5
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ I think you might find a note I wrote for one of my classes on Taylor series and polynomials instructive. $\endgroup$ – davidlowryduda Aug 23 '15 at 5:00
  • $\begingroup$ @mixedmath. Thanks for the link ! Very good material, indeed. $\endgroup$ – Claude Leibovici Aug 23 '15 at 6:15
  • $\begingroup$ Yes, very good material! Thank you! $\endgroup$ – ChaoSXDemon Aug 23 '15 at 6:29
1
$\begingroup$

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 $a$.

$\endgroup$
  • $\begingroup$ What's D and also what's 1-neighborhood? $\endgroup$ – ChaoSXDemon Aug 23 '15 at 5:03
  • $\begingroup$ 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 $\{ x \in \mathbb{R} \mid |x-a| < \varepsilon \}$. $\endgroup$ – Benicio Aug 23 '15 at 5:06
  • $\begingroup$ 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? $\endgroup$ – ChaoSXDemon Aug 23 '15 at 5:18
0
$\begingroup$

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.