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skapat

  • one year ago

Evaluate the following integral: cos((mpix)/L)cos((npix)/L) for (-L, L) where m≠n, and m,n are integers

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  1. skapat
    • one year ago
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    \[\cos \frac{ m Pix }{ L }\cos \frac{ n Pix }{ L }\]

  2. TuringTest
    • one year ago
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    Fourier?

  3. skapat
    • one year ago
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    what is fourier?

  4. TuringTest
    • one year ago
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    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.

  5. skapat
    • one year ago
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    what is alpha and what is beta? m and n? or L and -L?

  6. TuringTest
    • one year ago
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    \(\alpha\) and \(\beta\) are just the arguments of the cosines:\[\alpha=\frac{m\pi x}L\]\[\beta=\frac{n\pi x}L\]

  7. skapat
    • one year ago
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    so do we then take the integral of (1/2)[cos(alpha-beta)+cos(alpha+beta)]?

  8. TuringTest
    • one year ago
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    yes

  9. skapat
    • one year ago
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    but how would i integral the inside? (alpha-beta)?

  10. skapat
    • one year ago
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    sin(alpa)cos(beta)?

  11. TuringTest
    • one year ago
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    alpha and beta just represent any argument in the trig identity I am showing you substitute\[\alpha=\frac{m\pi x}L\]\[\beta=\frac{n\pi x}L\]

  12. skapat
    • one year ago
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    so wouldnt the integral be \[\sin( \alpha)\]\[\cos (\beta)\]

  13. skapat
    • one year ago
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    multiplied together that is?

  14. TuringTest
    • one year ago
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    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

  15. TuringTest
    • one year ago
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    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

  16. skapat
    • one year ago
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    ok, just a second, let me give that a try

  17. TuringTest
    • one year ago
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    also, because this integral is even we can write\[\int_{-L}^L\cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi x}L\right)dx=2\int_0^L\cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi x}L\right)dx\]

  18. skapat
    • one year ago
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    (1/2pi)(L((sin(pix(m-n)/L)/(m-n))+sin(pix(m+n)/L)/(m+n))

  19. skapat
    • one year ago
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    is that right?

  20. TuringTest
    • one year ago
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    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!

  21. skapat
    • one year ago
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    ok thank you for your patient and your help TurningTest

  22. skapat
    • one year ago
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    patience*

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