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

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• Subject: [mg40238] Re: [mg40202] Simplyfing Sum[Mod[f(k),y],{k,k0,n}] type expressions? (Newbie question)
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Thu, 27 Mar 2003 06:49:49 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```I looked again at your answer and noticed that I indeed misunderstood
it and it is not in fact independent of n. (It's just the way you wrote
it as f(t) where t does not depend on n, but f itself does!). In fact
some reason that you do not mention for wanting to express it in that
particular form.

Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/

On Wednesday, March 26, 2003, at 06:47  pm, Andrzej Kozlowski wrote:

> First of all, your answer, if I understood it correctly, seems
> obviously wrong.
> If G= d - 2n + 2 then t=  (n - 2 - G) mod 3 = -4 - d + 3 n mod 3 = 2 +
> 2d mod 3, which is independent of n. But the sum is clearly not
> independent of n.
> It is not clear to me what sort of answer you expect Mathematica to
> deliver. You can certainly do this:
>
> In[1]:=
> K = PolynomialMod[Sum[3*n - k - 4, {k, 2*n - 1, d}], 3]
>
> Out[1]=
> 2 + d^2 + 2*n + 2*n^2
>
> Now, if you want to get an answer involving G=  d - 2n + 2 you can do:
>
>  In[2]:=
> G = d - 2*n + 2;
>
> In[3]:=
> PolynomialReduce[K, G, Modulus -> 3]
>
> Out[3]=
> {{1 + d + 2*n}, 0}
>
>
> So your sum can be written (mod 3)  as (1+d+2n)G. This seems rather
>
> Andrzej Kozlowski
> Yokohama, Japan
> http://www.mimuw.edu.pl/~akoz/
> http://platon.c.u-tokyo.ac.jp/andrzej/
>
>
> On Wednesday, March 26, 2003, at 04:50  am, prodogoss 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!
>>