From: Wilbert Dijkhof Newsgroups: sci.electronics.cad,[...],sci.math Subject: Re: Math formulas Date: Tue, 05 Nov 1996 10:36:40 -0800 george wrote: > > :Algebraic isn't the same as closed-form. > :You know that polynomials of degree 3 are solvable, expressed in two ways: > :in roots or sin/arcsin functions. > ^^^^^^^^^^ > > Anybody know what they are? Or how to coax mathematica to give that > form..? > > -- > george > george@mech.seas.upenn.edu Begin with x^3+ax^2+bx+c=0 with x=(z-a)/3 gives: z^3-pz+q=0 with p(a,b,c) and q(a,b,c), the solutions are: z1 = 2sqrt(p/3)cos(1/3 arccosA) z2,3= 2sqrt(p/3)cos(2pi/3 +- arccosA) with A = (q/2)*(sqrt(p^3/27)) Wilbert ============================================================================== From: Wilbert Dijkhof Newsgroups: sci.electronics.cad,[...],sci.math Subject: Re: Math formulas Date: Tue, 05 Nov 1996 13:59:38 -0800 Wilbert Dijkhof wrote: > > george wrote: > > > > :Algebraic isn't the same as closed-form. > > :You know that polynomials of degree 3 are solvable, expressed in two ways: > > :in roots or sin/arcsin functions. > > ^^^^^^^^^^ > > > > Anybody know what they are? Or how to coax mathematica to give that > > form..? > > > > -- > > george > > george@mech.seas.upenn.edu > > Begin with x^3+ax^2+bx+c=0 with x=(z-a)/3 gives: > z^3-pz+q=0 with p(a,b,c) and q(a,b,c), the solutions are: > > z1 = 2sqrt(p/3)cos(1/3 arccosA) > z2,3= 2sqrt(p/3)cos(2pi/3 +- arccosA) Sorry, must be: z1 = -2sqrt... and z2,z3 = -2sqrt... > with A = (q/2)*(sqrt(p^3/27)) > > Wilbert Wilbert ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Newsgroups: sci.math Subject: Re: Math formulas Date: 5 Nov 1996 07:43:08 -0500 In article <55ljj1$hk6@netnews.upenn.edu>, george wrote: >:Algebraic isn't the same as closed-form. >:You know that polynomials of degree 3 are solvable, expressed in two ways: >:in roots or sin/arcsin functions. > >Anybody know what they are? Or how to coax mathematica to give that >form..? [I edited out all groups except sci.math] In a way; it is designed for equations with real coefficients, but sometimes one works with imaginary angles, and this can be translated into hyperbolic functions. Recall (or prove for yourself) a form of triple angle formulas: (cos(t))^3 = (3 * cos(t) + cos(3*t))/4 (sin(t))^3 = (3 * sin(t) - sin(3*t))/4 The procedure: reduce a general cubic equation to the form x^3 + p*x + q = 0 (p and q real, non-zero) Denote D = q^2 + 4 * p^3/27, and suppose D<=0, then one of the roots is x_1 = 2*sqrt(-p/3) * sin((1/3)*arcsin(-(3*q)*sqrt(-3/p)/(2*p)) and in the other two, add or subtract 2*pi/3 inside arcsin. If D>0 and p>0: the real root is x_1 = 2*sqrt(p/3) * sinh((1/3) * ln(z)) where z = 3*q*sqrt(3)/(2*p*sqrt(p)) + sqrt(1+27*q^2/(4*p^3)) (Don't you wish you hadn't asked?) For D>0 and p<0, there is a hyperbolic cosine formula - try it out. For D<=0, there is also a trisection formula using cosine (more often quoted in old handbooks), it is a nice exercise to develop it, as well as check out what I wrote. Numerical qualities of these formulas are not certain; it is more efficient to find the eigenvalues of the companion matrix using QR-method (as MATLAB does in its "roots" routine). Good luck, ZVK (Slavek).