Express a function as a power series if it respects some nice conditions

Question

Find all functions $f$ such that there is sequence $(a_n)_{n \in \mathbb{N}}$ such that : $$\forall x \in ]-1, 1[, \sum_{n = 0}^\infty a_nx^n = f(x)$$ $$\forall k \in \mathbb{N}, k \geq 2, f\left(\frac{1}{k}\right) = e^{-\sqrt{k}}$$

Here are my thoughts :

First I guess we need to use that :

$$e^{-\sqrt{x}} =\sum_{k= 0 }^\infty \frac{(-1)^k x^{k/2}}{k!}$$

So we need to find a sequnece such that :

$$\forall k \geq 2, \sum_{n = 0}^\infty\frac{ a_n }{k^n} = \sum_{n= 0 }^\infty \frac{(-1)^n k^{n/2}}{n!}$$

But from here I don't what to do. I know that two power series are equal if there is a segment on which there are equal. Here it's not the case, but these two power series have infinitely many points where they have the same value, but it's not sufficient to conclude that these power series are equal right ?

Moreover $0$ is an accumulation point of $\{1/k \mid k \geq 2\}$ so maybe it means that the two series are equal ? Since it means we can't find an open ball around $0$ such that the series are not equals, but the problem is that the theorem I know is only about a segment... So is what I am saying is true or they are nice counter-examples ?

Thank you !

Answer 1

There is no such function. Of course, the null function is no such function. On the other hand,$$a_0=f(0)=\lim_{k\to\infty}f\left(\frac1k\right)=\lim_{k\to\infty}e^{-\sqrt k}=0.$$So, there is a smallest $N\in\mathbb N$ such that $a_N\neq0$. But then$$\lim_{k\to\infty}\frac{\left\lvert f\left(\frac1k\right)\right\rvert}{\frac1{k^N}}=\lvert a_N\rvert\neq0,$$whereas$$\lim_{k\to\infty}\frac{e^{-\sqrt k}}{\frac1{k^N}}=\lim_{k\to\infty}\frac{k^N}{e^{\sqrt k}}=0.$$

Answer 2

Assume the existence of such $f$, which is not identically zero.

Since $f(0)=\lim_{k\to \infty}f(1/k)=\lim_{k\to \infty}e^{-\sqrt k}=0$, $f$ has a zero at $x=0$. Assume the zero is of order $(m-1)\in\mathbb N^+$.

From the series expansion, we know $$\lim_{x\to 0}\frac{f(x)}{x^m}=1$$

Thus, we expect $$\lim_{x\to 0}\frac{f(x)}{x^m}=\lim_{k\to\infty}k^m f(1/k) =\lim_{k\to\infty}k^m e^{-\sqrt k}\color{red}{=}1$$

However, there is no $m$ that can satisfy the red equality. Therefore a contradiction is created and we conclude that such $f$ does not exist.