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?
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Asaf Karagila
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tox123
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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: $$f(n)=f^{(n)}(0)$$ – Henry Lee May 05 '19 at 01:33

@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? – tox123 May 05 '19 at 01:48

They both mean the same thing and what you have showed is equal and shows what you mean – Henry Lee May 05 '19 at 01:51
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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: $a=e^{W(1)}\approx 0.318+1.337i$.
Alex R.
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Is this exclusively the only answer or are there other functions that satisfy my conditions? – tox123 May 05 '19 at 01:55


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