January 2020 - mathfun@mailman.xmission.com

Re: [math-fun] Cubic equations, ellipses and hyperbolic functions
by Henry Baker 31 Jan '20

31 Jan '20

At 08:35 PM 1/30/2020, Brad Klee wrote: >One issue with your model though: how do you set parameters to get >z^3=1? It sounds like circular Steiner ellipses would have c = 0, implying >only z^3=0. Can f(z) equal a constant? z^3=c^3 is soooo boring -- you can solve it by inspection. But I'd be interested if there is another parameterization that generalizes my parameterization to handle this degenerate case.

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[math-fun] Gimbal lock??
by Henry Baker 31 Jan '20

31 Jan '20

Perhaps it's time to learn quaternions?? >From comp.risks: Date: Fri, 10 Jan 2020 20:24:07 +0000 From: "Clive D.W. Feather" <clive(a)davros.org> Subject: Boeing 737s can't land facing west (FAA) "The FAA received reports earlier this year of three incidents of display electronic unit (DEU) software errors on Model 737 NG airplanes flying into runway PABR in Barrow, Alaska. All six display units (DUs) blanked with a selected instrument approach to a runway with a 270-degree true heading, and all six DUs stayed blank until a different runway was selected. [...] The investigation revealed that the problem occurs when this combination of software is installed and a susceptible runway with a 270-degree true heading is selected for instrument approach. Not all runways with a 270-degree true heading are susceptible; only seven runways worldwide, as identified in this AD, have latitude and longitude values that cause the blanking behavior." (Note that this is all 6 displays on each plane, not 2 displays on each of three planes.) The runways in question are: Runway 26, Pine Bluffs, Wyoming, USA (82V) Runway 28, Wayne County, Ohio, USA (KBJJ) Runway 28, Chippewa County, Michigan, USA (KCIU) Runway 26, Cavern City, New Mexico, USA (KCNM) Runway 25, Barrow, Alaska, USA (PABR) Runway 28, La Mina, La Guajira, Colombia (SKLM) Runway 29, Cheddi Jagan, Georgetown, Guyana (SYCJ) (The numbers are magnetic bearings, whereas the problem is apparently related to true bearing.) Original FAA notice: <http://rgl.faa.gov/Regulatory_and_Guidance_Library/rgad.nsf/0/3948342a978cc…>

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[math-fun] Cubic equations, ellipses and hyperbolic functions
by hbaker1 31 Jan '20

31 Jan '20

Cubic equations, ellipses and hyperbolic functions I contend that the most elegant form for the general cubic equation in the complex plane is the trigonometric form, with *complex* c,A,z: f(z) = (z-c*cos(A))*(z-c*cos(A+2pi/3))*(z-c*cos(A+4pi/3)), with roots located at positions r1,r2,r3 on the complex plane: r1 = c*cos(A) r2 = c*cos(A+2pi/3) r3 = c*cos(A+4pi/3) We assume that the centroid of the roots is at the origin, so if c,A,z are all complex, then this form is completely general. Expanding and simplifying, we get 4*f(z) = 4*z^3 - 3*c^2*z - c^3*cos(3*A) Solving f'(z)=0, we find that z = +-c/2, which are the *foci* of the *Steiner inellipse* of the triangle of the roots, while the roots r1,r2,r3 lie on the *Steiner (circum)ellipse* of the triangle of the roots, which ellipse is twice as big, hence its foci are at z = +-c. So what are the parameters a,b of the Steiner ellipse? a = c*cosh(ip(A)) b = c*sinh(ip(A)) where ip(A) is the *imaginary part* of the complex angle A. We have the familiar equation c^2=a^2-b^2 for this Steiner ellipse, as well as this less familiar equation for its eccentricity: e = sech(ip(A)) Note that the basic properties of the Steiner ellipse are *independent* of the realpart of A, and hence the roots can revolve about the origin along the Steiner elliptical orbit, with the orbital position(s) indicated by the realpart of complex angle A. Indeed, if we keep ip(A) constant, we can *animate* the roots in their Steiner orbit by making complex angle A a function of time t, e.g.: r1 = c*cos(A(t)), A(t) = t + i*ip(A) ("i" is the complex number i^2=-1.) The area of the triangle of the roots is the real number: Area = area(0,r1,r2)+area(0,r2,r3)+area(0,r3,r1) = 3*area(0,r1,r2) (0 is the centroid) = 3 * sqrt(3)/8 * sinh(2*ip(A)) * c*c' (c' = conjugate(c)) = 3^(3/2)/4 * sinh(ip(A))*cosh(ip(A)) * c*c' = 3^(3/2)/4 * c*sinh(ip(A)) * c'*cosh(ip(A)) = 3^(3/2)/4 * a * b' analogous to the formula for the area of an ellipse with semimajor axis a and semiminor axis b, and where 3^(3/2)/4 is the area of an equilateral triangle inscribed within the unit circle. Once again, note that the triangle area is *independent* of the location of the roots within their Steiner elliptical orbit. Since we're talking about polynomials, it is interesting to consider the *discriminant* of our cubic: (r1-r2)^2*(r2-r3)^2*(r3-r1)^2 = (27*sin(3*A)^2*c^6)/16 Note that the discriminant is *complex*, and is obviously *NOT* independent of the realpart of complex angle A. Q: Are there additional interesting parameters computable from the roots which are independent of the realpart of complex angle A ? Q: Which symmetric functions of the roots are independent of the realpart of complex angle A? Clearly, there are some, as r1+r2+r3=0, and r1*r2+r2*r3+r3*r1=-3*c^2/4.

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[math-fun] A Legendre Question
by Brad Klee 30 Jan '20

30 Jan '20

Cross-post from seqfan: http://list.seqfan.eu/pipermail/seqfan/2020-January/020454.html See also: https://pdfs.semanticscholar.org/3a41/3ed6779db174ff01c0debea7f41e12a215d0.… Maybe it is easy to implement a reduction algorithm, but I'm not quite to the point of trying yet. --Brad

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[math-fun] Neat Dragon question
by Bill Gosper 28 Jan '20

28 Jan '20

What is the longest line segment inscribable in the Dragon linking 0+0i to 1 + 0i via (1+i)/2? "Hint": If the segment be t z2 + (1-t)z1, with 0<t<1, then every point on it is (empirically, at least) a triple point of the spacefill whenever t is a dyadic rational! —rwg (This impacts OEIS A260747, etc.)

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[math-fun] For an anti-diversity argument read:
by Brad Klee 28 Jan '20

28 Jan '20

https://phys.org/news/2020-01-math-teacher-plea-pi-irrational.html Where a young student is implicitly denounced as stupid for using python to compute a curve that the math teacher can easily prove circular. I am guessing that the student's real "mistake" was using floating point numbers instead of rationals or arbitrary precision, but the article doesn't say. The student doesn't need derision from the teacher, but might be well-served with an explanation of machine epsilon, and also with pointers to python calculations using fractions: https://docs.python.org/2/library/fractions.html https://www.tutorialspoint.com/python-rational-numbers-fractions The student can search for rational configurations by any expedient algorithm. Eventually, arbitrarily-large sets of high-quality data can be derived, without ever mentioning congruent triangles. The teacher should complement the student's data set, and ask them to keep refining their analysis of the data. This sort of development can and will lead to a correct proof. There's nothing else I can do, except to reject this author's plea to "keep pi an irrational number". I look forward to the next generation of creative scientists, who will undoubtedly continue to find value in thinking of pi as something other than just "an irrational number". -- Brad

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[math-fun] New Heesch tilings
by ed pegg 27 Jan '20

27 Jan '20

I've put some new Heesch tilings found by Craig S Kaplan at the top of http://www.mathpuzzle.com/ --Ed Pegg Jr

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[math-fun] This will probably be incredibly stupid when we figure it out.
by Bill Gosper 26 Jan '20

26 Jan '20

But that might be so easy. Plot says that f[InverseFunction[#1 - f[#1] &][t]] - InverseFunction[#1 - f[#1] &][t] == -t for the several different functions f that I tried. Here's f@t:=Sin[4t] <http://gosper.org/crazysinx-x.png>. (The gold overprinted the blue.) How can those two subexpressions plot so crazily, yet differ merely by the identity function? It's hard not to suspect Mathematica of malfeasance. Strangely methodical malfeasance. Regardless of the vibrations, does anyone understand this identity? Does anyone understand the vibrations? —rwg PS, did everyone (but me) get my (almost trivial) solution to c[0]:=0, c[n+1] = 2 + √c[n] ?

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[math-fun] Mystery table 2
by Keith F. Lynch 25 Jan '20

25 Jan '20

Yes, this is loosely related to my previous mystery table. Yes, it's symmetrical, unlike my previous table. Yes, the numbers on the diagonal are all integers. This operation is defined for all pairs of positive real numbers, not just for integers, and it always returns a positive real number. I'd be glad to extend this table, or to give the value for any two positive real numbers, if anyone asks. 1 2 3 4 5 6 7 8 1: 1.000000 1.545362 2.293678 3.240906 4.248364 5.259193 6.267074 7.272747 2: 1.545362 2.000000 2.576905 3.343752 4.270644 5.262645 6.267506 7.272793 3: 2.293678 2.576905 3.000000 3.589622 4.367036 5.288115 6.272582 7.273628 4: 3.240906 3.343752 3.589622 4.000000 4.596536 5.380654 6.299598 7.279833 5: 4.248364 4.270644 4.367036 4.596536 5.000000 5.600888 6.389624 7.307724 6: 5.259193 5.262645 5.288115 5.380654 5.600888 6.000000 6.603882 7.395990 7: 6.267074 6.267506 6.272582 6.299598 6.389624 6.603882 7.000000 7.606068 8: 7.272747 7.272793 7.273628 7.279833 7.307724 7.395990 7.606068 8.000000

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[math-fun] Reccurence for Pi^2
by françois mendzina essomba2 25 Jan '20

25 Jan '20

Hi, a [n+1]=2^(2*n+5)*(3-sqrt(3)*sqrt(sqrt((3-2^((-2*n)-3)*a[n])^2/3+1)+1)); a [0]=8*(3-sqrt (3)) Best regards Envoyé depuis Yahoo Mail pour Android

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