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Problem

It can be shown that the function

$$ f(t) = \begin{cases} \dfrac{\pi}{2} + t & , -\pi < t < 0,\\ \dfrac{\pi}{2} - t & , 0 < t < \pi, \end{cases} $$

has Fourier series $FS_f(t) = \dfrac{4}{\pi} \sum^{\infty}_{n = 1, odd} \dfrac{\cos(nt)}{n^2}$.

Use integration to find which function has Fourier series $\dfrac{4}{\pi} \sum^{\infty}_{n = 1, odd} \dfrac{\sin(nt)}{n^3}$.


I understand that term-by-term integration of a $2\pi$-periodic Fourier series is the formula $\int^t_{-\pi} f(\alpha) \ d\alpha = \sum_{n = 1}^{\infty} \dfrac{a_n}{n}\sin(nt) - \sum_{n = 1}^{\infty} \dfrac{b_n}{n}(\cos(nt) - \cos(n\pi))$. However, I do not understand how to solve this problem.

I would greatly appreciate it if people could please take the time to explain the reasoning involved in solving this problem.

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  • $\begingroup$ You might consider rewriting your Fourier series as the Fourier transform of a function and then using the Fourier inversion theorem. You can use dominated convergence to swap the integral with the sum in your inverse Fourier transform. $\endgroup$ – Michael Lee May 21 '17 at 4:58
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    $\begingroup$ The result is a primitive function of $f(t)$ made of second degree polynomials. Take care to find the adequate constant of integration $\endgroup$ – Jean Marie May 21 '17 at 5:07
  • $\begingroup$ @JeanMarie But how do I solve such a problem in the first place? $\endgroup$ – The Pointer May 21 '17 at 5:09
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    $\begingroup$ a primitive of f(t) is $(\pi/2)t \pm t^2/2 + C$, that's all... $\endgroup$ – Jean Marie May 21 '17 at 5:11
  • $\begingroup$ @JeanMarie I knew there must have been some reason the Fourier series of $f$ was given. Silly me. My method would work in the general case where you might not have been given a hint like the one here. $\endgroup$ – Michael Lee May 21 '17 at 5:12
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Using the property that the function evaluated at $t=0$ is the average of the left and right piece-wise function evaluated at zero and $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 \, n + 1)^3} = \frac{\pi^3}{32}$$ then the following is obtained:

Integrating $$ f_{c}(t) = \begin{cases} \frac{\pi}{2} + t & , -\pi < t < 0,\\ \frac{\pi}{2} - t & , 0 < t < \pi, \end{cases} $$ becomes $$ f_{s}(t) = \begin{cases} c_{0} + \frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\ c_{1} + \frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi. \end{cases} $$ Now, using $f_{s}(0) = (f_{s}(0-) + f_{s}(0+))/2$, then $c_{1} = - c_{0}$ and $$ f_{s}(t) = \begin{cases} c_{0} + \frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\ -c_{0} + \frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi. \end{cases} $$ In order to determine $c_{0}$ evaluate the series at the point $t = \pi/2$ leading to $$f_{s}\left(\frac{\pi}{2}\right) = - c_{0} + \frac{\pi^2}{4} - \frac{\pi^2}{8} = - c_{0} + \frac{\pi^2}{8} $$ From the series it is determined that $$f_{s}(t) = \frac{4}{\pi} \, \sum_{n=0}^{\infty} \frac{\sin(2\,n +1)t}{(2\, n +1)^3}$$ and $$f_{s}\left(\frac{\pi}{2}\right) = \frac{\pi^2}{8}.$$ This yields $c_{0}=0$ and $$ f_{s}(t) = \begin{cases} \frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\ \frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi. \end{cases} $$

One can verify this by using \begin{align} f_{s}(t) &= \sum_{n=0}^{\infty} B_{n} \, \sin((2\, n +1) \, t) \\ B_{n} &= \frac{1}{\pi} \, \int_{- \pi}^{\pi} g(t) \, \sin((2 \, n +1) t) \, dt \\ g(t) &= \begin{cases} \frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\ \frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi. \end{cases} \end{align}

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