Let D be the differential operator d/dx and E the differential operator xD(xD - 1)(xD - 2) ... (xD - n). Find an expression of the form y = ∫1x g(t) dt for the solution of the differential equation Ey = f(x), with initial conditions y(1) = y'(1) = ... = y(n)(1) = 0, where f(x)is a continuous real-valued function on the reals.
The first step is to try to simplify Ey. Experimenting with small n, we soon suspect that Ey = xn+1y(n+1) and that is easily proved by induction.
So we seek a solution to xn+1y(n+1) = f(x), subject to y(1) = y'(1) = ... = y(n)(1) = 0. The trick is to use the Taylor expansion: y(x) = y(1) + (x-1)y'(1) + (x-1)2/2! y''(1) + ... + (x-1)ny(n)(1) + ∫1x g(t) dt, where g(t) = (x - t)nyn+1(t)/n! = (x - t)nf(t)/(n! tn+1).
24th Putnam 1963
© John Scholes
5 Feb 2002