Improving the Dirichlet's approximation theorem.

Question

Recall that Dirichlet showed the following:

For every real number $x$ and every $Q>1$, there exists an integer vector $(p,q)\in \mathbb Z^2$ such that $|xq-p|<1/Q$ and $0<q<Q$.

I wonder if the following is true:

For every real number $x$ and every $\epsilon>0$, there exists $Q_{\epsilon}$ such that for all $Q>Q_{\epsilon}$, there is an integer vector $(p,q)\in \mathbb Z^2$ such that $|xq-p|<\epsilon/Q$ and $0<q<Q$.

Of course this is trivially true when $x$ is rational, but I don't know what happens when $x$ is irrational (in particular when it is transcendental).

Note the Lerendre's theorem should be a special case of this by taking $\epsilon=1$ and we can always assume $Q>Q_{\epsilon} \ge 1$.

Update: As per the comments by rtybase below, by Liouville's theorem (on diophantine approximation), my statement is false if $x$ is algebraic. But what happens when $x$ is transcendental?

I thought it's Dirichlet's approximation theorem. Do you have any historical references pointing at Legendre? — rtybase, Oct 04 '20 at 18:40
In any case, that may not be valid for all the $x$. Specifically, algebraic numbers of degree $2$. — rtybase, Oct 04 '20 at 18:49
@rtybase To be safe, let me change it to Dirichlet. My source isn't very reliable... Thanks! — No One, Oct 04 '20 at 18:58
Anyway, you have the answer in my 2nd comment. Here and here are a few specific examples. — rtybase, Oct 04 '20 at 19:09
Yes, thanks! But now I am also curious about the transcendental case...Just updated my question. @rtybase — No One, Oct 04 '20 at 19:12
Look up "Badly approximable numbers"; basically this property you describe has to do with whether or not the partial quotients of the continued fraction expansion are bounded — MathematicsStudent1122, Oct 04 '20 at 19:17
https://en.wikipedia.org/wiki/Liouville_number#Irrationality_measure — Qiaochu Yuan, Oct 05 '20 at 21:06

Sophie · Answer 1 · 2020-10-05T22:18:01.697

1

Your conjecture is false. The best possible constant is $\epsilon = \frac 1 {\sqrt 5}$, which is tight for $\phi=\frac{1+\sqrt 5}2$ and its equivalent numbers under the equivalence relation:

$$y\sim x \iff y=\frac{ax+b}{cx+d}; ad-bc=1$$

You can prove that by showing that the best approximants are the convergents to the continued fraction, then noting that those for $\phi$ are ratios of Fibonacci numbers. Then proving that

$$\lim_{n\to\infty}F_n|\phi F_n-F_{n+1}|\to \frac 1{\sqrt 5}$$

Using the closed form for Fibonacci numbers.

If you exclude all the $x\sim\phi$ you can improve the constant to $\epsilon=2\sqrt 2$, and if you repeat the process you find the Lagrange numbers, which converge to 3.

As noted in the comments, the irrationality measure of all irrational algebraic numbers is 2 (Roth's theorem). As for transcendental numbers, almost all of them have irrationality measure 2 still.

edited Oct 05 '20 at 22:18

answered Oct 05 '20 at 22:06

Sophie

3,378

The way the question is asked is ... tricky. For example OP asks for "there is an integer vector", i.e. one or more. The $\frac{1}{\sqrt{5}}$ from Hurwitz's theorem is of the style "there are infinitely many", over that value it is still possible for "there exist only finitely many", which fits the "there is an integer vector" criteria. – rtybase Oct 05 '20 at 22:49
@rtybase they asked for an integer vector for every $Q>1$. You're right that I was a bit sloppy in establishing that my results hold with the $<1/Q$, as I substituted for $1/q$ which is weaker, but they're nearly equivalent. Another thing that can be problematic is if Q isn't an integer and an inequality that holds for $\lfloor Q\rfloor$ might not hold for Q. – Sophie Oct 05 '20 at 23:10

Improving the Dirichlet's approximation theorem.

1 Answers1

Linked