Functions $f, g$ are given. We know that we can expand them into power series around $x_0=0$, they also satisfy: $f(\frac{1}{k})=g(\frac{1}{k})$ for sufficiently large $k \in \mathbb{N} $. Prove that $f(x)=g(x)$ around $x_0=0 $. $$$$My solution: we know that $$f(x)=\sum a_nx^n, \\g(x)=\sum b_nx^n$$We also know that $$\lim_{k \rightarrow \infty} f(\frac{1}{k})=f(0),\\ \lim_{k \rightarrow \infty} g(\frac{1}{k})=g(0)$$ what with the fact in second line implies that $f(0)=g(0)$. $ \\ $And know I got stuck: I don't know exactly what means that$f, g$ are equal around $x_0=0$. I would even tell that those above is enough as there stands $f(0)=g(0)$ what to me seems desirable. Any ideas?
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This is answered here: http://math.stackexchange.com/a/35768/462 – Andrés E. Caicedo Apr 25 '13 at 16:14
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Consider the function $h(x)=f(x)-g(x)$. We know that $h(\frac 1k)=0$ for all sufficiently large $k$. But we also know that $h(x)$ has infinitely many derivatives at $0$. Suppose that there were some sequence of points $p_1, p_2, \ldots$ such that $|p_i|\rightarrow 0$ and $h(p_i)\neq 0$. Could you get a contradiction?
vadim123
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ee... a contradiction of what? You mean such a function that the statement is not true? – fdhd Apr 25 '13 at 16:34
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Correct; if no such sequence of points exists then there must be an interval where $h(x)=0$. – vadim123 Apr 25 '13 at 17:58