he denominator--but feel
that there must be a better.
Thanks,
Dan Schult
There is a method. Lets start with the given expression:
In [1]: var('a,b,r')
Out[1]: (a, b, r)
In [2]: e = (a+a*r+b*r)/(a+b)
In [3]: e
Out[3]:
a + a⋅r + b⋅r
─
a + b
First expand the expressio
It would be great if you could provide a concrete test case (with your
expected result) to work on. You are welcome to open a new issue for
this in our issue tracker.
I'll try I've simplified my actual problem to get a shorter answer.
The equations come from circuits and the result we're l
On Mar 15, 2010, at 2:37 PM, smichr wrote:
I think the thing to do is to use polynomial division which is going
to tell you exactly (as in a numerical division) what the whole and
remainder parts are:
n,d = eq.as_numer_denom(); s = eq.atoms(Symbol); w,r = div(Poly
(n,*s),Poly(d, *s))
print (
On Mar 27, 2010, at 2:00 PM, Toon Verstraelen wrote:
Ondrej Certik wrote:
On Sat, Mar 27, 2010 at 7:44 AM, Vinzent Steinberg
wrote:
I like mpmath's format:
**Arguments**
*arg1*
description
*arg2*
description
Note that each docstring must (should) contain examples of usa
ith === in docstrings doesn't play well with
sphinx, I guess because it expects it to be on a different level
then it ends up being on.
Aaron Meurer
On Mar 27, 2010, at 12:29 PM, Dan Schult wrote:
On Mar 27, 2010, at 2:00 PM, Toon Verstraelen wrote:
Ondrej Certik wrote:
On Sat, Mar 27,
