In this post we make use of Taylor polynomial expansions to efficiently solve one of 2018 Putnam competition problems.

Here is the problem statement.

Let $$f:\Bbb R \to \Bbb R$$ be an infinitely differentiable function satisfying $$f(0) = 0$$ and $$f(1) = 1$$, and $$f(x) \geq 0$$ for all $$x \in \Bbb R$$. Show that there exist a positive integer $$n$$ and a real number $$x$$ such that $$f^{(n)}(x)< 0$$.

We proceed by contradiction, thus assuming that $$f^{(n)}(x) \geq 0$$ for all real $$x$$ and all positive integers $$n$$.

1. Show that if there exists $$c < 0$$ such that $$f(c)>0$$, then the Mean Value Theorem implies the existence of a point $$x \in (c,0)$$ such that $$f'(x) < 0$$. Conclude that it must be $$f(x) = 0$$ for $$x \leq 0$$.
2. Use the above fact, and infinite differentiability, to conclude that $f^{(n)}(0) = 0$ for all $$n \in \Bbb Z^+$$.
3. Apply Taylor’s Theorem in the interval $$[0,1]$$ to show that $f(1) = \frac{f^{(n)}(\eta_n)}{n!},$for some $$\eta_n \in (0,1)$$. Conclude that for each positive integer $$n$$ there is a point $$\eta_n \in (0,1)$$ such that $f^{(n)}(\eta_n) = n!$
4. Use 3. and the fact that, for all real $$x$$, $$f^{(n+1)}(x) \geq 0$$ to prove that $f^{(n)}(1) \geq n!$
5. Apply again Taylor’s Theorem, in the interval $$[1,2]$$ this time, to write $f(2) = \sum_{k=0}^n \frac{f^{(k)}(1)}{k!} + \frac{f^{(n+1)}(\xi_n)}{(n+1)!}$ where we defined $$f^{(0)}(x) = f(x)$$ and where $$\xi_n \in (1,2)$$. By 4. and our hypothesis, we get the contradiction $f(2) \geq n, \ \ \ \forall n \in \Bbb Z^+.$

You might have noticed that there is nothing special in the hypothesis $$f(1) = 1$$. As a further exercise, you could replace this hypothesis with the weaker requirement for $$f$$ to be non-constant. (Hint: show that there must exist $$a>0$$ such that $$f(a) >0$$. Use the auxiliary function $$g(x) = \frac{f(ax)}{f(a)}$$…)