On 08/11/2011 07:04 PM, Tomo Lazovich wrote:
I just tried this in isympy and as long as you specify both real and
positive, it seems to work:
In [1]: r = Symbol('r', real=True, positive=True)
In [2]: sqrt(r*r)
Out[2]: r
In [3]: r = Symbol('r', real=True)
In [4]: sqrt(r*r)
Out[4]: │r│
On Thu, Aug 11, 2011 at 6:01 PM, Alan Bromborsky <[email protected]
<mailto:[email protected]>> wrote:
I have a symbol r which should be assumed real and greater than
zero. What I want is for sqrt(r*r) to evaluate to r. Can this be
currently done in sympy. I am trying to clean up some old code
that has a lot of hacks in it (curvilinear coordinates).
--
You received this message because you are subscribed to the Google
Groups "sympy" group.
To post to this group, send email to [email protected]
<mailto:[email protected]>.
To unsubscribe from this group, send email to
[email protected]
<mailto:sympy%[email protected]>.
For more options, visit this group at
http://groups.google.com/group/sympy?hl=en.
--
You received this message because you are subscribed to the Google
Groups "sympy" group.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/sympy?hl=en.
That worked fine. Now for another similar and more difficult problem.
Evaluate sqrt(r**2*cos(theta)**2) where -pi/2<theta<pi/2. This should
evaluate to r*cos(theta). Are symbol assumptions flexible enough yet
to handle this case (again curvilinear coordinates)?
--
You received this message because you are subscribed to the Google Groups
"sympy" group.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/sympy?hl=en.
