# I Convergence of Taylor series in a point implies analyticity

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1. Dec 10, 2016

### crick

Suppose that the Taylor series of a function $f: (a,b) \subset \mathbb{R} \to \mathbb{R}$ (with $f \in C^{\infty}$), centered in a point $x_0 \in (a,b)$ converges to $f(x)$ $\forall x \in (x_0-r, x_0+r)$ with $r >0$. That is
$$f(x)=\sum_{n \geq 0} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n \,\,\,\,\,\,\,\,\,\,\,\,\,\forall x \in (x_0-r,x_0+r)$$
(In particular, I do not know in advance anything about the analyticity of $f(x)$ in $(a,b)$, but only the convergence of the Taylor series in $x_0$.)

Does this imply that, chosen a $x_1 \in (x_0-r,x_0+r)$, the Taylor series of $f(x)$ centered in $x_1$ converges at least in $(x_1-r',x_1+r')$ with $r'=r-|x_0-x_1|$? That is
$$f(x)=\sum_{n \geq 0} \frac{f^{(n)}(x_1)}{n!} (x-x_1)^n \,\,\,\,\,\,\,\,\,\,\,\,\,\forall x \in (x_1-r',x_1+r')$$
If this is true, how can I prove it?

2. Dec 11, 2016

### stevendaryl

Staff Emeritus
Fooling around with the power series, I see that the conclusion is easily provable if you additionally know the following:
1. The series for $f(x)$ is absolutely convergent (so that you can reorder terms and get the same answer).
2. The series for $\frac{df}{dx}, \frac{d^2 f}{dx^2}, ...$ are also absolutely convergent.
As to point two--can you prove that if $f(x)$ is equal to its Taylor series at a point, then $f^{(n)}(x)$ is equal to its Taylor series at that point?

As to point one--I'm not sure whether you can get away without assuming that, or not.