Let $\operatorname{AGM}(x,y)$ be the arithmetic-geometric mean of $x$ and $y$. Given an error $\varepsilon>0$, a bound $b\in\mathbb R_+$ and a function $f:\mathbb R\rightarrow\mathbb R$ with $f(x)=O(\log x)$ and $f(x)=\Omega(\log\log x)$ for which rationals $\frac pq\in\mathbb Q_+$ with $0<p<b$ and $0<q<b$ is it possible to find $x=\frac{p'}{q'},y=\frac{p''}{q''}\in\mathbb Q$ with $\mathsf{\max}(|p'|,|q'|,|p''|,|q''|)<f(b)$ such that $$\Bigg|\frac pq-\operatorname{AGM}(x,y)\Bigg|<\varepsilon$$ holds?

Is there explicit methods to write such $x,y$ down?

The density of such representable $\frac pq$ should be tiny. Nevertheless are there special forms where this can be done. So are there special family of rationals?

  • 1
    $\begingroup$ What's AGM? ${}$ $\endgroup$
    – Wojowu
    May 16 '20 at 8:15
  • $\begingroup$ @Wojowu The arithmetic-geometric mean, I think. $\endgroup$ May 16 '20 at 11:08

This is just some comments.

Since $x,y>0$ for $AGM(x,y)$ to be defined, why the absolute values?

I will assume $f(\cdot)$ takes only integer values and that $p’,p’’,q’,q’’ \leq f(b)$

Let $F(b)$ be the set of fractions with numerator and denominator bounded by $b$. Then $|F(b)|<b^2$ and the largest members are integers until $\frac{b}2$. The smallest are their reciprocals. One would expect things to be densest near $1.$

Fix $b$ And let $c=f(b)$, then the $|F(c)|$ values of $x,y$ are all between $1/c$ and $c$ so $p/q$ better be between $1/c-\epsilon$ and $c+\epsilon$ . Then there are somewhat under $\binom{c^2}2$ values for $AGM(x,y)$ . Each determines an interval of radius $\epsilon$.

Certainly we can get all rationals in the $\epsilon$ neighborhoods of the members of $F(c)$ using $x=y.$ Those alone should allow all rationals in some interval , depending on $\epsilon$, and containing $1$ near its lower bound. Taking $x$ near $y$ would seem to allow a larger interval.

Past that I would suggest starting with $c=10$ or smaller and computing to see what the collection of intervals looks like.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.