May 2008 - mathfun@mailman.xmission.com

[math-fun] Earthquake magnitudes
by Henry Baker 14 May '08

14 May '08

I saw a press conference on the internet where the Chinese govt officials were saying that the 2008 Olympics were safe, because the facilities were constructed to survive an 8.0 earthquake. I also understand that the recent Sichuan quake was magnitude 7.9. If I understand the Richter scale, it is log_10(amplitude), so amplitude 8.0 should be 26% greater than amplitude 7.9. http://en.wikipedia.org/wiki/Richter_scale Is 26% much of a safety margin?

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Re: [math-fun] integration by parts for numerical computation
by James Propp 07 May '08

07 May '08

Thanks to Mike Stay for finding me an on-line copy of Klamkin's article "On the teaching of mathematics so as to be useful". The relevant excerpts are on pages 134-135 and page 159. The former excerpt reads as follows: "In many courses, once a method is established in theory for solving a class of problems that supposedly finishes the topic. However, actually obtaining a numerical answer to a specified accuracy may be an entirely different and difficult problem. For example, in evaluating definite integrals numerically, Simpson's rule may be derived and illustrated on several 'nice' functions. However, if one assigned an integral such as $\int_{0}^{100} x^x \ dx$ to be evaluated numerically to within 10%, almost all students will use Simpson's rule 'blindly' and be lucky to come within 500%. The difficulty here is that x^x is not a 'nice' function since it increases too rapidly. Similarly for the integral $\int_{10}^{\infty} e^{-x^2} \ dx$." The passage ends with a footnote leading the reader to an explicit calculation on page 159 (which would be a bit of a pain to include here, and would not be very readable even if I did). Jim Propp

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[math-fun] integration by parts for numerical computation
by James Propp 06 May '08

06 May '08

I seem to recall that Murray Klamkin once gave an example about how one can use integration by parts to get a better estimate of the integral of x^x from x=100 to x=101 than one would get by looking at the original integral. I may have the details wrong, though; can anyone recall how this goes? (In fact, I think I even know where Klamkin published it: it was in "On the teaching of mathematics so as to be useful" or "On the ideal role of an industrial mathematician and its educational implications". But I can't locate a world-readable copy of either article.) Thanks, Jim Propp

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Re: [math-fun] The value of PI
by James Propp 06 May '08

06 May '08

Henry Baker writes: >If you go through all the formulae in a large book -- e.g., Stegun, et al -- >I think that "2pi" shows up more often than "pi". Sooner or later, we're >bound to run up against a civilization that chose differently which to >commemorate with a name. There was a short piece (in the Mathematical Intelligencer, I think; perhaps the March 2007 issue?) that explored exactly this theme, and made the "modest proposal" that we should switch from using __ || = 3.14... to using --- ||| = 6.28... in our formulas. (An impassioned advocacy of this position would make a fun "heresy" at a college Pi Day.) Dan Asimov writes: >It is amusing to note that the sum of the V(n) for all even n is e^pi >(if one reasonably takes V(0) = 1, since the 0-ball is just a >point and 0-dimensional measure just counts points). Cute, and provocative. As long as we're on this subject, permit me to bring up a curiosity I may have mentioned before on this list: The formula V(n) = (2/n) pi^(n/2) / Gamma(n/2)) for the volume of the unit n-ball gives the familiar formulas V(3) = 4 Pi / 3 V(2) = Pi V(1) = 2 (and makes V(0) blow up), but the formula also gives V(-1) = 1 / Pi V(-2) = 0 V(-3) = - 1 / 2 Pi^2 V(-4) = 0 V(-5) = 3 / 4 Pi^3 with V(-2n) = 0 and V(2n-1) V(-2n-1) = (-1)^n / Pi^2. Do the numbers V(-1), V(-2), etc. have any kind of meaning, and is there any significance to the reciprocity between V(2n-1) and V(-2n-1)? In closing, here is a counterfactual self-referential mnemonic for pi I sent Doug Hofstadter about twenty years ago (I can't remember whether he used it in his column): Had \pi only a teeny bit smaller value, this sentence would be an excellent mnemonic for it. Jim Propp

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Re: [math-fun] The value of PI
by James Propp 06 May '08

06 May '08

Dan writes: >... >Plugging in n = 0 gives V(0) = 1. Right you are! Jim

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Re: [math-fun] The value of PI
by Dan Asimov 06 May '08

06 May '08

<< The formula V(n) = (2/n) pi^(n/2) / Gamma(n/2)) for the volume of the unit n-ball . . . . . . makes V(0) blow up >> Does it? (2/n) / Gamma(n/2) = 1 / ((n/2) Gamma(n/2)) = 1 / Gamma(1 + n/2) so the formula for V(n) becomes V(n) = pi^(n/2) / Gamma(1 + n/2). Plugging in n = 0 gives V(0) = 1. --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] Erdos-Turan conjecture
by Dan Asimov 02 May '08

02 May '08

Someone posted a question to sci.math.research on the Erdos-Turan conjecture: ---------------------------------------------------------- If for a subset S of the positive integers satisfies (*): (*) Its sum of reciprocals diverges. Then S contains arbitrarily long arithmetic progressions. ---------------------------------------------------------- The celebrated Green-Tao theorem proves this is true if S is the set of primes. Apparently it is not even known if (*) is sufficient to guarantee one single arithmetic progression of length 3 (!). Can anyone say what *is* known about this conjecture? --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] The value of PI
by Henry Baker 02 May '08

02 May '08

Some might prefer this version: (sqrt(%pi)*r)^n --------------- (n/2)! At 12:24 PM 5/1/2008, rwg(a)sdf.lonestar.org wrote: >hgb>Even more bizarre, IMHO, is the fact that in the volume formula for n >dimensions, it isn't pi itself, but sqrt(pi), that is important. > >That's sort of a notational artifact: > >(c41) sphere_volume(n,r) > > n/2 n > %pi r >(d41) --------- > n > (-)! > 2

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[math-fun] The value of PI
by Bernie Cosell 01 May '08

01 May '08

Pi shows up a lot in geometry. I was wondering when "we" (ahem) realized that it was just one underlying constant. For example, the area of a circle is 3.14... times the square of the radius and the volume of a sphere is 4.18... times the cube of the radius. It is already profound discovering that those ratios are fixed [and to be able to calculate a bunch of digits of them]; I don't know if he did or not, but I can' easily see Archimedes managing to calculate a bunch of digits of the 4.18... constant even as he did for the 3.14. one]. But when did mathematicians realize that those weren't two separate gnarly constants, but actually "reflections" [if you will] of a single underlying constant? I don't think the Greeks had enough math machinery to figure all that out, did they? [that is, that the ratio of those two constants is exactly 3:4. Or that the ratio of the radius of a circle to its curcumferenace is really just exactly twice the ratio of the square of the radius of the circle to its area; 3.14... and 6.28... will certainly *look* like 1:3 [not clear the 3:4 ratio is as obvious..:o)], but could they actually 'know' that enough to say that that was really "proven", in some sense? If not the greeks, then when did we figure that out? /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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Re: [math-fun] The value of PI
by Dan Asimov 01 May '08

01 May '08

<< Even more bizarre, IMHO, is the fact that in the volume formula for n dimensions, it isn't pi itself, but sqrt(pi), that is important. >> Not true. The "surface" (n-1)-measure of the (n-1)-sphere is S(n-1) = (2 pi^(n/2) / Gamma(n/2)) r^(n-1), and the "volume" n-measure of the n-ball is V(n) = (2/n) pi^(n/2) / Gamma(n/2)) r^n, but the sqrt(pi)'s that appear for n odd cancel in the numerator and denominator, so that all S(n-1)'s and V(n)'s for r = 1 are rational numbers times integer powers of pi. The power of pi is floor(n/2) for both S(n-1) and V(n), so we list the power of pi in parentheses after n, and just list the rational numbers in the lines below: n 0(0) 1(0) 2(1) 3(1) 4(2) 5(2) 6(3) 7(3) 8(4) ---------------------------------------------------- S(n-1) - 2 2 4 2 8/3 1 16/15 1/3 ---------------------------------------------------- V(n) 1 2 1 4/3 1/2 8/15 1/6 16/105 1/24 if my calculations are correct. It is amusing to note that the sum of the V(n) for all even n is e^pi (if one reasonably takes V(0) = 1, since the 0-ball is just a point and 0-dimensional measure just counts points). --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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