Can sexp(z) be periodic ?? tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 01/12/2015, 12:46 AM I wonder if sexp(z) can be periodic. In particular 2pi i periodic. A search for period(ic) gives too many results. SO forgive me if this has been asked/answered before. If I recall correctly a pseudoperiod of 2 pi i / L is known to be possible ( L is a fixed point of exp ) because of f(z) = lim exp^[n](L^(z-n)) and similar limits. Im unaware of limit formula's that give a periodic solution. Maybe pentation helps us out , although I want to avoid adding this function. Maybe the system of equations is overdetermined ?? f(0) = 1 f(z+a) = f(z) f(z+1) = exp(f(z)) On the other hand maybe it has uniqueness ? I think it has uniqueness if it has existance. Reason is f(z+theta(z)) = f(z+a+theta(z+a)) Implies that theta(z) is both '1' and 'a' periodic HENCE double periodic ; A nonconstant theta can thus not be entire here ! COMBINE THAT WITH THE RIEMANN MAPPING THEOREM AND nonconstant theta(z) cannot exist !!! ( I think ) regards tommy1729 sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 01/12/2015, 01:54 AM (This post was last modified: 01/12/2015, 02:11 AM by sheldonison.) (01/12/2015, 12:46 AM)tommy1729 Wrote: I wonder if sexp(z) can be periodic. In particular 2pi i periodic.... tommy1729 L=0.318131505204764 + 1.33723570143069i, and one can develop the standard Schroder equations about the fixed points. At the fixed point, $\lambda=L$, where $\lambda$ is the fixed point multiplier, since $\exp(L+\delta)=L\cdot(1+\delta)=L+\delta L$ The definition of the formal Schroder equation, which leads to a formal Taylor series is $S(L)=0$ $S(\exp(z)) = \lambda\cdot S( z)$ So then $\exp^{oz} = S^{-1}(\lambda^z)\;\;\;\text{period}=\frac{2\pi }{\ln(L)}=\frac{2\pi }{L}\;\approx \;4.4469+1.05794i$ The $S^{-1}(\lambda^z)$ super function is also entire. Of course, the Schroder function of 0,1,e,e^e, are all singularities... so this function needs a lot of work to become the real valued sexp(z) we use for Tetration, but it is the starting point... - Sheldon tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 01/14/2015, 01:19 PM (This post was last modified: 01/14/2015, 01:21 PM by tommy1729.) (01/12/2015, 01:54 AM)sheldonison Wrote: (01/12/2015, 12:46 AM)tommy1729 Wrote: I wonder if sexp(z) can be periodic. In particular 2pi i periodic.... tommy1729 L=0.318131505204764 + 1.33723570143069i, and one can develop the standard Schroder equations about the fixed points. At the fixed point, $\lambda=L$, where $\lambda$ is the fixed point multiplier, since $\exp(L+\delta)=L\cdot(1+\delta)=L+\delta L$ The definition of the formal Schroder equation, which leads to a formal Taylor series is $S(L)=0$ $S(\exp(z)) = \lambda\cdot S( z)$ So then $\exp^{oz} = S^{-1}(\lambda^z)\;\;\;\text{period}=\frac{2\pi }{\ln(L)}=\frac{2\pi }{L}\;\approx \;4.4469+1.05794i$ The $S^{-1}(\lambda^z)$ super function is also entire. Of course, the Schroder function of 0,1,e,e^e, are all singularities... so this function needs a lot of work to become the real valued sexp(z) we use for Tetration, but it is the starting point... But the Schroder function does not give a real-analytic sexp. I prefer not to " abuse " notation. $S(L)=0$ $S(k)=1$ $S(\exp(z)) = \lambda\cdot S( z)$ So then $\exp^{[z]}(k) = S^{-1}(\lambda^z)$ Im intrested in both periodic sexp's ; both real-analytic and not real-analytic. I was thinking about other limits forms , for instance including terms like exp(z) to " force " periodicity , but I have convergeance issues that cannot be solved by analytic continuation. regards tomm1729 « Next Oldest | Next Newest »

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