For complicated expression, the simplification does not seem to work out.
The beta_1 in the numerator and the denominator should have precisely
canceled out, however they do not.
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On Wed, Aug 21, 2019 at 3:10 AM David Bailey wrote:
>
> On 20/08/2019 23:35, Oscar Benjamin wrote:
>
> Hi David,
>
> Can you open an issue on Github please. That's a bug that should be fixed.
>
> The problem is, I tried to join Github a while back, and then needed a
> password reset. It said it
I've opened an issue:
https://github.com/sympy/sympy/issues/17473
On Wed, 21 Aug 2019 at 12:11, Oscar Benjamin wrote:
>
> On Wed, 21 Aug 2019 at 11:50, David Bailey wrote:
> >
> > On 20/08/2019 21:10, Aaron Meurer wrote:
> >
> > "TypeError: cannot determine truth value of Relational" generally
On Wed, 21 Aug 2019 at 11:50, David Bailey wrote:
>
> On 20/08/2019 21:10, Aaron Meurer wrote:
>
> "TypeError: cannot determine truth value of Relational" generally
> indicates a bug in SymPy. And yes, integrate() should always return
> unevaluated when it can't compute the integral.
>
> I
On 20/08/2019 21:10, Aaron Meurer wrote:
"TypeError: cannot determine truth value of Relational" generally
indicates a bug in SymPy. And yes, integrate() should always return
unevaluated when it can't compute the integral.
I imagine that symbolic integration may be littered by tricky problems
We should have, for example, `if (b < 0) == True or ...` here
File "C:\SymPyWorkbook\lib\site-packages\sympy\integrals\meijerint.py",
line 1684, in _meijerint_indefinite_1
if b < 0 or f.subs(x, 0).has(nan, zoo):
Kalevi Suominen
On Tuesday, August 20, 2019 at 11:08:59 PM UTC+3, David Bailey
On 20/08/2019 23:35, Oscar Benjamin wrote:
Hi David,
Can you open an issue on Github please. That's a bug that should be fixed.
The problem is, I tried to join Github a while back, and then needed a
password reset. It said it had sent me an email to reset my password,
but it never appeared
