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[2014] The angle fractal.
#1
If an initial value x converges to a fixpoint at an angle y , we can give that value pair(x,y) a corresponding color depending on the angle y.

We can make a plot that would then be the analogue of a fractal , the coloring based on the values y corresponding to the x's.

---

That is the main idea ...

Some comments : the derivative at the fixpoints x_0 needs to be positive real.

SO :

Arg ( f ' (x_0) ) = 0

0 < Abs ( f ' (x_0) ) < 1

Also the case is simplest when f(z) only has 2 conjugate fixpoints.

***


Therefore I seek :

Taking - without loss of generality ? - the fixpoints equal to +/- i .


g(z) = real entire = ??

f(z) :=

exp(g(z)) (z^2+1) + z

Arg ( f ' (i) ) = 0

0 < Abs ( f ' (i) ) < 1

And also f(z) ~ exp(z) z^A for some small real A.

Many solutions g(z) must exist , but which is best ?

And how do the " new fractals " look like ?

---

Also can the angle be given by an integral ?

Contour integral ?

z_0 = initial value
z_n = f ( z_(n-1) )

lim n -> +oo

[z_n - z_(n-1)] / [ | z_n - z_(n-1) | ]

= contour integral ( z_0 ) ???



regards

tommy1729
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#2
So f(z) is of the form

f(z) =
exp(z + fake_ln( (- e^(-z) z^3 - e^(-z) z + 1 ) / (z^2+1) )) + z

, such that the derivative at both the fixpoints is a real Q.

If Q lies between 0 and 1 that can give a nice " angle fractal ".

However the case Q > 1 is also very intresting and gives an analogue of sexp for the superfunction of f(z).

This should be worth an investigation !


regards

tommy1729
Reply


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