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        riddles >> putnam exam (pure math) >> Derivative goes to zero
        
(Message started by: Pietro K.C. on Sep 29^th, 2002, 10:43am)

Title: Derivative goes to zero
Post by Pietro K.C. on Sep 29^th, 2002, 10:43am

Suppose f(t) is a continuously once-differentiable strictly decreasing positive function for all t >= 0 . Must the derivative f'(t) = 0 as t approaches infinity? Why?
// modified post to include problem statement at top - willywutang 5/21/2008

Hint: [hideb]not necessarily.[/hideb]

Further hint: [hideb]for a function g(t) to have zero as a limit at infinity, it means that, given any r > 0, there exists an M > 0 such that |g(t)| < r for *EVERY* t > M. [/hideb]

Answer:[hideb]

consider the function f(t) = 1 + 1/(t+1); its derivative, f'(t) = -1/(t+1)², goes to zero as t->oo. But we can make that change without altering the overall behavior that f(t) > 0, and that is the main idea.

Suppose we pick a family of intervals (a_n,b_n) defined as:

a_n = 2ⁿ,
b_n = a_n + 2^-n,

and modify f inside them. Intuitively, even if we add -1 to f '(t) the intervals and change the value of f(t) at the right endpoint to make it still continuous, f(t) will remain positive for all t, because f(b_n) will be changed to f(b_n) - (1/2 + ... + 1/2ⁿ) > 0 since f(t) > 1 for all t.

So that is what we do. We define the function F(t) on the nonnegative reals as:

F(t) = f(t) if 0 <= t <= a₁
F(t) = f(t) - 1 + 1/2ⁿ⁺¹ if b_n <= t <= a_n+1

The rest of the argument is just mechanic, since it amounts to finding the appropriate modification of f inside the intervals that will retain continuity and differentiability. Since we know the value o F at a_n and b_n, plus the required value of the derivatives (which are less than 1) at these points, this amounts to fitting an appropriate cubic, which I shall NOT do, since I have no easy access to MATLAB or Mathematica right now. :)

Then, by construction, F is continuous, and by its definition we know that its derivative must be at least piecewise continuous. However, we constructed it so that it was in fact continuous. And by the mean value theorem, inside each interval (a_n,b_n) there must exist a point c_n with F'(c_n) < -1. Therefore we have a counterexample:

- F is continously once-differentiable;
- F is strictly decreasing for t >= 0;
- for every M > 0 there exists some t such that F'(t) < -1, so lim F'(t) is not zero.

[/hideb]