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 Growth rate of the recurrence x(n+1) = x(n) + (arcsinh( x/2 ))^[1/2] ( x(n) )? tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 04/29/2013, 11:29 PM Let x(0) = 2. I was wondering about the growth rate of the recurrence x(n+1) = x(n) + (arcsinh( x/2 ))^[1/2] ( x(n) ) ? Again ^[1/2] stands for half-iterate here. And arcsinh(x/2) is used as replacement for ln(x) because it is easy to find the half-iterate of 2sinh(x) ( because of the fixpoint at 0 ... and hence also the inverse of 2sinh being arcsin(x/2) ). Because of the big involvement of my friend mick surrounding these ideas I will partially name my conjecture after him. the tommy1729-mick Conjecture : For sufficiently large n : x(n) = 2 + n (f(n)+e(n)) (arcsinh( x/2 ))^[1/2](2) where -1 < e(n) < 1 and f(n) = A1 + A2 ln^(n) ln(n) + A3 ln(n) + A4 ln^(n) + A5 ln^[3/2](n) + A6 ln^[3/2](n) ln^(n) + A7 ln^[3/2](n)/ln^(n) + A8 sqrt(ln^[3/2](n)) + A9 [ ( int_2_n 1 / ln^[3/2](n) dn ) / ln(n) ] + A10 ln^[5/2](n) where A1,A2,... are real numbers. ( under the assumption that the integral from 2 till n of 1/ln^[3/2](n) dn divided by ln(n) is indeed < ln(n)/ln^(n) otherwise A9 is = 0. - I did not check that yet srr -) I do not know how to begin even trying here. ( I do know that f(n) cannot be elementary though , which was a motivation for the conjecture btw ) regards tommy1729 « Next Oldest | Next Newest »

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