Thread Rating:
  • 1 Vote(s) - 5 Average
  • 1
  • 2
  • 3
  • 4
  • 5
exp^[3/2](x) > sinh^[1/2](exp(x)) ?
#3
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
Reply


Messages In This Thread
RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by tommy1729 - 04/23/2015, 04:38 PM

Possibly Related Threads...
Thread Author Replies Views Last Post
  using sinh(x) ? tommy1729 99 109,845 03/12/2016, 01:20 PM
Last Post: tommy1729
  2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? Gottfried 5 5,012 09/11/2013, 08:32 PM
Last Post: Gottfried
  zeta and sinh tommy1729 0 1,833 05/30/2011, 12:07 PM
Last Post: tommy1729



Users browsing this thread: 1 Guest(s)