how do I verify that this converges uniformly to $f(x)$?

Question

I had to find the fourier series for $f(x)=|x|, -\pi \le x \le \pi$
I got the fourier series as $$f(x)=\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)$$ but now I have to verify that this converges uniformly to $f(x)$ and I have to evaluate the series when $x=0$. I do not know how to show convergence but for evaluating the series at $x=0$ all I have to do is find the first few terms of the sum and substitute $0$ for $x$?

Answer 1

You may find p. 4 here useful.

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Anyway,

$$|\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)|$$

$$\le \frac{\pi}{2}+|\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)|$$

$$\le \frac{\pi}{2}+\sum_{n=0}^{\infty}|\frac{4}{\pi(2n+1)^2}cos((2n+1)x)|$$

$$\le \frac{\pi}{2}+\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}|cos((2n+1)x)|$$

$$\le \frac{\pi}{2}+\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}$$

$$\le \frac{\pi}{2}+\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$$

Now

$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} < \infty$$

Hence

$$\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} < \infty$$

$\therefore$, by the Weierstrass M-test,

$$\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)$$

converges uniformly QED

Answer 2

HINT: Notice that the terms of the series drop off like $n^{-2}$. Can you use this fact to get a uniform (small) bound on $\sup_x|f(x)-S_N(x)|$, where $S_N(x)$ is the $N-$th partial sum?