Function $f(x)$, such that $\sum_{n=0}^{\infty} f(n) x^n = f(x)$

Consider a function $f(x)$. Define Taylor series $\sum_{n=0}^{\infty} f(n) x^n$. Is there such a function, other than constant $0$, that $\sum_{n=0}^{\infty} f(n) x^n = f(x)$?

The Taylor series of $f(x)$ at $0$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$. The series is unique, so $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty} f(n) x^n.$ This means that $f^{(n)}(0)=f(n) n!$. The constant function $0$ meets this condition. Are there others?

Add: There is an almost similar question here: Is there a function with the property $f(n)=f^{(n)}(0)$? It is not same, but looks a bit similar. Can it be used in any way?

• Not really. The link is about function(s) with the property $f(n)=f^{(0)}(0)$, which is different from the property $n!f(n)=f^{(0)}(0)$. – Danijel Apr 5 '14 at 17:26
• Correct me if I'm wrong but I think that this is impossible. For $f(1)$ to exist it must be that $f(n) \rightarrow 0$. But $\sum_n f(n) x^n$ is increasing in $x$ if any $f(n) > 0$. – Reinstate Monica Jul 10 '15 at 19:11
• Actually I'm an idiot. Obviously $f(n)$ may be negative. Then it seems $f(n)$ must change signs infinitely often or else the last nonzero term in the sum will dominate for sufficiently large $x$, and yet $f(n) \rightarrow 0$. – Reinstate Monica Jul 10 '15 at 19:16
• You need to add the further condition that the radius of convergence of the Taylor series is infinite, to avoid trivial examples like $f(x)=1/(1-x)$ for $x<1$ with $f(x)=1$ otherwise. – John Bentin Dec 10 '15 at 15:12
• This may be trivial, but it helps that for any function $f(0)=f(0)0!$, so we only need to concern ourselves with the derivatives – Yuriy S Mar 20 '16 at 8:45

1 Answer

Far from a solution, but a few thoughts:

Say that an entire function is $n$-fine if $f(0)=0$, $f'(0)=1$, and $f^{(k)}(0)=f(k)k!$ for $0\le k\le n$.

For $n\ge 2$, consider the polynomial $$p_n(x)=x^{n}(x-1)(x-2)\cdots(x-n+1)=(-1)^{n-1}(n-1)!x^{n}+O(x^{n+1}).$$ Then for $0\le k<n$, we have $p_n^{(k)}(0)=p_n(k)=0$, whereas $p_n(n)=n^n (n-1)!$ and $p_n^{(n)}(0)=(-1)^{n-1}n!(n-1)!$. In particular, $$p_n(n)n!-p_n^{(n)}(0)=(n^n+(-1)^n)n!(n-1)!\ne0.$$

Thus,

Lemma. If $f$ is $n$-fine, $n\ge 1$, then there exists a unique $c$ such that $f+cp_{n+1}$ is $(n+1)$-fine.

Given a 1-fine function $f_1$, in its most general form $f_1(x)=x+x^2g(x)$, we use the lemma to recursively define a sequence $\{f_n\}_n$ such that $f_n$ is $n$-fine and $f_{n+1}-f_n$ is a multiple of $p_{n+1}$.

The question is: Does $f=\lim_{n\to\infty}f_n$ exist? As $x^n\mid p_n$, the limit certainly exists as a formal power series. This gives us a linear map $g\mapsto f$ from entire functions (or even formal power series) to formal power series, and we are looking for a fixed-point of this map ...