February 2008 - mathfun@mailman.xmission.com

[math-fun] Re: another calculus question
by James Propp 08 Feb '08

08 Feb '08

Josh Zucker writes: >If f is continuous, isn't it automatically integrable? Oops, you're right! I mis-stated my problem. Let's try that again, replacing the word "continuous" by the word "defined": Suppose that f is defined on the interval [a,b] and F is an antiderivative of f on (a,b) (that is, if F is a function that is continuous on [a,b] and differentiable on (a,b) with F'(x) = f(x) for all x in (a,b)). Can we conclude that f is integrable on [a,b]? Jim

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FW: [math-fun] Re: calculus question
by Cordwell, William R 07 Feb '08

07 Feb '08

Well, if you pick your semicircle so that the center of the semicircle is always 1/2 of a radius to the left of the singular point, for example, then shrink the circle while keeping this constraint, it will, indeed, converge to the singular point, but the left bound of the circle is 3 times as far away from the singular point as the right. I'm fairly sure that I worked an example (long ago, when I was taking a class in Complex Analysis), and the answer was different than the regular PV result. The "purpose" is to illustrate that it matters how one constructs the bypass of the singular point. I admit, however, that I am unfamiliar with the definition of legitimate contour. Bill C. ------------------------------------------------------------------------------ Quoting "Cordwell, William R" <wrcordw(a)sandia.gov>: > The Principal Value of a complex integral also depends on how the > semi-circle (or half-square, or half-rectangle) is drawn to miss the > singularity. That is, the limit can still approach the singular point > from both sides, but unevenly, and this can give a different answer > from the case where the point is approached evenly from both sides. Not really. When the singularity is a point, it should be surrounded by a small circle centered on the point. Otherwise the limiting value as the excision is shrunk will not work out, although it would likely be possible to concoct wierd limits. To what purpose? Other than to make clever trickery? Then again, maybe there is a legitimate contour in which the singularity sits at the tip of a corner that is only part of 360 or 180 degrees. -hvm

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[math-fun] Re: calculus question
by James Propp 07 Feb '08

07 Feb '08

Dan Asimov writes: >... >This strategy, however, would certainly benefit from a theorem in between >the above definition of integrals and the calculation thereof, stating that >a large class of functions (e.g., piecewise continuous ones) actually *are* >integrable in the sense mentioned. Stewart includes exactly such a theorem. Andy Latto writes: >Pedagogically, I think it makes sense to do things the other way >around. This is a class for people who do not have a lot of experience >with mathematics, and the problem people have understanding >mathematics is the high level of abstraction involved. The way to >combat this is by being as concrete as possible at first, and only >introducing abstraction as needed. I'm in agreement with this. Dan asks: >Jim did, by the way, say this is an honors course. (Is it honors for >anyone, or mainly honors math majors?) Most of the students have declared an intention to major in physics or engineering. My goal is to get at least a few of them to decide that math is cooler. >Pedagogy aside, certainly one could cook up a function having the value >1 on all points of the form k/n for integers k,n such that >0 <= k/n <= 1, and the value 2 at other points. In fact, on the homework I define the function f(x) that's 1 when x is rational and 2 when x is irrational, and ask them to show that the function isn't integrable on [0,1]. >Let's call this method EWAP. So is EWAP integrable equivalent to APAP >integrable, and yielding the same values? That's a good and concise question of the mathematical issue (as opposed to the pedagogical issue) that I'm raising. >Conversely, there is a chance that a strange singularity can be created >near 0, say, of a function f:[0,1} -> R, such that f is EWAP integrable, >but not APAP integrable. I suspect not, but I don't see a proof off-hand. >Good question. Thanks! Stephen Gray writes: >When you go from equal intervals to unequal ones, it's important to specify >exactly how all the intervals approach zero width. Stewart's definition requires that the width of the widest interval (aka the "mesh" of the partition) must go to zero. >I imagine that the limit >is affected by how the limit is reached. One might worry that using a partition in which some intervals are much wider than others would mess up the answer, but this isn't the case, in the limit where the mesh goes to zero. For a situation in which Stephen's qualms *do* apply, see the "cylinder area paradox": one can define a sequence of polyhedral approximations to the cylinder for which the surface areas of the approximations do not converge on the true surface area of the cylinder, on account of the triangular faces of the polyhedron getting too long and skinny in shape, even though they're shrinking away to zero diameter. (Anyone know a good web reference for this? Freida Zames' article "Surface Area and the Cylinder Area Paradox" is only available via JSTOR, as far as I can tell.) Jim Propp

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[math-fun] moderator outage; new email address
by Schroeppel, Richard 07 Feb '08

07 Feb '08

I'm headed off to Maui for three weeks. Michael Kleber [michael.kleber(a)gmail.com] will be the temporary moderator. --- I've changed my email address. The new address is rcs(a)xmission.com. The old address (16 years), rcs(a)cs.arizona.edu, will forward for a while. I also have a "permanent" address at schroeppel(a)alum.mit.edu. Rich

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Re: [math-fun] calculus question
by Dan Asimov 07 Feb '08

07 Feb '08

Jim writes: << [W]hy would it make sense pedagogically to use sub-intervals of unequal widths as part of the _definition_ of the integral, and then state a _theorem_ that says that if a function is integrable then you might as well use equal-width sub-intervals (which is what Stewart does), rather than _define_ the integral using equal-width sub-intervals, and then state a _theorem_ that says that you can use more general partitions and still get Riemann sums that converge on the correct value, as long as the mesh goes to zero? >> Seems to me that to define the integral in terms of the same limit for a wide variety of partitions would remove any suspicion that the nice way it turns out is solely an artifice of using equal-length intervals. Then to specialize to equal-length partitions when *calculating* the integral makes sense, just for the sake of convenience. This strategy, however, would certainly benefit from a theorem in between the above definition of integrals and the calculation thereof, stating that a large class of functions (e.g., piecewise continuous ones) actually *are* integrable in the sense mentioned. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] calculus question
by Dan Asimov 06 Feb '08

06 Feb '08

Andy writes: << . . . [I wrote] << Conversely, there is a chance that a strange singularity can be created near 0, say, of a function f:[0,1} -> R, such that f is EWAP integrable, but not APAP integrable. >> The thing is, once the mesh size is less than 10^-100, the effect of the singularity at 0 is tiny, so the AWAP integral will still converge whenever the EWAP one does. . . . >> Only if you're assuming something about f (like continuity or boundedness), which I wasn't. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] calculus question
by Dan Asimov 06 Feb '08

06 Feb '08

Andy is certainly right that a definition of integral that uses equal-with intervals and, say, always left endpoints (EWLE) is easier to conceive of than the more general "any partition whose mesh approaches 0, and any points within the intervals" (APAP) definition. Pedagogically, there's no question that I'd introduce integrals, and confine all or almost all examples to the EWLE (or EWRE) type. But when first learning calculus I recall being a bit confused by which endpoint -- or was it the midpoint -- to evaluate the function at. When later (I think when using the ubiquitous text by G.B. Thomas) I learned the APAP definition, it was a relief not to have to be concerned about such details as the exact partition and where to evaluate the function. I would *state* but certainly not *prove* such a theorem in an intro course (other than to honors math majors, who might use a book like Apostol's). Jim did, by the way, say this is an honors course. (Is it honors for anyone, or mainly honors math majors?) ------------------------------------ Pedagogy aside, certainly one could cook up a function having the value 1 on all points of the form k/n for integers k,n such that 0 <= k/n <= 1, and the value 2 at other points. But this doesn't answer the interesting math question brought up by Jim: << Stewart says that a function f on the interval [a,b] is integrable with definite integral I if for all epsilon > 0 there exists a delta > 0 such that EVERY Riemann sum for f(x) on [a,b] whose mesh (the maximum of the widths of the sub-intervals) is less than delta has value within epsilon of I. >> [what I called APAP] << The student asked, would it be equivalent to require the sub-intervals to be of equal width (still allowing the sample-points to be arbitrary within the sub-intervals)? >> Let's call this method EWAP. So is EWAP integrable equivalent to APAP integrable, and yielding the same values? Obviously APAP implies EWAP since equal width is a special case of arbitrary partitions. Conversely, there is a chance that a strange singularity can be created near 0, say, of a function f:[0,1} -> R, such that f is EWAP integrable, but not APAP integrable. Good question. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Tower of Hanoi with random moves
by Guy Haworth 06 Feb '08

06 Feb '08

I recently modelled a 'Reference Fallible Player' playing a chess endgame with perfect information. Because their 'competence factor c' is not +infinity, they don't necessarily play the best moves even though they have perfect information. c = 0 corresponds to zero-skill, i.e., legal moves chosen at random. The state-space gives rise to a Markov System, from which all sorts of good things may be calculated. One can show how long a player with competence 'c' is likely to take to win a position from depth 'd', which in a way serves to characterize the difficulty of the endgame. The pdfs of the relevant papers used to be on a webpage about me on our University site, but due to an upgrade (?!) to a Content-Management System, I cannot currently point interested readers to those papers. My colleagues are still learning how to upload papers to the CMS without the 'website editor', a non-academic, vetoing their actions. However, if anyone wants to check these ideas out, they are welcome to ask for pdfs of the papers. Regards - Guy

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[math-fun] Re: calculus question
by James Propp 06 Feb '08

06 Feb '08

Just to clarify: I'm well aware that for many problems, a non-uniform partition is handy. For instance, one can integrate x^c on [0,1] for any exponent c>0 as Fermat did, by dividing the domain [0,1] into infinitely-many sub-intervals of the form [q^(i-1),q^i] (with 0 < q < 1), which gives an infinite Riemann sum whose terms are in geometric progression, and then take the limit as q->1. (I'll guide my students through this in the case c=2 on their next homework assignment.) So, the issue isn't whether there's benefit from using non-uniform partitions; I already know that there is. The issue is whether one should (a) DEFINE the integral using uniform partitions and then PROVE (or at least state) a theorem about non-uniform partitions, or (b) DEFINE the integral using non-uniform partitions and then DERIVE (as a trivial consequence) a theorem about non-uniform partitions. Stewart takes course (b); it seems to me that (a) is easier for students to digest, and I wonder if there are reasons to prefer (b) that I might not be aware of. Jim Propp

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[math-fun] calculus question
by James Propp 06 Feb '08

06 Feb '08

Hi, all. I'm teaching second semester freshman calc (honors), using Stewart's book "Essential Calculus" (which on the whole I like), and a student asked me a question that I don't know the answer to. First, some background: Stewart defines the definite integral in terms of Riemann sums where the widths of the sub-intervals are allowed to be different, and where the sample-points within those sub-intervals are allowed to vary (i.e., they aren't required to be the left endpoints, right endpoints, or mid-points of the sub- intervals). Stewart says that a function f on the interval [a,b] is integrable with definite integral I if for all epsilon > 0 there exists a delta > 0 such that EVERY Riemann sum for f(x) on [a,b] whose mesh (the maximum of the widths of the sub-intervals) is less than delta has value within epsilon of I. The student asked, would it be equivalent to require the sub-intervals to be of equal width (still allowing the sample-points to be arbitrary within the sub-intervals)? I suspect that the answer is yes (though I also suspect that a proof would be more suitable for an upper-division course than a freshman honors course). So my first question is, can anyone come up with a really simple way to see that the two notions of integrability are the same? (Assuming that my instinct is correct and they really are the same.) And my second question is, why would it make sense pedagogically to use sub-intervals of unequal widths as part of the _definition_ of the integral, and then state a _theorem_ that says that if a function is integrable then you might as well use equal-width sub-intervals (which is what Stewart does), rather than _define_ the integral using equal-width sub-intervals, and then state a _theorem_ that says that you can use more general partitions and still get Riemann sums that converge on the correct value, as long as the mesh goes to zero? Jim Propp

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