Transcript Document 7186866

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)
Trig Property in use here:
= cos3(α) – 3sin2(α)cos(α)
= 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
and
3st = A
(s3 - t3) = B
(Equation 1)
(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 AlKhayammi’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