Sum of a serie involving Fourier coefficients

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SUMMARY

The discussion centers on evaluating the sum of the series involving Fourier coefficients of the 2-periodic function \( u(t) = t \) for \( t = 2 \). The Fourier coefficients \( \hat{u}_k \) lead to the expression \( \sum_{k=-\infty}^{\infty} \hat{u}_k e^{\pi i k t} \), which simplifies to \( \sum_{k=-\infty}^{\infty} \hat{u}_k \). At \( t = 2 \), the function is discontinuous, and the Fourier series converges to the average of the left and right limits, yielding a final result of 1 for the sum of the series.

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Mathitalian
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Homework Statement



Let \hat{u}_k the Fourier coefficients of 2-periodic function u(t)=t with t\in [0,2). Evaluate the sum of the serie:

\sum_{k=-\infty}^{\infty}\hat{u}_k e^{\pi i k t} for t= 2


Ok, I think there is a trick that I don't know...

\sum_{k=-\infty}^{\infty}\hat{u}_k e^{\pi i k t} for t= 2

becomes

\sum_{k=-\infty}^{\infty}\hat{u}_k

What can i do now? Any help will be appreciated :)
 
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Mathitalian said:

Homework Statement



Let \hat{u}_k the Fourier coefficients of 2-periodic function u(t)=t with t\in [0,2). Evaluate the sum of the serie:

\sum_{k=-\infty}^{\infty}\hat{u}_k e^{\pi i k t} for t= 2


Ok, I think there is a trick that I don't know...

\sum_{k=-\infty}^{\infty}\hat{u}_k e^{\pi i k t} for t= 2

becomes

\sum_{k=-\infty}^{\infty}\hat{u}_k

What can i do now? Any help will be appreciated :)

Draw the graph of the periodic extension of ##f(t) = t## on [0,2). What happens at ##t = 2##? What do you know about the sum of the series there?
 
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Well.. u(t) is discontinue for t=2 so the Fourier serie converges to [u(2+)+ u(2-)]/2 where

u(2+)= \lim_{t\to 2^+}u(t)= 0
u(2-)= \lim_{t\to 2^-}u(t)= 2

so

\sum_{k=-\infty}^{\infty}\hat{u}_k= 1

Right?
 
Last edited:
Yes. That's the idea.
 
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Thank you so much LCKurtz :)
 

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