tommy's simple solution ln^[n](2sinh^[n+x](z)) - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: tommy's simple solution ln^[n](2sinh^[n+x](z)) (/showthread.php?tid=1144) tommy's simple solution ln^[n](2sinh^[n+x](z)) - tommy1729 - 01/16/2017 Consider f(z,x) = Lim(n --> oo)  ln^[n] ( 2sinh^[n+x] (z) ). This simple Function satisfies exp(f(z,x)) = f(z,x+1). So we have a simple superfunction that requires only the real iterations of 2sinh(z). Notice lim ( n --> oo) 2sinh^[n]( 2^(z-n)) is a superf for 2sinh. f(z,x) could be analytic for re(z) > 1. Also , is it really new ? Or is it the ( analytic continuation ? ) of the 2sinh method ? It sure is very similar. ---- Mick wondered if F^[n] ( g^[n] ) is analytic for f = sqrt and g = x^2 +1. --- Regards Tommy1729 RE: tommy's simple solution ln^[n](2sinh^[n+x](z)) - sheldonison - 01/17/2017 (01/16/2017, 01:29 PM)tommy1729 Wrote: Consider f(z,x) = Lim(n --> oo)  ln^[n] ( 2sinh^[n+x] (z) ). This simple Function satisfies exp(f(z,x)) = f(z,x+1). ... Or is it the ( analytic continuation ? ) of the 2sinh method ?consider ignoring z, and using the formal inverse Schroeder series (below) for putting 2sinh^[ox] into correspondence with 2^x.  Then your sexp2sinh/TommySexp function is exactly: $k=f^{-1}(1);\;\;\;\text{TommySexp}(x)=f(k+x)$ k=0.0678383660707522254065 This also happens to be the definition I personally used for your TommySexp function, but numerically, they are all exactly the same; infinitely differentiable but conjectured nowhere analytic.   TommySexp(-0.5)=0.498743364531671 Kneser's sexp(-0.5)= 0.498563287941114 Since f(x) is only defined at the real axis, the term analytic continuation has no meaning. "Mick wondered if F^[n] ( g^[n] ) is analytic for f = sqrt and g = x^2 +1",  yes it is.  So long as you restrict yourself to the region where |g^[n]|>>1 then it will converge.  Have Mick ask on Mathstack if he wants more details. Code:2sinh^[0] = formal2sinh_ischroeder(1) = 1.05804904330694441126 {formal2sinh_ischroeder=  x + +x^ 3*  1/18 +x^ 5*  13/5400 +x^ 7*  1193/14288400 +x^ 9*  219983/87445008000 +x^11*  225002297/3280062250080000 +x^13*  3624242332901/2095369366596105600000 +x^15*  294797208996087793/7208971629918239589408000000 +x^17*  532541776280711150089/581464560943620715682280960000000 +x^19*  4423796286922654904342141267/225896613039975363731770463347368960000000 ...}