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

Nov 25, 2016

The series is $= = 1 + {x}^{2} / 2 - {x}^{4} / 8 + {x}^{6} / 16. \ldots$

#### Explanation:

The binomial theorem is
${\left(a + b\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) {a}^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {a}^{n - 1} b + \left(\begin{matrix}n \\ 3\end{matrix}\right) {a}^{n - 2} {b}^{2} + \left(\begin{matrix}n \\ 4\end{matrix}\right) {a}^{n - 3} {b}^{3} + \ldots \ldots$

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

Normally $n \in \mathbb{N}$

But there is an extension for ${\left(1 + x\right)}^{k}$ where ∣x∣<1 and $k$ any number

Let's rewrite $f \left(x\right) = {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}$

So, ${\left(1 + {x}^{2}\right)}^{\frac{1}{2}} = 1 + \left(\frac{1}{2}\right) {x}^{2} + \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(\frac{1}{2}\right) {\left({x}^{2}\right)}^{2} + \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(\frac{1}{6}\right) {\left({x}^{2}\right)}^{3} + \ldots \ldots$

$= 1 + {x}^{2} / 2 - {x}^{4} / 8 + {x}^{6} / 16. \ldots .$

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