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[math-fun] Anti-sphere?
by Leo Broukhis 26 Oct '17

26 Oct '17

Take a sphere with volume V and radius R. Its surface area is minimal among all shapes with volume V. What will happen if we try the opposite - *maximizing* the surface area of a shape with a given V, while restricting the max curvature of the surface (or, equivalently, requiring that the radius of curvature at any point of the surface be no less than a given fraction of R)? Intuitively, for convex shapes, the sphere should be gradually flattening to a pancake rather than stretching to a string. For simply connected shapes and a particular value of the curvature limit, the shape should look like an erythrocyte, I think; but for greater curvatures my intuition fails me. For arbitrary multiply connected shapes, things should be even more interesting. For example, at what value of the curvature the optimum becomes 2-connected, etc. I hope that this problem been considered before, and the only thing I'm missing is the proper search term. Thanks, Leo

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Re: [math-fun] math-fun Digest, Vol 176, Issue 38
by françois mendzina essomba2 26 Oct '17

26 Oct '17

RE: We have noted that Gene's x + sin x HAKMEM recurrence Hi Professor Bill Gosper ... x + sin (x) is the first term obtained by the general formula of October 23, 2017, which I proposed ... There is an undervaluation of the potential of this formula which is rather of convergence (2 * n +1) and not 2 * m ... I realized this, it's a confusion due to the mechanism that gave birth to him whose jumps from one point to another are rather 2 * m Best regards... Le Jeudi 26 octobre 2017 17h17, "math-fun-request(a)mailman.xmission.com" <math-fun-request(a)mailman.xmission.com> a écrit : Send math-fun mailing list submissions to math-fun(a)mailman.xmission.com To subscribe or unsubscribe via the World Wide Web, visit https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun or, via email, send a message with subject or body 'help' to math-fun-request(a)mailman.xmission.com You can reach the person managing the list at math-fun-owner(a)mailman.xmission.com When replying, please edit your Subject line so it is more specific than "Re: Contents of math-fun digest..." Today's Topics: 1. Fabricating a glass/crystal sphere circa 1500 (Bill Gosper) 2. We have noted that Gene's x+sin x HAKMEM recurrence (Bill Gosper) 3. A new formula of the natural logarithm of a positive real and curious identities .. (Fran?ois Mendzina Essomba) 4. Re: Fabricating a glass/crystal sphere circa 1500 (Richard Howard) ---------------------------------------------------------------------- Message: 1 Date: Wed, 25 Oct 2017 18:05:05 -0700 From: Bill Gosper <billgosper(a)gmail.com> To: math-fun(a)mailman.xmission.com Subject: [math-fun] Fabricating a glass/crystal sphere circa 1500 Message-ID: <CAA-4O0FYVud=ymNZVHtP+kxX+AG4Mp1-kOenZf=AxQi0zc6dAA(a)mail.gmail.com> Content-Type: text/plain; charset="UTF-8" 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 >> > ------------------------------ Message: 2 Date: Thu, 26 Oct 2017 02:12:52 -0700 From: Bill Gosper <billgosper(a)gmail.com> To: math-fun(a)mailman.xmission.com Subject: [math-fun] We have noted that Gene's x+sin x HAKMEM recurrence Message-ID: <CAA-4O0EknOVRyUmDQaRqVTqPrpgzZrgDm6gGf6XnB0BHv5enqg(a)mail.gmail.com> Content-Type: text/plain; charset="UTF-8" converges cubically to odd multiples of ?. E.g., In[539]:= Series[x+Sin@x,{x,3?,3}] Out[539]= 3 ?+1/6 (x-3 ?)^3+O[x-3?]^4 Rohan notes that x-sin x converges cubically to the even multiples: In[523]:= Series[x-Sin@x,{x,2?,3}] Out[523]= 2 ?+1/6 (x-2 ?)^3+O[x-2 ?]^4 and x+cos x converges cubically to (2n+1/2)?, In[519]:= Series[Cos[x]+x,{x,5?/2,3}] Out[519]= 5 ?/2+1/6 (x-(5?)/2)^3+O[x-5 ?/2]^4 In[540]:= Series[Cos[x]+x,{x,?/2,3}] Out[540]=?/2+1/6 (x-?/2)^3+O[x-?/2]^4 and x-cos x converges to (2n+3/2)?: In[522]:= Series[x-Cos[x],{x,3?/2,3}] Out[522]= (3?)/2+1/6 (x-(3?)/2)^3+O[x-(3?)/2]^4 --rwg "... given the difficulty related to the calculation of sinuses." Indeed few things are as difficult as pushing stones up your nose. ------------------------------ Message: 3 Date: Thu, 26 Oct 2017 13:47:07 +0200 From: Fran?ois Mendzina Essomba <m_essob(a)yahoo.fr> To: math-fun(a)mailman.xmission.com, m_essob(a)yahoo.fr Subject: [math-fun] A new formula of the natural logarithm of a positive real and curious identities .. Message-ID: <ae6c7966-c2c9-8df8-9c8b-e6f2813e101b(a)yahoo.fr> Content-Type: text/plain; charset=utf-8; format=flowed Hello, I found this formula which makes it possible to make an approximation of the natural logarithm of a number. Formula 35 ln(x)=Limit(2^(n)*(x^(1/2^n)-1)/(x^(1/(2^(n+1)))),n=infinity,right); I have deduced the following identities: Formula 36 Limit(2^(n)*((x*y)^(1/2^n)-1)/((x*y)^(1/(2^(n+1)))),n=infinity,right)=Limit(2^(n)*(x^(1/2^n)-1)/(x^(1/(2^(n+1)))),n=infinity,right)+Limit(2^(n)*((y)^(1/2^n)-1)/((y)^(1/(2^(n+1)))),n=infinity,right); Formula 37 Limit(2^(n)*((x/y)^(1/2^n)-1)/((x/y)^(1/(2^(n+1)))),n=infinity,right)=Limit(2^(n)*(x^(1/2^n)-1)/(x^(1/(2^(n+1)))),n=infinity,right)-Limit(2^(n)*((y)^(1/2^n)-1)/((y)^(1/(2^(n+1)))),n=infinity,right); Formula 38 Limit(2^(n)*((product(x(i),i=1..infinity))^(1/2^n)-1)/((product(x(i),i=1..infinity))^(1/(2^(n+1)))),n=infinity,right)=sum(Limit(2^(n)*(x(i)^(1/2^n)-1)/(x(i)^(1/(2^(n+1)))),n=infinity,right),i=1..infinity); Formula 39 Limit(2^(n)*((product(x(i)/y(i),i=1..infinity))^(1/2^n)-1)/((product(x(i)/y(i),i=1..infinity))^(1/(2^(n+1)))),n=infinity,right)=sum(Limit(2^(n)*(x(i)^(1/2^n)-1)/(x(i)^(1/(2^(n+1)))),n=infinity,right),i=1..infinity) - sum(Limit(2^(n)*(y(i)^(1/2^n)-1)/(y(i)^(1/(2^(n+1)))),n=infinity,right),i=1..infinity); However, in the numerical results, formula 35 intrigues me about its convergence. It seems that for a given n, we get approximately the same number of exact decimals despite the size of the numbers. by example: X=55 ; u := ln(55) ~? 1048576/55*(55^(1/1048576)-1)*55^(2097151/2097152) 4.00733318523247091866270291119 = 4.00733318523490960132292657729 Exact number of decimal places : 11 X=5555555555555 u := ln(5555555555555) ~ 1048576/5555555555555*(5555555555555^(1/1048576)-1)*5555555555555^(2097151/2097152) 29.3458195440203748840441577653 = 29.3458195449780708532896646379 Exact number of decimal places : 9 Moreover, its convergence is not linear with n. Best Regards ------------------------------ Message: 4 Date: Thu, 26 Oct 2017 11:16:27 -0400 From: Richard Howard <rich(a)richardehoward.com> To: math-fun <math-fun(a)mailman.xmission.com> Subject: Re: [math-fun] Fabricating a glass/crystal sphere circa 1500 Message-ID: <CABkJZAZP9Mf5sMRC6MjLGYmRHB47KWpTOkKEGtCBb_DcUCUfwQ(a)mail.gmail.com> Content-Type: text/plain; charset="UTF-8" If you want to cheat and make a perfect parabola for a telescope mirror, it is hard to beat the spinning liquid trick. Mercury work well, if you want to only look up, but epoxy is more versatile (and less toxic). In the 60's I made a 120" f/1 mirror at Cal Tech out of epoxy poured on a roughly parabolic form and spun on air bearings. This mirror and its 60" brothers did the first IR sky surveys looking for dim stars, molecular clouds, etc. The parabola you get by spinning is not quite perfect enough for first-class optical imaging, but just fine for 2-10 micron IR. --R On Wed, Oct 25, 2017 at 9:05 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > 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 > >> > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > ------------------------------ Subject: Digest Footer _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ------------------------------ End of math-fun Digest, Vol 176, Issue 38 *****************************************

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[math-fun] A new formula of the natural logarithm of a positive real and curious identities ..
by François Mendzina Essomba 26 Oct '17

26 Oct '17

Hello, I found this formula which makes it possible to make an approximation of the natural logarithm of a number. Formula 35 ln(x)=Limit(2^(n)*(x^(1/2^n)-1)/(x^(1/(2^(n+1)))),n=infinity,right); I have deduced the following identities: Formula 36 Limit(2^(n)*((x*y)^(1/2^n)-1)/((x*y)^(1/(2^(n+1)))),n=infinity,right)=Limit(2^(n)*(x^(1/2^n)-1)/(x^(1/(2^(n+1)))),n=infinity,right)+Limit(2^(n)*((y)^(1/2^n)-1)/((y)^(1/(2^(n+1)))),n=infinity,right); Formula 37 Limit(2^(n)*((x/y)^(1/2^n)-1)/((x/y)^(1/(2^(n+1)))),n=infinity,right)=Limit(2^(n)*(x^(1/2^n)-1)/(x^(1/(2^(n+1)))),n=infinity,right)-Limit(2^(n)*((y)^(1/2^n)-1)/((y)^(1/(2^(n+1)))),n=infinity,right); Formula 38 Limit(2^(n)*((product(x(i),i=1..infinity))^(1/2^n)-1)/((product(x(i),i=1..infinity))^(1/(2^(n+1)))),n=infinity,right)=sum(Limit(2^(n)*(x(i)^(1/2^n)-1)/(x(i)^(1/(2^(n+1)))),n=infinity,right),i=1..infinity); Formula 39 Limit(2^(n)*((product(x(i)/y(i),i=1..infinity))^(1/2^n)-1)/((product(x(i)/y(i),i=1..infinity))^(1/(2^(n+1)))),n=infinity,right)=sum(Limit(2^(n)*(x(i)^(1/2^n)-1)/(x(i)^(1/(2^(n+1)))),n=infinity,right),i=1..infinity) - sum(Limit(2^(n)*(y(i)^(1/2^n)-1)/(y(i)^(1/(2^(n+1)))),n=infinity,right),i=1..infinity); However, in the numerical results, formula 35 intrigues me about its convergence. It seems that for a given n, we get approximately the same number of exact decimals despite the size of the numbers. by example: X=55 ; u := ln(55) ~ 1048576/55*(55^(1/1048576)-1)*55^(2097151/2097152) 4.00733318523247091866270291119 = 4.00733318523490960132292657729 Exact number of decimal places : 11 X=5555555555555 u := ln(5555555555555) ~ 1048576/5555555555555*(5555555555555^(1/1048576)-1)*5555555555555^(2097151/2097152) 29.3458195440203748840441577653 = 29.3458195449780708532896646379 Exact number of decimal places : 9 Moreover, its convergence is not linear with n. Best Regards

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[math-fun] We have noted that Gene's x+sin x HAKMEM recurrence
by Bill Gosper 26 Oct '17

26 Oct '17

converges cubically to odd multiples of π. E.g., In[539]:= Series[x+Sin@x,{x,3π,3}] Out[539]= 3 π+1/6 (x-3 π)^3+O[x-3π]^4 Rohan notes that x-sin x converges cubically to the even multiples: In[523]:= Series[x-Sin@x,{x,2π,3}] Out[523]= 2 π+1/6 (x-2 π)^3+O[x-2 π]^4 and x+cos x converges cubically to (2n+1/2)π, In[519]:= Series[Cos[x]+x,{x,5π/2,3}] Out[519]= 5 π/2+1/6 (x-(5π)/2)^3+O[x-5 π/2]^4 In[540]:= Series[Cos[x]+x,{x,π/2,3}] Out[540]=π/2+1/6 (x-π/2)^3+O[x-π/2]^4 and x-cos x converges to (2n+3/2)π: In[522]:= Series[x-Cos[x],{x,3π/2,3}] Out[522]= (3π)/2+1/6 (x-(3π)/2)^3+O[x-(3π)/2]^4 --rwg "... given the difficulty related to the calculation of sinuses." Indeed few things are as difficult as pushing stones up your nose.

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Re: [math-fun] Fabricating a glass/crystal sphere circa 1500
by Keith F. Lynch 24 Oct '17

24 Oct '17

Richard Howard <rich(a)richardehoward.com> 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. Nitpick: Flat surfaces also fit together in all orientations. I suppose a flat surface can be considered part of a sphere of infinite radius. Indeed, the method you describe is also used to make extremely flat mirrors, except that three surfaces are used, two at a time of which are ground against each other. I recently mentioned this in another forum in which we were discussing whether heliographs could be used for long distance communication on the Ringworld (an immense fictional solid ring around a sun-like star) without high technology. I argued that it should be possible to send something like Morse code over distances up to about 100 million kilometers if the recipient has good binoculars, even if the heliograph mirror is no larger than the Great Pyramid, a size that pre-technological cultures could obviously build things. > 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). How was the spherical gyro in Gravity Probe B, supposedly the roundest object ever made, constructed? My understanding is that it wasn't round enough, and the measurement of the frame dragging effect was below the noise floor, though some claim the noise could be compensated for. It's undisputed that it did measure the much larger space-time curvature effect. (When Magellan's crew returned to Europe after sailing consistently west, they thought that was because the world was round. They were mostly right, but if they had kept very close track of the curvature, they'd have noticed that they were short by about an inch. The missing inch was because of the space-time curvature around the Earth.) (Copernicus, Galileo, Kepler et al believed that the celestial sphere stood still and Earth rotated. They were mostly right, but if they had kept close track, they would have noticed that the celestial sphere rotates by about 0.04 seconds of arc per year due to frame dragging.) As for Rupert's drops, you can find high-speed videos on YouTube. Some show that the drops aren't harmed by being shot by a rifle on their thick part. Others show that the slightest tap on the narrow part makes them explode, and that the shock travels up the drop at several kilometers per second. Never watch this in person without safety goggles!

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[math-fun] Frame dragging (was Re: Fabricating a glass/crystal sphere circa 1500)
by Keith F. Lynch 24 Oct '17

24 Oct '17

Brent Meeker <meekerdb(a)verizon.net> wrote: > Keith F. Lynch wrote: >> (Copernicus, Galileo, Kepler et al believed that the celestial >> sphere stood still and Earth rotated. They were mostly right, but >> if they had kept close track, they would have noticed that the >> celestial sphere rotates by about 0.04 seconds of arc per year due >> to frame dragging.) > I don't understand that. The celestial sphere is defined as not > rotating and is empirically fixed by distant galaxies. That certainly wasn't the original definition. Before Copernicus it was generally believed that it rotated about once per day around the stationary Earth. According to relativity, with linear motion it's arbitrary which frame of reference is regarded as stationary. Not so with rotary motion. Numerous experiments can be done to determine whether the Earth turns or whether the celestial sphere, distant galaxies and all, revolve around the Earth. Or some of each. At Earth's north or south pole, set up a frictionless pendulum that's free to swing in any horizontal direction. Set it to swing towards and away from some distant galaxy that's on the horizon. If Earth is rotating, it will continue to swing towards and away from that distant galaxy as the Earth rotates under it. Anywhere on or near Earth, point a gyroscope's axis at some distant galaxy. See if it keeps pointing at that galaxy. Fire a cannonball in the northern hemisphere. See if its trajectory curves to the right, and if so by how much. On the equator, simultaneously launch two satellites into equatorial orbits, identical except that one is eastbound, the other westbound. See whether they're directly overhead when they collide. Send a radio signal in all horizontal directions at once. Measure when it returns to you from each direction. Deduce Earth's rotational rate from the magnitude of its equatorial bulge. The frame dragging effect, also known as Lense-Thirring precession, says that space in and near a rotating mass such as Earth will be dragged along with it somewhat. As such, those experiments ought to show that Earth is rotating slightly less than observations of distant galaxies would imply, and that those distant galaxies are all revolving slowly around the Earth in the opposite direction. Slowly in terms of arc seconds, quite rapidly in meters per second, given how distant those galaxies are. They're going many times faster than light. No, that doesn't contradict general relativity. GR says that that speed limit is strictly local. With the possible exception of Gravity Probe B, Earth frame dragging has never been detected, as the numbers are just too small. But it's certainly been detected around neutron stars and black holes. In _Incandescence_ by Greg Egan, a pre-technological civilization with no knowledge of astronomy (they live inside a hollow asteroid) works out general relativity from a few simple experiments and measurements and some careful reasoning. They have the advantage of being in close orbit of a massive black hole where GR effects are significant. (Well, it's an advantage in discovering GR, but a great disadvantage in all other ways, as it's a rather dangerous neighborhood.)

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[math-fun] Four constants (e, pi, gamma and phi) in a continuous cubic root.
by François Mendzina Essomba 24 Oct '17

24 Oct '17

*Hello, Four constants (e, pi, gamma and phi) in a continuous cubic root. Formula 33 a:=2/((3*sqrt(5))*((1/9)*((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2)))^(-1/3)-((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2))))) ; b:=(3/sqrt(5))*(((3*sqrt(5)*((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2))))-((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2)))^(1/3))/(1-9*((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2)))^(4/3))); r:=1/((((((((((((exp(-gamma)*Pi+1)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3)+ …) ; (phi)^(-1)=(1/a)*(b+r); An approximate value will be: (phi)^(-1)=(1/a)*(b+1/(((((((((((exp(-gamma)*Pi+1)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3)); Formula 34 continuous fraction phi:=sqrt(4*exp(-5*Pi)+4106118241)/(sqrt(5)*(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)-sqrt(4*exp(-5*Pi)+4106118241)/(2*sqrt(5))+64079/2)+…) ; An approximate value will be: phi:=sqrt(4*exp(-5*Pi)+4106118241)/(sqrt(5)*(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)-sqrt(4*exp(-5*Pi)+4106118241)/(2*sqrt(5))+64079/2)) ; Best regards...*

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Re: [math-fun] Random slice of a cube
by Keith F. Lynch 24 Oct '17

24 Oct '17

Allan Wechsler <acwacw(a)gmail.com> wrote: > Your lowbrow procedure excludes some planes that do intersect the > cube. Pick a point Q just "above" the center of one of the cube's > faces, (so it would have been excluded by your procedure) and erect > the plane through Q perpendicular to OQ. Now tilt the plane just a > little toward one of the four vertices of the chosen face, until the > plane nicks that vertex. There is a point Q' in this tilted plane > such that the tilted plane is perpendicular to OQ'; I contend that > this can be done so that Q' is still outside the cube, and hence the > tilted plane would have been excluded even though it intersects the > cube. My method was to pick a point Q at a distance less than the cube root of 3 from the cube's center, i.e. a point inside the sphere that circumscribes the cube. But now I'm wondering what *is* the locus of points such that a plane including Q and perpendicular to the line connecting Q to the center would intersect the cube. The locus obviously includes everything in the cube and nothing outside the circumscribing sphere. But it's not coincident with either the cube or the sphere. What is its shape, its volume, and its surface area? I'm assuming the cube has a side length of 2 and is centered on the origin. I'm pretty sure that parts of the shape are concave. A simpler problem is to reduce it by a dimension. What's the shape, area, and perimeter of the locus of points such that a line through the point and perpendicular to the line connecting the point to the center intersects the square with side length 2 and centered on the origin? Again, I'm pretty sure that parts of the shape are concave.

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Re: [math-fun] Fabricating a glass/crystal sphere circa 1500
by Bill Gosper 23 Oct '17

23 Oct '17

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 > > > Henry Baker wrote: > ----- > If you don't remove the air in such a column, you're likely to get > shot that is more raindrop-shaped than sphere-shaped. > ----- > > I wrote: > ----- > ... I'd try dropping molten glass some distance through the air to > make it form into a sphere, and have it land in water to make it > rigid. > > —Dan > > > Henry Baker wrote: > ----- > It's the year 1500, and some wealthy patron wants a transparent glass > and/or rock crystal sphere perhaps 6-8" in diameter. > > How would you fabricate it? > ... > ... > ----- > ----- > > _______________________________________________ > 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] Getting dense in a sphere
by Dan Asimov 23 Oct '17

23 Oct '17

Henry's question suggests the following math problem in the context of polishing a (large) BB: Suppose you are given a continuous curve A: [0,L] —> S^2 parametrized by arclength for some L > 0, where S^2 is the unit sphere. Let D(r, p) denote the closed disk of radius r about the point p of S^2, where r is measured along S^2. Then define gap_L(A) = sup{ r > 0 | For some p in S^2, D(r, p) does not intersect A([0,L])} For any L > 0, define mingap(L) = inf { gap(A) | A: [0,L] —> S^2 is parametrized by arclength }. Then what can be said about the values of mingap(L) ? What is its asymptotic behavior as L —> oo ??? —Dan Henry Baker wrote: ----- It's the year 1500, and some wealthy patron wants a transparent glass and/or rock crystal sphere perhaps 6-8" in diameter. How would you fabricate it? ... ... -----

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