I've a fourier series with a period = $2\pi$ that is even. f(t) = \begin{cases} 0 \text{, when: } 0<t<\pi-2 \\ \pi \text{, when: } \pi-2<t<\pi \end{cases} The functions trigonometric fourier series equals:

$2 + 2\sum_{k=1}^{\infty} (-1)^k\cos(kt)\cdot\frac{\sin(2k)}{k}$

What is the value of the series when $t = \pi-2$?

I tried to insert the value of t in the series and use Parevals formula and calculate the integral but I didn't manage to handle the integral. What should I do to calculate the sum of the series?

  • $\begingroup$ You don't need to use the series. What do you know about the value of a fourier series at a point of discontinuity of the function? $\endgroup$ – Paul Oct 2 '14 at 14:21

I don't think you need to derive a 'bonus sum' using Parseval's formula here - note that the function satisfies Dirichlet's conditions on the open interval $ (0,\pi) $ (prove this), and note that the function is discontinuous at $ t= \pi -2 $, and recall that the Fourier series converges to the average of the left and right limits at any point of discontinuity.

  • $\begingroup$ We haven't gone through Dirichlet's conditions so even if it might work I don't think I'm supposed to use it unfortunately :/ You don't happen to have another method to calculate this? :) $\endgroup$ – Superovan Oct 2 '14 at 14:39
  • $\begingroup$ The value of a fourier series at a discontinuity (jump) is its average value. $\endgroup$ – Paul Oct 2 '14 at 14:41
  • $\begingroup$ Ah so it seems. Thanks for the fast Response Paul, and sorry for not responding to your first comment (was going to). :) $\endgroup$ – Superovan Oct 2 '14 at 14:46
  • $\begingroup$ @user139094 - I guess if you haven't covered Dirichlet's conditions, stating what Paul said in your answer is fine $\endgroup$ – Eweler Oct 2 '14 at 14:51
  • $\begingroup$ @Eweler - I've been teaching engineers too long:) $\endgroup$ – Paul Oct 2 '14 at 14:55

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