Maths Olympiad Prep

Library / Putnam /26 of 60

Putnam

Algebra Difficulty 7.8 National Olympiad, round 2 Find the answer

Let dnd_n be the determinant of the n×nn \times n matrix whose entries, from
left to right and then from top to bottom, are cos⁡1,cos⁡2,…,cos⁡n2\cos 1, \cos 2, \dots, \cos n^2. Evaluate
lim⁡n→∞dn\lim_{n\to\infty} d_n.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The limit is 00; we will show this by checking that dn=0d_n = 0 for all n≥3n \geq 3.
Starting from the given matrix, add the third column to the first column; this does not change the
determinant. However, thanks to the identity
cos⁡x+cos⁡y=2cos⁡x+y2cos⁡x−y2\cos x + \cos y = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2},
the resulting matrix has the form
(2cos⁡2cos⁡1cos⁡2⋯2cos⁡(n+2)cos⁡1cos⁡(n+2)⋯2cos⁡(2n+2)cos⁡12cos⁡(2n+2)⋯⋮⋮⋱) \begin{pmatrix} 2 \cos 2 \cos 1 & \cos 2 & \cdots \\ 2 \cos (n+2) \cos 1 & \cos (n+2) & \cdots \\ 2 \cos (2n+2) \cos 1 & 2 \cos (2n+2) & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix}
with the first column being a multiple of the second. Hence dn=0d_n = 0.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.