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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?

A positive answer to this question has been obtained in Does there exist a function that generates itself?.

abacaba
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Valtteri
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    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 05 '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
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    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 08:45
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    What is the point of adding a bounty without clarifying the questions asked by @JohnBentin in the comments? – GEdgar May 07 '23 at 15:24
  • Probably one of the most annoying problems I've ever spent 2 hours on without any gain. – Alexander Conrad May 09 '23 at 20:24

2 Answers2

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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 ...

Hagen von Eitzen
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This is only a partial answer, but it seems to be an interesting observation, and it's too long for the comments, so I write it here. The point of this answer is to show that the real difficulty lies in ensuring that the required condition holds for all $x$, because it is extremely easy to get a construction if the condition is only to be satisfied on a bounded interval, even if you require smoothness everywhere on $\mathbb R$ (and analyticity almost-everywhere). In fact, there is so little difficulty in making such a construction that any values of the $f(n)$ that you pick by hand will work (modulo some issues about convergence, so e.g. you can't pick values that diverge too quickly).

Claim. For any sufficiently nice sequence $(a_n)_{n\geq 1}$ (e.g. boundedness suffices), it is possible to find a smooth function $f\colon\mathbb{R}\to\mathbb{R}$ satisfying $f(n)=a_n$ for positive integers $n$, as well as $\displaystyle f(x)=\sum f(n)x^n$ in a neighbourhood of zero.

To construct such a function we will exploit the fact that smooth functions are easy to bend around to our will, for example using bump functions. Let $$ \sigma_{a,b}(x):=\begin{cases}1,&|x|\leq a\\\dfrac{\exp(-\frac{b-a}{x-a})}{\exp(-\frac{b-a}{x-a})+\exp(-\frac{b-a}{b-x})},&a<|x|<b,\\0,&|x|\geq b\end{cases}$$ be a smooth transition function from $1$ on the interval $(-a,a)$ to $0$ for $|x|\geq b$ for any real numbers $0<a<b$.

Proof. Let $f_r\colon (1-\epsilon,\infty)\to\mathbb R$ be any smooth function that interpolates the $a_n$, i.e. $f_r(n)=a_n$ for all $n\geq1$, for some fixed $\epsilon>0$. I claim that we can construct a function $f\colon\mathbb R\to\mathbb R$ which restricts to $f_r$ on $(1-\epsilon,\infty)$ with the desired properties. To do so consider the function on a neighbourhood of $0$ defined by $g\colon (-r,r)\to\mathbb R, g(x)=\sum a_nx^n$ (where $r$ is the radius of convergence of this series). Let $\epsilon'>0$ be any real number smaller than $\min\{r,1-\epsilon\}$. Then, the function $$f(x)=g(x)\sigma_{\epsilon',(1-\epsilon)}(x)+f_r(x)$$ has the desired properties with the identity $f(x)=\sum f(n)x^n$ holding in the neighbourhood $(-\epsilon',\epsilon')$. (Of course, the functions $g(x)$ and $f_r(x)$ are understood to mean $0$ outside their domains of definition.) $\square$

More generally, it is easy following this method to "twist the function to do whatever we want" in the range $(k,k+1)$ for any integer $k$. The difficulty is to "get past the integer values without ruining the desired property". Unfortunately, I have no idea how to do that.

YiFan Tey
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