November 2012 - mathfun@mailman.xmission.com

[math-fun] union codes
by Warren Smith 21 Nov '12

21 Nov '12

if the code words are some set of orbits under some permutation group (Colth. used just the cyclic group) doesn't that explain Veit's observations & also explain how said code was found?

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[math-fun] Galileo
by Warren Smith 21 Nov '12

21 Nov '12

Galileo had, theoretically, no notion of calculus and Newton laws, so no way to realize circle was not "perfect." Experimentally, he had no clocks, he measured time using heartbeat, which is less precise.

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[math-fun] Fwd: Best proof that sqrt(2) is irrational?
by Bill Gosper 21 Nov '12

21 Nov '12

Hans H>Bill Gosper: "Years ago, Conway said, perhaps on this list, that he was working on a book of Extreme Proofs. His version of this one was to cut (or fold) a square of paper along the diagonal..." Michael Reid: "So it's possible that this proof is several millenia old." According to Miller & Montague, Stanley Tennenbaum discovered Conway's 'origami' proof in the 1950's: http://web.williams.edu/go/math/sjmiller/public_html/math/papers/irrational… According to < http://www.qedcat.com/proofs/geometricproof.html >, it may go back even further, mentioning a 'numerical' version in 'A Course of Modern Analysis' by Whittaker & Watson (1927). As I mentioned to Mike Hirschhorn, I think Conway's[?] origami is novel because you don't actually need to compute the quantities b-a and 2a-b. You can just say they're obviously differences of integers forming a smaller isosceles right triangle. --Bill On Tue, Nov 20, 2012 at 10:11 PM, Mike Hirschhorn <m.hirschhorn(a)unsw.edu.au>wrote: > Dear Bill, > > I do not know who first thought of the proof that if b^2=2a^2 then > (2a-b)^2=2(b-a)^2, leading to a contradiction, > though I found it many years ago, and it was accepted as a filler in the > American Mathematical Monthly till I found it in an article entitled > ``Irrationality without number theory'' in the Monthly of the previous > year, under the previous editor, and withdrew it. > > If you can find that article, you may find a reference to earlier work > which includes this proof. > > I found it by reversing the recurrence that gives the convergents to the > continued fraction for sqrt{2}. > > Best wishes, > > Mike. > > > ________________________________________ > From: mathfuneavesdroppers(a)googlegroups.com [ > mathfuneavesdroppers(a)googlegroups.com] on behalf of Bill Gosper [ > billgosper(a)gmail.com] > Sent: Wednesday, 21 November 2012 3:08 PM > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] Best proof that sqrt(2) is irrational? > > Hans Havermann> > > Dan Asimov: > > > Does anyone know who first thought of the following proof that > sqrt(2) is irrational? > > > It looks like Tom M. Apostol's proof, published as "Irrationality of The > Square Root of Two - A Geometric Proof" in the American Mathematical > Monthly, November 2000, pp. 841–842. > > Years ago, Conway said, perhaps on this list, that he was working on a > book of Extreme Proofs. > His version of this one was to cut (or fold) a square of paper along the > diagonal, making > an isosceles right triangle. Label the sides a,a,b. Then bisect-fold one > of the 45˚ angles. > > This creates an obvious right-angled b-a,b-a,2a-b. > --rwg > >

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Re: [math-fun] Best proof that sqrt(2) is irrational?
by Bill Gosper 21 Nov '12

21 Nov '12

Hans Havermann> Dan Asimov: Does anyone know who first thought of the following proof that sqrt(2) is irrational? It looks like Tom M. Apostol's proof, published as "Irrationality of The Square Root of Two - A Geometric Proof" in the American Mathematical Monthly, November 2000, pp. 841–842. Years ago, Conway said, perhaps on this list, that he was working on a book of Extreme Proofs. His version of this one was to cut (or fold) a square of paper along the diagonal, making an isosceles right triangle. Label the sides a,a,b. Then bisect-fold one of the 45˚ angles. This creates an obvious right-angled b-a,b-a,2a-b. --rwg

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[math-fun] Bettsian but not vertex-transitive
by Adam P. Goucher 21 Nov '12

21 Nov '12

Given some positive integer n, we consider a finite set S of points in R^n. We are interested in sets which are Bettsian (the set of distances from a point X in S to the other points in S is independent of the choice of X). I have shown that in R^2, all Bettsian sets are vertex-transitive. For R^4 and higher dimensions, I have counter-examples (Bettsian sets which are not vertex-transitive). At the time of writing, the problem is still open for R^3. http://cp4space.wordpress.com/2012/11/21/betts-revisited/ Sincerely, Adam P. Goucher

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Re: [math-fun] Best proof that sqrt(2) is irrational?
by Michael Reid 20 Nov '12

20 Nov '12

I've heard that the ancient Greeks had proofs of irrationality for sqrt(2), sqrt(3), sqrt(5), sqrt(6), ... up to some modest bound, and I think they used this type of reduction. So it's possible that this proof is several millenia old. (Does Apostol's article have any info?) For what it's worth, my favorite proof that sqrt(2) is irrational is: Consider the set S = { a + b sqrt(2) | a, b are integers } , and note that S is closed under multiplication. If sqrt(2) were rational, say p/q , then S would be contained in (1/q)Z . However, for large n , we have 0 < (2 - sqrt(2))^n < 1/q , a contradiction, since (2 - sqrt(2))^n is in S . Michael Reid > Does anyone know who first thought of the following proof that sqrt(2) is irrational? It's definitely my favorite. > > Suppose that the isosceles right triangle whose legs are 1 has hypotenuse = p/q with p,q in Z+. > > By magnifying it by a factor of q, we get the right triangle with sides q, q, p. > > Call this triangle ABC, with A at the right angle. > > Now draw arc of the circle O with center at B and radius q, from side AB to the hypotenuse. > Assume it hits the hypotenuse at point X. Draw a perpendicular to the hypotenuse at X, > extended to hit side AC, say at point Y. > > Then the segments CX and XY are of equal length p-q. And since the lines AY and XY are > tangents to the circle O from an external point (namely Y), they are equal length, > namely p-q, an integer. > > This means that the segment CY must be of length q-(p-q) = 2q-p, also an integer. > > Hence the triangle CXY is another, smaller, isosceles right triangle with integer sides. > > Repeating this process leads to an infinite regress of integer right triangles each > with smaller sides than the last one: contradiction. > > --Dan

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[math-fun] "The Loves of the Triangles"
by Henry Baker 19 Nov '12

19 Nov '12

FYI -- The Rime of the Ancient Mathematician. Believe it or not, the following is math porn ~1798. Yes, I think they twittered after reading this doggerel. http://uniformlyatrandom.wordpress.com/2009/12/14/the-loves-of-the-triangle… And first, the fair PARABOLA behold, Her timid arms, with virgin blush, unfold! Though, on one focus fixÂd, her eyes betray A heart that glows with loveÂs resistless sway, Though, climbing oft, she strive with bolder grace Round his tall neck to clasp her fond embrace, Still eÂer she reach it from his polishÂd side Her trembling hands in devious Tangents glide. Not thus HYPERBOLA:Âwith subtlest art The blue-eyed wanton plays her changeful part; Quick as her conjugated axes move Through every posture of luxurious love, Her sportive limbs with easiest grace expand; Her charms unveilÂd provoke the loverÂs hand:Â UnveilÂd except in many a filmy ray Where light Asymptotes oÂer her bosom play, Nor touch her glowing skin, nor intercept the day. Yet why, ELLIPSIS, at thy fate repine? More lasting bliss, securer joys are thine. Though to each fair his treacherous wish may stray, Though each in turn, may seize a transient sway, ÂTis thine with mild coercion to restrain, Twine round his struggling heart, and bind with endless chain. See: http://books.google.ca/books/download/Poetry_of_the_Anti_Jacobin.pdf pdf page 126 = book page 113. LOVES OF THE TRIANGLES. ARGUMENT OF THE FIRST CANTO. Warning to the Profane not to approachÂ-Nymphs and Deities of Mathematical MythologyÂ-Cyclois of a pensive turn-ÂPendulums, on the contrary, playfulÂ-and why?Â-Sentimental union of the Naiads and HydrostaticsÂ-Marriage of Euclid and Algebra Pulley the emblem of MechanicsÂ-Optics of a licentious DispositionÂ-distinguished by her Telescope and Green Spectacles.Â-Hyde Park Gate on a Sunday MorningÂ-CochneysÂ-Coaches.Â-Didictic PoetryÂ-NonsensiaÂ-Love delights in Angles or Corners-ÂTheory of Fluxions explainedÂ-Trochais, the Nymph of the WheelÂSmoke-Jach describedÂPersonification of elementary or culinary Fire.ÂLittle Jack HornerÂStory of CinderellaÂRectangle, a Magician, educated by Plato and MenecmusÂin love with Three Curves, at the same timeÂserved by Gins, or GeniiÂtransforms himself into a ConeÂThe Three Curves requite his PassionÂdescription of themÂParabola, Hyperbola, and EllipsisÂAsymptotesÂConjugated Axes.ÂlllustrationsÂRewbell, Barras, and Lepaux, the Three virtuous Direct orsÂMacbeth and the Three WitchesÂThe Three FatesÂThe Three Graces I &c.

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[math-fun] N-adic numbers
by Michael Reid 19 Nov '12

19 Nov '12

I haven't been following too closely, but let me add a few remarks. Suppose N > 1 is an integer. Multiplication on the ring of N-adic integers is continuous (as already discussed below). The purported issue: > If multiplication is continuous, you can then extend multiplication to > the n-adics by continuity. But continuity of multiplication depends on > the fact that | a | * | b | = | (a * b) | which only holds for n > prime. is not really an issue. Continuity does not require equality there; indeed it follows from the inequality | a | * | b | >= | a * b | , which holds trivially. In fact, continuity would follow from much less than that. ) So as I had read somewhere, there is indeed a natural multiplication on the N-adic integers which is continuous relative to the topology induced by the N-adic metric. ) ) Even though for N composite the N-adic integers do have zero-divisors. Not necessarily. If N is a power of a prime, say p^a (with a > 0 ) then the N-adic integers are the same as the p-adic integers, which have no zero-divisors. In general, if N = p_1^a_1 p_2^a_2 ... p_r^a_r is its prime factorization, then the ring of N-adic integers is isomorphic to the direct product of the q_i-adic integers , for i = 1, 2, ... r , where q_i = p_i^a_i , by the Chinese remainder theorem. So if N has (at least) two different prime factors, then there are zero-divisors. The construction can be done in much greater generality. Atiyah-MacDonald (in "Introduction to Commutative Algebra") define the a-adic completion of a commutative ring A having identity, where a is an ideal of the ring A . It's quite possible that these completions work equally well for non-commutative rings. Michael Reid

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[math-fun] a peculiar pair
by Bill Gosper 19 Nov '12

19 Nov '12

The eighth singular value: http://www.wolframalpha.com/input/?i=EllipticK[-112+-+80+Sqrt[2]+%2B+4+Sqrt… http://www.wolframalpha.com/input/?_=1353320405208&i=EllipticK[113+%2b+80+S… ^%281%2f4%29+Sqrt[%28+++2+Sqrt[2]+%2b+Sqrt[1+%2b+5+Sqrt[2]]%29%2f\[Pi]]+Gamma[1%2f8]+Gamma[3%2f8]&fp=1&incTime=true If this times out, try http://www.wolframalpha.com/input/?i={EllipticK[113+%2B+80+Sqrt[2]+-+4+Sqrt… ^%281%2F4%29+Sqrt[%28+++2+Sqrt[2]+%2B+Sqrt[1+%2B+5+Sqrt[2]]%29%2F\[Pi]]+Gamma[1%2F8]+Gamma[3%2F8]} Found by accident: http://www.wolframalpha.com/input/?i=EllipticK[-112+%2B+80+Sqrt[2]+%2B+4+I+… http://www.wolframalpha.com/input/?i={EllipticK[113-80Sqrt[2]-4I+Sqrt[2%28-… ^%28-I+ArcTan[1%2F23Sqrt[2]%287%2B6Sqrt[2]-3Sqrt[-7%2B17Sqrt[2]]%29]%29Gamma[1%2F8]Gamma[3%2F8]%29%2F%2816+2^%281%2F4%29Sqrt[%CF%80]%29} To view two side by side, I was able to open a second Alpha in a separate Firefox window. Beware: GMail may have inserted linebreaks. --rwg Also, Weisstein's Mathworld has more K special values than Borwein&Borwein. Also, that Byrd&Friedman Handbook of Elliptic Integrals appears to have only the small modulus Fourier series for am, whereas the pendulum formula requires http://www.wolframalpha.com/input/?i={JacobiAmplitude[u%2Cm]%2C+Sum[%28Sech… with briskly exponential convergence. So I still lack evidence that this useful formula is actually in use. (The HoldForm is to prevent Alpha timeout.) --rwg

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Re: [math-fun] math-fun Digest, Vol 117, Issue 15
by Tom Duff 18 Nov '12

18 Nov '12

On Fri, 16 Nov 2012, math-fun-request(a)mailman.xmission.com wrote: > If you try to provide a similar epsilon-delta proof that the series > ...9999.0 sums to -1, you will fail, and I can provide an epsilon-delta > proof that this series does *not* have a limit. Yes. Looking more closely (by a factor of 1+epsilon) though, the result you really have is that if the epsilon-delta definition is the correct definition of what it means to be a limit, then there is no limit. So either it's not the limit, or the definition is bogus. What we're really exploring here is whether the second option is the case. The interesting historical question is how we came to agree that e-d is the correct definition. There are multiple ways that Euler's intuitive approach to limits could have been formalized, and somehow we settled on this particular one, which fails to capture important parts of his intuition. This is historically interesting because in other cases, like the definition of the definite integral, there is still no consensus but we're mostly OK with that. We are taught the Riemann definition in introductory calc, but eventually we may want to work in more complicated spaces than Rn or handle functions that Riemann fails on where some other definition will win out. -- Tom Duff. Linux? Is that an OS, like Pentium?

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