### Route to Root

A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?

### Sixational

The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.

### LOGO Challenge - Sequences and Pentagrams

Explore this how this program produces the sequences it does. What are you controlling when you change the values of the variables?

# 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?