Consider the function f(z) = exp(z) + z.

f(z) has no finite fixpoints.

We do iterates ( fractional near the real line ) by moving the fixpoint at oo to a finite place.

Of other ways.

I Came to consider the following

f^[2](z) = exp(exp(z)) exp(z) + exp(z) + z.

The finite fixpoints are then

Exp(z) ( exp(exp(z)) +1 ) + z = z

=>

Exp(z) ( exp(exp(z)) + 1) = 0

=>

Z = ln( ln(-1) )

One solution is

ln( pi ) + pi / 2 i = A

And another the complex conjugate of A : B.

Now i wonder if we take the half- iterates of f^[2] from those 2 fixpoints ( A and B ) based on koenigs solutions ,

Call them

FA(z) and FB(z).

And then take the average of them :

C(z) = ( FA(z) + FB(z))/2

[ i call this merged fixpoints method ]

Then is C(x) , where x is real , close to f(x) ??

And how does C(z) behave on the complex plane ?

----

Similar questions for the merged fixpoints of g(z,t) = (z^2 + t) + z with t ≥ 1.

Notice g has only 2 fixpoints.

Plots would be nice too.

Regards

Tommy1729

f(z) has no finite fixpoints.

We do iterates ( fractional near the real line ) by moving the fixpoint at oo to a finite place.

Of other ways.

I Came to consider the following

f^[2](z) = exp(exp(z)) exp(z) + exp(z) + z.

The finite fixpoints are then

Exp(z) ( exp(exp(z)) +1 ) + z = z

=>

Exp(z) ( exp(exp(z)) + 1) = 0

=>

Z = ln( ln(-1) )

One solution is

ln( pi ) + pi / 2 i = A

And another the complex conjugate of A : B.

Now i wonder if we take the half- iterates of f^[2] from those 2 fixpoints ( A and B ) based on koenigs solutions ,

Call them

FA(z) and FB(z).

And then take the average of them :

C(z) = ( FA(z) + FB(z))/2

[ i call this merged fixpoints method ]

Then is C(x) , where x is real , close to f(x) ??

And how does C(z) behave on the complex plane ?

----

Similar questions for the merged fixpoints of g(z,t) = (z^2 + t) + z with t ≥ 1.

Notice g has only 2 fixpoints.

Plots would be nice too.

Regards

Tommy1729