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Are there non-trivial solutions $f:\Bbb R\to\Bbb R$ of the following "differential equation"?

$$f'(x)=f(x+1).$$

I consider $f(x)=0$ as trivial solution. I tried to express $f$ as a power series, i.e. $f(x)=\sum_n a_n/n!\cdot x^n$. This brought me to the following identity for the coefficients:

$$a_{n+1}=\sum_{k=0}^\infty \frac {a_{k+n}}{k!}.$$

This does not really help, as the right side contains infinitely many coefficients itself. And infinitely many must be non-zero. I expect that a possible solution looks something like this:

M. Winter
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    A complex one: let $\alpha \in \mathbb{C}$ be the solution to $\alpha = e^\alpha$, which is $\alpha \approx 0.318132 - 1.33724i$.

    Consider $f(z) = e^{\alpha z}$.

    $$f'(z) = \alpha e^{\alpha z}= e^\alpha e^{\alpha z} = e^{\alpha(z+1)} = f(z+1)$$

     – mechanodroid Sep 22 '17 at 13:36
  • It's not continuous, but can $e^{{x}}$ where ${x}=x-\lfloor x\rfloor$ qualify ? Since it is periodic, $f(x+1)=f(x)$ so we arrive at $f'=f$ which is simply exponential, (at the expense of small issues at the bounds). – zwim Sep 22 '17 at 13:41
  • @zwim and mechanodroid: Very nice ideas! Unfortunatly I am looking for differentiable and real valued solutions. – M. Winter Sep 22 '17 at 13:43
  • @mechanodroid: That equation has a lot of solutions. http://math.stackexchange.com/a/858985/8157 – Giuseppe Negro Sep 22 '17 at 13:58
  • @GiuseppeNegro It's not the same equation: this one is $z = e^z$, that one is $ze^z = a$. Regardless, it's just important that a solution exists. – mechanodroid Sep 22 '17 at 14:03
  • @mechanodroid $\alpha=W(-1)$, where $W$ is the Lambert W function. Choose branch appropriately. – Simply Beautiful Art Sep 22 '17 at 14:05
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    You can combine $e^{\alpha x}$ and $e^{\bar\alpha x}$ (which is also a solution) to get a real valued continuous function, e.g. $e^{(\Re\alpha) x}\cos((\Im\alpha)x)$ – Reinhard Meier Sep 22 '17 at 14:06
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    This is a "delay differential equation" ... see https://math.stackexchange.com/questions/514618/differential-delay-equations – GEdgar Sep 22 '17 at 14:08
  • Technically this is not a delay differential equation, as the derivative is determined by values to the right of $x$, not the left. – Paul Sep 22 '17 at 14:12
  • Have you tried a Laplace Transform? – Paul Sep 22 '17 at 15:16
  • @mechanodroid: Then see here https://math.stackexchange.com/q/859274/8157 – Giuseppe Negro Sep 22 '17 at 20:41
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    general solution https://math.stackexchange.com/a/4379608/532409 – Quillo Feb 11 '22 at 15:15

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