Solving Cubic Equations Ben Anderson Jeff Becker

Use provided ruler and compass to find 1/3 of the given angles. Warm-up Activity Use provided ruler and compass to find 1/3 of the given angles.

Explanation of Activity Many early solutions involved geometric representations. Later answers involved arithmetic without concepts of negatives & zero. Even as math evolved, solutions involving square roots of negative numbers eluded mathematicians.

Explanation of where the equation 4x3 - 3x – a = 0 comes from… Let θ = 3α, so cos(θ) = cos(3α) and let a = cos(θ), then… a = cos(θ) = cos(3α) = cos(2α + α) = cos(2α)cos(α) – sin(2α)sin(α) = [cos2(α)-sin2(α)]cos(α) – [2sin(α)cos(α)]sin(α) = cos3(α) – sin2(α)cos(α) – 2sin2(α)cos(α) = cos3(α) – 3sin2(α)cos(α) = cos3(α) – 3[1-cos2(α)]cos(α) = cos3(α) – 3cos(α) + 3cos3(α) = 4cos3(α) – 3cos(α) Thus, 4cos3(α) – 3cos(α) – a = 0. Finally, setting x = cos(α) gives us 4x3 - 3x – a = 0. Trig Property in use here: cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

Early Approaches to Solving the Cubic Equation Arabic and Islamic mathematicians Formulating the Problem - Islamic mathematicians, having read major Greek texts, noticed certain geometric problems led to cubic equations. - They solved various cubic equations in the 10th and 11th centuries using the Greek idea of intersecting conics.

Arabic and Islamic Mathematicians ’Umar ibn Ibrāhīm al-Khayyāmī - known in the West as Omar Khayyam - systematically classified and solved all types of cubic equations through the use of intersecting conics - publishes Treatise on Demonstrations of Problems of Al-jabr and Al-muqābala, devoted to solving the cubic equation - classifies various forms possessing a positive root and 14 other cases not reducible to quadratic or linear equations

Arabic and Islamic Mathematicians Sharaf al-Dīn al-Tūsī - born in Tus, Persia - He classified cubic equations into several groups that differed from the ones Omar conceived: 1. equations that could be reduced to quadratic ones, plus x3 = d. 2. eight cubic equations that always have at least one (positive) solution 3. types that may or may not have (positive) solutions, dependent upon values for the coefficients, which include: x3 + d = bx2, x3 + d = cx, x3 + bx2 + d = cx, x3 + cx + d = bx2, and x3 + d = b x2 + cx (his study of this third group was his most original contribution)

Italian Invention and Dispute algebra and arithmetic develop in 13th century in Italy through such publications as Leonardo of Pisa’s Liber Abbaci little progress is made toward solving the cubic equation until the 16th century when the Italian theater is set to stage a new production:

Solving the Cubic Equation: A Bad Sixteenth Century Italian Play List of Characters Scipione del Ferro, the dying scholar Antonio Maria Fiore, his student Niccolò Fontana (Tartaglia or “the Stammerer”), rival of Fiore Girolamo Cardano, new rival to Tartaglia Lodovico Ferrari, Cardano’s pupil Rafael Bombelli, the imaginary fool

Act I Scene i Ferro: Fiore, Fiore! Alas, I am dying, Fiore, and need to pass on my method for solving the cubic equation. Fiore (aside): An equation, he speaks? Though I am not a great mathematician, I will learn his method and someday reveal it to claim great fame.

Act I Scene ii Meanwhile, Tartaglia also knew a method for solving the cubic equation… Tartaglia: Though I know how to solve the cubic equation, I will not reveal how. This gives me the ability to challenge others with a type of problem they can’t solve. Rich patrons support me while I am defeating other scholars in public competitions! Fiore: I have heard your claim of knowing the solution to the cubic equation. I challenge you to a competition!

Act I Scene ii continued Tartaglia: Nave! Willingly I accept! Fiore (aside): Victory is within reach, for Tartaglia is known as “The Stammerer” and couldn’t act in a Bad Sixteenth Century Italian Play if he tried!

Act I Scene ii continued The contest ensues: Tartaglia: It appears you only know how to solve equations of the form x3 + cx = d. Fiore: That is strange, for you only know how to solve equations of the form x3 + bx2 = d. Tartaglia: Well, no matter. I have prepared fifty more unrelated homework problems for you. Fiore: Argh!!

Act I Scene ii continued Tartaglia wins the competition: Fiore: It appears my knowledge of mathematics does not extend beyond cubic equations, yet you have managed to find solutions to all my problems. Tartaglia: I forgive your ignorance. I will decline the prize for my victory, which included thirty banquets hosted by you, loser, for me, the winner, and all my friends. Fiore recedes into the obscurity of history.

Act II Scene i Cardano: Greetings. I am a great doctor, philosopher, astrologer, and mathematician. Please give me the solution to the cubic equation. Tartaglia: Okay, but you must swear never to reveal my secret. Cardano: I swear.

Act II Scene i continued Within the next decade, Cardano publishes the Ars Magna, containing complete solutions to solving any cubic equation. Included are geometric justifications for why his methods work. He includes a subtle footnote to Tartaglia that del Ferro had discovered the crucial solution before him, justifying his publication of del Ferro’s work. Cardano also includes a solution for the quartic, which his pupil Ferrari devised.

Act II Scene ii While Tartaglia is furious, Ferrari contacts him to challenge him to a competition. Tartaglia refuses until he is offered a professorship in 1548 on the condition that he defeats Ferrari in the contest. Ferrari: I know how to solve the general cubic and quartic equations. Tartaglia may not have read Cardano’s book on those equations, which contains a solutions manual.

Act II Scene ii continued Tartaglia: I don’t like books, especially when it’s other people’s bad writing, but I like math competitions and I assume I’ll win easily. This time I think I will accept my victory spoils! Tartaglia loses and remains resentful of Cardano for the rest of his life

Act II Scene ii Continued “This is not the end of the story.” Some expressions that resulted from Cardano’s method in equations of the form x3 = px + q didn’t make sense. For x3 = 15x + 4, Cardano’s method produces:

Act II Scene iii Bombelli: For the expression x3 = px + q, there is always a positive solution, regardless of the positive values of p and q. For x3 = 15x + 4, this would be x = 4. However, for many values of p and q, solving the equation gives square roots of negative numbers. Hence, I will legitimize these numbers by calling them “imaginary numbers,” making myself nothing short of a genius in my time!!!

Play Conclusion The next target, equations of degree 5, proves more difficult, turning a different chapter in history when abstract algebra rears its head.

Procedure for Solving Cubic Equations Beginning with an equation of the form: ax3 + bx2 + cx + d = 0 Substitute x = y – b/3a a(y – b/3a)3 + b(y – b/3a)2 + c(y – b/3a) + d = 0 and simplifying gives: ay3 – b2y/3a + cy + 2b3/27a2 – bc/3a + d = 0 Make equation into the form y3 + Ay = B y3 + (c/a - b2/3a2)y = (bc/3a2 – d/a - 2b3/27a3)

Procedure Continued Find s and t such that 3st = A (Equation 1) and (s3 - t3) = B (Equation 2) Fact: y = s – t is a solution to the cubic of the form y3 + Ay = B To find s and t, we solve Equation 1 in terms of s and substitute into Equation 2. (A/3t)3 – t3 = B Through algebra, we obtain: (A3/27t3) – t3 = B t6 + Bt3 – A3/27 = 0 Substitute u = t3 u2 + Bu – A3/27 = 0 Quadratic formula gives us value of u.

Procedure Concluded We then use this value of u to obtain t, which in turn is used to find s. Next, we use the fact (from the previous page) that y = s – t is a solution to the cubic and plug s – t into the original substitution of x= y – b/3a to find the first real root. To find the other roots (real or imaginary) of the equations, we use this solution to reduce the cubic equation into a quadratic equation by long division. At this point, we can use the quadratic formula to obtain the other roots of the equation.

Present Day Today there exists numerous cubic equation calculators that solve cubics at the click of a mouse. One such example is: www.1728.com/cubic.htm

Timeline 400 B.C. - Greek mathematicians begin looking at cubic equations 1070 A.D. - Al-Khayammi publishes his best work Treatise on Demonstrations of Problems of Aljabr and Al-muqābala Late 12th century – Sharaf continues Al-Khayammi’s work and adds new solutions 14th century - Algebra reaches Italy Early 16th century – del Ferro and Tartaglia discover how to solve certain cubics but keep their solutions secret

Timeline continued 1535 – Fiore challenges Tartaglia to a competition involving cubic equations, and Tartaglia wins. News of his victory reaches Cardano. 1539 - Tartaglia explains his partial solution to Cardano.

Timeline continued 1545 – Cardano produces a complete solution to cubic equations and publishes it in Ars Magna, which also includes Ferrari’s solution to the quartic. Late 16th century – Bombelli introduced the idea of using imaginary numbers in the solution to cubic equations

References Berlinghoff and Gouvea. Math Through the Ages. Katz, Victor J. A History of Mathematics. Cubic Equation Calculator. www.1728.com/cubic.htm Cubic Equations. en.wikipedia.org/wiki/Cubic_equation The “Cubic Formula”. http://www.sosmath.com/algebra/factor/fac11 /fac11.html