Asymptotics of the Weierstrass function

Question

It is known that the Weierstrass function

$$f (x) =\sum_{n\geq 1 } a^n \cos (b^n x ) , $$

with $0 < a < 1$, $ab>1$, is nowhere differentiable. But this should not prevent us from studying its asymptotic behaviors.

So, the question is, can we find a asymptotic formula for $f$ as $x\rightarrow 0 $?

Answer 1

no, the derivative tends to infinite at anywhere. and the condition you give may be hard to do inequality scaling, it should be $ab≥1+π$ instead

for $x\in[\pi\frac{j}{b^{n}},\pi\frac{j+1}{b^{n}}]$, and $n→∞$

we have $f^{'}(x)=\lim\limits_{n\to\infty}|\frac{f(\pi\frac{j+1}{b^{n}})-f(\pi\frac{j}{b^{n}})}{\pi\frac{j+1}{b^{n}}-\pi\frac{j}{b^{n}}}|≥\lim\limits_{n\to\infty}\frac{a^{n}b^{n}}{\pi}≥∞$

so we need to prove that $|{f(\pi\frac{j+1}{b^{m}})-f(\pi\frac{j}{b^{m}})}|≥a^m$ for every $m≥1$

it'll be quite long, you can skip it

here we go

$|{f(\pi\frac{j+1}{b^{m}})-f(\pi\frac{j}{b^{m}})}|$

$=|\sum\limits_{n=1}^{m-1}a^{n}(cos(b^{n-m}\pi(j+1))-cos(b^{n-m}\pi j))+a^{m}(cos(j+1)\pi-cos(j\pi))+\sum\limits_{n=m+1}^{∞}a^{n}(cos(b^{n-m}\pi(j+1))-cos(b^{n-m}\pi j))|$

$=|\sum\limits_{n=1}^{m-1}a^{n}(cos(b^{n-m}\pi(j+1))-cos(b^{n-m}\pi j))+a^{m}((-1)^{j+1}-(-1)^{j})|$

$≥||\sum\limits_{n=1}^{m-1}a^{n}(cos(b^{n-m}\pi(j+1))-cos(b^{n-m}\pi j))|-|a^{m}((-1)^{j+1}-(-1)^{j})||$

for any $x,y\in\mathbb{R}$, $ξ\in(x,y)$, we have $|\frac{cos(x)-cos(y)}{x-y}|=|cos^{'}(ξ)|=|-sin(ξ)|≤1$

so $|cos(x)-cos(y)|≤|x-y|$

$|\sum\limits_{n=1}^{m-1}a^{n}(cos(b^{n-m}\pi(j+1))-cos(b^{n-m}\pi j))|≤\sum\limits_{n=1}^{m-1}a^{n}|cos(b^{n-m}\pi(j+1))-cos(b^{n-m}\pi j))|$

$≤\sum\limits_{n=1}^{m-1}a^{n}|b^{n-m}\pi|$

$=\frac{\pi}{b^m}\sum\limits_{n=1}^{m-1}(ab)^{n}$

$=\frac{\pi}{b^m}\frac{ab(1-(ab)^{m-1})}{1-ab}$

$=\frac{\pi}{ab-1}\frac{(ab)^{m}-ab}{b^m}$

$\frac{\frac{\pi}{ab-1}\frac{(ab)^{m}-ab}{b^m}}{a^{m}(|(-1)^{j+1}-(-1)^{j})|}=\frac{\frac{\pi}{ab-1}\frac{(ab)^{m}-ab}{b^m}}{2a^{m}}$

$=\frac{\pi}{2(ab-1)}(1-(ab)^{1-m})≤\frac{\pi}{2(ab-1)}＜1$, for $ab≥1+π$

so $|{f(\pi\frac{j+1}{b^{m}})-f(\pi\frac{j}{b^{m}})}|≥2a^{m}-|\sum\limits_{n=1}^{m-1}a^{n}(cos(b^{n-m}\pi(j+1))-cos(b^{n-m}\pi j))|$

$\frac{\frac{\pi}{ab-1}\frac{((ab)^m-ab)}{b^m}}{a^m}≤\frac{\pi}{ab-1}≤1$

so $|{f(\pi\frac{j+1}{b^{m}})-f(\pi\frac{j}{b^{m}})}|≥2a^{m}-a^m＝a^m$

thus we complete the proof