Given the following, and assuming that $f(x)$ is infinitely differentiable: $$\frac{d^nf(x)}{dx^n}\Bigg|_{x=0}=f(n)$$ What functions $f$ could satisfy this equation? Do any functions of $f$ have a closed form, or if not does it have a form that is just a normal ODE form?

  • $\begingroup$ From what you've written it looks like: $$f(x=n)=f^{(n)}(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^{(n)}(0)$$ $\endgroup$ – Henry Lee May 5 at 1:33
  • $\begingroup$ @Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation? $\endgroup$ – tox123 May 5 at 1:48
  • $\begingroup$ They both mean the same thing and what you have showed is equal and shows what you mean $\endgroup$ – Henry Lee May 5 at 1:51

Let $f(x)=a^{x+1}$, where $a$ satisfies $\ln(a)=a$. Then $f^{(n)}(0)=\ln^n(a) a^{1}=a^{n+1}=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^{-W(-1)}\approx 0.318+1.337i$.

  • $\begingroup$ Is this exclusively the only answer or are there other functions that satisfy my conditions? $\endgroup$ – tox123 May 5 at 1:55
  • $\begingroup$ I can't see how $ln^n(a)a^1=a^{n+1}$... $\endgroup$ – Thehx May 5 at 1:55
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    $\begingroup$ @Thehx: look at the definition of $a$. $\endgroup$ – Alex R. May 5 at 2:08
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    $\begingroup$ oh god, this is beautiful. $\endgroup$ – Thehx May 5 at 2:11

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