tag:blogger.com,1999:blog-8231784566931768362.post9097345397054625000..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: 'Rigor Mathis' - A Calculus Paradox or...Dave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-8231784566931768362.post-81230458318298535892007-05-06T07:00:00.000-04:002007-05-06T07:00:00.000-04:00Here is another common problem that students get w...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:<BR/>The definite integral from 2 to 4 of 1/(x-3)^2.<BR/>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!Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-67108439219360728562007-04-21T13:05:00.000-04:002007-04-21T13:05:00.000-04:00I occasionally still think like one, or like a tea...I occasionally still think like one, or like a teacher.<BR/><BR/>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.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-23269900304061963632007-04-21T06:06:00.000-04:002007-04-21T06:06:00.000-04:00nice analysis, eric!how many of us have such a pro...nice analysis, eric!<BR/>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.<BR/><BR/>There is also the issue of thinking that the improper integral from -1 to 1 of 1/x can be rewritten as:<BR/>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.<BR/>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!Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-29237108289640075312007-04-20T23:11:00.000-04:002007-04-20T23:11:00.000-04:00Probably the best analogue is the problem of findi...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.<BR/><BR/>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:<BR/><BR/>f(x) = 0, x ≤ 0,<BR/>f(x) = (−1)^{n+1}/n, if n−1 < x ≤ n.<BR/><BR/>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?Anonymousnoreply@blogger.com