Math 数学中的求和问题

Math 数学中的求和问题,math,wolfram-mathematica,series,Math,Wolfram Mathematica,Series,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) 1/(1 - w^2) 鉴于严格等效: Simplify[(1 - w)^2 Sum[w^(k+kp) Mi

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)
1/(1 - w^2)
鉴于严格等效:

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

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) *)
但是


请解释为什么这些表达式是“严格等价的”,它只是一个数学陈述:对于k>=0和kp>=0:
Sum[If[l==lp,1,0],{l,0,k},{lp,0,kp}]=Min[k,kp]+1