Hello Chris,
Thanks a lot for your answer! It's very interesting. By now I just need the
bisection method: nsolve(eq,c2,(1,1000),solver='bisect'). But I will keep
in mind the continuation method for the future.
Cheers,
Camille
Le jeudi 26 juin 2014 18:36:58 UTC+2, Chris Smith a écrit :
>
> Above
Try only signsimp before substitution. Can you post code to generate one of
these expressions?
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# Yes: if greeting:
# No:if greeting == True:
# Worse: if greeting is True:
ok. but in the case of differentiating between None or False you have to use
the literal.
...is_foo cane use is True but for bool it's bad...
On Thursday, June 26, 2014 10:47:01 AM UTC-5, Aaro
>>> from sympy.utilities.iterables import permutations
>>> for i in permutations(range(3)):
... print i
...
(0, 1, 2)
(0, 2, 1)
(1, 0, 2)
(1, 2, 0)
(2, 0, 1)
(2, 1, 0)
;-) but if you help(permutations) you will see that we have just imported
permutations from itertools.
Also note that we
If you create a constant Poly, it picks the domain. You should look at
the source to see how it picks that.
Aaron Meurer
On Sat, Jun 21, 2014 at 10:48 AM, Christophe Bal wrote:
> Hello.
>
> I do not know if it is the case but I think that sympy should have a domain
> method for expressions. This
I think I'm losing track of your question here. is_foo can't use
BooleanTrue because it isn't symbolic. It will never return an
arbitrary boolean expression. Also, it uses three-valued logic,
whereas BooleanTrue is explicitly two-valued.
Aaron Meurer
On Fri, Jun 27, 2014 at 7:36 AM, Chris Smith
MatLab also have something similar to our UniversalSet
http://www.mathworks.in/help/symbolic/mupad_ref/universe.html.
On Saturday, 10 May 2014 23:33:12 UTC+5:30, Harsh Gupta wrote:
>
> This comes from the discussions on this PR
> https://github.com/sympy/sympy/pull/7462#issuecomment-42111992
>
>
But you can't create a constant Poly without giving the domain. Try
something like the following (and, again, trace the code to see if there is
a more direct way of learning the domain):
>>> Poly(1.2+var('x'))
Poly(1.0*x + 1.2, x, domain='RR')
>>> Poly(1+var('x'))
Poly(x + 1, x, domain='ZZ')
>>>
It's just that you have to explicitly use False or None to test an is_foo
result sometimes. And I'm so conditioned to use "is False" that when it
comes to testing the result of a relationship I do the same...but shouldn't
because (3>0) is True is False (because it's true, not True)and I note
For is_foo, you can use "is False", but for an inequality, it returns
a sympified boolean, because that is an example of a case where the
expression might not be true or false, but a symbolic expression (like
x > y). So a symbolic boolean makes sense in that case.
Aaron Meurer
On Fri, Jun 27, 201
You can't create it without giving a variable, but you can easily do
Poly(1, x).get_domain(). There's probably a more direct way, though.
Aaron Meurer
On Fri, Jun 27, 2014 at 2:15 PM, Chris Smith wrote:
> But you can't create a constant Poly without giving the domain. Try
> something like the f
There have been several questions about solving equations where it seems
that we haven't featured nsolve enough: nsolve is a great way to get a
numerical solution to an equation. You can use sympy to do the derivations
of some equation that you want to solve and then use nsolve to process it
fo
original question at
http://stackoverflow.com/questions/24393022/distance-from-point-to-complicated-curve
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I agree. We usually try to point people to numeric libraries for
numeric solutions, because they do it better. But there are advantages
to using SymPy, like being able to represent and manipulate the
equation symbolically. I think we should integrate with other numeric
solvers better (e.g., through
signsimp didn't seem to help. I'll try to get a gist of the expression up
tomorrow, it's so huge I haven't succesfully ran simplify on it. However, I
had the same thing happen with a readable expression. The issue was
something like:
>>> expr = sin(a)/tan(a)
>>> expr.subs(a, 0)
nan
>>> expr = e
Ahh...trisgsimp is like simplify in that it tries several things that are
known to generally work. If you know that it's trig-related because of
these sorts of ratios, you might just try rewriting
>>> cancel((sin(x)/tan(x)).rewrite(exp)).subs(x,0)
1
>>> w=Wild('w'); (sin(x)/tan(x)).replace(tan(w
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