TITLE: Plot of a ... Square?
QUESTION [30 upvotes]: Well there are equations which can plot a square like : $|x-y|+|x+y|=a$
But how about this equation: ? (At the end ... bear with me!)

[Here I have taken $a = 1$]
Plot of $$x^2 + y^2 = a^2$$

Plot of $$x^4 + y^4 = a^4$$

Plot of $$x^6 + y^6 = a^6$$

Plot of $$x^{100} + y^{100} = a^{100}$$

Since we can see that as the degree of the equation is increasing, the sharpness of the possible rounded square is also increasing...
So can we say that:

Plot of :
$$\lim\limits_{p \rightarrow \infty} \space (x^p + y^p = a^p)$$
is the plot of a square???

REPLY [4 votes]: Of course the equation
$$\max\bigl\{|x|, \>|y|\bigr\}=a$$
describes the circumference $\partial Q$ of a square simply and accurately. But you have the feeling that $\partial Q$ is also the "limit" of the sets
$$S_p:=\bigl\{(x,y)\>\bigm|\>|x|^p+|y|^p=a^p\bigr\}$$
when $p\to\infty$. There are several notions for limits of sequences of sets. For the problem at hand the notion of Hausdorff distance is most appropriate: For any two nonempty sets $A$, $B\in{\mathbb R}^n$ one defines
$$d(A,B):=\inf\bigl\{\epsilon>0\>\bigm|\>A\subset B_\epsilon\ \wedge\ B\subset A_\epsilon\bigr\}\ ,$$
whereby $A_\epsilon:=A+U_\epsilon(0)$ denotes the set of all points $y\in{\mathbb R}^n$ having distance $<\epsilon$ from some point $x\in A$.
In terms of this Hausdorff metric $d$ it is not difficult to show that
$$\lim_{p\to\infty}d(S_p,\partial Q)=0\ ;$$
whence $\lim_{p\to\infty} S_p=\partial Q$.