11/22/2021, 11:25 PM

The idea is to go experimental.

We approximate exp(s) by using (1+s/n)^n.

And perhaps later take n to +oo in the limit.

t(s) = (1 + erf(s))/2.

R_n(s) = n*(s^(1/n) - 1)

then

f_n(s) = (1 + (f_n(s-1) * t(s))/n )^n.

F_n(s) = lim m to +oo of R_n^[m] ( f_n(s + m) )

And then build the superfunctions from those.

And then ofcourse we can ask the analogue typical questions.

We could also test to which fixpoints it agrees.

(1 + z/n)^n = z has most of its zero's with negative real part and most on an almost circle for large n btw.

For n odd we have a real fixpoint and there are always zero's close to the pair of ln(z) = z.

Another benefit is ( for finite m at least ) we ( probably ) do not have log singularities but only root singularities.

I have many more ideas but to avoid making assumptions and speculations I will stop here.

Also this idea is something we ( top forum posters ) reached together and is partially very intuitive so I gave no name.

regards

tommy1729

We approximate exp(s) by using (1+s/n)^n.

And perhaps later take n to +oo in the limit.

t(s) = (1 + erf(s))/2.

R_n(s) = n*(s^(1/n) - 1)

then

f_n(s) = (1 + (f_n(s-1) * t(s))/n )^n.

F_n(s) = lim m to +oo of R_n^[m] ( f_n(s + m) )

And then build the superfunctions from those.

And then ofcourse we can ask the analogue typical questions.

We could also test to which fixpoints it agrees.

(1 + z/n)^n = z has most of its zero's with negative real part and most on an almost circle for large n btw.

For n odd we have a real fixpoint and there are always zero's close to the pair of ln(z) = z.

Another benefit is ( for finite m at least ) we ( probably ) do not have log singularities but only root singularities.

I have many more ideas but to avoid making assumptions and speculations I will stop here.

Also this idea is something we ( top forum posters ) reached together and is partially very intuitive so I gave no name.

regards

tommy1729