OK so

Since exp(g - h) =< g(exp) - h,

It follows that the 2sinh limit approaches iTS limit from below.

Therefore 2sinh^[1/2] < exp^[1/2].

Also these two are asymptotic.

Now notice that 3sinh method would be identical to 2sinh method !!

In the thread about base exp(2/5) i defined an f(x).

Using f would-be identical to 2sinh too !

And therefore Also f^[1/2] is asymptotic to both 2sinh and exp half-iterates !

But thats not all.

If half-f has no intersection with half-2sinh within ]0,e^e^e]

1) half-f => 2sinh

From 1) : the 2sinh analogue to compute half-f from half-2sinh gets the SAME function f as the analytic one Because of the Uniquenness of being both asympt and larger, THUS we have An 2sinh-type analytic FUNCTION !! ... And that property comes from investigating other functions !

Nice.

Regards

Tommy1729

Since exp(g - h) =< g(exp) - h,

It follows that the 2sinh limit approaches iTS limit from below.

Therefore 2sinh^[1/2] < exp^[1/2].

Also these two are asymptotic.

Now notice that 3sinh method would be identical to 2sinh method !!

In the thread about base exp(2/5) i defined an f(x).

Using f would-be identical to 2sinh too !

And therefore Also f^[1/2] is asymptotic to both 2sinh and exp half-iterates !

But thats not all.

If half-f has no intersection with half-2sinh within ]0,e^e^e]

1) half-f => 2sinh

From 1) : the 2sinh analogue to compute half-f from half-2sinh gets the SAME function f as the analytic one Because of the Uniquenness of being both asympt and larger, THUS we have An 2sinh-type analytic FUNCTION !! ... And that property comes from investigating other functions !

Nice.

Regards

Tommy1729