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