Let z_i = r_i/c, and define Chebyshev T_n polynomials such that T_n(cos(x)) = cos(n*x). Then T_n(r_1) + T_n(r_2) + T_n(r_3) = 0, unless n mod 3 = 0, when T_n(r_1) = T_n(r_2) = T_n(r_3). Function[{a}, Expand[Total[TrigReduce[ ChebyshevT[a, #]] & /@ {r1, r2, r3}]]] /@ Range[9] Out[] = {0, 0, 3 Cos[3 A], 0, 0, 3 Cos[6 A], 0, 0, 3 Cos[9 A],...} I think that these identities define the entire null ring, for example: Expand[Subtract[ 8 Total[ ChebyshevT[1, #]/4 & /@ {z1, z2, z3}]^2, (1/4)*Total[TrigReduce[ChebyshevT[2, #]] & /@ {z1, z2, z3}]]] Out[] = 3/4 + z1 z2 + z1 z3 + z2 z3 This does sound like a nice model for Cubics; though, an alternative approach is to define a cubic elliptic curve and intersect with a line. That is also elegant, especially when iterating through roots via the addition rules. In the recent thread on Legendre I allude to a lesser known model of triangle-symmetric elliptic curves. There is a clear analogy with Edwards's normal form, but I have yet to find any sort of proof that the triangular form is also a normal form. --Brad
On Jan 30, 2020, at 1:41 PM, hbaker1 <hbaker1@pipeline.com> wrote:
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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