k and n are positive integers. Let f(x) = 1/(xk - 1). Let p(x) = (xk - 1)n+1 fn(x), where fn is the nth derivative. Find p(1).
Solution
Answer: (-1)n+1 n! kn.
A routine differentiation.
Let pn(x) = (xk - 1)n+1fn(x). So fn = pn/(xk - 1)n+1. Differentiating, fn+1 = (pn' (xk - 1) - (n+1)kxk-1pn)/(xk - 1)n+2. Hence pn+1(1) = - (n+1)kpn(1). Also p1(1) = -k. Hence pn(1) = (-1)n n! kn.
© John Scholes
jscholes@kalva.demon.co.uk
11 Dec 2002
Last corrected/updated 16 Dec 02