[sympy] Help writing release notes

2016-12-02 Thread Aaron Meurer 
If anyone is free for the winter break and wants to help out, I could
really use some help getting together the release notes for the next
release https://github.com/sympy/sympy/wiki/Release-Notes-for-1.1.
Feel free to just edit that page.

We only need to list significant changes. Listing backwards compatible
changes and major changes is the most important. It's not a big deal
if the notes miss some minor changes.

The git command

git log --oneline 1.0..master --grep="Merge pull request"

will show all pull requests that have been merged since the release.
You can also get (roughly) the same thing here
https://github.com/sympy/sympy/pulls?utf8=%E2%9C%93=merged%3A%3E%3D2016-02-19%20is%3Apr%20

Aaron Meurer

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

2016-12-02 Thread Sébastien Labbé 
Indeed, I am using :

>>> import sympy
>>> sympy.__version__
'1.0'

Thank you,

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

2016-12-02 Thread Isuru Fernando 
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 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
>
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Re: [sympy] How to simplify -polylog(2, exp_polar(I*pi))/16 + pi**2/96 ?

2016-12-02 Thread Sébastien Labbé 
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

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

2016-12-02 Thread Isuru Fernando 
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 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 + pi**2/96
>
> because it seems doable:
>
> sage: from sympy.abc import n
> sage: from sympy import summation, oo
> sage: A = summation(1/((2*n+1)^2-4)^2, (n, 0, oo))
> sage: A._sage_()
> 1/64*pi^2
>
> This came up in https://ask.sagemath.org/question/35839/sage-
> incorrectly-evaluates-series/
>
> Sébastien
>
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[sympy] How to simplify -polylog(2, exp_polar(I*pi))/16 + pi**2/96 ?

2016-12-02 Thread Sébastien Labbé 
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 + pi**2/96

because it seems doable:

sage: from sympy.abc import n
sage: from sympy import summation, oo
sage: A = summation(1/((2*n+1)^2-4)^2, (n, 0, oo))
sage: A._sage_()
1/64*pi^2

This came up in 
https://ask.sagemath.org/question/35839/sage-incorrectly-evaluates-series/

Sébastien

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