Check the proof that f(x)=cos⁡({x})⋅cos⁡({3x}) is not periodicIf the function is periodic then:f(x)=f(x+T)cos⁡(x)⋅cos⁡(3x)=cos⁡(x+T)⋅cos⁡(3(x+T))Consider the function at 0:cos⁡(T)⋅cos⁡(3T)=1But this equation has only one solution at T=0 which contradicts the initial assumption that there exists a positive period.Or the other way:cos⁡(T)⋅cos⁡(3T)=1Let cos⁡(T)=1 and cos⁡(3T)=1, henceT=2πm3T=2πnSubstituting T in the second equation gives:2πm3=2πn3=nmBut m,n∈N and 3∈{R∖Q} which gives a contradiction.

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Check the proof that f(x)=cos⁡({x})⋅cos⁡({3x}) is not periodicIf the function is periodic then:f(x)=f(x+T)cos⁡(x)⋅cos⁡(3x)=cos⁡(x+T)⋅cos⁡(3(x+T))Consider the function at 0:cos⁡(T)⋅cos⁡(3T)=1But this equation has only one solution at T=0 which contradicts the initial assumption that there exists a positive period.Or the other way:cos⁡(T)⋅cos⁡(3T)=1Let cos⁡(T)=1  and  cos⁡(3T)=1, henceT=2πm3T=2πnSubstituting T in the second equation gives:2πm3=2πn3=nmBut  m,n∈N and 3∈{R∖Q} which gives a contradiction.

TonysennY2cp · Accepted Answer

As an alternative, we have thatcos⁡(x)⋅cos⁡(3x)=12cos⁡(x(3−1))+12cos⁡(x(3+1))and&amp;nbsp;cos⁡(x(3−1))&amp;nbsp;has period&amp;nbsp;2π3−1&amp;nbsp;while&amp;nbsp;cos⁡(x(3+1))+&amp;nbsp;has period&amp;nbsp;2π3+1and&amp;nbsp;∃k∈ℤ2π3−1=k2π3+1⇔3+13−1=k⇔(3+1)2=2k

Check the proof that \(\displaystyle{f{{\left({x}\right)}}}={\cos{{\left({\left\lbrace{x}\right\rbrace}\right)}}}\cdot{\cos{{\left({\left\lbrace\sqrt{{{3}}}{x}\right\rbrace}\right)}}}\) is not

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