23rd Putnam 1962

Problem A2

Let R be the reals. Find all f : K → R, where K is [0, ∞) or a finite interval [0, a), such that (1/k ∫0k f(x) dx )2= f(0) f(k) for all k in K.



Answer: for [0, ∞) f(x) = b/(cx + 1)2 with c non-negative. For [0, a), f(x) = b/(cx + 1)2 with c >= -1/a.

Put y = ∫0x f(t) dt. Then the given equation becomes y2/x2 = by', where b = f(0). Integrating, b/y = c + 1/x, or y = bx/(cx + 1). Differentiating gives f(x) = y' = b/(cx + 1)2.

f(-1/c) is undefined, so if K is the half-line, then c cannot be negative. If K is the finite interval [0, a), then we require c ≥ -1/a. Note that we can have equality because k must be strictly less than a. On the other hand, b can have any value.

Note that if we start by differentiating, then things get somewhat messy, albeit manageable. We get: x y' = A y3/2 - 2y, where A = 2/√ f(0). So we have to integrate 1/(A y3/2 - 2y). Differentiating the obvious log(A y3/2 - 2y) gives us the desired term plus 3/(2y). Thus we can integrate the equation to get: B + log x = log(A y3/2 - 2y) - 3/2 log y and after some simplification and extra care with signs we recover the result above.



23rd Putnam 1962

© John Scholes
5 Feb 2002