November 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Elementary triangle puzzle
by Michael Reid 17 Nov '11

17 Nov '11

Theorem. Any Heronian triangle can be embedded in the plane so its vertices have integer coordinates. Proof: First position the triangle so that A is at (0, 0) , B is at (a, 0) . It is easy to show that vertex C has rational coordinates. Consider the points as complex numbers; the vertices are all in Q(i) , the fraction field of Z[i] . It is well-known that Z[i] , which is the ring of Gaussian integers, is a PID, and even a Euclidean domain. Write the coordinate(s) of C as alpha / beta , in lowest terms. We have |AC|^2 = N(alpha / beta) = alpha alpha' / (beta beta') (where ' denotes complex conjugation, and N(z) is the norm of z , i.e. z z' ). Since |AC|^2 is (the square of) an integer, we have beta divides alpha alpha' . As it is relatively prime to alpha , we have beta divides alpha' , and thus beta' divides alpha , i.e. alpha = beta' gamma , for some gamma in Z[i] . Since beta is relatively prime to alpha , it follows that beta is also relatively prime to beta' . Also, |BC|^2 = N(a - alpha / beta) = a^2 - a (alpha / beta + alpha' / beta') + alpha alpha' / (beta beta') is an integer, so it follows that a (alpha / beta + alpha' / beta') = a (alpha beta' + alpha' beta) / (beta beta') is an integer. Note that beta is relatively prime to alpha beta' and therefore to alpha beta' + alpha' beta . Take complex conjugates to see that beta' is also relatively prime to alpha beta' + alpha' beta , and therefore beta beta' is relatively prime to alpha beta' + alpha' beta . However, beta beta' divides a (alpha beta' + alpha' beta) . This now implies that beta beta' divides a , so we may write a = b beta beta' , where b is a rational integer. Finally, rotate the triangle by multiplying by beta / beta' , which has magnitude 1 . The vertices of the triangle rotate to 0 , a beta / beta' = b beta^2 and alpha / beta' = gamma , which are Gaussian integers. Therefore the vertices of the rotated triangle have integer coordinates. QED Note that this gives a constructive way to find such an embedding. Michael Reid > > Yet another elementary program glitch caused misprinted vertex coordinates > --- for [a, b, c] = [5, 29, 30] the lattice vertices should have read > [0, 0], [3, 4], [21, -20] . > > The data list sent out subsequently was similarly garbled, and a > corrected version > will follow [embarrassed groan]. WFL > > On 11/17/11, Fred lunnon <fred.lunnon(a)gmail.com> wrote: > > As far as I can see, splitting a Heronian triangle (by its altitude) into > > two Pythagorean (right-angled, with rational edge-lengths, hence areas) > > constructs merely a pose with rational coordinates. > > > > For example, suppose one edge joins [0, 0] and [a, 0]; then the third > > vertex lies at [(a^2+b^2-c^2) / 2 a, 2 d / a], where d denotes the area. > > This will not be an integer unless some edge-length a happens to divide 2d > > . > > > > For triangles such as [a, b, c] = [5, 29, 30], where s, d = 32, 72, > > there is no such edge. However, the rotated pose with vertices > > [0, 0], [-4, 3], [-20, 21] > > still lies on the lattice. A similar apparently happy accident overtakes > > every > > case with s <= 200. > > > > Fred Lunnon > > >

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[math-fun] Elementary triangle puzzle
by Dan Asimov 17 Nov '11

17 Nov '11

Given an arbitrary triangle T in R^2, characterize p in R^2 such that the sum of distances S(p) := d(p, L_1) + d(p, L_2) + d(p, L_3) is minimized, where the L_j are the affine lines containing the sides of T. --Dan "Things are seldom what they seem." --W.S. Gilbert

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[math-fun] Paul Krugman's The Theory of Interstellar Trade
by Henry Baker 17 Nov '11

17 Nov '11

FYI -- Interstellar Economics explained... Krugman proves that interest rates on Earth & Asimov's Trantor are the same. http://krugman.blogs.nytimes.com/2008/03/11/economics-the-final-frontier/ March 11, 2008, 2:09 pm Economics: the final frontier Thirty years ago I was an oppressed assistant professor, caught up in the academic rat race. To cheer myself up I wrote Â well, see for yourself. Joshua Gans of the University of Melbourne scanned a copy of the thing I wrote Â back then academics did their work with typewriters, abacuses, and stone axes Â and was good enough to send me a copy. Ladies and gentlemen, I give you The Theory of Interstellar Trade. www.princeton.edu/~pkrugman/interstellar.pdf Some of this is explained in the 1-hour video: Doing Business in Interstellar Space - Professor John D. Barrow http://blip.tv/greshamcollege/doing-business-in-interstellar-space-professo… or http://www.youtube.com/watch?v=OUlWkx0wdME --- A similar set of arguments shows that if time travel exists, then interest rates here on Earth will tend to zero. If we utilize an analogy of the Fermi Paradox, then our current _zero interest rates_ are proof of the existence of time travel. http://en.wikipedia.org/wiki/Fermi_paradox

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Re: [math-fun] Silly question (?)
by Henry Baker 17 Nov '11

17 Nov '11

I've just become aware of some results that updates Tarski's decision procedure for real closed fields. If you add the (real) exponential function to the theory of real closed fields, the theory is probably decidable, but there remain gaps (how come this problem didn't make it onto the Millenium list?). This means in practise that you would be able to reason about _hyperbolic functions_ (sinh, cosh) in a decidable way. Reasoning about _circular functions_ (sin, cos) seems to allow the reasoning about the integers, due to their periodic nature, and thereby enter the undecidable domain. Thus, hyperbolic functions are _easier_ than circular functions. http://en.wikipedia.org/wiki/Tarski%27s_exponential_function_problem http://www.oocities.org/tamara_servi/thesis.html "Undecidable and Decidable Problems in Mathematics - Professor Angus MacIntyre" lecture video http://blip.tv/greshamcollege/undecidableanddecidableproblemsinmathematics_… At 07:51 PM 10/21/2011, Henry Baker wrote: >The theory of "real numbers" (or at least the theory of algebraic numbers) is easier than the theory of "integers". > >The theory of real order is decidable. >The theory of real addition & ordering is decidable. >The theory of real closed fields (+,*,<; i.e., polynomials with integral coefficients) is decidable. >The theory of integers with a single successor function ("1+") is decidable, but harder than that of real closed fields. This is like the integers, but you can only emulate addition/subtraction of constants, and cannot emulate multiplication of 2 arbitrary integers. > >The theory of integers (with addition & multiplication) is undecidable. > >So, in order to define the integers within the real numbers, you need to add enough additional structure to emulate finite chains (i.e., induction). Most of the emulations I've seen involve things like complex numbers, sines & cosines, so you emulate the integral periodic structure of sines/cosines. > >At 06:13 PM 10/21/2011, David Wilson wrote: >>How do you show that there is an integer greater than any given real number? >> >>The real question is, given the axioms for the real numbers (Dedekind-complete ordered field), how do you define the "real integers" within them. >> >>Two possibilities spring to mind. >> >> Definition A: The "real integers" are the minimal subset of R that contains 1 and is closed over negation and addition. >> >> Definition B: 1 is a "real integer", and there is a unique "real integer" in every half-open unit interval of R. >> >>Definition A seems more in the spirit of the number theory definition of Z. However, I do not immediately see how it answers my original question. >> >>Definition B immediately answers my original question. For any real number r, there is a unique integer on the half-open interval (r, r+1] which must be greater than r. >> >>Presumably definitions A and B are equivalent. Is the proof easy?

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Re: [math-fun] Heronian triangles can be made to have integer vertex coordinates
by Warren Smith 16 Nov '11

16 Nov '11

Lunnon: The parameterisation given by Buchholz (spelling!) --- which he ascribes to Carmichael, incidentally --- does not guarantee a primitive triangle with GCD(edge-lengths) = 1, only some triangle similar to it. --so let me understand your question. What you want is a proof that every PRIMITIVIZED heronian triangle is integer-posable? I.e. you conjecture any Bucholz triangle AFTER dividing by GCD(a,b,c) is integer-posable? Call that "revised Lunnon conjecture." To repeat the formulas so far Bucholz: a := (n*n + k*k)*m; b := (m*m + k*k)*n; c := (m + n)*(m*n - k*k); semiperim=s = (m + n)*m*n, area=d = k*m*n*(m + n)*(m*n - k*k) = k*m*n*c; constraints GCD(m,n,k) = 1; m > n > 0 & 0 < k <= sqrt( m*n*n/(m + 2*n) ) vertex coordinates proposed by WDS: A=(+n*(m-k)*(m+k), 0) B=(-m*(n-k)*(n+k), 0) C=(0, 2*k*m*n) a^2 = |B-C|^2 = |B|^2 + |C|^2, b^2 = |A-C|^2 = |A|^2 + |C|^2, c^2 = |B-A|^2. Well clearly G = GCD(a,b,c) = sqrt(GCD(a^2, b^2, c^2, area)) divides a, and b, and c, and 2*s=2*(m+n)*m*n; and G^2 divides area. If G divides 2*k*m*n=|C|, then G^2 divides |A|^2 and |B|^2 and |C|^2 and a^2 and b^2 and c^2 and area forcing G to divide a, b, c, A, B, and C. So "revised Lunnon" <== "G must divide 2*k*m*n." But we know G divides GCD(2*s, c, a+b). Hence G divides (m+n)*GCD(2*m*n, m*n-k*k, m*n+k*k). Hence G divides (m+n)*GCD(m*n, 2*k*k). But Lunnon-constraint implies that GCD(m*n,2*k*k)=GCD(m*n,2*k). Hence G divides 2*k*m*n. Therefore, the revised Lunnon conjecture, is also proven. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] strip mining to find nice provably irrational numbers?
by Warren Smith 16 Nov '11

16 Nov '11

It occurs to me that Apery's sensational and then-mysterious discovery of a proof that zeta(3) is irrational could have been found by computer by the following method for seeking numbers that are provably irrational. 1. consider all continued fractions of the form p1/(q1+p2/(q2+p3/(q3+...))) where pn is a polynomial of n with small-integer coefficients and low degree and qn is another, and we ignore continued fractions which do not converge fast enough. [Also, one can allow the pn to be sequences which are modular-cased polynomials, such as "n^2 if n odd, 3 if n even."] All such numbers should be provably irrational. 2. compute these numerically, and for each number you get, look it up in the Plouffe catalog of important real numbers. 3. if any number arises that was not previously known to be irrational then you have a proof it is irrational.* *Except you have to prove the continued fraction really does represent that number. But even if you cannot prove that you will have found a very cool conjectured identity. So... maybe you should try this. It ought to be dead easy for Plouffe to try. [Of course if the irrational has a rational ratio with a Plouffe-catalog number, that also suffices.] -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Heronian triangles can be made to have integer vertex coordinates
by Warren Smith 16 Nov '11

16 Nov '11

Re Fred Lunnon's Heronian triangle problem. Lunnon says the Heronian triangles = the triangles with all 3 sides a,b,c, and the area, being integers. Wikipedia http://en.wikipedia.org/wiki/Heronian_triangle says: Theorem: Given a Heronian triangle, one can split it into two right triangles, whose sidelengths form Pythagorean triples with rational entries. This proves any Heronian triangle can be made to have rational vertex coordinates. Therefore after scale-up by LCM(denominators of coordinates), it can be made to have all coordinates integer. Lunnon: Bucholtz' parameterisation should read a = (n^2 + k^2)m, b = (m^2 + k^2)n, c = (m + n)(m n - k^2), semiperim=s = (m + n)m n, area=d = k m n(m + n)(m n - k^2); constraints GCD(m,n,k) = 1; m > n > 0 & 0 < k <= sqrt( m n^2/(m + 2 n) ) enforce one (nonprimitive) triangle in each Heronian similarity class. The inverse mapping (modulo similarity) is given by k = d, n = s(s-b), m = s(s-a). The wikipedia Pythag proof shows the altitude of the triangle is given by altitude = 2*area / base and the two legs of the right triangles which both lie on the base are ( hypot1^2 - hypot2^2 + base^2 ) / (2*base) and ( hypot2^2 - hypot1^2 + base^2 ) / (2*base). Taking base=c=leg1+leg2 and hypot1=a and hypot2=b we find altitude = 2*k*m*n leg1 = m*(n^2-k^2) leg2 = n*(m^2-k^2) all of which are integers, proving Lunnon's "integer poseability" conjecture. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Erdos "minimize count of distinct distances" problem (pretty much) solved
by Warren Smith 16 Nov '11

16 Nov '11

this seems a breakthrough. The proof technique they use probably has many more applications but it seems mysterious why/how it works so well: http://arxiv.org/pdf/1011.4105v3

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Re: [math-fun] Giant planet was ejected from the solar system ?
by Steve Witham 15 Nov '11

15 Nov '11

>From: "Schroeppel, Richard" <rschroe(a)sandia.gov> > >Moreover ... galactic escape velocity (in our neck of the woods) > is a couple of hundred miles/second. >Ordinary planetary encounters (near misses) can change the velocities > of the players by at most twice the surface escape velocity of the opponent. > [Imagine a parabolic orbit that just skims the surface of the opponent.] > For Jupiter as the ejector, that's ~ 70 miles/sec. > Our hypothetical ejected planet would still be in a galactic orbit. Yes, sorry, I didn't mean that the planet would travel twice the diameter of the galaxy in a straight line, or that it would reach that distance from us. Also, by saying "truckin'" I mainly wanted to imply a long strange trip, see. >From: math-fun-bounces(a)mailman.xmission.com on behalf of Mike >Speciner [ms(a)alum.mit.edu] > >Remember that escape velocity is the velocity needed, at a particular >distance from the object (the sun in this case) to just make it to >"infinity" with zero velocity. The escapee slows down as it escapes. Yes, I took escape velocity as an order-of-magnitude for the range of left-over velocities "after" escaping. >And that's "affecting", not "effecting". That's the most embarrassing of all. "I shall effect repair." --NOMAD --Steve > >--ms > Wikipedia says escape velocity from the Sun's gravity starting at > Jupiter's > (current) orbit is 18.5 km/s. That's about one light year every sixteen > thousand years. > > Wikipedia also says the Solar System is 4.6 billion years old. Starting > when the Solar System was 600 million years old, and traveling at 18.5 > km/s, an object would travel about 244 thousand light years (if the Milky > Way weren't effecting its path). > > That's about twice the diameter of the Milky Way, or ten times the > distance from the Sun to the center of the galaxy. In that same period, > the Sun has orbited the center of the Galaxy 16 times. > > So I guess this means the ejected planet won't be found loitering > between the nearby stars. It's truckin'. > > --Steve

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Re: [math-fun] How to generate an infinite number of true-random bits in hardware
by Warren Smith 15 Nov '11

15 Nov '11

> this reminds me, there are noise diodes designed exactly for the purpose of generating noise. One supplier is noisecom.com (they were at the top of Google search). Their NC104 model provides 3 MHz bandwidth with noise output into a 2200ohm load of 3 microV/Hz^1/2, i.e. 5 milliV RMS if the full bandwidth is usable, or 1.6 mV if filtered down to 300 kHz. If the first amplifier has 7 pF input capacitance, then, together with the 2200 Ω load, the upper cutoff frequency is 10 MHz. --and to illustrate the problems you can get into with this "big amp" naive strategy, say you are aiming to amplify your 5 milliV by factor 200 to reach 1 volt. If you have merely around (7/200)pF = 0.035 pF of parasitic capacitance or more coupling the amp output back to the amp input, then that'd produce a bogus signal at the input comparable to the real signal, probably causing you to go into wild oscillation. That's a well known problem with high gain amps used near their bandwidth limit (at which the input is coming from maximum permitted impedance while the bogo-signal feedback gets minimum impedance). OK, you maybe can escape this fate using careful amp design and careful shielding... but why go there if you do not have to? I agree noise diodes, neons, etc (noisy things) are the way to go if you do want this approach, but to make it work you'd probably want to play it safe and operate far below the bandwidth limit.

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