Question

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)

而严格等同于：

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)

Answer 1

不是答案，但如果您使用

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

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) *)

但

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) *)

mathematica中的求和问题

1 个答案: