Find the Second coefficient of the series $\frac{1}{(1+x)^\frac{1}{4}} = \sum_{n=0}^\infty c_nx^n$ If $$\frac{1}{(1+x)^\frac{1}{4}} = \sum_{n=0}^\infty c_nx^n$$ then $c_2$ is....
I think I know how to attempt the problem but I'm not sure if I'm on the right track
I started out with the known series 
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
I then attempted to manipulate the left side so it would be the same as the equation above. Then depending on how the x term looks I could plug in a value for n to find the second coefficient of the series; however, I am having some trouble manipulating the problem due to the quarter power. 
So far I got up to here...
$$\frac{1}{1+x}=\sum_{n=0}^\infty x^n(-1)^n$$
I know that changing the $x$ term on the left side will change the $x$ term on the right side, but I'm unclear if I can change the whole denominator on the left side and how that would effect the right side since the quarter power is over the whole denominator. 
It is an old final so I do know the answer is $\frac{5}{32}$ but I don't know how to get that answer.  
 A: Have a Taylor expansion at $0$
$$f(x)=\frac{1}{(1+x)^\frac{1}{4}} =\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x-0)^n =\sum_{n=0}^\infty c_nx^n$$ 
then $c_2=\frac{f''(0)}{2!}$
$f''(x)=((1+x)^{-0.25})''=(-0.25 \times (1+x)^{-1.25})'=\frac{5}{16}(1+x)^{-\frac{9}{4}}$
Thus $c_2 = \frac{5}{32}$
A: There are two ways to do this. First, there is the formula
$$ (1 + x)^\alpha = \sum_{k = 0}^\infty {\alpha \choose k}x^k $$
where for an arbitrary complex number $\alpha$ and $k \in \mathbb N$
$$ {\alpha \choose k} = \frac{\alpha(\alpha - 1)(\alpha - 2) \cdots (\alpha - k + 1)}{k!}. $$
However, I suspect that the problem is instead asking you to do it by calculating derivatives. Recall that if $f(x)$ admits a power series representation at $0$ then
$$ f(x) = \sum_{k = 0}^\infty \frac{f^{(k)}(0)}{k!} x^k. $$
So in this case, the number you are looking for is
$$ c_2 = \frac12 \cdot \left. \frac{d^2}{dx^2} \frac{1}{(1 + x)^{1/4}} \right|_{x \leftarrow 0}. $$
A: Here is a simple and straightforward way which avoids any deep theorems of analysis. We have $$\left(\sum c_{n}x^{n}\right)^{4}(1 + x) = 1$$ or $$\left\{c_{0}^{2} + 2c_{0}c_{1}x + (2c_{0}c_{2} + c_{1}^{2})x^{2} + \cdots\right\}^{2}(1 + x) = 1$$ or $$\left\{c_{0}^{4} + 4c_{0}^{3}c_{1}x + \{2c_{0}^{2}(2c_{0}c_{2} + c_{1}^{2}) + 4c_{0}^{2}c_{1}^{2}\}x^{2} + \cdots\right\}(1 + x) = 1$$ and then we have by comparing coefficients $$c_{0}^{4} = 1, 1 + 4c_{0}^{3}c_{1} = 0, 4c_{0}^{3}c_{1} + 2c_{0}^{2}(2c_{0}c_{2} + c_{1}^{2}) + 4c_{0}^{2}c_{1}^{2} = 0\tag{1}$$ From the original equation $\sum c_{n}x^{n} = (1 + x)^{-1/4}$ we get $c_{0} = 1$ by putting $x = 0$. Then putting this value of $c_{0}$ in $(1)$ we get $$c_{1} = -1/4, -1 + 2(2c_{2} + 1/16) + 1/4 = 0$$ or $c_{2} = (1/2)(3/8 - 1/16) = 5/32$. 
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
{1 \over \pars{1 + x}^{1/4}} & =
\pars{1 + x}^{-1/4} =
1 + \pars{-\,{1 \over 4}}x\ +\
\overbrace{{1 \over 2!}
\pars{-\,{1 \over 4}}\pars{-\,{1 \over 4} - 1}}
^{\ds{-1/4 \choose 2}}\
\,x^{2} + \,\mrm{O}\pars{x^{3}}
\\[5mm] & =
1 - {1 \over 4}\,x +
\color{#f00}{5 \over 32}\,x^{2} + \,\mrm{O}\pars{x^{3}}
\end{align}
