Fourier Series Integral

  • Thread starter tigertan
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  • #1
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Hey there,

First time user of this forum.

I have a question regarding an integral i've been stuck on for the past few days. I would really appreciate any eye opener into this problem!

How do I find the Fourier coefficients for f(x)=xcos(x), x [-∏,∏]

So when calculating the coefficient bn I figure that the equation I'll be working with is (1/∏)∫xcos(x)sin(kx)dx (FROM -∏ to ∏).

I keep getting down to 1/2∏∫xsin((k+1)x)dx (FROM -∏ to ∏) + 1/2∏∫xsin((k-1)x)dx (FROM -∏ to ∏) . I some how always get a 0 for b1. What am I doing incorrectly??
 

Answers and Replies

  • #2
I like Serena
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Welcome to PF, tigertan! :smile:

I'm not getting 0 for b1.
So what did you do?
 
  • #3
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Am I right in breaking it up to 1/2∏∫xsin((k+1)x)dx (FROM -∏ to ∏) + 1/2∏∫xsin((k-1)x)dx (FROM -∏ to ∏)?

I then end up with 1/2∏[-(k+1)cos((k+1)x) + sin((k+1)x)](FROM -∏ to ∏) + 1/2∏[-(k-1)cos((k-1)x) + sin((k-1)x)](FROM -∏ to ∏)

This equals to 0 when I put b1
 
  • #4
I like Serena
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You lost a factor x in your integration.
It seems you did not properly apply integration by parts.

You should have an expression containing: x cos(k+1)x.
When you substitute the bounds -pi and +pi, these add up.
 

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