anonymous
  • anonymous
Evaluate the following integral: cos((mpix)/L)cos((npix)/L) for (-L, L) where m≠n, and m,n are integers
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\cos \frac{ m Pix }{ L }\cos \frac{ n Pix }{ L }\]
TuringTest
  • TuringTest
Fourier?
anonymous
  • anonymous
what is fourier?

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TuringTest
  • TuringTest
It's what you are leading up to studying, though apparently your teacher has not told you yet use\[\cos\alpha\cos\beta=\frac12[\cos(\alpha-\beta)+\cos(\alpha+\beta)]\]to get this integral manageable.
anonymous
  • anonymous
what is alpha and what is beta? m and n? or L and -L?
TuringTest
  • TuringTest
\(\alpha\) and \(\beta\) are just the arguments of the cosines:\[\alpha=\frac{m\pi x}L\]\[\beta=\frac{n\pi x}L\]
anonymous
  • anonymous
so do we then take the integral of (1/2)[cos(alpha-beta)+cos(alpha+beta)]?
TuringTest
  • TuringTest
yes
anonymous
  • anonymous
but how would i integral the inside? (alpha-beta)?
anonymous
  • anonymous
sin(alpa)cos(beta)?
TuringTest
  • TuringTest
alpha and beta just represent any argument in the trig identity I am showing you substitute\[\alpha=\frac{m\pi x}L\]\[\beta=\frac{n\pi x}L\]
anonymous
  • anonymous
so wouldnt the integral be \[\sin( \alpha)\]\[\cos (\beta)\]
anonymous
  • anonymous
multiplied together that is?
TuringTest
  • TuringTest
no, I have no idea where you are getting that :p \[\cos\alpha\cos\beta=\frac12[\cos(\alpha-\beta)+\cos(\alpha+\beta)]\]sub for \(\alpha\) and \(\beta\) as I have stated above twice before
TuringTest
  • TuringTest
you get\[\cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi x}L\right)=\frac12[\cos\left(\frac{\pi x}L(m-n)\right)+\cos\left(\frac{\pi x}L(m+n)\right)\]which you can integrate
anonymous
  • anonymous
ok, just a second, let me give that a try
TuringTest
  • TuringTest
also, because this integral is even we can write\[\int_{-L}^L\cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi x}L\right)dx=2\int_0^L\cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi x}L\right)dx\]
anonymous
  • anonymous
(1/2pi)(L((sin(pix(m-n)/L)/(m-n))+sin(pix(m+n)/L)/(m+n))
anonymous
  • anonymous
is that right?
TuringTest
  • TuringTest
It's hard for me to see if all those parentheses are in the right places, but it looks right I'm afraid I have to go. Here's a link to the problem you are solving: http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex1 Look at example 1 (the last part of it) for reference Good luck!
anonymous
  • anonymous
ok thank you for your patient and your help TurningTest
anonymous
  • anonymous
patience*

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