November 2017 - mathfun@mailman.xmission.com

[math-fun] square root approximating sequence ?
by Henry Baker 27 Nov '17

27 Nov '17

I'm looking for a simple approximation to sqrt(abs(b*c)) that goes something like this: b*c=b_1*c_1=b_2*c_2=...=(sign(b)*sqrt(b*c))*(sign(c)*sqrt(b*c)) where b_(i+1),c_(i+1) are rationally computed from b_i,c_i. In other words, we preserve the product while approximating the square root. What I'm really after is a sequence: [a b] [a b_1] [a b_2] [c d], [c_1 d], [c_2 d], ... such that the determinant is preserved, but each matrix is computed from the previous one by X.M.Y, where X,Y are 2x2 matrices rationally computed from the entries of M.

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[math-fun] Mobius motion
by James Propp 27 Nov '17

27 Nov '17

I'm planning to make a short video about hexaflexagons in which I'll demonstrate a kind of move I haven't seen described elsewhere: sliding the band "along" itself. This is not to be confused with flexing, which causes the band to slide "through" itself. "Along", "through" --- ordinary English prepositions aren't much help here. Is there any existing terminology for describing these kinds of motions of a band? To be clearer (or at least more techincal) about what I mean: Imagine we coordinatize the band with coordinates theta and y, where theta goes from 0 to 2pi and y goes from -1 to +1. These coordinates hide the details of the imbedding (and the three half-twists), but for present purposes that's desirable. If you imagine making each point (theta,y) travel in a circle of radius |y| lying in a plane locally perpendicular to the theta-axis, then what you're doing (more or less) is flexing the hexaflexagon. In contrast, what I'm talking about is moving the point (theta,y) along the band parallel to the theta-axis. I've gone through all Martin Gardner's writings, and watched a few YouTube videos about hexaflexagons, but haven't found anything along these lines. Jim Propp

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[math-fun] Geoff Tootill obituary: factored 2^18
by Henry Baker 27 Nov '17

27 Nov '17

FYI -- The Guardian doesn't usually screw up scientific/technical/mathematical things, but mathematically mature babies are rolling their eyes... Perhaps it was 2^18-1 ? 2^18+1 ? BTW, if the total instruction store only holds 32 instructions, how complex could the program be? How long could the longest-running program that *halts* run (in terms of program steps) ? https://www.theguardian.com/global/2017/nov/08/geoff-tootill-obituary Geoff Tootill obituary Computer engineer who developed 'Baby', the world's first stored-program computer Martin Campbell-Kelly Wednesday 8 November 2017 15.53 GMT ... In order to test the memory when it was constructed, Kilburn and Tootill designed an elementary computer, officially known as the Small-Scale Experimental Machine, but better known as "Baby". The computer could store just 32 instructions or numbers using a single cathode ray tube. The machine first worked in June 1948, taking 52 minutes to find the highest factor of 2^18, involving about 3.5m arithmetic operations. ---- Even less forgivable, *SlashDot* also screwed this one up: https://hardware.slashdot.org/story/17/11/26/197221/computer-pioneer-geoff-…

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[math-fun] The OEIS Foundation needs a new Treasurer
by Neil Sloane 26 Nov '17

26 Nov '17

Dear Math Fun: The OEIS Foundation is looking for a new treasurer, to start right away. Must be a US resident. Will be made a Trustee of the Foundation. Will not be paid (Trustees contribute their services "Pro Bono"). Main duties are to pay occasional bills and to process donations. Should not take more than an hour or two per week. By the way, the Trustees are a very distinguihed group: see http://oeisf.org/trustees.txt. Neil Sloane, njasloane(a)gmail.com

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[math-fun] More random polygons: fun puzzle
by James Propp 26 Nov '17

26 Nov '17

Given two parallelograms A,B in the plane and a translation vector v, let P(v) be the intersection of A+v with B. If we choose a random v such that P(v) is nonempty, how many sides on average do we expect P(v) to have? The set of such vectors v is a compact subset of R^2, so it's clear what probability measure to use: ordinary Lebesgue measure, rescaled. There is a nice answer to my question and at least one nice proof. I'll be curious to see proofs different from the one I found. I'm happy to answer requests for clarification, but no spoilers till Sunday, please. Jim Propp

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[math-fun] more fall-out
by Wouter Meeussen 26 Nov '17

26 Nov '17

E^(-(1/12) + Derivative[1, 0][Zeta][-1, 1/2 + n])*Glaisher*BarnesG[3/2 + n]*Gamma[1/2 + n]^(-(1/2) - n) == 1 since Product[ -i+j+n, {i,n}, {j, 1-i+n}]== Product[(2*k-1)!/(k-1)!,{k,1,n}] according to A086205 and thus, again Product[(2 k - 1)!/(k - 1)! /(2 k - 1)^(n - k + 1), {k, n}] == 2^((n - 1) n/2) (* why? proof? *) ok, I’ll stop this now ;-) wouter.

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[math-fun] The sin (pi / 18) of Ramanujan (generalization)
by François Mendzina Essomba 26 Nov '17

26 Nov '17

General formula giving $ p-1 $ sign of $ - $ and $ q $ sign of $ + $ with a period of $ p + q - 1 $ \\ latex-code. we will consider the two following cases: \\ case of $ p = 2k + 1 $ odd \begin{align} 2\sin\left(\dfrac{(2^{p+q-1}+2^{q+1}-3)\pi}{3\times 2^{p+q}-6}\right)&=2\cos\left(\dfrac{(2^{p+q}-2^{q+1})\pi}{3\times 2^{p+q}-6}\right)\\& =\sqrt{2-\sqrt{\cdots-\sqrt{2+\sqrt{\cdots+\sqrt{2}}}}} \end{align} case of $ p = 2k $ peer \begin{align} 2\sin\left(\dfrac{(2^{p+q-1}-2^{q+1}+3)\pi}{3\times 2^{p+q}+6}\right)&=2\cos\left(\dfrac{(2^{p+q}+2^{q+1})\pi}{3\times 2^{p+q}+6}\right)\\& =\sqrt{2-\sqrt{\cdots-\sqrt{2+\sqrt{\cdots+\sqrt{2}}}}} \end{align} We arbitrarily choose p and q, For example, if we want (04) sign of (-) and (05) sign of (+), we will have: p=4+1=5 ; q=5 ; \begin{align} 2\sin\left(\dfrac{(2^{5+5-1}+2^{5+1}-3)\pi}{3\times 2^{5+5}-6}\right)&=2\cos\left(\dfrac{(2^{5+5}-2^{5+1})\pi}{3\times 2^{5+5}-6}\right)\\& =\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}}}}}}} \end{align} we are getting: \begin{align} 2\sin(\dfrac{191\pi}{1022})=2\cos(\dfrac{320\pi}{1022})=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}}}}}}} \end{align} An approximate evaluation under maxima gives: 2*sin((191*Pi)/(1022))=sqrt(2-sqrt(2-sqrt(2-sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2-sqrt(2-sqrt(2-sqrt(2-sqrt(2))))))))))))));

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[math-fun] who can't count, doesn't
by Wouter Meeussen 25 Nov '17

25 Nov '17

https://math.stackexchange.com/questions/2498295/counting-semistandard-youn… funny fail in Mathematica 11 for the simple expression like (2^1 4^2 6^3 8^4 10^3 12^2 14^1) / (1^7 3^5 5^3 7^1) when presented like this: \frac{\prod _i^n \prod _j^{-i+n+1} (-i+j+n)}{\prod _k^n (2 k-1)^{-k+n+1}} it produces the quite unsavory \frac{2^{\frac{1}{24} \left(-12 n^2-12 n+1\right)} \pi ^{\frac{n}{2}+\frac{1}{4}} \Gamma \left(n+\frac{1}{2}\right)^{-n-\frac{1}{2}} e^{\zeta ^{(1,0)}\left(-1,n+\frac{1}{2}\right)+\frac{1}{24}} \prod _i^n \prod _j^{-i+n+1} (-i+j+n)}{\sqrt{A}} where A is the Glaisher constant (~ 1.282427..) having read the above link, I had naively expected just 2^(n(n-1)/2) . Bummer! Wouter.

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[math-fun] The plural of "dimension"
by Henry Baker 25 Nov '17

25 Nov '17

is dimentia? As in: "Space-time consists of one time dimension and three spatial dimentia". Or, "the vector space Q(r), r a root of x^5-6x+3, has 5 dimentia". Or, "Flatland: A Romance of Many Dimentia".

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[math-fun] a morning not very rich
by françois mendzina essomba2 24 Nov '17

24 Nov '17

hello... for all x, sum(((-1)^i*2^(2*n-i)*(cos(x))^(2*n-i)*cos(i*x))/(i!*(2*n-i)!),i,0,2*n-1)=0; FME...

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