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 C
Thanks a lot.
Christophe
2011/12/13 achrzesz
> sage: (cos(x)^3).reduce_trig()
> 1/4*cos(3*x) + 3/4*cos(x)
>
> On Dec 13, 7:45 pm, Christophe BAL wrote:
> > Hello,
> > what is the easiest way to linearize cos(x)^3 ?
>
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> T
sage: (cos(x)^3).reduce_trig()
1/4*cos(3*x) + 3/4*cos(x)
On Dec 13, 7:45 pm, Christophe BAL wrote:
> Hello,
> what is the easiest way to linearize cos(x)^3 ?
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Hello,
what is the easiest way to linearize cos(x)^3 ?
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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"
wrote:
> OK I found few things, I thought I'll wr
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
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