I would like to know how to calculate the derivative of a non-analytic smooth function?
Suppose $f:\mathbb R\rightarrow \mathbb R$ is in $\mathcal C^\infty\backslash \mathcal C^\omega$ and in particular has no Taylor series expansion at $x$. The right (left) derivative at point $x$ is:
\begin{equation} f'(x)=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\Big(f(x+\epsilon)-f(x)\Big) \end{equation}
The limit, due to smoothness, exists even though we can't expand, but there is no way to find the limit by means of a calculation or is there?
thanks a lot!
It's not clear to me what your asking but maybe this will help. The classic example of nonanalytic smooth function is $f : \mathbb{R} \to \mathbb{R}$ where $f(x) = e^{\frac{-1}{x}}$ on $(0, \infty)$ and $f(x) =0$ on $(-\infty, 0]$. In particular the function isn't analytic at $0$ but we are still able to calculate derivatives.