How do you use the binomial series to expand f(x)=1/(sqrt(4+x^2))?

Jul 3, 2018

The answer is $= \frac{1}{2} - \frac{1}{16} {x}^{2} + \frac{3}{256} {x}^{4} + \ldots . .$

Explanation:

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

$= \left(\begin{matrix}n \\ 0\end{matrix}\right) {a}^{n} {b}^{0} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {a}^{n - 1} b + \left(\begin{matrix}n \\ 2\end{matrix}\right) {a}^{n - 2} {b}^{2} + \ldots . .$

$= {a}^{n} + \frac{\left(n\right) \left(n - 1\right)}{1 \cdot 2} {a}^{2} b + \frac{\left(n\right) \left(n - 1\right) \left(n - 2\right)}{1 \cdot 2.3} {a}^{3} {b}^{2} + \ldots$

Where,

((n),(k))=(n!)/((n-k)!(k!))

Also,

${\left(1 + b\right)}^{n} = 1 + \frac{\left(n\right) \left(n - 1\right)}{1 \cdot 2} b + \frac{\left(n\right) \left(n - 1\right) \left(n - 2\right)}{1 \cdot 2.3} {b}^{2} + . .$

Here, we have

$\frac{1}{\sqrt{4 + {x}^{2}}} = \frac{1}{2 {\left(1 + {\left(\frac{x}{2}\right)}^{2}\right)}^{\frac{1}{2}}}$

Here

$n = - \frac{1}{2}$

$a = 1$

$b = {\left(\frac{x}{2}\right)}^{2}$

Therefore,

$\frac{1}{\sqrt{4 + {x}^{2}}} = \frac{1}{2} {\left(1 + {\left(\frac{x}{2}\right)}^{2}\right)}^{- \frac{1}{2}}$

$= \frac{1}{2} \left(\left(\begin{matrix}\frac{1}{2} \\ 0\end{matrix}\right) {1}^{\frac{1}{2}} {\left({x}^{2} / 2\right)}^{0} + \left(\begin{matrix}\frac{1}{2} \\ 1\end{matrix}\right) {1}^{\frac{1}{2} - 1} {\left({x}^{2} / 2\right)}^{2} + \left(\begin{matrix}\frac{1}{2} \\ 2\end{matrix}\right) {1}^{\frac{1}{2} - 2} {\left(\frac{x}{2}\right)}^{4} + \ldots \ldots\right)$

$= \frac{1}{2} \left(1 + \left(- \frac{1}{2}\right) {\left(\frac{x}{2}\right)}^{2} + \frac{\left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right)}{1 \cdot 2} {\left(\frac{x}{2}\right)}^{4} + . .\right)$

$= \frac{1}{2} - \frac{1}{16} {x}^{2} + \frac{3}{256} {x}^{4} + \ldots . .$