December 2014 - mathfun@mailman.xmission.com

[math-fun] The triangular squares
by Bill Gosper 07 Dec '14

07 Dec '14

Something for the youngsters: gosper.org/triangsq.pdf But you might see a Mathematica trick, e.g. --rwg

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Re: [math-fun] Symbolic derivatives in Mma
by Dan Asimov 07 Dec '14

07 Dec '14

Neil, I had seen that formula but it was a long time ago and I had forgotten what it was for. It appears to be spot-on to answer my question. Thanks!! --Dan << Dan, You about Faa di Bruno's formula, right? Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com << OK, I suppose Mathematica isn't going to answer my real question. Which is: Let u(x) be any infinitely differentiable function. Then if f(x) = exp(u(x)), its nth derivative f^(n)(x) (n >= 1) is a polynomial with positive integer coefficients in the derivatives u', u'', ..., u^(n), times exp(u(x)): f^(n)(x) = P_n(u', u'', ..., u^(n)) exp(u(x)) The polynomial P will be homogenous of (let's call it) "index" n if in each monomial, the factor u^(k) is assigned index k, and the indices of all factors are added. Using x_1,...,x_n as the variables in P_n, we'd have for instance P_3(x_1, x_2, x_3) = (x_1)^3 + 3 x_1 x_2 + x_3 Question: Is there a closed formula for P_n (without recursion) ? Do they have a specific name? What properties do they have? --Dan >> >>

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[math-fun] Marriages and schooling the children
by Warren D Smith 07 Dec '14

07 Dec '14

"How Game Theory Helped Improve New York City's High School Application Process" http://www.nytimes.com/2014/12/07/nyregion/how-game-theory-helped-improve-n… This article (pretty good press math piece) explains how the problem of matching children to schools in NY city is treated as a "stable marriage problem" -- schools & children both provide preference rank-orderings of the other -- repairing a totally brain-dead previous system. However, I suspect there is a better approach. Instead of children & schools providing rank-orderings, have them provide numerical RATINGS on a fixed allowed-score-range such as the real interval [0,100]. (Greater scores preferred.) Then find the pairing (associating of children--schools) that maximizes the sum of the used ratings, subject to valency constraints (e.g. each school can accomodate at most X children, where X depend on the school and is known). This is a max-weight "degree constrained subgraph" problem within a complete bipartite graph and is soluble in polynomial time using known algorithms (theory related to "maximum weight matching" graph problem). Also should be soluble using max-flow min-cut technique to find min-cost flows. Advantages: 1. ratings allow strengths of preference, not pretending all preferences same strength. 2. ratings easier to produce than rank orderings. 3. generically unique optimum solution found, unlike with stable marriages where there are generically many solutions and which is optimum is not known (and often not even defined). So why didn't New York City do it in this better way? 1. Since it really isn't better for some reason? 2. the article lauds on and on about how this system was invented by Nobel prize economists. Well, that was the problem. Stable marriage theory was done in substantial part by economists. Matching theory was done by mathematicians & computer scientists. If we then add as an axiom that Nobel Prize Economists are incapable of knowing anything about anything a non-economist did, we explain this. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Kinda nice anamorphic graffiti
by Marc LeBrun 07 Dec '14

07 Dec '14

Kinda nice anamorphic graffiti: http://io9.com/mind-boggling-optical-illusion-graffiti-looks-like-it-f-16673 50000

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[math-fun] infinite magic squares
by Warren D Smith 06 Dec '14

06 Dec '14

Adam Goucher's argument should indeed work not only for the sequence 1/N, but indeed any decreasing positive sequence with infinite sum; and also for e.g. constructing infinite magic cubes, 4-dimensional magic hypercubes etc. What I like about it is, it shows infinite magic is actually way easier to attain than finite magic (for that, you actually have to work).

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[math-fun] unexpected hangings redux
by Warren D Smith 06 Dec '14

06 Dec '14

I agree this is cute ("cute" & "annoying" being synonyms, apparently)... my question: does it have an actual use?

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[math-fun] Infinite magic squares
by Keith F. Lynch 06 Dec '14

06 Dec '14

An infinite magic square would look something like this: 1/2 1/4 1/8 1/16 1/32 1/64 ... 2/3 2/9 2/27 2/81 2/243 2/729 ... 3/4 3/16 3/64 3/256 3/1024 3/4096 ... 4/5 4/25 4/125 4/625 4/3125 4/15625 ... 5/6 5/36 5/216 5/1296 5/7776 5/46656 ... ... ... ... ... ... ... R/(R+1)^C No number would appear more than once, and every row, every column, and every main diagonal would sum to (i.e. converge to) the same sum. (I know what you're thinking: There can't be an infinite magic square since there's only one main diagonal. But I assert that it's valid if *every* main diagonal works, even if there's only one main diagonal.) In my arrangement above, no number appears more than once, and every row has the same sum. Unfortunately the columns and the main diagonal don't work. Indeed, the first two columns don't even converge. If such a magic square can't be found, or can be proven not to exist, perhaps a semi-magic square can be found. Maybe even by permuting each row of my above non-solution. For instance what about arrangements that always move the largest, left-most, term to the diagonal? What about magic squares that use 1/N for all N and use no other numbers? Consider: 1 1/2 1/6 1/7 1/15 1/16 ... 1/3 1/5 1/8 1/14 1/17 ... 1/4 1/9 1/13 1/18 ... 1/10 1/12 1/19 ... 1/11 1/20 ... 1/21 ... ... in which I just wrote each successive reciprocal on successive cells on successive diagonals, such that you could draw a curve through them in succession without lifting the pen. At least every row and every column converges, as does the main diagonal. Of course they converge to different values. (At least I assume no two sums are the same. Can anyone prove or disprove it?) Is it possible to make an arrangement in which every row and every column converge if you use *all* rationals between 0 and 1, rather than just the ones with a numerator of 1? If not, what about in higher dimensions? I expect it would be very easy to find an infinite magic square if negative values are allowed. So lets keep it positive.

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[math-fun] Seven resistor solution
by Keith F. Lynch 06 Dec '14

06 Dec '14

Lee Sallows <Lee.Sal(a)inter.nl.net> wrote: > ... I do have another in the form of a second resistor problem. > Only this time I'm not going to reveal whether it is unsolved > or not. > I have a set of 7 resistors, all distinct in value. You choose one. > Following which I take the remaining 6 resistors and connect them > together to form a two-terminal network whose total resistance is > the same as that of the resistor you selected. I can do this > whichever one you choose. What are these seven resistors? 1 = (p 2 3 6) 2 = (p 3 6) 3 = (s 1 (p 4 6 12)) 4 = (p 6 12) 5 = (s 1 2 (p 3 6)) 6 = (s 1 2 3) 12 = (s 1 2 3 6) Resistors not mentioned are lying around loose in the two-terminal network. Or, if you prefer, only one side is soldered to anything, and it doesn't matter what. Is that the same as your solution? Who first came up with this puzzle, and when? Thanks. I solved it by hand, but not before wasting time trying to figure out how to write a program to loop over all possible networks of six or fewer resistors.

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[math-fun] Tokieda's big rattleback
by Bill Gosper 05 Dec '14

05 Dec '14

In at least one online recording (that I can't locate) of T. Tokieda's toy talk (e.g., https://www.youtube.com/watch?v=f07KzjnL2eE) instead of his usual small plastic ones (with or without testudinal passengers) he had a large wooden (rosewood?) (outwardly) symmetrical one which he rubbed in his hair (ha ha) to reverse the preferential direction. Can someone suggest where to buy such a thing? Maybe not symmetrical. Maybe not wood. But fairly large? (I've heard of making them from old automobile bumpers. Maybe that's too large.) --rwg

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[math-fun] Do your folks have a big screen monitor? Do you have an iPhone?
by Bill Gosper 05 Dec '14

05 Dec '14

Could you give us a hint about your family's income? Why else would we ask you this ridiculous homework question? Saxon Math Student EBook 2: 8. a. Estimate the length of segment *AB* in inches. ⁎————————————————————————————⁎ *A* *B* b. Measure the length of segment *AB* to the nearest eighth of an inch. (Screenshot: gosper.org/saxongem.png . This is an electronic book.) --rwg

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