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

Aug 4, 2016

1/sqrt(2)sum_{n=0}^oo(-1)^n/(n!2^{2n} )(Pi_{k=0}^{n-1}(2k+1))x^(n+2)

#### Explanation:

Expanding for $x = 0$ the function $f \left(x\right) = {\left(x + 2\right)}^{- \frac{1}{2}}$ gives

f(x) = f(0) + f^'(0)/(1!)x+f^('')(0)/(2!)x^2+cdots + f^((n))(0)/(n!)x^n +cdots

but

$\frac{{d}^{n}}{{\mathrm{dx}}^{n}} f \left(x\right) = \left(- \frac{1}{2}\right) \left(- \frac{1}{2} - 1\right) \cdots \left(- \frac{1}{2} - n + 1\right) {\left(x + 2\right)}^{- \frac{1}{2} - n}$

or

$\frac{{d}^{n}}{{\mathrm{dx}}^{n}} f \left(x\right) = {\left(- 1\right)}^{n} {\Pi}_{k = 0}^{n - 1} \frac{2 k + 1}{2} {\left(x + 2\right)}^{- \frac{1}{2} - n}$

then

${f}^{\left(n\right)} \left(0\right) = {\left(- 1\right)}^{n} {\Pi}_{k = 0}^{n - 1} \left(\frac{2 k + 1}{2}\right) {2}^{\left(- \frac{1}{2} - n\right)}$
${f}^{\left(n\right)} \left(0\right) = \frac{1}{\sqrt{2}} {\left(- 1\right)}^{n} / {2}^{2 n} {\Pi}_{k = 0}^{n - 1} \left(2 k + 1\right)$

So the series for $f \left(x\right)$ is

f(x) = 1/sqrt(2)sum_{n=0}^oo(-1)^n/(n!2^{2n} )(Pi_{k=0}^{n-1}(2k+1))x^n

The series representation for ${x}^{2} {\left(x + 2\right)}^{- \frac{1}{2}}$ is given by

1/sqrt(2)sum_{n=0}^oo(-1)^n/(n!2^{2n} )(Pi_{k=0}^{n-1}(2k+1))x^(n+2)