11th Putnam 1951

R is the reals. Find an example of functions f, g : R → R, which are differentiable, not identically zero, and satisfy (f/g)' = f '/g' .

Solution

Suppose we take f = gh. Then we require h' = h + g/g' h'. If we take g/g' = 2, then h' = -h. So this suggests f(x) = e^-x/2, g(x) = e^x/2. Checking, we see that that works.

11th Putnam 1951

© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002