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Re: Problem rephrased: how to simplify summation of millions of exponentials symbolicly

• To: mathgroup at smc.vnet.net
• Subject: [mg50197] Re: Problem rephrased: how to simplify summation of millions of exponentials symbolicly
• From: kzhang at flashmail.com (Kezhao Zhang)
• Date: Fri, 20 Aug 2004 04:57:35 -0400 (EDT)
• References: <cg20h0$ok0$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

If (ai, bi) are located on a regular lattice on the 2D plane, you may 
consider to use discrete Fourier transfrom with the function Fourier[].

K. Z.

"networm" <networm8848 at yahoo.com> wrote in message news:<cg20h0$ok0$1 at smc.vnet.net>...
> Hi all,
> 
> I have a summation of exponentials:
> 
> SUM= 1+r(a1, b1)*exp(-j*(a1*u+b1*v))+r(a2, b2)*exp(-j*(a2*u+b2*v))+r(a3,
> b3)*exp(-j*(a3*u+b3*v))
>            + ...
>            + r(a1000000, b1000000)*exp(-j*(a1000000*u+b1000000*v))
> 
> where "j" is the imaginary sign.  a1, a2, ... a1000000, b1, b2, ... b1000000
> are known constants... (a's and b's) constitute some grids on the 2D plane,
> r(a, b) is the function defined on this 2D plane...
> r(a, b) is known as a look-up-table, but there is no closed-form expression
> for r(a, b)...
> u, v are frequency variable in 2D case.
> 
> Do you think it is possible to compute the close-form of the above SUM
> symbolically/analytically?
> 
> If not, is there any simple/efficient way to compute it ? 
> 
> I just need to compute this huge expression once, then if an simplified
> symbolic expression is found, it will save my subsequent numerical
> evaluations(that's going to tens of millions...)