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

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?

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 $$~\pi ~$$.

• Last term will change to $\cot \frac{x}{16}$ @Jean-ClaudeArbaut – cmi Aug 14 at 6:15
• Then what is happening? – cmi Aug 14 at 6:20
• 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 – cmi Aug 14 at 6:35
• Yes. ${}{}{}{}{}$ – Jean-Claude Arbaut Aug 14 at 6:47
• 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 – 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 $$16\pi$$.

• 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 – cmi Aug 14 at 7:05
• @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'. – Sangchul Lee Aug 14 at 7:08
• Yea that is what I was missing..Can you please tell me how continuity would make this possible..@Sangchul Lee – cmi Aug 14 at 7:25
• @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 $f$. – Sangchul Lee Aug 14 at 7:31
• By the way this is not a continuos function...Can you please elaborate little bit more? – 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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• Do you know Dirichilet function? Then what is the fundamental period of dirichilet function by your theorem?@mathdiscussion.com – cmi Aug 14 at 6:44
• 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. – 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 $$T=LCM(2\pi,\pi,2\pi, 4\pi, 8\pi, 16 \pi)=16 \pi$$