# General question on Taylor Series

The Taylor Series comes from an assumption that a function has an expression as power series. Given such assumption we can then say that the $n$-th derivative and evaluate them at $x = a$, it can give us the coefficient on the $n$-th term when the function converges on $a$. How do we know that every function has a power series?

• I think you might find a note I wrote for one of my classes on Taylor series and polynomials instructive. – davidlowryduda Aug 23 '15 at 5:00
• @mixedmath. Thanks for the link ! Very good material, indeed. – Claude Leibovici Aug 23 '15 at 6:15
• Yes, very good material! Thank you! – ChaoSXDemon Aug 23 '15 at 6:29

## 2 Answers

Verily, the story begins at the Taylor polynomials.

Let $a \in \mathbb{R}$; let $N(a)$ be a 1-neighborhood of $a$; let $f: N(a) \to \mathbb{R}$; and let $D^{n}f(a)$ exist. Then it can be shown that (try to prove it) the polynomial $p_{n}: x \mapsto f(a) + \sum_{1}^{n}D^{k}f(a)(x-a)^{k}/k!$ on $N(a)$ is the unique choice such that $$D^{k}p_{n}(a) = D^{k}f(a)$$ for all $0 \leq k \leq n$. And $p_{n}$ is called the Taylor polynomial of degree $n$ generated by $f$ at $a$.

The problem is about the remainder, or the error term $E_{n}: x \mapsto f(x) - p_{n}(x)$ on $N(a)$. Thanks to a theorem neighboring on the mean-value theorem for integrals, under suitable conditions we can write $E_{n}$ in the so-called Lagrange form $$\frac{D^{n+1}f(c)(x-a)^{n-1}}{(n+1)}.$$ And it can shown that, under suitable conditions, if there is some $M \geq 0$ such that $\sup_{x \in N(a)}|D^{k}f(x)| \leq M^{k}$ for all $k \geq 1$, then the error term vanishes as $k$ grows, and hence we can express $f$ as its Taylor polynomial to any degree at $a$.

• What's D and also what's 1-neighborhood? – ChaoSXDemon Aug 23 '15 at 5:03
• For convenience I write $D^{n}f$ for $f^{(n)}$, the $n$-th derivative of $f$. A 1-neighborhood of $a$ is a set of the form $\{ x \in \mathbb{R} \mid |x-a| < \varepsilon \}$. – Benicio Aug 23 '15 at 5:06
• To prove what you've asked ... noticed that if we take the n-th derivative all the terms that has (x-a)^(p) where p > 0 will become zero since we are saying x = a. This leaves us the nth coefficient. Since the n-th derivative is known to exist, this coefficient exists. Thus the polynomial? – ChaoSXDemon Aug 23 '15 at 5:18

We don't. For example, the Weierstrass function will have no power series that works everywhere on the real numbers. This is true for any non-differentiable function, e.g. $f(x) = |x|$ will need different power series for different domains, but no one single one will work...