July 2003 - mathfun@mailman.xmission.com

Re: [math-fun] NY State math exam flapment / Baseball seams
by Henry Baker 21 Jul '03

21 Jul '03

I'm currently in England, and had occasion to play with a cricket ball. The cricket ball does indeed have a seam around the equator. I was told that the center was cork. The cover looked like leather, but I couldn't be sure. Since the cricket ball is also pitched/"bowled", the aerodynamics of the seam would appear to be important. This is an interesting case of parallel evolution... At 11:01 AM 7/4/03 -0400, Bernie Cosell wrote: >On 4 Jul 2003 at 7:25, Henry Baker wrote: > >> I hope that you know enough about baseball to appreciate the >> elegance of the baseball's seam: >> >> http://www.sciencenews.org/sn_arc98/4_11_98/mathland.htm > >What property of the 'seam' problem for the baseball precludes just an >equator for the seam as a possible solution? Is there some "minimization >of necessary stretch" or other criterion that drives the more complicated >shape? > > /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] Calculating the standard deviation incrementally
by Henry Baker 17 Jul '03

17 Jul '03

I haven't been following this thread that carefully, so the following point may already have been made. The computation of the variance below involves the addition/subtraction of very large values, which can result in massive cancellations. Thus, though while the formulae work, you may have to carry very large precisions. In fact, even the simple summation of the values themselves is a serious problem, as I discovered in a real application many years ago, and which I passed on to my students in the following form: for i=0 to 1 by 0.000000001 do <loop> I believe that if i is single precision IEEE float, the increment is washed out well before hitting 1.0, which means that this loop is actually an infinite loop. (In the old days, computers were so slow that loops of this sort appeared to be infinite in any case! Large scientific calculations need the precision to go up with the speed of the computer, so that naive computations like this still "work".) A very good summation is to remember all the values, sort them by absolute value, and then add them in terms of increasing value. An almost-as-good method is to keep log2(n) levels of sums, and sum as a binary tree. I seem to recall that the CACM (Comms. of the ACM) had a number of articles about the "real-time" standard deviation problem back in the late 1960's - early 1970's time frame. The index of this publication is online and freely accessible at acm.org. At 10:17 PM 7/16/03 -0600, Roland Silver wrote: >Bernie Cosell, >All you need to do is compute on the fly the sum of squares S = x[1]^2 >+ ... + x[n]^2 of the empirical values, as well as the sum of values T >= x[1] + ... + x[n]. The mean M = T/n, and the variance V = ((x[1] - >M)^2 + (x[n] - M)^2)/(n-1). The SD is the square root of V. If you >expand the terms in the equation for V, you get V = (S -2*M*T + >n*M^2)/(n-1) = (S - n*M^2)/(n-1), which you can compute in one pass.

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[math-fun] new book, _Automatic Sequences_
by Jeffrey Shallit 17 Jul '03

17 Jul '03

Jean-Paul Allouche and I are pleased to announce the publication of our book (AS)^2, also known as Automatic Sequences: Theory, Applications, Generalizations by Cambridge University Press (2003), ISBN 0-521-82332-3. It is currently selling for US $50 on amazon.com. This book is about the class of sequences generated by finite automata, their generalizations, and applications to number theory and theoretical physics. The book has 571+xvi pages, 1600 citations to the literature, 460 exercises, 85 open problems, 1 musical score, and 2 jokes in the index. It will be of interest to number theorists and theoretical computer scientists. The web page for the book, http://www.math.uwaterloo.ca/~shallit/asas.html has a table of contents and other information. Jeffrey Shallit, Computer Science, University of Waterloo, Waterloo, Ontario N2L 3G1 Canada shallit(a)graceland.uwaterloo.ca URL = http://www.math.uwaterloo.ca/~shallit/

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Re: [math-fun] matrix exp
by asimovd＠aol.com 17 Jul '03

17 Jul '03

Harold (mcintosh(a)servidor.unam.mx ) writes: << An interesting intersection of these two trains of thought is the fact that the Lie Algebra of O4 is the direct product of the Lie Algebras of two O3 rotation groups >> And the rotation group SO(4) = S^3 x S^3 / (g,h) ~ (-g,-h) (where S^3 is the group of unit quaternions). Who would've thought that the rotation group SO(4) of something so round as the 3-sphere would be practically "square" (and its Lie algebra exactly square)? There's a very nice discussion of this in the Thurston notes on 3D geometry & topology that were sold by MSRI before the corresponding book got published. I'm not sure whether this discussion is in the book, too. --Dan

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[math-fun] matrix exp
by Shel Kaphan 17 Jul '03

17 Jul '03

Is anyone's symbolic math package able to take the matrix exponential of [ 0 a 0 -d ] [ -a 0 b 0 ] [ 0 -b 0 c ] [ d 0 -c 0 ] ? I set Macsyma to work on it, and it seems only to be trying to melt my laptop and burn the house down after long enough to suspect it isn't going to finish. If a result of that is available, I'd also like to know the resulting eigenvalues and eigenvectors. Thanks in advance, Shel

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[math-fun] Mathematicians under the Nazis
by Thane Plambeck 17 Jul '03

17 Jul '03

Perhaps not math-fun, but maybe math-interesting http://www.pupress.princeton.edu/titles/7558.html Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/ehome.htm

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[math-fun] Calculating the standard deviation incrementally
by Bernie Cosell 16 Jul '03

16 Jul '03

I thought *SURE* I knew an algorithm once, long ago, for calculating the SD "on the fly", but I can't remember any details of any such trickery and a quick Google search only came up with algorithms that I already know and that start with the obvious "calculate the mean..." Basically, I have a largish data set that it will only be convenient to read once and I'd like to be able to calculate [or at least estimate] the SD as the data comes by and avoid having to scroll it into a temp file and then do a second pass over the data from the temp file... Is there such an algorithm or am I just misremembering? Thanks! /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] Bessel functions for continued fractions
by Mike Stay 16 Jul '03

16 Jul '03

Here's something not directly related, but that might go somewhere (I'm certainly going to research it some more): For the sake of brevity below, F means GF(2^n). You can talk about functions that act on elements of F, but the elements of F can be treated as functions themselves, in a limited sense. Usually we think of them as polynomials in x with coefficients in {0,1}. That suggested to me that I could take the q-derivative of a single element of F rather than the q-derivative of a function acting on elements of F. Then you can identify an element e^x with the property that D((e^x)^k) = D(e^kx) = [k]e^kx, where [k] is the q-number (q^k - 1)/(q-1). So in addition to treating elements of F as polynomials, we've can treat them as trig functions, which all have nice continued fraction representations. Continued fractions lose uniqueness of the representation when there's no concept of magnitude (and therefore no "floor" function) but a lot of the theory still works. Of course, AES is just an overblown continued fraction, so it would be fun to see if there's some way to write it as, say, a sum of Bessel functions or something. >Gabor Hetyei <ghetyei(a)uncc.edu> writes > >> While browsing the internet, I came accross your remarkable formula > http://www.inwap.com/pdp10/hbaker/hakmem/cf.html#item99 > expressing the value of a continued fraction with partial quotients > increasing in arithmetic progression. > I would be interested in the known proof of this formula. Is there any > published paper or preprint on the issue that you could share with me or > point me to? > The reason of my interest is that I think it is possible to prove this > formula using combinatorial enumeration, and I wonder if this has already > been done or would be interesting to do. > >I'd been interested in the value of the continued fraction 1+1/2+1/3+1/4+... >since high school. In college, I came across the Dover reprint of the NIST >tables by Abramowicz & Stegun, and spent a fair amount of time studying the >special functions. I came across the Bessel-CF formula 9.1.73 for J, and >recognized its relevance to my puzzle. Then I found the recurrence for I >(first formula in section 9.6.26) and a litle manipulation does the rest. > >I was later able to find a direct proof using the power series for I, the >one with a product of factorials in the denominator. I don't recall details, >but it was pretty straightforward, just lining up terms to check the >recurrence. You need to add something about convergence, I don't recall >if I finished this step. > >I suspect "my" formula is actually well-known among continued fraction experts. > >I'm not a professional combinatorist, so I don't know if your work is new >or not. I'd think that anything relating combinatorics to Bessel functions >would count as interesting, since there's no obvious connection. (Although >I think they turn up in the formula for partitions into unequal parts.) > >There are some people I can ask. I'm forwarding this reply to them. > >Rich Schroeppel rcs(a)cs.arizona.edu > > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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[math-fun] Bessel functions for continued fractions
by Richard Schroeppel 16 Jul '03

16 Jul '03

Gabor Hetyei <ghetyei(a)uncc.edu> writes > While browsing the internet, I came accross your remarkable formula http://www.inwap.com/pdp10/hbaker/hakmem/cf.html#item99 expressing the value of a continued fraction with partial quotients increasing in arithmetic progression. I would be interested in the known proof of this formula. Is there any published paper or preprint on the issue that you could share with me or point me to? The reason of my interest is that I think it is possible to prove this formula using combinatorial enumeration, and I wonder if this has already been done or would be interesting to do. I'd been interested in the value of the continued fraction 1+1/2+1/3+1/4+... since high school. In college, I came across the Dover reprint of the NIST tables by Abramowicz & Stegun, and spent a fair amount of time studying the special functions. I came across the Bessel-CF formula 9.1.73 for J, and recognized its relevance to my puzzle. Then I found the recurrence for I (first formula in section 9.6.26) and a litle manipulation does the rest. I was later able to find a direct proof using the power series for I, the one with a product of factorials in the denominator. I don't recall details, but it was pretty straightforward, just lining up terms to check the recurrence. You need to add something about convergence, I don't recall if I finished this step. I suspect "my" formula is actually well-known among continued fraction experts. I'm not a professional combinatorist, so I don't know if your work is new or not. I'd think that anything relating combinatorics to Bessel functions would count as interesting, since there's no obvious connection. (Although I think they turn up in the formula for partitions into unequal parts.) There are some people I can ask. I'm forwarding this reply to them. Rich Schroeppel rcs(a)cs.arizona.edu

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[math-fun] Areas bounded by fractals
by R. William Gosper 16 Jul '03

16 Jul '03

sum(c_j e^(i j t)) draws a closed curve in the complex plane as 0<=t<=2 pi. By Green's thm, the enclosed area (weighted by winding number) is pi sum(j |c_j|^2), which is interestingly independent of the phase of c_j. Applying this to the Fourier series for Snowflake-like fractals arranged around a regular m-gon yields the identity inf 1 /===\ %pi (- + k) | | m 2 3 2 s %pi inf ( | | (1 - s tan(-----------))) %pi (------ + cot(---)) ==== | | n 2 m \ n = 1 (-2) s + 3 > --------------------------------- = ------------------------ . / 1 3 2 %pi ==== (- + k) sin (---) k = - inf m m The boundary dimension is 1/(1-lg(1+s^2)/2). The standard snowflake is m=3, s=1/sqrt(3), or m=6, s=-1/sqrt(3). This differs from my previous strange identities in that the (rapidly convergent) infinite product is squared. As s -> 1 (spacefilling), the sum converges extremely slowly. In the spacefilling (s=1) limit, this formula seems to ascribe half the spacefilled area to the interior. In other words, it gives an area midway between the s=0 m-gon and the s=1 2m-gram obtained by erecting isosceles right triangles on the sides of the m-gon. --rwg

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