Cos ax/sin pi*x

  • Thread starter pepe
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Main Question or Discussion Point

I' m trying to solve something as apparently simple like this

cos ax/sin pi*x

which appears solved in
https://archive.org/details/TheoryOfTheFunctionsOfAComplexVariable

in the page 157, exercise 9. second part.

I'm trying by fourier series, but by the moment I can't achieve it.

Thanks.
 

Answers and Replies

  • #2
mathman
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Your original statement is incomplete. You don't have an equation to solve. Also what are you solving for?
 
  • #3
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Thanks, reading a book on hidrogeology, there that equation was used and referred to Copson, I went there and found it, as an exercise with the solution, but it didn't appear how to reach the solution. I attach the complete equation (in fact, there are two). I'm trying to do it with Fourier series, because of the solution, and understanding as the hidrogeology book said that (-1)^n is in fact = cos n*pi

Thanks for your attention, and I can see that if I continue with this hobby I'll have to learn LaTex (I'm sorry I don't use it).
 

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  • #4
mathman
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Sorry - I can't help.
 
  • #5
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Sorry - I can't help.
Thanks anyway. I'm looking and looking in the internet, for a track to the solotion. But I don't reach it. Maybe nowadays maths afford than problems in a very different way, by means of computers, and not looling for "exact" solutions. Hope someone has the answer found in an old book. Thanks again.
 
  • #6
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looking instead of looling! My english is bad, but I make it even worse.
 
  • #7
HallsofIvy
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You keep saying "solve" but don't say what problem you are trying to solve. Are you trying to integrate [itex]\int \frac{cos(ax)}{sin(\pi x)}dx[/itex]?
 
  • #8
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Good morning (afternoon in Spain). Probably I'm not expressing well. What I mean is that in that book (Copson, 1935, pp.157. Exercice 39), Copson wrote that cos (ax)/sin pi*x was equal to the expression including a sumatory that I sent attached. But I have not found how to get that second expression from cos (ax)/sin pi*x. The sixth chapter of the book where that expression appears is dedicated to "The calculus of residues", maybe that can indicate the way to afford the exercise. Thanks.
 
  • #9
pasmith
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It's generally the case that an exercise at the end of a chapter will be solved by methods developed in that chapter. It may of course be possible to solve the exercise by other means, but to do so is, in the context of the textbook, to miss the point.

You are trying to show that if [itex]-\pi < \alpha < \pi[/itex] then
[tex]\frac{\cos \alpha z}{\sin \pi z} = \frac{1}{\pi z} + \frac{2z}{\pi}\sum_{n=1}^\infty (-1)^n \frac{\cos n\alpha}{z^2 - n^2}.[/tex] The Cauchy integral formula is a good place to start: for [itex]|z| < R[/itex] we have [tex]f(z) = \frac{1}{2\pi i}\int_{|\zeta| = R} \frac{f(z)}{z - \zeta}\,d\zeta.[/tex] You will need to take [itex]R \to \infty[/itex] and calculate the residues of [itex]\dfrac{\cos \alpha \zeta}{\sin \pi \zeta} \dfrac{1}{z - \zeta}[/itex] at [itex]\zeta = 0, \pm 1, \pm 2, \dots[/itex].
 
Last edited:
  • #10
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Thanks Mr. Pasmith! I'll begin to study it. Maybe I reach...
 

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