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Assume that there is a Taylor Series for a function $f(x)$ with centered at $x=a$, and converges for all $x$.

Then does the equation $f(b)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(b-a)^n, b \in \mathbb{R}$ hold true?

From my limited knowledge of Taylor series, it seems like this is the exact formulation of the Taylor Series, but in order for this equation to hold true, $f(x)$ has to have derivatives at this point. How do I determine if this function has derivatives at all? I know that it converges for all $x$. Is this information useful? I am not sure how to approach this question :/

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That statement is false, even if $f$ is differentiable everywhere. The standard example is$$f(x)=\begin{cases}e^{-1/x^2}&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}$$Then $(\forall n\in\Bbb N):f^{(n)}(0)=0$, and therefore$$(\forall b\in\Bbb R):\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}b^n=0\ne f(b)\text{ (unless $b=0$)}.$$

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If it's true then $f $ is called analytic. There are two standard settings, real and complex. In the first case, there are infinitely differentiable functions that are not real analytic (see José's answer). In the complex case, once differentiable (holomorphic), implies analytic.

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  • $\begingroup$ Analytic - is there also a synthetic function? $\endgroup$
    – Veritas
    Apr 11 at 0:22
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    $\begingroup$ Synthetic? Please explain. $\endgroup$
    – user403337
    Apr 11 at 0:24
  • $\begingroup$ Nevermind :D analytic/synthetic is a distinction in another subject $\endgroup$
    – Veritas
    Apr 11 at 0:25

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