March 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Nice chess problem
by Dan Asimov 06 Mar '11

06 Mar '11

That chess problem Richard posted reminds me of another backwards chess problem that I think I first saw in a Martin Gardner column of almost 40 years ago: This board shows a position from a possible game after the White king was accidentally knocked off the board immediately following a legal move. Lowercase = Black, uppercase = White. * * * * * * * * * * * * * * * * * * * * * * * * * r * b * * * * B * * * * * * * * * * * * * * * * * * * * * * * * * * k * * * * Where was the White king? Show a sequence of moves from an obviously plausible position that could lead to this one. (The board is oriented normally modulo a 180-degree rotation. But which side of the board is which, and whose turn is next, are suppressed.) --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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[math-fun] recent afterthoughts
by Bill Gosper 06 Mar '11

06 Mar '11

I closed 29nov with: --------------- Thought for today (simplfied by Julian) sum(sech(((%pi * n)/(2 * sqrt(3)))),n, - inf,inf) = ((3 * 2^(1/6) * Gamma((1/3))^3)/(2 * 3^(1/4) * (sqrt(3) + 1)^(5/2) * sqrt(5 * sqrt(3) - 4 * sqrt(2) - 3) * (%pi)^2)) inf ==== \ %pi n > sech(---------) = / 2 sqrt(3) ==== n = - inf 1/6 3 1 3 2 gamma (-) 3 ------------------------------------------------------------ 1/4 5/2 2 2 3 (sqrt(3) + 1) sqrt(5 sqrt(3) - 4 sqrt(2) - 3) %pi ------------------- This was subseqently *radically* simplified by Corey's denester: Sum[Sech[(n*Pi)/(2*Sqrt[3])], {n, -Infinity, Infinity}] == (3^(3/4)*(1 + 2*Sqrt[2] + Sqrt[3])*Gamma[1/3]^3)/(4*2^(5/6)*Pi^2) which also completely poofed the 8th roots from my tabulated eta(e^-(2 pi rt 19), 1/3 8 1/8 (d148) ((3 sqrt(57) + 1) - -------------------) = 1/3 (3 sqrt(57) + 1) 1/3 3/8 4 (3 sqrt(57) + 1) sqrt(57) - 5 2 (--------------------- + ------------------- - 2) sqrt(57) - 5 1/3 (3 sqrt(57) + 1) ------------------------------------------------------ 6 (as with the 19th singular value) and eta(e^-(2 pi rt 11) 1 ------------------------------------------------------------------ = 1/3 1/3 1/3 1/3 1/8 (2 (21 sqrt(33) + 283) + 2 (283 - 21 sqrt(33)) + 8) 1/3 1/3 2/3 1/4 7/8 2 (21 sqrt(33) + 283) 2 (sqrt(33) - 1) 2 3 (--------------------------- - ---------------------- + 1) sqrt(33) - 1 1/3 (21 sqrt(33) + 283) -------------------------------------------------------------------- 9 (Gammastorms omitted for brevity.) This leads one to ask why we tabulate singular values instead of etas. I can get the former from the latter, but vice versa? 23Feb:Victor (Miller)> Hi Simon, I would bet that your values are related to this: http://www.mathstat.carleton.ca/~williams/papers/pdf/220.pdf (rwg)>YOW! So evaluating etas (and thetas and lambdas and ...) is reduced to evaluating L series! Presumably, a solved problem. I had no idea. YOWed too soon. This (truly impressive) paper solves only a fraction of the problem: It only does integers, not rationals, and only 0 or -1 mod 4. And it only produces magnitudes, not phases. I thought it might be possible to get the phase directly from the infinite product, but how the heck do you get j 1/2 (1/2 + j) Pi Sum[(-1) ArcCot[E ], {j, 0, Infinity}] == 1/4 1 4 2 - (Pi - ArcTan[---------------]) ? 8 3 (3 + Sqrt[2]) Eta triples permit extension to rationals, but is there a science to finding them? I just mine the powerseries for empirical relations. 3Mar: Integrating a (simpler but harder to justify) variation on the W&W Mittag-Leffler derivation, 2 r 2 y y -2 Pi Sum[(-1) r (ArcTan[----] + ArcTanh[----]) Csch[Pi r], Pi r Pi r 2 {r, 1, Infinity}] == Integrate[x Csc[x] Csch[x], {x, 0, y}] Unintegrated: inf ==== r 1 2 \ r (- 1) csch(%pi r) csc(x) csch(x) = -- - 4 %pi x > -------------------- 2 / 4 4 4 x ==== %pi r - x r = 1 which probably also comes out of convolving the separate csc and csch expansions. Expanding at x=0 gives the infinite set of identities r 3 (-1) Csch[Pi r] Pi {Sum[----------------, {r, 1, Infinity}] == -(---), 3 360 r r 7 (-1) Csch[Pi r] 13 Pi Sum[----------------, {r, 1, Infinity}] == -(------), 7 453600 r r 11 (-1) Csch[Pi r] 4009 Pi Sum[----------------, {r, 1, Infinity}] == -(-----------)} 11 13621608000 r ... These are less like Simon's results than I thought--he has r^3 instead of r^-3. I was hoping this constant would be in ISC: r 1 Pi Sum[(-1) r Csch[Pi r] Log[1 - --], {r, 2, Infinity}] == 4 r 3 1 4 t Csch[Pi] Integrate[-(-) - ------------- + t Csc[t] Csch[t], {t, 0, Pi}] t 4 3 t Pi (1 - ---) 4 Pi There's lot's more fun to be had here. --rwg Also, it should be possible to let sides->oo in that spherical triangle area formula and get a path integral for the area bounded by a piecewise smooth curve.

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[math-fun] Nice chess problem
by Richard Guy 06 Mar '11

06 Mar '11

Recently Noam Elkies showed me the following cute chess problem. I don't know if it is original with him. Normal starting position for Chess. Remove Black's King's Knight, and White's King's Knight and King's and Queen's pawns. White has just played her 5th move in a normal play of Chess. What was it? R.

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[math-fun] why does the amazon river from space look like blood vessels?
by Gary Antonick 06 Mar '11

06 Mar '11

Hi all, I'm curious about what seems to be a branching pattern shared by blood vessels, the amazon river and oak trees, among other things. Can anyone point me in the right direction to learn a bit more about this? Here are some examples: - the amazon river<http://www.nasa.gov/centers/goddard/images/content/103371main_amazon-river.…> - blood vessels<http://www.istockphoto.com/file_thumbview_approve/6303204/2/istockphoto_630…> - lightning<http://www.istockphoto.com/file_thumbview_approve/3506509/2/istockphoto_350…> - an oak tree<http://www.telegraph.co.uk/news/uknews/4697449/Oak-tree-dating-back-250-yea…> What I find interesting are the odd details. An oak tree, lightning, and blood vessels, for example, seem to have these odd double-back knotted points. Things that — to me — do not look this this are.. - a winding river<http://www.istockphoto.com/file_thumbview_approve/4280336/2/istockphoto_428…> - veins in this leaf<http://www.istockphoto.com/file_thumbview_approve/3569643/2/istockphoto_356…> - a crack in ice<http://www.istockphoto.com/file_thumbview_approve/11913741/2/istockphoto_11…> I actually expected a crack in ice to look a lot more like lightning but it's completely different. These also seem to have their own groupings. The grand canyon<http://earthobservatory.nasa.gov/IOTD/view.php?id=6929> looks like frost<http://www.masterfile.com/stock-photography/image/600-02912692/Close-up-of-…>, for example. But I'm particularly curious about the blood vessels/amazon pattern. Cheers, Gary

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Re: [math-fun] advanced math 'Captchas'
by Dan Asimov 05 Mar '11

05 Mar '11

quad wrote: << It should be O(sqrt(x)log(x)), no? Not to mention that unmatched absolute value bar. >> That's one problem, yes. Also, the Gauss integral doesn't converge if its lower limit is 0. (Wikipedia mentions its "Cauchy principal value" but I don't know how that's defined in a case like this.) --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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Re: [math-fun] advanced math 'Captchas'
by Dan Asimov 05 Mar '11

05 Mar '11

"""quad""" wrote: << Can't forget about this Captcha: http://i.imgur.com/aAT52.jpg . . . >> It's a cute idea, but it misstates the Riemann hypothesis. --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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[math-fun] complex #'s, dot & cross product
by Henry Baker 05 Mar '11

05 Mar '11

Complex number algebra & vector algebra are usually taught as separate subjects, with little cross-over. However, I find the following operations to be extremely useful -- at least in programming & Maxima hacking -- when dealing with complex numbers. They are often more useful/perspicuous than the conjugate operation. dot(A,B) = dot(B,A) = realpart(A)*realpart(B)+imagpart(B)*imagpart(B) = (conjugate(A)*B+conjugate(B)*A)/2 = A.B cross(A,B) = -cross(B,A) = determinant([realpart(A),imagpart(A)],[realpart(B),imagpart(B)]) = realpart(A)*imagpart(B)-imagpart(A)*realpart(B) = (conjugate(A)*B-conjugate(B)*A)/2i = AxB 1.A = dot(1,A) = realpart(A) i.A = dot(i,A) = imagpart(A) 1xA = cross(1,A) = imagpart(A) Axi = cross(A,i) = realpart(A) (A.B)^2+(AxB)^2 = (A.A)*(B.B) For example, the vector A can be decomposed into the parallel & perpendicular components of B: A = (B.A)/(B.B)*B + i*(BxA)/(B.B)*B Trying to express this just with conjugates isn't particularly perspicuous. How come these vector operations on complex numbers aren't defined & used more often? (So what if complex numbers are only 2-dimensional; they're incredibly useful in that 2D domain.)

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[math-fun] advanced math 'Captchas'
by Henry Baker 05 Mar '11

05 Mar '11

FYI -- http://latimesblogs.latimes.com/technology/2011/03/web-security-against-hac… Technology The best Web security against hackers and spammers: advanced math! March 4, 2011 | 9:41 am If you don't want spam-bots or bad guys breaking into your site, make them do calculus. That's the approach the folks at a Croatia's Ru?er BoÂkovic' Institute are taking. Before you can log in to the research institute's Quantum Random Bit Generator service, you have to enter your name, password and affiliated organization, and then solve a math problem that would make most people run for the hills: http://www.irb.hr/en/geninfo/ http://random.irb.hr/signup.php [see article] [Captcha with calculus problem] The challenge is an extra-hard version of a regular CAPTCHA-type question (see image here), where users are prompted to retype a set of letters from a blurry image. And, like CAPTCHAs, users are given clemency if they can't get the first one right. "If you don't know the answer to this question," it says, "reload the page and you'll (probably) get another, easier question." How kind of you. It seems these scientists want to ward off ruffians who can't do advanced math. After all, the service they're offering is access to truly random numbers -- a difficult computer science feat on its own, and one that only responsible adults should have access to. But these math elitists may have a problem on their hands. As calculus teachers around the world are now discovering, the Internet will now do your math homework for you. Just go to WolframAlpha and pop in the problem, and boom, you'll have access to all the random numbers your heart could desire: http://www.wolframalpha.com/input/?i=d%2Fdx%28sin%282x+-+pi%2F2%29+%2B+6%29… Ha, ha, how do you like your little trick now, eggheads! Everyone knows the derivative of the sin of 2X minus pi over two plus six where x = 2 pi is zero! Isn't it? -- David Sarno

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[math-fun] Brownian motion
by Adam P. Goucher 05 Mar '11

05 Mar '11

I presume it is well-known that Brownian motion on Z^n is macroscopically isotropic, and therefore macroscopically indistinguishable from continuous Brownian motion on R^n? Basically, the position of a particle in each dimension at time t follows a binomial distribution, which asymptotically approaches a normal distribution. So, the probability density function in Z^n approaches a Gaussian 'hill' over n dimensions (easiest to visualise when n = 2). If the coordinates are x_1, x_2, x_3, ... , x_n, the probability density function is proportional to: e^-(x_1Â² + x_2Â² + x_3Â² + ... + x_nÂ²) = e^-dÂ² where d is the distance from the origin. Similarly, lattice gases can exhibit macroscopic isotropy, and thus cellular automata should be capable of approximating reaction-diffusion systems to arbitrary accuracy. Sincerely, Adam P. Goucher

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[math-fun] The hunt for the "true" 3D Mandelbrot - 4D?
by Stuart Anderson 04 Mar '11

04 Mar '11

The 3D object can be represented in 3D with the random generated buddhabrot algorithm; "Instead of selecting initial points on the real-complex plane one for each pixel, initial points are selected randomly from the image region. Then, each initial point is iterated using the standard mandelbrot function in order to test whether it escapes or not. Only those that do exit are then re-iterated. (The ones that don't escape -ie. which are believed to be within the Mandelbrot Set - are ignored)" http://www.superliminal.com/fractals/bbrot/bbrot.htm More intriguing than buddhabrots are the projections from 4D complex space into 3D, along with rotations and holographic views http://www.superliminal.com/fractals/bgram/ZrZiOut.htm cheers, Stuart

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