diff --git "a/stack-exchange/math_stack_exchange/shard_115.txt" "b/stack-exchange/math_stack_exchange/shard_115.txt" new file mode 100644--- /dev/null +++ "b/stack-exchange/math_stack_exchange/shard_115.txt" @@ -0,0 +1,5464 @@ +TITLE: $m+ni+k\lambda,\,\Re(\lambda),\Im(\lambda)\notin \mathbb{Q}$ is dense in $\mathbb{C}$! +QUESTION [6 upvotes]: As said in the comments below, it's needed to suppose $\{1,\Re(\lambda),\Im(\lambda)\}$ linearly independent over $\mathbb{Q}$, otherwise the result is false, according to Christian's example. + +This is an exercise of the book Funções de uma Variável Complexa, by Alcides Lins Neto and, well, let's see it before I explain my doubt: + +Consider $\lambda\in \mathbb{C}$ such that $\Re(\lambda)$ and $\Im(\lambda)$ are NOT rational numbers. Prove that the set + $$\{m+ni+k\lambda; m,n,k\in \mathbb{Z}\}$$ + is a dense subset of $\mathbb{C}$. + +This is exercise 19 of the very first section of exercises in the book. At exercise 18 is given a suggestion (which, I suppose, may also be used at the 19th) that is: + +If $a\notin\mathbb{Q}$ then the set $\{m+na;m,n\in \mathbb{Z}\}$ is dense in $\mathbb{R}$. + +Let's see what I've tried. +Let $z_0$ be any complex number and $\varepsilon>0$ be any "radius". In order to show that that set is dense in $\mathbb{C}$ is sufficient to show that there is an element of it in the square +$$R_\varepsilon(z_0)=\left\{u+iv\in \mathbb{C};|u-\Re(z_0)|<\varepsilon,\,\,\, |v-\Im(z_0)|<\varepsilon\right\}.$$ +It is sufficient because the collection of "open squares" is a basis for topology on $\mathbb{C}\cong\mathbb{R}^2$. +Now, using the suggestion, we have that there are $m_1,n_1,m_2,n_2\in \mathbb{Z}$ such that +$$\begin{array}{c}\Re(z_0)-\varepsilon0$ and $z\in \Bbb{C}$. We have to proof that +$$ +\exists m,n,k\in \Bbb{Z},\quad \text{such that}\quad \|m+ni+k\lambda-z\|<\varepsilon. +$$ +Define $d_1(-,-):\Bbb{C}^2\to [0,\sqrt{2}]$ to be the distance +of the reduced values, i.e., +if $z=a+bi$ and $w=c+di$, then +$$ +d_1^2(z,w)=(Fr(Re(z))-Fr(Re(w))^2+(Fr(Im(z))-Fr(Im(w)))^2, +$$ +Where $Fr$ denotes the fractional part of a real number: $x-Fr(x)\in\Bbb{Z}$ and +$0\le Fr(x)<1$ for all $x\in \Bbb{R}$. +We can assume that $z=a+bi$ with $a,b\in[0,1)$. We will prove that there exist +$n,m,k,r\in\Bbb{Z}$ and $c,d\in\Bbb{R}$ such that +$$ +d_1(k\lambda,r(c+di))=0\quad\text{and}\quad \|r(c+di)-(n+a-(m-b)i)\|<\varepsilon. +$$ +Let $t_1\ge\frac{\sqrt{2}}{\varepsilon}$. Then for $t\ge t_1^2$ +there exist $0\le j,j_1\le t$, $j_1\ne j$, such that +$d_1(j\lambda,j_1\lambda)<\varepsilon$. In fact, all values of +$Fr(Re(j\lambda))+iFr(Im(j\lambda))$ land in $[0,1]\times[0,1]$ and +we can partition the square into $t_1^2$ little squares of side length +$1/t_1$. So necessarily two of the $t_1^2+1$ elements have to be in the +same little square. The greatest distance between two points in a square is the length of its diagonal, in this case $\sqrt{2}/t_1\le \varepsilon$. +Eventually interchanging $j$ and $j_1$ we can assume that $Re(j\lambda)>Re(j_1\lambda)$. +Now consider $c=Fr(Re((j-j_1)\lambda))$ and define +$$ +d:=\begin{cases} Fr(Im((j-j_1)\lambda)),& \text{if +$Im((j-j_1)\lambda)>0$,}\\ Fr(Im((j-j_1)\lambda))-1,& \text{otherwise}\end{cases} +$$ + Then $d_1(r(c+di),r(j-j_1)\lambda)=0$ for all $r\in\Bbb{Z}$ and $\|c+di\|\le\varepsilon$. +Moreover, since $1,Re(\lambda),Im(\lambda)$ are l.i. over $\Bbb{Q}$, we have +$s:=\frac{d}{c}\notin \Bbb{Q}$. +Define $R:=b-sa$. Then there exist $n,m\in \Bbb{Z}$ such that $|ns+m-R|<\varepsilon/2$. +But then +$$ +\| (n+a)(1+si)-(n+a-(m-b)i)\|=\|(ns+m+as-b)i\|=\|ns+m-R\|<\varepsilon/2. +$$ +On the other hand, for each $u\in\Bbb{R}$ there exists $r\in\Bbb{Z}$ such that +$\|u(c+di)-r(d+ci)\|\le\varepsilon/2$, in particular, for $u=\frac{n+a}{c}(c+di)$ +we find an $r$ such that +$$ +\varepsilon/2\ge \|r(c+di)-u(c+di)\|=\|r(c+di)-(n+a)(1+si)\|, +$$ +hence $\|r(c+di)-(n+a-(m-b)i)\|<\varepsilon.$ +Since for $k=r(j-j_1)$ we have $d_1(k\lambda,r(c+di))=0$, this concludes the proof.<|endoftext|> +TITLE: How do you find the turning points of a polynomial without using calculus? +QUESTION [11 upvotes]: I have a polynomial $P(x) = -x^3+12x+3$, and I am asked to find the turning points of it, and hence state how many zeroes it has. Since this chapter is separate from calculus, we are expected to solve it without differentiation. +However, I'm not sure how I could solve this. I'm having trouble factorising it as well since the zeroes seem to be irrational. + +REPLY [22 votes]: You want to know for which $c$ it is the case that $P(x)+c$ has a double root. We could mess around with the discriminant of the cubic, but that's probably too much work. Instead, suppose $P(x) + c = -(x-a)^2(x-b)$, so that $$ + -x^3 + 12 x + 3 + c = - x^3 + (2a+b)x^2 -(a^2 + 2ab)x +a^2 b +$$ +From this, we read off $2a+b = 0$, $a^2 + 2ab = -12$, and $3+c = a^2 b$. From the first two, solutions $(a,b)$ are $(-2,4)$ and $(2,-4)$. We don't even need to solve for $c$ because the double root (the turning point) occurs at $x=a$, so the turning points are $(-2,P(-2)) = (-2, -13)$ and $(2,P(2)) = (2,19)$. + +REPLY [10 votes]: For a cubic, any turning point is a double root of a suitable translation on the Y-axis. i.e. we seek double roots of $P(x)+k = -x^3+12x+3+k$ for some value of $k$. So we must have for some $a, b$, +$$-x^3+12x+3+k = (x-a)(x-b)^2$$ +Equating coefficients, we get +$$a+2b=0,\quad 2ab+b^2=-12, \quad ab^2 = 3+k$$ +The first two gives you $b=\pm2$, which are the turning points! + +Here is another way with AM-GM. Note that from symmetry it is enough to find the positive turning point of $f(x) = -x^3+12x$. As $f(0) = 0, f(1) = 11, f(2\sqrt3) = 3$, we must have a local maximum in $[1, 2\sqrt3]$. +So we maximize $2f^2 = (2x^2)(12-x^2)(12-x^2)$ which is a product of three positive terms with a constant sum, which means the maximum is when the terms are all equal, viz. $2x^2=12-x^2 \implies x^2=4$.<|endoftext|> +TITLE: How to evaluate $\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}$? +QUESTION [5 upvotes]: How to evaluate $\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}$? +So I think we expand to $x^2$ since the lowest term for $\ln(1+x)$ is $x$ +Let $u=\arctan{(x)}$ +$\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}=\lim\limits_{x\to 0}\frac{1+u+\frac{u^2}{2}+o(u^2)-(x+\pi x^2+o())-1}{x^2+o()}=\lim\limits_{x\to 0}\frac{1+x+o(x^3)+\frac{x^2+o()}{2}+o(u^2)-(x+\pi x^2+o())-1}{x^2+o()}=\frac12-\pi$ +My answer is rather messy and likely incorrect. Could someone provide an easier way to solve such problem? + +REPLY [2 votes]: Let's try to simplify the expression first. We have +\begin{align} +L &= \lim_{x \to 0}\frac{e^{\arctan x} - xe^{\pi x} - 1}{(\log(1 + x))^{2}}\notag\\ +&= \lim_{x \to 0}\dfrac{e^{\arctan x} - xe^{\pi x} - 1}{\left(\dfrac{\log(1 + x)}{x}\right)^{2}\cdot x^{2}}\notag\\ +&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x + \arctan x - xe^{\pi x} - 1}{1^{2}\cdot x^{2}}\notag\\ +&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x - 1}{x^{2}} + \lim_{x \to 0}\frac{\arctan x - xe^{\pi x}}{x^{2}}\notag\\ +&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x - 1}{(\arctan x)^{2}}\cdot\frac{(\arctan x)^{2}}{x^{2}} + \lim_{x \to 0}\frac{\arctan x - xe^{\pi x}}{x^{2}}\notag\\ +&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x - 1}{(\arctan x)^{2}} + \lim_{x \to 0}\frac{\arctan x - x + x - xe^{\pi x}}{x^{2}}\notag\\ +&= \lim_{t \to 0}\dfrac{e^{t} - t - 1}{t^{2}} + \lim_{x \to 0}\frac{\arctan x - x}{x^{2}} + \lim_{x \to 0}\frac{1 - e^{\pi x}}{x}\text{ (putting }t = \arctan x)\notag\\ +&= \lim_{t \to 0}\dfrac{e^{t} - 1}{2t} + \lim_{x \to 0}\frac{\arctan x - x}{x^{2}} - \lim_{x \to 0}\frac{e^{\pi x} - 1}{\pi x}\cdot \pi \text{ (using LHR for first limit)}\notag\\ +&= \frac{1}{2} - \pi - \lim_{t \to 0}\frac{t - \tan t}{\tan^{2} t}\text{ (putting }t = \arctan x)\notag\\ +&= \frac{1}{2} - \pi - \lim_{t \to 0}\frac{t - \tan t}{t^{2}}\cdot\frac{t^{2}}{\tan^{2}t}\notag\\ +&= \frac{1}{2} - \pi - \lim_{t \to 0}\frac{t - \tan t}{t^{2}}\notag +\end{align} +The last limit for $(t - \tan t)/t^{2}$ as $t \to 0$ can be easily shown to be $0$ using the inequalities $$\sin t < t < \tan t$$ for $0 < t < \pi/2$. Hence the desired limit is $\dfrac{1}{2} - \pi$.<|endoftext|> +TITLE: Mirror Symmetry of Elliptic Curve +QUESTION [6 upvotes]: I'm a little bit unsure about the mirror symmetry statement for elliptic curves; specifically, how the flipping of the Kähler and complex moduli works. Perhaps I should say at the outset, the reason I've been thinking about this, is that I'm doing a computation involving a torus with parameter $\tau \in \mathbb{H}$, and my answer in invariant under $\tau \to \tau+1$, but not $\tau \to -1/\tau$. So I'm thinking that maybe I'm only using the Kähler structure, not the complex structure. +Of course, the complex structure is given by $\tau \in \mathbb{H}/\rm{PSL}(2, \mathbb{Z})$. I think for the Kähler structure, we choose a class $[\omega] \in H^{2}(X,\mathbb{C})$, which we can parameterize by Kähler parameter +$$t=t_{1} + i \, t_{2},\ t=\frac{1}{2\pi i } \int_{X} [\omega]$$ +We identify $t_{2}>0$ with the area of the torus. So unlike the complex structure, K��hler structures related by $\rm{PSL}(2,\mathbb{Z})$, aren't necessarily identical, correct? After all, one will have small area, the other large. +So I'm confused about how the mirror symmetry acts. If mirror symmetry is some mysterious equivalence of the torus under interchange of the complex and Kähler moduli, doesn't that imply that the space of equivalent Kähler structures is also $\mathbb{H}/\rm{PSL}(2, \mathbb{Z})$. Is this correct? +Thank you. + +REPLY [2 votes]: I know this question is now rather old, but I think it's worth putting up the reference. The place you'll want to look at is Section 2 of Dijkgraaf's Mirror Symmetry and Elliptic Curves. In summary, the extra symmetries you see on the complex moduli suggest an unexpected symmetry in the A-model structures of the mirror Elliptic curve. This is related to the prediction that generating series of Gromov-Witten invariants are something like modular forms. As you point out, that is not a symmetry you see on the classical level by comparing the areas of the Elliptic curves. +Of course, if you are using something which is not invariant under the modular group, then you have something which does not only depend on the complex structure of your Elliptic curve, but something more.<|endoftext|> +TITLE: The tangent space of the moduli space of connection? +QUESTION [5 upvotes]: I'm reading one of Floer's paper. (An Instanton-Invariant for 3-Manifold). Let $M$ be a $3$-manifold. A principal $SU_2$-bundle P over $M$ must be trivial. Fixed a trivialization $P \cong M \times SU_2$ and we can regard the space of connection as $su_2$-valued $1$-forms on $M$. Denote $L_1^4(\Omega^1(M)\otimes su_2)$ by $\mathcal{A}(M)$. The gauge group $L_1^4(M,SU_2)$ (denoted by $\mathcal{G}(M)$) atcs on $\mathcal{A}(M)$ by +$$ g(a) = g a g^{-1} + (dg) g^{-1}.$$ +Denote $\mathcal{A}(M)/\mathcal{G}(M)$ by $\mathcal{B}(M)$. For $a \in \mathcal{A}(M)$ such that $\operatorname{Fix}(a)$ is the center $\mathbb{Z}_2$. There is an identification for the tangent space +$$ T_{[a]} \mathcal{B}(M) = \{ \alpha \in L(\Omega(M) \otimes su_2) | d_a^\ast \alpha = 0 \}$$ +where $d_a$ extend $d$ on $M$ by the connection $a$ and $d_a^\ast$ is its $L^2$-adjoint. +My question is that how we can get the identification for the tangent space? Or is there any more detailed text about this type of objects? (Moduli spaces, Banach manifolds, etc.) +Thank you. + +REPLY [3 votes]: Whenever you have a quotient manifold $M/G$, $G$ some Lie group, if the tangent space to the orbit at $a$ is denoted $V_a$, you can identify the tangent space $T_{[a]}(M/G) \cong T_aM /V_a$. If $M$ further has a Riemannian metric, you can identify $V_a^\perp \cong T_a M /V_a$. +This case is no different. $\mathcal A$ is the space of connections on $E$, affine over the space $\Omega^1(\text{End}(E)) \cong \Omega^1(\mathfrak{su}(2))$ (because $E$ is trivial), and hence its tangent space at some point is $\Omega^1(\mathfrak{su}(2))$. The tangent space to the Lie group $\mathcal G$ at the identity is $\Omega^0(\mathfrak{su}(2))$. The Lie group action of $\mathcal G$ on $\mathcal A$ has derivative at $A$ equal to $\alpha \mapsto d_A \alpha$. So the image of this (the tangent space to the orbit) is equal to the image of $d_A$. So we can identify $T_{[A]} \mathcal B$ with $\Omega^1(\mathfrak{su}(2))/d_A\Omega^0(\mathfrak{su}(2))$. Now (as above, again) $\mathcal A$ has a Riemannian metric - $\langle \nu, \xi \rangle = \int \text{Tr}(\nu \wedge \ast \xi)$. (This is just to say it's the $L^2$ inner product with respect to the standard inner product on $\mathfrak{su}(2)$ - maybe up to a constant I'm forgetting). With respect to this inner product, the orthogonal complement of the image of $d_A$ is (more-or-less definitionally) the kernel of $d_A^*$. So we get our canonical identification $T_{[A]} \mathcal B = \text{ker } d_A^* \subset \Omega^1(\mathfrak{su}(2))$, as desired.<|endoftext|> +TITLE: Evaluate this infinite product: $\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)$ +QUESTION [8 upvotes]: $$\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)\;=\;\text{???}$$ + +I took the LCM and split the numerator as $(n+2)(n-2)$ and then took the product of the numerator and the denominator separately but I was not able to get the answer from that so can you please help me in what to do next. + +REPLY [4 votes]: A different solution. You can use the fact that +$$\sin(\pi x)=\pi x \prod_1^\infty \left(1-\frac{x^2}{n^2} \right).$$ +So your product is e qual to +$$\lim_{x \to 2} \frac{\sin(\pi x)}{\pi x (1-x^2)(1-x^2/4)}=1/6$$<|endoftext|> +TITLE: Pade approximant for the function $\sqrt{1+x}$ +QUESTION [5 upvotes]: I'm doing the followiwng exercise: + +The objective is to obtain an approximation for the square root of any given number using the expression +$$\sqrt{1+x}=f(x)\cdot\sqrt{1+g(x)}$$ +where g(x) is an infinitesimal. If we choose $f(x)$ as an approximation of $\sqrt{1+x}$, then we can calculate $g(x)$: +$$g(x)=\frac{1+x}{f^2(x)}-1$$ +$f(x)$ can be chosen as a rational function $p(x)/q(x)$, such that $p$ and $q$ have the same degree and it's Mclaurin series is equal to the Mclaurin series of the function $\sqrt{1+x}$ until some degree. Find a rational function $f(x):=p(x)/q(x)$, quotient of two linear polynomials, such that the McLaurin series of $p(x)-\sqrt{1+x}\cdot q(x)$ have the three first terms equal to $0$. + +How can I do this? Has something to be with the Pade approximant? +Any hint would be really appreciated. Thanks for your time. + +REPLY [2 votes]: You can work this out without MacLaurin, starting from +$$p^2(x)\approx(1+x)q^2(x).$$ +Then with +$$p(x)=(1+ax)^2,q(x)=(1+bx)^2,$$ +we have +$$1+2ax+a^2x^2\approx1+(2b+1)x+(b^2+2b)x^2+b^2x^3,$$ +and we identify +$$2a=2b+1,\\a^2=b^2+2b$$ +which gives the solution +$$\sqrt{1+x}\approx\frac{1+\frac34x}{1+\frac14x}.$$<|endoftext|> +TITLE: Continuity of Banach limit and existence of $\Lambda_0\in(\ell^\infty/c_0)^*$ such that $\Lambda=\Lambda_0\circ q_0$, with $q_0$ the quotient map +QUESTION [5 upvotes]: Let $\Lambda$ be any Banach limit on $\ell^\infty$, where $\ell^\infty$ denotes the space of bounded real sequences. A Banach limit is defined as a linear functional $\Lambda$ such that +$$ \Lambda(\tau x)=\Lambda(x), \forall x\in\ell^\infty$$ +$$ \liminf_{n\rightarrow\infty}x_n\leq\Lambda(x)\leq\limsup_{n\rightarrow\infty}x_n$$ +where we write $x=(x_n)_{n\in\mathbb{N}}$ for a sequence $x\in\ell^\infty$ and we define left translation on $\ell^\infty$ by $(\tau x)_n=x_{n+1},n=1,2,\dots$. +I would like to show that $\Lambda\in(\ell^\infty)^*$, which means that $\Lambda$ is a continuous, linear functional on $\ell^\infty$. Thus I need to show that $\Lambda$ is continuous. How do I do this? +Furthermore I wish to show that there exists a continuous, linear functional $\Lambda_0\in(\ell^\infty/c_0)^*$ such that $\Lambda=\Lambda_0\circ q_0$, where $$q_0:\ell^\infty\rightarrow\ell^\infty/c_0$$ is the quotient map and +$$c_0=\{(x_n)\in\ell^\infty\mid \lim_{n\rightarrow\infty}x_n=0\}$$ +I can't seem to get anywhere with these questions. Any help is greatly appreciated. + +REPLY [2 votes]: That $\Lambda$ is continuous follows directly from the second estimate: +$$ +-\|x\|_\infty\leq \liminf_{n\to\infty} \Lambda(x_n)\leq \Lambda(x)\leq \limsup_{n\to\infty}\Lambda(x_n)\leq\|x\|_\infty +$$ +Thus, $|\Lambda(x)|\leq \|x\|_\infty$ for all $x\in \ell^\infty$. +For your second claim define $\Lambda_0(x+c_0):=\Lambda(x)$. This is well-defined since $x-y\in c_0$ implies $\Lambda(x-y)=0$, i.e. $\Lambda(x)=\Lambda(y)$.<|endoftext|> +TITLE: Prove inequality $n\sqrt[n]{n!}-m\sqrt[m]{m!}\le\frac{(n−m)(n+m+1)}2.$ +QUESTION [6 upvotes]: Let $m,n\in\mathbb N$, $n>m$. Prove inequality + $$n\sqrt[n]{n!}-m\sqrt[m]{m!}\le\frac{(n−m)(n+m+1)}2.$$ + +My work so far: +$$\sqrt[n]{n!}=\sqrt[n]{1\cdot2\cdot...\cdot n}\le\frac{1+2+...+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}2.$$ +Then $n\sqrt[n]{n!}\le\frac{n(n+1)}{2}$ +$$\sqrt[m]{m!}=\sqrt[m]{1\cdot2\cdot...\cdot m}\le\frac{1+2+...+m}{m}=\frac{m(m+1)}{2m}=\frac{m+1}2.$$ +Then $m\sqrt[m]{m!}\le\frac{m(m+1)}{2}$ +And $$\frac{n(n+1)}{2}-\frac{m(m+1)}{2}=\frac{(n−m)(n+m+1)}2.$$ +But inequalities can be added, without subtract + +REPLY [5 votes]: The result follows immediately by writing +\begin{align} +n\sqrt[n]{n!} +&=n\sqrt[n]{m!\cdot(m+1)(m+2)\cdots n}\\ +&=n\sqrt[n]{(\sqrt[m]{m!})^m\cdot(m+1)(m+2)\cdots n}\\ +&\le m\sqrt[m]{m!}+(m+1)+(m+2)+\cdots+n\\ +&=m\sqrt[m]{m!}+\frac{(n-m)(n+m+1)}{2}. +\end{align}<|endoftext|> +TITLE: Show that any line in $\Bbb R^2$ is closed +QUESTION [5 upvotes]: Show that any line in $\Bbb R^2$ is closed under the usual topology +I think showing the complement is open is best way. +Let $ax+by+c=0$ be a line in $\Bbb R^2$ as a line it is the set of $(x,y) \in \Bbb R^2 $ such that it satisfies the above equation for constants $a,b,c$ for simplicity call the set $L$ +Take $(x,y) \in L^c$, the distance from $(x,y)$ to $L$ is $\displaystyle \frac{\mid ax+by+c \mid}{\sqrt{a^2+b^2}}$ +Let $\epsilon = \frac{\mid ax+by+c \mid}{\sqrt{a^2+b^2}}$ then take as $U$ the ball $((x,y), \epsilon)$ then $U \cap L = \emptyset$ so $L^c$ is open. + +REPLY [4 votes]: More generally +Any graph of a continuous function is closed. Indeed, let $$\Gamma=\{(x,f(x))\mid x\in \mathbb R\},$$ +the graph of a continuous function $f:\mathbb R\longrightarrow \mathbb R$. Let $\{(x_n,f(x_n))\}_n$ a sequence of $\Gamma$ that converge. Let denote $(a,b)$ it's limit. Then, by continuity of $f$, $$b=\lim_{n\to \infty }f(x_n)=f(a)$$ +and thus, $(a,b)\in\Gamma $. Therefore, $\Gamma$ is closed.<|endoftext|> +TITLE: Is the Fundamental Group of space with contractible universal cover torsion-free? +QUESTION [9 upvotes]: Some classmates and I were working on the following question - is the fundamental group of the Klein Bottle $K$ torsion-free? We have the following presentation: +$$\pi_1(K) = \langle a,b: aba = b \rangle.$$ +In trying to answer this problem, we came up with the following question that could resolve this problem, and seemed like an interesting claim in general: + +Is the Fundamental Group of a space with contractible universal cover torsion-free? + +We thought about it a bit and unfortunately did not find a counterexample or a proof. Is this true? If it's not, what is a good way to answer our original question about the Klein Bottle? + +REPLY [11 votes]: This is false in general; in fact every group $G$ is the fundamental group of a unique (up to weak homotopy) space with contractible universal cover, namely the classifying space $BG$, or equivalently the Eilenberg-MacLane space $K(G, 1)$. In particular, the Klein bottle is such a classifying space, as is the torus. +What's true is the following. + +Theorem: If $BG$ is a finite-dimensional CW complex, then $G$ is torsion-free. + +Proof. With the above hypotheses, the universal cover of $BG$ is also a finite-dimensional CW complex on which $G$ acts freely, hence for any subgroup $H$ of $G$, the quotient $BH$ of the universal cover by the induced action of $H$ is also a finite-dimensional CW complex, and in particular has finite cohomological dimension. But finite cyclic groups have infinite cohomological dimension, by the standard computation of their group cohomology. $\Box$<|endoftext|> +TITLE: Constructing a computably infinite tree with no computable infinite branches using PA +QUESTION [5 upvotes]: Define an infinite tree as any set of sequences closed under prefix restriction, i.e. any prefix restriction of a sequence in the set is also in the set, where a prefix restriction is a restritcion of the original sequence to an initial segment of it (the prefix). In our case, we are interested mainly in countable trees, so we may take trees to be sequences of natural numbers closed under restriction. A computable tree is a tree such that there is an algorithm for computing, for each sequence, if it belongs to the tree. Finally, a branch through a tree is an infinite sequence such that each finite prefix of this sequence is in the tree. (I hope these definitions are correct; I'm still getting used to working with trees) +It is well-known that it is possible to construct an infinite computable tree such that none of its branches are computable, e.g., the Kleene tree. However, I remember reading somewhere that another way of constructing such a tree would be to use Peano Arithmetic; something to the effect that the tree would consist of formulas of PA, and if one could compute every branch, this would imply the decidability of PA. Unfortunately, I don't remember where I read this and I couldn't reconstruct the proof by myself. Does this strategy sound plausible? If so, could anyone help me to reconstruct this proof? + +REPLY [8 votes]: Yes, this is a standard example. +Fix a computable listing of the sentences in the language of arithmetic, $\{\varphi_i: i\in\mathbb{N}\}$. Each binary sequence $\sigma\in 2^{<\omega}$ now corresponds to a set of sentences: $$[\sigma]=\{\varphi_i: \sigma(i)=1\}\cup\{\neg\varphi_i: \sigma(i)=0\}.$$ (Note that this set is finite, since $\sigma$ has finite length.) +Now, say that $\sigma\in 2^{<\omega}$ appears consistent with $PA$ if there is no proof of $0=1$, of length at most $\vert\sigma\vert$, from $\sigma$ together with the first $\vert\sigma\vert$-many axioms of $PA$ (using some standard enumeration of $PA$, here). Since this is a purely finitary property, the question "Does $\sigma$ appear consistent with $PA$?" can be answered effectively: the set $$T=\{\sigma\in 2^{<\omega}:\sigma\mbox{ appears consistent with $PA$}\}$$ is computable. +This $T$ is our tree. It's easy to see that it is a tree - that is, $\sigma\in T$ and $\tau\prec\sigma$ implies $\tau\in T$. To see that it has no infinite paths, note that if $f$ is a path through $T$ then $\{[\sigma]: \sigma\prec f\}$ is a complete consistent extension of $PA$, which is computable from $f$. So we are done. + +This is actually an example of a more general phenomenon. Say two disjoint r.e. sets $A, B\subseteq\omega$ are recursively inseparable if there is no computable $C$ with $A\subseteq C$ and $B\cap C=\emptyset$ (note that we could swap $A$ and $B$ and the definition would stay the same - replace $C$ by $\omega\setminus C$). Given any pair $A, B$ of recursively inseparable sets, we can form a tree $T_{A, B}$ of partial separations: $$T_{A, B}=\{\sigma\in 2^{<\omega}: \forall n<\vert\sigma\vert, [\sigma(n)=1\implies n\not\in B_{\vert\sigma\vert}]\wedge [\sigma(n)=0\implies n\not\in A_{\vert\sigma\vert}]\}$$ (where $D_s$ is the stage-$s$ approximation to $D$, if $D$ is an r.e. set (we assume a fixed recursive enumeration of $D$)). This $T_{A, B}$ is a recursive tree with no recursive path. +In the above example, we're using the recursively inseparable pair $\{$things $PA$ proves$\}$ and $\{$things $PA$ disproves$\}$.<|endoftext|> +TITLE: $\epsilon-\delta$ definition for limits involving $\infty$ +QUESTION [6 upvotes]: For a general limit the $\epsilon-\delta$ definition of a limit (the formal definition of a limit) states that +$$\lim_{x \ \to \ a} f(x) = L \Leftrightarrow \forall \epsilon > 0 \ (\exists \ \delta > 0 \ : \ (0<|x-a|<\delta \implies |f(x)-L| < \epsilon))$$ +However the $\epsilon -\delta$ definition of a limit changes for limits as $x \to +\infty$ and $x \to -\infty$, and it changes again for limits that evaluate to $+\infty$ and $-\infty$ + +1. Limit as $x \to +\infty$ +$$\lim_{x \ \to \ +\infty} f(x) = L \Leftrightarrow + \forall \ \epsilon>0\; (\exists \ \delta : (\;x>\delta\implies |f(x) - L|\leq\epsilon)) +$$ +2. Limit as $x \to -\infty$ +$$\lim_{x \ \to \ -\infty} f(x) = L \Leftrightarrow + \forall \ \epsilon>0\; (\exists \ \delta : (\;x<\delta\implies |f(x) - L|\leq\epsilon)) +$$ +3. Limit evaluating to +$\infty$ +$$\lim_{x \ \to \ a} f(x) = +\infty \Leftrightarrow \forall M > 0 \ (\exists \ \delta > 0 \ : \ (0<|x-a|<\delta \implies f(x) >M)$$ +4. Limit evaluating to -$\infty$ +$$\lim_{x \ \to \ a} f(x) = -\infty \Leftrightarrow \forall N < 0 \ (\exists \ \delta > 0 \ : \ (0<|x-a|<\delta \implies f(x) < N)$$ + +But why is that so? I understand that if you use the normal $\epsilon-\delta$ definition of a limit in these cases, you get contradictions that pop up like $0 < |x-\infty| < \delta \implies \delta > \infty$, which you cannot do anything further with. +While some of these differences might be subtle, it just seems counter-intuitive to change a formal and general definition. +I know that the fundamental idea behind the $\epsilon - \delta$ definition remains the same throughout all of these examples (that no matter how small we want to make our "error distance" ($\epsilon$) we can always find a "distance to our limit point" ($\delta$) that satisfies the definition of a limit) , but to get to that fundamental idea, the $\epsilon - \delta$ definition has to be subtly modified (and I'm not referring to modifications in notation) for each of these examples. +Or is it just a case that the $\epsilon - \delta$ definition for the general limit I put at the very start of this post is not as general as I thought? + +Furthermore, how does the $\epsilon - \delta$ definition of a limit change for these cases, (the formal definitions of these don't seem to be covered in any introductory Calculus textbook). +5. Limit as $x \to +\infty$ $= +\infty$ +$$ \lim_{x \ \to \ +\infty} f(x) = +\infty \Leftrightarrow \ ???$$ +6. Limit as $x \to -\infty$ $= -\infty$ +$$ \lim_{x \ \to \ -\infty} f(x) = -\infty \Leftrightarrow \ ???$$ +7. Limit as $x \to -\infty$ $= +\infty$ +$$ \lim_{x \ \to \ -\infty} f(x) = +\infty \Leftrightarrow \ ???$$ +8. Limit as $x \to +\infty$ $= -\infty$ +$$ \lim_{x \ \to \ +\infty} f(x) = -\infty \Leftrightarrow \ ???$$ + +REPLY [4 votes]: To understand this, you need to think of the intuition behind the $\epsilon$-$\delta$ definition. We want $\lim_{x\to a}f(x)=L$ if we can make $f(x)$ as close to $L$ as we like by making $x$ sufficiently close to $a$. Worded differently, we might say that: + +$\lim_{x\to a}f(x)=L$ if given any neighborhood $U$ of $L$, there is a neighborhood $V$ of $a$ such that elements of $V$ are mapped by $f$ to elements of $U$ (except possibly $a$ itself). + +In this context, a "neighborhood" of a point $p$ should be understood to mean "points sufficiently close to $p$". Let's make that precise by defining what we mean by "close". For $\epsilon>0$ (assumed, but not required, to be very small) define +$$B(x,\epsilon):=\{y\,:\,|x-y|<\epsilon\},$$ +the ball of radius $\epsilon$ about $x$. For our purposes, we say $U$ is a neighborhood of $x$ if $U=B(x,\epsilon)$ for some $\epsilon>0$. (The usual definition only requires that $U$ contains such a ball.) Assuming $\epsilon>0$ is very small, this agrees with our intuition of what closeness should mean. Now if we go back to our neighborhood "definition" of a limit, you should be able to think about it for a bit and convince yourself that it is equivalent to the usual definition. +How does this relate to the problem with infinity? Given that infinity is not a real number (and things like distance from infinity do not make sense), we must revise what it means to be "close" to infinity. So for $M>0$ (assumed this time to be very large) define +$$B(+\infty,M):=\{y\,:\,y>M\},\quad B(-\infty,M):=\{y\,:\,y<-M\},$$ +the neighborhoods of $\pm\infty$. Hopefully you can see why these make sense as definitions; a number should be close to infinity if it is very large (with the correct sign), so a neighborhood of infinity should contain all sufficiently large numbers. +Now we extend our neighborhood definition of limits to include the case where $a$ or $L$ can be $\pm\infty$. It is a similar exercise to before to verify now that the definition is still equivalent to the old one, only now we have in some sense unified somewhat.<|endoftext|> +TITLE: From correlation coefficient to conditional probability +QUESTION [9 upvotes]: In the best-selling book Thinking Fast and Slow (p. 205), Daniel Kahneman (a Nobel Prize winner in Economics) makes the following claim: 'Suppose you consider many pairs of firms. The two firms in each pair are generally similar, but the CEO of one of them is better than the other. How often will you find that the firm with the stronger CEO is the more successful of the two? . . . A correlation of .30 [between stronger CEO and more successful firm] implies that you would find the stronger CEO leading the stronger firm in about 60% of the pairs.' +How is the 'about 60%' computed under plausible assumptions? + +REPLY [4 votes]: For a randomly selected firm let $X$ be the CEO quality and $Y$ be the firm success (however these are measured). The author's assertion follows under the assumption that $(X,Y)$ has a bivariate normal distribution with correlation $\rho=0.3$. +If $(X_1,Y_1)$ and $(X_2,Y_2)$ are measured for independently selected firms, then the difference $(X_1-X_2,Y_1-Y_2)$ is also bivariate normal, with mean zero and the same correlation $\rho$. Kahneman's claim is that $P(Y_1>Y_2\mid X_1>X_2)\approx 0.6$. This follows from a fact(*) about bivariate normal variables: +If $(A,B)$ are bivariate normal with means $\mu_A$ and $\mu_B$ respectively, and correlation $\rho$, then +$$P(B>\mu_B\mid A>\mu_A)=\frac12 + \frac{\arcsin\rho}\pi.$$ +If $\rho=0.3$ the RHS works out to $0.59698668$. +(*) The fact can be deduced from this result.<|endoftext|> +TITLE: How to succeed in upper-level math +QUESTION [5 upvotes]: This question will essentially be more of a how-to plea or general help request. +I'm currently studying math and I'm at the point where I've transitioned into upper-division classes, most if not all of which are proof based. +To be blunt, I currently feel discouraged at the prospect of being able to succeed in "upper-level" classes. Last quarter I was able to get decent grades in my classes, but nothing like the success I enjoyed in lower-level math. My discouragement stems from the fact that in the past if I studied I did well. Now I feel as if I'm studying and not doing well at all. +Obviously this is a reflection of either my study methods or frequency with which I review the required material, but either way, I'm starting to become disillusioned with the good ol' saying that if you "practice, practice, practice" then you'll get better. +My question to whoever cares to comment, and believe me I greatly and truly appreciate all advice, is how did you transition into proof based classes and succeed in them? Obviously the material is different for each course, but in general how did you approach absorbing the material? While everyone is different, what are review methods that you found helped you to best understand the material and also retain your understanding? +I don't know if others have this same issue but I find that for proof based classes, since problems tend to all be different, I don't quite retain a sense of how to tackle problems even after completing an assignment whereas for classes such as calculus in which e.g. I had to learn integration, after so many integrals I had a general sense of how to solve them. Any advice on how to remedy this deficiency? +I apologize if this question is inappropriate in any way, but I'd love some perspective from those that have gone through a similar process. + +REPLY [3 votes]: This is a tough question with no easy answer, so I'll just write down some tips which may be helpful for proof-oriented courses. + +Read the proofs in your textbook (and maybe other sources), to the point where you can understand why each proof proves what it claims to prove. This means you understand why each sentence in the proof is correct, and how the paragraph(s) of the proof as a whole proves the claimed statement. When doing this, don't yet think about "How could I possibly come up with this proof?" (although eventually you will need to address this question). The first step to being able to write correct proofs of your own, is to be able to read other people's proofs and determine why they are correct (or incorrect!). +Study basic logic. Concepts like indirect proofs, the meaning of "if and only if", use of quantifiers, etc., should become second nature, so that you immediately recognize them when they are used in a proof. This is independent of whatever the actual content of the class is (linear algebra, group theory,...) Some logic textbooks use non-mathematical examples on familiar subjects for illustration ("If it's raining, then the ground is wet, but not conversely"), which might be helpful for separating logic from the new mathematical concepts in a class. Once you don't have to think about basic logic any more, you can concentrate more on the actual concepts in a class. +You say that in a proof-oriented class, "the problems tend to be all different". If this is the case, you need to be grouping the problems at a higher level. One possible approach is to group together proofs with a common goal. For example, in a linear algebra class, there may be many problems that ask you to prove that something is a subspace. The details of the "something" may look very different from one problem to the next, but there will be common steps in the proofs. (Some textbooks may explicitly point out these common steps.) These common steps will help you get started the next time you are asked to prove that something is a subspace. +When you're doing integration in a calculus class, you can compare your answer against someone else's pretty easily to determine whether they are the same. (Maybe you need to do a little algebra or use some trig identities, to put your answer in the same form.) This is not the case with proofs. It's possible for two different people to come up with quite different proofs, both of which are correct. That's why point 1 above is important. You need to be able to read a proof you wrote and determine whether it's correct. + +I hope this is helpful in some way.<|endoftext|> +TITLE: Can the partial sums of independent random variables with no normalization converge in distribution to a constant? +QUESTION [5 upvotes]: If $\{X_n , n\ge 1 \}$ is a sequence of independent random variables and $X_n$ is nondegenerate for at least one $n\ge1$, can there exist a finite constant c such that $S_n = \sum_{j=1}^n X_j \stackrel{d}{\rightarrow}c$ ? +(Here we are not dividing $S_n$ by any sequence of normalizing constants.) +I am struggling to find an example where this can hold. Could someone please provide an appropriate example, or a proof that no such constant $c$ can exist? + +REPLY [3 votes]: The answer is negative. +Note that convergence in distribution to a constant implies convergence in probability to that constant, so it suffices to prove we cannot have probability to a constant. Note that without loss of generality, we may suppose $c=0$ and that $X_1$ is non-degenerate. Set $Y=-\sum_{n=2}^\infty X_n$, where the limit is taken in probability. +Then the question is (more or less) equivalent to asking if $Y=X$, or if a sequence of random variables can converge in probability to a non-degenerate random variable independent from each of the summands. +Now, convergence in probability implies that a subsequence converges almost surely, so by grouping the $X_i$ we may suppose that the convergence to $Y$ is a.s. Then we have $Y=X$ a.s. for two independent random variables $Y$ and $X$ with $X$ non-constant, which is impossible.<|endoftext|> +TITLE: How to find $\lim\limits_{x\to 2}f(x)$ if $\lim\limits_{x\to 2}\frac{f(x)-5}{x-2}=100$? +QUESTION [6 upvotes]: How to find $\lim\limits_{x\to 2}f(x)$ if $\lim\limits_{x\to 2}\frac{f(x)-5}{x-2}=100$? +I suppose that we do not use L'Hopital rule here. Then $\lim\limits_{x\to 2}f(x)-5=0$, then $\lim\limits_{x\to 2}f(x)=5.$ +I highly doubt the answer, could someone help me understand? + +REPLY [9 votes]: First write $$\lim\limits_{x \to 2} f(x)=\lim\limits_{x \to 2}\left[\frac{f(x)-5}{x-2}(x-2)+5\right]$$ and since three limits exist $$=\lim\limits_{x \to 2}\frac{f(x)-5}{x-2}\lim\limits_{x \to 2}(x-2)+\lim\limits_{x \to 2}5=100\cdot0+5=5.$$ + +Proof of the statement that EdwardJiang used: If $$\lim\limits_{x \to a} \frac{f(x)}{g(x)}=b \text{ and } \lim\limits_{x \to a} g(x)=0$$ then $$\lim\limits_{x \to a} f(x)=\lim\limits_{x \to a} \frac{f(x)}{g(x)}g(x)=\text{(limits exist)}\lim\limits_{x \to a} \frac{f(x)}{g(x)}\lim\limits_{x \to a} g(x)=b\cdot 0=0.$$ (Maybe I am wrong but) I disagree about using this information without proof in this question. This question is actually about showing this as an example.<|endoftext|> +TITLE: Show that $\zeta'(0)=-\frac{1}{2}\ln(2\pi)$ +QUESTION [5 upvotes]: I started with the functional equation which was derived in class, +$$ +\zeta(s)=2^s\pi^{s-1}\sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s) +$$ +and took the logarithmic derivative of both sides to get +$$ +\frac{\zeta'(s)}{\zeta(s)}=\ln2+\ln\pi+\frac{\pi}{2}\frac{\cos(\frac{\pi s}{2})}{\sin(\frac{\pi s}{2})}+\psi(1-s)-\frac{\zeta'(1-s)}{\zeta(1-s)}. +$$ +Now I see that if I evaluate at $1$ or $0$ I will have an equation for $\zeta'(0)$ but I cannot solve either of the resulting equations. Can someone show me how to do this? + +REPLY [6 votes]: We may use the $\eta $ function as suggested by user1952009. Given $s\in(0,1)$, +$$ \eta(s) = \sum_{n\geq 1}\frac{(-1)^n}{n^s} $$ +fulfills $\lim_{s\to 0^+}\eta(s)=-\frac{1}{2}$ by Abel's lemma. On the other hand, +$$ \eta'(s) = \sum_{n\geq 1}\frac{(-1)^n \log n}{n^s}=\sum_{n\geq 1}\frac{(-1)^n}{n^s}\int_{0}^{+\infty}\frac{e^{-u}-e^{-nu}}{u}\,du $$ +by differentiation under the $\sum$ sign and Frullani's theorem. So we have: +$$ \lim_{s\to 0^+}\eta'(s) = \int_{0}^{+\infty}\frac{-\frac{1}{2}e^{-u}+\frac{1}{1+e^u}}{u}\,du=\frac{1}{2}\log\frac{\pi}{2}$$ +by Wallis' product and Frullani's theorem again. On the other hand, +$$ \zeta(s) = (2^{1-s}-1)\,\eta(s) $$ +so $\zeta(0)=-\frac{1}{2}$ and by switching to logarithmic derivatives we get: +$$ \frac{\zeta'(s)}{\zeta(s)} = \frac{\eta'(s)}{\eta(s)}-\frac{2^{1-s}\log(2)}{2^{1-s}-1}$$ +from which $\zeta'(0)=-\log\sqrt{2\pi}$ easily follows.<|endoftext|> +TITLE: Nested logarithm series +QUESTION [5 upvotes]: While working on problems from Spivak's Calculus, I came on one asking for the convergence/divergence of the series $$\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}.$$ +This is a straightforward comparison with the harmonic series to show divergence, but I ended up playing around with the idea a bit more and stumbling on the interesting fact that $$\sum_{n=2}^{\infty} \frac{1}{n^{1+\frac{1}{\log n}}}\,\,\text{diverges}\quad\text{and}\quad \sum_{n=2}^{\infty} \frac{1}{n^{1+\frac{1}{\log\log n}}}\,\,\text{converges}.$$ +The first is another limit comparison with the harmonic series; the second I tried to do by looking at $$\int_2^{\infty} \frac{1}{n}\cdot \frac{1}{n^{\frac{1}{\log \log n}}}\, dn$$ and substituting $u=\log n$ to give $$\int_2^{\infty}e^{-\frac{u}{\log u}}\, du,$$ which seems more tractable. +I have two questions: + +Can anyone show convergence of divergence of the second series by finishing up my attempt or through some other technique? +More generally, for the series $$\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{f(n)}}}$$ what can be said about the growth rates/other characterizations of the $f$ for which this converges? + +Any help would be much appreciated! + +REPLY [2 votes]: There was a request for a different approach from that discussed in the OP. Here, we use Cauchy's Condensation test on the series $S$ expressed as +$$S=\sum_{n=3}^\infty f(n) \tag 1$$ +where the summand $f(n)$ is given by +$$f(n)=\frac{1}{n^{1+\frac{1}{\log(\log(n)}}}$$ +Now, we analyze the sum $T$ expressed as +$$\begin{align} +T&=\sum_{n=3}^\infty 2^n f(2^n)\\\\ +&=\sum_{n=3}^\infty 2^n \frac{1}{2^{n\left(1+\frac{1}{\log(\log(2^n)}\right)}}\\\\ +&=\sum_{n=3}^\infty 2^{-\left(\frac{n}{\log(n\log(2))}\right)} \tag 2 +\end{align}$$ +Now, for any $\alpha >0$, the logarithm function satisfies the inequality SEE THIS ANSWER +$$\log(x)\le \frac{x^\alpha -1}{\alpha}\le \frac{x^\alpha}{\alpha} \tag 3$$ +Applying $(3)$ to $(2)$ reveals that for $0<\alpha <1$ +$$2^{-\left(\frac{n}{\log(n\log(2))}\right)}\le 2^{-\left(\frac{\alpha n^{1-\alpha}}{\log^\alpha(2)}\right)}$$ +Thus, the series in $(2)$ clearly converges by the comparison test. Take for example $\alpha =1/2$. Then, it is easy to show that +$$2^{-\left(\frac{ n^{1/2}}{2\sqrt{\log(2)}}\right)}\le \frac{48/\log^{3/2}}{n^{3/2}}$$ +and the series converges by comparing it to the series $\sum_{n=3}^\infty \frac{1}{n^{3/2}}$. +Finally, the condensation test guarantees that the series in $(1)$ also converges. And we are done!<|endoftext|> +TITLE: Find a Continuous Function with Cantor Set Level Sets +QUESTION [9 upvotes]: This was a problem from a class that I thought was really interesting. It asked to find function $f\in C[0,1]$ such that the sets $\{x:f(x)=c\}$ form a Cantor Set for all $0\leq c\leq 1$. I found a non-constructive proof of the existence of such functions, but would be curious if anyone could give a constructive example of such a function. +Definition: A set is a Cantor Set if it is uncountable, perfect, compact, nowhere-dense set. This makes it homeomorphic to "the Cantor set." + +REPLY [3 votes]: As the Cantor set is self-similar, it makes sense to try to construct a graph with some kind of self-similar structure. The only way the graph of a function can be strictly self-similar is if it is a line segment. Thus, we explore a slight generalization using a self-affine graph. I'll first present an example that is specifically designed to have the desired property and then comment on a couple of cases where such a structure arises quite naturally. +A rigged example +For our first example, we start with the Iterated Function System of six affine functions defined as follows: +$$ +\begin{align} + f_1(x,y) &= \left(\begin{array}{cc} 1/6 & 0 \\ 0 & 1/4 \end{array}\right) + \left(\begin{array}{c} x\\y\end{array}\right) \\ + f_2(x,y) &= \left(\begin{array}{cc} 1/6 & 0 \\ 0 & 1/4 \end{array}\right) + \left(\begin{array}{c} x\\y\end{array}\right) + + \left(\begin{array}{c} 1/6\\1/2\end{array}\right) \\ + f_3(x,y) &= \left(\begin{array}{cc} 1/6 & 0 \\ 0 & -1/4 \end{array}\right) + \left(\begin{array}{c} x\\y\end{array}\right) + + \left(\begin{array}{c} 2/6\\1\end{array}\right) \\ + f_4(x,y) &= \left(\begin{array}{cc} 1/6 & 0 \\ 0 & -1/4 \end{array}\right) + \left(\begin{array}{c} x\\y\end{array}\right) + + \left(\begin{array}{c} 3/6\\1/2\end{array}\right)\\ + f_5(x,y) &= \left(\begin{array}{cc} 1/6 & 0 \\ 0 & 1/4 \end{array}\right) + \left(\begin{array}{c} x\\y\end{array}\right) + + \left(\begin{array}{c} 4/6\\0\end{array}\right)\\ + f_6(x,y) &= \left(\begin{array}{cc} 1/6 & 0 \\ 0 & 1/4 \end{array}\right) + \left(\begin{array}{c} x\\y\end{array}\right) + + \left(\begin{array}{c} 5/6\\1/2\end{array}\right) +\end{align} +$$ +The geometric action of this IFS on the upwardly oriented unit square is illustrated in the following figure: + +This image shows the oriented initial set, the first two levels of approximation using rectangles, and a higher order approximation to the attractor, which is the graph of a continuous function. +Now, it's pretty easy to see that the intersection with the level $n$ approximation by rectangles with a horizontal line segment is just a collection of line segments and that these contain the corresponding intersection at level $n-1$. The intersection of all these is perfect, non-empty, and no where dense. Thus, it's a Cantor set. +This function maps $[0,1]\to[0,1]$ but it's easy to extend to $\mathbb R$, if desired. +Natural examples +There are some examples of functions with Cantor type level sets that occur naturally - i.e. constructed for some other purpose but happen to have the desired property. +Space-filling curves +The coordinate functions of Hilbert's space filling curve have Cantor type level sets. I proved this in my paper The Hausdorff Dimension of Hilbertʼs Coordinate Functions. I also published a more elementary exposition in The Math Mag. If you understood the previous example, you can see what's going on in figure 7 of that paper: + +The Takagi function +Finally, the Takagi function has this property at many points. More specifically, let +$\varphi(x)$ denote the distance from $x$ to the nearest integer and let +$$\tau(x) = \sum_{k=0}^{\infty} \frac{1}{2^k}\varphi(2^kx).$$ +Then, the maximum value of $\tau$ is $2/3$ and $\tau^{-1}(2/3)$ is a self-similar Cantor set with dimension $1/2$. The partial self-similar structure of the Takagi function implies that the level sets are Cantor sets for a dense set of points in $[0,2/3]$. +This an many other fun facts are proved in The Takagi function: a survey. It's not hard at all to see what's going on by examining the even partial sums of the series that generates $\tau$: + +The image shows +$$\tau(x) = \sum_{k=0}^{n} \frac{1}{2^k}\varphi(2^kx)$$ +for the first few even values of $n$. Note that the top is always a set of segments. As we move from $n=2$ to $n=4$, the one segment is replaced by two smaller sub-segments. The partial self-similar structure of the object suggests that this pattern repeats and, in the limit, the top of the object is a Cantor set. +Again referring to the partial self-similar structure, there is a dense set of points in $[0,2/3]$ whose inverse image is a Cantor set. This is formalized in Lemma 3.3 of the linked paper. Specifically, +the portion of the graph of $\tau$ above the interval +$[k/2^{2m},(k+1)/2^2m]$ is a +similar copy of the full graph reduced by the factor +$1/2^{2m}$ and shifted up by $\tau(k/2^{2m})$.<|endoftext|> +TITLE: Deducing the closed form for pentagonal numbers +QUESTION [5 upvotes]: Consider the sequence: +$0,1,5,12,22,35,51,70,92,117,145,176,\ldots$ +Find both a recurrence and a closed form for this sequence. + +I've done some research and found out that the majority of these (except 0) are "pentagonal numbers." We haven't learned about these yet, but further research did allow me to find a formula: +$A_n = ((3n^2)-n)/2$ +That is the closed form for this sequence, correct? Could someone possibly lead me as to how to get to that equation? +In terms of a recurrence, I'm somewhat lost. + +REPLY [3 votes]: Let $Sa_n=a_{n+1}$, then we can compute from the table that $(S-1)^3a_n=0$. The solution for this Linear Recurrence Relation is of the form +$$ +a_n=bn^2+cn+d\tag{1} +$$ +Using the known values for $n=0,1,2$, we get +$$ +\begin{align} +0&=d\\ +1&=b+c+d\\ +5&=4b+2c+d +\end{align}\tag{2} +$$ +Solving $(2)$ and substituting into $(1)$ gives +$$ +\begin{align} +a_n +&=\frac32n^2-\frac12n\\ +&=\frac{(3n-1)n}2\tag{3} +\end{align} +$$<|endoftext|> +TITLE: Is the absolute Galois Group of $\Bbb Q$ countable? +QUESTION [7 upvotes]: Is $\text{Gal} (\overline{\Bbb Q}/\Bbb Q)$ countable or uncountable? It seems like it should be countable (because the algebraic closure of $\Bbb Q$ is countable and there are countably many permutations of the irrational algebraic numbers, and a countable union of countable sets is countable. However, I've seen references that seem to imply it is uncountable. What is the answer, and why/how? + +REPLY [4 votes]: Let me elaborate on Lubin's comment: there are no countably infinite Galois groups. This follows from a general statement about topological spaces: If $X$ is compact, Hausdorff with no isolated points, then $X$ is uncountable. For the proof, see here: https://proofwiki.org/wiki/Compact_Hausdorff_Space_with_no_Isolated_Points_is_Uncountable/Lemma<|endoftext|> +TITLE: $n$-th derivative of Beta function +QUESTION [6 upvotes]: We pretty much know nothing about the high order derivatives of the Beta function. Well, we known for the example some recursive formulae for $\Gamma^{(n)}(1)$ as well as $\Gamma^{(n)}\left(\frac{1}{2}\right)$ but that is all. Yesterday , I encountered the following integral: +$$\int_{0}^{\infty} \frac{\log^n x}{\left ( 1+x^2 \right )^m}\, {\rm d}x$$ +As it can easily be seen , we have that: +$$\int_{0}^{\infty} \frac{\log x}{\left ( 1+x^2 \right )^m}\, {\rm d}x ={\rm B}^{(1)} \left ( \frac{1}{2}, m- \frac{1}{2} \right ) \implies \int_{0}^{\infty} \frac{\log^n x}{\left ( 1+x^2 \right )^m}\, {\rm d}x = {\rm B}^{(n)} \left ( \frac{1}{2}, m- \frac{1}{2} \right )$$ +The question is whether or not we can express that last $n$-th derivative in a closed form - most likely in an inductive form. I guess it will be difficult but with mathematics you never know. + +REPLY [2 votes]: Let us denote with +$$f(n, m)=\int_0^\infty \frac{\log^n x}{(1+x^2)^m}\, {\rm d}x$$ +It is quite straigtforward to note that: +$$f(n, 1)=\int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x = \left\{\begin{matrix} +0 &, &{\rm odd} \\ + 2n! \beta(n+1)&, &{\rm even} +\end{matrix}\right.$$ +Here $\beta$ denotes the Beta dirichlet function. +On the other hand +$$f(1,m)= \int_{0}^{\infty}\frac{\log x}{\left ( 1+x^2 \right )^m}= {\rm B}^{(1)} \left ( \frac{1}{2}, m - \frac{1}{2} \right )= \frac{\Gamma \left ( \frac{1}{2} \right ) \Gamma \left ( m- \frac{1}{2} \right ) \bigg[ \psi^{(0)}\left ( \frac{1}{2}\right) - \psi^{(0)} \left ( m- \frac{1}{2} \right ) \bigg]}{m^2 \Gamma(m)}$$ +Using gammatester's comment for the more general result, that is for $f(n,m)$ using Liebniz's general rule we have that: +$$f(n, m)= \int_{0}^{\infty} \frac{\log^n x}{\left ( 1+x^2 \right )^m}\, {\rm d}x = {\rm B}^{(n)} \left ( \frac{1}{2}, m- \frac{1}{2} \right )= \frac{1}{m^2 \Gamma(m)}\sum_{k=0}^{n} \binom{n}{k} \Gamma^{(k)} \left ( \frac{1}{2} \right )\Gamma^{(n-k)} \left ( m- \frac{1}{2} \right )$$ +Of course for $\Gamma^{(n)} \left(\frac{1}{2} \right)$ we have the following reduction formula: +$$\Gamma^{(n+1)}\left ( \frac{1}{2} \right )= -\left ( \gamma +2 \log 2 \right ) \Gamma^{(n)} \left ( \frac{1}{2} \right ) +n ! \sum_{k=1}^{n}\frac{(-1)^{k+1}}{\left ( n-k \right )!} \left ( 2^{k+1}-1 \right )\zeta(k+1) \Gamma^{(n-k)}\left ( \frac{1}{2} \right )$$<|endoftext|> +TITLE: $100$-th derivative of the function $f(x)=e^{x}\cos(x)$ +QUESTION [68 upvotes]: I've got this task I'm not able to solve. So i need to find the 100-th derivative of $$f(x)=e^{x}\cos(x)$$ where $x=\pi$. +I've tried using Leibniz's formula but it got me nowhere, induction doesn't seem to help either, so if you could just give me a hint, I'd be very grateful. +Many thanks! + +REPLY [2 votes]: I'm a little late to the game, but you can check your work in Wolfram Alpha with this query: +D[E^x Cos[x], {x, 100}] at x = pi + +which matches the answer posted by @choco_addicted. The syntax comes from the Wolfram Language page for the D function.<|endoftext|> +TITLE: $2^n + 3^n = x^p$ has no solutions over the natural numbers +QUESTION [14 upvotes]: A few weeks ago, I was asked to prove that $2^n + 3^n = x^2$ has no solutions over the positive integers. My proof was: +$2^n + 3^n \equiv (-1)^n \equiv \pm 1 \mod{3}\\\text{However, quadratic residue mod 3 is always 0 or 1. Therefore:}\\n = 2m\\\text{Now, working mod 5:}\\2^{2m} + 3^{2m} \equiv 4^m + 9^m \equiv 2 \times(-1)^m \equiv \pm 2 \mod{5}\\\text{However, quadratic residue mod 5 is }0, \pm 1\\\text{As such, } 2^n + 3^n \text{ is never a perfect square.}$ +Naturally, I questioned if the $2^n+ 3^n$ is ever a perfect cube, which a friend and I managed to prove was false, by working mod 7 and mod 9, and using Fermat's Last Theorem. This led us to consider the equation $2^n + 3^n = x^p,\,(n,x,p)\in \mathbb{N}^3,\,p\geq2$ and conjectured that there are no solutions. Obviously, if $p$ is composite and there are no solutions for any one of its factors then there are no solutions for p (e.g. $p = 4$ has no solutions because $x^4 = (x^2)^2$, so it is only necessary to consider the cases where $p$ is prime. +$p = 5$ was next, and we managed to partially prove it by working mod 11 and 25, and again, Fermat's Last Theorem, leaving only the case $n\equiv 6 \mod{10}$. +So my question is: is $2^n + 3^n$ ever a perfect power, and if not, how can this statement be proved? Alternatively, if this statement is false, are there any strategies other than using a computer to find a counterexample? + +REPLY [5 votes]: I'll provide a (relatively) simple proof (with the hard work hidden!) tailored for this problem - a similar argument works for more general equations such as, for example, +$$ +2^m \pm 3^n = x^p. +$$ +Suppose we have a solution to the equation +$$ +2^n+3^n=x^p, +$$ +where $n$ and $x$ are positive integers and $p$ is prime. Then, as noted by the proposer, we may, through congruence arguments, assume that $p \geq 5$. We may also suppose that $n \geq 5$. Consider the "Frey" curves +$$ +E_1 \; \; : Y^2 = X (X-3^n) (X+2^n), +$$ +if $n$ is odd, and +$$ +E_2 \; \; : Y^2 = X (X+x^p) (X+2^n), +$$ +if $n$ is even (we consider these two cases separately to get the smallest conductor we can). Then it can be shown (by Tate's algorithm) that each curve has conductor +$$ +N_{E_i} = 6 \prod_{q \mid x} q, +$$ +where the product is over prime $q$. Since we suppose that $p \geq 5$ (which guarantees that certain Galois representations that I haven't defined are irreducible), we may appeal to Ribet's level lowering results to conclude that these Frey curves correspond to certain weight $2$ modular forms of level $6$. This is a contradiction since the space of such forms turns out to be empty.<|endoftext|> +TITLE: Is a directed set countable, if for each element there are only finitely many smaller ones? +QUESTION [11 upvotes]: A directed set is a pair $(A,\leq)$ where $\leq$ is a reflexive, transitive relation such that for any $x,y\in A$ we have some $z$ such that $x,y\leq z$. (This comes up when dealing with categorical limits and topological nets). +In particular $(\mathbb{N},\leq)$ and $(\mathbb{R},\leq)$ are directed sets. +To help get comfortable with them, I imposed a "smallness" criteria: +Let's say a "finite-type" directed set is a directed set where every element has finitely many predecessors (smaller elements). + +My Guess: Finite-type directed sets are always countable. + +As before $(\mathbb{N},\leq)$ is an example, but now $(\mathbb{R},\leq)$ is too big and is a non-example. Another example is $(\mathbb{N}^2,\leq)$ where $(a,b)\leq (c,d)$ iff $(c,d)-(a,b)\in \mathbb{N}^2$ and it's higher dimensional analogues. However, I've personally been unable to equip $\mathbb{N}^\mathbb{N}$ with an appropriate finite-type directed set structure. +Is there a clean proof or counterexample regarding my guess? Or does this somehow end up touching upon foundational things such as the axiom of choice? + +REPLY [32 votes]: Let $X$ be any set, and let $A$ be the collection of finite subsets of $X$; $A$ is directed by $\subseteq$, and each member of $A$ has only finitely many predecessors in that order. However, if $X$ is infinite, then $|A|=|X|$, so the cardinality of $A$ can be as large as you like.<|endoftext|> +TITLE: Lebesgue measure of an intersection of a sequence of subsets +QUESTION [7 upvotes]: This is exercise 1.19 from "A User-Friendly Introduction to Lebesgue Measure and Integration" by Gail S. Nelson, and $m(E)$ is notation for Lebesgue measure of set $E$: +Let ${E_{k}}$ be a sequence of Lebesgue measurable sets with $E_{1}\supseteq E_{2}\supseteq E_{3}\supseteq\ldots$. Define the set $E$ to be $E=\cap_{k=1}^{\infty}E_{k}$. If $m(E_{1})<\infty$, show that $m(E)=\lim_{k\to\infty}m(E_{k})$. Show by example that this need not to be the case if we remove the assumption that $m(E_{1})<\infty$. +My attempt of proof: Since $E$ is subset of $E_{k}$ for each $k$, then $m(E)\leq m(E_{k})$ - this is according to a theorem stating that Lebesgue measure of a subset is less or equal than Lebesgue measure of its superset. Thus, it's certainly true that $m(E)\leq\lim_{k\to\infty}m(E_{k})$ too. If the limit is 0, then $m(E)$ is equal to the limit, as Lebesgue measure of any set is greater or equal to zero. However, now I don't see how to eventually prove that $m(E)$ cannot be less that the limit, in case when limit greater than 0. Also, for the second part of the question, I cannot find a counter-example for the case when $m(E_{1})=\infty$. + +REPLY [11 votes]: If $F_n:=E_n-E_{n+1}$ for $n=1,2,\dots$ then $E_k=E\cup\bigcup_{n=k}^{\infty}F_n$ and the sets in the union are disjoint, so that: +$$m(E_k)=m(E)+\sum_{n=k}^{\infty}m(F_n)$$ +So if $m(E_{k_0})<\infty$ for some $k_0$ then $m(E)<\infty$ and $\sum_{n=k_0}^{\infty}m(F_n)<\infty$ and consequently: $$\lim_{k\to\infty}m(E_k)=\lim_{k\to\infty}\left[m(E)+\sum_{n=k}^{\infty}m(F_n)\right]=m(E)+\lim_{k\to\infty}\sum_{n=k}^{\infty}m(F_n)=m(E)$$ + +Counterexample will only work if $m(E_n)=\infty$ for every $n$. You could take $E_n=[n,\infty)$.<|endoftext|> +TITLE: What exactly is an embedding in differential geometry? +QUESTION [7 upvotes]: I have seen several definitions of embedding in differential geometry. Let $f:M \to N$ be an injective smooth map between manifolds. Then I have heard that $f$ is an embedding if: + +$f_*$ is injective, that is, $f$ is an immersion +for any $p \in M$ there exists a neighbourhood $U$ of $p$ with $f^{-1}:f(U) \to M$ smooth +$f$ is diffeomorphic onto its image +$f$ is an immersion which is also homeomorphic onto its image + +My question is: are all these definitions equivalent, are some of them slightly incorrect, or are different definitions relevant for different purposes? Thanks. + +REPLY [6 votes]: The definition that I am familiar with is the last definition. For an embedding, the two properties give different ideas: + +A bijective immersion means that there is some topology that can be placed on the image so that the image is homeomorphic to the original manifold. +The condition of being homeomorphic with its image using the subspace topology says that the topology in the bullet above is actually the subspace topology (not some other weird one that differs from $N$'s topology). + +Now, some of what you write may be equivalent (certainly not the first one), and the diffeomorphism in the third one needs to be from the subspace topology. In general, an embedding means that the image is a regular submanifold of the codomain, so $f(M)$ looks like $M$ and also looks like a submanifold of $N$.<|endoftext|> +TITLE: possible norms of prime ideals in the class group of $K=\mathbb{Q}(\sqrt{-21})$ +QUESTION [6 upvotes]: I have an example in my notes where we try to compute the class group of the quadratic field $K=\mathbb{Q}(\sqrt{-21})$. My notes then proceed to evaluate the Minkowsk's bound< $\lambda(\sqrt{-21})$ which turns out to be 6. Therefore we know every ideal class has as a representative a prime ideal with a norm at most 6. My notes then claim that the only possible norms for those prime ideals must be 2 3 and 5. It's not clear to me why this should be true. I am aware that the norm of a prime ideal must be a prime power so I guess $4=2^2$ is a possible representative as well. Am I missing out something here does it have something to do with the fact that 2 is ramified in $\mathcal{O}_K$? +Thanks in advance + +REPLY [3 votes]: You're right that prime ideals can have prime powers for norms, but if a prime in $\mathbb Z$ does not ramify or split in an extension, its corresponding ideal is irrelevant to the class number computation. +I ask you to consider for a minute the case of 2 in $K = \mathbb Q(\sqrt{-35})$. The ideal $\langle 2 \rangle$ might be divisible by $\langle 2, 1 + \sqrt{-35} \rangle$. However, assigning $$\theta = \frac{1}{2} + \frac{\sqrt{-35}}{2},$$ which is an algebraic integer on account of the polynomial $x^2 - x + 9$, we readily see that $1 + \sqrt{-35} = 2 \theta$. Although $2 \nmid \theta$, obviously 2 does divide 2. +This suggests that 2 is prime, and not merely irreducible, in this domain. Of course $\mathcal O_{\mathbb Q(\sqrt{-35})}$ has class number greater than 1, since, for example, $\langle 3 \rangle = \langle 3, 1 - \sqrt{-35} \rangle \langle 3, 1 + \sqrt{-35} \rangle$. The norm of 2 is unnecessary for making this determination. +Now getting back to $\mathcal O_{\mathbb Q(\sqrt{-21})}$ and reassigning $$\theta = \frac{1}{2} + \frac{\sqrt{-21}}{2},$$ we see that $\theta$ is an algebraic number but not an algebraic integer since the relevant polynomial is $2x^2 - 2x + 11$. Thus $2 \mid (1 - \sqrt{-21})(1 + \sqrt{-21})$ but 2 divides neither of $1 \pm \sqrt{-21}$. +And since $(1 - \sqrt{-21}) \in \langle 2, 1 + \sqrt{-21} \rangle$ (and likewise $(1 + \sqrt{-21}) \in \langle 2, 1 - \sqrt{-21} \rangle$), we conclude $\langle 2 \rangle = \langle 2, 1 + \sqrt{-21} \rangle^2$, and hence $\mathcal O_{\mathbb Q(\sqrt{-21})}$ is not a prinicpal ideal domain and must have class number greater than 1. +It is easy to see that 3 ramifies with a corresponding non-principal ideal, and 25 has two distinct factorizations, but the norm of $\langle 5, 2 \pm \sqrt{-21} \rangle$ is still 5, right? I'm asking, so correct me if I'm wrong. +So, once again, the norm of 2 has turned out to be irrelevant to the class number computation. I figure that in a biquadratic domain like for $K = \mathbb Q(\root 4 \of 5)$ this might be much more fun, since the primes under the bound might ramify or split at the intermediate rings but not all the way down. Or would that be up? Depends on how you want to visualize it, I suppose.<|endoftext|> +TITLE: Componentwise Convergence in $\mathbb R^n$ +QUESTION [7 upvotes]: I came across the following question while preparing an exercise for basic analysis: +Suppose $d$ is some arbitrary metric on $\mathbb R^N$ and $(x^n) \subset \mathbb R^N$ converges to $x\in \mathbb R^N$ with respect to $d$. +Question: Does $(x^n)$ converge component-wise with respect to the usual norm? (And if not: Are there simple counter-examples?) +Up to now, I know the basics: + +The case is solved if $d$ is induced by a norm, because all norms on $\mathbb R^N$ are equivalent and $(x^n)$ converges componentwise iff it converges with respect to any $p$-norm. +The reverse of my question is not true. If for example $d$ is the discrete metric or the TGV metric, convergence in norm need not to imply convergence in metric. +If $\mathbb R^N$ is viewed as a topological space $(\mathbb R^N,\tau)$, then convergence with respect to $\tau$ need not to imply convergence in norm. (E.g. $\tau = \{\emptyset,\mathbb R^N\}$, which is not induced by a metric.) +All examples for metrics I considered so far (the usual examples) posess the desired property (and the proofs are easy, as these metrics are defined by using componentwise distances $|x_k-y_k|$). One could say, these metrics are just too normy, but I don't know, if there are not so normy examples or if my question can only be solved in a more abstract way. + +My question arises naturally, so maybe there is a simple answer I overlooked up to now. Thank you in advance for your help. + +REPLY [5 votes]: The answer is no. For instance, let $g:\mathbb{R}^N\to\mathbb{R}^N$ be any bijection at all, and define $d(x,y)=d_0(g(x),g(y))$, where $d_0$ is the Euclidean metric. Then $(x^n)$ converges to $x$ with respect to $d$ iff $g(x^n)$ converges to $g(x)$ with respect to $d_0$. Writing $f=g^{-1}$, your question thus becomes: if $f:\mathbb{R}^N\to\mathbb{R}^N$ is a bijection, then does $x^n\to x$ imply $f(x^n)\to f(x)$ (all with respect to the usual topology)? That is, is every bijection $\mathbb{R}^N\to\mathbb{R}^N$ continuous? +The answer is obviously no, since (for instance) a bijection could swap two points and be the identity everywhere else.<|endoftext|> +TITLE: If $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = 0$, can $\sum_0^\infty a_n$ be rational? +QUESTION [9 upvotes]: If a nonzero sequence of rationals +$$a_0, a_1 \dots a_n$$ +"decays fast" in the sense that $\lim_{n \rightarrow \infty} a_{n+1}/a_n = 0$, can the series converge to a rational number? That is, can $\sum_0^\infty a_n$ be rational? +The motivation is the idea that $e$ is irrational in some sense because its series $\sum \frac{1}{n!}$ grows too slowly. I was wondering if there was a way to make this precise. +(This is a restatement of this question, with thanks to Barry Cipra!) + +REPLY [3 votes]: Second attempt at a solution: +Yes, example: +\begin{eqnarray} a_{n} = \frac{1}{(2)_{n}} - \frac{1}{n!} \end{eqnarray} +where the Pochhammer symbol is $(\beta)_{n} = \beta(\beta+1) \cdots (\beta+n-1)$. Then, +\begin{eqnarray} \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=0 \text{ and } \sum_{n=0}^{\infty} a_{n} = -1. \end{eqnarray}<|endoftext|> +TITLE: Non-monadic adjunction +QUESTION [6 upvotes]: Could someone give some examples of a non-monadic adjunctions please? Possibly explaining why they are not monadic and how they contradict the monadicity theorem? +Thanks! + +REPLY [3 votes]: This might not be the answer you are looking for, but it could give you a little insight on what monadic really means. +Given a monad $(T,\eta,\mu)$ on a category $\mathcal C$, you can form the category $\operatorname{Adj}(\mathcal C,T)$ of adjunctions above $T$: + +its objects are the ajdunction $F: \mathcal C \rightleftarrows \mathcal D :U$ ($F$ on the left) such that the induced monad $(UF,{\rm unit},U\,{\rm counit}\,F)$ equals $(T,\eta,\mu)$ ; +a morphism from $F: \mathcal C \rightleftarrows \mathcal D :U$ to $F': \mathcal C \rightleftarrows \mathcal D' :U'$ is a functor $K \colon \mathcal D \to \mathcal D'$ such that $KF = F'$, $U'K = U$ and $K$ commutes with counits. + +This category $\operatorname{Adj}(\mathcal C,T)$ has a terminal object: the Eilenberg-Moore category of $T$. The unique morphism from any other object is the so called comparison functor. Hence, by definition, a adjunction $F: \mathcal C \rightleftarrows \mathcal D :U$ is monadic if and only if it is terminal in $\operatorname{Adj}(\mathcal C,UF)$. +It should make you realize that, in a sense, most adjunctions are non-monadic. For example, the previous category $\operatorname{Adj}(\mathcal C,T)$ always have an initial object: the Kleisli category of $T$. And this category is terminal only if every algebra over $T$ is free! Which is quite uncommon. It produces a lot of examples of non-monadic adjunctions $$Set \rightleftarrows \text{Free algrebas over T}$$ +for $T$ the monad of groups, monoids, modules over a non-field ring, etc. + +Note that the initial and terminal object can be the same (as in the example of Captain Lama), but still leaving some "room" for non monadic adjunctions in between.<|endoftext|> +TITLE: Catalan numbers formula derivation +QUESTION [7 upvotes]: I'm trying to follow a proof of the Catalan numbers being equal to $\frac{1}{n+1} {2n \choose n}$ from the recurrence relation $C_n = C_0C_{n-1}+C_1C_{n-2}+...+C_{n-2}C_{1}+C_{n-1}C_0$ +Now it's seen that the generating function satisfies $xf^2-f+1=0$ so $f=\frac{1-\sqrt{1-4x}}{2x}$ since the other root has a pole at 0 but I'm struggling to see how $f=\frac{1-\sqrt{1-4x}}{2x}$ can be expanded to obtain the necessary power series +I've seen Wikipedia's proof but I don't see how ${ \frac{1}{2} \choose n} = \frac{(-1)^{n+1}}{4^n(2n-1)} {2n \choose n}$ nor how this gets lost into ${2n \choose n}$ via plugging y=-4x and putting it into the expression $f=\frac{1-\sqrt{1-4x}}{2x}$ + +REPLY [3 votes]: Here as supplement to the answer of @MarcusM we show the validity of the binomial identity +\begin{align*} +\binom{\frac{1}{2}}{n}=\frac{(-1)^{n+1}}{4^n(2n-1)} \binom{2n}{n} +\end{align*} +In the following we use as definition of the binomial coefficient +\begin{align*} +\binom{\alpha}{n}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!} +\end{align*} + +We obtain + \begin{align*} +\binom{\frac{1}{2}}{n} +&=\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\cdots\left(\frac{1}{2}-n+1\right)}{n!}\\ +&=\frac{1(-1)(-3)\cdots(3-2n)}{2^nn!}\tag{1}\\ +&=\frac{(-1)^{n-1}}{2^nn!}(2n-3)!!\tag{2}\\ +&=\frac{(-1)^{n-1}}{2^nn!(2n-1)}(2n-1)!!\tag{3}\\ +&=\frac{(-1)^{n-1}}{2^nn!(2n-1)}\cdot\frac{(2n)!}{(2n)!!}\tag{4}\\ +&=\frac{(-1)^{n-1}}{2^nn!(2n-1)}\cdot\frac{(2n)!}{2^nn!}\tag{5}\\ +&=\frac{(-1)^{n+1}}{4^n(2n-1)}\cdot\frac{(2n)!}{n!n!}\tag{6}\\ +&=\frac{(-1)^{n+1}}{4^n(2n-1)}\binom{2n}{n} +\end{align*} + and the claim follows. + +Comment: + +In (1) we factor out $2^n$ +In (2) we factor out $(-1)^n$ use as convenient notation double factorial +\begin{align*} +(2n)!!&=(2n)(2n-2)(2n-4)\cdots 4\cdot 2\\ +(2n-1)!!&=(2n-1)(2n-3)(2n-5)\cdots 3\cdot 1\\ +\end{align*} +In (3) we multiply the expression with $\frac{2n-1}{2n-1}$ and use $(2n-3)!!(2n-1)=(2n-1)!!$ +In (4) we use the identity +\begin{align*} +(2n)!=(2n)!!(2n-1)!! +\end{align*} +In (5) we use the identity +\begin{align*} +(2n)!!=(2n)(2n-2)\cdots4\cdot2=2^n n! +\end{align*} +In (6) we do a small rearrangement<|endoftext|> +TITLE: What is the difference between |(d/dt)velocity| and (d/dt)|velocity| [Physical Interpretation] +QUESTION [7 upvotes]: According to me the modulus of $\frac {\text{d}v}{\text{d}t}$ gives the magnitude of acceleration without accounting for direction but I am stumbled on how to interpret $\left(\frac{\text{d}}{\text{d}t}\right)|velocity|$ + +REPLY [2 votes]: If we are dealing with motion in one dimension, then the velocity function is a scalar (i.e. $v: \mathbb{R} \to \mathbb{R}$), and it makes sense to talk about "positive velocity" and "negative velocity" (and likewise for acceleration). I'm going to assume that in what follows. +First, it may be helpful to know that $\frac{d}{dt} |t| = sgn(t)$, where $sgn(t)$ is the signum function: $sgn(t) = 1$ for $t>0$, $sgn(t) = -1$ for $t<0$, and $sgn(0)$ is undefined. You can also write this as $\frac{d}{dt} |t| = \frac{|t|}{t}$ for $ t\ne 0$. +By the chain rule, then, +$$\frac{d}{dt} |v(t)| = sgn(v(t)) \space a(t)$$ +That is, $\frac{d}{dt} |v(t)| $ tells you the acceleration times the sign of the velocity. +There are four cases to consider: + +If an object has positive velocity and positive acceleration, $\frac{d}{dt} |v(t)| $ is positive (and is equal to the acceleration). +If an object has positive velocity but negative acceleration, $\frac{d}{dt} |v(t)| $ is negative (and is equal to the acceleration, but not equal to the magnitude of the acceleration, which is of course positive). +If an object has negative velocity but positive acceleration, +$\frac{d}{dt} |v(t)| $ is negative (and opposite in sign to both the acceleration and the magnitude of the acceleration, which would be positive and equal to each other). +If an object has negative velocity and negative acceleration, $\frac{d}{dt} |v(t)| $ is positive (and is equal to the magnitude of the acceleration). + +Of the four cases above, in only the first and last is $\frac{d}{dt} |v(t)| $ equal to $|\frac{d}{dt} v(t) | $.<|endoftext|> +TITLE: $\sum \frac{a_n}{\ln a_n}$ converges $\implies \sum \frac{a_n}{\ln (1+n)}$ converges +QUESTION [10 upvotes]: Let $0 0$: so that $\ln\frac{1}{\tau_n} = \alpha\ln(n+1).$ +(Stop now if you didn't want the full solution, but only a hint.) + +In detail: As we'll see, any $\alpha > 1$ will do -- below, we set $\alpha \stackrel{\rm def}{=}2$. +Write +$$ +\frac{a_n}{\ln(n+1)} += \frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n < \frac{1}{(n+1)^2} \right\}} ++ \frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} +$$ +For the first term, we can write $$\frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n < \frac{1}{(n+1)^2} \right\}}\leq \frac{1}{(n+1)^2\ln(n+1)} \mathbb{1}_{\left\{ a_n < \frac{1}{(n+1)^2} \right\}} \leq \frac{1}{(n+1)^2\ln(n+1)}$$ +For the second, +$$\begin{align} +\frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} +&= +\frac{a_n}{\frac{1}{2}\ln((n+1)^2)} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} +\leq \frac{a_n}{\frac{1}{2}\ln \frac{1}{a_n}} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} \\ +&\leq \frac{2a_n}{\ln \frac{1}{a_n}} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} +\leq \frac{2a_n}{\ln \frac{1}{a_n}}. +\end{align}$$ +Together, we get +$$ +0 < \frac{a_n}{\ln(n+1)} \leq \frac{1}{(n+1)^2\ln(n+1)} + 2\frac{a_n}{\ln \frac{1}{a_n}} +$$ +which allows you to conclude by theorems of comparison.<|endoftext|> +TITLE: Plane tesselation, using stairs $n\times n$, is it possible? +QUESTION [6 upvotes]: The other day I was constructing new mathematical problems for my pupils and thought of something like this: + +Given the infinite sequence of "stairs" $n\times n$, constructed from $1\times1$ squares: +$\hspace{130pt}$ $\cdots$ +tesselate whole plane with it, using $\underline{\text{every}}$ "stair" and $\underline{\text{only once}}$. + +Unfortunately, neither I could prove it, nor my pupils. If you know a solution, please post it here, or any reference link will be much applied. But actually, if we weaken either of two initial conditions it becomes very easy. Let me demonstrate: +1. If we are allowed to use every stair, but not only once, then the solution comes from the idea of merging every $(2k-1)\times(2k-1)$ stair with $2k\times2k$ stair into square $2k\times2k$ like this: +$\hspace{110pt}$ $\cdots$ +Our plane tesselation is constructed like this: we stack $2\times2$ squares in an infinte vertical strip (roughly speaking $2\times\infty$), then to the right of it we put $4\times\infty$ infinte strip of stacked $4\times4$ squares, to the left we put $6\times\infty$, then to the right we put $8\times\infty$, then again to the left we put $10\times\infty$ and so on. Plane tesselation complete! +2. If instead we are allowed to use only once, but not every stair, then the solution comes from idea of squaring the square: just like in previous solution, merging every $(2k-1)\times(2k-1)$ stair with $2k\times2k$ stair give us a sequence of squares $2k\times2k$. +Plane tesselation constructed like this: we join some squares to make a bigger square $S_1$ with the help of squaring the square method, then we make new huge square $S_2$, consisting of obtained square $S_1$ as a smallest one, and others to complete construction and so on. +To point it out, we can solve the weaken problem, but is there a solution for a stronger one with both conditions kept? If you have an answer, post it here please, or at least give me a link. + +REPLY [4 votes]: Yes; the paper at http://www.maa.org/sites/default/files/pdf/pubs/SquaringThePlane.pdf shows that one can tile the plane with one of each size of square, exactly once. Thus, by doubling the side length of each square, one can do the same for one of each size of square with even side length. +One can create one square of each even side length by putting two consecutive stairs together, as explained in the post.<|endoftext|> +TITLE: How do I transpose an ellipse function to stretch the ellipse into curved space? +QUESTION [14 upvotes]: I'm working on an engineering project, using CAD software. +I can write simple parametric functions to draw an ellipse, with $\theta$ ranging from $0$ to $2\pi$ radians:$$x=3.75\cos\theta$$$$y=1.25\sin\theta$$ +But now I'm at a point where I need to bend the ellipse over a parabolic curve. +Here's a diagram demonstrating the curved space I'd like to bend/stretch my ellipse into. + +The white bottom-most curve is the parabolic curve, defined by the function: +$$y=-\frac{x(x+30)}{45}$$ +The other two green curves are defined as parallel (equidistant) to the parabola at all points, matching the flat rectangles above. +As you can see from the diagram, the vertical minor axis for the ellipse becomes slanted, making the top of the ellipse "longer" than the bottom of the ellipse - which follows the stretching of the enclosing space. +But the ellipse should still touch the top and bottom of the slanted minor axis, as well as touching the left-hand end of the center green curve, which is the transformed major axis for the ellipse. +So the question is, how do I transpose my simple ellipse functions to stretch the ellipse into this curved space? +In your answers, feel free to place the origin wherever is most convenient for your solution. I'll have to translate the solutions anyway, to move them into their proper place in my engineering diagrams. :) +Edit: I've been asked whether polar coordinates would be acceptable, rather than Cartesian. My CAD software does support polar coordinates, so for anyone considering answering this question, feel free to build your answers with polar coordinates if that would make things easier for you. + +REPLY [3 votes]: Others have mentioned that parameterizing a parabola by arc length is not possible with elementary functions. This is true, but it's not necessary. +First, we write the function for an ellipse centered at $(a,b)$ with semi-axes $a$ and $b$, respectively. +$$\frac{(x-a)^2}{a^2} + \frac{(y-b)^2}{b^2} = 1$$ +$$y = b \pm |b|\sqrt{1-\frac{(x-a)^2}{a^2}}.$$ +Instead of $x$ and $y$, we want the perpendicular height above the parabola, $h$, in terms of the distance along the parabola, $s$. +$$h = b \pm |b|\sqrt{1-\frac{(s-a)^2}{a^2}}.$$ +The arc length of the parabola $y = kx^2$ as a function of $x$ is given by +$$s(x) = \int_0^x\sqrt{1 + 4k^2x'^2}dx' = \frac{1}{2}x\sqrt{1+4k^2x^2} + \frac{1}{4k}\ln\left(\sqrt{1+4k^2x^2}+2kx\right).$$ +Next, we need the unit vector perpendicular to the tangent of the parabola (kindly provided by @Wouter), +$$\vec{n} = \frac{(-2kx, 1)}{\sqrt{4k^2 x^2 + 1}}.$$ +This will allow us to parameterize the ellipse as +$$(x_e, y_e) = (t, kt^2) + \vec{n}(t)h(t).$$ +The first part $(t, kt^2)$, is a point on the parabola. Adding the second part $\vec{n}(t)h(t)$ results in a point on the upper or lower half of the ellipse. I've switched $x$ for $t$ for clarity. Even though $x(t) = t$ in the parabola, the corresponding $x$-coordinate on the ellipse will be different. +So, the transformed ellipse is parameterized by +$$(x_e, y_e) = (t, kt^2) + \frac{(-2kt, 1)}{\sqrt{1+4k^2t^2}}\left(b \pm |b|\sqrt{1-\frac{(s(t)-a)^2}{a^2}}\right).$$ +For the full ugliness, let's substitute in $s(x)$: +$$(x_e, y_e) = (t, kt^2) + (-2kt, 1)\frac{1}{\sqrt{1+4k^2t^2}}\left(b \pm |b|\sqrt{1-\frac{\left(\frac{1}{2}t\sqrt{1+4k^2t^2} + \frac{1}{4k}\ln\left(\sqrt{1+4k^2t^2}+2kt\right)-a\right)^2}{a^2}}\right)$$ +with $t \in [0, \text{~}7.4]$ The value 7.4 comes from a guess-and-check numerical approximation that solves for $t$ so that the part inside the parentheses is non-negative (i.e., $0 \leq s(t) \leq 2a$). +Slightly easier to read (maybe): +$$x(t) = t + \frac{-2kt}{\sqrt{1+4k^2t^2}}\left(b \pm |b|\sqrt{1-\frac{\left(\frac{1}{2}t\sqrt{1+4k^2t^2} + \frac{1}{4k}\ln\left(\sqrt{1+4k^2t^2}+2kt\right)-a\right)^2}{a^2}}\right)$$ +$$y(t) = kt^2 + \frac{1}{\sqrt{1+4k^2t^2}}\left(b \pm |b|\sqrt{1-\frac{\left(\frac{1}{2}t\sqrt{1+4k^2t^2} + \frac{1}{4k}\ln\left(\sqrt{1+4k^2t^2}+2kt\right)-a\right)^2}{a^2}}\right)$$ +With $a = 3.75$, $b = 1.25$, and $k = -\frac{1}{45}$ (moving the vertex of the parabola to the origin), the graph looks like this: https://www.desmos.com/calculator/m8kxmdzbao<|endoftext|> +TITLE: When is $\mathbb{Z} [x]/f(x) $ a Dedekind domain? +QUESTION [5 upvotes]: Given a monic separable irreducible polynomial $f$ with integer coefficients, when $\mathbb{Z} [x]/f(x)$ is a Dedekind domain? And when it happens to be a Dedekind domain, how to know its class number? +The question is too broad. But do we have some useful criterion if we impose more conditions on $f$? + +REPLY [3 votes]: Edit (2016-05-05): I feel I should return to this question to give a better answer. What I had initially written (below the line) is, as far as I know, correct, but it's far from the best way of saying things. An excellent place for clear and detailed explanations on this question is Peter Stevenhagen's paper "The Arithmetic of Number Rings" in Algorithmic Number Theory: Lattices, Number Fields, Curves and Cryptography (Buhler & Stevenhagen eds., MSRI publications 44, Cambridge 2008): Stevenhagen discusses both the question of computing integral closure and that of computing the class number. But for the completeness of MSE, let just restate two things I might have said. +I explain below why $\mathbb{Z}[x]/(f)$ is normal (i.e., integrally closed, i.e., a Dedekind domain) iff $\mathbb{Z}_p[x]/(f)$ is normal (i.e., a discrete valuation ring — or perhaps the zero ring) for every $p$. I point out that this is automatically true for primes which don't divide the discriminant (I wrote "or the leading coefficient", that's redundant), so we need only worry about primes which do. I should have added: in fact, we need only worry about primes whose square divides the discriminant, because those that do not are automatically regular. So in particular, if the discriminant of $f$ is not a square, then $\mathbb{Z}[x]/(f)$ is normal. +The second thing I missed is part of the "Kummer-Dedekind" theorem. To state this, assume $f$ is monic (I didn't think too carefully about what happens if it is not, but essentially some primes disappear "at infinity"). Let $\bar f = \prod_{i=1}^r \varphi_i^{e_i} \in \mathbb{F}_p[x]$ be the factorization of $f$ mod $p$, where $\varphi_i$ are the distinct monic irreductible factors of $\bar f$, and let $g_i \in \mathbb{Z}[x]$ be an arbitrary monic lift of $\varphi_i \in \mathbb{F}_p[x]$; finally, let $f = q_i g_i + r_i$ be the division of $f$ by $g_i$ in $\mathbb{Q}[x]$ (which is, in fact, in $\mathbb{Z}[x]$ because $g_i$ is monic). Then: $\mathbb{Z}_p[x]/(f)$ is normal iff $e_i = 1$ or $r_i$ is not divisible by $p^2$ (in $\mathbb{Z}[x]$). +(Furthermore, if such is the case, then $(p)$ splits as the product of the prime ideals $(p,\;g_i\mathbin{\mathrm{mod}}f)$ with exponents $e_i$; and if such is not the case, then $\frac{1}{p}\cdot(q_i\mathbin{\mathrm{mod}}f)$ is an element of the field of fractions of $\mathbb{Z}[x]/(f)$ which is integral over the latter but does not belong to it.) +Again, at some deeper level this is equivalent to what I was saying below, but it's much clearer that way, and simple enough to test by hand for not-too-complicated polynomials $f$. + +The class number is a completely different question, so let me concentrate just on how to decide whether $\mathbb{Z}[x]/(f)$ is integrally closed (i.e., a Dedekind domain). +First, normality is a local condition: $\mathbb{Z}[x]/(f)$ is normal (i.e., integrally closed) iff $\mathbb{Z}_{(p)}[x]/(f)$ is normal for every prime $p$, where $\mathbb{Z}_{(p)}$ refers to the localization of $\mathbb{Z}$ at $p$ (i.e., the ring of rationals such that $p$ does not divide the denominator). Second, normality passes to the completion (the words "excellent ring" are a magic phrase here): we can, if we wish, replace $\mathbb{Z}_{(p)}[x]/(f)$ by $\mathbb{Z}_p[x]/(f)$ in the above. +Also, we don't need to test every prime $p$, only the finitely many which divide the discriminant of $f$ or its leading coefficient: the other primes are unramified and necessarily normal (see, e.g., Serre, Local Fields, I, §6, (i), prop. 15 and cor. 1 and 2, which is also relevant in what follows) (these unramified primes appear as $e=1$ in what follows). +So now let's concentrate on a single prime $p$. +Suppose $f$ for the time being that $f$ is monic (now that we're at a single prime $p$, it's not such a strong condition: we just want the leading coefficient not to be a multiple of $p$). If $\bar f = \prod_{i=1}^r \varphi_i^{e_i} \in \mathbb{F}_p[x]$ where $\varphi_i$ are the distinct irreductible factors of $\bar f$. By Hensel's lemma (since $\mathbb{Z}_p$ is complete) we can lift this factorization to one of $f = \prod_{i=1}^r f_i \in \mathbb{Z}_p[x]$ where $f_i$ reduces to $\varphi_i^{e_i}$ mod $p$, and the Chinese remainder theorem tells us that $\mathbb{Z}_p[x]/(f) = \prod_{i=1}^r \mathbb{Z}_p[x]/(f_i)$, which is normal iff every factor is, so we can assume that $f = f_i$ (i.e, $\bar f = \varphi^e$ where $\varphi\in\mathbb{F}_p[x]$ is irreducible and $e$ is the ramification). If $f$ isn't monic in the start of this paragraph, it should at least be primitive (i.e., gcd of coefficients $=1$), otherwise $\mathbb{Z}_p[x]/(f)$ certainly can't be integrally closed; even if it is primitive, some degree can be lost "at infinity": think of $f$ as a homogeneous polynomial in two variables instead, Henselize as such, and ignore the part at infinity. +So now we have $\mathcal{O} := \mathbb{Z}_p[x]/(f)$ which is a local noetherian domain of dimension $1$, whose maximal ideal $\mathfrak{P}=(p)+(\varphi)$ is the set of elements which are divisible by $\varphi$ after reduction mod $p$. So it is normal iff it is regular iff $\mathfrak{P}/\mathfrak{P}^2$ is $1$-dimensional as a vector space over $\mathcal{O}/\mathfrak{P} = \mathbb{F}_p[x]/(\varphi)$. At this stage, everything is quite explicit and it should be at least clear that the question is algorithmic (by doing linear algebra on $\mathbb{Z}_p$-modules, which is algorithmic); since every step thus far (factorization over finite fields, Hensel's lemma) was also algorithmic, and since I explained why we need to check only finitely many primes, the overall question is algorithmic. +A particularly important case is that when $\varphi = x$, in other words $f = x^e + a_1 x^{e-1} + \cdots + a_e$ with $a_i$ multiple of $p$ for $1\leq i \leq e$. Then $\mathfrak{P}$ consists of those $c_0 + c_1 \xi + \cdots + c_{e-1} \xi^{e-1}$ (where $\xi$ is the class of $x$ mod $f$) such that $c_0$ is a multiple of $p$, and $\mathfrak{P}^2$ is generated by $p^2$, $p\xi$ and $\xi^i$ for $2\leq i \leq e$. So the quotient $\mathfrak{P}/\mathfrak{P}^2$ is the $\mathbb{F}_p$-vector space of $(c_0,c_1)$ with $c_0 \in p\mathbb{Z}_p/p^2\mathbb{Z}_p$ and $c_1 \in \mathbb{Z}_p/p\mathbb{Z}_p$ quotiented out by the image of $\xi^e$ thither, which is $(-a_e,0)$. So it is $1$-dimensional iff $a_e$ is not a multiple of $p^2$, which is to say, $f$ is an Eisenstein polynomial. The general case will be similar, but more complicated to write down (if, instead of $\mathbb{Z}_p[x]/(f)$ we had a complete discrete valuation ring with an algebraically closed residue field, we could always reduce to the case just described).<|endoftext|> +TITLE: Easy partial fraction decomposition with complex numbers +QUESTION [6 upvotes]: There is an easy method to perform a partial fraction decomposition - described here, under the "Repeated Real Roots" title, for the coefficient A2. +The problem is - this method doesn't work in some cases, for example: +$$ +\frac{1}{(2 + j \omega)^2 \, (4 + j \omega)} + = \frac{A}{2 + j \omega} + + \frac{B}{(2 + j \omega)^2} + + \frac{C}{4 + j \omega} +$$ +The coefficients B,C get their proper value: +$$ +B = \frac{1}{4 + j \omega} \Big|_{\omega=2j} = \frac 12 +$$ +$$ +C = \frac{1}{(2 + j \omega)^2} \Big|_{\omega=4j} = \frac 14 +$$ +But the coefficient A get a false value: +$$ +A = \frac{d}{d \omega} \left( \frac{1}{4 + j \omega} \right)\Big|_{\omega=2j} + = \frac{-j}{4} +$$ +The true/proper value for A is $-\frac 14$. +I wonder what is the general formula for this kind of elegant/quick/easy trick to find the coefficients during partial fraction decomposition? + +REPLY [2 votes]: Here are the mechanics of the trick. Let's multiply your equation by $4+j\omega$: +$$ +\frac{1}{(2 + j \omega)^2 \, } + = \frac{(4 + j \omega)A}{2 + j \omega} + + \frac{(4 + j \omega)B}{(2 + j \omega)^2} + + C. +$$ +Now things nicely fall out if we look at $\omega = 4j$: +$$\frac{1}{(2-4)^2} = C.$$ +But ... the left side is just the original expression with the $(4+j\omega)$ "covered up" (since we multiplied everything by $4+j\omega$). Evaluating this covered-up expression at the root we covered up gives the coefficient. +Things don't all cancel away with a repeated root. Multiply by $2 + j\omega$: +$$ +\frac{1}{(2 + j \omega) \, (4 + j \omega)} + = A + + \frac{B}{(2 + j \omega)} + + \frac{(2 + j \omega)C}{4 + j \omega}. +$$ +There's still the $2 + j \omega$ in the denominator under $B$ so the "trick" fails and we have to find it some other way.<|endoftext|> +TITLE: Left-invariant vector fields on the circle $S^1$ +QUESTION [5 upvotes]: I'm trying to find the left-invariant vector fields on the circle $S^1$. + +If I understand correctly, $S^1$ is given the group structure of the multiplicative group of complex numbers on the unit circle in the complex plane, +and $S^1$ is made into a $C^\infty$ manifold by forming the maximal atlas from the four projection maps from open semi-circles of $S^1$ onto the $x$- and $y$-axes. +Given a point $p\in S^1$, if $p$ is on the upper or lower open semicircles, then a basis for $T_pS^1$ is given by $\{\partial/\partial \bar x \rvert_p\}$, where $\bar x$ is the projection onto the $x$-axis, +while if $p$ is on the left or right open semicircles, then a basis for $T_pS^1$ is given by $\{\partial/\partial \bar y \rvert_p\}$. +Given a vector field $X$ on $S^1$, $X$ is left-invariant provided that for all points $g,h\in S^1$, +$$ +(\ell_g)_{*,h}(X_h) = X_{gh} \text, +$$ +where $\ell_g\colon S^1 \to S^1$ is left-multiplication by $g$. +To obtain an arbitrary left-invariant vector field $X$ on $S^1$, I should be able to just pick an arbitrary tangent vector $A \in T_1S^1$ and see what it generates. +Let $A = a(\partial/\partial \bar y\rvert_1) \in T_1S^1$. +Then given a point $p\in S^1$, I can define +$$ +X_p = (\ell_p)_{*,1}(A). +$$ +However, I don't see how this is getting me any closer to concretely writing down what $X$ looks like. Any suggestions? + +REPLY [2 votes]: If you suppose that the circle $S^1$ is embedded in the complex plane $C$ you can represent an element of $S^1$ by $e^{it}, t\in R$. The tangent space $T_1S^1$ is the vertical line $x=0$. Let $(0,u)\in T_1S^1$, and $e^{it}\in S^1$, we denote by $l_{e^{it}}$ the left multiplication defined by $e^{it}$, thus you have $e^{it}(e^{iu})=e^{it}e^{iu}$. Remark that this map may be extended to $C$ in a linear map (the rotation of angle $t$), thus is equal to its differential, this implies that $l_{e^{it}}(0,u)=e^{it}(0,u)=(-u\sin(t),u\cos(t))$.<|endoftext|> +TITLE: Prove a polynomial identity +QUESTION [5 upvotes]: Define a sequence of polynomials in the following way: +$P_m(t)=\frac {1} {m!}\cdot t\cdot (t-1)\cdot...\cdot (t-m+1) $. +(Where $P_0(t)=1$). +I'm trying to prove the following identity: +$\frac d {dt} P_{m+1}(t) = \sum_{k=0}^{m} \frac {(-1)^{m-k}} {m-k+1} \cdot P_k(t)$ +Induction on $m$ doesn't seem to work here. +I've observed that if $f(x)=x^t$, then $P_m(t)=\frac {f^{(m)}(1)} {m!} $, but it didn't lead me anywhere. +Any ideas how to prove that? + +REPLY [3 votes]: Generating functions to the rescue! Your observation and Taylor series show us that +$$ +x^t = \sum_{m=0}^\infty P_m(t) (x-1)^m; +$$ +this converges uniformly for $\frac12\le x\le\frac32$ say and for $t$ in any bounded range. Therefore we may differentiate in $t$ term-by-term: +\begin{align*} +\sum_{m=0}^\infty \frac d{dt}P_m(t) (x-1)^m &= \frac d{dt} x^t \\ +&= x^t \log x \\ +&= \bigg( \sum_{m=0}^\infty P_m(t) (x-1)^m \bigg) \bigg( \sum_{m=1}^\infty \frac{(-1)^{m-1}}{m} (x-1)^m \bigg). +\end{align*} +Your identity then follows by comparing the coefficients of $(x-1)^{m+1}$ on both sides.<|endoftext|> +TITLE: Why is the group of unit upper triangular matrices solvable? +QUESTION [10 upvotes]: Let $GL_n(k)$ be the n by n general linear group over $k$, $B_n(k)$ be the subgroup of $GL_n(k)$ consisting of all upper triangular matrices, and $U_n(k)$ be the subgroup of $B_n(k)$ whose diagonal elements are all 1. +To show $B_n(k)$ is solvable, I'm proving it now by following steps: +1. $U_n(k)$ is a subgroup of $B_n(k)$. (done) +2. $U_n(k)$ is normal in $B_n(k)$. (done) +3. $U_n(k)$ is solvable. (question) +4. $B_n(k) / U_n(k)$ is also solvable. (not yet) +5. $B_n(k)$ is solvable. (by the below thm) +I'll use a theorem to verify $B_n(k)$ is solvable. +G is solvable if and only if H and G/H are solvable for some normal subgroup $H$ of $G$. +So, I have to prove both step3 and step4. But I have no idea about them. How to prove? Since my knowledge is not enough, I don't wanna show using Lie theory. +Thanks in advance. + +REPLY [15 votes]: You can prove $U_n$ solvable by induction on $n$. Consider $H_n$, the group of matrices of the form +$$ \begin{pmatrix} 1 & & & a_1 \\ & \ddots & & \vdots \\ & & 1 & a_{n-1} \\ & & & 1 \end{pmatrix} $$ +It is then easy to see that $H_n \triangleleft U_n$, that $H_n$ is abelian (in fact isomorphic to $k^{n-1}$), and that $U_n/H_n\cong U_{n-1}$.<|endoftext|> +TITLE: Diophantine equation: choosing the right modulus to prove an equation cannot be satisfied +QUESTION [7 upvotes]: I was looking at this problem, which asks to show that there are no $m,n \in \mathbb Z$ such that +$$3n^2+3n+7 = m^3.$$ +The result follows immediately from considering the equation modulo $9$ and enumerating the possibilities. But what if we somehow missed $9$ and instead considered other moduli? In other words, we would be aiming to find a $k \in \mathbb Z$ for which the sets $\{3n^2 + 3n + 7:n\in\mathbb Z\}$ and $\{m^3:m\in\mathbb Z\}$ are disjoint modulo $k$, thereby proving the impossibility of finding such $m,n$. +I've looked through all $k \in [2, 5000]$, and the winners are precisely the multiples of $9$. Why is this so? +The main question I have is this: + +Is it true that $\{3n^2 + 3n + 7:n\in\mathbb Z\}$ and $\{m^3:m\in\mathbb Z\}$ are disjoint modulo $k$ if and only if $k \equiv 0 \pmod 9$? + +REPLY [9 votes]: Here's a sketch. For all but finitely many primes $p$, your equation gives rise to an elliptic curve $\bmod p$. To see this very explicitly, rewrite it as +$$81n^2 + 81n + 27 \cdot 7 = 27m^3$$ +and substitute $z = 9n, x = 3m$ to get +$$z^2 + 9z + 27 \cdot 7 = x^3.$$ +Now complete the square to get +$$\left( z + \frac{9}{2} \right)^2 = x^3 - 27 \cdot 7 + \frac{81}{4} = x^3 - \frac{3^3 \cdot 5^2}{2^2}.$$ +Substituting $y = z + \frac{9}{2}$ produces an elliptic curve $\bmod p$ for all primes $p$ except possibly $p = 2, 3, 5$. +The Hasse-Weil bound implies that when $p \neq 2, 3, 5$ this curve has at least $p - 2 \sqrt{p}$ points on it, so for all primes $p \ge 7$ there is a solution $\bmod p$. (This is the least trivial step in the argument, and it really uses the fact that we have an elliptic curve and not just some polynomial.) Using Hensel's lemma will allow us to lift these to solutions $\bmod p^k$, and using the Chinese remainder theorem gives us solutions $\bmod n$ where $n$ isn't divisible by $2, 3, 5$. +Now the only moduli left to investigate are powers of $2, 3, 5$. For $2$ and $5$ you can find solutions by hand and use Hensel's lemma, which leaves only $3$.<|endoftext|> +TITLE: SVD - Decomposed Matrix Sizes +QUESTION [5 upvotes]: I had a question about SVD. Specifically about the size of matrices $U$, $\Sigma$ and $V$ decomposed from the $m\times n$ matrix $X$ using the formula +$$X = U \Sigma V^T $$ +Most of the the tutorial literature says that the resulting sizes are + +$U$ is $m \times m$ +$\Sigma$ is $m \times n$ +$V$ is $n \times n$ + +However, there have been quite few times when the sizes given are + +$U$ is $m \times n$ +$\Sigma$ is $n \times n$ +$V$ is $n \times n$ + +In other words instead of $\Sigma$ being the matrix with possibly different number of rows and columns, its $U$ with the different number of rows and columns. +The math works out, so why (and in what cases) is this less frequent version used? + +REPLY [2 votes]: Assume that $A$ is a "tall" matrix; that is $m \ge n$. Otherwise, we work with the transpose of $A$. The number of non-zero singular values, at most, is $n$. Consider your first version of the SVD which I call the full version. $A = U \Sigma V^t$. Often $m$ is large and $m >> n$ (such as in signal processing, statistics and data sciences). In that case, $U$ is a very large matrix. +The SVD can be written as +$$ +A = [U_1 \; U_2] \left[ \begin{array}{c} +\Sigma_1 \\ +O \end{array} \right] V^t \;\;\;,\;\;\; U=[U_1 \; U_2] +$$ +where $U_1$ is a $m \times n$ matrix, $U_2$ is a $m \times (m-n)$ and $\Sigma_1$ is a square diagonal matrix with entries in the non-increasing order. +Expanding the above, we get +$$ +A = U_1 \Sigma_1 V^t +$$ +which is sometimes called the "economical version". This is your second answer. Most of the time, $U_2$ is not required and hence the computation time and memory are saved by using the second version. However, note that $U_2^t A = 0$ (since $U_2^tU_1=0$) and if you require the left null space then you should compute $U_2$. However, in general, $U_2$ is not unique.<|endoftext|> +TITLE: $G_{\mathfrak a}(A)$ integral domain and $\bigcap \mathfrak a^n = 0$ implies $A$ is integral domain +QUESTION [5 upvotes]: This is Lemma 11.23 in Atiyah: + +For an ideal $\mathfrak a \subseteq A$, define $G_{\mathfrak a} (A) = \bigoplus _{n=0} ^\infty \mathfrak a^n / \mathfrak a^{n+1}$. + +The statement of the Lemma: + +Let $A$ be a ring, $\mathfrak a$ an ideal of $A$ such that $\bigcap_n \mathfrak a^n = 0$. Suppose that $G_{\mathfrak a} (A)$ is an integral domain. Then $A$ is an integral domain. +Proof: Let $x,y$ be nonzero elements of $A$. Then since $\bigcap \mathfrak a^n = 0$, there exist nonnegative integers $r,s$ such that $x \in \mathfrak a^r - \mathfrak a^{r+1}$ and $y \in \mathfrak a^s - \mathfrak a^{s+1}$. + +My question is why should the powers of $\mathfrak a$ cover everything in $A$ to begin with? Certainly if this were true, and the intersection of all the powers is zero, then the above follows. So it seems that this depends on something along the lines of + +$$\bigcup \mathfrak a^n = A.$$ + +Does it have to do with the $\mathfrak a$-adic topology where these powers of $\mathfrak a$ are the neighborhoods of $0$? + +REPLY [2 votes]: The numbers $r$ and $s$ are allowed to be $0$, in which case by convention we define $\mathfrak a^0=A$. This makes sense, because $\mathfrak a^0$ should be the ideal generated by all $0$-fold products of elements of $\mathfrak a$, and the only such product is the empty product, $1$. The ideal generated by $1$ is all of $A$. +Note that if you did not allow $r$ and $s$ to be $0$, this would almost always be false. Indeed, $\mathfrak a^n\subseteq a$ for all $n>0$ since $\mathfrak a$ is an ideal, so $\bigcup_{n>0}\mathfrak a^n$ is just $\mathfrak a$.<|endoftext|> +TITLE: A Quotient of Free Group +QUESTION [5 upvotes]: If $F$ is a free group on a finite set $S$, then the squares in $F$ generate a normal subgroup $N$ and $F/N$ is elementary abelian $2$-group of order $2^{|S|}$. +Let $F$ be free group on infinite set $S$, and $N$ the normal subgroup generated by squares in $F$. + +Q. $F/N$ is infinite direct sum or product of $|S|$ many copies of $\mathbb{Z}_2$? + +REPLY [2 votes]: You could conclude easily by saying that since the words are finite, then you must be in the direct sum and not the direct product as Quang Hoang points out. +But I think the most adequate point of view in general is that of universal properties (it's the most helpful in general cases) : $F$ has the universal property the any function $f:S\to G$ to a group $G$ extends uniquely to a morphism $\tilde{f}: F\to G$. Then the quotient $F/N$ has the property that any group morphism $\tilde{f}: F\to G$ such that $\tilde{f}(a)^2 = 1$ for all $a\in S$ factors uniquely to a morphism $\overline{f}:F/N\to G$. +So $F/N$ has the property that any function $f:S\to G$ such that $f(a)^2= 1$ for all $a\in S$ extends uniquely to a group morphism $\overline{f}: F/N\to G$. +This is exactly the universal property of $\bigoplus_S \mathbb{Z}/2\mathbb{Z}$, since such a function $f:S\to G$ is the same as giving a morphism $f_a: \mathbb{Z}/2\mathbb{Z}\to G$ for all $a\in S$. +So $F/N$ and $\bigoplus_S \mathbb{Z}/2\mathbb{Z}$ have the same universal property : they are canonically equal.<|endoftext|> +TITLE: What is wrong with my combinatorics method? +QUESTION [5 upvotes]: Suppose I want to select a team of $7$ from a pool of $10$ from $A$, $8$ from $B$ and $5$ from $C$. However, I want to make sure that I have at least one from each group. +My idea was to do the following $$10 \times 8 \times 5 \times \binom{20}{4}$$ +the $10$, $8$ and $5$ are to choose the one member from each team, who is 'guaranteed' a spot. The $\binom{20}{4}$ is to choose the remaining $4$ team members from the remaining pool of $20$ people. +The problem is that this value is larger than $\binom{23}{7}$, which should be the largest possible value. + +REPLY [4 votes]: I think you may have to use inclusion / exclusion for this one. Then you get +$${23\choose 7}-\left({13\choose 7}+{15\choose 7}+{18\choose 7}\right)+\left({{10\choose 7}+{8\choose 7}+0}\right)=205\ 310.$$ +So the middle terms are the cases where no one is from A, B, or C respectively, and the rightmost terms are the cases where no one is from B or C, A or C, or A or B respectively.<|endoftext|> +TITLE: Is there an easy way to see that this simple recurrence is 9-periodic? +QUESTION [27 upvotes]: In a colloquium talk yesterday, Robert Bryant pointed out that for all initial values $a_0, a_1 \in \mathbb{R}$, the sequence generated by the recurrence relation +$$ a_{n+1} = |a_n| - a_{n-1} $$ +turns out to be periodic with period 9, i.e., +$$ a_n = a_{n+9} = a_{n+18} = a_{n+27} = \cdots $$ +This is an interesting and unexpected fact, especially when you consider that to directly prove the periodicity (by writing down expressions for $a_2, \dots, a_9$ in terms of $a_0, a_1$), you would have to prove the following monstrous identity: +$$ a_0 = -\left| \left| a_1\right| -a_0\right| -\left| -\left| \left| \left| a_1\right| -a_0\right| -a_1\right| +\left| a_1\right| +\left| -\left| \left| a_1\right| -a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| +a_1\right| -a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| +\left| \left| \left| a_1\right| -a_0\right| +\left| -\left| \left| \left| a_1\right| -a_0\right| -a_1\right| +\left| a_1\right| +\left| -\left| \left| a_1\right| -a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| +a_1\right| -a_0\right| -\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| -a_1\right| -\left| a_1\right| -\left| -\left| \left| a_1\right| -a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| +a_1\right| +a_0\right| +a_1 $$ +Of course, a straightforward "brute-force" approach this problem would be to consider sequentially the cases $a_1 > a_0 > 0$, $a_0 > 0 > a_1$, etc. and show that the RHS whittles down to the LHS in each case. But due to the simplicity of the original recurrence relation, I wonder if there isn't a more elegant way (hopefully involving less casework) to see that this recurrence must be 9-periodic. + +REPLY [18 votes]: Let me begin with a couple of remarks. +The sequence $(a_n)$ being periodic with period $9$ means that $a_{n+9}=a_n$ for all integers $n\geq 0$. The condition $a_9=a_0$ is not sufficient for $(a_n)$ to have period $9$, but if also $a_{10}=a_1$, then the recurrence, which computes a term from the preceding two terms, does start repeating the first nine values $a_0$, $a_1$, $\ldots$, $a_8$ over an over again. +Note that the recurrence relation also works backwards: $a_{n-1}=\left|a_n\right|-a_{n+1}$, so that for a given terms $a_0$ and $a_1$ we can calculate, recursively, the terms $a_n$ for all integers $n$. This observation provides another sufficient test of periodicity, namely $a_{-9}=a_0$ and $a_{-8}=a_1$, and on top of that (almost) halves the number of cases that must be considered, since by going backwards the roles of the initial terms $a_0$ and $a_1$ get interchanged. +So, for example, after considering the case $0\leq a_0\leq a_1\leq 2a_0$ there is no need to consider the mirrored case $0\leq a_1\leq a_0\leq 2a_1$. +The cases themselves are not quite what you imagined; you did not actually work them through, or did you? +It suffices to consider five cases. Here they are, together with the terms $a_2$, $a_3$, $\ldots$, $a_8$, $a_9$, $a_{10}$ they produce: +Case $a_0,\,a_1\leq 0\,$: $-a_0\!-\!a_1$, $\,-a_0\!-\!2a_1$, $\,-a_1$, $\,a_0\!+\!a_1$, $\,-a_0$, $\,-2a_0\!-\!a_1$, $\,-a_0\!-\!a_1$, $\,a_0$, $\,a_1$. +Case $0\leq a_0\leq a_1\leq 2a_0\,$: $-a_0\!+\!a_1$, $\,-a_0$, $\,2a_0\!-\!a_1$, $\,3a_0\!-\!a_1$, $\,a_0$, $\,-2a_0\!+\!a_1$, $\,a_0\!-\!a_1$, $\,a_0$, $\,a_1$. +Case $0\leq 2a_0\leq a_1\,$: $-a_0\!+\!a_1$, $\,-a_0$, $\,2a_0\!-\!a_1$, $\,-a_0\!+\!a_1$, $\,-3a_0\!+\!2a_1$, $\,-2a_0\!+\!a_1$, $\,a_0\!-\!a_1$, $\,a_0$, $\,a_1$. +Case $a_0\leq0\leq a_1$, $a_0\!+\!a_1\geq 0\,$: $\,-a_0\!+\!a_1$, $\,-a_0$, $\,-a_1$, $\,a_0\!+\!a_1$, $\,a_0\!+\!2a_1$, $\,a_1$, $\,-a_0\!-\!a_1$, $\,a_0$, $\,a_1$. +Case $a_0\leq0\leq a_1$, $a_0\!+\!a_1\leq 0\,$: $\,-a_0\!+\!a_1$, $\,-a_0$, $\,-a_1$, $\,a_0\!+\!a_1$, $\,-a_0$, $\,-2a_0\!-\!a_1$, $\,-a_0\!-\!a_1$, $\,a_0$, $\,a_1$. +There are four other cases which we do not need to consider since they are mirror-symmetric to the cases two to five above. +How many cases are there? Five plus four. That's nine! +Let us enumerate the nine cases. +$(1)\,$ $\,x\leq 0\,$ and $\,y\leq 0\,$. +$(2)\,$ $\,0\leq x\leq y\leq 2x\,$. +$(3)\,$ $\,0\leq y\leq x\leq 2y\,$. +$(4)\,$ $\,0\leq 2x\leq y\,$. +$(5)\,$ $\,0\leq 2y\leq x\,$. +$(6)\,$ $\,x\leq 0\leq y\,$ and $\,x+y\geq 0\,$. +$(7)\,$ $\,y\leq 0\leq x\,$ and $\,x+y\geq 0\,$. +$(8)\,$ $\,x\leq 0\leq y\,$ and $\,x+y\leq 0\,$. +$(9)\,$ $\,y\leq 0\leq x\,$ and $\,x+y\leq 0\,$. +The cases are not mutually exclusive, but this does not affect the following result (prove it): + +The mapping $(x,y)\to(y,\,\left|y\right|\!-\!x)$ maps a part of the $xy$-plane that corresponds to one of the cases $(1)$--$(9)$ into a part that corresponds to some other case, in the following cyclic order: $(1)\to(6)\to(2)\to(5)\to(9)\to(8)\to(4)\to(3)\to(7)\to(1)$. + +We still have to prove that starting at any point, then applying the map nine times in a row, we will arrive back at precisely that point, and not only at some point of the part of the plane, corresponding to one of the cases, which contains the starting point. But now it suffices to consider only one case to which the starting point belongs: we choose a point that satisfies the condition of the case $(1)$, say, and then chase it all the way round, finding that we end at the chosen point. +Is this elegant enough? +[Added after the two comments about (non)elegance.] +I claim that we must know, in some way, the cases $(1)$--$(9)$ if we want to have a full an clear understanding of the $9$-periodic behavior of the sequence $(a_n)\,$: the nine cases are at the heart of the $9$-periodicity, they comprise its structure. +Denote by $T$ the mapping $\mathbb{R}^2\to\mathbb{R}^2$ that we have introduced above: $T(x,y) = (y,\,\left|y\right|\!-\!x)$. Then $T(a_{n-1},a_n)=(a_n,a_{n+1})$, which is precisely the reason why we introduced the transformation $T$. +The fact that the sequence $(a_n)$ can be recurrently produced backwards means that the transformation $T$ is invertible; and indeed, $T^{-1}(x,y) = (\left|x\right|\!-\!y,\,x)$. +The part of the plane $\mathbb{R}^2$ corresponding to the case $(1)$ is the lower-left quadrant (the third quadrant), denote it by $Q$. Now look at the parts $T^k Q$, $0\leq k\leq 8\,$: + +(The enumeration is by $k$ in $T^k Q$, not by the ad hoc case number. You may enjoy identifying the cases correponding to the $T^k Q$.)$\,$ +The transformation $T$ appears as a highly uneven piecewise-linear attempt at realizing the idea "rotate clockwise by two ninths of the full turn". +And with the figure in front of our noses we clearly see that $T^9$ is the identity transformation of the plane $\mathbb{R}^2$, which expresses in one breath the $9$-periodicity of all possible sequences $(a_n)$ generated by the given recurrence relation. +[Added some days later.] +It would be an extremely dull problem that could not inspire more problems. +The recurrence given in the present question is not of the dull kind, +it has an interesting cousin: +\begin{equation*} +a_{n+1} \,=\, \mathrm{sgn}(a_n) - a_{n-1}~, \qquad n\in\mathbb{Z}\,. +\end{equation*} +(The $\mathrm{sgn}(x)$ is the sign of a real number $x$: +it is $1$ if $x>0$, it is $0$ if $x=0$, and it is $-1$ if $x<0$.) +Like the original recurrence, this recurrence works also backwards: $a_{n-1}=\mathrm{sgn}(a_n)-a_{n+1}$. And like the original recurrence, each and every sequence $(a_n)$ it produces is periodic; the difference is that there is no common period, because there exist sequences that have arbitrarily large periods. So solving this recurrence offers much more fun. Another difference is that this time we do have a simple explanation of periodicity. Here's a hint. +Define the transformation $U\colon\mathbb{R}^2\to\mathbb{R}^2$ by $U(x,y):=(y,\,\mathrm{sgn}(y)\!-\!x)$. Then for every integer $m\geq 1$ the transformation $U$ induces a bijection $U_m\colon S_m\to S_m$, where $S_m$ is the (closed) set + +The bijection $U_m$ is not continuous, it has a discontinuity along the line segment $L_m$ with endpoints $(-m,0)$ and $(m,0)$. However, the restrictions of $U_m$ to the line segment $L_m$ as well as to each of the two connected components of $S_m\setminus L_m$ are isometries. The bijection $U_m$ induces permutations of $2$-cells in $S_m$ (the open unit squares), of $1$-cells (the open unit line segments, the sides of the squares), and of $0$-cells (the points, the vertices of the squares); this implies finite periods. You may enjoy determining the periods of sequences $(a_n)$ generated by particular points $(a_0,a_1)\in\mathbb{R}^2$.<|endoftext|> +TITLE: Integration and differentiation of Fourier series +QUESTION [11 upvotes]: I am interested in the properties of Fourier series under integration and differentiation, and I've noticed a "strange" phenomenon. +Suppose I have a Fourier series which I Integrate, and suppose that I can integrate it term by term. +$$f(x)=a_0+\sum_{n=1}^\infty a_n \cos(nx)+\sum_{n=1}^\infty b_n \sin(nx)$$ +$$\int f(x)dx=C+a_0 x+\sum_{n=1}^\infty \frac{a_n}{n} \sin(nx)-\sum_{n=1}^\infty \frac{b_n}{n} \cos(nx) $$ +Now because I want the integral to be represented as a Fourier series as well, I use the Fourier series for $x$: +$$x=\sum_{n=1}^\infty \frac{2}{\pi}\frac{(-1)^{n+1}}{n}\sin(nx)$$ +So I have: +$$\int f(x)dx=C+\sum_{n=1}^\infty \left(\frac{2a_0}{\pi}\frac{(-1)^{n+1}}{n}+\frac{a_n}{n}\right)\sin(nx)+\sum_{n=1}^\infty \frac{b_n}{n}\cos(nx)$$ +My problem is that upon differentiation (term by term), this series does not appear to give back the original Fourier series, as the constant term has been shifted into the cosine terms. +I think my mistake is the step I take converting the $x$ to a Fourier series, because it can't then be differentiated term by term (see this question). +Am I correct? And does this mean the integral of a Fourier series cannot be another Fourier series? + +REPLY [33 votes]: Fourier series (as with infinite series in general) cannot always be term-by-term differentiated. For general series we have the following theorem + +Theorem: Term-by-term differentiation: If a series $\sum f_k(x_0)$ converges at some point $x_0\in [a,b]$ and the series of derivatives $g(x) = \sum f_k'(x)$ converges uniformly on $[a,b]$ then the series $f(x) = \sum f_k(x)$ converges uniformly for all $x\in[a,b]$ and $f(x)$ is differentiable with $f'(x) = g(x)$. + +Your example fails this theorem dramatically as the derivative of the Fourier series of $x$ do not converge anywhere. +We also have some more specific results. The following theorem gives conditions for which this can be done for Fourier series: + +Theorem (Term-by-term differentiation of Fourier series): If $f$ is a piecewise smooth function and if $f$ is also continuous on $[-L,L]$, then the Fourier series of $f$ can be term-by-term differentiated if $f(-L) = f(L)$. + +The last condition is not satisfied when the Fourier series has a jump-discontinuity as $x=L$ so we in general we don't expect to be able to term-by-term differentiate a Fourier series that has a jump-discontinuity (thought the theorem does not rule it out). For your example of the Fourier series of $f(x) = x$ the first condition are satisfyed as $f$ is both smooth and continuous on $[-1,1]$ however $f(-1) \not= f(1)$ so the theorem does not apply. +For integration of Fourier series we have the following theorem + +Theorem (Term-by-term integration of Fourier series): The Fourier series of a piecewise smooth function $f$ can always be term-by-term integrated to give a convergent series that always + converges to the integral of $f$ for $x\in[-L,L]$. + +Note that the resulting series does not have to be a Fourier series. For example if we have a Fourier series $f(x) = a_0 + \ldots$ with $a_0\not= 0$ then $\int f(x){\rm d}x = a_0x+ \ldots$ and the presence of the $a_0x$ term makes this not be a Fourier series (though for this example one can probably expand $x$ in a Fourier series to get such a series). +The proofs of the theorems above can be found in any good textbook on Fourier series. You can also find them in the following course notes (by P. Laval).<|endoftext|> +TITLE: Which role does the $\frac{1}{24}$ in the Dedekind $\eta$-function play? +QUESTION [11 upvotes]: The Dedekind $\eta$-function is defined as $$\eta(z) = q^{\frac{1}{24}} \prod_{n = 1}^\infty (1 - q^n)^{-1}$$ +where $q = e^{2 \pi i z}$. +My question is: If I start with the Euler-product $\prod_{n = 1}^\infty (1 - q^n)^{-1}$, how do I come to the point where multiplication with $q^{\frac{1}{24}}$ makes sense? +Thanks! + +REPLY [9 votes]: The correct but unsatisfying answer is that, given $(q;q)_\infty=\prod_{n=1}^\infty(1-q^n)$, $1/24$ is the value for which $\eta(z)=q^{1/24}(q;q)_\infty$ satisfies $\eta(-1/z)=\sqrt{-iz}\eta(z)$, which leads to all the nice modular form properties. So really, the question you have is: why, conceptually, is $1/24$ the value that leads to that identity? +If you're familiar with Euler's pentagonal number theorem showing that only the terms with exponent $k(3k-1)/2$ are nonzero, then adding $1/24$ produces +$$\frac{3}{2}(k^2-2k/6+1/36)=\frac{3}{2}(k-1/6)^2$$ +which is a square, implying the eta function to be a sort of askew theta function and the modular relation follows from Poisson summation. +The silly answer (which I believe can be attributed to John Baez) is that while modular forms are defined on the upper half plane, many of the definitions, such as Eisenstein series, should work equally well for $-z$ and $z$; to do this with the Euler product, just replace $1-q^n$ with $q^{-n/2}-q^{n/2}$, or just multiply by $q$ to the power +$$-\frac{1}{2}\sum_{n=1}^{\infty}n=-\frac{\zeta(-1)}{2}=\frac{1}{24}.$$ +This use of the Riemann zeta function is not the actual value of the divergent sum, of course. +The value $1/24=-\zeta(-1)/2$ can also be "smuggled in" through its relation to $\zeta(2)=\pi^2/6$ just by looking at the asymptotics of the logarithmic derivative. Given the modular equation $\eta(-1/z)=\sqrt{-iz}\eta(z)$, take the derivative +$$\frac{1}{z^2}\eta'(-1/z)=\sqrt{-iz}\eta'(z)-\frac{i}{2\sqrt{-iz}}\eta(z)$$ +and, letting $F(z)=\eta'(z)/\eta(z)$ while dividing through by $\eta(-1/z)$, get +$$F(-1/z)=z^2 F(z) + \frac{z}{2}. $$ +If all we knew was $\eta(z)=q^c(q;q)_\infty$, then we'd still be able to get +$$F(z)=2\pi i\left(c-\sum_{n=1}^\infty \frac{n q^n}{1-q^n} \right) $$ +and for any $c$ to have the modular equation hold, $z^2 F(z)$ must approach $2 \pi i c$ as $z \rightarrow 0$. To find this limit, note +$$\sum_{n=1}^\infty \frac{n q^n}{1-q^n}=\sum_{m,n=1}^\infty n q^{mn} = \sum_{m=1}^\infty \frac{q^m}{(1-q^m)^2} $$ +and so +$$c=\frac{1}{2 \pi i}\lim_{z \rightarrow 0} z^2 F(z)=-\lim_{z \rightarrow 0} \sum_{m=1}^\infty \frac{z^2}{(1-e^{2 \pi i m z})^2}=-\sum_{m=1}^\infty \frac{1}{(2\pi i m)^2}=-\frac{\pi^2/6}{-4\pi^2}=\frac{1}{24}. $$ +The function $F$ and its quasi-modular relation are actually linked to the "Eisenstein series" $E_2$ but proving the relation from there is more complicated and implicitly involves reordering sums that aren't absolutely convergent.<|endoftext|> +TITLE: Geometrical Interpretation of Tensors (intuition) +QUESTION [9 upvotes]: How would one describe to a non-mathematician (an undergraduate physicist actually- so please do use mathematics-i am not even sure if they can be described without mathematics!) what do tensors represent geometrically? + +REPLY [2 votes]: We can add something to the already excellent answer given by Ben Grossmann. +The wonderful Gravitation by Misner,Thorpe and Wheeler (affectionately called MTW) on page 75 describes the most general (m,n) tensor as:" a linear machine with n input slots for n 1-forms and m input slots for m vectors, given the requested input, it puts out a real number..." +(1-forms are simply the linear operators that take a vector and give out a number). +Tensors are useful in physics because (roughly): +each observer is described by a different 4D basis in spacetime and +each measurement is described by the value of a specific component of a specific tensor in the observer's basis. +Tensor components in one basis are (linearly) related to the components in another basis, so if we know what one observer is measuring, we can find out what another observer will measure. +In short: measurements made by different observers are related in the same way that certain tensor components are related when passing from one basis to another basis<|endoftext|> +TITLE: Why is this proof that a circular cone is not a surface not rigorous? +QUESTION [5 upvotes]: In example $4.1.5$, page $73$ of Pressley's Elementary Differential Geometry, a "heuristic" argument is given to prove that the circular cone with vertex the origin and angle $\pi/4$, is not a surface. Here are the exact words: + +To see that it is not a surface, suppose that $\sigma: U \to S \cap W$ is a surface patch containing the vertex $(0,0,0)$ of the cone, and let $a\in U$ correspond to the vertex. We can assume that $U$ is an open ball with center $a$, since any open set $U$ containing $a$ must contain such an open ball. The open set $W$ must obviously contain a point $p$ in the lower half $S_{-}$ of $S$ where $z < 0$ and a point $q$ in the upper half $S^{+}$ where $z>0$; let $b$ and $c$ be the corresponding points in $U$. It is clear that there's a curve $\pi$ in $U$ passing through $b$ and $c$, but not passing through $a$. This is mapped by $\sigma$ into a curve $\gamma = \sigma \circ \pi$ lying entirely in $S$, passing through $p$ and $q$, and not passing through the vertex. (It is true that $\gamma$ will, in general, only be continuous, and not smooth, but this does not affect the argument.) This is clearly impossible. (Readers familiar with point set topology will be able to make this heuristic argument rigorous). + +Why is this argument not considered rigorous? Can someone give an outline of how a rigorous argument should be? + +REPLY [5 votes]: This question hasn't been answered and as I just happen to work out this example I will try to present a different idea to prove that it is not a surface. I follow Do Carmo's notation and ideas closely. +The problem is at the center of coordinates, that is at the point $(0,0,0)$. The usual parametrization for a cone of one sheet is the following, $z=\sqrt{x^2 +y^2}$ which is not differentiable at $(0,0,0)$. But could there be another parametrization such that the cone is a regular surface? Here comes to rescue the following result which is a converse to the usual result that the graph of a differentiable function is a regular surface. In this case the converse is local. + +Now if the cone where to be a regular surface, there should be a neighborhood $V$ of $(0,0,0)$ such that $V$ is the graph of a differentiable function which has one of those forms. If you haven't seen the proof of this proposition, what you actually use to construct the function is (not surprisingly) the inverse function theorem applied to the composition of the parametrization of the surface $X$ and the projection $\pi$ to one of the planes $\{z=0\} \hspace{0.2cm} \text {or} \hspace{0.2cm} \{y=0\} \hspace{0.2cm} \text {or} \hspace{0.2cm} \{x=0\} $. +In this case in particular of the cone, it can be seen that a neighborhood of $(0,0,0)$ whenever is projected to any other plane different than $\{z=0\}$, the projection can never be a bijection much less a diffeomorphism, so that you wouldn't be able to use the inverse function theorem. This tells us that the graph of the differentiable function ought to be of the type $z=f(x,y)$. This finishes the proof because now you know from the usual parametrization of the cone that $f(x,y)= \sqrt{x^2 +y^2}$ in a neighborhood of $(0,0,0)$. This function is not differentiable at $(0,0)$ so that you have arrived at a contradiction. This contradiction arises from supposing that the cone was a regular surface.<|endoftext|> +TITLE: Do these two permutations generate $A_n$? +QUESTION [7 upvotes]: Let $n$ be odd and not a multiple of $3$. Do the cycle $\sigma:=(1, 2, \dots, n)$ and any cycle of length $3$ generate $A_n$? + +REPLY [5 votes]: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$No. +$$ +\Span{(1,2,3,4,5,6,\dots,35), (1,6,11)} +$$ +is not the alternating group on $35$ letters. +The reason is that it is imprimitive. A block is +$$ +\Set{1,6,11,16,21,26,31} +$$<|endoftext|> +TITLE: Assuming $0 \leq a_{n+1} \leq c_n a_n + b_n$ (+ other conditions), show $a_n \to 0$ +QUESTION [5 upvotes]: In the paper "A primal-dual splitting method for convex optimization ..." (see here https://www.gipsa-lab.grenoble-inp.fr/~laurent.condat/publis/Condat-optim-JOTA-2013.pdf), Lemma 4.6 states the following: + +Let $(a_n)_n, (b_n)_n, (c_n)_n$ be sequences of nonnegative real numbers such that + +$0 \leq c_n < 1$ for all $n$, +$a_{n+1} \leq c_n a_n + b_n$ for all $n$, +$\sum_n (1 - c_n) = \infty$, +$b_n / (1-c_n) \to 0$. + +Then $a_n \to 0$. + +The paper cites the book "Introduction to Optimization" by Polyak as the source. It just quotes "Lemma 3" from that book, which is actually Lemma 3 of Section 2.2, page 45 (there are multiple Lemmas named "Lemma 3"). +Nevertheless, the book provides no proof. +Does anyone see how this can be obtained? It seems to be a (more or less) well-known lemma, so I could also imagine that there is some other source where this is proved. + +As to my own input: A few lemmas before the current one are proved by considering a "transformed" sequence (something like $(u_k - \alpha_k /(1 - q_k))_k$ comes to mind) which then satisfy a more "friendly" estimate. However, I don't see which transformation would be helpful here. + +REPLY [4 votes]: Given $\epsilon>0$, because of condition 4, there is $N_\epsilon$ such that for $n\ge N_\epsilon$ we have $b_n\le \epsilon (1-c_n)$ and so by the second condition +$$a_{n+1}\le c_na_n+\epsilon (1-c_n)=c_n(a_n-\epsilon)+\epsilon,$$ i.e., +with $\tilde a_n:=a_n-\epsilon$ we have +$$\tilde a_{n+1}\le c_n\tilde a_n$$ +for all $n>N_\epsilon$. We conclude that +$$\tag1\limsup \tilde a_n\le \tilde a_{N_\epsilon}\cdot \lim_{M\to\infty}\prod_{n=N_\epsilon}^M c_n.$$ +If infinitely many $c_n$ are $\le q<1$, the limit on the right clearly exists and equals $0$ (here we the first condition). +Hence we may assume (using $c_n<1$ from the first condition) that all $c_n$ with $n>N_\epsilon$ are so close to $1$ that we can approximate $\ln c_n\approx c_n-1$ (i.e., we find $\gamma>1$ with $\gamma(c_n-1)\le\ln c_n\le c_n-1$). +From $$\ln\prod_{n=N}^M c_n=\sum_{n=N}^M\ln c_n\le-\gamma \sum_{n=N}^M(1-c_n)$$ +we see uisng condition 3 that +$$\lim_{M\to\infty} \ln\prod_{n=N}^M c_n=-\infty$$, +hence +$$\lim_{M\to\infty} \prod_{n=N}^M c_n= 0$$ +and $(1)$ becomes $\limsup \tilde a_n\le 0$. So $\limsup a_n\le \epsilon$ and as $a_n\ge 0$ and $\epsilon$ was arbitrary, we conclude $$\lim a_n= 0.$$<|endoftext|> +TITLE: Proving the Ricci identity +QUESTION [9 upvotes]: I'm trying to prove the Ricci identity + +Let $Z^a$ be a vector field, $R^a_{\,bcd}$ the Riemann curvature tensor and $\nabla$ a torsion-free connection. Then: $\nabla_c\nabla_dZ^a-\nabla_d\nabla_cZ^a=R^a_{\,bcd}Z^b$. + +In particular, I want to start from the RHS. To do so, I've multiplied it by two arbitrary vector fields $X^c$ and $Y^d$, and used the definition of the Riemann tensor: +$$\begin{align}R^a_{\,bcd}Z^bX^cY^d&=\left(R(X,Y)Z\right)^a\\ +&=\left(\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\right)^a\\ +&=\left(\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\partial_{[X,Y]}Z\right)^a +\end{align}$$ +where in the last line I use the fact that the connection is torsion-free to move from covariant to partial derivative. Obviously, the next step is showing that the partial derivative of $Z$ with respect to $[X,Y]$ vanishes. +If they were basis vectors, I know that $[e_\mu,e_\nu]=0$, but what about the case of general vector fields? + +EDIT: second attempt + +$$\begin{align}R^a_{\,bcd}Z^bX^cY^d&=\left(R(X,Y)Z\right)^a\\ +&=\left(\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\right)^a\\ +&=\left(X^c\nabla_c\left(Y^d\nabla_dZ\right)-Y^d\nabla_d\left(X^c\nabla_cZ\right)-\nabla_{X^c\nabla_cY}Z+\nabla_{Y^d\nabla_dX}Z\right)^a\\ +&=\left(\left(X^c\nabla_cY\right)^d\nabla_dZ+X^cY^d\nabla_c\nabla_dZ-\left(Y^d\nabla_dX\right)^c\nabla_cZ-Y^dX^c\nabla_d\nabla_cZ\right.\\ +&\left.\qquad-\left(X^c\nabla_cY\right)^d\nabla_dZ+\left(Y^d\nabla_dX\right)^c\nabla_cZ\right)^a\\ +&=\left(X^cY^d\nabla_c\nabla_dZ-Y^dX^c\nabla_d\nabla_cZ\right)^a\\ +&=X^cY^d\left(\nabla_c\nabla_d-\nabla_d\nabla_c\right)Z^a +\end{align}$$ + +REPLY [10 votes]: This seems like mainly a question about what the abstract index notation is asking you to do. Let's look at a single term from the right hand: $\nabla_X\nabla_YZ$. Recall that in abstract index notation, two tensors juxtaposed signifies their tensor product. For example, if $V^a$ and $W^b$ are vector fields, $V^aW^b$ is supposed to mean the tensor field $V \otimes W$. Pairing of an upper and lower index signifies a trace: if $\varphi_a$ is a 1-form and $V^b$ is a vector field, $\varphi_aV^a =\operatorname{tr}(\varphi\otimes V) = \varphi(V)$. So the way we're supposed to think of this is first taking their tensor product and then tracing. Lastly, notice that $\nabla_VW = \operatorname{tr}(\nabla V \otimes W)$ when considered as a section of $T^* \otimes T \otimes T$, where we trace over the first and last slots. +Returning to the first term above, we then have: +\begin{align} +\nabla_X\nabla_YZ & = X^c\nabla_cY^d\nabla_dZ^b \\ +& = \nabla_X(\operatorname{tr}(Y \otimes \nabla_-Z)) \\ +& = \operatorname{tr}(\nabla_XY \otimes \nabla_-Z + Y \otimes \nabla_X\nabla_-Z)) \\ +& = Y^dX^c \nabla_c\nabla_dZ^b + (X^c\nabla_cY)^d\nabla_dZ^b \\ +& = \nabla^2_{X,Y}Z + \nabla_{\nabla_XY}Z. +\end{align} +And from here the rest shouldn't be bad.<|endoftext|> +TITLE: Show that 1 + $\lambda$ is an eigenvalue of $I + A$ +QUESTION [5 upvotes]: Show that if $\lambda$ is an eigenvalue of $A$, then 1+$\lambda$ is an eigenvalue of $I+A$. +What is the corresponding eigenvector? +What I have done so far (if it is correct at all...): +$(I+A)x=Ix+Ax=1x+Ax=1x+{\lambda}x=(1+\lambda)x$, thus, $1+{\lambda}$ is an eigenvector of $I+A$ +But how do I find the corresponding eigenvector just from the information above? + +REPLY [7 votes]: If $x$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$, then your computation of +$$(A + I)x = (\lambda + 1) x$$ +is exactly the definition that states $x$ is an eigenvector of $A + I$ corresponding to eigenvalue $\lambda + 1$.<|endoftext|> +TITLE: Is it true for $n > 2$ then there always exists a prime $\le n$ that does not divide $n$? +QUESTION [5 upvotes]: I was thinking of how to prove $\frac{n^n}{n!}$ is never an integer for $n > 2$. I think if I prove the above question, then this follows immediately. + +REPLY [16 votes]: Since $\gcd(n,n-1)=1$, no prime factor of $n-1$ divides $n$. And since $n>2$, there must exist a prime dividing $n-1$.<|endoftext|> +TITLE: Why only consider Dolbeault cohomology? +QUESTION [14 upvotes]: On a complex manifold we have the differential operators +$$\partial:A^{p,q}\to A^{p+1,q}$$ +$$\bar\partial:A^{p,q}\to A^{p,q+1}$$ +which both square to zero. Hence one can define cohomology groups +$$H^{p,q}_\partial=\frac{\ker\partial:A^{p,q}\to A^{p+1,q}}{\partial A^{p-1,q}}$$ +$$H^{p,q}_{\bar\partial}=\frac{\ker\bar\partial:A^{p,q}\to A^{p,q+1}}{\bar\partial A^{p,q-1}}$$ +But for some reason I only ever see people discussing the $H_{\bar\partial}^{p,q}$, i.e. Dolbeaut cohomology. What makes this more interesting the the $H_\partial^{p,q}$ groups? + +REPLY [14 votes]: $\newcommand{\dd}{\partial}$To supplement the existing (excellent) answers: If $E \to M$ is a holomorphic vector bundle, the transition functions of $E$ are "constant with respect to $\bar{\dd}$", so there's a natural notion of $E$-valued Dolbeault cohomology $H_{\bar{\dd}}^{p,q}(M, E)$. Unless the transition functions of $E$ are constant, however, there is no corresponding notion of $H_{\dd}^{p,q}(M, E)$.<|endoftext|> +TITLE: When Borel functions and Baire functions are equal? +QUESTION [6 upvotes]: Suppose $X$ is compact metric space. Let $A$ be the smallest set of complex functions containing all continuous functions such that: +If $f_n \in A$ are uniformly bounded and $f_n \to f$ pointwise then $f \in A$. +Is it true that $A$ is equal to the set of all Borel functions? If not, what is relation between these two? What assumptions are needed to get this? +Answers as well as references will be appreciated. + +REPLY [3 votes]: In Kechris' Classical Descriptive Set Theory, 11.6 and 11.7, it is proved that that for every metrizable $X$, the class of (real, complex) Borel functions is the smallest class containing the continuous functions and it is closed under taking pointwise limits. It is also noted that bounded real Borel functions are generated by bounded continuous functions and uniformly bounded pointwise limits. +Taking this two into account, the set you describe will result in the union of all continuous and all bounded complex Borel functions on $X$. This follows from the real case by considering real and imaginary parts, and that unbounded continuous functions can't be used in any uniformly bounded limit to get new functions.<|endoftext|> +TITLE: Finding the Gradient of a Tensor Field +QUESTION [6 upvotes]: Finding the Gradient of a Scalar Field +I understand that you can find the gradient of a scalar field, in an arbitrary number of dimensions like so : +$$grad(f) = \vec{\nabla}f = \left<\frac{\partial f}{\partial x_{1}}, \frac{\partial f}{\partial x_{1}},...,\frac{\partial f}{\partial x_{n}}\right> = \begin{bmatrix} + \frac{\partial f}{\partial x_{1}} \\ + \frac{\partial f}{\partial x_{2}} \\ + ... \\ + \frac{\partial f}{\partial x_{n}} +\end{bmatrix}$$ +where $f$ is a scalar function, $f: \mathbb{R^n} \to \mathbb{R}$. And as you can see this generalizes consistently to higher-dimensional scalar fields. All of this is straight out of Multivariable Calculus. +Finding the Gradient of a Vector Field +Furthermore finding the gradient of a Vector Field, is given by a Tensor i.e. given $f$ to be a vector function, $f : \mathbb{R^m} \to \mathbb{R^n}$ : +$$grad(\vec{f}) = \displaystyle + \nabla \vec{f} = T = +\begin{bmatrix} +\frac{\partial f_1}{\partial x_1} & +\frac{\partial f_1}{\partial x_2} & +... & +\frac{\partial f_1}{\partial x_m} \\ +\frac{\partial f_2}{\partial x_1} & +\frac{\partial f_2}{\partial x_2} & +... & +\frac{\partial f_2}{\partial x_m} \\ +... & +... & +... & +... \\ +\frac{\partial f_n}{\partial x_1} & +\frac{\partial f_n}{\partial x_2} & +... & +\frac{\partial f_n}{\partial x_m} \\ +\end{bmatrix}$$ +with $T$ denoting the tensor (a $n\ $x$\ n$ matrix of partial derivatives of $\vec{f}$'s scalar components, i.e. rank-$0$ tensor components, correct me if what I said in these brackets is wrong) which tells us how the vector field changes in any direction. + +How to find the Gradient of a Tensor Field? +But how do you find the gradient of a Tensor Field? I understand that to answer this question we may need to generalize the concept of a tensor field a bit further. +If I understand correctly, a scalar is a tensor of rank-$0$, a vector is a tensor of rank-$1$. Is it then fine to generalize scalar fields as tensor fields of rank-$0$, and vector fields as tensor fields of rank-$1$? If so then it means we've been finding the gradient of tensor fields (albeit of rank-0 being scalar fields) in our Multivariable courses all along, we just didn't know it. +By extending the logic behind the leap between taking the gradient of a Scalar Field, to taking the gradient of a Vector Field, is it then correct to say that : +The gradient of a Tensor field of rank-$n$ is a Tensor field of rank-($n+1$) ? + +REPLY [3 votes]: A tensor is just a multilinear, scalar-valued function. If I write $V \multimap W$, for the collection of linear functions from a vector space $V$ to a vector space $W$, then, over the reals, a rank-$n$ tensor is just $V^{\otimes n}\multimap\mathbb R$ where $V^{\otimes n}$ means the $n$-fold tensor product, e.g. $V^{\otimes 2} = V\otimes V$. A tensor field is just a smoothly indexed family of tensors. +For simplicity, I'll just talk about the manifold $\mathbb{R}^n$, but anywhere I explicitly write out $\mathbb{R}^n$ (as opposed to $V$), you could just as well use a submanifold of $\mathbb{R}^n$, e.g. a 1-dimensional curve in $\mathbb{R}^n$. A rank-$k$ tensor field on $\mathbb{R}^n$ is a (suitably smooth) function $\tau : \mathbb{R}^n \to (V^{\otimes k} \multimap \mathbb R)$ where $V$ is itself $\mathbb{R}^n$. Now say we write the directional derivative of $\tau$ in some direction $v \in V$ at $x \in \mathbb{R}^n$ as $D(\tau)(x; v)$. The result itself would be a rank-$k$ tensor, i.e. $D(\tau)(x; v) : V^{\otimes k} \multimap \mathbb R$. So what is the type of $D(\tau)$ itself. Well we know the type of $\tau$ and we know the type of $V$ and we know $D(\tau)(x; v)$ is linear in $v$ and non-linear in $x$. So we have +$$D(\tau) : \mathbb{R}^n \to (V \multimap (V^{\otimes k}\multimap\mathbb{R}))$$ but it is easy to show that $(V \multimap (V^{\otimes k}\multimap\mathbb{R})) \cong (V\otimes V^{\otimes k} \multimap \mathbb R) = (V^{\otimes k+1}\multimap \mathbb R)$. Which is to say $D(\tau)$ as a function of $x$ alone — essentially currying $D(\tau)$ — is a rank-$(k+1)$ tensor field. +So to answer your question, you find the gradient of a tensor field by viewing the directional derivative as a linear function of the direction. When you have a basis, as you do for $\mathbb{R}^n$, this linear function can be represented by a vector of partial derivatives (which are directional derivatives along the basis directions). A bit more concretely, +$$D(\tau)(x)_i = \frac{\partial\tau}{\partial x_i}(x)$$ +And yes, even high school, single variable calculus was doing this for the $n = 1$ case. It's just that $\mathbb{R}^{\otimes k} \cong \mathbb{R}$ for any $k$, and $(\mathbb{R}\multimap\mathbb{R}) \cong \mathbb{R}$.<|endoftext|> +TITLE: Orthogonal matrix norm +QUESTION [16 upvotes]: If $H$ is an orthogonal matrix, then $||H||=1$ and $||HA||=||A||, \forall A$-matrix (such that we can writ $H \cdot A$). +What norm is this about? + +REPLY [2 votes]: The original question was asking about a matrix H and a matrix A, so presumably we are talking about the operator norm. The selected answer doesn't parse with the definitions of A and H stated by the OP -- if A is a matrix or more generally an operator, (A,A) is not defined (unless you have actually defined an inner product on the space of linear operators, but if that is the case it may be surprising to the OP) +The operator norm of AH would usually be defined by +$$||HA||=\sup_{||x||=1} ||HAx||$$ +where $||.||$ is any norm, such as the norm induced by the inner product (the euclidean norm in the case of the dot-product) . +$$=\sup_{||x||=1} \sqrt{(HAx,HAx)}$$ +$$=\sup_{||x||=1} \sqrt{(H^*HAx,Ax)} $$ +(definition of adjoint) +$$=\sup_{||x||=1} \sqrt{(Ax,Ax)} $$ + (H is an orthogonal matrix) +$$=\sup_{||x||=1} ||Ax||$$ +(definition of operator norm) +$$=||A||$$<|endoftext|> +TITLE: Prove the positive definiteness of Hilbert matrix +QUESTION [9 upvotes]: This is so called Hilbert matrix which is known as a poorly conditioned matrix. +$$ +A = \left(\begin{matrix} + 1 & \frac{1}{2} & \frac{1}{3} & ... & \frac{1}{n} \\ + \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & ... & \frac{1}{n + 1} \\ + \vdots & \vdots & \vdots & \ddots & \vdots \\ + \frac{1}{n} & \frac{1}{n + 1} & \frac{1}{n + 2} & ... & \frac{1}{2n - 1} \\ + \end{matrix}\right) +$$ +The task is to prove that matrix A is positively definite. +A possible way to go is to look at the scalar product: +$$(f, g) = \int_0^1f(x)g(x)dx$$ +within the space of polynomials of degree not higher than $(n - 1)$. There our form $(\cdot, \cdot)$ is bilinear, symmetric and positively defined thus our polynomials' space with that form is, in fact, an euclidean space of degree $n$. +Now let's take the following basis: ${1, x, x^2, ..., x^\left(n - 1\right)}$. It's easy to see that matrix A is a Gram matrix for this basis. +But is there another approach? + +REPLY [11 votes]: Let $X=(x_i)_{1\leq i\leq n} \in \cal{M}_{n,1}(\mathbb{R}).$ We have +$$ +{}^tXAX=\sum_{1\leq i,j\leq n}\frac{x_ix_j}{i+j-1}=\sum_{1\leq i,j\leq n}x_ix_j\int_0^1t^{i+j-2}dt=\int_0^1\left(\sum_{i=1}^nx_it^{i-1}\right)^2dt>0 +$$ for $X\neq0$, giving the announced result since $A$ is symmetric.<|endoftext|> +TITLE: "Every open cover admits an open locally finite refinement" - can this refinement always be realized in terms of basis sets? +QUESTION [13 upvotes]: Let $X$ be a paracompact topological space, let $C$ be an open cover, and let $\mathscr B$ be a basis for the topology. Does there always exist a locally finite refinement that consists of basis sets? +The reason this is trickier than it sounds is that, while every set in the refinement is a union of basis sets, this is definitely not guaranteed to be a locally-finite sort of union, and taking a locally finite refinement of this lands you outside the world of basis sets... +Now, let me add that I don't quite remember the context that led me to consider this question, but I'm still interested in finding out the answer. + +REPLY [5 votes]: The property in question was introduced by Ralph Ford in his dissertation, 1963. A space is totally paracompact if every open basis has a locally finite subcover. +The property may be further differentiated, depending on whether you start your definition with "every open basis $\mathcal B$" (which I assume you mean, and which coincides with total paracompactness), or you say +"there is an open basis $\mathcal B$, etc", which requires consideration of some other conditions to make the latter property different from paracompactness, given that the topology itself is a basis. One variation is base-paracompactness, when you say there is a base $\mathcal B$ of cardinality the weight of the space such that every open cover has a localy-finite refinement with elements of this base. Another variation is if one says there is a base $\mathcal B$ such that every base $\mathcal B'\subseteq\mathcal B$ has a locally-finite subcover, this is called base-base paracompactness. +The comment by bof above provides an example of a paracompact space that is not totally paracompact and answers your question. I will put together some links to the literature where you can read more. +H. H. Corson, T. J. McMinn, E. A. Michael, J. Nagata, +Bases and local finiteness. +Notices Amer. Math. Soc. 6 (1959), 814 (abstract). +One result announced there: +The base of all bounded, open, +convex sets in a reflexive, infinite-dimensional Banach space is coarse (i.e. it has no locally-finite subcover). Also the irrationals have a coarse base. +H. H. Corson, Collections of convex sets which cover a Banach space. +Fund. Math. 49 (1960/1961), 143–145. +http://matwbn.icm.edu.pl/ksiazki/fm/fm49/fm49110.pdf +This paper has a proof of the result announced earlier. +R. Ford, Basic properties in dimension theory, +Dissertation, Auburn Univ., (1963). +He proved that small and large inductive dimension coincide for totally paracompact metric spaces. +http://topology.auburn.edu/tp/reprints/v25/tp25101.pdf +(the above is not a link to the dissertation itself, and the following is a link that I don't know if it works) +https://www.researchgate.net/publication/34359587_Basis_properties_in_dimension_theory +A.Lelek, On totally paracompact metric spaces. +Proc. Amer. Math. Soc. 19 (1968), 168–170 +https://www.jstor.org/stable/pdf/2036160.pdf +https://www.ams.org/journals/proc/1968-019-01/S0002-9939-1968-0219032-3/S0002-9939-1968-0219032-3.pdf +A. Lelek, Some cover properties of spaces +Fundamenta Mathematicae (1969) +Volume: 64, Issue: 2, page 209-218 +http://matwbn.icm.edu.pl/ksiazki/fm/fm64/fm64117.pdf +One of the properties considered is total paracompactness. +John O’Farrell, +Some methods of determining total paracompactness, +Dissertation, Auburn University (1982). +John O’Farrell, The Sorgenfrey line is not totally metacompact, +Houston J. Math 9 (1983), no.2, 271-273. +https://www.math.uh.edu/~hjm/vol09-2.html +The irrationals with the usual topology are not totally paracompact. The Sorgenfrey line is not totally paracompact. +Zoltan Balogh and Harold Bennett +Total Paracompactness of Real GO-Spaces +Proceedings of the American Mathematical Society +Vol. 101, No. 4 (Dec., 1987), pp. 753-760 +https://www.jstor.org/stable/2046684?seq=1#page_scan_tab_contents +https://www.ams.org/journals/proc/1987-101-04/S0002-9939-1987-0911046-2/S0002-9939-1987-0911046-2.pdf +Real GO-spaces here means take the reals as a set, with a generalized-ordered topology: Any topology stronger than the usual, that has a basis of order-convex sets. Some examples are the Sorgenfrey line and the Michael line. +Francisco Gallego Lupiañez +Total paracompactness and Banach spaces +Proc. Amer. Math. Soc. 103 (1988), 210-214 +http://www.ams.org/journals/proc/1988-103-01/S0002-9939-1988-0938670-6/ +(Seems very relevant, though I just discovered it with google.) +John E. Porter, Base-paracompact spaces +Topology and its Applications 128 (2003) 145–156 +https://core.ac.uk/download/pdf/81947055.pdf +Base-paracompactness is a less restrictive property, there is a base $\mathcal B$ of cardinality of the weight, such that every open cover has a locally-finite refinement with elements of $\mathcal B$. At present 6/26/2019, it is an open question if every paracompact space is base-paracompact. The introduction lists some literature about total paracompactness. +Strashimir G. Popvassilev, Base-family paracompactness +Houston Journal of Mathematics +Volume 32, No. 2, 2006 +http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.640.9733&rep=rep1&type=pdf +Base-family paracompactness is a related property, and the introductions lists some of the results and literature relevant to total paracompactness. +Strashimir G. Popvassilev, Base-cover paracompactness, +Proc. Amer. Math. Soc. 132 (2004), 3121-3130 +https://www.jstor.org/stable/4097277?seq=1#page_scan_tab_contents +http://www.ams.org/journals/proc/2004-132-10/S0002-9939-04-07457-X/home.html +Another related property and discussion of literature. +Strashimir G. Popvassilev +Base-base paracompactness and subsets of the Sorgenfrey line +Mathematica Bohemica (2012) +Volume: 137, Issue: 4, page 395-401 +http://mb.math.cas.cz/full/137/4/mb137_4_3.pdf +https://eudml.org/doc/246244 +Another related property and an open question. +Gary Gruenhage, Generalized metrizable spaces +In Recent progress of general topology III, +see section 20. Base paracompact +http://webhome.auburn.edu/~gruengf/papers/genmetricfinalb.pdf<|endoftext|> +TITLE: Find the value of $ab+ 2cb+\sqrt3 ac$? +QUESTION [5 upvotes]: Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$, $b^2+c^2=9$ and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$? +Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find $a,b,c$ + +REPLY [5 votes]: Here is a geometrical solution: + +by rewriting the equations as +$$\left\{ +\begin{aligned} +5^2&=a^2+b^2-2ab\cos\frac{5\pi}{6}\\ +3^2&=b^2+c^2-2bc\cos\frac{\pi}{2}\\ +4^2&=c^2+a^2-2ca\cos\frac{2\pi}{3} +\end{aligned} +\right. +$$ +and evaluating +$$4\left(\frac{1}{2}ab\sin\frac{5\pi}{6}+\frac{1}{2}bc\sin\frac{\pi}{2}+\frac{1}{2}ca\sin\frac{2\pi}{3}\right)=4S_{\Delta ABC}=24.$$ + +Sketch of a existence proof: +Since $3^2+4^2=5^2$, $\Delta ABC$ exists. Consider $\overset{\mmlToken{mo}{⏜}}{HC}$, i.e., the segment that lies between $AB$ and $CA$ of a circle, for which $BC$ is a diameter. For any $P$ on the arc, $\angle BPC=\pi/2$ while $\angle CPA$ increases continuously from $\pi/2$ to $\pi$ (from $H$ to $C$), so there exists $P$ so that $\angle CPA=2\pi/3$. +In general, if such configuration does exist, the point where three segments meet is the intersection of three arcs. Each arc results the given opening angle with respect to each side of the triangle and they sum up to $2\pi$.<|endoftext|> +TITLE: Evaluation of $\int_{0}^{10 \pi} ([\sec ^{-1}x]+[\cot^{-1} x])~\mathrm dx$ +QUESTION [8 upvotes]: Find the value of the integral +$$\int_{0}^{10 \pi} (\lfloor\sec ^{-1}x\rfloor+\lfloor\cot^{-1} x\rfloor)~\mathrm dx$$ +where $\lfloor . \rfloor$ denotes greatest integer function. +Could some help me with this? I cannot break greatest integer function here into different interval. Please provide some insight. + +REPLY [4 votes]: $\left\lfloor \sec^{-1}x\right\rfloor=\begin{cases} + 0 & \text{ for } x<\sec(1)\approx1.851\\ + 1 & \text{ for } x\ge\sec(1) + \end{cases}$ +$\left\lfloor\cot^{-1}x\right\rfloor=\begin{cases} + 1 & \text{ for } 0\le x<\cot(1)\approx0.642\\ + 0 & \text{ for } x\ge\cot(1) + \end{cases}$ +So +\begin{eqnarray} +\int_0^{10\pi}\left\lfloor \sec^{-1}x\right\rfloor+\left\lfloor\cot^{-1}x\right\rfloor\,dx +&=\int_0^{\cot(1)}1+0\,dx+\int_{\sec(1)}^{10\pi}0+1\,dx\\ +&=\cot(1)+10\pi-\sec(1) +\end{eqnarray} +Graph<|endoftext|> +TITLE: Why is continuity permissible at endpoints but not differentiability? +QUESTION [6 upvotes]: Differentiable at endpoints? +Does differentiation only work on open sets? +Admittedly, there are some questions and answers as to why a function defined on a closed interval is not differentiable on the interval's endpoints. But I find no answer as to why, in spite of this, continuity can be defined on an interval's endpoints. +For differentiability, the intuition is that the neighborhood of $x$ that allows $\lim_{x \rightarrow c}\frac{f(x) - f(c)}{x - c} = L\quad$ must (or is this our definition) be populated from both the left ($x < c\;$) and the right ($x > c\;$). Hence, differentiability isn't defined on endpoints of an interval. + +Why does such a requirement not exist for continuity? +A function $f$ is continuous at $c$ if $$ \forall \epsilon, \exists \delta, |x - c| < \delta \rightarrow |f(x) - f(c)| < \epsilon$$ +Continuity seems not to care whether points appear left or right of $c$ -- just that, if they exist in some chosen neighborhood, they satisfy the condition above. + +Here is an example of this "dichotomy" + +REPLY [4 votes]: Not all mathematicians forbid taking a derivative at the endpoint of an interval. +Consider the question Derivative at Endpoint, +which provides a passage from Rudin's Principles of Mathematical Analysis +according to which a function defined only on a closed interval may have derivatives at the endpoints of that interval. +You can also examine the answers to +derivative on endpoints. +The choice to allow or disallow derivatives at endpoints depends on technical details of how you define a derivative in the first place. +For example, if you use a definition that computes left and right derivatives separately and then requires them to be equal in order have and ordinary derivative, +differentiation is clearly impossible unless the function is defined on both sides of the point where you wish to differentiate it. +But it is not necessary to define differentiation that way. +With regard to the examples you cited in the question: +The fact that someone may say in some piece of mathematics that if +$f$ is differentiable on $(a,b)$ then a certain fact is true, +it does not mean that $f$ cannot be differentiable at $a$ or $b.$ +It merely means that we do not need $f$ to be differentiable at $a$ or $b$ +in order to reach the desired conclusion. +Putting unnecessary conditions in a theorem is generally a bad idea, since in those cases where the condition isn't met you are unable to use the theorem, +whereas if you had omitted the unnecessary condition you might find all the other conditions are satisfied and the theorem applies. +Conversely, when we say that +if some other fact is true then $f$ is differentiable on $(a,b),$ +we may mean only that sometimes $f$ can fail to have a derivative at +$a$ or $b$ even though the given fact is true. +It might even just mean that it was too much bother (and not worthwhile enough) +to prove that $f$ is differentiable at the endpoints. +It is certainly not warranted to conclude from such a statement that +$f$ cannot be differentiable at $a$ or $b.$<|endoftext|> +TITLE: Lebesgue measure-preserving differentiable function +QUESTION [10 upvotes]: Let $\lambda$ denote Lebesgue measure and let $f: [0,1] \rightarrow [0,1]$ be a differentiable function such that for every Lebesgue measurable set $A \subset [0,1]$ one has $\lambda(f^{-1}(A)) = \lambda (A)$. Prove that either $f(x) = x$ or $f(x) = 1- x$. +I will appreciate a hint or a solution that doesn't use ergodic theory as this is an old qual problem in measure theory. $f$ is bounded and continuous and so it is integrable and hence the hypothesis of Lebesgue's Differentiation Theorem are satisfied, but I haven't seen a way to use this or if it is even applicable. + +REPLY [7 votes]: First note $f$ is onto: Otherwise the range of $f$ is a proper subinterval $[a,b],$ which implies $\lambda(f^{-1}([a,b])) = \lambda([0,1]) = 1 > b-a = \lambda([a,b]),$ contradiction. +Suppose $f'(x) = 0$ for some $x\in [0,1].$ Then $f(B(x,r))\subset B(f(x),r/4)$ for small $r>0.$ ($B(y,r)$ denotes the open ball with center $y$ and radius $r$ relative to $[0,1].$) Thus for such $r,$ $f^{-1}(B(f(x),r/4))$ contains $B(x,r).$ But $B(f(x),r/4)$ has measure no more than $2\cdot (r/4) = r/2,$ while $B(x,r)$ has measure at least $r.$ That's a contradiction. Hence $f'(x) \ne 0$ for all $x\in [0,1].$ +By Darboux, $f'>0$ on $[0,1]$ or $f'<0$ on $[0,1].$ Let's assume the first case holds. Then $f$ is strictly increasing, hence a bijection on $[0,1].$ Thus if $A$ is an interval, $\lambda(A) = \lambda(f^{-1}(f(A)))= \lambda(f(A)).$ Since $f(0) = 0,$ this gives $x = \lambda([0,x]) = \lambda(f([0,x])) = f(x) - f(0) = f(x)$ for all $x\in [0,1].$<|endoftext|> +TITLE: The continuity of injectivity radius +QUESTION [6 upvotes]: Let $M$ be a Riemannian manifold. $r:M\to [0,+\infty]$ denotes the function assigns to $p\in M$ the injectivity radius $r_p$ of the exponential map $\exp_p$. + +Is this function $r$ is continuous or even smooth? If not, what additional conditions could guarantee the continuity of $r$? + +REPLY [3 votes]: [Edit: Asaf Shachar has pointed out a mistake in the last paragraph of the proof below. I'm no longer sure whether the stated claim is even true.] +If $M$ is complete, then $r$ is continuous. (I'm not sure of an example where $M$ is not complete and $r$ is not continuous, so it may also be that $r$ is continuous in this case as well.) +I'll give a proof. I use the phrase "$\gamma$ is a geodesic segment at $p$" to indicate that $\gamma$ is a unit speed geodesic map $\gamma: [0,L] \to M$, where $\gamma(0) = p$. Call $\gamma(L)$ is the endpoint of $\gamma$. $\gamma$ is minimizing if $d(\gamma(a), \gamma(b)) = |a-b|$ for all $a,b \in [0,L]$. +It's easy to see that the limit of a sequence of minimizing geodesic segments is a minimizing geodesic segment. +From this it follows easily that $r$ is upper semicontinuous, that is, that +$$ + \limsup_{p_n \to p} r(p_n) \leq r(p). +$$ +To see this, suppose otherwise; then there's some infinite sequence $p_n \to p$ with $r(p_n) > r(p) + \epsilon$. Fix a non-minimizing geodesic segment $\gamma$ at $p$ of length $r(p) + \epsilon$; we can write $\gamma$ as the limit of a sequence of geodesic segments $\gamma_n$ at $p_n$, all of length $r(p) + \epsilon$, and hence all minimizing; this leads to a contradiction. +Likewise, $r$ is lower semicontinuous. We wish to show +$$ + \liminf_{p_n \to p} r(p_n) \geq r(p). +$$ +Again suppose otherwise. Then there's an infinite sequence $p_n \to p$ with $r(p_n) < r(p) - \epsilon$. For each $n$ we can choose a non-minimizing geodesic segment $\gamma_n$ of length $r(p)$ at $p_n$. Let $\sigma_n$ be a minimizing geodesic segment from $p_n$ to $\gamma_n(r(p))$; by hypothesis, the length of $\sigma_n$ is no greater than $r(p) - \epsilon$. Passing to some subsequence of the $p_n$'s, we can assume that the $\gamma_n$ converge to some geodesic segment $\gamma$ at $p$ of length $r(p)$, and that the $\sigma_n$ converge to some geodesic segment $\sigma$ at $p$ of length $L \leq r(p) - \epsilon$. But $\sigma$ and $\gamma$ have the same endpoint, so $\gamma$ cannot be minimizing, a contradiction.<|endoftext|> +TITLE: Is this stronger statement of the squeeze theorem valid? +QUESTION [8 upvotes]: The squeeze theorem formally states that if $f,g$ and $h$ are real functions defined on an interval $I$ containing $c$ as a limit point and satisfy $g(x) \leq f(x) \leq h(x)$ for all $x \in I$ except possibly $c$, and furthermore, $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$. This is proven by choosing, for given $\epsilon$, the minimum of the two corresponding $\delta$s of $h$ and $g$ as the $\delta$ for $f$. +This statement, however, isn't as strong as can be. In particular, suppose $g$ and $h$, about every neighbourhood of $c$, "interchange" roles of being the lower and upper bounds of $f$. A concrete example would be $f(x) = 0$, $h(x) = x^2\sin\frac{1}{x}$ and $g = -h$ as $x \to 0$. I think that the conclusion of the squeeze theorem should be valid in this case regardless, because so long as we choose $\delta$ such that $g,h$ are within $\epsilon$ of $L$, $f$ will be within $\epsilon$ of $L$. +Essentially, I want to weaken the hypothesis so that rather than $\forall x \in I, c \neq x\left[g(x) \leq f(x) \leq h(x)\right]$, it states $\forall x \in I, c \neq x \left[g(x) \leq f(x) \leq h(x) \ \lor h(x) \leq f(x) \leq g(x)\right]$. + +Questions: Is my reasoning correct? If not, what mistake am I making? If so, why is the stronger statement not more common in calculus/analysis texts? + +REPLY [3 votes]: Your reasoning is right. +If, for all $x,$ either $g(x)\le f(x)\le h(x)$ or $h(x)\le f(x)\le g(x),$ then $$\min\{g(x),h(x)\}\le f(x)\le\max\{g(x),h(x)\}$$ for all $x.$ So all we have to do is show that, if $g(x)\to L$ and $h(x)\to L,$ then $\max\{g(x),h(x)\}\to L.$ (Of course it follows that $\min\{g(x),h(x)\}=-\max\{-g(x),-h(x)\}\to L$ as well.) +Observe that +$$\max\{g(x),h(x)\}=\frac{g(x)+h(x)+|g(x)-h(x)|}2\to\frac{L+L+|L-L|}2=L.$$<|endoftext|> +TITLE: Linearity of the right inverse of a surjective linear map +QUESTION [6 upvotes]: Suppose we have a surjective linear map $f:V\to V$ on an infinite-dimensional vector space $V$. We know that every surjective map has at least one right inverse. So I was wondering... I know not all right inverses are linear, for example, on the space of real sequences +$$(a_0,a_1,a_2,\cdots)\mapsto(1,a_0,a_1,a_2,\cdots)$$ +is a right inverse of the surjective linear map +$$(a_0,a_1,a_2,\cdots)\mapsto(a_1,a_2,\cdots)$$ but obviously isn't a linear map itself, but on the other hand $$(a_0,a_1,a_2,\cdots)\mapsto(0,a_0,a_1,a_2,\cdots)$$ is a linear right inverse. So I was wondering... does there always exist a linear right inverse of any linear surjective map on an infinite-dimensional vector space over an arbitrary field $K$? + +REPLY [3 votes]: Since you're assuming that surjective maps have a right inverse (which is equivalent to the axiom of choice), you know that every vector space has a basis. +Let $f\colon V\to W$ be a surjective linear map. Let $\mathscr{B}=\{w_\alpha:\alpha\in A\}$ be a basis of $W$; for each $\alpha\in A$, pick $v_\alpha$ such that $f(v_\alpha)=w_\alpha$ and define $g\colon W\to V$ by stipulating that +$$ +\text{for each $\alpha\in A$, $g(w_\alpha)=v_\alpha$} +$$ +By standard results, $g$ is well defined; moreover $f\circ g$ is the identity on $\mathscr{B}$, so it is the identity on $W$.<|endoftext|> +TITLE: Determining the presentation matrix for a module +QUESTION [5 upvotes]: I am trying to study some module theory using the book "Algebra" by Michael Artin (2nd Edition, to be precise), and I can't really fathom what is written in Section 14.5. + +Left multiplication by an $m \times n$ matrix defines a homomorphism + of $R$-modules $A: R^n \rightarrow R^m$. Its image consists of all + linear combinations of the columns of $A$ with coefficients in the + ring, and we may denote the image by $AR^n$. We say that the quotient + module $V=R^m/AR^n$ is presented by the matrix $A$. More generally, + we call any isomorphism $\sigma: R^m/AR^n \rightarrow V$ a + presentation of a module $V$, and we say that the matrix $A$ is a presentation matrix for $V$ if there is such an isomorphism. +For example, the cyclic group $C_5$ of order $5$ is presented as a + $\mathbb{Z}-$module by the $1\times 1$ integer matrix $[5]$, because + $C_5$ is isomorphic to $\mathbb{Z}/5\mathbb{Z}$. + +The above two paragraphs seem to be clear. + +We use the canonical map $\pi: R^m \rightarrow V=R^m/AR^n$ to + interpret the quotient module $V=R^m/AR^n$, as follows: +Proposition 14.5.1 +(a) $V$ is generated by a set of elements B $=(v_1, \cdots, v_m)$, the images of the standard basis elements of $R^m$. +(b) If $Y=(y_1, \cdots, y_m)^t$ is a column vector in $R^m$, the element B$Y=v_1y_1+\cdots + v_my_m$ of $V$ is zero if and only if + $Y$ is a linear combination of the columns of $A$, with coefficients + in $R$ - if and only if there exists a column vector $X$ with entries + in $R$ such that $Y=AX$. +Proof. The images of the standard basis generate $V$ because the map $\pi$ is surjective. Its kernel is the submodule $AR^n$. This + submodule consists precisely of the linear combinations of the columns + of $A$. + +First, could you expound on the proof please? The proposition is obvious when applied to the above example with $R^m=R^n=\mathbb{Z}$, $V=C_5$, and $A=[5]$; but it doesn't help me to carry out more detailed proof. + +If a module $V$ is generated by a set B $=(v_1, \cdots, v_m)$, we + call an element $Y$ of $R^m$ such that B$Y=0$ a relation + vector, or simply a relation among the generators. We may also + refer to the equation $v_1y_1+\cdots+v_my_m=0$ as a relation, meaning + that the left side yields $0$ when it is evaluated in $V$. A set $S$ + of relations is a complete set if every relation is a linear + combination of $S$ with coefficients in the ring. +Example 14.5.2 The $\mathbb{Z}-$module or an abelian group $V$ that is generated by three elements $v_1, v_2, v_3$ with the compete + set of relations +$$ 3v_1+2v_2+v_3=0\\ 8v_1+4v_2+2v_3=0\\ 7v_1+6v_2+2v_3=0\\ + 9v_1+6v_2+v_3=0 $$ +is presented by the matrix +$$ + A=\begin{bmatrix} + 3 & 8 & 7 & 9 \\ + 2 & 4 & 6 & 6 \\ + 1 & 2 & 2 & 1 \\ + \end{bmatrix}. $$ +Its columns are the coefficients of the (above) relations: $(v_1, v_2, + v_3)A=(0, 0, 0, 0)$. + +Here comes the second question. How did he find out immediately what matrix presents the group $V$? Artin gives an algorithm for determining the presentation matrix below this example, but since this example precedes it, the reader should be able to determine this matrix without the algorithm. So, in this case $R=\mathbb{Z}$, but what are $m$ and $n$? How do I find this matrix $A$ neatly? + +REPLY [4 votes]: I know this is an old question but it's not yet answered. I'm also just an algebra student, but I'll take a stab at this. +First, let's just make sure we're on the same page with what Artin is saying. We're given a ring $R$. Now, let's say we have a finitely generated $R$-module $M$ over this ring $R$. If there are no relations between the generators of the module, then the module is isomorphic to the free module $R^m$. +Now how do we use this to inform thinking about modules that $do$ have relations. Let's say we have a finitely generated module $M$ that can be finitely generated by a set of generators $B = (v_1, ... , v_m)$. Symbolically, the entire module can be thought of as $BR^m$, all $R-$ linear combinations of the generators. But notice something here with $BR^m$... it's that many of these elements are going to be zero. We need a way of capturing which of the elements of this are zero. To do this, we'll send all of the elements of $R^m$ which are the $coordinates$ of the zero elements of $M$ to zero. These are all the elements $AR^n$. See that the columns of the $A$ matrix are indeed the relation vectors for the module. Another way to think of this is that +$$ +BAR^n = 0 \text{ in the module} +$$ +Well that's precisely what quotienting is! If $R^m$ is the space of all the coordinate vectors by which we can compose the basis of generators, then we want to send all of the coordinate vectors that are relation vectors to zero! So we set $AR^n$ to zero. It's the $columns$ of $A$ which are the coordinate vectors i.e. relation vectors for $M$. So we get that $M \cong R^m/(AR^n)$. +After meditating on this for a little bit, Artin's rules for determining the presentation matrix should follow pretty directly.<|endoftext|> +TITLE: GCD of an empty set? +QUESTION [10 upvotes]: The empty sum and empty product have clear, widely accepted definitions. But I can't seem to find any such definition of an "empty GCD". +The GCD, I'm told, is associative. Hence, given a binary GCD function $\gcd(a, b)$, we can define the n-ary GCD function as $\gcd(a, b, c) \equiv \gcd(\gcd(a, b), c)$. +Extending this "below" the binary case, I'm pretty sure no-one would disagree that the unary case, $\gcd(a) = a$, makes sense. Then, can we say that $\gcd(a) = \gcd(a, \gcd())$? +If so, then we need a value $x = \gcd()$ such that $\forall a$, $\gcd(a, x) = a$. It seems to be an accepted convention that $\gcd(a, 0) \equiv a$, and I don't think there's any other candidate. +So, does it make sense to define $\gcd() \equiv 0$? +Motivation: I was implementing an n-ary gcd() function and got curious about whether I should require at least one argument. + +REPLY [5 votes]: Define $\gcd(S)$ to be natural number $g$ (if it exists) such that: + +$g \mid x$ for every $x \in S$. +For any $d$ such that $d \mid x$ for every $x \in S$, we have $d \mid g$. + +Then indeed $\gcd(\{\}) = 0$, and moreover it can be proven that $\gcd(S)$ exists for every set $S \subset \mathbb{Z}$. +Also $\gcd(\{0,0\}) = 0$. (This differs from the other common definition of $\gcd$ as the greatest common divisor in terms of size, which would simply stipulate that $\gcd(0,0)$ is undefined.) + +REPLY [5 votes]: Yes, the convention $\gcd \emptyset =0 $ makes sense. +Every integer divides all elements of $\emptyset$ thus the "greatest" among them is $0$, when "greatest" is understood with respect to the partial order given by divisibility, which is the appropriate notion of 'size' in this context.<|endoftext|> +TITLE: The number of zeros in the expansion of $n!$ in base $12$ +QUESTION [5 upvotes]: During an interview last year I was asked the following question: + +How many zeros appear at the end of $n!$ in base $12$, where $n$ is a positive integer? + +I applied the known Legendre formula for $p=3$ and then for $p=2$ (in the last case followed by a division by $2$). Then I realised that I couldn't compare the two series produced. +It turned out that the interviewer was an applied mathematician and considered the two series to be "equal". +So the question remained unanswered: + +Which of the produced series is greater for each value of $n$? + +REPLY [3 votes]: This is quite a nice problem, though a proper solution is pretty involved for a job interview. +Here's a compact way to repackage this: The largest power $k$ of a prime $p$ that divides $n!$ is the $p$-adic valuation $$v_p(n!) := \sum_{k = 1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor$$ of $n!$. Expanding this necessarily finite sum and applying a little middle school algebra shows that we can rewrite this as $$v_p(n!) = \frac{n - S_p(n)}{p - 1},$$ where $S_p(n)$ is the sum of digits of $n$ in base $p$. (So, if one already has access to the base-$2$ and -$3$ representations of $n$, this computation is very fast.) +Now, the largest power of $3$ that divides $n$ is $$\boxed{v_3(n!) = \tfrac{1}{2}(n - S_3(n))} ,$$ +and the largest power of $4$ that divides $n$ is $$\boxed{\left\lfloor \tfrac{1}{2} v_2(n!)\right\rfloor = \left\lfloor \tfrac{1}{2} (n - S_2(n))\right\rfloor}$$ (note that $S_2(n)$ is just the number of $1$'s in the binary expansion of $n$). Thus, the largest power of $12$ that divides $n!$ (and therefore the number of $0$'s at the end of the base-$12$ representation of $n!$) is +$$\color{#bf0000}{\boxed{Z_{12}(n) := \min \left\{\tfrac{1}{2}(n - S_3(n)), \tfrac{1}{2} (n - S_2(n))\right\}}} .$$ +If $n$ is a power $2^a$ of $2$, $a \geq 3$, then $S_2(n) = 1$, and so $$\left\lfloor \tfrac{1}{2} (n - S_2(n))\right\rfloor = \left\lfloor \tfrac{1}{2} (n - 1)\right\rfloor = \tfrac{1}{2} n - 1 > \left\lfloor \tfrac{1}{2} (n - S_3(n))\right\rfloor ,$$ that is, the number of factors of $3$ in $n!$ is smaller than the number of factors of $4$. On the other hand, if $n$ is a power of $3$, the reverse is true, so it is not the case that one factor or the other always dominates for sufficiently large $n$. +Remark This last fact is not equivalent to saying that each dominates the other with equal probability; it could be an interesting problem to compute, e.g., how often factors of $3$ dominate, asymptotically, i.e., the limit +$$\lim_{n \to \infty}\frac{\#\left\{\tfrac{1}{2}(n - S_3(n)) > \tfrac{1}{2} (n - S_2(n)) : 1 \leq n \leq N\right\}}{N} .$$ +From the above analysis, we can see immediately how to choose bases that have this critical behavior wherein some powers of the prime factors of the base are on even footing; the next smallest after $12$ is $45 = 3^2 \cdot 5$. The smallest base in which three prime factors are on mutually equal footing is $2^6 \cdot 3^3 \cdot 7 = 12096$.<|endoftext|> +TITLE: Homotopy connectedness +QUESTION [6 upvotes]: I am trying to prove the following statement: +Let $X,Y$ be two homotopy equivalent topological spaces. If $X$ is connected then $Y$ is connected. +So far, this is my attempt: If $X,Y$ are homotopy equivalent then exist a continuous function $f: X \rightarrow Y$ (homotopy equivalence). For the principal theorem of connectedness then $f(X)$ is connected in $Y$. Suppose $Y$ is not connected, then exists a separation (namely a couple of nonempty open sets $ \{U,V \}$ of $Y$) such that $Y=U \cup V$ (disjoint union). $f(X)$ is connected in $Y$ means that $f(X) \subset U$ or $f(X) \subset V$. Now I want to use the fact that exist unique map called homotopy inverse $g:Y \rightarrow X$ such that $ g \circ f \sim id_X$ (where $\sim$ stands for homotopy equivalent) and I want to find a contradiction, but nothing comes into my mind at the moment. Any hint? thank you in advance + +REPLY [8 votes]: So $f\circ g \equiv Id_Y$ i.e $H:Y\times I\to Y$ is a continuous map and $H(y,0)=f\circ g (y)$ and $H(y,1)=y$. $Img(f\circ g) \subset U$ (lets say). Then consider the path $\gamma (t)=H(v,t)$. This connect $U$ with a point in $V$. But $U,V$ are disjoint open sets so they cannot be connected by a path. (Contradiction).<|endoftext|> +TITLE: Why is Euler's number used as a base for logarithms? +QUESTION [15 upvotes]: Is there some special property of '$e$' which makes it suitable to be used as a base for logarithms? +Moreover, does the natural logarithm possess some advantage over the common logarithm? I don't understand why there is a need to choose an irrational number, '$e$', for a base. Isn't it much simpler to use 10 as a base? + +REPLY [2 votes]: Historically, I believe it was because it was much easier to compute. +How do you compute $10^{4.32}$? Maybe you can raise $10^{432}$, but then how do you extract that $100^{th}$ root? Maybe you use Newton's method or something, but this is a long process, and it is not "uniform": making a table of values this way would be horrible. +However, to approximate a table of values for $e^x$ is pretty easy: +$\begin{align*} +e^{0.1} &\approx \left((1+\frac{1}{10})^{10} \right)^{0.1}=1.1\\ +e^{0.2} &\approx 1.1^2 = 1.21\\ +e^{0.3} &\approx 1.1^3 = 1.331\\ +\vdots +\end{align*}$ +You can improve the accuracy systematically by using finer approximations, like $e^{0.01} \approx 1.01$. All you have to do to generate the table is multiply (which for these bases amounts to shifting and adding). +Napier used $e^{0.00000001}$ to make his table of logarithms. Handling the error in the terms (which get small fast, but add up over the course of a very long table) was highly nontrivial. Once he has his log tables, it allowed a huge number of computations to be done more easily (including the creation of tools like the slide rule).<|endoftext|> +TITLE: Showing that the cardinal $\mu=\min\{\lambda: \kappa^{\lambda} > \kappa\}$ is regular +QUESTION [8 upvotes]: I've been thinking about this question, but to no avail and I've got to ask. +How to show that for $\kappa\geq\aleph_0,$ $\mu=\min\{\lambda: \kappa^{\lambda} > \kappa\}$ is regular? +If I wanted a contradiction, it would suffice to prove $\kappa^{cf(\mu)}>\kappa$, but I don't see how this could be true. + +REPLY [10 votes]: Let $\kappa$ be infinite and $\mu=\min\{\lambda\mid \kappa^\lambda>\kappa\}$ (this cardinal exists, since already $\kappa^\kappa>\kappa$). If $\mu$ is singular, write $\mu=\sum_{\alpha<\theta}\mu_\alpha$ where $\theta={\rm cf}(\mu)<\mu$ and the $\mu_\alpha$ form a strictly increasing sequence of regular cardinals. We have + $$ \kappa^\mu=\kappa^{\sum_\alpha \mu_\alpha}=\prod_\alpha \kappa^{\mu_\alpha}. $$ +Now, by definition of $\mu$, $\kappa^{\mu_\alpha}=\kappa$ for all $\alpha$, and therefore the product above reduces to $\prod_\alpha\kappa=\kappa^\theta$. Again, since $\theta<\mu$, then $\kappa^\theta=\kappa$, and we are done. +The key equality is of course $\kappa^{\sum_\alpha \mu_\alpha}=\prod_\alpha \kappa^{\mu_\alpha}$, which is proved directly: The sum $\sum_\alpha\mu_\alpha$ is the cardinality of the disjoint union $\bigcup_\alpha \mu_\alpha\times\{\alpha\}$, and we can identify any function $\phi$ from this union to $\kappa$ with the sequence of functions $(\phi_\alpha\mid \alpha<\theta)$ given by $\phi_\alpha=\phi\upharpoonright \mu_\alpha\times\{\alpha\}$. This is actually a bijection between ${}^{\bigcup_\alpha \mu_\alpha\times\{\alpha\}}\kappa$ and the Cartesian product $\prod_\alpha {}^{\mu_\alpha\times\{\alpha\}}\kappa$.<|endoftext|> +TITLE: Why can this cosine sum function show all primes less than $N^2$? +QUESTION [10 upvotes]: I constructed this cosine sum that puts all primes within N on line y=1, and its zeros show the sieve by primes less than N. For $x0$ and $-j-t$ may be combined to give $2\cos(\cdot\cdot\cdot)$ while sine being an odd function implies that the $j+t>0$ and $-j-t<0$ contributions will cancel. Thus, we may rewrite our formula as: +$$=\sum_{k=2}^m\frac{1}{k^2}\left[\sum_{j+t=0}\cos\left(\frac{2\pi x}{k}(j+t)\right)+2\sum_{j+t>0}\cos\left(\frac{2\pi x}{k}(j+t)\right)\right]$$ where $1-k\leq j,t\leq k-1$ in the above summations. A simple counting argument for when different values of $j+t$ are equal then gives: +$$=\sum_{k=2}^m\frac{1}{k^2}\left[k+2\sum_{j=1}^{k-1}(k-j)\cos\left(\frac{2\pi xj}{k}\right)\right]$$ and multiplying through by one of the factors of $\frac{1}{k}$ therefore indeed gives $f_m(x)=P(m,x)$. +Now, we will apply this useful identity to analyze the function: +If $x$ is an integer, then $\sin(\pi x/k)$ will be zero exactly when $k|x$, and $\lim_{x\to kt}\left(\frac{\sin(\pi x)}{\sin(\pi x/k)}\right)^2=k^2$, where $t\in\mathbb{Z}$. Every other term with $k\nmid x$ ($x$ still an integer) will have $\sin(\pi x)=0$ and $\sin(\pi x/k)$ nonzero. That is, for $x$ an integer, the only non-zero contributions come from divisors of $x$, and each of these will contribute $\frac{1}{k^2}\cdot k^2=1$. So, $f_m(x)$ gives the number of distinct divisors of $x$ which lie in between $2$ and $m$. This is why for $x$ a prime number between $2$ and $m$, the function gives $1$; for $x$ a prime number between $m$ and $m^2$, the function gives $0$; and why the function acts as a prime sieve for $x$ larger than $m^2$. +In this form, it is fairly easy to see that the period of $f_m(x)$, as it is the sum of these $\sin$ terms, is $\operatorname{lcm}(2,3,...,m)$.<|endoftext|> +TITLE: Solution to exponential congruence +QUESTION [6 upvotes]: Is there a clever solution to the congruence without going through all the values of x up to 58?$$2^x \equiv 43\pmod{59}$$ +Can I somehow use the fact that $2^4 \equiv -43\pmod{59}$ ? + +REPLY [2 votes]: Somebody posted a very nice solution using quadratic residues, but here is an easier, and slightly more "dirty" solution. +Note that $$2^{29} \equiv (2^6)^{4} \times 32 \equiv 5^4 \times 32 \equiv 35 \times 32 \equiv 7 \times 160 \equiv 7 \times 42 \equiv 294 \equiv -1 \pmod {58}$$ +Also note $$2^2 \equiv 4 \pmod {58}$$ +This implies that $2$ is a primitive root. Since $$2^{33} \equiv 2^{29} \times 2^{4} \equiv -16 \equiv 43 \pmod {58}$$ +We have that $33+58k$ is the only possible solution where $k$ is an integer.<|endoftext|> +TITLE: Why is $v/\|v\|$ not a unit vector? +QUESTION [10 upvotes]: I have a homework question that is seriously stumping me. It was a true/false statement. +The statement is + +"If $v$ is any vector in an inner product space $V$, then $v/\|v\|$ is a unit vector" + +and according to my professor, the statement is false. +Isn't dividing a vector by its length the very definition of normalizing a vector? +Edit: Thanks everyone! I didn't even consider the null vector. + +REPLY [3 votes]: I consider your professor's question to be badly stated. +The expression + +If $v$ is any vector in an inner product space $V$, then $v/\|v\|$ is a unit vector. + +can be translated into symbols as +$$\forall v \in V: \left\| \frac{v}{\|v\|} \right\| =1. \tag{1}\label{first}$$ +If your professor has true/false as possible answers then either \eqref{first} is true or its negation, which is +$$ \exists v \in V: \left\| \frac{v}{\|v\|} \right\| \ne 1. \tag{2}\label{second}$$ +Now note that \eqref{second} isn't true either since it "breaks" if you try to plug in $0\in V$ (for one can't divide by zero). +So while one can see what was meant by the question and that your professor meant false to be the correct answer, I, being a student, am always somewhat peeved when I see something like this on an exam.<|endoftext|> +TITLE: Every loop space $(\Omega Y,w_0)$ has the structure of an $H$-group. +QUESTION [7 upvotes]: The most important example of an $H$-group is the loop space +$(\Omega Y,w_0)$ of any pointed space $(Y,y_0)$. Let +$\mu:\Omega Y\times \Omega Y\to \Omega Y; \;\; \mu(\alpha,\beta)=\alpha \star\beta$, +where $\alpha \star\beta$ is the product of two loop $\alpha$ and $\beta$. To show $\mu$ is continuous, we use Theorem of exponential correspondence and Pasting lemma on $\Omega Y\times \Omega Y\times[0,\frac{1}{2}]$ and $\Omega Y\times \Omega Y\times[\frac{1}{2},1]$ as closed subsets of $\Omega Y\times \Omega Y\times I$. In fact we show that +$\Omega Y\times \Omega Y\times I\mathop \to \limits^{\mu\times 1}\Omega Y\times I\mathop \to \limits^{E}Y;\;\; (\alpha,\beta,t)\mapsto \alpha\star\beta(t),$ +is continuous. $\mu$ is an H-multiplication. that is, the constant map $e:\Omega Y\to\Omega Y$ whose value is the constant loop $w_0$ is an $H$-unit (i.e. $\mu\circ (e,1)\simeq 1 \simeq \mu\circ (1,e)$. +We want a homotopy $H:\Omega Y\times I\to \Omega Y$ from $\mu\circ (1,e)$ to $1$. To do this, fixed $t\in I$ we define an +$H_t:\Omega Y\times \{t\}\times I\to Y$ by +$ +H_t(w,t,s)= +\begin{cases} +w(\frac{2s}{t+1})& 0\leq s\leq \frac{t+1}{2}\\ +y_0 & \frac{t+1}{2}\leq s\leq 1 +\end{cases} +.$ +In a similar way, Pasting lemma on $\Omega Y\times\{t\}\times [0,\frac{t+1}{2}]$ and $\Omega Y\times\{t\}\times [\frac{t+1}{2},1]$ implies that $H_t$ is continuous for every $t\in I$. But now how we can show that $\bar H:\Omega Y\times I\times I\to Y$ by +$ +H(w,t,s)= +\begin{cases} +w(\frac{2s}{t+1})& 0\leq s\leq \frac{t+1}{2}\\ +y_0 & \frac{t+1}{2}\leq s\leq 1 +\end{cases} +$ is continuous? + +[1] Theorem of exponential correspondence: If $X$ is a locally compact + hausdorff space and $Y$ and $Z$ are topological spaces, a map $g: Z\to Y^X$ is continuous if and only if $E\circ (g \times 1): Z\times X\to Y$ is continuous. + +[1] E. H. Spanier, Algebraic topology, New York, McGraw-Hill, 1966. + +REPLY [5 votes]: This is just the pasting lemma: you are pasting two continuous maps, one on $\Omega Y\times \{(s,t)\in I\times I:0\leq s\leq\frac{t+1}{2}\}$ and one on $\Omega Y\times\{(s,t)\in I\times I:\frac{t+1}{2}\leq s\leq 1\}$. These are closed subsets of $\Omega Y\times I\times I$ whose union is the whole space, and the continuous maps agree on the intersection, so they paste to give a continuous map on all of $\Omega Y\times I\times I$.<|endoftext|> +TITLE: Find the number of tuples consisting of $0, 1$ and $3$ +QUESTION [6 upvotes]: How can I find the number of tuples $(k_1, k_2, ...,k_{26})$ such that each $k_i$ equals $0, 1$ or $3$ and $k_1 + k_2 + ... + k_{26} = 15$. +I can reduce this problem to finding the coefficient of $x^{15}$ in expression $(1 + x + x^3)^{26}$, but I do not now any simple way to do it. + +REPLY [2 votes]: Okay, so you have options: Lets say you have $a$ 1s and $b$ 3s, and the remainder are zeros. So you have that $a+3b = 15$. This gives the following collection of $(a,b)$ pairs as solutions: +$$ (15,0), (12, 1), (9,2), (6,3), (3,4), (0,5),$$ +or equivalently, $(15-3k, k)$ for $0 \leq k \leq 5$. +Now from a pair $(a,b)$ you obtain solutions to the original equation by choosing, from the 26 positions, $a$ places to put a 1, and from the remaining $26-a$ positions, $b$ places to put a $0$. In other words, you have contribution of ${26 \choose a}{26-a \choose b}$ from the pair $(a,b)$. This gives a total of +$$ \sum_{k=0}^{5} {26 \choose 15 -3k}{26-3k \choose k}.$$ +This gives the same result as in the answer of Markus.<|endoftext|> +TITLE: calculate arbitrary points from a plane equation +QUESTION [5 upvotes]: I understand how one can calculate a plane equation (ax+by+cz=d) from three points but how can you go in reverse? +How can you calculate arbitrary points from a plane equation? + +REPLY [7 votes]: From your comment I finally understood what you are looking for: +If you have a plane defined by $a x + b y + c z = d$ then you also have the following properties: + +Plane normal direction: +$$\hat{n} = \begin{pmatrix} + \frac{a}{\sqrt{a^2+b^2+c^2}} \\ + \frac{b}{\sqrt{a^2+b^2+c^2}} \\ + \frac{c}{\sqrt{a^2+b^2+c^2}} \end{pmatrix}$$ +Point on plane closest to the origin (position of plane) +$$ \vec{r} = \begin{pmatrix} + \frac{a d}{a^2+b^2+c^2} \\ + \frac{b d}{a^2+b^2+c^2} \\ + \frac{c d}{a^2+b^2+c^2} \end{pmatrix} $$ +Distance of plane from the origin +$$ r = \frac{d}{\sqrt{a^2+b^2+c^2}} $$ +Directions along the plane (not unit vectors), and perpendicular to $\hat{n}$. +$$ \begin{align} + \hat{e}_1 & = \begin{pmatrix} c-b \\ a-c \\ b-a \end{pmatrix} & + \hat{e}_2 & = \begin{pmatrix} a (b+c)-b^2-c^2 \\ b (a+c) -a^2-c^2 \\ c (a+b)-a^2 - b^2 \end{pmatrix} \end{align} $$ + +You can verify that $\hat{e}_1 \cdot \hat{n} =0$, $\hat{e}_2 \cdot \hat{n}=0$ and $\hat{e}_1 \cdot \hat{e}_2 =0$, where $\cdot$ is the dot (inner) product. +Confirmation via GeoGebra + +NOTES: Please edit the equation to make it clear you are looking for the plane properties when given a plane in equation form.<|endoftext|> +TITLE: Is there an integral that proves that $\sin \tan 1\lt 1$? +QUESTION [5 upvotes]: I recently noted that this inequality is unbelievably sharp: +$$\sin \tan 1\lt 1$$ +Is there some sort of integral that can prove that this is true? +This question might be of some use: +Prove: $\sin (\tan x) \geq {x}$ +I also noted, after seeing some of the comments, that an integral of $$\pi/2-\tan 1$$ +would also greatly help + +REPLY [5 votes]: Hint: + +$\sin(x) \lt 1$ unless $x = \frac{\pi}{2} + 2n\pi$ for integer $n$<|endoftext|> +TITLE: Properties of sin(x) and cos(x) from definition as solution to differential equation y''=-y +QUESTION [5 upvotes]: I recently came across the interesting definition of the sine function as the unique solution to the Initial Value Problem +$$y'' = -y$$ +$$y(0) = 0, y'(0) = 1$$ +(My first question would be why this solution is unique, but I found that proofs of uniqueness and existence are too complicated for me to understand) +Taking existence and uniqueness for granted still leaves a lot of questions about the properties of sine like: + +$y$ is periodic with smallest period $T$ (defined to be $2\pi$) +$ -1 \leq y(x) \leq 1 $ for all $x \in \mathbb{R}$ +The value of sine for certain values related to the period: $ y(\frac{\pi}{6}) = \frac{1}{2}, y(\frac{\pi}{3}) = \frac{\sqrt3}{2}$ +All the various trigonometric identities + +How would one go about proving these properties using only the differential equation definition? I have no idea where to start. Usually these properties (at least for me) are intuitively understood through the unit circle but that doesn't help here. + +REPLY [9 votes]: The first thing you are going to want is the cosine function. If $y=s(x)$ is the solution to $y^{\prime\prime}(x)+y(x)=0$, $y(0)=0$, $y^{\prime}(x)=1$, then let $c(x)=y^{\prime}(x)$. Then $$c^{\prime\prime}(x)+c(x)=s^{\prime\prime\prime}(x)+s^{\prime}(x)=\frac d{dx}(s^{\prime\prime}(x)+s(x))=0$$ +So $c(x)$ is a solution to the same differential equation with $c(0)=s^{\prime}(0)=1$ and $c^{\prime}(0)=s^{\prime\prime}(0)=-s(0)=0$. It is independent of $c(x)$ because if $ac(x)+bs(x)=0$, then at $x=0$ we see that $a=0$ and since $ac^{\prime}(x)+bs^{\prime}(x)=0$, then at $x=0$, $b=0$, so there is no nontrivial linear combination of the two solutions that is identically $0$. Then we can say the general solution to the differential equation $y^{\prime\prime}(x)+y(x)=0$ is +$$y(x)=c_1c(x)+c_2s(x)$$ +We can see that $c^{\prime}(x)=s^{\prime\prime}(x)=-s(x)$ and we can prove the pythagorean identity because +$$\frac d{dx}(c^2(x)+s^2(x))=-2c(x)s(x)+2s(x)c(x)=0$$ +so $c^2(x)+s^2(x)=\text{constant}=c^2(0)+s^2(0)=1^2+0^2=1$. The pythagorean identity establishes an upper and lower bound for s(x). Then we can write down the differential equation for $s(x+a)$ because +$$\frac d{dx}s(x+a)=s^{\prime}(x+a)\frac d{dx}(x+a)=s^{\prime}(x+a)=c(x+a)$$ +$$\frac {d^2}{dx^2}s(x+a)=c'(x+a)=-s(x+a)$$ +Since it satisfies the same differential equation, we can match it to the general solution. $s(0+a)=s(a)=c_1% and s^{\prime}(0+a)=c(a)=c_2$. Thus $s(x+a)=s(a)c(x)+c(a)s(x)$ Similarly you can show that $c(x+a)=c(a)c(x)-s(a)s(x)$. The periodicity seems to be much harder. Initially $s(x)$ is zero and then it increases because its derivative is positive. $c(x)$ starts out at $1$ but it decreases because its derivative is negative. Do the two function ever cross? Well, if they do at $\kappa$, then +$$c^2(\kappa)+s^2(\kappa)=2c^2(\kappa)=1$$ +So $c(\kappa)=s(\kappa)=\frac1{\sqrt2}$. Now, for $0 +TITLE: Laplace transforms of powers of cosine (solved!) +QUESTION [8 upvotes]: During the past several hours, while studying the Laplace transform, I've started wondering what +\begin{equation} +\mathcal{L} \{ \cos^n(at)\}(s) +\end{equation} +would look like – since it won't appear on your typical transform table. After some thinking and tedious pattern-finding and rearranging (and typing on Wolfram Alpha), I've come up with the following (interesting, yet messy) two formulae: +(1) For odd $n$'s ($n=2k+1, k \in \mathbb{N}$), +\begin{equation} +\mathcal{L}\{\cos^{2k+1} (at)\}(s) = \frac{1}{2^{2k}}\sum_{j=0}^{k} \frac{\binom{2k+1}{k-j}s}{ (2j+1)^2 a^2 + s^2} \tag{1} +\end{equation} +(2) For even $n$'s ($n=2k, k \in \mathbb{N}$), +\begin{equation} +\mathcal{L}\{\cos^{2k} (at)\}(s) = \frac{\binom{2k}{k}}{2^{2k}s} + \frac{1}{2^{2k-1}}\sum_{j=1}^{k} \frac{\binom{2k}{k-j}s}{(2j)^2 a^2 + s^2} \tag{2} +\end{equation} +So, + +First of all, are these statements even correct? +If they are, how can I prove it using the definition of the transform? I already have a few ideas in my mind, which involve the exponential definition of cosine, "complexifying" the Laplace integral (but isn't $s$ already complex?), applying the inverse Laplace transform on each of the series terms, or heck, maybe even the Maclaurin expansion of cosine or some trig identities... it'd be lovely to read some suggestion from you all! +I'm also looking at merging the two formulae into a single one that works for every $n$, but the major obstacle seems to be the $ \frac{\binom{2k}{k}}{2^{2k}s} $ term in the even-$n$ formula: where does its weird binomial coefficient come from (1/2 of what the sum's general binomial coefficient would predict)? +A similar approach has been tried for $ \sin^n (at) $, with similar results — messy sums which only work for one or the other parity of $n$. How do the two relate to each other, and possibly to the (albeit trivial) transform of $\exp^n{(at)}=\exp{n(at)}$? And how about $ \sinh^n (at) $ and $ \cosh^n (at) $? + +Thanks in advance, hope this has provided some interesting input + +EDIT. Following Alex R.'s suggestion, I've tried using +\begin{equation} +\cos^n (at) = \frac{\left(e^{iat} + e^{-iat}\right)^n}{2^n} +\end{equation} +The Laplace transform then becomes: +\begin{equation} +\mathcal{L}\{\cos^{n} (at)\}(s) = \int_{0}^{\infty}2^{-n}\left(e^{iat} + e^{-iat}\right)^n e^{-st} dt \\ += 2^{-n} \int_{0}^{\infty} \sum_{j=0}^{n} \binom{n}{j} e^{(iat)j} e^{(-iat)(n-j)} e^{-st} dt \\ += 2^{-n} \sum_{j=0}^{n} \binom{n}{j} \int_{0}^{\infty} e^{-\big(s+ia(2j-n)\big)t} dt \\ += 2^{-n} \sum_{j=0}^{n} \binom{n}{j} \frac{1}{-\big(s+ia(2j-n)\big)} \left[\lim_{\beta \to \infty} e^{-\big(s+ia(2j-n)\big)\beta} - e^{-\big(s+ia(2j-n)\big)(0)}\right] +\end{equation} +\begin{equation} += 2^{-n} \sum_{j=0}^{n} \binom{n}{j} \frac{[-1]}{-\big(s+ia(2j-n)\big)} = 2^{-n} \sum_{j=0}^{n} \binom{n}{j} \frac{1}{s+ia(2j-n)} +\tag{3} \end{equation} +I'm not very satisfied with what I've got — not even sure it's right. And besides that, I can't figure out how it connects with the two formulae given above. Please let me know what you think + +EDIT 2. We could actually check the result in $(3)$ by the use of the Fourier-Mellin integral below: +\begin{equation} +f(t)=\mathcal{L}^{-1}\{\hat{f}(s)\} = \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} e^{st} \hat{f}(s) ds +\end{equation} +where $\gamma \in \mathbb R$ is greater than the real part of all singularities of $\hat{f}(s)$. Applying this formula to the sum in $(3)$, we get +\begin{equation} +\begin{split} +f(t) &= \frac{2^{-n}}{2\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} e^{st} \sum_{j=0}^{n} \binom{n}{j} \frac{1}{s+ia(n-2j)} ds \\ +&= \frac{2^{-n}}{2\pi i} \sum_{j=0}^{n} \binom{n}{j} \int_{\gamma-i\infty}^{\gamma+i\infty} \frac{e^{st}}{s+ia(n-2j)}ds +\end{split} +\end{equation} +All the poles of our sum-defined $\hat{f}(s)$ lie on the imaginary axis, so we can choose any $\gamma > 0 $. +Let us now consider the integral alone. We may now close the contour with a semicircle lying on the left side of the plane and call it $\Gamma$. +\begin{equation} \newcommand{\Res}{\mathop{\text{Res}}} +\oint_\Gamma \frac{e^{st}}{s+ia(n-2j)}ds = 2\pi i \lim_{s\to -ia(n-2j)} e^{st} = 2\pi i e^{-ia(n-2j)t} \tag{4} +\end{equation} +which can be demonstrated to be equal to the integral over the line $\gamma+iq, \quad -\infty < q <\infty$. +We can plug $(4)$ into our formula for $f(t)$ +\begin{equation} +f(t) = \frac{2^{-n}}{2\pi i} \sum_{j=0}^{n} \binom{n}{j} (2\pi i e^{-ia(n-2j)t}) = 2^n \sum_{j=0}^{n} \binom{n}{j} e^{-(n-j)(iat)}e^{j(iat)} +\end{equation} +which of course we recognize as being the binomial expansion of +\begin{equation} +f(t) = \left(\frac{e^{iat} + e^{-iat}}{2}\right)^n = \cos^n(at) +\end{equation} +I will soon try to prove $(1)$ and $(2)$ by the same means. + +FINAL EDIT. I just had to expand the result in $(3)$ to obtain +\begin{equation} +\bbox[5px,border:2px solid red]{ +\mathcal{L} \{ \cos^n(at)\}(s) = \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j} \frac{s+(n-2j)a}{s^2+(n-2j)^2a^2}} \tag{5} +\end{equation} +(a big duhhh on my behalf!) +We can try out $(5)$ with even or odd $n$'s and it will work, connecting beautifully to the formulae in $(1)$ and $(2)$. The painful asymmetry in $(2)$ can now be explained: it was actually caused by the fact that, if $n=2k$ and $k$ is integer, then the $n$th row in Pascal's triangle contains an odd number of terms – of which $\binom{2k}{k}$ is the central one. Since $\binom{n}{q} = \binom{n}{n-q}$, every other term besides that middle one were appearing twice in the sum, and so they could be reduced to $2\binom{n}{q}$. The asymmetry in $(2)$ was just concealing the underlying symmetry expressed in $(5)$. +Finding an expression for the Laplace transform of the powers of sine was trickier, but possible: +\begin{equation} +\mathcal{L} \{ \sin^n(at)\}(s) = \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j} (-1)^{j+\lfloor\frac{n}{2}\rfloor} \frac{s+(n-2j)a}{s^2+(n-2j)^2a^2} \tag{6} +\end{equation} +It works in a similar fashion for the hyperbolic functions: +\begin{equation} +\mathcal{L} \{ \cosh^n(at)\}(s) = \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j} \frac{s+(n-2j)a}{s^2-(n-2j)^2a^2} \tag{7} +\end{equation} +\begin{equation} +\mathcal{L} \{ \sinh^n(at)\}(s) = \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j} (-1)^j \frac{s+(n-2j)a}{s^2-(n-2j)^2a^2} \tag{8} +\end{equation} + +REPLY [4 votes]: Perhaps the dirtiest but most straightforward way would be to write $\cos^n(kx)=\left(\frac{e^{ikx}+e^{-ikx}}{2}\right)^n$, use the binomial theorem on this and then you'll have an easy sum of integrals of an exponential.<|endoftext|> +TITLE: Is there a way to write an infinite set that contains only irrational numbers without integer multiples? +QUESTION [21 upvotes]: Is there a way to write an infinite set that contains only irrational numbers without integer multiples? +The infinite set must not contain integer multiples of any other members of that set. For example,$\pi$ is a member, but we cannot have $2\pi, 3\pi$, and so on. Same applies for any other irrational number in the set. +Also, that infinite set must be equinumerous to $\mathbb{N}$ (natural numbers). This seems intuitive to me, as there are many ways to line up infinite sets with $\mathbb{N}$. But I am having trouble thinking of such an infinite sets regarding only irrationals. +Thanks. + +REPLY [24 votes]: What about: all irrationals in $[2,3]$ ... None of them is an integer multiple of another. +Oops, that's uncountable. +How about all numbers in $[2,3]$ that are rational multiples of $\sqrt{2}$? Again, none of them is an integer multiple of another.<|endoftext|> +TITLE: Deriving the Normalization formula for Associated Legendre functions: Stage $1$ of $4$ +QUESTION [8 upvotes]: The question that follows is needed as part of a derivation of the Associated Legendre Functions Normalization Formula: $$\color{blue}{\displaystyle\int_{x=-1}^{1}[{P_{L}}^m(x)]^2\,\mathrm{d}x=\left(\frac{2}{2L+1}\right)\frac{(L+m)!}{(L-m)!}}$$ where for each $m$, the functions $${P_L}^m(x)=\frac{1}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L$$ are the associated Legendre functions on $[−1, 1]$. +The question in my textbook asks me to + +Show that $$\begin{align}\require{enclose}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L&=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\end{align}$$ Where $L\,\text&\,m\, \text{are constants}$ and $0\leq m\leq L$. + + +Hint: Write $$(x^2-1)^L=(x-1)^L(x+1)^L$$ and find the derivatives by Leibniz' rule. + +So this is what I have tried: +$$\begin{align}\require{enclose}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L&=\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}(x^2-1)^L +\\&=\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}(x-1)^L(x+1)^L\quad\quad\longleftarrow\bbox[#F8A]{\text{Using the Hint}} +\\&=(x-1)^L\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}(x+1)^L ++(L-m)\frac{\mathrm{d}}{\mathrm{d}x}\left(x-1\right)^L\frac{\mathrm{d}^{L-(m+1)}}{\mathrm{d}x^{L-(m+1)}}(x+1)^L +\\&\phantom{Abcde}+\frac{(L-m)(L-[m+1])}{2}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\left(x-1\right)^L\frac{\mathrm{d}^{L-(m+2)}}{\mathrm{d}x^{L-(m+2)}}(x+1)^L+\ldots\,.\end{align}$$ +But this could go on forever and I have no idea how to evaluate (or simplify) terms like $$\frac{\mathrm{d}^{L-(m+2)}}{\mathrm{d}x^{L-(m+2)}}(x+1)^L\,.$$ +Is there any chance someone could please give me some hints or advice on how to show that +$$\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L=\color{#180}{\fbox{$\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L$}}$$ by starting at one side of the equation and showing that it is equal to the other side? +Best Regards. + +REPLY [2 votes]: Note: Please find here a former variation of my current answer. OP was asking to republish it to make the connection with some of OPs comments and his related question plausible. +Both answers are more or less the same. Originally I've decided to apportion it into two somewhat smaller pieces to enhance readability. The current answer was only written since OP was explicitely asking for a one sided approach. + +Here we transform the left hand side to get a convenient representation. Then we transform the right hand side to obtain the same representation and equality is shown. + +The following is valid for $0\leq m\leq L$ + \begin{align*} +\frac{d^{L-m}}{dx^{L-m}}\left(x^2-1\right)^L +=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L +\end{align*} + +We apply the Leibniz Product rule +\begin{align*} +(f\cdot g)^{(n)}=\sum_{j=0}^n\binom{n}{j}f^{(j)}g^{(n-j)} +\end{align*} +according to the hint to $(x^2-1)^L=(x+1)^L(x-1)^L$. + +We obtain from the LHS + \begin{align*} +\frac{d^{L-m}}{dx^{L-m}}&\left(x^2-1\right)^L\\ +&=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L-m-j}}{dx^{L-m-j}}(x-1)^L\tag{1}\\ +&=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(m+j)!}(x-1)^{m+j}\tag{2}\\ +&=(L-m)!\sum_{j=0}^{L-m}\binom{L}{j}\binom{L}{m+j}(x+1)^{L-j}(x-1)^{m+j}\tag{3} +\end{align*} + +Comment: + +In (1) we apply the Leibniz product rule +In (2) we do the differentiation which is not too hard since we have powers of linear factors only +\begin{align*} +\frac{d^j}{dx^j}(x+1)^L&=L(L-1)(L-2)\cdots(L-j+1)(x+1)^{L-j}\\ +&=\frac{L!}{(L-j)!}(x+1)^{L-j} +\end{align*} +In (3) we factor out $(L-m)!$ from $\binom{L-m}{j}$ and rearrange the factorials to get nice binomial coefficients. + + +And now the RHS + \begin{align*} +&\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L\\ +&\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m +\sum_{j=0}^{L+m}\binom{L+m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L+m-j}}{dx^{L+m-j}}(x-1)^L\tag{1}\\ +&\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m +\sum_{j=m}^L\binom{L+m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(j-m)}!(x-1)^{j-m}\tag{5}\\ +&\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m +\sum_{j=0}^{L-m}\binom{L+m}{j+m}\frac{L!}{(L-j-m)!}(x+1)^{L-j-m}\frac{L!}{j!}(x-1)^j\tag{6}\\ +&\quad=\frac{(L-m)!}{(L+m)!} +\sum_{j=0}^{L-m}\binom{L+m}{j+m}\frac{L!}{(L-j-m)!}(x+1)^{L-j}\frac{L!}{j!}(x-1)^{j+m}\tag{7}\\ +&\quad=(L-m)!\sum_{j=0}^{L-m}\binom{L}{j}\binom{L}{m+j}(x+1)^{L-j}(x-1)^{m+j}\tag{8} +\end{align*} + and the claim follows. + +Comment: + +In (5) we do it similarly to (2). Note, that we also shrink the range of the index variable to $m\leq j\leq L$, since for all other values of $j$ the differentiation give zero. +In (6) we shift the index $j$ to start from $0$ and substitute in the expression $j\rightarrow j+m$ +In (7) we apportion $(x^2-1)^m=(x+1)^m(x-1)^m$ +In (8) we cancel $(L+m)!$ and reorder the factorials similarly to (3)<|endoftext|> +TITLE: When does convergence in quotient space $X / {\sim}$ induce convergence in $X$ +QUESTION [6 upvotes]: Let $X$ be a topological space, $\sim$ an equivalence relation and $q : X \to X/{\sim}$ the quotient map that induces on $X/{\sim}$ the quotient topology. +Assume that $p_n \to p$ in $X / {\sim}$. Under what possibly weak assumptions on $X$ and $\sim$ can we deduce that there is $x_n \in p_n$ and $x \in p$ such that $x_n \to x$ in $X$? + +REPLY [5 votes]: You can't expect this to happen even in some very nice cases. For instance, suppose $X=[0,1]$ and $\sim$ is the equivalence relations whose equivalence classes are $\{0,1\}$ and $\{x\}$ for each $x\in(0,1)$. Then in $X/\sim$, the sequence $\{1/2\},\{1/3\},\{2/3\},\{1/4\},\{3/4\},\{1/5\},\{4/5\},\dots$ converges to $\{0,1\}$. However, there is only one possible choice of a representative of each equivalence class, and the sequence $1/2,1/3,2/3,1/4,3/4,\dots$ you get does not converge in $X$. +Here is the most general condition I can find in which it does work. Suppose $X$ is first-countable and $\sim$ is an equivalence relation such that the closure of any saturated set is saturated (for instance, this is true if the equivalence classes of $\sim$ are the orbits of a group of homeomorphisms of $X$). Fix any point $x\in p$, and let $U_1\supseteq U_2\supseteq U_3\supseteq\dots$ be a descending local base at $x$. For each $N$, let $S_N=\{n:p_n\cap U_N=\emptyset\}$. Suppose $S_N$ is infinite for some $N$. Let $A=\overline{\bigcup_{n\in S_N}p_n}$. Then $A$ is the closure of a saturated set and hence saturated, so its image in $X/{\sim}$ is also closed. Since $S_N$ is infinite and $p_n\to p$, this means the image of $A$ contains $p$, and so $p\subseteq A$. But this is impossible, since $U_N$ is a neighborhood of $x$ that witnesses that $x\not\in A$. +Thus $S_N$ is finite for all $N$. We can now choose a sequence of points $x_n\in p_n$ such that $x_n\in U_N$ whenever $N\leq n$ and $n\not\in S_N$. By the finiteness of $S_N$, this sequence is eventually in each $U_N$, and hence converges to $x$. + +To connect this argument to Henno Brandsma's answer, note that the closure of any saturated set is saturated iff the quotient map is an open map. Indeed, the closure of any saturated set is saturated iff the interior of any saturated set is saturated. If this holds and $U\subseteq X$ is open, then the interior of the saturation of $U$ is saturated and contains $U$ and hence is the entire saturation of $U$. That is, the saturation of $U$ is open, so its image in the quotient is open. Conversely, if the quotient map is open and $S\subseteq X$ is saturated, the saturation of the interior of $S$ is open and hence equal to the interior of $S$ since $S$ is saturated. That is, the interior of $S$ is saturated.<|endoftext|> +TITLE: Definition of primary ideal +QUESTION [5 upvotes]: I am confused with the definition of a primary ideal. The definition states that if $R$ is a commutative ring then $I$ is called a primary ideal of $R$ is the following condition holds. If $xy\in I$ then either $x\in I$ or $y^n\in I$ for some $n$. My confusion lies in the following case. Suppose we have that $xy\in I$ but both $x$ and $y$ are not in $I$ and further $y^n \notin R\ \forall n$. Then by definition we have that $I$ is not primary. However if there exists $k$ such that $x^k\in R$ then the element $yx$ does not lead us to a contradiction. But since our ring is commutative $yx=xy$. So what I don't understand is the following this definition might show me that an ideal turns out to be primary depending on how I write the elements (i.e $xy$ or $yx$) which simply doesn't make sense. So what am I saying wrong here? +Thanks in advance. + +REPLY [3 votes]: You can look at the definition from the opposite side: + +$I$ is primary when, for all $x,y\in R$, if $xy\in I$ and no power of one of the elements belongs to $I$, then the other element belongs to $I$. + +You get a symmetric definition also looking at the quotient ring. The ideal $I$ is primary if and only if every zero divisor in $R/I$ is nilpotent. + +In particular, the nilradical of $R/I$ contains all zero divisors, so if you kill all nilpotent elements, you get a domain, which is to say that the radical ideal of $I$ is a prime ideal.<|endoftext|> +TITLE: Link between the negative pell equation $x^2-dy^2=-1$ and a certain continued fraction +QUESTION [5 upvotes]: Consider the generalized continued fraction +$$F(x)=(x-1)-\cfrac{(x+1)}{x+\cfrac{(-1)(5)} {3x+\cfrac{(1)(7)}{5x+\cfrac{(3)(9)}{7x+\cfrac{(5)(11)}{9x+\ddots}}}}}$$ +I experimentally discovered that at certain cases it converges to quadratic irrationals $\frac{a+b\sqrt{d}}{c}$ where $a,b,c$ and$ d\gt1$(square free) are integers. Here's a list of some of its closed forms +$$F(1)=\frac{11+5\sqrt{5}}{3}$$ +$$F(2)=-\sqrt{2}$$ +$$F(3)=\frac{120-26\sqrt{13}}{69}$$ +$$F(4)=\frac{109-25\sqrt{5}}{33}$$ +$$F(5)=\frac{349-29\sqrt{29}}{71}$$ +$$F(6)=\frac{258-35\sqrt{10}}{39}$$ +$$F(7)=\frac{3610-212\sqrt{53}}{429}$$ +$$F(8)=\frac{485-51\sqrt{17}}{47}$$ +$$F(9)=\frac{8841-425\sqrt{85}}{717}$$ +$$F(10)=\frac{1604-143\sqrt{26}}{111}$$ +$$F(11)=\frac{5988-1250\sqrt{5}}{359}$$ +$$F(12)=\frac{6105-481\sqrt{37}}{321}$$ +$$F(13)=\frac{32395-1211\sqrt{173}}{1509}$$ +$$F(14)=\frac{1754-625\sqrt{2}}{73}$$ +$$F(16)=\frac{16891-1105\sqrt{65}}{573}$$ +$$F(17)=\frac{27935-879\sqrt{293}}{863}$$ +$$F(18)=\frac{12840-779\sqrt{82}}{363}$$ +$$F(20)=\frac{12471-707\sqrt{101}}{299}$$ +$$F(22)=\frac{26330-1403\sqrt{122}}{543}$$ +$$F(23)=\frac{82390-2132\sqrt{533}}{1583}$$ +$$F(26)=\frac{16036-765\sqrt{170}}{253}$$ +To my surprise, the discriminants $d$ of the quadratic irrationals are only numbers for which the negative pell equation $x^2-dy^2=-1$ is solvable,see OEIS A031396. +Question:Can the link between the continued fraction and pell equation be proven? + +REPLY [4 votes]: Given your continued fraction $F(x)$, it seems it has a general closed-form. Define, +$$d=x^2+4\tag1$$ +then, +$$F(x) = x-1-\frac{(x+1)\big(-x^3+12x+d\sqrt{d}\big)}{6(3x^2-4)}\tag2$$ +hence, +$$G(x)=\frac{-x^3+12x+d\sqrt{d}}{6(3x^2-4)}=\cfrac{1}{x+\cfrac{(-1)(5)} {3x+\cfrac{(1)(7)}{5x+\cfrac{(3)(9)}{7x+\cfrac{(5)(11)}{9x+\ddots}}}}}\tag3$$ +Plugging in $x$, you'll recover all your values, including the missing $F(15)$ and $F(19)$. +If $d$ as defined by $(1)$ is supposed to be square-free $d'$, then the negative Pell equation, +$$p^2-d'q^2 = -1\tag4$$ +always has a solution. + +Even $x$: If $x=2v$, then $d'=v^2+1$, and $p =v,\;q = 1$. +Odd $x$: Then $d'=x^2+4$, and $p=\frac{x(x^2+3)}{2},\;q=\frac{x^2+1}{2}$ while $p,q$ are integers since $x$ is odd.<|endoftext|> +TITLE: How to prove a regular pentagon is formed by knotting a rectangular strip of paper? +QUESTION [10 upvotes]: I found this interesting problem from a friend (From Arthur Engel's-Problem Solving Strategies book). +The method to begin the problem is as follows- + +Step 1.Take a rectangular strip of paper +Step 2.We all know how to make a knot.Do the same (make a knot) using the strip of paper and fold it along the creases to form a polygon. +(If you are having a problem to make the knot please go to this link to see how to do it. See here ) +Step 3.Prove that the polygon $abcde$ formed is regular. + + +What I could make out from the thing- +When I open the fold again I get this- + +Note:-e is mentioned by dotted lines with pencil since e lies in the opposite side of the paper. +Now,my intuition somehow suggest that $a$ and $b$ must be mutually parallel and $d$ and $e$ must also be parallel.(Since then they would become congruent trapeziums). +But unfortunately I have no clue on how to prove this or how to prove that they are equal to $c$. +Any help or response is highly appreciated and thanks a lot in advance!! + +REPLY [5 votes]: Not sure if it's a complete answer, but an attempt to check a converse statement... +Consider the following illustration: + +The quadrilateral $ABCD$ is a rhombus (the other $D$ on the right is the length of the diagonal, sorry for the confusion between letters) +We have that +\begin{equation} +d=\frac{1}{\sin\theta},\qquad D=\frac{1}{\cos\theta} +\end{equation} +and +\begin{equation} +a = \frac{1}{2}\sqrt{D^2+d^2},\qquad b = a-2d\cos\theta +\end{equation} +Let's suppose that we want a regular polygon. Then we have to impose that the segments $DE$ and $EB$ are of equal length. Since $\overline{DE}=d$ and $\overline{EB}=b$, that means imposing $d=b$: +\begin{equation} +d=b\Longrightarrow \frac{1}{\sin\theta}\left(\frac{1}{2\cos\theta}-\frac{2}{\tan\theta}-1\right)=0 +\end{equation} +i.e. +\begin{equation} +\left(\frac{1}{2\cos\theta}-\frac{2}{\tan\theta}-1\right)=0 +\end{equation} +The only solution to this equation in the interval $[0,\pi/2]$ is actually $\theta=2\pi/5$.<|endoftext|> +TITLE: Generators for a matrix group +QUESTION [9 upvotes]: Let's denote $\Gamma_0(4)$ the subgroup of $\mathrm{SL}_2(\mathbb Z)$: +$$\Gamma_0(4):=\left\{\begin{pmatrix} a &b\\ 4c&d \end{pmatrix}\in \mathrm{SL}_2(\mathbb Z)\right\}.$$ +We also define $A$ and $B$ in $\Gamma_0(4)$ as follow: +$$\gamma_1:=\begin{pmatrix} 1 &1\\ 0&1 \end{pmatrix},$$ +$$\gamma_2:=\begin{pmatrix} 1 &0\\ 4&1 \end{pmatrix}.$$ +We already know (with sophisticated arguments using group actions) that $\Gamma_0(4)/\{\pm 1\}$ is generated by $\gamma_1$ and $\gamma_2$: +$$\Gamma_0(4)/\{\pm 1\}=\langle \gamma_1,\gamma_2\rangle.$$ +I would like to prove it with elementary arguments if possible. +Do you have any leads/solutions to this problem? + +REPLY [2 votes]: Since I managed to find the answer, I am going to write it here in case it is useful for someone. + +Let's defined the group $G=<\gamma_1,\gamma_2>$ generated by: +$$ \displaystyle \gamma_1=\begin{pmatrix} 1 & 0\\ 4 & 1 \end{pmatrix} \text{ and } \displaystyle \gamma_2=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}.$$ +First, we compute ${\gamma_1}^n$ and ${\gamma_2}^n$ by induction +$$\forall n \in \mathbb Z\quad {\gamma_1}^n = \begin{pmatrix} 1 & 0\\ 4n & 1 \end{pmatrix} \text{ and } {\gamma_2}^n = \begin{pmatrix} 1 & n\\ 0 & 1 \end{pmatrix}.$$ +Let $\gamma = \begin{pmatrix} a & b\\ 4c & d \end{pmatrix} \in \Gamma_0(4)$. +Everything we will be doing here works if we are dividing by non zero numbers. If it is not the case, there is nothing to prove. + +Let $q$ and $r$ be the quotient and the rest of the euclidien division of $a$ by $4c$. +We have $a=4cq+r$, so $r=a-4cq$. +Then we have +$$ {\gamma_2}^{-q}\gamma=\begin{pmatrix} a-4cq & b-dq \\ 4c & d \end{pmatrix} = \begin{pmatrix} r & b'\\ 4c & d \end{pmatrix}. $$ +Let $q'$ and $r'$ be the quotient and the rest of the euclidien division of $c$ by $a$. +We have $c=aq'+r'$, so $4r'=4c-4aq'$. +Then we have +$$ {\gamma_1}^{-q'}\gamma=\begin{pmatrix} a & b \\ 4c-4aq' & d-4bq' \end{pmatrix} = \begin{pmatrix} a & b \\ 4r' & d' \end{pmatrix}.$$ + +We now apply a slightly modified version of Euclid's algorithm (I can add the details if someone's interested) so the algorithm works on $\mathbb Z$. +We then obtain a sequence of matrixes of the form: +$$ \begin{pmatrix} a & b \\ 4c & d \end{pmatrix} \rightarrow +\begin{pmatrix} r & \star \\ 4c & \star \end{pmatrix} \rightarrow +\begin{pmatrix} r & \star \\ 4r' & \star \end{pmatrix} \rightarrow +\ \cdots \ \rightarrow +\begin{pmatrix} \pm1 & \star \\ 0 & \star \end{pmatrix}. $$ +This is beacause, $\gamma\in \text{PSL}_2(\mathbb Z)$, so $ad-4bc=1$. +So with Bezout's theorem, $a\wedge c=1$. +We work in the quotient by $\{\pm 1\}$, which allow us to consider only the positive case. +Then, +$$ \exists \delta \in G,\quad \delta\gamma= \begin{pmatrix} 1 & b' \\ 0 & d' \end{pmatrix}.$$ +And $\delta\gamma \in \text{PSL}_2(\mathbb Z)$ so $\det(\delta\gamma)=1$, so $d'=1$. +So, +$$ \delta\gamma= \begin{pmatrix} 1 & b' \\ 0 & 1 \end{pmatrix}.$$ +Then, +$${\gamma_2}^{-b'}\delta\gamma= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$ +So $\gamma\in G$. +So $\Gamma_0(4) \subset G$. +Finally, +$$\Gamma_0(4) / \{\pm 1\}= G.\qquad \square$$<|endoftext|> +TITLE: A nice and hard colouring problem +QUESTION [10 upvotes]: This question is a generalization of a problem recently appeared in a Italian mathematical competition. + +$A$ and $B$ are two coprime integers, both greater than $2$. A + non-constant colouring $$ c:I=\{1,2,3,\ldots,n\}\mapsto \{0,1\}$$ has the + following properties: + +if $u,v\in I$ and $|u-v|=A$, $c(u)=c(v)$; +if $u,v\in I$ and $|u-v|=B$, $c(u)=c(v)$. + +What is the maximum value of $n$, as a function of $A$ and $B$? + +In the original problem, $A=11$ and $B=16$. If we assume that $c(a)=1$ (let we say that $a$ is black) then $a+11,a+22,a+33,\ldots$ are black, too. By assuming that $n>22$, then $a+6,a+17,a+28,\ldots$ are black, too. We may continue this way till a number in the range $a,\ldots,a+10$ is still white, then deduce the maximum value of $n$. +We may also notice that if $AB_1-BA_1=1$, with $1\leq A_11\leq r\leq A-1$, and $r$ prime to $A$. +Imagine an "abstract" regular $A$-gon with vertex set $V\sim{\mathbb Z}/A$. If we draw the $A$ chords $\{k,k+r\}$ we obtain a cyclic graph $\Gamma$. Omitting an edge from $\Gamma$ leaves a connected graph $\Gamma'$, but if we omit $\geq2$ edges from $\Gamma$ the resulting graph on $V$ is no longer connected. +The set $[n]:=\{1,2,\ldots, n\}$ splits into $A$ remainder classes modulo $A$. All members of the same class are colored with the same color; hence each class has its color. Some of these classes are related by a $B$-jump and therefore are assigned the same color. The possible $B$-jumps are $$1\to1+B, \quad 2\to 2+B, \quad \ldots, \quad n-B\to n\ .$$ +The first $A$ of these jumps lead to different edges of $\Gamma$. +If $n-B\leq A-2$, or $n\leq A+B-2$, then not all classes modulo $A$ are connected by $B$-jumps, hence a nonconstant coloring of $[n]$ of the desired kind is possible. If, however, $n-B\geq A-1$, or $n\geq A+B-1$, then $\geq A-1$ different chords in $\Gamma$ are realized, which then implies that all remainder classes modulo $A$ will get the same color.<|endoftext|> +TITLE: Intuitive explanation of extended binomial coefficient +QUESTION [7 upvotes]: We all are familiar with the following formula - +$$\dbinom{n}r = \dfrac{n!}{(n-r)! \space r!} \space\space \space ; \space \space n>r$$ +This is the binomial formula where $n$ and $r$ are positive integers ($r$ can be $0$). However, recently I have been introduced to the Extended binomial theorem which increases the domain of $n$ to any real number. + +Suppose $n \in \mathbb{R}$ then for every $x$ satisfying $|x| < 1$, we have $$ (1+x)^{n} = \sum_{r=0}^{\infty} \dbinom{n}r x^r \space \space \space \space ; \space \space r = 0,1,2...$$ and the extended binomial coefficient is given as $$\dbinom{n}r = \dfrac{n!}{(n-r)! \space r!}$$ + +This is the extended binomial theorem. I do understand the intuition behind the (so as to say) regular binomial coefficient. In simplest language, $\dbinom{n}r$ basically means number of ways to choose $n$ different objects taken $r$ at a time. But in the extended binomial theorem, $n$ can be any real number and $n +TITLE: Why does $z^{-1}$ not have an anti derivative? +QUESTION [5 upvotes]: I had been given the question as shown in the following image, with the answer also given. + +Surely however the anti derivative of $z^{-1}$ would be $\log(z)$ ? I have seen a similar question asked, however I didn't see the answer to this specific question. + +REPLY [6 votes]: $\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}$Because the integral of the reciprocal about the unit circle is non-zero, the value of the integral +$$ +w = \int_{1}^{z} \frac{d\zeta}{\zeta} +\tag{1} +$$ +depends not only on $z$, but on the path from $1$ to $z$. You can denote the value of the integral $w = \log z$, but doing so does not make $\log$ single-valued. The green paths below depict two "lifts" of the unit circle, starting at $(1, 0)$ and tracing counterclockwise, ending at $(1, 2\pi i)$; tracing clockwise, ending at $(1, -2\pi i)$. +For each non-zero $z$, there are infinitely many distinct values of $\log z$. Fixing a path $\gamma$ from $1$ to $z$ fixes a value of +$$ +\log z = \int_{\gamma} \frac{d\zeta}{\zeta}. +$$ +Every other choice differs by an added integer multiple of $2\pi i$, corresponding to a path that winds about the origin some number of times before following $\gamma$ to $z$. +There's no perfectly adequate picture because the graph of the relation $w = \log z$ naturally sits in the space of ordered pairs $(z, w)$ of complex numbers (a real four-dimensional space), but the projection +$$ +(z, w) \mapsto (\Re z, \Im z, \Re w + \Im w) +$$ +(which perhaps is not as widely-known as it should be), captures the qualitative "parking garage" behavior of the Riemann surface of $\log$.<|endoftext|> +TITLE: Cosemisimple Hopf algebra and Krull-Schmidt +QUESTION [9 upvotes]: A cosemisimple Hopf algebra is one which is the sum of its cosimple sub-cobalgebras. Is it clear that a comodule of a cosemisimple Hopf algebra always decomposes into irreducible parts? Moreover, will this decomposition obey Krull-Schmidt, by which I mean will the type and multiplicity of the irreducible comodules appearing be the same in any decomposition. +I am sure that this should be the case but I can't see how to prove it. One thing that confuses me is the prospect of infinite multiplicity in the case of an infinite dimensional comodule. + +REPLY [2 votes]: Regarding your first question, I think the following definition and theorem settles the answer to the affirmative: +Definition: A coalgebra $C$ is called right cosemisimple (or right completely reducible coalgebra) if the Category $M^C$ is a semisimple Category i.e. if every right $C$-comodule is cosemisimple. +Similarly, left cosemisimple coalgebras are defined by the semisimplicity of the Category of left comodules. +Theorem: Let $C$ be a coalgebra. The following assertions are equivalent: + +$C$ is a right cosemisimple coalgebra +$C$ is a left cosemisimple coalgebra +$C=C_0$ +Every left (right) rational $C^*$-module is semisimple + +where $C_0=Corad(C)$ i.e. the coradical of $C$, which is the sum of all its simple subcoalgebras. +For a proof of the above you can see for example Hopf algebras-an introduction, Dascalescu-Nastasescu-Raianu, Ch.3, p.118-119, Theorem 3.1.5. (However this is more or less standard material, you can also find it in Sweedler's book on Hopf algebras, etc). +Now, regarding your second question, I think you can use the above theorem together with the correspondence between $C$-comodules and rational $C^*$-modules to investigate the situation closer.<|endoftext|> +TITLE: Adjoint of a bounded linear operator is bounded +QUESTION [7 upvotes]: Suppose $X$ and $Y$ are normed spaces over $\mathbb{R}$ and suppose $T: X \rightarrow Y$ is a bounded linear map. I want to prove that the adjoint map $T^\star : Y^\star \rightarrow X^\star$ is bounded as a linear map also, but I get super confused when trying to deal with the norms on the dual spaces since there are four norms, to deal with and I am unsure whether it is the operator norm of $T^\star$ or the dual norms i need to take? +I suppose the question is that since $\Vert T(x) \Vert_X \leq M \Vert x \Vert_Y$ it should be possible to find a $K$ s.t. $\Vert T^\star(l) \Vert_X^{\star} \leq K \Vert l \Vert_Y^{\star}$, but expanding these seems a little unfeasible. + +REPLY [5 votes]: Let $A \in L(X,Y)$, $\tilde y \in Y^*$, $x \in X$. Then from $\left(A^*(\tilde y)\right)(x) := \tilde y (A(x))$ you get +$$|A^*(\tilde y)(x)|= |\tilde y (A(x))|≤\|\tilde y\|_{Y^*} \|A(x)\|_Y ≤ \|\tilde y\|_{Y^*} \|A\|_{L(X,Y)} \|x\|_X$$ +For $A^*$ to be bounded you must consider the following expression +$$\|A^*\|_{L(Y^*,X^*)}=\sup_{\tilde y \in Y^*, \ \|\tilde y\|_{Y^*}≤1} \left\{\|A^*(\tilde y)\|_{X^*}\right\}$$ +Note that you have the following inequality, which follows from the first one +$$\|A^*(\tilde y)\|_{X^*}=\sup_{x \in X, \ \|x\|_X≤1}\{|A^*(\tilde y)(x)|\}≤\sup_{x \in X, \ \|x\|_X≤1}\{\|\tilde y\|_{Y^*} \|A\|_{L(X,Y)}\ \|x\|_X \}$$ +Putting it together gives you +$$\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$$ +Note further that the construction gives you $\|A^{**}\|_{L(X^{**},Y^{**})}≤\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$. If you restrict $A^{**}$ to the subspace given by the embedding of $X$ into $X^{**}$ you get precisely $A$ again (composed with the embedding of $Y$ into $Y^{**}$). So $\|A^{**}\|_{L(X^{**},Y^{**})}≥\|A\|_{L(X,Y)}$ also holds, and the two inequalities are equalities, so also $\|A^*\|=\|A\|$. + +REPLY [4 votes]: the definition of the adjoint of a bounded operator $T : X \to X$ is +$$ \text{for every }\ x \in X, y \in X^* : \qquad y(T(x)) = (T^* y)(x)$$ +then note that for any $x \in X$ :$$\|x\|_{X} = \max_{y \in X^*, \ \|y\|_{X^*} =1} |y(x)|$$ +hence $$\|T\| = \max_{\|x\|_X=1} \|Tx\| = \max_{\|x\|_X=1}\max_{\|y\|_{X^*}=1} |y(Tx)| = \max_{\|y\|_{X^*}=1} \max_{\|x\|_X=1}|y(Tx)|$$ $$ = \max_{\|y\|_{X^*}=1} \max_{\|x\|_X=1}|(T^*y)(x)| = \max_{\|y\|_{X^*}=1} \|T^*y\|_{X^*} = \|T^*\|$$ +(this can easily be extended to the case $T : X \to Y$, if the adjoint exists)<|endoftext|> +TITLE: Surjective exponentials for algebraically closed fields +QUESTION [12 upvotes]: The existence of the exponential on $\mathbb{C}$ has a very basic, yet very strong consequence : $(\mathbb{C}^*,\cdot)$ is a quotient of $(\mathbb{C},+)$. This question is concerned with fields $K$ such that $K^*$ is a quotient of $K$ ; that is, with the existence of a surjective group morphism $K\to K^*$. I will refer to such morphisms as "exponentials" on $K$. (I know that the notion of exponential fields exists, but I only consider the surjective case.) +The existence of such a map is not benign : since the additive group $K$ is $q$-divisible for all $q\neq char(K)$, it implies that $K^*$ is also $q$-divisible. In characteristic $0$, this actually implies that $K$ is algebraically closed (I suspect that in characteristic $p$ this implies that $K$ is separably closed, but I'll focus on zero characteristic for now). EDIT : That was false, but it doesn't really matter since Eric's answer gives a construction when $K^*$ is divisible. +Question 1: It's a basic fact that algebraically closed fields of characteristic zero and with the same transcendance degree over $\mathbb{Q}$ are isomorphic. So $\mathbb{C}_p \simeq \mathbb{C}$ for all $p$, and thus $\mathbb{C}_p$ admits (at least one) exponential map. On the other hand, isomorphisms $\mathbb{C}\to \mathbb{C}_p$ are highly non-constructible objects, so this does not give us any clue about what an exponential on $\mathbb{C}_p$ may look like. + +Can such an exponential map $\mathbb{C}_p\to \mathbb{C}_p^*$ be explicitly defined ? In particular, does it exist without the axiom of choice (or with only a weak version) ? + +Question 2: If we put the axiom of choice back on : + +For any countable cardinal $\kappa$, do algebraically closed fields of transcendance degree $\kappa$ over $\mathbb{Q}$ admit a surjective exponential ? (In particular, what about $\overline{\mathbb{Q}}$ ?) + +We already know from $\mathbb{C}$ that this is true for $\kappa = 2^{\aleph_0}$ so since "algebraically closed fields of characteristic zero with a surjective exponential" may be countably axiomatized in first-order logic, Löwenheim-Skolem implies that there are models of all infinite cardinality, so in particular question 2 has a positive answer for all uncountable $\kappa$, and for some countable $\kappa$. But since all countable $\kappa$ give models of the same cardinality $\aleph_0$, we cannot use that remark to answer the question. + +REPLY [7 votes]: I have no idea how to answer question 1, but in the presence of the axiom of choice, everything is very simple. If $K$ is an algebraically closed field of characteristic $0$, then the additive group of $K$ is isomorphic to a direct sum of $|K|$ copies of $\mathbb{Q}$ (this is trivial when $K$ is uncountable; when $K$ is countable note that it must have infinite degree over $\mathbb{Q}$ to be algebraically closed). The torsion subgroup of the multiplicative group (i.e., the roots of unity) is $\mathbb{Q}/\mathbb{Z}$, and since $\mathbb{Q}/\mathbb{Z}$ is injective, the multiplicative group splits as a direct sum $\mathbb{Q}/\mathbb{Z}\oplus A$ for some group $A$. The group $A$ is torsion-free and divisible and hence a $\mathbb{Q}$-vector space, and its dimension is easily seen to be $|K|$ (this is trivial when $K$ is uncountable; when $K$ is countable note that the prime integers are an infinite linearly independent set). +Thus $K^*$ is a direct sum of $\mathbb{Q}/\mathbb{Z}$ and $|K|$ copies of $\mathbb{Q}$. There is clearly a surjective homomorphism from $K$ to this direct sum, since there is a surjective homomorphism $\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$ and $|K|$ is infinite. In fact, there is a surjective homomorphism whose kernel is cyclic, just like in the case of $\mathbb{C}$. +Note that all that this argument needs from the axiom of choice is for $K$ to be well-orderable. So if you know $K$ is well-orderable without the axiom of choice, you get a surjective exponential without choice. In particular, for instance, $\overline{\mathbb{Q}}$ has a surjective exponential which you can actually write down (in terms of explicit bases for $\overline{\mathbb{Q}}$ and $\overline{\mathbb{Q}}^*$), though this will of course still not be very "natural". +Note also that it is not actually true that $K^*$ being divisible implies $K$ is algebraically closed. Indeed, this just says $K$ is closed under taking radicals, but since not all polynomials can be solved by radicals, this will not generate all roots of polynomials. The above discussion, however, applies perfectly well to any field of characteristic $0$ such that $K^*$ is divisible (with the modification that the group of roots of unity might be a subgroup of $\mathbb{Q}/\mathbb{Z}$ rather than all of $\mathbb{Q}/\mathbb{Z}$; however, as a divisible subgroup, it will be a direct summand of $\mathbb{Q}/\mathbb{Z}$, and hence still a quotient of $\mathbb{Q}$). +Finally, I would note that almost no field $K$ of characteristic $p>0$ has a surjective exponential. Indeed, every element of $K$ is annihilated by $p$, so this would mean $x^p=1$ for all $x\in K^*$. This implies $x=1$ for all $x\in K^*$, so the only such field is $K=\mathbb{F}_2$.<|endoftext|> +TITLE: In how many ways can the letters of word $PERMUTATIONS$ be arranged if there are always 4 letters between P and S? +QUESTION [5 upvotes]: In how many ways can the letters of word $PERMUTATIONS$ be arranged if there are always $4$ letters between $P$ and $S$? + +Now there are $12$ blank spaces, which we have to fill by the letters of the given word. +That is :- +[][][][][][][][][][][][] +it is given that two places are always filled with P and S such that there are 4 blank spaces between them. +{[P][][][][][S]}[][][][][][] +$Therefore$, the number of arrangements that can be made such that there are 4 letter between $P$ and $S$ is $P(10,4)$ +$p(10, 4) = \frac{10!}{(10 - 4)!} =\frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040 \tag 1$ +Also {[P][][][][][S]} can be arranged in different ways with respect to {[P][][][][][S]}[][][][][][]. +$Therefore$ the number arrangements will be $p(7, 7)$ +$p(7, 7,) = 7! = 5040 \tag 2$ +$Therefore$ the the total number of arrangements for the given question will be +$5040 \times 5040 \text{ (results of 1 and 2)}\\ += 25401600 \tag3$ +but there are two T's in the word permutation +$Therefore$ we divde the result of 3 by 2 to get the real answer. +That is , $$\frac{25401600}{2} = 12700800$$ +So I am getting $12700800$ as the answer but it is wrong the given answer is $25401600$. How ? +Thanks and sorry for the formatting. + +REPLY [6 votes]: You can arrange the $10$ letters other than P and S in a row in ${10!\over2}$ ways (divide by $2$ because of the double T). There are $11$ slots between these letters and at the ends of the row. Write P (or S) in slot $k\in[1\>..\>7]$, and $S$ (or P) in slot $k+4\in[5\>..\>11]$. This allows for $2\cdot7$ choices in all. The total number of arrangements therefore comes to $7\cdot10!=25\,401\,600$.<|endoftext|> +TITLE: Prove that $\exists x_0, x_1\in (0,1)$, such that $\frac{f'(x_0)}{x_0}+\frac{f'(x_1)}{x_1^2}=5$ +QUESTION [10 upvotes]: Let $f:[0,1]\to\mathbb{R}$ be a differentiable function, such that + $f(0)=0$ and $f(1)=1$. Prove that there exist different $x_0, x_1\in +(0,1)$, such that $$\frac{f'(x_0)}{x_0}+\frac{f'(x_1)}{x_1^2}=5$$ + +I have thought of possibly using Cauchy's theorem, so that we would only need to prove that there exist $k, l\in (0,1)$ such that $f(k)=k^2$ and $f(l)=l^3$, but I don't know how to prove these. Any hint? +Edit 1: Apparently my thought here is wrong, so any ideas? + +REPLY [3 votes]: Consider the function $g(x) = \frac{1}{3}x^3$. Then as it's differentiable on $(0,1)$ and continuous on $[0,1]$, by Cauchy Mean Value Theorem we have $\exists c \in (0,1)$, s.t.: +$$\frac{f(1) - f(0)}{g(1) - g(0)} = \frac{f'(c)}{g'(c)} \implies \frac{f'(c)}{c^2} = 3$$ +Similarly now taking $h(x) = \frac{1}{2}x^2$, by Cauchy Mean Value Theorem $\exists d \in (0,1)$, s.t.: +$$\frac{f(1) - f(0)}{h(1) - h(0)} = \frac{f'(d)}{h'(d)} \implies \frac{f'(d)}{d} = 2$$ +Hence the proof.<|endoftext|> +TITLE: How to prove that the wedge product is the determinant (Spivak's Claim) +QUESTION [6 upvotes]: In the book I am using now, Spivak's A comprehensive introduction to differential geometry volume 1 I have a question on page 205. Because he says the following +I think this is equivalent to say the following Let $\{e_i\}$ be the standard basis for $\mathbb{R}^{n}$ and let $\{e_i^{\ast}\}$ be dual basis then I want to conclude that $e_1^{\ast} \wedge e_2^{\ast} \wedge ... \wedge e_n^{\ast}$ is the determinant tensor because I have seen that is books or courses this is given as the definition but for me in dimension 2 is little bit different, +$$v_1^{\ast} \wedge v_2^{\ast}=v_1^{\ast} \otimes v_2^{\ast}-v_2^{\ast} \otimes v_1^{\ast}$$ +I know that somehow this could be interpret as the determinant but I can't figure out how can I prove the equivalence of the definitions and what Spivak is claiming in that part of the book. +I was trying to figure out this knowing how the wedge product $e_1^{\ast} \wedge e_2^{\ast} \wedge ... \wedge e_n^{\ast}$ evaluate on the standard basis, and using skew symmetry and multi-linearity, but I'm still puzzled. +Thanks a lot in advance. + +Attempts + + +I was trying to do something of the following sort + +We take $T \in \Lambda^{k}(V^{k})$ and $\{v_i\}$ a basis for V, then we can form k-tuples but we discard the ones that are $v_i\times ... \times v_i$ since this is zero so $T$ is uniquely determined by $T(v_1,...,v_k)=c$. But then +$$c(v_1^{\ast} \wedge v_2^{\ast} \wedge ... \wedge v_n^{\ast})$$ +is in $\Lambda^{k}(V^{k})$ and $c(v_1^{\ast} \wedge v_2^{\ast} \wedge ... \wedge v_n^{\ast})(v_1,...,v_k)=c$ so that implies that $T=c(v_1^{\ast} \wedge v_2^{\ast} \wedge ... \wedge v_n^{\ast})$ +But I don't know if this is right or how to perform this in my case. + +The reason because of I was trying to use the evaluation in the standard basis is because I have the following lemma + +Lemma: Let $A: V \to V$ be a linear map. Then if $w_i=Av_i$, $v_1,v_2 \in V$ then +$$v_1^{*} \wedge v_2^{*}(w_1,w_2)=detA (v_1^{*} \wedge v_2^{*})(v_1,v_2)$$ + +May be another way to prove this is to prove that that the wedge product satisfies the definition of determinant given in the book Second Year Calculus by Bressoud but by definition the wedge product is linear and the conditions that $e_1^{\ast} \wedge e_2^{\ast} \wedge ... \wedge e_n^{\ast}(e_1,...,e_n)=1$ holds (I don't know how to prove that) the same for the condition that $e_1^{\ast} \wedge e_2^{\ast} \wedge ... \wedge e_n^{\ast}(e_1,e_2,...,e_2,e_n)=0$ +So I don't know if this is a good way to proceed. + +REPLY [5 votes]: As $\dim \Lambda^k (V^n) = \binom{n}{k}$, if $k = n$, then clearly $\dim \Lambda^n (V^n) = 1$. If $\lambda \in \mathcal L(V,V)$ (set of all linear transformations) then $\lambda^* : \Lambda^n (V^n) \to \Lambda^n (V^n)$ is a linear transformation between one-dimensional vector spaces, and so it must be a multiplication by an element in $\mathbb R$. Thus there is a unique number (call it) $\det (\lambda) \in \mathbb R$ such that $$\lambda^*(\omega) = \det (\lambda) \omega $$ +where $\omega \in \Lambda^n(V^n)$. +Now if $A = (a^{i}_{j})$ is the matrix of $\lambda \in \mathcal L (V,V)$ and taking $\{e_1,\ldots, e_n\}$ some basis with dual $\{e^1,\ldots, e^n\}$ (I'm using $e^i$ instead $e_i^*$ just to avoid typing) then we have +$$\begin{align}\lambda^*(e^1 \wedge \ldots \wedge e^n) (e_1, \ldots, e_n) &= (e^1\wedge \ldots \wedge e^n)\lambda(e_1, \ldots, e_n)\\&= (e^1 \wedge \ldots \wedge e^n) (\lambda e_1, \ldots, \lambda e_n)\\&= \det (e^i (\lambda e_j)) = \det A\end{align}$$ +since $\lambda (e_i) = \sum a^i_j e_j$. On the other hand, $$\lambda^* (e^1 \wedge \ldots \wedge e^n) = (\det \lambda )(e^1 \wedge \ldots \wedge e^n)$$ +and as $(e^1 \wedge \ldots \wedge e^n) (e_1,\ldots, e_n) = 1$ we get that $\det (\lambda) = \det A$. +Edit: Here is a Lemma + +Let $\alpha^1 , \ldots, \alpha^k$ elements of $\Lambda^1(V^n)$ and $v_1, \ldots, v_k$ vectors in $V$. Then $$(a^1 \wedge \ldots \wedge \alpha^k)(v_1, \ldots, v_k)=\det A$$ + +where $A = (a^i_j)$ is the $k \times k$ matrix whose $ij-$th entry is $a^i_j = \alpha^i (v_j)$. +Proof: (idea) Show that $$\alpha \wedge (\beta \wedge \gamma) = \frac{(k_1 + k_2 + k_3)!}{k_1!k_2!k_3!} \mathrm {Alt} \, (\alpha \otimes (\beta \otimes \gamma))$$ +by inductive application you get $$a^1 \wedge \ldots \wedge \alpha^k = k! \mathrm {Alt} (\alpha^1 \otimes \ldots \otimes \alpha^k)$$ +Thus $$a^1 \wedge \ldots \wedge \alpha^k(v_1, \ldots, v_k) = \sum_{\sigma} \mathrm{ sign} (\sigma)\,\,\alpha^1(v_{\sigma(1)}) \cdots \alpha^k(v_{\sigma(k)}) = \det A$$ +where $\sigma \in S_k$ is a permutation. +Note: Here $\alpha \in \Lambda^{k_1}(V^n)$, $\beta \in \Lambda^{k_2}(V^n)$, $\gamma \in \Lambda^{k_3}(V^n)$.<|endoftext|> +TITLE: Difference in use between $d$, $\partial$, $\operatorname d$, $\varDelta$ and $D$ for derivatives. +QUESTION [6 upvotes]: While reading different sources on implicit differentiation (and thereafter differentiation in general), I came across many different "d's" being used for (or similar to) the familiar +$$\frac{dy}{dx}$$ +The variant $d$ appears in the majority of cases, and also with integrals: +$$\int x~dx$$ +Now, it appears like the convention on Wikipedia is to use $d$ or $\operatorname d$ for "normal" and $\partial$ for partial (which is also its name in Tex) derivatives. Is there an additional difference between $d$ and $\operatorname d$ (except the amount of characters needed), and are there any general conventions? + +REPLY [6 votes]: I regard $d$ and $\rm d$ as the same. In general, I think of it like this: + +If $f: \Bbb R \to \Bbb R$, the derivative at $x$ is $f'(x)$ or $\frac{{\rm d}f}{{\rm d}x}(x)$; +If $f: \Bbb R^n \to \Bbb R$, the $k-$th partial derivative at $x$ is $\frac{\partial f}{\partial x^k}(x)$, or $D_kf(x)$, or $D_{x^k}f(x)$, or $\partial_kf(x)$ or $\partial_{x^k}f(x)$; +If $f: \Bbb R^n \to \Bbb R^m$, the total derivative (Fréchet derivative, if you like it) at $x$ is $Df(x)$. In particular, we can call the partial derivative $\frac{\partial f}{\partial x^k}(x)$, which will be a vector whose components are the partial derivatives of the components, following the above item. If $f: \Bbb R^n \times \Bbb R^m \to \Bbb R^k$ we can also use $D_xf(x,y)$ and $D_yf(x,y)$ to denote blocks of $Df(x,y)$. This is useful sometimes when doing things with the inverse/implicit function theorems. +If $f: \Bbb R^n \to \Bbb R^m$, the differential of $f$ at $x$ is the map ${\rm d}f_x: T_x\Bbb R^n \to T_{f(x)}\Bbb R^m$ given by ${\rm d}f_x(v_x) = (Df(x)(v))_{f(x)}$. Some people write ${\rm d}_xf$ instead of ${\rm d}f_x$. +The exterior derivative of $k$-form $\omega$ is denoted simply by ${\rm d}\omega$.<|endoftext|> +TITLE: Do endomorphisms of infinite-dimensional vector spaces over algebraically closed fields always have eigenvalues? +QUESTION [6 upvotes]: Let $V$ be a vector space over an algebraically closed field $K$ and let $f:V\to V$ be an endomorphism. If $V$ is finite-dimensional, we know that the characteristic polynomial $\chi_f$ has a zero $\lambda$, which is an eigenvalue of $f$. However, does a similar argument hold if $\dim V=\infty$? + +REPLY [7 votes]: The answer is no. Consider the space of infinite real sequences and let $T(x)=(0,x_1,x_2,\dots)$. Suppose $T(x)=\lambda x$. +If $\lambda=0$, then $x$ has to be zero, so $0$ is not an eigenvalue. +If $\lambda \neq 0$, then $0=\lambda x_1$, so $x_1=0$, but then $x_1=0=\lambda x_2$. Continuing by induction we again conclude that $x=0$. So $\lambda$ is not an eigenvalue either. +So $T$ has no eigenvalues. +Note that there is no contradiction with the finite dimensional case, because the finite dimensional analogue of this "forward shift" operator must necessarily "forget" about $x_n$, which means that $(0,0,\dots,1)$ will be an eigenvector with eigenvalue zero. + +REPLY [4 votes]: No, the shift homomorphism defined on $(e_1,...,e_n,...)$ by $f(e_i)=e_{i+1}$ does not have an eigenvalue.<|endoftext|> +TITLE: Convergence of moments of a sequence of random variables +QUESTION [5 upvotes]: I encountered this problem in my study of time series. It seemed trivial at first but I don't see the finishing move to complete the proof. The problem is as follows. +Let $(X_n)_n$ be a sequence of random variables and $X$ also a random variable on some probability space. Suppose that $E\left[\lvert X_n - X\rvert^p\right]\to 0$ as $n$ goes to infinity and $E\left[\lvert X\rvert^p\right]< \infty$ for some $p \geq 1$. Show that $E[X_n^k] \to E[X^k]$ for every $0 +TITLE: Linear transform of a strictly stationary time series +QUESTION [5 upvotes]: First, let me clarify what I mean by a strictly stationary time series. Let $(X_t)_{t\in \mathbb{Z}}$ be a sequence of random variables on some probability space. If it holds that +$$(X_t, X_{t+1},\ldots,X_{t+h}) \stackrel{d}{=} (X_s, X_{s+1},\ldots,X_{s+h})$$ +for every $s,t \in \mathbb{Z}$ and every $h \in \mathbb{Z}_{\geq 0}$. +Next, I will explain what I mean by linear transform. Given a sequence of reals $(\psi_j)_{j\in \mathbb{Z}}$ such that $\sum_j\lvert\psi_j\rvert < \infty$, linear transform of $(X_t)$, which we denote by $(Y_t)$, is defined as +$$Y_t = \sum_j\psi_jX_{t-j}$$ +Now suppose that the sum above is well-defined, i.e. $\sum_j\psi_jX_{t-j}$ is finite almost surely. ($E\lvert X_t\rvert < \infty$ would be sufficient for this but we don't assume that.) +How do I then show that $(Y_t)$ is also strictly stationary? Intuitively speaking, +$$\sum_j\psi_jX_{t-j} \stackrel{d}{=} \sum_j\psi_jX_{s-j}$$ +since $(X_t)$ is strictly stationary. But shifting a finite sequence is not quite the same as ``shifting an infinite sequence". How do I make this rigorous? + +REPLY [3 votes]: Fix some $s,t \in \Bbb Z$. Note that since $(X_t, X_{t+1},\ldots,X_{t+h}) \stackrel{d}{=} (X_s, X_{s+1},\ldots,X_{s+h})$ for all $h$, we have that $$(X_{t+k})_{k \in \Bbb Z} \stackrel{d}{=} (X_{s+k})_{k \in \Bbb Z}$$ since both processes have the same finite-dimensional distributions. If you are not familiar with this fact from elementary measure theory, see for example Theorem 23 here (there $\Xi$ denotes the space of functions endowed with product $\sigma$-algebra) or alternatively Proposition 3.1 here. +Now define the map $\psi: \Bbb R^{\Bbb Z} \to \Bbb R^{\Bbb Z}$ by sending $(x_j)_{j \in \Bbb Z} \mapsto (\sum_j \psi_j x_{k-j})_{k \in \Bbb Z}$ and note that it is measurable on the product $\sigma$-algebra (since the individual components are). Fix some $s,t \in \Bbb Z$, and since $(X_{t+k})_{k \in \Bbb Z} \stackrel{d}{=} (X_{s+k})_{k \in \Bbb Z}$ we easily get that $$(Y_{t+k})_{k\in\Bbb Z}=\psi\big((X_{t+k})_{k \in \Bbb Z}\big) \stackrel{d}{=} \psi\big((X_{s+k})_{k \in \Bbb Z}\big)=(Y_{s+k})_{k\in\Bbb Z}$$ which precisely means stationarity of $Y$.<|endoftext|> +TITLE: Understanding the proof of Möbius inversion formula +QUESTION [8 upvotes]: I am trying to understand one step in the proof of the Möbius inversion formula. +The theorem is + +Let $f(n)$ and $g(n)$ be functions defined for every positive integer $n$ satisfying $$f(n) = \sum_{d|n}g(d)$$ + Then, g satisfies $$g(n)=\sum_{d|n}\mu(d) f(\frac{n}{d})$$ + +The proof is as follows: +We have $$\sum_{d|n}\mu(d)f(\frac{n}{d}) = \sum_{d|n}\mu(\frac{n}{d})f(d) = \sum_{d|n}\mu(\frac{n}{d}) \sum_{d'|d}g(d') = \sum_{d'|n}g(d')\sum_{m|(n/d')}\mu(m)$$ +Where in the last term, the inner sum on the right hand side is $0$ unless $d'=n$. +My question is: how do we get the last equation? I don't really understand how the author interchanges the summation signs. +Thanks for any help! + +REPLY [7 votes]: First, considering the sum +\begin{align*} +\sum_{d|n}\mu(\frac{n}{d})\sum_{m|d}g(m), +\end{align*} +let's take a look into the indices +$$ +n=\frac{n}{d}d=\frac{n}{d}\frac{d}{m}m=khm, +$$ +with +$$ +\frac{n}{d}=k, \frac{d}{m}=h. +$$ +Thus, we have +\begin{align*} +\sum_{d|n}\mu(\frac{n}{d})\sum_{m|d}g(m) +&=\sum_{dk=n}\mu(k)\sum_{hm=d}g(m)\\ +&=\sum_{khm=n}\mu(k)g(m)\\ +&=\sum_{mkh=n}g(m)\sum_{kh=\frac{n}{m}}\mu(k)\\ +&=\sum_{m|n}g(m)\sum_{k|\frac{n}{m}}\mu(k)\\ +&=\sum_{m|n}g(m)[\frac{m}{n}]=g(n). +\end{align*} +If you are reading Apostol, then in his book's convention, we can see that we're in fact doing the convolution +$$ +f*\mu=(g*U)*\mu=g*(U*\mu)=g*I=g, +$$ +with $U(n)=1$ and $I(n)=[\frac{1}{n}].$ +The tricky part here is that there are really three arithmetic functions convoluting with each other, namely, $f$, $g$ and $U.$<|endoftext|> +TITLE: Evaluation of $ \int_{0}^{\pi}\ln(5-4\cos x)\,dx$ +QUESTION [6 upvotes]: Evaluation of $\displaystyle \int_{0}^{\pi}\ln(5-4\cos x)dx = \int_{0}^{\pi}\ln(5+4\cos x)dx$ + +$\bf{My\; Try::}$ Let $\displaystyle I(a,b) = \int_{0}^{\pi}\ln(a+b\cos x)dx$ +Then $$\frac{d}{db}(a,b) = \frac{d}{db}\left[\int_{0}^{\pi}\ln(a+b\cos x)dx\right]db$$ +So $$I'(a,b) = \int_{0}^{\pi}\frac{\cos x}{a+b\cos x}dx = \frac{1}{b}\int_{0}^{\pi}\frac{(a+b\cos x)-a}{a+b\cos x}dx$$ +So we get $$I'(a,b) = \frac{\pi}{b}-\frac{a}{b}\int_{0}^{\pi}\frac{1}{a+\cos x}dx$$ +Using half angle formula $\displaystyle \tan x = \frac{1-\tan^2 x/2}{1+\tan^2 x/2}$ +so we get $$I'(a,b) = \frac{\pi}{b}-\frac{a}{b}\int_{0}^{\pi}\frac{\sec^2 x/2}{(a+b)+(a-b)\tan^2 x/2}dx$$ +Now Put $\tan x/2= t\; $ Then $\displaystyle \sec^2 \frac{x}{2}dx = 2dt$ +So we get $$I'(a,b) = \frac{\pi}{b}-\int_{0}^{\infty}\frac{1}{(a+b)+(a-b)t^2}=\frac{\pi}{b}-\frac{a\pi}{2b\sqrt{a^2-b^2}}$$ +So we get $$I(a,b) = \pi\ln|b|-\frac{\pi a}{2}\left[-\frac{\ln|b^2-a^2|+2\ln |b|}{2a^2}\right]$$ +So we put $a = 5$ and $b=4$ +We get $$I(5,4) = \pi\cdot \ln (5)-\frac{\pi}{2}\left[\frac{-\ln(9)+2\ln(5)}{2\cdot 5}\right]$$ +So We get $$I(5,4) = \left[\frac{18\ln(5)+\ln(9)}{20}\right]\cdot \pi$$ +I did not understand where i have done mistake, Help me +Thanks + +REPLY [2 votes]: This approach seems really good, but the first mistake was in +$$\int_0^{\pi}\frac1{a+b\cos x}dx=\frac1a\int_0^{\pi}\frac1{1+e\cos x}dx$$ +Where $e=\frac ba$. If you used the eccentric anomaly at this point, +$$\sin E=\frac{\sqrt{1-e^2}\sin x}{1+e\cos x}$$ +So that $$dE=\frac{\sqrt{1-e^2}}{1+e\cos x}dx$$ +You would get +$$\int_0^{\pi}\frac1{a+b\cos x}dx=\frac1a\int_0^{\pi}\frac{dE}{\sqrt{1-e^2}}=\frac{\pi}{\sqrt{a^2-b^2}}$$ +So you were off by a factor of $2$ at this point. Next, if we let $b=a\sin\theta$, then +$$\begin{align}\int\frac{db}{b\sqrt{a^2-b^2}}&=\frac1a\int\csc\theta\,d\theta=-\frac1a\ln\left|\csc\theta+\cot\theta\right|+C_1(a)\\ +&=-\frac1a\ln\left(\frac ab+\frac{\sqrt{a^2-b^2}}b\right)+C_1(a)\\ +&=-\frac1a\ln\left(a+\sqrt{a^2-b^2}\right)+\frac1a\ln b+C_1(a)\end{align}$$ +So you would arrive at +$$I(a,b)=\pi\ln\left(a+\sqrt{a^2-b^2}\right)+C(a)$$ +At $b=0$ we get $\pi\ln a=\pi\ln2a+C(a)$ so $C(a)=-\pi\ln2$. Then +$$I(5,4)=\pi\ln8-\pi\ln2=\pi\ln4$$ +Numerical quadrature confirms this result.<|endoftext|> +TITLE: Convergence properties of $z^{z^{z^{...}}}$ and is it "chaotic" +QUESTION [9 upvotes]: $\DeclareMathOperator{\Arg}{Arg}$ +Let $z \in \mathbb{C}.$ Let $b = W(-\ln z)$ where $W$ is the Lambert W Function. Define the sequence $a_n$ by $a_0 = z$ and $a_{n+1} = {a_0}^{a_n}$ for $n \geq 1$, that is to say $a_n$ is the sequence $z, z^z, z^{z^z} ...$ +I am trying to classify the sequence $a_n$ as convergent or divergent (i.e. not convergent) according to $|b|$. My approach to this involves defining $2$ additional sequences: $b_n = \ln a_n$ and $c_n = b_n - b$. Then $b_{n+1} = \ln \left(z^{e^{b_n}}\right) = e^{b_n} \ln z$ and $c_{n+1} = e^{(b + c_n)} \ln z - b= (e^b \ln z) e^{c_n} - b = b e^{c_n} - b = b (e^{c_n} - 1)$. Suppose $b_n \to b \implies c_n \to 0 \implies e^{c_n} - 1 \sim c_n \implies c_{n+1} \sim bc_n$. From here there are $3$ cases to consider: +If $|b| > 1$ then $c_n$ is increasing without bound, so it clearly cannot converge except if $c_0 = 0$ which would imply $c_n = 0$ for every $n$ which is not the case because $b_0 =\ln z$ so $b_1 = e^{\ln z}\ln z \ne b_0$ so $c_1 \ne c_0$, a contradiction $\implies b_n$ is divergent $\implies a_n$ is divergent. +If $|b| < 1$ then $c_n$ is decreasing towards $0 \implies b_n$ is convergent $\implies a_n$ is convergent. +If $|b| = 1$ then $c_n \to c \ne 0 \implies b_n$ is divergent $\implies a_n$ is divergent. +I would like to give credit to user1952009. The method involving the sequences $b_n$ and $c_n$ is due to them. +Edit 1: After reading Yiannis Galidakis's answer several times, I have decided it is mostly satisfactory. Having said that, I feel I must clarify my position on chaotic sequences. When I said the concept of a chaotic sequence was nonsense, that was an exaggeration. What I meant to say was the definition given in the previous post didn't really make sense to me. The one given here is detailed and precise enough that I can understand what it actually means. +However, I do not believe the statements "$a_n$ is chaotic" and "$a_n$ does not converge" are incompatible. Setting aside the question of whether $a_n$ is chaotic or not, I feel I have given a plausibility argument (although not a rigorous proof) that $a_n$ is convergent if and only if $c_n \to 0$ which happens if and only if $|b| < 1$. +On a different note, I have been experimenting numerically with $a_n$ for at least $3$ years. I believe (but cannot prove) that whenever $z$ is not purely real and $|b| = |W(-\ln z)| > 1$ the sequence $a_n$ +actually has a $k$-cycle for some $k \in \mathbb{N}$. It appears that $k = 3$ for all sufficiently large $|z|$ and that $k \to \infty$ as $|b| \to 1$. +Similarly, whenever the sequence "shoots off" too a neighborhood of $\infty$ I have found that $a_n \sim$ {$..., \infty, 0, 1, z, z^z ...$} I will admit however that I cannot predict when the sequence will "shoot of" to $\infty$. +At this point I have $2$ main questions: is my convergence analysis correct? Are there any values of $z$, not purely real, such that $a_n$ is unbounded, i.e, the sequence "blows up" to $\infty$ and never "comes back down." +Apologies for a somewhat lengthy post. +Edit 2: In Addendum 2 in Yiannis Galidakis's answer, it is stated that if $c \in \mathbb{C}$ and $|\Re(c)| > \exp(\exp(-1))$ and $\Arg{c}=2\pi\alpha$ with $\alpha\in\mathbb{R}\setminus\mathbb{Q}$ then the iterated exponential will be unbounded. This is not quite correct. The part about irrational rotation is fine, but it is not enough to have $|\Re(c)| > \exp(\exp(-1))$. Consider $c = 2\exp(e\pi i/14) \approx 1.6393177+1.1457037i$. Clearly $c$ satisfies both conditions. However $|W(-\ln c)| \approx .890512$. Hence the iterated exponential is convergent, hence it is bounded. A few other counterexamples are given by $c = k\exp(e\pi i/14)$ for $k \in \mathbb{R}$ and $1.76253 \le k \le 2.34896$ + +REPLY [5 votes]: This is an extended response to Yiannis Galidakis's question. +I have been experimenting with the value $c$ they gave in the previous post. It is in case 3b since $|t| = |W(-\ln c)| = 1$ and there is no $n \in \mathbb{N}$ such that $t^n = 1$. I have evaluated the sequence $a_n$ for $0 \leq n \leq 10^8$. As in Gottfried Helm's analysis, I have found no evidence that the terms are contracting towards some limit. In fact, it appears the sequence is frequently "slingshot" to a neighborhood of $\infty$, and that the terms are being "slingshot" progressively farther out as $n \to \infty$. As an example, the largest such "slingshot" has $|a_n| = 17.1$ for $n \leq 10^6$. For $n \leq 10^8$ this increases to about $25$. In all such cases, I have observed that $a_n \sim$ {$...,\infty, 0, 1, z, z^z, ...$}. For these reasons I believe $a_n$ is divergent (i.e. not convergent) in case 3b. +As to whether $a_n$ is chaotic or not, I am not completely sure. This may be a bad analogy, but, the way I have come to think of it, when we say a sequence or other mathematical object/phenomenon is chaotic, what we're really saying is the condition number of some function is very high relative to some other function. By this definition, I believe the sequence $a_n$ can be chaotic in all cases, but is especially chaotic in cases 3a and 3b. For one thing, an arbitrarily small perturbation in $a_0 = z$ can leave the sequence in the same case or shift it into any of the others. Also, using the formula $cond(f(x)) = \frac{xf'(x)}{f(x)}$ we see that $a_n$ is very ill-conditioned when it is "slingshot" near $\infty$. +In spite of these issues, I do not lose much sleep over it. As far as I know, there are $4$ ways to alleviate the difficulty of an ill-conditioned problem. + +Supercomputers +Arbitrary precision calculators +A very sophisticated algorithm +A lot of time and effort + +Of course there are some problems where even all $4$ of these are not enough. As someone once said "Computers and calculators are not replacements for mathematical thought. They are just replacements for some kinds of mathematical labor, either numerical or symbolic. There are, and always will be, mathematical problems that can’t be solved by a calculator or computer, regardless of its size and speed."<|endoftext|> +TITLE: Why is this determinant positive? +QUESTION [11 upvotes]: I have seen that the $k$-dimensional volume of an parallelepiped in $\mathbb{R}^n$, i.e., $$P(v_1, \ldots, v_k) = \{t_1v_1 + \dotsb + t_kv_k : 0 \le t_i \le 1 \}$$ is $\sqrt{\det(T^{\top}T)}$, where $T$ is the $n\times k$ matrix with columns $v_1, \ldots, v_k$. +How do we know that $\det(T^{\top}T)$ is non-negative? + +REPLY [19 votes]: $T^{\top}T$ is positive semidefinite, so all the eigenvalues are non-negative. +The determinant of $T^{\top}T$ is the product of the eigenvalues; hence, it is non-negative.<|endoftext|> +TITLE: Existence of an $n\times n$ real matrix $A$ such that $A^2=-I$. +QUESTION [8 upvotes]: Let $A$ be a $n\times n$ real matrix $A$ such that $A^2=-I$. Such an $A$ cannot be, + +Orthogonal. +Invertible. +Skew-symmetric. +Symmetric. +Diagonalizable. + +I tried to figure out the answer by looking at the [possible] determinant of $A$, using the fact $\det(AB)=\det(A)\det(B)$. So $\det(A^2)=(\det(A))^2=(-1)^n\det(I)$. If $n$ is even, then $\det(A)=\pm1$ and if odd then $\det(A)=\pm i$. There I stuck, if $n$ is even, I can try some guessing, but when $n$ is odd, the determinant becomes complex and I have no logic to go forward. Is there any other approach that might be more correct? How can I do this? Help me out. + +REPLY [2 votes]: I experimented with rotation matrices and powers and found the following solution for $\small n=4$ where $\small w = \sqrt{1/3}$ : + 0 w -w w +A= -w 0 w w + w -w 0 w + -w -w -w 0 + +Then + -3*w^2 . . . + . -3*w^2 . . + A^2= . . -3*w^2 . = - I + . . . -3*w^2 + +And also, + +$\small A$ is invertible +$\small A^{-1} = A^T$ (or $\small A \cdot A^T=I $ ) which is a condition for +orthogonal/orthonormal matrices. +Also $\small A^T = - A$ which means, $\small A$ is skew-symmetric. +Using complex numbers, a diagonalization is possible and gives the diagonal $\small D = [1,1,-1,-1] \cdot \sqrt{-3} $<|endoftext|> +TITLE: Eigenvalues of block matrix of order $m+1$ +QUESTION [5 upvotes]: How to find eigenvalues of following matrix? +$\begin{bmatrix} +mkI-A & -A & -A & \cdots & -A\\ +-A & kI-A & O & \cdots & O\\ +-A & O & kI-A & \cdots & O\\ +\vdots & \vdots & \vdots & \ddots & \vdots\\ +-A & O & O & \cdots & kI-A\\ +\end{bmatrix}_{m+1}$ +Where, +$A$ is any square matrix of order $n$, +$O$ is zero matrix of order $n$ +$I$ is identity matrix of order $n$, +$k \in \mathbb{N}$ +As well as one eigenvalue of above matrix is zero. +I think if we can convert above matrix in terms of either Kronecker product or Kronecker sum of two matrices then we can find eigenvalues of above matrix by taking multiplication or addition of two matrices respectively. +The other way is might be if we can convert above matrix in block diagonal matrix then we can find eigenvalues easily. + +REPLY [3 votes]: At first, to summarize this problem, let replace diagonal factor matrices as: +$$ +\begin{bmatrix} +X \\ +Y +\end{bmatrix} += +\begin{bmatrix} +\lambda-mk & 1 \\ +\lambda-k & 1 +\end{bmatrix} +\begin{bmatrix} +I \\ +A +\end{bmatrix} +$$ +where $\lambda$ is eigenvalue. Therefore, we can suppose the following determinant: +$$ +\begin{aligned} +\det +\begin{bmatrix} +X & A & A & \cdots & A & A \\ +A & Y & O & \cdots & O & O\\ +A & O & Y & \cdots & O & O \\ +\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ +A & O & O & \cdots & Y & O \\ +A & O & O & \cdots & O & Y +\end{bmatrix} +\end{aligned} +$$ +Calculation determinant has an important property that each rows can be added or reduced from another row. Thus, top row can be subtracted from bottom row to extract the diagonal factor as $\det(Y)$. Then the block matrix becomes smaller from $(m+1)n$ square to $mn$ square: +$$ +\begin{aligned} +\det +\begin{bmatrix} +X-AY^{-1}A & A & A & \cdots & A & O \\ +A & Y & O & \cdots & O & O\\ +A & O & Y & \cdots & O & O \\ +\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ +A & O & O & \cdots & Y & O \\ +A & O & O & \cdots & O & Y +\end{bmatrix} +=& +\det +\begin{bmatrix} +X-AY^{-1}A & A & A & \cdots & A \\ +A & Y & O & \cdots & O \\ +A & O & Y & \cdots & O \\ +\vdots & \vdots & \vdots & \ddots & \vdots \\ +A & O & O & \cdots & Y +\end{bmatrix} +\det(Y) +\end{aligned} +$$ +Likewise, we repeat this operation until the block matrix becomes $n$ orders matrix. Eventually, the below determinant is obtained: +$$ +\det(Y)^m\det(X-mAY^{-1}A) +$$ +Now, we attempt to factorize the second determinant $\det(X-mAY^{-1}A)$ using this consideration. Then the whole result is shown below: +$$ +\det(\lambda I-(kI+A))^{m-1} +\det\biggl(\lambda I-\cfrac{B+\sqrt{C}}{2}\biggr) +\det\biggl(\lambda I-\cfrac{B-\sqrt{C}}{2}\biggr) +$$ +where: +$$ +\begin{cases} +B=-2A+k(m+1)I \\ +\\ +C=4mA^2-5k(m+1)A+k^2(1+m)^3I +\end{cases} +$$ +In conclusion, this $n(m+1)$ square matrix has $3n$ kinds eigenvalues. However, $C=O$ case is $2n$ kinds exceptionally because of multiple solution.<|endoftext|> +TITLE: Is my proof correct? If $f$ has a finite number of discontinuities on $[a, b]$, then it is integrable on $[a, b]$ +QUESTION [5 upvotes]: Question: +Suppose a function $f(x)$ over the interval $[a, b]$ is bounded and has only a finite number of discontinuous points on $[a, b]$. I intend to prove that it must be integrable on $[a, b]$. +Is my proof below correct? +My Answer (Is it correct?): +Since $f(x)$ has only a finite number of discontinuous points, call these points $p_1, p_2, ..., p_n$. Now let $r$ be some number greater than $0$ such that for all $\epsilon > 0$, $2r < \frac{\epsilon}{4M}$. Now let +$M_i =$ sup$_{x \in [p_i - r, p_i +r]}f(x)$ and $m_i =$ inf$_{x \in [p_i - r, p_i +r]}f(x)$. Then $M_i - m_i \leq 2M$. +Now then $f$ must be continuous on each interval $[a, p_1 - r], [p_1 + r, p_2 - r], ...,[p_n + r, b]$ and thus $f$ must be integrable on each one of these intervals. Therefore there exist partitions $P_1, P_2,..., P_{N+1}$ of each of these intervals such that $U(P_k, f) - L(P_k, f) < \frac{\epsilon}{2(N + 1)}$. $U(P_k, f)$ and $L(P_k, f)$ are the upper and lower sums of $f$ over their respective partitions. +Let $P$ be the partition given by the $P_1 \cup P_2 \cup ... \cup P_{N+1}$. +Then $U(P, f) - L(P, f) = U(P_1, f) - L(P_1, f) + U(P_2, f) - L(P_2, f) + ... U(P_{N + 1}, f) - L(P_{N + 1}, f) + M_1 - m_1 + M_2 - m_2 + ... M_N - m_N < 2M(\frac{\epsilon}{4M}) + \frac{\epsilon}{2(N + 1)}(N + 1) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. +Thus for all $\epsilon > 0$, there exists a partition $P$ of $[a, b]$ such that $U(f, P) - L(f, P) < \epsilon$. +Therefore $f(x)$ is integrable over $[a, b]$. +Is this proof correct ? + +REPLY [4 votes]: You have the correct approach. Just clean up a few details. +You are trying to show that given any $\epsilon > 0$, there exists a partition $P_\epsilon$ such that $U(P_\epsilon,f) - L(P_\epsilon,f) < \epsilon$. +First define $M$, +$$M := \sup_{x \in [a,b]} |f(x)|.$$ +Then it follows that for $i = 1, 2, \ldots, n$ we have +$$M_i - m_i \leqslant 2M,$$ +and the contribution to the difference in upper and lower sums from intervals $[p_i - r, p_i +r]$ is +$$\sum_{i=1}^n(M_i - m_i)[p_i + r - (p_i-r)] \leqslant 4Mnr.$$ +Don't choose $r < \epsilon/4M$ for any $\epsilon$. For a given $\epsilon > 0,$ choose a particular $r = r(\epsilon)$ such that $r < \epsilon/ (8Mn)$ and $[p_i-r,p_i+r] \subset (a,b).$ +Choose partitions $P_k$ of the other $n+2$ intervals needed to cover $[a,b]$ such that +$$U(P_k,f)-L(P_k,f) < \frac{\epsilon}{2(n+2)}.$$ +Also you are proving this for $p_i \in (a,b)$. The same argument applies with some slight modification if $p_1 = a$ or $p_n = b.$<|endoftext|> +TITLE: How to show that $(X(t)-\lambda t)^2 - \lambda t$ is a martingale, where $X(t)$ is a Poisson Process? +QUESTION [5 upvotes]: I am trying to show that $(X(t)-\lambda t)^2 - \lambda t$ is a martingale, where $X(t)$ is a Poisson process with rate $\lambda$. So far, what I have done is: +\begin{align*} +E\left((X(t)-\lambda t)^2 - \lambda t| \mathcal{F}_s\right)&= E\left(X(t)^2-2\lambda t X(t) + \lambda^2 t^2 - \lambda t| \mathcal{F}_s\right) \\ +&=E\left((X(t)-X(s))^2| \mathcal{F}_s\right) + E\left(2X(t)X(s)| \mathcal{F}_s\right) - E\left(X(s)^2| \mathcal{F}_s\right) - E\left(2X(t)\lambda t| \mathcal{F}_s\right) \\ +& + E\left(\lambda^2 t^2| \mathcal{F}_s\right) -\lambda t\\ +&=E\left((X(t)-X(s))^2\right) + 2X(s)E\left(X(t)| \mathcal{F}_s\right) - X(s)^2 -\lambda t - 2 \lambda t E\left(X(t)| \mathcal{F}_s\right)+ \lambda^2 t^2\\ +&= 2\lambda t - 2\lambda tX(s) + 2X(s)\left(X(s)+ \lambda(t-s)\right) -\lambda t - 2\lambda t\left(X(s) + \lambda (t-s)\right)+ \lambda^2 t^2\\ +\end{align*} +I am not sure why my result is not working out. Does anyone have any ideas what I am doign wrong? Thanks! + +REPLY [3 votes]: I think everything you're doing is correct, but perhaps it would benefit from exploiting some cancellations so things work out more smoothly. +In your second line, with six terms, I claim if you take the first, fourth, fifth, and sixth terms and sum them, you get +$$\boxed{ \lambda^2 s^2 -\lambda s -2\lambda t X_s}.$$ +I used this grouping because lots of things cancel. I also used the fact if $Y$ is a Poisson random variable with mean (parameter) $\mu$, $E(Y^2)=\mu^2+\mu$. +This leaves us with evaluating the conditional expectation of +$$2X_sX_t - X_s=X_s(2X_t-X_s)=X_s(2(X_t-X_s) + X_s)=2X_s(X_t-X_s)+X_s^2$$ +Conditioning, and noting that the conditional expectation of $X_t-X_s$ is $\lambda(t-s)$ yields +$$\boxed{2X_s\lambda(t-s)+X_s^2}.$$ +Summing the boxed expressions gives +$$(X_s-\lambda s)^2 - \lambda s,$$ +which shows the martingale property holds.<|endoftext|> +TITLE: (Finitely many minimal primes) + ($R_M$ domain for all maximal $M$) $\Rightarrow$ ($R$ = product of domains) +QUESTION [5 upvotes]: We are doing a homework problem for our commutative algebra class, which asks us to prove: + +Let $R$ be a commutative ring with $1$ containing finitely many minimal prime ideals $P_1, \dots, P_n$. If in addition $R$ satisfies that $R_M$ (the localization of $R$ at $M$) is a domain for all maximal ideals $M \subset R$, then $$ R \cong \frac{R}{P_1} \times \dots \times \frac{R}{P_n}. $$ + +If we could show that $R$ is Artinian, then we would be done by a result from class. But we think we don't have enough to show this. +Any help would be appreciated. Thanks. + +REPLY [5 votes]: If $i \neq j$, we deduce that $P_i+P_j$ is not contained in any maximal ideal: Assume $P_i+P_j \subset M$, then $R_M$ has at least two minimal primes - namely $P_i$ and $P_j$ - and is thus not a domain. +So $P_i+P_j=R$ for $i \neq j$. This is precisely what we need for the Chinese Remainder Theorem, hence we are done. +Note that $\bigcap_{1 \leq i \leq n} P_i$ is the nil-radical of $R$, but $R$ is reduced, since it is locally reduced. +By the way: Your conclusion, that there only finitely many maximal ideals, is plain wrong. For example, $R=\mathbb Z$ clearly satisfies all the assumptions and has infinitely many maximal ideals.<|endoftext|> +TITLE: I have a answer to a question about trace. Is there an easier answer to this question? +QUESTION [5 upvotes]: Let $A\in M_n(\mathbb{C})$. Show that +$$tr\left(\frac{A+A^*}{2}\right)\leq tr((A^*A)^{1/2}).$$ +My answer: +It is easy to see that +$$tr\left(\frac{A+A^*}{2}\right)=\text{Re}(tr(A))\qquad and\qquad tr((A^*A)^{1/2})=\sum_{i=1}^n\sigma_i(A),$$ +where $\sigma(A)$ is the singular value of $A$. +On the other hand, based on Polar decomposition theorem, there exists a unitary matrix $P$ such that $A=P((A^*A)^{1/2})$. +Now, according to [Theorem 8.7.6, Page 551, R. Horn, C. Johnson, Matrix analysis, Second edition], we have: +$$\begin{eqnarray} + \text{Re}(tr(A))&=\text{Re}(tr(P((A^*A)^{1/2})))\\ +&\leq\sum_{i=1}^n \sigma_i(P)\sigma_i ((A^*A)^{1/2}). + \end{eqnarray} $$ +Since $P$ is unitary, $\sigma_i(P)=1$. We have also $\sigma_i ((A^*A)^{1/2})=\sigma_i(A)$. So, +$$ \text{Re}(tr(A))\leq\sum_{i=1}^n \sigma_i(A).$$ +Therefore, we get +$$tr\left(\frac{A+A^*}{2}\right)\leq tr((A^*A)^{1/2}).$$ + +Now, is there an easier answer to this question? + +REPLY [2 votes]: You’ve presented a nice clean proof of that trace inequality, and one could hardly expect to find a shorter argument. The proof of Theorem 8.7.6 given in Horn and Johnson does, however, rely on some results for doubly stochastic matrices (such as Birkhoff’s theorem) that are not exactly obvious. +In the first edition of Matrix Analysis, though, Horn and Johnson have a fairly straightforward exercise that provides an even stronger result than the trace inequality that you presented. Namely, if $A$ is a complex $n\times n$ matrix, $\lambda_1 \ge \ldots \ge \lambda_n$ are the ordered eigenvalues of $\frac{1}{2}\left[{A + A^*} \right]$, and $\sigma_1 \ge \ldots \ge \sigma_n$ are the ordered singular values of $A$, then $\lambda_i \left( {\frac{1}{2}\left[ {A + A^*} \right]} \right) \le \sigma_i \left( A \right)$ for $1 \le i \le n.$ This result, which immediately implies the trace inequality, is outlined in exercise 15 on page 454 (section 7.4) of the first edition of Matrix Analysis. +The outline consists of simply showing that $$\frac{1}{2}y^* \left( {A + A^* } \right)y = {\mathop{\rm Re}\nolimits} (y^* Ay) \le \left\| {Ay} \right\|_2$$ for a Euclidean unit vector $y$, and then applying the Courant-Fischer theorem to $\frac{1}{2}\left[{A + A^*}\right]$ on the left side of the inequality, and the singular-value analog of the Courant-Fischer theorem to the right side of the inequality. If not easier, it is at least arguably more elementary (and stronger) than the proof using Theorem 8.7.6 from Horn and Johnson's book.<|endoftext|> +TITLE: Castelnuovo-Mumford regularity and exact sequence. +QUESTION [5 upvotes]: In a question on MathOverflow it is said that: + +It is known that given a short exact sequence of finitely generated graded modules over a polynomial ring over a field:$$0 \to M'' \to M \to M' \to 0$$ then $\operatorname{reg}M\leq\max(\operatorname{reg}M',\operatorname{reg}M'').$ + +Unfortunately no proof is given there of the above result. Is this result obvious? Please help me to prove this. + +REPLY [6 votes]: Yes, it is obvious. Take the long exact sequence of cohomology groups. Anytime, the cohomology of the two outer terms vanishes, the cohomology of the inner term will vanish, too.<|endoftext|> +TITLE: Is $\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots$ an irrational number? +QUESTION [25 upvotes]: Obviously: +$$\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\cdots=0.1111\dots=\frac{1}{9}$$ +is a rational number. +Now, if we make terms with demoninators in the form: +$$q_n=\sum_{k=0}^{n} 10^k$$ +Then the sum will be: +$$\sum_{n=1}^{\infty}\frac{1}{q_n}=\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots=0.1009181908362007\dots$$ +The decimal expansion of this number appears to be non-periodic. + +How can we prove/disprove that this number is irrational? + +Edit +This number is at OEIS: http://oeis.org/A065444 + +REPLY [2 votes]: This is not the answer to the question of irrationality. This a piece of information about an interesting closed form of the series. +$$q_n=\sum_{k=0}^n 10^k = \frac{10^{n+1}-1}{9}$$ +$q_0=1\quad;\quad q_1=11\quad;\quad q_2=111\quad...$ +$\sum_{n=0}^m \frac{1}{q_n}=\frac{1}{1}+\frac{1}{11}+\frac{1}{111}+... \quad$ ($m+1$ terms). +$$\sum_{n=0}^m \frac{1}{q_n}=\sum_{n=1}^m \frac{9}{10^{n+1}-1}= +\frac{9}{\ln(10)}\left(\psi_{10}(m+2)-\psi_{10}(1)\right)-9(m+1)$$ +$\psi_q(x)$ is the q-digamma function. The leading terms of the asymptotic series are: +$$\psi_{10}(x)\sim (x-\frac{1}{2})\ln(10)-\ln(9)$$ +$$\sum_{n=0}^{m\to\infty} \frac{1}{q_n}\sim +\frac{9}{\ln(10)}\left( (m+2-\frac{1}{2})\ln(10)-\ln(9)-\psi_{10}(1)\right)-9(m+1)$$ +After simplification : +$$\sum_{n=0}^{\infty} \frac{1}{q_n}=\frac{9}{2} +-\frac{9}{\ln(10)}\left( \ln(9)+\psi_{10}(1)\right)$$ +$\psi_{10}(1)\simeq-1.32759401026424207 $ +$\frac{9}{2} +-\frac{9}{\ln(10)}\left( \ln(9)+\psi_{10}(1)\right)\simeq 1.100918190836200736 $ +Note that $\sum_{n=0}^{\infty} \frac{1}{q_n}$ includes the first term$=1$<|endoftext|> +TITLE: Is this a valid argument for proving that a sum of reciprocals is irrational? +QUESTION [10 upvotes]: Suppose we have a strictly increasing sequence of natural numbers. +Suppose that the sum of the reciprocals of the elements converges. +And suppose that the elements have infinitely many prime factors. +Does this imply that the sum of the reciprocals of the elements is irrational? + +My only thought is - yes: + +We can never find a common denominator for all elements +So there is no way to represent the sum as a simple fraction + +My motivation comes from this question on $\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\dots$ +I'm not even sure that $11,111,1111,\dots$ have infinitely many prime factors. +But I was still wondering if this argument (assuming it's correct) could be used. +Thanks + +REPLY [10 votes]: In the general case, this argument is false. +You can use the Sylvester sequence, defined this way : $a_0 = 2$ and $a_{n+1} = \prod \limits_{k=0}^n a_k + 1$. +Hence $a_0 = 2$, $a_1 = 3$, $a_2 = 7$, $a_3 = 43$, $a_4 = 1807$... +You can prove that $a_{n+1} = a_n(a_n - 1) + 1$, and also by induction that $\sum \limits_{k=0}^n \frac{1}{a_k} = 1 - \frac{1}{a_{n+1} - 1}$. +So $\sum \limits_{k=0}^{+\infty} \frac{1}{a_k} = 1$, and the $a_k$ are all coprime (and $1$ is rational)<|endoftext|> +TITLE: Is there a way to cut a an ellipsoid with a plane such that it gives a circle? +QUESTION [5 upvotes]: I'm trying to answer this + +In $\Bbb {R^3} $ consider the ellipsoid: + $2x^2+3y^2+4z^2=1$ It exists a subspace of dimension 2 which intersection with the ellipsoid is a circle. Justify any answer. + +I know that I must search for a plane since these are the only linear subspaces of $\Bbb {R^3} $ of dimension 2, but I don't know what else to do to check if such a plane exists. any hint would be much appreciated. Thanks! + +REPLY [3 votes]: Let $a,b,c$ be any 3 positive numbers such that $0 < a < b < c$. +Let $\mathcal{E}$ and $\mathcal{S}$ be the ellipsoid and sphere defined by +$$\mathcal{E} = \{ (x,y,z) : ax^2+by^2 + cz^2 = 1 \} +\quad\text{ and }\quad +\mathcal{S} = \{ (x,y,z) : b(x^2+y^2+z^2) = 1 \}$$ +For any point $(x,y,z) \in \mathcal{E} \cap \mathcal{S}$, we have +$$ax^2 + by^2 + cz^2 = 1 = b(x^2+y^2+z^2) +\implies (c-b)z^2 = (b-a)x^2 +\iff z = \pm k x +$$ +where $k = \sqrt{\frac{b-a}{c-b}}$. The RHS is the equation for a pair of planes +$P_{\pm} = \{ (x,y,z) : z = \pm k x \}$. +This implies +$$\mathcal{S} \cap \mathcal{E} \subset \mathcal{P}_{+} \cup \mathcal{P}_{-}$$ +Converse, for any point $(x,y,z) \in \mathcal{S} \cap (\mathcal{P}_{+} \cup \mathcal{P}_{-})$, a similar argument tell us $ax^2 + by^2 + cz^2 = 1$. +This means we also has +$$\mathcal{S} \cap (\mathcal{P}_{+} \cup \mathcal{P}_{-}) \subset \mathcal{E}$$ +Combine these two piece of information, we get +$$\mathcal{S} \cap \mathcal{E} += \mathcal{S} \cap ( \mathcal{P}_{+} \cup \mathcal{P}_{-} ) += ( \mathcal{S} \cap \mathcal{P}_{+} ) \cup (\mathcal{S} \cap \mathcal{P}_{-})$$ +Since the intersection of a sphere and a plane is a circle, we find +$\mathcal{S} \cap \mathcal{E}$ is the union of two circles +$\mathcal{S} \cap \mathcal{P}_{+}$ and $\mathcal{S} \cap \mathcal{P}_{-}$. +We can repeat essentially the same argument to other combinations of $\mathcal{S}$, $\mathcal{E}$ and $\mathcal{P}_{\pm}$. At the end, we have something like +$$\mathcal{E} \cap \mathcal{P}_{+} = \mathcal{S} \cap \mathcal{P}_{+} +\quad\text{ and }\quad +\mathcal{E} \cap \mathcal{P}_{-} = \mathcal{S} \cap \mathcal{P}_{-}$$ +Substitute $(a,b,c)$ by $(2,3,4)$, we find the intersection of +the ellipsoid $$\{ (x,y,z) : 2x^2 + 3y^2 + 4z^2 = 1 \}$$ with the two planes $z = \pm x$ are two circles of radius $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{3}}$.<|endoftext|> +TITLE: Find 10 commuting $2\times 2$ matrices of the same order +QUESTION [5 upvotes]: Prove that there exists 10 distinct real $2\times 2$ matrices which are pairwise commuting and all of the same finite order. +Here, the order of matrix A is the smallest integer $k > 0$ such that $A^k = I.$ Also by 'pairwise commuting', I mean that if $\{A_1, \cdots, A_{10}\}$ are the ten matrices, then $A_i A_j = A_j A_i$ for any $i,j = 1,2,\cdots, 10$ and $i\neq j.$ +Linear algebra is not my forte yet, I haven't got a clue where to begin. The matrices commute so they all have the same eigenvectors... so letting $A_1$ be a matrix of just these eigenvectors, $A_2 = A_1 + I$ would have the same eigenvectors, thus $A_2A_1 = A_1A_2.$ However I am not sure how to deal with making them the same order. + +REPLY [4 votes]: First Hint. Define $$R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos\theta\end{bmatrix}$$ +Then $R(\theta)$ represents the linear transform that rotates vectors counter-clockwise by an angle $\theta$. +Second Hint. $11$ is prime.<|endoftext|> +TITLE: Evaluate $\int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx$ +QUESTION [8 upvotes]: Evaluate $$\int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx$$ i used substitution $x=\tan^2 y$ so $dx=2\tan y \sec^2 y dy$ so the integral becomes +$$I=\int\frac{2\cos 2y\: \tan y\: \sec^2 y \:dy}{\sqrt{\tan^2 y+\tan^4 y+\tan^6 y}}=\int\frac{2\cos 2y \:\sec^2 y\: dy}{\sqrt{1+\tan^2 y+\tan^4 y}}$$ I was stuck here + +REPLY [5 votes]: Let $$I = \int\frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx =\int\frac{(1-x^2)}{(x^2+2x+1)\sqrt{x+x^2+x^3}}dx$$ +Now again Reaaranging we get $$I = -\int\frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}dt$$ +Now put $\displaystyle x+\frac{1}{x}+1=u^2\;,$ Then $\displaystyle \left(1-\frac{1}{x^2}\right)dt = 2udu$ +So we get $$I = -2\int\frac{1}{u^2+1}du = -2\tan^{-1}(u)+\mathcal{C} = -2\tan^{-1}\left(\sqrt{x+\frac{1}{x}+1}\right)+\mathcal{C}.$$<|endoftext|> +TITLE: It there an example of sum and product of continuous functions is not continuous? +QUESTION [7 upvotes]: The sum and product of two continuous functions is continuous + +I can prove this easily when the space is metrizable, but I don't get it when the space is non-metrizable. Is there a counterexample of this? or it is true for all topological spaces those have a binary operation? +edit: It it true even if the space has a non-contionuous binary operation? + +REPLY [12 votes]: If $X$ is any space and $f,g: X \to \mathbb R$ are any two continuous functions, then $(f,g): X \to \mathbb R \times \mathbb R$ is continuous by definition (or construction) of the product, and $+: \mathbb R^2 \to \mathbb R$ is a continuous function. Then $f+g = {+} \circ (f,g)$, and a composition of continuous functions is continuous. +Similarly for $\times$, of course. +Edit: or replace $\mathbb R$ with $\mathbb C,\mathbb Q$, or any topological ring -- that is, a ring so that $+,-,\times$ are continuous.<|endoftext|> +TITLE: How can I show that $G$ is non abelian of order 20? +QUESTION [5 upvotes]: Problem says: + +Let $G=\langle x,y|x^4=y^5=1,x^{-1}yx=y^2\rangle$. Show that $G$ is + nonabelian group of order 20. + +To show it, I tried to turn $x^ny^m$ into $y^kx^l$ for some $k,l$. Since I have $yx=xy^2$, $xy^m=(xy^2)y^{m-2}=\cdots =y^kxy^{m-2k}$. So, if $m$ is even, $xy^m=y^{m/2}x$. If $m$ is odd, then $xy^m=y^{(m-1)/2}xy$. So, I have 12 distinct elements of $G$, $1,x,x^2,x^3,y,y^2,y^3,y^4,xy,xy^3,yx,y^2x$. How could I do find more here? And I want to the general way to solve this type of problem. + +REPLY [4 votes]: A slightly more concrete presentation of the argument from Hagen von Eitzen's answer. +Consider the group (verify that it is subgroup of a suitable matrix group) +$$ +H=\left\{\left(\begin{array}{cc}a&b\\0&1\end{array}\right)\bigg\vert\ a,b\in\Bbb{Z}_5,a\neq0\right\}. +$$ +It has 20 elements. The matrix +$$ +Y=\left(\begin{array}{cc}1&1\\0&1\end{array}\right) +$$ +has order five. The matrix +$$ +X=\left(\begin{array}{cc}3&0\\0&1\end{array}\right) +$$ +has order four. I am leaving it to you to verify these and the remaining relation $X^{-1}YX=Y^2$ (observe that $3=2^{-1}$). Anyway, the relations are there meaning that we have a homomorphism $f$ from your group $G$ to this group $H$ such that $f(x)=X$ and $f(y)=Y$. To complete the proof you need to show that $X$ and $Y$ generate a group of order $20$, i.e. all of $H$. Then you can polish off: + +$G$ has at most $20$ elements as per the calculations using the relations. +$G$ has at least $20$ elements because it has a homomorphic image $H$ with $20$ elements. +Ergo, $G$ has exactly $20$ elements and is also isomorphic to $H$ (in particular as $H$ is non-abelian, so is $G$).<|endoftext|> +TITLE: How do outer products differ from tensor products? +QUESTION [11 upvotes]: From my naive understanding, an outer product appears to be a particular case of a tensor product, but applied to the "simple" case of tensor products of vectors. However, in quantum mechanics there appears to be a distinction between the two. For example, given two state vectors ("kets"), $\lvert\psi\rangle$ and $\lvert\phi\rangle$, an outer product of the two is given by $$\lvert\psi\rangle\langle\phi\rvert$$ and this corresponds to a representation of some operator acting on vectors in this Hilbert space. +However, if we consider the two kets, $\lvert\psi\rangle$ and $\lvert\phi\rangle$ as representing two single-particle states, then the tensor product of their two single-particle Hilbert spaces gives a two-particle state of the form $$\lvert\psi\rangle\otimes\lvert\phi\rangle\equiv\lvert\psi\rangle\lvert\phi\rangle$$ which gives a new two-particle state in the tensor-product Hilbert space (of the two single-particle Hilbert spaces). +Is the distinction that the outer product maps two vectors in a given Hilbert space to an operator acting on that same Hilbert space, whereas, the tensor product maps two vectors from two different Hilbert spaces to a vector in the (tensor) product Hilbert space? + +REPLY [7 votes]: It's basically the same, it's mainly a matter of implicit identifications. +Namely, in a Hilbert space $H$ there is a natural isomorphism $H\to H^*$ (which is the bra-ket duality), so $H\otimes H\simeq H\otimes H^*$. +Now there is a map $H\otimes H^*\to L(H)$ which corresponds to what you call the outer product. +So for two kets $|\psi\rangle$ and $|\phi\rangle$ in $H$, you can either see $|\psi\rangle\otimes |\phi\rangle$ as an element of $H\otimes H$ and interpret that as a kind of two-particles state, or you can look at its canonical image in $H\otimes H^*$, which the physicists note $|\psi\rangle \langle\phi|$ (which makes sense since $\langle\phi|$ is the element of $H^*$ corresponding to $|\phi\rangle$), and then interpret it as an operator in $L(H)$.<|endoftext|> +TITLE: Combinatorics problem involving n-dimensional space +QUESTION [6 upvotes]: Consider a set of more than $\frac {2^{n+1}} {n}$ points $(n>2)$, chosen from the $2^n$ points of the $n$-dimensional space which have the coordinates $\{ \pm1, \pm1, ..., \pm1 \}$. +Show that there are $3$ points in the set that form an equilateral triangle. +I don't know how to start, as I am new in combinatorics problems. + +REPLY [4 votes]: Using the pigeonhole principle (which the problem formulation clearly suggests to do), one can show that moreover there is an equilateral triangle with the side equal to $2\sqrt{2}$. +Consider the set $V$ of points of the form $\{\pm 1,\dots, \pm 1\}$ (these are vertices of an $n$-hypercube). Call two of them neighbors if they differ in exactly one coordinate (in other words, they are connected by an edge). Note that the distance between any two different points from $V$ with a common neighbor is $2\sqrt{2}$. +Now each of the chosen points has $n$ neighbors. So in total there are more than $2^{n+1}$ neighbors to the chosen points. Since there are $2^n$ points in $V$, by the pigeonhole principle, some element of $V$ is a neighbor to at least three of the chosen points. By the above remark, these three points form the desired equilateral triangle.<|endoftext|> +TITLE: Does there exist a complex function which is differentiable at one point and nowhere else continuous? +QUESTION [17 upvotes]: Let $f\colon\mathbb{C}\to\mathbb{C}$. We know that if $f^{\prime}(a)$ exists for some $a\in\mathbb{C}$ then $f$ is continuous at $a$. This is because, from the definition of the derivative, $$f(z)-f(a)=[f^{\prime}(a)+\varepsilon(z)](z-a)$$ for some $\varepsilon$ such that $\varepsilon(z)\to0$ as $z\to a$, and hence $f(z)\to f(a)$. +I think we can interpret this geometrically by saying that, for $z$ close to $a$, if we take the vector from $a$ to $z$, rotate it counterclockwise about its tail by $\mathrm{Arg}\,{(f^{\prime}(a))}$ and scale its length by $\lvert f^{\prime}(a)\rvert$, then we get approximately the vector from $f(a)$ to $f(z)$ (it's only approximate, because of the $\varepsilon$ error term, but for $z$ close to $a$ this $\varepsilon$ is small). +However, consider the following heuristic argument: let $z_{1}$ and $z_{2}$ both be "close to" $a$ (and therefore close to each other). The above interpretation then suggests that the vector from $f(a)$ to $f(z_{1})$ should be approximately the same as the vector from $f(a)$ to $f(z_{2})$, and therefore $f(z_{2})-f(z_{1})$ should be small in magnitude. This seems to suggest that $f$ should be continuous not just at $a$, but also in some small interval around $a$. This seems too good to be true. +Hence my question: + +Does there exist a function $f\colon\mathbb{C}\to\mathbb{C}$ such that $f^{\prime}(a)$ exists at some point $a\in\mathbb{C}$ but $f$ is continuous nowhere else? + + +Update, added later: A quick thanks to those who responded. I have since thought through this with a friend and we've found a problem with the heuristic argument: + +Properly interpreted, the heuristic argument is saying that if $z_{1}$ and $z_{2}$ are close to $a$, then $f(z_{1})$ is close to $f(z_{2})$. However, to have continuity at $z_{1}$, say, we need to be able to make $f(z_{2})$ arbitrarily close to $f(z_{1})$ by choosing $z_{2}$ to be sufficiently close to $z_{1}$ --- therein lies the problem. Suppose $\lvert z_{3}-z_{1} \rvert < \lvert z_{2}-z_{1} \rvert$, i.e., suppose $z_{3}$ is even closer to $z_{1}$ than $z_{2}$ is. Then, for $f$ to be continuous, we need to be able to ensure that $f(z_{3})$ is closer still to $f(z_{1})$ (than $f(z_{2})$ is); in general, we can't ensure this. + +If I get round to it, I might try to work through the details in a specific case and see if there are any other problems. If I get anywhere, I'll post the results as an additional answer. + +REPLY [30 votes]: Let $g \colon \mathbb{C} \rightarrow \mathbb{C}$ be a function that is discontinuous everywhere and bounded. Consider $f(z) := z^2 g(z)$. Then $f$ is continuous only at $z = 0$, and $f$ is differentiable at $z = 0$ as +$$ \lim_{z \to 0} \frac{f(z) - f(0)}{z} = \lim_{z \to 0} z g(z) = 0.$$<|endoftext|> +TITLE: Limit of recursive sequence $n^2q_n=1+(n-1)^2q_{n-1}+2(n-2)q_{n-2}$ +QUESTION [10 upvotes]: When looking at this riddle, I came across the following sequence for the frequency of sampled integers between 1 and $n$ in a without replacement/without neighbour sampling: +$$q_1=1,\quad q_2=1/2,\quad q_n=\frac{1}{n^2}+\frac{(n-1)^2}{n^2}q_{n-1}+\frac{2(n-2)}{n^2}q_{n-2}\quad (n>2)$$ +and I wonder if there is a generic mathematical approach to computing an analytic solution for $$\lim_{n\to\infty} q_n$$Assessing the numerical limit with an R code like +q=rep(1,1e7) +for (n in 3:1e7) q[n]=(1+2*q[n-2]+(n-1)*q[n-1])/n +q[1e7]/1e7 + +led me to $0.432332...$ And a probabilistic reasoning indicates that $(n>1)$ $$\frac{1}{3}\le q_n\le \frac{1}{2}$$ + +REPLY [19 votes]: Let $a_n = (n+1)q_{n+1}$. We have $a_0 = a_1 = 1$ and +$$(n+1)a_n = 1 + na_{n-1} + 2a_{n-2}\quad\text{ for }\quad n > 1\tag{*1}$$ +Let $f(z) = \sum\limits_{n=0}^\infty a_n z^n$, multiply $(*1)$ by $z^n$ and start to sum from $n = 2$, we get +$$\begin{align} +& \left(z\frac{d}{dz} + 1 \right)(f(z) - 1 - z) = \frac{z^2}{1-z} + \left(z\frac{d}{dz}\right)(z(f(z)-1)) + 2z^2f(z)\\ +\iff & +zf' + f - 1 -2z = \frac{z^2}{1-z} + z^2 f' + zf - z + 2z^2 f\\ +\iff & +z(1-z)f' + (1-z-2z^2)f = \frac{1}{1-z} +\end{align} +$$ +Solving the ODE give us +$$f(z) = \frac{1-e^{-2z}}{2z(1-z)^2} = \frac{A}{(1-z)^2} + \frac{B}{(1-z)} + g(z)\tag{*2}$$ +where +$\displaystyle\;\begin{cases} +A &= \frac{1-e^{-2}}{2}\\ +B &= \frac{1-3e^{-2}}{2}\\ +\end{cases}\; +$ +and $g(z) = \sum\limits_{n=0}^\infty g_n z^n$ +is some function analytic over all $\mathbb{C}$. +Expanding $(*2)$ as a power series and compare coefficients of $z^n$ on both sides, we get +$$(n+1)q_{n+1} = a_n = (n+1)A + B + g_n$$ +Since $g(z)$ is entire, its coefficients of taylor expansion $g_n$ is bounded. +As a result, +$$ +\begin{align} +\lim_{n\to\infty} q_{n+1} +&= A + \lim_{n\to\infty} \frac{B + g_n}{n+1} = A = \frac{1 - e^{-2}}{2}\\ +&\approx 0.432332358381693654053000252513757798296184227045212059265 +\end{align} +$$<|endoftext|> +TITLE: Is the Epsilon-Delta definition of a limit not precise enough? +QUESTION [5 upvotes]: Consider the function f: {-1, +1} -> R defined by +$f(x)= \arcsin (\frac{1+x^2}{2x})$. +Due to the following two inequalities : +(i) $1+x^2 \geq 2x$ +(ii)$1+x^2 \geq -2x$ , +the function can only be defined at $x=1$ and $x=-1$. I have learnt that the epsilon delta definition only includes those values of $x$ which are in the domain of $f(x)$. But in this case, the function isn't defined on either side of x=1. +So this is my question: Is it correct to say that the limit as $x$ approaches $1$ of $f(x$) is $\frac{\pi}{2}$ ? +Can the above question be given a definitive "yes" or "no" answer, or must it(unfortunately) vary from person to person? +If the latter, is the "precise" definition of a limit not precise enough? +How can the answer be proved or disproved using the epsilon delta definition? +I have also read that functions are by default continuous at isolated points. Can I conclude from the definition of continuity (the limit equals the value of the function evaluated at the point) that the limit must exist? +Note : Forgive my ignorance but I do not know a thing about topology. I'm looking for a detailed answer but in simple terms, preferably written in the language of calculus. +Thanks for the help. + +REPLY [5 votes]: The bottom line is: + +Limits are only defined on limit points. Continuity is only defined on the domain. + +If we have a point $p$ on the domain such that $p$ is a limit point, then continuity at $p$ is equivalent to the limit being the value. However, if we have a point not on the domain, continuity does not make sense (although we can enlarge the domain and define it at the point in order to force continuity, but I digress), and if we have a point on the domain which is not a limit point, then limit does not make sense. +The problem is explictly the following: +The definition of limit is: Let $x_0$ be a limit point of the domain. Then, we say $f(x) \stackrel{x \to x_0}{\to} L$ if for all $\epsilon>0$, there exists $\delta>0$ such that $0<|x-x_0|< \delta \implies |f(x)-L| < \epsilon$. +If $x_0$ is not a limit point, there exists a $\delta$ such that there is no $x$ with $|x-x_0|<\delta$. Therefore, any value of $L$ can be the limit. If we want to be really forceful and withstand the non-uniqueness of limit on metric spaces, then we can indeed let this be the case: every point on the codomain is a limit of $f(x)$ as $x \to x_0$ if $x_0$ is not a limit point. Then, for instance, it would hold that $f$ is continuous at $x_0$ if and only if $f(x) \stackrel{x \to x_0}{\to} f(x_0)$ even if $x_0$ is not a limit point, but that would be a very degenerate case. +What we conclude is that the case of $x_0$ not being a limit point results in a nuisance of non-uniqueness. Continuity is different: +The definition of continuity at $x_0$ is: $f$ is continuous at $x_0$ if for all $\epsilon>0$ there exists $\delta>0$ such that $|x-x_0|<\delta \implies |f(x)-f(x_0)| < \epsilon$. +There is no vacuous problem. Rather, the definition gives immediately that if $x_0$ is not a limit point, then $f$ is automatically continuous at $x_0$ (go through the definition and verify this). There is a certain degree of triviality, but no vacuity. +Summing up, your function $f$ is continuous, but limits on the points of the domain are not well-defined (at least, not unique when viewed under the usual definition). +PS: I acknowledge that there may be different definitions of continuity. However, the one used here is the special case of the definition used in topology, which is by far the most useful and well-behaved.<|endoftext|> +TITLE: A problem in group theory_dsom +QUESTION [5 upvotes]: Let $H$ be a group of integers modp, under addition, where $p$ is a prime number. Suppose that $n$ is an integer satisfying $1 \leq n \leq p$, and let G be the group $ H \times H \times \cdots \times H$ (n factors). Show that $G$ has no automorphism of order $p^2$. + +REPLY [4 votes]: $H$ is in fact a field, hence $G$ can be viewes as a vector space over $H$, and every automorphism of the group $G$ is also a linear automorphism of the vector space $G$. As such it can be described by an $n\times n$ matrix $A$. If $G$ has order $p^2$ then this matrix has an eigenvalue (in an algebraic closure of $H$) that is a primitive $p^2$th root of unity. The degree of its minimal polynomial is $\phi(p^2)=p\cdot(p-1)$ and this larger than the degree $\deg(A-\lambda)$ of the characteristic polynomial - unless $p=n=2$. But the automorphism group of $C_2\times C_2$ is well-known and equals $S_3$, which does not contain a $C_4$.<|endoftext|> +TITLE: Do finite morphisms preserve dimensions? +QUESTION [5 upvotes]: If $f: Y\to Z$ is a finite, surjective morphism of normal integral schemes (of finite type over a field) and $y$ is a prime divisor of $Y$, is then also $z= f(y)$ of codimension 1? +We have an inclusion $\mathcal{O}_{Z,z} \to \mathcal{O}_{Y,y}$ and the latter is a DVR, i.e. it has dimension 1. Does the integrality of both local rings imply that $\operatorname{codim}z =1$? +More generally, is there a relationship between $\dim y, \dim f(z)$ and $\dim f^{-1}(z'), \dim z'$ for $z'\in Z$ arbitrary? + +REPLY [4 votes]: Answer to your first question: We prove the more general statement that if $f: Y \to Z$ is a finite surjective morphism of integral schemes, then $\dim y = \dim f(y)$ (in the case that $Y,Z$ are also of finite type over a field this will also yield that $codim \; y = codim \; f(y)$). +Claim: $\dim Y = \dim Z$, the reason being that if $Spec B$ is any affine open of $Z$, then $f^{-1}(Spec B) = Spec A$ with $B \hookrightarrow A$ being an integral extension hence $dim Spec A = dim Spec B$, now cover $Y$ with affine opens and recall that a scheme has dimension $n$ if and only if it has an affine cover where each affine in the cover has dimension $\le n$ and equality is achieved for at least one, from taking preimages of such a cover of $Z$ we now see that $\dim Z = \dim Y$. +Now we have that the morphism $y \to f^{-1}(z) \to f(y) = z$ is a finite surjective morphism of integral schemes (being the composition of a closed embedding together with a projection from a base change of a finite morphism) hence it follows from the claim that $\dim y = \dim z$ and since codimension is the difference of dimensions for integral schemes of finite type over a field it then follows that $codim y = codim z$. +Answer to your second question: +If $f: Y \to Z$ is any morphism of locally Noetherian schemes, and $p \in Y$, $q = f(p)$ then we always have +$$(Eq \; 1) \; \; \; \; codim_Y p \le codim_Z q + codim_{f^{-1}(q)}p$$ (here $codim$ of a point is interpeted as the codimension of the closure in the appropriate space). Now for integral schemes of finite type over a field (and more generally for Catenary schemes if I am not mistaken?) this translates to +$$ (Eq \; 2) \; \; \; \; \dim Y - \dim \overline{\{p\}} \le \dim Z - \dim \overline{\{q\}} + codim_{f^{-1}(q)}p$$ +If the morphism is flat we have equality, and of course if the morphism is finite then the second term on the right hand side can be removed.<|endoftext|> +TITLE: Combinatorics: choose 5 out of 10 colored balls +QUESTION [5 upvotes]: I usually don't have any problems thinking about combinatorics but this problems answer doesn't seem correct. +There are $5$ black balls, $1$ red, $1$ green, $1$ blue, $1$ yellow and $1$ white. +In how many ways can you pick $5$ balls? +The answer is $2^5$ but why? +I'm comparing it with a lock with $4$ integers, which has the number of combinations of $10^4$. +Anyone got any explanations? + +REPLY [3 votes]: For the set of ways to pick balls, there is either a red or none; and if there are no red balls, then there must be a black ball. Similarly for all other non-black colours. +Consider the five pairs of balls: red and black, green and black, blue and black, yellow and black, and white and black. +The number of ways to pick $5$ balls is the same as the number of ways to pick one ball from each of the five pairs: $2^5$.<|endoftext|> +TITLE: Riddle similar to the 100 prisoners riddle, but different +QUESTION [5 upvotes]: The riddle goes like this: +$\qquad$ There are $100$ prisoners standing in line, each with a number on their back. The numbers are all different, and range from $1$ to $101$ (i.e. one number is missing). The person in the back of the line can see all 99 prisoners ahead of him's numbers. The prisoner in the front of the line sees no numbers. Starting with the prisoner in the back, each prisoner must shout the number on their own back, but each number can be said aloud only once. +$\qquad$The first prisoner clearly only has a $50$% chance of guessing his number correctly. However, once he has guessed his number correctly, the remaining 99 prisoners have a $100$% chance of correctly guessing the number on their backs. +What is their strategy to achieve this? +Update: +I have been given a hint that the solution "does not just involve modulo sums." There is more to the solution than just a modular trick. + +REPLY [2 votes]: The back prisoner imagines another behind with the missing number. He then guesses the number that will make the list an even permutation of the numbers $1$ to $101$. The person in front of him (assuming the first guess is correct) knows all the numbers but two and can select the one that makes the permutation even. Each prisoner in turn has just two choices (assuming the previous shouts are correct) and one of those will make the permutation even. +Second solution: back person shouts the number on the front person. As he has failed to state his number correctly, there is no requirement that the others are correct.<|endoftext|> +TITLE: Automorphisms of infinite abelian groups +QUESTION [13 upvotes]: It is well-known that the map $\operatorname{Aut}$ from the class of groups to itself has fixed points. For $n \neq 2$ or $6$, $\operatorname{Aut}(S_n) \cong S_n$, $\operatorname{Aut}(D_4) \cong D_4$ and if $G$ is a finite non-abelian simple group, then $\operatorname{Aut}(\operatorname{Aut}(G)) \cong \operatorname{Aut}(G)$. The question arises if there is a fixed point among the non-trivial abelian groups. It is relatively easy to see that finitely generated abelian groups do not qualify, so if such a group exists, it has to be infinitely generated. Note that for topological automorphisms of topological groups fixed points are known: for instance $\operatorname{Aut}(\mathbb{R} \times \mathbb{Z}/2\mathbb{Z}) \cong \mathbb{R} \times \mathbb{Z}/2\mathbb{Z}$. But we assume no additional structure imposed. + +REPLY [11 votes]: The automorphism group, as an additive group, of the $p$-adic integers $\mathbb{Z}_p$ is the same as its group of units, which is isomorphic to $\mathbb{Z}_p\times\mathbb{Z}/(p-1)\mathbb{Z}$ for odd primes $p$. So $\mathbb{Z}_3\times\mathbb{Z}/2\mathbb{Z}$ is isomorphic to its own automorphism group.<|endoftext|> +TITLE: Possibilities for a group $G$ that acts faithfully on a set of objects with two orbits? +QUESTION [5 upvotes]: A group $G$ acts faithfully on a set $X$ of 5 objects. The action has + two orbits: one of size 2, and one of size 3. What are the + possibilities for the group $G$? + +I think I should apply the orbit-stabilizer theorem. I believe a main hint is the fact that this group has precisely two orbits but I'm not sure how to use that information. What's a good way to go about listing the possibilities for this group $G$? + +REPLY [3 votes]: First determine the order of the group. Remember that it is a subgroup of $S_{5}$. The size of an orbit must divide the size of the group, which in turn must divide the order of $S_{5}$; so this means that $|G|$ is a multiple of $6$ which also divides $120$. +On the surface that looks like there are a lot of options, but consider that any $\sigma \in G$ can be decomposed into its' action on each orbit, so really you can consider $G$ a subgroup of $S_{2} \times S_{3}$, which has order $2\cdot 6 = 12$. Notice this also means you can describe the possibilities for $G$ by classifying the transitive subgroups of $S_{2}$ and $S_{3}$. +[EDIT] My hint really only makes two of the three possibilities apparent, namely $\mathbb{Z}_{2} \times \mathbb{Z}_{3}$ (order 6) and $\mathbb{Z}_{2} \times S_{3}$ (order 12). But as noted by Derek Holt, there is actually a third possibility, namely a second group of order 6. I would suggest to consider how $G = S_{3}$ can act on the orbits $\{1,2\}$ and $\{3,4,5\}$ in an appropriate way, by considering the action of the two generators of the group (i.e. consider $S_{3} = \langle \alpha, \beta \mid \alpha^{3} = \beta^{2} = e,\ \beta \alpha \beta = \alpha^{-1} \rangle$).<|endoftext|> +TITLE: The lattice points in the real cone of some semigroups are just the integer cone of that semigroup. +QUESTION [8 upvotes]: I'm trying to solve an exercise in Fulton's book on toric varieties, and have reduced it to the following: + +Let $M$ be a lattice of rank $n$ with $M \otimes \mathbb{R} = V$, and $S$ be a finitely generated semigroup of $M$, such that the following conditions hold: +i) S is saturated. (If $v \in M, n \in \mathbb{Z}_{> 0}$ with $nv \in S$, then $v \in S$.) +ii) S generates M as a group. +Then the lattice points inside the real cone on $S$ are simply the integer cone on $S$. + +In other words, given a saturated semigroup $S=\langle v_1,\dots,v_k\rangle_{\mathbb{Z}_{\geq 0}}$ with $S=\langle v_1,\dots,v_k\rangle_{\mathbb{Z}} = M$, then $\langle v_1,\dots,v_k\rangle_{\mathbb{R}_{\geq 0}}\cap M = S$. This seems geometrically obvious, but I can't find a way to prove it. +I can prove the result if $\langle v_1,\dots,v_k\rangle_{\mathbb{R}_{\geq 0}}\cap M=\langle v_1,\dots,v_k\rangle_{\mathbb{Q}_{\geq 0}}\cap M$ using the saturatedness of $S$, the problem is showing that any positve real linear combination of the $v_i$ lying in the lattice can also be written as a positive rational combination. +Thinking about the different ways to represent the same vector, we see that finding $r=(r_1,\dots,r_k)$ such that $\sum r_iv_i = v$ is equivalent to saying that $r$ is a solution to the matrix equation: +$$\bigg(v_1 \dots v_k \bigg)x=v $$ +The solutions to this equation are either an empty set, of an affine subspace of $\mathbb{R}^k$. But if $v$ is in the intersection we're lookinh at, we know that the solution space contains a "positive" point, i.e. a point of $(\mathbb{R}_{\geq 0})^k$, and an "integer" point, i.e. a point of $\mathbb{Z}^k$. I'd like to use this somehow to show that it also contains a positive rational point, which is easy enough if the positive point lies in the strict interior of the set of positive points, but seems like it could be impossible in some examples, such as if the solution space is an affine line running tangent to the set of positive points, so I'm not sure if this approach will work... + +REPLY [2 votes]: The question you ask (which is a variation on the theme of Gordan's lemma) appears exactly (and with almost exactly the same notations) as the "conversely" part of proposition 1.1 on page 3 in Tadao Oda's Convex Bodies and Algebraic Geometry (An Introduction to the Theory of Toric Varieties) (Springer 1985). +The key point is that if $v$ belongs to the convex cone $\rho$ generated by $v_1,\ldots,v_k$, then in fact there are $n$ among the $v_i$ (where $n = \dim\langle v_1,\ldots,v_k\rangle_{\mathbb{R}}$), say w.l.o.g. $v_1,\ldots,v_n$, linearly independent over $\mathbb{R}$, such that in fact $v$ belongs to the convex cone $\rho'$ generated by $v_1,\ldots,v_n$. This is Carathéodory's theorem (the version on Wikipedia doesn't say exactly the same thing, but the proof given there works). But now the coefficients $r_1,\ldots,r_n$ of $v \in M\cap\rho'$ on the $v_1,\ldots,v_n$ are uniquely determined, so the question is trivial: they are rational because $v \in M$ (and $v_1,\ldots,v_n$ generate $M\otimes\mathbb{Q}$ over the rationals) and nonnegative because $v \in \rho'$. And since $S$ is saturated, we can get nonnegative integer coefficients. +Edit: Following OP's comment, let me spell out the argument why $v_1,\ldots,v_n$ generate $M\otimes\mathbb{Q}$. They are in $M\otimes\mathbb{Q}$ and linearly independent over $\mathbb{R}$ (this follows from the proof of Carathéodory's theorem, although people usually forget to put this in the statement for some reason); so in particular, they are linearly independent over $\mathbb{Q}$. But $M\otimes\mathbb{Q}$ has dimension $n$ as a $\mathbb{Q}$-vector space (because the rank of $M$ as a free $\mathbb{Z}$-module is the rank of $M\otimes\mathbb{Q}$ as a $\mathbb{Q}$-vector space or of $M\otimes\mathbb{R}$ as an $\mathbb{R}$-vector space, and that is what we are calling $n$). So they are a basis of $M\otimes\mathbb{Q}$. This explains why the coefficients of $v$ on $v_1,\ldots,v_n$ are unique (and rational).<|endoftext|> +TITLE: How would you show that field automorphisms fix prime subfields? +QUESTION [8 upvotes]: Suppose $K$ is a prime subfield of $E$, then if $\phi$ is an automorphism from $E$ to $E$, we have for all $x \in K$, $\phi(x) = x$. +I feel like this is just the definition of a field automorphism, but my book says this should be proven as an exercise. + +REPLY [10 votes]: This is because the prime subfield is generated as a field by $1$. Since you have no choice but to send $1$ to itself, the prime subfield remains fixed as well.<|endoftext|> +TITLE: Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible. +QUESTION [7 upvotes]: This is what I'm trying to prove: + +Let $f(x)\in\mathbb{Q}[x]$ and $\deg(f(x))>1$. Prove that $f(x)$ is irreducible in $\mathbb{Q}[x]$ iff its reciprocal polynomial $f^*(x)$ is irreducible in $\mathbb{Q}[x]$. + +Note: The reciprocal polynomial $f^*(x)=x^nf(1/x) \in \mathbb{Q}[x]$, where $n=\deg f$. +So my thoughts are to prove the contrapositive, i.e. $f^*(x)$ is reducible iff $f(x)$ is reducible. +So to prove the ($\Rightarrow$) direction I assume that $f^*(x)=g(x)h(x)$, so I get that $g(x)h(x)=x^nf(1/x)$ which implies $f(1/x)=(1/x^n)g(x)h(x)$. I want to say this somehow makes $f(x)$ reducible but I am unable to proceed as $f(1/x)\notin \mathbb{Q}[x] $. I thought about substituting $y=1/x$, which gives $f(y)=y^ng(1/y)h(1/y)$, but I am unable to show that this is irreducible in $\mathbb{Q}[y]$. If I could, then it would be irreducible in $\mathbb{Q}[x]$ since $\mathbb{Q}[y]\cong \mathbb{Q}[x]$ +Any suggestions or hints will be appreciated. + +REPLY [7 votes]: Observe that $(f^*)^*=f$ if and only if $f$ has nonzero constant term, and $(gh)^*=g^*h^*$ for all $g,h$. +Assume $f$ is irreducible but $f^*=gh$ is not, where $g,h$ are nonconstant. Then since $f^*$ has nonzero constant term, so do $g$ and $h$, hence $g^*,h^*$ are both nonconstant, and since $f=(f^*)^*=g^*h^*$ we have a contradiction. +The other implication actually isn't true! For a counterexample, let $h(x)$ be irreducible with nonzero constant term and let $f(x)=xh(x)$. Then $(f^*)^*=h(x)$ and $((f^*)^*)^*=f^*$, so $f^*$ is irreducible even though $f$ is not.<|endoftext|> +TITLE: Proof outer measure satisfies monotonicity: $A \subseteq B \implies m^*(A) \leq m^*(B)$ +QUESTION [6 upvotes]: Theorem: $$A \subseteq B \implies m^*(A) \leq m^*(B)$$ +Proof Attempt: +By definition, $m^*(B) = \inf\{\sum\limits_{k=1}^\infty |J_k||\{J_k\} \text{ is a cover of B }\}$, $m^*(A) = \inf\{\sum\limits_{k=1}^\infty |I_k||\{I_k\} \text{ is a cover of A }\}$ +Then since $A \subset B \implies \bigcup_{k =1}^\infty I_k \subseteq \bigcup_{k=1}^\infty J_k$, then $\sum\limits_{k=1}^\infty |I_k| \leq \sum\limits_{k=1}^\infty |J_k|$, so we have $m^*(A) \leq m^*(B)$ +Can someone check if everything checks out? + +REPLY [4 votes]: I am just a little bit unsighted by your proof, hence I want to give you a proof which is roughly the same, but a little more sharpened so that you can see how get rid of small discrepancies in proofs. +So $m^*(B) = \inf \{ \sum_1^n|J_k|, \{ J_k\} \text{ is an open interval cover of B}\}$. +Similarly, $m^*(A) = \inf \{ \sum_1^n|J_k|, \{ J_k\} \text{ is an open interval cover of A}\}$. +Note that $A \subset B$,whence if $\{J_k\}$ covers $B$, it also covers $A$. Hence, the following is true: +$$ +\{ \sum_1^n|J_k|, \{ J_k\} \text{ an open interval cover of B}\} \subset \{ \sum_1^n|J_k|, \{ J_k\} \text{ an open interval cover of A}\} +$$ +Now, the infimum of the subset of a family is larger than the infimum of the family, because the family could contain an element that is not in the subset but smaller than every element in the subset. +Hence, it follows that the infimum of the left side, which is $m^*(B)$, is greater than the right side, which is $m^*(A)$. +EDIT: As other have pointed out, there was something wrong in your proof. However, you were thinking in the right direction, and hopefully this proof will show you how to put those efforts in a more rigorous direction.<|endoftext|> +TITLE: Show the Cauchy-Riemann equations hold but f is not differentiable +QUESTION [5 upvotes]: Let $$f(z)={x^{4/3} y^{5/3}+i\,x^{5/3}y^{4/3}\over x^2+y^2}\text{ if }z\neq0 +\text{, and }f(0)=0$$ + Show that the Cauchy-Riemann equations hold at $z=0$ but $f$ is not differentiable at $z=0$ + +Here's what I've done so far: +$\quad$As noted above, there are two cases: $z=0$ and $z\neq0$. The Cauchy-Riemann equations hold at $\quad z\neq0$ because $f(x)$ is analytic everywhere except at the origin. +$\quad$Now we consider $z=0$. In this case, +$$f_x(0,0) = \lim_{x\to 0} \frac{0 + 0}{x^2+0}=0$$ +$\quad$and +$$f_y(0,0) = \lim_{y\to 0} \frac{0 + 0}{0+y^2}=0$$ +$\quad$So, we see that $\frac{\delta u}{\delta x}=0=\frac{\delta v}{\delta y}$ and $\frac{\delta u}{\delta y}=0= -\frac{\delta v}{\delta x}$. Therefore, the Cauchy-Riemann equations $\quad$hold at $z=0$. +From here, I'm a little confused about how to show that $f$ is not differentiable at $z=0$. I instinctively believe that $f$ is not continuous at $z=0$, which would imply that $f$ is not differentiable at $z=0$. However, I'm not sure how I can prove this. +Any help would be greatly appreciated. Thank you!! +EDIT: +I now see that $f$ is continuous at $z=0$, but I'm not sure how to prove that it is not differentiable. I know I should use $\lim_{z\to 0} \frac{f(z) - f(0)}{z-0}$ to determine differentiability, and that I should get two different answers when I approach the limit from two different paths (for example: $x=0$, $y=0$, or $x=y$), but I'm not sure how to evaluate $\lim_{z\to 0} \frac{f(z) - f(0)}{z-0}$ for $x=0$, $y=0$ or $x=y$. +As before, any help would be much appreciated. Thanks! + +REPLY [2 votes]: HINT: Are the limits as $x \to 0$ and $y \to 0$ the same? Because if the limit does not exist and equal the same value for EVERY direction of approach to the origin, then the limit does not exist there. +EDIT: The same argument would work with directional derivatives at the origin; if any two are not equal, then the function (considered as a function on R^2), can not be differentiable and hence whether or not the Cauchy-Riemann equations hold is insufficient. +However, do note that even if all directional derivatives exist and are equal, this is still not sufficient to prove the differentiability at the origin; it is only a necessary, but not a sufficient condition.<|endoftext|> +TITLE: Show that $\Bbb R^{2n+1}$ is not a division algebra over $\Bbb R$ for $n>0$ +QUESTION [7 upvotes]: This is an exercise from Hatcher's Algebraic Topology (exercise 2.B.8). Here is the problem statement: + +Show that, for $n>0$, $\Bbb R^{2n+1}$ is not a division algebra over $\Bbb R$ by showing that if it were, then for nonzero $a \in \Bbb R^{2n+1}$ the map $S^{2n} \to S^{2n}$ defined by $x \mapsto \dfrac{ax}{|ax|}$ would be homotopic to $x \mapsto -\dfrac{ax}{|ax|}$, but these maps have different degrees. + +I found a duplicate of this question here where a suggestion was to find a path between $a$ and $-a$ that does not pass through $0$. I am struggling to do this and I would appreciate if someone could give me some direction here. I don't have much experience with algebras or division algebras, so I may be missing something simple. +I am also somewhat confused about how to find the degrees of the maps $x \mapsto {ax\over |ax|}$ and $x \mapsto {-ax\over |ax|}$. As I understand it a map $S^{n} \to S^{n}$ induces a homomorphism $H_n(S^n) \to H_n(S^n)$, and $H_n(S^n)$ is isomorphic to $\Bbb Z$, so the induced homomorphism is of the form $x \mapsto nx$ for some $n \in \Bbb Z$. Then that $n$ is the degree of the original map. I'm not sure how to actually find the induced maps on homology in this problem (or if there is another way to find the degree). +Edit: As anomaly has pointed out, the second part of my question is simpler than I thought. + +REPLY [5 votes]: Too long for a comment: +It is nice to prove that there are no nontrivial odd-dimensional real division algebras using algebraic topology. However, it should be noted that there are, of course, far more elementary ways of proving such a statement. One of my favorites is the following (I think I learnt this one from Alain Hillion). +Suppose that the real vector space $\mathbf R^{m}$ carries the structure of a real division algebra, where $m$ is odd. We prove that $m=1$. Let $a\in \mathbf R^m$. Consider the multiplication-on-the-left-by-$a$ map from $\mathbf R^m$ into itself, which we denote again by $a$. Its characteristic polynomial $\det(\lambda I_m-a)$ is a polynomial of degree $m$ with real coefficients. Since $m$ is odd, it has a real root $\lambda$. This means that that there is a nonzero $v\in\mathbf R^m$ such that $(a-\lambda)v=0$. Since $\mathbf R^m$ is a division algebra and $v\neq0$, one has $a-\lambda=0$, i.e., $a=\lambda\in\mathbf R\cdot 1$ in $\mathbf R^m$. This proves that $\mathbf R^m=\mathbf R\cdot 1$, i.e., $m=1$.<|endoftext|> +TITLE: Can this puzzle be solved without brute force? +QUESTION [8 upvotes]: Consider positive integers $a$ and $b$, where $a \ge b$ and the sum $\frac{a+1}{b}+\frac{b+1}{a}$ is also an integer. Find the sum of all $a$ values less than $1000$ that meet this criteria. + +For example, if $a=3$ and $b=2$, then +$$\frac{a+1}{b}+\frac{b+1}{a} = \frac{3+1}{2}+\frac{2+1}{3} = 3$$ +which is an integer and provides one legal value of $a$. The problem asks for all such values of $a \lt 1000$. I’m assuming that a particular $a$ suffices if there exists even one $b \le a$ that satisfies the condition. +This problem can be solved by simply writing a computer program that double-loops through the positive integers less than 1000, testing each case, however I’m wondering if there is a more general/abstract way to solve it. +Here are some thoughts … I’m not sure how helpful they are: + +$\frac{a+1}{b}+\frac{b+1}{a}$, which is an integer, can be rewritten as $\frac{a}{b}+\frac{b}{a}+\frac{1}{b}+\frac{1}{a}$. +either the quotients $(a+1) \div b$ and $(b+1) \div a$ are each also integers, or their fractional parts add up to $1$. + +REPLY [3 votes]: We use a now-standard technique called Vieta jumping, which has a fascinating and surprisingly recent history. +Suppose $a^2 + a + b^2 + b = kab$ for some integer $k$. Then, writing this as a polynomial in $a$ with integer coefficients: +$$a^2 - (kb-1) a + (b^2+b) = 0,$$ +we see that if $(a,b)$ is one of the roots of the above quadratic, then the other one is $c = (b^2+b)/a$, which is clearly positive. Furthermore, since the two roots sum to an integer, namely $kb-1$, then $c$ is necessarily a positive integer. +Now let's consider sizes. If $a>b$, then $a \ge b+1$ so $c = b(b+1)/a \le b$. So if $(a,b)$ is one solution to the problem, then either $a=b$ or $(b,c)$ is a smaller solution, with the same value of $k$. Since all values remain positive, they can only get smaller finitely many times, so eventually we must reach a solution where $a=b$. +The case where $a=b$ is very easy: we must have $k = 2 + 2/a$, so the only solutions of this type are $(1,1)$ with $k=4$ and $(2,2)$ with $k=3$. +Finally, since every solution eventually reduces to either $(1,1)$ or $(2,2)$ by the reduction process, we can recover all solutions by running the process in reverse, starting from $(1,1)$ and $(2,2)$: +$$(1,1) \leftarrow (2,1) \leftarrow (6,2) \leftarrow (21,6) \leftarrow \cdots $$ +$$(2,2) \leftarrow (3,2) \leftarrow (6,3) \leftarrow (14,6) \leftarrow \cdots $$ +The values grow exponentially, so it doesn't take long for either sequence to exceed $1000$.<|endoftext|> +TITLE: Which real numbers have $2$ possible decimal representations? +QUESTION [5 upvotes]: I know that all positive and negative whole numbers have $2$ possible decimal representations. For example, $1+1+1+1$ could be represented as $4$ or as $3.99999...$ (I believe $4.000..1$ isn't a thing, right?). +All terminating decimal numbers have $2$ representations as well, for example $1.5$ is the same as $1.4999..$ However, I can't really see how you would make a second representation for non-terminating decimals (especially irrational numbers) and zero. + +REPLY [3 votes]: A decimal representation represents the number that the corresponding series converges to. For a number $x$ whose standard representation doesn't terminate, using any digit other than the one in its standard representation at any point in the series would make it impossible for the series to converge to $x$: If a greater digit is used, the partial sum is greater than $x$, and the remaining terms are all positive and hence can't make the series converge to $x$; and if a smaller digit is used in the $n$-th digit after the decimal point, the result is less than $x$ by more than $10^{-n}$, and the remaining terms can add up to at most $10^{-n}$, so again the series can no longer converge to $x$. Thus each digit is uniquely determined by $x$ unless the standard representation terminates.<|endoftext|> +TITLE: Any Video Lectures Of An MIT, Harvard, Stanford, UC Berkeley, Yale, or Princeton Analysis Course Based On Baby Rudin? +QUESTION [6 upvotes]: I'm learning analysis from the book Principles of Mathematical Analysis by Walter Rudin, third edition. +This book, popularly known as Baby Rudin, is being used for analysis courses at such elite places as the MIT, Harvard, Stanford, UC Berkeley, Yale, and Princeton. Am I right? +Now is there any video lecture analysis course based on Baby Rudin available on the Internet from any of the above-mentioned institutions? +Which other text(s) treat the same material as Baby Rudin and so can be used in combination with it? + +REPLY [5 votes]: I would not limit yourself just to those universities if you want good analysis lectures. I would highly recommend looking into Francis Su's lectures from Harvey Mudd College. His is a one-semester course and he ends roughly when they get into single-variable differentiation. A first glance at MIT's Open Courseware doesn't show any video lectures in analysis, though you may possibly find other materials (possibly written lectures) on their site.<|endoftext|> +TITLE: Deriving the asymptotic distribution of a two-stage estimator +QUESTION [5 upvotes]: Suppose $X_i \stackrel{iid}{\sim}F_\theta$, where $F_\theta$ is a probability distribution parameterized by a finite-dimensional vector $\theta$. +Let $\hat{\theta}_n$ denote the maximum likelihood estimator of $\theta$, i.e. +$$ +\hat{\theta}_n = \arg \max_\theta \frac{1}{n}\sum_{i=1}^n f(X_i;\theta) +$$ +where $f(x;\theta)$ is the density of $X$. +Suppose we are interested in determining the asymptotic properties of $Z_{i,n} = g(X_i,\hat{\theta}_n)$, where $g$ is a continuously differentiable function. +What is the asymptotic distribution of $\bar Z_n := \frac{1}{n}\sum_{i=1}^n Z_{i,n}$? +My attempt: +Taking a Taylor expansion, we obtain +$$ +Z_{i,n} = g(X_i,\theta) + \frac{\partial g}{\partial y}(X_i,\theta) (\hat{\theta}_n - \theta) + o_P(||\hat{\theta}_n - \theta||) +$$ +from which it follows that +$$ +\sqrt{n}(\bar Z_n - EZ) = \sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n g(X_i,\theta) - EZ\right) + \left(\frac{1}{n}\sum_{i=1}^n\frac{\partial g}{\partial y}(X_i,\theta)\right) \sqrt{n} (\hat{\theta}_n - \theta) + o_P(||Y_n - \theta||) +$$ +where $EZ := Eg(X_i,\theta)$. +I can apply the CLT to get the asymptotic distribution of $\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n g(X_i,\theta) - EZ\right)$ and I can use the Delta Method to obtain the asymptotic distribution of $\left(\frac{1}{n}\sum_{i=1}^n\frac{\partial g}{\partial y}(X_i,\theta)\right) \sqrt{n} (\hat{\theta}_n - \theta)$. However, I know marginal convergence in distribution does not imply joint convergence in distribution, so I can't immediately use the continuous mapping theorem to determine the distribution of the sum of the two terms. How should I proceed? + +REPLY [2 votes]: Under some hypotheses which are assumed for asymptotic normality of MLE, it is possible. +Sketch: Denote $u(x,\theta) = D_\theta \log f(x,\theta)$, $U(X,\theta) = \sum_{k=1}^n u(X_k,\theta)$. Then we know from the proof of asymptotic normality of MLE that +$$ +\theta_n -\theta \sim n^{-1}I(\theta)^{-1}U(X,\theta), +$$ +where $I(\theta) = - \mathsf E_\theta [D_\theta u(X,\theta)]$ is the Fisher matrix. Indeed, + $$-U(X,\theta) = U(X,\hat\theta_n) - u(X,\theta) \sim D_\theta U(X,\theta)(\hat \theta_n-\theta), n\to\infty,$$ $\mathsf P_\theta$-a.s. in view of consistency of $\hat \theta_n$; on the other hand, +$$ +D_\theta U(X,\theta) = \sum_{k=1}^n D_\theta u(X_k,\theta)\sim - n I(\theta), n\to\infty, +$$ +$\mathsf P_\theta$-a.s. by LLN. +Hence it is easy to identify the required joint asymptotic behavior. Namely, +$$ +\left(\sqrt{n}\Big(\frac1n \sum_{k=1}^n g(X_k,\theta)- \mathsf E_\theta g(X_1,\theta)\Big), \sqrt{n} (\theta_n-\theta)\right) +$$ +converges in distribution to a centered normal vector with the covariance matrix +$$c_{i,j}(\theta) = \begin{cases} a_{i-1,j-1}(\theta), & i,j\ge 2\\ +\mathsf E_\theta [g(X_1,\theta)v_j(X_1,\theta)], & i=1,j\ge 2\\ +\mathsf V_\theta\big(g(X_1,\theta)\big), & i=j=1, +\end{cases}$$ +where $a_{i,j}(\theta)$ are elements of $I(\theta)^{-1}$, $v_j$ is the $j$th coordinate of $v(x,\theta) = I(\theta)^{-1}u(x,\theta)$. (Note that $\mathsf E_\theta v(X_1,\theta) = 0$, so in fact $\mathsf E_\theta = \operatorname{cov}_\theta$ in the second line.) +Note that instead of $\mathsf E_\theta[Z]$ I've used $\mathsf E_\theta[g(X_1,\theta)]$, for it is infinitely easier to compute. (Just like one uses $\theta$ instead of $\mathsf E_\theta[\theta_n]$ when writing the asymptotic normality of $\theta_n$.)<|endoftext|> +TITLE: Combinatorics problem that deals with trigonometric functions +QUESTION [5 upvotes]: If $m$ and $p$ are positive integers and $m \geq p$, then show that + $${m \choose 0}+{m \choose p}+{m \choose 2p}+{m \choose 3p}+\cdots$$ + has value $${2^m \over p}\left(1+\sum_{k=1}^{\left \lfloor {p-1 \over 2} \right \rfloor}{{\left( \cos{k\pi \over p}\right)}^m \cos{mk\pi \over p}}\right)$$ + +I have no idea how to approach this problem. I cannot find a way to express combinatorial arguments using trigonometric functions. It'd be great if someone can give me any hints. + +REPLY [3 votes]: With $\zeta=e^{2\pi\mathrm i/p}$ as Gerry suggested, +\begin{align} +\sum_{l=0}^{\left\lfloor\frac mp\right\rfloor}\binom m{lp} +&=\sum_{j=0}^m\binom mj\frac1p\sum_{k=0}^{p-1}\zeta^{jk}\\ +&=\frac1p\sum_{k=0}^{p-1}\sum_{j=0}^m\binom mj\zeta^{jk}\\ +&=\frac1p\sum_{k=0}^{p-1}\left(1+\xi^k\right)^m\\ +&=\frac{2^m}p\sum_{k=0}^{p-1}\left(\cos\frac{k\pi}p\right)^m\mathrm e^{mk\pi\mathrm i/p}\\ +&=\frac{2^m}p\left(1+2\sum_{k=1}^{\left\lfloor\frac{p-1}2\right\rfloor}\left(\cos\frac{k\pi}p\right)^m\cos\frac{mk\pi}p\right)\;, +\end{align} +so I think you were missing a factor of $2$ before the sum.<|endoftext|> +TITLE: Real numbers as element of a universe +QUESTION [5 upvotes]: defn. A universe is a set $U$ such that: + +$x\in u\in U\Rightarrow x\in U$ +$u\in U$ and $v\in U$ imply $\left\{u,v\right\}, \langle u,v\rangle, u\times v\in U$ +$x\in U\Rightarrow \mathcal{P}(x)$ and $\cup x\in U$ +$\omega\in U$ +if $f:x\longrightarrow y$ is a surjective mapping with $x\in U$ and $y\subseteq U$ then $y\in U$ + +Questions: +1) is the set $\mathbb{R}$ of real numbers an element of $U$? +2) if yes, how can be proved? + +REPLY [12 votes]: Let me outline a proof. As previously stated in a comment, the only nontrivial part is to actually define what $\Bbb R$ is. In set theory, it is common to just let $\Bbb R := \mathcal P(\omega)$ and in this case, we immediately have that $\Bbb R \in U$. However, this definition is not very common amongst non-set theorists, and I shall therefore outline a different approach, using Dedekind cuts (as drhab suggested). +First of all, let us define $\Bbb Z := \{ (a,b) \in \omega \times \omega \mid a = 0 \vee b = 0 \}$. We may think of $(a,b)$ as the integer $a-b$. Usually, the part "$a = 0 \vee b = 0$" is omitted and one simply defines an equivalence relation $\sim$ on $\omega \times \omega$ by letting $(a,b) \sim (c,d)$ iff $a+d = c+b$ (which in our indented interpretation just means that $a-b = c-d$. However, we can't write this down until we actually defined $-$. Note that in contrast $+$ is defined, as the usual ordinal addition.) +Clearly $\Bbb Z \subseteq \omega \times \omega$ and hence $\Bbb Z \in \mathcal P(\omega \times \omega)$. Since, $\omega, \omega \times \omega$ and $\mathcal P(\omega \times \omega)$ are elements of $U$, we now get that $\Bbb Z \in U$ as well, by condition 1. +Next, we define $\Bbb Q = \{ (p,q) \in \Bbb Z \times \omega \mid q \neq 0 \wedge (p,q) = 1 \}$. Here $(p,q) = 1$ means that $p$ and $q$ are coprime. I leave it to you to formalize this. (There are two cases, depending on whether $p = (a,0)$ or $p = (0,b)$ and both are trivial.) Again, one usually defines $\Bbb Q$ simply as $\Bbb Z \times (\omega \setminus \{0\})$ and takes the obvious quotient of this, as in the case of $\Bbb Z$. Repeating the argument above, we now get that $\Bbb Q \in U$. +This allows us to define $\Bbb R$: We say that $C \subseteq \Bbb Q$ is a cut iff $C \not \in \{ \emptyset, \Bbb Q \}$, $C$ is downward closed, i.e. for every $p \in C$ and every $q \in \Bbb Q$ with $q < p$, we have that $q \in C$, and $C$ has no $<$-maximal element. (Note that we didn't define the order $(\Bbb Q; <)$ and we actually don't need to do this - formally. We may just work with our intuitive notion of $<$ on $\Bbb Q$. It is, however, not difficult to define $<$ as a set and to show that ${<} \in U$.) We now let $\Bbb R := \{ C \subseteq \Bbb Q \mid C \text{ is a cut} \}$.Our intended interpretation is that $C$ represents that unique real $x$ such that $\{ y \in \Bbb Q \mid y < x \} = C$. Since $\mathbb R \in \mathcal P(\mathcal P(\Bbb Q))$ and $\Bbb Q \in U$, we may apply condition 1 and 3 to conlude that $\Bbb R \in U$. +So far, we only defined a set $\Bbb R$ and it is not entirely obvious that this set has anything in common with our intuitive definition of $\Bbb R$. It is however not difficult to define sets $+, {\,\cdot\,}, {<} \in U$ such that $(\Bbb R, +, {\,\cdot\,}, {<}) \in U$ is provably isomorphic to our intuitive notion of $\Bbb R$ in the sense that $(\Bbb R, +, {\,\cdot\,}, {<})$ is a complete, totally ordered field.<|endoftext|> +TITLE: Proving error bound on Simpson's Rule, Numerical Integration +QUESTION [8 upvotes]: The approximation from "Simpson's Rule" for $\int_a^b f(x)\, dx$ is, +\begin{equation} +S_{[a,b]}f = \bigg[\frac{2}{3}f\Big(\frac{a+b}{2}\Big) + \frac{1}{3}\Big(\frac{f(a) + f(b)}{2}\Big)\bigg](b-a). +\end{equation} +If $f$ has continuous derivatives up to order three, prove that +\begin{equation} +\bigg|\int_a^b f(x)\, dx - S_{[a,b]}f\bigg| \leq C(b-a)^4 \max_{[a,b]} |f^{(3)}(x)|, +\end{equation} +where $C$ does not depend on $f.$ +I've seen many other error estimations for Simpson's Rule, but this one has given me some trouble. I've attempted using some type of majorizing interpolation, but my result was complicated and I think it is over thinking it. Any hints or solutions would be appreciated. + +REPLY [11 votes]: Let $F$ be an anti-derivative of $f$, $F'=f$. W.l.o.g. $a+b=0$, set $x=(b-a)/2$ then we are interested in the error expression +$$ +g(x)=F(x)-F(-x)-\frac{x}3(f(x)+4f(0)+f(-x)). +$$ +This has derivatives +\begin{alignat}{2} +g'(x)&=\frac23(f(x)-2f(0)+f(-x))&&-\frac x3 (f'(x)-f'(-x)) +\\ +g''(x)&=\frac13(f'(x)-f'(-x))&&-\frac x3(f''(x)+f''(-x)) +\\ +g'''(x)&=&&-\frac x3(f'''(x)-f'''(-x)). +\end{alignat} +Consequently, by the extended mean value theorem +$$ +\frac{g(x)}{x^4}=…=\frac{g'''(x_3)}{24x_3}=-\frac{f'''(x_3)-f'''(-x_3)}{72} +$$ +which gives an error bound of +$$ +|g(x)|\le\max_{s\in[a,b]}|f'''(s)|·\frac{x^4}{36}=\max_{s\in[a,b]}|f'''(s)|·\frac{(b-a)^4}{576}. +$$ + +Of course, the more common error term is obtained via +$$ +\frac{g(x)}{x^5}=\frac{g'''(x_3)}{60x_3^2}=-\frac1{90}·\frac{f'''(x_3)-f'''(-x_3)}{2x_3}=-\frac1{90}·f^{(4)}(x_4). +$$ + +$x_3$ above is the middle point in the third iterative application of the mean value theorem. As $g(0)=g'(0)=g''(0)=g'''(0)=0$ you get for $m\ge 3$ $$\frac{g(x)}{x^m}=\frac{g'(x_1)}{mx_1^{m-1}}=\frac{g''(x_2)}{m(m-1)x_2^{m-2}}=\frac {g'''(x_3)}{m(m-1)(m-2)x_3^{m-3}}$$ with $0 +TITLE: Orthogonal Factorization Systems and functoriality +QUESTION [5 upvotes]: In definition 1.1 of these notes on factorization systems, (III) calls the factorization functorial if given the solid diagram described, there's a unique horizontal arrow making both squares commute. Why does this deserve the name 'functorial'? + +REPLY [6 votes]: Let $[n]$ denote the category $0\to 1\to\cdots\to n$. Given a factorization system $(L, R)$ in a category $\mathcal{C}$, we can choose a map $F_0\colon\operatorname{Ob}\mathcal{C}^{[1]}\to\operatorname{Ob}\mathcal{C}^{[2]}$ which maps an object $x\xrightarrow{f} y$ in $\operatorname{Ob}\mathcal{C}^{[1]}$ to an object $x\xrightarrow{e} E\xrightarrow{m} y$ in $\mathcal{C}^{[2]}$ such that $e\in L$, $m\in R$ and $f=me$. +If your factorization system is functorial, this extends to a functor $F\colon\mathcal{C}^{[1]}\to\mathcal{C}^{[2]}$ mapping a morphism $(u,v)$ in $\mathcal{C}^{[1]}$ to a morphism $(u,\omega,v)$ in $\mathcal{C}^{[2]}$ where $\omega$ is the unique morphism required in (III), making the respective diagram commutative. The uniqueness guaranties that $F$ is indeed a functor, i.e. that given two morphism $(u,v),(u',v')\in\mathcal{C}^{[1]}$, we have +$$ +F((u,v))\circ F(u',v') = (u,\omega,v)\circ (u',\omega',v') = (u\circ v',\omega\circ\omega',v\circ v') = F((u\circ u',v\circ v'))$$ +Also note that even though the choice of $F_0$ is not unique, (III) guarantees that it is unique up to isomorphisms. To prove this, take two factorizations $(e, m)$ and $(e',m')$ of the same morphism $f$ and consider the diagrams +$$ +\require{AMScd} +\begin{CD} + x@>\operatorname{id}>> x @>\operatorname{id}>> x\\ +@V{e}VV @V{e'}VV @V{e}VV \\ +E@>\omega>>E' @>\omega'>> E \\ +@V{m}VV @V{m'}VV @V{m}VV\\ + y@>\operatorname{id}>> y @>\operatorname{id}>> y +\end{CD} +$$ +and +$$ +\begin{CD} + x@>\operatorname{id}>> x\\ +@V{e}VV @V{e}VV \\ +E@>\omega'\circ\omega>> E \\ +@V{m}VV @V{m}VV\\ + y@>\operatorname{id}>> y +\end{CD} +$$ +By (III): $\omega'\circ\omega$ is unique, hence $\omega'\circ\omega = \operatorname{id}$. Thus our factorization functor $F$ is determined up to isomorphisms, and it makes sense to talk about the associated factorization functor.<|endoftext|> +TITLE: Calculate $\int_0^{\infty}\frac{1}{(x+1)(x-2)}dx$ using residues +QUESTION [6 upvotes]: I'm supposed to calculate +$$\int_0^{\infty}\frac{1}{(x+1)(x-2)}dx$$ +using residues. The typical procedure on a problem like this would be to integrate a contour going around an upper-half semicircle of radius $R$, and come back through the real axis, taking two indents on the path at the points $z=-1$, $z=2$, say of radius $\rho_1,\rho_2$ respectively. Then the total integral around the path is $0$ and I can calculate the limits as $\rho_1,\rho_2\to0$ and $R\to\infty$ using the known formulas/theorems. However, this leaves me with $\int_{-\infty}^{\infty}\frac{1}{(x+1)(x-2)}dx$, instead of $0$ to $\infty$. And the function is not even so I can't just take half of the whole integral. +Does anybody know a way around this problem? Taking a path going a quarter around the circle and back down to the origin seems unnecessarily complicated, and I'm not even sure that would work here. + +REPLY [3 votes]: You use, as Jack said, a keyhole contour with bumps above and below the pos. real axis at $z=2$. Thus, if $C$ is that keyhole contour with the bumps of radius $\epsilon$, we consider +$$\oint_C dz \frac{\log{z}}{(z+1)(z-2)}$$ +which is equal to (assuming we have taken the radius of the large circular arc to go to $\infty$) +$$PV \int_0^{\infty} dx \frac{\log{x}}{(x+1)(x-2)} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log{(2+\epsilon e^{i \phi}})}{(3+ \epsilon e^{i \phi})(e^{i \phi})} \\ + PV \int_{\infty}^0 dx \frac{\log{x}+i 2 \pi}{(x+1)(x-2)} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\log{(2+\epsilon e^{i \phi})+i 2 \pi}}{(3+ \epsilon e^{i \phi})(e^{i \phi})} $$ +The contour integral is equal to $i 2 \pi$ times the residue at the pole $z=e^{i \pi}$. Thus, +$$-i 2 \pi PV \int_0^{\infty} \frac{dx}{(x+1)(x-2)} - i 2 \pi \frac{\log{2}}{3} + \frac{2 \pi^2}{3} = i 2 \pi \frac{i \pi}{-3}$$ +Therefore we now say that +$$PV \int_0^{\infty} \frac{dx}{(x+1)(x-2)} = - \frac{\log{2}}{3} $$<|endoftext|> +TITLE: Can the fractal dimension of a surface be less than 2? +QUESTION [8 upvotes]: I have two surfaces represented as raster images with heights as grayscale values. +One is natural landscape elevations; the other is just distance from a line. +I have computed Minkowsky D = 2 - H where, for surface area A and cell size s at aggregation level k: +$$ A_k = s_k^{-H} \quad \mathrm{rewritten} \quad \log A_k = -H \log s_k$$ +and H is found by linear regression. +D for the elevation surface is 2.31, but D for the distance surface comes out as 1.81. Does this indicate an error in computation? a wrong choice of dimension formula? Is there some meaning associated with a fairly simple 2D surface having a fractal dimension less than 2? +code snippet + +REPLY [2 votes]: A locally Lipschitz map can't increase Hausdorff dimension. In particular, affine projections are Lipschitz maps. So if your surface $S$ lies over a nonempty open set $U$ in the $xy$ plane, in that the image of $S$ under the projection $(x,y,z) \to (x,y)$ contains $U$, then the Hausdorff dimension of $S$ must be at least the Hausdorff dimension of $U$, namely $2$.<|endoftext|> +TITLE: if $\left | x_{n+1}-\frac{x_{n}^2}{x_{n-1}} \right |\leq 1$, show that $(\frac{x_{n+1}}{x_{n}}) $ convergent +QUESTION [13 upvotes]: Let a real positive number sequence $(x_{n})$ such that $\left | x_{n+1}-\frac{x_{n}^2}{x_{n-1}} \right |\leq 1$ and $\sqrt{x_{1}}\geq \sqrt{x_0+1}$. Show that $(\frac{x_{n+1}}{x_{n}}) $ convergent. +I have tried to define a sequence $(y_{n})$ such that $y_{n}=\frac{x_{n+1}}{x_{n}}$. So the preceding inequality equivalent to : +$\left |x_{n}( \frac{x_{n+1}}{x_{n}}-\frac{x_{n}}{x_{n-1}}) \right |\leq 1$ +$\equiv \left |y_{n}-y_{n-1}\right |\leq \frac {1}{x_{n}}$ +I want to show that $(y_{n})$ is contractive sequence, and then it will be a cauchy sequence if and only if convergent sequence. But i stuck at this part. +Please give me some clue, thank you. + +REPLY [4 votes]: What about the following sequence: +$$1,2,3,\frac{7}{2},\frac{37}{12},\frac{865}{504},1,1,\frac{1}{2},1,1,\frac{1}{2},1,1,\frac{1}{2},...$$ +I believe it satisfies the given conditions, but clearly $\frac{x_{n+1}}{x_n}$ doesn't converge. I think the problem statement might be wrong.<|endoftext|> +TITLE: Finite sum $\sum_{n=2}^N\frac{1}{n^2}\sin^2(\pi x)\csc^2(\frac{\pi x}{n})$ +QUESTION [27 upvotes]: I was looking for a closed form but it seemed too difficult. Now I'm seeking help to simplify this sum. The 50 bounty points or more will be awarded for any meaningful simplification of this sum. +I found this function that has very interesting property to show if $n \nmid x$. And the sum of this function from $2$ to $N$ can show all the primes within $N$ on 1, and primes beyond $n$ and within $N^2$ as zeros, and prime sieve beyond that +$$ +f(n,x)=\frac{1}{n}F_n\left(\frac{2\pi}{n}x\right)=\frac{1}{n^2}\left(\frac{1-\cos(2\pi x)}{1-\cos\left(\frac{2\pi}{n}x\right)}\right)=\frac{1}{n^2}\left(\frac{\sin(\pi x)}{\sin\frac{\pi x}{n}}\right)^2 +$$ +This is a prime test function and I try to find a closed form formula to simplify the calculation of $$\sum_{n=2}^Nf(n,x)=\sum_{n=2}^N\frac{1}{n^2}\left(\frac{\sin(\pi x)}{\sin\left(\frac{\pi x}{n}\right)}\right)^2=\sum_{n=2}^N\frac{1}{n^2}\sin^2(\pi x)\csc^2\left(\frac{\pi x}{n}\right)$$ +Is there a close form formula (i.e. no or few terms of summation) for this sum? It's known that $$\int_{2}^{N} f(t,x)dt=\sin^2(\pi x)\int_{2}^{N}\frac{1}{t^2}\csc^2\left(\frac{\pi x}{t}\right)dt=\left. \left(\frac{1}{\pi x}\sin^2(\pi x)\cot\left(\frac{\pi x}{t}\right) \right)\right|_{t=2}^{t=N}$$ +Does the Euler–Maclaurin Summation Formula help here? +Here's the example graphic of $\sum_{n=2}^{50}f(n,x)$ shows all primes within $2500$. + +REPLY [2 votes]: An idea: +Note the identity +$$\sin n \theta=U_{n-1}\left( \cos \theta \right) \sin \theta ,$$ +where $U_i$ is the $i^{th}$ Chebyshev orthogonal polynomial of the second kind. This gives +$\sin^2 \pi x=\sin^2 \left(n \frac{\pi x}{n}\right)=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right) \sin^2 \frac{\pi x}{n},$ +and then +$\sin^2 \pi x \csc^2 \frac{\pi x}{n}=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right) \sin^2 \frac{\pi x}{n} \csc^2 \frac{\pi x}{n}=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right),$ +so your sum becomes +$\sum \limits_{n=2}^N \frac{1}{n^2} U_{n-1}^2\left( \cos \frac{\pi x}{n} \right).$ +There may be an identity on the Chebyshev polynomials to further simplify this sum.<|endoftext|> +TITLE: Solve these functional equations: $\int_0^1\!{f(x)^2\, \mathrm{dx}}= \int_0^1\!{f(x)^3\, \mathrm{dx}}= \int_0^1\!{f(x)^4\, \mathrm{dx}}$ +QUESTION [19 upvotes]: Let $f : [0,1] \to \mathbb{R}$ be a continuous function such that + $$\int_0^1\!{f(x)^2\, \mathrm{dx}}= \int_0^1\!{f(x)^3\, \mathrm{dx}}= \int_0^1\!{f(x)^4\, \mathrm{dx}}$$ + Determine all such functions $f$. + +So far, I've managed to show that $\int_0^1\!{f(x)^t\, \mathrm{dx}}$ is constant for $t \geq 2$. Help would be appreciated. + +REPLY [4 votes]: By Holder's Inequality, +$$\left(\int_{[0,1]}ff^2dx\right)^2\leq \int_{[0,1]}f^2dx \int_{[0,1]}f^4 dx=\left(\int_{[0,1]}f^3dx\right)^2$$ +Equality occurs in this inequality iff $f=af^2$ for some $a\in\mathbb{R}$ i.e., Linearly dependence of $f,f^2$.. +$f=af^2$ implies $f^2=af^3$. Considering the integrals and their equality, we see that $f=f^2$. +Which in turn implies that $f(x)=0$ or $1$ for $x\in [0,1]$<|endoftext|> +TITLE: Does the Lie derivative commute with $\partial$? +QUESTION [11 upvotes]: It is well-known that on a smooth manifold $M$, the Lie derivative commutes with the exterior derivative, i.e. +$${\cal L}_Xd\alpha=d{\cal L}_X\alpha$$ +for any vector field $X$ and differential form $\alpha$. +If $M$ is a complex manifold, is there a similar result for the partial derivative +$${\cal L}_X\partial\alpha=\partial{\cal L}_X\alpha?$$ +(Edit: By "similar" I mean maybe it does not hold in this form but there is nevertheless an analogous statement?) + +REPLY [2 votes]: It's true if the complex structure is invariant under the vector field X. In that case: $\mathcal{L}_X \partial a = \mathcal{L}_X (1 - \imath J) da = (1 - \imath J)\mathcal{L}_X da = \partial \mathcal{L}_X a$. (Note: This is written with $a$ being a scalar. For a $(p,q)$ form you just need to pick a suitable projector.)<|endoftext|> +TITLE: Find limit recursion of sequence $x_{n+1} = \frac{x_n+ n x_{n-1}}{n+1} $ +QUESTION [5 upvotes]: Prove sequence +$$x_{n+1} = \frac{x_n+ n x_{n-1}}{n+1} $$ +$$x_0 = 0, x_1 = 1 $$ +converges and find it's limit + +My attempt + +Let's prove $0 \le x_n \le 1$: +$x_n \ge 0 $ (obvious) +By induction +if $x_n \le 1$ and $x_{n-1} \le 1$ then: $$ \frac{x_n+ n x_{n-1}}{n+1} \le \frac{1+ n}{n+1}=1 \implies x_{n+1}<=1$$ +Let's prove convergence + +$$\lim_{n \to \infty }{x_{n+1}} = \lim_{n \to \infty }{ \frac{x_n + n x_{n-1}}{n+1} } = \lim_{n \to \infty }{ (\frac{x_n}{n+1} } + \frac{x_{n-1}}{1+\frac{1}{n}}) = x_{n-1}$$ +So, sequence converges. +Question: I'm right so far and how to find the limit? +Thanks + +REPLY [6 votes]: Hint: +Check that $\displaystyle \forall n\geq 0, x_{n+1}-x_n=\frac{(-1)^n}{n+1}$ +The series $\displaystyle \sum (x_{n+1}-x_n)$ is therefore convergent, and so is the sequence $(x_n)$, say $x_n\to l$ +Furthermore, $\displaystyle l= \sum_{k=0}^\infty (x_{k+1}-x_k) =\sum_{k=0}^\infty \frac{(-1)^k}{k+1} = \log 2$. + +Edit: from a simulation on Mathematica, here is the plot of the first few terms of the sequence, giving intuition about its behavior:<|endoftext|> +TITLE: Distance of bch code +QUESTION [5 upvotes]: I want to show that a binary , narrow-sense BCH code of length $n=2^m-1 $ and designed distance $ \delta=2t+1 $ has minimum distance $ \delta$ , given that $ \sum_{i=0}^{t+1} \binom{2^m-1}{i}> 2^{mt} $. +We know that: +A BCH code with designed distance $ \delta $ has minimum distance at least $ \delta $. +So we have that $ d \geq \delta $. +In order to show that the minimum distance is $\delta $,we have to find a codeword with Hamming-weight $\delta $. +How can we use the given inequality in order to show the existence of such a codeword? +EDIT: Also how can we show that the distance of the code has to be odd? + +REPLY [3 votes]: Hints: + +The sum $\sum_{i=0}^{t+1}\binom n i$ counts the number of points at a distance $\le t+1$ from any given point $\in\Bbb{F}_2^n$. +The dimension of your BCH code is at least $n-mt$, so it has at least $2^{n-mt}$ distinct words. +Compare with the derivation of the sphere packing bound. + + +[Added after Evinda spotted a gap] We can conclude that that minimum distance $d$ of this code is thus at most $2t+2$. But we can show that $d$ is always odd. This is because if $d$ were $2t+2$, then this would also be the minimum distance of the extended BCH code of length $n+1=2^m$. But that code has a transitive automorphism group, so it has words of minimum weight $2t+2$ such that the extended bit is also $1$. The claim follows. +Alternatively we can see this using finite field arithmetic. The existence of a word of weight $d=2t+2$ means that there exists a subset $S\subseteq \Bbb{F}_{2^m}\setminus\{0\}$ of size $|S|=d$ such that for all integers $1\le j\le 2t$ we have +$$ +\Sigma(S,j):=\sum_{x\in S}x^j=0. +$$ +Because $d$ is an even number it follows that $\Sigma(S,0)=0$ also. Let $s\in S$ be a fixed element. Let $S+s=\{x+s\mid x\in S\}$, so $0=s+s\in S+s$. The binomial formula tells us that +$$ +\Sigma(S+s,j)=\sum_{x\in S}(x+s)^j=\sum_{k=0}^j\binom j k s^{k-j} \Sigma(S,j)=0 +$$ +for all $j$ in the range $1\le j\le 2t$. This means that the set $S+s\setminus \{0\}$ is the support of a word of this BCH-code also. But it has only $2t+1$ elements.<|endoftext|> +TITLE: Where does the gap come from? +QUESTION [24 upvotes]: Can anyone tell me please where does the gap come from? +Thanks and sorry if the question is not exactly relevant, I just didn't know where else to ask. + +REPLY [27 votes]: Figure out the area of the little gap in the middle: + +(you can probably guess it, but it's easy to calculate). +Try it for yourself before peeking at the explanation at the bottom. +From there it's easy to see where the area for the square comes from. + +Total area of the 13 × 5 region = 65 squares. +Total area of the two unmarked rectangular parts = (5 × 3) × 2 = 30 +Total area of blue triangles = (5 × 2 × ½) × 2 = 10 +Total area of red triangles = (8 × 3 × ½) × 2 = 24 +Remaining area = 65 - 30 - 10 - 24 = 1 square. +If it's still not clear, rotate the upper two triangles each about the center of its own diagonal (which leaves the upper edge of the gap unchanged), you get the triangles from your two diagrams overlaid on each other in the correct positions: + +It takes the extra square to make up for the area of that skinny parallelogram-shaped gap between the regions covered by the two arrangements of shapes.<|endoftext|> +TITLE: What is a sample of a random variable? +QUESTION [6 upvotes]: I've tried to learn probability, one way or another, many times in the last 50 years, and finally settled on the Kolmogorov approach, where a random variable isn't described as something like "a roll of a die," but rather as "a measurable function from a measure space $X$ to the reals, where the (fixed) measure on $X$ has the property that $\mu(X) = 1$." +That's something concrete and mathematical that I can work with, even if my students hate it. :) +Now I'm finding that I don't really know what a "sample" is. I'm reading a paper that says "Let us consider some dataset $X = \{ x^{(i)}\}_{i=1}^N$ consisting of $N$ i.i.d. samples of some continuous or discrete random variable $x$." And I realized that I didn't know what that meant. +I went down the hall and asked a guy who does lots of probability stuff, and usually helps me clarify things, and he said "a sample...well, sampling is a process that produces outputs distributed ... well, ... it's a single value, not a whole function like the random variable $x$." (I'm being a little unfair here; I think what he said was more coherent than this, but I couldn't make sense of it.) +My question: Can someone pleaes tell me what a sample is, in formal terms? (By that I mean that it should be some familiar mathematical object -- a set, or a function on some well-described domain, or a relation, or ... you get the idea. It should not be something like "an algorithm" or "a process" or anything like that.) +Perhaps a followup question is this: Can someone recommend a book or other source from which I can read things like this? I thought that Feller might help, but didn't find it very satisfactory the last time I looked. + +REPLY [6 votes]: ^I think @adjan's answer and @Ian's answer above are other correct interpretations. +Generally, "sample" is more of a concept from statistics then probability. +If you have a "population" which assumes a certain property with probability defined by distribution $P$, then a sample is a vector (of arbitrary length, but ideally specified beforehand) of independent copies of a random variable $X$ distributed according to distribution $P$. +To be technical, two copies $X_1, X_2$ of an independent random variable $X$ are actually two random variables on the product space $\Omega \times \Omega$ which have the product measure (i.e. $\mathbb{E}X_1X_2 = \mathbb{E}X_1 \mathbb{E}X_2$). If $\pi_1, \pi_2$ are the canonical projections and $i_1, i_2: \Omega \to \Omega \times \Omega$ are any two injections such that $\pi_1 \circ i_1 = id_{\Omega}, \pi_2 \circ i_2 = id_{\Omega}$, then $X_1 \circ i_1 \sim X$ and $X_2 \circ i_2 \sim X$. +Sampling a predetermined, fixed size $n$ of independent "samples" from the total population is really just considering the product space $\Omega^n$ with random variables $X_1, \dots, X_n$ endowed with the product measure, thus $X_1 \circ i_1 \sim \dots \sim X_n \circ i_n \sim X$. +If we don't want to fix in advance the number of samples we take, then sampling is just considering the product space $\Omega^{\mathbb{N}}$ with random variables $X_1, \dots, X_n, \dots$ endowed with the product measure, thus $X_1 \circ i_1 \sim \dots \sim X_n \circ i_n \sim \dots \sim X$. Such a construction is possible in the countably infinite case by the Ionescu Tulcea extension theorem, for example, even in many cases when the conditions of Kolmogorov's extension theorem do not apply. +It is also worth noting that we don't necessarily need to assume that the "copies" of $X$ are independently distributed; using an entirely analogous definition, it would still be possible for $X_1 \circ i_1 \sim \dots \sim X_n \circ i_n \sim \dots \sim X$ without having the product measure on $\Omega^{\mathbb{N}}$ (here I am drawing an implicit distinction between "the" product measure on the product of measure spaces and an arbitrary measure on such a space, which need not equal the canonical such measure.) +Then a "statistic" is any (measurable) function of this sample vector in the finite case or sample sequence in the infinite case. +Because in practice every experiment conducted in real life only takes finitely many variables, references to sample vectors (of an arbitrary, unspecified size $n$) rather than sample sequences will often be more common in statistics-oriented literature, although the greater generality afforded (in some sense) by the limiting case $n \to \infty$ is used very frequently in the pure probability literature (e.g. when discussing random walks or results like the CLT or Law of the Iterated Logarithm). However in the pure probability literature such infinite samples are usually termed "a sequence of identical (independently distributed) random variables" rather than a "sample", even though any sequence of identical random variables is the same type of mathematical object as the unbounded version of the theoretical model of "samples" used by statisticians. Indeed, this correspondence is why the terms "sampling with replacement" and "sampling without replacement" turn up in the pure probability literature. Since the central object of study of statistics are vectors of random variables, this explains why the theory of vector-valued random variables is more prominent or prevalent in the statistical literature than other fields. +For example, we could have a population of particles with velocities distributed as $N(0,1)$. Then a sample would correspond to a vector $(X_1, \dots, X_n)$ with each $X_i$ having the distribution $N(0,1)$. +"In practice" this is the theoretical model for choosing $n$ particles from the population and noting what their velocity is. +Given this sample, we can calculate statistics, for example, the "sample mean" $(X_1 + \dots X_n)/n$. +However, this is in general different from the "mean of the population" (which is 0, the expectation of the normal distribution), since it is a random variable and hence assumes different values randomly depending on which particles we choose. In this case, however, the statistics has the optimal property that its expectation is equal to the population mean (i.e. $E[(X_1 + \dots X_n)/n]=EX$.<|endoftext|> +TITLE: Would such a function be of any importance (primality test)? +QUESTION [5 upvotes]: While experimenting with some Maths, I came up with a really cool function. +Let's call this function $\space \beta \space$. Which is a function of a real variable $\space r \space $. +Here is the function: +$$\beta (r) = \tan \Bigg({\pi \over 2 } - {{4 \pi \Gamma(r)} \over r} \Bigg)$$ +Where $\space \Gamma(r) \space $ is the Gamma Function. +Here is why this function is cool. +Let's take $\space \beta (z) \space$ where $\space z \space$ is a positive integer. +Well, $\space \beta (z) \space$ is only well defined IF AND ONLY IF $\space z \space$ is a prime GREATER than $\space 2$ (I have a proof for this, I didn't just assume it). +So when $\space z \space$ is not prime, $\space \beta (z) \space$ is undefined. +Would this serve ANY importance at all? Is it worth mentioning? +Either way, I found it pretty cool and I hope you do too :). + +REPLY [3 votes]: If the question is about the usefulness of this function as a primality test (as I assumed in the comments and the OP appeared to agree with me), then it's probably even less useful than the Wilson's theorem on its own. +The primality tests are only needed for really big numbers. First, this one involves computing the factorial of large numbers. And second, for large numbers the function becomes larger (in absolute value), which makes it harder to distinguish the finite case from the infinite one. +It's more convenient to use the reciprocal function: +$$\frac{1}{\beta(r)}=\cot \left(\frac{\pi}{2}-\frac{4\pi \Gamma(r)}{r} \right)=\tan \left(\frac{4\pi \Gamma(r)}{r} \right)$$ +Then it will be zero for every integer $r$ unless $r$ is prime. +But! Its absolute value will steadily decrease with increasing $r$, thus making it harder and harder to determine if a particular large number is prime. See the list plot up to $r=100$:<|endoftext|> +TITLE: Find the eigenvalues of a 3 x 3 matrix +QUESTION [8 upvotes]: I have a question on determining eigenvalues for a given matrix A: +$$ A= + \begin{bmatrix} + 2 & 1 & 2 \\ + 0 & 2 & -1 \\ + 0 & 1 & 0 \\ + \end{bmatrix} +$$ +Here's what I have so far: +\begin{align} +& det(A-\lambda I_3) = 0\\ +& =det \begin{bmatrix} + 2-\lambda & 1 & 2 \\ + 0 & 2-\lambda & -1 \\ + 0 & 1 & 0-\lambda \\ + \end{bmatrix} = 0 \\ +& = (2-\lambda) \begin{vmatrix} + 2-\lambda & -1 \\ + 1 & 0-\lambda \\ + \end{vmatrix} \\ +& = (2-\lambda)(-2\lambda+\lambda^2+1) \\ +& = (2-\lambda)(\lambda-1)^2 = 0\end{align} +So my eigenvalues are $2$ and $1$. I am almost postitive this is correct. Here's my confusion/question. I know that the determinant of an upper triangular matrix is the product of the terms along the diagonal. +Some of my classmates used a different approach where from the beginning they row reduced $A$ into an upper triangular matrix by eliminating the $1$ from the $3$rd row, then subtracted $\lambda I$ and then instead of solving the determinant like I did they just multiplied the diagonal terms together. +The problem I see with this is that you can obtain different matrices depending on how you choose to row reduce, which means you can get different eigenvalues. Is it possible to solve for the eigenvalues by using the approach of making $A$ into an upper triangular matrix first? (So that you can solve the determinant by multiplying the diagonal entries) I hope that makes sense. Thanks! + +REPLY [4 votes]: The answer to your question is no. That is, your friends are wrong. In general, applying row-reduction to your matrix will change its eigenvalues. +We can see what goes wrong if we take this approach slightly differently. In particular: +$$ +\pmatrix{ + 2 & 1 & 2 \\ + 0 & 2 & -1 \\ + 0 & 1 & 0 \\ +} \to +\pmatrix{ + 2 & 1 & 2 \\ + 0 & 1 & 0 \\ + 0 & 2 & -1 \\ +} \to +\pmatrix{2&1&2\\0&1&0\\0&0&-1} +$$ +however, $-1$ is not an eigenvalue of $A$. +What's even more extreme: if we keep row-reducing, we can go from $A$ to $I$. It is certainly not the case, however, that $A$ has only the eigenvalue $1$. +If you insist on calculating the determinant via row-operations, subtract $\lambda I$ first: +$$ +\det(A - \lambda I) = +\det \pmatrix{ + 2 - \lambda & 1 & 2 \\ + 0 & 2 - \lambda & -1 \\ + 0 & 1 & -\lambda \\ +} = \\ +\det\pmatrix{ +1&0&0\\ +0&1&0\\ +0&-\frac{1}{2 - \lambda} & 1 +}^{-1} +\det \pmatrix{ + 2 - \lambda & 1 & 2 \\ + 0 & 2 - \lambda & -1 \\ + 0 & 0 & -\lambda + \frac{1}{2 - \lambda} \\ +} = \\ +1 \cdot \det \pmatrix{ + 2 - \lambda & 1 & 2 \\ + 0 & 2 - \lambda & -1 \\ + 0 & 0 & -\lambda + \frac{1}{2 - \lambda} \\ +} +$$<|endoftext|> +TITLE: Uses of step functions +QUESTION [5 upvotes]: My highschool teacher has informally told us about what continuity is and used step functions as an example of a discontinuous function. +The Wikipedia page for it links to a lot of other kind of step functions, such as the Heaviside step function. +What are some uses of these step functions to mathematicians at higher levels (or even in highschool)? + +REPLY [5 votes]: Two of the most frequently used step functions are the "floor" and "ceiling" functions. +The floor function takes a real number and returns the largest integer(whole) number less then or equal the real number. For a positive real number this is equivalent to cutting off all the numbers after the decimal place. +The ceiling function returns the smallest integer number greater then the number given. +They are useful in rounding off and often arise in cases where there is some sort of calculation that results in a real number but only the whole part is meaningful.<|endoftext|> +TITLE: Wallpaper groups for the hyperbolic plane +QUESTION [9 upvotes]: I would be grateful if someone could direct me to a reference that classifies the equivalent of the wallpaper groups (and the frieze groups and the point groups, if possible) for the hyperbolic plane, i.e., the symmetry groups of isometries acting properly discontinuously on the hyperbolic plane. +The Euclidean case is classical, well-known and easy to find proofs for. The spherical case is also well-known, and normally described in the same terms (see, for example, the Wikipedia article List of spherical symmetry groups). But it seems rather more difficult to find a place where the same ideas for the hyperbolic plane are described in the same way (although I'm sure this has been done), as rotations, translations, reflections, and possibly glide reflections, acting on some sort of fundamental region, as seen in Wikipedia's article on the wallpaper groups, for example. +There are several ways of approaching hyperbolic symmetries, including groups of reflections in the sides of $(l,m,n)$ hyperbolic triangles with $1/l+1/m+1/n<1$, (see e.g. Wikipedia's article), or tessellations of $p$-sided polygons, $q$ of which meet at each vertex, with $(p-2)(q-2)>4$; presumably each hyperbolic wallpaper group appears as a symmetry group of one of these tessellations, and is a subgroup of some triangle group; is there a reference that demonstrates this? + +REPLY [8 votes]: What you are asking for is the theory of hyperbolic 2-dimensional orbifolds. It's a very big theory. If you do not add additional hypotheses, it becomes somewhat unreasonable to expect a good description of the theory. Even if you add enough hypotheses to tame the question, it still requires a lot of mathematics to even describe the classification. +Let me tame the question a bit by adding some hypotheses and putting some restrictions on the classification, just in order to give a taste of the big picture. +First, I'll add the hypothesis that the group $\Gamma$ is finitely generated. Next, I'll add a stronger hypothesis of co-compactness, meaning $\Gamma$ has a compact fundamental domain. Co-finite area is also a good hypothesis, as said in my comment, but is trickier to describe correctly, although not really any deeper mathematically. So I am focussing on discrete, co-compact groups $\Gamma < \text{Isom}(\mathbb{H}^2)$ (which, as a consequence, side-steps any questions about analogues of Frieze groups). +The basic level is classification up to isomorphism: the group $\Gamma$ is determined by its quotient orbifold $M_\Gamma = \text{H}^2 / \Gamma$, in the sense that $\Gamma,\Gamma' < \text{Isom}(\mathbb{H}^2)$ are isomorphic if and only if their quotient hyperbolic orbifolds $M_\Gamma,M_{\Gamma'}$ are "orbifold equivalent", meaning that there is a homeomorphism $M_\Gamma \to M_{\Gamma'}$ which preserves the orbifold structure. +On this basic level, the classification in the hyperbolic case is identical (in theory) to the Euclidean and spherical cases: the groups $\Gamma$ with above hypotheses care classified by the closed hyperbolic orbifolds; the 17 wallpaper groups are classified up to isomorphism by the 17 closed Euclidean orbifolds up to orbifold equivalence; and there is a similar spherical classification. +There are infinitely many distinct closed hyperbolic 2-orbifolds $\mathcal{O}$ up to equivalence, and hence infinitely many groups $\Gamma$ up to isomorphism. It is possible to completely list all closed hyperbolic 2-orbifolds $\mathcal{O}$ by keeping track of certain invariants: + +the topological type of the surface $S$ underlying $\mathcal{O}$ (topological Euler characteristic $\chi(S)$, orientability of $S$, number of boundary components of $S$); +the numbers of cone points of all angles in the interior of $S$; +the cycles of dihedral points going around each boundary component of $S$ (keeping track of angles of the dihedral points in the cycle, and taking appropriate care of orientation of boundary components depending on whether $S$ itself is orientable). + +Of course one must also be sure to discard from this list the spherical, Euclidean, and bad 2-orbifolds, but that can be done using the sign of the orbifold Euler characteristic $\chi(\mathcal{O})$, which can be computed by a formula using the invariants listed above, with the following outcome: + +spherical or bad is equivalent to $\chi(\mathcal{O})>0$; +Euclidean is equivalent to $\chi(\mathcal{O})=0$; +hyperbolic is equivalent to $\chi(\mathcal{O})<0$. + +The link provided in the first line of the answer gives a complete list of the Euclidean, spherical, and bad cases; everything else is hyperbolic. + +You have asked for references. One reference is anything about Conway's orbifold notation. You might also look up any references on Fuchsian groups, although this will cover only the cases where $\Gamma$ preserves orientation (in particular, no reflections). Also, the literature on Fuchsian groups tends to focus even more strictly on the case where the groups have no finite order elements (no rotations), but still one can find some references without that restriction.<|endoftext|> +TITLE: Is a map that preserves the hyperbolic distance biholomorphic? +QUESTION [5 upvotes]: Let $\lVert z \rVert_w = \frac{|z|}{1 - |w|^2}$ be the hyperbolic distance in $\mathbb{D}$, and let the hyperbolic metric be $d(z, w) = \inf_\gamma \int_0^1 \lVert \gamma'(t) \rVert_{\gamma(t)} \, dt$. Automorphisms of $\mathbb{D}$ preserve this metric. I'm looking to prove the converse, that is, if $f \colon \mathbb{D} \to \mathbb{D}$ preserves $d$ then it is a biholomorphism from $\mathbb{D}$ to itself, or $\overline{f}$ is. +If $f$ is assumed $C^1$ it is mostly not so hard - you can get that at every point either $f$ is holomorphic or its conjugate is by considering $\psi_{f(0)} \circ f$, which preserves the origin and must also preserve the ordinary absolute value, hence is isotropic hence either $f$ or $\overline{f}$ is holomorphic at the origin, and since $f$ was arbitrary this is true of all points. (And hopefully this gives that one of them is globally holomorphic! But I haven't figured out that part yet.) +If nothing is assumed of $f$, is the result even true? I see why $f$ must be continuous, but does preservation of the hyperbolic metric imply $C^1$-ness? + +REPLY [5 votes]: Yes, this is true. The proof is not specific to hyperbolic metric: one can argue the same way about the Euclidean metric on the plane, or in higher dimensions. +Step 1: Compose $f$ with a Möbius transformation that sends $f(0)$ to $0$. This reduces the problem to the case $f(0)=0$. +Step 2: Since $f$ is an isometry, every circle around the origin is mapped to self. By composing $f$ with a rotation, we reduce the problem to the case $f(1/2)=1/2$. +Step 3: Consider the (hyperbolic) circles with centers $0,1/2$ passing through $i/2$. Like any two distinct circles, they have at most two points in common: specifically, $i/2$ and $-i/2$. Hence, $f(i/2)$ must be one of these points. By composing $f$ with conjugation, the problem reduces to the case $f(i/2)=i/2$. +Step 4. The same reasoning as in step 3 shows that $f(z)$ is either $z$ or $\bar z$ for every $z$. And since $d(f(z), i/2) = d(z,i/2)$, it must be $f(z)=z$.<|endoftext|> +TITLE: Why is complex analysis so nice? And how is it connected/motivating for algebraic topology? +QUESTION [21 upvotes]: This is very much a soft question, but after seeing Cauchy's integral formula in lecture today I was really struck by how neat complex analysis is. I don't understand how all of these amazing analytic properties (global extrapolations from local properties/holomorphic implies infinitely holomorphic) can come from just algebraically adjoining the square root of -1. +When I asked my professor about this, he said it was a function of the complementary relationship between complex analysis and algebraic topology and didn't really expand on that. +Even not knowing much algebraic topology, this connection does seem clear in some ways (the importance of simple connectedness for Cauchy's theorem and dealing with so many paths/using code words for homotopy). However, I am still not sure what it is about the complex plane that lends itself to this special link, especially when it comes to functions. $\mathbb{R}^{2}$ is topologically equivalent (maybe just point set topology questions?) from what I understand and it definitely isn't as nice. +I would appreciate any sort of discussion or direction towards references (especially for someone that hasn't learned much topology formally-Hatcher is a difficult text for me to grapple with on my own) and I hope this is interesting to other people. + +REPLY [2 votes]: I'm going to narrow your question to the following first paragraph: + +I don't understand how all of these amazing analytic properties (global extrapolations from local properties/holomorphic implies infinitely holomorphic) can come from just algebraically adjoining the square root of -1. + +There are three questions hidden here: Why $-1$? Why the square root (as opposed to, say, the cube root)? And how do these produce "these amazing analytic properties"? +The other answer and the question John Kyon linked in comments give excellent answers to the third question. To quickly summarize, holomorphisms are nice, because Cauchy's formula and shifting contours give us the implication $\text{integrable}\Rightarrow\text{differentiable}$. We get Cauchy's formula and shifting contours from the Cauchy-Riemann equations, and the CR-equations arise because we want the derivative of a $\mathbb{C}\to\mathbb{C}$ function at a given point to itself be an element of $\mathbb{C}$. +But this leaves the first two questions more mysterious. We can generalize the construction of $\mathbb{C}$ quite substantially: given a commutative ring $R\leq\mathbb{R}$ and an $R$-algebra $A$, we can ask about the functions $A\to A$ with derivatives given by the multiplication action of $A$ on itself. For example, we could always look at numbers of the form $a+b\sqrt{-2}$ instead of $a+b\sqrt{-1}$. Of course, it turns out that those numbers are just $\mathbb{C}$ again…but can you be sure this isn't just a bad example? Why isn't there just as nice a theory for these other algebras? Why don't we hear about them? +The answer is that there are (essentially) no other algebras. We need to have some sort of underlying complete field in order to define derivatives. So we need to start with $R=\mathbb{R}$ above. But then abstract algebra tells us that, since $\mathbb{R}$ is a field, any commutative, finite-dimensional $\mathbb{R}$-algebra is a direct sum (as $\mathbb{R}$-algebras) of field extensions of $\mathbb{R}$. So the $A$ we wanted to analyze above is built out of objects like $\mathbb{C}=\mathbb{R}(\sqrt{-1})$…or $\mathbb{R}(\sqrt{-2})$. +So what sort of numbers can we adjoin to $\mathbb{R}$ to get something bigger than $\mathbb{R}$? By a Galois-theoretic argument (see Dummit and Foote, section 14.6), the answer is precisely "square roots of negative numbers." Moreover, those additions give all of $\mathbb{C}$, by comparing the dimensions as $\mathbb{R}$-vector spaces. So if we create an algebra by adjoining a different square root, we still get $\mathbb{C}$, but with a weird coordinatization that makes no geometric sense. We might as well use the coordinatization that gives good geometry!<|endoftext|> +TITLE: What are the group objects in the category of finite sets and bijections, and its functor category? +QUESTION [8 upvotes]: An object $G$ in a category $\mathcal{C}$ is called a group object if, given any object $X$ in $\mathcal{C}$, there is a group structure on the morphisms $\operatorname{hom}\left(X,G\right)$ such that $X\mapsto \operatorname{hom}\left(X,G\right)$ is a (contravariant) functor from $\mathcal{C}$ to $\text{Grp}$. + +Group objects of the category $\text{Set}$ are groups in the usual sense. Similarly, group objects of the category $\text{FinSet}$ are finite groups. +Assuming that good (non-tautological) descriptions exist, + +What are the group objects of the category $\text{FinBij}$, the category whose objects are finite sets, and whose morphisms are bijections? +What are the group objects of the functor category of $\text{FinBij}$, the category whose objects are functors of $\text{FinBij}$, and whose morphisms are natural transformations? + +REPLY [11 votes]: There is no group object: since you only have bijections in your category, for all G there exists an X such that hom(x,G) is empty, so not a group in any way at all.<|endoftext|> +TITLE: Expressing the minimum function in terms of the absolute value in a symmetric manner (generalized to more variables) +QUESTION [9 upvotes]: It is well known that: +$$\max(a,b) = \frac12(a+b)+\frac12|a-b|$$ +and similarly: +$$\min(a,b) = \frac12(a+b)-\frac12|a-b|.$$ +In fact, they are equivalent since $\max(a,b) = -\min(-a,-b)$. We can try generalizing these representations to more arguments. One obvious approach is to do the following: +$$\max(a,b,c) = \max(a,\frac12|b+c|+\frac12|b-c|)$$ +and then again use the previous expression. However, the final expression is quite messy and not obviously symmetric under permuting the $3$ variables. Is there a neater expression? +I expect there should be some form in the general case of $n$ variables as a sum over the permutations in $S_n$ with some sort of sign but it is not entirely clear... +Edit: This question answers the $n=3$ case. I am still interested in the general $n$ case or alternative representations for $n=3$. + +REPLY [2 votes]: If we apply the three powers of the permutation $(a \ b \ c)$ to the "obvious" recursive expression and add, we get $$ + \max(a, \max(b, c)) + \max(b, \max(c, a)) + \max(c, \max(a, b)) \text{.} +$$ +Clearly all three of these terms have the same value, so one-third of them is the value sought. Applying the definition in terms of absolute values and simplifying a bit, we get $$ +\frac{1}{12} (4 a + 4 b + 4 c + |a - b| + |a - c| + |b - c| + + \left| 2 a - b - c - | b - c | \right| + \left| a - 2 b + c + | a - c | \right| + \left|a + b - 2 c + |a - b| \right| ) \text{,} +$$ which demonstrates the symmetry you were wanting. +Note that we input that symmetry by hand. There might be a way to get there from the $\max(a,\max(b,c))$ form, but I bet it requires a lot of "insight".<|endoftext|> +TITLE: Limit with x approaching infinity +QUESTION [5 upvotes]: The problem says: + +If $$\lim_{x\to +\infty} \left\lbrack\frac{ax+1}{ax-1}\right\rbrack^x=9$$, determine $a$. + +It appears to be a case of $\left\lbrack\frac{\infty}{\infty}\right\rbrack^\infty$. How can I solve such a case? + +REPLY [6 votes]: This is definetely a very interesting question. Here is what I think is a good way to think about it. First rearrange the expression (omitting limits for clarity): +$$\left( \frac{a x + 1 - 2 + 2 }{a x -1} \right)^x = \left( 1 + \frac{ 2 }{a x -1} \right)^x $$ +Notice that, in the limit of $x \rightarrow ∞ $, the denominator reduces to $a x$. We rewrite the equation as: +$$\lim_{x\to \infty} \left( 1 + \frac{ 2 }{a x -1} \right)^x = 9 $$ +or by noting what we have said about the denominator: +$$\lim_{x\to \infty} \left( 1 + \frac{ 2 }{a x} \right)^x = 9 $$ +The LHS is exactly the definition of $e^{2/a}$, hence you can say that: +$$e^{2/a} = 9$$ +from which you can easily find that: +$$a = \frac{2}{\ln{9}} = \frac{1}{\ln{(3)}}$$ +I hope this helps you!<|endoftext|> +TITLE: Minimum number of marked squares on $n × n$ board +QUESTION [6 upvotes]: Came across this question: +Consider an $n × n$ square board, where $n$ is a fixed even positive integer. The board is divided into $n^2$ unit squares. We say that two different squares on the board are adjacent if they have a common side. +$N$ unit squares on the board are marked in such a way that every square (marked +or unmarked) on the board is adjacent to at least one marked square. Determine the smallest possible value of $N$. +I've found an upper bound of $N = \frac {n^2}2$. But is this the smallest possible value of $N$? If so, how can I prove it? +I've included an illustration of my upper bound for $n=2$ and $n=4$ below. + +Edit +@jwsiegel has linked to the supposed solution here: http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln993.html +It gives the solution +$$N = \frac{n(n+2)}{4}$$ +I don't understand how this solution is correct. The instructions for marking squares is given as: + +Color alternate squares black and white (like a chess board). +Look first at the odd-length white diagonals. In every other such diagonal, mark alternate squares (starting from the border each time, so that r+1 squares are marked in a diagonal length 2r+1). + +That's it. I have followed these instructions (as I understand them) for $n=4$ and $n=6$ below, using an X to indicate a marked square: + +The number of marked squares fits the given solution, i.e. $N = 6$ for $n = 4$ and $N = 12$ for $n = 6$, but NO marked square is adjacent to any other marked square! How then is this a valid solution? +I can only see 3 possibilities: + +I have marked incorrectly, or +The people who made this solution count a common corner as a common side between 2 squares, or +The solution is incorrect. + +Please explain what I'm missing. + +REPLY [2 votes]: jwsiegel has already provided both an asymptotic answer to the original question, and a link to a closed-form solution. I will address the Juan's question in the comments on jwsiegel's answer and Jens's follow-up question in the original post. +In answer to Juan's question, "For even n, this is better, but does this work if n is odd?": +Nothing in jwsiegel's asymptotic result makes any assumption on the parity of $n$. His answer provides a covering of the infinite plane which satisfies the rules given in the original post. When this marking is restricted to an $n \times n$ board, you get a marking of the board in which every interior square is adjacent to a marked board, with roughly a forth of the squares marked (exactly 1/4 if $n$ is a multiple of 4). To then cover the boundary as well, you need to add no more than $4(n-1)$ additional squares. When $n$ approaches infinity, this contribution becomes negligible, so we get that, for $n$ large, his covering of the board requires about a quarter of the squares to be marked. Since each square is adjacent to at most 4 other squares, it follows that this result is asymptotically optimal. +It is, in principle, possible to precisely calculate how many squares are marked by this procedure, but there is no reason to expect that it will be optimal for any fixed $n$. What the asymptotic solution does do is provide guidance for what a closed-form solution must satisfy, namely, that $$\lim_{n \to \infty} \frac{N}{n^2} = \frac{1}{4}.$$ +Now, for OP's question. You forgot this part: "In every other such diagonal..." With that in mind, here are the cases you wrote up, corrected. Let the upper-left hand corner be white, as in your pictures. I've indicated the black squares using an ascii box; hopefully that should make it pretty clear. +$n=4$: + X | ■ | | ■ +--------------- + ■ | | ■ | X +--------------- + | ■ | | ■ +--------------- + ■ | X | ■ | + +$n=6$: + X | ■ | | ■ | X | ■ +----------------------- + ■ | | ■ | | ■ | +----------------------- + | ■ | X | ■ | | ■ +----------------------- + ■ | | ■ | | ■ | X +----------------------- + X | ■ | | ■ | | ■ +----------------------- + ■ | | ■ | X | ■ | + +Observe that every black square is adjacent to exactly one marked white square. Now, as you observed, of course, no white square is adjacent to any marked square. But if we apply the exact same procedure to mark the black squares, then we obtain a marking of the entire board such that every square is adjacent to exactly one marked square. Then the claims are that (1) this board has $n(n+2)/4$ marked squares, and (2) this number of marked squares is optimal. Note that this closed form solution absolutely does depend on assuming that $n$ is even.<|endoftext|> +TITLE: Finding number of functions from a set to itself such that $f(f(x)) = x$ +QUESTION [6 upvotes]: The questions states that $f: A\rightarrow A$ is a function which satisfies $f(f(x)) = x.$ +We have to find the number of such functions with $A = \left\{1,2,3,4\right\}$. +The given condition clearly states that the function is the inverse of itself. Since the set $A$ was given to be so small, I first tried to think of some familiar functions that could fit. I came up with $f(x) = x$ and $f(x) = 5-x$. But the answer was nowhere near $2$. +I think there has to be a general method to find the no. of functions that satisfy a given condition instead of hit and trial and counting on fingers. But I'm not able to figure out what it is. + +REPLY [2 votes]: This can me asked as a graph theory problem: how many graphs with vertices $\{a,b,c,d\}$ exist such that each vertex lies in exactly one cycle of length at most $2$? +Case 1: All cycles are of length 1. This is equivilant to the identity function, which we know is unique. However from a graph theoretic point of view this is equal to $\binom{4}{4}$ because we are choosing 4 vertices to have cycle length one (notice that order does not matter because cycles start and end in the same place, and if two functions have the same cycles regardless of order they are equal). $\binom{4}{4} = 1$, consistent with what we know about the identity. +Case 2: One cycle is of length 2, and 2 are of length one. We know that the two of length one are chosen by $\binom{4}{2}$ and the reaming two vertices are forces to be in the cycles of length two (or are 'chosen' by $\binom{2}{2}$). +Case 3: 2 cycles of length 2, which is the same amount of choices as above, since the two remaining ones are forced to be the cycle. We have to divide by the number of ways there are to arrange 2 $2$-cycles, which is $2!$ +Summing the amounts of functions from the three cases we get $10$.<|endoftext|> +TITLE: How to prove Lagrange multiplier theorem in a rigorous but intuitive way? +QUESTION [35 upvotes]: Following some text books, the Lagrange multiplier theorem can be described as follows. +Let $U \subset \mathbb{R}^n$ be an open set and let $f:U\rightarrow \mathbb{R}, g:U\rightarrow \mathbb{R}$ be $C^1$ functions. Let $\mathbf{x}_{0} \in U, c=g(\mathbf{x}_0)$, and let $S$ be the level set of $g$ with value $c$. Suppose that $\nabla g(\mathbf{x}_0) \neq 0$. +If the function $f|_S$ has a local extremum at $\mathbf{x}_0$ then there is a $\lambda \in \mathbb{R}$ so that $\nabla f(\mathbf{x}_0)=\lambda g(\mathbf{x}_0)$ +Some textbooks give a strict proof using implicit function theorem (see proof). The proof is very clear, however I can't build some intuition from the proof. Can someone help me prove this theorem in a rigorous but more intuitive way? Thanks. + +REPLY [5 votes]: Here's a different way to see that Lagrange multipliers are "intuitively" the correct way to go. To first order, the change in $f$ due to a change in $\vec r$ is simply:$$\delta f = f(\vec r + \delta\vec r) - f(\vec r) \approx \nabla f\cdot\delta\vec r.$$This has the obvious interpretation for $f$ above, "go in the direction of $\nabla f$ to increase $f$," but it also has an interpretation for constraints: to first order, the paths allowed by a constraint $g(\vec r) = c$ must be the ones such that $\nabla g \cdot \delta \vec r = 0,$ otherwise you would increase $c$ to first-order by following that tangent vector. That is, if you obey the constraint, we must have $\delta g = 0.$ +Now we simply combine these two. At the constrained extremum of $f$ on $S,$ it must be the case that for all $\delta\vec r$ such that $\nabla g \cdot \delta \vec r = 0,$ we also have $\delta f = \nabla f\cdot\delta\vec r = 0,$ otherwise we can follow $\pm \delta\vec r$ to increase/decrease $f$, proving that this wasn't either sort of extremum on $S$ after all. So just to phrase the very condition for a constrained maximum, we would have to write it as $$\forall \vec a . (\nabla g \cdot \vec a = 0) \rightarrow (\nabla f \cdot \vec a = 0)$$ +But that "for-all" is 100% sufficient to prove that they are proportional! Effectively, you can't have two linearly independent vectors $\vec a, \vec b$ such that the nullspaces of $(\vec a \cdot)$ and $(\vec b \cdot)$ are the same. Let's just prove this last aspect to complete the proof. +Consider $\vec p = \vec a - \alpha \vec b$ for $\alpha = \vec a \cdot \vec b / (\vec b \cdot \vec b)$, this satisfies by construction $\vec b \cdot \vec p = 0$ and therefore we must have $\vec a \cdot \vec p = 0$ due to the shared nullspace. But when you chug through the algebra of this it means ultimately that $(\vec a \cdot \vec a)(\vec b \cdot \vec b) = (\vec a \cdot \vec b)^2.$ That guarantees the = sign in a triangle inequality looking like either of $$|\vec a \pm \vec b|^2 = |\vec a|^2 + |\vec b|^2 \pm 2 (\vec a \cdot \vec b) ~\le~ (|\vec a| + |\vec b|)^2 = |\vec a|^2 + |\vec b|^2 + 2 |\vec a||\vec b|.$$ We know that those equals signs only hold when the vectors are degenerate, that is, $\vec a = \lambda \vec b.$<|endoftext|> +TITLE: Why is the Barycenter operation in Hadamard spaces Lipschitz continuous? +QUESTION [9 upvotes]: I am looking into exercise 9.2.22 of "A course in metric geometry" by Burago-Burago-Ivanov. +For a Hadamard space $H$ (a complete simply connected metric space of nonpositive curvature in the Alexandrov sense) +and a fixed point $p \in H$ the function $f(x) = (d(p,x))^2$ is strongly convex, where $d$ denotes the metric of $H$. +Therefore for a fixed collection of points $p_1, \ldots, p_n$ the function +$$ f(x) = \sum_{i=1}^n (d(p_i,x))^2 $$ +is also strictly convex and therefore has a unique point $y$ where $f$ attains its maximum. The Point $y$ is called the Barycenter of $p_1, \ldots ,p_n$. +The Statement I want to prove is the following: +The Barycenter operation is Lipschitz continuous, i.e if $q_1, \ldots q_n$ is another collection of points and $y'$ is their barycenter then one has a relation of the form +$$ d(y,y') \leq L \cdot \left( d(p_1,q_1) + \ldots + d(p_n,q_n) \right) $$ +I suspect that i need some statement of the form: If strongly convex functions $f,g$ satisfy $\Vert f-g\Vert_{\infty}< C $ then their Minima $y,y'$ satisfy $ d(y,y')< CL$. Does such a statement exist? + +REPLY [2 votes]: (1) In a Hardmard space $(X ,|\ |)$, define a convex set +$K$ wrt $|\ |$ Then then we have $$ f : X\rightarrow \mathbb{R},\ +f(x)=|xK|\ ({\rm or}\ |xK|^2) +$$ Note that $f$ attains a minimum at $x_0$ and $x_0$ is unique So +we have +$$ \pi : X\rightarrow K,\ \pi (x)=x_0 $$ +Then $\pi$ is distance nonexpanding +Proof : Note that $$\angle (xx_0,y_0x_0),\ \angle (yy_0,x_0y_0) > +\frac{\pi}{2}$$ +If ${\rm geod}_{[ab]}(t)$ is unit speed geodesic from $a$ to $b$, then +we have first variation formula, since $X$ is Hadamard space, $$ +|{\rm geod}_{[x_0x]} (t)- {\rm geod}_{[y_0y]} (\tau ) | = |x_0y_0| +-t\cos\ \angle (xx_0,y_0x_0) -\tau \cos\ \angle (yy_0,x_0y_0) + +o(t+\tau ) +$$ Hence we complete the proof +(2) Now we will apply to our case : If we have $(H,d)$, +then define +$$ (X,|\ |)= \underbrace{(H,d)\times \cdots \times (H,d)}_{n\ times} +$$ where $$ |(p_i)(p_i')| := \sqrt{ \sum_i d(p_i,p_i')^2 } $$ +Then we have a convex set $K:=\Delta H$ And let $f((p_i)):=|(p_i)K|=\inf_{x\in H}\ \sqrt{\sum_i d(p_i,x)^2} +$ +By (1) since $\pi$ is distance nonexpanding so $$ \pi:X\rightarrow K,\ \pi((x_i))=(a_i),\ a_i=x_0,\ \pi((y_i))=(b_i),\ b_i=y_0$$ $$\sqrt{n}d(x_0,y_0) +\leq \sqrt{\sum_i d(x_i,y_i)^2} $$ +This complete the proof<|endoftext|> +TITLE: Rings in which $ab=0$ implies $axb=0$ +QUESTION [5 upvotes]: I'm sure there must be some standard term for (not necessarily commutative) rings $R$ in which $ab=0$ implies $(\forall x)\, axb=0$ (for example, this is the case if $R$ is commutative or is a domain). What is this term? +Additionally, or alternatively, what about (two-sided) ideals $I$ such that $ab\in I$ implies $(\forall x)\, axb\in I$, i.e., ideals quotienting by which gives a ring as I just said? Do they have a name? +Edit: I should probably also mention the stronger condition that $ab=0$ implies $ba=0$: such rings are called "reversible" (Cohn, "Reversible Rings", Bull. London Math. Soc. 31 (1999), 641–648). Clearly, commutative rings and domains are reversible, and reversible rings satisfy the property I'm looking for a name for (because in a reversible ring, if $ab=0$ then $ba=0$ so $bax=0$ so $axb=0$). + +REPLY [5 votes]: The most modern term for this is + +$R$ satisfies the $SI$ condition + +This is seen in papers by Greg Marks, which I have found to be the most thoughtful and comprehensive recent papers discussing these things. For example , see A taxonomy of 2-primal rings. +Earlier works used the following terms: + +$R$ is zero-insertive (zi) +$R$ satisfies the insertion of factors property (IFP) +$R$ is semi-commutative (conflicts with different usages in the literature)<|endoftext|> +TITLE: Is $\sigma$-finiteness unnecessary for Radon Nikodym theorem? +QUESTION [5 upvotes]: Let $(X,\mathfrak{M},\mu)$ be a $\sigma$-finite measure space and $\lambda:\mathfrak{M}\rightarrow [0,\infty]$ be a measure. If $\lambda\ll \mu$, then there exists a measurable $f:(X,\mathfrak{M})\rightarrow [0,\infty)$ such that $d\lambda= fd\mu$. + +The above is the Radon Nikodym theorem stated in wikipedia. However, both texts Rudin and Folland prove the Radon Nikodym theorem under assumption that $\lambda$ is $\sigma$-finite. How do I prove the Wikipedia vesion of Radon Nikodym therem? Is there any reference? + +REPLY [8 votes]: EDIT: It seems that OP does not allow $f$ to take $\infty$, then the $\sigma$ finiteness is needed. If we allow function to take $\infty$, we do not need the $\sigma$ finiteness condition. I'll leave the answer below as an extension. + +Let $(X,\mathcal A)$ is a measurable space, and $\mu,\nu$ are two measures on it, with $\nu \ll \mu$. We need $\sigma$ finiteness for $\mu$, and we need $\mu$ to be positive measure. However we do NOT need $\sigma$ finiteness for $\nu$, and actually $\nu$ could be of any measure, according to an exercise in the book of A course in abstract analysis by J.B. Conway; (Page 113, Exercise 3). I put a solution of mine to that exercise below: +Let $(X,\mathcal A)$ is a measurable space, and $\mu$ is a positive $\sigma$-finite measure on $(X,\mathcal A)$, and $\nu$ is any measure on $(X,\mathcal A)$ (could be positive, signed, or complex measure), and $\nu \ll \mu$. Then Radon-Nikodym still holds. +Now, I'm gonna provide a proof given that we've already proved Radon-Nikodym Theorem for $\sigma$-finite positive measure of $\mu$ and $\sigma$-finite signed measure $\nu$, where $\nu \ll \mu$. +Proof: Step 1, we consider the case that $\mu$ is $\sigma$-finite positive measure, and $\nu$ is signed measure. Firstly, let $\lambda$ be a positive finite measure, we will prove two conclusions (*1) and (*2). +Given $n\ge 1$, let $\{P_n,N_n\}$ be the Hahn decomposition for the signed measure of $\nu-n\lambda$, where $P_n$ are positive sets, and $N_n$ are negative sets. Thus we have $(\nu-n\lambda)(N_n) \le 0 \Rightarrow \nu(N_n) \le n\lambda(N_n)$. Let $P=\cap_{n=1}^\infty P_n, N=\cup_{n=1}^\infty N_n$. Consider $\nu$ restricted to set $N$, $\nu|_N(E)=\nu(E\cap N)$, we have $$\forall n \ge 1, \nu|_N(N_n)=\nu(N_n\cap N)=\nu(N_n) \le n\lambda(N_n) < \infty$$, and thus $\nu|_n$ is $\sigma$-finite (*1). +$\forall A \in \mathcal A$ and $A \subset P$, we have $\forall n \ge 1, A \subset P_n$, and thus $\forall n \ge 1, (\nu-n\lambda)(A)\ge 0 \Rightarrow \nu(A) \ge n\lambda(A)$. This means (*2): $$\nu(A)=\begin{cases}0, \mu(A) = 0 \\ \infty, \mu(A) > 0\end{cases}$$ +Now $\mu$ is $\sigma$-finite, and thus we could find a series of increasing measurable sets $F_i \uparrow X$, s.t. $\forall i \in \mathbb N, \mu(F_i) < \infty$. Let $\mu_i=\mu|_{N_i}$, then $\mu_i$ are finite positive measures. We apply (*1) to every $\mu_i$, and get a series of $\sigma$-finite signed measure $\nu_i=\nu|_{N_i}$, where $N_i=\cup_{n=1}^\infty N_{i,n}\subset F_i$, and it's obviously that $\nu_i \ll \mu_i$. +We could apply Radon-Nikodym Theorem for $\sigma$-finite positive $\mu$, and $\sigma$-finite signed measure $\nu$, to find $\mu_i$-measurable function $f_i$, s.t. $$\forall A \in \mathcal A, A \subset N_i, ~\nu_i(A)=\int_Af_i~d\mu_i$$ +Now notice that if $i \le j$, then $F_i \subset F_j$, and thus $\mu_i(N_{i,n})=\mu(N_{i,n} \cap F_i)=\mu(N_{i,n} \cap F_j)=\mu_j(N_{i,n})$. So we have: +$$(\nu-n\mu_i)(N_{i,n}) \le 0 \Rightarrow (\nu-n\mu_i)(N_{i,n}) \le 0$$ This means that $N_{i,n} \subset N_{j,n}$, and $P_{i,n} \supset P_{j,n}$, and therefore $N_i \uparrow N^*,~P_i \downarrow P^*, ~ N^* \cap P^* = \emptyset, ~N^* \cup P^*= X$. Thus on the set of $N_i$, we have $f_i = f_j, a.e.(\mu_i)$. Now we can finally define our function: +$$f(x)=\begin{cases}f_i(x), x \in N_i \\ \infty, x\in P^*\end{cases}$$ +And we have $$\nu(A \cap N_i)=\nu_i(A)=\int_Af~d\mu_i=\int_{A\cap N_i}f~d\mu$$ +Let $i \to \infty$, we have $$\nu(A\cap N^*)=\int_{A\cap N^*}f~d\mu$$ +Next, let's consider set $A \cap P^*$. Obviously, $\forall i,~ A\cap P^*\subset P_i$, and we apply (*2) to know that: +When $\mu(A \cap P^*)=0$, and recall definition of Lebesgue integration that if $a_i=0, \mu(E_i) = \infty$, we define $a_i\mu(E_i) = 0$, thus we have +$$\nu(A\cap P^*)=0=\int_{A \cap P^*}f~d\mu$$ +When $\mu(A\cap P^*) > 0$, we have +$$\nu(A\cap P^*)=\infty=\int_{A\cap P^*}f~d\mu$$ +Finally in summary, we have: +$$\forall A \in \mathcal A, ~ \nu(A) = \int_{A\cap P^*}f~d\mu + \int_{A \cap N^*}f~d\mu = \int_Af~d\mu$$ +Now we've approved Radon-Nikodym for $\mu$ to be $\sigma$-finite positive measure and $\nu$ to be any signed measure +Note here we assume the signed measure would take $\infty$, some other definition would let signed measure to take $-\infty$ - but it cannot take both $\infty$ and $-\infty$. For the case where signed measure would take $-\infty$, proof is similar. +Step 2, we consider the case that $\mu$ is $\sigma$-finite positive measure, while $\nu$ is complex measure. We write $\nu=\nu_R + i\nu_I$, where $\nu_R,\nu_I$ are both signed measures (to be precise, we know that they have to be finite as well), and we apply conclusion in Step1 to $\nu_R$ and $\nu_I$ respectively and add them together - we then get the Radon-Nikodym Theorem for $\mu$ is $\sigma$-finite positive measure, and $\nu$ is complex measure. +Thus, now we've proved Radon-Nikodym Theorem for the case when $\mu$ being a positive $\sigma$-finite measure on $(X,\mathcal A)$, and $\nu$ being any measure on $(X,\mathcal A)$ (could be positive, signed, or complex measure), with $\nu \ll \mu$. Q.E.D<|endoftext|> +TITLE: What is the difference between partial order relations and equivalence relations? +QUESTION [5 upvotes]: From googling it, I understood that a relation is both a partial order relation and an equivalence relation when they are reflexive, symmetric and transitive. So it appears they are the same. But as far I know, they are not supposed to be the same. +What is the difference between partial order relations and equivalence relations? + +REPLY [8 votes]: It looks like you misread slightly: partial orders and equivalence relations are both reflexive and transitive, but only equivalence relations are symmetric, while partial orders are antisymmetric. A relation $R$ is symmetric if $aRb$ implies $bRa$, while a relation $R$ is antisymmetric if $aRb$ and $bRA$ implies $a=b$. +For example, the relation "has the same parity as" is symmetric (and in fact an equivalence relation) on the integers, since if $a$ has the same parity as $b$ then $b$ has the same parity as $a$. On the other hand, it is not antisymmetric, because $2$ has the same parity as $4$ and $4$ has the same parity as $2$, even though $2\neq 4$. +The relation $\leq$ on integers, in contrast, is antisymmetric (and in fact a partial order), since if $a\leq b$ and $b\leq a$ then $a=b$. But it is not symmetric, since (for instance) $2\leq 4$ but $4\not\leq 2$.<|endoftext|> +TITLE: Corollary to Lemma of Nakayama +QUESTION [5 upvotes]: In Matsumura's Commutative Algebra there is the following Corollary to the Lemma of Nakayama: + +Let $A$ be a ring, $M$ an $A$-module, $N$ and $N'$ submodules of $M$, and $I$ an ideal of $A$. Suppose that $M=N+IN'$, and that either: + +$I$ is nilpotent +$I\subseteq \mathrm{rad}(A)$ and $N'$ is finitely generated. + +Then $M=N$. + +In both cases the idea is to prove that $M/N=I(M/N)$ and I am struggling with that. +We have that $M/N=(N+IN')/N=IN'/(IN'\cap N)$ and I do not know how to take it from here. + +REPLY [2 votes]: Since $M=N+IN'\subseteq N+IM$ you get $M=N+IM$. Can you continue from here? +Since $M=N+IN'\subseteq N+N'$ you get $M=N+N'$. Can you see now why $M/N$ is finitely generated?<|endoftext|> +TITLE: Show $x_n := \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1}$ is increasing in n for $p \in (0,1)$ +QUESTION [6 upvotes]: I need to show that $x_n := \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1}$ is increasing in $n$ for $p \in (0,1)$. My attempts have involved trying to show +$\begin{eqnarray*} \frac{1-p^{n+2}}{1-p^{n+1}} \frac{n+1}{n+2} &>& \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1} \\ +\left(1 + \frac{p^{n+1}}{\sum_{k=0}^n p^k} \right) \frac{n+1}{n+2} &>& \left(1 + \frac{p^{n}}{\sum_{k=0}^{n-1} p^k} \right) \frac{n}{n+1} \\ +\frac{n+1}{n+2} - \frac{n}{n+1} &>& \left(\frac{p^{n}}{\sum_{k=0}^{n-1} p^k} \right)\frac{n}{n+1}-\left(\frac{p^{n+1}}{\sum_{k=0}^n p^k} \right) \frac{n+1}{n+2}\\ +\frac{n+1}{n+2} - \frac{n}{n+1} &>& p^n \left[\left(\frac{1}{\sum_{k=0}^{n-1} p^k} \right)\frac{n}{n+1}-\left(\frac{p}{\sum_{k=0}^n p^k} \right) \frac{n+1}{n+2} \right]\end{eqnarray*}$ +but this does not seem to simplify. +I have also attmpted to treat $x_n$ as a continuous function in $n$ and differentiated (this leaves a mess so I won't put it up here). +Sadly, I seem to be unable to get my required result from either. +EDIT: It might be helpful to note that $x_n = \frac{\sum_{k=0}^n p^k}{\sum_{k=0}^{n-1} p^k}\frac{n}{n+1}$ +EDIT 2: I can show that $x_n < 1$ for all $n \in \mathbb{N}$ and that $x_n \to 1$. But I don't think this suffices... +EDIT 3: If we can show that +$\frac{\sum_{k=0}^{n+1}p^k}{n+2} \cdot \frac{\sum_{k=0}^{n-1}p^k}{n} \geq \left(\frac{\sum_{k=0}^{n+1}p^k}{n+2} \right)^2$, we are done. But I don't know how to show this. + +REPLY [3 votes]: Writing $x_n = \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}$, we see: +$$x_n>x_{n-1}\iff \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}>\frac{1-p^{n}}{1-p^{n-1}}\frac{n-1}{n}$$ +$$\iff(1-p^{n-1})(1-p^{n+1})n^2>(1-p^n)^2(n^2-1)$$ +$$\iff (1-p^n)^2 > n^2[(1-p^n)^2-(1-p^{n-1})(1-p^{n+1})]=n^2[p^{n-1}+p^{n+1}-2p^n]$$ +$$\iff\frac{(1-p^n)^2}{p^n}>n^2\left(\frac{1}{p}+p-2\right)=n^2\frac{(p-1)^2}{p}$$ +$$\iff \frac{1-p^n}{p^{n/2}}>n\frac{1-p}{p^{1/2}}\iff\frac{p^{n/2}}{p^{1/2}}\frac{1-p^{-n}}{1-p^{-1}}>n$$ +$$\iff p^{\frac{n-1}{2}}(1+p^{-1}+p^{-2}+\cdots+p^{-(n-1)})>n$$ +$$\iff \frac{1+p^{-1}+p^{-2}+\cdots+p^{-(n-1)}}{n}>p^{-\frac{n-1}{2}}$$ +which is true by the AM-GM inequality.<|endoftext|> +TITLE: Is it possible for $R \oplus M$ and $R \oplus N$ to be isomorphic to each other if $M$ and $N$ are not isomorphic? +QUESTION [7 upvotes]: Suppose $M$ and $N$ are non-isomorphic $R$-modules (where $R$ is a commutative ring with a unit element).Can we conclude that $R \oplus M \not\simeq R \oplus N$ ? If not in this most general setup, is there an affirmative answer in some special circumstances ? + +Motivation : I am reading Proposition 9(1.4) in Dale Husemoller's book 'Fibre Bundles' which reads : + +If $u,v : \theta^1 \to \xi^k$ are two monomorphisms of the trivial line bundle over $B$ into $\xi^k$ (a k-dimensional vector bundle over $B$) , such that $n \leq ck-2$, then coker $u$ and coker $v$ are isomorphic over $B$. + +Using the fact that short exact sequences of finite dimensional vector bundles over a paracompact space split, and replacing vector bundles over $B$ by $R$-modules, I get the question that I am asking here. +Essentially I am just trying to see if the theorem about vector bundles can be generalized to an arbitrary abelian category. + +REPLY [5 votes]: Two $R$-modules $M$ and $N$ are said to be "stably isomorphic" (see Stacks for example) if $M \oplus R^n$ and $N \oplus R^n$ are isomorphic for some $n \in \mathbb{N}$. It is possible for two non-isomorphic modules to be stably isomorphic. If $M'$ and $N'$ are such modules, let $n$ be the smallest (necessarily positive) integer such that $M' \oplus R^n \cong N' \oplus R^n$; then $M = M' \oplus R^{n-1}$ and $N = N' \oplus R^{n-1}$ satisfy your condition. +In The K-book: an introduction to algebraic K-theory, Weibel gives an example (Chapter I, Example 1.2.2) of a stably free module which is not free. The ring is $R = \mathbb{R}[x,y,z]/(x^2+y^2+z^2=1)$ and the module is $P = \ker \sigma$, where $\sigma : R^3 \to R$ is $(P,Q,R) \mapsto xP+yQ+zR$. Then $P \oplus R \cong R^3 = R^2 \oplus R$ as $R$-modules, but $P$ is not isomorphic to $R^2$. (Crazily enough, the reason comes from algebraic topology: it's the hairy ball theorem!) +In fact if you go through the definition, you will see that two finitely generated projective modules $M$, $N$ over $R$ are stably isomorphic iff they are equal in $K_0(R)$, the zeroth algebraic K-theory group of $R$. (See also the K-book, Chapter 2.)<|endoftext|> +TITLE: What comes after diagram chasing? +QUESTION [7 upvotes]: An early edition of Lang's algebra textbook gives the famous exercise to + +Take any book on homological algebra, and prove all the theorems without + looking at the proofs given in that book. + +Here are some of theorems you would find in such a book: the snake lemma, the five lemma, the nine lemma, the many unnamed lemmas for constructing long exact sequences in algebraic topology (e.g. Mayer-Vietoris). +The only way I know to prove these theorems, and which I have read in homological algebra books, is by a diagram chase. +I have nothing against diagram chasing -- it's rather fun once you figure out how it works. But it also feels rather mechanical and devoid of insight. + +Is there any systematic way to think about/prove/remember theorems in homological algebra like the ones that I cited without having to resort to diagram chasing? +Are the theorems of the type I mentioned small enough in number to be subsumed by some general theory which explains them? +Is homological algebra anything more than a mindless subject used for its applications in other areas of math? + +A satisfactory answer would allow us to (for instance) look at the hypotheses of the five lemma and then see immediately that the vertical arrow is an isomorphism. +Note: I am aware of the formalism of abelian categories, though not yet a competent user. It seems to me that abelian categories generalize the setting of homological algebra without clarifying its content. For instance, it seems like most people, when pressed to prove theorems of the type I mentioned in an abelian category, use Mitchell's Embedding Theorem to justify performing a diagram chase. + +REPLY [7 votes]: I highly recommend reading through Ravi Vakil's notes on spectral sequences. For convenience I'll outline the general points relating spectral sequences to diagram lemmas. +Diagram lemmas usually begin with a double complex. That is a grid of maps + +composing to $0$. Here the blue bullet denotes the $(0,0)$ position (or the "origin") of the grid. This grid is called the "$0$th page" of the spectral sequence, or $E_0$. +Spectral sequences supply us with two different ways of studying this grid. We can think "vertically" or "horizontally". Each perspective supplies us with an algorithm. +The "horizontal" perspective is an algorithm where we start by replacing each object in $E_0$ with its corresponding homology obtained from the horizontal arrows. This gives us a new grid of objects, $_{>}E_1$. Additionally, this new grid is supplied with differentials that now point "up". We continue this process, taking homology and inheriting a differential to obtain new "pages". The direction of the differential on the $_{>}E_r$ page is $_{>}d_r:E_r^{p,q}\to E_r^{p-r+1,q+r}$. Thus the differentials "spiral out in a counter-clockwise fashion" + +On the other hand, we can think "vertically" and obtain a similar algorithm. Take homology to obtain new grids $_{\hat{}}E_r$, starting with the vertical maps. The differentials in this case "spiral out in a clockwise fashion" + +The idea of spectral sequences roughly states: + +The two collections of grids $_{>}E_r$ and $_{\hat{}}E_r$ eventually stabilize to the same grid $E_{\infty}$. + +This gives us a surprisingly powerful tool for studying diagram lemmas. For example, let's prove the 5-lemma. Suppose we have a diagram + +with exact rows where $\alpha$, $\beta$, $\delta$, and $\varepsilon$ are isomorphisms. +The first page of the horizontal homology $_{>}E_1$ is + +where the $\bullet$'s are the only objects that are not necessarily zero. The next page $_{>}E_2$ is + +The maps here are omitted because they are all necessarily zero! From here we see that the spectral sequences converges to $E_{\infty}=_{>}E_2$ since the only possibly nonzero objects have differentials entering and exiting zero objects from page two onward. +Now, the first page of the vertical homology $_{\hat{}}E_1$ is + +The vertical pages stabilize immediately. Thus we see that $_{\hat{}}E_1\simeq _{>}E_2\simeq E_\infty$. In particular, $\ker\gamma\simeq\DeclareMathOperator{cok}{cok}\cok\gamma\simeq0$. Hence $\gamma$ is an isomorphism. +The moral here is: if we accept the theory of spectral sequences as a black box, then many diagram lemmas are quite natural.<|endoftext|> +TITLE: Extension of real analytic function to a complex analytic function +QUESTION [31 upvotes]: I just learned that real analytic functions (by real analytic, I mean functions $f: \mathbb{R} \to \mathbb{R}$ which admit a local Taylor series expansion around any point $p \in \mathbb{R}$) cannot be extended to complex entire function always. I believe functions with this extension property are called real entire functions in some books, and a function which is real analytic, but not real entire is $f(x) = \frac{1}{1 + x^2}$. Clearly, with this example, the problem with any extension happens around $\pm i$. See also this MSE post. +My question is, do real analytic functions admit extensions to a complex analytic function even locally? That is, given a real analytic function $f : \mathbb{R} \to \mathbb{R}$, can we find a complex analytic function $g : \Omega \to C$, such that $g|_\mathbb{R} = f$, and $\Omega$ contains a strip $\mathbb{R} \times (-\varepsilon, \varepsilon)$ around the real axis? + +REPLY [10 votes]: As David C. Ullrich and Jason pointed out, the answer is not always affirmative. Let me quote a Theorem that gives a partial affirmative answer to your question, with somewhat different assumptions. + +Theorem: Assume that $f$ satisfies the Fourier inversion formula, i.e. $f(x) = \int_{\mathbb{R}}\hat{f}(\xi)e^{2\pi i x\xi}\,d\xi$ and that $$|\hat{f}(\xi)| \le Ae^{-2\pi a|\xi|}$$ for some constants $a,A > 0$. Then $f$ is the restriction to $\mathbb{R}$ of a function $f(z)$ holomorphic in the strip $S_b = \{z \in \mathbb{C}: |\text{Im}(z)| < b\}$, for any $0 < b < a$. + +I think this is an interesting result since it also tells how much you can push the $\epsilon$ is you question. You can find a proof of this theorem and further discussions on the possibility of extending functions on the real line to complex-analytic functions in Complex Analysis by E.M. Stein and R. Shakarchi.<|endoftext|> +TITLE: How many divisors does $111...1$ have? +QUESTION [5 upvotes]: Let $A=\underbrace{11..1}_{2010}$. How many divisors does $111...1$ have? +Original problem: Prove that $τ(A)>50$ (or $τ(A)<50$) + +My work so far: + +If $\tau(A) -$ the number of divisors of $A$ and $A=p_1^{\alpha_1}\cdotp_2^{\alpha_2}\cdot...\cdot p_n^{\alpha_n}$, then $\tau(A)=(\alpha_1+1)(\alpha_2+1)\cdot...\cdot(\alpha_n+1)$ + +If $a=\underbrace{11..1}_{k}$ and $b=\underbrace{11..1}_{l}$ and $l|k$ then $b|a$ + +$2010=2\cdot 3\cdot5\cdot67$. Then $\tau(A)\ge 2^4=16$ + +REPLY [3 votes]: A partial answer: note that +$$ +A = \frac{\overbrace{99\cdots9}^{2010}}{9} = \frac{10^{2010} - 1}{10-1} +$$ +that is, $A = \frac{x^{2010} - 1}{x-1}$ with $x = 10$. You can exploit the many factorizations of $x^n - 1$, since $n$ isn't prime. For example: +$$ +x^{2010} - 1 = (x^{67} - 1)(1 + x^{67} + x^{2\cdot 67} + \cdots x^{29 \cdot 67}) +$$ +This nicely confirms some of your results so far.<|endoftext|> +TITLE: proof on trace trick +QUESTION [13 upvotes]: I wanted to find the maximum likelihood estimator for $\mathbf{\Sigma}$ in the multivariate gaussian. +I was anticipating the solution would be a bit involved and messy, if not 'brute-forced', but I was surprised to find an elegant and clever shortcut known as the trace trick. +For a vector $\mathbf{x}$ and a matrix $\mathbf{A}$, $\mathbf{x}^T\mathbf{Ax} = tr (\mathbf{x}^T\mathbf{Ax}) = tr (\mathbf{xx}^T\mathbf{A}) = tr (\mathbf{Axx}^T)$. +I am not sure how this can be true. Any insight/rough sketch of proof would be great. + +REPLY [17 votes]: The first equality is just using the fact that $x^TAx$ is a real number, so when seen as a $1\times 1$ matrix, it's equal to its trace! +The other equalities use the fact that $tr(AB)=tr(BA)$.<|endoftext|> +TITLE: Evaluating $\int_{0}^{3} \sqrt{1+x}\: dx$ using Limit of a Sum approach +QUESTION [5 upvotes]: Evaluate $\int_{0}^{3} \sqrt{1+x}\: dx$ using Limit of a Sum approach. +Using the formula $$\int_{a}^{b} f(x)\:dx=(b-a) \times \lim_{n \to \infty} \frac{1}{n} \times \sum_{k=1}^{n}f\left(a+\frac{(b-a)k}{n}\right)$$ we have +$$I=\int_{0}^{3} f(x)\:dx=(3-0) \times \lim_{n \to \infty} \frac{1}{n} \times \sum_{k=1}^{n}f\left(0+\frac{(3-0)k}{n}\right)=3 \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f\left(\frac{3k}{n}\right)$$ So +$$I=3\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} \sqrt{1+\frac{3k}{n}}$$ Now if we expand the summation we get limits of the form +$$\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{3}{n}}+\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{6}{n}}+\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{9}{n}}+\cdots+\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{3n}{n}}$$ +But each limit is clearly zero and hence $I=0$. I know the answer is wrong but what is my mistake? + +REPLY [3 votes]: You have to choose the partition points $x_k$ such that $\sqrt{1+x_k}$ becomes manageable. Therefore put +$$x_k:=u_k^2-1\quad(0\leq k\leq n)$$ whereby the $$u_k:=1+{k\over n}\quad(0\leq k\leq n)$$ +are equally spaced. The $x_k$ are then inequally spaced between $0$ and $3$, but in any case the differences $x_k-x_{k-1}$ tend to $0$ when $n\to\infty$. A typical Riemann sum is then given by +$$\eqalign{R_n&:=\sum_{k=1}^n \sqrt{1+x_k}\>(x_k-x_{k-1})=\sum_{k=1}^n u_k(u_k+u_{k-1})(u_k-u_{k-1})\cr &={1\over n}\sum_{k=1}^n\left(1+{k\over n}\right)\left(2+{2k-1\over n}\right)={14\over3}+{3\over2n}-{1\over 6n^2}\ .\cr}$$ +Note that we have computed the last sum exactly, using the formulas for $\sum_{k=1}^n k^\alpha$ when $\alpha\in\{0,1,2\}$. It follows that +$$\int_0^3\sqrt{1+x}\>dx=\lim_{n\to\infty} R_n={14\over3}\ .$$<|endoftext|> +TITLE: Showing that this is a martingale.(4.13 in Øksendals SDE) +QUESTION [6 upvotes]: This is an exercise from Øksendals stochastic differential equations, where I get stuck. It is exercise number 4.13.(I simplified the notation a bit.) + +I have that X is an Itô-process where: +$dX_T=u(\omega,t)dt+dB_t$, $t \in [0,T]$, and I assume u is bounded. +I define the Itô-process Z, where: +$dZ_t = -u_t dB_t -\frac{1}{2}u_t^2dt$, $Z_0 = 0$ +I am supposed to show that $X_te^{Z_t}$ is a + $\mathcal{F}_t$-martingale, $\mathcal{F}_t$ is the filtration generated by the Brownian + motion. + +Using Itô formula for the multidimensional case(I omit all the algebra) I get that: +$d(X_te^{Z_t})=[e^{Z_t}(1-X_tu_t)]dB_t$. +Hence $X_te^{Z_t}=X_0+\int_0^te^{Z_t}(1-X_tu_t)dB_t$. +From the theory of the construction of the Itô process I know that this is a martingale if: +$E[\int_0^t(e^{Z_t}\{1-u_tX_t\})^2dt]<\infty$. +But how do I show this? +What I am able to get is: +$|Z_t| \leq K+ |\int_0^tu_TdB_t|$, so since u is bounded we know that $|\int_0^tu_tdB_t|\in L_2(\omega \times [0,T])$. So $Z_t \in L_2(\Omega \times [0,T])$. +And by the boundedness of u it also follows that $(1-u_tX_t)\in L_2(\Omega \times [0,T])$. +But how do I show that $e^{Z_t} \in L_2(\omega \times [0,T])$, and $e^{Z_t}u_tX_t \in L_2(\omega \times [0,T])$? The $L_2$ vector-space isnt closed under multiplication, and we can not just take the exponential and expect to still be there? +Any hints? + +REPLY [3 votes]: Let's show that $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] < \infty $ +By Cauchy-Schwarz inequality: +$\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] \leq \mathbb{E}[\int_0^t e^{4 Z_s}ds]^{\frac{1}{2}} \mathbb{E}[\int_0^t (1-X_su_s)^{4}ds]^{\frac{1}{2}} $ +Because $u$ is bounded, it is clear that $\mathbb{E}[\int_0^t (1-X_su_s)^{4}ds] < \infty$. +Hence it is left to show that $\mathbb{E}[\int_0^t e^{4 Z_s}ds] < \infty$. +It is enough to show that $\mathbb{E}[e^{4 Z_t}] <\infty$ for all $t \in [0,T]$. +$e^{4 Z_t} = \exp( -\int_0^t 4u_s dB_s - \frac{1}{2}\int_0^t ({4u_s})^{2} ds)\exp(6 \int_0^t u_s^{2} ds)=M_t \exp(6 \int_0^t u_s^{2} ds)$. (*) +$M_t$ is the Doléans-Dade exponential (applied to the process $-\int_0^t 4u_s dB_s$) and you can test the Novikov condition: $\mathbb{E}[\frac{1}{2} \exp(\int_0^T (4u_s)^{2} ds)] < \infty $ which is true because $u$ is bounded. +Hence by the Novikov condition $M_t$ is a martingale and $\mathbb{E}[M_t]< \infty$. +Using the fact that $u$ is bounded, we conclude by (*) that $\mathbb{E}[e^{4 Z_t}] < \infty$ and that your process is a martingale.<|endoftext|> +TITLE: True or False: there is a space $X$ such that $S^1$ is homeomorphic to $X\times X$ +QUESTION [5 upvotes]: I had an exam this morning, one of the questions asked about the truth of the statement + +There is a space $X$ such that $S^1$ is homeomorphic to $X\times X$. + +I said that this was false and this was my reasoning... was I correct? (Note, an earlier part of the question asked me to prove $\pi_1(Y\times Z)=\pi_1(Y)\times\pi_1(Z)$, I took this as a hint.) +Suppose for the sake of contradiction that such an $X$ exists. Then $\Bbb{Z}\cong G\times G$ where $G\cong\pi_1(X)$. Let $\Phi$ be the isomorphism. +Let $\Phi(1)=(g_1,g_2)\in G\times G$. Then at least one of $g_1,g_2$ is nonzero, since $\Phi(0)=(0,0)$ (the identity on $G\times G$, and $\Phi$ is injective). Without loss of generality $g_1\ne 0$. Then which element $n\in\Bbb{Z}$ has $\Phi(n)=(kg_1,0)$? Since $$kg_1=\underbrace{g_1+\dots +g_1}_k\in G,$$ +and $\Phi$ is surjective, such an $n$ should exist. But this is not possible since $\Phi(n)=(kg_1,0)$ implies (since $\Phi$ is a homomorphism) that $kg_1=ng_1$ and $ng_2=0$, and since $k$ was arbitrary and this need not hold if $k\ne 0$. + +REPLY [6 votes]: Your idea is OK, using homotopy groups. +Suppose $\mathbb{Z} = G \times G$ for some (Abelian) group $G$, which must be infinite. Then note that $\{0\} \times G$ and $G \times \{0\}$ are infinite subgroups of $G \times G$ that intersect only in the unit element $\{(0,0)\}$. By the isomorphism that supposedly exists, such subgroups should also exist in the integers. +But any subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ and any two of those have infinite intersection (at least $nm\mathbb{Z}$ for $n\mathbb{Z}$ and $m\mathbb{Z}$), so such subgroups cannot exist. +Using your original approach: suppose $h: \mathbb{Z} \rightarrow G \times G$ is an isomorphism, and $h(1) = (g_1,g_2)$, so that $h(n) = (ng_1, ng_2)$. As $(0,g_2)$ should be an image, $h(k) = (0,g_2)$ for some $k$, but it also equals $(kg_1, kg_2)$ so $kg_1 = 0$, and $kg_2 = g_2$, the latter implies $k=1$ (and then $g_1 = 0$ from the first equation), or $g_2 = 0$. In either case the image of $\mathbb{Z}$ can only be a subset of $G \times \{0\}$ or $\{0\} \times G$, both of which are never all of $G \times G$.<|endoftext|> +TITLE: Why this set is dense in $C_0(\mathbb{R})$? +QUESTION [14 upvotes]: Let $C_0=\{f~|~ f:\mathbb{R}\to\mathbb{R}, \textrm{$f$ is continuous},\lim\limits_{\vert x\vert \to\infty}f(x)=0\}$ +$A=\{f~|~f(x)=p(x)e^{-x^2}, \textrm{$p(x)$ is polynomials}\}$ +Why $A$ is dense in $C_0$? +The topology induced by the supremum metric $d(f,g)=\sup\limits_{x\in\mathbb{R}}|f(x)-g(x)|$. + +REPLY [5 votes]: Well, functional analysis provides very natural set up for dealing with this question. +We will prove the same for $$A_{c(x)}:= \left \{p(x) e^{-c(x)} \right\}$$ +Where $c(x)$ is a fixed even degree polynomial with positive leading coefficient and of degree $\ge 2$ and $p(x)$ is any polynomial. +Note that we need to prove that $A_{c(x)}$ is dense in $C_0 (\mathbb R)$. Else, by the Hahn-Banach Theorem, there would exist a nonzero functional $\mathcal{K}$ which would be zero on $A_{c(x)}$. +By Riesz–Markov–Kakutani representation theorem (referring to Rudin, Real and Complex analysis), $\ \mathcal{K}(f) = \int_{\mathbb R} f \, \text{d} \mu$ for some real measure $\mu$. Thus $\int_{\mathbb R} p(x) e^{-c(x)} \, \text{d} \mu = 0$ for every polynomial $p(x)$. Writing $\mu$ as $\mu_1 - \mu_2$ where $\mu_1, \mu_2$ are positive finite measures we have $\int_{\mathbb R} p(x) e^{-c(x)} \, \text{d} \mu_1 = \int_{\mathbb R} p(x) e^{-c(x)} \, \text{d} \mu_2$. +Define the measure $\nu_i := \int_{\mathbb R} e^{-c(x)} \text{d} \mu_i $ for $i= 1,2$. Then we have $\int_{\mathbb R} p(x) \text{d} \nu_1 = \int_{\mathbb R} p(x) \text{d} \nu_2 $ for every polynomial $p(x)$. (*) + +Now Define $ \Phi_i(z) := \int _{\mathbb R} e^{zt} \, \text{d} \nu_i (t)$ for $z \in \mathbb C$ and $i=1,2$. It is easy to check that $\Phi_1, \Phi_2$ are entire functions (Here you actually use that $c(x)$ is in fact an even degree polynomial of degree $\ge 2$ with positive leading coefficient.). + +Now by (*) we have $\Phi_1 ^{(n)} (0) = \Phi_2 ^{(n)} (0) $ for all $n \ge 0$. So their power series representations are identical, hence they are identical as functions, i.e., $\Phi_1(z) = \Phi_2(z)$. +So in particular, for any $s \in \mathbb R$, $\Phi_1( i s) = \Phi_2 (is)$, i.e., $ \int _{\mathbb R} e^{ist} \, \text{d} \nu_1 (t) = \int _{\mathbb R} e^{ist} \, \text{d} \nu_2 (t)$. This implies that $\nu_1 = \nu_2$ (by Inversion theorem), which implies $\mu_1 = \mu_2$. +Therefore $\mu = 0$, so the functional $\mathcal K$ must be zero, contrary to our assumption. This completes the proof.<|endoftext|> +TITLE: With the exception of $\mathbb{Z}$, every infinite abelian group contains a subgroup isomorphic to $\mathbb{Z}^2$? +QUESTION [5 upvotes]: With the exception of $\mathbb{Z}$, every infinite abelian group + contains a subgroup isomorphic to $\mathbb{Z}^2$. + +Is this statement true? I don't have much experience working with non-finite groups so any advice would be appreciated, thanks! + +REPLY [8 votes]: This is false. You could consider for a (non trvial) finite abelian group $G$ the group $\mathbb{Z} \times G $. +But there are also infinite abelian groups that do not contain a copy of $\mathbb{Z}$ even. Take for example the product of infinitely many copies of $\mathbb{Z}/2 \mathbb{Z}$. +Also note that $(\mathbb{Q},+)$ does not contain a copy of $\mathbb{Z}^2$ (yet $(\mathbb{R},+)$ does). + +REPLY [5 votes]: No, consider $\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ : it's infinite, not isomorphic to $\mathbb{Z}$ (it has torsion) and does not contain $\mathbb{Z}^2$.<|endoftext|> +TITLE: Do non-second-countable spaces have "small" non-second-countable subspaces? +QUESTION [8 upvotes]: If $X$ is any space which is not second-countable, can one find a subspace $Y \subseteq X$ with $|Y| \leq \aleph_1$ which is also not second-countable? (Recall that a topological space $X$ is second-countable if it has a countable base.) +Note that if we require $|Y| \leq \aleph_0$ there are obvious counterexamples. For example, take $X$ to be the ordinal space $\omega_1$. Since $X$ has uncountably many isolated points, it is not second-countable, but every countable subspace is second-countable. +Looking through the non-second-countable spaces in π-base seems to suggest this is possible, but perhaps the spaces listed there are "too nice". + +REPLY [4 votes]: The answer is yes. +There is special terminology and known results along these lines. +If every space with property $P$ has a subspace of size (at most) $\aleph_1$ with property $P$ then one says that property $P$ reflects in size $\le\aleph_1$ spaces. +For example non-metrizability (for compact spaces) reflects in size $\le\aleph_1$ spaces, i.e. +every non-metrizable (compact Hausdorff) space has a non-metrizable subspace of size $\aleph_1$ (a result of Alan Dow). (Note that in this context $\aleph_1$ and $\omega_1$ are often used interchangeably, he writes +$\omega_1$ in his paper. Similarly $\aleph_0$, $\omega$ and $\omega_0$ are used interchangeably.) +Recall that $w(X)$ denotes the weight of $X$, i.e. the smallest infinite cardinality of an open base for the topology of $X$. +The answer to your question is yes, see +Having a small weight is determined by the small subspaces, +Authors: A. Hajnal and I. Juhász +Journal: Proc. Amer. Math. Soc. 79 (1980), 657-658. +http://www.ams.org/journals/proc/1980-079-04/S0002-9939-1980-0572322-2/ +Abstract: We show that for every cardinal $ \kappa > \omega $ and an arbitrary topological space X if we have $ w(Y) < \kappa $ whenever $ Y \subset X$ and $ \vert Y\vert \leqslant \kappa $ then $ w(X) < \kappa $ as well. M. G. Tkačenko proved this for $ {T_3}$ spaces in [2]. We also prove an analogous statement for the $ \pi $-weight if $ \kappa $ is regular. +Here the property that "reflects" is "having weight $\ge\omega_1$". +If $w(X)\ge\omega_1$ then $w(Y)\ge\omega_1$ for some $Y\subseteq X$ with $|Y|\le\omega_1$. Indeed, if this were not the case, then taking $\kappa=\omega_1$ in the above theorem, we would have that every subspace $Y\subseteq X$ with $|Y|\le\omega_1$ satisfies $w(Y)<\omega_1$ +(i.e. $w(Y)\le\omega_0$, $Y$ has countable weight, $Y$ is second-countable), and then the conclusion would be that $w(X)<\omega_1$ (i.e. $w(X)\le\omega_0$, $X$ has countable weight, $X$ is second-countable). +For reflection of non-metrizability (with a proof using elementary submodels) see: +An empty class of nonmetric spaces, Alan Dow, +Proc. Amer. Math. Soc. 104 (1988), 999-1001. +http://www.ams.org/journals/proc/1988-104-03/S0002-9939-1988-0964886-9/ +And here is a more-or-less random selection about reflecting other properties +(just a few sample links, out of many more). +On Dow's reflection theorem for metrizable spaces +Jerry E. Vaughan +TOPOLOGY PROCEEDINGS Volume 22, 1997, 351-360 +http://topology.auburn.edu/tp/reprints/v22/tp22123.pdf +Reflecting point-countable families, Zoltan T. Balogh +Proc. Amer. Math. Soc. 131 (2003), 1289-1296 +http://www.ams.org/journals/proc/2003-131-04/S0002-9939-02-06621-2/ +Reflecting Lindelöf and converging $\omega_1$-sequences +Alan Dow, Klaas Pieter Hart +arXiv:1211.2764v2 [math.GN] http://arxiv.org/abs/1211.2764 +(Submitted on 12 Nov 2012 (v1), last revised 17 Jul 2013 (this version, v2)) +Reflecting Lindelöfness +James E. Baumgartner, Franklin D. Tall, +Topology and its Applications +Volume 122, Issues 1–2, 16 July 2002, Pages 35–49 +http://www.sciencedirect.com/science/article/pii/S0166864101001353 +Ideal reflections +Paul Gartside, Sina Greenwood, and David Mcintyre +TOPOLOGY PROCEEDINGS +Volume 27, No. 2, 2003, Pages 411-427 +http://topology.auburn.edu/tp/reprints/v27/tp27203.pdf<|endoftext|> +TITLE: How do I go about solving this? +QUESTION [6 upvotes]: I have tried substitution, but it is not working for me. +$$ +\int_0^\pi \frac{dx}{\sqrt{(n^2+1)}+\sin(x)+n\cos(x)}=\int_0^\pi \frac{n dx}{\sqrt{(n^2+1)}+n\sin(x)+\cos(x)}=2 +$$ +General form of this integral is +$$ +\int_0^\pi \frac{dx}{\sqrt{(n^2+m^2)}+m\sin(x)+n\cos(x)}=\frac{2}{m} +$$ + +REPLY [2 votes]: This seems so easy in polar form. Let $1=r\cos\phi$, $n=r\sin\phi$, then $r=\sqrt{1+n^2}$, $\phi=\tan^{-1}n$ and then let $x+\phi=y+\frac{\pi}2$ +$$\begin{align}\int_0^{\pi}\frac{dx}{\sqrt{1+n^2}+\sin x+n\cos x}&=\int_0^{\pi}\frac{dx}{\sqrt{1+n^2}(1+\sin(x+\phi))}\\ +&=\int_{-\frac{\pi}2+\phi}^{\frac{\pi}2+\phi}\frac{dy}{\sqrt{1+n^2}(1+\cos y)}\\ +&=\left.\frac{\sin y}{\sqrt{1+n^2}(1+\cos y)}\right|_{-\frac{\pi}2+\phi}^{\frac{\pi}2+\phi}\\ +&=\frac1{\sqrt{1+n^2}}\left(\frac{\cos\phi}{1-\sin\phi}+\frac{\cos\phi}{1+\sin\phi}\right)\\ +&=\frac1{\sqrt{1+n^2}}\left(\frac1{\sqrt{1+n^2}-n}+\frac1{\sqrt{1+n^2}+n}\right)\\ +&=2\end{align}$$ +The second integral, after the substitution $y=x-\phi$ becomes +$$\begin{align}\int_0^{\pi}\frac{n\,dx}{\sqrt{1+n^2}+n\sin x+\cos x}&=\int_0^{\pi}\frac{n\,dx}{\sqrt{1+n^2}(1+\cos(x-\phi))}\\ +&=\int_{-\phi}^{\pi-\phi}\frac{n\,dy}{\sqrt{1+n^2}(1+\cos y)}\\ +&=2\end{align}$$<|endoftext|> +TITLE: Linear Algebra with functions +QUESTION [15 upvotes]: Basically my question is - How to check for linear independence between functions ?! +Let the group $\mathcal{F}(\mathbb{R},\mathbb{R})$ + Be a group of real valued fnctions. +i.e $\mathcal{F}(\mathbb{R},\mathbb{R})=\left\{ f:\mathbb{R}\rightarrow\mathbb{R}\right\} $ +Let 3 functions $f_{1},f_{2},f_{3}$ + be given such that +$\forall x\in\mathbb{R}\,\,\,f_{1}=e^{x},\,\,f_{2}=e^{2x},\,\,f_{3}=e^{3x}$ +$W=sp(f_{1},f_{2},f_{3})$ + what is $dim(W)$ + ? +How to approach this question ? (from a linear algebra perspective) +I know that $\forall x\in\mathbb{R}\,\,\,W=\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}$ +And to get the dimension I need to find the base of $W$ +so I need to check whether the following holds true : +$\forall x\in\mathbb{R}\,\,\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\,\Leftrightarrow\,\alpha,\beta,\gamma=0$ +However when $x=0$ + I get $\alpha+\beta+\gamma=0$ + which leads to infinite amount of solutions. +How to approach this question ? + +REPLY [5 votes]: You need to show the three vectors are linearly independent. In this case I would use this trick; so that you don't need to worry about them being functions and the equality to hold for every value of $x$. +If you consider $D: \mathcal{F} \rightarrow \mathcal{F}$, the derivative operator, is an endomorphism in $\mathcal F$ (i.e. a linear map from $\mathcal{F}$ to itself). The derivatives of the three functions are +$$ +Df_1=De^x=e^x=f_1 +$$ +$$ +Df_2=De^{2x}=2e^{2x}=2f_2 +$$ +$$ +Df_3=De^{3x}=3e^{3x}=3f_3 +$$ +So $f_1,f_2,f_3$ are eigenvectors of $D$, with eigenvalues $\lambda_1=1, \lambda_2=2, \lambda_3=3$, respectively. Since $f_1, f_2, f_3$ are eigenvectors with distinct eigenvalues of the same endomorphism $D$, they are linearly independent so they form a base for $W$ and $\text{dim}W=3$.<|endoftext|> +TITLE: Linear algebra for modern differential geometry( and other types of modern geometry, like analytic, complex and algebraic) +QUESTION [6 upvotes]: I wish to study real and complex analysis(for example, Pugh "Real Mathematical Analysis" and Cartan "Elementary theory of analytic functions of one and several complex variables") and modern differential and riemannian geometry(for example, Jeffrey Lee "Manifolds and differential geometry"). +I would like to chose a book on theoretical linear algebra to cover prerequisites for these. +That is, I don't need a linear algebra for applications in science. I don't need it for its own sake or history of mathematics. I want to study modern geometry and analysis - real and complex analysis, analysis on manifolds, differential geometry, riemannian geometry, complex algebraic and analytic geometry. +I came across a few books on my search, but I'm not sure if they are the best option for my goals. +Axler "Linear Algebra Done Right" - this book seems to have a different goal in mind. It avoid determinants as well as works only over $\mathbb{R}$ and $\mathbb{C}$. Do you need linear algebra over an arbitrary field $F$ for study of smooth manifolds? +Roman "Advanced Linear Algebra". Isn't it an overkill for my goals? What do you think of this book? +Any other recommendations will be considered, of course. I would prefer modern books over likes of Hoffman-Kunze or Halmos, but if you know any old gems, feel free to tell. + +REPLY [4 votes]: The topics that you want to study use mostly the very essential ideas from Linear Algebra. Yes, over $\mathbb{R}$ and $\mathbb{C}$ is all you need. Since you seem to be a theoretically minded person with interest in geometry related subjects, I would recommend Gelfand's "Lectures on Linear Algebra". Strang's textbook is excellent, but probably not the style you are looking for. +My area of research is differential geometry (moving surfaces) and my favorite subjects to teach are Linear Algebra and Tensor Analysis. My advice would be to not treat Linear Algebra merely a means to studying the topics that you listed. Linear Algebra is the language and the framework for dealing with all branches of applied mathematics where physics, algebra and geometry meet. I would therefore recommend to study Linear Algebra thoroughly and, in fact, to give it several passes following a few different sources. +In case you want to learn my personal perspective on Linear Algebra and Differential Geometry, check out the Linear Algebra course on Lemma (http://lem.ma) and my Tensor Calculus videos on YouTube. +When Niels Henrik Abel was asked how he acquired his expertise, we replied "By studying the masters and not their pupils." Just one of my reactions to your desire to read a modern textbook.<|endoftext|> +TITLE: Proving that the tangent to a convex function is always below the function +QUESTION [12 upvotes]: Consider a real-valued convex function $f$ defined on an open interval $(a,b) \subset \mathbb{R}$. Let $x,y \in (a,b)$. I want to prove that +\begin{equation} +f((1-\lambda)x + \lambda y) \leq (1-\lambda)f(x) + \lambda f(y) , \lambda \in [0,1]\implies f(y) + f'(y)(x-y) \leq f(x) +\end{equation} +No other assumptions (regarding the second derivatives) is allowed. +While both facts are really obvious and clear, I am unable to prove this without making additional assumptions. + +REPLY [13 votes]: Presumably you're allowed to assume that $f$ is differentiable. +The Idea: If you draw a picture you convince yourself that if, say, $y0.$$ +Since $(1-\lambda)(x-y)>0$ you can divide both sides of the second-last inequality by it and you get +$$\frac{f((1-\lambda)x+\lambda y)-f(y)}{[(1-\lambda)x+\lambda y]-y} +\le\frac{f(x)-f(y)}{x-y}.$$Now if $\lambda\to 1$ then $(1-\lambda)x+\lambda y\to y$, so the left side approaches $f'(y)$, and you get $$f'(y)\le\frac{f(x)-f(y)}{x-y}.$$Since $x-y>0$ you can multiply both sides by $x-y$, and you get $$f'(y)(x-y)\le f(x)-f(y).$$ +(If $a +TITLE: Can a separable isogeny of elliptic curves have an inseparable dual? +QUESTION [8 upvotes]: Let $\phi: E_1\to E_2$ be an isogeny of elliptic curves over a field $K$ of characteristic $p>0$. Suppose that $\phi$ is separable and let $\hat{\phi}: E_2\to E_1$ denote the dual isogeny. Then $\hat{\phi}$ satisfies several nice properties. Two of these that I am interested in are: + +$\deg\hat{\phi} = \deg\phi$; +$\hat\phi\circ\phi = [\deg\phi]$ on $E_1$ and $\phi\circ\hat\phi = [\deg\phi]$ on $E_2$. + +Note that if $\deg\phi$ is divisible by $p$ then both compositions $\hat\phi\circ\phi$ and $\phi\circ\hat\phi$ are inseparable. +But is it possible for $\phi$ to be separable while $\hat\phi$ is inseparable? Or, by the above remark, can $\phi$ be separable whilst having degree divisible by the characteristic $p$? These properties, plus others in Silverman's book, don't tell me how the inseparability degrees of an isogeny and its dual are related. My intuition (and hope!) is that these two things are impossible, but I am wary of strange counterexamples in characteristic $p$. + +REPLY [8 votes]: Just so this has an answer, let me expand upon what I said in the comments slightly. +So, let us begin with the following trivial observation: + +Observation: If $f:E\to E'$ is an isogeny of elliptic curves over $k$, with $\mathrm{char}(k)=p>0$, then $f$ and $\widehat{f}$ are separable if and only if $p\nmid\deg(f)$. + +Indeed,this follows immediately from the follow facts: in a tower $E/K/F$ the extension $E/F$ is separable if and only if $E/K$ and $E/F$ is separable, $[n]$ is separable if and only if $p\nmid n$, and $f\circ\widehat{f}=[n]$. +Thus, your question comes down to whether or not there even exist separable isogenies $f:E\to E'$ with $p\mid \deg(f)$. The answer is yes if and only if $E$ is ordinary. +To see this most clearly, note that every isogeny $f:E\to E'$ is of the form $E\to E/K$ (quotient map) for $K\subseteq E$ a finite subgroup (scheme) and that $\ker f=K$ so that $\deg(f)=|K|$. Moreover, $f$ is separable if and only if it's étale (separable=generically étale=étale for group schemes since one can translate the generic étaleness everywhere) if and only if $K$ is étale (over $\text{Spec}(k)$). +Now, we may as well assume that $\deg(f)=p^r$ for some $r$ (since this is the only case of real interest to us) in which case we see that $K\subseteq E[p^r]$. Thus, your whole question comes down to whether or not $E[p^r]$ has an étale subgroup scheme. Moreover, it's easy to deduce from the fact that $E[p^r]$ is an iterative extension of $E[p]$ that this is true for some $r$ if and only if it's true for $r=1$. Thus, the existence of an isogeny $f$ which is separable but for which $\widehat{f}$ is inseparable is equivalent to the question of whether or not $E[p]$ has an étale subgroup scheme. +But, this is precisely equivalent to whether or not $E$ is ordinary or supersingular. Namely, depending on your definitions, $E$ is defined to be ordinary if $E[p]$ has a non-trivial étale subgroup (and supersingular otherwise). If you, instead, use the definition that $E$ is ordinary if and only if $|E[p](\overline{k})|>1$ note merely that a finite scheme $X/k$ is étale if and only if $\#(X_{\overline{k}})=\dim(\mathcal{O}_X(X))$. Using this it's easy to see that this definition of ordinary agrees with the one I said above. +To summarize, the only interesting isogenies in this question are of the form $f:E\to E'$ with $\deg(f)=p^r$ for some $r\geqslant 1$, and are of the form $E/\to E/K$ (quotient map). Moreover, this $f$ is separable with inseparable dual if and only if $K$ is an étale subgroup scheme of $E[p^r]$, which exist if and only if $E$ is ordinary.<|endoftext|> +TITLE: Showing that $\sup\{|f(x)-f(y)|, x,y\in X\}= \sup \ f - \inf \ f$ +QUESTION [7 upvotes]: I need to show that, for $f:X\to \mathbb{R}$ bounded, we have: +$$\sup\{|f(x)-f(y)|, x,y\in X\}= \sup f - \inf f$$ +Well, I know that +$$\sup\{|f(x)-f(y)|, x,y\in X\}\ge |f(x)-f(y)|$$ but in what this helps? I really have no idea in how to prove this one + +REPLY [4 votes]: Fix any $x_0\in X$ and $y_0\in X$. Then, one has that $$f(x_0)-f(y_0)\leq|f(x_0)-f(y_0)|\leq\sup_{x,y\in X}\{|f(x)-f(y)|\},$$ since every real number is less than or equal to its absolute value. Rearranging yields: $$f(x_0)\leq\sup_{x,y\in X}\{|f(x)-f(y)|\}+f(y_0).$$ This is true for any $x_0\in X$, so taking supremum on the left-hand side yields $$\sup_{x\in X}f(x)\leq\sup_{x,y\in X}\{|f(x)-f(y)|\}+f(y_0).$$ Another rearrangement yields: $$\sup_{x\in X}f(x)-\sup_{x,y\in X}\{|f(x)-f(y)|\}\leq f(y_0),$$ and taking infimum on the right-hand side results in $$\sup_{x\in X}f(x)-\sup_{x,y\in X}\{|f(x)-f(y)|\}\leq\inf_{x\in X} f(x),$$ or $$\sup_{x\in X}f(x)-\inf_{x\in X} f(x)\leq\sup_{x,y\in X}\{|f(x)-f(y)|\}.$$ + +As for the other direction, fix again arbitrary $x_0\in X$ and $y_0\in X$. If $f(x_0)\geq f(y_0)$, then $$|f(x_0)-f(y_0)|=f(x_0)-f(y_0)\leq \sup_{x\in X} f(x)-f(y_0)\leq\sup_{x\in X}f(x)-\inf_{x\in X}f(x).$$ If, on the other hand, $f(x_0) +TITLE: Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $ +QUESTION [5 upvotes]: The following is an exercisein complex analysis: + +Use contour integrals with $-\pi/2<\operatorname{arg} z<3\pi/2$ to compute + $$ +I:=\int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx. +$$ + +I don't see why the branch $-\pi/2<\operatorname{arg} z<3\pi/2$ would work. Let +$$ +f(z)=\frac{z^{1/3}\log z}{z^2+1}. +$$ +Denote $\Gamma_r={re^{it}:0\leq t\leq \pi}$. One can then consider a contour consisting of $\Gamma_R$, $\Gamma_r$, and $[-R,-r]\cup[r,R]$. The integral along $[r,R]$ will give $I$. But without symmetries, how could one deal with the integral along $[-R,-r]$? + +REPLY [2 votes]: To first answer the question in the OP regarding the applicability of a branch cut along the negative imaginary axis, we note that we can certainly evaluate the integral of $\frac{z^a \log(z)}{z^2+1}$ over the proposed contour as +$$\begin{align}\oint_C \frac{z^a \log(z)}{z^2+1} &=\int_r^R \frac{x^a\log(x)}{x^2+1}\,dx-\int_{r}^{R}\frac{e^{i\pi a}x^a(\log(x)+i\pi )}{x^2+1}\,dx\\\\ +&+\int_{0}^\pi \frac{R^ae^{ia\phi}(\log(R)+i\phi)}{R^2e^{i2\phi}+1}\,iRe^{i\phi}\,d\phi\\\\&-\int_{0}^\pi \frac{r^ae^{ia\phi}(\log(r)+i\phi)}{r^2e^{i2\phi}+1}\,ire^{i\phi}\,d\phi\end{align}$$ +Proceeding, for $0 +TITLE: $|f| $ is Lebesgue integrable , does it implies $f$ is also? +QUESTION [6 upvotes]: If $ f $ is Lebesgue integrable then $|f|$ is Lebesgue integrable but does the converse of the result is also true? + +REPLY [6 votes]: No, it does not: If $E$ is (bounded and) non-measurable, then $f = 2\chi_{E} - 1$ is everywhere equal to $\pm1$, so $|f| \equiv 1$, but $f$ is non-measurable, hence not integrable.<|endoftext|> +TITLE: Commutativity of a ring from idempotents. +QUESTION [10 upvotes]: In a ring $R$ with unity, every element can be written as product of finitely many idempotents. Can one show that the ring is commutative? + +REPLY [11 votes]: Lemma 1: The only unit is the identity. + +If $e$ is the leftmost idempotent in a factorization of a unit $u$, then $(1-e)u=0$ implies $1-e=0$ and $e=1$. After finitely many steps we have proven $u=1$. + +Lemma 2: The ring is reduced, that is, it has no nonzero nilpotent elements. The only nilpotent element is $0$. + +If $x$ is nilpotent, then $1-x$ is a unit, and that must be $1$ by the first lemma. Thus $x=0$. + +Lemma 3: Idempotents in a reduced ring commute with all elements. + +For any $r\in R$, both $er-ere$ and $re-ere$ square to zero, so they are both zero, and $er=ere=re$. + +Proof of the original post: + +everything is a product of idempotents, all of which commute with each other, so the ring is commutative. Moreover, every product of commuting idempotents is idempotent, so we can even conclude the ring is a Boolean ring.<|endoftext|> +TITLE: Evaluation of an integral associated with integral kernel of resolvent of Laplacian +QUESTION [5 upvotes]: I came across evaluating the following sort of integral when I was considering the integral kernel for resolvent of Laplacian $(I-\Delta)^{-1}$: +$$ +K(x)=\int_0^{\infty}\frac{\exp(-t-\frac{|x|^2}{4t})}{t^{\frac{n}{2}}}dt +$$ +However, what I did know is that in the 1-dimensional case, i.e. $n=1$, the operator $(1-\partial_x^2)^{-1}$ has an integral kernel representation of the form: +$$ +(1-\partial_x^2)^{-1}f=\int_{-\infty}^{\infty}\frac{1}{2}e^{-|x-y|}f(y)dy +$$ +i.e. the integral kernel is$$ +K(x)=\frac{1}{2}e^{-|x|} +$$ +My question is, how can one evaluate the first integral for the case $n=1$ to obtain exactly the integral kernel in the third equation? + +REPLY [5 votes]: Plugging in $n=1$, the integral becomes +$$ K(x) = \int_0^{\infty} t^{-\frac{1}{2}} e^{-t-\frac{x^2}{4t}} \,dt.$$ +Taking a Fourier transform, we have +$$\mathcal{F}K(y) = \frac{1}{\sqrt{2\pi}} \int_0^{\infty} t^{-\frac{1}{2}} e^{-t}\int_{-\infty}^{\infty} e^{-ixy} e^{-\frac{x^2}{4t}}\,dx\,dt.$$ +The inner integral is the Fourier transform of the Gaussian with a dilation. In general, if $\mathcal{D}_{\alpha}f(x) = f(\alpha x)$ for nonzero $\alpha$, then a change of variable gives +$$\mathcal{F}\mathcal{D}_{\alpha}f(y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ixy}f(\alpha x)\,dx = \frac{1}{|\alpha|\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ix\frac{y}{\alpha}} f(x)\,dx.$$ +Which is nothing more than $\frac{1}{|\alpha|}\mathcal{D}_{\alpha^{-1}} \mathcal{F}f(y)$, meaning if you dilate in one domain, you get the opposite dilation in the other domain (with a scale factor). In our case $\alpha = (2t)^{-1/2}$. Moreover, the Fourier transform of a Gaussian ($\exp(-x^2/2)$) is another Gaussian since the Gaussian is an eigenfunction of the Fourier transform, so we get that +$$ \mathcal{F}K(y) = \int_0^{\infty} t^{-\frac{1}{2}}e^{-t} \cdot (2t)^{\frac{1}{2}} e^{-ty^2}\,dt = 2^{\frac{1}{2}}\int_0^{\infty} e^{-(1+y^2)t}\,dt. $$ +Letting $u = (1+y^2)t$, we get that $du = (1+y^2)\,dt$ and so +$$ \mathcal{F}K(y) = \frac{\sqrt{2}}{1+y^2} \int_0^{\infty} e^{-u}\,du = \frac{\sqrt{2}}{1+y^2}.$$ +The Fourier transform is injective and the only function whose Fourier transform is $\frac{\sqrt{2}}{1+y^2}$ is $\sqrt{\pi} e^{-|x|}$. +I will leave it to you to argue that $K$ (as given in the first line) is in $L^1(\Bbb R)$ and that you can interchange the integrals as I did in the second line so that the use of the Fourier transform is justified. +Edit: Our constants do not quite match up. My guess is that your initial definition of $K$ has a constant factor out front which should take care of the factor of $\sqrt{\pi}$. +My reason for using the Fourier transform here is twofold. The main problem in your integral is the $e^{-\frac{x^2}{4t}}$ part. This is a Gaussian in $x$ - not $t$. So my thought was along the lines of: in the $t$-integral, the $\frac{1}{t}$ part is very problematic and hard to work with. Moreover the dilation property of the Fourier transform means that if I did a Fourier transform, the $\frac{1}{t}$ would appear in the numerator afterward which is infinitely easier to handle. Coupled with the Gaussian being an eigenfunction of the Fourier transform, it was clear that the Fourier transform was the way to go.<|endoftext|> +TITLE: Right Triangle's Proof +QUESTION [7 upvotes]: A right triangle has all three sides integer lengths. One side has length 12. What are the possibilities for the lengths of the other two sides? Give a proof to show that you have found all possibilities. +EDIT: I figured out that there are a total of 4 combinations for a side with length 12. +$$a^2= c^2 -b^2$$ +factoring the right side gives +$$a^2 = (c+b)(c-b)$$ +so from there I just looked for values of c and b which would make the equation true. And I got: $(37,35), (13,5), (15,9), (20,16)$. +My only problem now is proving that these are all the possibilities. I have an intuition as to why that's true but I don't know how to go about a full proof. I'm guessing I need to make use of the fact that all sides are to be integer lengths and that $12$ itself cannot be equal to $c$. +I know that if I were to try values for values for $b$ and $c$ whose absolute value difference is greater than 8, then the equation would not hold true. +Ex: +$(37,35)$ has a difference of $2$ and works. +$(13,5)$ has a difference of $8$ and works. +$(15,9)$ has a difference of $6$ and works. +$(20,16)$ has a difference of $4$ and woks. +But if I were to pick any pair with an absolute difference greater than 8 it would not work. Can I just prove this is true with a couple of examples? Or do I need a full generic proof ? If so, how would I go about it? + +REPLY [5 votes]: If $$144=(c+b)(c-b)$$ we know that $c+b$ is a divisor of $144$. So there are only finitely many pairs of $(c,b)$ we have to investigate. $(c+b,c-b)$ are from the set $$\{(1,144),(2,72),(3,48),(4,36),(6,24),\ldots,(144,1)\}$$. But we know that $c+b \gt c-b$ and that both $c+b$ and $c-b$ are even. Because $12\cdot 12=144$ we have $c-b<12$ and $c+b>12$. The only even divisors of $144$ are $\{2,4,6,8\}$. So your solutions are the only ones. +But the side of length $12$ can be the hypotenuse $c$. So you should check the equation $$12^2=a^2+b^2$$ too. Again only finitely many pairs are to check. You can assume that $a \ge b $ and therefore $a \ge \sqrt{\frac{144}{2}} = 8.\cdots$. So you have only to check $a \in \{9,10,11\}$.<|endoftext|> +TITLE: Solution of functional equation $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$ +QUESTION [8 upvotes]: If $x,y\in \mathbb{R}$ and $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$ and $f'(0)=0\;,$ Then $f(x)$ is + +$\bf{My\; Try::}$ Using $$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+h\sqrt{f(x)}-f(x)}{h}$$ +Now Put $x=y=0$ in $$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}\;,$$ We get $f(0)=0$ +So we get $f(0)=0$ +So $$f'(x) = \sqrt{f(x)}+\lim_{h\rightarrow 0}\frac{f(h)}{h}=\sqrt{f(x)}$$ +So $$\int\frac{f'(x)}{\sqrt{f(x)}}dx = 1\int dx\Rightarrow 2\sqrt{f(x)}=x+c$$ +Now Put $x=\;,$ We get $c=0$ +So we get $2\sqrt{f(x)}=x\Rightarrow 4f(x)=x^2\Rightarrow \displaystyle f(x)=\frac{x^2}{4}$ +Can we solve it some short way, If yes then please explain here, Thanks + +REPLY [5 votes]: The above answers didn't see, that, by squaring, an incorrect solution is added. Indeed, there is no nontrivial solution: +In order for the functional equation to be well defined for each $x,y\in\mathbb R$, we need $f\geq 0$. Let's assume that there is $z\in\mathbb R$ with $f(z)>0$. +Because $f(0) = f(0)+f(0)$, we have $f(0)=0$, hence $z\neq 0$. +Then, for $x=z,~y=-z$, we obtain +$$f(0) = f(z)+f(-z)-z\sqrt{f(z)}.$$ +On the other hand, for $x=-z,~y=z$, we obtain +$$f(0)=f(-z)+f(z)+z\sqrt{f(-z)}.$$ +Subtracting those equations, we get +$$0 = z\sqrt{f(-z)}+z\sqrt{f(z)}.$$ +Since $z\neq 0$, this yields $0=\sqrt{f(-z)}+\sqrt{f(z)}\geq \sqrt{f(z)}>0$, contradiction. +An other possibility is, to see that the only nontrivial solution could be $x^2/4$ as shown above and then check that it is indeed not a solution to the equation.<|endoftext|> +TITLE: Angles in Hilbert's axioms for geometry +QUESTION [5 upvotes]: In Hilbert's axioms for geometry, the following elements are presented as undefined (meaning "to be defined in a specific model"): point, line, incidence, betweenness, congruence. +In the euclidean plane $\mathbb{E}^2$, for example, we define a point as an element of $\mathbb{R}^2$, a line as a set $\{(x, y) \in \mathbb{R}^2 \;|\; ax+by=c\}$ for some $a, b, c \in\mathbb{R}$ with $(a, b) \neq(0, 0)$ and so on, which is all fine. +My question is: what about angles? +Even the most rigorous textbooks I've found treat angles as a natural thing, but to me, it seems something hard to explain formally. How can we actually define an angle? I'm not talking about sines or cosines (in $\mathbb{E}^2$ for example, we could define the cosine of the angle between vectors $u$ and $v$ as $cos(\theta):=\frac{}{||u||.||v||}$), but really about THE angle between two lines, the real number $\theta$ itself. I haven't found any convincing definition so far. +Since angles are so important in geometry, I don't understand why they are not presented in Hilbert's axioms, or at least clearly defined somewhere along the way. +Thanks! + +REPLY [2 votes]: First I'll talk about Hilbert axioms. +Angles themselves are not an undefined term. Angle is defined with the notion of line and betweenness. Namely an angle is an unordered pair $\{A,B\}$ where $A$ and $B$ are halflines having the same origin (we also exclude the situation when $A$ and $B$ are contained in the same line) and halfline is a set $\{b:B(oba)\vee a=b\vee B(oab)\}$ for some $a\neq o$ ($o$ is called the origin of the angle). +What is an undefined term is the congruence relation between angles, which tells us which angles are in a sense 'equal', but nothing is said about real numbers assigned to the angles. +However there is a notion of measure which correlates with every angle a positive real number in such way that congruent angles have equal measures and roughly speaking the measure is additive (strict definition requires some other definitons). +The following theorem is true: +Let $\{A,B\}$ be the angle and $x_0>0$. There exists exactly one measure of angles $\psi$ such that $\psi(\{A,B\})=x_0$. +We usually use this theorem with $\{A,B\}$ being the right angle and $x_0=\frac{\pi}{2}$ or $x_0=90$. +The proof is somewhat unconstructive because it doesn't give the formula for the measure (it uses the continuity of the real line). You can find the proof, all lemmas and definitions and generally much more detail in the book K. Borsuk, W. Szmielew, Foundations of geometry. Unfortunately in english edition they changed the system of axioms greatly in comparison with original Hilbert's axioms and for example they removed the congruence relation between angles from undefined terms. However the part about measure of angles stays the same and doesn't depend on axioms that much. +Now let's go to our euclidian model $\mathbb{R}^2$. Once we define the notions of line and betweenness we also have the notion of angle. Next we define the notion of congruence between angles in such way: +The angle $\{A,B\}$ with origin $o$ is congruent to the angle $\{A_1,B_1\}$ with origin $o_1$ if and only if there exist $a\in A, b\in B, a_1\in A_1, b_1\in B_1$ such that $|oa|=|o_1a_1|, |ob|=|o_1b_1|, |ab|=|a_1b_1|$. +Once we prove all axioms we know that all theorems which can be derived from the axioms are true in the model, in particular the theorem about measure of angles.<|endoftext|> +TITLE: Is there a fundamental theorem of algebra for matrices? +QUESTION [19 upvotes]: The fundamental theorem of algebra says we can do this ($z\in\mathbb{C}$ of course) +$$\sum_{k=0}^n a_kz^k= a_n\prod_{k=1}^n (z-\omega_k)=0$$ +for some set $\{\omega_k \in\mathbb{C}\}_{k=1,2,\ldots , n}$. Is there a similar answer for square matrices with complex entries (i.e. for linear operators $A_k:\mathbb{C}^n \to\mathbb{C}^n,\,\,(k=0,1,2,\ldots, n))$? That is, can we do such a factorization +$$\sum_{k=0}^n A_kz^k= A_n\prod_{k=1}^n (z\mathbb{1}-\Omega_k)=0$$ +where $\mathbb{1}$ is the identity operator and $\Omega_k : \mathbb{C}^n\to\mathbb{C}^n$ are linear operators? Certainly we're restricted here by the non-commutativity of matrices in general, so the order of this decomposition would matter. But, maybe there are some types of matrices that admit commutative "root" matrices? +Sorry if my notation is awkward, and also if the answer to this is considered common knowledge. + +REPLY [2 votes]: When $n>1$, to write $\sum_{k=0}^n A_kz^k=0$ is generically non-sense. Indeed, that implies that the $(A_k)_k$ are linearly dependent over $\mathbb{C}$. More precisely, the previous equation is equivalent to say that $n^2$ complex polynomials have a common complex root, that is generically false when we consider only $2$ polynomials !! +EDIT. The correct question can be written as follows: let $(A_k)_{0\leq k\leq n})\in M_n(\mathbb{C})$; our problem: (*) can we factorize $\sum_{k=0}^n z^kA_k$ in the form $A_n \Pi_{i=1}^n (zI-X_i)$ where the $X_i\in M_n(\mathbb{C})$ ? Of course, there are counter-examples; mathematicians have a great time when they find such results; the best known example is probably linked to the equation $X^2=U$; the mathematician people immediately says that this equation may have no solutions; yet, everybody knows that, for a generic $U$, this equation has always $2^n$ solutions ! Why not say it first? Mathematics are beautiful when they work. +Proposition. Assume that $n\leq 3$. For generic $(A_i)_i$, (*) has always $6$ simple solutions when $n=2$ and $1680$ simple solutions when $n=3$. +Remarks. 1. "Generically" is considered in the sense "Zariski open dense set" or in the probabilistic sense (the entries of the $(A_i)$ are iid complex that follow a normal law; the result is true with probability $1$). The results and definitions used or cited in this post are in my paper [1] http://arxiv.org/pdf/1304.2506.pdf (published in Linear Algebra and its Applications). + +Clearly, we can also consider equations of degree $p$ in $M_n(\mathbb{C})$. +I wrote a proof only when $n\leq 3$; I think that the result can be generalized when $n>3$. + +Proof. Since $A_n$ is generic, it is invertible and we may assume that it is $I$. +Case 1. $n=2$. $U,V$ are known generic matrices and the unkowns $X,Y$ satisfy $z^2I-Uz+V=(zI-X)(zI-Y)$, that is $U=X+Y,V=XY$; then $X^2-XU+V=0,Y=U-X$. The first previous equation is an unilateral equation of degree $2$ (a Riccati one) and (generically) admits $6$ solutions in $X$ and then, $6$ solutions in $(X,Y)$. +Case 2. $n=3$. $U,V,W$ are known generic matrices and the unkowns $X,Y,Z$ satisfy $z^3I-Uz^2+Vz-W=(zI-X)(zI-Y)(zI-Z)$, that is $U=X+Y+Z,V=XY+YZ+XZ,W=XYZ$; then $Z=U-X-Y$ and putting $K=XY,L=X+Y$, we obtain the system $V=K+L(U-L),W=K(U-L)$ in the unknowns $K,L$. Since $Z=U-L$, from $V(U-L)=W+L(U-L)^2$ we deduce (1) $Z^3-UZ^2+VZ-W=0$. Moreover, we can apply the previous result concerning $Z$ to the transpose of the equality (*); we obtain (2) $X^3-X^2U+XV-W=0$. +Equations (1) and (2) are unilateral and, consequently, have $\binom{3^2}{3}=84$ simple solutions $Z$ or $X$ (cf. Theorem 1 in [1]). Let $Z$ be a solution of (1). Then $X+Y=U-Z,XY=V-UZ+Z^2$ and $Y$ is solution of a unilateral equation of degree $2$. In all probability, $U-Z,V-UZ+Z^2$ are generic (and independent) and there are $\binom{2\times 3}{3}=20$ solutions $Y$. Finally, we obtain $20\times 84=1680$ solutions in $X,Y,Z$.<|endoftext|> +TITLE: How many times $n$ must you play a game in which you have a $1/N$ chance of winning to have a better than 50% chance of winning at least once +QUESTION [6 upvotes]: I am having difficulty approaching the above problem, and would like a hint. I tried doing an inclusion exclusion argument: Let $A_{i}$ be winning the game on the i'th try, then by inclusion exclusion we have, for winning at least 1 game in n tries: +$$P\left(\bigcup_{i=1}^{n}A_i\right)=\frac{n}{N}-{n\choose 2}\frac{1}{N^2}+{n\choose 3}\frac{1}{N^2}+.....\pm \frac{1}{N^{n}} +$$ +unless I am missing something, I don't see how I could evaluate numerically the probability without knowing the value of $N$ +edit: We have spent a fair amount of time on stirling's formula, so maybe this would be a way to at least make sense of the choose function portion? + +REPLY [6 votes]: using André's suggestion, we want +$\left(1-\frac{1}{N}\right)^{n}\leq .5 \Rightarrow n\ln\left(1-\frac{1}{N}\right)\leq \ln(.5)\Rightarrow \frac{\ln\left(1-\frac{1}{N}\right)}{\ln(.5)}\leq n$ so for any $n\geq \left\lceil\frac{\ln(1-\frac{1}{N})}{\ln(.5}\right\rceil$ we will have a greater than 50% chance of winnning.<|endoftext|> +TITLE: Proving Fatou type lemma +QUESTION [6 upvotes]: Let $f_1, f_2, \cdots$ and $f$ be nonnegative lebesgue integrable functions on $\mathbb{R}$ such that $$\lim_{n \to \infty}\int_{-\infty}^y f_n(x)dx = \int_{-\infty}^y f(x)dx \; \; \text{ for each $y \in \mathbb{R}$}$$ and $$\lim_{n \to \infty}\int_{-\infty}^{\infty} f_n(x)dx = \int_{-\infty}^{\infty} f(x)dx $$ +Then I want to prove that $ \liminf_{n \to \infty} \int_{U}f_n(x)dx \geq \int_{U}f(x)dx\:$ for any open set $U$ of $\mathbb{R}.$ +I think this can done by proving $\lim_{n \to \infty}f_n(x) = f(x) $ and then using Fatou's lemma. But I am not able to prove $\lim_{n \to \infty}f_n(x) = f(x) $ . +Thank you in advance. + +REPLY [4 votes]: This is a particular instance of the Portmanteau theorem. +Proof of the statement. Let $F_n(x) = \int_{-\infty}^{x} f_n(t) \, dt$ and $F(x) = \int_{-\infty}^{x} f(t) \, dt$. Then for all test functions $\varphi \in C_c^{\infty}(\Bbb{R})$, integration by parts and the bounded convergence theorem shows that +\begin{align*} +\int_{\Bbb{R}} \varphi(x) f_n(x) \, dx +&= -\int_{\Bbb{R}} \varphi'(x) F_n(x) \, dx \\ +&\xrightarrow[n\to\infty]{} -\int_{\Bbb{R}} \varphi'(x) F(x) \, dx += \int_{\Bbb{R}} \varphi(x) f(x) \, dx +\end{align*} +Now let $U$ be open. Then there exists a sequence of test functions $(\varphi_n) \subset C_c^{\infty}(\Bbb{R})$ such that $0 \leq \varphi_n \leq 1$ and $\varphi_n \uparrow \mathbf{1}_U$ as $n\to\infty$. Thus for any fixed $m$, we have +\begin{align*} +\int_{\Bbb{R}} \varphi_m (x) f(x) \, dx +&= \liminf_{n\to\infty} \int_{\Bbb{R}} \varphi_m (x) f_n(x) \, dx \\ +&\leq \liminf_{n\to\infty} \int_{\Bbb{R}} \mathbf{1}_U(x) f_n(x) \, dx \\ +&= \liminf_{n\to\infty} \int_{U} f_n(x) \, dx. +\end{align*} +Now taking $m \to \infty$ and utilizing the monotone convergence theorem shows the claim. //// +Next, let us discuss a counterexample of pointwise convergence. Let +$$ f_n(x) = \begin{cases} 1 + \cos(2n\pi x),& 0 \leq x \leq 1 \\ 0,& \text{otherwise} \end{cases}, \qquad f(x) = \mathbf{1}_{[0,1]}(x) $$ +and define $(F_n)$ and $F$ as before. Then it is easy to check that $F_n$ converges to $F$ pointwise, but $f_n(x)$ does not converge if $x \in [0, 1] \setminus \Bbb{Q}$.<|endoftext|> +TITLE: What exactly is an "analytic function"? +QUESTION [7 upvotes]: This is for real analysis so I'm not worried about complex analytic functions. +The definition in my book just says: "A function f(x) which is represented by a power series with a positive radius of convergence is said to be 'real analytic at the origin', or simply 'analytic'." +To me it seems like this definition ends prematurely. Like when I was reading it I expected there to be an "if" and then a list of qualifications. +I guess I just don't really understand the rationale behind creating this definition in the first place. What would even be the point of representing a function by a power series that didn't have a positive radius of convergence? If the power series doesn't converge for any values of x, it's not really representative of anything, right? +I feel like this is actually obvious and I'm just overthinking it, but is this definition literally just giving a name to functions that can be represented as power series that actually converge for x values in some radius? There was an exercise earlier in the chapter before analytic functions were defined that asked us to find an infinite power series which represented a function; was that function analytic as well? +Thanks in advance and sorry if this is a silly question. + +REPLY [3 votes]: Why do we define any classification? So we can develop a language that can quickly say whether or not a structure exhibits a certain property. +Some functions can't be represented by a power series while others can. We have this definition to put functions into this analytic or not analytic category -- we don't have to say "$f$ can be expanded into a power series", we just have to say "$f$ is analytic." Another thing, a lot of the time we can have equivalent definitions of the same thing. This way we can avoid confusion by calling it by one name. +Of course, we make classifications only when they are useful. In this case, we can get a lot of very useful properties from the fact that a function is analytic: for example, smoothness follows from analyticity.<|endoftext|> +TITLE: Proof of an inequality in $\mathbb{C}$ +QUESTION [9 upvotes]: Let $z\in \mathbb{C}, n \geq 2$. Show this complex inequality +$$|z^n-1|^2\le |z-1|^2\left(1+|z|^2+\dfrac{2}{n-1}\Re{(z)}\right)^{n-1}$$ +For $n=2$ the inequality is easy to prove: +$$|z^2-1|^2\le |z-1|^2\left(1+|z|^2+2\Re{(z)}\right)$$ +$$\Longleftrightarrow |z+1|^2\le 1+|z|^2+2\Re{(z)}$$ +$$\Longleftrightarrow z+\overline{z}\le 2\Re{(z)}$$ +which is in fact an equality, thus is exact. +for $n=3$. +$$\Longleftrightarrow |z^2+z+1|^2\le (1+|z|^2+\Re{(z)})^2$$ +$$\Longleftrightarrow(z^2+z+1)(\overline {z^2}+\overline{z}+1)\le |z|^4+2|z|^2+1+\Re^2{z}+2\Re{(z)}+2|z|^2\cdot\Re{(z)}$$ +$$\Longleftrightarrow |z|^4+|z|^2+|z|^2(z+\overline{z})+1+(z+\overline{z})+(z^2+(\overline{z})^2)\le |z|^4+2|z|^2+1+\Re^2{z}+2\Re{(z)}+2|z|^2\cdot\Re{(z)}$$ +$$\Longleftrightarrow |z|^2+(\Re{(z)})^2\ge z^2+(\overline{z})^2=(z+\overline{z})^2-2|z|^2$$ +$$\Longleftrightarrow 3|z|^2\ge 3(\Re{(z)})^2$$ +It is clear +Is it true for a general $n$? + +REPLY [4 votes]: Firstly: $$\frac{z^n-1}{z-1}=z^{n-1}+z^{n-2}+\cdots1 \hspace{2cm}(1)$$ +Let $z_k, \; k=0,\cdots,n-1$ be the roots of unity, i.e. the roots of $z^n-1=0$, and take $z_0=1$. Using Vieta's relations for the sum of the roots (https://en.wikipedia.org/wiki/Vieta's_formulas), one has $\sum\limits_{i=0}^{n-1} z_i=0$. Using that $z_0=1$, we get $\sum\limits_1^{n-1}z_i=-1$. With $z_k = \cos\theta_k+I\sin\theta_k$, and separating the real and imaginary parts, we get: $$\sum\limits_{k=1}^{n-1}\cos\theta_k=-1$$ $$\sum\limits_{k=1}^{n-1}\sin\theta_k=0$$ +Equation $(1)$ can be written now as: $$\frac{z^n-1}{z-1}=(z-z_{n-1})(z-z_{n-2}\,)\cdots(z-z_1)$$, hence: $$\vert\frac{z^n-1}{z-1}\vert^2=\prod\limits_{k=1}^{n-1}\vert z-z_k\vert^2$$ +We use now the inequality between the geometric and arithmetic means (and $z=x+ I y$): +$$\left( \prod\limits_{k=1}^{n-1}\vert z-z_k\vert^2 \right)^{\frac{1}{n-1}}\le\frac{\sum\limits_{k=1}^{n-1}\vert z-z_k\vert^2}{n-1}=\frac{\sum\limits_{k=1}^{n-1}(1+x^2+y^2-2x\cos\theta_k-2y\sin\theta_k)}{n-1}=1+x^2+y^2+2\frac{x}{n-1},$$ +where we used the above-derived equalities for the sum of $\sin$ and $\cos$. +Putting it all together: +$$\vert\frac{z^n-1}{z-1}\vert^2 \le (1+\vert z\vert^2+2\frac{\Re(z)}{n-1})^{n-1}$$. +We can re-state the problem also as: +"For an arbitrary point in a plane, the squared product of distances to the vertices of any regular polygon inscribed in the unit circle (with $n$ vertices, that contains the point $(1,0)$, which is not to be included in the product), is at most equal to the power $n-1$ of the sum of the squared distance to the point $(-\frac{1}{n-1},0)$ and the constant $1-(\frac{1}{n-1})^2$ ". +The minimum of the right-hand side term reads: $\left[ 1-(\frac{1}{n-1})^2\right]^{n-1}$. Notably is the limit of it, for $n \to \infty$.<|endoftext|> +TITLE: Does rationality of $\cosh(nx)$ and $\cosh((n+1)x)$ imply rationality of $\cosh(x)$? +QUESTION [6 upvotes]: Suppose that $x\in\mathbb{R}^+$ and $n\in \mathbb{N}$. If $\cosh(nx)$ and $\cosh((n+1)x)$ are rational, can we show that $\cosh(x)$ is rational too? +I guess the following equalities should be useful: +$$\begin{eqnarray} +\cosh(x) &=& \cosh((n+1)x−xn)\\ +&=& \cosh((n+1)n).\cosh(nx) − \sinh((n+1)x).\sinh(nx)\\ +&=& \cosh((n+1)n).\cosh(nx) − \sqrt{\cosh^2((n+1)x) − 1}.\sqrt{\cosh^2(nx) − 1}. +\end{eqnarray}$$ + +REPLY [4 votes]: Suppose you have an $u>1$ such that both $u^n+u^{-n}$ and $u^{n+1} + u^{-n-1}$ are rational. +Then $u^n$ satisfies the equation with rational coefficients $(T - u^n)(T-u^{-n}) = T^2 - (u^n + u^{-n})T + 1 = 0$, so $u^n$ is algebraic and its only possible conjugate (over $\Bbb Q$) is $u^{-n}$. +Similarly, $u^{n+1}$ is algebraic and its only possible conjugate is $u^{-n-1}$. +Now, $u = u^{n+1}/u^n$, so $u$ is algebraic and its only possible conjugates are among $u^{-1},u^{2n+1},u^{-2n-1}$. +If $u^{2n+1}$ really is a conjugate of $u$, then so is $u^{(2n+1)^2}$. But since $u^{(2n+1)^2}$ is larger than any of $u,u^{-1},u^{2n+1},u^{-2n-1}$, this is impossible. +Therefore, $u$ has only one possible conjugate ($u^{-1}$), and so $u+u^{-1}$ is rational. +Taking $u = \exp x$, this shows that if $\cosh (nx)$ and $\cosh((n+1)x)$ are rational, $\cosh x$ is rational. + +You can easily generalize this to show that if $n,m$ are positive integers and $\cosh (nx)$ and $\cosh((n+m)x)$ are rational, then $\cosh(mx)$ is rational + +In fact, since we can view $\cosh(nx)$ as polynomials in $\cosh(x)$, we can interpret $\Bbb Q(\cosh(nx),\cosh((n+1)x)$ as a subfield of $\Bbb Q(\cosh x)$. Luroth's theorem then shows that this subfield is $\Bbb Q(\cosh x)$ itself, so there is a rational fraction $f_n$ in two variables with rational coefficients such that $\cosh(x) = f_n(\cosh(nx),\cosh((n+1)x)$. +This immediately implies that for almost all $x \in \Bbb C$, $\cosh(x) \in \Bbb Q(\cosh(nx),\cosh((n+1)x)$ (there is a finite number of exceptions where the denominator and numerator of $f_n$ vanish and $\cosh x$ may be in a bigger field).<|endoftext|> +TITLE: Should a metric always map into $\mathbf{R}$? +QUESTION [5 upvotes]: Typically you see the definition of a metric as a function which maps $X\times X\to\mathbf{R},$ but does this always have to be the case? +Motivating example: When you complete $\mathbf{Q}$ with the Archimedean metric you think of it as mapping into $\mathbf{R},$ but if you were to choose the $p$-adic metric instead it can only take values in $\{0\}\cup p^\mathbf{Z}$. These are elements of $\mathbf{Q}_p$ as well as $\mathbf{R}.$ The difference is then that either your metric maps into an ordered field or one where you can't define any ordering (but you still know that the distances can have different values). +Does not mapping a metric into $\mathbf{R}$ always lead to problems? And are the issues of the example above avoidable? + +REPLY [4 votes]: It depends on what you want for your "metric". It makes perfect sense as far as the definition goes to replace $\mathbb{R}$ by any totally ordered abelian group. But of course the properties won't be the same... +First note that when you suggest to interpret $|x|_p = p^{-v_p(x)}$ as an element of $\mathbb{Q}_p$, this has the pretty annoying inconvenient that if $x_n\to 0$ in $\mathbb{Q}_p$ then $|x|_p\to \infty$, so this interpretation does not give the $p$-adic topology. +Then let's assume that you define a metric as a map $X\times X \to \Gamma$ where $\Gamma$ is a totally ordered abelian group, with the usual axioms for metrics. This does give you a topology on $X$ by defining open balls as usual, and taking them as a basis for the topology. +But in general this topology need not be secound countable : this strongly uses the fact that in $\mathbb{R}$ there is a sequence $(\varepsilon_n)$ such that $\varepsilon_n>0$ and every $x>0$ satisfies $x<\varepsilon_n$ for some $n$ (take $\varepsilon_n = \frac{1}{n}$). So if you want to keep this very crucial property of metric spaces, you need to assume that such a sequence $\varepsilon_n$ exists in $\Gamma$. +More importantly, if you want that $X$ is the union of the open balls $B(a,n)$ with $n\in \mathbb{N}$ for any $a\in X$, then you need that $\Gamma$ is archimedian (ie for any $\varepsilon>0$ and any $x\in \Gamma$ there in $n\in \mathbb{N}$ such that $n\varepsilon>x$). But that actually implies that $\Gamma$ can be embedded in $\mathbb{R}$ as an ordered group. So if you want that property for metric spaces, then you may assume $\Gamma = \mathbb{R}$ without loss of generality.<|endoftext|> +TITLE: How to evaluate $\int_0^\infty \frac{e^{-x}+x-1}{x(e^{2x}-e^{-2x})}dx$? +QUESTION [5 upvotes]: We are unable to verify this this equality +$$ +4\int\limits_0^\infty \frac{e^{-x}+x-1}{x\left(e^{2x}-e^{-2x}\right)}\;\mathrm{d}x=\gamma+\ln\frac{16\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}\;. +$$ +where $\gamma$ is Euler's constant which is approximately $0.5772156\dots$, and $\Gamma(n)$ is the gamma function and the particular value for $\Gamma\left(\frac{1}{4}\right)=3.625690\dots$. + +REPLY [2 votes]: Another approach. We have $$\sum_{k\geq0}\frac{\left(-1\right)^{k}x^{k}}{k!}=e^{-x} + $$ hence $$e^{-x}-1+x=\sum_{k\geq2}\frac{\left(-1\right)^{k}x^{k}}{k!} + $$ and so $$I=4\int_{0}^{\infty}\frac{e^{-x}-1+x}{x\left(e^{2x}-e^{-2x}\right)}dx=4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k!}\int_{0}^{\infty}\frac{x^{k-1}}{e^{2x}-e^{-2x}}dx + $$ now note that $$ \begin{align} +\int_{0}^{\infty}\frac{x^{k-1}}{e^{2x}-e^{-2x}}dx= & \int_{0}^{\infty}x^{k-1}\sum_{m\geq0}e^{-\left(2m+1\right)2x}dx \\ = & \sum_{m\geq0}\int_{0}^{\infty}x^{k-1}e^{-\left(2m+1\right)2x}dx + \\ = + & \frac{\left(k-1\right)!}{2^{k}}\sum_{m\geq0}\frac{1}{\left(2m+1\right)^{k}} + \\ = + & \frac{\left(k-1\right)!}{2^{k}}\left(1-\frac{1}{2^{k}}\right)\zeta\left(k\right) + \end{align}$$ and so $$I=4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k2^{k}}\zeta\left(k\right)-4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k4^{k}}\zeta\left(k\right) + $$ and now we can use the identity, for $x\geq0, + $ $$\sum_{k\geq2}\frac{\left(-x\right)^{k}}{k}\zeta\left(k\right)=x\gamma+\log\left(\Gamma\left(x+1\right)\right) + $$ so finally we have $$I=\gamma+\log\left(\frac{16\pi^{2}}{\Gamma\left(\frac{1}{4}\right)^{4}}\right). + $$<|endoftext|> +TITLE: Sum over fourth power of the sine +QUESTION [5 upvotes]: I am considering the sum +$$ +A_m = \sum_{j=0}^m \sin^4\left(\frac{j}{m}\cdot\frac{\pi}{2}\right). +$$ +I think that for $m>1$ it holds +$$ +A_m = \frac{3m+4}{8}, +$$ +but I can't really get to it. + +REPLY [9 votes]: Start by linearizing $\sin^{4}(x)$ : +$$ \forall x \in \mathbb{R}, \, \sin^{4}(x) = \frac{1}{8}\Big( \cos(4x) - 4\cos(2x) + 3 \Big). $$ +This leads to : +$$ \sum_{j=1}^{m} \sin^{4}\Big( \frac{j\pi}{2m} \Big) = \frac{1}{8}\Bigg( \sum_{j=1}^{m} \cos\Big( \frac{2j\pi}{m} \Big) - 4 \sum_{j=1}^{m} \cos\Big( \frac{j\pi}{m} \Big) + 3m \Bigg). $$ +The sums $\displaystyle \sum_{j=1}^{m} \cos\Big( \frac{2j\pi}{m} \Big)$ and $\displaystyle \sum_{j=1}^{m} \cos\Big( \frac{j\pi}{m} \Big)$ can be computed using complex numbers. For $z \in \mathbb{C}$, $\Re(z)$ denotes the real part of $z$. Indeed, note that : +$$ \sum_{j=1}^{m} \cos \Big( \frac{2j\pi}{m} \Big) = \Re\Bigg( \sum_{j=1}^{m} e^{2ij\pi / m} \Bigg). $$ +and, since $\exp(2i\pi /m) \neq 1$ for all $m$, we have : +$$ \sum_{j=1}^{m} e^{2ij\pi / m} = e^{2i\pi / m} \frac{1 - \big( e^{2i\pi / m} \big)^{m} }{1 - e^{2i\pi / m} } = 0. $$ +Similarily : $\displaystyle \sum_{j=1}^{m} \cos\Big( \frac{j\pi}{m} \Big) = -1$. +As a consequence : +$$ \sum_{j=1}^{m} \sin^{4}\Big( \frac{j\pi}{2m} \Big) = \frac{1}{8}(3m+4). $$<|endoftext|> +TITLE: Why is the determinant of the 0x0 matrix equal 1? +QUESTION [16 upvotes]: I don't understand why the determinant of the 0x0 matrix equals 1. + +REPLY [15 votes]: The only linear map $\{0\}\to\{0\}$ is the identity map and the identity map has determinant $1$, because this is a required property of the determinant. +From a slightly different point of view, since the basis of $\{0\}$ is empty, the representative matrix of the unique endomorphism is the empty matrix (which is invertible and its inverse is, of course, the empty matrix). Invertible matrices should have nonzero determinant, shouldn't they?<|endoftext|> +TITLE: If $A$ is normal and upper triangular then it is diagonal +QUESTION [14 upvotes]: Let $A$ be a normal matrix in Mat$_{n\times n}(\mathbb C)$, if $A$ is upper triangular then it is diagonal + +(Normal means $AA^*=A^*A$, where $A^*$ is the conjugate transpose of $A$) +If I consider the diagonal of $AA^*$, let denote $(a_{ij})=A$ and $(â_{ij})_{i,j}=AA^*$ then, since $AA^*=A^*A$ +$â_{ii}=\sum\limits_{k=1}^na_{ik}\overline{a}_{ik}=\sum\limits_{k=1}^n\overline{a_{ki}}{a}_{ki}$ +$\implies\sum\limits_{k=1}^n|a_{ik}|^2=\sum\limits_{k=1}^n|a_{ki}|^2$. +If I take $i=n$ then it follows that $a_{in}=0, \forall 1\le i\le n-1$ and continuing in this manner the upper diagonal entries are zero, Is this correct ? +Can I show it in another way, because in a previous exercise I had to show that ''If A is normal and nilpotent then $A=0$'' so using this can I decompose $A$ into diagonal and nilpotent matrix, then show that the nilpotent part is zero ? + +REPLY [11 votes]: Let's prove this by induction. Suppose $A$ is an $n\times n$ upper-triangular, normal matrix. +If $n=1$, this is trivial. +If $n=2$, we have +$$ A = \begin{pmatrix} +a & b\\ +0 & c +\end{pmatrix}$$ +So +\begin{align*} +0 = AA^* - A^*A &= \begin{pmatrix} +a & b\\ +0 & c +\end{pmatrix}\begin{pmatrix} +\overline{a} & 0\\ +\overline{b} & \overline{c} +\end{pmatrix}- \begin{pmatrix} +\overline{a} & 0\\ +\overline{b} & \overline{c} +\end{pmatrix}\begin{pmatrix} +a & b\\ +0 & c +\end{pmatrix} \newline +&= \begin{pmatrix} +|a|^2 + |b|^2 & b\overline{c}\\ +\overline{b}c & |c|^2 +\end{pmatrix} - \begin{pmatrix} +|a|^2 & \overline{a}b \\ +a\overline{b} & |b|^2+|c|^2 +\end{pmatrix} \newline +&= \begin{pmatrix} +|b|^2 & b\overline{c}-\overline{a}b\\ +\overline{b}c -a\overline{b} & |b|^2 +\end{pmatrix}_. +\end{align*} +Therefore $|b|^2 = 0$. Thus $b = 0$ and $A$ is diagonal. +Now assume the result holds for $n-1$. +Then $$A = \begin{pmatrix} +a & B\\ +0 & C +\end{pmatrix}$$ +where $B$ is an $1\times (n-1)$ matrix and $C$ is an $(n-1)\times(n-1)$ upper-triangular matrix. +Since $A$ is normal, +\begin{align*} +0 = A^*A-AA^* &= \begin{pmatrix} +\overline{a} & 0\\ +B^* & C^* +\end{pmatrix}\begin{pmatrix} +a & B\\ +0 & C +\end{pmatrix} - \begin{pmatrix} +a & B\\ +0 & C +\end{pmatrix}\begin{pmatrix} +\overline{a} & 0\\ +B^* & C^* +\end{pmatrix} \newline +&= \begin{pmatrix} +|a|^2 & \overline{a}B\\ +aB^* & B^*B + C^*C +\end{pmatrix} - \begin{pmatrix} +|a|^2 + BB^* & BC^*\\ +CB^* & CC^* +\end{pmatrix} \newline +&= \begin{pmatrix} +-|B|^2 & \overline{a}B-BC^*\\ +aB^*-CB^* & B^*B + C^*C-CC^* +\end{pmatrix}_. +\end{align*} +Thus $|B| = 0$ and so $B=0$. Hence $C^*C-CC^* = 0$, i.e. $C$ is normal. By the hypothesis of induction, $C$ is diagonal. Finally, +$$ A = \begin{pmatrix} +a & 0\\ +0 & C +\end{pmatrix}_.$$ +So $A$ is diagonal.<|endoftext|> +TITLE: Can the boy escape the teacher for a regular $n$-gon? +QUESTION [14 upvotes]: This is related to Prove that the boy cannot escape the teacher +Suppose there is a boy in the center of a regular $n$-gon. The teacher is on the edge of the $n$-gon (but cannot leave the edge) and wants to capture the boy. At the beginning he is on a vertex. The teacher is $v(n)$ times faster than the boy. Which is the maximum $v(n)$ such that the boy can escape? (By escaping means he reaches the edge of the $n$-gon and the teacher is not there) +From the linked question we know $3 \le v(4) < 6$, and for $n= \infty$ (a circle) then I know a strategy such that $v(\infty) = \pi + 1$ suffice; I don't know if this is optimal. I also put convergence in the tags because my wild hypothesis is that the maximum $v(n)$ will converge to some value and it would be interesting to know which one! +Any cool way to solve this? + +REPLY [5 votes]: See all the details here; they are too long for a MSE answer. In summary, suppose the pool occupies a regular $n$-gon $P$. Let $\theta=\pi/n$, and let +$K$ be the largest integer between $0$ and $n$ such that +$$ + \sin(K\theta)-(K+n)\tan\theta\cos(K\theta) +$$ +is negative. Equivalently, $K$ is the floor of the unique root of +$\tan(x\theta)-(x+n)\tan\theta$ in $[1,n/2)$. Then +$$ + \cos((K+2)\theta)\leq\frac{2\sin(K\theta)}{(K+n)\tan\theta}-\cos(K\theta)<\cos(K\theta) +$$ +so we can define +$$ + \alpha=\frac12\left(K\theta+\cos^{-1}\left( + \frac{2\sin(K\theta)}{(K+n)\tan\theta}-\cos(K\theta)\right)\right), +$$ +Then the cutoff speed is $\lambda=1/\cos\alpha$. Explicitly, for speeds +$<\lambda$ the student wins regardless of starting positions. For speeds +$>\lambda$ the teacher wins provided the student starts in a regular +$n$-gon $Q$, where $Q$ is obtained from $P$ by rotating by $K\theta$ and +scaling by +$$ + s=\frac{\cos\alpha}{\cos(\alpha-K\theta)} +$$ +about the center of $P$. +As for a circular pool, the student must consider two cases in which the +teacher chases clockwise or anti-clockwise, and find the optimal path in +each. In fact from each vertex of $Q$ there are two equally optimal paths in each direction, as shown below (follow the link for other values of $n$). + +For a square we have $K=1$, $\alpha\approx1.397$ and $\lambda\approx5.789$. +For large $n$, $\theta$ is small and $n\tan\theta\approx\pi$. +Thus $\alpha\approx K\theta\approx\mu$, where $\mu$ satisfies +$$ + \tan\mu=\mu+\pi. +$$ +This agrees with the cutoff for a circular pool.<|endoftext|> +TITLE: You have to estimate $\binom{63}{19}$ in $2$ minutes to save your life. +QUESTION [43 upvotes]: This is from the lecture notes in this course of discrete mathematics I am following. +The professor is writing about how fast binomial coefficients grow. + +So, suppose you had 2 minutes to save your life and had to estimate, up to a factor of $100$, the value of, say, $\binom{63}{19}$. How would you do it? I will leave this (hopefully intriguing!) question hanging and maybe come back to the topic of efficiently estimating binomial coefficients later. + +Any ideas/hints on how to do it? + +REPLY [2 votes]: So we want to estimate: +$$\frac{63!}{19!44!} $$ +We do this by doing a geometric average approximation, replacing each factorial by it's geometric average, and the following approximations $9.5^{19} \approx 9^{20}$ and since it's a matter of life and death we expose we know the 10% and 5% interest tables by heart (useful since $22 = 20\cdot 1.10$ and $31.5 = 30\cdot 1.05$ so the exponents will copy over to the "percentage" part): +$$\frac{31.5^{63}}{ 9.5^{19} \cdot 22^{44}} = \frac{22\cdot 3^{63}\cdot10^{63}}{66\cdot 2^{44}\cdot10^{44}\cdot 3^{2\cdot 20}} = \frac{1}{3}\frac{3^{23}10^{19}}{2^{44}}$$ then cancelling by using $2^3 \approx 3^2 \approx 10$: +$$ \approx \frac{10^{19}}{2^{11}} \approx 2.5\cdot 10^{15}$$<|endoftext|> +TITLE: An integral related with the Riemann $\zeta$ function +QUESTION [6 upvotes]: I have to prove that: +$$ +\forall s>1,\qquad\int_0^\infty \sum_{k=1}^{\infty}\frac{1}{(k^s+1)^x+k^s}dx=\zeta(s). +$$ +I how do I find the closed form for this sum? +$$ +\sum_{k=1}^{\infty}\frac{1}{(k^s+1)^x+k^s} +$$ + +REPLY [3 votes]: Just to complete Ron Gordon's excellent answer, notice that: +$$ I(a,b)=\int_{0}^{+\infty}\frac{dx}{e^{x\log a}+b} = \frac{1}{\log a}\int_{0}^{+\infty}\frac{dz}{e^z+b}=\frac{1}{\log a}\int_{1}^{+\infty}\frac{dt}{t(t+b)}$$ +and: +$$ \int_{1}^{+\infty}\frac{dt}{t(t+b)}=\int_{0}^{1}\frac{du}{1+bu}=\frac{1}{b}\int_{0}^{b}\frac{dv}{1+v}=\frac{\log(1+b)}{b}$$ +through the substitutions $x=\frac{z}{\log a}$, $z=\log t$, $t=\frac{1}{u}$ and $u=\frac{v}{b}$. That leads to: +$$\int_{0}^{+\infty}\sum_{k\geq 1}\frac{dx}{(k^s+1)^x+k^s}=\sum_{k\geq 1}\frac{\log(1+k^s)}{k^s \log(1+k^s)} = \zeta(s)$$ +as wanted.<|endoftext|> +TITLE: $R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective, then is $R$ Noetherian? +QUESTION [8 upvotes]: Let $R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective, then is $R$ a Noetherian ring ? + +REPLY [12 votes]: Here's a counterexample. Let $R=\mathbb{Z}[2^{1/3},2^{1/9},2^{1/27},\dots]\subset\mathbb{R}$. Then $R$ is not Noetherian, since the ideal $(2^{1/3},2^{1/9},2^{1/27},\dots)$ is not finitely generated. But the only ring-homomorphism $f:R\to R$ is the identity. Indeed, any such $f$ must be the identity on $\mathbb{Z}$, and then $f\left(2^{1/3^n}\right)$ must be $2^{1/3^n}$ for each $n$ since $2^{1/3^n}$ is the only $3^n$th root of $2$ in $R$. + +More generally, any non-Noetherian subring $R$ of $\overline{\mathbb{Q}}$ works. Indeed, if $a\in R$, then it is a root of a nonzero polynomial $p(x)$ with integer coefficients. If $a\neq 0$, $p(x)$ can be taken to have nonzero constant term. The constant term of $p(x)$ is then in the ideal generated by $a$, since $p(a)=0$ and every other term of $p(a)$ is in the ideal generated by $a$. Thus every nontrivial ideal in $R$ contains a nonzero integer. It follows that any homomorphism $f:R\to R$ must be injective. +(In fact, any homomorphism $f:R\to R$ must also be surjective: it must send roots of any $p(x)\in\mathbb{Z}[x]$ to roots of $p(x)$, and there are only finitely many such roots. An injection from a finite set to itself is surjective, so the image of $f$ must contain all the roots of $p(x)$ in $R$.)<|endoftext|> +TITLE: How much proof is needed in such paper (Maths related)? +QUESTION [5 upvotes]: I'm writing a paper (report) regarding Euler's Number $\space e \space$ (even though he didn't discover it). +Within this paper, I show that: +$${d\over dx} {e^x} = {e^x}$$ +**NOTE: ** This is not what the whole paper is about. +However, the proof uses the fact that: +$$\space {d\over dx}\ln f(x) = {f'(x)\over f(x)}$$ +Do I need to prove this first? +Or can I just leave it as prerequisite knowledge before reading the paper? +**P.S. - ** I understand this isn't exactly a Maths question in a sense. But it is Maths related and it takes a Mathematician to answer. + +REPLY [4 votes]: You haven't really proven anything. You have used something that needs proving, to "prove" that the derivative of $\mathrm e^x$ is itself. It's really just a circular argument. +I would suggest that you prove it from first principles: +$$\mathrm f'(x) = \lim_{h \to 0} \left(\frac{\mathrm f(x+h)-\mathrm f(x)}{h}\right)$$ +In your case, $\mathrm f(x) = \mathrm e^x$ and so +\begin{eqnarray*}\frac{\mathrm d}{\mathrm d x}\mathrm e^x &=& \lim_{h \to 0}\left(\frac{\mathrm e^{x+h}-\mathrm e^x}{h}\right) \\ \\ +&=& \lim_{h \to 0}\left(\frac{(\mathrm e^{h}-1)\mathrm e^x}{h}\right) \\ \\ +&=& \mathrm e^x \cdot \lim_{h \to 0}\left(\frac{\mathrm e^{h}-1}{h}\right) +\end{eqnarray*} +To prove your statement from first principles you need only show that +$$\lim_{h \to 0}\left(\frac{\mathrm e^{h}-1}{h}\right)=1$$<|endoftext|> +TITLE: What is the domain of the successor function? +QUESTION [5 upvotes]: Before defining the natural numbers, how do you define the successor function? $S(x)=x\cup\{x\}$ is a function on _ into _? +In two textbooks I've seen, the authors introduce the function (before the naturals) by saying "for every set $x$, we define the successor $S(x)$...". But doesn't that imply that the domain of $S$ contains all sets? +Kinda funny that "what is the domain of the successor function?" is a title that does not meet this website's standards. (I guess if you click submit enough times...) + +REPLY [3 votes]: We can define the successor function on the class of all sets. In +fact, we may define $S \colon V \to V, x \mapsto x \cup \{x\}$, where +$V$ is the proper class of all sets. Note that this can be done within +ZFC: While $S$ does not exists as an object in ZFC, we have that $S = +\{ (x, x \cup \{x\}) \mid x \in V \}$ is a definable class and those +can be regarded as "virtual classes" in ZFC. Also note that for any +set $D$, the function $S \restriction D \colon D \to S"D, x \mapsto x +\cup \{x \}$ is a set and hence an object in ZFC. + +edit: I should probably remark that, while virtual classes +are a handy way to deal with proper classes in ZFC, there are some +limitations to this approach. For example, statements of the form "For +every virtual class ..." can in general not be formalized within ZFC +and there may be classes that are not virtual, i.e. are not a +collection of sets that satisfy a given, fixed formula. For example, +the modeling relation $ \models := \{ (\phi, x) \mid \phi(x) \text{ is +true } \}$ is provably not a virtual class.<|endoftext|> +TITLE: Can you determine the length of a curve by the lengths of its projections onto planes? +QUESTION [5 upvotes]: If $\,\Gamma \subset \mathbb R^n$ is $1$–rectifiable, then its Hausdorff measure is equal to its integral geometric measure. That is, +$$\displaystyle\mathcal H^1\left(\Gamma\right) = \int_{G\left({1,\mathbb R}^n\right)} \int_K \operatorname{Card}\left(\left\lbrace {y \in \Pi_K}^{-1}\left(\left\lbrace x\right\rbrace \right)\right\rbrace \right)\, {d\hspace{0.125ex}\mathcal H}^1\left(x\right)\, d \hspace{0.125ex}\Theta_{{1,\mathbb R}^n}\left(K\right),$$ +where $\operatorname{Card}(S)$ means the number of points in ${S,\Pi}_K$ denotes orthogonal projection onto ${K,\mathcal H}^1$ denotes the one–dimensional Hausdorff measure, $G\left({1,\mathbb R}^n\right)$ denotes the Grassmanian of unoriented lines through the origin in $\mathbb R^n$, and $\Theta_{{1,\mathbb R}^n}$ is the unique (up to suitable constant) finite Borel measure on $G\left({1,\mathbb R}^n\right)$ which is invariant under the action of the orthogonal group. +I would like to know if the following is true: +$$\displaystyle\mathcal H^1\left(\Gamma\right) = \int_{G\left({2,\mathbb R}^n\right)} \int_V \operatorname{Card}\left(\left\lbrace {y \in \Pi_V}^{-1}\left(\left\lbrace x\right\rbrace \right)\right\rbrace \right)\, {d\hspace{0.125ex}\mathcal H}^1\left(x\right)\, d \hspace{0.125ex}\Theta_{{2,\mathbb R}^n}\left(K\right),$$ +The more general question where the numbers $1$ and $2$ are replaced by $j$ and $k$ with $j +TITLE: 21 points on circumference of a circle must have at least 100 pairs separated by 120+ degrees. +QUESTION [5 upvotes]: Prove that at least 100 of the arcs determined by the pairs of these points subtend an angle not exceeding 120 degrees at the center. How do I prove this? Induction? Help please. Thanks. + +REPLY [2 votes]: Form a graph with 21 vertices corresponding to the points and edges between pairs of points more than 120 degrees apart. +This graph is triangle-free. +The number of edges in a triangle-free graph on 21 vertices is (uniquely) maximized for the bipartite graph on 10 + 11 vertices which has 110 edges. Therefore, there are at least ${21 \choose 2} - 110 = 210 - 110 = 100$ pairs of vertices not joined by an edge. Those pairs correspond to pairs of points at most 120 degrees apart. +This also explains what the maximal arrangements of points look like. All of them can be formed by taking two disjoint closed 120 degree arcs on the circle, and placing $10$ and $11$ points respectively inside the two open intervals that are the complement of those arcs.<|endoftext|> +TITLE: Painting the plane red and blue: Is it possible for each unit circumference to contain exactly $n$ blue points? +QUESTION [23 upvotes]: I recently stumbled upon the following problem: + +Consider the plane: You may color each point either red or blue. + Is there a way to color it such that each unit circumference (centred anywhere) contains exactly one blue point? And two? + +I solved it relatively easy: the "one" case has no solution, and the "two" case is solved placing the blue points in a set of parallel lines, at a distance of two from their neighbours. +Of course I couldn't resist the temptation to consider the $n$ case: For $n$ even, the solution is easily extended considering lines at a distance of $4/n$. + +My question is: any help for the odd $n$ case? + +If there is any justice in the world, there should be no solution, but I honestly don't know where to start. +EDIT: I have had this idea: if I have a solution for an $m$ case, and another for the $n$ case, such that they do not intersect (that is, whenever a point is blue in a solution it is not in the other) then simply superimposing the solutions yields a valid $m+n$ solution. It also stands to reason that if a $k$ solution exists, and contains an $m$ solution, then removing it yields a $k-m$ solution. +So all that I need to prove is that an odd $k$ solution must contain a $k-1$ solution, and therefore also a $1$ solution, which I have shown cannot exist, and therefore also the generic odd solution cannot exist. +For reference: I know geometry up to the basics of manifolds, analysis up to (but not including) Lebesgue measure and integration, and some group theory. +Here is the proof for the $n = 1$ case: +Obviously, there must be at least one blue point: otherwise any circumference would contain no blue points and break the condition. +Consider the circle centred in that blue point: it must have one blue point. We therefore have two blue points at a distance of 1, therefore there is a circle that contains both, and violates the condition. +Thus, there cannot be a successful configuration of blue points, QED. +I also have this other proof: +Take one blue point. +Consider all circles containing that point. +Those circles cannot have any other blue point, because they must have exactly one. +All those circles cover a disc of radius 2 centered in the blue point, and said disc cannot contain any other blue points. +Therefore the circle centered in the blue point contains no blue points and breaks the condition, QED. + +REPLY [2 votes]: Assuming the axiom of choice, there is nothing really preventing a solution with $n=3$ to exist (unlike the $n=1$ case where a single blue point forced some red all over the place) +Biject the set of unit circles onto the smallest ordinal with $\Bbb R$'s cardinality, $\alpha$, and do a transfinite induction on $\alpha$. +Let $\beta \in \alpha$, and suppose we have selected blue points for all the circles number $\gamma$ with $\gamma \in \beta$ without creating any circle with unit circumference that have four blue points. +If circle $\beta$ already has three blue points, we don't have to do anything. +If not, we will have to select between one and three points on it, while not introducing any circle with four blue points. +For any triplet of blue points, if their circle has unit circumference, then we can't add anymore blue point on them. But in total, there are always at most $\beta^3 <\alpha$ circles who already have three blue points, and each of those circles can only intersect circle number $\beta$ at at most two points, so those cannot cover our circle completely. +So we will always have room to choose the few necessary blue points without making any circle with four blue points.<|endoftext|> +TITLE: Reflection with respect to a parabola +QUESTION [8 upvotes]: I know how to find a reflection with respect to one of the axis or with respect to the origin, but let's say I want to find the reflection with respect to a parabola, how do I do it? +Let's say we have the standard $y = x^2$, in this case each point needs to be reflected by projecting them onto the parabola by finding the point whose tangent line at that point of the parabola and the line from the point to the parabola are perpendicular. Then the reflection would the point on the other side. So (0,1) goes to (0,-1). Is there a way to find a general formula? +I am thinking that since the parabola is $y = x^2$, I should be able to map the axis y=0 to $y=x^2$, and then any pair of points (x,y) and (x,-y), symmetric with respect to the y axis, would remain symmetric with respect to the parabola. But what is this mapping? +EDIT: +Is there a better way to define this reflection by only considering the x>0 half-plane, and adding the constraint that the reflected points must still be in the x>0 half-plane? Similarly to how the area under the y=0 axis for x>0 is reflected into the area over y=0 and x>0 for a reflection about the y=0 axis, I also would like the reflection for the area x>0 under the parabola to be reflected in a point x>0 inside the parabola. I am looking at a deformation of the plane that takes the y=0 line onto the parabola and drags all the points with it. Symmetric points after the deformation should match those before the deformation. + +REPLY [4 votes]: I will add the brutal (force) numerical solution ( in MATLAB). +% Author - Chenguang Zhang +% create a circle to be mirrored +theta = linspace(0, 2*pi, 50); +radius = 0.2; +as = 0.0 + radius * cos(theta); +bs = 0.5 + radius * sin(theta); + + +figure(1) +hold on +xx = linspace(-3,3,101); +plot(xx, xx.*xx, 'b-', 'LineWidth', 2) +plot(as, bs, 'ro') + +for ii = 1:length(theta) + a = as(ii); + b = bs(ii); + troots = roots([1 0 0.5-b -0.5*a]); + for i = 1:length(troots) + t = troots(i); + if(imag(t) ~= 0) + continue; + end + x = t; + y = t*t; + p_reflect = [2*x - a, 2*y - b]; + plot(p_reflect(1), p_reflect(2), 'bo') + end +end + +An image from multiple runs with different circles location and shape is below.<|endoftext|> +TITLE: The image of the transpose of $A^T$ is the orthogonal complement of its kernel +QUESTION [9 upvotes]: Suppose $V$ is a finite dimensional vector space over $\mathbb{K}$. Let A be a linear map. I am trying to prove that $$ImA^T=(KerA)^{\perp}$$ +I know one direction: $ImA^T \subset (KerA)^{\perp}$ but I don't know how to show the other direction. Can anyone help me? + +REPLY [6 votes]: If $x\notin{\rm im}(A^T)$, note that ${\rm im}(A^T)={\rm im}(A^T)^{\perp\perp}$ because $V$ is finite dimensional. So there exists +$x'\in{\rm im}(A^T)^\perp$ such that $ \langle x,x'\rangle\ne0$. +In fact, we have $x'\in\ker (A)$ because $A^TAx'\in{\rm im}(A^T)$, which implies +$$\langle Ax', Ax'\rangle=\langle x',A^TAx'\rangle=0\quad\Longrightarrow\quad +Ax'={\it 0}.$$ +Thus $x\notin \ker (A)^\perp$. + +REPLY [2 votes]: We need to show that $(Ker A)^\perp\subseteq ImA^T$. Taking orthogonal complements, this is equivalen tto $(ImA^T)^\perp\subseteq Ker A$. +Take $x\in (Im A^T)^\perp$. Since $V=Ker A\oplus(Ker A)^\perp$, we can write $x=y+z$, with $y\in Ker A$ and $z\in (Ker A)^\perp$. Then for all $v\in V$, +\begin{align*} +0&=\langle x,A^Tv\rangle=\langle y,A^Tv\rangle+\langle z,A^Tv\rangle=\langle Ay,v\rangle+\langle Az,v\rangle=\langle Az,v\rangle +\end{align*} +where the last equality follows from $y\in Ker A$. Using the above with $v=Az$ yields $Az=0$, and so $z\in KerA\cap(KerA)^\perp=0$, so $z=0$ and $x=y\in KerA$.<|endoftext|> +TITLE: On the linear combination of $\pm 1$ random variables +QUESTION [6 upvotes]: Let $X_1,\dots, X_n$ be i.i.d symmetric $\pm 1$ random variables, i.e. $X_j$ takes values in $\{-1,1\}$ with +$$\mathbb{P}(X_j=1)=\mathbb{P}(X_j=-1)=\frac{1}{2}.$$ +Let $a_1,\dots,a_n\in\mathbb{Z}$, define +$$X=\sum_{j=1}^n a_jX_j.$$ +How can we prove that +$$\mathbb{P}(X=0)=\frac{1}{2\pi}\int_0^{2\pi}\prod_{j=1}^n\cos(a_jt)dt.\quad(*)$$ +I noticed that the integrant in the right hand side of $(*)$ is the characteristic function of the random variable $X$, so I am thinking about using the inverse formula: +$$\frac{\mathbb{P}(X=a)+\mathbb{P}(X=b)}{2}+\mathbb{P}(a +TITLE: Finding the kernel of an epimorphism onto $S_3$? +QUESTION [6 upvotes]: Let $\Lambda$ denote the group with presentation $\langle a,b \mid + abab^{-1}a^{-1}b^{-1}\rangle$. We define the following epimorphism + from $\Lambda$ onto $S_3$: +$\theta: \Lambda(a,b) \rightarrow S_3$ using $a \mapsto (12)$ and $b + \mapsto (23)$. We have $(12)(23)(23)^{-1}(12)^{-1}(23)^{-1}=e$ + thus indicating that our homomorphism factors through the quotient + modulo the normal subgroup $H$ of $F$ generated by + $abab^{-1}a^{-1}b^{-1}$ and, by definition, $\Lambda=F/H$. Find an + explicit, finite presentation for the kernel, $\kappa$, of the + epimorphism + +I know the kernel of $\theta$ is the set of all elements of $\Lambda$ that are mapped to $e\in S_3$ and is a normal subgroup of $\Lambda$. I'm a little new to finding kernels so I'm not sure how to proceed. What's a good way to find the kernel of this epimorphism? + +REPLY [2 votes]: There is an algorithm known as the Reidemeister-Schreier algorithm for computing presentations of subgroups of finitely presented groups. I haven't got time to explain it right now and in any case you might do better to google it - there are lots of descriptions online. +Examples like this, where the index of the subgroup is small, can easily be done by hand. For subgroups of larger index, you can sue GAP or Magma. I was lazy and did your example by computer and found the presentation of the kernel: +$$\langle x,y,z \mid y^{-1}x^{-1}yzxz^{-1}, y^{-1}zxyx^{-1}z^{-1} \rangle,$$ +where $x=a^2$, $y=b^2$, and $z=ab^2a^{-1}$.<|endoftext|> +TITLE: Integral (Tanh and Normal) +QUESTION [15 upvotes]: I am trying to evaluate the following: The expectation of the hyperbolic tangent of an arbitrary normal random variable. +$\mathbb{E}[\mathrm{tanh}(\phi)]; \phi \sim N(\mu, \sigma^2)$ +Equivalently: +$\int_{-\infty}^{\infty} \mathrm{tanh}(\phi)\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{1}{2\sigma^2}(\phi-\mu)^2) d\phi$ +I've resorted to Wolfram Alpha, and I can sometimes (!) get it to evaluate the integral for $\sigma^2 = 1$. It gives: +$\exp(\mu) - 1$ for negative $\mu$ and $1-\exp(-\mu)$ for positive mu. +I have no idea how it got this, but it seems plausible as I've done some simulations (i.e. lots of draws from the normal distribution and then the mean of the tanh of the draws) and the formula it gives for the $\sigma^2 = 1$ case seems pretty close. +I cannot get it to evalute anything with a different variance. Mathematica also seems unwilling to compute the integral or the Fourier transform (which I thought might be a way forward) but this is tricker than the ones I know how to deal with. +Thanks! + +REPLY [2 votes]: $\def\f{\varphi} +\def\s{\sigma} +\def\m{\mu} +\def\p{\pi}$Let +$$f(\f) = \frac{1}{\sqrt{2\p}\s} e^{-(\f-\m)^2/(2\s^2)}.$$ +We have +\begin{align*} +I &= \int_{-\infty}^\infty f(\f)\tanh\f\, d\f += \int_0^\infty f(\f)\tanh\f\, d\f ++ \int_{-\infty}^0 f(\f)\tanh\f\, d\f \\ +&= \int_0^\infty (f(\f)-f(-\f))\tanh\f\, d\f. +\end{align*} +But +\begin{align*} +\tanh\f +&= \frac{1-e^{-2\f}}{1+e^{-2\f}} += (1-e^{-2\f})\sum_{k=0}^\infty (-1)^k e^{-2k\f} \\ +&= 1+2\sum_{k=1}^\infty (-1)^k e^{-2k\f}. +\end{align*} +It is a straightforward exercise to show that +\begin{align*} +J_k &= \int_0^\infty (f(\f)-f(-\f)e^{-2k\f} d\f \\ +&= \frac12 e^{-2k(\m-k\s^2)} +\left( +1+\mathrm{erf} +\left(\frac{\m-2k\s^2}{\sqrt2 \s}\right) +\right) +- \frac12 e^{2k(\m+k\s^2)} +\mathrm{erfc}\left(\frac{\m+2k\s^2}{\sqrt2\s}\right). +\end{align*} +One can show for large $k$ that $J_k$ is of order $1/k^2$. +Let +$$I_n = J_0 + 2\sum_{k=1}^n (-1)^k J_k.$$ +Thus, $I = \lim_{n\to\infty}I_n.$ +The zeroth approximation is +$$I_0 = J_0 = \mathrm{erf}\left(\frac{\m}{\sqrt{2}\s}\right).$$ +(This amounts to approximating $\tanh\f$ by $\mathrm{sign}(\f)$ in the defining integral.) +Below we plot $I_0$, $I_1 = J_0 - 2J_1$, and $\hat I$, an independent numerical approximation to $I$, as functions of $\m$ for $\s=1$. +On this scale $I_2$ would be indistinguishable from $\hat I$. + +Figure 1. $I_0$, $I_1$, and $\hat I$ (in red, orange, and black, respectively) vs $\m$ for $\s=1$. +Addendum: Interchanging summation and integration +Let $g(\f) = f(\f)-f(-\f)$. +Note that +$$g(\f)\sum_{k=1}^\infty (-1)^k e^{-2k \f}$$ +converges pointwise to +$$-\frac{g(\f)}{1+e^{2\f}}$$ +on $(0,\infty)$. +Also, +$$\left| +g(\f)\sum_{k=1}^n (-1)^k e^{-2k \f} +\right| +\le \left|g(\f)\right|.$$ +Since $|g(\f)|$ is integrable, interchanging summation and integration is justified by the dominated convergence theorem.<|endoftext|> +TITLE: Helmholtz decomposition of a vector field on surface +QUESTION [6 upvotes]: Does it make sense to do Helmholtz decomposition of a vector field defined on a surface or on a manifold? I am mostly interested in the surface case. I was trying to find a reference for this and found only a handful of them mostly from electromagnetics literature. e.g. +Scharstein, Robert W. "Helmholtz decomposition of surface electric current in electromagnetic scattering problems." System Theory, 1991. Proceedings., Twenty-Third Southeastern Symposium on. IEEE. +On the related note, is there a well defined curl operator for a (tangential) vector field defined on a surface? + +REPLY [3 votes]: The presence of a curl term in the Helmholtz decomposition is very much a three-dimensional phenomenon. While it is reasonable to define the curl of a vector field tangent to a surface, what you get is either a field normal to the sphere or a scalar field, so it's not going to help us decompose tangent vector fields. +The Hodge decomposition tells us that any vector field on a surface can be decomposed as $$ X = \nabla f + J \nabla g + \sum_{i=1}^m c_iY_i$$ where $f,g$ are functions, $J$ is rotation by $\pi /2$ in the tangent plane, $c_i$ are some real coefficients determined by $X$ and the $Y_i$ are some finite collection of curl-free, divergence-free vector fields (independent of $X$) which describe the relevant features of the topology of the surface. +Very roughly, there is one field $Y_i$ going around each independent "loop" in the surface - so a plane or a sphere would have none, an infinite cylinder would have one and a torus would have two. If you're interested in some search terms to learn the details: the $Y_i$ are vector fields dual to a basis of the space of harmonic one-forms, which provide representatives of every cohomology class. +In three-dimensional manifolds you get the more familiar +$$ X = \nabla f + \operatorname{curl} V + \sum_{i=1}^m c_iY_i. $$ +In higher dimensions, scalar fields and vector fields no longer suffice to describe the whole de Rham complex, so it's necessary to switch to talking about differential forms.<|endoftext|> +TITLE: (Elegant) proof of an inequality: $h(x) \geq 1- (1-\frac{x}{1-x})^2$, where $h$ is the binary entropy function +QUESTION [14 upvotes]: I am looking for the most concise and elegant proof of the following inequality: +$$ +h(x) \geq 1- \left(1-\frac{x}{1-x}\right)^2, \qquad \forall x\in(0,1) +$$ +where $h(x) = x \log_2\frac{1}{x}+(1-x) \log_2\frac{1}{1-x}$ is the binary entropy function. Below is a graph of the two functions. + +Of course, an option would be to differentiate $1,2,\dots,k$ times, and study the function this way — it may very well work, but is not only computationally cumbersome, it also feels utterly inelegant. (For my purposes, I could go this way, but I'd rather not.) +I am looking for a clever or neat argument involving concavity, Taylor expansion around $1/2$, or anything — an approach that would qualify as "proof from the Book." + +REPLY [2 votes]: $$h(x) = x \log_2\frac{1}{x}+(1-x) \log_2\frac{1}{1-x}\ge1- \left(1-\frac{x}{1-x}\right)^2$$ +If we let $x=\frac{1+y}{2}$ and push through the algebra, the claim is equivalent to: +$$1-\frac{1}{2}\log_2(1-y^2)-\frac{y}{2}\log_2\left(\frac{1+y}{1-y}\right)\ge1-\frac{4y^2}{(1-y)^2}$$ +which can be rearranged to: +$$\frac{8y^2}{(1-y)^2}\ge \log_2(1-y^2) + y\log_2\left(\frac{1+y}{1-y}\right)$$ +Now, using the well-known $\ln u\le u-1$: +$$\ln v^2\le v^2-1\implies \ln v \le \frac{v^2-1}{2}$$ +and taking $v=\frac{1+y}{1-y}$, this becomes: +$$\ln\left(\frac{1+y}{1-y}\right)\le \frac{2y}{(1-y)^2}$$ +Noting that $\log_2(1-y^2)\le0$, we can see +$$\log_2(1-y^2) + y\log_2\left(\frac{1+y}{1-y}\right)\le y\frac{1}{\ln2}\frac{2y}{(1-y)^2}=\frac{2}{\ln2}\frac{y^2}{(1-y)^2}\le \frac{8y^2}{(1-y)^2}$$ +and we are done for $y>0$. +For $y<0$ it's a little trickier, because the other $\log$ term does come into play - still working out a nice argument for how to bound things appropriately.<|endoftext|> +TITLE: How to show that a continous function $f:\mathbb{R}^m \to \mathbb{R}$ has a maximum? +QUESTION [6 upvotes]: My task is this: +Suppose $f:\mathbb{R}^m \to \mathbb{R}$ is a positive, continous function such that $\lim_{\mid \textbf{x}\mid \to \infty} f(\textbf{x}) = \textbf{0}$. +Show that $f$ has a maximum. +I am not sure exactly how to show this since the domain of $f$ is all of $\mathbb{R}^m$. According to the extreme value theorem, we need $A\subset \mathbb{R}^m$ to be closed, bounded and that $f:A\to\mathbb{R}$ to be continous (which it is over the entire $\mathbb{R}^m$). I am thinking that if we could somehow split up the domain into closed sets and take the union, one could probably show that $f$ had a maximum, but I am not sure if that's the right approach. Any help would be more than welcome! + +REPLY [2 votes]: Two ways: +First: +Compactifying $\mathbb{R}^m$ with one point and defining $f(\infty)=0$ makes this function continuous on a compact set, hence achieves a maximum somewhere. Since it is positive on $\mathbb{R}^m$ and $0$ on $\infty$, it must be on $\mathbb{R}^m$ and not on $\infty$. Therefore, the restriction back to $\mathbb{R}^m$ has a maximum. +Second: +Take $\epsilon:=f(0)/2$. Let $K$ be a compact set so big that out of it, $f(x)<\epsilon$ (this is possible by hypothesis). Since $f$ is smaller than $\epsilon$ outside of $K$, and $\epsilon$ is smaller than $f(0)$, what matters is all inside $K$. But $K$ is compact. + +REPLY [2 votes]: Here is the idea, which I leave to you to make rigorous. If $f$ is identically $0$, then the conclusion is clear, so suppose there is some $\alpha \in \mathbb{R}^{n}$ such that $L := f(\alpha) > 0$. Since $f(x) \to 0$ as $|x| \to \infty$, there exists an $r > 0$ such that $|x| > r$ implies $f(x) < L/2$; clearly, $r > \alpha$ since $f(\alpha) = L > L/2$. Thus, the maximum of $f$ (if it exists) must occur in the closed ball of radius $r$. This is a compact subset of $\mathbb{R}^{n}$, and so $f$ attains its maximum on this subset. +Edit: I suppose by positive you probably meant $f(x) > 0$ for all $x \in \mathbb{R}^{n}$, in which case one can avoid this business at the beginning and just choose $\alpha = 0$. Since the above shows that the claim still holds even if positive is replaced with the (slightly) weaker hypothesis of non-negative, I'll leave it as is.<|endoftext|> +TITLE: How many integers from $43523$ to $93107$ contain at least one digit $7$ +QUESTION [5 upvotes]: How many integers from $43523$ to $93107$ contain the digit $7$ at least once? +I know that if we had $43000$ to $93000$, we would subtract integers that do not contain digit $7$ from the total number: $50000 - (5\times9\times 9\times 9 \times 9)$ +But how to do it with the given numbers? + +REPLY [7 votes]: First idea: if you want the number of something from $42523$ through $93107$, it may be easier to count the ones from $0$ through $93107$ and subtract the ones from $0$ through $42522$. Let $N(n)$ be the number of numbers up to $n$ that contain at least one $7$. You want $N(93107)-N(43522)$. It helps to prefix the numbers with zero so all numbers have five digits. Second idea: $N(93107)$ consists of all the numbers that start with $7$, which is $70000$ through $79999$, plus all the others that contain at least one $7$. For the ones that start with $0$ through $8$, that is $8$ times as many as the four digit numbers that contain at least one $7$. This suggests a recursive algorithm. I leave the problem that you only want some of the numbers that start with $9$ to you. The idea is the same.<|endoftext|> +TITLE: Number of ways of visiting N places +QUESTION [7 upvotes]: A tourist wants to visit $N$ cities, each numbered from $1$ to $N$, but he wants to visit them in a weird order. +A weird order is such in which no city numbered $i$ is the $i$-th to visit in his travel itinerary. +He wants to visit each city exactly once. +What are the number of possible weird orders for his travel? + +REPLY [11 votes]: Considering that the sequence in which he visits these $n$ cities, is $$a_1, a_2, \dots, a_n$$ then $a_i \neq i$. This is a derangement, whose formula is given by $$!n=\left[\frac{n!}{e}\right]$$ where $[x]$ is the nearest integer function and $!n$ is the number of derangements for an sequence with $n$ elements. +This sequence also appears in the Online Encyclopedia of Integer Sequences, which also provides you with recurrence relations, namely $$!n=(n-1)[!(n-1)+!(n-2)]$$for the number of derangements. + +REPLY [3 votes]: Hint look up dearrangements. $D_n=N!(1-1\frac{1}{2!}...+\frac{(-1)^n}{n!})$<|endoftext|> +TITLE: Kunen exercise III.8.21 +QUESTION [7 upvotes]: Let $f: \omega_1\to \mathbb{R}$ be one-one. Let $g:[\omega_1]^2\to 2$ be such that for any $\alpha<\beta<\omega_1$, $g(\{\alpha, \beta\})$ is $0$ when $f(\alpha) +TITLE: My formula for sum of consecutive squares series? +QUESTION [8 upvotes]: I stumbled upon a specific series, who's Sum of squares of consecutive integers equals the sum of squares of the continuation of that consecutive integers. +For exmaple, this first number in the series results in: $3^2 + 4^2 = 5^2$. +The second results in: $10^2 + 11^2 + 12^2 = 13^2 + 14^2$. +The thrid: $21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2$. etc. +The formula for the series is: + +By using; $1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$ it is easy to prove that both side are the same. +Have I found something new? Useful? +I have a feeling that this is the only series with this specific property (consecutive sum of squares equal continuation of consecutive sum of squares), but I am unsure how to prove that. +Any comments would be appreciated. + +REPLY [3 votes]: Here are the examples of $\sum_{j=a}^{b} j^2 = \sum_{j=b+1}^{c} j^2$ where the limits are less than $1000$. +There are examples that do not follow your pattern. But I don't know whether there are other counterexamples beyond $1000$. + sum a b c L R + 25 3 4 5 2 1 true + 365 10 12 14 3 2 true + 2030 21 24 27 4 3 true + 7230 36 40 44 5 4 true + 11900 18 34 42 17 8 false + 19005 4 38 48 35 10 false + 19855 55 60 65 6 5 true + 42419 12 50 63 39 13 false + 45955 78 84 90 7 6 true + 94220 105 112 119 8 7 true + 176460 136 144 152 9 8 true + 308085 171 180 189 10 9 true + 379525 60 110 135 51 25 false + 508585 210 220 230 11 10 true + 802010 253 264 275 12 11 true + 963295 16 142 179 127 37 false + 1217450 300 312 324 13 12 true + 1254539 67 159 198 93 39 false + 1789515 351 364 377 14 13 true + 2558815 406 420 434 15 14 true + 3572440 465 480 495 16 15 true + 4884440 528 544 560 17 16 true + 6556305 595 612 629 18 17 true + 8657445 666 684 702 19 18 true +11265670 741 760 779 20 19 true +14467670 820 840 860 21 20 true +18359495 903 924 945 22 21 true<|endoftext|> +TITLE: Can you use both sides of an equation to prove equality? +QUESTION [31 upvotes]: For example: + +$\color{red}{\text{Show that}}$$$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$ + +In high school my maths teacher told me + +To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. + +So starting from the LHS: $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{4\cos^2(x)-2}{\cos^2(x)}=4-2\sec^2(x)$$ $\large\fbox{}$ +At University, my Maths Analysis teacher tells me + +To prove a statement is true, you must not use what you are trying to prove. + +So using the same example as before: +LHS = $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{2\Big(2\cos^2(x)-\left[\sin^2(x)+\cos^2(x)\right]\Big)}{\cos^2(x)}=\frac{2(\cos^2(x)-\sin^2(x))}{\cos^2(x)}=\bbox[yellow]{2-2\tan^2(x)}$$ +RHS =$$4-2\sec^2(x)=4-2(1+\tan^2(x))=\bbox[yellow]{2-2\tan^2(x)}$$ +So I have shown that the two sides of the equality in $\color{red}{\rm{red}}$ are equal to the same highlighted expression. But is this a sufficient proof? +Since I used both sides of the equality (which is effectively; using what I was trying to prove) to show that $$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$ +One of the reasons why I am asking this question is because I have a bounty question which is suffering from the exact same issue that this post is about. + +EDIT: +Comments and answers below seem to indicate that you can use both sides to prove equality. So does this mean that my high school maths teacher was wrong? +$$\bbox[#AFF]{\text{Suppose we have an identity instead of an equality:}}$$ $$\bbox[#AFF]{\text{Is it possible to manipulate both sides of an identity to prove that the identity holds?}}$$ +Thank you. + +REPLY [5 votes]: Unfortunately your high-school teacher (and some of the other answers) is wrong. It is false that proving something of the form "$A = B$" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side, not to say necessary. +Not necessary +Let us first deal with the false misconception that you can only go in one direction. Suppose you have proven the following, where $A,B,C$ are any expressions: +  $A = C$. +  $B = C$. +Then you can use the second sentence to substitute $C$ for $B$ in the first to obtain: +  $A = B$. +This is logically valid because "$=$" means "is exactly the same as". +Not always possible +Let us now consider an example where it is simply impossible to manipulate from one side to the other to prove an equality! +$\def\zz{\mathbb{Z}}$ +$\def\qq{\mathbb{Q}}$ +$\def\rr{\mathbb{R}}$ +Theorem: +  Take any $x \in \rr$ such that $x^2 \le 0$. Then $x = 0$. +Proof: +  $x = 0$ or $x > 0$ or $x < 0$.   [by trichotomy] +  If $x > 0$: +    $x^2 = x \times x > 0$.   [by positive multiplication] +    This contradicts $x^2 \le 0$. +  If $x < 0$: +    $0 < -x$.   [by subtraction] +    $x^2 = (-x) \times (-x) > 0$.   [by positive multiplication] +    This contradicts $x^2 \le 0$. +  Therefore $x = 0$. +Here "positive multiplication" denotes "multiplying by a positive real", which preserves the inequality sign. Similarly subtraction preserves an inequality. The above theorem cannot be proven directly by algebraic manipulation from one side "$x$" to the other "$0$", simply because the only given condition is an inequality. +Similarly the construction of $\sqrt{2}$ in elementary real analysis, as shown below, does not permit a proof by algebraic manipulation. +Theorem: +  Let $S = \{ r : r \in \qq_{\ge 0} \land r^2 \le 2 \}$. +  Then $S$ is non-empty and has an upper bound in $\rr$. +  Let $x = \sup_\rr(S)$. +  Then $x^2 = 2$. +Proof: +  [Exercise! Or see a good textbook like Spivak's.] +General identities +To address the new sub-question in blue, it suffices to generalize the above theorem to the cube-root of arbitrary real numbers, namely: + +$\sup( \{ r : r \in \qq \land r^3 \le x \} )^3 = x$ for any $x \in \rr$. + +This is an identity but cannot be proven by direct manipulation. You may not be satisfied with this counter-example, but in higher mathematics this kind of identity is in fact the usual kind that we are interested in, rather than identities that can be proven by algebraic manipulation (which are usually considered trivial). Here are some more examples: + +$\lceil \frac{m}{n} \rceil = \lfloor \frac{m-1}{n}+1 \rfloor$ for any $m \in \zz$ and $n \in \zz_{>0}$. +$\sum_{k=0}^{n-1} 2^k = 2^n-1$ for any $n \in \zz_{>0}$. + +Nevertheless, if you desire an arithmetic identity, the question becomes much more interesting and depends heavily on what you mean by "arithmetic" and what axioms you are allowed to use. Tarski's problem asked whether an arithmetic identity concerning positive integers can be proven using only the basic high-school identities about them, and Wilkie gave an explicit and simple identity that cannot be so proven, basically because any proof needs to use subtraction and hence negative integers.<|endoftext|> +TITLE: A bipartite graph like $G(X,Y)$ such that $|X|=|Y|=k$ and $\delta(G) \gt \frac {k}{2}$ is Hamiltonian. +QUESTION [5 upvotes]: A bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint sets $X$ and $Y$ (that is, $X$ and $Y$ are each independent sets) such that every edge connects a vertex in $X$ to one in $Y$. +A Hamiltonian cycle, also called a Hamiltonian circuit, Hamilton cycle, or Hamilton circuit, is a graph cycle (i.e., closed loop) through a graph that visits each node exactly once. +A Hamiltonian graph is a graph which contains a Hamiltonian cycle. +Now Prove that: A bipartite graph like $G(X,Y)$ such that $|X|=|Y|=k$ and $\delta(G) \gt \frac {k}{2}$ is Hamiltonian. +Note: I tried to prove this question the same as proving Dirac's Theorem but it didn't work! I don't reach the contradiction. Also I tried to prove it using Ore's theorem but it doesn't fit the condition of using Ore's theorem. + +REPLY [3 votes]: As in Dirac, build to a maximal counterexample. That is, let $G$ be a bipartite graph with $\delta(G) > \frac{k}{2}$ and no Hamiltonian cycle. If any edge can be added to the graph that preserves all these properties, then do so. +Since $G$ is now a maximal counterexample, it has a Hamiltonian path $x_0 \cdots x_n$ (without loss of generality, $x_0 \in X$). Applying the pigeonhole principle to the high minimum degree (I leave the details to you), there are vertices $u \in X$ and $v \in Y$ such that + +$x_0 \sim v$, +$x_n \sim u$, and +$u \sim v$. + +The cycle $u v x_0 \cdots x_n u$ is Hamiltonian, which is a contradiction.<|endoftext|> +TITLE: How can mathematics work in wildly different set theories? +QUESTION [7 upvotes]: There are several set theories, e.g. ZFC and NF, which often have different axioms or are even outright contradictory. And yet most of other branches of mathematics, e.g. abstract algebra or topology, even though they use the concept of set, can work in any of these theories. How is this possible? +Why isn't abstract algebra based on ZFC be different than that based on NF, even though they both have different set axioms? + +REPLY [4 votes]: It is not strictly true that mathematics turns out the same in both theories. For instance, since NF refutes Choice, and it's unclear if something like the Ultrafilter Lemma is consistent, the Completeness Theorem for first order logic is not known to be provable in NF or any of its extensions. There are categories $\mathcal{C},\mathcal{D}$ that are "small" in the traditional sense, but for which the functor category $\mathcal{D}^\mathcal{C}$ is not an exponential. Not every group can be the quotient of a free group. And moving to NFU, which can support Choice and easily be extended to match ZFC in consistency strength, doesn't fix all of these examples. +But note that there's a theory called KF which is axiomatized by extensionality, pairing, union, power set, and a scheme of stratified $\Delta_0$ separation. KF+"There is a Dedekind infinite set" (we'll call this "KFI") is pretty weak, but strong enough to talk about a pretty surprising amount of arithmetic and analysis. Now observe that KFI+"There is a universal set" is NF; and KFI+Regularity+Full Separation is, more or less, Zermelo. The fact that both theories are extensions of a common theory which is adequate to a good deal of ordinary mathematics should make it clear why they agree so well on the basics, even when the extensions are incompatible.<|endoftext|> +TITLE: On those integers $n>1$ such that any commutative ring with identity having exactly $n$ ideals is a PIR +QUESTION [7 upvotes]: Convention : All rings are commutative with unity unless stated otherwise. By ideals we will mean to include $\{0\}$ and $R$ also. +Let us call an integer $n>1$ a "principal number" if any ring $R$ having exactly $n$ ideals is a PIR (from this answer we know that for any $n>1$ there exists a corresponding PIR, here we want all the corresponding rings be PIR). I know that any ring $R$ having $5$ or less ideals is a PIR, so certainly any $n \le 5$ is a principal number. +My question is: + +Are there infinitely many principal numbers? Are there infinitely many positive integers which are not principal number? Can we in any way characterize the principal numbers? + +Please help. Thanks in advance. + +REPLY [4 votes]: To show that there are infinitely many positive integers which are not principal it suffices to exhibit infinitely many positive integers $n$ such that there exists a non-principal ring with $n$ ideals. This is straightforward and can be done in many ways; here's one. +Let $R = \mathbb{F}_q[x, y]/(x^2, xy, y^2)$. This ring is not principal because $m = (x, y)$ is not principal, and it has exactly $q + 4$ ideals, as follows. Beside the unit ideal $(1)$, these ideals correspond to subspaces of $\text{span}(x, y)$, of which there are $q + 3$: $1$ zero-dimensional, $q + 1$ one-dimensional, and $1$ two-dimensional. +This means that $n = q + 4$ is not principal whenever $q$ is a prime power. This sequence begins +$$6, 7, 8, 9, 11, 12, 13, 15, 17, 20, \dots$$ +By taking the product of the rings above with the rings $\mathbb{F}_q[x]/x^k$ we can furthermore show that $n$ isn't principal whenever it is divisible by a number of the form $q + 4$, $q$ a prime power, which I believe more or less already implies that almost all positive integers aren't principal (in the sense that the principal positive integers have density $0$). +Various modifications of this construction can be done but I haven't been able to get as fine a control on the resulting number of ideals as I would like. I suspect it might even be true that every positive integer $n \ge 6$ is non-principal but am pretty far from showing this. Right now the smallest two numbers I don't know the status of are $n = 10, 19$.<|endoftext|> +TITLE: Let $f: [0, 1] \to \mathbb{R}$ s.t $f(0)=f(1)=0$ then measure of $A = \{h \in [0, 1] \mid \exists x \text{ such that }f(x+h) =f(x)\} \geq 1/2$. +QUESTION [5 upvotes]: Let $f:[0,1]\to\mathbb R$ be a continuous function s.t. $f(0)=f(1)=0$. Let $$A = \{h \in [0, 1] \mid \exists x \text{ such that }f(x+h) =f(x)\}.$$ Show that set $A$ has Lebesgue measure $\geq 1/2$. + +I have no clue to prove the claim. Can anyone give me some detail hints? Thank you in advance. + +REPLY [2 votes]: First let's show that $A$ is Borel. Given $\epsilon>0$, let $A_\epsilon=\left\{h\in[0,1]:\exists x\in[0,1-h]\text{ such that }|f(x)-f(x+h)|<\epsilon\right\}$. Since $f$ is continuous, we can consider only rational numbers in the definition of $A_\epsilon$, that is, if $\left\{q_n:n=1,2,\ldots\right\}$ denote the rationals in $[0,1]$, we have +\begin{align*} +A_\epsilon&=\left\{h\in[0,1]:\exists n\text{ s.t. }q_n+h\leq 1\text{ and }|f(q_n)-f(q_n+h)|<\epsilon\right\}\\ +&=\bigcup_{n=1}^\infty\left\{h\in[0,1-q_n]:|f(q_n)-f(q_n+h)|<\epsilon\right\} +\end{align*} +Each set in the union in the RHS is Borel (because $h\mapsto f(q_n+h)$ is continuous), so each $A_\epsilon$ is Borel, and thus $A=\bigcap_{k=1}^\infty A_{1/k}$ is Borel. +Now extend $f$ to $[0,2]$ by setting $f(x)=f(x-1)$ for each $x\in[0,2]$. Let $M\in[0,1]$ be a point of maximum for $f$ and $m\in[0,1]$ be a point of minimum for $f$ (so they are also points of maximum and minimum for the extension of $f$). +Given $h\in[0,1]$, we have $f(M+h)\leq f(M)$ and $f(m+h)\geq f(m)$, so by Intermediate Value Theorem there exists a point $x\in[0,1]$ between $M$ and $m$ such that $f(x+h)=f(x)$. We have two cases: + +If $x+h\in[0,1]$, then $h\in A$. +If $x+h\not\in[0,1]$, i.e., $x+h\in[1,2]$. Denote $y=x+h-1$, so that $y\in[0,1]$, $y+(1-h)=x\in[0,1]$, and $f(y+(1-h))=f(x)=f(x+h)=f(x+h-1)=f(y)$, therefore $1-h\in A$. + +Anyway, for each $h\in[0,1]$ we have either $h\in A$ or $1-h\in A$. Denoting by $\chi_B$ the characteristic function of a set $B$, this implies that $\chi_{[0,1]}(h)\leq \chi_A(h)+\chi_{A}(1-h)$. Integrating this over $[0,1]$ yields $1<2\lambda(A)$ (to integrate the last term use invariance of $\lambda$ under the map $h\mapsto(1-h)$), which is what we wanted. + +If $f$ is positive, then we indeed have $A=[0,1]$. Indeed, in this case, for each $h\in[0,1]$ we have $f(0)\leq f(h)$ and $f(1-h)\geq f(1)$, so again there is some point $x$ between $0$ and $1-h$ for which $f(x)=f(x+h)$, so $h\in A$. +An example to see that the inequality $\lambda(A)\geq 1/2$ is optimal in the general case is $f(x)=\sin(2\pi x)$, in which case $A=[0,1/2]$.<|endoftext|> +TITLE: Does the series converge +QUESTION [5 upvotes]: We know that all series of the following form diverge: \begin{equation} + S_k = \sum_{n=\left\lceil \mathrm{e}^k \right\rceil}^\infty \frac{1}{n (\ln n) (\ln \ln n)\dots(\ln^k n)} +\end{equation} +where the notation $\ln^k$ is function composition. Additionally, this series is interesting because for large $k$, it diverges very slowly. +Then does the following series converge? \begin{equation} + S = \sum_{n=1}^\infty \frac{1}{f(n)} +\end{equation} +where $f(n)$ is defined recursively by the monotonic function \begin{equation} +f(n) = \begin{cases} + n & \text{if }n \le \mathrm{e} \\ + n \cdot f{\left(\ln n\right)} & \text{otherwise} +\end{cases} +\end{equation} +The first few terms of this series are \begin{equation} +1 + \frac{1}{2} + \frac{1}{3 \ln 3} + \frac{1}{4 \ln 4} + \dots + \frac{1}{16 (\ln 16) (\ln \ln 16)} + \dots +\end{equation} +and the maximum power of $\ln$ in the denominator gradually increases. +This series should eventually grow slower than all $S_k$ (it is in some kind of sense a limit) because the terms will eventually drop below those of $S_k$ for any particular $k$. But at the same time it seems to grow faster than all series of the form \begin{equation} + T_k = \sum_{n=\left\lceil \mathrm{e}^k \right\rceil}^\infty \frac{1}{n (\ln n) (\ln \ln n)\dots(\ln^{k-1} n){(\ln^k n)}^2} +\end{equation} which are known to converge, albeit also very slowly. + +REPLY [6 votes]: Since $f$ is monotonic, we can use the integral comparison. For any fixed $u \geqslant e$, we have +$$\int_u^{e^u} \frac{dx}{f(x)} = \int_u^{e^u} \frac{1}{f(\ln x)} \frac{dx}{x} = \int_{\ln u}^u \frac{dt}{f(t)}.$$ +Since $\int_1^e \frac{dx}{f(x)} > 0$, we have +$$\int_1^{\infty} \frac{dx}{f(x)} = \sum_{k = 0}^{\infty} \int_{x_k}^{x_{k+1}} \frac{dx}{f(x)} = \sum_{k = 0}^{\infty} \int_1^e \frac{dx}{f(x)} = +\infty,$$ +where $x_0 = 1$ and $x_{k+1} = e^{x_k}$ for all $k$. Thus the integral, and hence the series diverges. Very very slowly.<|endoftext|> +TITLE: Absolute value graph sketching: $||x-1|-1|$ +QUESTION [9 upvotes]: Where would you start if you were told to plot: +$$||x-1|-1|$$ +I looked at just $f(x) = |x-1|$ and noticed that the two equations are: $\pm (x-1)$ for $x \geq 1$ and $x < 1$. Extrapolating then: $\pm(\pm(x-1)-1)$, but how would I know which regions which equation would work for? + +REPLY [31 votes]: 1) $y=|x|$ + +2) $y=|x-1|$ + +3) $y=|x-1|-1$ + +4) $y=||x-1|-1|$<|endoftext|> +TITLE: Problem about a compact operator $T:l^p\rightarrow l^p$ +QUESTION [6 upvotes]: I have to solve this problem. + +Let $\{\lambda_n\}$ be a sequence of real number such that $\lim_{n\rightarrow\infty}\lambda_n=0$ and consider the operator $T:l^p\rightarrow l^p$, $1\leq p\leq \infty$, defined by +$$T(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_n x_n,\ldots\} +$$ + Prove that $T$ is compact. + +My attempt is the following. +I have considered the operator $$ +T_N:l^p\rightarrow l^p,\quad T_N(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_N x_N,0,0,\ldots\}. +$$ +$T_N$ is clearly linear and it is also compact. In fact given any bounded sequence $\{x^{(n)}\}_{n\geq 1}$ in $l^p$, the sequence $\{T_Nx^{(n)}\}_{n\geq 1}$ is bounded in $\mathbb{R}^N\subset l^p$. Then, since every bounded sequence in $\mathbb{R}^N$ has a convergent subsequence, it follows that $T_N$ is compact. +Now, I have to prove that $T$ is compact. For this I proved that $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$. In fact we have: +$$ +||T_N-T||=\sup_{||x||=1}||(T_N-T)x||=\sup_{||x||=1}||T_Nx-Tx||=\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}} +$$ +and, given $\epsilon>0$, there exists $\bar n$ such that $|\lambda_n|<\epsilon$, for every $n>\bar n$. So if we take $N>\bar n$, we get +$$ +\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}}\leq\epsilon\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|x_i|^p\right)^{\frac{1}{p}}\leq\epsilon +$$ +Therefore $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$ implies that $T$ is compact. +My questions are: +(1) is the proof of the compactness of the operator $T_N$ correct? +(2) is the conclusion correct? +Thanks + +REPLY [4 votes]: Yes, it is correct. The operators $T_N$ are of finite rank (hence compact) and converge to $T$. Thus, $T$ is compact as well.<|endoftext|> +TITLE: Left Invariant vector field on SO(3) +QUESTION [5 upvotes]: I have a Lie group, namely on SO(3), i.e. $SO(3,\mathbb{\mathbb{R}})=\left\{ A\in GL\left(3,\mathbb{R}\right)\mid A^{T}A=\mathbb{1},\,\det\left(A\right)=1\right\}$. +I have a Left action $L_g$ and I know that by definition a left invariant vector field is a vector field such that +$$(L_g)_{\ast}(X_e)=X_g$$ +I worked out few thing with the tangent space at the identity and I arrived at formulating the idea that this should be a left invariant vector field on $SO(3)$: +$$L_{\mathbf{1}}=y\frac{\partial}{\partial z}-z\frac{\partial}{\partial x}$$ +But now how to I effectively demonstrate that is left invariant. Can anybody show me the computation that once in a lifetime must be seen? Or at least give a refernce that actually shows the computation technique? + +REPLY [4 votes]: This kind of formula only works when you have the action of the lie group $G = SO(3)$ on some manifold $M$, in this case, the euclidean space $\mathbb{R}^3$. Let then $G$ act on $M$ by $(g,x) \mapsto gx$. +A left-invariant vector field $V$ in $G$ induces a vector field on $M$ such that the flow in $M$ is: $${\gamma(t,x) = \exp(tV)x,}$$ where $\exp: \mathfrak{g} \rightarrow G$ is the exponential map. Define the induced vector field $\tilde{V}$ on $M$ by: $${\tilde{V} = \frac{d}{dt}\exp(tV)x|_{t=0}.}$$Then we have defined a map $\mathfrak{g} \rightarrow \mathfrak{X}(M)$ by $V \mapsto \tilde{V}$. +In your case, take an element of $\mathfrak{so}(3)$ (the lie algebra of $SO(3)$, e.g.: the tangent space at the identity), namely: +$$X = \begin{bmatrix} +0 & 0 & 0\\ +0 & 0 & -1\\ +0 & 1 & 0 +\end{bmatrix}$$ +(You can see that this is an element of $\mathfrak{so}(3)$ because it is an antissymetric matrix). +Then your induced vector field is $\tilde{X} = -z\frac{\partial}{\partial y} + y \frac{\partial}{\partial x}$.<|endoftext|> +TITLE: Significance of the Riemann hypothesis to algebraic number theory? +QUESTION [5 upvotes]: Of course, the truth of the Riemann hypothesis is a central question in analytic number theory. Does its truth/falsehood have important consequences in purely algebraic number theory as well? Moreover, are there any known methods of studying or "attacking" the Riemann hypothesis that are more algebraic than analytic? + +REPLY [10 votes]: You can find here http://www.claymath.org/sites/default/files/sarnak_rh_0.pdf +a short but very enlightening account on RH and GRH by P. Sarnak , « Problem of the Millennium : The Riemann hypothesis », 2004 (building on E. Bombieri’s official presentation of the problem). Your question , as well as many others on this subject, could be placed under the common headline « What is so interesting about the zeroes of the Riemann Zeta function ? » (as termed by Karmal, April 24). I can see that a large majority of the answers concentrate on applications to the distribution of primes, which is natural since Riemann himself started the subject, but one has the right to marvel at e. g. how GRH can lead to an information on the arithmetic of elliptic curves (Serre’s result recalled by Sarnak op. cit.). Even more wonderful is the parallel with the Zeta function of a curve (Weil’s theorem recalled by Jake) and more generally, of a smooth projective variety (Weil’s conjectures, proved by Deligne and others) over a finite field. This is not just an analogy, but a manifestation of the unity of mathematics : « What was really eye-catching [in the Weil conjectures], from the point of view of other mathematical areas, was the proposed connection with algebraic topology. Given that finite fields are discrete in nature, and topology speaks only about the continuous, the detailed formulation of Weil (based on working out some examples) was striking and novel. It suggested that geometry over finite fields should fit into well-known patterns relating to Betti numbers, the Lefschetz fixed-point theorem and so on. The analogy with topology suggested that a new homological theory [étale cohomology] be set up applying within algebraic geometry. This took two decades (it was a central aim of the work and school of Alexander Grothendieck) building up on initial suggestions from Serre » (Wikipedia). +Modifying slightly your question, let us now concentrate on the “significance of the zeroes of the Zeta function to algebraic number theory”. Away from RH, the deep arithmetic properties surrounding the so called trivial zeroes are all the more astonishing in view of the opposition discrete vs. analytic. Looking at the special values $\zeta (n)$, $n \in \mathbf Z$ and using the functional equation, one can easily show that $\zeta (0) = 1/2$, and thus restrict to the integers $n$ of the form $-2m$ or $1 – 2m$, $m$ > 0. Then: +1) The special value $\zeta (1 - 2m)$ = - $B_{2m}$ /$2m$, where $B_k$ is the $k$-th Bernouilli number, is a plain rational value. But when a number theorist falls upon a rational number, he inevitably asks whether the numerator and the denominator could not be the orders of some finite groups. Astonishingly, this is the case, see below. +2) $\zeta (- 2m) = 0$ . These trivial zeroes are simple and the special values $\zeta (- 2m)^* $ are by definition the first non zero terms in the Taylor expansion at $s = \zeta (- 2m)$ . +The general formula for all the special values then reads: for all $n \ge 1$, $\zeta (- n)^* $ is equal, up to a sign and a power of 2, to $R_n (\mathbf Q)$. # $K_{2n}(\mathbf Z)$ / # torsion $K_{2n+1}(\mathbf Z)$. Here the $K_i (\mathbf Z)$ are the Quillen K-groups associated to the ring $\mathbf Z$ and $R_n(\mathbf Q)$ is the $n$-th Borel regulator. This was the so called Lichtenbaum conjecture (around 1973), now a theorem. +Comments: Quillen’s definition of the K-groups of a ring comes from algebraic topology – a further manifestation of the unity of mathematics. Note that the K- groups of $\mathbf Z$ are in no way easy to compute. For any number field $F$, one has the K-groups of the ring of integers $O_F$; one can define also the Borel regulators of $F$, higher analogues of the Dedekind regulator which appears in the classical expression of $\zeta_F (0)^* $; and one can extend the Lichtenbaum conjecture to $F$ to get a higher analogue of the class number formula. The conjecture has been proved for abelian fields. Its depth can be judged by simply looking at the list of tools which come into play: the values of $\zeta_F(s)$ at positive integers expressed in terms of polylog functions; the Main Conjecture of Iwasawa theory on p-adic L-functions proved by Mazur and Wiles; the so called Milnor-Bloch-Kato conjecture relating the K-theory of fields to Galois cohomology, proved by Voevodsky; the Quillen-Lichtenbaum conjecture relating the the K-theory of rings of integers to étale cohomology, proved by Vv. and others. Note that a far reaching generalization of the Lichtenbaum conjecture has been proposed by Bloch-Kato concerning the special values of “motivic” L-functions – another strong indication, if still needed, of the unity of mathematics. A convenient introductory book is “The Bloch-Kato conjecture for the Riemann Zeta function”, Coates-Raghuram-Saikia-Sujatha ed., London Math. Soc. LN 418, 2015 .<|endoftext|> +TITLE: Probability of number of people who know a rumor +QUESTION [10 upvotes]: Suppose that among a group of $n$ people, some unknown number of people $K$ know a rumor. If someone knows the rumor, there is a probability $p$ that they will tell it to us if we ask. If they don't know the rumor they will always say they don't know it. +If I go around and ask each person if they know about the rumor, and $M$ people say they do, what does that tell me about the number of people who actually know the rumor? +In particular, what is the distribution $P(K=k|M=m)$ in terms of $P(K)$? +Edit: +I've been able to show that +$P(K=k|M=m)=\frac{b(m,k,p)P(K=k)}{\sum_{j=m}^{n-1}b(m,j,p)P(K=j)}$ +where $b(m,k,p)$ is the binomial density function (probability of $m$ successes out of $k$ trials with probability $p$ of success). Is it possible to take this any further? + +REPLY [3 votes]: For the expected value/mean, we can ignore the people who haven't heard the rumor entirely. You should find that $M=pK$, which makes sense as the people who haven't heard of the rumor won't add anything to the number of people who tell you they have heard the rumor. This means that $K\approx\frac M p$. Unfortunately, this is only the mean, and not a distribution. Now, what we are going to do is rewrite $M$ as a Binomial Distribution, so $P(X=M)={K\choose M}\cdot p^M\cdot(1-p)^{K-M}$, where $M$ is the number of people who answer yes and $K$ is the total number of people who know the rumor. Now, we want to fix $M$ and let $K$ vary. First, we know that $M\leq K$, so $P(K=x)=0\quad(\forall x +TITLE: Can second degree polynomials generate as many as we wish prime numbers in the way described? +QUESTION [15 upvotes]: While I was getting in my pyjamas, a few minutes ago, the Euler polynomial $n^2+n+41$ came into my mind. As you know, this polynomial is famous because the set $\{f(0),f(1),...f(39)\}$ consists of prime numbers only, so this polynomial takes for the first $40$ of its values on the set $\mathbb N_0$ only prime numbers. +So the natural question to ask is: + +Is it true that for every $m \in \mathbb N$ there exists second degree real polynomial of a real variable $P$ and a number $k(m) \in \mathbb N$ such that all numbers in the set $\{P(k(m)), P(k(m)+1),...,P(k(m)+m-1)\}$ are prime numbers? + +REPLY [2 votes]: If you allow non-monic polynomials the question is not very arithmetic in nature. +Any class of integer functions closed under the operations $f(n) \to Af(n)+B$ with integer $A,B$, is guaranteed to have this property for any set that contains long arithmetic progressions (such as the primes, by the Green/Tao theorem as people have mentioned). This includes degree $k$ polynomials for any $k$. +For monic non-linear polynomials, as long as $f(n)$ does not assume all values mod $p$ for some prime, then for all $k$ there is an integer $A$ such that $(f(x) + A)$ has no congruence obstruction to having a string of $k$ consecutive prime values. According to the the Hypothesis H conjectures (Schinzel, Bateman-Horn) such strings then occur infinitely often.<|endoftext|> +TITLE: The three unsolved problems of antiquity +QUESTION [11 upvotes]: In Sidelights on the Cardan-Tartaglia Controversy (Apr., 1938) by Martin A. Nordgaard in the National Mathematics Magazine, Vol. 12, No. 7, pp. 327-364, it is written on the first page + +The solution of the cubic had presented itself to the human mind as + an intellectual problem already in the fifth century B. C.; it became + a scientific need in Archimedes' calculation on floating bodies in the + third century B. C.; it confronted the Arab astronomers in the Middle + Ages. And now it was solved! The first of "the three unsolved + problems of antiquity" to be solved. + +But which are the other two of "the three unsolved problems of antiquity"? What people usually seem mean when they refer to "the three unsolved problems of antiquity" are the trisecting of the angle, the doubling of the cube (or the Delian problem) and the squaring of the circle (just use Google to verify). Evidently Nordgaard refers to these as one. The other problem could then very well be the question of the necessity of Euclid's fifth postulate. +Is there any other place were this terminology is used in this manner, or is it just an imagination of Nordgaard? Are my suspicions correct? Or does he refer to different problems? Was the solution to the cubic indeed such a "big deal" to the ancients? + +REPLY [10 votes]: Here I provide some text passages which rather indicate that the three great/unsolved problems of antiquity + +duplication of the cube (Delian problem) +trisection of the angle +squaring the circle + +were not considered as three connected problems by the ancients. + +These three problems are discussed in B.L. van der Waerdens Science Awakening Vol. 1. Most of the following information is from chapter VII, the Alexandrian Era (330 - 200 B.C.) + +duplication of the cube: + +(p. 124) ... The pure cubic $x^3=V$, in geometric form, poses the problem of contructing a cube of given volume. The ancients were concerned with this problem as well; a special case is the famous duplication of the cube, ... +(p.136) ... Hippocrates (ca. 470 - ca. 410 B.C.) must also have occupied himself with the duplication of the cube, more exactly with the problem of constructing a cuve whose volume has a given ration (e.g. $1:2$) to that of a given cube. +It is an established tradition (see Eutocius in Archimedis Opera III, 104) that Hippocrates was the first to recognize that the problem would be solved if one could construct two mean proportionals $x$ and $y$ between two given line segments $a$ and $b$ whose ratio is the given ratio, + \begin{align*} +a:x=x:y=y:b\tag{1} +\end{align*} + For, as we have developed before, if (1) holds then the cube $a^3$ would have to $x^3$ the same ratio as $a$ has to $b$. ... + +Section The duplication of the cube +... I shall discuss the history of the problem in some detail, known to us chiefly from Eutocius' commentary on Archimedes. A few of the main points will be touched upon now. We have already seen that Hippocrates of Chios reduced the general problem of increasing the volume of a cube in a given ratio to that of constructing two mean proportionals between two given line segments. Archytas of Taras (428 - 347 B.C.) discovered an extremely ingenious three-dimensional solution of this problem. Later, solutions were given by Eudoxus, Menaechmus, Plato (?), Eratosthenes, Nicomedes, Appolonius, Heron, Philon of Byzantium, Diocles, Sporus and Pappus. Obviously great importance was attributed to this problem. +In the supplement to the Laws of Plato, known as the Epinomis, counted as one of Plato's works but not published until after his death, the problem of the two mean proportionals is indeed considered as the problem of solid geometry. + +$$ $$ + +About Archytas of Taras: + +... The versatility of this remarkable South-Italian Dorian is unparalleled even in that era. By means of an extremely ingenious space construction, he obtained a solution of the famous Delian problem, the duplication of the cube. + + +On page 150,151 we can read about the really ingenious construction. See here the corresponding graphic Figure (1) which is also shown in v.d. Waerdens book. + +Later on, when v.d. Waerden discusses the possible origins of the sources we can read about two legends. + +... Fortunately, a quotation in Theon of Smyrna gives the source of the second legend +In his work entitled Platonics, Eratosthenes relates that, when God announced to the Delians through an oracle that, in order to be liberated from the pest, they would have to make an altar, twice a great as the existing one, the architects were much embarassed in trying to find out how a solid cube could be made twice as great as another one. They went to consult Plato, who told them that the god had not given the oracle because he needed a doubled altar, but that it had been declared to censure the Greeks for their indifference to mathematics and their lack of respect for geometry. +The same story, in almost the same words, is found in Plutarch (de Ei apud Delphos, 386 E). Elsewhere (de genio Socratis 579 CD), he adds that Plato referred the Delphians to Eudoxus and to Helicon of Cyzicus for the solution of the problem. ... + + + +Trisection of the angle: +We start with Nicomedes (280 - 210 B.C.) + +(p. 235) ... Chronologically, Nicomedes comes between Eratosthenes and Apollonius, since he criticised Eratosthenes' duplication of the cube, while Appolonius refers somewhere to the cochloid of Nicomedes. Apparently he belonged to the same group of Alexandrian mathematicians as Eratosthenes and Apollonius. +... Pappus states that Nicomedes described an instrument, with which the curve can be drawn. ... +... Nicomedes took great pride in the discovery of this curve. He applied his neusis-construction to two famous problems, viz. the trisection of the angle and the duplication of the cube. + ... The underlying idea of the trisection will be seen to be the same as that of the trisection of Archimedes. +But Nicomedes only uses a neusis between two straight lines, while Archimedes calls for a neusis between a circle and a straight line. Pappus shows that the neusis of Nicomedes can also be obtained by the intersection of a circle and a hyperbola. Nicomedes used the intersection of his cochloid with a straight line, every bit as simple, since a cochloid is defined and drawn more easily than a hyperbola. + + +It's interesting to note how the ancients considered the use of different techniques and means to solve geometric problems. + +About constructions by means of compasses and straight edge by the ancients +(p. 264) The idea, sometimes expressed, that the Greeks only permitted constructions by means of compasses and straight edge, is inadmissible. It is contradicted by the numerous constructions, which have been handed down, for the duplication of the cube and the trisection of the angle. +It is true however, that such constructions were considered to be more elementary, and somewhere Pappus says that, whenever a construction is possible by means of straight edge, more advanced means should not be used. See in this connection A.D. Steele, Quellen und Studien, B 3, p.287. + + +Some hundred years later Pappus of Alexandria (290 - 350 A.D.) did a great unifying work + + +(p. 286) ... The reader will gradually have gotten some idea of his principal work, the Mathematical Collection, ... In this work, Pappus brought together everything in the work of his predecessors which interested him: on higher plane curves, on the quadrature of the circle, the duplication of the cube, the trisection of the angle, the method of analysis, etc. +Wherever explanations of or supplements to the works of the great geometers seemed to him necessary, he formulated them as lemmas. This gives us a great deal of varied information about what was contained in the lost works of Euclid and of Apollonius. But furthermore, Pappus supplemented and extended in several points the work of his predecessors. + + + +The third of the problems: + +Squaring the circle: + +(p. 130) ... It has already been mentioned that, while he was in prison, Anaxagoras, (510 - 428 B.C.) occupied himself with the quadrature of the circle. +Altogether this problem was very popular towards the end ot the 5th century. The comic poet Aristophanes even made a joke about it. In The Birds, he introduces the astronomer Meton, who says: ... Meton's quadrature was only a jest. +But, although not accurate, the quadrature of the circle of the sophist Antiphon was intended to be serious. ... He inscribes some polygon in the circle, perhaps a triangle or a square. By bisecting the arcs, he obtains a polygon of double the number of sides, a hexagon or an octagon. +He believed that, by continuing in this way, he would ultimately obtain a polygon, whose sides, on account of their small size, would coincide with the circumference. Since for any polygon, one can construct a square of equal area, and since the polygon that has been obtained is to be considered as equal to the circle, one can construct a square as equal to the circle, he concluded that the circle can also be squared. +(p.131) ... The most famous among the geometers of the 5th century, Hippocrates of Chios (not to be confused with the still more famous physician Hippocrates of Cos), also occupied himself with the quadrature of the circle. +... We can get some idea of the way in which he attacked this problem from a statement which Simplicius, the most learned and most reliable among the commentators of Aristotle, gets from his teacher Alexander of Aphrodisias. According to this Alexander, Hippocrates began with an isosceles right triangle and proved that the sum of the areas of the two lunules, formed by the semicircles on the right sides and by the semi-circle circumscribed about the triangle, is equal to the area of the triangle. ... + + +Another contributions from Dinostratus and Nicomedes + + +(p. 191) ... In the large compendium of Pappus, which must have been written in the time of the emperor Diocletian (284 - 305), it is mentioned that Dinostratus and Nicomedes used for the quadrature of the circle a curve, which for that reason was called the quadratrix. ... + + + +Conclusion: + +The impression when looking at this information is that different ingenious geometers have attacked these problems with different techniques. Sometimes as we could see by Nicomedes one technique, the Neusis construction could be used to attack two of the three problems. But there is no evidence, that the ancients have considered the three problems as related. +It seems to me that nowadays for our convenience these problems are put together, since these are famous examples of what can't be solved by means of compasses and straight edge alone. A typical example is chapter 10 of this introduction to Galois theory by Richard Koch. +I would like to point to an instructive answer of a related question by DavidSpeyer who transforms the problem to a more familiar one, namely to +the standard five function calculator: $+$, $-$, $\times$, $\div$, $\sqrt{ \ }$. So the question was: using these keys, and the ability to type in integers or recopy numbers that you have computed before, what can you compute? Can you compute $\sqrt[3]{2}$, or $\cos (20^{\circ})$?. +A typical convenient representation for us nowadays is presented by Jennifer Doung and Daniel Szara in The 3 unsolved problems of Greek Mathematics.<|endoftext|> +TITLE: A limit involving $\cot$ that seemingly shouldn't exist +QUESTION [7 upvotes]: According to Wolfram Alpha, $$\lim_{x \to \infty} \frac{x - \cot x}{x} =1.$$ +But does the limit even exist? Isn't $\frac{x - \cot x}{x}$ unbounded near $x= n \pi$ for all $n \in \mathbb{N}$? +Assuming that the limit doesn't actually exist, what might explain why Wolfram Alpha thinks that it does exist? + +REPLY [4 votes]: When we ask Mathematica (10.4) to compute a power series for $f(x) = \frac{x - \cot(x)}{x}$ around $x = \infty$ (of order 10), we get $$ 1 + \cot(x)\left(\frac{-1}{x} + O\left(x^{-12}\right) \right) + O\left(x^{-11}\right) \text{.} $$ If we imagine that cotangent was a very nice function (heh), we'd say it is getting crushed by $-1/x$ as $x \rightarrow \infty$ and similarly for the big-O residuals. This just leaves the "$1$" in the limit. (This property of Mathematica's Series[] function to use some simple transcendental functions rather than expand them into the series is frequently irritating.) +If we ask Mathematica to evalute the limit, it stares at us blankly +In: Limit[(x - Cot[x])/x, x -> \infty ] + +Out: Limit[(x - Cot[x])/x, x -> \infty ] + +If we ask Wolfram Alpha to do expand $f$ in a series around $\infty$, it stares at us blankly +Series[(x - Cot[x])/x,{x,\infty,2}] + +(no series expansion available) + +So I can't guarantee that Alpha is using the above expansion to arrive at the wrong limit. But I'd bet a dollar that it does.<|endoftext|> +TITLE: $\mathbb{Z} [\sqrt{2}]$ is an integral domain +QUESTION [5 upvotes]: We know that $(\mathbb{Z} [\sqrt{2}],+,\cdot)$ is an integral domain. +Someone can prove it easily if he says that is a subring of $(\mathbb{R} ,+,\cdot)$ . +Can we find a different proof, more analytical? +How can we show that $$\forall x,y\in \mathbb{Z} [\sqrt{2}], x\ne 0, y\ne 0\implies xy\neq 0$$ + +REPLY [8 votes]: In these cases, considering the conjugate can help. The conjugate of $a+b\sqrt{2}$ is $a-b\sqrt{2}$. Now, if $(a+b\sqrt{2})(c+d\sqrt{2})=0$, also +$$ +(a+b\sqrt{2})(a-b\sqrt{2})(c+d\sqrt{2})(c-d\sqrt{2})=0 +$$ +and therefore +$$ +(a^2-2b^2)(c^2-2d^2)=0 +$$ +Since the integers form a domain, we conclude $a^2-2b^2=0$ or $c^2-2d^2=0$. The irrationality of $\sqrt{2}$ tells us that either $a=b=0$ or $c=d=0$.<|endoftext|> +TITLE: Positiveness of energy of differential equation +QUESTION [6 upvotes]: Let $x(t) : [0,T] \rightarrow \mathbb{R}^n$ be a solution of a differential equation + $$ +\frac{d}{dt} x(t) = f(x(t),t). +$$ + In addition we have functions $E :\mathbb{R}^n \rightarrow \mathbb{R}$ and $h:\mathbb{R}^{n+1}\rightarrow \mathbb{R}$ such that + $$ +\frac{d}{dt}{E(x(t))} = h(x(t),t) E(x(t)) +$$ + for any solution $x(t)$ of the differential equation. Can we show that if $E$ is positive at the time $0$ then it is positive at all times? i.e. + $$ +E(x(0))>0 \Longrightarrow E(x(t)) \qquad 00 \Leftrightarrow f(0)=0)$, which gives: $$f(t) \leqslant f(0) \times \int_0^t h(x(s),s) f(s) ds=0 $$ +By definition of $f$, this implies that $E(x(t))$ is positive. Of course, here all the $\leqslant$ could be replaced by $=$ since you have equalities. Note that the derivative of $1_{E(x(t),t)\leqslant 0}$ has a meaning with the notion of weak derivative; its "derivative" is a dirac at the point "$E(x(t),t)=0$". \ No newline at end of file