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 ?
 A: Consider solutions of the form $f(x)=\exp(\lambda x)$.  Then 
$\lambda=\exp(\lambda)$.
This equation has no real roots.  But it does have complex conjugate roots, we will see there are infinitely many of them actually.  Render one such pair of roots as $\alpha\pm\beta i$.  Then by superposition we may construct real solutions
$f(x)=\exp(\alpha x)\cos(\beta x)$
$f(x)=\exp(\alpha x)\sin(\beta x)$
and with infinitely many pairs of roots for this transcendental equation we can get infinitely many linearly independent solutions.  These exponential-trigonometric components in turn may be superposed to give a series solution of the form:
$f(x)=\Sigma_{k=1}^{\infty}(\exp(\alpha_k x)(a_k\cos(\beta_k x)+b_k\sin(\beta_k x)))$
I have a hunch that this would be a complete set of such solutions.
Can we find any roots analytically?  It appears not, to me.  If we had $f(x+\pi/2)$ on the right side we could pick off $\lambda=\pm i$ getting the usual trigonometric functions, but that does not work without the $\pi/2$.
But we can prove the existence of the roots.  Render the parts of $\lambda$ and $\exp(\lambda)$ equal:
$\alpha=\exp(\alpha)\cos(\beta)$
$\beta=\exp(\alpha)\sin(\beta)$
Then from the Pythagorean identity
$\alpha^2+\beta^2=\exp(2\alpha)$
$\beta=\pm\sqrt{\exp(2\alpha)-\alpha^2}$
The real part equation is now an equation for $\alpha$ alone:
$\alpha=\exp(\alpha)\cos(\sqrt{\exp(2\alpha)-\alpha^2})$
Now we see what happens: As $\alpha$ increases without bound the cosine function on the right oscillates ever faster and with an amplitude far exceeding $\alpha$ itself.  So the above relation has infinitely many fixed points in this limit proving the claim of infinitely many roots for the complex parameter $\lambda$.
A: 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 $\xi$ !
