Slog(Exponential Factorial(x))

[tommy1729]

EF(x) = exponential factorial(x) = x^(x-1)^(x-2)^...^3^2^1.
What happens if you do the tetration logarithm of the exponential factorial function. (I am thinking tetration logarithm base the Tetra-Euler Number.) How can the tetration logarithm of the exponential factorial function be approximated?
slog(e4,EF(1)) = 0.
slog(e4,EF(2)) ~ .636.
slog(e4,EF(3)) ~ 1.612.
slog(e4,EF(4)) ~ 2.693.
slog(e4,EF(5)) ~ 3.703.
Numbers worked out with the Kneser method.
This is my 16th thread! 
I conjecture that slog(k,EF(x))-x approaches a number, as x grows larger and larger. For any k greater than eta.

I think your conjecture is wrong.

For a fixed x :

slog(x^x^... n times ) is about n + constant , for large n.

You might want to look at base change formula and base change constant.

slog( 3^... n times ) is a converging sequence.

because ln ln ... ( 3^ 3^ ... ) converges.

but the base change constant increases as the value x increases.

the base of the slog is not so relevant if it is above e and so is x.

SO for sufficiently large  x :

 x < slog_b(EF(x)) < x + basechange(b,x)

Now since x^(x-1)^(x-2)^... is much closer to x^x^x than b^b^b I think slog_b(EF(x)) is closer to  x + basechange(b,x) than to x.

therefore slog_b(EF(x)) - x is probably a strictly increasing function of x.

Maybe slog_e4 ( EF(x) ) - x - 3/4 * basechange(e4,x) converges ...

This is not a formal proof ofcourse.

If we find something like slog_e4 ( EF(x) ) - x - 3/4 * basechange(e4,x) converges or similar we could use that to construct a C^oo solution to EF(x).




regards

tommy1729