July 2008 - mathfun@mailman.xmission.com

[math-fun] Fw: Beauty of Math
by Harvey Dubner 17 Jul '08

17 Jul '08

A relative sent me the following message. I had never seen this collection of number equations before. Harvey Dubner ----- Original Message ----- To: Harvey(a)dubner.com Subject: Beauty of Math Absolutely amazing! Beauty of Math! 1 x 8 + 1 = 9 12 x 8 + 2 = 98 123 x 8 + 3 = 987 1234 x 8 + 4 = 9876 12345 x 8 + 5 = 98765 123456 x 8 + 6 = 987654 1234567 x 8 + 7 = 9876543 12345678 x 8 + 8 = 98765432 123456789 x 8 + 9 = 987654321 1 x 9 + 2 = 11 12 x 9 + 3 = 111 123 x 9 + 4 = 1111 1234 x 9 + 5 = 11111 12345 x 9 + 6 = 111111 123456 x 9 + 7 = 1111111 1234567 x 9 + 8 = 11111111 12345678 x 9 + 9 = 111111111 123456789 x 9 +10= 1111111111 9 x 9 + 7 = 88 98 x 9 + 6 = 888 987 x 9 + 5 = 8888 9876 x 9 + 4 = 88888 98765 x 9 + 3 = 888888 987654 x 9 + 2 = 8888888 9876543 x 9 + 1 = 88888888 98765432 x 9 + 0 = 888888888 Brilliant, isn't it? And look at this symmetry: 1 x 1 = 1 11 x 11 = 121 111 x 111 = 12321 1111 x 1111 = 1234321 11111 x 11111 = 123454321 111111 x 111111 = 12345654321 1111111 x 1111111 = 1234567654321 11111111 x 11111111 = 123456787654321 111111111 x 111111111=12345678987654321 -------------------------------------------------------------------------------- Create a Home Theater Like the Pros. Watch the video on AOL Home.

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[math-fun] Playing with c = (a*b) mod (a+b)
by Eric Angelini 14 Jul '08

14 Jul '08

Hello MathFun & SeqFans, Start S = 2, 9,... Next term is (2*9)mod(2+9) 18 mod 11 = 7 S = 2, 9, 7,... Next term is (9*7)mod(9+7) 63 mod 16 = 15 S = 2, 9, 7, 15,... Next term is (7*15)mod(7+15) 105 mod 22 = 17 S = 2, 9, 7, 15, 17,... Next term is (15*17)mod(15+17) 255 mod 32 = 31 S = 2, 9, 7, 15, 17, 31,... Next term is (17*31)mod(17+31) 527 mod 48 = 47 S = 2, 9, 7, 15, 17, 31, 47,... Does this end in a loop? If yes, is there another start which doesn't? Best, É.

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[math-fun] Celestial Geometry
by Michael Kleber 11 Jul '08

11 Jul '08

I have a question in celestial geometry whose answer is surely well-known -- but not to me. Do any of you know this? I want to know the function whose inputs are my latitude on the earth (eg 0=equator, +-90=poles) and the latitude the sun is over today (0 on equinox, +-23.5 on the solstices), and whose output is what direction I need to look in to see the sun at sunrise/sunset (0=due east/west, -90=due south -- which would be the case if it's the winter solstice and I'm at the arctic circle). Surely all almanac-writers have a button on their pocket calculators which does this... --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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[math-fun] (no subject)
by rwg＠sdf.lonestar.org 08 Jul '08

08 Jul '08

If you split Ramanujan's 1/pi = 4F3[1/64] (search for Ramanujan in www.tweedledum.com/rwg/idents.htm) into two 3F2s, Mma 6.0 recognizes one of them, producing the identity expand(Hyper_F[P,Q]([1,1,3]/2,[1,1],1/64)=16/(21*%Pi) + (64*Elliptic_Kc(1/2*(1-3*Sqrt(7)/8))^2)/(21*%Pi^2)) 3 1 1 1 hyper_f ([-, -, -], [1, 1], --) = 3, 2 2 2 2 64 2 1 3 sqrt(7) 64 elliptic_kc (- - ---------) 2 16 16 ------------------------------ + ------ 2 21 %pi 21 %pi ~ 1.01336493939236. If we can find one more, we might get the whole contiguous family, since 3F2[z] obeys four term recurrences. Embarrassments: Mma 6.0 can't do FullSimplify[ FunctionExpand[ HypergeometricPFQ[{1, a, b}, {c, 3 - c + b + a},1]]] nor even the special case HypergeometricPFQ[{1,1,1},{5,5}/2,1] = 9 pi - 27, even though the general sum telescopes. My enhanced Macsyma also fails unless you first convert to Sigma notation (makesum). --rwg PS, overall, I'm favorably impressed with Mma 6.0, although I did lose a session when something in the help system cleared all my definitions.

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Re: [math-fun] Playing with c = (a*b) mod (a+b)
by Eric Angelini 08 Jul '08

08 Jul '08

Thanks to Chuck, Jack and Maximilian : http://www.cetteadressecomportecinquantesignes.com/ModuloPlay.htm Best, É. -----Message d'origine----- De : Eric Angelini [mailto:Eric.Angelini@kntv.be] Envoyé : lundi 7 juillet 2008 19:03 À : math-fun; seqfan(a)ext.jussieu.fr Objet : Playing with c = (a*b) mod (a+b) Hello MathFun & SeqFans, Start S = 2, 9,... Next term is (2*9)mod(2+9) 18 mod 11 = 7 S = 2, 9, 7,... Next term is (9*7)mod(9+7) 63 mod 16 = 15 S = 2, 9, 7, 15,... Next term is (7*15)mod(7+15) 105 mod 22 = 17 S = 2, 9, 7, 15, 17,... Next term is (15*17)mod(15+17) 255 mod 32 = 31 S = 2, 9, 7, 15, 17, 31,... Next term is (17*31)mod(17+31) 527 mod 48 = 47 S = 2, 9, 7, 15, 17, 31, 47,... Does this end in a loop? If yes, is there another start which doesn't? Best, É.

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[math-fun] Li and the Riemann Hypothesis
by victor miller 07 Jul '08

07 Jul '08

It seems that Terry Tao has found a bad flaw: http://terrytao.wordpress.com/2008/02/07/structure-and-randomness-in-the-pr…, and then there's the following comment. Victor ---------- Forwarded message ---------- From: Robert Guralnick <guralnic(a)usc.edu> Date: Thu, Jul 3, 2008 at 1:48 PM Subject: Re: [math-fun] proof of riemann hypothesis ? (fwd) I had lunch with Sarnak today -- he looked at the paper and said that there are many problems with the proof. Bob

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[math-fun] d/dx dinosaur
by Jason 06 Jul '08

06 Jul '08

Calculus humor, but with some swearing and immature stick figure violence: http://abstrusegoose.com/26

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Re: [math-fun] Xian-Jin Li's announced proof of Riemann Hypothesis
by Dan Asimov 03 Jul '08

03 Jul '08

Optimist (please feel free to identify yourself) wrote: << . . . Interestingly enough, Li is a former Ph.D. student (Purdue, 1993) of Louis de Branges, who has been somewhat notorious of late for announcing his own "proof" of RH back in 2004, which as I understand it was never accepted by experts. See http://genealogy.math.ndsu.nodak.edu/id.php?id=16641 This initially gave me pause, . . . >> Despite what I think is more than one supposed proof of RH by de Branges that was not accepted by the mathematical community, he is also known for proving the Bieberbach conjecture, a longstanding and important unsolved problem in complex function theory since 1916, in 1984. --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] f(0,0)-f(0,1)-f(1,0)+f(1,1)
by James Propp 03 Jul '08

03 Jul '08

Is there an easy way to show that if f(x,y) is twice-differentiable on [0,1]x[0,1] with continuous second-order partial derivatives, then there must exist a point at which (d/dx)(d/dy) f(x,y) equals f(0,0)-f(0,1)-f(1,0)+f(1,1)? (I wonder if forums like math-fun, and sci.math in its heyday, are bad for a mathematician's brain. The above problem is probably a fairly standard exercise in advanced calculus, and in the old days, I would've had to wait until coffee hour to ask a colleague, by which time my own subconscious would have dredged up the needed facts and supplied the answer. Nowadays, I can just post the question and wait for someone in the 24-hour-a-day virtual coffee hour to post an answer. So my brain doesn't get as much exercise. Then again, my subconscious has more chew-toys to play with, thanks to *other* people's questions!) Jim Propp

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Re: [math-fun] proof of riemann hypothesis ?
by Dan Asimov 02 Jul '08

02 Jul '08

P.S. Interestingly, while Li's putative proof of RH depends on a 2000 theorem of Bombieri, this theorem of Bombieri depends on a 1997 theorem of (the very same) Li. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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