Question

From uniform convergence. anyone please help.

Answer 1

the Weierstrass function function ic.

Answer 2

Hmm???

Answer 3

@Zarkon..

Answer 4

Maybe this is a step in the right direction? For the series to be uniformly convergent, you must have for any \(\epsilon>0\) and \(x\) in the domain some \(N\) such that
\[\large\left|S_p(x)-S_r(x)\right|<\epsilon\]
where \(S_k(x)\) denotes the \(k\)th partial sum.

You have
\[\large\begin{align*}\left|\sum_{n=1}^pb^n\cos(a^n\pi x)-\sum_{n=1}^rb^n\cos(a^n\pi x)\right|&=\left|\sum_{n=r+1}^pb^n\cos(a^n\pi x)\right|\\\\
&\le\sum_{n=r+1}^p\left|b^n\cos(a^n\pi x)\right|\\\\
&=\sum_{n=r+1}^pb^n\left|\cos(a^n\pi x)\right|\\\\
&\le\sum_{n=r+1}^pb^n\\\\
&=\frac{b\left(b^n-b^m\right)}{b-1}<\epsilon
\end{align*}\]
so you know that whatever \(N\) you end up choosing, it will be independent of \(x\).

Answer 5

That last line should read
\[\large\frac{b(b^p-b^r)}{b-1}\]

Answer 6

Oh and there's slight error in the definition I wrote above for uniform convergence. It should say for any \(\epsilon\) there exists \(N\) for all \(x\), etc. (The order matters, I believe...)

Answer 7

@SithsAndGiggles Cauchy's criterion??

Answer 8

Am I remembering things incorrectly? Sorry, it's been a while since I've taken a course in real analysis.