Hello everyone !
Time to get serious.
Another infinite composition method.
This time I took care of unneccessary complications such a branch points, singularities etc.
Periodic points remain a topic however.
A sketch of the idea :
Let oo denote real infinity.
Basically it combines these 3 :
1) https://math.eretrandre.org/tetrationfor...p?tid=1320
2) https://math.eretrandre.org/tetrationfor...p?tid=1326
3) And most importantly the following f(s) ;
Tommy's Gaussian method :
f(s) = exp(t(s) * f(s-1))
t(s) = (erf(s)+1)/2
Notice that t(s - oo) = 0 and t(s + oo) = 1 for all (finite complex) s.
In particular
IF 2 + Re(w)^4 < Re(w)^6 < Im(w)^6 - Re(w)^2
THEN t(w)^2 is close to 0 or 1.
Even more so when Im(w)^2 is small compared to Re(w)^2.
The continued fraction for t(s) gives a good idea how it grows on the real line ; it grows at speeds about exp(x^2) to 0 or 1.
A visual of t(w) would demonstrate that it converges fast to 0 or 1 in the (resp) left and right triangle of an x shaped region.
That x shape is almost defined by Re(w)^2 = Im(w)^2 thus making approximately 4 90 degree angles at the origin and having only straith lines.
Therefore we can consistantly define for all s without singularities or poles ( hence t(s) and f(s) being entire ! )
f(s) = exp( t(s) * exp( t(s-1) * exp( t(s-2) * ...
thereby making f(s) an entire function !
Now we pick a point say e.
And we can try the ideas from
1) https://math.eretrandre.org/tetrationfor...p?tid=1320
2) https://math.eretrandre.org/tetrationfor...p?tid=1326
to consistently define
exp^[s](e)
and then by analytic continuation from e to z ;
exp^[s](z).
We know this analytic continuation exists because f(s) is entire and for some appropriate q we must have exp^[q](e) = z.
By picking the correct branch we also got the slog function.
It should be as simple as ( using small o notation )
lim n to +oo , Re( R ) > 0 ;
exp^[R](z) = ln^[n] ( f( g(z) + n + R) ) + o( t(-n+R) )
and ofcourse using the appropriate brances of ln and g.
regards
tommy1729
Tom Marcel Raes
Time to get serious.
Another infinite composition method.
This time I took care of unneccessary complications such a branch points, singularities etc.
Periodic points remain a topic however.
A sketch of the idea :
Let oo denote real infinity.
Basically it combines these 3 :
1) https://math.eretrandre.org/tetrationfor...p?tid=1320
2) https://math.eretrandre.org/tetrationfor...p?tid=1326
3) And most importantly the following f(s) ;
Tommy's Gaussian method :
f(s) = exp(t(s) * f(s-1))
t(s) = (erf(s)+1)/2
Notice that t(s - oo) = 0 and t(s + oo) = 1 for all (finite complex) s.
In particular
IF 2 + Re(w)^4 < Re(w)^6 < Im(w)^6 - Re(w)^2
THEN t(w)^2 is close to 0 or 1.
Even more so when Im(w)^2 is small compared to Re(w)^2.
The continued fraction for t(s) gives a good idea how it grows on the real line ; it grows at speeds about exp(x^2) to 0 or 1.
A visual of t(w) would demonstrate that it converges fast to 0 or 1 in the (resp) left and right triangle of an x shaped region.
That x shape is almost defined by Re(w)^2 = Im(w)^2 thus making approximately 4 90 degree angles at the origin and having only straith lines.
Therefore we can consistantly define for all s without singularities or poles ( hence t(s) and f(s) being entire ! )
f(s) = exp( t(s) * exp( t(s-1) * exp( t(s-2) * ...
thereby making f(s) an entire function !
Now we pick a point say e.
And we can try the ideas from
1) https://math.eretrandre.org/tetrationfor...p?tid=1320
2) https://math.eretrandre.org/tetrationfor...p?tid=1326
to consistently define
exp^[s](e)
and then by analytic continuation from e to z ;
exp^[s](z).
We know this analytic continuation exists because f(s) is entire and for some appropriate q we must have exp^[q](e) = z.
By picking the correct branch we also got the slog function.
It should be as simple as ( using small o notation )
lim n to +oo , Re( R ) > 0 ;
exp^[R](z) = ln^[n] ( f( g(z) + n + R) ) + o( t(-n+R) )
and ofcourse using the appropriate brances of ln and g.
regards
tommy1729
Tom Marcel Raes
