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 [2014] The secondary fixpoint issue. tommy1729 Ultimate Fellow Posts: 1,421 Threads: 346 Joined: Feb 2009 06/15/2014, 06:31 PM 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 sheldonison Long Time Fellow Posts: 664 Threads: 23 Joined: Oct 2008 06/15/2014, 08:07 PM (This post was last modified: 06/15/2014, 09:05 PM by sheldonison.) (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. - Sheldon tommy1729 Ultimate Fellow Posts: 1,421 Threads: 346 Joined: Feb 2009 06/15/2014, 08:17 PM (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. Thanks for the reply though. regards tommy1729 « Next Oldest | Next Newest »

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