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.?
Age
16 to 18
Challenge level
A sequence of numbers $x_1, x_2, x_3, \ldots$ , starts with $x_1 = 2$, and, if you know any term $x_n$, you can find the next term $x_{n+1}$ using the formula: $$x_{n+1} = \frac{1}{2}\bigl(x_n + \frac{3}{x_n}\bigr)$$ 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.?
A sequence of numbers $x_1, x_2, x_3, ... ,$ starts with $x_1 = 2$, and, if you know any term $x_n$, you can find the next term $x_n+1$ using the formula $x_{n=1} = \frac{1}{2} \left( x_n + \frac{3}{x_n} \right)$.
Solution by Andaleeb of Woodhouse Sixthform College, London.
For the iteration $$x_{n+1} = \frac{1}{2} \left( x_n + \frac{3}{x_n} \right)$$
\begin{eqnarray} \\ x_1 &=& 2,\; x_2 =
1.75,\; x_2 = 1.732142857,\; x_4 = 1.73205081 \\ x_5 &=&
1.732050808,\; x_6 = 1.732050808,\; x_7 = 1.732050808 \\ x_8
&=& 1.732050808;\end{eqnarray}
We notice that when $x_n = 1.732050808$, so is $x_{n+1}$. Squaring
these terms we get $x_1^2 = 4, x_2^2 = 3.0625, ... , x_5^2 = 3$ and
the rest of the other terms are the same!! This implies that when
$x_n \approx \sqrt{3}$ so is $x_{n+1}$ and the values of $x_n$ tend
to the limit $\sqrt{3}$. This special property can easily be
proven. Assume that the limit exists, so $x_{n+1} = x_n = x$, then
solve the equation $$X = \frac{1}{2}\left(X + \frac{3}{N} \right)$$
If we test it for $N = 3$, we see that $x_{29} = 1.44224957$, which
is what the calculator gives for the cube root of 3. Testing it for
$N = 8$, we get $x_1 = 2$, which is the right answer. By
experimentation you can soon discover for yourself that it is not
safe to assume that the same method works finding fourth roots
using the iteration formula. $$x_{n+1} = \frac{1}{2} \left( x_n +
\frac{N}{x_n^3} \right)$$ There is work to do to show that the
iteration $x_{n+1} = F(x_n)$ converges to a limit $L$ if and only
if $-1 < F'(L) < 1$.