purely periodic continued fractions
We know that periodic continued fractions represent quadratic irrationals; this article characterizes purely periodic continued fractions. We will use freely the results on convergents^{} to a continued fraction^{}.
Theorem 1.
(Galois) A quadratic irrational $t$ is represented by a purely periodic simple continued fraction if and only if $t\mathrm{>}\mathrm{1}$ and its conjugate^{} $s$ under the transformation $\sqrt{d}\mathrm{\mapsto}\mathrm{-}\sqrt{d}$ satisfies $$.
Proof.
Suppose first that $t$ is represented by a purely periodic continued fraction
$$t=[\overline{{a}_{0},{a}_{1},\mathrm{\dots},{a}_{r-1}}].$$ |
Note that ${a}_{0}\ge 1$ since it appears again in the continued fraction. Thus $t>1$. The ${r}^{\mathrm{th}}$ complete convergent is again $t$, so that we have
$$t=\frac{{p}_{r-2}+t{p}_{r-1}}{{q}_{r-2}+t{q}_{r-1}}$$ |
so that
$${q}_{r-1}{t}^{2}+({q}_{r-2}-{p}_{r-1})t-{p}_{r-2}=0$$ |
Consider the polynomial^{} $f(x)={q}_{r-1}{x}^{2}+({q}_{r-2}-{p}_{r-1})x-{p}_{r-2}$. $f(t)=0$, so the other root of $f(x)$ is the conjugate $s$ of $t$. But $f(-1)=({p}_{r-1}-{p}_{r-2})+({q}_{r-1}-{q}_{r-2})>0$ since the ${p}_{i}$ and the ${q}_{i}$ are both strictly increasing sequences^{}, while $$. Thus $s$ lies between $-1$ and $0$ and we are done.
Now suppose that $t>1$ and $$, and let the continued fraction for $t$ be $[{a}_{0},{a}_{1},\mathrm{\dots}]$. Let ${t}_{n}$ be the ${n}^{\mathrm{th}}$ complete convergent of $t$, and ${s}_{n}=\overline{{t}_{n}}$. Thus ${s}_{0}=s$. Then
$$t={t}_{0}={a}_{0}+\frac{1}{{t}_{1}}$$ |
so that
$${s}_{0}=\overline{{t}_{0}}={a}_{0}+\frac{1}{\overline{{t}_{1}}}={a}_{0}+\frac{1}{{s}_{1}}$$ |
and thus
$$ |
so that $$. Inductively, we have $$ for all $n\ge 0$. Suppose now that the continued fraction for $t$ is not purely periodic, but rather has the form
$$t=[{a}_{0},{a}_{1},\mathrm{\dots},{a}_{k-1},\overline{{a}_{k},{a}_{k+1},\mathrm{\dots},{a}_{k+j-1}}]$$ |
for $k\ge 1$. Then ${t}_{k}={t}_{k+j}$ and so
$${t}_{k-1}-{t}_{k+j-1}=\left({a}_{k-1}+\frac{1}{{t}_{k}}\right)-\left({a}_{k+j-1}+\frac{1}{{t}_{k+j}}\right)={a}_{k-1}-{a}_{k+j-1}$$ |
But ${a}_{k-1}\ne {a}_{k+j-1}$, otherwise ${a}_{k-1}$ would have been the first element of the repeating period. Thus ${t}_{k-1}-{t}_{k+j-1}$ is a nonzero integer and thus ${s}_{k-1}-{s}_{k+j-1}$ is as well. But $$, which is a contradiction^{}. Thus $k=0$ and the continued fraction is purely periodic. ∎
References
- 1 A.M. Rockett & P. Szüsz, Continued Fractions, World Scientific Publishing, 1992.
Title | purely periodic continued fractions |
---|---|
Canonical name | PurelyPeriodicContinuedFractions |
Date of creation | 2013-03-22 18:04:44 |
Last modified on | 2013-03-22 18:04:44 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 6 |
Author | rm50 (10146) |
Entry type | Theorem |
Classification | msc 11Y65 |
Classification | msc 11A55 |