1. ## Maclaurin Series

I have to find the Maclaurin series of this:
f(x)=x^2/(sqrt(2+x))
I'm not really sure where to start. Any help would be appreciated.

2. $f(x) = x^2 \left(2+x\right)^{-\frac{1}{2}} = 2^{-\frac{1}{2}}x^2 {\color{blue}\left(1 + \frac{x}{2}\right)^{-\frac{1}{2}}}$

Here, we can use binomial series:
${\color{blue} \left(1+\frac{x}{2}\right)^{-\frac{1}{2}}} = 1 \ + \ \left(-\frac{1}{2}\right)\frac{x}{2} \ + \ \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!} \left(\frac{x}{2}\right)^2 \ + \ \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!} \left(\frac{x}{2}\right)^3 \ + \ \cdots$ ${\color{white}.} + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots \left(-\frac{1}{2} - n + 1\right)}{n!} \left(\frac{x}{2}\right)^n + \cdots$ ..... $\text{if } \left| \frac{x}{2} \right| < 1$

Now simplify ...

You could've went down the route by taking derivatives. You would still get the same coefficients (since they are unique).

3. Originally Posted by o_O
$f(x) = x^2 \left(2+x\right)^{-\frac{1}{2}} = 2x^2 {\color{blue}\left(1 + \frac{x}{2}\right)^{-\frac{1}{2}}}$

Here, we can use binomial series:
${\color{blue} \left(1+\frac{x}{2}\right)^{-\frac{1}{2}}} = 1 \ + \ \left(-\frac{1}{2}\right)\frac{x}{2} \ + \ \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!} \left(\frac{x}{2}\right)^2 \ + \ \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!} \left(\frac{x}{2}\right)^3 \ + \ \cdots$ ${\color{white}.} + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots \left(-\frac{1}{2} - n + 1\right)}{n!} \left(\frac{x}{2}\right)^n + \cdots$ ..... $\text{if } \left| \frac{x}{2} \right| < 1$

Now simplify ...

You could've went down the route by taking derivatives. You would still get the same coefficients (since they are unique).
This is the preferred method, but when o_O suggested if there was another method I was intrigued and I found one. Lets give it a go...it is completely impractical but interesting

$\int\frac{dx}{\sqrt{1+x^2}}=\text{arcsinh}(x)$

And it can easily be shown that \begin{aligned}\text{arcsinh}(x)&=\ln\left(x+\sqrt {1+x^2}\right)\\
&=\frac{1}{2}\ln\left(\left(x+\sqrt{1+x^2}\right)^ 2\right)\\
&=\frac{1}{2}\ln\left(1+2x^2+2x\sqrt{1+x^2}\right) \end{aligned}

So we know that

$\forall{x}\backepsilon|x|<1~~\ln(1+x)=-\sum_{n=0}^{\infty}\frac{(-1)^nx^{n}}{n}$

So

$\forall{x}\backepsilon\left|2x^2+2x\sqrt{1+x^2}\ri ght|<1~~\frac{1}{2}$ $\ln\left(1+2x^2+2x\sqrt{1+x^2}\right)=\frac{-1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n\left(2x^2+2x\sqrt{1+x^2}\right)^{n}}{n}$

So

$\forall{x}\backepsilon\left|2x^2+2x\sqrt{1+x^2}\ri ght|<1$ $\frac{1}{\sqrt{1+x^2}}=\frac{-1}{2}\sum_{n=1}^{\infty}(-1)^n\left(2x^2+2x\sqrt{1+x^2}\right)^{n-1}\left(2x^2+2x\sqrt{1+x^2}\right)'$

And finally

$\forall{x}\backepsilon\left|2x+2\sqrt{x+x^2}\right |<1~$ $\frac{1}{\sqrt{1+x}}=\frac{-1}{2}\sum_{n=1}^{\infty}(-1)^n\left(2x+2\sqrt{x+x^2}\right)^{n-1}\left(2x^2+2x\sqrt{1+x^2}\right)'\bigg|_{x=\sqrt {x}}$

I might have made a computational error..its late