Hello Dear
how can I calculate the integral point for this curve:
[0,0,0,-3568202637461440265241263457,0]
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The spring layout is "accidentally" finding the normal subgroup
{+/-1}. If we look at a smaller subgraph, the problem should be
apparent.
G= SL(2, ZZ)
S, T= G.gens(); ST= S*T
L= [S^i*ST^j for i in range(4) for j in range(3)]
els= Set([ a*b for a in L for b in L])
gr= G.cayley_graph(generators = [
i think i see what is going on. By reducing the number of vertices
plotted, it appears that the "double edges" are really little clusters
of four edges!
On 13 mar, 21:27, Pierre wrote:
> Dear Tom,
>
> Thanks for your answer! I get the empty set, too. I really wonder what
> is going on with the pi
Dear Tom,
Thanks for your answer! I get the empty set, too. I really wonder what
is going on with the picture though... if one cannot "rely on the
picture", then it pretty much defeats the purpose when it comes to
Cayley graphs, doesn't it?
and i mean, there *is* a double arrow on some edges.
th
I tracked down a confusing bug in some code I was writing in Sage down to
the fact that pow(a,b,m) applied to Sage Integers does not give an Integer
result, but rather an object with parent the ring of integers mod m (a
simple call to either int() or .lift() fixed my code).
I like this design d
This is clever idea.
It may be also relatively easy to "separate" this 4 ( or 2in example)
variables and then solve 4 recurrence relations directly.
But it looks like pure algorithmic, so I would like to ask if someone
implement that.
Thank You for Your help!
K
On Mar 13, 3:36 pm, David Joyner w
On Tue, Mar 13, 2012 at 10:27 AM, Kakaz wrote:
> Thanks for answer.
>
> Yes, I did. I found similar question but it was about rsolve and
> second order relation.
> I am looking for rsolve or similar solver in context of several
> variables ( ei. a(n), b(n) in example above and 4 variables in my ca
Thanks for answer.
Yes, I did. I found similar question but it was about rsolve and
second order relation.
I am looking for rsolve or similar solver in context of several
variables ( ei. a(n), b(n) in example above and 4 variables in my case
I want to solve).
I have read this:
http://groups.googl
I think it is in sympy (included in Sage).
I vaguely remember the question has been asked before on
this list but I don't remember the exact answer. Did you look through the
sage-support archive?
On Tue, Mar 13, 2012 at 9:35 AM, Kakaz wrote:
> Of course it should be:
> a(n+1) = A*a(n) + B*b(n)
Of course it should be:
a(n+1) = A*a(n) + B*b(n)
b(n+1) = C*a(n) + D*b(n)
On Mar 13, 2:32 pm, Kakaz wrote:
> Hi all!
> I would like to ask - is there possibility in Sage ( or Maxima) to
> solve first order recurrence relation given by linear system with
> several variables?
> For example:
>
> a(n
Hi all!
I would like to ask - is there possibility in Sage ( or Maxima) to
solve first order recurrence relation given by linear system with
several variables?
For example:
a(n+1) = A*a(n) + B*b(n)
b(n) = C*a(n) + D*b(n)
Have I find solution of linear system at first an then solve
"separated" rel
Pierre,
Don't rely on the picture!
sage: U = set(gr.edges())
sage: V = set(gr.reverse().edges())
sage: U.intersection(V) #for me, this is the empty set
On Tue, Mar 13, 2012 at 3:26 AM, Pierre wrote:
> Hi,
>
> I've been playing with Cayley graphs in Sage (thanks to whoever
> implemented this!)
Hi,
I've been playing with Cayley graphs in Sage (thanks to whoever
implemented this!) I got funny results on one example, and I'd like to
understand.
I've tried SL(2, ZZ):
sage: G= SL(2, ZZ)
sage: S, T= G.gens(); ST= S*T
sage: L= [S^i*ST^j for i in range(4) for j in range(3)] #S has order
4, ST
That's good news. It's a long way until usage of Sage becomes standard
in France, unfortunately. I hear that Lyon has made the transition
from Maple and has been confronted with a lot of resistance from
teachers who did not want to change their old habits...
Here in Strasbourg we were about to swi
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