• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 How many methods have this property ? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 05/22/2014, 08:43 AM How many C^oo ( or analytic ) solutions to tetration satisfy : For all bases larger than eta for which the method works : Property : sexp_a(x) > sexp_b(x) For all a>b and x>1. I used convention sexp_a(1) = a , sexp_b(1) = b here. I could not find it here ? I assume this property fails for base change and kneser. I think it works for Cauchy contour method (kouznetsov?) and 2sinh method (tommy1729). Maybe this has been said long ago. regards tommy1729 sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 05/22/2014, 04:56 PM (This post was last modified: 05/22/2014, 09:16 PM by sheldonison.) (05/22/2014, 08:43 AM)tommy1729 Wrote: How many C^oo ( or analytic ) solutions to tetration satisfy : For all bases larger than eta for which the method works : Property : sexp_a(x) > sexp_b(x) For all a>b and x>1. I used convention sexp_a(1) = a , sexp_b(1) = b here. I could not find it here ? I assume this property fails for base change and kneser. I think it works for Cauchy contour method (kouznetsov?) and 2sinh method (tommy1729). Maybe this has been said long ago. regards tommy1729 I think all tetration methods have your property: Quote:Property : sexp_a(x) > sexp_b(x) For all a>b and x>1. There are anomalies, but the definition is different. The typical Kneser method anomalies involve sexp(x)=sexp(y), x<>y, a>b, but counter intuitively, sexp_a(x+0.5) sexp_10(2.3637006685629+0.5) The solutions which avoid this anomaly are coo, but not analytic, like the basechange method, which derives all tetrations for all bases from a single master base. Originally, for the basechange solution we used cheta, for exp(1/e), but this is really arbitrary. Then if f is the arbitrary chosen superfunction for the arbitrarily chosen master base, we can define a family of tetrations for all bases>exp(1/e), which avoid the anomaly. For example, f(z) could be 2sinh^z, or f(z) could be sexp(z) for any base. $k(a)=\lim_{n \to \infty} f^{-1}(a\uparrow\uparrow n)-n$ $\text{sexp}_a(z) = \lim_{n \to \infty} \log_a^{on}f(z+k(a)+n)$ - Sheldon « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post the distributive property tommy1729 17 13,197 06/06/2014, 12:09 PM Last Post: MphLee [Update] Comparision of 5 methods of interpolation to continuous tetration Gottfried 30 27,885 02/04/2014, 12:31 AM Last Post: Gottfried Borel summation and other continuation/summability methods for continuum sums mike3 2 5,041 12/30/2009, 09:51 PM Last Post: mike3 Iterations of ln(mod(ln(x^(1/x))) and number with property Left a[4]n=exp (-(a^(n+1)) Ivars 0 2,354 03/30/2008, 06:41 PM Last Post: Ivars Applying the iteration methods to simpler functions bo198214 11 12,036 11/13/2007, 12:39 AM Last Post: andydude singularities (at least diagonalization-based methods) Gottfried 0 2,214 09/27/2007, 03:29 PM Last Post: Gottfried

Users browsing this thread: 1 Guest(s)