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[math-fun] Geoff Tootill obituary: factored 2^18
by Henry Baker 26 Nov '17

26 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 23 Nov '17

23 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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[math-fun] Cooking a turkey with potential energy
by Henry Baker 23 Nov '17

23 Nov '17

FYI -- I wonder if this would work from MIT's Green Building? www.ohio.edu/mechanical/thermo/Intro/Chapt.1_6/energy/CookingPE.pdf >From the Journal of Irreproducible Results, 1987.

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[math-fun] Robust convex closure
by Allan Wechsler 22 Nov '17

22 Nov '17

Suppose you want to distribute a small given finite number of points around the origin of the Cartesian plane so that the origin is in the interior of the convex closure of the points. It's easy - you can't do it with two points, but you can do it with three. Now suppose you want to ensure that the origin will still be inside the convex closure even if one of the points is removed by an evil entity. I have pretty much convinced myself that you can't do it with four points, though I would appreciate a clear proof. You can do it with five; a regular pentagon centered on the origin clearly works. Now what if the evil entity is allowed to remove any two points? How many points do you need to be safe? I think the answer is six (a regular hexagon works). Seven points defends against three removals, but I think you might need nine to defend against four. I suspect the answer in general is that you need more than twice as many points as the evil entity is allowed to remove, but I don't have a proof.

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