I have a question on whether the functions following have a discontinuity, and if not, what are the points where two functions meet.

First, the piece wise equation : \begin{align*} f(x)= \begin{cases} \sin(x), &\text{ if } 0 \leq x \leq 2\pi;\\ 0, &\text{ if } x<0 \text{ or } x>2\pi. \end{cases} \end{align*}

What are the points called at $0$ and $2\pi$ ?. Are they a discontinuity, as I expect that they are not since there is no jump.

The second piece wise equation is : \begin{align*} f(x)= \begin{cases} \sin(x), &\text{ if }0 \leq x \leq \pi/2;\\ 0, &\text{ if } x<0;\\ 1, &\text{ if } x>\pi/2. \end{cases} \end{align*}

What is the point called at $\pi/2$? Is it a discontinuity, and if not, how would they be described ?

The reason for this question is that I am examining DSP filters – and I need to ensure that I use the correct terminology when documenting results, and also, to ensure that I understand the theory. There have been statements that the first equation does have discontinuities, but I examined the definition, and I am unsure.

Thanks and regards, Code_X.

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    $\begingroup$ I've only constructed the pictures mentally, so may not have this right. You are right that the first function is continuous. But at the crucial points it's not differentiable. The second function is continuous at $0$ but not differentiable there. At $\pi/2$ it is continuous and differentiable but the derivative is not differentiable. $\endgroup$ – Ethan Bolker Aug 25 '18 at 14:13
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    $\begingroup$ Thanks for confirming that there are no discontinuities. I sort of understood the differentiation aspect - thanks for confirming this too. (Apologies for not using the math text option). Does yourself, or anyone else know what the names are for where the two functions join ??? (thanks). $\endgroup$ – Code_X Aug 25 '18 at 14:27
  • $\begingroup$ @mattstokes I believe you messed up the formulas. $\endgroup$ – Don Thousand Aug 25 '18 at 14:43
  • $\begingroup$ Just changed after taking Matts lead. thanks. $\endgroup$ – Code_X Aug 25 '18 at 14:48
  • $\begingroup$ @Code_X nice! Also put a \ in front of the sin in the formula in order to make it not italicized. $\endgroup$ – Don Thousand Aug 25 '18 at 14:52

I don't know if the points where the components of piece wise functions join have a name, but checking whether they are discontinuities, or non-differentiable points is not difficult.

Lets begin with discontinuities. In order for the point $p$ not to be a discontinuity, $$\lim_{x\to p^+} f(x) = \lim_{x\to p^-} f(x) = f(p)$$

In the case of $p=0$, for your first piece-wise function, since $\sin(x)$ is continuous, $$f(0) = \lim_{x\to0^+} f(x)$$

So all that needs to be shown is that $\lim_{x\to0^-} f(x)=f(0)$, which is obvious since $$\forall x<0, f(x)=0$$

For the second piecewise function, the exact same logic applies since the functions at this point are the same.

You can apply similar logic to the second point joining point of the piece-wise function, but of course, the cases of both functions are a tad bit different there. See how far you can get!

On the other hand, if we find a point to be continuous (i.e., $f(0)$), we can also check whether we can find the derivative of the point, or if the point is a non-differentiable point on the function.

We do that with the following formula :$$f'(p)=\lim_{h\to 0}\frac{f(p+h)-f(p)}{h}$$ If this exists, then it is differentiable. This exists iff $$\lim_{h\to 0^+}\frac{f(p+h)-f(p)}{h}=\lim_{h\to 0^-}\frac{f(p+h)-f(p)}{h}$$

In the case of $p=0$, when $h>0$, $f(p+h)=sin(h)$ and $f(p)=sin(p)=0$, so $$\lim_{h\to 0^+}\frac{f(p+h)-f(p)}{h} = \lim_{h\to 0^+}\frac{\sin(h)}h = 1$$

However, when $h<0$, $f(p+h)= 0$ and $f(p) = 0$, so $$\lim_{h\to 0^-}\frac{f(p+h)-f(p)}{h} = \frac{0-0}h = 0$$ Since these one sided limits are not equal, the function is not differentiable at $p=0$. Again, you can use this logic on the other point where the piece-wise functions join.

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    $\begingroup$ many thanks for the answer, much appreciated. I have upvoted the answer, shame we cannot say thanks too. Thanks for the help on differentiation too :O) $\endgroup$ – Code_X Aug 25 '18 at 15:17
  • $\begingroup$ @Code_X No problem! Nice question $\endgroup$ – Don Thousand Aug 25 '18 at 15:22
  • $\begingroup$ @Code_X If you could give me the accepted answer too (by hitting the green check mark) I would appreciate that! $\endgroup$ – Don Thousand Aug 25 '18 at 19:38
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    $\begingroup$ Apologies- i thought that was the indicator for an up vote. Thanks for your help on this, much appreciated. $\endgroup$ – Code_X Aug 25 '18 at 20:38

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