using sinh(x) ?

[sheldonison]

After that, do you agree that converting from your f(y) to g(y) by iterating natural logarithms would appear to be exactly the same? ... The value I'm using for k is 0.067838366.

\( \operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (
\operatorname{superfunc}(z+n+k)) \)

no , i use iterations of exp(x) !

you only use logaritms and the superfunction of 2sinh(x) ...

i use logaritms , superfunction of 2sinh(x) and exp(x).

numerically it might not be a big difference for large values ( since 2sinh(x) and exp(x) are close for large x ) , but for small x or properties it might be a huge difference ... maybe that is why you get 0.92715... ?

i dont know if i need a correcting 'k'.

and btw as for the diff-eq , i made a correction to that post (typo mainly but important ) , maybe it will help.

regards

tommy1729

The correcting factor is
k=\( \lim_{n \to \infty}\operatorname{superfunc}^{-1}\exp^{[n]}(0) \)

If you try your equations to calculate TommySexp(0.5) with x=0, y=0.5, and k=3

let g(x,y) be the y'th iterate of exp(x) evaluated at x.

let f(x,y) be the y'th iterate of 2 sinh(x) evaluated at x.

then g(x,y) = lim k -> oo
log log log ... (k times ) [f( g(x,k) ,y)]

then g(0,0.5) = (for k=3)
log log log (3 times) [f( g(0,3), 0.5)]

g(0,3) is \( \exp^{[3]}(0) \)=3814279.104....

Calculating f( g(3,0) ,0.5) is a a somewhat complex operation, and I'd like to understand if you would calculate that expression the same way I would. My approach would be to use the inverse superfunction, then add 0.5, then calculate the superfunction.

\( x=\exp^{[3]}(0) ; y=0.5 \)
\( \text{f(x,y)=}\operatorname{superfunc}(\operatorname{superfunc}^{-1} (x) + y) \)

Update or alternatively, iterate
\( \text{f(x,y)=}\lim_{n \to \infty}\operatorname{2sinh}^{[n]}((\operatorname{2sinh^{-1 [n times]}(x))*2^y) \)
\( \text{f( g(0,3) ,0.5)=}\lim_{n \to \infty}\operatorname{2sinh}^{[n]}((\operatorname{2sinh^{-1 [n times]}(g(0,3)))*\sqrt 2) \)

The inverse superfunction gives 3.0678383... which is 3 plus the normalization constant, then adding 0.5 ... then calculate the superfunction of 3.5678383.... 8.10306E+77

Finally, take the logarithm three times, and you get (for Tommy's k=3)
TommySexp(0.5) =~ f(0,0.5) =~ 0.4987433...

Numerically, this is exactly what the limit equations I've posted give.
- Sheldon