Let ∑0∞ xn(x - 1)2n / n! = ∑0∞ an xn. Show that no three consecutive an are zero.
Let p(x) = x(x - 1)2, and f(x) = ep(x). Then f(x) has the expansion given. Differentiating f ' = ep(x)p'(x) = f p' (*). Now since p is a polynomial of degree 3, we have p(n) = 0 for n > = 4. So differentiating (*) n times we get: f(n+1) = nC0 f(n) p' + nC1 f(n-1) p'' + nC2 f(n-2) p'''. So if f(n)(0) = f(n-1)(0) = f(n-2)(0), then f(m)(0) = 0 for all m ≥ n - 2. In other words if three consecutive an are zero then all subsequent an are zero, and hence f(x) is a finite polynomial, which is impossible.
33rd Putnam 1972
© John Scholes
27 Jan 2001