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...)