Putnam 2000, Problem B-3

Let $f(t) = \sum_{j=1}^N a_j \sin(2\pi jt)$, where each $a_j$ is real and $a_N$ is not equal to 0$. Let $N_k$ denote the number of zeroes (including multiplicities) of $\frac{d^k f}{dt^k}$. Prove that $N_0 \leq N_1 \leq N_2 \leq \cdots$ and $\lim_{k \to \infty} N_k = 2N$.  
[Editorial clarification: only zeroes in $[0,1)$ should be counted.]

Solution:
Put \(f_k(t) = \frac{d^k}{dt^k}\). Recall Rolle's theorem: if \(f(t)\) is differentiable, then between any two zeroes of \(f(t)\) there exists a zero of \(f'(t)\). This also applies when the zeroes are not all distinct: if \(f\) has a zero of multiplicity \(m\) at \(t = x\), then \(f'\) has a zero of multiplicity at least \(m - 1\) there.