A nonstandard cubic equation

Transcription

1 MATH-Jan-05-0 A nonstandard cubic euation J S Markoitch PO Box West Brattleboro, VT 050 Dated: January, 05 A nonstandard cubic euation is shown to hae an unusually econoical solution, this solution incorporates an angle that seres as the euation s discriinant A nonstandard cubic euation is shown to hae an unusually econoical solution, this solution incorporates an angle that seres as the euation s discriinant This euation is treated as four special cases: In Sec I the euation has just two independent constants and In Sec II the euation has three independent constants,, and k, and represents the general case In Sec III the euation has just a single independent constant, but is especially interesting as it possesses the siple approxiate solution x + In Sec IV the euation again has one independent constant, but ust be a perfect cube The solution to the standard cubic euation is gien in Appendix A I THE CUBIC EQUATION WITH TWO CONSTANTS We begin with a theore proiding the solution to the nonstandard cubic euation haing just two constants Theore Define the cubic euation + x + + x =, haing positie constants and, and the ariable x ero out x fro the aboe euation to define and let and W = + sin θ = W = + sin θ sin θ Electronic address: jsarkoitch@gailco Then x = + 5 soles E Proof We will expand E into the standard cubic euation, identify its coefficients, and then sole it by using its classical solution This solution will then be siplified by a series of substitutions until Es and 5 are recoered The standard cubic euation ax + bx + cx + d = 0 a = 6 has this solution x = p = c b = b 7 + bc d r = b 7 r, 7 8 see Appendix A for proof When E is expanded we get or so that x + x + x + + x + x + = x + 6x + 9 x + =, x + 6x + 9 x + = 0 This produces coefficients of for E 6 a = b = 6 c = 9 d = = W 9

2 Substituting these alues into E 8 gies p = = + 7 r = 6 which siplifies to p = 6 9 = r = + 0 Substituting these coefficients into E 7 gies x = or x = Factoring out gies x = + + Because the alues in the aboe two outer radicals are reciprocals of each other, it follows that letting + allows E to be rewritten x = u + u But this euation is identical to E 5 except that u has replaced It follows that E 5 our goal holds proided that, 5 which is to say if + sin θ sin θ 6 But this is easily shown: Obsere that E gies = W This allows reoing fro E by substituting W to get W + W = W + W = W + W W + = W + W W = W + W W We now need to eliinate and W by substituting sin θ A glance at E shows that this reuires rewriting the aboe euation using powers of W So, we diide the aboe nuerator and denoinator by W to get W + W W and rearrange ters so that + W + W W 7 Now we can eliinate powers of W by substituting powers of sin θ as defined by E This gies which factors into so that + sin θ + sin θ sin θ + sin θ sin θ + sin θ + sin θ + sin θ sin θ Finally, we substitute into E to recoer E 5,,

3 Reark If θ = 0 then E has two distinct real roots Reark If 0 < θ < π/ then E has one real and two coplex roots Reark If θ is purely iaginary then E has three distinct real roots Note: sin iθ = i sinh θ Reark As a side issue, note the use of W in the siple alternate expression for d in E 9 II THE CUBIC EQUATION WITH THREE CONSTANTS It is possible to odify E slightly by joining x with a new real constant k, so as to create a general ersion of E In + k + x + + k + x = and are again positie constants, but the expression k +x now seres in the role earlier sered by x alone Hence, E 5 becoes k + x = +, so that the solution to E is x = + k, W, θ, and are defined as in Es Note that this use of k does not affect the usefulness of θ as the discriinant Euation produces coefficients of a = b = 6 + k c = 9 + k + k d = + 9 k + 6k + k = W + k[c kb ka] for E 6, k = 0 recoers E 9 III THE CUBIC EQUATION WITH ONE CONSTANT Now suppose that ceases to be an independent constant, but instead deries fro the constants and M = M M M = +, + M M, but now 9 Then, a surprisingly siple, but accurate, approxiate solution to E becoes possible: naely, x M In the theore that follows the extreely sall size coputed for ɛ is not proof of the accuracy of the aboe approxiate solution but the proof does help explain why the approxiation is so accurate Theore Let [ ] M y + M y [ M M ] + M M y =, M, and and M are positie constants such that M = +, Then 9M 7 + 9M 8 8M 7 Reark 5 Inforally speaking, the absolute alue for ɛ euals the difference between the alue for produced by E when x = M, ersus that produced by E Moreoer, as E 7 akes clear, for eer larger M the necessarily sall alue for ɛ shrinks rapidly Proof Substituting y, as defined by E, into E gies M M + M M M M + M M This expands and siplifies to 7M 0 + 9M 5 8M + 6M 5 + 9M 8 M M = 7M 0 + 9M 5 5M 9 + 9M 8M + 7M 9 + 8M 9 8M 8

4 TABLE I: Values produced by E when is deterined by E Values are coputed for the two sallest for which is a perfect cube The alues in the first row derie fro E Cubed expression Suared expression 9 a a Minial case Cobining large and sall ters separately gies 7M 9 7M 0 + 7M 9 + 9M 5 + 9M 8M 9 But the large ters of the aboe nuerator su to 0; that is to say, gien E 5, it follows that 7M 9 7M 0 + 7M 9 = M 9 M 9 M = M 9 M 9 = 0 So, the effects of M and M in E 8 alost copletely cancel What does not cancel is this relatiely sall aount 9M 5 + 9M 8M 0 This fraction, which has only coparatiely sall powers of M in its nuerator, gies 9M 7 + 9M 8 8M Reark 6 In the nuerator of E 9 all large ninth and tenth powers of M, which ight otherwise contribute greatly to approxiation error, copletely cancel; this leaes only the uch saller fourth and fifth powers of M as the ajor sources of error It follows fro Es 5, 6, and that ɛ IV THE CUBIC EQUATION WITH ONE CONSTANT AND A PERFECT CUBE If = 9, then E 5 gies M = 0, so that E gies = = = 706 TABLE II: Values produced by E when is deterined by E Values are coputed for the two sallest for which is a perfect cube M W / x sin θ 9 a b 09 c d 0006 a Minial case b Approxiately 0 = See Es and 5 c So, cos θ = θ d Approxiately 7 = Because = 9 is a perfect cube this ay be rewritten = = 706 With a perfect cube, E can likewise be rewritten So, substituting the aboe alues for and into E gies 0 = = 706 It is these alues which appear in the first rows of Tables I and II Because = 9 is the sallest positie nuber for which is a perfect cube it follows that = 9 and = 706 represent a inial case All of this shows that at the outset we ight hae chosen as a different starting point this logical alternatie to E + x + + x =, n n = And, finally, note that for the aboe and, E produces x , a alue ery close to the approxiate alue for x gien by E, naely x M

5 5 APPENDIX A: THE SOLUTION TO THE STANDARD CUBIC EQUATION Theore The standard cubic euation ax + bx + cx + d = 0 a = A has the solution x = + proided that 7 + p = c b = b 7 + bc d r = b Proof We introduce y as follows x = y r and substitute r as defined by E A to get x = y b Substituting into E A gies 7 r A A A y b + b y b + c y b + d = 0 This expands and siplifies to y + py = 0 A5 with p and fro E A neatly replacing all instances of b, c, and d Note the absence of a y ter: the point of this substitution We introduce z as follows y = z p A6 z and ake Vieta s substitution into E A5 to get z p + p z p = 0 z z This expands and neatly siplifies to z p 7 z = 0 A7 We turn this into a uadratic euation in z by ultiplying through by z to get z z p 7 = 0 A8 the point of Vieta s substitution The standard uadratic forula then gies ± p 7 z = = ± 7 A9 We are now close to recoering E A, which we hae to reasseble fro the trail of parts we left behind Essentially, we need to roll back the y r and z p z substitutions ade earlier We proceed in reerse order by eliinating z p z first Fro E A9 we know that z = + 7 A0 The inner radical we arbitrarily gie a plus sign, but a inus sign would lead to identical results We now introduce this identity p = into which we substitute z fro E A0 to get p = z 7 By oing z to the left, we then also know that p z = A 7 Substituting the aboe alues for z and p z A6 gies y = into E, A undoing Vieta s substitution Finally, we undo the first substitution by plugging this y into E A to recoer E A Reark 7 The discriinant of E A can be shown to be = 8abcd b d + b c ac 7a d a = Copare this against the econoy of the discriinant θ, discussed in Rearks,, and By playing a central role in the solutions to Es and, the siple discriinant θ shows these two euations to be at least in this liited respect ore fundaental than E A