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[math-fun] Is mathematical truth "real" ?
by Dan Asimov 11 Apr '18

11 Apr '18

If this common discussion subject among mathematicians has come up in math-fun previously, it must have been a while ago: Is mathematical truth real? I was reminded of this question when I happened to sit next to an esteemed molecular biologist who has strong opinions on the matter: He thinks math is "all in the brain" of humans who think about it. I could not convince him that math has any kind of independent existence — though I certainly believe it myself. What do other think about this? I would say that mathematical truth is *at least* as real of a thing as physical truth. —Dan

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Re: [math-fun] Who's going to G4G13?
by Keith F. Lynch 11 Apr '18

11 Apr '18

Not me, though I'd like to. Please have the people in charge ask the people who run the annual World Science Fiction Convention on how the latter manage to put on an event that lasts twice as long for half the price with no membership limit.

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Re: [math-fun] Miss those IBM/DEC blinking lights?
by Henry Baker 11 Apr '18

11 Apr '18

I'm surprised that auto LED headlights don't already broadcast their vehicle VIN# (maybe they do?). Apparently, one of the countries in the Middle East is now requiring some sort of RFID IFF for automobiles. Sure makes the CIA drone program a lot more accurate... I've heard that Disney is starting to replace some of its nighttime *fireworks* displays with *LED drone fleets* displays. I would imagine that some of these temporal streaks could be useful for these drone displays. At the last several CES's (Consumer Electronics Shows), I talked with vendors making bicycle LED displays, and none of them were very imaginative. I did see a T-shirt with a bit-mapped display, but it *wasn't programmable* !! What weren't they thinking? How come I still can't buy a T-shirt with a bitmapped display on it? -- it might be great for political protests, etc. It might be fun to hack the news cameras by synchronizing the T-shirt display with a GPS-derived 60Hz signal; with proper synchronization, it might be possible to have the TV news cameras pick up completely different content from that perceived by human observers! I seem to recall that someone at MIT had developed an LED display for bicycle spokes that was time-multiplexed in such a way as to provide a full circular bit-mapped display -- kinda like those WWII radar displays. At 01:28 AM 4/10/2018, Bill Gosper wrote: >It should also be possible to Morse-code messages by rapidly blinking LED auto taillights to leave dotted and dashed afterstreaks, if you know how to look.

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[math-fun] The mango lassi problem (and meta-problem)
by James Propp 11 Apr '18

11 Apr '18

I had dinner last night with Alan Frank (the inventor of the muffin problem that was discussed on math-fun a few years back and which Bill Gasarch will be talking about at the MIT Combinatorics Seminar tomorrow), and we encountered a problem about sharing that might turn out to be a sequel to the muffin problem, except that I don't quite know what the right problem is. (I solved the special case I needed to solve, but I'm not sure what the right general setting is.) Dining with me and Alan were three kids. I ordered two mango lassis to split between myself and the three kids, along with two extra glasses, so that each of the four lassi-drinkers would get half a serving. The two glasses containing (identical amounts of) mango lassi were the same shape as one another, and the two empty glasses were the same shape as one another, but the empty glasses were NOT the same shape as the other two glasses. Not a problem! I took a full glass of lassi and divided it evenly between the two initially empty glasses, and then I took the remaining full glass and divided the lassi it contained evenly between it and the glass I'd drained in the first step. But what if the two empty glasses had NOT been the same shape as each other? (We may assume that each is big enough to hold a full serving of mango lassi.) PROBLEM: Show that even if the two empty glasses were different shapes from one another (and from the identical full glasses), one can still divide the lassi into four equal portions. The only allowed operation is to equalize the amount of lassi in two glasses of the same shape by equalizing the height of the lassi. (One might quibble that in practice you can only do this approximately, but for purposes of this puzzle, ignore that nicety.) META-PROBLEM: What's the best way to alter and/or extend this problem, where "best" means "most fun"? Jim Propp Jim Propp

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[math-fun] Viete for inverse sinusoidal functions
by françois mendzina essomba2 10 Apr '18

10 Apr '18

(2/sqrt(2*x/(sqrt(x^2+1))+2))*(2/sqrt(sqrt(2*x/(sqrt(x^2+1))+2)+2))*(2/sqrt(sqrt(sqrt(2*x/(sqrt(x^2+1))+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2*x/(sqrt(x^2+1))+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2*x/(sqrt(x^2+1))+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2*x/(sqrt(x^2+1))+2)+2)+2)+2)+2)+2))*... = sqrt(x^2+1)*acot(x); (2/sqrt(2/(sqrt(-x^2+1))+2))*(2/sqrt(sqrt(2/(sqrt(-x^2+1))+2)+2))*(2/sqrt(sqrt(sqrt(2/(sqrt(-x^2+1))+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2/(sqrt(-x^2+1))+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(-x^2+1))+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(-x^2+1))+2)+2)+2)+2)+2)+2))*... = sqrt(1/x^2-1)*atanh(x); (2/sqrt(2*x/(sqrt(x^2-1))+2))*(2/sqrt(sqrt(2*x/(sqrt(x^2-1))+2)+2))*(2/sqrt(sqrt(sqrt(2*x/(sqrt(x^2-1))+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2*x/(sqrt(x^2-1))+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2*x/(sqrt(x^2-1))+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2*x/(sqrt(x^2-1))+2)+2)+2)+2)+2)+2))*... = sqrt(x^2-1)*acoth(x); (2/sqrt(2*x+2))*(2/sqrt(sqrt(2*x+2)+2))*(2/sqrt(sqrt(sqrt(2*x+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2*x+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2*x+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2*x+2)+2)+2)+2)+2)+2))*... = acosh(x)/sqrt(x^2+1); (2/sqrt(2*sqrt(x^2+1)+2))*(2/sqrt(sqrt(2*sqrt(x^2+1)+2)+2))*(2/sqrt(sqrt(sqrt(2*sqrt(x^2+1)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2*sqrt(x^2+1)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2*sqrt(x^2+1)+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2*sqrt(x^2+1)+2)+2)+2)+2)+2)+2))*... = asinh(x)/x; (2/sqrt(2*sqrt(-x^2+1)+2))*(2/sqrt(sqrt(2*sqrt(-x^2+1)+2)+2))*(2/sqrt(sqrt(sqrt(2*sqrt(-x^2+1)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2*sqrt(-x^2+1)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2*sqrt(-x^2+1)+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2*sqrt(-x^2+1)+2)+2)+2)+2)+2)+2))*... = asin(x)/x; (2/sqrt(2*x+2))*(2/sqrt(sqrt(2*x+2)+2))*(2/sqrt(sqrt(sqrt(2*x+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2*x+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2*x+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2*x+2)+2)+2)+2)+2)+2))*... = acos(x)/sqrt(1-x^2);

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[math-fun] Infinite product of the type of François Viet for atan (x)
by françois mendzina essomba2 10 Apr '18

10 Apr '18

Looking at the first infinite product of pi found by François Viet, I deduced an infinite product of atan(x) (2/sqrt(2/(17*sqrt(5))+2))*(2/sqrt(sqrt(2/(17*sqrt(5))+2)+2))*(2/sqrt(sqrt(sqrt(2/(17*sqrt(5))+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2/(17*sqrt(5))+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2/(17*sqrt(5))+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2/(17*sqrt(5))+2)+2)+2)+2)+2)+2)) * ... =(17*sqrt(5)*atan(38))/38; (2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2)/5+2)+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2)/5+2)+2)+2)+2)+2))* (2/sqrt(sqrt(sqrt(sqrt(sqrt(2)/5+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2)/5+2)+2)+2))*(2/sqrt(sqrt(sqrt(2)/5+2)+2))*(2/sqrt(sqrt(2)/5+2)) * ... =(5*sqrt(2)*atan(7))/7 ; And in a general way (2/sqrt(2/(sqrt(x^2+1))+2))*(2/sqrt(sqrt(2/(sqrt(x^2+1))+2)+2))*(2/sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2)+2)+2))* ... = sqrt(1/x^2+1)*atan(x) ; An approximation of order 6 will be (2/sqrt(2/(sqrt(x^2+1))+2))*(2/sqrt(sqrt(2/(sqrt(x^2+1))+2)+2))*(2/sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2)+2)+2)) = sqrt(1/x^2+1)*atan(x);

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Re: [math-fun] Who's going to G4G13?
by Richard Hess 10 Apr '18

10 Apr '18

I’m on my way now. Dick Sent from my iPhone > On Apr 10, 2018, at 9:41 AM, Scott Kim <scott(a)scottkim.com> wrote: > > I'll be going. -- Scott > >> On Tue, Apr 10, 2018 at 8:57 AM, Allan Wechsler <acwacw(a)gmail.com> wrote: >> >> No Jenny or Allan this year, I'm afraid. We will miss seeing you and >> Hilarie :( >> >>> On Tue, Apr 10, 2018 at 11:55 AM, Tomas Rokicki <rokicki(a)gmail.com> wrote: >>> >>> Susie and I will be there! >>> >>>> On Tue, Apr 10, 2018 at 8:02 AM, <rcs(a)xmission.com> wrote: >>>> >>>> Hilarie & I will be at Gathering for Gardner. >>>> Who else is going? >>>> >>>> Rich >>>> >>>> >>>> >>>> _______________________________________________ >>>> math-fun mailing list >>>> math-fun(a)mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>> >>> >>> >>> >>> -- >>> -- http://cube20.org/ -- http://golly.sf.net/ -- >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Who's going to G4G13?
by rcs＠xmission.com 10 Apr '18

10 Apr '18

Hilarie & I will be at Gathering for Gardner. Who else is going? Rich

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[math-fun] Draft of April 2018 blog post
by James Propp 10 Apr '18

10 Apr '18

I’m writing an essay called “Who Knows Two?” (to be made public a week from today) and would appreciate comments: https://mathenchant.wordpress.com/?p=2247&shareadraft=5acb8145805a2 Is my description of Frege’s work on number correct? What about my description of intuitionism? Is it okay that the article meanders so much, or should I restructure the opening to make it clearer where I’m going? Is there stuff I should cut or shorten (e.g., the Endnote about misreading Latin declension tables)? Or is there stuff I should add? (The piece is already on the long side, but if you know some really juicy things about primitive roots, I’ll consider adding them, or at least adding a reference.), As always, I hope you’ll sign your comments so that I can ask follow-up questions if need be. Thanks, Jim Propp

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[math-fun] The asymptotic muffin problem
by James Propp 10 Apr '18

10 Apr '18

It occurred to me after Bill Gasarch's recent MIT Combinatorics Seminar talk that there's a problem related to the muffin problem that has just a single real parameter, which plays the same role as m/n but can be irrational; call it p (for portion). We have infinitely many muffins and infinitely many noshers. Each muffin gets divided into pieces, and each piece is given to one of the noshers, in such a fashion that each piece gets eaten and each nosher gets p muffins. Let F(p) be the infimum of the sizes of the pieces. What can we say about F(p)? For instance, is it periodic with period 1? Or could F(p+1) be smaller than F(p) for some p? Note that when p is rational, F(p) = inf f(m,n) over all pairs m,n with m/n = p. (Or is it? This seemed obvious when I wrote it, but now I’m not so sure. There are two inequalities in this claim, and right now I’m not seeing how to prove either of them.) Can anyone compute F(p) for some quadratic irrational p, such as (-1+sqrt(5))/2? (That's probably asking for too much; I can’t even determine F(2/n) for all odd n.) Jim Propp

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