[seqfan] Re: Extending signature sequences for transcendental numbers
Robert Israel
israel at math.ubc.ca
Tue Jun 22 00:50:13 CEST 2010
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
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