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[math-fun] "optimal control" theory question
by Henry Baker 13 Jul '18

13 Jul '18

I seem to recall someone from NASA telling me about some famous theorem of control theory that goes something like this: Suppose you have a system with multiple degrees of freedom, e.g., x,y,z in 3-space, and you have sensors capable of sensing the coordinate positions, but the *precisions of the different dimensions are different*. I.e., you might be able to get x,y positions to within one meter, but the z position might have a precision only to 5 meters. The classical optimal control algorithms will tend to operate in such a way that *most of the position uncertainty will be *forced* into the most imprecise coordinate. Example: the usual GPS system is more accurate in the lat&lon dimensions than it is in the *elevation* dimension. So you construct an optimal control system to fly an airplane using GPS. The system operates beautifully to get your plane to land on the correct runway, but unfortunately, it nearly always crashes the plane due to the altitude uncertainty. In other words, the optimal control strategy is indeed "tail-stuffing", whereby small uncertainties are moved into degrees of freedom which are less observable, and hence will be magnified to the point where they are. Thus, a banker who can manipulate the location of credit risk can manipulate it into degrees of freedom that are less observable, and furthermore, any optimal algorithm would do precisely the same.

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Re: [math-fun] Skripals' anagram
by Dan Asimov 12 Jul '18

12 Jul '18

SKRIPALS' anagram: Avian constants: LARKS' PIS Places Noah moors the boat: ARK SLIPS Just the things for photographing the small mammals: PIKA SLRS Very handy for those times you need to file down your necktie: SILK RASP. —And

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[math-fun] Rohan notices that integer 60º triangles come in pairs:
by Bill Gosper 12 Jul '18

12 Jul '18

{7 | 8 | 3, 7 | 8 | 5, 13 | 15 | 7, 13 | 15 | 8, 19 | 21 | 5, 19 | 21 | 16, 31 | 35 | 11, 31 | 35 | 24, . . ., 859 | 989 | 429, 859 | 989 | 560, 889 | 999 | 295, 889 | 999 | 704,. . . Sometimes very close pairs: 181 | 209 | 104, 181 | 209 | 105, . . ., 2521 | 2911 | 1455, 2521 | 2911 | 1456, . . . --rwg

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[math-fun] How to make Bill Gosper fall off his chair
by Joerg Arndt 12 Jul '18

12 Jul '18

Enjoy: https://jjj.de/tmp-math-fun/ This one is on the (3.4.6.4)-grid (aka, apparently: "Rhombitrihexagonal tiling", see https://en.wikipedia.org/wiki/Rhombitrihexagonal_tiling ) Best regards, jj

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Re: [math-fun] "Stochastic" integration (?)
by Dan Asimov 11 Jul '18

11 Jul '18

I had almost never thought about stochastic integrals, and the idea of a probabilistic vector field is cool. ----- the speed at one moment is NOT independent of the speed at the next moment ----- I'm not sure if I'm interpreting this right. One way might be (let's say for curves {A(t)} in the complex plane with A(0) = 0, A'(0) = 1: Then something like Brownian motion defines a measure mu_t, for each t > 0, on all possible continuous paths p : [0, t] —> C so that certain probabilistic consistency equations are satisfied. This could lead to really interesting possibilities. —Dan

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[math-fun] Skripals' anagram
by Bill Gosper 11 Jul '18

11 Jul '18

"COUNTERSPIES" ↔︎ "PERSECUTIONS" --rwg

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[math-fun] "Stochastic" integration (?)
by Henry Baker 10 Jul '18

10 Jul '18

I find myself driving quite a bit on freeways, and I'm curious about how my car GPS can calculate a reasonable ETA (Estimated Time of Arrival) for me. I learned calculus in high school & college, but I don't recall learning about how to integrate a *probabilistic* variable -- i.e., a variable which is subject to some probability distribution. So here's the thing: suppose I'm driving at a speed whose values come from some distribution, and I want to integrate this speed over some amount of time to compute another probability distribution about how much distance is being covered within that time, or to estimate a distribution of my arrival time at a particular destination. (Let's assume that time itself moves precisely, but the car speed varies.) Now an integral is merely a fancy *sum*, and we know how to estimate the sum of *independent* random variables. But in the case of my car, the speed at one moment is NOT independent of the speed at the next moment! The acceleration, maybe, but not the speed. More likely, even the acceleration isn't independent from moment to moment, but the "jerk" (dacceleration/dt) might very well be. So now we have an even bigger problem: we need to go back an integrate the jerk(t) for each moment of time t. But wait, there's more! If I'm using a program like Waze, which has a probability distribution for the speed on each segment of each highway, then I should be somehow computing an overall distribution for my trip time from some sort of integration of all of these little contributions. If these individual distributions were independent, then presumably my result would be some sort of convolution of all of the little distributions. But the distributions aren't independent. (Even worse: Waze knows that the speed distributions are time-dependent -- e.g., based upon time of day. But I don't even know how to do the simpler problem.) So what now?

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Re: [math-fun] Power Station Problem
by Richard Hess 08 Jul '18

08 Jul '18

So happy to see your approach, Fred. Just what I was looking for. Thanks so much. Dick Hess Sent from my iPhone > On Jul 8, 2018, at 2:26 AM, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > > Z-shaped power station problem. > > > *--------------* > /| 1 /| > / | / | > / | s / | s > / | / 1 | > / *---/----------* > a / / / / > / 1/ /c / > / / / / d > / / / / > // / / > /// > * P : 3 inner angles = u [View without proportional spacing] > > > Via elementary trigonometry, > > cos(u) = t , > cos(2 u) = 2 t^2 - 1 , > cos(3 u) = 4 t^3 - 3 t ; > > now via cosine rule, 6 consistent(!) equations in 5 variables: > > (A) s^2 = a^2 + 1 - 2 a t , > (B) s^2 = c^2 + d^2 - 2 c d t , > (C) 1 = a^2 + c^2 - 2 a c (2 t^2 - 1) , > (D) 1 = 1 + d^2 - 2 d (2 t^2 - 1) , > (E) 1 + s^2 = 1 + c^2 - 2 c t , > (F) 1 + s^2 = a^2 + d^2 - 2 a d (4 t^3 - 3 t) ; > > subtracting, > > D => 0 = d ( d - 4 t^2 + 2) , > E - B => 1 = -2 c t - d^2 + 2 c d t , > F - A => 1 = d^2 - 2 a d (4 t^3 - 3 t) - 1 + 2 a t , > > whence > d = 4 t^2 - 2 , > c = (1 + d^2)/2 t (d - 1) = (16 t^4 - 16 t^2 + 5)/2 t(4 t^2 - 3) , > a = (16 t^4 - 16 t^2 + 2)/2 t(16 t^4 - 20 t^2 + 7) ; > > substituting into C , > 64 t^8 - 176 t^6 + 168 t^4 - 63 t^2 + 7 = 0 ; > and the right-hand side = T_7(t) (t^2-1)/t , where T_n denotes > Chebyshev polynomial. > > There is essentially only one solution with acute angles: u = pi/14 . *** > > Fred Lunnon > > > >> On 7/7/18, James Propp <jamespropp(a)gmail.com> wrote: >> Note the rarely-spotted-in-the-wild triple-negative in “I'm still not >> convinced there aren't solutions >> that use the non-cyclic order.” >> >> Jim Propp >> >>> On Saturday, July 7, 2018, Allan Wechsler <acwacw(a)gmail.com> wrote: >>> >>> Mike Speciner's trivial case does not satisfy the extra constraint about >>> the distance to the nearest vertex. If you start with that case, though, >>> and budge the viewpoint off center along one of the orthogonal axes of the >>> square, you can then fudge the aspect ratio so that the viewing angles are >>> still equal. Keep doing that, moving the viewpoint further and further off >>> center and adjusting the aspect ratio as required. Eventually you will hit >>> a point where Hess's extra constraint is satisfied: this is, I think, the >>> solution that Hess intended. All of these cases involve scanning the >>> vertices in cyclic order. I'm still not convinced there aren't solutions >>> that use the non-cyclic order. >>> >>>> On Sat, Jul 7, 2018 at 4:31 PM, Mike Speciner <ms(a)alum.mit.edu> wrote: >>>> >>>> There is the trivial case of a square with the viewpoint at the center. >>>> >>>> >>>> >>>>> On 07-Jul-18 16:12, Allan Wechsler wrote: >>>>> >>>>> Standing at the viewpoint, as you scan from left to right, you must be >>>>> enumerating the vertices of the rectangle in one of two kinds of order: >>>>> "U" >>>>> order, going around the perimeter of the rectangle, or "Z" order, >>>>> traversing one of the rectangle's diagonals. >>>>> >>>>> I am guessing that Richard Hess's problem involves "U" order, and that >>> the >>>>> viewpoint is on one of the axes of symmetry of the rectangle. >>>>> >>>>> But it is not obvious to me that there isn't another solution, >>>>> utilizing >>>>> "Z" order. Probably the extra condition, that the distance to the >>> closest >>>>> vertex from the viewpoint is equal to the long dimension of the >>> rectangle, >>>>> eliminates this class of solutions, but I haven't been able to prove >>>>> it. >>>>> >>>>> On Sat, Jul 7, 2018 at 3:40 PM, Richard Hess <rihess(a)cox.net> wrote: >>>>> >>>>> The conditions of the problem uniquely determine the ratio of side >>> lengths >>>>>> of the rectangle as .512858431636277... >>>>>> >>>>>> If t=tan(theta)^2 then define >>>>>> r = (3-t)/8/(1-t) >>>>>> s = 1-r >>>>>> then rectangle ratio = .25/sqrt(rs) >>>>>> >>>>>> >>>>>> Sent from my iPhone >>>>>> >>>>>>> On Jul 7, 2018, at 6:39 PM, James Propp <jamespropp(a)gmail.com> wrote: >>>>>>> >>>>>>> Dick, >>>>>>> >>>>>>> Can you tell us the conjectural aspect ratio of the rectangle? (I’m >>>>>>> assuming that it’s unique, or that it takes on only finitely many >>>>>>> values, >>>>>>> related to the cosines and sines of multiples of 90/7 degrees.) >>>>>>> >>>>>>> Jim Propp >>>>>>> >>>>>>>> On Saturday, July 7, 2018, Richard Hess <rihess(a)cox.net> wrote: >>>>>>>> >>>>>>>> Imagine a power station with towers of negligible (=0)width built at >>>>>>>> the >>>>>>>> four corners of a rectangle on a flat plane. At a certain viewing >>>>>>>> >>>>>>> point, P, >>>>>> >>>>>>> on the plane, the bases of the four towers are equally spaced in >>> viewing >>>>>>>> angle by an angle, theta. P is at a different distance from each >>> corner >>>>>>>> >>>>>>> and >>>>>> >>>>>>> the distance from P to the closest tower is equals the length of the >>>>>>>> >>>>>>> long >>>>>> >>>>>>> side of the rectangle. For this case theta equals 90/7 degrees to 15 >>>>>>>> >>>>>>> places >>>>>> >>>>>>> of accuracy but I’m unable to prove equality. >>>>>>>> >>>>>>>> Any takers in finding a proof? >>>>>>>> >>>>>>>> Dick Hess >>>>>>>> >>>>>>>> Sent from my iPhone >>>>>>>> >>>>>>>> _______________________________________________ >>>>>>>> math-fun mailing list >>>>>>>> math-fun(a)mailman.xmission.com >>>>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>>>> >>>>>>> _______________________________________________ >>>>>>> math-fun mailing list >>>>>>> math-fun(a)mailman.xmission.com >>>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>>> >>>>>> >>>>>> _______________________________________________ >>>>>> math-fun mailing list >>>>>> math-fun(a)mailman.xmission.com >>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>> >>>>>> _______________________________________________ >>>>> math-fun mailing list >>>>> math-fun(a)mailman.xmission.com >>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>> >>>> >>>> >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Redundant trinomials --- are they really necessary?
by Fred Lunnon 08 Jul '18

08 Jul '18

I need an existence proof --- it's the missing link in a theorem establishing full-plane integer zero-free number walls over finite characteristic ( |F_2 excepted). CONJECTURE: Given any r > 1 , there exists over |F_2 a trinomial f(X) = X^s + X^t + 1 = g(X) h(X) , where g(X) is irreducible of degree r . This topic turns out to have been studied previously, since it is of interest in elliptic curve public-key cryptography. Such "redundant trinomials" are used to do arithmetic in binary finite fields |F^(2^r) lacking a trinomial generating polynomial: the idea is to calculate instead in a larger ring using trinomial f(X) with factor some field generator g(X) , afterwards reducing the result to the field. Of course, engineers don't care greatly about proving that such a trinomial exists for every field degree: they're just content that some have been found for every degree up to 10,000, comfortably exceeding anything currently required. I haven't uncovered a great deal about this in the literature: Christophe Doche (2005) <doche(a)ics.mq.edu.au> "Redundant Trinomials for Finite Fields of Characteristic 2 " http://web.science.mq.edu.au/~doche/redundant.pdf is cited by most of the rest; Ryul Kim, Wolfram Koepf <koepf(a)mathematik.uni-kassel.de> "Divisibility of Trinomials by Irreducible Polynomials over |F_2 " https://arxiv.org/pdf/1311.1366.pdf includes results about cofactor h(X) , along with baffling typos. If required I can post lists of sample trinomials and cofactors, currently for prime r < 4096 (in 3 hrs), also all r < 1024 (in 12 mins). Fred Lunnon

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Re: [math-fun] Power Station Problem
by Richard Hess 07 Jul '18

07 Jul '18

I could hardly fail to disagree with you less. Dick Hess Sent from my iPhone > On Jul 7, 2018, at 10:37 PM, James Propp <jamespropp(a)gmail.com> wrote: > > Note the rarely-spotted-in-the-wild triple-negative in “I'm still not > convinced there aren't solutions > that use the non-cyclic order.” > > Jim Propp > >> On Saturday, July 7, 2018, Allan Wechsler <acwacw(a)gmail.com> wrote: >> >> Mike Speciner's trivial case does not satisfy the extra constraint about >> the distance to the nearest vertex. If you start with that case, though, >> and budge the viewpoint off center along one of the orthogonal axes of the >> square, you can then fudge the aspect ratio so that the viewing angles are >> still equal. Keep doing that, moving the viewpoint further and further off >> center and adjusting the aspect ratio as required. Eventually you will hit >> a point where Hess's extra constraint is satisfied: this is, I think, the >> solution that Hess intended. All of these cases involve scanning the >> vertices in cyclic order. I'm still not convinced there aren't solutions >> that use the non-cyclic order. >> >>> On Sat, Jul 7, 2018 at 4:31 PM, Mike Speciner <ms(a)alum.mit.edu> wrote: >>> >>> There is the trivial case of a square with the viewpoint at the center. >>> >>> >>> >>>> On 07-Jul-18 16:12, Allan Wechsler wrote: >>>> >>>> Standing at the viewpoint, as you scan from left to right, you must be >>>> enumerating the vertices of the rectangle in one of two kinds of order: >>>> "U" >>>> order, going around the perimeter of the rectangle, or "Z" order, >>>> traversing one of the rectangle's diagonals. >>>> >>>> I am guessing that Richard Hess's problem involves "U" order, and that >> the >>>> viewpoint is on one of the axes of symmetry of the rectangle. >>>> >>>> But it is not obvious to me that there isn't another solution, utilizing >>>> "Z" order. Probably the extra condition, that the distance to the >> closest >>>> vertex from the viewpoint is equal to the long dimension of the >> rectangle, >>>> eliminates this class of solutions, but I haven't been able to prove it. >>>> >>>> On Sat, Jul 7, 2018 at 3:40 PM, Richard Hess <rihess(a)cox.net> wrote: >>>> >>>> The conditions of the problem uniquely determine the ratio of side >> lengths >>>>> of the rectangle as .512858431636277... >>>>> >>>>> If t=tan(theta)^2 then define >>>>> r = (3-t)/8/(1-t) >>>>> s = 1-r >>>>> then rectangle ratio = .25/sqrt(rs) >>>>> >>>>> >>>>> Sent from my iPhone >>>>> >>>>>> On Jul 7, 2018, at 6:39 PM, James Propp <jamespropp(a)gmail.com> wrote: >>>>>> >>>>>> Dick, >>>>>> >>>>>> Can you tell us the conjectural aspect ratio of the rectangle? (I’m >>>>>> assuming that it’s unique, or that it takes on only finitely many >>>>>> values, >>>>>> related to the cosines and sines of multiples of 90/7 degrees.) >>>>>> >>>>>> Jim Propp >>>>>> >>>>>>> On Saturday, July 7, 2018, Richard Hess <rihess(a)cox.net> wrote: >>>>>>> >>>>>>> Imagine a power station with towers of negligible (=0)width built at >>>>>>> the >>>>>>> four corners of a rectangle on a flat plane. At a certain viewing >>>>>>> >>>>>> point, P, >>>>> >>>>>> on the plane, the bases of the four towers are equally spaced in >> viewing >>>>>>> angle by an angle, theta. P is at a different distance from each >> corner >>>>>>> >>>>>> and >>>>> >>>>>> the distance from P to the closest tower is equals the length of the >>>>>>> >>>>>> long >>>>> >>>>>> side of the rectangle. For this case theta equals 90/7 degrees to 15 >>>>>>> >>>>>> places >>>>> >>>>>> of accuracy but I’m unable to prove equality. >>>>>>> >>>>>>> Any takers in finding a proof? >>>>>>> >>>>>>> Dick Hess >>>>>>> >>>>>>> Sent from my iPhone >>>>>>> >>>>>>> _______________________________________________ >>>>>>> math-fun mailing list >>>>>>> math-fun(a)mailman.xmission.com >>>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>>> >>>>>> _______________________________________________ >>>>>> math-fun mailing list >>>>>> math-fun(a)mailman.xmission.com >>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>> >>>>> >>>>> _______________________________________________ >>>>> math-fun mailing list >>>>> math-fun(a)mailman.xmission.com >>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>> >>>>> _______________________________________________ >>>> math-fun mailing list >>>> math-fun(a)mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>> >>> >>> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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