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I am looking for examples of functions $f:\mathbb R \to \mathbb R$ where $f'(x)=f(f(x))$ for all $x$. The only example I can find is the trivial one where f is identically 0.

  • Where did this problem come up? Are you looking for all such functions or just some nontrivial examples? – Carl Schildkraut Jul 05 '18 at 04:04
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    You can get some complex functions of the form $\alpha x^\beta$ that satisfy this condition. – Theo Bendit Jul 05 '18 at 04:24
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    @Carl it's a curiosity of mine. I just want some examples. I can't find any. Determining all such functions would be nice but perhaps too ambitious. I'm not sure. –  Jul 05 '18 at 04:24
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    @Theo Indeed, I also found this. But I am just interested in real valued functions. –  Jul 05 '18 at 04:26
  • Find $f$ is equivalent to solve the constrained variational problem: $$ \min_{y=f(x)} \int_{\mathbb{R}^{2}}(f'(x)-f(y))^2 dx dy $$ Good luck! ;) – Alex Silva Jul 05 '18 at 10:02
  • @Alex Is this an open problem? –  Jul 05 '18 at 10:12
  • @user1488 I have no idea! I have just reformulated the problem to drop the composition. – Alex Silva Jul 05 '18 at 10:18

1 Answers1

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This has been solved here:

Result: for every $a > 0$, there exists a unique differentiable function $f : \mathbb{R} \mapsto \mathbb{R}$ such that $f(-a)=-a$ and $f' = f \circ f$.

charmd
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