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Suppose that the power series $$\sum b_nx^n$$ converges for $|x|$ less than or equal to $1$.

Suppose that for some $s$ greater than $0$, $p(x)=0$ for all $|x|$ less than $s$.

How to show that $b_n=0$ for all $n$ greater than equal to $1$?

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Based on the context of learning, if the textbook said you can differentiate a power series term by term inside circle of the convergence, then just differentiate it. You can prove $c_1 =0 $ and do it recursively.

Otherwise, you need to prove you can differentiate a converged power series term by term and still have a converged power series.

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  • $\begingroup$ Which you can show easily and indirectly by showing that you can integrate $\sum _{n>0}n a_n x^{n-1}$ term by term, because the sequence for the derivative converges uniformly on any closed set inside the open circle of convergence. $\endgroup$ Feb 10 '16 at 5:59
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You don't say what $n$ runs over in the sum. Let's say from $1$ to $\infty$.

Factor out $x$. You thus have $x\sum_{n=1}^\infty b_n x^{n-1}$. The latter series has the same radius of convergence as the former by the formula in Baby Rudin 3.39. Divide by $x$. You thus have $\sum_{n=1}^\infty b_n x^{n-1}=0$ for $|x|<s$. Let $x=0$. You thus have $b_1=0$.

Suppose you've proved $b_1, b_2, \ldots, b_m=0$. We want to show $b_{m+1}=0$. We have $p(x)=\sum_{n=m+1}^\infty b_n x^n=0$ for $|x|<s$. Factor out $x$ again, divide by $x$, and let $x=0$. We get $b_{m+1}=0$ as desired.

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