Prove that, if \(x\) is so small that its cube and higher powers can be neglected, \[ \sqrt{\frac{1+x}{1-x}} = 1 + x + \frac{x^2}{2}. \]
The binomial theorem says that, if \(r\) is any real number and \(\big|x\big| < 1\), then \[ (1+x)^r = 1 + rx + \frac{r(r-1)}{2!} x^2 + \frac{r(r-1)(r-2)}{3!} x^3 + \dotsb. \] Since \(x\) is close enough to zero so that its cube and higher powers can be neglected, this can be shortened to say that \[ (1+x)^r \approx 1 + rx + \frac{r(r-1)}{2!} x^2. \]
Now, as we can write \[ \sqrt{\frac{1+x}{1-x}} = (1+x)^{1/2} \times (1-x)^{-1/2} \] we can therefore use the binomial theorem to say that \[\begin{align*} \sqrt{\frac{1+x}{1-x}} &\approx \left( 1 + \tfrac{1}{2}x + \frac{\tfrac{1}{2}\left( -\tfrac{1}{2} \right)}{2!}x^2 \right) \times \left( 1 + \left(-\tfrac{1}{2}\right)(-x) + \frac{-\tfrac{1}{2}\left( -\tfrac{3}{2} \right)}{2!}(-x)^2 \right) \\ &= \left( 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 \right) \times \left( 1 + \tfrac{1}{2}x + \tfrac{3}{8}x^2 \right) \\ &\approx 1 + \left( \tfrac{1}{2} + \tfrac{1}{2} \right) x + \left( \tfrac{3}{8} + \tfrac{1}{4} - \tfrac{1}{8} \right) x^2\\ &= 1 + x + \frac{x^2}{2}, \end{align*}\]as required.
By taking \(x = \dfrac{1}{9}\), prove that \(\sqrt{5}\) is approximately equal to \(\dfrac{181}{81}\).
Since \(x = \dfrac{1}{9}\) is close to zero, the expansion that we proved in the first part of the question is reasonable.
The left-hand side becomes \[ \sqrt{\frac{1+x}{1-x}} = \sqrt{\frac{10/9}{8/9}} = \sqrt{\frac{10}{8}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \] while the right-hand side becomes \[ 1 + x + \frac{x^2}{2} = 1 + \frac{1}{9} + \frac{1}{162} = \frac{162 + 18 + 1}{162} = \frac{181}{162}. \] That is, \[ \frac{\sqrt{5}}{2} \approx \frac{181}{162} \] so on multiplying through by \(2\), \[ \sqrt{5} \approx \frac{181}{81} \] as we were aiming to show.
In fact, \(\left| \dfrac{181}{81} - \sqrt{5} \, \right| \approx 0.0015\). The approximation is good!
It is, however, far from being the “best” fraction approximation for \(\sqrt{5}\) using small denominators. We have \[\begin{align*} \left| \dfrac{38}{17} - \sqrt{5} \, \right| &\approx 0.0008,\\ \left| \dfrac{161}{72} - \sqrt{5} \, \right| &\approx 0.00004, \end{align*}\]which are far better. These are found from the continued fraction expansion of \(\sqrt{5}\).
