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Re: Multiple sum with iterators that cannot equal
*To*: mathgroup at smc.vnet.net
*Subject*: [mg19590] Re: Multiple sum with iterators that cannot equal
*From*: Paul Abbott <paul at physics.uwa.edu.au>
*Date*: Sat, 4 Sep 1999 01:34:26 -0400
*Organization*: University of Western Australia
*References*: <7pl714$cgd@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Eric Spahr wrote:
> 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.
Two suggestions:
[1] If there is no singularity if i==j then you could use
\!\(\[Sum]\+\(i = 1\)\%n\(\[Sum]\+\(j = 1\)\%n w\_i\ w\_j\) -
\[Sum]\+\(i = 1\)\%n w\_i\%2\)
[2] Alternatively,
\!\(\[Sum]\+\(i = 1\)\%n w\_i\ \((\[Sum]\+\(j = 1\)\%\(i - 1\)w\_j +
\[Sum]\+\(j = i + \
1\)\%n w\_j)\)\)
____________________________________________________________________
Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia
Nedlands WA 6907 mailto:paul at physics.uwa.edu.au
AUSTRALIA http://physics.uwa.edu.au/~paul
God IS a weakly left-handed dice player
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