The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Robin Chapman

Posts: 412
Registered: 5/29/08
Re: Help with identity
Posted: May 2, 2013 4:41 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 02/05/2013 00:34, Mike Trainor wrote:
> On Fri, 26 Apr 2013 14:31:57 +0100, Robin Chapman
> <> 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.

No we don't. Put z = cos(y) + i sin(y) = exp(iy).
What is z^n in terms of y?

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.