Re: lagrange and continued fractions

From: Robert Israel (israel_at_math.ubc.ca)
Date: 12/17/04 

Date: 17 Dec 2004 07:45:02 -0500

In article <cpt1d4$14l$1@dizzy.math.ohio-state.edu>,
Loic <teyssier@math.univ-lille1.fr> wrote:

> I was wondering about the following: take two continued fractions
>[a0;a1,a2,...] and [b0;b1,b2,...] (converging to irrationals x1 and
>x2), such that

> (1) there exists N : an=bn for n>=N
> (2) sum(ak-bk)=0

> If (2) then the respective convergents pn/qn and cn/dn lie on the
>same "horizontal" line in the Stern-Brocot tree (for n>=N). If (1)
>then Lagrange showed that x2=f(x1) where f(x)=(ax+b)/(cx+d) with
>integer coefficients (if I remember correctly).

> Question 1: is there en "explicit" form for f (for instance
>knowing ak,bk for k<N)? By any chance, would f be the same one that
>sends pN/qN on cN/dN? (I bet not...)

It is. Let y_k = [a_k; a_{k+1}, a_{k+2}, ...] and
z_k = [b_k; b_{k+1}, b_{k+2}, ...], so x1 = y_0 and x2 = z_0.
Let s_k be the Mobius transformation u -> a_k + 1/u
and t_k the Mobius transformation u -> b_k + 1/u.
Then y_k = s_k(y_{k+1}) and z_k = t_k(z_{k+1}).
By assumption y_N = z_N. Thus
x1 = s_0 o ... o s_{N-1} o t_{N-1}^{-1} o ... o t_0^{-1} x2.
Since the Mobius transformations form a group,
   T = s_0 o ... o s_{N-1} o t_{N-1}^{-1} o ... o t_0^{-1}
is a Mobius transformation (and easily computed from
a_0, ..., a_{N-1}, b_{N-1},...,b_0). Of course this doesn't depend
on a_N, a_{N+1}, ..., so it would be the same if a_N = a_{N+1} = ... = 0,
and thus T(c_N/d_N) = p_N/q_N.

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