# Tetration Forum

Full Version: [2014] The secondary fixpoint issue.
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Let exp(x) =/= x , exp(exp(x)) = x , Im(x) > 0.
Thus x is a secondary fixpoint in the upper half-plane.

Let y be a number in the upper half-plane such that sexp(y) = x and sexp is analytic in the upper half-plane.

Then by assuming the existance of y and the validity of sexp(z+1) = exp(sexp(z)) everywhere we get :

sexp(y+1) =/= sexp(y) , sexp(y+2) = sexp(y).

thus for y = a + b i and R any real , we have that :

sexp(R + b i) is a nonconstant analytic PERIODIC function in R.

And then by analytic continuation , sexp(z) is a nonconstant analytic periodic function in the upper half-plane !!

But that cannot be true !!

Same applies to n-ary fixpoints !!

Seems unlikely that sexp contains none of these n-ary fixpoints ?!

And as for the functional equation f(x+1) = exp(f(x)) + 2pi i that is on another branch. So that does not seem to help.

Where is the mistake ??

I posted this before , hoping the seeming paradox is understood now.

Also note that the "paradox" is not limited to 2nd ary fixpoints , 3rd ary fixpoints etc but also V-ary fixpoints for any V > 1 !

regards

tommy1729
(06/15/2014, 06:31 PM)tommy1729 Wrote: [ -> ]....
Seems unlikely that sexp contains none of these n-ary fixpoints ?!

And as for the functional equation f(x+1) = exp(f(x)) + 2pi i that is on another branch. So that does not seem to help.

Conjecture: if $\text{sexp}(z)=L$, than for some positive integer n, there is a non-zero integer m such that $\text{sexp}(z-n)=L+2m\pi i$

This would apply for L equals any fixed point of exp(z) and any finite value of z. This conjecture would apply to both the Kneser solution, and the secondary fixed point solution. I think it can be proven by showing for these two solutions, that sexp(z-n)<>L for large finite values of n.
(06/15/2014, 08:07 PM)sheldonison Wrote: [ -> ]
(06/15/2014, 06:31 PM)tommy1729 Wrote: [ -> ]....
Seems unlikely that sexp contains none of these n-ary fixpoints ?!

And as for the functional equation f(x+1) = exp(f(x)) + 2pi i that is on another branch. So that does not seem to help.

Conjecture: if $\text{sexp}(z)=L$, than for some positive integer n, there is a non-zero integer m such that $\text{sexp}(z-n)=L+2m\pi i$

This would apply for L equals any fixed point of exp(z). This conjecture would apply to both the Kneser solution, and the secondary fixed point solution. I think it can be proven by showing for these two solutions, that sexp(z-n)<>L as n goes to infinity.

Wait a minute , we have sexp(R + oo i) = L or conj(L) for all real R and all real oo !?

I thought we all agreed on that for years ?

But R + oo i - n is also of the form R + oo i.

??
Now Im even more confused.