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