Maybe the solution at z=0 to f(f(z))=-z+z^2
#5
Im not sure how these relate but 3 ideas cross my mind.

( for every petal seperately : )

1)

 the "realifier" mentioned by at minimum yourself and Bo. 

2) 

how about the 1-periodic theta functions present in the super functions ? 

3)

 vector calculus to " see " what is really going on and even perhaps solve it. Together with series expansions that might make more sense.

I mentioned fourier and L-series in the past.

Mittag-Leffler perhaps.

It seems taylor might not be the best here.


I assume you considered fractional calculus too. And the julia equation.

***

Also i had the idea

f^[t](x) = f_1(t) x + f_2(t) x^2 + f_3(t) x^3 + ...

Understanding these 3 functions seems key to me.

If those f_n agree with one of the iterates computed by the carleman matrix methods ( picking the correct roots ) then we are close to having the semi-group isom.

Notice f_1(t) also has the semi-group iso for both f_1(t) = 1 and for f_1(t) = (-1)^t.

In fact around a fixpoint f_1 defines bijectively a fitting theta function.


However the issues are not resolved ; radius 0 , divergeance , bad series expansions ? etc

But the truncated taylor function f_1(t) x + f_2(t) x^2 + f_3(t) x^3 should work well in the limit formula's I assume.

further ideas start to resemble the ideas around exp(x) - 1 ....

Ironically I think the branching thing is resolved in a way similar to your own ideas of that " realifier " picking branches... if we got the correct solution ofcourse.

I want to point out that dynamics on a plane is alot like liquid flow.

Not sure if that is of any help.

just my 50 cent.


regards

tommy1729
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RE: Maybe the solution at z=0 to f(f(z))=-z+z^2 - by tommy1729 - 01/20/2023, 08:33 PM

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