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What is the value of n $\in \mathbb{Z} $ for which the function $\displaystyle f(x) = \frac{\sin nx} { \sin \biggl( \frac{x}{n} \biggr) } \text { has } 4\pi $ as period?

Also could it be possible to solve this if we need $x\pi$ as period ?I am interested in learning the general approach for this particular type of the problem.

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Edit your question again please – Bryan Yocks Nov 30 '10 at 19:44
@ Bryan Yocks : Done! – Quixotic Nov 30 '10 at 19:45
Thanks – Bryan Yocks Nov 30 '10 at 19:46
up vote 3 down vote accepted

With $n=2$, $\sin(2x)$ has period $\pi$ and $\sin(x/2)$ has period $4\pi$ so their ratio must have a period of $4\pi$ since the latter period is an integral multiple of the former.

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so their ratio must have a period of 4π Could you elaborate how? I am not sure how to find the period of the function of the form $f(x)\cdot f(z)$ or $f(z)/f(x)$. – Quixotic Nov 30 '10 at 19:54
We could also add $n= -2 \textrm{ and } n = \pm 1.$ – Derek Jennings Nov 30 '10 at 19:55
@Debanjan: The first numerator repeats it's output values after every $\pi$ units. Hence, it also repeats after every $4\pi$ units. The denominator repeats it's values every $4\pi$ units. So their ratio must repeat after every $4\pi$ units. In general if you have two functions, one with period $m$ and the other with period $n$ their product or ratio (eseentially any function you can manufacture out of only the two) will have period which is lcm$(m,n)$ – Timothy Wagner Nov 30 '10 at 19:58
@Derek: Yes. I thought the OP just needed one value. – Timothy Wagner Nov 30 '10 at 19:58
@Debanjan: First: When you write "$f(x)/g(z)$", you are writing a function of two variables. Is that what you mean? I sincerely doubt it. Second: don't think, check! It's simple enough to plug in and check. – Arturo Magidin Nov 30 '10 at 21:17

You want $$\frac{\sin n(x + 4\pi)}{\sin \frac{x + 4\pi}{n}} = \frac{\sin nx}{\sin \frac{x}{n}}.$$

This is equivalent with $$\sin \frac{x + 4\pi}{n} = \sin \frac{x}{n}.$$

Therefore $\frac{x}{n} = \frac{x + 4\pi}{n} + 2k\pi$ or $\frac{x}{n} = \pi - \frac{x + 4\pi}{n} + 2k\pi$ for some $k \in \mathbb{Z}$. In the first case $x = x + 4 \pi + 2k \pi n$ and thus $n = \frac{4}{2k}$. In the second case $x = \pi n - x - 4\pi + 2k\pi n$ and thus $n = \frac{2x + 4\pi}{2k\pi + \pi}$, which is impossible since this should hold for every $x \in \mathbb{R}$.

Thus $n = \pm 1$ or $\pm 2$.

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