Nowhere-analytic generic smooth function
= Nowhere-analytic generic smooth function
The generic derivative bound implies that at every rational $q\in(0,1)$,
$$
\limsup_m\left|\frac{f^{(m)}(q)}{m!}\right|^{1/m}=\infty.
$$
The <Cauchy-Hadamard theorem> therefore gives radius zero to the Taylor series at every such $q$. If a Taylor series represented $f$ on a neighborhood of any point, that neighborhood would contain a rational point at which $f$ was real analytic, a contradiction.