How does one expand $\sqrt{r^2+a^2}-r$ to $\frac{a^2}{2r}-\frac{a^4}{8r^3}+\frac{a^6}{16r^5}$? The problem and solution are attached as photos below. 
I understand all parts of the solution except for the line above "By inspection..." where $V(r,\theta=0)$ is expanded. I tried a taylor expansion, taylor expansion + binomial, etc, to no avail, but maybe I don't know what I'm doing. Help is much appreciated!
Remember we assume r>~a in part b. The solution comes from a trusted source. 


 A: One first off factors out the factor of $r$ to have $r\sqrt{1 + (a/r)^2}$ and then Taylor-expands $y = \sqrt{x}$ about $x=1.$ The derivatives involve multiplying $(-1/2)$ by $(-3/2)$ by $(-5/2)$ by ... so we need to use a double-factorial ($m!! = 1\cdot 3\cdot 5 \cdot \dots \cdot m$) for the numerator, $2^n$ for the denominator. Choosing $(-1)!! = 1$ for simplicity, we find that $y^{(n)}(x) = (-1)^{n+1}~x^{-n+1/2}~(2n-3)!!~/ 2^n$ and hence:$$\begin{array}{rl}\sqrt{1 + \delta x} =& 1+\sum_{n=1}^\infty (-1)^{n+1}\frac{(2n-3)!!}{2^n~n!}~(\delta x)^n\\
 =& 1 + \frac12~\delta x - \frac18 \delta x^2 + \frac 1{16}\delta x^3 - \frac 5{128} \delta x^4 + \frac7{256} \delta x^5 -\dots \end{array}$$
and putting these both together by multiplying through the series by $r$ yields that expression as long as $-1 < (a/r)^2 < 1.$ 
There is a somewhat simpler expression for this but it requires knowing that $(-1/2)! = \sqrt{\pi},$ which you might not have encountered yet.
A: It's the usual binomial theorem $$(1+x) ^{n} =1+nx+\frac{n(n-1)}{2!}x^{2}+\frac{n(n-1)(n-2)}{3!}x^{3}+\dots\tag{1}$$ which is valid for all $n$ and all $x$ with $|x|<1$. Now $$\sqrt{r^{2}+a^{2}}-r=r\{\sqrt{1+(a/r)^{2}}-1\}=r\{(1+(a/r)^{2})^{1/2}-1\}\tag{2}$$ Since $r>a$ the expression $(a/r) ^{2}$ is less than $1$ and we can apply binomial theorem (by setting $x=(a/r) ^{2},n=1/2$ in $(1)$) to get $$(1+ (a/r)^{2})^{1/2}=1+\frac{a^{2}}{2r^{2}}-\frac{a^{4}}{8r^{4}}+\frac{a^{6}}{16r^{6}}+\cdots $$ Subtracting $1$ from the above expression and multiplying the result by $r$ we get the desired formula for the expression in equation $(2)$.
A: Taylor expansion is indeed the key here, but you have to be mindful of what variable you are expanding! Here, depending on how far $(r)$ you are away from the disk, the smaller the disk is to you and less its extended structure matters so $a$ is 'small' for your purposes. We then should Taylor expand around $a=0$.
You don't want to Taylor expand around $r=0$ because that would be assuming that $r$ is small but it's given in the problem that $r>a$. 
So we say
\begin{equation}
f(a) = \sqrt{r^2+a^2} = f(0) + f'(0)~a + \frac{1}{2}f''(0)~a^2+ \frac{1}{3!}f'''(0)~a^3 + \frac{1}{4!}f^{(4)}(0)~a^4 + \theta(a^5)
\end{equation}
Now the derivatives of $\sqrt{r^2+a^2}$ are
\begin{align}
\frac{\partial}{\partial a}\sqrt{a^2+r^2} = \frac{a}{\sqrt{r^2+a^2}}~,\\
\frac{\partial^2}{\partial a^2}\sqrt{a^2+r^2} = \frac{1}{\sqrt{r^2+a^2}}-\frac{a^2}{(r^2+a^2)^{3/2}}~,\\
\frac{\partial^3}{\partial a^3}\sqrt{a^2+r^2} = - \frac{3a r^2}{(r^2+a^2)^{5/2}}~,\\
\frac{\partial^4}{\partial a^4}\sqrt{a^2+r^4} = \frac{12 a^2 r^2}{(r^2+a^2)^{7/2}} - \frac{3r^4}{(r^2+a^2)^{7/2}}\\
...
\end{align}
so evaluating these at $a=0$ (only the 2nd and 4th derivatives are non-zero at $a=0$) and plugging into our Taylor expansion we get
$$
\sqrt{r^2+a^2} = \frac{a^2}{2r} - \frac{a^4}{8r^3} + \theta(a^5)~~.
$$
I'll leave it as an exercise to you to calculate the 5th and 6th derivatives with respect to $a$ to see if you can match the expression from your solution.
