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Orthogonal Properties for Sine Don't Hold if Pi is involded?

  • Thread starter mmmboh
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Orthogonal Properties for Sine Don't Hold if Pi is involded??

Normally I know [tex]
\int_{-L}^L \sin \frac{n x}{L} \sin \frac{\m x}{L} ~ dx = 0\mbox{ if }n\not =m , \ =L \mbox{ if }n=m
[/tex] but apparently this doesn't work for [tex]
\int_{-L}^L \sin \frac{\pi n x}{L} \sin \frac{\pi m x}{L} ~ dx
[/tex]

I am trying to find fourier coefficients, and my integral is [tex]\int_{0}^{L/2} \sin \frac{2\pi x}{L} \sin \frac{\pi n x}{L} ~ dx
[/tex]

For even n that aren't 2, the integral is 0, if n=2, then the integral is L/2, but if n is odd, then the integral doesn't equal 0 (it's actually a fairly complex answer) even thought it looks like it should by orthogonal properties ...why doesn't this work?

The integrals I am doing take very long to do, is there a property that would allow me to shorten them?
 
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Answers and Replies

  • #2
LCKurtz
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It does work. If [itex]n \ne m[/itex] are positive integers then

[tex]\int_0^L\ \sin(\frac{n\pi x}{L})\sin(\frac{m\pi x}{L})\, dx = 0[/tex]

Sturm-Liouville theory would guarantee this without working it out.
 
  • #3
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Sorry the limits are actually 0 to L/2, in this case it doesn't work, is there another property that would help evaluate the integral quicker or do I have to go through the whole thing every time?
 
  • #4
LCKurtz
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The set of functions

[tex]\left \{ \sin\frac{2n\pi x}{L}\right\}[/tex]

are orthogonal on [0, L/2].
 
  • #5
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Is there anything like that for [tex]
\left \{ \sin\frac{(2n)\pi x}{L}\right\ \left \sin\frac{(2n+1)\pi x}{L}\right\}
[/tex]?
 
  • #6
LCKurtz
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Is there anything like that for [tex]
\left \{ \sin\frac{(2n)\pi x}{L}\right\ \left \sin\frac{(2n+1)\pi x}{L}\right\}
[/tex]?
I doubt you will get zero, but why don't you just integrate it and see?
 

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