MATH 2400 LECTURE NOTES: POLYNOMIAL AND RATIONAL FUNCTIONS PETE L. CLARK Contents 1. Polynomial Functions 1 2. Rational Functions 6 1. Polynomial Functions Using the basic operations of addition, subtraction, multiplication, division and composition of functions, we can combine very simple functions to build large and interesting (and useful!) classes of functions. For us, the two simplest kinds of functions are the following: Constant functions: for each a R there is a function C a : R R such that for all x R, C a (x) = a. In other words, the output of the function does not depend on the input: whatever we put in, the same value a will come out. The graph of such a function is the horizontal line y = a. Such functions are called constant. The identity function I : R R by I(x) = x. function is the straight line y = x. The graph of the identity Recall that the identity function is so-called because it is an identity element for the operation of function composition: that is, for any function f : R R we have I f = f I = f. Example: Let m, b R, and consider the function L : R R by x mx + b. Then L is built up out of constant functions and the identity function by addition and multiplication: L = C m I + C b. Example: Let n Z +. The function m n : x x n is built up out of the identity function by repreated multiplication: m n = I I I (n I s altogether). The general name for a function f : R R which is built up out of the identity function and the constant functions by finitely many additions and multiplications is a polynomial. In other words, every polynomial function is of the form (1) f : x a n x n +... + a 1 x + a 0 for some constants a 0,..., a n R. 1
2 PETE L. CLARK However, we also want to take at least until we prove it doesn t make a difference a more algebraic approach to polynomials. Let us define a polynomial expression as an expression of the form n i=0 a ix i. Thus, to give a polynomial expression we need to give for each natural number i a constant a i, while requiring that all but finitely many of these constants are equal to zero: i.e., there exists some n N such that a i = 0 for all i > n. Then every polynomial expression f = n i=0 a ix i determines a polynomial function x f(x). But it is at least conceivable that two different-looking polynomial expressions give rise to the same function. To give some rough idea of what I mean here, consider the two expressions f = 2 arcsin x + 2 arccos x and g = π. Now it turns out for all x [ 1, 1] (the common domain of the arcsin and arccos functions) we have f(x) = π. (The angle θ whose sine is x is complementary to the angle φ whose cosine is x, so arcsin x + arccos x = θ + φ = π 2.) But still f and g are given by different expressions : if I ask you what the coefficient of arcsin x is in the expression f, you will immediately tell me it is 2. If I ask you what the coefficient of arcsin x is in the expression π, you will have no idea what I m talking about. This is the poly- One special polynomial expression is the zero polynomial. nomial whose ith coefficient a i is equal to zero for all i 0. Every nonzero polynomial expression has a degree, which is a natural number, the largest natural number i such that the coefficient a i of x i is nonzero. Thus in (1) the degree of f is n if and only if a n 0: otherwise the degree is smaller than n. Although the zero polynomial expression does not have any natural number as a degree, it is extremely convenient to regard deg 0 as negative, i.e., such that deg 0 is smaller than the degree of any nonzero polynomial. This means that for any d N the set of polynomials of degree at most d includes the zero polynomial. We will follow this convention here but try not to make too big a deal of it. Let us give some examples to solidify this important concept: The polynomials of degree at most 0 are the expressions f = a 0. The corresponding functions are all constant functions: their graphs are horizontal lines. (The graph of the zero polynomial is still a horizontal line, y = 0, so it is useful to include the zero polynomial as having degree at most zero.) The polynomials of degree at most one are the linear expressions L = mx + b. The corresponding functions are linear functions: their graphs are straight lines. The degree of L(x) is one if m 0 i.e., if the line is not horizontal and 0 if m = 0 and b 0. Similarly the polynomials of degree at most two are the quadratic expressions q(x) = ax 2 + bc + c. The degree of q is 2 unless a = 0. We often denote the degree of the polynomial expression f by deg f.
MATH 2400 LECTURE NOTES: POLYNOMIAL AND RATIONAL FUNCTIONS 3 Theorem 1. Let f, g be nonzero polynomial expressions. a) If f + g 0, then deg(f + g) max(deg f, deg g). b) We have deg(fg) = deg f + deg g. c) We have deg(f g) = deg f deg g. Proof. a) Suppose that and f(x) = a m x m +... + a 1 x + a 0, a m 0 g(x) = b n x n +... + b 1 x + b 0, b n 0 so that deg g = m, deg g = n. Case 1: m > n. Then when we add f and g, the highest order term will be a m x m, since the polynomial g only smaller powers of x. In particular the degree of f + g is m = max(m, n). Case 2: m < n. Similarly, when we add f and g, the highest order term will be a n x n, so the degree of f + g is n = max(m, n). Case 3: Suppose m = n. Then (f + g)(x) = (a m + b m )x m +... + (a 1 + b 1 )x + (a 0 + b 0 ). Thus the degree of f + g is at most m. It will be exactly m unless a m + b m = 0, i.e., unless b m = a m ; in this case it will be strictly smaller than m. b) If f and g are as above, then the leading term of f g will be a m b n x m+n, and since a m, b n 0, a m b n 0. Thus deg fg = m + n. c) Unlike the first two parts, this result is not actually very useful, so we leave the proof as an exercise for the interested reader. (To get the flavor of it, note that if f = x m, g = x n then f g = f(x n ) = (x n ) m = x mn. The general case is no harder than this but involves more notation.) From an algebraic perspective, the following result is the most important and fundamental property of polynomials. Theorem 2. (Polynomial Division With Remainder) Let a(x) be a polynomial expression and b(x) be a nonzero polynomial expression. There are unique polynomial expressions q(x) and r(x) such that (i) a(x) = q(x)b(x) + r(x) and (ii) deg r(x) < deg b(x). Note that Theorem 2 is directly analogous to Integer Division with Remainder. In fact the only difference in the statement comes when we want to express the fact that the remainder polynomial r(x) is smaller than the divisor polynomial b(x). The set of polynomial expressions doesn t come with a natural less than relation, so we interpret smaller in terms of the degree. Moreover the proof of Theorem 2 is analogous to that of Integer Division with Remainder: namely in junior / high school algebra we learned an explicit algorithm, the usual long division of polynomials, which when followed yields the quotient q(x) and the remainder r(x). What we probably didn t see in junior / high school is a proof that this algorithm always works and that the quotient and remainder polynomials are unique. We leave the writeup of this as an extra credit exercise. This has many important and useful consequences; here are some of them.
4 PETE L. CLARK Theorem 3. (Root-Factor Theorem) Let f(x) be a polynomial expression and c a real number. The following are equivalent: (i) f(c) = 0. ( c is a root of f. ) (ii) There is some polynomial expression q such that as polynomial expressions, f(x) = (x c)q(x). ( x c is a factor of f. ) Proof. We apply the Division Theorem with a(x) = f(x) and b(x) = x c, getting polynomials q(x) and r(x) such that f(x) = (x c)q(x) + r(x) and r(x) is either the zero polynomial or has deg r < deg x c = 1. In other words, r(x) is in all cases a constant polynomial (perhaps constantly zero), and its constant value can be determined by plugging in x = c: f(c) = (c c)q(c) + r(c) = r(c). The converse is easier: if f(x) = (x c)q(x), then f(c) = (c c)q(c) = 0. Corollary 4. Let f be a nonzero polynomial of degree n. Then the corresponding polynomial function f has at most n real roots: i.e., there are at most n real numbers a such that f(a) = 0. Proof. By induction on n (why not?). Base case (n = 0): If f has degree 0 then it is a nonzero constant function, so it has no roots at all. Induction Step: Let n N, suppose that every polynomial of degree n has at most n real roots, and let f(x) be a polynomial of degree n + 1. If f(x) has no real root, great. Otherwise, there exists a R such that f(a) = 0, and by the Root-Factor Theorem we may write f(x) = (x a)g(x). Moreover by Theorem 1, we have n + 1 = deg f = deg(x a)g(x) = deg(x a) + deg g = 1 + deg g, so deg g = n. Therefore our induction hypothesis applies and g(x) has m distinct real roots a 1,..., a m for some 0 m n. Then f has either m + 1 real roots if a is distinct from all the roots a i of g or m real roots if a = a i for some i, so it has at most m + 1 n + 1 real roots. Lemma 5. Let f = n i=0 a ix i be a polynomial expression. Suppose that the function f(x) = n i=0 a ix i is the zero function: f(x) = 0 for all x R. Then a i = 0 for all i, i.e., f is the zero polynomial expression. Proof. Suppose that f is not the zero polynomial, i.e., a i 0 for some i. Then it has a degree n N, so by Corollary 4 there are at most n real numbers c such that f(c) = 0. But this is absurd: f(x) is the zero function, so for all (infinitely many!) real numbers c we have f(c) = 0. Theorem 6. (Uniqueness Theorem For Polynomials) Let f = a n x n +... + a 1 x + a 0, g = b n x n +... + b 1 x + b 0 be two polynomial expressions. The following are equivalent: (i) f and g are equal as polynomial expressions: for all 0 i n, a i = b i. (ii) f and g are equal as polynomial functions: for all c R, f(c) = g(c). (iii) There are c 1 < c 2 <... < c n+1 such that f(c i ) = g(c i ) for 1 i n + 1.
MATH 2400 LECTURE NOTES: POLYNOMIAL AND RATIONAL FUNCTIONS 5 Proof. (i) = (ii): This is clear, since if a i = b i for all i then f and g are being given by the same expression, so they must give the same function. (ii) = (iii): This is also immediate: since f(c) = g(c) for all real numbers c, we may take for instance c 1 = 1, c 2 = 2,..., c n+1 = n + 1. (iii) = (i): Consider the polynomial expression h = f g = (a n b n )x n +... + (a 1 b 1 )x + (a 0 b 0 ). Then h(c 1 ) = f(c 1 ) g(c 1 ) = 0,..., h(c n+1 ) = f(c n+1 ) g(c n+1 ) = 0. So h is a polynomial of degree at most n which has (at least) n + 1 distinct real roots. By Corollary 4, h must be the zero polynomial expression: that is, for all 0 i n, a i b i = 0. Equivalently, a i = b i for all 0 i n, so f and g are equal as polynomial expressions. In particular, Theorem 6 says that if two polynomials f(x) and g(x) look different i.e., they are not coefficient-by-coefficient the same expression then they are actually different functions, i.e., there is some c R such that f(c) g(c). Finally, we want to prove an arithmetic result about polynomials, the Rational Roots Theorem. For this we need another number-theoretic preliminary. Recall that we say positive integers a and b are coprime (or relatively prime) if they are not both divisible by any integer d > 1 (equivalently, they have no common prime factor). Theorem 7. (Generalized Euclid s Lemma) Let x, y, z Z +. Suppose that x, y are coprime and that yz is divisible by x. Then z is divisible by x. Extra Credit Exercise: a) Prove Theorem 7 using the Fundamental Theorem of Arithmetic (uniqueness of prime factorizations). b) Explain why Theorem 7 is indeed a generalization of Euclid s Lemma. (Hint: let x = p be prime. What does it mean that p, y are coprime?) Theorem 8. (Rational Roots Theorem) Let a 0,..., a n be integers, with a 0, a n 0. Consider the polynomial P (x) = a n x n +... + a 1 x + a 0. Suppose that b c is a rational number, written in lowest terms, which is a root of P : P ( b c ) = 0. Then a 0 is divisible by b and a n is divisible by c. Proof. Well, we know 0 = P ( b c ) = a b n n c n +... + a b 1 c + a 0. Multiplying through by c n clears denominators, giving Rewriting this equation as a n b n + a n 1 b n 1 c +... + a 1 bc n 1 + a 0 c n = 0. a n b n = c( a n 1 b n 1... a 0 c n 1 ) shows that a n b n is divisible by c. But since b and c have no prime factors in common and b n has the same distinct prime factors as does b, b n and c have no prime factors
6 PETE L. CLARK in common and are thus coprime. So Theorem 7 applies to show that a n is divisible by c. Similarly, rewriting the equation as a 0 c n = b( a n b n 1 a n 1 b n 2 c... a 1 c n 1 ) shows that a 0 c n is divisible by b. As above, since b and c are coprime, so are b and c n, so by Theorem 7 a 0 is divisible by b. In high school algebra the Rational Roots Theorem is often employed to generate a finite list of possible rational roots of a polynomial with integer coefficients. This is nice, but the same result can be put to much more impressive use. For instance, taking a n = 1 and noting that 1 is divisible by c iff c = ±1 we get the following result. Corollary 9. Let a 0,..., a n 1 Z, and consider the polynomial P (x) = x n + a n 1 x n 1 +... + a 1 x + a 0. Suppose c is a rational number such that P (c) = 0. Then c is an integer. So what? Well, here s what. Let p be any prime number, n 2 any integer, and consider the polynomial P (x) = x n p. By Corollary 9, if c Q is such that P (c) = 0, then c Z. But if c Z is such that P (c) = 0, then c n = p. But this means that the prime number p is divisible by the integer c, so c = ±1 or c = ±p. But (±1) n = ±1 and (±p) n = ±p n, so c n = p is impossible. So there is no rational number c such that c n = p: that is, n p is irrational. This is a doubly infinite generalization of our first irrationality proof, that 2 is irrational, but the argument is, if anything, shorter and easier to understand. (However, we did use and prove, by induction the Fundamental Theorem of Arithmetic, a tool which was not available to us at the very beginning of the course.) Moral: polynomials can be useful in surprising ways! 2. Rational Functions A rational function is a functtion which is a quotient of two polynomial functions: f(x) = P (x) Q(x), with Q(x) not the zero function. Other than this definition and the remark that a natural domain of a rational function is the set of all real numbers except the finitely many roots of Q(x), I do not believe that I said anything about rational functions at this point in the course. This is a happy coincidence, since I have nothing in this section of my notes!