Brute force tetration A_k(s) is analytic !

[tommy1729]

Ok I put my money on the uncompleted gamma function "G" :

\( A_k(s)=exp(A_k(s-1) - G(-ks)) \)

We can solve this by infinite composition : \( A_k(s)=exp( - G(-ks) + exp( - G(-k(s-1)) + ...) \)
( yes to be formal we need a " z parameter " too, but that is a detail )

Now consider the functions \( A_5(s),A_6(s),A_7(s),... \)

The limit k to +oo gives us A_oo(s) = A(s).

Now A(s) satisfies for Re(s) > 0 :

\( A(0)=LIM.exp(-G(-oo)+exp(-oo+0))=LIM.exp(0+exp(-oo))=exp(0)=1. \)

and 

\( A(s+1)=exp(A(s)+0)=exp(A(s)) \)

I see no clear reason why this should fail.

So this deserves some attention.

In particular I assume G in the imaginary direction.

***

Btw I want to mention that the functional equation f(s+1) = exp(f(s) + s) does a poor job at approximating tetration when you consider the derivative ;

we do not get f ' (x) = f(x) ln(f(x)) ln(ln(f(x))) * ... but f ' (x) = f(x) * (1 + ln(f(x)) * ( 1 + ln ln ( f(x) )...

The uncompleted gamma function method described above however does ( in the limit ) give f ' (x) = f(x) ln(f(x)) ln(ln(f(x))) * ... and does not suffer the periodicity problem of the exp method solution.

Hence I am betting on this uncompleted gamma trick.


regards

tommy1729

Tom Marcel Raes