November 2017 - mathfun@mailman.xmission.com

[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 23 Nov '17

23 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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[math-fun] Never so soon
by Hans Havermann 21 Nov '17

21 Nov '17

Yesterday, Cliff Pickover's twitter feed presented a bit from Pickover's 2005 "A Passion for Mathematics" which references Guy's 1994 "Unsolved Problems in Number Theory" (2nd ed.) E15, a recursion of Göbel, wherein is stated that x(43) of the sequence is not an integer. The sequence is A003504: http://oeis.org/A003504 There's a different offset in the OEIS version, so A003504(44) is now the first one that is not an integer. Pickover in his book felt the need to add something to the problem so, noting that A003504(44) = 5.4093*10^178485291567, he stated that this number "is so large that humanity will *never* be able to compute all of its digits". I had a go on my four-year-old Mac Pro with 64 GB RAM and was only able to compute A003504(42) with its 44621322894 decimal digits. That suggests that when the next iteration of the Mac Pro, with 256 GB RAM, comes out in 2018 it should be able to calculate the number. But I know that there are personal computer setups out there right now that enjoy 256 GB RAM, so I emailed Cliff with a "never is now". :) I'm curious to find out what the fractional part of the number will turn out to be. I think it'll be some integer divided by 43.

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[math-fun] Computational geometry software
by James Propp 20 Nov '17

20 Nov '17

Anyone know of software (preferably in Mathematica) that computes the intersection of polytopes, or at least polygons? I'd like to explore some more random-nonempty-intersection problems, and it'd be very helpful to be able to do numerical experiments in a flexible setting. For instance, how many sides on average does the intersection of P, Q+v, and R+w have, conditioned on the event that the intersection is nonempty, where P, Q, and R are fixed parallelograms, and v and w are random translation vectors? (The usual caveats and clarifications about "random translation vectors" apply.) The answer might be that it depends on P, Q, and R, or it might be 4 (no matter what), or it might be some other number; maybe one of you can see a non-experimental approach, but I don't, so doing experiments is the next step for me. But I don't trust myself to write correct infrastructural code for handling polygons. Jim Propp PS: I'm still mulling over Veit's two puzzles about averages. I'm guessing that a key tool for solving his first puzzle is "V-E+F=2", or rather "V-E+F is negligible when V, E, and F are large", but I don't see what the answer is.

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Re: [math-fun] Computational geometry software
by Henry Baker 20 Nov '17

20 Nov '17

I suspect that Jim wants to do exact (or at least very high precision) calculations. Some computational geometry problems merely want the *sign* of a determinant, but they need quite high (quad?) precision in order to make sure that that sign is correct. At 08:36 AM 11/20/2017, Andy Latto wrote: >If you want to do a lot of this sort of computation, I'll bet this is >the sort of thing graphics processors are optimized to do fast. > >Andy > >On Mon, Nov 20, 2017 at 10:01 AM, James Propp <jamespropp(a)gmail.com> wrote: >> Anyone know of software (preferably in Mathematica) that computes the >> intersection of polytopes, or at least polygons? >> >> I'd like to explore some more random-nonempty-intersection problems, and >> it'd be very helpful to be able to do numerical experiments in a flexible >> setting. >> >> For instance, how many sides on average does the intersection of P, Q+v, >> and R+w have, conditioned on the event that the intersection is nonempty, >> where P, Q, and R are fixed parallelograms, and v and w are random >> translation vectors? (The usual caveats and clarifications about "random >> translation vectors" apply.) The answer might be that it depends on P, Q, >> and R, or it might be 4 (no matter what), or it might be some other number; >> maybe one of you can see a non-experimental approach, but I don't, so >> doing experiments is the next step for me. But I don't trust myself to >> write correct infrastructural code for handling polygons. >> >> Jim Propp >> >> PS: I'm still mulling over Veit's two puzzles about averages. I'm guessing >> that a key tool for solving his first puzzle is "V-E+F=2", or rather "V-E+F >> is negligible when V, E, and F are large", but I don't see what the answer is.

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[math-fun] pi number
by françois mendzina essomba2 19 Nov '17

19 Nov '17

u_(n+1)=sqrt(2+2*u_(n))/2; u_(0)=-1; a_(n)=2^(k*(2*n+1))*sum((k!/((i!)*(k-i)!))*(-1)^(i)*u_(n)^i,i,0,k); n--- infinity; a_(n)---Pi^(2*k);

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[math-fun] Fwd: Conway Life emulating Conway Life
by Brent Meeker 19 Nov '17

19 Nov '17

NEAT! Brent -------- Forwarded Message -------- Hi, Here is a wonderful video showing an emulation of the game of life, programmed in the game of life. It illustrates well the notion of emulation. In fact the program is supposed to emulate itself, so there is a nice "inception" effect, also: https://www.youtube.com/watch?v=xP5-iIeKXE8 <https://www.youtube.com/watch?v=xP5-iIeKXE8> I have to go, Have a good day, Bruno http://iridia.ulb.ac.be/~marchal/ <http://iridia.ulb.ac.be/%7Emarchal/>

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[math-fun] Fabricating a glass/crystal sphere circa 1500
by Bill Gosper 19 Nov '17

19 Nov '17

I watched a transit of Venus through an ~8" reflector made by Aerobie|Aeropress inventor Alan Adler, who precisely strained spherical mirrors. The editor of Sky & Telescope was very enthused, but the technique never caught on, apparently because CNC grinding of paraboloids got cheap. Alan is now using his extra mirrors under homemade tungsten tops that spin for twenty minutes. What is the optimal shape to trade air drag for moment of inertia? It turns out to be hard to model air drag. --rwg On 2017-10-23 14:37, Richard Howard wrote: > Large telescope mirrors start as extremely accurate sections of a sphere > made by carefully randomized grinding of two surfaces against each other. > The only two surfaces that fit together in all orientations are spheres. > > Starting with a rough glass sphere and a rough hemispherical hole in a > plate, continued random grinding produces a perfect sphere (and a perfect > hemispherical hole). > > BTW, the thermal approaches have the difficulty of dealing with the thermal > expansion of glass. A 8" sphere could take many months to anneal without > shattering. Corning museum has a nice video of making of the largest glass > paperweight (~13" diameter and over 109 lbs). > > --R > > On Mon, Oct 23, 2017 at 1:58 PM, Eugene Salamin via math-fun < > math-fun(a)mailman.xmission.com> wrote: > >> It was perhaps 58 years ago, in our High School auditorium, the speaker >> gave us a demo of this prestressed glass phenomenon. It was in the shape >> of a chemist's Florence flask, the rounded kind. He first used the flask >> to hammer a nail, no problem. Then he dropped in a speck of carborundum. >> It's harder than glass, and produces a tiny nick on impact. The flask just >> exploded. >> >> -- Gene >> >> >> On Monday, October 23, 2017, 1:29:12 PM PDT, Bill Gosper < >> billgosper(a)gmail.com> wrote: >> >> Interesting 16/sec ball bearing production process: >> https://www.youtube.com/watch?v=19duYMdiXi0 >> Falling raindrops form cabochons: >> https://www.youtube.com/watch?v=a9CRrGvQwe0 >> 1961|2 freshman chemistry lecturer demoed Prince Rupert's Drops, and then >> something >> more interesting. (Prince <othername>'s Drops? Google fails me.) >> Hollow, thick-walled glass blobs, open at one end. He hammered nails with >> one, >> then wrapped it in a towel, dropped in some carborundum crumbs, BAM! --rwg >> >> On 2017-10-22 11:57, Dan Asimov wrote: >> > Henry got that right — in fact, something quite interesting happens, >> > as Wikipedia says: >> > >> > ----- >> > Prince Rupert's Drops (also known as Dutch or Batavian tears) are >> > toughened glass beads created by dripping molten glass into cold >> > water, which causes it to solidify into a tadpole-shaped droplet with >> > a long, thin tail. These droplets are characterized internally by very >> > high residual stresses, which give rise to counter-intuitive >> > properties, such as the ability to withstand a blow from a hammer or a >> > bullet on the bulbous end without breaking, while exhibiting explosive >> > disintegration if the tail end is even slightly damaged. In nature, >> > similar structures are produced under certain conditions in volcanic >> > lava. >> > ----- >> > >> > more at https://en.wikipedia.org/wiki/Prince_Rupert's_Drop. >> > >> > —Dan >> >

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Re: [math-fun] Miscounting in the Cheddar Gorge ...
by Leo Broukhis 18 Nov '17

18 Nov '17

This was converted from 37.0 Celsius. Leo On Nov 12, 2017 22:23, "Andy Latto" <andy.latto(a)pobox.com> wrote: My favorite precision fail is the one for human body temperature. It's generally quoted in the US as 98.6 degrees Fahrenheit, but this was obtained by converting the value of 37 degreed Celsius, so a number with a precision of a Celsius degree was converted to one that people think has a precision of a tenth of a Fahrenheit degree. Andy On Sun, Nov 12, 2017 at 4:15 PM, Keith F. Lynch <kfl(a)keithlynch.net> wrote: > Guy Haworth <g.haworth(a)reading.ac.uk> wrote: >> The Cheddar Gorge is 10,000,001 years old this year. > >> I know this because last year ... > > That's a precision fail, not a counting fail. It's perfectly valid to > add one year to the age of something each year, no matter how old it > is. Except that the age of very old things is seldom known to the > nearest year. And gorges, unlike, say, individual dinosaurs, don't > have precise birthdays anyway. (That joke is more commonly told about > dinosaurs, with a museum guard answering that a particular specimen is > is 100,000,003 years old, or whatever.) > > Also, for very high precision or for extremely long distances or > times, the age depends on the trajectory of the clock. > > For instance how old is Mount Everest? I haven't been able to find > that number anywhere. Wikipedia says it's 8848 meters tall and that > the Himalayas are growing by 5 millimeters per year. That would imply > an age of 1769600 years. If, at the instant the mountain started its > climb from sea level someone had placed a perfect clock where it would > ride up the peak and another perfect clock where it would remain at > sea level, today the two clocks would disagree by 24 seconds as to the > age of the mountain, due to general relativity. > > You'd get far more variance on how long ago the Big Bang was, since > there are so many choices of "reasonable" trajectories for a clock that > formed that very instant and conveniently lands in from of you just when > you want to know the age of the universe. > > Precision fails are very common. For instance: > > Naturally, this means that it's ridiculously hot on the planet's > surface, with temperatures reaching 2000 degrees Celsius (3632 > Fahrenheit), ... > > https://www.outerplaces.com/science/item/16744- astronomers-planet-titanium-atmosphere > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- Andy.Latto(a)pobox.com _______________________________________________ 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] Phi, Pi and Gamma
by françois mendzina essomba2 18 Nov '17

18 Nov '17

(Phi)*(Pi)=(1/2)*Gamma(1/10)*Gamma(9/10) ;

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