The sequence \(x_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(x_0 = 1\) and by
\[x_{n+1} = \frac{x_n + 2}{x_n + 1}\]
for \(n \geqslant 0\).
Explain briefly why \(x_n \geqslant 1\) for all \(n\).
Show that \(x_{n+1}^2 - 2\) and \(x_n^2 - 2\) have opposite sign, and that
\[\left|x_{n+1}^2 - 2\right| \leqslant \tfrac{1}{4}\left|x_n^2 - 2\right|\,.\]
Show that
\[2 - 10^{-6} \leqslant x_{10}^2 \leqslant 2\,.\]
The sequence \(y_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(y_0 = 1\) and by
\[y_{n+1} = \frac{y_n^2 + 2}{2y_n}\]
for \(n \geqslant 0\).
Show that, for \(n \geqslant 0\),
\[y_{n+1} - \sqrt{2} = \frac{(y_n - \sqrt{2})^2}{2y_n}\]
and deduce that \(y_n \geqslant 1\) for \(n \geqslant 0\).
Show that
\[y_n - \sqrt{2} \leqslant 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^n}\]
for \(n \geqslant 1\).
Using the fact that
\[\sqrt{2} - 1 < \tfrac{1}{2}\,,\]
or otherwise, show that
\[\sqrt{2} \leqslant y_{10} \leqslant \sqrt{2} + 10^{-600}\,.\]