# Does convergence to $f(x)$ of Taylor series centered in $x_0$ in a interval $I$ imply that $f(x)$ is analytic in all $I$?

A function $f: (a,b) \subset \mathbb{R} \to \mathbb{R}$ with $f \in C^{\infty}(a,b)$ is said to be analytic in $x_0\in (a,b)$ if $\exists \delta >0$ such that the Taylor series of $f(x)$ centered in $x_0$ converges in $(x_0-\delta,x_0+\delta)$ and its sum is $f(x)$.

My question is:

Suppose that I already know that the Taylor series of a function $f\in C^{\infty}$ centered in $x_0$ $$\sum_{n \geq 0} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n$$ converges $\forall x \in (x_0-\xi,x_0+\xi)$ for some $\xi>0$ and that $$f(x)=\sum_{n \geq 0} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n \,\,\,\,\,\,\,\,\,\,\,\,\,\forall x \in (x_0-\xi,x_0+\xi)$$

Can I conclude that $f(x)$ is analytic in all the interval $(x_o-\xi,x_0+\xi)$?

Attempt: I think that the convergence of Taylor series centered in a particular $x_0 \in I$ in all $I$ does not imply the fact that $f$ is analytic in all an interval $I$. Is this correct?

For example, chosen one particular $x_0 \in \mathbb{R}$ I think that the following implication is incorrect.

$$e^{x}= \sum_{n \geq 0} \frac{e^{x_0}}{n!} (x-x_0)^n \,\,\,\,\,\forall x \in \mathbb{R} \implies e^{x} \,\,\,\mathrm{is}\,\,\, \mathrm{analytic} \,\,\, \mathrm{in} \,\,\, \mathbb{R}$$

While it should be correct to say that

$$e^{x}= \sum_{n \geq 0} \frac{e^{x_0}}{n!} (x-x_0)^n \,\, ,\,\,\forall x_0 \in \mathbb{R}\,\, ,\,\,\forall x \in \mathbb{R} \implies e^{x} \,\,\,\mathrm{is}\,\,\, \mathrm{analytic} \,\,\, \mathrm{in} \,\,\, \mathbb{R}$$

It is much easier to forget about $f$, and just start with a formal power series $\sum_{n\in\Bbb N}c_n(x-x_0)^n$ in $x-x_0$, which converges for $x$ in an open interval $I$ containing $x_0$. You are asking if, for a given $x_1\in I$, the function it converges to is again given by a power series in $(x-x_1)$ in a neighbourhood of$~x_1$.
The answer is yes. Given the hypotheses, the radius$~r$ of convergence of the power series with coefficients $(c_n)_{n\in\Bbb N}$ is such that $|x-x_0|<r$ for all $x\in I$. In particular this holds for $x_1$, and the function is given by a power series around $x_1$ with radius of convergence at least $r-|x_1-x_0|$, and where the two series both converge, they converge to the same value. So contrary to what you think in the last part of your question: once a power series converges everywhere in some open interval, the function that it converges to is automatically analytic in every point of that interval (and whenever a function$~f$ is given by a power series in$~x-x_0$, the series is in fact the Taylor series of$~f$ around$~x_0$).
To see the concrete relation between the coefficients, and why the power series give the same values where they both converge, one can treat them somewhat like formal power series, in that separate powers of shifted versions $x$ are treated separately; the condition that one is within the radius of convergence will ensure absolute convergence of the sums, and justify the necessary interchanging of summations. So if $\sum_{n\in\Bbb N}c_n(x-x_0)^n$ is to be written as a power series $\sum_{n\in\Bbb N}b_n(x-x_1)^n$ it suffices to put $d=x_1-x_0$ so that $x-x_0=(x-x_1)+d$ and then expand the former sum and gather like powers of $x-x_1$. One gets $$b_k=\sum_{n\geq k}\binom nkc_nd^{n-k},$$ which infinite sums converge because the factors $c_nd^{n-k}$ decrease exponentially ($d$ is inside the radius of convergence for $(c_n)_{n\in\Bbb N}$) while the binomial coefficients grow less than exponentially. It should come as no great surprise that if we now express $\sum_{k\in\Bbb N}b_k(x-x_1)^k$ back as a power series in $x-x_0$, we get our original series: $$\sum_{k\in\Bbb N}(x-x_0-d)^k\sum_{n\geq k}\binom nkc_nd^{n-k} =\sum_{k\in\Bbb N}\left(\sum_{m=0}^k\binom km(x-x_0)^m(-d)^{k-m}\right)\sum_{n\geq k}\binom nkc_nd^{n-k} \\= \sum_{m\in\Bbb N}(x-x_0)^m\left(\sum_{k\geq m}\binom km(-d)^{k-m} \sum_{n\geq k}\binom nkc_nd^{n-k}\right) \\= \sum_{m\in\Bbb N}(x-x_0)^m\left( \sum_{n\geq m}c_nd^{n-m}\sum_{k=m}^n\binom nk\binom km(-1)^{k-m}\right) \\= \sum_{m\in\Bbb N}(x-x_0)^m\left( \sum_{n\geq m}c_nd^{n-m}\delta_{m,n}\right) =\sum_{m\in\Bbb N}(x-x_0)^mc_m,$$ wher I used the known binomial coefficient identity $\sum_{k=m}^n\binom nk\binom km(-1)^{k-m}=\delta_{m,n}$. That is, whenever $x$ is such that we are within the radius of convergence of both series, they converge to the same value.
• Thanks for the answer! If I may ask, could you explain a little further the proof of this fact? How $f(x)=\sum_{n \geq 0} c_n (x-x_0)^n$ $\forall x \in (x_0-r,x_0+r)\implies f(x)=\sum_{n \geq 0} b_n (x-x_1)^n$ $\forall x \in (x-(r-|x_1-x_0|),x+(r-|x_1-x_0|))$ with $x_1 \in (x_0-r,x_0+r)$? In particular it must be $(c_n)_{n \in \mathbb{N}}=(\frac{f^{(n)}(x_0)}{n!})_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}=(\frac{f^{(n)}(x_1)}{n!})_{n \in \mathbb{N}}$ but, from the reasoning in the answer, I can't see how this relation between $c_n$ and $b_n$ is respected – Gianolepo Dec 10 '16 at 14:30
• I've expanded to show the relation between the sequences $c_n$ and $b_n$. Note that I never use a description of coefficients in terms of derivatives; that is not useful here. – Marc van Leeuwen Dec 10 '16 at 21:08