What is a necessary and sufficient condition for a Taylor series to exist?

Question

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?

Do you know the notion of analytic functions? You will know this only if you have studied complex variables subject. — Mhenni Benghorbal, Jan 04 '15 at 09:23
See Why don't taylor series represent the entire function? for some counter examples. — Martin R, Jan 04 '15 at 09:25
@Mhenni, what is gained by rephrasing the question as 'what is a necessary and sufficient condition for a function to be analytic?' — Titus, Jan 04 '15 at 09:26
You might be interested in Theorem 8.4, Rudin's PMA, 3rd ed. — hrkrshnn, Jan 04 '15 at 09:31
@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 — Quantum spaghettification, Jan 04 '15 at 09:35

score 10 · Accepted Answer · answered Jan 04 '15 at 10:18

10

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.

answered Jan 04 '15 at 10:18

Ittay Weiss

81,796

If all the derivatives are continuous, does this tell us any more e.g. there are some points where it converges? – Quantum spaghettification Jan 04 '15 at 10:22
not necessarily. – Ittay Weiss Jan 04 '15 at 10:23
Is there any condition it must satisfy for it to converge at a point, other then $a$. – Quantum spaghettification Jan 04 '15 at 10:30
tautologically, being analytic. – Ittay Weiss Jan 04 '15 at 10:31

score 0 · Answer 2 · answered Jan 04 '15 at 09:39

0

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.

answered Jan 04 '15 at 09:39

Frog

265

4

But this only applies if we already know that $f$ can be expressed as a power series at some point. – Matt Samuel Jan 04 '15 at 09:42

What is a necessary and sufficient condition for a Taylor series to exist?

2 Answers2