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Topic: This is hard.
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H.Verweij

Posts: 3
Registered: 12/12/04
This is hard.
Posted: Jul 7, 1996 2:42 PM
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Hello,

My name is Matthew and I need your help. I am struggling with the
summation of a double series:

Let alpha_n=(2n+1) pi/a;
Let beta_m =(2m+1) pi/b;
Both a and b are lengths, so the units of alpha_n and beta_m are m^-1.
Let gamma^2 be a complex number which real part is zero; the unit of
gamma is m^-1.
C and ab*C_0 have the same dimensions.

The series:

(1) C=64*C_0/(ab) sum of n=0 to infinity sum of m=0 to infinity of

{ (alpha_n)^2 + (beta_m)^2 } divided by { (alpha_n)^2 * (beta_m)^2 }

times 1 over {(alpha_n)^2 +(beta_m)^2-gamma^2 }.

It's a real hard one. I am studying for my Ph.D at Twente University in
Enschede, the Netherlands, so I went to the mathematics department over
here. (I am a physicist myself.) They solved it partially, removing one
summation. They got:

(2) C/C_0= ab * tan (gamma a/2) over (gamma a /2) +

16 a^4 gamma^2 over (pi^5) times the sum of n=0 to infinity of

tan{ b/2 sqrt(gamma^2-pi^2*(2n+1)^2/a^2) }over

(2n+1)^2 * [ gamma^2 a^2/pi^2 - (2n+1)^2 ]^(3/2).

If you manage to obtain these sums in a closed mathematical form, I would
be appreciate it. Since I am not into any of the mathematical software
packages, I did not try them. It might help to look upon (1) as a Laurent
series, with poles at gamma^2=(alpha_n)^2+(beta_m)^2.

I studied two cases: small and large gamma. For gamma small, the series
is easy summable and it should say that C=ab*C_0. In the case of large
gamma, I obtained:

C=ab*C_0 times [tan(gamma a/2) over (gamma a/2) + tan (gamma b/2) over
(gamma b/2) ];

Therefore, it might be that (1) is some weird product of tangenses.

I appreciate your effort. If you succeed, you can email me at
M.W.denOtter@ct.utwente.nl . Response guaranteed.

Good luck!

Matthew.








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