TITLE: Functions that are their Own nth Derivatives for Real $n$
QUESTION [20 upvotes]: Consider (non-trivial) functions that are their own nth derivatives.  For instance
$\frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x$
$\frac{\mathrm{d}^2}{\mathrm{d}x^2} e^{-x} = e^{-x}$
$\frac{\mathrm{d}^3}{\mathrm{d}x^3} e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2}) = e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2})$
$\frac{\mathrm{d}^4}{\mathrm{d}x^4} \sin x = \sin x$
$\cdots$
Let $f_n(x)$ be the function that is it's own nth derivative.  I believe (but I'm not sure) for nonnegative integer $n$, this function can be written as the following infinite polynomial:
$f_n(x) = 1 + \cos(\frac{2\pi}{n})x + \cos(\frac{4\pi}{n})\frac{x^2}{2!} + \cos(\frac{6\pi}{n})\frac{x^3}{3!} + \cdots + \cos(\frac{2t\pi}{n})\frac{x^t}{t!} + \cdots$
Is there some sense in which this function can be extended to real n using fractional derivatives?  Would it then be possible to graph $z(n, x) = f_n(x)$, and would this function be smooth and continuous on both $n$ and $x$ axes?  Or would it have many discontinuities?

REPLY [2 votes]: Sorry to give yet another answer that does not address the issue of fractional $n$ [it seems that fractional derivatives are not such a familiar topic to many research mathematicians; certainly they're not to me], but:
There is a little issue here which has not been addressed.  By the context of the OP's question, I gather s/he is looking for real-valued functions which are equal to their $n$th derivative (and not their $k$th derivative for $k < n$).  Several answerers have mentioned that the set of solutions to $f^{n} = 0$ forms an $n$-dimensional vector space.  But over what field?  It is easier to identify the space of such complex-valued functions, i.e., $f: \mathbb{R} \rightarrow \mathbb{C}$: namely, a $\mathbb{C}$-basis is given by $f(x) = e^{2 \pi i k/n}$ for $0 \leq k < n$.  But what does this tell us about the $\mathbb{R}$-vector space of real-valued solutions to this differential equation?
The answer is that it is $n$-dimensional as a $\mathbb{R}$-vector space, though it does not have such an immediately obvious and nice basis.  
Let $W$ be the $\mathbb{R}$-vector space of real-valued functions $f$ with $f^{(n)} = 0$ and $V$ the $\mathbb{C}$-vector space of $\mathbb{C}$-valued functions $f$ with $f^{(n)} = 0$.  
There is a natural inclusion map $W \mapsto V$.  Phrased algebraically, the question is whether the induced map $L: W \otimes_{\mathbb{R}} \mathbb{C} \rightarrow V$ is an isomorphism of $\mathbb{C}$-vector spaces.  In other words, this means that any given $\mathbb{R}$-basis of $W$ is also a $\mathbb{C}$-basis of $V$.  This is certainly not automatic.  For instance, viewing the Euclidean plane as first $\mathbb{R}^2$ and second as $\mathbb{C}$ gives a map 
$\mathbb{R}^2 \rightarrow \mathbb{C}$ which certainly does not induce an isomorphism upon tensoring with $\mathbb{C}$, since the first space has (real) dimension $2$ but the second space has (complex) dimension $1$.
For more on this, see Theorem 1.6 of
http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf
It turns out that this is actually a problem in Galois descent: according to Theorem 2.14 of the notes of Keith Conrad already cited above, the map $L$ is an isomorphism iff there exists a conjugate-linear involution $r: V \rightarrow V$, i.e., i.e., a map which is self-inverse and satisfies, for any $z \in \mathbb{C}$ and $v \in V$, $r(zc) = \overline{z} r(c)$.
But indeed we have such a thing: an element of $V$ is just a complex-valued function $f$, 
so we put $r(f) = \overline{f}$.  Note that this stabilizes $V$ since the differential equation $f^{(n)}) = 0$ "is defined over $\mathbb{R}$": or more simply, the complex conjugate of the $n$th derivative is the $n$th derivative of the complex conjugate.  Thus we have "descent data" (or, in Keith Conrad's terminology, a G-structure) and the real solution space has the same dimension as the complex solution space.
It is a nice exercise to use these ideas to construct an explicit real basis of $W$.