CODEWITHSUNDEEP
mathwolfram-mathematicaseries
bernard
bernard

Reputation: 161

summation issue in mathematica

I'm puzzled by this behavior of mathematica. The two following expressions should return the same result:

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]}]]

returns:

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

whereas the strictly equivalent:

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]}]]

returns:

1/(1 - w^2)

Upvotes: 0

Views: 99

Answers (1)

b.gatessucks
b.gatessucks

Reputation: 1242

Not an answer, but if you use

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

then :

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) *)

but

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) *)

Upvotes: 1

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