April 2006 - mathfun@mailman.xmission.com

[math-fun] conjugation of algebraic numbers
by Schroeppel, Richard 21 Apr '06

21 Apr '06

A curiosity: sqrt 6 + sqrt 10 + sqrt 15 is NOT a conjugate of - sqrt 6 - sqrt 10 - sqrt 15. The minimal polynomial of the first expression is X^4 - 62 X^2 - 240 X - 239. (Yet another property of 239.) On the other hand, sqrt 2 + sqrt 3 + sqrt 5 IS a conjugate of - sqrt 2 - sqrt 3 - sqrt 5. and the minimal polynomial is degree 8. Rich

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[math-fun] Scooby Dougall's thm
by R. William Gosper 21 Apr '06

21 Apr '06

Dougall's thm: a hyper_f ([a, - + 1, b, c, d, n - d - c - b + 2 a + 1, - n], 7, 6 2 a [-, - b + a + 1, - c + a + 1, - d + a + 1, - n + d + c + b - a, n + a + 1]) 2 (a + 1) (- c - b + a + 1) (- d - b + a + 1) (- d - c + a + 1) n n n n = -------------------------------------------------------------------. (- b + a + 1) (- c + a + 1) (- d + a + 1) (- d - c - b + a + 1) n n n n I found an early 70s (pre matrix-calculus) printout from a secret Xerox device (a LASER printer!) implying a hyper_f ([a, b, c, d, - n, n - d - c - b + 2 a, - + 1], 7, 6 2 a [- b + a + 1, - c + a + 1, - d + a + 1, - n + d + c + b - a + 1, n + a + 1, -], 1) 2 = - (a + 1) (- c - b + a + 1) (- d - b + a + 1) (- d - c + a + 1) n n - 1 n - 1 n - 1 ((d + c + b - a) n (n - d - c - b + 2 a) + (c + b - a) (d + b - a) (d + c - a)) /((- b + a + 1) (- c + a + 1) (- d + a + 1) (- d - c - b + a) ). n n n n (Tested through n=5). The latter cannot be a variable-rename of the former, because the lower parameters minus the upper parameters total 2 in the former and 4 in the latter. Presumably, there's a q-extension. Can this be "new"? We'd need several more before we'd have enough to seed the contiguity relations and evaluate the whole mesh. --rwg

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RE: [math-fun] Square thirds
by Torgerson, Mark D 20 Apr '06

20 Apr '06

I'm still stuck on the trivial way of thirding the square. If you use pencil and paper, sure two horizontal lines divide the thing into three pieces that are the same. But y'all have been talking, open sets, closed sets, point sets, etc. Down to that level of detail I don't see how to trivially make equal thirds. Actually, I don't see how to do it at all. If the trivial way is to make two equally spaced cuts, the middle third shares two sides, the other two share one. How are the points on the cuts assigned? Mark -----Original Message----- From: math-fun-bounces+mdtorge=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Wednesday, April 19, 2006 2:38 PM To: dasimov(a)earthlink.net; math-fun Subject: Re: [math-fun] Square thirds On 4/19/06, dasimov(a)earthlink.net <dasimov(a)earthlink.net> wrote: > A RELATED PROBLEM that some people could've predicted I'd ask is: > > In what ways can the square torus be decomposed into n congruent > pieces for any positive n (other than the obvious n parallel annuli > decomposition [whose slopes, btw, can be any fixed rational number]) ? > > --Dan > Well, hexagonal and triangular tilings as well, obviously: including the famous dissection into 7 hexagons, each adjacent at an edge to all the others. Fred Lunnon _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Re: Square thirds
by dasimov＠earthlink.net 20 Apr '06

20 Apr '06

Mark T. wrote: >If you use pencil and paper, sure two horizontal lines divide the thing >into three pieces that are the same. But y'all have been talking, open >sets, closed sets, point sets, etc. Down to that level of detail I don't >see how to trivially make equal thirds. Actually, I don't see how to do >it at all. If the trivial way is to make two equally spaced cuts, the >middle third shares two sides, the other two share one. How are the >points on the cuts assigned? Jim P. replied: << There's no way to do the dissection if you play the game that way (and if the pieces have polygonal interiors and boundaries composed of open, half-open, and closed intervals). Each region of this sort has an "Euler measure" (aka combinatorial Euler characteristic), which you can compute by decomposing it into points, open intervals, and open 2-cells, and then computing V-E+F, where V counts the points, E counts the open intervals, and F counts the open 2-cells. This quantity is always a whole number; it doesn't depend on how you decompose a region, and it's invariant under congruence. . . . >> I seem to recall once showing that *none* of the 5 Platonic solids (thought of as polyhedral surfaces) could be used for an exact down-to-the-last-point partition into congruent connected pieces by using its face interiors as disjoint open sets and then adjoining part of each's boundary. Using radial projection, these can all be considered as being about partitions of the usual sphere. In fact I'm aware of only aware of ONE way to partition the sphere into conguruent connected pieces each of which is an open set with an interval of its boundary adjoined. (Or infinitely many ways if the boundary is allowed to be a finie union of intervals.) (Apologies if this has come up in this venue before.) --Dan

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[math-fun] Re: Square thirds
by James Propp 20 Apr '06

20 Apr '06

>I'm still stuck on the trivial way of thirding the square. > >If you use pencil and paper, sure two horizontal lines divide the thing >into three pieces that are the same. But y'all have been talking, open >sets, closed sets, point sets, etc. Down to that level of detail I don't >see how to trivially make equal thirds. Actually, I don't see how to do >it at all. If the trivial way is to make two equally spaced cuts, the >middle third shares two sides, the other two share one. How are the >points on the cuts assigned? > >Mark There's no way to do the dissection if you play the game that way (and if the pieces have polygonal interiors and boundaries composed of open, half-open, and closed intervals). Each region of this sort has an "Euler measure" (aka combinatorial Euler characteristic), which you can compute by decomposing it into points, open intervals, and open 2-cells, and then computing V-E+F, where V counts the points, E counts the open intervals, and F counts the open 2-cells. This quantity is always a whole number; it doesn't depend on how you decompose a region, and it's invariant under congruence. For the closed square, the most natural decomposition gives 4-4+1, or 1. And since this is not divisible by 3, there's no way to "exactly" decompose the square into 3 congruent pieces. Klain and Rota's book "Introduction to Geometric Probability" talks about Euler measure. The people who have explored this notion are (roughly in chronological order) Hadwiger, Lenz, McMullen, Rota, Schanuel, Morelli, Chen, and myself. Jim Propp

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Re: [math-fun] equality of floating point numbers.
by Henry Baker 20 Apr '06

20 Apr '06

The problem of overloading operators that _appear_ to have certain properties, but don't, is one that I've complained about for years [http://home.pipeline.com/~hbaker1/ObjectIdentity.html] Your example of non-transitive "equals" is but the tip of the iceberg. Unfortunately, there doesn't seem to be any way to provide the object-oriented system with a _proof_ of the properties that are desired, so that these properties can be utilized by a compiler, for example. My current pet peeve is the lack of any mathematical elegance behind so-called "sync" operators -- that operation that "syncs" your PDA with your computer, or "syncs" your iPod with your computer. In my mind, "sync" is a lattice operation, e.g., "meet" or "join". Two important properties of such an operation is that it is _symmetric_, in that "sync"-ing should pass information in both directions so that both machines have exactly the same information; and it is _idempotent_, in that "sync"-ing twice shouldn't do any more than "sync"-ing once. Unfortunately, I've run across many so-called "sync" operations that were neither symmetric nor idempotent, and I wouldn't want to call them "sync" operations. At 06:56 AM 4/20/2006, Michael Kleber wrote: >Franklin T. Adams-Watters wrote: > >> >In a programming context, non-transitively can happen by accident in >> >an object-oriented language, if one creates two classes that know >> >how to compare themselves to a third, but not to each other. E.g., >> >Rational, Float, and Integer; 2/1 = 2, 2.0 = 2, but 2/1 != 2.0. > >...and Ray Tayek responded: > >> yes, but that works by overloading ==. > >But Ray, your original question was based on comparing floating-point >numbers using |a-b|<epsilon, which is itself redefining the equality >operator! The "real" equality operation on floating-point values is >one that requires perfect matching, which is flawed in the opposite >direction: too *few* things are equal, not too many. > >(When the "true" answer to some computation is not perfectly >representable by your favorite floating-point scheme, two >mathematically equivalent but computationally distinct evaluation >techniques might result in non-identical answers. For instance, >C tells me that (1/10)^2 != 1/100.) > >The problem here isn't with equality, it's with the associative and >distributive laws for the arithmetic operations that generated the >numbers in the first place. > >--Michael Kleber > >-- >It is very dark and after 2000. >If you continue you are likely to be eaten by a bleen.

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Re: [math-fun] equality of floating point numbers.
by dasimov＠earthlink.net 20 Apr '06

20 Apr '06

Ray Tayek writes: << hi, reading junit recipies by martin fowler. talking about equality as an equivalence relation on objects. if one of the fields of the objects that is in included in the equals method has a floating point number, then one uses some idea of "close enough" (i.e. within some epsilon). this is clearly reflexive and symmetric, but not transitive. he conjectures that this idea of "close enough" is the *only* case (in the space of equals methods for object) of having a equals relation that is symmetric and reflexive and not transitive. does anyone have a counter example? >> It's hard to know what is meant here by "an equals relation", since in math the concept of an "equivalence relation" is defined by possessing reflexivity, symmetry, and transitivity. It's easy to come up with a relation having the first two properties but not transitivity: Let the set on which the relation is defined be {A, B, C}. Let the entire relation be defined by A~A, B~B, C~C, A~B, B~A, B~C, and C~B. Then ~ is easily seen to be reflexive and symmetric, but not transitive (A~B and B~C but not A~C). This minimal example can be transplanted to any set S havng at least 3 elements, such as all real numbers: Pick 3 elements from S, call them A,B,C, and define ~ among A,B,C just as above. Let all other elements X satisfy X ~ X but no other "~" statement. Then you again have reflexivity and symmetry but not transitivity (since only one counterexample is needed). I don't know what it would mean to say this is or is not a "close enough" example. --Dan

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Re: [math-fun] equality of floating point numbers.
by Ray Tayek 20 Apr '06

20 Apr '06

At 08:14 PM 4/19/2006, you wrote: >Ray Tayek writes: > ><< >...equality >as an equivalence relation on objects. i.. some idea of "close >enough" (i.e. within some >epsilon).... >he conjectures that this idea of "close enough" is the *only* case ... >does anyone have a counter example? > >> > >It's hard to know what is meant here by "an equals relation", >since in math the concept of an "equivalence relation" is >defined by possessing reflexivity, symmetry, and transitivity. yes. the topic is an equals() method that *should* be tranistive and is not. >It's easy to come up with a relation having the first two >properties but not transitivity: ... sure >I don't know what it would mean to say this is or is not a "close enough" >example. my reading of it is that there is always something like abs(x1-x0)<epsilon and he conjectere that there are no other examples. thanks --- vice-chair http://ocjug.org/

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[math-fun] Re: Bernoulli numbers
by R. William Gosper 20 Apr '06

20 Apr '06

>Dear William Gosper, I'm looking for a reference for the identity involving the Bernoulli numbers. Please look at (34) at http://mathworld.wolfram.com/BernoulliNumber.html It is said that it is yours formula, but a reference is not given. This identity is equivalent to an (alternating) zeta relation. It's interesting if you proved your formula using this re lation, or if you used only properties of the Bernoulli numbers. Thanks in advance, Sergey. Dear Sergey, thank you for your inquiry. I can't locate that identity in the Math-Fun archive, where I usually announce things. In fact, I often flame there against Bernoulli number identities, simply because they can almost always be generalized to Bernoulli polynomials in new variables. E.g., (34) is the special case x=y=1/2 of n ==== \ n > ( ) B (x) B (y) = / i n - i i ==== i = 0 n (y + x - 1) B (y + x - 1) - (n - 1) B (y + x - 1), n - 1 n which follows immediately from multiplying the generating functions, so I don't know why I'd've bothered with such a routine announcement. Similarly, the preceding (32) and (33) can be greatly generalized by multisecting the expansion (c150) (sum(binomial(n,k)*t^k*bernpoly(x,k),k,0,n),%% = hypersimp(%%)) n ==== \ k n 1 (d150) > binomial(n, k) t bernpoly(x, k) = t bernpoly(x + -, n) / t ==== k = 0 E.g., hexasecting gives ==== \ (d180) 6 > binomial(n, 6 k) bernpoly(x, 6 k) = / ==== k = 0 %i %pi n -------- 2 7 %pi n %i sqrt(3) 1 2 (realpart(2 %e cos(-------) bernpoly(x + ---------- + -, n)) 6 2 2 2 %i %pi n - ---------- 3 %i sqrt(3) 1 n - 1 + realpart(- n %e (x + ---------- - -) )) + bernpoly(x + 1, n) 2 2 n + (- 1) bernpoly(x - 1, n) plus the five others for 6k+1, 6k+2,... . --Bill Gosper

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Re: [math-fun] cubic thirds
by Michael Reid 19 Apr '06

19 Apr '06

sorry to be late replying to this thread. > Polyominoes of order 3 do not exist > I. N. Stewart and A. Wormstein > > Journal of Combinatorial Theory, Series A > Volume 61, Issue 1 , September 1992, Pages 130-136 [ ... ] > So you can't do it with a polyomino. see also "Trisecting a rectangle", Samuel J. Maltby, Journal of Combinatorial Theory, Series A, v. 66, no. 1, April 1994, pp. 40-52. abstract: In this paper it is shown that it is impossible to dissect a rectangle into three congruent pieces unless those pieces are also rectangles. doi: http://dx.doi.org/10.1016/0097-3165(94)90049-3 math reviews: http://www.ams.org/mathscinet-getitem?mr=95g:52022 which seems to answer the original question asked. i note that it must be understood that the pieces are connected, for otherwise there are other (easy) dissections. this leaves open the case of dissection into 5, 7 or 9 congruent pieces. mike

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