July 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Veit's disks, again
by Dan Asimov 13 Jul '11

13 Jul '11

Ed wrote: << See http://www2.stetson.edu/~efriedma/cirRcir/ That's a very interesting set of solutions. Modifying it a bit will allow more asymmetry: n circles with the largest possible sum of radii packed inside a convex unit area. >> This may be related to the sausage conjecture. From a write-up on the Web, at < http://cambmathsociety.org/Math-Library1/Popular%20and%20historical/Szpiro%…'s%20conjecture%20(Wiley,%202003)(ISBN%200471086010)(306s).pdf >: << . . . the "sausage conjecture." It is related to Kepler’s conjecture and was proposed by that Hungarian grandmaster of discrete geometry, Laszlo Fejes-Tóth. To this day it remains unsolved. To illustrate, let us observe Fumiko, a shop assistant in Tokyo who boxes up gifts for the customers in the sports department. The question she faces is, what is the best way to pack balls? Which boxing method wastes the least amount of space? For example, is it more efficient to pack four tennis balls—or Ping-Pong balls, or golf balls—into a square box, or an elongated box or a pyramidal box? How about six balls, or three, or twentyfive? We will call the long boxes sausage-boxes and all others cluster-boxes. The results are quite surprising. For up to fifty-five balls the sausage seems to represent the best box. It may be a little awkward to load such a thing into the trunk of the car, but as far as wasted space is concerned it’s the best. If you want to pack additional balls something strange happens, however. For fiftysix balls or more, clusters are better than sausages. Fejes-Tóth was so shocked and confused when he found out that he called the switch from sausages to clusters the “sausage catastrophe.” Nobody is quite sure at exactly what number of balls optimality changes from sausage to cluster. It is believed that the switch occurs somewhere between fifty and fifty-six balls. But at exactly what number this happens is still an open question. If you become bored with regular tennis you may prefer to play the game in four dimensions. As you might expect, the problem does not become any easier. As long as you play four-dimensional tennis with less than about seventy- five thousand balls, it is better to store them in sausage-boxes. For more than 375,769, cluster-boxes are best. Somewhere in between the switch takes place. Where exactly? Nobody knows. Surprisingly, in still higher dimensions the situation calms down again. Fejes-Tóth claimed that in sufficiently high dimensions, sausages are always best, no matter how many balls you want to store. This contention was proved rigorously for all dimensions greater than forty-two. But Fejes-Tóth conjectured that in any dimension higher than five, sausages are always best. As of this writing, the claim is still awaiting a proof. By the way, can you guess what happens in two dimensions? It’s real simple: clusters are always superior to sausages.5 And in one dimension, sausages are the only option. If the surface area of the box’s walls are of primary concern and not the volume, then there is another conjecture in store. The “spherical conjecture” states that the optimum, wall-minimizing shape of the box for tennis balls is roughly spherical if the dimension is large. Whether this is true, and for which dimensions the conjecture should hold, is anybody’s guess at this stage. >> --Dan Sometimes the brain has a mind of its own.

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[math-fun] Polyomino question
by David Wilson 13 Jul '11

13 Jul '11

For polyominoes A and B, let A ~ B if A and B each tile the same nonempty polyomino. Is A ~ B an equivalence?

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[math-fun] Veit's disks, again
by Bill Gosper 12 Jul '11

12 Jul '11

rwg>BtW, the mnemonic "1,4" encodes a seed configuration from which the entire >solution follows almost immediately. >Veit Elser> Since fourteen is the first number where symmetry is completely> broken, (http://pratikpoddarcse.blogspot.com/2010/05/veit-elsers-formidable-14.html) >ve> that would make it my favorite number. >> My place of work was founded on a principle expressed in fourteen> words.>> Veit>> On Jun 1, 2011, at 5:30 AM, Bill Gosper wrote:>>> Veit>>>> Is your five disk packing of the unit disk maximizing sum(radii)>>>> (http://gosper.org/HTMLFiles/5disks.gif) an example of either of>>>> these asymmetries? How about your semisecret fourteen disk solution,>>>> which has no symmetry at all? (http://milou.msc.cornell.edu/images/>>>> seems to be down.)>>>> --rwg Whoa, what about twelve? http://gosper.org/HTMLFiles/12disks.gif I thought this was the "six dimes and three quarters around three nickels" I Mobius-transformed to make Twubblesome Twelve, but it seems to have been Mobius transformed already! It appears to retain bilateral symmetry, and the transformation should be described by a single real parameter. If I can reconstruct my "six dimes" formulae, it should become a Calculus I problem with an exact algebraic solution. --rwg

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[math-fun] tables
by Alex Bellos 11 Jul '11

11 Jul '11

Hi - I've just received an email from a reader who saw in a Cologne museum a document called Fundamentalis Tabula Arithmetica from 1638. It is a table of multiples for all the numbers up to 100 and also the tables for these five numbers: 106, 117, 256, 318 and 365 Its pretty obvious why 256 and 365 are included. Might anyone have any thoughts as to why 106, 117 and 318 (3x106) are also included? Alex

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[math-fun] zeta and Abel-Plana
by Bill Gosper 10 Jul '11

10 Jul '11

Another one for http://www.cs.cmu.edu/~adamchik/articles/catalan/catalan.htm: 8*Catalan + Pi^2 == 12 + 256*Integrate[y/((-1 + E^(2*Pi*y))*(1 + 16*y^2)^2), {y, 0, Infinity}] More generally, a^s*HurwitzZeta[s, a] == (-1 + 2*a + s)/(2*(-1 + s)) + 2*a*Integrate[(Cos[t]^(-2 + s)*Sin[s*t])/(-1 + E^(2*a*Pi*Tan[t])), {t, 0, Pi/2}] so two more for the Catalan list: (1/16)*(8*Catalan + Pi^2) == 3/4 + (1/2)*Integrate[Sin[2*t]/(-1 + E^((1/2)*Pi*Tan[t])), {t, 0, Pi/2}] , (9/16)*(-8*Catalan + Pi^2) == 5/4 + (3/2)* Integrate[Sin[2*t]/(-1 + E^((3/2)*Pi*Tan[t])), {t, 0, Pi/2}] . Note the rapidity of convergence of the HurwitzZeta integral, especially if you boost a by an integer. Is this how Odlyzko et al do zeta numerics miles from the real axis? --rwg

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[math-fun] Google+
by Michael Kleber 09 Jul '11

09 Jul '11

Hi funsters -- I've been enjoying Google+, and it looks like it is becoming easier to invite people to join it. (It's only been around for a week and they are rate-limiting new people getting on, but that seems to be opening up a bit.) If you'd like an invitation, let me know. (Please reply to me directly, though; I'm sure all of math-fun doesn't want to see it.) --Michael michael.kleber(a)gmail.com -- Forewarned is worth an octopus in the bush.

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[math-fun] innumeracy at JPL?
by Mike Speciner 09 Jul '11

09 Jul '11

From http://www.jpl.nasa.gov/news/news.cfm?release=2011-204 : "The dust is minus 429 to minus 416 degrees Fahrenheit (about minus 221 to 213 Celsius) -- colder than Pluto, which is about minus 400 degrees Fahrenheit (204 degrees Celsius). " This is not the first time they've blown the conversion between F and C. But the last time, they used the usual conversion formula for temperatures, but applied it to a temperature difference. Can you see what they did this time? I assume they started with the article in Science, which quoted dust temperatures of 17 to 23K and slightly misconverted those numbers to Fahrenheit (for domestic readers?), then grossly misconverted the results to Celsius.

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Re: [math-fun] bridge and randomness
by Dan Asimov 08 Jul '11

08 Jul '11

I agree with whoever expressed skepticism over players' ability to consciously figure out such a subtlety. On the other hand, for a game like bridge that's probably too complicated to completely analyze, it seems possible that top-players-to-be intuitively explore strategy space, and gravitate toward whatever style of play tends to improve their success. Even if that occurs on a unconscious level. --Dan Michael Kleber wrote: << ... Persi has a nearly unlimited number of great anecdotes on the subject, so actually I have no problem believing Bernie's (2), though. Bernie Cosell wrote: << ... 2) bridge masters seemed to occasionally make counter-probabilistic plays [playing finesses and for suit splits, etc], but as a corollary to (1), he discovered that their play was actually correct for the *actual* probabilities [due to the inadequate shuffles]. ... >> >> Sometimes the brain has a mind of its own.

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Re: [math-fun] bridge and randomness
by Dan Asimov 08 Jul '11

08 Jul '11

Interesting. But I think it would be difficult if not impossible to infer with confidence the reason that a top bridge player made a surprising play. In bridge you're not supposed to have any secret communication with your partner, even if it's limited to secret meanings of bids. But this rule seems impossible to enforce, especially for partners that change their codes often enough. --Dan << 1) that the 'standard' method of shuffling wasn't very good and you needed to do additional shuffles to really randomize the deck. 2) bridge masters seemed to occasionally make counter-probabilistic plays [playing finesses and for suit splits, etc], but as a corollary to (1), he discovered that their play was actually correct for the *actual* probabilities [due to the inadequate shuffles]. I found it pretty impressive that the deviation from "true randomness" almost certainly wasn't all that much, but it was enough that the top bridge players could detect the deviation and took advantage of it. >> Sometimes the brain has a mind of its own.

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[math-fun] Slanted coverage
by Bill Gosper 08 Jul '11

08 Jul '11

In those ubiquitous little tables of area and volume formulae in handbooks and textbooks, there's often a picture of a skewed cone next to the right cone to emphasize that they have the same volume formula. Did you ever wonder why there's never the skewed cone area formula? For the special case with the apex directly above a point on the circumference, I get pi h r 3F2[-1/2,1/4,3/4; 1/2, 1; -4 r^2/h^2], which doesn't even converge for h < 2 r. The area element for the general x,y,z parametric surface is just Out[85]= (1,0) (0,1) Norm[Cross[({x, y, z}) , ({x, y, z}) ]] (I don't recall this from 18.03.) E.g., for the torus with radii r and R, In[116]:= % /. {x :> (Cos[#2]*(R + r*Sin[#1]) &), y :> (Sin[#2]*(R + r*Sin[#1]) &), z :> (r*Cos[#1] &)} Out[116]= Norm[Cross[({Cos[#2] (R + r Sin[#1]) & , Sin[#2] (R + r Sin[#1]) & , (1,0) r Cos[#1] & }) , ({Cos[#2] (R + r Sin[#1]) & , (0,1) Sin[#2] (R + r Sin[#1]) & , r Cos[#1] & }) ]] In[117]:= % /. f_[L_List] :> f /@ L /. Abs[w_] -> w Out[117]= Sqrt[Power[-(-r Sin[#1] & ) (Cos[#2] (R + r Sin[#1]) & ) + (0 & ) (r Cos[#1] Sin[#2] & ), 2] + Power[-(0 & ) (r Cos[#1] Cos[#2] & ) + (-r Sin[#1] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2] + Power[(r Cos[#1] Cos[#2] & ) (Cos[#2] (R + r Sin[#1]) & ) - (r Cos[#1] Sin[#2] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2]] In[119]:= % /. (f_ &) :> (f &)[t, u] Out[119]= 2 2 2 2 Sqrt[r Cos[u] Sin[t] (R + r Sin[t]) + 2 2 2 2 r Sin[t] (R + r Sin[t]) Sin[u] + 2 2 2 (r Cos[t] Cos[u] (R + r Sin[t]) + r Cos[t] (R + r Sin[t]) Sin[u] ) ] In[120]:= FullSimplify[%] Out[120]= 2 2 Sqrt[r (R + r Sin[t]) ] In[125]:= Assuming[R > r > 0, Integrate[%120, {t, 0, 2*Pi}, {u, 0, 2*Pi}]] Out[125]= 2 4 Pi r R Did Archimedes know this? --rwg

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