Arithmetic-geometric mean for rationals?

Question

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?

Answer 1

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.