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I was wondering if there is a general way to solve the functional equation

$$(\exists k)(f^\prime(x) = f(x+k))$$

I know that this is true for certain functions:

$$(e^{cx})^\prime = e^{c(x+\frac{\ln c}{c})}$$ $$\sin(x)^\prime=\sin(x+\tfrac{\pi}{2})$$

but I was wondering if these functions take a general form or are related to the eigenvalues of the second derivative, because both of these are.

k_g
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    You can take any $C^\infty$ function you like on $[0,k]$ (given that it satisfy some conditions at $x=0$ and $x=k$ to make sure the procedure we are about to do gives us a differentiable function) and use the defining relation to define the function on $[k,2k]$ and then on $[2k,3k]$ and so on. This procedure gives you a function that satisfy your equation (at least on $[0,\infty)$). – Winther Jan 04 '15 at 06:35
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    ...it turns out this question has been asked before, see for example this MSE question which explains the same procedure as I hinted to above. See also this. – Winther Jan 04 '15 at 06:44
  • Oh thanks. I tried searching but I couldn't find anything... I guess I need to figure out how to search math properly. – k_g Jan 04 '15 at 07:16
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    It is not very easy to search here unless you know exactly what to search for (and even then its not always easy to find it). I tried combinations of 'functional equation' and 'derivative' and then after 3 pages of results I finally found it. – Winther Jan 04 '15 at 07:22

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