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