Putnam 2002, Problem A-1

Let $k$ be a fixed positive integer. The $n$-th derivative of $\frac{1}{x^k-1}$ has the form $\frac{P_n(x)}{(x^k-1)^{n+1}}$ where $P_n(x)$ is a polynomial. Find $P_n(1)$.

Solution:
By differentiating \( \frac{P_n(x)}{(x^k - 1)^{n+1}} \), we find that  
\[ P_{n+1}(x) = (x^k - 1)P_n'(x) - (n+1)kx^{k-1}P_n(x); \]  
substituting \( x = 1 \) yields \( P_{n+1}(1) = -(n+1)kP_n(1) \). Since \( P_0(1) = 1 \), an easy induction gives \( P_n(1) = (-k)^n n! \) for all \( n \geq 0 \).  

Note: one can also argue by expanding in Taylor series around 1. Namely, we have  
\[ \frac{1}{x^k - 1} = \frac{1}{k(x-1) + \cdots} = \frac{1}{k}(x-1)^{-1} + \cdots, \]  
so  
\[ \frac{d^n}{dx^n} \frac{1}{x^k - 1} = \frac{(-1)^n n!}{k(x-1)^{-n-1}} \]  
and  
\[ P_n(x) = (x^k - 1)^{n+1} \frac{d^n}{dx^n} \frac{1}{x^k - 1} \]  
\[ = (k(x-1) + \cdots)^{n+1} \left( \frac{(-1)^n n!}{k}(x-1)^{-n-1} + \cdots \right) \]  
\[ = (-k)^n n! + \cdots. \]