Nixon-Banach-Lambert-Raes tetration is analytic , simple and “ closed form “ !!

[tommy1729]

I found a mistake and therefore this edit :


...

consider tetration (base e) for \( Re(s) > 2 \).

We start with real s and then extend by analytic continuation. 

Anyways it thus makes sense to consider this :

For \( s > 2 \) : 

\( \lim_{n\to\infty} \log \log \cdots (n\,\text{times})\cdots \log \phi(s+n) = \phi(s) + r_n(s) \)

And then by recursion and induction we get

\( r_0(s) = 0 , r_1(s) = s \)


\( r_{n+1}(s) = ln(\phi(s+1) + r_n(s+1)) - \phi(s) \)

by using log(a + b) = log(a) + log(1 + b/a) we get 


\( r_{n+1}(s) = ln(\phi(s+1)) + ln(1 + (r_n(s+1)/\phi(s+1)) ) - \phi(s) \)

( notice up to here we could have done the analogue with the sinh method and in fact we could try the things below too. But for phi things work out nicer. )


by using the fundamental equation of phi : \( \phi(s+1) = e^{s+\phi(s)} \)
  we can simplify without any problems ;


\( r_{n+1}(s) = s + \phi(s) + ln(1 + (r_n(s+1)/\phi(s+1)) ) - \phi(s) \)

and finally 

\( r_{n+1}(s) = s  + ln(1 + (r_n(s+1)/\phi(s+1)) ) \)



NOW we can therefore consistently define

\( \lim_n r_n(s) = v(s) = V(s) \)

We arrive at 

\( V(s) = s  + ln(1 + (V(s+1)/\phi(s+1)) ) \)

Define 

\( V(s+1) = V(s) + R(s) \)

Notice this equation contains functions that are analytic almost everywhere. [1]
And remember the inverse of a locally analytic function is also locally analytic.[2]

Now from the concept of relativity we know we only need to solve this equation for V and that is defined and analytic for most complex numbers too by [1] and [2].


...


SO we are left to solve this :

\( V  =  s  + ln(1 + (V+R(s))/(ts) ) \)

Using the Lambert-W function we get 

\( V = - LAMBERT-W( -\exp(-ts - s - R(s)) ts) - ts - R(s) \)

But ts is just \phi(s+1) so 


\( V = -W( -\exp(- \phi(s+1) - s - R(s)) \phi(s+1) ) - \phi(s+1) - R(s) \)

So we have a closed form using phi, R(s) and lambert-W :

\( \lim_{n\to\infty} \log \log \cdots (n\,\text{times})\cdots \log \phi(s+n) = \phi(s) - W( -\exp(- \phi(s+1) - s - R(s)) \phi(s+1) ) - \phi(s+1) - R(s) \)

Notice this also gives us a way to compute the difference operator over phi , although I am uncertain how practical it is.

But let us continue

\( \phi(s+1) = e^{s+\phi(s)} \) therefore 



\( V = -W( -\exp(- \phi(s+1) + \phi(s) - R(s)) ) - \phi(s+1) - R(s) \)


and thus 

\( \lim_{n\to\infty} \log \log \cdots (n\,\text{times})\cdots \log \phi(s+n) = \phi(s) - \phi(s+1) - R(s) -W( -\exp(- \phi(s+1) + \phi(s) - R(s)) \)

So if we take \( M(s) =  - \phi(s+1) + \phi(s) \) then we get 


\( \lim_{n\to\infty} \log \log \cdots (n\,\text{times})\cdots \log \phi(s+n) = M(s) - R(s) - W( -\exp(M(s) - R(s)) ) \)


Ofcourse branches of logarithms , inverse functions of phi or M - R and ofcourse lambertW are not considered in this sketchy overview here.
So be careful with those issues.


***

This replaces alot of attention and mystery towards M(s) - R(s) and the correct branch of LambertW.

I think R(s) is close to 1 for all Re(s) > 2 with small imaginary parts but I need to investigate.

Another thing is that at first sight there are way too many " negative terms " in the solution.

WE CANNOT ACCEPT " POSITIVE = NEGATIVE ". Not even for a C^oo solution.
( reminds me of the famous 1+2+3+4+... = -1/2 )

I hope this is resolved by the understanding of M,R and the branches of LambertW.
Maybe we can rewrite the expression without negative terms ??

I mean M(s) is negative for s > 2 right ??

Did I make a sign mistake ?
Or another mistake ?

Or does the equation give other solutions than the one given here that cannot even be reached by the branches of the lambertW ?
How about Banach then ?? 

So many questions.

What do you think ??



regards

tommy1729