*Do either (1) or (2):*

(1) f is a real-valued function defined on the reals with a continuous second derivative and satisfies f(x + y) f(x - y) = f(x)^{2} + f(y)^{2} - 1 for all x, y. Show that for some constant k we have f ''(x) = ± k^{2} f(x). Deduce that f(x) is one of ±cos kx, ±cosh kx.

(2) a_{i} and b_{i} are constants. Let A be the (n+1) x (n+1) matrix A_{ij}, defined as follows: A_{i1} = 1; A_{1j} = x^{j-1} for j ≤ n; A_{1 (n+1)} = p(x); A_{ij} = a_{i-1}^{j-1} for i > 1, j ≤ n; A_{i (n+1)} = b_{i-1} for i > 1. We use the identity det A = 0 to define the polynomial p(x). Now given any polynomial f(x), replace b_{i} by f(b_{i}) and p(x) by q(x), so that det A = 0 now defines a polynomial q(x). Prove that f( p(x) ) is a multiple of &prodc; (x - a_{i}) plus q(x).

**Solution**

(1) Putting y = 0 gives f(0)^{2} = 1, so f(0) = ±1. Differentiating wrt y gives f '(x+y) f(x-y) - f(x+y) f '(x-y) = 2 f(y) f '(y). Putting y = 0 gives f '(0) = 0. Differentiating wrt x gives f ''(x+y) f(x-y) = f(x+y) f ''(x-y). Putting x = y = z/2 gives f ''(z) = h f(z), where h = ± f ''(0). If h is positive, we may put h = k^{2} and integrate, using f(0) = ±1, f '(0) = 0 to get f(x) = ± cos kx. If h is negative, we may put h = - k^{2} and integrate to get f(x) = ± cosh kx.

(2)

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002