### 11th Putnam 1951

**Problem B2**

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