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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 $a_n=0$. – Davide Giraudo Dec 12 '11 at 9:35
@Davide: Why not make this into an answer? – robjohn Dec 12 '11 at 10:46
Actually, you need radius of convergence $>0$, right? For $f(x_n)$ to make sense? – GEdgar Dec 12 '11 at 14:10

We can follow the hint and show the result by induction on $k$. Since $f(x_n)=0$ for all $n\geq 1$, $f$ is continuous at $x_0$ and $x_n\to x_0$ we have $f(x_0)=0$. But $f(x_0)=a_0$, so $a_0=0$. If we have shown that $a_0=\ldots=a_{n-1}=0$ for $n\geq 1$, we show that $a_n=0$. We have $f(x)=(x-x_0)^nf_{(n)}(x)$, so for all $j$, $0=f(x_j)=(x_j-x_0)f_{(n)}(x_j)$, and since $x_j\neq x_0$, we get $f_{(n)}(x_j)=0$ for all $j\geq 1$. $f_{(n)}$ has the same radius of convergence as $f$, and is continuous so $a_n=f_{(n)}(x_0)=0$.

If you don't want to use induction, we can suppose that the conclusion is wrong. Let $k_0:=\min\{k\geq 0, a_k\neq 0\}$. Then $f(x)=(x-x_0)^{k_0}f_{(k_0)}(x)$, and we conclude by the same way as in the use of induction hypothesis (so the two ideas are in fact the same).

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