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 one year ago
Evaluate the following integral: cos((mpix)/L)cos((npix)/L) for (L, L) where m≠n, and m,n are integers
 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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skapat
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos \frac{ m Pix }{ L }\cos \frac{ n Pix }{ L }\]

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1It'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.

skapat
 one year ago
Best ResponseYou've already chosen the best response.0what is alpha and what is beta? m and n? or L and L?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1\(\alpha\) and \(\beta\) are just the arguments of the cosines:\[\alpha=\frac{m\pi x}L\]\[\beta=\frac{n\pi x}L\]

skapat
 one year ago
Best ResponseYou've already chosen the best response.0so do we then take the integral of (1/2)[cos(alphabeta)+cos(alpha+beta)]?

skapat
 one year ago
Best ResponseYou've already chosen the best response.0but how would i integral the inside? (alphabeta)?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1alpha and beta just represent any argument in the trig identity I am showing you substitute\[\alpha=\frac{m\pi x}L\]\[\beta=\frac{n\pi x}L\]

skapat
 one year ago
Best ResponseYou've already chosen the best response.0so wouldnt the integral be \[\sin( \alpha)\]\[\cos (\beta)\]

skapat
 one year ago
Best ResponseYou've already chosen the best response.0multiplied together that is?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1no, 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
 one year ago
Best ResponseYou've already chosen the best response.1you get\[\cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi x}L\right)=\frac12[\cos\left(\frac{\pi x}L(mn)\right)+\cos\left(\frac{\pi x}L(m+n)\right)\]which you can integrate

skapat
 one year ago
Best ResponseYou've already chosen the best response.0ok, just a second, let me give that a try

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1also, because this integral is even we can write\[\int_{L}^L\cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi x}L\right)dx=2\int_0^L\cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi x}L\right)dx\]

skapat
 one year ago
Best ResponseYou've already chosen the best response.0(1/2pi)(L((sin(pix(mn)/L)/(mn))+sin(pix(m+n)/L)/(m+n))

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1It'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!

skapat
 one year ago
Best ResponseYou've already chosen the best response.0ok thank you for your patient and your help TurningTest
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