2013th derivative of rational function I am struggling to find $f^{(2013)}(0)$ for $$f(x) = \frac{1}{1 + x + x^3 + x^4}$$ I know that I should use power series, and following a hint I rewrote the equation as the following:
$$1 = (1 + x + x^3 + x^4)(\sum_{n = 0}^{\infty} a_n x^n) = \sum_{n = 0}^{\infty} a_n x^n + \sum_{n = 0}^{\infty} a_n x^{n + 1} + \sum_{n = 0}^{\infty} a_n x^{n + 3} + \sum_{n = 0}^{\infty} a_n x^{n + 4}$$
I can work out the value of $a_0$ easily, but I'm not really sure how to solve for the remaining coefficients and hence find the derivatives... I know that the pattern is supposed to be, but I'm completely at a loss. Any hints as to how to proceed?
 A: Following the hint and letting $f(x)=\sum_{n}a_n x^n$, you have
$$
\begin{eqnarray}
1&=&\left(1+x+x^3+x^4\right) f(x) \\ &=& \left(1+x+x^3+x^4\right)\sum_{n}a_n x^n \\ &=& \sum_na_nx^n+\sum_na_nx^{n+1}+\sum_na_nx^{n+3}+\sum_na_nx^{n+4} \\
&=&\sum_na_nx^n+\sum_na_{n-1}x^n+\sum_na_{n-3}x^{n}+\sum_na_{n-4}x^n \\
&=&\sum_n\left(a_n+a_{n-1}+a_{n-3}+a_{n-4}\right)x^{n}.
\end{eqnarray}
$$
Equating identical powers of $x$ tells you that $a_0=1$ and $a_{n}=-a_{n-1}-a_{n-3}-a_{n-4}$ for $n>0$.  Direct calculation of this recurrence relation shows that $a_{2013}=-672$.  (There is also a closed-form solution, but that doesn't seem necessary here.)  Since $f^{(2013)}(0)=2013!\times a_{2013}$, you find
$$
f^{(2013)}(0)=-672\cdot2013! \approx -1.91\times 10^{5781}.
$$
A: Does it help that $1+x+x^2 +x^3 + x^4 = \frac{1-x^5}{1-x}$? You can then separate the fraction and apply the extended binomial theorem.
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
$\ds{\fermi\pars{x} = \frac{1}{1 + x + x^3 + x^4} = {1 \over \pars{1 + x^{3}}\pars{1 + x}}\,,\quad \fermi^{\pars{2013}}\pars{0}:\ ?}$

\begin{align}
\fermi\pars{x}
&=
\sum_{\ell = 0}^{\infty}\pars{-1}^{3\ell}x^{3\ell}
\sum_{\ell' = 0}^{\infty}\pars{-1}^{\ell'}x^{\ell'}
=
\sum_{\ell = 0}^{\infty}\sum_{\ell' = 0}^{\infty}\pars{-1}^{\ell + \ell'}
\sum_{n = 0}^{\infty}x^{n}\delta_{n,3\ell + \ell'}
\\[3mm]&=
\sum_{n = 0}^{\infty}x^{n}\sum_{\ell = 0}^{\infty}\pars{-1}^{\ell}
\sum_{\ell' = 0}^{\infty}
\pars{-1}^{\ell'}\delta_{\ell', n - 3\ell}
=
\sum_{n = 0}^{\infty}x^{n}\bracks{\sum_{\ell = 0}^{\infty}
\pars{-1}^{\ell}\pars{-1}^{n - 3\ell}}_{n - 3\ell \geq 0}
\\[3mm]&=
\sum_{n = 0}^{\infty}{\fermi^{\pars{n}}\pars{0} \over n!}\,x^{n}
\quad\mbox{where}\quad
\fermi^{\pars{n}}\pars{0}
=
\pars{-1}^{n}\,n!\!\!\!\!\!\!\!\sum_{\ell = 0 \atop{\vphantom{\LARGE A}n - 3\ell\ \geq\ 0}}^{\infty}1
\end{align}

\begin{align}
\fermi^{\pars{2013}}\pars{0}
&=
\pars{-1}^{2013}\,2013!\!\!\!\!\!
\sum_{\ell = 0 \atop {\vphantom{\LARGE A}2013 - 3\ell\ \geq\ 0}}^{\infty}
\!\!\!\!\!\!\!\!\!\!\!1
=
-\pars{2013!}\sum_{\ell = 0}^{671}1
\end{align}
$$
\color{#0000ff}{\large\fermi^{\pars{2013}}\pars{0} = -672 \times \pars{2013!}}
$$
