TITLE: Are there any function which's derivate is the scaled one of the original?( $f'(x) = f(cx)$ )
QUESTION [7 upvotes]: which function could satisfy the following, for a certain $c\ne1$
$f'(x) = f(cx)$
...beyond the trivial $f=0$
i've been thinking about it for a while.
for a simpler case:
$f'(x)= f(x+c)$
i've found $e^{xe^v}$ where $v$ is the solution of $c=\frac{v}{e^v}$

REPLY [2 votes]: We get that $$f^{(k)}(x)=c^{\frac{k(k-1)}{2}}f(c^kx).$$
From this we get that $$f(x)=\sum_{k=0}^{\infty}\frac{c^{\frac{k(k-1)}{2}}}{k!}f(0)x^k.$$