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 \(n\).

- 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 \(x \leq 0\). - Use the above fact, and infinite differentiability, to conclude that \[f^{(n)}(0) = 0\] for all \(n \in \Bbb Z^+\).
- 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 \[f^{(n)}(\eta_n) = n!\]
- Use 3. and the fact that, for all real \(x\), \(f^{(n+1)}(x) \geq 0\) to prove that \[f^{(n)}(1) \geq n!\]
- 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 \(g(x) = \frac{f(ax)}{f(a)}\)…)*