53rd Putnam 1992 Problem A4 Let R be the reals. Let f : R → R be an infinitely differentiable function such that f(1/n) = n 2 /(n 2 +1) for n = 1, 2, 3, ... Find the value of the derivatives of f at zero: f (k) (0) for k = 1, 2, 3, ... . Solution
: the odd derivatives are zero; f (2n) (0) = (-1) n . Clearly one possibility is f(x) = 1/(1 + x 2 ). This has the power series expansion 1 - x 2 + x 4 - ... , so in this case f (k) (0) = 0 for k odd and (-1) n for k = 2n. Suppose g(x) is another such function. Define h by h(x) = f(x) - g(x). Then h is infinitely differentiable and zero at x = 1/1, 1/2, 1/3, ... . We show that all derivatives of h are zero at x = 0. Suppose otherwise. Take h (k) (0) as the non-zero derivative of lowest order. Then for some ε > 0, h (k) (x) is non-zero for all 0 ≤ x < ε. Take n sufficiently large that 1/n < ε. By the (nth order) mean value theorem, h(1/n) = h (k) (ε')/k! for some ε' < 1/n < ε. But the lhs is zero and the rhs is non-zero. Contradiction. So all the derivatives of h are zero at x = 0, and hence the derivatives of g at x = 0 are the same as those of 1/(1+x 2 ). 53rd Putnam 1992 © John Scholes jscholes@kalva.demon.co.uk 7 Jan 2001