Can we find a good rational approximation for $\sqrt{5}$?

Can we find a good rational approximation for $\sqrt{5}$?

Question

Prove that, if $x$ is so small that its cube and higher powers can be neglected, $\sqrt{\frac{1+x}{1-x}} = 1 + x + \frac{x^2}{2}.$ By taking $x = \dfrac{1}{9}$, prove that $\sqrt{5}$ is approximately equal to $\dfrac{181}{81}$.

Suggestion

Can we rewrite the left-hand side as a product of functions that we do know how to expand?

Solution

The binomial theorem says that, if $r$ is any real number and $\big|x\big| < 1$, then $(1+x)^r = 1 + rx + \frac{r(r-1)}{2!} x^2 + \frac{r(r-1)(r-2)}{3!} x^3 + \dots b.$ 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.

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 $\sqrt{5}$.