k and n are positive integers. Let f(x) = 1/(x^{k} - 1). Let p(x) = (x^{k} - 1)^{n+1} f^{n}(x), where f^{n} is the nth derivative. Find p(1).

**Solution**

Answer: (-1)^{n+1} n! k^{n}.

A routine differentiation.

Let p_{n}(x) = (x^{k} - 1)^{n+1}f^{n}(x). So f^{n} = p_{n}/(x^{k} - 1)^{n+1}. Differentiating, f^{n+1} = (p_{n}' (x^{k} - 1) - (n+1)kx^{k-1}p_{n})/(x^{k} - 1)^{n+2}. Hence p_{n+1}(1) = - (n+1)kp_{n}(1). Also p_{1}(1) = -k. Hence p_{n}(1) = (-1)^{n} n! k^{n}.

© John Scholes

jscholes@kalva.demon.co.uk

11 Dec 2002

Last corrected/updated 16 Dec 02