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Re: Splitting sums in mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg126859] Re: Splitting sums in mathematica
- From: Ray Koopman <koopman at sfu.ca>
- Date: Thu, 14 Jun 2012 05:29:31 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <jqusrg$rnm$1@smc.vnet.net> <jr9kqr$3m8$1@smc.vnet.net>
On Jun 13, 2:00 am, "james.a.gordon1" <james.a.gord... at googlemail.com>
wrote:
> Hi Koopman,
>
> Thanks for the quick reply, I believe your suggestions solve some
> of my problem. To further clarify my question, I have an equation
> that looks like this
>
> F = (X - Y1 - Y2 - Y3 + Z)^2
>
> X = Sum_p Sum_q Sum_r [a_p*a_q*a_r]
>
> Yi = Sum_p Sum_q!=p b_p*b_q, where different X_i have different
> terms in the sums
>
> Z = Sum_p Sum_q!=p Sum_r!=q!=p
>
> In other words I end up with a large number of terms, each of which
> can have as many as 6 nested sums, all of which need to be expanded
> resulting in hundreds of terms. I want mathematica to algebraically
> expand these nested sums for me rather than me having to type up
> the expanded form myself. I believe with your solution I would still
> need to manually keep track of the indices for each expansion.
>
> Once all the nested sums are expanded, I will make some
> substitutions of the a, a^2, a^3...a^6 terms, then simplify.
> At no point to I need to calculate this with real values,
> I just need an algebraic expression.
>
> Thanks again
>
> James
I don't understand what you're trying to do -- how the Yi differ
from one another, or what it means to "make some substitutions
of the a, a^2, a^3...a^6 terms", as opposed to substituting for
individual terms. Give us a small example, using Mathematica syntax.
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