# How do you use the binomial series to expand sqrt(z^2-1)?

Jul 13, 2016

$\sqrt{{z}^{2} - 1} = i \left[1 - \frac{1}{2} {z}^{2} - \frac{1}{8} {z}^{4} - \frac{1}{16} {z}^{6} + \ldots\right]$

#### Explanation:

I'd quite like a double check because as a physics student I rarely get beyond ${\left(1 + x\right)}^{n} \approx 1 + n x$ for small x so I'm a bit rusty. The binomial series is a specialised case of the binomial theorem which states that

${\left(1 + x\right)}^{n} = {\sum}_{k = 0}^{\infty} \left(\begin{matrix}n \\ k\end{matrix}\right) {x}^{k}$

With ((n),(k)) = (n(n-1)(n-2)...(n-k+1))/(k!)

What we have is ${\left({z}^{2} - 1\right)}^{\frac{1}{2}}$, this is not the correct form. To rectify this, recall that ${i}^{2} = - 1$ so we have:

${\left({i}^{2} \left(1 - {z}^{2}\right)\right)}^{\frac{1}{2}} = i {\left(1 - {z}^{2}\right)}^{\frac{1}{2}}$

This is now in the correct form with $x = - {z}^{2}$

Therefore, the expansion will be:

$i \left[1 - \frac{1}{2} {z}^{2} + \frac{\frac{1}{2} \left(- \frac{1}{2}\right)}{2} {z}^{4} - \frac{\frac{1}{2} \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right)}{6} {z}^{6} + \ldots\right]$

$i \left[1 - \frac{1}{2} {z}^{2} - \frac{1}{8} {z}^{4} - \frac{1}{16} {z}^{6} + \ldots\right]$