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Topic: Iterating the function f(x,y) = (sin(y)+x,sin(sin(y)+x)+y)
Replies: 12   Last Post: Jan 5, 2014 12:59 PM

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Thomas Nordhaus

Posts: 433
Registered: 12/13/04
Re: Iterating the function f(x,y) = (sin(y)+x,sin(sin(y)+x)+y)
Posted: Jan 4, 2014 7:25 PM
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Am 04.01.2014 23:38, schrieb quasi:
> quasi wrote:
>> Richard Clark wrote:
>>> Thomas.Nordhaus wrote:
>>>> Thomas.Nordhaus wrote:
>>>>> Richard Clark wrote:
>>>>>>
>>>>>> Take the function
>>>>>>
>>>>>> f(x,y) = (sin(y)+x, sin(sin(y)+x)+y)
>>>>>>
>>>>>> This has fixed points at (n.pi,m.pi) for integer values of
>>>>>> n and m. These are alternately centres and saddle points.
>>>>>>
>>>>>> If you start an iteration of the function ...

>>>>>
>>>>> f is area-preserving, so the dynamics is pretty well
>>>>> described by KAM-theory. I don't see many surprises - but
>>>>> still a fun map to iterate

>>>>
>>>> Actually, your map f lifts to a map on the torus if you look
>>>> at x and y modulo 2*Pi. The map then only has two centres at
>>>> (0,Pi) and (Pi,0) and two saddles at (0,0) and (Pi,Pi). Still
>>>> fun to play around with.

>>>
>>> Yes, it can be reduced mod 2*pi but it's worth using the
>>> original to demonstrate the random walk.
>>>
>>> I don't think the map is area preserving - As evidence for this
>>> I would point out that two points close to (0,0), say (0.1,0)
>>> and (0.11,0), can end up very far apart if we iterate the
>>> function enough times.

>>
>> The function f is definitely area preserving on R^2 -- check the
>> Jacobian.

>
> Of course, in this context, area preserving means _locally_
> area preserving.
>
> The function f is definitely not globally area preserving.
>
> After all, f is periodic.


No, f isn't periodic - it is a diffeomorphism! You can solve the equations

(1) X = sin(y) + x, (2) Y = sin(sin(y) + x) + y uniquely for x and y:
(2),(1) => y = Y - sin(X) and then (1) => x = X - sin(Y-sin(X)). Therefore:

f^(-1)(X,Y) = (X - sin(Y-sin(X)) , Y - sin(X)).

So f ia locally and globally area-preserving.


>
> quasi
>



--
Thomas Nordhaus



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