On 1 Lut, 00:34, William Stein wst...@gmail.com wrote:
2010/1/31 Kakaz kazimierz.k...@gmail.com:
I have several worksheets in my sage, and when I have some new ideas I
create other one, somethimes just for fun. So there are worksheets
named: Idea 1, Matrices, 'FFT, FFT3 etc. After a
Hi Flavio,
You can have a trac account if you like -- it's only as an anti-spam
measure that we require a login. Just email Michael Abshoff (address
on the trac front page). It's possible to create new tickets for
feature requests; these should be allocated to the milestone sage-
wishlist.
That
Thanks,
I think I'll do both: ask for an account, so that I can report bugs,
and post about my feature request here (I'll open a new thread, so
that the subject will be informative)
Flávio
On 7 abr, 16:20, davidloeffler dave.loeff...@gmail.com wrote:
Hi Flavio,
You can have a trac account
sage: matrix(SR, 3, 3, [[21,17,6],[-5,-1,-6],[4,4,16]]).exp()
[ (13*e^16 - e^4)/4 (13*e^16 - 5*e^4)/4 (e^16 - e^4)/2]
[ (e^4 - 9*e^16)/4 (5*e^4 - 9*e^16)/4 (e^4 - e^16)/2]
[ 4*e^16 4*e^16e^16]
this does not work for 4x4 matrices, try
Hello,
sage: r = matrix(SR, 4, 4, [[21,17,6,8], [-5,-1,-6,-3], [4,4,16,2],
[2,3,-4,-1]])
sage: r.exp()
.
This is happening since Maxima is failing to do the computation for
reasons that I don't know. I suppose it wouldn't be too difficult to
write our own matrix exponentiation.
Hi Georg,
There is currently support for taking the matrix exponential of a
symbolic matrix already in Sage since it is using Maxima in the
background. I suppose that this should be extended to other types of
matrices.
sage: matrix(SR, 3, 3, [[21,17,6],[-5,-1,-6],[4,4,16]]).exp()
[ (13*e^16
Thank you for the fast answer Mike,
What functionality did you envision having in a symmetric matrix class?
In general (not specific to the hermitian (symmetric) property)
exponentiation,
determinate,
elementary matrix operations:
- changing rows(colums)
- multiplication of specific
A variation of this, which would be useful in some elliptic curve
calculations, would be a function
RR(x).nearby_rational_whose_denominator_is_a_perfect_square().
For either problem, is there a better solution than going through the
continued fraction convergents until one is found which has the
