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Fourier factor frustration

  • Thread starter doive
  • Start date
  • #1
15
0

Homework Statement



I have conflicting notes and even textbooks about where i should and shouldn't have factors of 2 when doing fourier series. I just want to check once and for all i'm doing it right and not introducing erroneous factors of 2 or omitting any


Homework Equations


[tex]
\tilde{f}(x)= \frac{a_0}{2} + \sum^{\inf}_{n=1} [a_n cos(\frac{n \pi x}{L}) + b_n sin(\frac{n\pi x}{L})]
[/tex]

[tex]
a_n = \frac{1}{L} \int^{L}_{-L} cos \frac{n\pi x}{L} f(x) dx
[/tex]

[tex]
b_n = \frac{1}{L} \int^{L}_{-L} sin(\frac{n\pi x}{L}) f(x) dx
[/tex]

[tex]
a_0 = \frac{1}{2L} \int^{L}_{-L} f(x) dx
[/tex]

where the interval is [-L,L] and the period is 2L

The Attempt at a Solution



Is the above formulation correct (i'm particularly unsure about the 1/2L associated with a0)
 
Last edited:

Answers and Replies

  • #2
vela
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Just plug in the series for f(x) into the integrals and see if the coefficients work out correctly.
 
  • #3
jbunniii
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Is the above formulation correct (i'm particularly unsure about the 1/2L associated with a0)
You have an extra factor of 1/2 associated with [itex]a_0[/itex]. To see that this is the case, set [itex]f(x) = 1[/itex]. Then according to your formulas, [itex]a_0 = 1[/itex] and the Fourier series is

[tex]\tilde{f}(x) = 1/2[/tex].

There should be a 1/2 either on the defining integral for [itex]a_0[/itex] or on the [itex]a_0[/itex] term in the series expansion, but not both.

You can check your other coefficients in a similar way by setting

[tex]f(x) = \cos(n\pi x/L)[/tex]

or

[tex]f(x) = \sin(n \pi x/L)[/tex]
 
  • #4
15
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thank you guys!

I had tried checking it but couldn't be certain it wasn't my dodgy arithmetic!
 

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