On Monday, April 1, 2013 2:44:36 AM UTC-4, Ben Goodrich wrote:
>
> g = y * ((1-(z*y)**2)**(-2+K/2 + eta) * gamma(S(1)/2 *
> (-1+K)+eta))/(sqrt(pi) * gamma(-1+K/2+eta)) * (2 * (y**2)**eta
> (1-y**2)**(S(1)/2 * (-4+K)))/beta(S(1)/2+eta,-1+K/2)
> f = simplify(integrate(g, (y, 0, 1
Hi,
When I input
y = symbols("y", positive=True)
z = symbols("z", real=True)
K = symbols("K", integer=True, positive=True)
eta = Symbol("eta", positive=True)
g = y * ((1-(z*y)**2)**(-2+K/2 + eta) * gamma(S(1)/2 *
(-1+K)+eta))/(sqrt(pi) * gamma(-1+K/2+eta)) * (2 * (y**2)**eta
(1-y**2)**(S(1)/2 *
Hi Tom,
On Saturday, April 21, 2012 10:42:42 AM UTC-4, Tom Bachmann wrote:
> Instead of spamming further I will now really wrap this up and submit a
> pull request.
>
Thanks for working on this. In the example below, there seems to be a
problem with integer constants in rational expressions.
On Oct 9, 11:07 pm, Ben Goodrich wrote:
> I'll open an Issue for it.
Now issue 2073
http://code.google.com/p/sympy/issues/detail?id=2073
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On Oct 9, 11:07 pm, Ben Goodrich wrote:
> > I tried running the limit through Maple, and it couldn't do it (it just
> > returned an unevaluated form), and Wolphram Alpha times out. Any idea what
> > the answer should be?
>
> I conjecture the limit is 1
for c in
Hi Aaron,
On Oct 9, 10:32 pm, "Aaron S. Meurer" wrote:
> I tried running the limit through Maple, and it couldn't do it (it just
> returned an unevaluated form), and Wolphram Alpha times out. Any idea what
> the answer should be?
I conjecture the limit is 1 based on its similarity to some oth
Hi,
I didn't see this problem exactly when I searched through the Issues,
but I wanted to check that it was not a known problem first.
Thanks,
Ben
In [32]: c,n,r = symbols('cnr')
In [33]: limit( (( 2*n*(n-r+1)/(n + r*(n-r+1)) )**c + (r-1)*( n*(n-r
+2)/(n + r*(n-r+1)) )**c - n)/(n**c - n), n, oo
Sorry. You had said that already, but it did not stick in my brain. --
Ben
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Hi Mateusz,
This works with the high-level code but does not work with the low-
level code in polys9:
from sympy.polys.monomialtools import monomial_lex_key as O_lex
from sympy.polys.groebnertools import *
from sympy.polys.polytools import basic_from_dict
gens = [x,y]
u = 1
f = sdp_from_dict(Po
Hi Chris,
On Apr 19, 3:48 pm, smichr wrote:
> > FF = sdp_from_dict(Poly(F, *gens, domain = 'QQ').as_dict(), O_lex)
>
> > or specify any option after the *gens without getting the above syntax
> > error.
>
> > I am using Python 2.5.
>
> See Mateusz's 11:50 comment above. It's the same reason you r
es instead. You can probably fix it by doing
>
> dict([ (tuple(map(int, monom)), coeff) for monom, coeff in M[0,0] ])
>
> (though I haven't tried it).
>
> Aaron Meurer
> On Apr 19, 2010, at 1:43 PM, Ben Goodrich wrote:
>
>
>
> > Hi Mateusz,
>
> &g
Hi Mateusz,
I think I understand but ...
On Apr 19, 12:13 pm, Mateusz Paprocki wrote:
> They look identical, but they're not. As you use F.as_dict(), then in
> both cases c_i are of type Integer. However, keys of _50 and _51 are
> different. In _50 M_i are tuples of Integer instances and in _51
Hi,
Okay, I got the domain thing figured out, but now I am running into a
separate problem when the polynomial in sdp form is inside of a
matrix.
All this is fine:
from sympy.polys.monomialtools import monomial_lex_key as O_lex
from sympy.polys.groebnertools import *
from sympy.polys.polytools i
Hi Mateusz,
On Apr 18, 6:48 pm, Mateusz Paprocki wrote:
> Hi,
>
> > fgh = Poly(dict(FGH), domain = 'QQ', *gens)
>
> the above should read
>
> Poly(dict(FGH), *gens, domain='ZZ')
>
> (or domain='QQ' if you used the rational field in previous steps).
I tried that way previously, but it results in
Hi Mateusz,
On Apr 18, 2:06 pm, Ben Goodrich wrote:
> Mateusz's suggestions are working well for me, just ugly. And it is
> still slow and RAM-hungry when you convert a big sdp polynomial back
> into basic SymPy form at the end, but maybe your work on expand / core
> will hel
Hi Chris,
On Apr 18, 12:01 pm, smichr wrote:
> Hi Ben,
>
> I tried the original expressions that you posted and this takes 9
> minutes on my pretty modest machine with branch 1766 (on which I've
> reworked the expand quite a bit). Try
>
> numer = (first*second - third**2).expand(mul=1)
>
> and s
Hi Mateusz,
On Apr 16, 6:43 pm, Mateusz Paprocki wrote:
> This is all correct, because you are using top-level quo() function with
> FGH, which is a low-level polynomial representation.
Ah, now I understand. Thanks.
Ben
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Hi Mateusz,
On Apr 16, 4:10 pm, Mateusz Paprocki wrote:
> > What exactly is gens here? I thought it was supposed to be a list that
> > contains all the symbols in the polynomials, but then I eventually run
> > into this error
>
> > /tmp/mattpap-sympy-polys-cd30a32/sympy/polys/polytools.py in
> >
Hi Mateusz
On Apr 16, 11:46 am, Mateusz Paprocki wrote:
> SymPy can compute things like F*G - H**2 fast, unfortunately not on the
> user level. Below you will find a short tutorial how to do it fast but
> messy way.
>
> Lets assume you have polynomial-like expressions F, G, H. For my own
> conven
Hi Akshay,
On Apr 16, 11:51 am, Akshay Srinivasan
wrote:
> Forgot to expand the terms in the program,
>
> [neptune ~/bench] time g++ -lcln -lginac RAMspeed.cc -o RAMspeed
>
> real 0m57.911s
> user 0m55.526s
> sys 0m0.780s
> [neptune ~/bench] time ./R
Hi Mateusz,
On Apr 16, 11:46 am, Mateusz Paprocki wrote:
> SymPy can compute things like F*G - H**2 fast, unfortunately not on the
> user level. Below you will find a short tutorial how to do it fast but
> messy way.
Thank you for the tutorial and all your work on the polynomials. I
will try my
Hi again,
Things are a bit better now, but I still can't get as far as I'd like with
the polys8 branch (as of a few days ago). I didn't rigorously benchmark the
attached, but just a rough impression from looking at top implies that
thing = first * second - third.sqr()
where first, second, and th
On Apr 9, 7:05 pm, Ondrej Certik wrote:
> On Fri, Apr 9, 2010 at 1:07 PM, Ben Goodrich wrote:
> > So far, that is only using a small amount of RAM, but it is still
> > calculating. I will report back when it finishes or errors.
>
> Yep, report back.
>
> Ondrej
Oka
On Apr 9, 3:09 pm, Ondrej Certik wrote:
>
> And if this helps, we can then try to think how to make this
> automatic, so that you can just write:
>
> first * second - third**2
>
> and it will be the same fast.
>
> Ondrej
Okay, I had already assumed that if first * second - third**2 was
inside quo
On Mar 21, 4:13 pm, Ben Goodrich wrote:
> 3) Square both the numerator and denominator to finally get
>
> (((1 - Sigma_12**2/(tau_11**2*tau_22**2)) * (Sigma_34/(tau_33*tau_44)
> - Sigma_13*Sigma_14/(tau_11**2*tau_33*tau_44)) - (Sigma_23/
> (tau_22*tau_33) - Sigma_12*Sigma_13/(t
Hi,
I would like to write a function that rewrites an expression to
eliminate square root signs. I have read through the simplify module,
but I don't quite see how to do this. Is there something in the new
polys module that would help me?
Here is an example. I can assume that everything is real i
Is this a known issue? I see it on 0.7.0-git too. -- Ben
Python 2.5.5 console for SymPy 0.6.6
These commands were executed:
>>> from __future__ import division
>>> from sympy import *
>>> x, y, z = symbols('xyz')
>>> k, m, n = symbols('kmn', integer=True)
>>> f, g, h = map(Function, 'fgh')
Docum
Thanks Chris and Mateusz. I reopened issue 1472
http://code.google.com/p/sympy/issues/detail?id=1472
with a patch to substitute the one-liner and add a test.
Ben
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Hi,
The following code:
Lambda = Matrix(5, 2, lambda i, j: Symbol("Lambda_%d%d" % (i, j)))
Upsilon = eye(2)
Upsilon[0,1] = Upsilon[1,0] = Symbol("Upsilon_01")
Sigma = Lambda * Upsilon * Lambda.transpose()
Sigma == Sigma.transpose() # WTF
False
disputes the fact that Sigma is necessarily symmetri
On Jun 8, 8:08 pm, Ben Goodrich wrote:
> Thank you. I will do that. -- Ben
This is now issue 1465 at
http://code.google.com/p/sympy/issues/detail?id=1465
which has my patch. In my opinion, the previous behavior is
inconsistent with the definition of a Jacobian, so I hope the patch
can
Thank you. I will do that. -- Ben
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Can someone tell me if there is a subtle reason why the jacobian
function in matrices.py is restricted to only work when the Jacobian
is square?
The example in the documentation is
from sympy import symbols, sin, cos
rho, phi = symbols("rho phi")
X = Matrix([rho*cos(phi), rho*sin(phi)])
Y = Matr
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