p(x) is a polynomial of degree < 1992 such that p(0), p(1), p(-1) are all non-zero. The 1992th derivative of p(x)/(x^{3} - x) = f(x)/g(x) for polynomials f(x) and g(x). Find the smallest possible degree of f(x).

**Solution**

Answer: 3984.

We may write p(x)/(x^{3} - x) = q(x) + a/x + b/(x-1) + c/(x+1), where q(x) has degree less than 1992, and (since 0, 1, and -1 are not roots of p(x) ) a, b, c are all non-zero. When we differentiate 1992 times, q(x) disappears and we get a non-zero constant times (a/x^{1993} + b/(x-1)^{1993} + c/(x+1)^{1993}). Hence f(x) is a multiple of a(x^{2}-1)^{1993} + bx^{1993}(x+1)^{1993} + cx^{1993}(x-1)^{1993}. We can zero the highest power 2·1993 = 3986 by taking a + b + c = 0. We can zero the next highest power 3985 by taking b = c. That also zeros all the other odd powers. Now the coefficient of x^{3984} is -1992a + 1993.1992 b = 1992(1993 + 2)b which is non-zero. So the smallest possible degree of f(x) is 3984.

© John Scholes

jscholes@kalva.demon.co.uk

7 Jan 2001