7th Putnam 1947

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Problem B1

Let R be the reals. f : [1, ∞) → R is differentiable and satisfies f '(x) = 1/(x2 + f(x)2) and f(1) = 1. Show that as x → ∞, f(x) tends to a limit which is less than 1 + π/4.

 

Solution

Clearly f '(x) is always positive. But f(1) = 1, so f(x) > 1 for all x. Hence f '(x) < 1/(x2 + 1) for all x. Hence f(x) = 1 + ∫1x f '(t) dt < 1 + 1 + ∫1x 1/(1 + t2) dt = 1 + (tan-1t)|1x = 1 + tan-1x - π/4 < 1 + π/2 - π/4 = 1 + π/4. Since f '(x) is positive, f(x) is monotone increasing. It is bounded above by 1 + π/4, so it must tend to a limit less than 1 + π/4.

 


 

7th Putnam 1947

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