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I used half my Sunday for trying to proof the following, but I couldn't find the answer. Can you help me?

Let $f(x)=\sum_{k=0}^{\infty}a_k(x-x_0)^k$ a power series with radius of convergence $\ge 0$. Show: $f(x_n)=0$ for a sequence of points $\{x_n\}$ with $x_n \rightarrow x_0, x_n \neq x_0 \implies a_k = 0$ $\forall k \in \mathbb{N}$.

Hint: Define for $j \in \mathbb{N}$ $$f_{(j)}(x):= \sum_{k=0}^{\infty}a_{j+k}(x-x_0)^k$$ and use induction to prove for all $j \in \mathbb{N}$: $f_{(jn)}(x_n)=0$ for all $n \in \mathbb{N}$ and $a_j=0$.

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    For $k=0$, it's just the continuity of a function defined by a power series. If $a_0=\ldots=a_{n-1}=0$, then use the fact that $f(x)=(x-x_0)^nf_{(n)}(x)$ to conclude $a_n=0$.2011-12-12
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    @Davide: Why not make this into an answer?2011-12-12
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    Actually, you need radius of convergence $>0$, right? For $f(x_n)$ to make sense?2011-12-12

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