Improper integrals debate

Here’s a simple Calc 1 problem:

Evaluate  $\int_{-1}^1 \frac{1}{x}dx$

Before you read any of my own commentary, what do you think? Does this integral converge or diverge?

image from illuminations.nctm.org

Many textbooks would say that it diverges, and I claim this is true as well. But where’s the error in this work?

$\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_a^{1}\frac{1}{x}dx\right]$

$= \lim_{a\to 0^+}\left[\ln(a)-\ln(a)\right]=\boxed{0}$

Did you catch any shady math? Here’s another equally wrong way of doing it:

$\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_{2a}^{1}\frac{1}{x}dx\right]$

$= \lim_{a\to 0^+}\left[\ln(a)-\ln(2a)\right]=\boxed{\ln{\frac{1}{2}}}$

This isn’t any more shady than the last example. The change in the bottom limit of integration in the second piece of the integral from a to 2a is not a problem, since 2a approaches zero if does. So why do we get two values that disagree? (In fact, we could concoct an example that evaluates to ANY number you like.)

Okay, finally, here’s the “correct” work:

$\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^-}\left[\int_{-1}^{a}\frac{1}{x}dx\right]+\lim_{b\to 0^+}\left[\int_b^{1}\frac{1}{x}dx\right]$

$= \lim_{a\to 0^-}\left[\ln|a|\right]+\lim_{b\to 0^+}\left[-\ln|b|\right]$

But notice that we can’t actually resolve this last expression, since the first limit is $\infty$ and the second is $-\infty$ and the overall expression has the indeterminate form $\infty - \infty$. In our very first approach, we assumed the limit variables $a$ and $b$ were the same. In the second approach, we let $b=2a$. But one assumption isn’t necessarily better than another. So we claim the integral diverges.

All that being said, we still intuitively feel like this integral should have the value 0 rather than something else like $\ln\frac{1}{2}$. For goodness sake, it’s symmetric about the origin!

In fact, that intuition is formalized by Cauchy in what is called the “Cauchy Principal Value,” which for this integral, is 0. [my above example is stolen from this wikipedia article as well]

I’ve been debating about this with my math teacher colleague, Matt Davis, and I’m not sure we’ve come to a satisfying conclusion. Here’s an example we were considering:

If you were to color in under the infinite graph of $y=\frac{1}{x}$ between -1 and 1, and then throw darts at  the graph uniformly, wouldn’t you bet on there being an equal number of darts to the left and right of the y-axis?

Don’t you feel that way too?

(Now there might be another post entirely about measure-theoretic probability!)

What do you think? Anyone want to weigh in? And what should we tell high school students?

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**For a more in depth treatment of the problem, including a discussion of the construction of Reimann sums, visit this nice thread on physicsforums.com.

Great NCTM problem

Yesterday I presented this problem from NCTM’s facebook page:

Solve for all real values of $x$:

$\frac{(x^2-13x+40)(x^2-13x+42)}{\sqrt{x^2-12x+35}}$

Don’t read below until you’ve tried it for yourself.

Okay, here’s the work. Factor everything.

$\frac{(x-8)(x-5)(x-7)(x-6)}{\sqrt{(x-5)(x-7)}}=0$

Multiply both sides by the denominator.

$(x-8)(x-5)(x-7)(x-6)=0$

Use the zero-product property to find $x=5,6,7,8$. Now check for extraneous solutions and find that $x=5$ and $x=7$ give you $\frac{0}{0}\neq 0$ and $x=6$ gives $x=\frac{0}{\sqrt{-1}}=\frac{0}{i}=0$. This last statement DOES actually hold for $x=6$ but we exclude it because it’s not in the domain of the original expression.The original expression has domain $(-\infty,5)\cup(7,\infty)$. We could have started by identifying this, and right away we would know not to give any solutions outside this domain. The only solution is $x=8$.

Does this seem problematic? How can we exclude $x=6$ as a solution when it (a) satisfies the equation and (b) is a real solution? This is why we had such a lively discussion.

But this equation could be replaced with a simpler equation. Here’s one that raises the same issue:

Solve for all real values of x:

$\frac{x+5}{\sqrt{x}}=0$

Same question: Is $x=-5$ a solution? Again, notice that it DOES satisfy the equation and it IS a real solution. So why would we exclude it?

Of course a line is drawn in the sand and many people fall on one side and many fall on the other. It’s my impression that high-school math curriculum/textbooks would exclude $x=-5$ as a solution.

Here’s the big question: What does it mean to “solve for all real values of x“? Let’s consider the above equation within some other contexts:

Solve over $\mathbb{Z}$:

$\frac{x+5}{\sqrt{x}}=0$

Is $x=-5$ a solution? No, I think we must reject it. If we try to check it, we must evaluate $\frac{0}{\sqrt{5}}$ but this expression is undefined because $\sqrt{5}\notin\mathbb{Z}$. Here’s another one:

Solve over $\mathbb{Z}_5$:

$\frac{x+5}{\sqrt{x}}=0$

Is $x=-5$ a solution? No. Now when we try to check the solution we get $\frac{0}{\sqrt{5}}=\frac{0}{\sqrt{0}}=\frac{0}{0}$ which is undefined.

The point is that, if we go back to the same question and ask about the solutions of $\frac{x+5}{\sqrt{x}}=0$ over the reals, and we check the solution $x=-5$, we must evaluate $\frac{0}{\sqrt{-5}}$ which is undefined in the reals.[1]

So in the original NCTM question, we must exclude $x=6$ for the same reason. When you test this value, you get $\frac{0}{i}$ on the left side which YOU may think is 0. But this is news to the real numbers. The reals have no idea what $\frac{0}{i}$ evaluates to. It may as well be $\frac{0}{\text{moose}}$.

There’s a lot more to say here, so perhaps I’ll return to this topic another time. Special thanks to all the other folks on facebook who contributed to the discussion, especially my dad who helped me sort some of this out. Feel free to comment below, even if it means bringing a contrary viewpoint to the table.

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[1] This last bit of work, where we fix the equation and change the domain of interest touches on the mathematical concept of algebraic varieties, which I claim to know *nothing* about. If someone comes across this post who can help us out, I’d be grateful!

For Reals

A few people have pointed me to this mathy web comic:

Thanks to smbc-comics for a great mathy web comic! (http://www.smbc-comics.com/comics/20120517.gif)

I’m not sure how often discrete mathematics uses the phrase “for reals”….I would think “for natural numbers” would be more appropriate, don’t you?

Inverse functions and the horizontal line test

I have a small problem with the following language in our Algebra 2 textbook. Do you see my problem?

Horizontal Line Test

If no horizontal line intersects the graph of a function f more than once, then the inverse of f is itself a function.

Here’s the issue: The horizontal line test guarantees that a function is one-to-one. But it does not guarantee that the function is onto. Both are required for a function to be invertible (that is, the function must be bijective).

Example. Consider $f:\mathbb{R}\to\mathbb{R}$ defined $f(x)=e^x$. This function passes the horizontal line test. Therefore it must have an inverse, right?

$f(x)=e^x$

Wrong. The mapping given is not invertible, since there are elements of the codomain that are not in the range of $f$. Instead, consider the function $f:\mathbb{R}\to (0,\infty)$ defined $f(x)=e^x$. This function is both one-to-one and onto (bijective). Therefore it is invertible, with inverse $f^{-1}:(0,\infty)\to\mathbb{R}$ defined $f(x)=\ln{x}$.

This might seem like splitting hairs, but I think it’s appropriate to have these conversations with high school students. It’s a matter of precise language, and correct mathematical thinking. I’ve harped on this before, and I’ll harp on it again.

Happy Birthday to Bolzano

In honor of Bernard Bolzano, I share this old-but-good rap, from Dr. Sawin of Fairfield University: