[math-fun] Cubic eqn w/3 real roots: trig soln
[I hope I transcribed the formulae correctly.] The usual trig solution for a monic real cubic eqn with 3 real roots has the high school teacher pulling an obscure trig identity out of a hat in deus ex machina fashion, leaving the high school student feeling inadequate, helpless, irritable, and (s)he probably then tunes out of the rest of the presentation & starts texting on his/her cellphone. Here's a version of a cubic solution that a student familiar with complex numbers should feel very good about. We first eliminate the quadratic x^2 term in the usual fashion by shifting the center of mass of the roots to the origin. Finding the center of mass doesn't require solving the equation -- it is merely -1/3 the coefficient of the quadratic term. It is pretty easy to motivate this step -- we almost always translate to the origin whenever easily possible! That having been done, we are left with the so-called "incomplete" cubic: x^3+px+q = 0. Since the 3 roots r1,r2,r3 are real, they can take on any real values, so long as r1+r2+r3=0 (remember, we shifted the center of mass to the origin). All three could be equal (hence zero), or two could be equal (hence equal to minus half the third root). We assume that all three roots are distinct, and will check back later to make sure that everything still works for the other cases. Here is the modest bunny we will pull out of the hat: we conjecture that all three _real_ roots are the _real parts_ of the 3 vertices of an equilateral triangle in the complex plane centered on the origin. We first check the dimensionality: our triangle has only two real parameters: the angle a with the x-axis (real axis) and the radius r of the circumcircle. But our incomplete cubic also has only two real parameters, p,q, so we are still in business. We next realize that the three vertices of the equilateral triangle are the three cube roots of a complex number: z^3 = (r*(cos(a)+i*sin(a)))^3 = r^3*(cos(3a)+i*sin(3a)) [de Moivre] Now let z=x+iy, so z^3 = (x+iy)^3 = (x^3-3xy^2) - i*(y^3-3yx^2) = r^3*(cos(3a)+i*sin(3a)) Since we are projecting onto the x-axis, we focus only on the real part: x^3-3xy^2 = r^3*cos(3a) But how to get rid of y^2 ? We have the equation of the circumcircle: x^2+y^2=r^2, so y^2=r^2-x^2, so our real part equation now reads: x^3-3x(r^2-x^2) = r^3*cos(3a) = x^3-3xr^2+3x^3 = 4x^3-3xr^2 = r^3*cos(3a) or, dividing by 4, x^3-3xr^2/4-r^3*cos(3a)/4 We can now identify p = -3r^2/4, q = -r^3*cos(3a)/4 We first solve for r, r^2 = -4p/3 r = sqrt(-4p/3) So p must be negative in order to have 3 real roots. We then solve for a: cos(3a) = -4q/r^3 a = acos(-4q/r^3)/3 So |4q/r^3|<=1 in order to have 3 real roots. So our 3 real roots are: r1 = r*cos(a) r2 = r*cos(a+2pi/3) r3 = r*cos(a-2pi/3) So let us now check our work. (x-r1)(x-r2)(x-r3) = x^3 - (r1+r2+r3)x^2 + (r1r2+r1r3+r2r3)x - r1r2r3 We already know that r1+r2+r3=0, but if you want to check the trig, be my guest! -r1r2r3 = -r^3*cos(a)*cos(a+2pi/3)*cos(a-2pi/3) [after a lot of trig!] = -r^3*cos(3a)/4 r1r2+r1r3+r2r3 = r^2(cos(a)cos(a+2pi/3)+cos(a)cos(a-2pi/3)+cos(a+2pi/3)cos(a-2pi/3)) [more trig!] = r^2(-3/4) = -3r^2/4 So our equation is now x^3 - 0x^2 + (-3r^2/4)x - r^3*cos(3a)/4 = x^3-3xr^2/4-r^3*cos(3a)/4 Sure, there is a lot of trig involved in checking the solution, but since we have already motivated the solution & have it in hand, the trig involved in checking it is less stressful (although still tedious w/o Macsyma or equivalent). We do have to be careful in mapping 3 given real numbers (sum=0) to the three different angles a,a+pi/3,a-pi/3, to make sure that the mappings are consistent with the parameterization. We can now see that the case of all 3 roots being equal is handled by setting the parameter r=0, while the case of 2 roots being equal is handled by the parameter a=0 or a=pi, depending upon whether the double root is to the right or left, respectively, of the single root. An interesting fact that pops out of this characterization is a relatively tight bound, |ri|^2<r^2=-4p/3, on the size of the roots.
Quoting Henry Baker <hbaker1@pipeline.com>:
The usual trig solution for a monic real cubic eqn with 3 real roots has the high school teacher pulling an obscure trig identity out of a hat in deus ex machina fashion, leaving the high school student feeling inadequate, helpless, irritable, and (s)he probably then tunes out of the rest of the presentation & starts texting on his/her cellphone.
Those cell phones are really a plague on contemporary civilization comparable to the biblical locusts. I have always admired the trig solution, and it is not so obscure if you think of multiple angle formulas (not to mention Tchebycheff polynomials. Did Tchebycheff know that was what he was doing? One hopes so.) The cubic version is but a modest extension of the double angle formula, but how many trig or geometry courses get that far? I once got into very hot water (fired, in fact) for showing a similar derivation to a *college* trig class. Now that cell phones have graphics capabilities (and certainly more familiar computers), one can graph the coefficients of the polynomial as functions of the roots. For a (monic) quadratic polynomial, the remaining coefficients are the sum and the product of the roots, so you get a family of lines r1 + r2 = (coef of x), and a family of hyperbolas r1 x r2 = const. The pairwise crossings show the roots in a very symmetric way: straddling the mean, real when the curves actually intersect, horrendous numerical instability at double roota because of the tangency, and so on. I once heard while an undergraduate that there were tables of roots of cubic equations, although I am not sure whether Jahnke and Emde had them. Or even graphs. Well, following the quadratic analogy, monicize the equation and then make the sum of the roots 1, not 0, by the usual shift, and then graph r1 x r2 + r2 x r3 + r3 x r1 = (coefficient of x), and r1 x r2 x r3 = (const) (another hyperbola) using plane trilinear coordinate graph paper. Again it is possible to read the roots off the intersections. This is much too complicated for a course, but I have used it to amuse my numerical analysis students. I'd have to go look up the reference, but the british Mathematical Gazette (I think) had a nice article matching the triangle you mention to the critical points of the cubic. Centroid = point of inflection = center of 180 degree rotational symmetry, for example. But one really wonders what students can be expected to endure. -hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
From: "mcintosh@servidor.unam.mx" <mcintosh@servidor.unam.mx> To: math-fun <math-fun@mailman.xmission.com>; Henry Baker <hbaker1@pipeline.com> Cc: math-fun@mailman.xmission.com; mcintosh@servidor.unam.mx Sent: Sun, December 6, 2009 9:59:13 PM Subject: Re: [math-fun] Cubic eqn w/3 real roots: trig soln ... The cubic version is but a modest extension of the double angle formula, but how many trig or geometry courses get that far? I once got into very hot water (fired, in fact) for showing a similar derivation to a *college* trig class. ... ________________________________ Which college? Give us a forewarning, so that we can avoid the place. -- Gene
Quoting Eugene Salamin <gene_salamin@yahoo.com>:
Which college? Give us a forewarning, so that we can avoid the place. -- Gene
This happened long ago, so the less said the better; things did get sorted out after a while and after a fashion. As an interesting side comment, they had a calculus professor who adamantly opposed df(x)/dx, in favor of D_x f; supposedly to prevent students from cancelling the d's. Poor Leibnitz, who went to such trouble to create an appealing notation. But at least it has stood the test of time. There was a joke (at Newton's expense?) x^(dot) = dx/dt. Cancel the d's in the fraction, cancel the x's from both sides of the equation, and clear the fraction to get t^(dot) = 1, which, amazingly enough, is true. I don't recall who told us that - maybe a Physics professor, or just a myopic small boy who couldn't see emperors' vestiments. It is the sad fate of mathematics to suffer practicioners who never got the point (if there is any). Geometry teachers who won't let the students draw pictures, a recent proposal by some british teachers to drop the quadratic equation from the algebra course because it was "too hard" for the little darlings. As for my problem in retrospect, I think the class was mainly composed of premed students who were more familiar with biology and chemistry than mathematics and physics, and someone probably went complaining to the faculty when a complex number was mentioned. Anyway in a subsequent job we had a seminar presided by a really good statistician from whom I learned about finite differences, heard that exp(k Delta) generated a Newton's series in the style of exp(k d/dx) generating MacLaurin's series, and altogether lost my former loathing of arithmetic and numerical analysis. The teacher really can make a difference! "If Allah places an obstacle in the path of an animal, it is a clear indication that He desires that the animal take another path" -hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
participants (3)
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Eugene Salamin -
Henry Baker -
mcintosh@servidor.unam.mx