MHB Define a discontinuous sine function?

[samir]

Hi!

I want to define a sine function that is discontinuous at multiples of $\pi$. The multiplier is to be an integer.

How can I do that?

I am thinking about something like this:

$$f(x)=\begin{cases}sin(x) & x \in \Bbb{R} \\ \text{undefined} & x=n \cdot \pi | n \in \Bbb{Z} \text{ and } x \in \Bbb{R}\end{cases}$$

Is this a valid statement?

Not only does it need to be discontinuous, but also undefined at these multiples. I believe it is the only way to have such a function be discontinuous at those multiples. Let me know if you know of another way.

I tried to graph this on my calculator but it didn't work. I think I need to define the condition for undefined in terms of x. Can I do that?

It seems like my calculator wants me to define $n$ first before I use it. But how do you define a variable that holds an element of the infinite set of integers? I'm not so sure. If I could set the data type to integers, that would get me half way there. Since it is a programmable calculator it should be possible. This is more of a product usage question, so I will leave that off for now. What I really want to know is if my definition of such function above is mathematically correct? I can deal with the calculator and graphing later.

 

[Opalg]

A function can only be continuous or discontinuous at points where it is defined, because the definition of continuity/discontinuity involves the value of the function at that point. So your function need to be defined at multiples of $\pi$, but its value there should be different from the value of the sine function at those points (which is $0$). So you should amend your definition to say that $f(x) = \sin x$ when $x$ is not a multiple of $\pi$, and $f(x) = 1$ (say, or $17.629$, or anything at all other than $0$), when $x$ is a multiple of $\pi$.

 

[Evgeny.Makarov]

I would write
$$
f(x)=\begin{cases}\sin(x), & x \ne\pi n\text{ for any }n\in\mathbb{Z}\\ \text{undefined}, & \text{otherwise}\end{cases}
$$
The conditions in the two branches should be mutually exclusive. The conditions $x\in\Bbb R$ and $x=\pi n$ for some $n\in \Bbb Z$ are not mutually exclusive. Further, the vertical bar is only used in the set-builder notation, e.g., $\{x\in\Bbb R\mid x>0\}$ (and in this case it is written using the LaTeX command [m]\mid[/m] to create correct spacing on both sides). It does not have a universally accepted meaning of "such that" in other contexts. Finally, it is superfluous to write $x=n \cdot \pi | n \in \Bbb{Z} \text{ and } x \in \Bbb{R}$ because $x=\pi n$ and $n\in\Bbb Z$ imply $x\in\Bbb R$.