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

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

• All that matters is that the power series has a positive radius of convergence, which is necessary in order for $p$ to be analytic. Aug 13, 2016 at 8:25
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 $b_n=0$.
• If $p(x)$ is convergent only at $0$ and $p(0)=0$ then we could have that $b_n\not=0$ for $n>0$. Tale for example $\sum_{n\geq 1} n!x^n$. Aug 14, 2016 at 6:00