Olympiad Maths Prep

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Problem 1040

National olympiad, first round
Algebra Difficulty 6.0 Prove it

[ Periodicity and Aperiodicity ] [ Examples and Counterexamples. Constructions ]

Are there two functions with the smallest positive periods of 2 and 6, respectively, such that their sum has the smallest positive period of 3?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let, for example, f(x)=cos2/3πx+cosπx,g(x)=cosπxf(x)=\cos 2 / 3 \pi x+\cos \pi x, g(x)=-\cos \pi x, then their sum h(x)h(x) is cos2/3πx\cos 2 / 3 \pi x.

The smallest positive period of the function g(x)g(x) is 2π:π=22 \pi: \pi=2, and the smallest positive period of the function h(x)h(x) is 2π:2π/3=32 \pi: 2 \pi / 3=3.

We will prove that the smallest positive period of the function f(x)f(x) is 6. Indeed, the number 6 is a multiple of 3 and 2, so it is both a period of the function h(x)h(x) and a period of the function g(x)g(x), which means it is also a period of their difference.

Suppose that a positive number TT is a period of the function f(x)f(x). Then

cos2/3πT+cosπT=cos0+cos0=2cos2/3πT=cosπT=1cos2/3πT=cosπT=12/3πT=2πk,πT=\cos ^{2} / 3 \pi T+\cos \pi T=\cos 0+\cos 0=2 \Leftrightarrow \cos ^{2} / 3 \pi T=\cos \pi T=1 \Leftrightarrow \cos ^{2} / 3 \pi T=\cos \pi T=1 \Leftrightarrow 2 / 3 \pi T=2 \pi k, \pi T= 2πn,k,nN2 \pi n, k, n \in \mathbf{N}. Therefore,

T=3k=2nT=3 k=2 n. This means that TT is an integer that is a multiple of both 2 and 3, so it is at least 6.

## Answer

There exist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.