July 2015 - mathfun@mailman.xmission.com

[math-fun] Polynomials as electric potentials
by Warren D Smith 03 Jul '15

03 Jul '15

Actually, the whole view of a "monic polynomial" P(z) as P(z)=PRODUCT_j (z-r_j) where r_j are the roots hence log P(z) = SUM_j log(z-r_j) hence log|P(z)| = SUM_j log|z-r_j| = potential energy of movable negative unit charge located at z in the coulomb fields of positive unit charges fixed at the r_j, is often quite useful. For one example, suppose you want to make a rootfinding algorithm. "Newton's iteration" is equivalent to: you assume the whole electric field -P'(z)/P(z) at z is due to just one root R, and solve for its location; then go there (z:=R) and do it again. Some other iterations can also be thought of electrostatically. One can also create "simultaneous iterations" in this way which attempt to converge to all N roots of P(z) at once. Place N movable negative unit charges in the complex plane. They will repel each other. There will also be attractive forces from the N fixed roots r_j of P(z), which are positive unit charges. The min-energy locations of the N movable charges are co-located with the N roots. Minimizing energy solves for the roots. Basically any iteration which keeps decreasing energy in a sensible way must converge. You can at each guess-location compute the electric field arising from both the fixed and movable charges, pretend it is due to just a constant number of them, e.g. to 1 positive and 1 negative charge, and go to the location of the positive... or take a smaller step in that direction (keep decreasing step size until cause energy decrease; one can devise a "safety rule" of that sort to provably assure convergence). By fast polynomial algorithm methods the simultaneous iteration can be implemented to run in O(N*(logN)^2) flops per simultaneous-adjust. A famous theorem of electrostatics assures that there exists no energy local-min for the movable charges aside from the global min we want. These schemes can be shown to converge ultimately in a manner such that the number of correct decimals grows geometrically, if all roots are isolated. Convergence to multiple roots is slower, geometric, although you can devise variant algorithms which do not care about multiplicities up to say, 4, and you can remove all multiple roots ahead of time by solving for the roots not of P(z) but rather of Q(z)=P(z)/GCD(P(z), P'(z)) which has only single roots. I invented and programmed this class of rootfinding methods in the 1980s. They work pretty well if you use your own floating point arithmetic (not machine's) with larger exponent field so as to remove worries about numerical overflow with high-degree polynomials.

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[math-fun] Haptic hearing
by James Propp 03 Jul '15

03 Jul '15

At one of the earlier Gatherings for Gardner, James Randi spoke about a case in which a claimed paranormal ability turned out to be genuine (and in hindsight therefore not "paranormal" after all, except in the most restrictive and literal sense): the case of a physician who could "hear" vinyl LPs by touch alone, picking up vibrations by running his fingers along the grooves. Does anyone else remember this? Does anyone have any pointers to articles about this case? I am especially curious to know whether this talent, like echolocation, is one that many people can develop with practice. Jim Propp

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[math-fun] "There is no revere on a motorcycle"
by Bill Gosper 02 Jul '15

02 Jul '15

--Claude Shannon. http://www.digitaltrends.com/cars/missions-hot-new-160hp-electric-motorcycl… Some of the pix show it on an elaborate kickstand-looking thing. Seems to me you could just turn the handle bars, hold it straight up, and get off, if it had one of those Segway servomechanisms. --rwg

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[math-fun] Draft of "The Lessons of a Square-Wheeled Trike"
by James Propp 02 Jul '15

02 Jul '15

I've finished a draft of installment #1 (as opposed to #0) of "Mathematical Enchantments", and I'd welcome comments. Jim Propp

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Re: [math-fun] Discrimant (physical meaning)
by Henry Baker 02 Jul '15

02 Jul '15

Very interesting idea, Warren. Thanks! At 02:33 PM 7/1/2015, Warren D Smith wrote: >> If you also want to know not merely about the absolute value, >> but also about the phase angle... i.e. about Im(Log(discriminant)), >> that's kind of stupid physically since it is not >> rotation invariant.... but it's sort of like the quadrupole moment of >> the charge configuration, but not exactly. > >--Actually, a cheap statement about a=arg(discriminant) >for a monic polynomial P(z) of degree N, is that >exp(i*a/(N-1)) * P(z * exp(-i*a/((N-1)*N) ) ) >is an "improved" version of P(z) having same degree, >still monic, but now with pure-real discriminant >with same old absolute value. > >So it tells you how much you need to rotate in order >to orient P(z) "correctly." Any pure-real polynomial P(z) >already was oriented correctly without you needing to do anything. > >It is intuitively somewhat like quadrupole moment as I'd said, e.g. >a pure-real polynomial automatically has quadrupole moment pointing in the >real (or pure imaginary) direction.

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Re: [math-fun] Geometric significance of algebraic discriminant ?
by Henry Baker 02 Jul '15

02 Jul '15

Very interesting direction (!) to pursue. Thanks, Adam, Dan, Gene! At 03:09 PM 7/1/2015, Adam P. Goucher wrote: >Suppose the polynomial is monic and irreducible, and has a root x. > >Then the field K := Q[x], viewed as a vector space over Q, has a basis of {1, x, x^2, ..., x^(n-1)}. > >We can endow the vector space with an inner product, namely (y, z) = tr(yz*), which gives rise to a natural metric on the field (which does not dependent on the choice of x used to generate K). > >Then the discriminant is the squared volume of the parallelepiped generated by {1, x, x^2, ..., x^(n-1)}. > >Sincerely, Adam P. Goucher > >> Sent: Wednesday, July 01, 2015 at 9:19 PM >> From: "Dan Asimov" <asimov(a)msri.org> >> >> A math friend pointed out the following (slightly edited): >> >> ----- >> The roots-to-coefficients map, taking (r1,r2,...,rn) to the vector of elementary symmetric functions in the ri, (\sigma1,\sigma2,...,\sigma_n), has the square root of the discriminant as the determinant of its derivative/Jacobian matrix. [This may be a little imprecise.] >> >> Also, two triangles whose vertices are viewed as complex numbers, (z1,z2,z3) and (w1,w2,w3), are similar with zj going to wj for j=1,2,3 iff \det(1 1 1 \\ z1 z2 z3 \\ w1 w2 w3) = 0. Especially, taking w1=z2, w2=z3, w3=z1 gives a polynomial condition for (z1,z2,z3) to be equilateral. >> ----- >> >> ÂDann >> >> On Jun 30, 2015, at 3:05 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> >> I recall learning a bunch of algebra about the discriminant, which becomes zero when there are coincident roots. >> >> https://en.wikipedia.org/wiki/Discriminant >> >> Has someone come up with geometric insights about this particular formula ? >> >> In the case of a quadratic, the formula is (x1-x2)^2, but this isn't the real number |x1-x2|^2. Perhaps the norm of the discriminant (DD*) is more important? >> >> What about the discriminant of the cubic ? Shouldn't this say something interesting about the triangle in the complex plane formed by the roots?

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[math-fun] What to say to Fermat-ists
by James Propp 02 Jul '15

02 Jul '15

Recently someone wrote to me saying (apropos of FLT) About a year ago an idea presented itself to me such that I know it is > correct. I can "prove" 90% of it and demonstrate 100% of it. However I am > not a mathematician and a formal proof is not something I can accomplish. > Would you be willing to examine my paper? At the least, you can use it as > a bad example. If it's not laughable, then I'd be happy to ask you to be > co-author. > Here's what I wrote to him; comments? --------------------------------------------------------------------- I hope you'll forgive me if I treat the matter statistically. (Nobody is a mere statistic in his/her own eyes, but in the eyes of other people, we all inevitably are.) Nearly all attempts at proofs of FLT by mathematicians and non-mathematicians alike over the past few centuries have failed. The difference is that the mathematicians' failures have sometimes (not often, but sometimes) been educational: the flawed argument covered important infinite classes of possible counterexamples and ruled them out, or the flaws in the proof, when brought to light, uncovered important facts about whole numbers. Amateurs' failures, in contrast, when brought to light, uncovered facts that are well-known to number theorists, so the only ones who benefited from the bringing-to-light were the amateurs themselves, not the mathematical community at large, and not the bringer-to-light (typically a sympathetic mathematician or a graduate student who was given the proof by the more senior mathematician to whom it had been sent, as a way of helping the graduate student sharpen his/her skills). None of which rules out the possibility that you are the rare exception: an amateur with an approach that is both new and valid. In fact, I would say that, given what I know about you, the chance of that unlikely event is not one in a billion (the probability I'd assign if you were just a random mathematical amateur) but one in a million, since you've already shown an ability to come up with new and valid ideas in another field. Still, my time is finite, like my money, and I don't play the lottery with my money, so I don't want to invest a lot of time into a project that, from my point of view, is unlikely to lead to anything. But: you've been generous with me, so I'm willing to spend an hour taking a look at what you've done and giving you my feedback. --------------------------------------------------------------------- It occurs to me that there might be some way to be pseudo-quantitative and do a cost-benefit calculation to quantify just how much time I should spend reading his draft of a proof of FLT. Jim Propp

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[math-fun] Geometric significance of algebraic discriminant ?
by Henry Baker 01 Jul '15

01 Jul '15

I recall learning a bunch of algebra about the discriminant, which becomes zero when there are coincident roots. https://en.wikipedia.org/wiki/Discriminant Has someone come up with geometric insights about this particular formula ? In the case of a quadratic, the formula is (x1-x2)^2, but this isn't the real number |x1-x2|^2. Perhaps the norm of the discriminant (DD*) is more important? What about the discriminant of the cubic ? Shouldn't this say something interesting about the triangle in the complex plane formed by the roots?

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[math-fun] Discrimant (physical meaning)
by Warren D Smith 01 Jul '15

01 Jul '15

Log|discriminant| = Re(Log(discriminant)) is proportional to the Coulomb energy of unit charges (for energy formula appropriate for 2 dimensions) placed at the roots of the (assumed monic) polynomial. If you also want to know not merely about the absolute value, but also about the phase angle... i.e. about Im(Log(discriminant)), that's kind of stupid physically since it is not rotation invariant.... but it's sort of like the quadrupole moment of the charge configuration, but not exactly.

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Re: [math-fun] Geometric Discriminant
by Henry Baker 01 Jul '15

01 Jul '15

This is all true, but what interpretation/meaning should we give to the square of the Vandermonde determinant? For example, what meaning would this number have in the case that the three roots of a cubic are the vertices of a triangle? If the area of the triangle is zero, the discriminant is still non-zero, so if the discriminant is a function of the area, then it must somewhat complicated. I was wondering if the discriminant could be some function of other properties of the triangle (of the roots). At 12:43 PM 7/1/2015, rcs(a)xmission.com wrote: >Geometric Interpretation of the Discriminant > >Following up Gene's note on Discriminant = square >of Vandermonde determinant ... >The determinant is +- the area/volume/measure of a >parallelogram/piped/? with origin 0 and each row >(or each column) as a generating edge. The >determinant of the matrix > >[ 1 2 ] >[ 3 4 ] > >is -2, which is also the area of the parallelogram >00 - 12 - 46 - 34. (The +- depends on your sign >convention, and is extinguished if you square the >Vandermonde determinant.) >The determinant of a V matrix is 0 whenever two roots >of the related polynomial are equal.

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