[seqfan] Re: Extending signature sequences for transcendental numbers

[Robert Israel]

On Mon, 21 Jun 2010, Kerry Mitchell wrote:

> I've extended the idea of the signature sequence for transcendental
> numbers.  To get the signature sequence of x, form y values:
>
> y = a0 + a1x
>
> for positive integers a0 and a1.  Sort the list by the value of y and the
> sequence of a0's is the signature sequence.  If x is irrational, then the
> sequence is a fractal sequence.
>
> Transcendental numbers are not solutions to any polynomial equation with
> integer coefficients.  I used that idea to create a series of sequences:
>
> S1(x) is the sequence of a0's when sorting y = a0 + a1x.  (S1 is the
> standard signature sequence.)
> S2(x) is the sequence of a0's when sorting y = a0 + a1x + a2x^2.
> S3(x) is the sequence of a0's when sorting y = a0 + a1x + a2x^2 + a3x^3.
> etc.
> Sn(x) is the sequence of a0's when sorting y = a0 + a1x + a2x^2 + ... +
> anx^n.
>
> Then, I defined T(x) to be the limit of Sn(x) as n goes to infinity.
>
> For the few values I've investigated, it looks like T(x) is well defined and
> is similar to, but different from, the regular signature sequence S1(x).

Unless I misunderstand you, if 0 < x < 1 every entry in T(x) will be 1. 
It makes no difference whether x is transcendental.  It's just that
k x^n < 1 for all k < x^(-n), so at least the first floor(x^(-n)) entries 
of Sn(x) will be 1.

Robert Israel                                israel at math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada