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^[2](n) ln(n) + A3 ln(n) + A4 ln^[2](n) + A5 ln^[3/2](n) + A6 ln^[3/2](n) ln^[2](n) + A7 ln^[3/2](n)/ln^[2](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^[2](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

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^[2](n) ln(n) + A3 ln(n) + A4 ln^[2](n) + A5 ln^[3/2](n) + A6 ln^[3/2](n) ln^[2](n) + A7 ln^[3/2](n)/ln^[2](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^[2](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