I'd say this is a bug. The problem it it's evaluating the inner sum to
an expression that has terms like (w_1_2_3 - 1) in both the numerator
and denominator without cancelling them. You can work around it by
evaluating and simplifying the inner summand separately:
>>> 7*Sum((binomial(5, w_1_2_3)*Sum(binomial(0, -w_1_2_3 + w_1_3)*binomial(2,
>>> w_1_2_3 - w_1_3 - 2)*binomial(13, 11 - w_1_2_3), (w_1_3, w_1_2_3,
>>> 5)).doit()).simplify(), (w_1_2_3, 0, 5)).doit()/868017280
0
Aaron Meurer
On Wed, Aug 3, 2022 at 12:26 PM Hadas Biran <hadas.mor...@gmail.com> wrote:
>
> I'm trying to evaluate a long expression with lots of summations and n choose
> k's. For some reason, the following expression (which is a part of the long
> expression) returns NaN.
>
> from sympy import *
> w_1_2_3 = symbols('w_1_2_3')
> w_1_3 = symbols('w_1_3')
> w_2_3 = symbols('w_2_3')
> w_1_2 = symbols('w_1_2')
> exprr = 7*Sum(binomial(5, w_1_2_3)*Sum(binomial(0, -w_1_2_3 +
> w_1_3)*binomial(2, w_1_2_3 - w_1_3 - 2)*binomial(13, 11 - w_1_2_3), (w_1_3,
> w_1_2_3, 5)), (w_1_2_3, 0, 5))/868017280
> exprr.doit() -> NaN
>
> However, if I evaluate each of the terms in the first sum separately, I get 0
> for each of them (as should be):
>
> exprr2 = binomial(5, w_1_2_3)*Sum(binomial(0, -w_1_2_3 + w_1_3)*binomial(2,
> w_1_2_3 - w_1_3 - 2)*binomial(13, 11 - w_1_2_3), (w_1_3, w_1_2_3, 5))
> for t in range(0, 6):
> res = exprr2.subs(w_1_2_3, t).doit()
> print(t, res)
> -> 0 0
> 1 0
> 2 0
> 3 0
> 4 0
> 5 0
>
> What is going on here? Why do I get NaN for the overall term?
>
> I also posted a question in
> StackOveflow-https://stackoverflow.com/questions/73222278/sympy-doit-returns-nan-instead-of-zero
>
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