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
