Sum of a serie involving Fourier coefficients

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Mathitalian
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Homework Statement



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

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


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

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

becomes

[tex]\sum_{k=-\infty}^{\infty}\hat{u}_k[/tex]

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

Homework Statement



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

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


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

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

becomes

[tex]\sum_{k=-\infty}^{\infty}\hat{u}_k[/tex]

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

[tex]u(2+)= \lim_{t\to 2^+}u(t)= 0[/tex]
[tex]u(2-)= \lim_{t\to 2^-}u(t)= 2[/tex]

so

[tex]\sum_{k=-\infty}^{\infty}\hat{u}_k= 1[/tex]

Right?
 
Last edited:
Yes. That's the idea.
 
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