As mentioned before my 2sinh method comes from my lost notebook.
Most of the notebook was number theory but there were also things about tetration.
I recall a big part of the notebook.
So its time to post yet another C^oo solution to tetration.
Due to lack of time and the huge amount of pages I cannot post all right now in a single post.
There are 2 ways to compute the tetration (method) and they are equivalent.
Method 1 assumes analyticity and analytic continuation.
(it uses Taylor series)
Method 2 depends strongly on convergeance assumptions.
The method is basicly a way to solve the equation
f ' (x) = Continuum product f(x)
Or f ' (x) ~ f(x) f(x-1) f(x-2) ... if you like.
It then follows f(x) should be tetration (base e) or at least close to it.
I will focus on method 2 here.
It is based on a real sequence a_n.
a_0 a_1 ...
Forgive any typos.
***
Method 2
k is a postive integer.
The larger k , the better the approximation.
So lim k -> +oo is the solution.
Let sexp(1) = sexp ' (1) = 1
Let a_(n+m) > a_n (a_n is strictly increasing).
g(k) ~ k
a_0 = 1
a_1 = 1 +1/g(k)
a_2
a_3
...
a_k = exp(a_0) = e
a_(k+1) = exp(a_1) = exp(1+1/g(k))
k(a_(k+1)-a_k) = a_1 a_k
k(a_(km + q) - a_(km + q -1)) = a_(km + q - 1)[k(a_(k(m-1) + q) - a_(k(m-1) + q -1))]
where n = km + q
and q = n mod k
then
sexp(1+x) = lim a_(x/k)
to be continued ...
regards
tommy1729
Most of the notebook was number theory but there were also things about tetration.
I recall a big part of the notebook.
So its time to post yet another C^oo solution to tetration.
Due to lack of time and the huge amount of pages I cannot post all right now in a single post.
There are 2 ways to compute the tetration (method) and they are equivalent.
Method 1 assumes analyticity and analytic continuation.
(it uses Taylor series)
Method 2 depends strongly on convergeance assumptions.
The method is basicly a way to solve the equation
f ' (x) = Continuum product f(x)
Or f ' (x) ~ f(x) f(x-1) f(x-2) ... if you like.
It then follows f(x) should be tetration (base e) or at least close to it.
I will focus on method 2 here.
It is based on a real sequence a_n.
a_0 a_1 ...
Forgive any typos.
***
Method 2
k is a postive integer.
The larger k , the better the approximation.
So lim k -> +oo is the solution.
Let sexp(1) = sexp ' (1) = 1
Let a_(n+m) > a_n (a_n is strictly increasing).
g(k) ~ k
a_0 = 1
a_1 = 1 +1/g(k)
a_2
a_3
...
a_k = exp(a_0) = e
a_(k+1) = exp(a_1) = exp(1+1/g(k))
k(a_(k+1)-a_k) = a_1 a_k
k(a_(km + q) - a_(km + q -1)) = a_(km + q - 1)[k(a_(k(m-1) + q) - a_(k(m-1) + q -1))]
where n = km + q
and q = n mod k
then
sexp(1+x) = lim a_(x/k)
to be continued ...
regards
tommy1729