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 exp^[3/2](x) > sinh^[1/2](exp(x)) ? tommy1729 Ultimate Fellow Posts: 1,421 Threads: 346 Joined: Feb 2009 04/23/2015, 04:38 PM (This post was last modified: 04/23/2015, 04:44 PM by tommy1729.) 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 « Next Oldest | Next Newest »

 Messages In This Thread exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by tommy1729 - 04/21/2015, 12:09 PM RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by tommy1729 - 04/21/2015, 01:21 PM RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by tommy1729 - 04/23/2015, 04:38 PM RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by sheldonison - 04/24/2015, 12:40 AM RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by tommy1729 - 04/23/2015, 04:54 PM RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by tommy1729 - 04/24/2015, 02:07 PM RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by sheldonison - 04/24/2015, 02:59 PM RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by tommy1729 - 10/26/2015, 01:07 AM

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