June 2009 - mathfun@mailman.xmission.com

Re: [math-fun] Fractal puzzles
by Dan Asimov 24 Jun '09

24 Jun '09

There's also a possible intermediate case, where as the finite stages approach the limiting fractal in question, there *is* self-identification in the limit, yet the limiting fractal is nevertheless a simple closed curve. (E.g., in the way that identifying each x in [1,2] in the reals R with both y(x) = 2-x in [0,1], and z(x) = 4-x in [2,3], leads to a quotient space that is still topologically equivalent to R.) --Dan I wrote: << For this fractal, of dimension = log_5(9), it's not immediately obvious to me that it is a simple closed curve. Each of the finite stages that approach the limit is indeed a simple closed curve, but it seems possible that the limiting process might lead to some self-identification. I'd probably lean towards betting that it's *not* a simple closed curve. Anyone know the answer? >> _____________________________________________________________________ "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] Fractal puzzles
by Dan Asimov 24 Jun '09

24 Jun '09

For this fractal, of dimension = log_5(9), it's not immediately obvious to me that it is a simple closed curve. Each of the finite stages that approach the limit is indeed a simple closed curve, but it seems possible that the limiting process might lead to some self-identification. I'd probably lean towards betting that it's *not* a simple closed curve. Anyone know the answer? --Dan RWG wrote: << b) What is the dimension of the boundary of the 2D quincunx fractal, the fixed point of surrounding one by four in a square, then dividing by 2+i? >> _____________________________________________________________________ "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] Fractal puzzles
by Dan Asimov 24 Jun '09

24 Jun '09

The rotation by atan(1/2) should ensure that the sequence of finite stages converges to a geometric figure, but it doesn't concern how I calculated the Hausdorff dimension. --Dan Rich wrote: << Does the rotation from the /(2+i) make a difference here? --Rich I wrote: << For the fractal boundary I describe below -- and for the one described by Bill G. and Alan, if they're all the same -- I get Hausdorff dimension = log_5(9) (~ 1.3652+). >> _____________________________________________________________________ "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] Fractal puzzles
by Dan Asimov 24 Jun '09

24 Jun '09

For the fractal boundary I describe below -- and for the one described by Bill G. and Alan, if they're all the same -- I get Hausdorff dimension = log_5(9) (~ 1.3652+). --Dan << So, does this make it the same as starting with a plus sign of 5 adjacent squares, and taking 5 congruent copies snuggled together and normalized -- then iterate to the limit? (Actually the boundary of this limit.) If so, it's a fractal I was studying just a couple of months ago. I'll go review what I had about it. >> _____________________________________________________________________ "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] Fractal puzzles
by Dan Asimov 24 Jun '09

24 Jun '09

Thanks, Alan. So, does this make it the same as starting with a plus sign of 5 adjacent squares, and taking 5 congruent copies snuggled together and normalized -- then iterate to the limit? (Actually the boundary of this limit.) If so, it's a fractal I was studying just a couple of months ago. I'll go review what I had about it. =-Dan << I'm pretty sure I understand the intended construction. We'll construct a sequence of sets of complex numbers; the limit of this sequence will be (well-defined and) the intended fractal. Let Q[0] contain only 0. Then for any nonnegative integer i, let Q[i+1] = (Q[i] + {0, 1, -1, i, -i}) / (2i + 1). Here, if A and B are sets, the set A+B is intended to mean {a+b | a in A and b in B}. Each Q is roughly cross-shaped; RWG observes (very tersely) that dividing by 2i+1 rotates and shrinks each such cross by just enough that five crosses can snuggle together to make a meta-cross. I think this fractal is in Mandelbrot; I cannot dig up my copy at the moment. >> _____________________________________________________________________ "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] credit card numbers
by Dan Asimov 24 Jun '09

24 Jun '09

That's happened to me at least twice -- when I needed to cancel a card due to loss or theft, and the replacement card differed only in the last 4 digits. Which suggested to me that *something* was coded in the first 8, but what I can't say. --Dan << I learned (through entirely licit means) that a friend of mine who lives not far from me has the same first eight digits on her Visa card as mine. >> _____________________________________________________________________ "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] Fractal puzzles
by Dan Asimov 23 Jun '09

23 Jun '09

RWG wrote: << A private conversation raised the following: a) continuous maps from 1D onto a 2D patch visit countably many points at least three times. What is the minimum dense revisitation count for continuous volume fillers, 1D onto 3D? It was seven for that "Treano" function I discussed a few months ago. >> Whst is known about the dense visitation count from 1D onto 2D ? << b) What is the dimension of the boundary of the 2D quincunx fractal, the fixed point of surrounding one by four in a square, then dividing by 2+i? >> Could you please elaborate on this? I don't understand the construction. (I do know that a quincunx is the arrangement of 5 points, or dots, as the center and corners of a square.) _____________________________________________________________________ "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] The hexadeciaml pi formula; Catalan &al.
by rcs＠xmission.com 23 Jun '09

23 Jun '09

Simon Plouffe was one of the discoverers of the nice pi formula inf 4 2 1 1 -k pi = sum (---- - ---- - ---- - ----) 16 k=0 8k+1 8k+4 8k+5 8k+6 which allows computing individual bits of pi without computing the preceding part. It seems to me that there should be another such formula, in which the mod 8 residues are rearranged a little. Perhaps with 8k+3 and 8k+7 replacing 1 & 5, or maybe just 1&3 instead of 1&5, or with 8k+2 in place of 8k+6. Has anyone seen variations like this? ----- The sums of 1/(2n+1)^k, for even k, give rational multiples of pi^k, and the alternating sums, for odd k, also give rational multiples of pi^k. So 1 - 1/27 + 1/125 - ... = pi^3 / 32. Perhaps there's some relationship between the leftovers: gamma, Catalan, zeta(odd), etc? Rich

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Re: [math-fun] capture the cat
by Dan Asimov 16 Jun '09

16 Jun '09

I find I can win virtually every time on the website, except when I make a clearly unwise blunder. (Probably having played a lot of Hex doesn't hurt any.) The mathematical question on an infinite board seems to be * From which starting positions (darkened cells and position of the cat) can the cat be captured? * In particular: Can the cat be captured even if no cells are darkened? The game is similar in spirit to Conway's Angel & Devil game. --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] capture the cat
by Thane Plambeck 16 Jun '09

16 Jun '09

I'd welcome insight into whether this cat-catching game is indeed always winnable, as the site seems to claim http://www.members.shaw.ca/gf3/circle-the-cat.html -- Thane Plambeck tplambeck(a)gmail.com http://thaneplambeck.typepad.com/

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