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Find a period of the following function. $$x(t) = 2\cos(3\pi t) + 7\cos(9t).$$

After finding the individual time period how to proceed?

Michael Rozenberg
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Sajal
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    What makes you think that's periodic? – dxiv Aug 03 '17 at 07:18
  • cause cos is a periodic function........? – Sajal Aug 03 '17 at 07:20
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    @Sajal The sum of two periodic functions doesn't need to be periodic... – 5xum Aug 03 '17 at 07:23
  • @Sajal That's not enough for an intuition, let alone a proof. In fact, your $x(t)$ isn't periodic, see for example [The sum of two continuous periodic functions is periodic if and only if the ratio of their periods is rational?](https://math.stackexchange.com/questions/1356802/the-sum-of-two-continuous-periodic-functions-is-periodic-if-and-only-if-the-rati). – dxiv Aug 03 '17 at 07:23
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    Dear @Sajal, welcome to Mathematics SE. For the future, include your question into the body, to help people read and understand your problem, and use the mathematics formatting, please. If you need some hint how to do it, have a look [here](https://math.stackexchange.com/help/notation). Moreover, it would be nice from you if you provide also some attempt you already tried to follow. – Ender Wiggins Aug 03 '17 at 07:27
  • The periods of the terms have no common multiple so the period is $\infty$ – lasen H Aug 03 '17 at 07:33
  • @lasenH Thank you sir. – Sajal Aug 03 '17 at 07:40
  • @lasenH `so the period is ∞` Sorry, but that makes no sense. A periodic function has a *finite* period by definition. If you meant to say that $\,x(t)\,$ is not periodic, then that's not the right way to say it. – dxiv Aug 03 '17 at 07:49

1 Answers1

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$T$ names a period of $f$ if $T>0$ and for all $x$ from domain $f$ we need $f(x+T)=f(T)$.

Also, if there is $T'>0$ with previous properties then $T'\geq T$.

Now, let $T$ is a period of our function.

Thus, for all $x$ we have: $$f(x+T)=f(x),$$ where $f(x)=2\cos3\pi x+7\cos9x$ or $$2\cos3\pi(x+T)+7\cos9(x+T)=2\cos3\pi x+7\cos9x.$$ Let $x=0$.

Thus, $$2\cos3\pi T+7\cos9T=9,$$ which gives $\cos3\pi T=1$ and $\cos9T=1$, which gives $$T=\frac{2}{3}k,k\in\mathbb Z$$ and $$T=\frac{2\pi}{9}m, m\in\mathbb Z,$$ which gives $\pi=\frac{3k}{m}$, which is absurd.

Thus, our function has no period.

Michael Rozenberg
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  • This is just amazing.. Very nice explanations.. This is counter-intuitive if you try to plot and deduce from the plot because it seems to be periodic. – Ahmad Bazzi Aug 03 '17 at 08:27