4th Putnam 1941

Define f(x) = ∫₀^x ∑_i=0^n-1 (x - t)ⁱ / i! dt. Find the nth derivative f ⁽ⁿ⁾(x).

Solution

Straightforward.

Note that x appears both in the integrand and in the limits, so a little care is needed. Write g_r(x, t) = ∑_i=0^r-1 (x - t)ⁱ / i! so that f(x) = ∫₀^x g_n(x, t) e^nt dt. By definition f '(x) = lim_{δx -> 0} (∫₀^x+δx g_n(x+δx, t) e^nt dt - ∫₀^x g_n(x, t) e^nt dt) / δx = e^nx g_n(x, x) + ∫₀^x g_n'(x,t) e^nt dt, where the ' denotes the partial derivative wrt the first variable. But g_n(x, x) = 1, and g_r'(x, t) = g_r-1(x, t), so f '(x) = e^nx + ∫₀^x g_n-1(x, t) e^nt dt. Hence by an easy induction f ⁽ⁿ⁾(x) = e^nx( 1 + n + n² + ... + n^n-1 ).

4th Putnam 1941

© John Scholes
jscholes@kalva.demon.co.uk
15 Sep 1999