May 2012 - mathfun@mailman.xmission.com

[math-fun] sqrt(pi)
by Bill Gosper 24 Mar '13

24 Mar '13

Pairwise combining the terms of A&S 6.5.31 (upper Gamma CF), then adding last week's lower gamma CF: In[185]:= Equal[Gamma[ z], ((y^z*((1/(cfk[n*(z - n), -z + y + 2*n + 1, List[n, Infinity]] - z + y + 1)) + (1/(z + cfk[n*y, z - y + n, List[n, Infinity]] - y))))/(E^y))] z 1 Gamma[z] == (y (------------------------------------------- + -y + z + cfk[n y, n - y + z, {n, Infinity}] 1 y -----------------------------------------------------------)) / E 1 + y - z + cfk[n (-n + z), 1 + 2 n + y - z, {n, Infinity}] where cfk:=ContinuedFractionK and y>0 can be chosen to optimize convergence. Specializing, In[186]:= % /. z -> 1/2 /. y -> 1 Out[186]= 1 Sqrt[Pi] == (-------------------------------------- + 1 1 -(-) + cfk[n, -(-) + n, {n, Infinity}] 2 2 1 ------------------------------------------) / E 3 1 3 - + cfk[(- - n) n, - + 2 n, {n, Infinity}] 2 2 2 In[192]:= N[List @@ %186 /. cfk[num_, den_, _] :> ContinuedFractionK[num, den, {n, 9}]] Out[192]= {1.77245, 1.77244} This is the first non-asymptotic sum or CF I've seen for sqrt(pi). But the sum behind the lowergamma cf has been there all along, so I must not have been paying attention. --rwg

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[math-fun] "quasicrystal"
by Bill Gosper 07 Jun '12

07 Jun '12

I'm intrigued by Jack Holloway's recursive pentagram snowflake design, Fig 31 in the Minskys & Trinskys book<http://www.blurb.com/bookstore/detail/2172660>. However, circularizing and sampling obscure the pixel structure, and how, if at all, it relates to Fig 30. As usual, Julian comes to the rescue, with a "period 5" trinsky in which orbital x+y stays constant mod 2. The "black squares" (checkerboard) subset of this bitmap strongly resembles Fig 31, except the bounding pentagons have vertex angles of π/2 and 3π/4 instead of 3π/5. So I linearly crunched it to have "rhombic pixels" and wrote http://www.tweedledum.com/rwg/rhombicp5.svg The .svg is so you can scale it up without antialiasing. Unfortunately, my Apple Firefox scales it right off the screen, and refuses to scroll it back! Safari scrolls, but won't let you zoom very far. And Preview won't load svgs at all! And, as of this AM, gosper.org delivers source text instead of graphics for it. So to mega zoom it, Neil suggested downloading inkscape<http://inkscape.org/>, which seems to work great. Can one speak of the limit of this pattern in the large? The pentagram outlines grow without bound and get "infinitely thin", yet seem to fill space. And what is a formula for their limiting shape? Julian (quoted without permission) explains the original image as follows: (d_t means the trinsky d multiplier, etc.) It's the Trinsky equivalent of p5d1, which is d=g=1, e=2*sin^2(π/5). The Minsky-Trinsky equivalence is has d_t=g_t=d_m, e_t=e_m/2, has x about twice as large to compensate. This is what causes the checkerboarded images–for a given Trinsky (x,y), the equivalent Minsky point is ((x+⸤d_t*y⸥)/2,y) (sorry about the floors I haven't found any decent floor characters to use), and the checkerboard pattern comes from whether (x+⸤d_t*y⸥)/2 is an integer or not. The "checkerboard" pattern is only when the Trinksy d=g=1, in which case x+y is constant mod 2: x'=x-y=c mod 2, y'=y+⸤e*x⸥, x''=x'-y'=c-y' mod 2. A granularity of 1/2 would not produce checkerboarded images–it would produce four normal (i.e. at coordinates (ax+b,cy+d)) images (or two, alternating in either columns or row but not both, if only one granularity=1/2). --rwg

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Re: [math-fun] Shouryya Ray's equation
by Dan Asimov 01 Jun '12

01 Jun '12

There's a long debate in Wikipedia, or maybe Wikimedia, on whether to delete an article on Shouryya Ray. The consensus seems to be that there is no verifiable source, other than the media yammering to itself and copying each other's yammerings. There are also assertions that the equation in question was already discovered in the late 1800's. My guess is that the teenager actually did rediscover some equation that he did not know was not new, and the world press -- with a collective IQ below room temperature -- has just parroted what was on his science fair project. --Dan << An Indian teenager in Dresden supposedly came up with a new closed form solution to a trajectory in the presence of air resistance. Supposedly, the solution involved cos, sqrt and _asinh_ (this last one is what intrigued me). However, I haven't been able to find his actual paper after consulting with Google. Does anyone here have the reference or link ? >> ________________________________________________________________________________________ It goes without saying that .

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[math-fun] combinatorial programming contest
by James Buddenhagen 01 Jun '12

01 Jun '12

Since there are clearly wizards here (both mathematicians and programmers) I thought some here might be interested in the just-started programming contest here: http://infinitesearchspace.dyndns.org/ . It is just for fun. The object is to find the largest square grid that can be colored with C colors so that corners of sub-rectangles are not all the same color, and to do this for various values of C (each value a sub-contest). You just register and then start submitting your best results. Since this is somewhat mathematical and for fun, I hope it is not inappropriate to mention here.

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[math-fun] Shouryya Ray's equation
by Henry Baker 31 May '12

31 May '12

An Indian teenager in Dresden supposedly came up with a new closed form solution to a trajectory in the presence of air resistance. Supposedly, the solution involved cos, sqrt and _asinh_ (this last one is what intrigued me). However, I haven't been able to find his actual paper after consulting with Google. Does anyone here have the reference or link ?

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Re: [math-fun] combinatorial programming contest
by Dan Asimov 31 May '12

31 May '12

Good point! I withdraw my objection. --Dan Ed wrote: << This is equivalent to a torus grid. If its a rectangle on a torus grid, it's a rectangle on the plain grid. >> ________________________________________________________________________________________ It goes without saying that .

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Re: [math-fun] combinatorial programming contest
by Dan Asimov 30 May '12

30 May '12

Nice question! Though I wish it were a bit more symmetrical: same question for the largest square *torus* grid. --Dan << Since there are clearly wizards here (both mathematicians and programmers) I thought some here might be interested in the just-started programming contest here: http://infinitesearchspace.dyndns.org/ . It is just for fun. The object is to find the largest square grid that can be colored with C colors so that corners of sub-rectangles are not all the same color, and to do this for various values of C (each value a sub-contest). You just register and then start submitting your best results. Since this is somewhat mathematical and for fun, I hope it is not inappropriate to mention here. >> ________________________________________________________________________________________ It goes without saying that .

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[math-fun] integrals for 2F1
by Bill Gosper 30 May '12

30 May '12

Besides the traditional definite In[13]:= FunctionExpand[Integrate[t^(b - 1)*(1 - t)^(c - b - 1)/(1 - z*t)^a, {t, 0, 1}], 0 < z < 1] Out[13]= ConditionalExpression[( Gamma[b] Gamma[-b + c] Hypergeometric2F1[a, b, c, z])/Gamma[c], Re[b] < Re[c] && (Re[z] < 1 || z \[NotElement] Reals) && Re[b] > 0] (it ignored the 0 < z < 1), we have the indefinite Hypergeometric2F1[A, B, C, z] == E^Integrate[ContinuedFractionK[(A + n)*(B + n), (C + n - (1 + A + B + 2*n)*t)/Sqrt[(1 - t)*t], {n, 0,∞}]/Sqrt[(1 - t)*t], {t, 0, z}] --rwg Testing this is hard: As of 8.04, ContinuedFractionK is numerically and symbolically unusable.

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Re: [math-fun] math-fun Digest, Vol 111, Issue 28
by Henry Baker 29 May '12

29 May '12

Perhaps "black holes" should be called "red holes"; after all, as time slows down close to the event horizon, all emitted light becomes red-shifted into the long wave radio spectrum; Mary no longer "sees" Joe, so much as hears him on her radio. Don't count photons; count energy. The number of photons actually goes up dramatically as the frequency goes down. Photons merely count the number of Planck lengths in a single wavelength of "light". Photons are merely the smallest units of a "transaction" in the "transactional interpretation of quantum theory". At 05:17 PM 5/28/2012, Warren Smith wrote: >> From: Eugene Salamin <gene_salamin(a)yahoo.com> >> If we examine the situation before the black hole has formed, and after it has evaporated, we need not be concerned about the peculiarities of general relativity. Baryonic matter goes in; mostly photons (and gravitons also) come out. This is the "Joe point of view" wherein Joe has been destroyed in the black hole. This must be correct, so what is wrong with the "Mary point of view" in which Joe lingers forever just above the event horizon? > >> In Joe's rest frame, he emits a finite number of photons before crossing the horizon, and for him, crossing the horizon is not particularly eventful. > >--not necessarilly. Joe could emit an infinite number of photons in a >finite amount of his time with a finite summed energy (e.g. photon n >has energy=1/n^2)... at least if >we had magic fake-photons which were point particles not waves so we did not > need to worry about their redshifts or antenna effects. >Not that this quibble terribly matters. So if photons were "points" >not waves and we did not need to worry about their redshifts, then >this would enable Joe to stay >visible to Mary the whole time. > >>Mary detects these photons within a finite amount of her proper time, and after the last one, there are no more. Photon number is conserved under propagation and Doppler shift in general relativity. How could Mary test the hypothesis that Joe never actually fell into the black hole? >She might send in a beam of light and look for the scattering off of Joe. > >--the usual story is that does not work because after some short >finite amount of Mary-time, she is unable to emit a photon capable of >catching up to Joe. It works >until that moment comes, however. Furthermore, after that moment, the >light Mary had shined on Joe before, still is in transit toward Joe or >back to Mary... hence she would keep on seeing Joe using that old >light, forever, except for the fact there are only a finite number of >photons in the beam since light is not infinitely divisible. >And eventually the last one catches >up to Joe, then flies back to Mary, for her last glimpse of Joe. > >>But on a space-time diagram, when the probe light reaches the event horizon, Joe is already inside. >If the light does catch up with and scatter off Joe, the scattered >light is trapped inside the horizon, and never gets back to Mary. > >> Where does the notion of "lingering just outside the horizon" come from? > >--it comes from the picture where Joe keeps emitting an infinite >number of magic fake-photons constantly, causing Joe to remain >visible to Mary eternally, meaning "He never fell in." > >Before 1969 or so, the literature often called black hole s"frozen >stars" to connot the fact they were "frozen eternally just outside >their event horizon." >But then Wheeler coined the phrase "black hole" and it stuck -- he >laughed off complaints this name was "undignified" and "obscene" :) > >> In the elementary textbook description for the Schwszschild black hole, we have coordinates r and t that are the usual radius and time far from the black hole.In these coordinates, Joe's orbit r(t) approaches the Schwarzschild radius as t approaches infinity, and so he seems to linger there forever. But just because we name something "t" doesn't make it into a time. > >--very true. Indeed t is not time, and r is not length. You can >integrate ds along >radial or t-directed lines though, to figure out lengths and times... >But t really happens to be Mary's time-coordinate, and r for her >really is a length coordinate (assuming Mary located at large r >value). > >> Inside the horizon, r becomes timelike, and t becomes spacelike, while on the horizon itself, the metric in these coordinates becomes singular. > >> The black hole singularity is not so much "at the center" as it is in the future. > >--true. > >>Joe is compelled to fall into the singularity for the same reason that we are all compelled to move into the future, whatever that reason may be. > >--true again. As Wheeler once put it (which struck me as rather poetic wording, >it has an appealing air of finality) >"Time stops. That is the lesson of the Schwarzschild black hole."

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Re: [math-fun] math-fun Digest, Vol 111, Issue 28
by Warren Smith 29 May '12

29 May '12

> From: Eugene Salamin <gene_salamin(a)yahoo.com> > If we examine the situation before the black hole has formed, and after it has evaporated, we need not be concerned about the peculiarities of general relativity. Baryonic matter goes in; mostly photons (and gravitons also) come out. This is the "Joe point of view" wherein Joe has been destroyed in the black hole. This must be correct, so what is wrong with the "Mary point of view" in which Joe lingers forever just above the event horizon? > In Joe's rest frame, he emits a finite number of photons before crossing the horizon, and for him, crossing the horizon is not particularly eventful. --not necessarilly. Joe could emit an infinite number of photons in a finite amount of his time with a finite summed energy (e.g. photon n has energy=1/n^2)... at least if we had magic fake-photons which were point particles not waves so we did not need to worry about their redshifts or antenna effects. Not that this quibble terribly matters. So if photons were "points" not waves and we did not need to worry about their redshifts, then this would enable Joe to stay visible to Mary the whole time. >Mary detects these photons within a finite amount of her proper time, and after the last one, there are no more. Photon number is conserved under propagation and Doppler shift in general relativity. How could Mary test the hypothesis that Joe never actually fell into the black hole? She might send in a beam of light and look for the scattering off of Joe. --the usual story is that does not work because after some short finite amount of Mary-time, she is unable to emit a photon capable of catching up to Joe. It works until that moment comes, however. Furthermore, after that moment, the light Mary had shined on Joe before, still is in transit toward Joe or back to Mary... hence she would keep on seeing Joe using that old light, forever, except for the fact there are only a finite number of photons in the beam since light is not infinitely divisible. And eventually the last one catches up to Joe, then flies back to Mary, for her last glimpse of Joe. >But on a space-time diagram, when the probe light reaches the event horizon, Joe is already inside. If the light does catch up with and scatter off Joe, the scattered light is trapped inside the horizon, and never gets back to Mary. > Where does the notion of "lingering just outside the horizon" come from? --it comes from the picture where Joe keeps emitting an infinite number of magic fake-photons constantly, causing Joe to remain visible to Mary eternally, meaning "He never fell in." Before 1969 or so, the literature often called black hole s"frozen stars" to connot the fact they were "frozen eternally just outside their event horizon." But then Wheeler coined the phrase "black hole" and it stuck -- he laughed off complaints this name was "undignified" and "obscene" :) > In the elementary textbook description for the Schwszschild black hole, we have coordinates r and t that are the usual radius and time far from the black hole.In these coordinates, Joe's orbit r(t) approaches the Schwarzschild radius as t approaches infinity, and so he seems to linger there forever. But just because we name something "t" doesn't make it into a time. --very true. Indeed t is not time, and r is not length. You can integrate ds along radial or t-directed lines though, to figure out lengths and times... But t really happens to be Mary's time-coordinate, and r for her really is a length coordinate (assuming Mary located at large r value). > Inside the horizon, r becomes timelike, and t becomes spacelike, while on the horizon itself, the metric in these coordinates becomes singular. > The black hole singularity is not so much "at the center" as it is in the future. --true. >Joe is compelled to fall into the singularity for the same reason that we are all compelled to move into the future, whatever that reason may be. --true again. As Wheeler once put it (which struck me as rather poetic wording, it has an appealing air of finality) "Time stops. That is the lesson of the Schwarzschild black hole."

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