# Is there a closed form, or cleaner way of writing a function satisfying $\frac{d^nf(x)}{dx^n}|_{x=0}=f(n)$ for all $n$?

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?

• 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 5 at 1:33
• @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 5 at 1:48
• They both mean the same thing and what you have showed is equal and shows what you mean – 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: $$a=e^{-W(-1)}\approx 0.318+1.337i$$.
• I can't see how $ln^n(a)a^1=a^{n+1}$... – Thehx May 5 at 1:55
• @Thehx: look at the definition of $a$. – Alex R. May 5 at 2:08