- mathfun@mailman.xmission.com - mailman.xmission.com

[math-fun] Abramowitz and Stegun on the web
by victor＠idaccr.org 21 Aug '03

21 Aug '03

The following may be of interest to members of this list. A mirror site for the scanned version of Abramowitz and Stegun exists at: http://jove.prohosting.com/~skripty/ -- Victor S. Miller | " ... Meanwhile, those of us who can compute can hardly victor(a)idaccr.org | be expected to keep writing papers saying 'I can do the CCR, Princeton, NJ | following useless calculation in 2 seconds', and indeed 08540 USA | what editor would publish them?" -- Oliver Atkin

4 3

[math-fun] What are these polytopes called?
by asimovd＠aol.com 19 Aug '03

19 Aug '03

Given a regular n-simplex, it contains (n+1)-choose-(k+1) k-simplices. Define T(n,k) as the convex hull of the barycenters of all these k-simplices. These polytopes T(n,k) are highly regular, but I haven't heard them referred to. Do they have standard names? E.g., T(3,0) ~ T(3,2) is a tetrahedron; T(3,1) is an octahedron. Combinatorially, T(n,k) should be identical to T(n,n-k-1) by duality. Clearly any lower-dimensional face of an T(n,k) is an T(m, j) itself for m < n. First interesting case seems to be T(4,1) ~ T(4,2), bounded by 10 vertices, 30 edges, 30 triangles, and 10 3-cells (5 tetrahedra and 5 octahedra). Define a polytope of any dimension to be "Archimedean" if its isometry group carries any vertex into any other. Then clearly each T(n,k) is Archimedean in this sense. QUESTION: For sufficiently high n, what are the (non-regular) Archimedean polytopes of dimension n other than all the T(n,k)'s ? (A guess: All others are obtained from the n-cube and its dual (the "n-cross-polytope") in a manner like (but not the same as) the way the T(n,k)'s are obtained from the n-simplex.) --Dan

3 2

[math-fun] Re: Gosper circular disk problem: right radius for tray?
by R. William Gosper 18 Aug '03

18 Aug '03

Amazing, Rod! I accidentally posted a completely forgotten set scaled to the 7.5" inner diameter of a Model A10 Aerobie. How did you deduce the 7.5" value with enough conviction to spend that much effort solving it? Without revealing too much of your solution, I also eagerly ask: How many different partial solution candidates did you have to test by hand? Were you able to determine that the solution is unique? How would you describe the "nice symmetry" you found? --Bill Gosper

1 0

[math-fun] Mathematical Musician and/or Musical Mathematician
by Henry Baker 15 Aug '03

15 Aug '03

I learned of this fellow through my high school (Walnut Hills, Cinn, OH) alumni rag http://www.mctague.org/carl/

1 0

[math-fun] Magic Sqaures
by Richard Schroeppel 15 Aug '03

15 Aug '03

I've found a curious transformation for Magic Squares of order 4. I didn't see it in my copy of Winning Ways, so perhaps it's new. The squares use the numbers from 1-16, and the magic sum is 34. Multiply each element in the square by a small number K, and reduce mod 17. This is a permutation on the 16 non-zero residues mod 17, so the numbers are still 1-16 in some order. The sum of each row/column/diagonal is still 0 mod 17, so it might be 17, 34, or 51. 17 and 51 are less likely, having a smaller number of decompositions into A+B+C+D (with 0<A<B<C<D<17) than 34 does. So sometimes, the new object will also be a magic square. For K=2, doubling: Mark a number white if it's < 8.5, and black if it's > 8.5. Then if each line of the magic square has two black and two white cells, doubling works. (But it might not work twice.) This seems to imply a curious result: There are a lot of other groups of cells that add up to the magic constant -- the four corners, the four center cells, etc. If each line of the square has two black & two white cells, then the other groups must also. K= -1 always works, and is the standard "subtract from 17" isomorphism. K=3 works when the total "overage" of a row is 4. Overage is the number of multiples of 17 dropped in the modular reduction, floor[N/5.66], 0 if N<=5, 1 for 6-11, 2 for N>=12. Again, rows with overage/=4 are less common. Under one way of counting, there are 880 MSQ4s. 880 = 16*55. A similar story should work for Magic Squares of order 6, since 37 is prime. But the chances are smaller, since more lines have to work, and the central tendency of the line sums is weaker. For order 5, 26 isn't prime, but odd K should work (except 13). ----- Counting MSQ6s. I looked at partitioning {1...36} into six sets of six numbers that add to 111. (Then we could count all the MSQ6s for a sample of the partitions, and estimate the MSQ6 total. Counting a sample of one million would give a 3-place estimate of MSQ6s.) 36! = 3.72 * 10^41. There are binom(36 6) = 1947792 ways of picking six of 36. Of these, 32134 sum to 111. A ratio of 1/60.6. One estimate of MSQ6 is 36! / 60.6^12 /192 = 788 quadrillion. The model is you dump the 36 numbers into the square and see what the chance is that the lines add up. The count of independent constraints is 12: 5 rows, 5 columns, 2 diagonals. One row and one column are redundant, since if 5 rows work, the sixth must also, etc. The 192 is for the group-of-the-square, and some more complicated permutations of the cells that preserve lines. (Another coincidence: same size as the groups for 4^3 and 5^3 TicTacToe.) 4739 of the 32134 potential row-sets contain "1". These are potential first sets of a partition. Extending to second rows, I count 8739938 pairs of (first,second). I'm assuming the second row must contain the smallest element that's missing from the first row. The ratio, 1844 combinations per first-row, is (very roughly) 1/e of 4739. Moving to third rows, I find 4957935915 combinations, again with the smallest-unused-element restriction. The ratio triples/pairs is 567, very roughly a 1/e drop from 1844. My program fills in one element at a time, so it's running out of steam about now. 100 minutes for the triple count. But it looks like the approach of listing all possible rows as bit patterns, and filtering the list recursively, could get to row 5. (Row 6 must work, since it's dependent.) My estimate for the total number of partitions is roughly 4 trillion. Attempting to guess the number of MSQ6s per partition, I wind up with a guess of 600 quadrillion total MSQ6s. This matches the simple probability method above. If we need 1 nsec to count each magic square, that's about 20 years. Just over the edge of practicality. But the edge moves. Rich rcs(a)cs.arizona.edu

1 0

RE: [math-fun] surprisingly hard knapsack puzzle
by Meeussen Wouter (bkarnd) 13 Aug '03

13 Aug '03

hi Bill, what are the radii of the 12? W. -----Original Message----- From: R. William Gosper [mailto:rwg@spnet.com] Sent: maandag 11 augustus 2003 22:12 To: math-fun(a)mailman.xmission.com Subject: [math-fun] surprisingly hard knapsack puzzle Last month I had the pleasure to visit the Microsoft Theory Group. Mike Sinclair and Gary Starkweather in a neighboring group generously helped me laser-cut a puzzle consisting of an 8" circular tray and twelve unequal circular disks ranging from .86" to 2.8". To my amazement, *no one* has yet succeeded in fitting all twelve disks back into the tray, although Christian Borgs, a Microsoft physicist, solved two slightly easier versions, one of which, unfortunately, gave him the answer to the harder version. I believe the solution to the current version is unique, but my first "cut" neglected to allow for the .006" diameter of the laser beam, and this small amount of slop permitted at least three unintended solutions. The current version is fairly pleasant to see and work, and draws small crowds when worked in public. Thus it invites frequent and prolonged attacks, but protects itself with the "too many cooks" effect. This puzzle is also fairly resistant to computer solution--I'm not even sure how to quantify the combinatorics of the search. --rwg _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

2 1

Re: [math-fun] Calculus I problem
by reid＠math.arizona.edu 13 Aug '03

13 Aug '03

just saw this message from rwg several weeks back (june 22, to be precise): > with a nice shortcut and an interesting answer: > Q: A cylindrical water tank, radius 1, rests on its side, skewered > by a coaxial bar of radius r. Choose r to maximize capacity if > the water level may not exceed the top of the bar. > > A: .retal etaitnereffid od gniog tsuj er'uoy ecnis )2^x-1(trqs largetni > rotces eht od ot deen t'nod uoy taht si tuctrohs ehT .)2^ip+1(trqS/1 or do it geometrically ... the maximum occurs when the circumference of (the cross section of) the bar equals the length of the chord formed by the cross section of the top of the water. mike

1 0

[math-fun] Decades ago, David R. Barton wrote Macsyma's polynomial decomposer:
by R. William Gosper 13 Aug '03

13 Aug '03

(c18) (ratsubst(x^2+3,x,x^3-27),%% = polydecomp(%%,x)) 6 4 2 3 2 (d18) x + 9 x + 27 x = [x - 27, x + 3] But it won't recognize the numerator of a composition of rational functions: (c20) (num(ratsubst(x+3/x,x,x^3-27)),%% = polydecomp(%%,x)) 6 4 3 2 (d20) x + 9 x - 27 x + 27 x + 27 = 6 4 3 2 [x + 9 x - 27 x + 27 x + 27] Anyone know how to do this? I can't even remember how Barton's program works, despite attending his elegant presentation at IBM Yorktown (SYMSAC 76?). --rwg

1 0

FW: Re: [math-fun] distinct and square-free factorizations
by James Tanton 13 Aug '03

13 Aug '03

Let D_k (n) = number of partitions of n with any number repeated at most k times. (Thus D_1(n) = D(n).) O_k (n) = number of partitions of n into numbers none of which is a multiple of k. ( Thus O_2(n) = O(n).) Then D_k (n) = O_(k+1) (n) for all n. The argument using generating functions to establish D(n) = O(n) easily generalises to establish this, but the kids (ages 11 - 17) at the Boston Math Circle recently came up with a nice fun proof of this in terms of representing numbers base k. - JT > [Original Message] > From: R. William Gosper <rwg(a)spnet.com> > To: <math-fun(a)mailman.xmission.com> > Date: 5/18/2003 4:11:57 PM > Subject: Re: [math-fun] distinct and square-free factorizations > > mlb>In fact, recalling this, I'd hoped to retrieve some collective memories! > > acw>> Gosper showed me an absurdly intricate _analytic_ proof of this, and > >>bewailed the lack of a _combinatorial_ proof; > > Here are my memories. Unfortunately, the TeX sources appear to to have been > abandoned at Xerox PARC. Shortly after the announcement of the factorization > of 2^2^2^3+1, I sent a(n unrelated) letter to the the usual gang of maniacs > (Dick Askey, George Andrews, George Gasper (I think), Rich&Hilarie, etc.) > that began: "Dear Thai food lovers and Rich," > answering Hilarie's challenge to find a bijection between the *complements* > of the odd partitions and distinct partitions. I.e., pair the not-all-differents > with the not-all-odds. A very young Peter Weyhrauch and I came up with one, > and then a PARC coworker, Ted Kaehler, came up with a different one! I > then found a generalization of the theorem to residue classes mod k, of which > #(some evens) = #(some duplicates) was the case k=2. This led to a nonobvious > generating function identity. > It would be nice to locate a hardcopy of that letter. > --rwg > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ========================= James Tanton mathcircle(a)earthlink.net P.O. Box 2344 Acton MA 01720 978 635 8036 FIND *MATH CIRCLE* ON THE WEB AT: www.themathcircle.org

2 1

[math-fun] A complex numbers ...
by Jon Perry 12 Aug '03

12 Aug '03

What is the term for the relation of (a+bi) and (c+di), both complex numbers, where, given a+bi, (a+bi)(c+di) = x or = iy i.e., is either pure real or pure imaginary? I came up with the 'realization' or the 'imagination'. Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

1 0