Let ∑_{0}^{∞} x^{n}(x - 1)^{2n} / n! = ∑_{0}^{∞} a_{n} x^{n}. Show that no three consecutive a_{n} are zero.

**Solution**

Let p(x) = x(x - 1)^{2}, and f(x) = e^{p(x)}. Then f(x) has the expansion given. Differentiating f ' = e^{p(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 a_{n} are zero then all subsequent a_{n} are zero, and hence f(x) is a finite polynomial, which is impossible.

© John Scholes

jscholes@kalva.demon.co.uk

27 Jan 2001