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 exp^[3/2](x) > sinh^[1/2](exp(x)) ? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/21/2015, 12:09 PM Although it might appear like a random question to some , this is a rather important question : Prove or disprove : For any real x and exp^[3/2] computed with the 2sinh method : exp^[3/2](x) > sinh^[1/2](exp(x)) *** I assume that both the first and second derivative of sinh^[1/2](x) for x >0 are positive. The answer to these will lead to more progress ! And possibly some counterintuitive things ! regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/21/2015, 01:21 PM We can simplify this by looking in the interval x E [0,e^e] for the curves exp^[1/2] and sinh^[1/2]. This suggests that it is true. But that very likely gives counter-intuitive results. I Will clarify later. Formal proofs and plots are still Desired and appreciated. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 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 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/23/2015, 04:54 PM Bye contrast the half-iterates of x^2 and x^2 - 1/4 intersect infinitely often. Regards Tommy1729 sheldonison Long Time Fellow Posts: 631 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 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/24/2015, 02:07 PM Sheldon, that is exactly as expected. The reason you get this result is because you used kneser instead of my 2sinh method. If you would have used my method you should get the same result as me. Intresting numerics through. Regards Tommy1729 sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 04/24/2015, 02:59 PM (04/24/2015, 02:07 PM)tommy1729 Wrote: Sheldon, that is exactly as expected. The reason you get this result is because you used kneser instead of my 2sinh method. I assumed Kneser's exp^[1/2](z), since it is analytic in the complex plane - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 10/26/2015, 01:07 AM I doubt 3sinh method would be identical to 2sinh method. Need to reconsider Some things. Regards Tommy1729 « Next Oldest | Next Newest »

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