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

  • $\begingroup$ All that matters is that the power series has a positive radius of convergence, which is necessary in order for $p$ to be analytic. $\endgroup$
    – florence
    Aug 13, 2016 at 8:25
  • $\begingroup$ Please make the title more informative. $\endgroup$
    – Did
    Aug 13, 2016 at 8:48
  • $\begingroup$ Why do we need positive radius of convergence?@florence@Did $\endgroup$
    – Learnmore
    Aug 13, 2016 at 9:06

1 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 $b_n=0$.

  • $\begingroup$ Why do we need positive radius of convergence? $\endgroup$
    – Learnmore
    Aug 14, 2016 at 5:18
  • $\begingroup$ What happens if we take radius=0 $\endgroup$
    – Learnmore
    Aug 14, 2016 at 5:19
  • $\begingroup$ 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 $. $\endgroup$
    – Robert Z
    Aug 14, 2016 at 6:00

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