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Topic: Continuous iteration (not tetration)
Replies: 12   Last Post: Feb 8, 2009 11:52 PM

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matt271829-news@yahoo.co.uk

Posts: 2,136
Registered: 1/25/05
Re: Continuous iteration (not tetration)
Posted: Feb 8, 2009 11:52 PM
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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)




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