Solving $f'(x)=f(x+1)$ I was wondering if it was possible to find functions $f$ such that
$$
f'(x)=f(x+1)
$$
for all $x \in \mathbb{R}$. The only thing i've found so far is that it implies
$$
f^{\left(n\right)}\left(x\right)=f\left(x+n\right)
$$
is there any way to solve this ?

By the Fourier transform,
$$2i\pi\xi F(\xi)=e^{2i\pi\xi}F(\xi).$$
Then $F(\xi)$ is only nonzero for the roots of $2i\pi\xi=e^{2i\pi\xi}$ and the spectrum is discrete. But the equation has no real roots in $\xi$ !
