
#1
Feb1513, 08:19 PM

P: 30

So I know that sec(x) has period 2Pi, and it's even so I don't need to figure out coefficients for bn.
Let's take the limits of the integral to go from 3/2 Pi to 1/2 Pi. How do I integrate sec(x) sin(nx) dx?! Am I on the right path? PS: I know that this doesn't satisfy the Dirichlet Theorem, but the textbook I'm using still says to compute it. 



#2
Feb1613, 12:10 AM

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Seems to me that integral is zero for n even and indeterminate for n odd.




#3
Feb1613, 12:23 AM

P: 2,265

well since [itex]\sec(x) = 1/\cos(x) [/itex] then i think you can express the cosine functions, both on the top and on the bottom as exponentials with imaginary argument. i think you can get to an integral that looks like
[tex] c_n = \frac{1}{2 \pi} \int_{\pi}^{+\pi} \frac{2}{e^{i (n+1) x}+e^{i (n1) x}} \ dx [/tex] i think you can find an antiderivative of that. might still come out as indeterminate. but that's the integral you have to solve. 



#4
Feb1613, 12:36 AM

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Fourier Series of Secant 



#5
Feb1613, 02:58 AM

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#6
Feb1613, 03:27 AM

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mathskier indicated the wrong integrand. It should be sec x cos(nx), which is the integrand I used.




#7
Feb1613, 04:59 AM

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