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

Nov 15, 2016

(1+x^2)^(-1/2)=sum_(k=0)^oo (-1)^k (Pi_(j=1)^k(1/2+j-1))/(k!)x^(2k)

for $\left\mid x \right\mid \le 1$

#### Explanation:

We know that

(1+y)^n = 1+ny+(n(n-1))/(2!)y^2+cdots+(Pi_(j=1)^k(n+j-1))/(k!) y^k+cdots

so here

(1+x^2)^(-1/2)=sum_(k=0)^oo (-1)^k (Pi_(j=1)^k(1/2+j-1))/(k!)x^(2k)

The five first terms are

${\left(1 + {x}^{2}\right)}^{- \frac{1}{2}} \approx 1 - {x}^{2} / 2 + \frac{3 {x}^{4}}{8} - \frac{5 {x}^{6}}{16} + \frac{35 {x}^{8}}{128}$

Of course this series is convergent only for $\left\mid x \right\mid \le 1$

Nov 15, 2016

The series is $= 1 - {x}^{2} / 2 + \frac{3 {x}^{4}}{8} - \frac{5 {x}^{4}}{16}$

#### Explanation:

The binomial theorem is ${\left(a + b\right)}^{n}$
$= {a}^{n} + n \cdot \frac{{a}^{n - 1} b}{1} + \frac{n \left(n - 1\right) {a}^{n - 2} {b}^{2}}{1 \cdot 2} + \frac{n \left(n - 1\right) \left(n - 2\right) {a}^{n - 3} {b}^{3}}{1 \cdot 2 \cdot 3}$

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

So, $a = 1$

$b = {x}^{2}$

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

Therefore, ${\left(1 + {x}^{2}\right)}^{- \frac{1}{2}} = 1 + \left(- \frac{1}{2}\right) {x}^{2} + \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) {x}^{4} / \left(1 \cdot 2\right) + \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) {x}^{6} / \left(1 \cdot 2 \cdot 3\right)$

$= 1 - {x}^{2} / 2 + \frac{3 {x}^{4}}{8} - \frac{5 {x}^{6}}{16}$