4
$\begingroup$

Take the following code:

Simplify[
  Sum[
    Sum[
      ((-1)^(1 + q1 + q2) 2^(-2 q1 - 2 q2) q2 Theta^(1 + 2 q1 + 2 q2))/((1 + 2 q1)! (2 q2)!), 
      {q1, 0, Infinity}], 
    {q2, 0, Infinity}]]

Simplify[
  Sum[
    ((-1)^(1 + q1 + q2) 2^(-2 q1 - 2 q2) q2 Theta^(1 + 2 q1 + 2 q2))/((1 + 2 q1)! (2 q2)!), 
    {q1, 0, Infinity}, {q2, 0, Infinity}]]

I am basically summing the same coefficients, but in two different ways: one with two summations, the other with a summation over two indices. The results are the following:

1/2 Theta Sin[Theta/2]^2

-(1/4) Theta Cos[Theta]

As you can see they dont match. Why is it so?

To give some interpretation, it is actually the Taylor coefficient of sin(theta/2) * q1) * Taylor_coefficient(cos(theta/2)) * q2 * (-2 q2) that I integrate. Said differently, I am summing

-2 q2 Coeff[Sin[q1]] Coeff[Cos[q2]]

over q1 and q2 :

$\endgroup$
2
  • $\begingroup$ Last point is not clear. Can you please write more precisely how you get the series, either in mathematical form or a respective derivation with MA. The equation in orange box has no Theta. $\endgroup$
    – yarchik
    Dec 3, 2019 at 13:59
  • $\begingroup$ You could substitute numerical values to find which gives the correct answer. Remember that it may not be mathematically legitimate to interchange the 2 limiting processes. $\endgroup$
    – mikado
    Dec 3, 2019 at 17:57

1 Answer 1

3
$\begingroup$

Not an answer, just a long comment...

One can simplify this case to this:

expr = ( 2^(q1 + q2) q1)/((1 + 2 q1)! ( 2 q2)!);

Sum[expr, {q1, 0, Infinity}, {q2, 0, Infinity}, Method -> "IteratedSummation"] -
Sum[expr, {q1, 0, Infinity}, {q2, 0, Infinity}] // FullSimplify

(* 1/4 *)

Which more clearly specifies the discrepancy in two methods.

$\endgroup$
2
  • $\begingroup$ To understand: is it then a a problem of interversion of two series (thus a mathematical problem) behind ? I tried to look at what means "iterated univariate summation" but I didn't find something interesting yet. $\endgroup$
    – StarBucK
    Dec 3, 2019 at 16:39
  • $\begingroup$ @StarBucK I suspect it's a bug in Automatic method Wolfram uses to calculate this double sum. IteratedSummation is a method of just symbolically summing one sum and then another, like in your question Sum[Sum[...],...] $\endgroup$
    – swish
    Dec 3, 2019 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.