# Continuity of $\sqrt[\leftroot{0}\uproot{0}3]{\sin(\theta)}$ on interval $[-\pi, \pi]$

I'm doing a problem where my task is to determine whether a given function is continuous, piecewise continuous or piecewise smooth on interval $[-\pi, \pi]$. According to my book the function:

$$\sqrt[\leftroot{0}\uproot{0}3]{\sin(\theta)}$$

is continuous on that interval. Here is a graph of the function:

I can see that the function is continuous on the interval $[0, \pi]$ and not defined on the interval $(-\pi, 0)$. I have understood that continuity means that we can draw the graph "without lifting the pencil" at any point. But if I would draw the function $\sqrt[\leftroot{0}\uproot{0}3]{\sin(\theta)}$ on interval $[-\pi, \pi]$ then I would not have any gap on the interval $[0, \pi]$, but I would have a gap between the point $-\pi$ and the interval $[0, \pi]$. So is this then continuous or not on the interval $[-\pi, \pi]$?

• You only look at continuity on the interval $[0,\pi]$, and the point $-\pi$ is outside this interval. Mar 6, 2014 at 7:47
• +1 @Svinepels Thank you for your help. So the function is NOT then continuous on the interval $[-\pi, \pi]$? Because the question is: Is the given function continuous n the interval $[-\pi, \pi]$? =) Mar 6, 2014 at 7:48
• Any function is not continuous where it is undefined. Mar 6, 2014 at 7:49
• Obviously, the extension of the cube root as in $\sqrt[3]{x}=\text{sign}(x)\sqrt[3]{|x|}$ is intended in the task. So that $\sqrt[3]{-8}=-2$. -- And the function is Hölder-continuous with Hölder index $1/3$. Mar 6, 2014 at 7:51
• Your graph is not correct, since $\sin$ is defined on $\mathbb R$ and $\sqrt[3]{x}$ is defined on $\mathbb R$, so the composition $f(\theta)=\sqrt[3]{\sin(\theta)}$ is defined on $\mathbb R$. For example, $f(-\frac{\pi}{2})=\sqrt[3]{-1}=-1$. In addition, the claim that a continuous function is "drawn without lifting the pencil" (which is by the way, not a mathematical statement) is true only if f is continuous and defined on an interval. Mar 6, 2014 at 7:53

The cube root function is continuous in ${\Bbb R}$, $\sin$ is continuous in ${\Bbb R}$. Your function is continuous because is composition of two continuous functions.
• By the way I'm curious, I have tried four different online graphers and they all give the same graph. I have used as my input the function $\sin(x)^{1/3}$, shouldn't this be the cubic root of $\sin(x)$? Mar 6, 2014 at 8:07
• Fractional powers of negative numbers are problematic for numerical computing. The computer can't tell the difference between $\alpha = 1/3$ and $\beta = \frac {33333333333333333}{100000000000000000}$. But a negative number to the power $\beta$ should be imaginary since $\beta$ has an even denominator. Mar 6, 2014 at 8:13
• On the other hand, for symbolic computation, many computer algebra systems prefer the "principal branch" of non-integer powers, defined by $z^\alpha = \exp(\alpha \text{Log}(z))$. Thus in Maple, $(-1)^{1/3}$ is not $-1$ but $e^{i\pi/3}$. Mar 6, 2014 at 8:16