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 exp^[3/2](x) > sinh^[1/2](exp(x)) ? sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 04/24/2015, 12:40 AM (This post was last modified: 04/24/2015, 04:34 PM by sheldonison.) (04/23/2015, 04:38 PM)tommy1729 Wrote: .... Therefore 2sinh^[1/2] < exp^[1/2].I use Koenig's for 2sinh^[1/2] and Kneser for exp^[1/2] $\;\;\alpha^{-1}(\alpha(z)+0.5)\;\;$ where $\alpha(z)$ is the Abel or slog function s2inh^[1/2](6) = 20.0717649 exp^[1/2](6) = 20.0860615 at x=6, exp^[1/2](x) is bigger s2inh^[1/2](20) ~= 399.098221 exp^[1/2](20) ~= 398.247512 at x=20, 2sinh^[1/2](x) is bigger. The first crossing where the two functions are equal to each other for positive reals occurs at x~=9.49535513; the second crossing where the two functions are equal occurs at x~=54.3741864, which is ~=exp^[1/2](9.49). The third crossing where the two functions are equal occurs at ~=13297.7591 which is near exp(9.49); the fourth crossing occurs near 4.11537228*10^23, which is almost exactly exp(54.3741864). This pattern of the two functions intersecting each other repeats infinitely.... To explain the pattern, you might look at $\theta(z)=\alpha(\text{sexp}(z))-z\;\;$ where$\alpha(z)$ is the Abel function for 2sinh. $\theta(z)$ converges very quickly (super-exponentially) to a 1-cyclic function as z increases, rather than converging to a constant. Wherever $\theta(z+0.5)-\theta(z)<0$ the 2sinh^[1/2](sexp(z)) function is larger than exp^[1/2](sexp(z)). Here is a graph of $\theta(z+0.5)-\theta(z)\;$ The zeros correspond exactly to the zeros above.     - Sheldon « 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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