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 On replacing 2sinh(2 x) with F(x) ... tommy1729 Ultimate Fellow Posts: 1,668 Threads: 368 Joined: Feb 2009 02/20/2013, 10:37 PM It came to my mind to replace 2sinh(b x) with another similar realanalytic function F and then perform the same iterations as with the sinh method. The problem is the fixpoints. If I use 2sinh(2 x) to construct a superfunction for exp(2x) then 2sinh(2 L) =/= L. But we have the advantage that 2sinh(2 x) has only one fixpoint so entire iterations for 2sinh(2 x) exists. If however we have F(L)=L then that might make things easier ( analytic continuation ). But there is fear for other fixpoints such as 0 and/or L* resulting in a nonentire and nonunique superfunction of L(x). However if we consider a real superfunction of F(x) and we only have the fixpoints L and L* and both are repelling then it might be intresting to investigate this. Notice here that L and L* means the 2 conjugate fixpoints of the function e^bx. Note however that analytic continuation will still be neccessary. tommy1729 Ultimate Fellow Posts: 1,668 Threads: 368 Joined: Feb 2009 02/20/2013, 10:50 PM However I fear such a function does not exist. The reason is F(x) must be so different from exp and sinh. It cannot be of the form exp(x)(x-a)(x-b) and it cannot be of the form exp(x) - exp(ax) - exp(bx). And at the same time F(x) must be entire and have only 2 fixpoints. It seems the Weierstrass factorization theorem implies it must be of a form it cannot be. Also a Dirichlet series for F(x) seems impossible. tommy1729 Ultimate Fellow Posts: 1,668 Threads: 368 Joined: Feb 2009 02/21/2013, 11:37 PM I came to realise that I am close to a proof that any formula of the form lim n-> oo ln^[n](f^[x](exp^[n])) for f^[1] entire , cannot be analytic. For certain the probability is at most 1/R where R is the cardinality of the continuum. « Next Oldest | Next Newest »

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