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Is this THE equation for parabolic fix ?
Since jan 2015 I am very intrested in parabolic fixpoints again.
I take the fixpoint at 0 for simplicity.

Maybe I havent read the right books, but I feel a large gap in understanding of the subject.

On the other hand - as usual - I was not able to find the answers anywhere.

Anyway I came to the idea of working with analogue ideas of the hyperbolic fixpoint methods.

First I used the limit ideas.
This lead to the results here :

But I do not want just limits now , I want equations and series expansions.

The problem with Taylor is that it has zero radius of convergeance.

I considered the so-called Tommy-Kouznetsov expansions and the tommy equation and so forth.

But they do not seem to work either.

So apart from the limit formula's - and their explaination - and some perturbations I know almost nothing for sure.

Since trying to find the "right equation" AND the right (associated) converging expansions (as I tried in the thread tommy equation) might be hard because we want 2 things at the same time , I consider to return to Taylor series and just find the "right equation".

( It seems easier to find the right equation , then it is to find the right expansion WITHOUT the equation , since the last does not seem to make sense )

SO we return to the Taylor series with nonzero radius.

SO what is THE equation for parabolic fixpoints ?

For hyperbolic it turned out to be

g( f(x) ) = f(A x)

Notice that near the fixpoint 0 the function g behaves like A x and the iterates behave like A^x.

Notice A^x is the super of A x.

The analogue should then be

g ( f(x) ) = f( q(A,x) )

Now there are 2 ideas about what q should be based on the above.
( and those 2 might be both correct and equivalent ! )

1) q(A,x) is the Ath iteration of a nonlinear polynomial truncation of g(x).
We then solve the equation by using the Taylor expansion of q(A,x).

2) For simplicity assume g grows to 0 at a rate 1/sqrt(x).
( yes this idea resembles )

To compute what t(x) = 1/sqrt(x) is the super of we use the traditional :

t(x) is super of t( t^[-1](x) + 1).

so here 1/sqrt(x) is the super
of 1/sqrt(1/x^2 + 1)) = x/sqrt(x^2 + 1).

Now let K(x) be the Taylor of x/sqrt(x^2 + 1).

Then we get the equation

g( f(x) ) = f( K(x) )

In both cases [ 1) and 2) ] it is assumed that g can be solved in terms of a Taylor series with nonzero radius.

( I estimate the radius between 0.5 and 1. )

From this we can then compute the analytic superfunctions.

I noticed that this probably has an analogue in terms of linear algebra ;

G( F(x) ) = F ( T(x) )

is isomorphic to G * F = F * T where G,F,T are carleman matrices.

Im skeptical about what I wrote , so I wonder what you guys think about it.

If the parabolic indeed works similar to the hyperbolic case I find that an intresting " meta-math result " ; by which I mean an intresting fundamental generalization.



" Truth is whatever does not go away when we stop believing in it "

Messages In This Thread
Is this THE equation for parabolic fix ? - by tommy1729 - 03/18/2015, 09:49 PM

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