October 2014 - mathfun@mailman.xmission.com

[math-fun] cells in arrangements of H hyperplanes in D-space: the nice counting formula
by Warren D Smith 19 Oct '14

19 Oct '14

Generically, and maximally, H hyperplanes partition D-dimensional space into F(H,D) regions where F(H,D) = the unique polynomial(H) of degree=D whose value agrees with 2^H for each H=0,1,2,3,...,D.

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[math-fun] BAHFest tonight
by James Propp 19 Oct '14

19 Oct '14

Science-fun isn't the same thing as math-fun, but still, if any of you funsters are in the area of Cambridge, MA and don't have plans for the evening, you might want to attend the Festival of Bad Ad Hoc Hypotheses: see bahfest.com for more info. I believe the event is sold out, but I've got a couple of e-tickets that I'd be willing to part with (seeing as how I'm not going to be able to attend anyway). Email me if interested. Jim Propp (jamespropp(a)gmail.com)

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Re: [math-fun] Circles in circles
by Warren D Smith 19 Oct '14

19 Oct '14

Edwards wrongly replaces k^3 by k^2 in his derivation, but this does not matter since k cancels out in numerator & denominator below. (So do the factors of 2 and pi.) When Edwards integrates, one way to try to do it (not his) is to compute int(r=0..a-x) / int(r=0..infinity) another way is to compute int(x=0..a-r) / int(x=0..infinity). My point is, probability = a ratio, numerator counts ways to be happy, denominator counts all ways. I'm not necessarily claiming either of these two ways is actually valid. Anyhow, the latter is what Edwards does, except he OMITS the denominator! Also, Edwards makes an implicit assumption that the tuple (P, x) is uniformly distributed within the outer circle. This is nowhere justified and is in fact wrong... since... the denominator he omits obviously has to have some nonobvious nonuniform probability distribution for x, to avoid infinite nonsense results. So: Edwards made plenty of errors, his derivation is garbage.

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[math-fun] counting orderings of points in D-space
by Warren D Smith 19 Oct '14

19 Oct '14

To generalize Wilson's problem to D-space, combining previously posted ideas, if there are N points inside a D-dimensional ball, N large, D fixed, then the number of possible orderings of the N points along the X-coordinate (when the ball is rotated) is upper bounded by: if D=1: 2. if D=2: N*(N-1). if D>2: Any point-pair defines a "great-circle" (hyperplane intersect sphere) from bisector hyperplane. The arrangement defined on the sphere-surface by these H=N*(N-1)/2 hyperplanes has F faces, and the generic value of F upper bounds the number of orderings. And obviously F(H,D) is upper bounded by F(H,D) <= F(H-1,D-1) + F(H-1,D) which allows us to compute a table of upper bounds starting from F(H,2) = 2*H and F(1,D) = 2. (This is, of course, the Pascal triangle recurrence...) Note that F(H,D) with H large and D fixed is bounded by a degree=(D-1) polynomial in H, you can prove this inductively. Its leading (highest degree) term should have coefficient=4/D! when D>1.

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[math-fun] counting orderings of points in D-space -- problem reposted to election science forum
by Warren D Smith 19 Oct '14

19 Oct '14

I have re-posted D.W.Wilson's ordering problem (my version of it, with some notes) here: https://groups.google.com/forum/#!forum/electionscience under the heading puzzle: honest rank-orderings in D dimensions. (Since it's really a political science problem :) Unfortunately "election scientists" tend to be less smart than math-funsters, so they probably won't be able to do much, but there are some smart ones.

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Re: [math-fun] Lockheed Martin Claims Fusion Breakthrough That Could Change World Forever
by Warren D Smith 18 Oct '14

18 Oct '14

> From: Eugene Salamin <gene_salamin(a)yahoo.com> > If there are profits to be had, then the investors would expect to share in > them. That is what usually motivates investors. Perhaps they are expecting > a grant, rather than an investment, from the taxpayers. > -- Gene --well, cold fusioneers Pons & Fleischman got the state of Utah or some govt body anyhow to donate $5 million dollars to build them "the National Cold Fusion Institute." Then the Univ. of Utah spent $1 million securing itself the patents, then P & F got $20 M from Toyota and Japan over 8 years. And in the case of Lockheed, if some investor gives them, say, $5 billion, they spend it on research, and then they say "sorry, no fusion plant" then this may not count as a "profit" but somebody got money and employment, while somebody else lost money.

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Re: [math-fun] Lockheed Martin Claims Fusion Breakthrough That Could Change World Forever
by Warren D Smith 18 Oct '14

18 Oct '14

It seems to me, that this is exactly like the previous "cold fusion" hoax, except for the fact that the cold fusion guys gave far more details than the Lockheed guys, hence were far more convincing. (I will say, though, I predicted from day 1 those guys were wrong.) The reason LM is seeking partners, rather than financing this revolutionary breakthru development themselves so they can reap 100% of the huge profits is (I predict): because it is a scam, and the "profits" they plan to reap are entirely from the gullible outside investors. And who would those gullible investors be? Lockheed appears to be aiming at the biggest-pocketed, stupidest investors in the world... the US government. And if you truly want stupid, I suggest US congress committees on scirnce & technology. Those guys are unbelievably stupid. Will they fall for it?

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Re: [math-fun] unsolved magic square problem
by Lee Sallows 17 Oct '14

17 Oct '14

Hi all, and thanks for your responses to the magic square problem. First, I knew it was an unsolved problem because I invented it myself but couldn't solve it. But I'm no big shakes as a mathematician so recently I tried it out on my friend Hendrik Lenstra, who also tried it out on some of his students, none of whom came up with a solution. The problem arose as follows. Suppose we take an ordinary n x n magic square and replace the numbers with resistors of same ohmic value. Then connecting the n resistors in any row, column or diagonal in series yields the same total resistance in each case. I asked myself whether a set of resistors could be found that yield a constant total resistance when connected in PARALLEL. That was easy, my example can be seen here: http://www.leesallows.com/index.php?page_menu=Magic . The next step was obvious: can we find a set of resistors that work both in series and in parallel? Keith's solution is cool, my congratulations. Rich's clarifications were helpful, thanks. Richard's query about the reciprocal of 0 surely doesn't apply here? But Keith's solution is hopefully only a good start rather than the best that can be done. It makes me think I should have stipulated a solution in reals. Moreover, I'm disappointed in myself for not coming up with his solution myself. The notion of using roots of unity in a magic square had crossed my mind before, as seen here: http://www.leesallows.com/index.php?page_menu=Magic . And as he says, the reciprocal = complex conjugate is an obvious come-on. Btw, recall that a multimagic square is defined as one in which replacing all numbers with their squares, cubes, etc, preserves magic. So this problem is addressing the case of a negative exponent. A such Keith's solution ought to be recorded on Christian Boyer's most excellent site http://www.multimagie.com/indexengl.htm. So, is there a solution in reals? Lee At 08:20 AM 10/16/2014, you wrote: >Keith is using a little shorthand. >The elements are really w^k, with w an Nth root of 1. >His upper left corner entry is w^0, i.e. 1. >His main trick is that reciprocals automatically work, if >the row or column sum is real, since reciprocals of >roots of unity are also conjugates. > >One possibly useful idea for finding/proving that >certain w^k sums are 0 is to note that the sum must be >a multiple of the corresponding cyclotomic polynomial, >of degree phi(n). The cyclotomic polys often have both >positive & negative coefficients, and for N>=105, may have >some |coeff| >= 2, which will complicate trying to create >poly multiples with only 0,1 coeffs. When N is even, >the trick of converting -w^k into w^(k+N/2), which Keith >uses in his square, helps a lot. > >Rich > >----------- >Quoting rkg <rkg(a)cpsc.ucalgary.ca>: > >>Reciprocal of zero ?? R. >> >>On Wed, 15 Oct 2014, Keith F. Lynch wrote: >> >>>Lee Sallows <Lee.Sal(a)inter.nl.net> wrote: >>>>An unsolved problem that MathFunsters may like to ponder is as >>>>follows: >>> >>>I'm curious how you know whether it's unsolved. >>> >>>>Does there exist an n x n magic square using n^2 distinct numbers >>>>that remains magic when every number is replaced by its reciprocal? >>> >>>>Or failing that, how about a semimagic square (magic on rows and >>>>columns only)? >>> >>>I've found the latter, but not yet the former. Here's my semi-magic >>>square: >>> >>>0 18 1 19 2 20 >>>12 30 13 31 14 32 >>>24 6 25 7 26 8 >>>3 21 4 22 5 23 >>>15 33 16 34 17 35 >>>27 9 28 10 29 11 >>> >>>The numbers are the exponents of cos(pi/18) + i sin(pi/18), the >>>"first" 36th root of 1. All 36 36th roots appear once each. The >>>semi-magic sum is zero. The semi-magic sum of the reciprocals is >>>also zero. >>> >>>I chose to work with numbers of unit absolute value since all such >>>numbers have their complex conjugate for their reciprocal, hence any >>>(semi-)magic square consisting only of such numbers is guaranteed to >>>remain a (semi-)magic square if all the numbers are replaced by their >>>reciprocals. >>> >>>I conjectured that for a finite set of unit-absolute-value numbers >>>that includes 1 to sum to 0, it must contain all the Nth roots of 1 >>>for some integer N>1, and analogously if it doesn't include 1. If so, >>>the order of a semi-magic square must be the product of at least two >>>distinct primes. Note that on each row, numbers are next to their >>>negative, i.e. their antipodes on the unit circle, and that on each >>>column, numbers are grouped in threes, forming equilateral triangles >>>on the unit circles. >>> >>>I did the above entirely by hand, no computers. >>> >>>To make a true magic square by this method, the order of the magic >>>square would have to be divisible by three distinct primes. So I'd >>>try order 30, using the 900th roots of 1, with groupings by 2 on the >>>rows, by 3 on the columns, and by 5 on the diagonals. Tricky. And >>>completely infeasible by a brute-force search. >>> >>>But is the conjecture correct? It's not obvious to me, perhaps >>>because I'm not a visual thinker. So I tested it with a >>>semi-brute-force computer search. I found what appear to be >>>exceptions, starting with the 30th roots of 1. Unless there's a >>>glitch in my program, 0 1 7 13 19 20 add up to zero, as do 0 6 7 17 18 >>>24. (Again, these are powers of cos(pi/15) + i sin(pi/15), or points >>>on a regular 30-gon.) Can anyone confirm or refute this exception >>>to my conjecture? If the exception is valid, what is the rule for >>>finding numbers of unit absolute value that add up to zero? Thanks. >>> >>>If my conjecture is wrong, it may be possible to make a smaller magic >>>square than order 30. >>> >>>I just did a Google search on "0 1 7 13 19 20," and I got >>> >>> On the chromatic number of integral circulant graphs - ScienceDirect >>> >>> Vertices from C={0,1,7,13,19,20} C = { 0 , 1 , 7 , 13 , 19 , 20 } >>> generate a clique in the given integral circulant graph, ... >>> >>>so maybe it is a thing. >>> >>>_______________________________________________ >>>math-fun mailing list >>>math-fun(a)mailman.xmission.com >>>https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >>_______________________________________________ >>math-fun mailing list >>math-fun(a)mailman.xmission.com >>https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > >_______________________________________________ >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] Good qubits made in silicon
by Warren D Smith 16 Oct '14

16 Oct '14

http://arxiv.org/abs/1407.1950 http://arxiv.org/abs/1402.7140 These sounded promising at least before I noticed that all their experiments were done at about 0.05 kelvin temperature and apparenly they've never managed to build a q-gate, only single qubits. You get a crystal of ultrapure silicon-28 (spin-0 isotope of Si). You put in a phosphorus-31 dopant atom. Different spin states of the dopant atom nucleus & electron store two qubits. Atomic state-transition spectral frequencies can be used to stimulate transitions, thus controlling the state of the qubit. Now you can create an electric field near that dopant atom by charging some metal gate. This "Stark shifts" its spectral frequencies. That way, by charging only the gate you want, you can control only the qubit you want and leave the others alone. Instead of using a phosphorus dopant atom, you can use an "artificial atom" (system with state-transitions) made from a "quantum dot." Can systems with many interacting qubits (via q-gates) thus be built? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] unsolved magic square problem
by Keith F. Lynch 16 Oct '14

16 Oct '14

Lee Sallows <Lee.Sal(a)inter.nl.net> wrote: > An unsolved problem that MathFunsters may like to ponder is as > follows: I'm curious how you know whether it's unsolved. > Does there exist an n x n magic square using n^2 distinct numbers > that remains magic when every number is replaced by its reciprocal? > Or failing that, how about a semimagic square (magic on rows and > columns only)? I've found the latter, but not yet the former. Here's my semi-magic square: 0 18 1 19 2 20 12 30 13 31 14 32 24 6 25 7 26 8 3 21 4 22 5 23 15 33 16 34 17 35 27 9 28 10 29 11 The numbers are the exponents of cos(pi/18) + i sin(pi/18), the "first" 36th root of 1. All 36 36th roots appear once each. The semi-magic sum is zero. The semi-magic sum of the reciprocals is also zero. I chose to work with numbers of unit absolute value since all such numbers have their complex conjugate for their reciprocal, hence any (semi-)magic square consisting only of such numbers is guaranteed to remain a (semi-)magic square if all the numbers are replaced by their reciprocals. I conjectured that for a finite set of unit-absolute-value numbers that includes 1 to sum to 0, it must contain all the Nth roots of 1 for some integer N>1, and analogously if it doesn't include 1. If so, the order of a semi-magic square must be the product of at least two distinct primes. Note that on each row, numbers are next to their negative, i.e. their antipodes on the unit circle, and that on each column, numbers are grouped in threes, forming equilateral triangles on the unit circles. I did the above entirely by hand, no computers. To make a true magic square by this method, the order of the magic square would have to be divisible by three distinct primes. So I'd try order 30, using the 900th roots of 1, with groupings by 2 on the rows, by 3 on the columns, and by 5 on the diagonals. Tricky. And completely infeasible by a brute-force search. But is the conjecture correct? It's not obvious to me, perhaps because I'm not a visual thinker. So I tested it with a semi-brute-force computer search. I found what appear to be exceptions, starting with the 30th roots of 1. Unless there's a glitch in my program, 0 1 7 13 19 20 add up to zero, as do 0 6 7 17 18 24. (Again, these are powers of cos(pi/15) + i sin(pi/15), or points on a regular 30-gon.) Can anyone confirm or refute this exception to my conjecture? If the exception is valid, what is the rule for finding numbers of unit absolute value that add up to zero? Thanks. If my conjecture is wrong, it may be possible to make a smaller magic square than order 30. I just did a Google search on "0 1 7 13 19 20," and I got On the chromatic number of integral circulant graphs - ScienceDirect Vertices from C={0,1,7,13,19,20} C = { 0 , 1 , 7 , 13 , 19 , 20 } generate a clique in the given integral circulant graph, ... so maybe it is a thing.

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