MathDB
Putnam 2000 B3

Source:

September 6, 2011
Putnamtrigonometrylimitcalculusderivativealgebrapolynomial

Problem Statement

Let f(t)=j=1Najsin(2πjt)f(t) = \displaystyle\sum_{j=1}^{N} a_j \sin (2\pi jt), where each aja_j is areal and aNa_N is not equal to 00. Let NkN_k denote the number of zeroes (including multiplicites) of dkfdtk\dfrac{d^k f}{dt^k}. Prove that N0N1N2 and limkNk=2N. N_0 \le N_1 \le N_2 \le \cdots \text { and } \lim_{k \rightarrow \infty} N_k = 2N. [Only zeroes in [0, 1) should be counted.]