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Topic: Help with identity
Replies: 15   Last Post: May 9, 2013 7:13 AM

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 Mike Trainor Posts: 28 Registered: 4/21/13
Re: Help with identity
Posted: May 1, 2013 8:50 PM
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On Wed, 01 May 2013 19:34:16 -0400, Mike Trainor
<mtrainor@hotmail.com> wrote:

>On Fri, 26 Apr 2013 14:31:57 +0100, Robin Chapman
><R.J.Chapman@ex.ac.uk> wrote:
>

>>On 26/04/2013 13:10, Mike Trainor wrote:
>>

>
>>> It comes down to doing the integral of
>>>
>>> cos(ny)/(cosh(x) - cos(y))
>>>
>>> from 0 to 2 pi, for integer n, where x => 0.

>>
>>How about integrating z^{n-1}/(cosh(x) - (z+1/z)/2)
>>over the unit circle in C?

>
>Thanks, Robin, from bringing back 25+ year old
>memories ... have not done this kind of work in
>a while. Funny about the cosh(x) terms as it
>simplifies the terms.
>
>I do have a question as my memory is shot and
>I cannot figure it out. I see why you would have
>z^(n-1) and not z^n as the 'dy' becomes
>dz/(i*z). Now, there are simple poles at
>z = exp(+/- x), and only the z = exp(-x) lies
>within the contour as x > 0 in my case, at least.
>That gives the cosech(x) term I need in the
>answer.
>
>But, I have a question. What above the
>z^n term in the numberator that comes due
>to the numerator that should be there from
>the cos(ny) part? That messes up things as
>the residues now have exp(-n*x) and,
>unforturnately, exp(n*x). Other than that,
>it all works out.
>
>I have yet to get to the basement, locate and
>pull out my grad school textbooks .... I guess
>the answer is there.
>
>Thanks for the pointer and would appreciate it
>if you can give me a hint.
>
>tia
>mt

And, I might as well add that I have not gotten
near the n-th order pole at z = 0 :-)

mt

Date Subject Author
4/25/13 Mike Trainor
4/25/13 David C. Ullrich
4/26/13 Mike Trainor
4/26/13 Robin Chapman
5/1/13 Mike Trainor
5/1/13 Mike Trainor
5/2/13 Robin Chapman
5/2/13 Robin Chapman
5/2/13 Mike Trainor
5/2/13 Mike Trainor
4/26/13 David C. Ullrich
4/29/13 Mike Trainor
5/2/13 RGVickson@shaw.ca
5/8/13 AP
5/8/13 Virgil
5/9/13 Mike Trainor

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