For each integer n > 0 show that there is a polynomial p(x) such that p(2 cos x) = 2 cos nx.
Solution
Induction on n. Obvious for n = 1. For n = 2, we have 2 cos 2x = 4 cos2x - 2 = (2 cos x)2 - 2, so it is true for n = 2. Now suppose it is true for n and n+1. Then 2 cos(n+1) cos x = cos(n+2)x + cos nx, so 2 cos(n+2) x = (2 cos(n+1)x)(2 cos x) - (2 cos nx), and so it is true for n+2.
© John Scholes
jscholes@kalva.demon.co.uk
6 March 2004
Last corrected/updated 6 Mar 04