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Farey Approximation

Age 16 to 18 Short Challenge Level:

In 1816 the British geologist John Farey defined the Farey sequence $F_n$ as the list, written in increasing order, of all the rational numbers between $0$ and $1$ that have only the numbers $1, 2, 3, ... , n$ as denominators. We can do the same thing for rational numbers between any two positive numbers. For example we can consider sequences betweeen $1$ and $2$ where we have
\[ \eqalign{F_1&=\frac{1}{1}, \frac{2}{1}\cr F_2&=\frac{1}{1}, \frac{3}{2}, \frac{2}{1}.} \]

What would $F_3$ and $F_4$ be in this case?

For the two positive rational numbers $\frac{b}{d}$ and $\frac{a}{c}$ the mediant is defined as $\frac{a+b}{c+d}$. The mediant has the nice property that it is always in between the two fractions giving rise to it: if $0< \frac{b}{d} < \frac{a}{c}$ then $\frac{b}{d} < \frac{a+b}{c+d} < \frac{a}{c}$. You might want to verify this for yourself.

Clearly each Farey sequence $F_{n+1}$ must contain all of the terms of $F_{n}$, along with some new terms. Mediants also have the nice property that each 'new' term in the Farey sequence $F_{n+1}$ is the mediant of two consecutive terms in $F_n$. Why not test this out on $F_2, \dots, F_5$?

Farey sequences have all sorts of interesting properties. One is that they give an interesting way of approximating rational numbers.

In order to find a rational approximation to an irrational number using Farey fractions you need to pick the interval between Farey fractions that contains the target number and narrow the interval at each step. If the target number is between $\frac{b}{d}$ and $\frac{a}{c}$ then at the next step you have to decide which of the two intervals
\[ \Big[\frac{b}{d}, \frac{a+b}{c+d}\Big], \quad \Big[\frac{a+b}{c+d}, \frac{a}{c}\Big]. \]

contains the target number.

Carry out this numerical approximation process for $\sqrt 2$, and for $\pi$, to get rational approximations that are correct to 4 decimal places. You may like to use a spreadsheet. Compare your results to the approximations obtained by using a straightforward interval-halving method. What do you notice?