The Physics Philes, lesson 29: Math, Discontinuously

The Physics Philes, lesson 29: Math, Discontinuously

In which discontinuity is removed, x-intercepts are determined, and physics is promised.

Happy New Year’s Eve, and I hope you all had a wonderful Newton-mas! Thanks for tolerating my half-assed post from last week. But I’m back and better than (actually, probably the same as) ever! Today I’m going to finish the discussion of continuity, as well as the Intermediate Value Theorem. 

First, let’s talk about removable and nonremovable discontinuity.

Removable discontinuity is discontinuity that can be eliminated by redefining a finite number of points. Uh…what? This is basically just filling in the holes in a function. Still, what? Maybe it will become ore clear with an example. Let’s look at function f(x).

Picture 1Since the limit of this function is f(x) = -9/11, there is point discontinuity at x = -1/3. But, if we redefine f(x) to be f(-1/3) = -9/11, then the limit equals the function value when x = -1/3 and f(x) is continuous there. The new function would then be:

Picture 2So here’s my question: What is the difference between this function, in which we removed the discontinuity, and a completely different function? Does that make sense? This doesn’t seem like a “fixed” function, just a very similar but different function. Can anyone illuminate this?

If there can be removable discontinuity, it stands to reason that there can be nonremovable discontinuity. Nonremovable discontinuity is basically what it sounds like. There is no way to redefine a finite number of points to repair the discontinuity. This type of discontinuity occurs when a function has no general limit at a given x-value. Infinite and jump discontinuity are forms of nonremovable discontinuity. In the previous example, there is a vertical asymptote at x = -4. That means that no general limit exists at that point, thus it’s a nonremovable discontinuity.

This makes more sense to me than removable discontinuity. It’s pretty straight forward. Now it’s time for me to try to explain the Intermediate Value Theorem. Emphasis on try.

Here is the technical definition of this theorem: If a function f(x) is continuous on the closed interval [a,b], then for every real number d between f(a) and f(b) there exists a c between a and b such that f(x) = d.

Ooooooookay. What does this mean? I think it means that between any two x-values on a graph, we can know the graph reaches a certain height. (Is that right? Please tell me if it’s not right.)

This theorem does not claim to say where your function reaches that value or how many times it does so. It only claims that every height a function reaches on a specific x-interval boundary will be output at least once by some x within that interval.

Ah…I’m still not sure I understand. Let’s try to work though this. Explain why function g(x) = x^2 + 3x – 6 must have a root (x-intercept) on the closed interval [1,2].

Let’s just plug in 1 and 2 into the function:

g(1) = x^2 + 3x – 6

= 1^2 + 3(1) – 6

= 1 + 3 – 6

= -2

g(2) = x^2 + 3x – 6

= 2^2 + 3(2) – 6

= 4 + 6 – 6

= 4

If you’ll notice, 0 (which is an x-intercept) lies between -2 and 4. So in a mathy kind of way we’ve shown that the function has an x-intercept on the closed interval [1,2].

That’s it for discontinuity and the Intermediate Value Theorem. Ta-da!

You know, I’ve been focusing on math for a while now. I miss the physics. So next week you can look forward to a discussion of a physics topic to be determined later.

Featured image credit: Mikey Angels

Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.

4 Comments

  1. The difference would be that you eliminated the hole with a piecewise assembly thing. Other than that point, it isn’t different from the function with the hole.

  2. They are two different functions. They are however, only different at one point. If you’re going to remove a discontinuity you’ll need to change the function. (It was discontinuous, so it can’t become continuous without being changed.) The point is that a removable discontinuity requires a tiny amount of change. You can make it continuous by changing the function at only finitely may points. (This is tiny in comparison to the number of points in an interval of any length, which is one of the larger infinities.)

    Here’s some thoughts on the intermediate value theorem. First some pedantry: when you state it, the last line should be f(c)=d, since that equation is giving a condition that the number ‘c’ has to satisfy.

    This theorem is a special case of the fact that if you take a connected space, and feed it into a continuous function, what comes out will still be connected. (Continuous functions might bend or stretch, but they can never tear.)

    So you start with a connected set, such as the interval [a,b] and feed it into a continuous function f. Then you know the following things about what is spit out: it is connected, it is a subset of the real number line, and it contains the numbers f(a) and f(b). From these you can tell immediately that it must contain every number between f(a) and f(b).

    In terms of the graph of the function, the graph of a continuous function can’t get from one height to another without passing through every height in between.

    • “In terms of the graph of the function, the graph of a continuous function can’t get from one height to another without passing through every height in between.”

      This made the Intermediate Value Theorem make sense (at least, some sense) to me. Thank you :)

  3. Another example of a discontinuity that isn’t a jump or an infinite discontinuity is that of the function defined by f(x)=sin(1/x) for x not 0. In this case, the function does not blow-up near 0 (so it’s not an infinite discontinuity), and neither the limit from the left nor the right exist (so it’s not a jump per se). The problem is that we vary x near zero, the value of sin(1/x) varies wildly between -1 and 1, and there is no single value which deserves to be called the limit. Since the limit does not exist, there is no hope of defining f(0)=c to be some value and have the resulting function, f(x)=sin(1/x) if x is not 0 and f(0)=c, be continuous at 0.

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