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Solving Cubic Equations

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					Solving Cubic Equations




      Ben Anderson
       Jeff Becker
            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)      Trig Property in use here:
       = cos3(α) – 3[1-cos2(α)]cos(α) cos(x+y) = cos(x)cos(y) - sin(x)sin(y)
       = 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.
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

				
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