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我对mathematica的这种行为感到困惑。以下两个表达式应返回相同的结果:

Simplify[(1 - w)^2 Sum[w^(k+kp) Sum[If[l == lp, 1, 0], {l, 0, k}, {lp, 0, kp}],
{k,0, \[Infinity]}, {kp, 0, \[Infinity]}]]

返回:

(-1 - w + w^3)/(-1 + w^2)

而严格等价的:

Simplify[(1 - w)^2 Sum[w^(k+kp) Min[k, kp],{k,0,\[Infinity]},{kp,0,\[Infinity]}]
       + (1 - w)^2 Sum[w^(k+kp)           ,{k,0,\[Infinity]},{kp,0,\[Infinity]}]]

返回:

1/(1 - w^2)
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1 回答 1

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不是答案,但如果你使用

$$\sum_{k=0}^{\infty} \sum_{l=0}^{k} = \sum_{l=0}^{\infty} \sum_{k=l}^{\infty} $$

然后 :

sum1 = Simplify[(1 - w)^2 
         Sum[w^(k + kp) Sum[KroneckerDelta[l, lp] , {l, 0, k}, {lp, 0, kp}], 
             {k, 0, Infinity}, {kp, 0, Infinity}]]
(* (-2 + w + w^2 - w^3)/(-1 + w^2) *)

sum2 = Simplify[(1 - w)^2 
         Sum[KroneckerDelta[l, lp] 
             Sum[w^(k + kp), {k, l, Infinity}, {kp, lp, Infinity}] , 
         {l, 0, Infinity}, {lp, 0, Infinity}]]
(* 1/(1 - w^2) *)
于 2013-04-24T15:54:23.343 回答