April 2008 - mathfun@mailman.xmission.com

Re: [math-fun] And I suppose heavy objects fall faster than light ones?
by Eugene Salamin 30 Apr '08

30 Apr '08

If the centrifuge spins at angular velocity omega, the potential energy of a molecule of mass m in the centrifugal potential is V(r) = - (1/2) m omega^2 r^2. In a gas at temperature T, the equilibrium density varies as exp(- V/(r)/kT), k being Boltzmann's constant. The two molecules U235F6 and U238F6 distribute independently, and since they have slightly different mass, the gas near the axis is slightly enriched, and the gas near the periphery is slightly depleted, in U235. Since the peripheral velocity is v = omega R, R the radius of the object, the separatory power is proportional to the square of the peripheral velocity of the rotor, and independent of its radius. Original Message ---- From: Dan Asimov <dasimov(a)earthlink.net> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Tuesday, April 29, 2008 6:50:56 PM Subject: Re: [math-fun] And I suppose heavy objects fall faster than light ones? Mike Speciner wrote: << [quote of passage from NY Times article that Mike evidently considers erroneous] >> Mike, I don't recall if I ever understood just how centrifuges work. Would you be so kind as to explain? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ

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[math-fun] And I suppose heavy objects fall faster than light ones?
by Mike Speciner 30 Apr '08

30 Apr '08

>From http://www.nytimes.com/2008/04/29/science/29nuke.html Why do the contents separate? As Newton explained in his second law of motion, the more massive the object, the greater the tug. In the lurching car, an adult feels the force more than a child. In the centrifuge, heavy objects feel it more than light ones and, if possible, they move more vigorously toward the outer wall.

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[math-fun] Article in "Science" re learning math via concrete examples vs. via abstractions
by Dan Asimov 28 Apr '08

28 Apr '08

Did anyone else see this article: < http://www.nytimes.com/2008/04/25/science/25math.html > ? I'm curious what people's reaction to it was. The original article is not hard to find by googling, though in the interest of probity I shan't post a link. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] minimum area convex lattice polygons
by Chris Landauer 27 Apr '08

27 Apr '08

hihi, all - i was a bit cryptic in my message - the problem was one from this list in december 2003, to find, for each n, the minimum area convex n-gon in the 2-d integer lattice so convex polygon in the plane, all of whose vertices have integer coordinates i announced earlier results up to n=128 in december 2006, but quit after n=256 results seemed strange i also found the minimum L1-perimeter convex lattice polygons up to n=512 (corroborating the n<=256 results with a completely different program) so i have a question for you all - when i write up the results, where do they go (aside from to oeis of couree)? more soon, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation cal(a)aero.org

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[math-fun] minimum area convex attice polygons, yet again
by Chris Landauer 27 Apr '08

27 Apr '08

i have further results in the minimum area convex lattice polygon search, up to n=512 i said i was done with it after n=128, but i wasn't - i ran the same old programs up to n=256 from january to march last year, but did not trust the results, so i left them dangling these further results use incremental algorithms that are much smaller and way much faster than the ones i used last year and the years before (i wrote and ran these new ones in march and april this year, after dreaming about the problem for some unknown reason, on the fastest two of the old machines and on two newer slightly faster ones) i have minimum area and minimum L1-perimeter results for 3<=n<=512, with adequate margin to believe them (margin 10 for perimeter and 9 for area) margin is the difference between the maximum edge L1-length in a polygon and the a priori search bound b, which is used to constrain the program to consider only edges (d, n) with d > 0, n >= 0, gcd(d, n) = 1, and d+n < b (with the convention that gcd(0,x) = x for any integer x), and their three other rotations by multiples of pi/2 (-n, d) (-d, -n) (n, -d) i had to impose some such bound, since all SL(2,Z)-images of a convex lattice polygon are convex lattice polygons with the same area (but not the same edge lengths and perimeter), so the search would otherwise be infinite my minimum area program looks for polygons with minimum area, then minimum maximum edge L1-length among those with the same area, then minimum L1-perimeter among those of the same area and maximum edge length (this ordering is different from my earlier search criterion, but it is more consistent with the search bound, and it almost always finds the same polygon anyway) there is exactly one exception for n<=512, at n=275 my minimum perimeter program looks for polygons with minimum perimeter, then minimum maximum edge L1-length among those with the same perimeter, then minimum L1-area among those of the same perimeter and maximum edge length (this ordering is also different from my earlier search criterion, but it is more consistent with the search bound, and it frequently finds the same polygon, but not always) there are many differences for n<=512 (over a hundred), but all are small i am still making the as yet unproven assumption that for odd n, there is a minimum area polygon and a minimum perimeter polygon for which the two polyhalves are both minimal (corners are defined by the quadrant containing the edges, and polyhalves are two successive corners) and have almost the same number of vertices (by analogy with the case of even n, in which it was proven that there is a minimum area polygon with two identical minimal polyhalves, one rotated by pi radians and translated to fit) the reason i believe this assertion is that for n<=49 odd, several earlier versions of these programs found many such examples, in every case, without making that assumption i thought that if my program can search for area, then it can also search for perimeter - it is a simple coding change - but perimeter, as it turns out, is much simpler i have devised a construction that seems to generate almost minimum perimeter polygons for all n, giving a pretty close upper bound on area (within 3.6% for n<=256, except for 4.3% for n=24) and perimeter (off by exactly 1 or 2 for all odd n<=256, and exact when n is even) this construction provides a close upper bound for area, and transforms the search for minimum area (well, really only for a good bound) to a computation with number-theoretic sums (mainly combinations of the euler phi function) all of these computations are simpler with L1-lengths, since the L1-perimeter of a convex polygon is the same as the perimeter of the circumscribed axis-parallel rectangle (though i chose it because the values are always integers in this program search, and because it directly relates to the edge bound) on a related note, the asymptotic shape of the minimum area polygon cannot be easily defined, since there are always infinitely many of them (all SL(2,Z)-images), and the one from the papers (by Baranyi and others) is not the one my program finds, which is the one with minimum maximum edge length - that one seems to tend towards a near circle (i expect that the one from the papers is an easily found SL(2,Z)-image of the one my program finds) curiosities: (0) the maximum edge length for the minimum area polygon suddenly decreases after a run of generally increasing values the minimum maximum edge length is not monotonic - it has two major recessions 17 to 13 at n=171 to 178 27 to 21 at n=417 to 418 (i have made pictures of this) (1) the minimum area polygon with the two different orderings differs for exactly one case for n<=512, namely at n=275 there is a minimum area 376330.5 polygon for n=275 with a smaller perimeter 2936 but longest edge 20 than the perimeter 2938 for a polygon with longest edge 19 (2) the minimum perimeter polygon with the two different orderings differs for many values of n the new ordering never has a maximum edge length more than one different from the old ordering, though it frequently delays the onset of larger maximum edge lengths when compared to the old ordering as n increases (3) intermediate files for b=031 get up to 12GB, for b=036 get up to 42GB, and for b=041 are expected to get up to 89GB (they are already at 65GB for p=186) it should be noted that these intermediate files are produced in such a way that there are always two of them - the old one that is being read, and the new one that is being produced, so the size required is twice the individual file size (the program scripts actually remove the old file after it has been read completely) it should also be noted that the b=041 runs are not expected to change the results in any way, only the margin of the area search (these runs are taking over five hours each at p=186) (4) intermediate files for b=028 that keep all the edge information (so the programs can determine the minimal polygons, not just their parameters) are expected to get up to 130GB (these runs are taking over six hours each at p=089, so we expect them to take a very long time) these last few expectations will take a few weeks to corroborate or refute, since the polygon parameter search that has already finished does not provide the edges of the best polygon, only its parameters (this simplification is a major part of the speedup) if anyone would like the files that show the results, i can send them or maybe send them out to the list (they are each about 550 lines of ascii text) more soon, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation cal(a)aero.org

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[math-fun] Fwd: Sad news about Peter Pleasants
by Fred lunnon 24 Apr '08

24 Apr '08

In case there may be any of his old acquaintences here who have not heard. WFL ---------- Forwarded message ---------- From: K Matthews <keithmatt(a)gmail.com> Date: Apr 23, 2008 9:04 AM Subject: Sad news about Peter Pleasants To: fred(a)cs.nuim.ie 6.00pm Brisbane, Wednesday 23rd April Dear Fred, Peter Pleasants died on Sunday morning, losing a battle with T-cell lymphoma. The funeral will be held tomorrow and will be a small family affair. He was doing mathematics in hospital up till about two weeks ago, until complications from the last bout of strong chemo, plus the reintroduction of his stem cells, made further work too difficult. Ann's email address is pleasantsann(a)yahoo.com.au if you wish to email her. Best regards Keith Matthews

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[math-fun] Peano polygons
by Bill Gosper 23 Apr '08

23 Apr '08

If Peano maps the unit interval onto the complex unit square, we get the canonical diagrams by stepping by powers of 1/4, connecting points offset by 1/6 of a step: Graphics[Polygon[{Re[#],Im[#]}&/@Join[{1,0},Table[Peano[(k+1/6)/256],{k,0,255}]]]] gives http://gosper.org/peano256+1536s.gif . With zero offset, joining every fifth point gives http://gosper.org/peano1024o5+0.gif, which by startling coincidence is the Foo Dynasty ideogram for "My hovercraft is full of eels." --rwg ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ

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Re: [math-fun] Peano polygons
by Dan Asimov 22 Apr '08

22 Apr '08

Gosper's http://gosper.org/peano256+1536s.gif is especially striking with the figure and the background competing for attention and coming out about even. I think it would look particularly great surrounded with a gray (neutral) frame. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Help!
by rcs＠xmission.com 19 Apr '08

19 Apr '08

This is from Gosper; it was somehow trapped by the mail filtering software. I've edited it slightly. --rich ------------ Date: Sat, 19 Apr 2008 12:15:02 -0700 (PDT) Subject: Re: [math-fun] Help! From: Bill Gosper <rwmgosper(a)yahoo.com> To: math-fun <math-fun(a)mailman.xmission.com>, Bill Gosper <rwmgosper(a)yahoo.com>, rkg(a)cpsc.ucalgary.ca All: I sent this (retrieved from my outbox) yesterday, and Yahoo somehow poofed it: > Hi Richard, at least for k=3, z needs a factor of (k^n-1)^(2*k^n-1). > I haven't checked k>3. I think this differs slightly from Gareth's > solution, so we'll need a third opinion. > --Bill Macsyma has subsequently confirmed this result for general k and n, and they can't both be right... --rwg Fred: [Come to think of it, I could have done without them inventing fireworks as well ...] But then we wouldn't have space travel: During the first and second stage flights of the vehicle, if a serious irretrievable fault should occur and the deviation of the flight attitude of the vehicle exceeds a predetermined value, the attitude self-destruction system will make the vehicle self-destroyed. --Long March 3 User's Manual Ministry of Astronautics, People's Republic of China (1985) --- On Fri, 4/18/08, Richard Guy <rkg(a)cpsc.ucalgary.ca> wrote: <I clipped out a copy of RKGs reply to Gareth McCaughan --rich>

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[math-fun] Help!
by Richard Guy 19 Apr '08

19 Apr '08

This isn't much fun, but some of you with access to algebra manipulation systems may be able to help. Etienne Garnier has pointed out what must be a mistake in D13 of UPINT. I'm not able to access the relevant paper or even a review thereof, so could someone check what value of z (if any) should replace the one given below in the solution of the equation x1^x1 * x2^x2 * ... * xk^xk = z^z ? [the exponent of k in z is less than that of x_1 and there is no factor k^n-1 at all] Many thanks in anticipation. R. Claude Anderson conjectured that the equation $w^wx^xy^y=z^z$ has no solutions with $1<w<x<y<z$, but Chao Ko & Sun Qi had earlier found an infinity of counterexamples to a generalization of the conjecture to any number of variables: x_1 = k^{k^n(k^{n+1}-2n-k)+2n}(k^n-1)^{2(k^n-1)} x_2 = k^{k^n(k^{n+1}-2n-k)}(k^n-1)^{2(k^n-1)+2} x_3 = ... = x_k = k^{k^n(k^{n+1}-2n-k)+n}(k^n-1)^{2(k^n-1)+1} z = k^{k^n(k^{n+1}-2n-k)+n+1} ??????? Chao Ko & Sun Qi, On the equation $\prod^k_{i=1}x_i^{x_i}$, {\it J.\ Sichuan Univ.}, {\bf2}(1964) 5--9. [I may even have got the title of the paper wrong, since an ``= z^z'' seems to be missing.]

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