I've attached the patch to the issue's page - it's just a one-line
change :)
Akshay
On Jan 18, 8:58 pm, "Ondrej Certik" wrote:
> On Sun, Jan 18, 2009 at 3:04 AM, Akshay Srinivasan
>
> wrote:
>
> > Hi Ondrej,
> > I think I've fixed it. As you said, it was something in the printing/
> > str.py.
Thanks.
I am computing something like
Prob(A + B) = Prob(A) + Prob(B) - Prob(A*B)
for a complicated discrete distribution where '+' means union and '*'
means intersection. This is simple enough for a small number of sets,
but there might be millions of sets and the above formula gets more
compl
Hello Alan,
Thanks for your answer. Apparently all I need from the exterior
algebra
can be done with '^'. If so, I can still use, as you suggest, a
diferent metric in the setup. (I will probably use a euclidien
metric)
and then use the geometric product ('*') to compute the dual: "-I*a".
I will
Hello Alan,
Thanks for your answer. Apparently all I need from the exterior
algebra
can be done with '^'. If so, I can still use, as you suggest, a
diferent metric in the setup. (I will probably use a euclidien
metric)
and then use the geometric product ('*') to compute the dual: "-I*a".
I will
Hello Alan,
Thanks for your answer. Apparently all I need from the exterior
algebra
can be done with '^'. If so, I can still use, as you suggest, a
diferent metric in the setup. (I will probably use a euclidien
metric)
and then use the geometric product ('*') to compute the dual: "-I*a".
I will
Hello Alan,
Thanks for your answer. Apparently all I need from the exterior
algebra
can be done with '^'. If so, I can still use, as you suggest, a
diferent metric in the setup. (I will probably use a euclidien
metric)
and then use the geometric product ('*') to compute the dual: "-I*a".
I will
Hello Alan,
Thanks for your answer. Apparently all I need from the exterior
algebra
can be done with '^'. If so, I can still use, as you suggest, a
diferent metric in the setup. (I will probably use a euclidien
metric)
and then use the geometric product ('*') to compute the dual: "-I*a".
I will
luis wrote:
> Hi,
>
> Thanks,
>
> Non, I don't want a Minkowsky metric. I am simply doing exterior
> algebra in R^4 (or R^n) and in such case I need a null metric
> to have the wedge product in Lambda(R^n).
>
> As you can see in
> http://en.wikipedia.org/wiki/Clifford_algebra#As_quantization_of_
Hi,
Thanks,
Non, I don't want a Minkowsky metric. I am simply doing exterior
algebra in R^4 (or R^n) and in such case I need a null metric
to have the wedge product in Lambda(R^n).
As you can see in
http://en.wikipedia.org/wiki/Clifford_algebra#As_quantization_of_exterior_algebra
the exterio
luis wrote:
> Hi,
>
>
> Thanks. Since I want to do exterior algebra, I am using a null
> metric.
> Also, I am not in 3-dim and so I don't have a cross product.
>
> Let's suppose we have the following code:
>
> ###
> #!/bin/env python
> from sympy.galgebra.GAsympy import *
> set
Hi,
Thanks. Since I want to do exterior algebra, I am using a null
metric.
Also, I am not in 3-dim and so I don't have a cross product.
Let's suppose we have the following code:
###
#!/bin/env python
from sympy.galgebra.GAsympy import *
set_main(sys.modules[__name__])
MV.s
luis wrote:
> Hi,
>
> Is there some way in 'galgebra.GAsympy'
> to apply the Hodge-* operator?
>
> Thanks,
>
> Luis
>
> PS. I am using MV.setup(basis,metric) with a null metric.
> >
>
>
See http://en.wikipedia.org/wiki/Geometric_algebra. I think the Hodge
dual of a vector, v, is just the geo
Hi,
Is there some way in 'galgebra.GAsympy'
to apply the Hodge-* operator?
Thanks,
Luis
PS. I am using MV.setup(basis,metric) with a null metric.
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Hi Reckoner,
On Sat, Jan 17, 2009 at 12:51 PM, Reckoner wrote:
>
>
> I know python has a set() object, but I'm wondering if sympy has set
> algebra built into it (e.g. union, intersection) for symbolic objects.
> Note that I'm interested in manipulating the sets themselves and not
> necessarily
On Tue, Jan 13, 2009 at 12:41 PM, Ondrej Certik wrote:
>>> If you did add everything, then you need to use "git rebase -i" and
>>> rewrite the history to remove the unwanted things. I can do that for
>>> you, if it's too dificult --- just push everything you have to github,
>>> I'll have a look a
On Sun, Jan 18, 2009 at 3:04 AM, Akshay Srinivasan
wrote:
>
> Hi Ondrej,
> I think I've fixed it. As you said, it was something in the printing/
> str.py.
Excellent, when you are ready, send us a patch please (either to
sympy-patches, or attach it to the issues).
> In doing that I came across t
On Sat, Jan 17, 2009 at 11:25 PM, Luke wrote:
>
> If I have a symbolic expression like:
> expr = a*sin(x) + b*cos(y) - c*tan(z)
>
> I can get each individual additive term, sort of like:
> In[1]: terms(expr)
> Out[1]: [a*sin(x), b*cos(y), -c*tan(z)]
In [2]: var("a b c")
Out[2]: (a, b, c)
In [3
Luke wrote:
> I would like everything to be symbolic. If you have two reference
> frames, say A and B, you can only take the dot product of a vector
> expressed using the A coordinates with a vector expressed in the B
> coordinates if you know the rotation matrix that relates the two
> reference
Hi Ondrej,
I think I've fixed it. As you said, it was something in the printing/
str.py.
In doing that I came across this,
>>(x/sin(x)).args()
(x,sin(x))
>>(cos(x)/sin(x)).args()
(1/sin(x),cos(x))
Shouldn't
>>(cos(x)/sin(x)).args()
give
(cos(x),1/sin(x))
??
Akshay (Neptune)
On Jan 18, 12:06
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