May 2012 - mathfun@mailman.xmission.com

Re: [math-fun] More grist for the mill re galaxies and gravity theory
by Henry Baker 07 May '12

07 May '12

So we're missing "grey matter" ?? (Sorry, couldn't resist.) At 07:13 PM 5/6/2012, Warren Smith wrote: >So the whole "dark matter" problem is not solved, but major >pieces of stupidity allegedly have been revealed.

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[math-fun] More grist for the mill re galaxies and gravity theory
by Warren Smith 07 May '12

07 May '12

1. http://arxiv.org/pdf/astro-ph/0507619v1.pdf 2. http://arxiv.org/pdf/astro-ph/0602519.pdf Newton gravity might naively be expected to work well with no need to use full blown general relativity (GR) for galaxies. But the first paper claims that is not true! Actually they say you DO need to use GR, and they do in a small-G approximate form, and find much less "dark matter" is needed to explain a bunch of galaxies. In fact in some cases none needed? But the second paper claims the first paper had errors but agrees GR is needed and says GR reduces amount of dark matter needed by 33% vs Newton. This however still leaves us needing a huge amount dark matter. So the whole "dark matter" problem is not solved, but major pieces of stupidity allegedly have been revealed.

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[math-fun] arxiv paper about the Elsevier boycott
by Joerg Arndt 06 May '12

06 May '12

By Douglas N. Arnold and Henry Cohn, titled "Mathematicians take a stand" http://arxiv.org/abs/1204.1351 Best regards, jj

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Re: [math-fun] The "math pages"
by Dan Asimov 04 May '12

04 May '12

Hmm, interesting. I was on the math faculty at the Minneapolis campus of U of M 1973-1977 (though on leave at the IAS '75-'77). I don't recall his name, but can recall very few students' names from 37+ years ago. I did, however, have a lot of engineering students in my classes, including several large calculus lectures with about 250 students in each, in 1973, probably the year he was a freshman. So there is at least a modest chance he was in one of my classes. I wonder if I'd recognize his photo. --Dan Hans wrote: << Kevin S. Brown received a BS (1977) and MS (1978) in mechanical engineering at the University of Minnesota. . . . . . . >> ________________________________________________________________________________________ It goes without saying that .

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[math-fun] The "math pages"
by Warren Smith 04 May '12

04 May '12

http://www.mathpages.com/ contains an interesting collection of essays about a lot of mathy subjects. According to a hard to find note, they were written by "Kevin S. Brown." I can't vouch for them all, and there also are some broken links unfortunately,... but some seem highly insightful, and the amount of work it took to write all these was very large. Also, you can buy book version -- he self-publishes books.

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Re: [math-fun] xkcd
by Bill Gosper 03 May '12

03 May '12

Dan A> What's enigmatic about this kind of very simple expression for a near-integer is whether it is a coincidence, or whether it begs for an explanation. To explain or not to explain: that is the question. For example, it's fairly well known that exp(pi*sqrt(163)) is an integer up to < 10^(-12), and this can be explained by the q-expansion of the j-invariant. This explanation also holds for the near-integers exp(pi*sqrt(67), up to < 10^(-5), exp(pi*sqrt(43), up to < 10^(-3), exp(pi*sqrt(19), up to < 1/4. Wikipedia also mentions this amazing discovery of D. Bailey & J. Borwein: Integral_{x=0 to oo} (cos(2x) * Prod_{n=1 to oo} cos(x/n)) dx = pi/8, up to < 10^(-42). Wikipedia claims this is just a coincidence, but I'd call that chutzpah. --Dan Me too. To numerically investigate this, the infinite product requires drastic acceleration. A completely inadequate but mildly interesting acceleration follows. Starting with the middle form in Out[153]= {Sqrt[Product[Sinc[x/(k-1/2)], {k, n}]^3/Product[Cos[x/k], {k, n}]], Product[Cos[x/k], {k, 2*n}], Product[Sinc[x/(k-1/2)], {k, 2*n}]} Convert to the third form by the usual differentiate - Mittag-Leffler - sumswap, giving a negligible speedup. Combine weighted powers to get the first form, and take half as many terms to reflect the doubled complexity. It still wins: In[157]:= Table[Evaluate[%153 /. x -> .69], {n, 3^Range[5]}] Out[157]= {{0.66406, 0.685026, 0.669198}, {0.660881, 0.668976, 0.663342}, {0.660477, 0.663315, 0.661396}, {0.66043, 0.661393, 0.660748}, {0.660425, 0.660748, 0.660532}} --rwg

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[math-fun] Major slurp & burp
by Warren Smith 03 May '12

03 May '12

They claim to have observed a star being eaten by a supermassive black hole. http://www.nature.com/nature/journal/vaop/ncurrent/full/nature10990.html http://www.sciencedaily.com/releases/2012/05/120502132958.htm

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[math-fun] Personalized fireworks -- your face in the sky
by Henry Baker 02 May '12

02 May '12

FYI -- Didn't Arthur C. Clarke invent something like this back in the 1950's ? http://www.firebox.com/product/5314/Personalised-Fireworks Always wanted to see your face in lights? Now you can! With Personalised Fireworks you can plaster your smiling phizzog right across the night sky. A jaw-dropping climax to any fireworks display, this bespoke box will ignite up to 200 brilliant white rockets all at once, to create a startling likeness of your face that can be seen from up to 30km away. Just set up the self-contained box Â making sure itÂs facing the right direction Â light the single fuse and retreat to the minimum safe distance (approx. 200m). In a blinding explosion that will completely ionize the paper box, your personalised display will launch every single rocket into the sky simultaneously, producing the equivalent of full stadium lighting for up to 5 seconds. Upload your photo The software begins to plot out the detail Your face is in the sky So how does it work? Well, all we need is your favourite profile picture. The Chinese boffins behind this miraculous invention will then identify the most distinctive parts of your face using advanced mapping software. Selecting from a ÂpaletteÂ of over 5000 unique rockets, a fully automated manufacturing process will then decide how best to illustrate them as an explosion. It will then optimise their position in your Personalised Fireworks box, making it a convenient and portable size. Planning to launch multiple personalities at your next display? No problem. Each box features your original picture on the front and your choice of charming message ( , , , or ). Which also makes it the perfect gift at weddings, Bar Mitzvahs, birthdays or wakes. So whether itÂs to wow the crowds on November 5th [Guy Fawkes Day], signal for help, or just guide friends to your campsite at the next festival, Personalised Fireworks will sure to blow up in your face. In a good way.

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Re: [math-fun] geometry puzzle
by Michael Reid 02 May '12

02 May '12

> > Actually, it's easy to do for the regular hexagon, so the last sentence > > should probably read "n = 5 and n >= 7". > > It can be done for any n = 6m, by the following procedure: The "it" here refers to a dissection into congruent pieces with the center of the regular n-gon in the interior of one of the pieces.

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Re: [math-fun] geometry puzzle
by Michael Reid 02 May '12

02 May '12

> Variation: Dissect a disk into congruent pieces, so that a small region > around the center is within one of the pieces. -- Rich This is also included in the problem from UPIG that I previously mentioned. From my 1991 edition, problem C6 "Cutting up squares, circles and polygons", at the bottom of page 87: "Stein asks whether it is possible to partition the unit circle into congruent pieces so that the center is in the interior of one of the pieces?[sic] It need not be on the boundary of all the pieces, as Figure C8 [on page 89] shows. Is it true that the pieces must have a diameter of at least one? The first question is also of interest for the regular n-gon, n >= 5 ." Actually, it's easy to do for the regular hexagon, so the last sentence should probably read "n = 5 and n >= 7". For the equilateral triangle and square, there are such dissections into 4 and 3 congruent pieces, respectively. What's the fewest number of pieces in such a dissection of the regular hexagon? Is there any update on any of these in a later edition of the book? Michael Reid

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