Warning: I haven't thought this through, but the ideas might be useful.
It looks like you'd call something a 3-cycle if for some x_0 we had
f(f(f(x_0))) == x_0, right? Then we should be able to do this
numerically, with some caveats:
# can't remember the slick way, so brute force
def iter_apply(f0, n):
fs = [f0()]
for i in xrange(n-1):
last = fs[-1]
fs.append(f0(last))
return fs
def find_pcycles(f0, n):
fs = iter_apply(f0, n)
req = fs[-1] == x # defining equation of the cycle
roots = req.roots(ring=RR)
for root, mult in roots:
yield [fi(x=root) for fi in fs]
which gives
sage: f(x) = x^2-1.76
sage:
sage: p1 = list(find_pcycles(f, 1))
sage: p1
[[-0.917744687875782], [1.91774468787578]]
sage: f(p1[0][0])
-0.917744687875783
sage: f(p1[1][0])
1.91774468787578
sage: list(find_pcycles(f, 2))
[[0.504987562112089, -1.50498756211209], [-0.917744687875783,
-0.917744687875782], [-1.50498756211209, 0.504987562112089],
[1.91774468787578, 1.91774468787578]]
sage: list(find_pcycles(f, 3))
[[1.33560128916886, 0.0238308036295167, -1.75943209279837],
[1.27545967679486, -0.133202612870350, -1.74225706392451],
[-0.917744687875783, -0.917744687875782, -0.917744687875783],
[-1.74225706392451, 1.27545967679486, -0.133202612870348],
[-1.75943209279837, 1.33560128916886, 0.0238308036295143],
[-0.133202612870349, -1.74225706392451, 1.27545967679486],
[0.0238308036295150, -1.75943209279837, 1.33560128916886],
[1.91774468787578, 1.91774468787578, 1.91774468787578]]
Note that this says [-0.917744687875783, -0.917744687875782,
-0.917744687875783] is a 3-cycle, which it kind of is..
Anyway, maybe playing off of something like that would help?
Doug
--
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
URL: http://www.sagemath.org
