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**

Answer: 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}).

© John Scholes

jscholes@kalva.demon.co.uk

7 Jan 2001