Friday, April 20, 2007

'Rigor Mathis' - A Calculus Paradox or...

Update: I've added another 'paradox' in the comments. With the AP Calculus (BC) Exam looming, AP teachers may want to share this with their students for review.

[The following AP level question is designed for upper level students.]

Why do mathematicians have to be so rigid, um, I mean, rigorous?

Here's an AP Calculus problem brought to my attention this morning by one of our outstanding Calculus teachers, Mr. D. He found this in an AP Review book.

[Rather than play with the symbols, I'll 'write it out']:

The definite integral of sec2(x) from x = 0 to x = 3pi/4 is?
It was multiple choice and the answer given was -1.

I shared this problem with my AP group later on in the morning and I asked them why that answer makes no sense. R.J. immediately replied, "The answer can't be negative since sec2(x) is never negative."
Of course, but let's work it out!

[I intentionally did it incorrectly at first]:
By the Fundamental Theorem, the integral equals tan(3pi/4) - tan(0) = -1!

What's going on here! I was gratified that one of my students recognized that the function sec2(x) has an infinite discontinuity at x = pi/2, so the original integral is improper. When we integrate from 0 to pi/2, then from pi/2 to 3pi/4, and apply the rigorous limit definition of an improper integral, we see that the integral diverges! If anything, the 'area' is infinite or unbounded.

Using these kinds of 'paradoxical' examples and asking students to 'FIND THE ERROR' is a wonderful device many educators use to deepen student understanding of mathematics and demonstrate the need to be rigorous!

Now why isn't the definite integral of 1/x from -1 to 1 equal to zero, since the region in the first quadrant 'clearly cancels' the part in the 3rd quadrant?? Hmmm... I'll bet some of you could explain this and find many other such 'paradoxes'!


Eric Jablow said...

Probably the best analogue is the problem of finding the definite integral of x dx from −∞ to ∞. If you let the lower and upper limits head to −∞ and ∞ at the same rate, you get zero. If you let one limit head to infinity faster than the other, you get something completely different. If you claim that the integral is zero, then cutting the region below the curve apart and rearranging it changes the area.

Now, consider a conditionally convergent series such as the alternating harmonic series, ∑_{n=1}^{∞} (−1)^{n+1}/n. Draw the graph of the appropriate step function whose “area” this should be:

f(x) = 0, x ≤ 0,
f(x) = (−1)^{n+1}/n, if n−1 < x ≤ n.

Does the region between the graph and the x-axis have an area? You can make infinitely-many vertical cuts, and rearrange the cut regions, to get any area you want. That's Riemann's theorem. Do you want areas to work that way?

Dave Marain said...

nice analysis, eric!
how many of us have such a profound understanding of this beautiful topic. In the past I had the time to show how one could rearrange the terms of a conditionally convergent series to make any sum you want but time seems to diminish every year.

There is also the issue of thinking that the improper integral from -1 to 1 of 1/x can be rewritten as:
lim (b-->0-)(def int of 1/x from b -1 to b) + lim(b-->0+)(def int of 1/x from b to 1). Using the same variable 'b' in both is dangerous and wrong. Students might argue that one could combine the two proper integrals first, producing zero, regardless of the limit, but, of course, the limit does not exist in either case! Thus, one cannot use the 'limit of a sum equals the sum of the limits' argument at all.
Your geometric analysis of cutting up the areas and rearranging the pieces is profound. I often get the sense that you have never stopped being a research mathematician!

Eric Jablow said...

I occasionally still think like one, or like a teacher.

You'll notice that you can't get that sort of behavior when all the summands are positive, or when all the signed areas are positive. The Cauchy principal value, which is the limit you discussed, is rarely used, but it does show up occasionally. You also get some odd contour integrals in complex analysis.

Dave Marain said...

Here is another common problem that students get wrong by not recognizing the integral is improper. Wuth the AP Calculus exam looming, you may want to discuss it:
The definite integral from 2 to 4 of 1/(x-3)^2.
If the student attempts to integrate over the entire interval, without splitting it up, one obtains -2 as the result, which, of course, makes no sense for a function which is never negative. Since the definite integral from 2 to 3 diverges, the entire integral diverges. It might be instructive to come up with a collection of these!