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 Post subject: Non periodic function!
PostPosted: Thu Jul 07, 2016 12:16 pm 

Joined: Wed Nov 11, 2015 12:47 pm
Posts: 13
Prove that \(\sin\left(x^3\right)\) is a non-periodic function.


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PostPosted: Thu Jul 07, 2016 12:17 pm 

Joined: Mon Nov 09, 2015 11:52 am
Posts: 76
Location: Limassol/Pyla Cyprus
Let \(f(x) = \sin(x^3)\), suppose it is periodic, and let \(T\) be a period of \(f\). Since \(f\) is differentiable it follows that \(T\) is a period of \(f'\) as well. Observe that \(f'(x) = 3x^2 \cos(x^3)\) is continuous and so is bounded on \([0,T]\). Since \(T\) is a period of \(f'\), it follows that \(f'\) is bounded. But this is absurd as \(f'(a_n) \to \infty\) when \(a_n = \sqrt[3]{2n\pi}\). So \(f\) is not periodic.


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PostPosted: Thu Jul 07, 2016 12:18 pm 

Joined: Tue Nov 10, 2015 5:35 pm
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After Demetres' answer, let us generalize this problem:

Find all polymomials \(p(x)\) such that \(\sin (p(x))\) is a periodic function.


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PostPosted: Thu Jul 07, 2016 12:19 pm 

Joined: Mon Nov 09, 2015 11:52 am
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Location: Limassol/Pyla Cyprus
I claim that \( \sin(p(x))\) is periodic if and only if \(p\) is linear (or constant). The "if" part is obvious. For the "only if" part we work similarly as my answer above. If it was periodic then its derivative \(p'(x)\cos(p(x))\) would also be periodic and since it is continuous it would also be bounded. But this is not the case. For each \(K,\) since \(p'\) is a non-constant polynomial we can find \(N\) such that \(|p'(x)| \geqslant K\) for each \(x \geqslant N\). Furthermore we can find \(x \geqslant N\) such that \(|\cos(p(x))| = 1\) and therefore for that \(x\) we would have \(|p'(x)\cos(p(x))| \geqslant K\). This contradicts the boundedness of the derivative and therefore the periodicity of the function.


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