How to show $f$ is a constant zero function...

Question

Power series $f(x)=\sum\limits_{n=0}^\infty$ $a_n x^n$ with the radius of convergence $R>0$
And the sequence $(b_k)$ satisfies
$R>b_1>b_2>..., $ $\lim_{k\to\infty} b_k=0$
Need to show that if $f(b_k) =0$ for all $k\geq 1$, $f$ is a constant zero function.

So I decided to use contraposition. I assumed that there exists at least one nonzero component of$(a_n)$, namely $a_p$.
Then the series $f(b_k)$ become $a_p$$b_k$ for all $b_k$, which is nonzero.
Am I right? Can it be the end of the proof?

Answer 1

Almost, you cannot totally neglect the higher terms. If $p$ is minimal with $a_p\ne 0$, then $f(x)=x^p\cdot\sum_{n=0}^\infty a_{n+p}x^n$. The function $x\mapsto \sum_{n=0}^\infty a_{n+p}x^n$ is continuous in $|x|<R$ and is nonzero at $x=0$, hence in a neighbourhood of $0$, i.e. for almost all $b_k$.

Answer 2

$f$ is expressed as a power series in $(-R,R)$ and hence it is $C^\infty$ in $(-R,R)$.

Hence, if $x_n\to x\in(-R,R)$, then $f(x_n)\to f(x)$.

In particular, as $b_n\to 0$ and $f(b_n)=0$, then $$ f(0)=\lim_{n\to\infty}f(b_n)=0. $$ Next, using Rolle's Theorem, we can find $c_n\in (b_{n+1},b_n)$, such that $f'(c_n)=0$, and as $c_n\to 0$ and $f'$ is continuous at $x=0$, we obtain as above that $f'(0)=0$.

Repeating this argument, we show next that $f''(0)=0$, and inductively, $f^{n}(0)$, for all $n\in\mathbb N$. But, if $$ f(x)=\sum_{n=0}^\infty a_nx^n, $$ then $$ a_n=\frac{f^{(n)}(0)}{n!}=0, $$ and hence $f\equiv 0$ in $(-R,R)$.