Review question

# If $x_1 = (x_0 + 2)/(x_0 + 1)$, can we show $\sqrt{2}$ lies between $x_0$ and $x_1$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8404

## Solution

If $x_0$ is any positive rational number and $\begin{equation*} x_1 = \frac{x_0 + 2}{x_0 + 1}, \quad x_2 = \frac{x_1 + 2}{x_1 + 1}, \quad x_3 = \frac{x_2 + 2}{x_2 + 1}, \end{equation*}$

prove that

1. $\sqrt{2}$ lies between $x_0$ and $x_1$, and between $x_2$ and $x_3$,
We can reason as follows: \begin{align*} x_{n+1} < \sqrt{2} &\iff \frac{x_n + 2}{x_n + 1} < \sqrt{2} \\ &\iff (x_n + 2) < \sqrt{2}(x_n + 1) \\ &\iff (1 - \sqrt{2})x_n < (\sqrt{2} - 2) \\ &\iff x_n > \frac{2 - \sqrt{2}}{\sqrt{2} - 1} = \sqrt{2} \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \sqrt{2}. \end{align*}

Since the same argument applies with $<$ and $>$ reversed, we’ve shown that $x_n$ and $x_{n+1}$ always lie on opposite sides of $\sqrt{2}$.

Thus $x_{0}$ and $x_1$, and $x_{2}$ and $x_3$, always lie on opposite sides of $\sqrt{2}$.

1. $|x_1^2 - 2| < |x_0^2 - 2|$,
We can write \begin{align*} |x_1^2 - 2| &= \left| \left( \frac{x_0 + 2}{x_0 + 1} \right)^2 - 2 \right|\\ &= \left| \frac{(x_0 + 2)^2 - 2(x_0 + 1)^2}{(x_0 + 1)^2} \right|\\ &= \frac{|x_0^2 + 4x_0 + 4 - 2x_0^2 - 4x_0 - 2|}{|x_0 + 1|^2} \\ &= \frac{|2 - x_0^2|}{|x_0 + 1|^2} \\ &= \frac{|x_0^2 - 2|}{|x_0 + 1|^2} \\ &< |x_0^2 - 2|. \end{align*}

The final inequality follows since $x_0 + 1 > 1$.

1. $|x_2^2 - 2| < \frac{1}{9} |x_0^2 - 2|$.
The same method as above shows that $\begin{equation*} |x_2^2 - 2| = \frac{|x_1^2 - 2|}{|x_1 + 1|^2} = \frac{|x_0^2 - 2|}{|x_0 + 1|^2 |x_1 + 1|^2}. \end{equation*}$ We can write $\begin{equation*} |x_1 + 1|^2 = \left( \frac{x_0 + 2}{x_0 + 1} + 1 \right)^2 = \frac{(x_0 + 2 + x_0 + 1)^2}{(x_0 + 1)^2} = \frac{(2x_0 + 3)^2}{(x_0 + 1)^2}. \end{equation*}$ Hence, $\begin{equation*} |x_2^2 - 2| = \frac{|x_0^2 - 2|}{|x_0 + 1|^2} \times \frac{(x_0 + 1)^2}{(2x_0 + 3)^2}= \frac{|x_0^2 - 2|}{(2x_0 + 3)^2} < \tfrac{1}{9} |x_0^2 - 2|, \end{equation*}$

with the inequality holding as $x_0 > 0$ which implies that $(2x_0+3)^2>9$.

By taking $x_0 = \dfrac{7}{5}$, or otherwise, show that $\begin{equation*} \frac{41}{29} < \sqrt{2} < \frac{99}{70} \end{equation*}$
From the first part of the question, the iteration goes from one side of $\sqrt{2}$ to the other, and back again, and so on. Since $x_0 = 1.4 < \sqrt{2}$, we know that $\begin{equation*} \sqrt{2} < x_1 = \frac{x_0 + 2}{x_0 + 1} = \frac{7 + 10}{7 + 5} = \frac{17}{12} \end{equation*}$ and that $\begin{equation*} \sqrt{2} > x_2 = \frac{x_1 + 2}{x_1 + 1} = \frac{17 + 24}{17 + 12} = \frac{41}{29} \end{equation*}$ and, finally, that $\begin{equation*} \sqrt{2} < x_3 = \frac{x_2 + 2}{x_2 + 1} = \frac{41 + 58}{41 + 29} = \frac{99}{70}. \end{equation*}$

…and, without using tables, show that each of $\dfrac{41}{29}$ and $\dfrac{99}{70}$ differ from $\sqrt{2}$ by less than $5 \times 10^{-4}$.

With $x_0 = \dfrac{7}{5}$, we saw that $x_2 = \dfrac{41}{29}$ and $x_3 = \dfrac{99}{70}$. We saw from the very first part of the question that $x_2$ and $x_3$ lie on opposite sides of $\sqrt{2}$, and so $\begin{equation*} |x_2 - \sqrt{2}| < |x_2 - x_3| = \left| \frac{41}{29} - \frac{99}{70} \right| = \left| \frac{41 \times 70 - 99 \times 29}{2030} \right| = \frac{1}{2030} < 5 \times 10^{-4}. \end{equation*}$

The same inequalities above also apply when $x_2$ and $x_3$ are interchanged, which proves the result.

Where did our initial iteration come from? We have $\begin{equation*} f(x) = x^2 - 2 = 0 \iff x^2 + x - 2 = x \iff x(x+1) - 2 = x \iff x = \frac{x+2}{x+1} = g(x). \end{equation*}$

Thus we have in action here the $x = g(x)$ iteration, or the rearrangement method, where we start with $f(x) = 0$, and move to $x = g(x)$.