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

TuringTest
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what is alpha and what is beta? m and n? or L and L?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but how would i integral the inside? (alphabeta)?

TuringTest
 3 years 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\[\alpha=\frac{m\pi x}L\]\[\beta=\frac{n\pi x}L\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0multiplied together that is?

TuringTest
 3 years 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
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, just a second, let me give that a try

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

anonymous
 3 years 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
 3 years 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok thank you for your patient and your help TurningTest
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