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What is the period of $\sin 2\theta + \sin \frac{\theta}{2}$? The period of the first term is $\pi$ and that of the second is $4\pi$. Does that mean that the period of the whole is $4\pi$?

Simon Kuang
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1 Answers1

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It is equal to the smallest $>0$ generator of $2\mathbf{Z} \cap \frac{1}{2}\mathbf{Z}$ as $\mathbf{Z}$-module. So what is the smallest $a\in\mathbf{Q}_{+}^{\times}$ such that $2\mathbf{Z} \cap \frac{1}{2}\mathbf{Z} = a \mathbf{Z}$ ?

Olórin
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  • How about $2$?? – Simon Kuang Jan 30 '15 at 00:53
  • The Fourier transform of $f(\theta) = \sin(2 \theta) + \sin(\theta/2)$ is $i \sqrt{2 \pi } \delta (1-2 r)+i \sqrt{\frac{\pi }{2}} \delta (r-2)-i \sqrt{\frac{\pi}{2}} \delta (r+2)-i \sqrt{2 \pi } \delta (2 r+1)$, which gives you the answer immediately. – David G. Stork Jan 30 '15 at 00:54
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    @DavidG.Stork Is this a joke or ? – Olórin Jan 30 '15 at 00:57
  • @DavidG.Stork Do you mean $f(\theta) = sin (2\theta)\ldots$? – Simon Kuang Jan 30 '15 at 00:58
  • @SimonKuang So your guess is $2$ ? Can you prove it then ? – Olórin Jan 30 '15 at 00:58
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    @RobertGreen: I was following your solution style. – David G. Stork Jan 30 '15 at 01:01
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    Simon asked a relatively simple question, and I think it's a question that is best answered without the use of Fourier transforms or complicated terms like "generator" or "Z-module." This is a question that a person in high school trigonometry should be answering. Also, David, I don't think your sarcasm helps the problem here. – nukeguy Jan 30 '15 at 01:01
  • @nukeguy Note that OP asked THE period, and not A period. For A period, I would have answered $4\pi$ with this proof : "check it !". But to show that $4\pi$ is THE period, the only way to do it is, definition of THE period (smallest > 0 generator of the group of "periods"), to do what I have done. A period of $g$ is a $T$ s.t. $g(x+t) = g(x) \forall x$, and such $T'$ form a sub-group of $\mathbf{R}$, etc etc... and THE period of $g$ will be the generator of this subgroup, if it has one... – Olórin Jan 30 '15 at 01:07
  • From the Fourier transform we can see that the lowest frequency corresponds to $r = 2$, and others are integral portions of that. Thus the period $T$ is $2 \times 2\pi = 4\pi$. – David G. Stork Jan 30 '15 at 01:12
  • @DavidG.Stork This is not a proof, but anyway, that's not the point. – Olórin Jan 30 '15 at 01:14
  • @SimonKuang BTW you was right for $2$, so that you get that the period you're looking for is... $2 \times (2\pi)$. ;-) Beware of something, saying "the period" and "a period" of a periodic function are very different things, as I'm sure you'll see later in your studies. – Olórin Jan 30 '15 at 01:16