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MathGroup Archive 2003

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Re: Simplyfing Sum[Mod[f(k),y],{k,k0,n}] type expressions? (Newbie question)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg40224] Re: [mg40202] Simplyfing Sum[Mod[f(k),y],{k,k0,n}] type expressions? (Newbie question)
  • From: Dr Bob <drbob at bigfoot.com>
  • Date: Wed, 26 Mar 2003 02:43:58 -0500 (EST)
  • References: <200303251950.OAA11794@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

You haven't told Mathematica that n and d are integers, so the sum isn't as 
simple as you pretend.  However, the following may help:

Simplify[Block[{Sum, rule, assumption = {n, d} â?? Integers},
    rule = Sum[Mod[expr_, x_Integer], iter_] :> Mod[Sum[expr, iter], x]; 
Sum[Mod[3n - k - 4, 3], {k, 2n - 1, d}]
      /. rule], assumption]

Bobby

On Tue, 25 Mar 2003 14:50:20 -0500 (EST), prodogoss 
<prodogoss at btinternet.com> wrote:

> I'm dealing with a problem that requires lots of summations of (x mod
> y) type expressions. I had hoped that Mathematica (which I've just
> started using) could simplify these but it doesn't seem to work.
>
> E.g., why won't (can't?) Mathematica simplify the expression:
>
> Sum[Mod[3n-k-4, 3],{k, 2n-1,d}] ?
>
>
> By *hand* I can show that if
>
> S = SUM((3n-k-4) mod 3, k = 2n-1 to d) and
>
> G = d - 2n + 2, then S = f(t) where t = (n - 2 - G) mod 3, and
>
> (f(0), f(1), f(2)) = ({1, 0, 0}, {0, 1, 0}, {0, 0, 1})(G-1, G, G)
>
> ie 3x1 matrix = (3rd order unit matix) x (3x1 matrix)
>
> Hence, given values of d & n, S can be expressed in such a way that
> its calculation is possible without having to labourously evaluate S
> for each step, k.
>
>
> I had heard so much about Mathematica and thought it would excel at
> handling problems like this involving summations of Mod[] etc.? Is
> there another way of doing this? Any help would be great!
>
> Thanks in advance!
>
>



-- 
majort at cox-internet.com
Bobby R. Treat



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