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Let $f:[a,b]\rightarrow \mathbb R$. We say that $f$ is $\mathbb R$-analytic if for each $x_0 \in [a,b]$ there is $R(x_0)>0$ and power series $\sum_{k=0}^\infty c_k(x_0)(x-x_0)^k$ convergent for $x\in (x_0-R(x_0), x_0+R(x_0))$ such that $f(x)=\sum_{k=0}^\infty c_k(x_0)(x-x_0)^k$ for $x\in [a,b] \cap (x_0-R(x_0), x_0+R(x_0))$.

I wish to show elementary, without the Cauchy formula and complex analysis's theorems, that if $f:[a,b]\rightarrow \mathbb R$ is analytic then there is $R>0$, non depending on $x_0$, such that for each $x_0 \in [a,b]$ there is a power series $\sum_{k=0}^\infty c_k(x_0)(x-x_0)^k$ converegent for $x\in (x_0-R, x_0+R)$ such that $f(x)=\sum_{k=0}^\infty c_k(x_0)(x-x_0)^k$ for $x\in [a,b] \cap (x_0-R, x_0+R)$.

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up vote 1 down vote accepted

Use compactness of $[a,b]$: the given intervals $I_{x_0}=\big(x_0-R(x_0),x_0+R(x_0)\big)$ for all $x_0$ definitely cover $[a,b]$, so finitely many can be chosen that already covers it. Then take the minimum diameter of all the intersections of those finitely many intervals (even the number of all possible intersections is finite).

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You have to be a little more careful, since the the diameter of the intersection of these intervals can be much smaller than the smallest $R$. However, there are only finitely many of those intersections, so you still get a minimum size. – Lukas Geyer Oct 29 '12 at 12:50

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