p(x) is a polynomial with real coefficients such that p(x) > 0 for x ≥ 0. Show that (1 + x)np(x) has non-negative coefficients for some positive integer n.
Solution
A product of two polynomials with non-negative coefficients also has non-negative coefficients and any real polynomial can be factorised into linear and quadratic factors, so it is sufficient to prove the result for p(x) linear and quadratic.
If p(x) = ax + b, then b = p(0) > 0. Also p(-b/a) = 0, so -b/a ≤ 0 and hence a ≥ 0. So p itself has non-negative coefficients, and hence (1 + x)np(x) has non-negative coefficients for all positive n. Thus the result is true for p(x) linear.
Now suppose p(x) is quadratic, p(x) is positive for x zero and x large, so the coefficients of x2 and x0 must be positive. There is no loss of generality in taking the coefficient of x2 to be 1. If the coefficient of x is also positive, then we are home. So assume it is negative. Thus p(x) = x2 - bx + c, with b, c > 0. Also b2 < 4c or p(b/2) < 0. So the coefficients of (1+x)np(x) are c, nc - b, n - b, 1 which are all positive for any sufficiently large n, and nCk-2 - nCk-1 b + nCk c for k = 2, 3 ... , n.
Ignoring the positive factor n!/(k! n-k+2!), we get k2(1+b+c) - k(n(b+2c)+2b+3c+1) + c(n2+3n+2). Put A = 1+b+c, B = b+2c, C = 2b+3c+1. So we have Ak2 + (Bn+C)k + c(n2+3n+2) (*). Obviously A > 0. But b2 < 4c, so (b+2c)2 < 4(1+b+c)c or B2 < 4Ac. Hence for all sufficiently large n, (Bn+C)2 < 4Ac(n2+3n+2) and so (*) is positive for all k, as required.
Thanks to Kamil Duszenko
© John Scholes
jscholes@kalva.demon.co.uk
2 Dec2003
Last corrected/updated 2 Dec 2003