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Theory of Equations
Lesson 2
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. TI-85/ are trademarks of Texas Instruments Inc.
1.4. POLYNOMIAL ARITHMETIC 13
1.4 Polynomial Arithmetic
In this chapter a polynomial will be a function
p(x) = a 0 + a 1 x + a 2 x^2 + · · · + anxn
where the numbers a 0 , a 1 , etc. are called the coefficients. We will attempt a more formal and precise definition in Chapter 5. The symbol x is simply a place holder, known as the ”variable.” The coefficients may be real or complex numbers, the set of all polynomial functions with real coefficients will be denoted R[x], the set of all polynomial functions with complex coefficients will be denoted C[x]. A polynomial p(x) ∈ R[x] can be thought of either as a real-valued function of a real variable or a complex-valued function of a complex variable. Of course if p(x) ∈ C[x] has some imaginary coefficients we will generally only be able to think of p(x) as a complex- valued function of a complex variable. Specification of the coefficients is enough to specify a polynomial but the converse statement, i.e. that the function determines the coefficients, should not be taken as obvious. It is a theorem that will be proven later in this chapter. We “evaluate” a polynomial by substituting an actual number (real or complex) for the variable x. There are good ways and not so good ways to do this. We will discuss the good ways in Chapter 2. Of course if you are using Maple the computer can worry about the details. We add and multiply polynomial functions like functions in calculus, i.e. pointwise. Thus if p(x), q(x) are polynomials, the sum is the polynomial f (x) given by f (x) = p(x) + q(x) for each x, and the product is g(x) given by g(x) = p(x) ∗ q(x) for each x. Thus if p(x), q(x) are polynomials with p(3) = 7 and q(3) = − 5 then (p + q)(3) = 2 and (p ∗ q)(3) = − 35. We can also subtract polynomials, but we cannot divide as the quotient of two polynomial functions is no longer a polynomial function. This arithmetic would not be particularly useful if it were not for the fact that one can fairly easily determine the coefficients of the sum and product from the summands and factors. It is an exercise with the associative, commutative and distributive laws of R and C to show that if
p(x) = a 0 + a 1 x + a 2 x^2 + · · · + anxn q(x) = b 0 + b 1 x + b 2 x^2 + · · · + bmxm
then p(x) + q(x) = f (x) and p(x) ∗ q(x) = g(x) where
f (x) = c 0 + c 1 x + · · · + ckxk g(x) = d 0 + d 1 x + · · · + dm+mxm+n
1.4. POLYNOMIAL ARITHMETIC 15
Note that you must type 2*xˆ2 not 2xˆ2 even though Maple returns 2 x^2. This isn’t actually fair but Maple is not as smart as you are and so needs help interpreting the expression 2xˆ. To evaluate a polynomial, say f , at a number, say c, use the procedure
subs(x=c, f);
Do not set x:= c because this will destroy x as a variable and then f will simply be the number f (c) and you will not be able to recover the polynomial f. c need not be an actual number, for in- stance c might be
2 and the result of subs(x=sqrt(2),f);, for f = x^3 − 2 x^2 + 5 above would be 2
2 + 1, to get a decimal value use evalf(subs(x=sqrt(2),f)); c may even be a variable ex- pression, for example subs(x=h+2, f); returns h^3 +4h^2 +4h+5. Now you can define the sum h(x) and product p(x) by
h := f + g; p := f*g;
Don’t be surprised if the results are
h := (x^3 − 2 x^2 + 5) + (x^2 − x + 3) p := (x^3 − 2 x^2 + 5)(x^2 − x + 3)
There are procedures normal, simplify, expand, sort and collect which may change the algebraic form of your func- tions without changing the functions. Try them
h := normal(h); h := simplify(h); p := expand(p); p := sort(p); p := collect(p,x);
For getting polynomials into standard form p(x) = a 0 + a 1 x + a 2 x^2 + · · · + anxn^ the most reliable of the above is sort(collect(p,x)). The x is the name of the variable that you want collected.
1.5 Degree
The degree of a polynomial p(x) = a 0 + a 1 x + a 2 x^2 + · · · + anxn^ is the largest index i for which the coefficient ai 6 = 0. Thus the polynomial p(x) = 1 + x^2 has degree 2. Constant polynomials have degree 0 with one exception: the constant polynomial 0, which has no non-zero coefficients does not have a degree. Alternatively, we can think of the 0 polynomial as having degree −∞; this will make later results true so long as we accept the convention that m + (−∞) = −∞ for any integer m or m = −∞. We will denote the degree of a polynomial p(x) by deg p(x).
Theorem 1.5.1 Let p(x) = a 0 + a 1 x + a 2 x^2 + · · · + anxn^ and q(x) = b 0 + b 1 x + · · · + bmxm_. Assume_ deg p(x) = n and deg q(x) = m. Then
i) deg(p(x) + q(x)) ≤ max(n, m)
ii) deg(p(x) ∗ q(x)) = m + n if p(x) 6 = 0, q(x) 6 = 0
In particular we note that in the second formula that because of the definition of degree an, bm are not 0; however ai, bj are 0 for i > n, j > m so the coefficient dm+n of p(x) ∗ q(x) is simply anbm which is not zero by the axiom I1. We thus note that for non-zero polynomials p(x), q(x), the product p(x) ∗ q(x) has a degree and hence is also non-zero. Thus the polynomial rings R[x], C[x] also satisfy axiom I1 and are thus integral domains.
On the other hand deg(p(x) ∗ q(x)) ≥ deg p(x) so if deg p(x) ≥ 1 there is no way the product p(x) ∗ q(x) = 1 since deg 1 = 0. Hence the property in axiom F1 fails most of the time so division of polynomials in the sense of division of real numbers is not possible.
Maple Implementation
You can find the degree of the polynomial f by using the function degree(f,x), the leading coefficient (the coefficient of the highest power of x with a non-zero coefficient) by lcoeff(f,x), the trail- ing coefficient (the coefficient of the smallest power of x with non- zero coefficient, usually the constant term if this term is not zero) by tcoeff(f,x). To find the coefficient of xk^ use coeff(f,x,k). In all of these x is the name of the variable.
The relation of divides is implemented by divide(p,g). MAPLE interprets this correctly by returning the word “TRUE” if the remainder is 0 and “FALSE” otherwise.
Although we will have occasion to use arbitrary polynomials f (x) as divisors, in most of our applications of the division theorem we will use the special case of f (x) = x − c where c is a real or complex number. The reason is because of the following important theorem:
Theorem 1.6.2 (The Remainder Theorem) Let p(x) be a polynomial and c ∈ C_. Di- viding_ p(x) by the degree 1 polynomial x−c gives as remainder the constant polynomial p(c).
Proof: We have p(x) = (x − c)q(x) + r(x) by the Division Algorithm where r(x) = 0 or deg r(x) < deg(x − c) = 1; in either case r(x) is a constant. Evaluating both sides at x = c gives p(c) = (c − c)q(c) + r(x) = r(x).
1.7 Factors and Roots.
Given a real or complex polynomial p(x), a complex number c is called a zero or a root of p(x) if p(c) = 0. The following theorem is one of the most important results on polynomials. This result was first stated by the philosopher- mathematician Rene Descartes in Chapter 3 of his book La Geometrie published in 1637. The idea may have been known previously, but Descartes was the first to write polynomials as we do today and thus the first person capable of appropriately phrasing this result. La Geometrie is famous as the book in which analytic geometry first appears; however, Chapter 3 is actually the modern beginning of the Theory of Equations.
Theorem 1.7.1 (The Factor Theorem) The complex number c is a root of p(x) if and only if (x − c) is a factor of p(x).
Proof: By the Remainder Theorem p(x) = (x − c)q(x) + p(c) and thus p(x) = (x − c)q(x) if and only if p(c) = 0.
For the polynomial p(x) = x^2 − 2 x + 1 the number c = 1 is a root so x − 1 is a factor. But in this case, the other factor is also x − 1 so p(x) = (x − 1)^2. In this case we say c = 1 is a root of multiplicity 2.
1.7. FACTORS AND ROOTS. 19
More generally, we say c is a root of p(x) of multiplicity k if (x − c)k^ divides p(x) but (x − c)k+1^ does not. In other words, c is a root of p(x) of multiplicity k if p(x) = (x − c)kq(x) and c is not a root of q(x). A consequence of the Factor Theorem is
Theorem 1.7.2 If p(x) 6 = 0 has degree n , then p(x) has at most n roots, counted according to multiplicity.
Proof: Use induction on n. The theorem is obvious for n = 0, 1. Suppose the theorem is true for all polynomials of degree less than n and deg p(x) = n. Possibly p(x) has no roots, which does not violate the conclusion of the theorem, but if p(x) has a root of multiplicity k then p(x) = (x − c)kq(x) where deg q(x) = n − k ≤ n. By induction q(x) has roots counted according to multiplicity of at most n − k But by the integral domain property, roots of p(x) are either c or roots of q(x) so the total multiplicity is less than or equal to k + (n − k) = n.
We can now prove a result which implies one we took for granted earlier:
Theorem 1.7.3 If p(x) = a 0 + a 1 x + a 2 x^2 + · · · + anxn^ and q(x) = b 0 + b 1 x + · · · + bmxm^ are two polynomials with p(x) = q(x) for all real (or complex) values of x then deg p(x) = deg q(x) and aj = bj for all j_. In other words if_ p(x) = q(x) as functions, their degrees and coefficients are equal.
Proof: Suppose there is some j with aj 6 = bj , then f (x) = p(x) − q(x) has degree k where k ≥ j ≥ 1 (note a 0 = p(0) = q(0) = b 0 by hypothesis). By Theorem 1.7.2 f (x) has at most k roots, but since p(x) = q(x) for all x, f (x) has infinitely many roots. This contradiction shows our original supposition to be wrong.
Theorem 1.7.2 says p(x) of degree n has at most n roots, but if we are willing to consider complex roots the following theorem is actually true:
Theorem 1.7.4 (Fundamental Theorem of Algebra) If p(x) ∈ C[x] has degree n ≥ 1 then p(x) has exactly n complex roots, counted according to multiplicity.
We will devote Chapter 3 to a discussion of proofs of this theorem and related subjects. In the meantime we can use this when necessary. Since the days of Al Khwarizmi mathematicians have devoted much thought and energy to the problem of actually finding the roots of polynomials. We will devote much of Chapters 2 and 4 to a discussion of the various methods used. The designers
