Re: [sympy] How to simplify -polylog(2, exp_polar(I*pi))/16 + pi**2/96 ?

Indeed, I am using : >>> import sympy >>> sympy.__version__ '1.0' Thank you, -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To

Re: [sympy] How to simplify -polylog(2, exp_polar(I*pi))/16 + pi**2/96 ?

It does give me the correct answer in the latest master, but I get the same incorrect answer in sympy-1.0. It was fixed here, https://github.com/sympy/sympy/pull/10799 Isuru Fernando On Fri, Dec 2, 2016 at 6:48 PM, Sébastien Labbé wrote: > Ok, thank you. I see: > > >>> from

Re: [sympy] How to simplify -polylog(2, exp_polar(I*pi))/16 + pi**2/96 ?

Ok, thank you. I see: >>> from sympy import * >>> -polylog(2, exp_polar(I*pi))/16 + pi**2/96 -polylog(2, exp_polar(I*pi))/16 + pi**2/96 >>> exp_polar(I*pi) exp_polar(I*pi) Then should not this gives me pi**2/64 ? >>> -polylog(2, -1)/16 + pi**2/96 pi**2/192 -- You received this message because

Re: [sympy] How to simplify -polylog(2, exp_polar(I*pi))/16 + pi**2/96 ?

This is a known bug, https://github.com/sympy/sympy/issues/8404 A._sage_() works because exp_polar(I*pi) is -1 in sage and is unevaluated in SymPy. Isuru Fernando On Fri, Dec 2, 2016 at 2:55 PM, Sébastien Labbé wrote: > Can SymPy tell me that A = 1/64*pi**2 by simplifying

[sympy] How to simplify -polylog(2, exp_polar(I*pi))/16 + pi**2/96 ?

Can SymPy tell me that A = 1/64*pi**2 by simplifying the polylog in below example? >>> from sympy.abc import n >>> from sympy import summation, oo >>> A = summation(1/((2*n+1)^2-4)^2, (n, 0, oo)) >>> A -polylog(2, exp_polar(I*pi))/16 + pi**2/96 >>> A.simplify() -polylog(2, exp_polar(I*pi))/16 +