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Anyone ever come across a real non-textbook example of a graph with a hole in it?

In Precalc, you get into graphing rational expressions, some of which reduce to a non-rational. The cancelled factors in the denominator still identify discontinuity, yet can't result in vertical asymptotes, but holes.

Thanks!

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    $\begingroup$ Do you mean that $\frac{2x}{x} =2$ if $x\neq 0$ and has a 'hole' at $x=0$? $\endgroup$
    – S.D.
    Commented Nov 30, 2011 at 13:58
  • $\begingroup$ Well, that equation is not really a graph in the form y=f(x). I am more talking visually, when you graph some rational equation, and you get a hole from the cancelled factors that can't be vertical asymptotes. I guess I am just looking for some real life example of a rational expression vs. something contrived in a textbook. Who would ever graph a rational expression? Does that ever come up outside of a precalc textbook? $\endgroup$
    – JackOfAll
    Commented Nov 30, 2011 at 17:32
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    $\begingroup$ At the risk of immodesty, I think my answer about the speed of a car below is the simplest explanation that you'll find, and it ties it instantly to a serious subject. So please don't miss it before you leave this thread behind you. $\endgroup$ Commented Nov 30, 2011 at 17:49
  • $\begingroup$ What would be quite interesting to me is a function that a. has more than one removable discontinuity, and b. "naturally" shows up in applications. $\endgroup$ Commented Nov 30, 2011 at 17:54

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Yes, this actually came up in a loan calculator I was asked to code. Given whole number $n > 0$ and real $r>0$, there is a formula for the geometric sum, $$ 1 + r + r^2 + r^3 + \cdots + r^{n} = \frac{1 - r^{n+1}}{1-r}.$$ This works fine when $r \neq 1$. However, it failed for $r=1$ (program crashes, and all that happy stuff), because there is a hole in the function $S(r) = \frac{1 - r^{n+1}}{1-r}$ at $r=1$. It is a removable discontinuity, but the computer didn't know that. It had to be hard-coded that: $$ 1+ r + r^2 + r^3 + \cdots + r^{n} = n+1, \quad \textrm{when $r = 1$}. $$

Hope this helps!

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  • $\begingroup$ There is an integral formula $\int x^t\,dx = x^{t+1}/(t+1)+C$ for $t\ne -1$ but $\int x^{-1}\,dx = \ln x + C$. I don't know if the OP would call this "real life" though. $\endgroup$
    – GEdgar
    Commented Nov 30, 2011 at 14:59
  • $\begingroup$ Minor point: Could you have just hardcoded a 2 where r=1? $\endgroup$
    – JackOfAll
    Commented Nov 30, 2011 at 17:30
  • $\begingroup$ @JackOfAll: There's also a dependence on $n$. So for example, when $n=4$ and $r=1$, we have $1 + r +r^2 + r^3 + r^4 = 1 + 1 + 1 + 1 + 1 = 5$. $\endgroup$
    – Shaun Ault
    Commented Nov 30, 2011 at 18:52
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The sine cardinal function, $\dfrac{\sin\,x}{x}$. It turns up often enough in signal processing and a number of other applications.

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  • $\begingroup$ However, I think one usually adds to the definition that the value is 1 at 0, which makes the hole disappear. $\endgroup$ Commented Nov 30, 2011 at 15:37
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    $\begingroup$ Yes; that the sine cardinal has to be patched that way makes it similar to Shaun's example, no? $\endgroup$ Commented Nov 30, 2011 at 15:38
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A car goes 60 miles in 2 hours. So 60 miles/2 hours = 30 miles per hour.

But how fast is the car going at a particular instant? It goes 0 miles in 0 hours. There you have a hole!

It is for the purpose of removing that hole that limits are introduced in calculus. Then you can talk about instantaneous rates of change (such as the speed of a car at an instant), which is the topic of differential calculus.

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  • $\begingroup$ How would this translate to a graph? $\endgroup$
    – JackOfAll
    Commented Dec 1, 2011 at 1:33
  • $\begingroup$ @JackOfAll : Look at the curve $y=x^3$ and ask: When $x=1$ and $y=1$, then $y$ is changing how many times as fast as $x$ is changing? That's like the instantaneous speed of a car. You can say the change in $x$ is $x-1$, and the change in $y$ is $x^2-1$, so the change in $y$ over the change in $x$ is $(x^2-1)/(x-1)$. This has a "hole" at $x=1$. If you factor the numerator and do the cancellation, you get $x+1$, and when $x=1$, this is $2$. So at that point, $y$ is changing $2$ times as fast as $x$ is changing. $\endgroup$ Commented Dec 1, 2011 at 1:53
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I guess the derivative of the absolute value (on the reals) comes up in certain "actual" applications. It is undefined at $0$, and no way of plugging the hole makes it continuous. Which doesn't prevent one from defining arbitrarily a value of for instance $0$ at $0$, but it seems better to just leave the hole.

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  • $\begingroup$ One does tend to encounter the sign function (and its close relative the unit step function) in a lot of applications... $\endgroup$ Commented Nov 30, 2011 at 15:48
  • $\begingroup$ @J.M.: But the sign and step functions usually don't have a hole in their definition, just one or more discontinuities. I was trying to answer the question... $\endgroup$ Commented Nov 30, 2011 at 16:03
  • $\begingroup$ In that case, I don't know what the "derivative of the absolute value (on the reals)" looks like to you... $\endgroup$ Commented Nov 30, 2011 at 16:04
  • $\begingroup$ @J.M.: It looks just like the sign function, except that it has a hole in its graph at $0$. $\endgroup$ Commented Nov 30, 2011 at 16:06
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    $\begingroup$ @J.M. It's closely related to the sign function, but it's unhelpful to define an arbitrary value for "the derivative of [any non-differentiable function]" $\endgroup$
    – Random832
    Commented Nov 30, 2011 at 17:11

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