0

I want to ask if there is a condition that is necessary and sufficient to prove that a function $F(x)$ can be can be decomposed into power series ? If yes, tell me the condition please, thanks in advance!

Hptunjy Prjkeizg
  • 171

1 Answers1

1

For functions $F : \mathbb{R} \to \mathbb{R}$, being $C^{\infty}$ is, as you noted, necessary but not sufficient.

The nice frame in which to study functions with convergent Taylor series, is that of complex analysis. Because for a function $F : \mathbb{C} \to \mathbb{C}$, as soon as it is differentiable everywhere, its Taylor series (in every point!) converges to $F$ globally. Here one could say differentiability is necessary and sufficient.

Of course, this tells you nothing in the case you're not familiar with complex functions. For example, there's no way you could know that the complex sine function is differentiable if you don't know what complex sine function and complex derivative are. But it can give you some intuition (and theorems) about which real functions cannot be expressed globally as a power series.

  • The function $\ln : ]0, +\infty[ \to \mathbb{R}$ can never have a power series converging on all of its domain, because it can not be extended to a complex function differentiable at $0$, since $\lim_{x \to 0^+} \ln x = -\infty$. The same argument holds for any function which can only be defined (in a smooth way) on a strict subset of $\mathbb{R}$.
  • The function $\mathbb{R} \to \mathbb{R} : x \mapsto \frac{1}{1+x^2}$ is smooth, but also cannot be extended to a differentiable function $\mathbb{C} \to \mathbb{C}$, since (intuitively) such a function diverge as $x \to i$ or $x \to -i$. That's why this real function might have a Taylor series in every point, but never a globally converging one.

If you don't have the tools of complex analysis, here are some helpful statements which might come in handy when checking whether a function $F: \mathbb{R} \to \mathbb{R}$ can be written as a global Taylor series:

  • If for some $x \in \mathbb{R}$ the Taylor series around $x$ does not converge to $F$ outside $x$, then $F$ does not have a global Taylor expansion.
  • If for some $x \in \mathbb{R}$ the Taylor series around $x$ does converge for some $y \neq x$, then this series will globally converge to $F$ if and only if it globally converges, if and only if $F$ has a globally convergent Taylor series around some point.
Bib-lost
  • 3,890
  • first of all, thank you @bib-lost for your answer, so to prove that a fuction can be decomposed into power series, all i gotta do is, do the dormula $\sum \frac{f^{(n)}(0)}{n!}x^n$ and see if it converges to f or not ? – Hptunjy Prjkeizg May 18 '17 at 16:24
  • Jup! And in fact, it's sufficient that it converges to $f$ for some $x \neq 0$ and converges to some value for all $x$. – Bib-lost May 18 '17 at 18:40
  • but the answer to my question is that the necessary and sufficient condition for a function $f : \mathbb{R} \to \mathbb{R}$ to be decomposed into power series, is that $f$ has to be extendable to complex function that is differentiable at $D_f$ right ? – Hptunjy Prjkeizg May 18 '17 at 21:31
  • Yes, it is a necessary and sufficient condition for a function to have a global power series decomposition, that it can be extended to a globally differentiable complex function. But it is entirely possible to show that a function $\mathbb{R} \to \mathbb{R}$ can be written as a power series without knowing anything about complex functions. For example, on Proofwiki you can find a proof that $e^x$ has a power series representation without using anything about complex numbers: https://proofwiki.org/wiki/Power_Series_Expansion_for_Exponential_Function – Bib-lost May 19 '17 at 11:03
  • You're welcome! Don't forget you can mark an answer as 'accepted' if you feel it answered your question. – Bib-lost May 22 '17 at 12:33
  • yea sorry, i forgot XD just done it ty again – Hptunjy Prjkeizg May 24 '17 at 18:55