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I am trying to decide when a function can be written as a Taylor series. I think it exists if the following condition is met:

For a Taylor series of $f(x)$ about the point $a$

In the region $R$ containing both $x$ and $a$, the function $f(x)$ is single-valued with an infinite number of continuous derivatives that all exist.

Is this both a necessary and sufficient condition? If not then what is?

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  • $\begingroup$ Do you know the notion of analytic functions? You will know this only if you have studied complex variables subject. $\endgroup$ Jan 4, 2015 at 9:23
  • $\begingroup$ See Why don't taylor series represent the entire function? for some counter examples. $\endgroup$
    – Martin R
    Jan 4, 2015 at 9:25
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    $\begingroup$ @Mhenni, what is gained by rephrasing the question as 'what is a necessary and sufficient condition for a function to be analytic?' $\endgroup$
    – Titus
    Jan 4, 2015 at 9:26
  • $\begingroup$ See my How to Answer where you can find a nice fact. $\endgroup$ Jan 4, 2015 at 9:29
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    $\begingroup$ @MhenniBenghorbal Yes I do know what an analytic function but it is simply defined as a function that has a Taylor series... so as Titus does not change the question $\endgroup$ Jan 4, 2015 at 9:35

2 Answers 2

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Any function which is infinitely differentiable at a point $a$ has a Taylor series at that point. Whether or not the Taylor series converges at any point other than $a$ is a different issue. But for the existence of a Taylor series all you need is the coefficients to exist, and these only require knowing the derivatives of the function at that point, so this is your sufficient condition. It is of course also necessary since if the function has a Taylor series, then the coefficients contain all higher derivatives at the point.

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  • $\begingroup$ If all the derivatives are continuous, does this tell us any more e.g. there are some points where it converges? $\endgroup$ Jan 4, 2015 at 10:22
  • $\begingroup$ not necessarily. $\endgroup$ Jan 4, 2015 at 10:23
  • $\begingroup$ Is there any condition it must satisfy for it to converge at a point, other then $a$. $\endgroup$ Jan 4, 2015 at 10:30
  • $\begingroup$ tautologically, being analytic. $\endgroup$ Jan 4, 2015 at 10:31
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Theorem 8.4, Principles of Mathematical analysis says:

Suppose $$ f(x) = \sum\limits_{n=0}^{\infty} c_n x^n,$$ the series converging in $|x|<R$. If $-R<a<R$, then $f$ can be expanded in a power series about point $a$ which converges in $|x-a|<R-|a|$, and $$f(x)= \frac{f^{(n)}(a)}{n!}(x-a)^n \quad (|x-a|<R-|a|)$$

We can consider this as a necessary and sufficient condition.

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    $\begingroup$ But this only applies if we already know that $f$ can be expressed as a power series at some point. $\endgroup$ Jan 4, 2015 at 9:42

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