Show that $b_n=0$ for all $n$.

Question

Suppose that the power series $p(x)=\sum b_nx^n$ converges for $|x|\le 1$.Suppose that for some $\delta >0$, $p(x)=0$ for $|x|<\delta $.

Show that $b_n=0$ for all $n$.

My effort:The series $p(x)=\sum b_nx^n$ represents an analytic function and the power series has uncountably many zeros since $p(x)=0$ in $B(0,\delta)$.

Hence $p(x)=0$ which in turn $\implies b_n=0$

Is the solution okay ?

I have some doubts as I have not used the hypotheses that $p(x)=\sum b_nx^n$ converges for $|x|\le 1$.Please help

Answer 1

By hypothesis the power series is convergent $p(x)=\sum b_nx^n$ in $[-1,1]$ and therefore it has a positive radius of convergence. Hence $\displaystyle n! b_n=\frac{d^np(0)}{dx^n}$ for all $n\geq 0$.

Moreover you know that $p(x)=0$ for $|x|<\delta$ then $\frac{d^np(0)}{dx^n}=0$ for all $n\geq 0$, which implies that $b_n=0$.