I've been trying to solve this problem from Abbott's Understanding Analysis for hours and can't seem to get the last piece of the proof.
Let $g(x) = \sum_{n=1}^\infty b_n x^n$ be a power series that converges for all $x \in (-R,R)$. Let $x_n \rightarrow 0$, while $x_n \neq 0$, and $g(x_n) = 0$ for every $n$. Show that $g(x)$ must be identically zero on $(-R,R)$.
I know that $g(0) = 0$ since $g$ is continuous and the set of zeros of a continuous function is closed. I know that $g'(0) = b_1$, and I need to show that this is zero. I know that if $f(x) = 0$ on an interval, then $f'(x) = 0$ on that interval as well, but I can't seem to find a parallel of that result to this problem.
Best Answer
Because we have convergence on a interval around zero, convergence is uniform for both the series and the series of derivatives, and so we can differentiate term by term.
You have
$$ b_1=g'(0)=\lim_{h\to0}\frac{g(h)-g(0)}{h}=\lim_k\frac{g(x_k)-g(0)}{x_k}=0. $$ We now have that $$g(x)=x\sum_{n=2}^\infty b_n x^{n-1}.$$ Since $x_n\ne0$ and $g(x_n)=0$, we get that $$ g_2(x_k)=\sum_{n=2}b_n(x_k)^{n-1}=0 $$ for all $k$. Since $g_2(0)=0$, we may apply the above reasoning again, to obtain $b_2=0$. Now we can repeat the argument to obtain $b_3=0$, $b_4=0$, etc.