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I have a question that I'm trying to solve: enter image description here

I've been asked to show that the Fourier series is: $$f(t)=\frac{\pi}{4}-\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\frac{1}{n^2}\left ( 1-\left ( 1-in\pi \right )\left ( -1 \right )^n \right )e^{int}$$

Now, I've been working to solve the Fourier coefficient and I have arrived at:

$$c_{n}=-\frac{1}{2\pi}\left ( \frac{1}{n^2}\left ( 1-\left ( 1-in\pi \right )\left ( -1 \right )^n \right ) \right )$$

I can see by plugging it into the standard form of the Fourier series that I should have $$f(t)=-\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\frac{1}{n^2}\left ( 1-\left ( 1-in\pi \right )\left ( -1 \right )^n \right )e^{int}$$

But I'm at a complete loss for where the $\frac{\pi}{4}$ comes from in the solution I'm supposed to be showing. Can anyone explain this to me? It's not really shown in my text where these leading terms arise from. The worked examples never state the full Fourier series, just the coefficient, and subsequent sections just show these leading terms without explanation.

Also, all these sums have a $n\neq 0$ under them, but I don't know how to LaTeX that...

C. Wolfe
  • 184

1 Answers1

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The $\frac{\pi}{4}$ term represents $c_{0}$ in the series.

$c_{0}$ is the average value of the function over one period which, here, is from $-\pi$ to $\pi$.

To find the average value of a function, use this formula: $$\frac{1}{b-a}\int_{a}^{b} f(t) dt$$

Take the piecewise function in each part: $$\frac{1}{0-(-\pi)}\int_{a}^{b} -t\ dt=\frac{1}{\pi}\left [ \frac{t^2}{2} \right ]_{-\pi}^{0}=\frac{\pi}{2}$$

$$\frac{1}{\pi-0}\int_{a}^{b} 0\ dt=\frac{1}{\pi}\left [ 0 \right ]_{-\pi}^{0}=0$$

Then take the average of the two symmetrical pieces: $$\frac{\frac{\pi}{2}+0}{2}=\frac{\pi}{4}$$

This is now term $c_{0}$ which appears in front of the Fourier series.

C. Wolfe
  • 184