I have updated the Linear Algebra Quick Reference card to more closely match
version 4.8 and to catch up on 2.5 years worth of changes. You can find it (and
others) at:
http://wiki.sagemath.org/quickref
Rob
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe
When you factor out relations you are factoring out the normal
subgroup generated by the relations, which is (or may be) bigger than
just the subgroup they generate. Does that make sense?
John Cremona
On Dec 13, 3:12 am, syd.lavas...@gmail.com syd.lavas...@gmail.com
wrote:
OK I found few
Hello,
what is the easiest way to linearize cos(x)^3 ?
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more options, visit this group at
http://groups.google.com/group/sage-support
sage: (cos(x)^3).reduce_trig()
1/4*cos(3*x) + 3/4*cos(x)
On Dec 13, 7:45 pm, Christophe BAL projet...@gmail.com wrote:
Hello,
what is the easiest way to linearize cos(x)^3 ?
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
Thanks a lot.
Christophe
2011/12/13 achrzesz achrz...@wp.pl
sage: (cos(x)^3).reduce_trig()
1/4*cos(3*x) + 3/4*cos(x)
On Dec 13, 7:45 pm, Christophe BAL projet...@gmail.com wrote:
Hello,
what is the easiest way to linearize cos(x)^3 ?
--
To post to this group, send email to
It makes lots of sense. This happens because we identifies the words
in the Relation set with identity, hence by definition for each word r
\in R and element a \in G we have:
a r a^-1 = a 1 a^-1 = 1 and hence a r a^-1 \in R too.
Thank you very much for the clarification
On 13 דצמבר, 11:21, John
