Re: how to simplify a summation of exponentials symbolicly/analytically?

From: Magnus Sundberg (nisse_at_manpower.sg)
Date: 08/17/04 

Date: Tue, 17 Aug 2004 10:12:08 +0200

Hi!

I doubt it is possible to do this analytically unless the a's and b'a have
some fixed relationship (such as equidistant integers), where you could
transform your sum into a summation over powers of some constant.

Also, in matlab, if u and v are not scalars it is difficult/costly to avoid
a loop unless you have vast amounts of memory.

Fortunately, with the new just-in-time compile (from matlab version 6.5 and
later), a well designed loop over the summation will not be that costly in
terms of execution speed - with the added benefit of much more readable code
compared to a fully vectorized solution (which some wizard will undoubtedly
suggest).

(Of course if u and v are scalars you can just write SUM =
1+sum(exp(-j*(a*u+b*v)) where a and b are vectors of your coefficients).

HTH,
Magnus

"networm" <networm8848@yahoo.com> wrote in message
news:cfsced$rv4$1@news.Stanford.EDU...
> Hi all,
>
> I have a summation of exponentials:
>
> SUM= 1+exp(-j*(a1*u+b1*v))+exp(-j*(a2*u+b2*v))+exp(-j*(a3*u+b3*v))
> + ...
> + exp(-j*(a1000000*u+b1000000*v))
>
> where "j" is the imaginary sign. a1, a2, ... a1000000, b1, b2, ...
b1000000
> are known constants... 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 ? In Matlab and
> Maple...
>
>

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