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[math-fun] Has someone repealed the laws of physics?
by Mike Speciner 31 Jul '06

31 Jul '06

Science News reports in "Out of Sight" [15 July 2006, vol 170 no 3 p 42]: "Electromagnetic radiation typically exerts a repulsive pressure on objects, so it should, in theory, attract metamaterials. Also, the well-known Doppler shift, in which the frequency of electromagnetic radiation from an object increases as the object approaches, is expected to reverse for metamaterials. Experiments have yet to confirm those two predictions." IMHO, attractive pressure from EM radiation would violate conservation of energy and momentum, while reverse Doppler shift would violate relativity. Am I confused, or should Science News be embarassed? --ms

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[math-fun] asinh (hyperbolic arc sine) identities ?
by Henry Baker 31 Jul '06

31 Jul '06

Hyperbolic arc sine function is an interesting function which is 1-1, linear at the origin and logarithm towards +-oo. I'm looking for interesting identities involving this function. In particular identities for things like asinh(sinh(x)+c), c constant asinh(sinh(x)*c), c constant asinh(sinh(x)+sinh(y)) asinh(asinh(x)*asinh(y)) Also, identities that allow asinh(x) to be computed recursively in terms of itself (shifts up or down). If these don't work so well, then identities involving acosh or atanh, but neither of these functions are particularly interesting due to their reduced domains.

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[math-fun] MathPath Math Camp
by Henry Baker 29 Jul '06

29 Jul '06

Congratulations to Funster Conway on "MathPath" summer math camp for kids! It got a first-page article in today's (Sat 7/29) Wall Street Journal. Here's the MathPath web site: http://www.mathpath.org/ Here's the entrance exam (not easy!) in pdf format: http://www.mathpath.org/Quiz/quiz.2006.pdf ---- I had great fun in the 1960's with the NSF Summer Science Training Program math course when I was in high school, and it sounds like these camps are filling this important void.

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Re: [math-fun] Calculating pi
by dasimov＠earthlink.net 26 Jul '06

26 Jul '06

Jason writes: << . . . Let's see if I can write it as a formula instead of an algorithm: n n 4 ___ ___ --- \ \ pi ~ n*n * / / 1 if sqrt(x*x+y*y) < n --- --- x=0 y=0 Does that look about right? . . . >> Yes. In fact one of the old classic math books (Geom. & the Imagination? What is Mathematics? Math. & the Imagination?) uses this observation to obtain the series pi/4 = 1 - 1/3 + 1/5 - 1/7 + . . . . --Dan

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[math-fun] The mathemagician and pied puzzler: a collection in tribute to Martin Gardner
by Thane Plambeck 25 Jul '06

25 Jul '06

Funsters might be interested to know that I just finished a little project for Tom Rodgers and AK Peters, Ltd. to put the entire 1999 book The mathemagician and pied puzzler: a collection in tribute to Martin Gardner online, as separate PDF files, one per article (or one big one, if you have the patience to download it). You can find it here http://g4gardner.pbwiki.com Please let me know if there are any problems with it -- Thane Plambeck http://www.plambeck.org/ehome.htm

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[math-fun] Buried Cable problem
by Michael Kleber 24 Jul '06

24 Jul '06

A week or two ago I heard for the first time the following enjoyable brain teaser: --------- A company has just buried a 10-mile-long cable containing 120 individual wires. Unfortunately, the project was overseen by a summer intern, and the sad result is that the wires are entirely unlabelled. Your job is to fix this mess: label each wire with the same name on both ends. The tools you have available are (1) a battery, (2) a lightbulb, and (3) an unlimited amount of patch cord. And (4) your feet, and no other form of transportation. You start at one end of the cable, and can do as much as you want there, but then you need to walk down to the other end and repeat. In how few trips down the length of the cable can you label the wires? --------- I have a second problem for follow-up, but I can only present it to people who have solved the first problem (for general n, of course, not just n=120). So tune in tomorrow for my tale of the ups and downs in solving this one, and perhaps help me out of my end state (down). --Michael Kleber p.s.: An interesting observation is that with n=2 the problem is unsolvable; there is no way to tell the two wires apart. -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

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[math-fun] Blender help
by Bernie Cosell 22 Jul '06

22 Jul '06

My wife isn't exactly sure where to turn. She's been struggling mightily with Blender and losing at every turn. We bought the [800 page] book, but it really doesn't address the problems she's running into and she's going nuts constantly fighting with the program. As I mentioned a few months back, what she wants to do is "geometric" stuff [for example, a currrent hack: two intersecting tetrahedra and then be able to highlight and play with the octahedron that shows up in their intersection]. I'd have thought that that kind of thing would be fairly easy [extrapolating frm the similar 2-D stuff I've done with both Illustrator and Visio], but apparently not. The types of things that are making her go bonkers: 1) the edges. Since Blender deals with 3D objects, you need to have some kind of 'strut' to use for the edges. She decided on using triangular-cross-section struts. But then to make that work out properly, she apparently had to hand-compute the intersections between the structs [at the odd angles that they show up in in polyhedra] and by-hand "mitre" the struts so that they fit "just right" with no points sticking out. Does that sound right? [she and I ended up struggling to remember high-school trig and were throwing around lots of sqrt(3)s and such, but I can't imagine that an "artist type" who would want some kind of polyhedron or the like would even know HOW to do the 8-decimal- place calculations, much less want to bother to be hand-trimming each strut. She/I/We *must* be missing something. 2) Faces. Apparently, in order to do faces she needed to compute-by- hand the coordinates of the vertices and then fit a face to it, but apparently once she does that the vertices are "gone" and if she needs to remove or redo the face, she has to start over with the vertex calculations... [not to mention just having to *compute* the coordinates of the vertices of a polyhedron feels very wrong to me]. My guess is that she's missing some key intution on how to use blender for this type of application [partly extrapolating from my experience with "2D" tools, where I don't do any of that stuff: I use skews and rotates and copying/pasting and such and let the program do all the heavy-trig-lifting; I use layers to allow me to "scribble" and test fitting pieces in over stuff that was already done, but that allows me to easily wipe it away if I screw it up]. The problem is that we don't know where to turn to try to figure out what's going on -- virtually all of the Blender info is aimed at *animation* [which is not what my wife is interested in - what she wanted/needed was a 3D modeling/drawing program] and *artists* [rather than geometers or engineers or the like]. Do any of you know of any Blender resources that might help shed some light on this kind of use of the program? THANKS!!! /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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[math-fun] Coffin for sale
by Michael Kleber 21 Jul '06

21 Jul '06

If there's anyone out there who, like I did, wants to own a hard copy of Stewart Coffin's book _The Puzzling World of Polyhedral Dissections_ (available free on line), permit me to point out that abebooks.com has just listed a copy for the unusually low price of $27.50. http://www.abebooks.com/servlet/SearchResults?tn=The+Puzzling+World+of+Poly… --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

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[math-fun] Question-sequence probability
by Guy Haworth 20 Jul '06

20 Jul '06

There are 9! sequences for the 9 questions. So there are (9!-1) 'other' sequences. There are 9*8/2 = 36 pairs to flip if one flip is required. So there are 1+36 sequences 0 or 1 'flip' away from the 1-9 sequence. Therefore the probability = 37/(9!-1) = 37/362,879, just over 1 in 10,000 g

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Re: [math-fun] How to Compute This Probability?
by Daniel Asimov 20 Jul '06

20 Jul '06

Robert Baillie asks: << . . . the same 9 topics, in the same order, except with items 6 and 9 reversed. Let's assume that there are only these 9 topics to choose from. What is the probability that . . . ? >> I'll answer the question I think was intended: Question: Given a random permutation of {1,...,9}, what is the probability that it moves at most 2 symbols? The # of perms. moving 0 symbols is 1. No perm. can move only 1 symbol. The number of pairs of symbols is 9*8/2 = 36. For each pair, the number of perms. interchanging the pair but moving no other symbols is 1. Thus the answer is 37/9! = 37/362880 = 1.01962+ x 10^(-4) or about 1 out of 10,000. ***************************************************** Statistics tries to address questions of the sort that can't really be answered, like "What is the probability that two permutations of {1,...,9} that differ in only 2 places were actually chosen independently?" The basic method is to define the "null hypothesis" as the assumption that the two perms. were in fact chosen independently. Then (before examining the data) an "N% confidence interval" is chosen to include events whose probability -- *assuming the null hypothesis* -- is at least N%. N is typically selected to be 5 or 1. The decision procedure is then to accept or reject the null hypothesis "at the N% confidence level" according as the event in question falls inside the N% confidence interval or not. This method was originally designed to deal with events having a normal distribution. In a different situation like this one, defining the "confidence interval" can be a bit of an art. Let d(k) be the number of derangements (perms without any fixed points) of k elements. According to A000166 of Neil Sloane's OEIS, d(n) for n = 0,...,9 is given by n | 0 1 2 3 4 5 6 7 8 9 d(n)| 1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496* The number of permutations of {1,...,9} moving exactly k elements, 0 <= k <= 9 is then #(k) = C(9,k)*d(k). My calculator tells me this sequence is k | 0 1 2 3 4 5 6 7 8 9 #(k) | 1 0 36 168 1134 5544 22260 66744 133497 133496 9-k | 9 8 7 6 5 4 3 2 1 0 (Here 9-k is the number of elements FIXED by #(k) different perms. of {1,...,9}.) Now, how to define an N% confidence interval? I'd start with the average number of fixed elements -- call it Favg -- and move outward to get an interval # fixed elements F lies in the interval I(t) = [Favg-t, Favg+t] for some real number t. The corresponding probabilities are of course #(9-F) for each F, and the P(t) := Prob(F lies in I(t)) = sum{F in I(t)} of #(9-F). Thus choose your confidence level of N% = c, say, and find the largest t such that P(t) <= 1-c. Call it t(c). Then I(t(c)) is your confidence interval at the N% level of confidence. --Dan __________________ * Had I checked, I'd have learned that this is part of A098825 of the OEIS *before* doing the calculation!

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