The Cubic Equation. Urs Oswald. 11th January 2009

Transcription

1 The Cubic Equation Urs Oswald th January 009 As is well known, equations of degree u to 4 can be solved in radicals. The solutions can be obtained, aart from the usual arithmetic oerations, by the extraction of roots. In the case of the quadratic equation, this has a very concrete meaning. Even if the coefficients are arbitrary comlex numbers, the solutions can always be calculated by the extraction of roots from nonnegative real numbers. This can, if necessary, even be done by hand. It is therefore imortant to emhasize that, already in the case of the cubic equation with only real coefficients, solvability in radicals means much less. Whenever there are three distinct real solutions, calculating them involves finding a solution to the equation z = u for some comlex, nonreal u. To be quite recise, this is not necessary if coefficients as well as solutions are rational. Most of the time this cannot be done by just extracting roots from nonnegative real numbers. Therefore, while the question of whether or not a certain equation can be solved in radicals has had a rofound influence on the develoment of mathematics, it is not of any decisive imortance to the actual search for solutions. Arbitrary comlex coefficients As any cubic equation z az bz c = 0 can be transformed to the form z z q = 0 by the substitution of z with z a, solving the cubic boils down to solving equation. The identity connects with the system { u v u v q = u v q uv u v uv =, u v = q. The idea of letting z = u v aarently goes back to the Italien mathematician Tartaglia 499 or However, while to Tartaglia all numbers were real and even nonnegative, we shall admit arbitrary comlex numbers. Theorem i If u, v is a solution of, then u v is a solution of. ii If u v is a solution of and u v = 0, then u, v is a solution of. iii If z is a solution of, then there is a solution u, v of such that z = u v. Aart from order, u and v are uniquely determined: for some ε such that ε = z. u = z ε, v = z ε

2 Proof: i and ii are immediate consequences of. iii The conditions u v = z and u v = uniquely determine aart from order u, v as the solutions of the equation w z w = 0. Therefore u = z ε, v = z ε for some ε such that ε = z. By ii, u, v is a solution of. q.e.d. The first equation of imlying u v =, imlies that u, v are the solutions of the equation w q w = 0. These solutions are q δ and q δ for some δ such that δ =, where q =. Therefore, imlies { uv =, u = q δ and v = q δ for some δ such that δ =. 4 The converse is trivially true, hence we get: Lemma. The equation systems and 4 are equivalent. Next we rove: Theorem Let δ, u be such that δ =, u = q δ, where u 0, and let v = u. Then i v = q δ, ii u v is a solution of. Proof: δ = = q imlies that = q δ = q δ q δ = u q δ. From the definition of v we get v = = u u q δ = q u δ. So u, v is a solution of 4 and, by Lemma., of. By i of Theorem, u v is a solution of. q.e.d. This theorem is, in a way, an instruction on how to obtain a solution z : Find a δ such that δ = can be done by extraction of roots from nonnegative real numbers. Find a solution u to the equation u = q δ by de Moivre s formula. Calculate v = u. Let z = u v. We let ζ be one of the nonreal solutions of the equation x =, let s say As z = z z z, ζ satisfies ζ = i. 5 ζ ζ = 0. 6 If u, v is a solution of, then obviously, ζ k u, ζ k v and ζ k v, ζ k u also are k Z. We now rove the converse: Theorem If u, v and u, v are both solutions of, then for some k {0,, }. u, v = ζ k u, ζ k v or u, v = ζ k v, ζ k u

3 Proof: We first assume that = 0. Then, e.g., u = v = 0. From the second equation of follows v = q = u. So u = ζ k v for some k {0,, }. Trivially, v = ζ k u. We let k = k mod. Then ζ k = ζ k and ζ k = ζ k, therefore u, v = ζ k v, ζ k u, where k {0,, }. All other cases are treated similarly. Now we assume 0. Then Lemma. imlies that for some δ such that δ = q, u = q δ, and either u = q δ or v = q δ. In the first case, u = u, and therefore u = ζ k u for some k {0,, }. The first equation of imlies that u v = u v =, therefore ζ k u v = u v. As 0, u 0, and we conclude that ζ k v = v, and therefore v = ζ k v. In the second case, v = u, and therefore v = ζ k u for some k {0,, }. As before, u v = u ζ k u = u v =, and u ζ k = v, u = ζ k v. q.e.d. Theorem 4 Let z, z, z be the not necessarily distinct solutions of, and let u, v be any solution of. Then for k = 0,,, the values ζ k u ζ k v are a ermutation of the solutions z, z, z. Proof: By iii of Theorem, each of the solutions z, z, z is of the form u v, where u, v is a solution of. By Theorem, u, v = ζ k u, ζ k v or u, v = ζ k v, ζ k u, and therefore u v = ζ k u ζ k v for some k {0,, }. q.e.d. Examle The following icture illustrates how the solution of the equation z 6 iz 6 7 i = 0 is reduced to the solution of the simler equations u = q δ = 8 8 i, v = q δ = i, where δ = 65 9 i, δ = = i. We can take as a start u = 4 4 i and v = i. u = 4 4 i v = i u u v z z v v z u We then get u, u by twice rotating u by π v by π twice multilying v by ζ. twice multilying u by ζ, and v, v by twice rotating

4 Theorem 5 If z, z and z are the solutions of, then Therefore = 08 z z z z z z. i 0 if and only if all solutions are distinct. ii If z, z, z are all distinct and real, then is also real, and < 0. Proof: In view of Theorem 4 we may assume that z = u v, z = ζu ζ v, z = ζ u ζ v for some solution u, v of. We find z z = ζ u ζ v = ζ u ζ v = ζ u ζ v, z z = ζ u ζv = ζ ζu v = ζ ζ u v and therefore, alying ζ ζ = ζ 4 ζ ζ = ζ ζ = =, z z z z = ζ u ζ v ζ u v = ζ ζ u ζu v u v ζ v = ζ ζ u ζ u v ζ v = ζ ζ u u v v = u u v v. We further have z z = ζ ζu ζ ζ v = ζ ζu v and therefore z z z z z z = u u v v ζ ζu v = ζ ζu v. By Lemma., u, v is also a solution of 4, therefore u = q δ, v = q δ and u v = δ for some δ such that δ =. Thus finally, z z z z z z = 9 4 = 08. q.e.d. All coefficients real While the above results hold for arbitrary comlex coefficients, we now retrict ourselves to real values of and q. Then = q also is real, and there are three ossibilities: I. = 0 vanishing discriminant, II. > 0 classical case, III. < 0 irreducibel case. We will show that in cases I and III, there are only real solutions, whereas in case II, there are two nonreal solutions which then are conjugate comlex.. The case of the vanishing discriminant Let ρ be the uniquely determined real number such that ρ = q. Then = q = 0 imlies ρ 6 = 0, = ρ 6, = ρ. Thus we have = ρ, q = ρ, and becomes x ρ x ρ = 0. Obviously, one of the solutions is z = ρ, and after slitting off the factor x ρ, for z, z we get the equation x ρ x ρ = 0 having the solutions ρ and ρ. Theorem 6 If, q are real and = 0, then has the solutions q, q. q. 4

5 . The classical case If and q are real, and > 0, equation has one real and two nonreal solutions, the latter being conjugate comlex. We start with the following Lemma. If z is a solution, then is equivalent to x z x z x z = 0, and = z 08 4 z = 08 z, where = z 4 is the discriminant of the quadratic factor. Proof: From z z q = 0, we get q = z z and therefore = q = 4 z z 7 = 08 7z 6 54z 4 7 z 4 = 08 z 4 9z 4 6z, thus = z 4 z. For the discriminant of the quadratic factor, we find = z 4z = 08 z 4 = z 4. Hence = 08 z. q.e.d. We define two real numbers u and v, letting u = q, v = q. 7 Theorem 7 i u, v is a solution of. ii z = u v is a solution of. iii z is the only real solution of. The other solutions are z, z = u v ± u v i. Proof: i u v = q q = q = q q =, hence u v =. u v = q q = q. ii As u, v is a solution of, u v is a solution of by i of Theorem. iii By Lemma., z, z are the solutions of the quadratic equation x z x z = 0 having the discriminant = z 4. On the other hand, from ii we get z u v = u v u v = 4u v = 4, hence z 4 = u v and = z 4 = u v. Therefore z, z = z ± u v i, where z = u v. q.e.d. Of course, Theorem 7 yields the same result as Theorem 6 if = 0. Examle Alying Theorem 7 to the equation x x = 0, we obtain z =. 9 9 Now this seems rather surrising, because obviously, one of the solutions is z =. This is the only real solution, as x x = x x x, the second factor having no real zeroes. So we are forced to conclude that =. 9 9 If there are rational numbers a and b such that a b = 9, then a b = 9, and the left side reduces to ab a b = a. We therefore look for b such that b = 9. The conditions for b are { 8 6 b = 8b = 9. b 4 This system has a unique solution: b = 6. So the sum of the two third roots reduces to 6 =. 6 5

6 We might have reached this insight easier with the shortcut furnished by iii of Theorem. Knowing that z =, we get ε = = 6, u = 6, v = 6. For u v we get and finally, alying iii of Theorem 7, z, z = ± 7 i. Examle x x = 0. By Theorem 7, we get = q = 08 = 8 69 and z = Any rational solution would have to be an integer, and any integer solution would have to be a divisor of q =, but and are not solutions. So there are no rational solutions. Consequently, there is no use looking for rational r, s such that r s 69 = If there were, this would imly r s 69 = 8 69, and we would have the rational solution z = r s 69 r s 69 = r. Theorem 8 Let z, z, z be the solutions of the equation x x q = 0 where z and b are real numbers such that b > 0 and x, = z ± b i. Then if z = u v, u, v = z ± b. Proof: The equation corresonding to the given solutions is x b 4 z x q = 0. From art iii of Theorem follows u, v = z ± ε, where ε = z = z 4 b z 4 = b. q.e.d. Examle 4 x 6x 0 = 0. We get = 08. As z = is a solution, alying Theorem 7 ii, we conclude that =. Part iii of Theorem yields u, v = ± ε, where ε = z = =. Hence u, v = ±. We indeed find ± = 0 ± 6. Slitting off the factor x, we find x, = ± i. Using Theorem 8, we again find u, v = ± = ±.. The irreducibel case If, q are real, then = q δ = i. Then q δ = q i = < 0 imlies < 0, =, and = q, hence q. We let q δ = 0. 8 Therefore the equation u = q δ has three distinct solutions u, u, u such that u k = 0. We let v k = u k and x k = u k v k i =,,. Then by Theorem, z, z, z are solutions of. From u k v k = = and u ku k = u k = = = u kv k, we conclude that v k = u k and x k = u k u k = Reu k = u k cosargu k. 6

7 But argu k = arg q δ k π, and arg q δ = arg q i, where > 0. Therefore ϕ = arg q δ = arccos q q δ = arccos q by 8. By i of Theorem 5, all solutions are distinct. We can also see this directly, as 0 < ϕ < π, and therefore 0 < ϕ π < ϕ π < π, 4π < ϕ 4π < 5π. As a consequence, < π, ϕ < cos π < ϕ < cos 4π < < cos ϕ <. 9 This imlies that Reu, Reu, Reu, and therefore z, z, z are all distinct. Theorem 9 If < 0, the equation has three distinct real solutions: x k = cos arccos q k π k =,,. We might add that our deduction of course garanties that in this result, arccos is always defined. More directly, as < 0, the following are equivalend: > q >= q = 0, < < 0, > =<, hence we have: Lemma. If < 0, then > = 0 if and only if < q > =<..4 Another, most elegant formulation In view of the result of Theorem 9, we let z = w. 0 Then transforms into 8 w w q = 0. As = sgn, this is equivalent to 8 w 6 sgn w = q and, if 0, to Therefore, is equivalent to I. 4w w = C, if > 0, II. 4w w = C, if < 0. 4w sgnw = C, where C = q = q. These equations match in form with the identities 4 sinh ϕ sinh ϕ = sinh ϕ, 4 cosh ϕ cosh ϕ = cosh ϕ, 4 cos ϕ cos ϕ = cos ϕ, whereby we have to kee in mind that the ranges of sinh, cosh, cos are R, {x R/x }, and {x R/ x }, resectively. Comaring with, we get 7

8 Theorem 0 i w = sinh arsinh C if > 0, { cosh ii w = arcosh C if < 0 and C cosh arcosh C if < 0 and C, iii w k = cos arccos C k π, where k =,, if < 0 and C. By Lemma., i and ii corresond to the classical case, while iii is the result in the irreducibel case Theorem 9. There is, however, a difference. While the deduction of Theorem 9 is based on comlex numbers, the deduction of art iii of Theorem 0, based on the formula cos ϕ = 4 cos ϕ cos ϕ, lies wholly within the realm of the real numbers. In order to confirm that cases i and ii of Theorem 0 lead to the result of Theorem 7, we first rove Lemma. q C C if > 0, = C C if < 0 and C. Proof: If > 0, then C = q =, therefore C C = q. If < 0 and C, then C = q C = q = q. q.e.d. q = = 0, therefore C Theorem If > 0, or < 0 and C, then i and ii of Theorem 0 both lead to z = w = q q. Proof: i. arsinh C = ln C C, hence w = sinh arsinh C = By Lemma., C C = q. q, therefore w = q, C = C q q q C C = C C q =, and z = w = q ii. The roof runs analogously to that of case i, using that for C, arccos C = ln C C and thus cosh arccos C = C C, C C where, by Lemma., C C = q. If C, the equation 4w w = C has to be relaced by the equivalent equation 4 w w = C. q.e.d.. Begun: osurs@dell98:/home/osurs/cubic/cubic.tex th January 009 8