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Suppose that $f:\mathbb{R}\to\mathbb{R}$ is analytic at $x=0$, and $T(x)$ its Taylor series at $x=0$, with radius of convergence $R>0$. Is it true that $f(x)=T(x)$ whenever $|x|<R$ ?

José Carlos Santos
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ashpool
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2 Answers2

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Yes, it is true. It follows from the definition of analytic function that the equality $f(x)=T(x)$ holds on some interval $(-r,r)$. But then you can use the identity theorem: since $(-r,r)$ has accumulation points (actually, each of its points is an accumulation point), and since $(-R,R)$ is connected, $f=T$.

José Carlos Santos
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  • Thanks. But the Wikipedia article you linked seems to say that the identity theorem does not apply to real-valued functions. – ashpool Sep 22 '19 at 09:25
  • I think I found the answer to my question here (https://math.stackexchange.com/q/739476/4341), and the proof seems highly nontrivial (and not very satisfying). – ashpool Sep 22 '19 at 09:33
  • The article is written in the context of complex analysis, but, yes, it also holds for real analytical functions. And the proof is exactly the same as in the context of complex analytical functions. – José Carlos Santos Sep 22 '19 at 09:37
  • José, would you check if my answer is correct? – Botond Sep 22 '19 at 11:19
  • @Botond I have no idea what the last sentence means. – José Carlos Santos Sep 22 '19 at 11:30
  • I wanted to express that $T=f$ on $(-r, r)$. But for all $x$ in $(-R, R)$, we can find an $r$ so that $x<r<R$, which means that $T=f$ on $(-R,R)$. – Botond Sep 22 '19 at 11:32
  • From what you did, it follows that the equality $f(x)=T(x)$ holds on some interval $(-r,r)$ (assuming that $f$ is analytic), but I don't see how you jump from that to the assertion that the equality holds for every $x\in(-R,R)$. – José Carlos Santos Sep 22 '19 at 11:37
  • I think I fixed it now. The $r \in (0, R)$ is only needed to get that $f^{(n)}(0)=n! a_n$. – Botond Sep 22 '19 at 11:46
  • How do you know that $\bigl(\forall x\in(-R,R)\bigr):f(x)=T(x)$? – José Carlos Santos Sep 22 '19 at 11:48
  • In general, nothing guarantees that. But let $y \in (-R, R)$. Then we have that $f$ is defined at $y$ and $\sum_{n \in \mathbb{N}}a_n y^n = f(y)$. Which means that $\sum_{n \in \mathbb{N}}a_n y^n =T(y)$, so $T$ is convergent at $y$ and $T(y)=f(y)$ (because if $a_n=b_n \forall n \in \mathbb{N}$, then $a_n \to x \Leftrightarrow b_n \to x$) – Botond Sep 22 '19 at 11:54
  • How do you know that $\sum_{n=0}^\infty a_ny^n=f(y)$? – José Carlos Santos Sep 22 '19 at 11:57
  • You are right, $R$ is the radius of convergence of the Taylor series, not of the series that defines $f$. Than you! – Botond Sep 22 '19 at 11:59
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No. For a counterexample, try

$$f(x) = \begin{cases}e^x & \text{ if } x < 1 \\ 0 & \text{ if } x \geq 1\end{cases}$$

Then $R = \infty,$ but $T(x) = e^x \neq f(x)$ for $x > 1.$

Brian Moehring
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