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MathGroup Archive 1999

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Re: Multiple sum with iterators that cannot equal

  • To: mathgroup at smc.vnet.net
  • Subject: [mg19392] Re: [mg19365] Multiple sum with iterators that cannot equal
  • From: "Andrzej Kozlowski" <andrzej at tuins.ac.jp>
  • Date: Mon, 23 Aug 1999 13:57:09 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

One simple way is as follows. Define a function, say f, by

f[i_, i_] := 0

f[i_, j_] := Subscript[w, i]*Subscript[w, j] /; i != j

and  evaluate

Sum[f[i, j], {i, 1, n}, {j, 1, m}]

This is of course a pretty general method which can be used to find sums
over idices satisfying various conditions.
--
Andrzej Kozlowski
Toyama International University
JAPAN
http://sigma.tuins.ac.jp
http://eri2.tuins.ac.jp


----------
>From: Eric Spahr <rspahr at worldnet.att.net>
To: mathgroup at smc.vnet.net
>To: mathgroup at smc.vnet.net
>Subject: [mg19392] [mg19365] Multiple sum with iterators that cannot equal
>Date: Sat, Aug 21, 1999, 4:09 AM
>

> Using  multiple summmation signs with condition 'i "not
> equal" j', evaluation stops immediately as 'i' starts as
> 1 as does 'j'.  My formula requires that the 'i's' and
> 'j's' together:
>
>  Sum[Sum[f,{j,1,n}],{i,1,n}] ignore cases where i = j.
>
>  Or as copied from my notebook:
>
> \!\(\(\(\[Sum]\+\(i = 1\)\%n\[Sum]\+\(j = 1\)\%n\)\+\(i
> != j\)\) \(w\_i\)
>     w\_j\)
>
> How do I solve this without some cumbersome loops.
>
> Many Thanks
>
> Eric Spahr
> Cottage Technology rspahr at worldnet.att.net
>
> 


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