September 2013 - mathfun@mailman.xmission.com

[math-fun] Muffins
by rcs＠xmission.com 17 Sep '13

17 Sep '13

Richard Chatwin <rchatwin1(a)yahoo.com> has joined Math-Fun. He's interested in the Muffin Problem from 2009. He's reviewing our work, and may be contacting some of us about it. Rich

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[math-fun] maximum quadrilateral area
by Bill Gosper 16 Sep '13

16 Sep '13

WDS> With given sides occurs when the quad is inscribable in (i.e. inscribed in) a circle (and this indeed does not depend on the order of the sides due to Brahmagupta's area formula). It is also interesting that the "max area happens when quad inscribable" theorem is an "absolute" theorem, i.e. it works in both euclidean and nonEuclidean geometries. And the fact the area then is independent of the side-ordering also is "absolute" (which should be obvious). (I'm recounting all this from memory, since I recall proving this theorem a long time ago. I was going to put it in a paper I never finished.) ---------------------------- So, is the solid angle formed by four dihedrals maximal when it inscribes in a circular cone? (Actually, I should still have the max constraints. But I think the cone part was way messy.) --rwg 3 Sep 2008: "Everybody" knows that, given any triangle, the sum of the angle tangents equals their product. This "Brahmagupts" into Given any quadrilateral, the sum of the angle tangents equals their product times the sum of the cotangents. Can't be new. But probably newer than Brahmagupta, since, even for trapezoids, it just says 0=0. --rwg Am I the only one who hasn't seen In[650]:= FullSimplify[Integrate[Log[1/z]^n, {z, 0, 1}], n \[Element] Integers && n > 0] Out[650]= n! until today? It's just a change of variables in Euler's integral, but why isn't it preferred, or even mentioned, in e.g., A&S? Dicey for complex n?

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[math-fun] curves winding their way around dense subsets
by Cordwell, William R 16 Sep '13

16 Sep '13

Hello, There was some interesting discussion about rational points being dense in the unit circle (corresponding to all the Pythagorean triples). It is straightforward to show that there are many circles that contain no rational points. So, is there any theory about (smooth, closed, whatever modifier you like) curves in the plane that completely miss a dense subset? Thanks, Bill

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Re: [math-fun] The Bernoulli Manifesto
by Bill Gosper 16 Sep '13

16 Sep '13

VMiller>How about "Bernewli" numbers? Sent from my iPhone On Sep 15, 2013, at 16:57, Eugene Salamin <gene_salamin(a)yahoo.com> wrote: It would cause much less confusion if a new name is associated with the new definition, e.g. a "new-Bernoulli" number. -- Gene Naah. Just call the present ones Bernoldies. --rwg

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[math-fun] The Bernoulli Manifesto
by Hans Havermann 16 Sep '13

16 Sep '13

Back in 2004, Peter Luschny wrote an open letter to Donald E. Knuth about his conviction that the proper definition of the Bernoulli number B(1) should be 1/2, not -1/2. Professor Knuth declined to embrace the idea. http://www.luschny.de/math/zeta/OpenLetter.pdf Peter has now put up a detailed response to Knuth, calling it The Bernoulli Manifesto. http://luschny.de/math/zeta/The-Bernoulli-Manifesto.html

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[math-fun] The Bernoulli Manifesto
by Bill Gosper 15 Sep '13

15 Sep '13

Hans Havermann>Back in 2004, Peter Luschny wrote an open letter to Donald E. Knuth about his conviction that the proper definition of the Bernoulli number B(1) should be 1/2, not -1/2. Professor Knuth declined to embrace the idea. http://www.luschny.de/math/zeta/OpenLetter.pdf Peter has now put up a detailed response to Knuth, calling it The Bernoulli Manifesto. http://luschny.de/math/zeta/The-Bernoulli-Manifesto.html -------------- Wow, I expected DEK's only defense of the status quo would be "It's too late to change." But they've already changed at least once. Whittaker and Watson's nth Bernoulli number was |B_2n|. Date: Sun, 20 Feb 1994 02:54-0800 From: Bill Gosper <rwg(a)NEWTON.macsyma.com> Subject: Bernoulli numbers and polynomials To: conway(a)math.Princeton.EDU Cc: math-fun(a)cs.arizona.edu In-Reply-To: <19931017045155.6.RWG(a)TSUNAMI.macsyma.com> Message-Id: <19940220105422.2.RWG(a)SWEATHOUSE.macsyma.com> Date: Sat, 16 Oct 1993 21:51 PDT From: Bill Gosper <rwg@NEWTON> . . . JHC> We won't win on B^1 for a very long time. Which leads to my modest proposal: * Outlaw Bernoulli numbers altogether! * This is founded on my previously stated . . . religious conviction that any formula involving Bernoulli numbers can be elegantly generalized to instead use Bernoulli polynomials of a new variable. Then our only concession is always having to write B(x+1). (Like Gamma(x+1).) Several years ago I wrote these sentiments to Rich and Hilarie, and included some polynomialized formulas traditionally given with mere Bernoulli numbers. One I'd especially like to recover, ascribed to Rademacher, was posed as a challenge by N.J.A. Sloane in reponse to the polys-only proposal. Found it. 26 Nov 1980, (claimed) cc: D&E Lehmer, R Askey, N Sloane, L Guibas. But I think we already knew this: k k k k k k y y - (y + 1) y - 2 (y + 1) + (y + 2) B (y) = -- + ------------- + -------------------------- + . . . k 1 2 3 Anyway, it directly extends B (a) 1-s zeta(s, a) = ------- s - 1 to negative s. (And indirectly to conjectures like 3 zeta(3) - --------- = 2 4 pi 0 1 1 2 2 3 ----- log(1) + ----- (log(1) - log(2)) + ----- (log(1) - 2 log(2) + log(3)) 2 3 4 3 4 5 4 5 6 + . . ., later confirmed.) Other gossip therein: 7 zeta(3) --------- inf 2 /===\ 1/12 - n 8 pi | | e e | | ---------------------- = ---------- | | 1 2 (n - 1/3) n 5/36 n = 1 (1 - ---) 2 2 n and denom(B ) = 2 Fib(20). 40 --------------- rwg And I do NOT advocate τ over 2π.

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[math-fun] rationals hard to approximate by rationals
by Warren D Smith 14 Sep '13

14 Sep '13

Consider a rational number (in reduced form) X=A/B. Suppose I tell you B, and I ask you to "find A such that X has only 1s and 2s in its partial fraction expansion." (Even better, "...and with the fewest 2s.") For example if B=2^32. You could by brute force just try all 2^32 possible A (actually only 2^31 if we insist A=odd) computing the continued fraction of each A/B, Is there a faster method? How fast can you get? For which (if any) B's is the problem unsolvable? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Gyroalgebra?
by Mike Stay 13 Sep '13

13 Sep '13

I was recently introduced to the notion of a "gyrovector" in the context of special relativity. Addition of velocities in the open ball G = {v s.t. ||v|| < c} is not associative, but there's a deformation of associativity that holds, namely a⊕(b⊕c) = (a⊕b)⊕gyr[a, b](c), where gyr[a, b] is an automorphism of the groupoid (G, ⊕). Does anyone know if there's a notion of "multiplication" of gyrovectors to get a "gyroalgebra"? I expect the multiplication wouldn't be associative, either. Googling "gyroalgebra" gives a single hit, and there it's used to mean calculating velocities with gyrovectors, not the multiplicative structure. -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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[math-fun] insects invented gears
by Warren D Smith 13 Sep '13

13 Sep '13

http://www.cam.ac.uk/research/news/functioning-mechanical-gears-seen-in-nat…

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Re: [math-fun] tribia
by Bill Gosper 13 Sep '13

13 Sep '13

PVS numbers? http://en.wikipedia.org/wiki/Pisot–Vijayaraghavan_number http://en.wikipedia.org/wiki/Salem_number WFL On 9/11/13, Bill Gosper < billgosper(a)gmail.com> wrote: DanA> When I try to evaluate that exact sum in Mma, it gives me only numbers that are to > 20 decimal places equal to 0. (Adding only 1000 or 10000 terms, asking for 10 digits' precision.) I'd guess only a set of measure 0 of real numbers x would have their Sum[Round[x^n]-x^n,{n,0,∞}] <> 0. But it's easy to believe that many algebraic numbers don't. What's your secret? --Dan On 2013-09-10, at 8:11 PM, Bill Gosper wrote: Let t:=1/3 (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3)), the tribonacci constant. Then, empirically, Sum[Round[x^n]-x^n,{n,0,∞}] Gaa, I meant Round[t^n]-t^n (Mac OS crashed as I prepared to send.) ==-0.80851211604688125... -2 + ((3 Sqrt[3] - Sqrt[11])^(1/3) + (3 Sqrt[3] + Sqrt[11])^(1/3))/(2^(2/3) Sqrt[3]) In[2]:= NSum[Round[GoldenRatio^k] - GoldenRatio^k, {k, 288}, NSumTerms -> 288, WorkingPrecision -> 99] Out[2]= 0.618033988749894848204586834365638117720309179806089199107257598931385927496544945583310480117151672 In[3]:= (% + 1/2)^2 Out[3]= 1.2500000000000000000000000000000000000000000000007297116525363033564609 5964404522000200488900285933 This can't be new. Anybody? --rwg _________________________________ Thanks, Fred! So, defining roundoff[x]:=FractionalPart[x+1/2]-1/2 (or Round[x]-x), why has nobody thought to Sum[roundoff[Pisot^k]]? At first I thought it's because of the modern habit of norming everything. But Sum[Abs[roundoff[Pisot^k]]] is quadratic when Pisot is, so somebody should have noticed. This is and endless source of identities. It seems that, if it converges, Sum[roundoff[Pisot1^k]^p1*roundoff[Pisot2^k]^p2.../r^k] is algebraic with the same degree as Pisot1 Pisot2 ... . E.g., NSum[(FractionalPart[(1/3 (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3)))^k + 1/2] - 1/2)^2/2^k, {k, 288}, NSumTerms -> 288, WorkingPrecision -> 288]; In[31]:= RootApproximant[%] Out[31]= Root[-215317 + 2949880 #1 + 1899072 #1^2 + 326144 #1^3 &, 1] In[29]:= RootApproximant[NSum[(FractionalPart[(1/ 3 (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3)))^ k + 1/2] - 1/2)*(FractionalPart[GoldenRatio^k + 1/2] - 1/2), {k, 288}, NSumTerms -> 288, WorkingPrecision -> 105]] Out[29]= Root[ 41 + 567 #1 - 144 #1^2 - 313 #1^3 - 8 #1^4 + 55 #1^5 + 11 #1^6 &, 2] --rwg

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