Download Ancient and Modern Algebra: Exact Solutions of Polynomial Equations - Prof. Barry Dayton and more Study notes Mathematics in PDF only on Docsity!
Theory of Equations
Lesson 9
by
Barry H. Dayton
Northeastern Illinois University
Chicago, IL 60625, USA
www.neiu.edu/˜bhdayton/theq/
These notes are copyrighted by Barry Dayton, 2002. The PDF files are freely available on the web and may be copied into your hard drive or other suitable electronic storage devices. These files may be shared or distributed. Single copies may be printed for personal use but must list the website www.neiu.edu/˜bhdayton/theq/ on the title page along with this paragraph.
“Maple” and “MAPLE” represent registered trademarks of Waterloo Maple Inc.
Chapter 4
Ancient and Modern Algebra
THE EXACT SOLUTION OF POLYNOMIAL EQUATIONS
In this chapter we turn our attention to exact solutions of polynomial equations. The Babylonians used tables of values of n^3 + n to find numerical solutions of easy cubic equations, however the Greeks around the time of Euclid became obsessed with exactness. Largely due to the dominant influence of Euclid’s Elements there was a great interest in exact solutions up to the proof by Abel and Galois of the impossibility of such solutions in general. In the first few sections of this chapter we review some of the history of exact solutions. A good overview of this history is included in B.L van der Waerden’s A History of Algebra.
4.1 Solutions of Quadratic Equations
Although Euclid, in keeping with the philosophy of his time, rejected all numbers he still indirectly considered quadratic equations. His version of the solution of the quadratic equation ax + x^2 = b^2 is the geometric theorem:
If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole (with the added straight line) and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line.
Essentially this is completing the square.
4.1. SOLUTIONS OF QUADRATIC EQUATIONS 113
Figure 4.1: Abu Kamil’s solution to x^2 + 10x = 39
to N L, the whole line OM is the required line z. This is expressed in the following way:
z =
a +
aa + bb
This is essentially the first statement in more or less modern language of the general quadratic equation. (See Figure 4.2). Incidentally, Descartes justification of his formula is algebraic, the point of his ge- ometry is to establish that the number
aa/4 + bb exists, i.e. it is the length of the hypotenuse of the triangle 4 LM N. Although later in his book, Descartes allows negative coefficients and negative roots, at this point in his book, the first chapter on analytic geometry, only positve coefficients and roots are considered. We now know that the complete solution to all quadratic equations ax^2 + bx + c = 0 is given by the general quadratic equation
x =
−b ±
b^2 − 4 ac 2 a
This solution is valid even when a, b and c are complex numbers and when the quantity inside the square root sign is positive, negative or imaginary.
114 CHAPTER 4. ANCIENT AND MODERN ALGEBRA
Figure 4.2: Descartes’ Solution of z^2 = az + b^2
4.2 Omar Khayyam and Viete
SOLUTION OF THE CUBIC
Omar Khayyam, the Persian poet and mathematician of the second half of the 11th cen- tury, was one of the first mathematicans to consider the cubic equation. Like Descartes 6 centuries later he avoided associating x^2 with area and x^3 with volume, but unlike Descartes who disassociated the two concepts, Omar Khayyam got around this prob- lem with the statement “Every time we shall say in this book ‘a number is equal to a rectangle’, we shall understand by the ‘number’ a rectangle of which one side is unity, and the other a line equal in measure to the given number...”
Figure 4.3: Omar Khayyam’s solution of Cubic (Modern version)
116 CHAPTER 4. ANCIENT AND MODERN ALGEBRA
- del Ferro did not publish his solution but it was known to some of his pupils, no- tably Antonio Maria Fiore, who in 1535 challenged a young man nicknamed Tartaglia to a contest in solving the equation “the first power plus the cube equal to a number.” Tartaglia managed to solve the problem the night before the contest and defeated Fiore.
The distinguished physician Cardano, who was at that time engaged in writing an elementary book on mathematics, heard about Tartaglia’s success and in 1539 tried to persuade Tartaglia to expain his method. Tartaglia had his own intentions to publish and first refused but later was induced to give his method to Cardano, but even then only in obscure verses, and only after Cardano swore a solemn oath not to reveal the method until Tartaglia had published.
Cardano mastered and expanded Tartaglia’s method and became impatient to pub- lish. In 1543 Cardano received permission to inspect del Ferro’s papers and found the original solution at which point he published his famous Ars Magna (it appeared in 1545). Tartaglia was furious and the ensuing debate between Tartaglia, Cardano and Cardano’s former houseboy and student Ferrari lasted over 30 years. Ferrari, who had been present when Tartaglia gave Cardano the method, discovered how to solve the biquadratic (fourth degree) using the solution to the cubic. Cardano included this in his book however with the qualification:
Although a long series of rules might be added and a long discourse given about them, we concude our detailed consideration with the cubic, others being merely mentioned, even if generally, in passing. For as the first power refers to a line, the square to a surface, and the cube to a solid body, it would be very foolish to go beyond this point. Nature does not permit it.
Cardano used a rhetorical notation (no symbols) and did not allow negative coef- ficients. Thus his book has separate chapters for the 13 different cases of cubics that Cardano identifies. A nice translation of Ars Magna has been given by T.R. Witmer and published under the title The Great Art.
The one case where the cubic has three real roots was called the irreducible case. Cardano could not solve this case since the real roots could be found only by taking the square roots of imaginary numbers. Rafael Bombelli in 1572 mastered enough complex arithmetic to solve this problem. Viete, who did not accept negative numbers, much less imaginary ones, invented his trigonometric method to handle this case.
4.4. ALGEBRAIC SOLUTION OF THE CUBIC 117
4.4 Algebraic solution of the Cubic
The modern algebraic solution of the cubic is essentially that described by Cardano with some improvements by Viete. However, since we can now take square and cube roots of negative and imaginary numbers we need only one case to handle all cubics, including the “irreducible case.” We give an exposition of the method combining that of Birkhoff and MacLane and Uspensky. Given a cubic polynomial equation (we can assume monic)
x^3 + ax^2 + bx + c = 0 (4.1)
the first step is to do a change of variables to eliminate the x^2 term. This can be done using Horner’s process, see theorem 2.5.1. In particular, letting y = x + a/ 3 we get the equation y^3 + py + q = 0 (4.2)
where p = f ′(−a/3) = b − a^2 / 3 and q = f (−a/3) = c − ba/3 + 2a^3 / 27. If we can solve Equation 4.2 for y then the solutions for x are given by x = y − a/ 3. Thus for the rest of this section we will assume the equation to be solved is Equation 4.2. The trick (due to Viete) is to make the substitution in Equation 4.2 of
y = w −
p 3 w
From the binomial theorem and algebra we get
w^3 −
p^3 27 w^3
or (w^3 )^2 + qw^3 − p^3 /27 = 0 (4.4)
which is quadratic in w^3 and can be solved by the general quadratic equation as
w^3 = −q/ 2 ±
q^2 /4 + p^3 / 27
Thus we have two possible values for w^3 , and, unless one of these is 0, three possible cube roots for value of w^3. However, upon substituting in Equation 3 we find only 3 separate values of y i.e. the three cube roots of just one of the solutions of Equation 4.4. In fact, we only need to calculate one cube root. To see this, it is known that the two roots α, β of the quadratic equation x^2 +bx+c = (x − α)(x − β) = 0 satisfy αβ = c. Thus the two roots (i.e. values of w^3 ) of Equation
4.4. ALGEBRAIC SOLUTION OF THE CUBIC 119
We summarize Cardano’s solution as follows, while not quite as compact as the quadratic equation, it is just as mechanical.
To solve x^3 + ax^2 + bx + c = 0
Let p = b − a^2 / 3 , q = c − ba 3 + 2 a
3 27 Note that if^ a^ = 0^ then^ p^ =^ b^ and q = c.
Let A be any cube root of −q/ 2 ±
q^2 /4 + p^3 / 27 , where the sign ± is chosen so that A 6 = 0. Let B = −p/(3A) and γ = −1+
√ 3 i
Then the solutions are
x 1 = A + B −
a 3 x 2 = γA + γ^2 B −
a 3 x 3 = γ^2 A + γB −
a 3
Maple Implementation
To obtain Cardano’s solutions for a cubic f simply use the com- mand
solve(f, x); But unless Maple sees an obvious simplification the solution will be quite literally the one above.
In the following two exercises do all work by hand, do not use Maple, except, perhaps, as a check.
Exercise 4.4.1 [30 points] Use the method of this section to find the 3 complex roots of x^3 + 6x − 20 = 0. Give an exact answer.
Exercise 4.4.2 [40 points] An open wooden box (without a top) is in the shape of a cube with each outer edge 10 inches long. If the volume is 500 cubic inches what is the thickness of the wood (assuming uniform thickness). Give an exact, not decimal, answer! This problem is taken from Uspensky.
120 CHAPTER 4. ANCIENT AND MODERN ALGEBRA
We remark that the number inside the radical sign is (− 4 p^3 − 27 q^2 )/(−108) and the numerator has a special meaning. Suppose that Equation 4.2 has the three complex roots z 1 , z 2 , and z 3. Then (y − z 1 )(y − z 2 )(y − z 3 ) = y^3 + py + q and so we get
z 1 + z 2 + z 3 = 0 z 1 z 2 + z 1 z 3 + z 2 z 3 = p −z 1 z 2 z 3 = −q
Now let D = (z 1 − z 2 )^2 (z 1 − z 3 )^2 (z 2 − z 3 )^2. A straight forward multiplication using the three formulas above shows that D = − 4 p^3 − 27 q^2 , i.e. the numerator under the radical sign. D is called the discriminant of Equation 4.2. This should be compared with the quadratic case where if z 1 , z 2 are the two roots of x^2 + bx + c = 0 then (z 1 − z 2 )^2 = b^2 − 4 c. As in the quadratic case, the discriminant in the cubic case gives us information on the number of imaginary roots:
Theorem 4.4.1 Let D = − 4 p^3 − 27 q^2 be the discriminant of the equation y^3 +py +q = 0 where p and q are real. If D > 0 this equation has 3 distinct real roots, if D < 0 this equation has one real root and 2 imaginary roots. If D = 0 this equation has multiple real roots.
Note that in the case of 3 real roots that q^2 /4+p^3 / 27 < 0 so A, B in the solution are imaginary numbers. So imaginary numbers may be required to find the real solutions of a real equation.
4.5 Solution of the Biquadratic Equation
The method still in use is essentially the method due to Ferrari. The first step, as in the case of the cubic is to simplify the equation, by a linear change of variable using 2.5. to get rid of the x^3 term. Thus we may assume our equation is
y^4 + py^2 + qy + r = 0 (4.6)
The trick is to add uy^2 + u^2 / 4 to both sides of the equation where u is, for now, unknown. Thus we get
y^4 + uy^2 + u^2 /4 = (u − p)y^2 − qy + (u^2 / 4 − r)
or (y^2 + u/2)^2 = (u − p)y^2 − qy + (u^2 / 4 − r) (4.7)
122 CHAPTER 4. ANCIENT AND MODERN ALGEBRA
However, unless there is a major simplification, Maple will not give Ferrari’s solution, often returning only RootOf(f,x). Generally Maple feels that the Ferrari solution is so complicated that you don’t want to see it. If you insist on seeing the full solution, give the com- mand
_EnvExplicit := true;
before executing solve. You’ll be sorry you did.
