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skapat
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Evaluate the following integral: cos((mpix)/L)cos((npix)/L) for (L, L) where m≠n, and m,n are integers
 one year ago
 one year ago
skapat Group Title
Evaluate the following integral: cos((mpix)/L)cos((npix)/L) for (L, L) where m≠n, and m,n are integers
 one year ago
 one year ago

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skapat Group TitleBest ResponseYou've already chosen the best response.0
\[\cos \frac{ m Pix }{ L }\cos \frac{ n Pix }{ L }\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
Fourier?
 one year ago

skapat Group TitleBest ResponseYou've already chosen the best response.0
what is fourier?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

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

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

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

skapat Group TitleBest ResponseYou've already chosen the best response.0
sin(alpa)cos(beta)?
 one year ago

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

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you get\[\cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi x}L\right)=\frac12[\cos\left(\frac{\pi x}L(mn)\right)+\cos\left(\frac{\pi x}L(m+n)\right)\]which you can integrate
 one year ago

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

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

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

skapat Group TitleBest ResponseYou've already chosen the best response.0
is that right?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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!
 one year ago

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