January 2006 - mathfun@mailman.xmission.com

Re: [math-fun] Digital silliness
by Michael Reid 10 Jan '06

10 Jan '06

> For a number n, let f(n) be the set of numbers gotten by splitting n^2 at > the 0 digits. For example > > 29648^2 = 879003904 > > so f(29648) = { 4, 39, 879 } > > Let S be the smallest set of numbers containing 2 and fixed by f. > What is the largest element of S? perhaps you mean "if s in S , then f(s) is a subset of S ". it is not obvious to me that S is finite. it is easy to exhibit arbitrarily large integers that do not have 0 as a decimal digit, and whose squares also do not have any 0's . of course this does not show that your S is infinite, but raises that issue as a possibility. is there reason to believe (i.e. an heuristic argument) that S is finite? or better yet, a proof? mike

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[math-fun] more hyjynx
by R. William Gosper 10 Jan '06

10 Jan '06

A datum, probably long known, apropos the computational complexity of the Gamma fcn: 3 1 6 9 5 1 6 8 1 6 10 hyper_2f1 (-, -, - ) hyper_2f1 (-, -, - ) hyper_2f1(-, -, -- ) 7 7 7 7 7 7 7 7 7 7 1 %pi 588 (-)! cos(---) 7 14 = ------------------, %pi so (n/d)! is probably easy, but with effort proportional to d. (Or maybe d^2). ------- My POWERSERIES function got a double sum instead of Clausen's 4F3[z] for not knowing the well-poised, 3-balanced (sesquiSaalschutzian?), terminating 4F3[1] in the inner sum. "SesquiPfaffian" is more just, but also more confusing. I got ambitious and derived the whole 4F3[a,b,c,d;e,f,g] (and 3F2[z]) matrix system, and extracted the general, noterminating case: 1 hyper_f ([a, b, c, c - b - a + -], 4, 3 2 1 [c - a + 1, c - b + 1, b + a + -], 1) = 2 1 (b + a - -)! (c - a)! (c - b)! (sqrt(%pi) (c - 2 b - 2 a)! 2 1 1 /((a - -)! (b - -)! (c - 2 a)! (c - 2 b)! (c - b - a)!) 2 2 - hyper_f ([1, c - 2 a + 1, c - 2 b + 1, c - b - a + 1], 4, 3 3 [c + 1, c - 2 b - 2 a + 2, c - b - a + -], 1) 2 c + 1 1 /((a - 1)! (b - 1)! (----- - b - a) c! (c - b - a + -)!)) 2 2 Notice the rhs 4F3 getting divided by infinity exactly when an upper parameter on the left is a nonpositive integer. Also note the weird oo - oo when (c+1)/2 = a+b. As usual, the matrices provide the complete system of identities contiguous to the above. The algebra gets very heavy in transforming the general 4x4 system (which has eight dimensions), but there's a way to exploit the symmetry in a,b,c,d and e,f,g. You can actually substitute a=b=c=d=S and later recover the original generality with an extension of the following trick: Supposing there were only two variables, a and b: Replace every f(S)^3 with f(a) f(b) f((a+b)/2), then f(S)^2 with f(a) f(b) and finally f(S) with f((a+b)/2). This isn't rigorous, but elementary theology seems to find the right symmetric functions. A rigorous way to fight malignant algebroma is to get even more ambitious and derive the q-extension of the 4F3 system. Normally, the addition of a new variable makes matters nonlinearly worse, but q-extending (n + d + c + b + a) (n + e + c + b + a) (n + e + d + b + a) (n + e + d + c + a) (n + e + d + c + b) to n + d + c + b + a n + e + c + b + a (1 - q ) (1 - q ) n + e + d + b + a n + e + d + c + a (1 - q ) (1 - q ) n + e + d + c + b (1 - q ) actually *reduces* the expanded form from 247 terms to 31 terms, with five 5-term exponents and 26 sixes. All the nonlinear terms (n^5, a^4, ...) disappear, but are recoverable with l'Hospital's rule. Unfortunately, 4F3 is just shy of the breakeven point. But at least I now have the 4phi3 system too. --rwg (-: Mesotonic economist comes-into emoticons :-) PS, I'm dabbling in sudoku as a programming exercise. Our program (brute-force backtrack) typically finds one or two "redundant" clues in published puzzles, but of course removing them renders the puzzle both unsymmetrical and insoluble without (perhaps humanly infeasible) lookahead.

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Clarification Re: [math-fun] Bott periodicity
by Daniel Asimov 10 Jan '06

10 Jan '06

In "Explication du texte" I inadvertently switched notation from g to f. g and f are both supposed to be the same reflection in v-w. Sorry about that. --Dan

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Re: [math-fun] Bott periodicity
by Daniel Asimov 10 Jan '06

10 Jan '06

Fred Lunnon wrote: << Which reminds me that I recently acquired a copy of the august tome J.H.Conway, D.A Smith "On Quaternions and Octonions" and promptly became mired at Lemma 3 on page 6. I can prove this result easily, but I'm hanged if I can follow their (3 line) proof --- can another reader clarify? Fred Lunnon >> I've finally found my copy of Conway & Smith (which I highly recommend, although you could spend a lifetime trying to absorb all the goodies therein). At last I know what Fred is talking about. The proof is flawless, if laconic. ----------------------------------------------------------------------------------------------------- Lemma 3 (page 6, "On Quaternions and Octonions", John Conway and Derek Smith, A.K. Peters, 2003): Every element g of O(n) that fixes a k-dimensional subspace can be written as a product of at most n-k reflections. ----------------------------------------------------------------------------------------------------- [Note: O(n) is the group of all isometries of R^n that fix the origin. This book calls it GO_n, but the rest of the world calls it O(n) or O_n.] ----------------------------------------------------------------------------------------------------- Proof as given in the book: Take a vector v that is *not* fixed by g, say v -> w. Then reflection in v-w restores w to v while fixing any vector u fixed by g. ------------------------------------------------------------------------------------------------------ Explication du texte: We're doing induction on the codimension n-k. First assume n-k = 0. Then g fixes all of R^n and we're done, having used 0 reflections. Now assume Lemma proved for all codimensions < n-k. Let r denote the reflection in v-w. Now compose r with f to get rf, satisfying rf(v) = r(w) = v. It's easy to check that rf fixes the same k-dimensional subspace fixed by f. But rf also fixes the subspace generated by v, so the dimension of rf's fixed subspace greater than k and its codimension is less than n-k. Hence by the induction hypothesis, rf is the product of at most n-k-1 reflections: rf = r_1 r_2 . . . r_q for q <= n-k-1. Applying r on the left of each side of this equation, we get (since r^2 = identity) that f is the prouct of at most n-k reflections. qed ------------------------------------------------------------------------------------------------------- --Dan

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[math-fun] Grocery store geometry...
by Henry Baker 10 Jan '06

10 Jan '06

http://www.canstruction.org/2005winners.html CanstructionÂ®, trademarked by the Society of Design Administration (SDA) and held in conjunction with chapters of the American Institute of Architects (AIA) and allied professional organizations, is an annual charity competition where design firms test their mettle to see who can build the most astounding structure made entirely from full cans of food. At the close of the competition all of the food used in the structures is donated to local food banks to feed those most in need. Structures, ranging in size from 1,000 to 13,000 cans, are created by stacking a variety of can sizes and shapes using the product labels as the color pallet. Structurally self-supporting, the only other materials to be used are 1/4Â leveling, cardboard, tape, rubber bands and wire. This past year competitions took place in 45 cities across North America. Local winners went on to compete via photography and were judged at the AIA/SDA national convention.

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Re: [math-fun] Bott periodicity
by Daniel Asimov 09 Jan '06

09 Jan '06

Gareth wrote: << [I wrote:] > Er, guys, would one of you be so kind as to state what theorem is being > discussed here? The answer: Any isometry of R^n that fixes a k-dimensional subspace is the product of at most n-k reflections. >> OK, I'll give it a shot: The proposition is equivalent to saying that any isometry of R^p with no nontrivial fixed subspace is the product of at most p reflections. Any isometry of R^n preserves mutually orthogonal subspaces: of dim = 2, and another one of dim = 1 (if n is odd). (This follows readily from complexification.) The cases of 1 dimension, and one reflection in 2D, are trivial. And a rotation of R^2 of angle theta is the product of a reflection about the x-axis, and one about the line at angle theta/2. QED. --Dan

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[math-fun] Sudoku Clue - minimum figure
by Guy Haworth 09 Jan '06

09 Jan '06

Apologies if this is an 'old one' and I haven't been paying attention. The question of the minimum number of clues for Sudoku (for a unique solution) is circulating. Apparently, 17 is ok but there are no known puzzles with 16 clues and a unique solution. Anything known in this area? g

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Re: [math-fun] Bott periodicity
by Daniel Asimov 08 Jan '06

08 Jan '06

Gareth wrote: << [I wrote:] > Er, guys, would one of you be so kind as to state what theorem is being > discussed here? The answer: Any isometry of R^n that fixes a k-dimensional subspace is the product of at most n-k reflections. >> OK, I'll give it a shot: The proposition is equivalent to saying that any isometry of R^p with no nontrivial fixed subspace is the product of at most p reflections. Any isometry of R^n preserves mutually orthogonal subspaces: of dim = 2, and another one of dim = 1 (if n is odd). (This follows readily from complexification.) The cases of 1 dimension, and one reflection in 2D, are trivial. And a rotation of R^2 of angle theta is the product of a reflection about the x-axis, and one about the line at angle theta/2. QED. --Dan

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Re: [math-fun] Bott periodicity
by Gareth McCaughan 08 Jan '06

08 Jan '06

I think Dan intended this e-mail, which went just to me, to go to the list: > Er, guys, would one of you be so kind as to state what theorem is being > discussed here? The answer: Any isometry of R^n that fixes a k-dimensional subspace is the product of at most n-k reflections. -- g

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Re: [math-fun] Bott periodicity
by Daniel Asimov 08 Jan '06

08 Jan '06

Er, guys, would one of you be so kind as to state what theorem is being discussed here? Thanks, Dan ------------------------------------------------------- Gareth wrote: << On Sunday 08 January 2006 14:41, Fred Lunnon wrote: > Which reminds me that I recently acquired a copy of the august tome > J.H.Conway, D.A Smith "On Quaternions and Octonions" > and promptly became mired at Lemma 3 on page 6. > I can prove this result easily, but I'm hanged if I can follow their > (3 line) proof --- can another reader clarify? Fred Lunnon I'll have a go. Let a be any element of GO(n) fixing a k-dimensional subspace. If it's the identity, we're done. Otherwise, pick something it doesn't fix; say it takes v to w. Let b be reflection in v-w. Then ba is still in GO(n); it fixes all the same things a did; and it also fixes v. So it fixes (at least) a (k+1)-dimensional subspace, so by induction on n-k we're done. How do we know that ba fixes all the same things a did? Because anything a fixes is orthogonal to v-w (Why? Because an isometry preserves inner products; so if au=u and av=w then (u,v) = (au,av) = (u,w), so (u,v-w)=0.) and is therefore fixed by b as well as by a. Is that any better? >>

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