What is the fundamental period of the function $$ f(x) = \sin x + \tan x + \tan\frac{x}{2} + \tan\frac{x}{4} + \tan\frac{x}{8} + \tan\frac{x}{16}~ .$$

I know that $16\pi$ is one period but how can I determine the fundamental period?

Can anyone please help me to find out it's fundamental period?

My friend was telling me that it's fundamental period will also be $~16\pi~$. Because $~16\pi~$ is the L.C.M of all periods of the periodic functions in the expression.

But I can not understand this argument because the well known function $~|\sin x | + |\cos x|~$ is a periodic function with period $~\frac{\pi}{2}~$ where as $~|\sin x |~$ and $~|\cos x|~$ are of period $~\pi ~$.

  • $\begingroup$ Last term will change to $\cot \frac{x}{16}$ @Jean-ClaudeArbaut $\endgroup$ – cmi Aug 14 at 6:15
  • $\begingroup$ Then what is happening? $\endgroup$ – cmi Aug 14 at 6:20
  • $\begingroup$ Ohh sorry.. I did not see properly.. It will be $\infty$ which is different from $f(0)$ .So $8\pi$ Can not be fundamental period.Thus similarly for any number in your list which is less than $16\pi$ can not be the fundamental period.. Is this what you are trying to say?@Jean-ClaudeArbaut $\endgroup$ – cmi Aug 14 at 6:35
  • $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$ – Jean-Claude Arbaut Aug 14 at 6:47
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    $\begingroup$ I am having a network issue.. I was trying to edit the post..But it asked to change the title. Then I saw that a new post had been posted already.. I was trying to delete that.. But I could not..@GerryMyerson $\endgroup$ – cmi Aug 14 at 7:38

It is easy to check that $f(x + 16\pi) = f(x)$ for all $x \in \mathbb{R}$. Since $f$ is continuous on its domain (or even continuous as function $\mathbb{R} \to \mathbb{R}\cup\{\infty\}$) and non-constant, it follows that the fundamental period of $f$ is of the form $16\pi/n$ for some positive integer $n$.

Now if $n$ is a positive integer for which $16\pi/n$ is a period of $f$, then we must have $f(16\pi/n) = 0$.

  • If $n > 32$, then it is easy to see that each summand of $f(16\pi/n)$ is positive, and so, $f(16\pi/n) > 0$.

  • So it suffices to check that $f(16\pi/n) \neq 0$ for each $2 \leq n \leq 32$. This is the trickiest part, and to be honest, I do not see any clear argument for this. (Although we can easily remove $n = 2, 4, 8, 16, 32$ out of options, all the other values still deserve to be investigated.) But any CAS is capable of computing those values, and it turns that none of them are zero.

  • Therefore the only possible choice is $16\pi$.

  • $\begingroup$ So continuity is the main thing. If the function was not continuous then , we could not have said that the fundamental period will belong to [$\frac{16\pi}{n}$:$n \in \mathbb N$ ]..Am i right?@Sangchul Lee $\endgroup$ – cmi Aug 14 at 7:05
  • $\begingroup$ @cmi, That is right. Continuity is one such condition that prevents from having 'infinitesimal periods' (or more precisely, arbitrarily small periods), thus enabling the notion of 'fundamental period'. $\endgroup$ – Sangchul Lee Aug 14 at 7:08
  • $\begingroup$ Yea that is what I was missing..Can you please tell me how continuity would make this possible..@Sangchul Lee $\endgroup$ – cmi Aug 14 at 7:25
  • $\begingroup$ @cmi, For any function $f$ on $\mathbb{R}$, it is easy to check that the set of all periods $$P=\{a\in\mathbb{R}:f(x+a)=f(x)\text{ for all }x\in\mathbb{R}\}$$ forms an additive subgroup of $\mathbb{R}$. Then as we can confirm from this posting, we have a trichotomy: (1) $G=\{0\}$, (2) $G$ is dense in $\mathbb{R}$, or (3) $G$ is of the form $G=T\mathbb{Z}$ for some $T>0$. But if $f$ is periodic, non-constant and continuous, then neither of (1) nor (2) occurs, and so, only (3) is possible. Then of course, $T$ is the fundamental period of $f$. $\endgroup$ – Sangchul Lee Aug 14 at 7:31
  • $\begingroup$ By the way this is not a continuos function...Can you please elaborate little bit more? $\endgroup$ – cmi Aug 14 at 7:31

In the given function 16π is the fundamental period. The reason is if "T " is one of period then $ \frac{T}{n} $ where n$ {\in N } $ can be a fundamental period if $\\ f(x\,+\,T)\,=f(x) $ . Generally it happens in function which transform from one to the other given e.g |Sin((π/2)+x)| = | Cos x | and vise versa there fundamental period get refused to π/2.

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  • $\begingroup$ Do you know Dirichilet function? Then what is the fundamental period of dirichilet function by your theorem?@mathdiscussion.com $\endgroup$ – cmi Aug 14 at 6:44
  • $\begingroup$ Remember that I am talking with respect to the given problem only other wise there is only one definition of periodic function if there exist any T>0 such that f(x+T)=f(x) then f(x) periodic and least such positive T is fundamental period. This hold for all functions except constant function in which case any real number is one of its period whereas fundamental period is not well defined. $\endgroup$ – mathsdiscussion.com Aug 14 at 7:17

If $T_1,T_2,T_3,...$ are the periods of individually added functions and if the ratios of these periods are rational the period of the the whole function is given by the $LCM(T_1,T_2,T_3...)$ If $T$ is period of $g(x)$, the period of $g(ax)$ is $T/a$. So the period of the whole given function is $$T=LCM(2\pi,\pi,2\pi, 4\pi, 8\pi, 16 \pi)=16 \pi$$


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