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+--- Thread: Can sexp(z) be periodic ?? (/showthread.php?tid=950)
Can sexp(z) be periodic ?? - tommy1729 - 01/12/2015
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
RE: Can sexp(z) be periodic ?? - sheldonison - 01/12/2015
(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...
RE: Can sexp(z) be periodic ?? - tommy1729 - 01/14/2015
(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