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Re: nested sums

  • To: mathgroup at smc.vnet.net
  • Subject: [mg125254] Re: nested sums
  • From: Severin Pošta <severin at km1.fjfi.cvut.cz>
  • Date: Fri, 2 Mar 2012 07:49:45 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201202291226.HAA28458@smc.vnet.net> <jinjbt$ctp$1@smc.vnet.net>

Sure, but that's not what I am asking for, I wanted to ask if such check 
can be somehow performed automatically?
Thanks, S.

On 3/1/2012 11:34 AM, Bob Hanlon wrote:
> Sum[x^a y^b z^c, {a, 0, Infinity}, {b, 0, a}, {c, 0, a - b}]
>
> -(1/((-1 + x)*(-1 + x*y)*(-1 + x*z)))
>
>
> Bob Hanlon
>
> 2012/2/29 Severin Po=C5=A1ta<severin at km1.fjfi.cvut.cz>:
>> I have noticed that
>>
>> Sum[x^a y^b z^c, {a, 0, Infinity}, {b, 0, Infinity}, {c, 0, a - b}]
>>
>> produces "wrong" result
>>
>> (-y + z - y z + x y z)/((-1 + x) (-1 + y) (y - z) (-1 + x z))
>>
>> because a-b is not checked to be nonnegative in the nested sums. This is
>> probably by design (?).
>> I wonder if I can achieve such check to be perfomed.
>>
>> This does not work:
>>
>> Sum[x^a y^b z^c If[a - b>= 0, 1, 0], {a, 0, Infinity}, {b, 0,
>> =C2  Infinity}, {c, 0, a - b}]
>>
>> This also does not work:
>>
>> Sum[x^a y^b z^c Piecewise[{{1, a - b>= 0}}], {a, 0, Infinity}, {b, 0,
>> =C2  =C2 Infinity}, {c, 0, a - b}]
>>
>>
>> It is also interesting that Sum[...,{},{}] is not equivalent to
>> Sum[Sum[...,{}],{}]. The following produces correct result:
>>
>> s= Sum[x^a y^b z^c Piecewise[{{1, a - b>= 0}}], {c, 0, a - b}];
>> s1=Sum[s, {b, 0, Infinity}];
>> Sum[s1,{a, 0, Infinity}]
>>
>>
>> -(1/((-1 + x) (-1 + x y) (-1 + x z)))
>>
>>
>> Interesting.
>>
>> S.
>>
>




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