December 2018 - mathfun@mailman.xmission.com

[math-fun] Draft of December 31 blog post
by James Propp 28 Dec '18

28 Dec '18

I've written something different for this month: a (mostly) tongue-in-cheek promotion for an upcoming mathematical holiday called "Thirdsday". (The holiday is bogus; the math is real.) You can read the current draft at http://mathenchant.org/043-share.pdf . Suggestions are welcome, especially suggestions for mathematically interesting occurrences of the number 1/3. (I'm saving "the volume of a pyramid is 1/3 times the area of the base times the height" and "the integral of x^2 from x=0 to x=1 is 1/3" for the next Thirdsday, just over a decade from now.) Jim Propp

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Re: [math-fun] Amusing Permutation Representations of Group Extensions
by Dan Asimov 28 Dec '18

28 Dec '18

Nice pictures in that paper! (I know nothing about group representations, and don't even know the definition of a wreath product.) Funny, just last night I was musing about how for any finite group G the fewest number N = N(G) of points on which it can act faithfully is an interesting invariant. Which is just finding G as a subgroup of the permutation group S_N. For instance, what is this number for the rotation groups of the regular polyhedra: T (order 12), O (order 24), I (order 60) ? Each group acts faithfully on the face centers, edge midpoints, and also the face centers of the dual polyhedron (for O and I). For the octahedron and icosahedron, and for the edges of the tetrahedron, antipodal points get rotated to antipodal points, so we can divide all these numbers by 2 except for the 4 face centers of the tetrahedron. Hence T permutes 4 or 3 = 6/2 points, O permutes 4 = 8/2 or 6 = 12/2 or 3 = 6/2 points, and I permutes 10 = 20/2 or 15 = 30/2 or 6 = 12/2 points. But wait! We can't represent T or O faithfully on 3 points! S_3 is only order 6, while these groups are bigger. Which of these 8 cases are faithful (represent all group elements as distinct permutations)? Can we do better than 4 for T (clearly not), 6 for O, and 12 for I ? —Dan Joerg Arndt schrieb: ----- I thought some folks here would enjoy the paper https://arxiv.org/abs/1812.08475 Abstract: Wreath products of finite groups have permutation representations that are constructed from ... -----

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[math-fun] formulae
by Bernie Cosell 28 Dec '18

28 Dec '18

There are often posts to math-fun that obviously include formulae. Such as today's <math>e^{i \pi} + 1 = 0</math> That looks like an HTML tag, but the messages don't come through as HTML [to me at least: Content-Type: text/plain; charset="utf-8"] so if I try to open a message in my browser it is a mess. Do you folks all read math-fun with email clients that magically handle the <math> tag in a plain text message? Is there some trick to being able to view the forumlae? thanks /Bernie\ Bernie Cosell bernie(a)fantasyfarm.com -- Too many people; too few sheep --

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[math-fun] Amusing Permutation Representations of Group Extensions
by Joerg Arndt 28 Dec '18

28 Dec '18

I thought some folks here would enjoy the paper https://arxiv.org/abs/1812.08475 Abstract: Wreath products of finite groups have permutation representations that are constructed from the permutation representations of their constituents. One can envision these in a metaphoric sense in which a rope is made from a bundle of threads. In this way, subgroups and quotients are easily visualized. The general idea is applied to the finite subgroups of the special unitary group of (2Ã2)-matrices. Amusing diagrams are developed that describe the unit quaternions, the binary tetrahedral, octahedral, and icosahedral group as well as the dicyclic groups. In all cases, the quotients as subgroups of the permutation group are readily apparent. These permutation representations lead to injective homomorphisms into wreath products. Comments: 42 figures included to blow your mind Best regards, jj

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[math-fun] Permutation question
by David Wilson 28 Dec '18

28 Dec '18

Let P be a permutation of {1, 2, ..., n}. Let the longest initial decreasing prefix of P have k elements. For example, let n = 5 and P = (5, 3, 2, 4, 1) Then the longest decreasing prefix of P is (5, 3, 2), so k = 3 and P_k = 2. In terms of n, what is the expected value of k? of P_k?

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Re: [math-fun] playing card color guessing game
by Dan Asimov 27 Dec '18

27 Dec '18

Although I believe this: ----- Hmm, strategy seems clear: guess the majority color remaining, if any ----- , I'm a lot less sure of the rarish case when it's a tie: Is it wiser to, say, alternate your guesses when it's a tie (to pick up, you know, a second-order advantage, now that you have maxed out your first-order advantage), or is it entirely irrelevant (to your expected gain) what you pick at those times? Actually, now that I'm writing it, I kind of sort of *think* that it is completely obvious that it's entirely irrelevant. On the other hand, suppose that instead of only considering the *expected* value of this game we view it as a 2-player thang. You are playing against an opponent — each playing in isolation from the other — but with the winner being *not* necessarily the person who is ahead by the end. Instead, what if the winner is defined to be *the one who was *leading* more*. Where a scorekeeper has a running tally of each player's progress through the game. Each player gets one metapoint for each completed t'th stage during the ordinary game, 1 <= t <= 52, at which they are strictly ahead of the other. If at the end of the game *leading more* (i.e., having the most metapoints) were the only thing that mattered, would that affect strategy at ties? Now: Onward, to the third-order effects. —Dan

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[math-fun] 3 was not Dedekind's lucky number
by Bill Gosper 27 Dec '18

27 Dec '18

DedekindEta[1/2 + I] == ((-1)^(1/24) 2^(11/16) (1/4)!)/((-1 + Sqrt[2])^(1/4) \[Pi]^(3/4)) == (2^(5/16) ((2 + Sqrt[2]) (I + Sqrt[3]))^(1/4) (1/4)!)/\[Pi]^(3/4) but DedekindEta[1/3 + I] == (1/(\[Pi]^(3/4))) 2 (1/81 (15 + 8 Sqrt[3]) + (2 I ((6 + 7 I) + (7 + 2 I) Sqrt[3]))/( 9 (611 + 344 Sqrt[3] + 77 Sqrt[2 (98 + 55 Sqrt[3])])^(1/3)) + 2/81 (1 + I Sqrt[3]) (611 + 344 Sqrt[3] + 77 Sqrt[2 (98 + 55 Sqrt[3])])^(1/3))^(1/12) (1/4)! Do these (often dreadful) algebraic factors always come out in radicals? —rwg

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[math-fun] Palindrome Coin (was Re: New Mersenne; GIMPS Coin? Why not GIMPS for POW?)
by Keith F. Lynch 27 Dec '18

27 Dec '18

"Adam P. Goucher" <apgoucher(a)gmx.com> wroe: > Henry Baker <hbaker1(a)pipeline.com> wrote: >> Why not use GIMPS as proof-of-work for a "coin" that is only minted >> ~1X per year? > The main reason is that proof-of-work solutions should be hard to > find and easy to verify. > Mersenne primes are hard to verify. I sympathize with the desire that the proof-of-work do something mathematically interesting. I propose finding dual palindromes. Preferably binary-ternary, as those seem to be the rarest. It's difficult to discover that 24014998963383302600955162866787153652444049 is a palindrome in both binary and (standard) ternary, but easy to confirm that it is one once someone makes the claim. Before anyone claims that that's not interesting, look up the tasks that Bitcoin mining solves. They're *designed* to be as ugly, boring, random, and unsymmetrical as possible, lest someone find an elegant shortcut and corner the market.

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Re: [math-fun] playing card color guessing game
by Dan Asimov 26 Dec '18

26 Dec '18

Hmm, strategy seems clear: guess the majority color remaining, if any. So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum): E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc. —Dan Allan Wechsler wrote: ----- I know how to analyze this, but I haven't tried to actually do the work. But I'm going to make a wild-ass stab based on an intuition. Is it $32.87? On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck(a)gmail.com> wrote: > the usual 52-card playing deck is shuffled and then revealed one-by-one to > a player who guesses the color (red or black) of each card prior to its > being revealed. the player earns one dollar for each card whose color he > guesses correctly; there is no penalty for being wrong. what is a fair > entry price to play? ----=-

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[math-fun] playing card color guessing game
by Thane Plambeck 26 Dec '18

26 Dec '18

the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play?

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