
Re: Continuous iteration (not tetration)
Posted:
Feb 8, 2009 11:52 PM


On Feb 8, 11:51 pm, mike3 <mike4...@yahoo.com> wrote: > On Feb 7, 11:29 pm, Robert Israel > > <isr...@math.MyUniversitysInitials.ca> wrote: > > mike3 <mike4...@yahoo.com> writes: > > > Also, one thing that I noticed here is that the seemingly "chaotic" > > > orbits > > > of points under the z^2  2 map, when graphed continuously using the > > > formulas > > > you gave, look like simple "sine"like waves whose frequency increases > > > exponentially. > > > Is this something interesting? > > > With z = 2 cos(t), you get f(z) = 2 cos(2 t), and iterating > > f^[n](z) = 2^(n) cos(2^n t). Is that what you mean? > > I'm referring to the graph: set z = 0.45, for example. Then > plot f^n(z) against n, where n varies continuously from, say, > 0 to 5, using the formulas given for the continuous iteration. > The result will be a "sine" wave whose frequency increases > exponentially. Rotate n discretely: 0, 1, 2, 3, 4, 5, and the > result is an erratically jumping sequence of numbers. > > But I guess the above is why this would happen, no? As it > really _is_ a sinewave of exponentially growing frequency. > > And I think your formula should be f^n(z) = 2 cos(2^n t), > as the one with 2^n doesn't seem to give the proper > values. Set n = 1, you get 1/2 cos(2 t) not > 2 cos(2 t). :)
Right... as you probably realise, the "cos(2^n t)" formulation is one case of the general solution I posted for type 2  specifically, the case when alpha = (2*a*x + b + sqr((2*a*x + b)^2  16))/4 is complex. The complex exponentials unravel to give
f^n(x) = (4*cos(2^n*theta)  b)/(2*a)
where theta = arg(alpha)

