Editing Non-analytic smooth function (section)

==A smooth function that is nowhere real analytic==
[[File:Smooth non-analytic function.png|thumb|Approximation of the smooth-everywhere, but nowhere-analytic function mentioned here. This partial sum is taken from {{math|''k'' {{=}} 2<sup>0</sup>}} to 2<sup>500</sup>.|right]]

A more [[Pathological (mathematics)|pathological]] example is an infinitely differentiable function which is not analytic ''at any point''. It can be constructed by means of a  [[Fourier series]] as follows. Define for all <math>x \in \mathbb{R}</math>
:<math>F(x):=\sum_{k\in \mathbb{N}} e^{-\sqrt{2^k}}\cos(2^k x)\ .</math>

Since the series <math>\sum_{k\in \mathbb{N}} e^{-\sqrt{2^k}}{(2^k)}^n</math> converges for all <math>n \in \mathbb{N}</math>, this function is easily seen to be of class  C<sup>∞</sup>,  by a standard inductive application of the [[Weierstrass M-test]] to demonstrate [[uniform convergence]] of each series of derivatives.

We now show that <math>F(x)</math> is not analytic at any [[dyadic rational]] multiple of π, that is, at any <math>x := \pi \cdot p \cdot 2^{-q}</math> with <math>p \in \mathbb{Z}</math> and <math>q \in \mathbb{N}</math>.  Since the sum of the first <math>q</math> terms is analytic, we need only consider <math>F_{>q}(x)</math>, the sum of the terms with <math>k>q</math>.  For all orders of derivation <math>n = 2^m</math> with <math>m \in \mathbb{N}</math>, <math>m \geq 2</math> and <math>m > q/2</math> we have

:<math>F_{>q}^{(n)}(x):=\sum_{k\in \mathbb{N}\atop k>q} e^{-\sqrt{2^k}} {(2^k)}^n\cos(2^k x)  = \sum_{k\in \mathbb{N}\atop k>q} e^{-\sqrt{2^k}} {(2^k)}^n \ge  e^{-n} n^{2n}\quad  (\mathrm{as}\; n\to \infty)</math>

where we used the fact that <math>\cos(2^k x) = 1</math> for all <math>2^k > 2^q</math>, and we bounded the first sum from below by the term with <math>2^k=2^{2m}=n^2</math>. As a consequence, at any such <math>x \in \mathbb{R}</math>

:<math>\limsup_{n\to\infty} \left(\frac{|F_{>q}^{(n)}(x)|}{n!}\right)^{1/n}=+\infty\, ,</math>
so that the [[radius of convergence]] of the [[Taylor series]] of  <math>F_{>q}</math>  at  <math>x</math>  is  0  by the [[Cauchy-Hadamard theorem#Statement of the theorem|Cauchy-Hadamard formula]]. Since the set of analyticity of  a function  is an open set, and since dyadic rationals are [[Dense set|dense]], we conclude that  <math>F_{>q}</math>, and hence <math>F</math>, is nowhere analytic in <math>\mathbb{R}</math>.