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Given the function $$f(x)=\begin{cases} 1, \space\space\space\space 0\leq |x|\leq 1/4 \\ -1, \space 1/4< |x|\leq 1/2 \end{cases}$$

I am asked to expand the function $f(x)$ as a series of cosine. ( I am studying Fourier series). Knowing it is an even function, I have expanded it and I have $$Sf(x)=4\sum_{k=0}^{\infty}\frac{(-1)^k}{\pi (2k+1)}\cos(2\pi x(2k+1))$$Now I am asked to calculate $$\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)}$$ and $$\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$ How do I calculate it?

user
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codingnight
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2 Answers2

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For the first one, take $x=0$. Note that $Sf(0)=f(0)=1$ if you apply any pointwise convergence criterion.

For the second one, apply Parseval's identity.

Eclipse Sun
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1) Set $x=0$: $$\begin{array}{} 1=f(0)=S f(0)=\frac4\pi\sum\limits_{k=0}^{\infty}\frac{(-1)^k}{2k+1} \Rightarrow \sum\limits_{k=0}^{\infty}\frac{(-1)^k}{2k+1}=\frac\pi4. \end{array} $$

2)Integrate over $x$ from $0$ to $\frac14$:

$$\begin{array}{} \frac14&=\int\limits_0^{1/4}f(x)dx=\int\limits_0^{1/4}Sf(x)dx\\ &=\frac4\pi\sum\limits_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\left[\frac{\sin(2\pi(2k+1)x)}{2\pi(2k+1)}\right]_0^{1/4} =\frac{2}{\pi^2}\sum\limits_{k=0}^{\infty}\frac{1}{(2k+1)^2}\\ &\Rightarrow \sum\limits_{k=0}^{\infty}\frac{1}{(2k+1)^2}=\frac{\pi^2}8. \end{array} $$

user
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