If \(\dfrac{b}{a}\) is small enough for powers of \(\dfrac{b}{a}\) higher than the third to be neglected, show that
\[\begin{equation*}
(a+b)^\frac{1}{2} - (a-b)^\frac{1}{2} = a^\frac{1}{2} \left( \frac{b}{a} + \frac{b^3}{8a^3} \right).
\end{equation*}\]
If \(a > 0\), we can write \[(a+b)^\frac{1}{2} = \sqrt{a} \left( 1 + \frac{b}{a} \right)^{1/2}.\]
The binomial theorem says that, if \(\big |x \big| < 1\), \[\begin{equation*} (1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!} x^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!} x^3 + \dotsb \end{equation*}\] and so, if \(\dfrac{b}{a}\) is small enough, \[\begin{align*} (a+b)^\frac{1}{2} &= \sqrt{a} \left( 1 + \frac{b}{a} \right)^{1/2} \\ &= \sqrt{a} \left( 1 + \frac{1}{2} \frac{b}{a} + \frac{\frac{1}{2} \times \left(-\frac{1}{2} \right)}{2!} \left( \frac{b}{a} \right)^2 + \frac{\frac{1}{2} \times \left( -\frac{1}{2} \right) \times \left( -\frac{3}{2} \right)}{3!} \left( \frac{b}{a} \right)^3 + \dotsb \right) \\ &\approx \sqrt{a} \left( 1 + \frac{b}{2a} - \frac{b^2}{8a^2} + \frac{b^3}{16a^3} \right). \end{align*}\] Similarly, assuming that \(a > 0\) and \(\dfrac{b}{a}\) is small enough we will get the same result as above replacing \(b\) with \(-b\), \[\begin{align*} (a-b)^\frac{1}{2} &\approx \sqrt{a} \left( 1 + \frac{-b}{2a} - \frac{(-b)^2}{8a^2} + \frac{(-b)^3}{16a^3} \right)\\ &= \sqrt{a} \left( 1 - \frac{b}{2a} - \frac{b^2}{8a^2} - \frac{b^3}{16a^3} \right). \end{align*}\] As a result, \[\begin{align*} (a+b)^\frac{1}{2} - (a-b)^\frac{1}{2} &\approx \sqrt{a} \left( \left( 1 + \frac{b}{2a} - \frac{b^2}{8a^2} + \frac{b^3}{16a^3} \right) - \left( 1 - \frac{b}{2a} - \frac{b^2}{8a^2} - \frac{b^3}{16a^3} \right) \right) \\ &= \sqrt{a} \left( \frac{b}{a} + \frac{b^3}{8a^3} \right). \end{align*}\]Use this result to deduce a rational approximation to \(\sqrt 5 - \sqrt 3\).
Let’s put \(a + b = 5\) and \(a - b = 3\). Now adding gives \(2a = 8 \implies a = 4, b = 1\). Thus \(a > 0\) and \(\dfrac{b}{a}\) is suitably small.
From our work above, \[\begin{align*} \sqrt 5 - \sqrt 3 = (a+b)^\frac{1}{2} - (a-b)^\frac{1}{2} &\approx \sqrt{a} \left( \frac{b}{a} + \frac{b^3}{8a^3} \right) \\ &= \sqrt{4} \left( \frac{1}{4} + \frac{1}{8 \times 64} \right) \\ &= \frac{1}{2} + \frac{1}{256} \\ &= \frac{129}{256}\\ & \approx 0.5039... \end{align*}\]The approximation is an accurate one: \(\sqrt 5 - \sqrt 3= 0.5040...\)
Deduce, or obtain otherwise, a rational approximation to \(\sqrt 5 + \sqrt 3\).
Again, the approximation is a good one: \(\sqrt 5 + \sqrt 3 = 3.9681...\)
