$f(x)=\sum_{k=0}^\infty c_kx^k$ converges for $|x|<R$ with $R>0$, $\exists x_n\ne0$ s.t. $x_n\to0$, $f(x_n)=0$ $\forall n$, then $c_k=0$ $\forall k$.

Question

I meet a problem:

The power series $f(x)=\sum_{k=0}^\infty c_kx^k$ converges for $|x|<R$ with $R>0$. If there exists a sequence $x_n\ne0$ such that $x_n\to0$ and $f(x_n)=0$ for all $n$, then $c_k=0$ for all $k\in\mathbb{N}$.

I have no clue at all. Any hint or help will be appreciated.

Answer 1

Supposing that not all the $c_k$ are $0$, let $n_0=\min{\{n\geq 0, c_n\neq 0\}}$

Then $f(x)=\sum_{k=n_0}^\infty c_kx^k=x^{n_0}(c_{n_0}+\sum_{k=n_0+1}^\infty c_kx^{k-n_0})$

The power series $g(x)=\sum_{k=n_0+1}^\infty c_kx^{k-n_0}$ has the same radius as $f$ and $g(0)=0$.

$g$ being continuous at $x=0$, there is a neighborhood $V$ of $0$ such that $$\forall x\in V\setminus \{0\}, c_{n_0}+g(x) = c_{n_0}+\sum_{k=n_0+1}^\infty c_kx^{k-n_0}\neq 0$$

Hence $\forall x\in V\setminus \{0\}, f(x)\neq 0$, a contradiction.

Answer 2

Hint: $f$ is continuous for $|x|<R$ - conclude that $c_0=f(0)=0$. It is also differentiable there, and $f'(x)=\sum_{n=0}^\infty (n+1)c_{n+1}x^n$. - Use Rolle to find a sequence of zeroes of $f'$ converging to $0$.

Answer 3

Completion of Hagen von Eitzen's method:

Since $f(x)=\sum_{k=0}^\infty c_kx^k$ converges for $|x|<R$ with $R>0$, $f$ is continuous and infinitely differentiable in $(-R,R)$, and for any $j\in\mathbb{N}$, $$ f^{(j)}(x)=\sum_{k=j}^\infty \frac{k!}{(k-j)!}c_kx^{k-j}=\sum_{k=0}^\infty \frac{(k+j)!}{k!}c_{k+j}x^k $$ has the same radius with $f$, and thereby is also continuous in $(-R,R)$.

By the continuity, $c_0=f(0)=\lim_{n\to\infty}f(x_n)=0$.

Notice that $f(0)=f(x_1)=0$. By Rolle's mean value theorem, there exist a point $\xi_1\in(0,x_1)$ such that $f'(\xi_1)=0$. Choose $x_{n_2}\in(0,\xi_1)$ in the sequence $\{x_n\}$, then $f(0)=f(x_{n_2})=0$. Also by Rolle's theorem, there exist a point $\xi_2\in(0,x_{n_2})$ such that $f'(\xi_2)=0$. Obviously $\xi_2<\xi_1$. Keep on this way, we can get a strictly decreasing sequence $\{\xi_n\}$ with $\xi_n\ne0$ $(\forall n)$, $\xi_n\to0$ $(n\to\infty)$, and $f'(\xi_n)=0$ $(\forall n)$. By the continuity, $c_1=f'(0)=\lim_{n\to\infty}f'(\xi_n)=0$.

By the same argument and the induction, one can show that for any $j\in\mathbb{N}$, $$ c_j=\frac{1}{j!}f^{(j)}(0)=0. $$ QED.