How many methods have this property ?

[tommy1729]

How many C^oo ( or analytic ) solutions to tetration satisfy :

For all bases larger than eta for which the method works :

Property : sexp_a(x) > sexp_b(x)
For all a>b and x>1.

I used convention sexp_a(1) = a , sexp_b(1) = b here.

I could not find it here ?

I assume this property fails for base change and kneser.
I think it works for Cauchy contour method (kouznetsov?) and 2sinh method (tommy1729).

Maybe this has been said long ago.

regards

tommy1729

[sheldonison]

How many C^oo ( or analytic ) solutions to tetration satisfy :

For all bases larger than eta for which the method works :

Property : sexp_a(x) > sexp_b(x)
For all a>b and x>1.

I used convention sexp_a(1) = a , sexp_b(1) = b here.

I could not find it here ?

I assume this property fails for base change and kneser.
I think it works for Cauchy contour method (kouznetsov?) and 2sinh method (tommy1729).

Maybe this has been said long ago.

regards

tommy1729

I think all tetration methods have your property:

Property : sexp_a(x) > sexp_b(x)
For all a>b and x>1.

There are anomalies, but the definition is different. The typical Kneser method anomalies involve sexp(x)=sexp(y), x<>y, a>b, but counter intuitively, sexp_a(x+0.5)<sexp_b(y+0.5), where we would expect greater than. The anomalies usually involve really big numbers. For example, consider tetration base a=10, and base b=E, for x=3.5, 2.3637006685629
sexp_e(3.5) =~ sexp_10(2.3637006685629) =~ 6.1500942853737 E77
but sexp_e(4) > sexp_10(2.3637006685629+0.5)

The solutions which avoid this anomaly are coo, but not analytic, like the basechange method, which derives all tetrations for all bases from a single master base. Originally, for the basechange solution we used cheta, for exp(1/e), but this is really arbitrary. Then if f is the arbitrary chosen superfunction for the arbitrarily chosen master base, we can define a family of tetrations for all bases>exp(1/e), which avoid the anomaly. For example, f(z) could be 2sinh^z, or f(z) could be sexp(z) for any base.

\( k(a)=\lim_{n \to \infty} f^{-1}(a\uparrow\uparrow n)-n \)

\( \text{sexp}_a(z) = \lim_{n \to \infty} \log_a^{on}f(z+k(a)+n) \)
- Sheldon