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
- \(\sqrt{2}\) 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 \(\sqrt{2}\).
- \(|x_1^2 - 2| < |x_0^2 - 2|\),
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 \(x_0 + 1 > 1\).
- \(|x_2^2 - 2| < \frac{1}{9} |x_0^2 - 2|\).
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 \(x = g(x)\).
