May 2003 - mathfun@mailman.xmission.com

RE: [math-fun] knight problems from Elkies and Stanley
by Andy Latto 29 May '03

29 May '03

Like many proofs by contradiction, I find that this proof that there is no knight's tour on a 4xn chessboard hides a bit of what's really going on. Here's a more direct version of the same proof that I find more perspicuous. Call the squares in the top and bottom rows of a 4xn board the "extreme" squares. Since you can't get in one knight's move from one extreme square to another, the maximum density of extreme squares in a knight's tour is 50%, and this is only attained if every other square in the tour is extreme. Since half the squares on the board are extreme, a tour that includes every square must alternate extreme with non-extreme squares. But a knight's tour alternates white and black squares, so a tour that alternates in this way cannot reach both black and white extreme squares. QED > Stop reading here, if you so desire. Otherwise: first let's prove the > seemingly unrelated fact that, on a standard 8x8 chessboard, the > most knights you can put on with no two attacking each other is 32. > It's easy to see how to attain 32: cover all the white or all > the black > squares with knights; each knight move changes color, so these 32 > are clearly non-attacking. Now to show that these are the unique > two configurations that attain this maximum: we already know that > there is a knight's tour on the 8x8 board (a picture of > Euler's solution > is included in the article, for example). Given any two squares which > are used consecutively in the knight's tour, at most one of them can > be covered by a knight from a "no two attacking" configuration, of > course. > Therefore the only way to pack 32 knights is to place one on > every other > square in the knight's tour -- which means all the black or > all the white > squares, since again, knight moves alternate colors. Nice > proof of the > wrong theorem. But now apply the same proof to the 4xn board, and > we see that if a knight's tour existed, then the only way to fit 2n > nonattacking > knights on a 4xn board would be to use all the squares of a > single color. > But that's manifestly not the case: on a 4xn board, cover the > entire top > row and the entire bottom row with knights; 2n of 'em, no pair > attacking. QED.

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[math-fun] knight problems from Elkies and Stanley
by Thane Plambeck 29 May '03

29 May '03

I found the following in a review by Philip D. Straffin (College Mathematics Journal, May 2003) of Noam Elkies and Richard Stanley's "The Mathematical Knight," The Mathematical Intelligencer 25:1 (Winter 2003) 22-34: ******** "...there are also discussions and problems for the reader (with answers at the end) that lead in less familiar directions. Here are some of my favorites: 1) Determine the knight distance from (0,0) to (m,n) on an infinite board 2) If you plot the 4! shortest knight paths from (0,0) to (6,0) you get an elegant projection of the edges of the 4-dimensional hypercube. Explain..." ***** It's hard not to pick up pencil and pad when you read stuff like that! Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/home.htm

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[math-fun] pennies, handy constants, more Prentice Hall
by R. William Gosper 29 May '03

29 May '03

"My" seventh grader was asked to tabulate the ways to make 40 cents change without pennies, so I had couldn't resist showing him (c72) (1/(1-q^5)/(1-q^10)/(1-q^25),%%=taylor(%%,q,0,40)) 1 (d72)/T/ ---------------------------- = 5 10 25 (1 - q ) (1 - q ) (1 - q ) 5 10 15 20 25 30 35 40 1 + q + 2 q + 2 q + 3 q + 4 q + 5 q + 6 q + 7 q + . . . and then allowing for pennies by dividing by 1-q : (c73) %/(1-q) 1 (d73)/T/ ------------------------------------ = 5 10 25 (1 - q) (1 - q ) (1 - q ) (1 - q ) 2 3 4 5 6 7 8 9 10 1 + q + q + q + q + 2 q + 2 q + 2 q + 2 q + 2 q + 4 q 11 12 13 14 15 16 17 18 + 4 q + 4 q + 4 q + 4 q + 6 q + 6 q + 6 q + 6 q 19 20 21 22 23 24 25 26 + 6 q + 9 q + 9 q + 9 q + 9 q + 9 q + 13 q + 13 q 27 28 29 30 31 32 33 + 13 q + 13 q + 13 q + 18 q + 18 q + 18 q + 18 q 34 35 36 37 38 39 40 + 18 q + 24 q + 24 q + 24 q + 24 q + 24 q + 31 q + . . . But dividing by 1-q is the "undifference" operator, replacing the nth coefficient by the running sum of the coefficients from the 0th thru nth. Obviously! To allow for pennies, make change for anything, then pennies quantum sufficit. Then I illustrated the situational nature of the question by revoking the pennies, but including the 2, 3, and 20 cent pieces from 19th century coinage: (c74) d72/(1-q^2)/(1-q^3)/(1-q^20); 1 (d74)/T/ -------------------------------------------------------- = 2 3 5 10 20 25 (1 - q ) (1 - q ) (1 - q ) (1 - q ) (1 - q ) (1 - q ) 2 3 4 5 6 7 8 9 10 11 1 + q + q + q + 2 q + 2 q + 2 q + 3 q + 3 q + 5 q + 4 q 12 13 14 15 16 17 18 19 + 6 q + 6 q + 7 q + 9 q + 9 q + 10 q + 12 q + 12 q 20 21 22 23 24 25 26 + 17 q + 15 q + 19 q + 20 q + 22 q + 27 q + 27 q 27 28 29 30 31 32 33 + 30 q + 34 q + 35 q + 44 q + 42 q + 49 q + 52 q 34 35 36 37 38 39 40 + 56 q + 65 q + 66 q + 72 q + 79 q + 82 q + 97 q + . . . However, there were also half-cents. Dividing by 1-sqrt(q) gives ... + 1046 q^40 + ... . Restoring pennies gives an amazing 10135 ways to make 40 cents. Quick: How many ways if we just exclude half-cents? Answer: 1046 again! It doesn't matter if we fill out a lesser amount with pennies or half cents. Two new mathematical truths offered as a contribution to education to be used only by the subscribers to Math-Fun free of cost by paying any royalties whatever on the same, provided it is accepted and adopted by the official action of the membership: It has been found that the relationship of Fahrenheit to Centigrade is not C = (F - 32) 5/9, as previously supposed, but rather C = (F - 32) sin (3 pi/16), which is approximately C = (F - 32) .555570 and more elegantly C = (F - 32) sqrt(2-sqrt(2-sqrt(2)))/2. Moreover, the ratio of feet per second to miles per hour, previously believed to be 22/15 (= 1.4666...), is in reality, pi/(pi-1) (= 1.4669...) when one remembers to allow for the curvature of the Earth. The remainder of this contribution is quotes from that 7th grade Prentice Hall "astronomy" book, with my comments and questions in [brackets]. p17. You are completely uprepared for the spectacle that awaits you when you put on your glasses. To your left you notice a mass of brilliant stars that shine mainly with X-rays. [Gee, then how come the Hertzsprung-Russell diagram (p34?) only goes up to blue on the temperature scale?] p18, fig 1-8 Each galaxy in this cluster of galaxies might hold several trillion or more stars. [10^12? Are there hundredfold bigger galaxies than Andromeda out there?] p32 fig1-24 The brightness of these stars in theTrifid Nebula, as seen from Earth, is called absolute brighness. What term is used for the actual brighness of a particular star? [HUH??] p33: The Hertzsprung-Russell diagram is the single most important diagram astronomers use today. On the Hertzsprung-Russell diagram, the surface temperature of stars is plotted along the horizontal axis. The absolute magnitude, of actual brightness, of stars is plotted along the vertical axis. [It's a scatter plot with about twenty stars, all named, and grouped into either supergiants, giants, main sequence, or dwarfs. But there is absolutely no clue that this is anything but a poop sheet on these particular twenty stars. Why or how is this diagram is used? Later, in the section] Measuring star distance [they "explain" parallax and redshift, but don't mention H-R:] To determine the distance to a star more than 100 light years away, astronomers use the brightness of the star. They plug the star's apparent magnitude and its absolute magnitude into a complicated formula. The formula provides a close approximation of the distance to that star. [Well, if the formula is complicated enough, there's no telling what it might reveal. About redshift, they say] As you have read, light from a star moving away from the Earth has a red shift in its spectrum. Astronomers measure the amount of red shift in a star's spectrum and use complex formulas ... [The book earlier mentions that farther objects look dimmer and redder, but apparently this is not useful in "Measuring star distance".] [The parallax illustration has the star at about 1 AU, i.e., about a 90 degree swing. No mention of how wildly out of scale this drawing is, with the nearest star's actual parallax about 1/3 arc second, and all others less. But that would require numbers. And not just ordinary numbers. Big numbers (shudder).] [Even photographs can be misleadingly out of scale. A (necessarily) overexposed photo of Sirius shows Sirius B nearly submerged in the glare of Sirius A. I'll bet even the teacher would agree that B must be barely outside A's corona, when in fact, the photo only establishes which is brighter, not which is larger. The book blithely depicts brighter stars with bigger dots, and nowhere explains how small the images of stars should really be.] p42 [End of section problems include:] List and describe the four main layers of the sun. [Diagram p39 "Layers of the sun", shows corona, chromosphere, photosphere "about 550 km thick", and core, drawn with < 1/3 solar radius. This leaves 96% of the sun's volume unnmamed! Then they ask for the names of four kinds of solar storm. The book lists flares, prominences, and sunspots, but damned if I can find a fourth. The teacher probably couldn't either, since she omitted this set from the homework.] p25 According to the [big bang] theory, all the matter and energy in the universe was [a lucky mistake] once concentrated into a single place. This place, of course, was extremely hot and dense. Then some 15 to 20 billion years ago, an explosion--the big bang-- shot the concentrated matter and energy in all directions. The fastest moving matter traveled farthest away. Energy, too, began moving away from the area of the big bang. [I.e., there was already a big, empty universe, and a bang went off in the middle.] p27 Quasars If the universe is expanding, then objects near the edge of the universe are the oldest objects in the universe. Put another way, these objects took longer to reach their current position than objects closer to the center of the universe did. [HUH??] p46 In a massive star, gravity continues to pull together the carbon atoms in the core. When the core is squeezed so tightly that the heat given off reaches 600,000,000 deg C, tbe carbon atoms begin to fuse together to form new and heavier elements such as oxygen and nitrogen. The star has begun to become a factory for the production of heavy elements. The core of the massive star is so hot that fusion continues until the heavy element iron forms. But not even the tremendous heat of a massive star can cause iron atoms to fuse together. [Wouldn't tremendous heat *fission* heavier elements? What sort of heat would fission Pb, e.g.? Maybe a big ball of lead would burn via fission?] [According to, e.g., http://csep10.phys.utk.edu/guidry/violence/supernovae-info.html, iron production is so weakly exothermic as to only transpire in the final week before supernova, if at all. Doesn't most iron come from the supernova itself?] Supernovas By the time most of the nuclear fusion in a massive star stops [why?], the central core [is there a peripheral core?] is mainly iron.[?] Although the process is not well understood, the iron atoms begin to absorb energy. Soon this energy is released, as the star breaks apart in a tremendous explosion called a supernova. [Oh, is THAT what happens! No wonder Geritol cures tired blood.] --rwg PS, July 2 is the 36th anniversary of X-ray astronomy, inadvertently initiated by US spy satellites. Absent gamma ray bursts, what's the next best evidence for black holes?

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Re: [math-fun] Signatures of irrational numbers
by lkmitch＠att.net 28 May '03

28 May '03

My bad. A fractal sequence needs to be infinitive, which means that every positive integer shows up an infinite number of times. Here's what Wolfram has to say about them: http://mathworld.wolfram.com/FractalSequence.html Kerry -- lkmitch(a)att.net www.fractalus.com/kerry > > What is a fractal sequence? The sequence 1,1,1,1,... also possesses > the above mentioned property, but would not reasonably be called > fractal. >

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Re: [math-fun] Signatures of irrational numbers
by Henry Baker 28 May '03

28 May '03

I think that the constant function could be construed as "self-similar", but also boring. At 12:32 PM 5/28/03 -0700, Eugene Salamin wrote: >--- lkmitch(a)att.net wrote: >> ... >> And the signature begins: 1, 2, 1, 3, 2, 4, 1, 3, 5. If you strike >> the first >> occurence of every integer in the sequence, you get the original >> sequence back, >> which makes this a fractal sequence. >> ... > >What is a fractal sequence? The sequence 1,1,1,1,... also possesses >the above mentioned property, but would not reasonably be called >fractal.

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[math-fun] Small world phenomenon
by Henry Baker 27 May '03

27 May '03

>From the May 2003 MAA College Math Journal: "John Baez, an expert on mathematical physics at the University of California at Riverside, is Joan Baez's younger brother!" John Baez's home page: http://www.math.ucr.edu/home/baez/ He writes a column: "This Week's Finds in Mathematical Physics", which is very cool. He has a very nice paper on Octonions on his web site. http://www.math.ucr.edu/home/baez/Octonions/

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[math-fun] John Baez
by asimovd＠aol.com 27 May '03

27 May '03

Someone wrote: << >From the May 2003 MAA College Math Journal: "John Baez, an expert on mathematical physics at the University of California at Riverside, is Joan Baez's younger brother!" >> For what it's worth, John Baez has told me that he's Joan's cousin. --Dan "not his son" Asimov

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[math-fun] Puzzle
by David Wilson 25 May '03

25 May '03

Puzzle/sequence/programming exercise inspired by vague question on Mathpuzzle site. Given k, what is the fewest number of terms needed to get the best possible approximation to pi (or e) as a sum of (positive) fractions with denominators <= k? For example, if k = 8, the closest you can come to pi is 377/120 (Fib/Fac!), and this can be done with at best three terms (2/3 + 3/5 + 15/8).

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[math-fun] infinitary Zeta
by wouter meeussen 24 May '03

24 May '03

hi all, could anyone point me to a low-level text about Zeta functions? I'd like to play with the idea that since Sum( k=1..inf; mu(k)/k^w) = 1/Zeta(w) , an infinitary iZeta(w) could be defined by iZeta(w) = 1/Sum( k=1..inf; imu(k)/k^w) where imu(k) is the infinitary Moebius function A064179 (same as mu(k), except its zero's are turned into +1 or -1 depending on the sum of the binary digits over all prime exponents of k being even or odd). What properties of Zeta would be conserved in this iZeta ? Is iZeta(1) = inf ??? Wouter.

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[math-fun] wearable math
by James Propp 23 May '03

23 May '03

For a number of years I've been creating mathematical tee-shirts and selling them at cost. My latest effort of this sort is now available, for roughly $13.50 per shirt plus postage. (I won't know the price per shirt until I know the number of shirts that are ordered, since there are some fixed costs that don't depend on the number of shirts and will be distributed equally among all buyers.) This shirt represents work done during the past two years by participants in R.E.A.C.H. (Research Experiences in Algebraic Combinatorics at Harvard). The students did most of the work on the design of the shirt, and are still tweaking the details, but you can get a good idea of what the shirts will look like by going to http://www.math.wisc.edu/~propp/reach/newshirt.html Shirts will be white, and available in sizes S, M, L, X, and X-squared. If you're interested in ordering a shirt, please place your order with REACH-member Gabriel Carroll (gcarroll(a)fas.harvard.edu) Gabriel can also answer any questions you may have. (Don't reply to me, as I'll be incommunicado until June.) Be sure to include the address you want the shirts sent to. We'll continue to take orders throughout the summer, but we may run out of supplies. We'll be placing the order with the printer on or shortly after June 1. So, to make sure you get all the shirts you want, please send in your order by MAY 31. Jim Propp

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