TITLE: 2013th derivative of rational function
QUESTION [7 upvotes]: 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?

REPLY [8 votes]: 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
$$
f^{(2013)}(0)=-672\cdot2013! \approx -1.91\times 10^{5781}.
$$