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