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$\sin^2(x)$ has period $\pi$ but it seems to me $\sqrt{\sin x}$ is not periodic since inside square root has to be positive and when it is negative, it is not defined.

Does it creates problem for periodicity? Can we say square root of the periodic function need to be periodic? Thanks for your help.

enter image description here

This is graph of the $\sqrt{\sin x}$ above.

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    $\begingroup$ Well, it's periodic where it is defined. That's something! $\endgroup$
    – lulu
    Commented Mar 30, 2019 at 13:21
  • $\begingroup$ You could ask the same question of $\tan(x)$, because it's not defined at odd multiples of $\pi/2$. $\endgroup$ Commented Mar 31, 2019 at 13:43

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The function $\sqrt{\sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $\sin x$ is negative, you have $$\sqrt{\sin x} = \mbox{ undefined } =\sqrt{\sin(x+2\pi)}.$$

Or you can extend to the complex numbers and you'll have periodicity everywhere.

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  • $\begingroup$ Now we just need to choose how to define $\sqrt{z}$ in general, but that's doable. $\endgroup$
    – J.G.
    Commented Mar 31, 2019 at 13:21
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$f$ is periodic, if there is a positive real number $p\ $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:

$$f \text{ is periodic }\iff\exists p>0: \forall x \in D(f): x+p \in D(f)\ \land\ f(x) = f(x+p)$$

For the function $g = \sqrt f$ of a periodic function $f$ obviously $$x \in D(g) \iff x + k \in D(g)$$

(because $x \in D(g) \iff f(x) \ge 0 \iff f(x+k)=f(x) \ge 0 $)

and

$$f(x) = f(x+p),$$

so the square root of a periodic function is a periodic one, too.

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  • $\begingroup$ Can we conclude that $f$ and $g$ necessarily have same fundamental period? $\endgroup$
    – S.H.W
    Commented Mar 31, 2020 at 7:38
  • $\begingroup$ @S.H.W, it has neither relation to the question, nor to my answer. The matter was “periodic or not”. $\endgroup$
    – MarianD
    Commented Mar 31, 2020 at 11:19
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If a function $f$ is periodic, then there exists $k\ne0$ such that $f(x)=f(x+k)$ for all $x\in\mathbb{R}$. So if $g$ is its square root, we have $$g(x)=\sqrt{f(x)}=\sqrt{f(x+k)}=g(x+k).$$ And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.

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  • $\begingroup$ Can we conclude that $f$ and $g$ necessarily have same fundamental period? $\endgroup$
    – S.H.W
    Commented Mar 31, 2020 at 7:48
  • $\begingroup$ Yes, I'd say we can. $\endgroup$
    – AugSB
    Commented Mar 31, 2020 at 14:00
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The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do $$ \begin{align} f(x+T) &= \sqrt{\sin(x+T)} \\ &= \sqrt{\sin(x)\cos(T)+\cos(t)\sin(T)} \tag{1} \end{align} $$ We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2\pi$ sec, then Eq.(1) becomes $$ \begin{align} f(x+T) &= \sqrt{\sin(x)\cos(T)+\cos(t)\sin(T)} \\ &= \sqrt{\sin(x)(1)+\cos(t)(0)} \\ &= \sqrt{\sin(x)} \\ &= f(x) \tag{2} \end{align} $$ Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2\pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2\pi=6.28$ sec.

enter image description here

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