Split (1)

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A1 =   1 1 0 −1 1 0 2 5 4    A2 =  3 1 2 0 0 5 2 0 4    A3 =  3 1 0 −1 1 2   1 0 0 5We are developing a method in the forthcoming discussion. As with the discussion in Section 8.4 when we developed the ﬁrst algorithm to ﬁnd matrix inverses, we ask that you indulge us. 6The reader is encouraged to stop and think this through. 8.5 Determinants and Cramer’s Rule 621 If we expand det (A1) along the ﬁrst row, we get det (A1) = det = det −1 5 1 4 −1 5 1 4 − det 0 5 0 4 + 2 det 0 −1 1 0 Amazingly, this is none other than the C11 cofactor of A. The reader is invited to check this, as well as the claims that det (A2) = C12 and det (A3) = C13.7 (To see this, though it seems unnatural to do so, expand along the ﬁrst row.) Cramer’s Rule tells us x11 = det (A1) det(A) = C11 det(A), x21 = det (A2) det(A) = C12 det(A), x31 = det (A3) det(A) = C13 det(A) So the ﬁrst column of the inverse matrix X is:   =   x11 x21 x31                   C11 det(A) C12 det(A) C13 det(A) = 1 det(A)     C11 C12 C13 Notice the reversal of the subscripts going from the unknown to the corresponding cofactor of A. This trend continues and we get   =   x12 x22
x32 1 det(A)     C21 C22 C23   =   x13 x23 x33 1 det(A)     C31 C32 C33 Putting all of these together, we have obtained a new and surprising formula for A−1, namely A−1 = 1 det(A)   C11 C21 C31 C12 C22 C32 C13 C23 C33   To see that this does indeed yield A−1, we ﬁnd all of the cofactors of A C11 = −9, C21 = −2, C31 = C12 = 10, C22 = C13 = 7 8, C32 = −15 2, C23 = −1, C33 = −3 And, as promised, 7In a solid Linear Algebra course you will learn that the properties in Theorem 8.7 hold equally well if the word ‘row’ is replaced by the word ‘column’. We’re not going to get into column operations in this text, but they do make some of what we’re trying to say easier to follow. 622 Systems of Equations and Matrices A−1 = 1 det(A)   C11 C21 C31 C12 C22 C32 C13 C23 C33   = −   1 13 −9 −2 7 8 −15 10 2 −1 −3   =    2 13 − 7 9 13 − 10 − 2 13 13 − 8 13 1 13    13 15 13 3 13 To generalize this to invertible n × n matrices, we need another deﬁnition and a theorem. Our deﬁnition gives a special name to the cofactor matrix, and the theorem tells us how to use it along with det(A) to ﬁnd the inverse of a matrix. Deﬁnition 8.15. Let A be an n × n matrix, and Cij denote the ij cofactor of A. The adjoint of A, denoted adj(A) is the
matrix whose ij-entry is the ji cofactor of A, Cji. That is adj(A) =      C11 C21 C12 C22...... C1n C2n... Cn1... Cn2...... Cnn      This new notation greatly shortens the statement of the formula for the inverse of a matrix. Theorem 8.9. Let A be an invertible n × n matrix. Then A−1 = 1 det(A) adj(A) For 2 × 2 matrices, Theorem 8.9 reduces to a fairly simple formula. Equation 8.2. For an invertible 2 × 2 matrix, a b c d −1 = 1 ad − bc d −b a −c The proof of Theorem 8.9 is, like so many of the results in this section, best left to a course in Linear Algebra. In such a course, not only do you gain some more sophisticated proof techniques, you also gain a larger perspective. The authors assure you that persistence pays oﬀ. If you stick around a few semesters and take a course in Linear Algebra, you’ll see just how pretty all things matrix really are - in spite of the tedious notation and sea of subscripts. Within the scope of this text, we will prove a few results involving determinants in Section 9.3 once we have the Principle of Mathematical Induction well in hand. Until then, make sure you have a handle on the mechanics of matrices and the theory will come eventually. 8.5 Determinants and Cramer’s Rule 623 8.5.3 Exercises In Exercises 1 - 8, compute the determinant of the given matrix. (Some of these matrices appeared in Exercises 1 - 8 in Section 8.4.) 1. B = 12 −7 −5 3 3. Q = x x2 1 2x     5. F = 73 4 −3 6 2 i −1 k j 5 0 9 −4 −2   2. C = 6 15 14 35         
 4. L = 6. G = 8. H = 1 x3 3 x4     ln(x) x3 1 − 3 ln(x) x4   2 3 3 11 4 19 − 1 2 3 1 2 −2 0 0 0 −3 0 8 7 16 0 4 1 −5 1     In Exercises 9 - 14, use Cramer’s Rule to solve the system of linear equations. 3x + 7y = 26 5x + 12y = 39 9. 11. 13.    x + y = 8000 0.03x + 0.05y = 250 x + y + z = 3 2x − y + z = 0 −3x + 5y + 7z = 7 In Exercises 15 - 16, use Cramer’s Rule to solve for x4. 15.    x1 − x3 = −2 2x2 − x4 = 0 x1 − 2x2 + x3 = 0 −x3 + x4 = 1 16. 10. 2x − 4y = 5 10x + 13y = −6 12 6x + 7y = 3 14.       3x + y − 2z = 10 4x − y + z = 5 x − 3y − 4z = −1 4x1 + x2 = x2 − 3x3 = 10x1 + x3 + x4 = 4 1 0 −x2 + x3 = −3 624 Systems of Equations and Matrices In Exercises 17 - 18, ﬁnd the inverse of the given matrix using their determinants and adjoints. 17. B = 12 −7 −5 3 183 4 −3 6 2 19. Carl’s Sasquatch Attack! Game Card Collection is a mixture of common and rare cards. Each common card is worth $0.25 while each rare card is worth $0.75. If his entire 117 card collection is worth $48.75, how many of each kind of card does he own
? 20. How much of a 5 gallon 40% salt solution should be replaced with pure water to obtain 5 gallons of a 15% solution? 21. How much of a 10 liter 30% acid solution must be replaced with pure acid to obtain 10 liters of a 50% solution? 22. Daniel’s Exotic Animal Rescue houses snakes, tarantulas and scorpions. When asked how many animals of each kind he boards, Daniel answered: ‘We board 49 total animals, and I am responsible for each of their 272 legs and 28 tails.’ How many of each animal does the Rescue board? (Recall: tarantulas have 8 legs and no tails, scorpions have 8 legs and one tail, and snakes have no legs and one tail.) 23. This exercise is a continuation of Exercise 16 in Section 8.4. Just because a system is consistent independent doesn’t mean it will admit a solution that makes sense in an applied setting. Using the nutrient values given for Ippizuti Fish, Misty Mushrooms, and Sun Berries, use Cramer’s Rule to determine the number of servings of Ippizuti Fish needed to meet the needs of a daily diet which requires 2500 calories, 1000 grams of protein, and 400 milligrams of Vitamin X. Now use Cramer’s Rule to ﬁnd the number of servings of Misty Mushrooms required. Does a solution to this diet problem exist? 11 −7 −3 15 9 6 1 −5 9 6 −7 11 24. Let R =, and a) Show that det(RS) = det(R) det(S) (b) Show that det(T ) = − det(R) (c) Show that det(U ) = −3 det(S) 25. For M, N, and P below, show that det(M ) = 0, det(N ) = 0 and det(P ) = 02 −4 −6 9 8 7   8.5 Determinants and Cramer’s Rule 625 26. Let A be an arbitrary invertible 3 × 3 matrix. (a) Show that det(I3) = 1. (See footnote8 below.) (b) Using the facts that AA−1 = I3 and det(AA−1) = det(A) det(A−1), show that det(A−1) = 1 det(A) The purpose
of Exercises 27 - 30 is to introduce you to the eigenvalues and eigenvectors of a matrix.9 We begin with an example using a 2 × 2 matrix and then guide you through some exercises using a 3 × 3 matrix. Consider the matrix C = 6 15 14 35 from Exercise 2. We know that det(C) = 0 which means that CX = 02×2 does not have a unique solution. So there is a nonzero matrix Y with CY = 02×2. In fact, every matrix of the form Y = − 5 2 t t is a solution to CX = 02×2, so there are inﬁnitely many matrices such that CX = 02×2. But consider the matrix X41 = 3 7 It is NOT a solution to CX = 02×2, but rather, 3 7 6 15 14 35 CX41 = = 123 287 = 41 3 7 In fact, if Z is of the form Z = 3 7 t t then CZ = 6 15 14 35 3 7 t t = 123 7 t 41t = 41 3 7 t t = 41Z for all t. The big question is “How did we know to use 41?” We need a number λ such that CX = λX has nonzero solutions. We have demonstrated that λ = 0 and λ = 41 both worked. Are there others? If we look at the matrix equation more closely, what 8If you think about it for just a moment, you’ll see that det(In) = 1 for any natural number n. The formal proof of this fact requires the Principle of Mathematical Induction (Section 9.3) so we’ll stick with n = 3 for the time being. 9This material is usually given its own chapter in a Linear Algebra book so clearly we’re not able to tell you everything you need to know about eigenvalues and eigenvectors. They are a nice application of determinants, though, so we’re going to give you enough background so that you can start playing around with them. 626 Systems of Equations and Matrices we really wanted was a nonzero solution to (C − λI2)X = 02×2 which we know exists if and only if the determinant of C − λI2 is zero.10 So we computed det(C − λI2) = det 6 −
λ 15 14 35 − λ = (6 − λ)(35 − λ) − 14 · 15 = λ2 − 41λ This is called the characteristic polynomial of the matrix C and it has two zeros: λ = 0 and λ = 41. That’s how we knew to use 41 in our work above. The fact that λ = 0 showed up as one of the zeros of the characteristic polynomial just means that C itself had determinant zero which we already knew. Those two numbers are called the eigenvalues of C. The corresponding matrix solutions to CX = λX are called the eigenvectors of C and the ‘vector’ portion of the name will make more sense after you’ve studied vectors. Now it’s your turn. In the following exercises, you’ll be using the matrix G from Exercise 6 11 4 19 27. Show that the characteristic polynomial of G is p(λ) = −λ(λ − 1)(λ − 22). That is, compute det (G − λI3). 28. Let G0 = G. Find the parametric description of the solution to the system of linear equations given by GX = 03×3. 29. Let G1 = G − I3. Find the parametric description of the solution to the system of linear equations given by G1X = 03×3. Show that any solution to G1X = 03×3 also has the property that GX = 1X. 30. Let G22 = G − 22I3. Find the parametric description of the solution to the system of linear equations given by G22X = 03×3. Show that any solution to G22X = 03×3 also has the property that GX = 22X. 10Think about this. 8.5 Determinants and Cramer’s Rule 627 8.5.4 Answers 1. det(B) = 1 3. det(Q) = x2 5. det(F ) = −12 2. det(C) = 0 4. det(L) = 1 x7 6. det(G) = 0 7. det(V ) = 20i + 43j + 4k 8. det(H) = −2 9. x = 39, y = −13 11. x = 7500, y = 500 13. x = 1, y
= 2, z = 0 15. x4 = 4 17. B−1 = 3 7 5 12 18. F − 10. x = 41 66, y = − 31 33 12. x = 76 47, y = − 45 47 14. x = 121 60, y = 131 60, z = − 53 60 16. x4 = −1    19. Carl owns 78 common cards and 39 rare cards. 20. 3.125 gallons. 21. 20 7 ≈ 2.85 liters. 22. The rescue houses 15 snakes, 21 tarantulas and 13 scorpions. 23. Using Cramer’s Rule, we ﬁnd we need 53 servings of Ippizuti Fish to satisfy the dietary requirements. The number of servings of Misty Mushrooms required, however, is −1120. Since it’s impossible to have a negative number of servings, there is no solution to the applied problem, despite there being a solution to the mathematical problem. A cautionary tale about using Cramer’s Rule: just because you are guaranteed a mathematical answer for each variable doesn’t mean the solution will make sense in the ‘real’ world. 628 Systems of Equations and Matrices 8.6 Partial Fraction Decomposition This section uses systems of linear equations to rewrite rational functions in a form more palatable to Calculus students. In College Algebra, the function f (x) = x2 − x − 6 x4 + x2 (1) is written in the best form possible to construct a sign diagram and to ﬁnd zeros and asymptotes, but certain applications in Calculus require us to rewrite f (x) as f (x) = x + 7 x2 + 1 − 1 x − 6 x2 (2) If we are given the form of f (x) in (2), it is a matter of Intermediate Algebra to determine a common denominator to obtain the form of f (x) given in (1). The focus of this section is to develop a method by which we start with f (x) in the form of (1) and ‘resolve it into partial fractions’ to obtain the form in (2). Essentially, we need to reverse the least common denominator process. Starting with the form of f (x) in (1), we begin by factoring the denominator x2 − x −
6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) We now think about which individual denominators could contribute to obtain x2 x2 + 1 as the least common denominator. Certainly x2 and x2 + 1, but are there any other factors? Since x2 + 1 is an irreducible quadratic1 there are no factors of it that have real coeﬃcients which can contribute to the denominator. The factor x2, however, is not irreducible, since we can think of it as x2 = xx = (x − 0)(x − 0), a so-called ‘repeated’ linear factor.2 This means it’s possible that a term with a denominator of just x contributed to the expression as well. What about something like x x2 + 1? This, too, could contribute, but we would then wish to break down that denominator into x and x2 + 1, so we leave out a term of that form. At this stage, we have guessed x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) =? x +? x2 +? x2 + 1 Our next task is to determine what form the unknown numerators take. It stands to reason that since the expression x2−x−6 is ‘proper’ in the sense that the degree of the numerator is less than x4+x2 the degree of the denominator, we are safe to make the ansatz that all of the partial fraction resolvents are also. This means that the numerator of the fraction with x as its denominator is just a constant and the numerators on the terms involving the denominators x2 and x2 + 1 are at most linear polynomials. That is, we guess that there are real numbers A, B, C, D and E so that x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) = A x + Bx + C x2 + Dx + E x2 + 1 1Recall this means it has no real zeros; see Section 3.4. 2Recall this means x = 0 is a zero of multiplicity 2. 8.6 Partial Fraction Decomposition 629 However, if we look more closely at
the term Bx+C term B Hence, we drop it and, after re-labeling, we ﬁnd ourselves with our new guess: x2. The x which means it contributes nothing new to our expansion. x has the same form as the term A, we see that Bx+C x2 = Bx x2 + C x2 = B x + C x2 x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) = A x + B x2 + Cx + D x2 + 1 Our next task is to determine the values of our unknowns. Clearing denominators gives x2 − x − 6 = Ax x2 + 1 + B x2 + 1 + (Cx + D)x2 Gathering the like powers of x we have x2 − x − 6 = (A + C)x3 + (B + D)x2 + Ax + B In order for this to hold for all values of x in the domain of f, we equate the coeﬃcients of corresponding powers of x on each side of the equation3 and obtain the system of linear equations    (E1) A + C = (E2) B + D = (E3) (E4) 0 From equating coeﬃcients of x3 1 From equating coeﬃcients of x2 A = −1 From equating coeﬃcients of x B = −6 From equating the constant terms To solve this system of equations, we could use any of the methods presented in Sections 8.1 through 8.5, but none of these methods are as eﬃcient as the good old-fashioned substitution you learned in Intermediate Algebra. From E3, we have A = −1 and we substitute this into E1 to get C = 1. Similarly, since E4 gives us B = −6, we have from E2 that D = 7. We get x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) = − 1 x − 6 x2 + x + 7 x2 + 1 which matches the formula given in (2). As we have seen in this opening example, resolving a rational function into partial fractions takes
two steps: ﬁrst, we need to determine the form of the decomposition, and then we need to determine the unknown coeﬃcients which appear in said form. Theorem 3.16 guarantees that any polynomial with real coeﬃcients can be factored over the real numbers as a product of linear factors and irreducible quadratic factors. Once we have this factorization of the denominator of a rational function, the next theorem tells us the form the decomposition takes. The reader is encouraged to review the Factor Theorem (Theorem 3.6) and its connection to the role of multiplicity to fully appreciate the statement of the following theorem. 3We will justify this shortly. 630 Systems of Equations and Matrices Theorem 8.10. Suppose R(x) = than the degree of D(x) and N (x) and D(x) have no common factors.a is a rational function where the degree of N (x) less N (x) D(x) If α is a real zero of D of multiplicity m which corresponds to the linear factor ax + b, the partial fraction decomposition includes A1 ax + b + A2 (ax + b)2 +... + Am (ax + b)m for real numbers A1, A2,... Am. If α is a non-real zero of D of multiplicity m which corresponds to the irreducible quadratic ax2 + bx + c, the partial fraction decomposition includes B1x + C1 ax2 + bx + c + B2x + C2 (ax2 + bx + c)2 +... + Bmx + Cm (ax2 + bx + c)m for real numbers B1, B2,... Bm and C1, C2,... Cm. aIn other words, R(x) is a proper rational function which has been fully reduced. The proof of Theorem 8.10 is best left to a course in Abstract Algebra. Notice that the theorem provides for the general case, so we need to use subscripts, A1, A2, etc., to denote diﬀerent unknown coeﬃcients as opposed to the usual convention of A, B, etc.. The stress on multiplicities is to help us correctly group factors in the denominator
. For example, consider the rational function 3x − 1 (x2 − 1) (2 − x − x2) Factoring the denominator to ﬁnd the zeros, we get (x + 1)(x − 1)(1 − x)(2 + x). We ﬁnd x = −1 and x = −2 are zeros of multiplicity one but that x = 1 is a zero of multiplicity two due to the two diﬀerent factors (x − 1) and (1 − x). One way to handle this is to note that (1 − x) = −(x − 1) so 3x − 1 (x + 1)(x − 1)(1 − x)(2 + x) = 3x − 1 −(x − 1)2(x + 1)(x + 2) = 1 − 3x (x − 1)2(x + 1)(x + 2) from which we proceed with the partial fraction decomposition 1 − 3x (x − 1)2(x + 1)(x + 2) = A x − 1 + B (x − 1) Turning our attention to non-real zeros, we note that the tool of choice to determine the irreducibility of a quadratic ax2 + bx + c is the discriminant, b2 − 4ac. If b2 − 4ac < 0, the quadratic admits a pair of non-real complex conjugate zeros. Even though one irreducible quadratic gives two distinct non-real zeros, we list the terms with denominators involving a given irreducible quadratic only once to avoid duplication in the form of the decomposition. The trick, of course, is factoring the 8.6 Partial Fraction Decomposition 631 denominator or otherwise ﬁnding the zeros and their multiplicities in order to apply Theorem 8.10. We recommend that the reader review the techniques set forth in Sections 3.3 and 3.4. Next, we state a theorem that if two polynomials are equal, the corresponding coeﬃcients of the like powers of x are equal. This is the principal by which we shall determine the unknown coeﬃcients in our partial fraction decomposition. Theorem 8.11. Suppose anxn + an−1xn−1 + · · · + a2x
2 + a1x + a0 = bmxm + mm−1xm−1 + · · · + b2x2 + b1x + b0 for all x in an open interval I. Then n = m and ai = bi for all i = 1... n. Believe it or not, the proof of Theorem 8.11 is a consequence of Theorem 3.14. Deﬁne p(x) to be the diﬀerence of the left hand side of the equation in Theorem 8.11 and the right hand side. Then p(x) = 0 for all x in the open interval I. If p(x) were a nonzero polynomial of degree k, then, by Theorem 3.14, p could have at most k zeros in I, and k is a ﬁnite number. Since p(x) = 0 for all the x in I, p has inﬁnitely many zeros, and hence, p is the zero polynomial. This means there can be no nonzero terms in p(x) and the theorem follows. Arguably, the best way to make sense of either of the two preceding theorems is to work some examples. Example 8.6.1. Resolve the following rational functions into partial fractions. 1. R(x) = x + 5 2x2 − x − 1 4. R(x) = 4x3 x2 − 2 Solution. 2. R(x) = 5. R(x) = 3 x3 − 2x2 + x x3 + 5x − 1 x4 + 6x2 + 9 3. R(x) = 6. R(x) = 3 x3 − x2 + x 8x2 x4 + 16 1. We begin by factoring the denominator to ﬁnd 2x2 − x − 1 = (2x + 1)(x − 1). We get x = − 1 2 and x = 1 are both zeros of multiplicity one and thus we know x + 5 2x2 − x − 1 = x + 5 (2x + 1)(x − 1) = A 2x + 1 + B x − 1 Clearing denominators, we get x+5 = A(x−1)+B(2x+1) so that x+5 = (A+2B
)x+B −A. Equating coeﬃcients, we get the system A + 2B = 1 −A + B = 5 This system is readily handled using the Addition Method from Section 8.1, and after adding both equations, we get 3B = 6 so B = 2. Using back substitution, we ﬁnd A = −3. Our answer is easily checked by getting a common denominator and adding the fractions. x + 5 2x2 − 2x + 1 632 Systems of Equations and Matrices 2. Factoring the denominator gives x3 − 2x2 + x = x x2 − 2x + 1 = x(x − 1)2 which gives x = 0 as a zero of multiplicity one and x = 1 as a zero of multiplicity two. We have 3 x3 − 2x2 + x = 3 x(x − 1)x − 1)2 Clearing denominators, we get 3 = A(x − 1)2 + Bx(x − 1) + Cx, which, after gathering up the like terms becomes 3 = (A + B)x2 + (−2A − B + C)x + A. Our system is    A + B = 0 −2A − B + C = 0 A = 3 Substituting A = 3 into A + B = 0 gives B = −3, and substituting both for A and B in −2A − B + C = 0 gives C = 3. Our ﬁnal answer is 3 x3 − 2x2 + x − 1)2 3. The denominator factors as x x2 − x + 1. We see immediately that x = 0 is a zero of multiplicity one, but the zeros of x2 − x + 1 aren’t as easy to discern. The quadratic doesn’t factor easily, so we check the discriminant and ﬁnd it to be (−1)2 − 4(1)(1) = −3 < 0. We ﬁnd its zeros are not real so it is an irreducible quadratic. The form of the partial fraction decomposition is then 3 x3 − x2 + x = 3 x (x2 − x + 1) = A x + Bx + C x2 − x + 1 Proceeding as usual, we clear
denominators and get 3 = A x2 − x + 1 + (Bx + C)x or 3 = (A + B)x2 + (−A + C)x + A. We get    A + B = 0 −A + C = 0 A = 3 From A = 3 and A + B = 0, we get B = −3. From −A + C = 0, we get C = A = 3. We get 3 x3 − x2 + x = 3 x + 3 − 3x x2 − x + 1 4. Since 4x3 x2−2 isn’t proper, we use long division and we get a quotient of 4x with a remainder of 8x. That is, 4x3 x2−2 so we now work on resolving 8x x2−2 into partial fractions. The quadratic x2 −2, though it doesn’t factor nicely, is, nevertheless, reducible. Solving x2 −2 = 0 x2−2 = 4x + 8x 8.6 Partial Fraction Decomposition 633 √ gives us x = ± enables us to now factor x2 − 2 = x − 2, and each of these zeros must be of multiplicity one since Theorem 3.14 2. Hence, 2 x + √ √ 8x x2 − 2 = x − √ √ 8x or 8x = (A + B)x + (A − B) √ 2. Clearing fractions, we get 8x = A x + We get the system √ A + B = 8 2 = 0 (A − B) √ From (A − B) Hence, A = B = 4 and we get 2 = 0, we get A = B, which, when substituted into A + B = 8 gives B = 4. 4x3 x2 − 2 = 4x + 8x x2 − 2 = 4x +. At ﬁrst glance, the denominator D(x) = x4 + 6x2 + 9 appears irreducible. However, D(x) has three terms, and the exponent on the ﬁrst term is exactly twice that of the second. Rewriting + 6x2 + 9, we see it is a quadratic in disguise and factor D(x) = x2 + 32
D(x) = x22. Since x2 + 3 clearly has no real zeros, it is irreducible and the form of the decomposition is x3 + 5x − 1 x4 + 6x2 + 9 x3 + 5x − 1 (x2 + 3)2 = When we clear denominators, we ﬁnd x3 + 5x − 1 = (Ax + B) x2 + 3 + Cx + D which yields x3 + 5x − 1 = Ax3 + Bx2 + (3A + C)x + 3B + D. Our system is Cx + D (x2 + 3)2 Ax + B x2 + 3 = +    A = 1 B = 0 5 3A + C = 3B + D = −1 We have A = 1 and B = 0 from which we get C = 2 and D = −1. Our ﬁnal answer is x3 + 5x − 1 x4 + 6x2 + 9 = x x2 + 3 + 2x − 1 (x2 + 3)2 6. Once again, the diﬃculty in our last example is factoring the denominator. In an attempt to get a quadratic in disguise, we write x4 + 16 = x22 + 42 = x22 + 8x2 + 42 − 8x2 = x2 + 42 − 8x2 634 Systems of Equations and Matrices and obtain a diﬀerence of two squares: x2 + 42 and 8x2 = 2x √ 22. Hence, x4 + 16 = x2 + 4 − 2x √ 2 x2 + 4 + 2x √ 2 = x2 − 2x √ 2 + 4 x2 + 2x √ 2 + 4 The discrimant of both of these quadratics works out to be −8 < 0, which means they are irreducible. We leave it to the reader to verify that, despite having the same discriminant, these quadratics have diﬀerent zeros. The partial fraction decomposition takes the form 8x2 x4 + 16 = √ x2 − 2x √ 8x2 2 + 4 x2 + 2x √ 2 + 4 = 2 + 4 + (C
x + D) x2 − 2x x2 − 2x Ax + B √ √ 2 + 4 or + Cx + D √ x2 + 2x 2 + 4 2 + 4 We get 8x2 = (Ax + B) x2 + 2x 8x2 = (A + C)x3 + (2A √ 2 + B − 2C √ 2 + D)x2 + (4A + 2B √ 2 + 4C − 2D √ 2)x + 4B + 4D which gives the system    √ 2A 4A + 2B 2 + B − 2C √ 2 + 4C − 2D √ 4B + 4D = 0 √ We choose substitution as the weapon of choice to solve this system. From A + C = 0, we get A = −C; from 4B + 4D = 0, we get B = −D. Substituting these into the remaining two equations, we get −2C √ √ 2 − D − 2C √ 2 + 4C − 2D √ 2 + D = 8 2 = 0 or −4C − 2D √ √ −4C −4D 2 = 8 2 = 0 We get C = − √ 2 so that A = −C = √ 2 and D = 0 which means B = −D = 0. We get 8x2 x4 + 16 = √ x x2 − 2x 2 √ − 2 + 4 √ x x2 + 2x 2 √ 2 + 4 8.6 Partial Fraction Decomposition 635 8.6.1 Exercises In Exercises 1 - 6, ﬁnd only the form needed to begin the process of partial fraction decomposition. Do not create the system of linear equations or attempt to ﬁnd the actual decomposition. 1. 3. 5. 7 (x − 3)(x + 5) m (7x − 6)(x2 + 9) A polynomial of degree < 9 (x + 4)5(x2 + 1)2 2. 4. 6. 5x + 4 x(x − 2)(2 − x) ax2 + bx + c x3(5x + 9)(3x2 + 7
x + 9) A polynomial of degree < 7 x(4x − 1)2(x2 + 5)(9x2 + 16) In Exercises 7 - 18, ﬁnd the partial fraction decomposition of the following rational expressions. 7. 9. 11. 13. 15. 17. 2x x2 − 1 11x2 − 5x − 10 5x3 − 5x2 −x2 + 15 4x4 + 40x2 + 36 5x4 − 34x3 + 70x2 − 33x − 19 (x − 3)2 −7x2 − 76x − 208 x3 + 18x2 + 108x + 216 4x3 − 9x2 + 12x + 12 x4 − 4x3 + 8x2 − 16x + 16 8. 10. 12. 14. 16. 18. −7x + 43 3x2 + 19x − 14 −2x2 + 20x − 68 x3 + 4x2 + 4x + 16 −21x2 + x − 16 3x3 + 4x2 − 3x + 2 x6 + 5x5 + 16x4 + 80x3 − 2x2 + 6x − 43 x3 + 5x2 + 16x + 80 −10x4 + x3 − 19x2 + x − 10 x5 + 2x3 + x 2x2 + 3x + 14 (x2 + 2x + 9)(x2 + x + 5) 19. As we stated at the beginning of this section, the technique of resolving a rational function into partial fractions is a skill needed for Calculus. However, we hope to have shown you that it is worth doing if, for no other reason, it reinforces a hefty amount of algebra. One of the common algebraic errors the authors ﬁnd students make is something along the lines of Think about why if the above were true, this section would have no need to exist. 8 x2 − 9 = 8 x2 − 8 9 636 Systems of Equations and Matrices 8.6.2 Answers 1. 3. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18 7x − 6 A x + 4 A x + 2x x2 − 1 + + Bx + C x2 + 9 C B (x + 4)3 + (x
+ 4)2 + C B (4x − 1)2 + 4x − x + 4)4 + E (x + 4)5 + Dx + E x2 + 5 + F x + G 9x2 + 16 2. 4x − 2)2 + B x2 + F x + G x2 + 1 C x3 + + D 5x + 9 Hx + I (x2 + 1)2 + Ex + F 3x2 + 7x + 9 −7x + 43 3x2 + 19x − 14 11x2 − 5x − 10 5x3 − 5x2 = = 5 3x − x2 − 4 5(x − 1) −2x2 + 20x − 68 x3 + 4x2 + 4x + 16 = − 9 x + 4 + 7x − 8 x2 + 4 −x2 + 15 4x4 + 40x2 + 36 = 1 2(x2 + 1) − 3 4(x2 + 9) = − −21x2 + x − 16 6 3x3 + 4x2 − 3x + 2 x + 2 5x4 − 34x3 + 70x2 − 33x − 19 (x − 3)2 − 3x + 5 3x2 − 2x + 1 = 5x2 − 4x + 1 + 9 x − 3 − 1 (x − 3)2 x6 + 5x5 + 16x4 + 80x3 − 2x2 + 6x − 43 x3 + 5x2 + 16x + 80 = x3 + x + 1 x2 + 16 − 3 x + 5 −7x2 − 76x − 208 x3 + 18x2 + 108x + 216 = − 7 x + 6 + 8 (x + 6)2 − 4 (x + 6)3 −10x4 + x3 − 19x2 + x − 10 x5 + 2x3 + x = − 10 x + 1 x2 + 1 + x (x2 + 1)2 4x3 − 9x2 + 12x + 12 x4 − 4x3 + 8x2 − 16x + 16 = 1 x − 2 + 4 (x − 2)2 + 3x + 1 x2 + 4 2x2 + 3x + 14 (x2 + 2x + 9)(x2 + x + 5) = 1 x2 + 2
x + 9 + 1 x2 + x + 5 8.7 Systems of Non-Linear Equations and Inequalities 637 8.7 Systems of Non-Linear Equations and Inequalities In this section, we study systems of non-linear equations and inequalities. Unlike the systems of linear equations for which we have developed several algorithmic solution techniques, there is no general algorithm to solve systems of non-linear equations. Moreover, all of the usual hazards of non-linear equations like extraneous solutions and unusual function domains are once again present. Along with the tried and true techniques of substitution and elimination, we shall often need equal parts tenacity and ingenuity to see a problem through to the end. You may ﬁnd it necessary to review topics throughout the text which pertain to solving equations involving the various functions we have studied thus far. To get the section rolling we begin with a fairly routine example. Example 8.7.1. Solve the following systems of equations. Verify your answers algebraically and graphically. x2 + y2 = 4 4x2 + 9y2 = 36 x2 + y2 = 4 4x2 − 9y2 = 36 1. 2. Solution: x2 + y2 = 4 y − 2x = 0 x2 + y2 = 4 y − x2 = 0 3. 4. 1. Since both equations contain x2 and y2 only, we can eliminate one of the variables as we did in Section 8.1. (E1) x2 + y2 = 4 (E2) 4x2 + 9y2 = 36 Replace E2 with −−−−−−−−−−→ −4E1 + E2 (E1) x2 + y2 = 4 5y2 = 20 (E2) From 5y2 = 20, we get y2 = 4 or y = ±2. To ﬁnd the associated x values, we substitute each value of y into one of the equations to ﬁnd the resulting value of x. Choosing x2 + y2 = 4, we ﬁnd that for both y = −2 and y = 2, we get x = 0. Our solution is thus {(0, 2), (0, −2)}. To check this algebraically, we need to show that both points satisfy both of the original equations. We leave it to the reader to verify this. To check our answer graph
ically, we sketch both equations and look for their points of intersection. The graph of x2 + y2 = 4 is a circle centered at (0, 0) with a radius of 2, whereas the graph of 4x2 + 9y2 = 36, when written in the standard form x2 4 = 1 is easily recognized as an ellipse centered at (0, 0) with a major axis along the x-axis of length 6 and a minor axis along the y-axis of length 4. We see from the graph that the two curves intersect at their y-intercepts only, (0, ±2). 9 + y2 2. We proceed as before to eliminate one of the variables (E1) x2 + y2 = 4 (E2) 4x2 − 9y2 = 36 Replace E2 with −−−−−−−−−−→ −4E1 + E2 (E1) x2 + y2 = 4 (E2) −13y2 = 20 638 Systems of Equations and Matrices Since the equation −13y2 = 20 admits no real solution, the system is inconsistent. To verify this graphically, we note that x2 + y2 = 4 is the same circle as before, but when writing the second equation in standard form, x2 4 = 1, we ﬁnd a hyperbola centered at (0, 0) opening to the left and right with a transverse axis of length 6 and a conjugate axis of length 4. We see that the circle and the hyperbola have no points in common. 9 − y2 y 1 −1 −3 −2 −1 1 2 3 x −3 −2 −1 y 1 −1 1 2 3 x Graphs for x2 + y2 = 4 4x2 + 9y2 = 36 Graphs for x2 + y2 = 4 4x2 − 9y2 = 36 √ 3. Since there are no like terms among the two equations, elimination won’t do us any good. We turn to substitution and from the equation y − 2x = 0, we get y = 2x. Substituting this √ into x2 + y2 = 4 gives x2 + (2x)2 = 4. Solving, we ﬁnd 5x2 = 4 or x = ± 2 5 5. Returning √ 5 when x = 2 5 to the
equation we used for the substitution, y = 2x, we ﬁnd y = 4 5, so one solution is. We. Similarly, we ﬁnd the other solution to be leave it to the reader that both points satisfy both equations, so that our ﬁnal answer is 2. The graph of x2 + y2 = 4 is our circle from before and the graph of y − 2x = 0 is a line through the origin with slope 2. Though we cannot verify the numerical values of the points of intersection from our sketch, we do see that we have two solutions: one in Quadrant I and one in Quadrant III as required. While it may be tempting to solve y − x2 = 0 as y = x2 and substitute, we note that this system is set up for elimination.1 (E1) x2 + y2 = 4 y − x2 = 0 (E2) Replace E2 with −−−−−−−−−−→ E1 + E2 (E1) x2 + y2 = 4 y2 + y = 4 (E2) From y2 + y = 4 we get y2 + y − 4 = 0 which gives y = −1±. Due to the complicated 2 nature of these answers, it is worth our time to make a quick sketch of both equations to head oﬀ any extraneous solutions we may encounter. We see that the circle x2 + y2 = 4 intersects the parabola y = x2 exactly twice, and both of these points have a positive y value. Of the two solutions for y, only y = −1+ is positive, so to get our solution, we substitute this 2 17 17 √ √ 1We encourage the reader to solve the system using substitution to see that you get the same solution. 8.7 Systems of Non-Linear Equations and Inequalities 639 into y − x2 = 0 and solve for x. We get 2+2 2 17, −1+ 2 17, − −2+2 2 17, −1+ 2 17 √ −1+ 2 17 = ± √ √ −2+2 2 17. Our solution is, which we leave to the reader to verify. y 1 −3 −2 −1 1 2 3 x −3 −2 −1 y 1 −1 1 2 3 x Graphs for x2 + y2 = 4 y −
2x = 0 Graphs for x2 + y2 = 4 y − x2 = 36 A couple of remarks about Example 8.7.1 are in order. First note that, unlike systems of linear equations, it is possible for a system of non-linear equations to have more than one solution without having inﬁnitely many solutions. In fact, while we characterize systems of nonlinear equations as being ‘consistent’ or ‘inconsistent,’ we generally don’t use the labels ‘dependent’ or ‘independent’. Secondly, as we saw with number 4, sometimes making a quick sketch of the problem situation can save a lot of time and eﬀort. While in general the curves in a system of non-linear equations may not be easily visualized, it sometimes pays to take advantage when they are. Our next example provides some considerable review of many of the topics introduced in this text. Example 8.7.2. Solve the following systems of equations. Verify your answers algebraically and graphically, as appropriate. x2 + 2xy − 16 = 0 y2 + 2xy − 16 = 0 1. y + 4e2x = 1 y2 + 2ex = 1 2.    3. z(x − 2) = x yz = y (x − 2)2 + y2 = 1 Solution. 1. At ﬁrst glance, it doesn’t appear as though elimination will do us any good since it’s clear that we cannot completely eliminate one of the variables. The alternative, solving one of the equations for one variable and substituting it into the other, is full of unpleasantness. Returning to elimination, we note that it is possible to eliminate the troublesome xy term, and the constant term as well, by elimination and doing so we get a more tractable relationship between x and y (E1) x2 + 2xy − 16 = 0 (E2) y2 + 2xy − 16 = 0 Replace E2 with −−−−−−−−−−→ −E1 + E2 (E1) x2 + 2xy − 16 = 0 y2 − x2 = 0 (E2) 640 Systems of Equations and Matrices √ √ 4 3 or x = ± 4 We get y2 − x2 = 0 or y = ±x.
Substituting y = x into E1 we get x2 + 2x2 − 16 = 0 so √ 3 that x2 = 16 3. On the other hand, when we substitute y = −x into E1, we get √ 3 x2 − 2x2 − 16 = 0 or x2 = −16 which gives no real solutions. Substituting each of x = ± 4 3 into the substitution equation y = x yields the solution. We leave it to the reader to show that both points satisfy both equations and now turn to verifying our solution graphically. We begin by solving x2+2xy−16 = 0 for y to obtain y = 16−x2 2x. This function is easily graphed using the techniques of Section 4.2. Solving the second equation, y2 + 2xy − 16 = 0, for y, however, is more complicated. We use the quadratic formula to x2 + 16 which would require the use of Calculus or a calculator to graph. obtain y = −x ± Believe it or not, we don’t need either because the equation y2 + 2xy − 16 = 0 can be obtained from the equation x2 + 2xy − 16 = 0 by interchanging y and x. Thinking back to Section 5.2, this means we can obtain the graph of y2 + 2xy − 16 = 0 by reﬂecting the graph of x2 + 2xy − 16 = 0 across the line y = x. Doing so conﬁrms that the two graphs intersect twice: once in Quadrant I, and once in Quadrant III as required4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 The graphs of x2 + 2xy − 16 = 0 and y2 + 2xy − 16 = 0 2. Unlike the previous problem, there seems to be no avoiding substitution and a bit of algebraic unpleasantness. Solving y + 4e2x = 1 for y, we get y = 1 − 4e2x which, when substituted into the second equation, yields 1 − 4e2x2 + 2ex = 1. After expanding and gathering like terms, we get 16e4x − 8e2x + 2ex = 0. Factoring gives us 2ex 8e3x − 4ex + 1 = 0, and since 2ex = 0 for any real x, we
are left with solving 8e3x − 4ex + 1 = 0. We have three terms, and even though this is not a ‘quadratic in disguise’, we can beneﬁt from the substitution u = ex. The equation becomes 8u3−4u+1 = 0. Using the techniques set forth in Section 3.3, we ﬁnd u = 1 2 8u2 + 4u − 2. We is a zero and use synthetic division to factor the left hand side as u − 1 2 √ use the quadratic formula to solve 8u2 + 4u − 2 = 0 and ﬁnd u = −1± 5. Since u = ex, we 4 = − ln(2). As now must solve ex = 1 for ex = −1± has no real solutions. We are 4 2 and ex = −1± 4, we ﬁrst note that −1−. From ex = 1 4 < 0, so ex = −1− 2, we get x = ln.7 Systems of Non-Linear Equations and Inequalities 641 left with ex = −1+ 4 accompanying y values for each of our solutions for x. For x = − ln(2), we get, so that x = ln. We now return to y = 1 − 4e2x to ﬁnd the 5 5 √ −1+ 4 √ y = 1 − 4e2x = 1 − 4e−2 ln(2) = 1 − 4eln For x = ln √ −1+ 4 5, we have y = 1 − 4e2x 2 ln = 1 − 4e √ 5 −1+ 4 √ 2 5 2 5 ln −1+ 4 √ −1+ 4 √ 3− 8 5 = 1 − 4e = 1+ 2 5 √ 5 −1+ 4, −1+ 2 √ 5 (0, − ln(2)), We get two solutions,. It is a good review of the properties of logarithms to verify both solutions, so we leave that to the reader. We are able to sketch y = 1 − 4e2x using transformations, but the second equation is more diﬃcult and we resort to the calculator. We note that to graph y2 + 2ex = 1, we need to graph both
the positive and negative roots, y = ± 1 − 2ex. After some careful zooming,2 we get ln √ The graphs of y = 1 − 4e2x and y = ± √ 1 − 2ex. 3. Our last system involves three variables and gives some insight on how to keep such systems organized. Labeling the equations as before, we have 2The calculator has trouble conﬁrming the solution (− ln(2), 0) due to its issues in graphing square root functions. If we mentally connect the two branches of the thicker curve, we see the intersection. 642 Systems of Equations and Matrices    z(x − 2) = x E1 E2 yz = y E3 (x − 2)2 + y2 = 1 The easiest equation to start with appears to be E2. While it may be tempting to divide both sides of E2 by y, we caution against this practice because it presupposes y = 0. Instead, we take E2 and rewrite it as yz − y = 0 so y(z − 1) = 0. From this, we get two cases: y = 0 or z = 1. We take each case in turn. Case 1: y = 0. Substituting y = 0 into E1 and E3, we get E1 z(x − 2) = x E3 (x − 2)2 = 1 Solving E3 for x gives x = 1 or x = 3. Substituting these values into E1 gives z = −1 when x = 1 and z = 3 when x = 3. We obtain two solutions, (1, 0, −1) and (3, 0, 3). Case 2: z = 1. Substituting z = 1 into E1 and E3 gives us E1 (1)(x − 2) = x E3 (1 − 2)2 + y2 = 1 Equation E1 gives us x − 2 = x or −2 = 0, which is a contradiction. This means we have no solution to the system in this case, even though E3 is solvable and gives y = 0. Hence, our ﬁnal answer is {(1, 0, −1), (3, 0, 3)}. These points are easy enough to check algebraically in our three original equations, so that is left to the reader. As
for verifying these solutions graphically, they require plotting surfaces in three dimensions and looking for intersection points. While this is beyond the scope of this book, we provide a snapshot of the graphs of our three equations near one of the solution points, (1, 0, −1). Example 8.7.2 showcases some of the ingenuity and tenacity mentioned at the beginning of the section. Sometimes you just have to look at a system the right way to ﬁnd the most eﬃcient method to solve it. Sometimes you just have to try something. 8.7 Systems of Non-Linear Equations and Inequalities 643 We close this section discussing how non-linear inequalities can be used to describe regions in the plane which we ﬁrst introduced in Section 2.4. Before we embark on some examples, a little motivation is in order. Suppose we wish to solve x2 < 4 − y2. If we mimic the algorithms for solving nonlinear inequalities in one variable, we would gather all of the terms on one side and leave a 0 on the other to obtain x2 + y2 − 4 < 0. Then we would ﬁnd the zeros of the left hand side, that is, where is x2 + y2 − 4 = 0, or x2 + y2 = 4. Instead of obtaining a few numbers which divide the real number line into intervals, we get an equation of a curve, in this case, a circle, which divides the plane into two regions - the ‘inside’ and ‘outside’ of the circle - with the circle itself as the boundary between the two. Just like we used test values to determine whether or not an interval belongs to the solution of the inequality, we use test points in the each of the regions to see which of these belong to our solution set.3 We choose (0, 0) to represent the region inside the circle and (0, 3) to represent the points outside of the circle. When we substitute (0, 0) into x2 + y2 − 4 < 0, we get −4 < 4 which is true. This means (0, 0) and all the other points inside the circle are part of the solution. On the other hand, when we substitute (0, 3) into the same inequality, we get 5 < 0 which is false. This means (0, 3) along with all other points outside the circle are not part of the solution.
What about points on the circle itself? Choosing a point on the circle, say (0, 2), we get 0 < 0, which means the circle itself does not satisfy the inequality.4 As a result, we leave the circle dashed in the ﬁnal diagram. y 2 −2 2 x −2 The solution to x2 < 4 − y2 We put this technique to good use in the following example. Example 8.7.3. Sketch the solution to the following nonlinear inequalities in the plane. 1. y2 − 4 ≤ x < y + 2 Solution. 2. x2 + y2 ≥ 4 x2 − 2x + y2 − 2y ≤ 0 1. The inequality y2 − 4 ≤ x < y + 2 is a compound inequality. It translates as y2 − 4 ≤ x and x < y + 2. As usual, we solve each inequality and take the set theoretic intersection to determine the region which satisﬁes both inequalities. To solve y2 − 4 ≤ x, we write 3The theory behind why all this works is, surprisingly, the same theory which guarantees that sign diagrams work the way they do - continuity and the Intermediate Value Theorem - but in this case, applied to functions of more than one variable. 4Another way to see this is that points on the circle satisfy x2 + y2 − 4 = 0, so they do not satisfy x2 + y2 − 4 < 0. 644 Systems of Equations and Matrices y2 − x − 4 ≤ 0. The curve y2 − x − 4 = 0 describes a parabola since exactly one of the variables is squared. Rewriting this in standard form, we get y2 = x + 4 and we see that the vertex is (−4, 0) and the parabola opens to the right. Using the test points (−5, 0) and (0, 0), we ﬁnd that the solution to the inequality includes the region to the right of, or ‘inside’, the parabola. The points on the parabola itself are also part of the solution, since the vertex (−4, 0) satisﬁes the inequality. We now turn our attention to x < y + 2. Proceeding as before, we write x − y − 2 < 0 and focus our attention on x − y − 2 = 0, which is the line y = x − 2.
Using the test points (0, 0) and (0, −4), we ﬁnd points in the region above the line y = x − 2 satisfy the inequality. The points on the line y = x − 2 do not satisfy the inequality, since the y-intercept (0, −2) does not. We see that these two regions do overlap, and to make the graph more precise, we seek the intersection of these two curves. That is, we need to solve the system of nonlinear equations (E1) y2 = x + 4 y = x − 2 (E2) Solving E1 for x, we get x = y2 − 4. Substituting this into E2 gives y = y2 − 4 − 2, or y2 − y − 6 = 0. We ﬁnd y = −2 and y = 3 and since x = y2 − 4, we get that the graphs intersect at (0, −2) and (5, 3). Putting all of this together, we get our ﬁnal answer below. y 3 y y −5−4 x 2 3 4 5 x −5−4 2 3 4 5 x −3 −3 −3 y2 − 4 ≤ x x < y + 2 y2 − 4 ≤ x < y + 2 2. To solve this system of inequalities, we need to ﬁnd all of the points (x, y) which satisfy both inequalities. To do this, we solve each inequality separately and take the set theoretic intersection of the solution sets. We begin with the inequality x2 + y2 ≥ 4 which we rewrite as x2 + y2 − 4 ≥ 0. The points which satisfy x2 + y2 − 4 = 0 form our friendly circle x2 + y2 = 4. Using test points (0, 0) and (0, 3) we ﬁnd that our solution comprises the region outside the circle. As far as the circle itself, the point (0, 2) satisﬁes the inequality, so the circle itself is part of the solution set. Moving to the inequality x2 − 2x + y2 − 2y ≤ 0, we start with x2 − 2x + y2 − 2y = 0. Completing the squares, we obtain (x − 1)2 + (y − 1)2 = 2, which is 2. Choosing (1, 1)
to represent the inside of the a circle centered at (1, 1) with a radius of circle, (1, 3) as a point outside of the circle and (0, 0) as a point on the circle, we ﬁnd that the solution to the inequality is the inside of the circle, including the circle itself. Our ﬁnal answer, then, consists of the points on or outside of the circle x2 + y2 = 4 which lie on or √ 8.7 Systems of Non-Linear Equations and Inequalities 645 inside the circle (x − 1)2 + (y − 1)2 = 2. To produce the most accurate graph, we need to ﬁnd where these circles intersect. To that end, we solve the system (E1) x2 + y2 = 4 (E2) x2 − 2x + y2 − 2y = 0 We can eliminate both the x2 and y2 by replacing E2 with −E1 + E2. Doing so produces −2x − 2y = −4. Solving this for y, we get y = 2 − x. Substituting this into E1 gives x2 + (2 − x)2 = 4 which simpliﬁes to x2 + 4 − 4x + x2 = 4 or 2x2 − 4x = 0. Factoring yields 2x(x − 2) which gives x = 0 or x = 2. Substituting these values into y = 2 − x gives the points (0, 2) and (2, 0). The intermediate graphs and ﬁnal solution are below. y 1 −3 −2 −1 2 x −3 −2 −1 2 x −1 −2 −3 −1 −2 −3 x2 + y2 ≥ 4 x2 − 2x + y2 − 2y ≤ 0 Solution to the system. 646 Systems of Equations and Matrices 8.7.1 Exercises In Exercises 1 - 6, solve the given system of nonlinear equations. Sketch the graph of both equations on the same set of axes to verify the solution set. x2 − y = 4 x2 + y2 = 4 x2 + y2 = 16 9x2 − 16y2 = 144 1. 4. 2. 5. x2 + y2 = 4 x2 − y = 5 x2 + y2
= 16 16 x2 = 1 9 y2 − 1 1 x2 + y2 = 16 16x2 + 4y2 = 64 x2 + y2 = 16 x − y = 2 3. 6. In Exercises 9 - 15, solve the given system of nonlinear equations. Use a graph to help you avoid any potential extraneous solutions. 7. 10. x2 − y2 = 1 x2 + 4y2 = 4 (x − 2)2 + y2 = 1 x2 + 4y2 = 4 √ x + 1 − y = 0 x2 + 4y2 = 4 x2 + y2 = 25 y − x = 1 8. 11. 13. y = x3 + 8 y = 10x − x2 14. x2 − xy = 8 y2 − xy = 8 x + 2y2 = 2 x2 + 4y2 = 4 9. 12. 15.    x2 + y2 = 25 x2 + (y − 3)2 = 10 x2 + y2 = 25 4x2 − 9y = 0 3y2 − 16x = 0 16. A certain bacteria culture follows the Law of Uninbited Growth, Equation 6.4. After 10 minutes, there are 10,000 bacteria. Five minutes later, there are 14,000 bacteria. How many bacteria were present initially? How long before there are 50,000 bacteria? Consider the system of nonlinear equations below    1 If we let u = 1 x and v = 1 y then the system becomes 4u + 3v = 1 3u + 2v = −1 This associated system of linear equations can then be solved using any of the techniques presented earlier in the chapter to ﬁnd that u = −5 and v = 7. Thus x = 1 We say that the original system is linear in form because its equations are not linear but a few substitutions reveal a structure that we can treat like a system of linear equations. Each system in Exercises 17 - 19 is linear in form. Make the appropriate substitutions and solve for x and y. 5 and.7 Systems of Non-Linear Equations and Inequalities 647 4x3 + 3 3x3 + 2 √ √ y = 1 y = −1 17.
18. 4ex + 3e−y = 1 3ex + 2e−y = −1 19. 4 ln(x) + 3y2 = 1 3 ln(x) + 2y2 = −1 20. Solve the following system    √ x2 + √ 3x2 − 2 −5x2 + 3 √ y + log2(z) = 6 y + 2 log2(z) = 5 y + 4 log2(z) = 13 In Exercises 21 - 26, sketch the solution to each system of nonlinear inequalities in the plane. 21. 23. 25. x2 − y2 ≤ 1 x2 + 4y2 ≥ 4 (x − 2)2 + y2 < 1 x2 + 4y2 < 4 x + 2y2 > 2 x2 + 4y2 ≤ 4 22. 24. 26. x2 + y2 < 25 x2 + (y − 3)2 ≥ 10 y < y > 10x − x2 x3 + 8 x2 + y2 ≥ 25 y − x ≤ 1 27. Systems of nonlinear equations show up in third semester Calculus in the midst of some really cool problems. The system below came from a problem in which we were asked to ﬁnd the dimensions of a rectangular box with a volume of 1000 cubic inches that has minimal surface area. The variables x, y and z are the dimensions of the box and λ is called a Lagrange multiplier. With the help of your classmates, solve the system.5    2y + 2z = λyz 2x + 2z = λxz 2y + 2x = λxy xyz = 1000 28. According to Theorem 3.16 in Section 3.4, the polynomial p(x) = x4 + 4 can be factored into the product linear and irreducible quadratic factors. In this exercise, we present a method for obtaining that factorization. (a) Show that p has no real zeros. (b) Because p has no real zeros, its factorization must be of the form (x2 +ax+b)(x2 +cx+d) where each factor is an irreducible quadratic
. Expand this quantity and gather like terms together. (c) Create and solve the system of nonlinear equations which results from equating the coeﬃcients of the expansion found above with those of x4 + 4. You should get four equations in the four unknowns a, b, c and d. Write p(x) in factored form. 29. Factor q(x) = x4 + 6x2 − 5x + 6. 5If using λ bothers you, change it to w when you solve the system. 648 Systems of Equations and Matrices 8.7.2 Answers √ 1. (±2, 0), ± y 3, −1 2 1 −2 −1 1 2 x −1 −2 −3 −4 3. (0, ±4) y 4 3 2 1 2. No solution y 2 1 −2 −1 1 2 x −1 −2 −3 −4 4. (±4, 0) y 4 3 2 1 −4−3−2−1 −1 1 2 3 4 x −6−5−4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 −2 −3 −4 √ ± 4 5. 7 5, ± 12 √ 5 y 2 √ 6. 1 + 7, −1 + y √ 7, 1 − √ 7, −1 − √ 7 4 3 2 1 −4−3−2−1 −1 1 2 3 4 x −2 −3 −4 √ 15 5, ± √ 7. ± 2 10 5 √ 104−3−2−1 −1 1 2 3 4 x −2 −3 −4 8. (0, 1) 9. (0, ±1), (2, 0) 11. (3, 4), (−4, −3) 12. (±3, 4) 13. (−4, −56), (1, 9), (2, 16) 14. (−2, 2), (2, −2) 15. (3, 4) 16. Initially, there are 250000 49 ≈ 5102 bacteria. It will take 5 ln(49/5) ln(7/5) ≈ 33.92 minutes for the colony to grow to 50,000 bacteria. 8.7 Systems of Non-Linear Equations and Inequalities 649 √ 17
. − 3 5, 49 20. (1, 4, 8), (−1, 4, 8) 21. x2 − y2 ≤ 1 x2 + 4y2 ≥ 4 y 2 1 −2 −1 1 2 x −1 −2 23. (x − 2)2 + y2 < 1 x2 + 4y2 < 4 y 1 −1 1 2 x 25. x + 2y2 > 2 x2 + 4y2 ≤ 4 y 1 −1 18. No solution 19. e−5, ± √ 7 22. x2 + y2 < 25 x2 + (y − 3)2 ≥ 10 y 4 3 2 1 −5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 −5 24. y > 10x − x2 x3 + 8 y < y 16 9 −4−3−2−1 1 2 x −56 26. x2 + y2 ≥ 25 5 −3 −1 1 3 5 x −3 −5 650 Systems of Equations and Matrices 27. x = 10, y = 10, z = 10, λ = 2 5 28. (c) x4 + 4 = (x2 − 2x + 2)(x2 + 2x + 2) 29. x4 + 6x2 − 5x + 6 = (x2 − x + 1)(x2 + x + 6) Chapter 9 Sequences and the Binomial Theorem 9.1 Sequences When we ﬁrst introduced a function as a special type of relation in Section 1.3, we did not put any restrictions on the domain of the function. All we said was that the set of x-coordinates of the points in the function F is called the domain, and it turns out that any subset of the real numbers, regardless of how weird that subset may be, can be the domain of a function. As our exploration of functions continued beyond Section 1.3, we saw fewer and fewer functions with ‘weird’ domains. It is worth your time to go back through the text to see that the domains of the polynomial, rational, exponential, logarithmic and algebraic functions discussed thus far have fairly predictable domains which almost always consist of just a collection of intervals on the real line. This may lead some readers to believe that the only important functions in a College Algebra text have domains which consist
of intervals and everything else was just introductory nonsense. In this section, we introduce sequences which are an important class of functions whose domains are the set of natural numbers.1 Before we get to far ahead of ourselves, let’s look at what the term ‘sequence’ means mathematically. Informally, we can think of a sequence as an inﬁnite list of numbers. For example, consider the sequence 1 2, − 3 4, 9 8, − 27 16,... (1) As usual, the periods of ellipsis,..., indicate that the proposed pattern continues forever. Each of the numbers in the list is called a term, and we call 1 8 the ‘third term’ and so forth. In numbering them this way, we are setting up a function, which we’ll call a per tradition, between the natural numbers and the terms in the sequence. 4 the ‘second term’, 9 2 the ‘ﬁrst term’, − 3 1Recall that this is the set {1, 2, 3,...}. 652 Sequences and the Binomial Theorem 1 2 n a(n − 27 16...... In other words, a(n) is the nth term in the sequence. We formalize these ideas in our deﬁnition of a sequence and introduce some accompanying notation. Deﬁnition 9.1. A sequence is a function a whose domain is the natural numbers. The value a(n) is often written as an and is called the nth term of the sequence. The sequence itself is usually denoted using the notation: an, n ≥ 1 or the notation: {an}∞ n=1. 2, a2 = − 3 Applying the notation provided in Deﬁnition 9.1 to the sequence given (1), we have a1 = 1 4, a3 = 9 8 and so forth. Now suppose we wanted to know a117, that is, the 117th term in the sequence. While the pattern of the sequence is apparent, it would beneﬁt us greatly to have an explicit formula for an. Unfortunately, there is no general algorithm that will produce a formula for every sequence, so any formulas we do develop will come from that greatest of teachers, experience. In other words, it is time for an example. Example 9.1.1
. Write the ﬁrst four terms of the following sequences. 1. an = 5n−1 3n, n ≥ 1 3. {2n − 1}∞ n=1, k ≥ 0 2. bk = (−1)k 2k + 1 1 + (−1)i i 4. ∞ i=2 5. a1 = 7, an+1 = 2 − an, n ≥ 1 6. f0 = 1, fn = n · fn−1, n ≥ 1 Solution. 32 = 5 1. Since we are given n ≥ 1, the ﬁrst four terms of the sequence are a1, a2, a3 and a4. Since the notation a1 means the same thing as a(1), we obtain our ﬁrst term by replacing every occurrence of n in the formula for an with n = 1 to get a1 = 51−1 3. Proceeding similarly, we get a2 = 52−1 33 = 25 27 and a4 = 54−1 2. For this sequence we have k ≥ 0, so the ﬁrst four terms are b0, b1, b2 and b3. Proceeding as before, replacing in this case the variable k with the appropriate whole number, beginning with 0, we get b0 = (−1)0 2(3)+1 = − 1 7. (This sequence is called an alternating sequence since the signs alternate between + and −. The reader is encouraged to think what component of the formula is producing this eﬀect.) 2(0)+1 = 1, b1 = (−1)1 5 and b3 = (−1)3 3, b2 = (−1)2 9, a3 = 53−1 2(1)+1 = − 1 2(2)+1 = 1 34 = 125 81. 31 = 1 9.1 Sequences 653 3. From {2n − 1}∞ n=1, we have that an = 2n − 1, n ≥ 1. We get a1 = 1, a2 = 3, a3 = 5 and a4 = 7. (The ﬁrst four terms are the ﬁrst four odd natural numbers. The reader is encouraged to examine whether or not this pattern continues indeﬁnitely.) 4. Here, we are using the letter i as a counter, not as the imaginary
unit we saw in Section 3.4. 2 and a5 = 0. Proceeding as before, we set ai = 1+(−1)i, i ≥ 2. We ﬁnd a2 = 1, a3 = 0, a4 = 1 i 5. To obtain the terms of this sequence, we start with a1 = 7 and use the equation an+1 = 2−an for n ≥ 1 to generate successive terms. When n = 1, this equation becomes a1 + 1 = 2 − a1 which simpliﬁes to a2 = 2−a1 = 2−7 = −5. When n = 2, the equation becomes a2 + 1 = 2−a2 so we get a3 = 2 − a2 = 2 − (−5) = 7. Finally, when n = 3, we get a3 + 1 = 2 − a3 so a4 = 2 − a3 = 2 − 7 = −5. 6. As with the problem above, we are given a place to start with f0 = 1 and given a formula to build other terms of the sequence. Substituting n = 1 into the equation fn = n · fn−1, we get f1 = 1 · f0 = 1 · 1 = 1. Advancing to n = 2, we get f2 = 2 · f1 = 2 · 1 = 2. Finally, f3 = 3 · f2 = 3 · 2 = 6. Some remarks about Example 9.1.1 are in order. We ﬁrst note that since sequences are functions, we can graph them in the same way we graph functions. For example, if we wish to graph the sequence {bk}∞ k=0 from Example 9.1.1, we graph the equation y = b(k) for the values k ≥ 0. That is, we plot the points (k, b(k)) for the values of k in the domain, k = 0, 1, 2,.... The resulting collection of points is the graph of the sequence. Note that we do not connect the dots in a pleasing fashion as we are used to doing, because the domain is just the whole numbers in this case, not a collection of intervals of real numbers. If you feel a sense of nostalgia, you should see Section 1.21 − 3 2 1 2 3 x Graphing y = bk = (−1)k 2
k + 1, k ≥ 0 Speaking of {bk}∞ k=0, the astute and mathematically minded reader will correctly note that this technically isn’t a sequence, since according to Deﬁnition 9.1, sequences are functions whose domains are the natural numbers, not the whole numbers, as is the case with {bk}∞ k=0. In other words, to satisfy Deﬁnition 9.1, we need to shift the variable k so it starts at k = 1 instead of k = 0. To see how we can do this, it helps to think of the problem graphically. What we want is to shift the graph of y = b(k) to the right one unit, and thinking back to Section 1.7, we can accomplish this by replacing k with k − 1 in the deﬁnition of {bk}∞ k=0. Speciﬁcally, let ck = bk−1 where k − 1 ≥ 0. We get ck = (−1)k−1 2k−1, where now k ≥ 1. We leave to the reader to verify that {ck}∞ k=0, but the former satisﬁes Deﬁnition 2(k−1)+1 = (−1)k−1 k=1 generates the same list of numbers as does {bk}∞ 654 Sequences and the Binomial Theorem 9.1, while the latter does not. Like so many things in this text, we acknowledge that this point is pedantic and join the vast majority of authors who adopt a more relaxed view of Deﬁnition 9.1 to include any function which generates a list of numbers which can then be matched up with the natural numbers.2 Finally, we wish to note the sequences in parts 5 and 6 are examples of sequences described recursively. In each instance, an initial value of the sequence is given which is then followed by a recursion equation − a formula which enables us to use known terms of the sequence to determine other terms. The terms of the sequence in part 6 are given a special name: fn = n! is called n-factorial. Using the ‘!’ notation, we can describe the factorial sequence as: 0! = 1 and n! = n(n − 1)! for n ≥ 1. After 0
! = 1 the next four terms, written out in detail, are 1! = 1 · 0! = 1 · 1 = 1, 2! = 2 · 1! = 2 · 1 = 2, 3! = 3 · 2! = 3 · 2 · 1 = 6 and 4! = 4 · 3! = 4 · 3 · 2 · 1 = 24. From this, we see a more informal way of computing n!, which is n! = n · (n − 1) · (n − 2) · · · 2 · 1 with 0! = 1 as a special case. (We will study factorials in greater detail in Section 9.4.) The world famous Fibonacci Numbers are deﬁned recursively and are explored in the exercises. While none of the sequences worked out to be the sequence in (1), they do give us some insight into what kinds of patterns to look for. Two patterns in particular are given in the next deﬁnition. Deﬁnition 9.2. Arithmetic and Geometric Sequences: Suppose {an}∞ n=k is a sequencea If there is a number d so that an+1 = an + d for all n ≥ k, then {an}∞ arithmetic sequence. The number d is called the common diﬀerence. n=k is called an If there is a number r so that an+1 = ran for all n ≥ k, then {an}∞ n=k is called a geometric sequence. The number r is called the common ratio. aNote that we have adjusted for the fact that not all ‘sequences’ begin at n = 1. Both arithmetic and geometric sequences are deﬁned in terms of recursion equations. In English, an arithmetic sequence is one in which we proceed from one term to the next by always adding the ﬁxed number d. The name ‘common diﬀerence’ comes from a slight rewrite of the recursion equation from an+1 = an + d to an+1 − an = d. Analogously, a geometric sequence is one in which we proceed from one term to the next by always multiplying by the same ﬁxed number r. If r = 0, we can rearrange the recursion equation to get an+1 = r, hence the name ‘common ratio.’ Some an sequences are arithmetic,
some are geometric and some are neither as the next example illustrates.3 Example 9.1.2. Determine if the following sequences are arithmetic, geometric or neither. arithmetic, ﬁnd the common diﬀerence d; if geometric, ﬁnd the common ratio r. If 1. an = 5n−1 3n, n ≥ 1 3. {2n − 1}∞ n=1 2. bk = (−1)k 2k + 1, k ≥ 0 4, − 27 16,... 2We’re basically talking about the ‘countably inﬁnite’ subsets of the real number line when we do this. 3Sequences which are both arithmetic and geometric are discussed in the Exercises. 9.1 Sequences 655 Solution. A good rule of thumb to keep in mind when working with sequences is “When in doubt, write it out!” Writing out the ﬁrst several terms can help you identify the pattern of the sequence should one exist. 1. From Example 9.1.1, we know that the ﬁrst four terms of this sequence are 1 27 and 125 81. To see if this is an arithmetic sequence, we look at the successive diﬀerences of terms. We ﬁnd that a2 − a1 = 5 27. Since we get diﬀerent numbers, there is no ‘common diﬀerence’ and we have established that the sequence is not arithmetic. To investigate whether or not it is geometric, we compute the ratios of successive terms. The ﬁrst three ratios 9 and a3 − a2 = 25 9 = 10 27 − 5 3 = 2 9 − 1 9, 25 3, 5 a2 a1 = 5 9 1 3 = 5 3, a3 a2 = = 5 3 and a4 a3 = 25 27 5 9 125 81 25 27 = 5 3 suggest that the sequence is geometric. To prove it, we must show that an+1 an = r for all n. an+1 an = 5(n+1)−1 3n+1 5n−1 3n = 5n 3n+1 · 3n 5n−1 = 5 3 This sequence is geometric with common ratio r = 5 3. 3, 1 2. Again, we have Example 9.1.1 to thank for providing the
ﬁrst four terms of this sequence: 15. Hence, the sequence is not 5. Since there is no 1, − 1 3 and b2 − b1 = 8 3 and b2 arithmetic. To see if it is geometric, we compute b1 b1 b0 ‘common ratio,’ we conclude the sequence is not geometric, either. 7. We ﬁnd b1 − b0 = − 4 5 and − 1 = − 1 = − 3 3. As we saw in Example 9.1.1, the sequence {2n − 1}∞ n=1 generates the odd numbers: 1, 3, 5, 7,.... Computing the ﬁrst few diﬀerences, we ﬁnd a2 − a1 = 2, a3 − a2 = 2, and a4 − a3 = 2. This suggests that the sequence is arithmetic. To verify this, we ﬁnd an+1 − an = (2(n + 1) − 1) − (2n − 1) = 2n + 2 − 1 − 2n + 1 = 2 This establishes that the sequence is arithmetic with common diﬀerence d = 2. To see if it is geometric, we compute a2 3. Since these ratios are diﬀerent, we conclude the a1 sequence is not geometric. = 3 and a3 a2 = 5 4. We met our last sequence at the beginning of the section. Given that a2 − a1 = − 5 4 and a3 − a2 = 15 8, the sequence is not arithmetic. Computing the ﬁrst few ratios, however, gives us 2, a3 a2 = − 3 2. Since these are the only terms given to us, we assume that a2 a1 the pattern of ratios continue in this fashion and conclude that the sequence is geometric. 2 and a4 a3 = − 3 = − 3 We are now one step away from determining an explicit formula for the sequence given in (1). We know that it is a geometric sequence and our next result gives us the explicit formula we require. 656 Sequences and the Binomial Theorem Equation 9.1. Formulas for Arithmetic and Geometric Sequences: An arithmetic sequence with ﬁrst term a and common diﬀerence d is given by an = a + (n − 1)d,
n ≥ 1 A geometric sequence with ﬁrst term a and common ratio r = 0 is given by an = arn−1, n ≥ 1 While the formal proofs of the formulas in Equation 9.1 require the techniques set forth in Section 9.3, we attempt to motivate them here. According to Deﬁnition 9.2, given an arithmetic sequence with ﬁrst term a and common diﬀerence d, the way we get from one term to the next is by adding d. Hence, the terms of the sequence are: a, a + d, a + 2d, a + 3d,.... We see that to reach the nth term, we add d to a exactly (n − 1) times, which is what the formula says. The derivation of the formula for geometric series follows similarly. Here, we start with a and go from one term to the next by multiplying by r. We get a, ar, ar2, ar3 and so forth. The nth term results from multiplying a by r exactly (n − 1) times. We note here that the reason r = 0 is excluded from Equation 9.1 is to avoid an instance of 00 which is an indeterminant form.4 With Equation 9.1 in place, we ﬁnally have the tools required to ﬁnd an explicit formula for the nth term of the sequence given in (1). We know from Example 9.1.2 that it is geometric with common ratio r = − 3 2. The ﬁrst term is a = 1 for n ≥ 1. After a touch of simplifying, we get an = (−3)n−1 for n ≥ 1. Note that we can easily check our answer by substituting in values of n and seeing that the formula generates the sequence given in (1). We leave this to the reader. Our next example gives us more practice ﬁnding patterns. 2 so by Equation 9.1 we get an = arn−1 = 1 − 3 2 n−1 2n 2 Example 9.1.3. Find an explicit formula for the nth term of the following sequences. 1. 0.9, 0.09, 0.009, 0.0009,... 2. 2 5, 2. 1, − 2 7, 4 13, − 8 19,... Solution
. 1. Although this sequence may seem strange, the reader can verify it is actually a geometric for 10n, n ≥ 1. There is more to this sequence than meets the sequence with common ratio r = 0.1 = 1 n ≥ 0. Simplifying, we get an = 9 eye and we shall return to this example in the next section. 10. With a = 0.9 = 9 10, we get an = 9 1 10 n−1 10 2. As the reader can verify, this sequence is neither arithmetic nor geometric. In an attempt to ﬁnd a pattern, we rewrite the second term with a denominator to make all the terms appear as fractions. We have 2 7,.... If we associate the negative ‘−’ of the last two −3, 2 1, 2 terms with the denominators we get 2 −7,.... This tells us that we can tentatively sketch out the formula for the sequence as an = 2 where dn is the sequence of denominators. dn 4See the footnotes on page 237 in Section 3.1 and page 418 of Section 6.1. 9.1 Sequences 657 Looking at the denominators 5, 1, −3, −7,..., we ﬁnd that they go from one term to the next by subtracting 4 which is the same as adding −4. This means we have an arithmetic sequence on our hands. Using Equation 9.1 with a = 5 and d = −4, we get the nth denominator by the formula dn = 5 + (n − 1)(−4) = 9 − 4n for n ≥ 1. Our ﬁnal answer is an = 2 9−4n, n ≥ 1. 3. The sequence as given is neither arithmetic nor geometric, so we proceed as in the last problem to try to get patterns individually for the numerator and denominator. Letting cn and dn denote the sequence of numerators and denominators, respectively, we have an = cn. After dn some experimentation,5 we choose to write the ﬁrst term as a fraction and associate the negatives ‘−’ with the numerators. This yields 1 19,.... The numerators form the sequence 1, −2, 4, −8,... which is geometric with a = 1 and r = −2, so we get cn =
(−2)n−1, for n ≥ 1. The denominators 1, 7, 13, 19,... form an arithmetic sequence with a = 1 and d = 6. Hence, we get dn = 1 + 6(n − 1) = 6n − 5, for n ≥ 1. We obtain our formula for an = cn dn 6n−5, for n ≥ 1. We leave it to the reader to show that this checks out. = (−2)n−1 13, −8 1, −2 7, 4 While the last problem in Example 9.1.3 was neither geometric nor arithmetic, it did resolve into a combination of these two kinds of sequences. If handed the sequence 2, 5, 10, 17,..., we would be hard-pressed to ﬁnd a formula for an if we restrict our attention to these two archetypes. We said before that there is no general algorithm for ﬁnding the explicit formula for the nth term of a given sequence, and it is only through experience gained from evaluating sequences from explicit formulas that we learn to begin to recognize number patterns. The pattern 1, 4, 9, 16,... is rather recognizable as the squares, so the formula an = n2, n ≥ 1 may not be too hard to determine. With this in mind, it’s possible to see 2, 5, 10, 17,... as the sequence 1 + 1, 4 + 1, 9 + 1, 16 + 1,..., so that an = n2 + 1, n ≥ 1. Of course, since we are given only a small sample of the sequence, we shouldn’t be too disappointed to ﬁnd out this isn’t the only formula which generates this sequence. For example, consider the sequence deﬁned by bn = − 1 2 n − 5, n ≥ 1. The reader is encouraged to verify that it also produces the terms 2, 5, 10, 17. In fact, it can be shown that given any ﬁnite sample of a sequence, there are inﬁnitely many explicit formulas all of which generate those same ﬁnite points. This means that there will be inﬁnitely many correct answers to some of the exercises in this section.6 Just because your answer doesn’t match ours doesn’t mean it
’s wrong. As always, when in doubt, write your answer out. As long as it produces the same terms in the same order as what the problem wants, your answer is correct. Sequences play a major role in the Mathematics of Finance, as we have already seen with Equation 6.2 in Section 6.5. Recall that if we invest P dollars at an annual percentage rate r and compound the interest n times per year, the formula for Ak, the amount in the account after k compounding periods, is Ak = P 1 + r, k ≥ 1. We now spot this as a geometric sequence with ﬁrst term P 1 + r. In retirement planning, it is seldom n the case that an investor deposits a set amount of money into an account and waits for it to grow. Usually, additional payments of principal are made at regular intervals and the value of the investment grows accordingly. This kind of investment is called an annuity and will be discussed in the next section once we have developed more mathematical machinery. 1 + r n and common ratio n2 + 25 2 n3 − 31 4 n4 + 5 k−1 k n 5Here we take ‘experimentation’ to mean a frustrating guess-and-check session. 6For more on this, see When Every Answer is Correct: Why Sequences and Number Patterns Fail the Test. 658 Sequences and the Binomial Theorem 9.1.1 Exercises In Exercises 1 - 13, write out the ﬁrst four terms of the given sequence. 1. an = 2n − 1, n ≥ 0 3. {5k − 2}∞ k=1 5. xn n2 ∞ n=1 2. dj = (−1) j(j+1) 2, j ≥ 1 4. 6. ∞ n=0 ∞ n2 + 1 n + 1 ln(n) n n=1 7. a1 = 3, an+1 = an − 1, n ≥ 1 8. d0 = 12, dm = dm-1 100, m ≥ 1 9. b1 = 2, bk+1 = 3bk + 1, k ≥ 1 11. a1 = 117, an+1 = 1 an, n ≥ 1 10. c0 = −2, cj = cj-1 (j + 1)(j + 2), j ≥ 1 12.
s0 = 1, sn+1 = xn+1 + sn, n ≥ 0 13. F0 = 1, F1 = 1, Fn = Fn-1 + Fn-2, n ≥ 2 (This is the famous Fibonacci Sequence ) In Exercises 14 - 21 determine if the given sequence is arithmetic, geometric or neither. If it is arithmetic, ﬁnd the common diﬀerence d; if it is geometric, ﬁnd the common ratio r. 14. {3n − 5}∞ n=1 16. 1 3, 1 6, 1 12, 1 24,... 15. an = n2 + 3n + 2, n ≥ 1 3 1 5 17. n−1∞ n=1 18. 17, 5, −7, −19,... 19. 2, 22, 222, 2222,... 20. 0.9, 9, 90, 900,... 21. an = n! 2, n ≥ 0. In Exercises 22 - 30, ﬁnd an explicit formula for the nth term of the given sequence. Use the formulas in Equation 9.1 as needed. 22. 3, 5, 7, 9,... 25. 1, 2 3, 1 3, 4 27,... 23. 1,... 26. 1, 1 4, 1 9, 1 16,...,... 24. 1, 2 3, 27. x, − 4 5 x3 3,, 8 7 x5 5, − x7 7,... 9.1 Sequences 659 28. 0.9, 0.99, 0.999, 0.9999,... 29. 27, 64, 125, 216,... 30. 1, 0, 1, 0,... 31. Find a sequence which is both arithmetic and geometric. (Hint: Start with an = c for all n.) 32. Show that a geometric sequence can be transformed into an arithmetic sequence by taking the natural logarithm of the terms. 33. Thomas Robert Malthus is credited with saying, “The power of population is indeﬁnitely greater than the power in the earth to produce subsistence for man. Population, when unchecked, increases in a geometrical ratio. Subsistence increases only in an arith
metical ratio. A slight acquaintance with numbers will show the immensity of the ﬁrst power in comparison with the second.” (See this webpage for more information.) Discuss this quote with your classmates from a sequences point of view. 34. This classic problem involving sequences shows the power of geometric sequences. Suppose that a wealthy benefactor agrees to give you one penny today and then double the amount she gives you each day for 30 days. So, for example, you get two pennies on the second day and four pennies on the third day. How many pennies do you get on the 30th day? What is the total dollar value of the gift you have received? 35. Research the terms ‘arithmetic mean’ and ‘geometric mean.’ With the help of your classmates, show that a given term of a arithmetic sequence ak, k ≥ 2 is the arithmetic mean of the term immediately preceding, ak−1 it and immediately following it, ak+1. State and prove an analogous result for geometric sequences. 36. Discuss with your classmates how the results of this section might change if we were to examine sequences of other mathematical things like complex numbers or matrices. Find an explicit formula for the nth term of the sequence i, −1, −i, 1, i,.... List out the ﬁrst four terms of the matrix sequences we discussed in Exercise 8.3.1 in Section 8.3. 660 Sequences and the Binomial Theorem 9.1.2 Answers 1. 0, 1, 3, 7 3. 3, 8, 13, 18 5. x, x2 4, x3 9, x4 16 7. 3, 2, 1, 0 9. 2, 7, 22, 67 11. 117, 1 117, 117, 1 117 13. 1, 1, 2, 3 14. arithmetic, d = 3 16. geometric, r = 1 2 18. arithmetic, d = −12 20. geometric, r = 10 2. −1, −1, 1, 1 4. 1, 1, 5 3, 5 2 6. 0, ln(2) 2, ln(3) 3, ln(4) 4 8. 12, 0.12, 0.0012, 0.000012 10. −2, − 1 3, − 1 36, − 1 720 12. 1, x + 1, x2 + x +
1, x3 + x2 + x + 1 15. neither 17. geometric, r = 1 5 19. neither 21. neither 22. an = 1 + 2n, n ≥ 1 23. an = − 1 2 n−1, n ≥ 1 24. an = 2n−1 2n−1, n ≥ 1 25. an = n 3n−1, n ≥ 1 26. an = 1 n2, n ≥ 1 27. (−1)n−1x2n−1 2n−1, n ≥ 1 28. an = 10n−1 10n, n ≥ 1 29. an = (n + 2)3, n ≥ 1 30. an = 1+(−1)n−1 2, n ≥ 1 9.2 Summation Notation 661 9.2 Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a deﬁnition, which, while intimidating, is meant to make our lives easier. Deﬁnition 9.3. Summation Notation: Given a sequence {an}∞ satisfying k ≤ m ≤ p, the summation from m to p of the sequence {an} is written n=k and numbers m and p p n=m an = am + am+1 +... + ap The variable n is called the index of summation. The number m is called the lower limit of summation while the number p is called the upper limit of summation. In English, Deﬁnition 9.3 is simply deﬁning a short-hand notation for adding up the terms of the sequence {an}∞ n=k from am through ap. The symbol Σ is the capital Greek letter sigma and is shorthand for ‘sum’. The lower and upper limits of the summation tells us which term to start with and which term to end with, respectively. For example, using the sequence an = 2n − 1 for n ≥ 1, we can write the sum a3 + a4 + a5 + a6 as 6 (2n − 1) = (2(3) − 1) + (2(4) − 1) + (2(5) − 1) + (2(6) − 1) n=3 = 5 + 7 +
9 + 11 = 32 The index variable is considered a ‘dummy variable’ in the sense that it may be changed to any letter without aﬀecting the value of the summation. For instance, 6 (2n − 1) = n=3 6 k=3 (2k − 1) = 6 j=3 (2j − 1) One place you may encounter summation notation is in mathematical deﬁnitions. For example, summation notation allows us to deﬁne polynomials as functions of the form f (x) = n k=0 akxk for real numbers ak, k = 0, 1,... n. The reader is invited to compare this with what is given in Deﬁnition 3.1. Summation notation is particularly useful when talking about matrix operations. For example, we can write the product of the ith row Ri of a matrix A = [aij]m×n and the jth column Cj of a matrix B = [bij]n×r as Ri · Cj = n k=1 aikbkj 662 Sequences and the Binomial Theorem Again, the reader is encouraged to write out the sum and compare it to Deﬁnition 8.9. Our next example gives us practice with this new notation. Example 9.2.1. 1. Find the following sums. (a) 4 k=1 13 100k (b) 4 n=0 n! 2 (c) 5 n=1 (−1)n+1 n (x − 1)n 2. Write the following sums using summation notation. (b) 1 − (a) 1 + 3 + 5 +... + 117 1 3 1 4 (c) 0.9 + 0.09 + 0.009 +... 0 zeros + −... + 1 117 1 2 + − Solution. 1. (a) We substitute k = 1 into the formula 13 100k and add successive terms until we reach k = 4. 4 k=1 13 100k = 13 1002 + 13 1001 + 13 1003 + = 0.13 + 0.0013 + 0.000013 + 0.00000013 = 0.13131313 13 1004 (b) Proceeding as in (a), we replace every occurrence of n with the values 0 through 4. We recall
the factorials, n! as deﬁned in number Example 9.1.1, number 6 and get: 4 n=0 n! 2 + + + 2! 2 2 · 1 2 = 4! 3 + 12 1! 2 1 2 1 2 = = 0! 2 1 2 1 2 = 17 = + 4 · 3 · 2 · 1 2 (c) We proceed as before, replacing the index n, but not the variable x, with the values 1 through 5 and adding the resulting terms. 9.2 Summation Notation 663 5 n=1 (−1)n+1 n (x − 1)n = (−1)1+1 1 (x − 1)1 + (−1)2+1 2 (x − 1)2 + (−1)3+1 3 (x − 1)3 + (−1)1+4 4 = (x − 1) − (x − 1)4 + (x − 1)2 2 + (−1)1+5 5 (x − 1)3 3 (x − 1)5 − (x − 1)4 4 + (x − 1)5 5 2. The key to writing these sums with summation notation is to ﬁnd the pattern of the terms. To that end, we make good use of the techniques presented in Section 9.1. (a) The terms of the sum 1, 3, 5, etc., form an arithmetic sequence with ﬁrst term a = 1 and common diﬀerence d = 2. We get a formula for the nth term of the sequence using Equation 9.1 to get an = 1 + (n − 1)2 = 2n − 1, n ≥ 1. At this stage, we have the formula for the terms, namely 2n − 1, and the lower limit of the summation, n = 1. To ﬁnish the problem, we need to determine the upper limit of the summation. In other words, we need to determine which value of n produces the term 117. Setting an = 117, we get 2n − 1 = 117 or n = 59. Our ﬁnal answer is 1 + 3 + 5 +... + 117 = 59 (2n − 1) n=1 (b) We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives ‘−’
with the numerators to get 1 1 + −1 2 + 1 3 + −1 4 +... + 1 117 The numerators, 1, −1, etc. can be described by the geometric sequence1 cn = (−1)n−1 for n ≥ 1, while the denominators are given by the arithmetic sequence2 dn = n for n ≥ 1. Hence, we get the formula an = (−1)n−1 for our terms, and we ﬁnd the lower and upper limits of summation to be n = 1 and n = 117, respectively. Thus 117 = 117 n=1 (−1)n−1 n (c) Thanks to Example 9.1.3, we know that one formula for the nth term is an = 9 10n for n ≥ 1. This gives us a formula for the summation as well as a lower limit of summation. To determine the upper limit of summation, we note that to produce the n − 1 zeros to the right of the decimal point before the 9, we need a denominator of 10n. Hence, n is 1This is indeed a geometric sequence with ﬁrst term a = 1 and common ratio r = −1. 2It is an arithmetic sequence with ﬁrst term a = 1 and common diﬀerence d = 1. 664 Sequences and the Binomial Theorem the upper limit of summation. Since n is used in the limits of the summation, we need to choose a diﬀerent letter for the index of summation.3 We choose k and get 0.9 + 0.09 + 0.009 +... 0. 0 · · · 0 n − 1 zeros 9 = n k=1 9 10k The following theorem presents some general properties of summation notation. While we shall not have much need of these properties in Algebra, they do play a great role in Calculus. Moreover, there is much to be learned by thinking about why the properties hold. We invite the reader to prove these results. To get started, remember, “When in doubt, write it out!” Theorem 9.1. Properties of Summation Notation: Suppose {an} and {bn} are sequences so that the following sums are deﬁned. p n=m p n=m p (an ± bn) = p p bn an ±
n=m n=m c an = c p n=m an, for any real number c. j an + p an = an, for any natural number m ≤ j < j + 1 ≤ p. n=m n=m n=j+1 p an = p+r n=m n=m+r an−r, for any whole number r. We now turn our attention to the sums involving arithmetic and geometric sequences. Given an arithmetic sequence ak = a + (k − 1)d for k ≥ 1, we let S denote the sum of the ﬁrst n terms. To derive a formula for S, we write it out in two diﬀerent ways S = S = (a + (n − 1)d) + (a + (n − 2)d) +... + (a + d) + a +... + (a + (n − 2)d) + (a + (n − 1)d) (a + d) + a If we add these two equations and combine the terms which are aligned vertically, we get 2S = (2a + (n − 1)d) + (2a + (n − 1)d) +... + (2a + (n − 1)d) + (2a + (n − 1)d) The right hand side of this equation contains n terms, all of which are equal to (2a + (n − 1)d) so we get 2S = n(2a + (n − 1)d). Dividing both sides of this equation by 2, we obtain the formula 3To see why, try writing the summation using ‘n’ as the index. 9.2 Summation Notation 665 S = n 2 (2a + (n − 1)d) If we rewrite the quantity 2a + (n − 1)d as a + (a + (n − 1)d) = a1 + an, we get the formula S = n a1 + an 2 A helpful way to remember this last formula is to recognize that we have expressed the sum as the product of the number of terms n and the average of the ﬁrst and nth terms. To derive the formula for the geometric sum, we start with a geometric sequence ak = ark−1, k ≥ 1, and let S once again
denote the sum of the ﬁrst n terms. Comparing S and rS, we get S = a + ar + ar2 +... + arn−2 + arn−1 rS = ar + ar2 +... + arn−2 + arn−1 + arn Subtracting the second equation from the ﬁrst forces all of the terms except a and arn to cancel out and we get S − rS = a − arn. Factoring, we get S(1 − r) = a (1 − rn). Assuming r = 1, we can divide both sides by the quantity (1 − r) to obtain S = a 1 − rn 1 − r If we distribute a through the numerator, we get a − arn = a1 − an+1 which yields the formula S = a1 − an+1 1 − r In the case when r = 1, we get the formula times = n a Our results are summarized below. 666 Sequences and the Binomial Theorem Equation 9.2. Sums of Arithmetic and Geometric Sequences: The sum S of the ﬁrst n terms of an arithmetic sequence ak = a + (k − 1)d for k ≥ 1 is S = n k=1 ak = n a1 + an 2 = n 2 (2a + (n − 1)d) The sum S of the ﬁrst n terms of a geometric sequence ak = ark−1 for k ≥ 1 is 1. S = 2. S = n k=1 n k=1 ak = a1 − an+1 1 − r = a 1 − rn 1 − r, if r = 1. ak = n k=1 a = na, if r = 1. While we have made an honest eﬀort to derive the formulas in Equation 9.2, formal proofs require the machinery in Section 9.3. An application of the arithmetic sum formula which proves useful in Calculus results in formula for the sum of the ﬁrst n natural numbers. The natural numbers themselves are a sequence4 1, 2, 3,... which is arithmetic with a = d = 1. Applying Equation 9.2(n + 1) 2 2 = 5050. So, for example, the sum of the ﬁrst 100 natural numbers5
is 100(101) An important application of the geometric sum formula is the investment plan called an annuity. Annuities diﬀer from the kind of investments we studied in Section 6.5 in that payments are deposited into the account on an on-going basis, and this complicates the mathematics a little.6 Suppose you have an account with annual interest rate r which is compounded n times per year. We let i = r n denote the interest rate per period. Suppose we wish to make ongoing deposits of P dollars at the end of each compounding period. Let Ak denote the amount in the account after k compounding periods. Then A1 = P, because we have made our ﬁrst deposit at the end of the ﬁrst compounding period and no interest has been earned. During the second compounding period, we earn interest on A1 so that our initial investment has grown to A1(1 + i) = P (1 + i) in accordance with Equation 6.1. When we add our second payment at the end of the second period, we get A2 = A1(1 + i) + P = P (1 + i) + P = P (1 + i) 1 + 1 1 + i The reason for factoring out the P (1 + i) will become apparent in short order. During the third compounding period, we earn interest on A2 which then grows to A2(1 + i). We add our third 4This is the identity function on the natural numbers! 5There is an interesting anecdote which says that the famous mathematician Carl Friedrich Gauss was given this problem in primary school and devised a very clever solution. 6The reader may wish to re-read the discussion on compound interest in Section 6.5 before proceeding. 9.2 Summation Notation 667 payment at the end of the third compounding period to obtain A3 = A2(1 + i) + P = P (1 + i) 1 + 1 1 + i (1 + i) + P = P (1 + i)1 + i)2 During the fourth compounding period, A3 grows to A3(1+i), and when we add the fourth payment, we factor out P (1 + i)3 to get A4 = P (1 + i)1 + i)2 + 1 (1 + i)3 This pattern continues so that at the end of the kth
compounding, we get Ak = P (1 + i)k−1 + i)2 +... + 1 (1 + i)k−1 The sum in the parentheses above is the sum of the ﬁrst k terms of a geometric sequence with a = 1 and r = 1 1+i. Using Equation 9.2, we get 1 + 1 1 + i + 1 (1 + i)2 +... + 1 (1 + i)k−1 = 1     Hence, we get 1 − 1 (1 + i)1 + i) 1 − (1 + i)−k i Ak = P (1 + i)k−1 (1 + i) 1 − (1 + i)−k i P (1 + i)k − 1 i = If we let t be the number of years this investment strategy is followed, then k = nt, and we get the formula for the future value of an ordinary annuity. Equation 9.3. Future Value of an Ordinary Annuity: Suppose an annuity oﬀers an annual interest rate r compounded n times per year. Let i = r n be the interest rate per compounding period. If a deposit P is made at the end of each compounding period, the amount A in the account after t years is given by A = P (1 + i)nt − 1 i The reader is encouraged to substitute i = r n into Equation 9.3 and simplify. Some familiar equations arise which are cause for pause and meditation. One last note: if the deposit P is made a the beginning of the compounding period instead of at the end, the annuity is called an annuitydue. We leave the derivation of the formula for the future value of an annuity-due as an exercise for the reader. 668 Sequences and the Binomial Theorem Example 9.2.2. An ordinary annuity oﬀers a 6% annual interest rate, compounded monthly. 1. If monthly payments of $50 are made, ﬁnd the value of the annuity in 30 years. 2. How many years will it take for the annuity to grow to $100,000? Solution. 1. We have r = 0.06 and n = 12 so that i = r n = 0.06 12 = 0.005. With P =
50 and t = 30, A = 50 (1 + 0.005)(12)(30) − 1 0.005 ≈ 50225.75 Our ﬁnal answer is $50,225.75. 2. To ﬁnd how long it will take for the annuity to grow to $100,000, we set A = 100000 and solve for t. We isolate the exponential and take natural logs of both sides of the equation. 100000 = 50 (1 + 0.005)12t − 1 0.005 10 = (1.005)12t − 1 (1.005)12t = 11 ln (1.005)12t = ln(11) 12t ln(1.005) = ln(11) ln(11) t = 12 ln(1.005) ≈ 40.06 This means that it takes just over 40 years for the investment to grow to $100,000. Comparing this with our answer to part 1, we see that in just 10 additional years, the value of the annuity nearly doubles. This is a lesson worth remembering. We close this section with a peek into Calculus by considering inﬁnite sums, called series. Consider the number 0.9. We can write this number as 0.9 = 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 +... From Example 9.2.1, we know we can write the sum of the ﬁrst n of these terms as 0. 9 · · · 9 n nines =.9 + 0.09 + 0.009 +... 0. 0 · · · 0 n − 1 zeros 9 = n k=1 9 10k Using Equation 9.2, we have 9.2 Summation Notation 669   k=1   1 − n 9 10 9 10k = 1 10n+1 1 10 It stands to reason that 0.9 is the same value of 1 − 1 10n+1 as n → ∞. Our knowledge of exponential 10n+1 → 1. We have expressions from Section 6.1 tells us that just argued that 0.9 = 1, which may cause some distress for some readers.7 Any non-terminating decimal can be thought of as an inﬁ
nite sum whose denominators are the powers of 10, so the phenomenon of adding up inﬁnitely many terms and arriving at a ﬁnite number is not as foreign of a concept as it may appear. We end this section with a theorem concerning geometric series. 10n+1 → 0 as n → ∞, so 1 − 1   = 1 − 1 10n+1 1 − 1 Theorem 9.2. Geometric Series: Given the sequence ak = ark−1 for k ≥ 1, where \|r\| < 1, a + ar + ar2 +... = ∞ ark−1 = k=1 If \|r\| ≥ 1, the sum a + ar + ar2 +... is not deﬁned. a 1 − r The justiﬁcation of the result in Theorem 9.2 comes from taking the formula in Equation 9.2 for the sum of the ﬁrst n terms of a geometric sequence and examining the formula as n → ∞. Assuming \|r\| < 1 means −1 < r < 1, so rn → 0 as n → ∞. Hence as n → ∞, n k=1 ark−1 = a 1 − rn 1 − r → a 1 − r As to what goes wrong when \|r\| ≥ 1, we leave that to Calculus as well, but will explore some cases in the exercises. 7To make this more palatable, it is usually accepted that 0.3 = 1 3 so that 0.9 = 3 0.3 = 3 1 3 = 1. Feel better? 670 Sequences and the Binomial Theorem 9.2.1 Exercises In Exercises 1 - 8, ﬁnd the value of each sum using Deﬁnition 9.3. 1. 5. 9 (5g + 3) g=4 4 i=1 1 4 (i2 + 1) 2. 6. 8 k=3 1 k 100 (−1)n n=1 3. 7. 5 j=0 5 n=1 2j (n + 1)! n! 4. 8. 2 (3k − 5)xk k=0 3 j=1 5! j! (5 − j)! In Exercises 9 - 16, rewrite the sum using summation notation. 9. 8 + 11
+ 14 + 17 + 20 10 11. x − x3 3 + x5 5 − x7 7 13 15 16 + 1 25 − 1 36 12. 1 + 2 + 4 + · · · + 229 14. − ln(3) + ln(4) − ln(5) + · · · + ln(20) 16. 1 2 (x − 5) + 1 4 (x − 5)2 + 1 6 (x − 5)3 + 1 8 (x − 5)4 In Exercises 17 - 28, use the formulas in Equation 9.2 to ﬁnd the sum. 17. 20. 10 n=1 5n + 3 n 10 n=1 1 2 18. 21. 20 n=1 2n − 1 n 5 n=1 3 2 19. 22. 15 k=0 3 − k k 5 k=0 1 4 2 23. 1 + 4 + 7 +... + 295 24. 4 + 2 + 0 − 2 −... − 146 25. 1 + 3 + 9 +... + 2187 26 256 27 256 28. 10 n=1 −2n + n 5 3 In Exercises 29 - 32, use Theorem 9.2 to express each repeating decimal as a fraction of integers. 29. 0.7 30. 0.13 31. 10.159 32. −5.867 9.2 Summation Notation 671 In Exercises 33 - 38, use Equation 9.3 to compute the future value of the annuity with the given terms. In all cases, assume the payment is made monthly, the interest rate given is the annual rate, and interest is compounded monthly. 33. payments are $300, interest rate is 2.5%, term is 17 years. 34. payments are $50, interest rate is 1.0%, term is 30 years. 35. payments are $100, interest rate is 2.0%, term is 20 years 36. payments are $100, interest rate is 2.0%, term is 25 years 37. payments are $100, interest rate is 2.0%, term is 30 years 38. payments are $100, interest rate is 2.0%, term is 35 years 39. Suppose an ordinary annuity oﬀers an annual interest rate of 2%, compounded monthly, for 30 years. What should the monthly payment be to have $100,000 at
the end of the term? 40. Prove the properties listed in Theorem 9.1. 41. Show that the formula for the future value of an annuity due is A = P (1 + i) (1 + i)nt − 1 i 42. Discuss with your classmates what goes wrong when trying to ﬁnd the following sums.8 (a) ∞ k=1 2k−1 ∞ (1.0001)k−1 (b) k=1 ∞ (−1)k−1 (c) k=1 8When in doubt, write them out! 672 Sequences and the Binomial Theorem 9.2.2 Answers 1. 213 5. 17 2 9. 13. 5 (3k + 5) k=1 5 k=1 k + 1 k 17. 305 21. 633 32 25. 3280 29. 7 9 2. 341 280 6. 0 10. 14. 8 k=1 20 k=3 (−1)k−1k (−1)k ln(k) 18. 400 22. 26. 30. 1365 512 255 256 13 99 3. 63 7. 20 11. 15. 4 k=1 6 k=1 (−1)k−1 x2k−1 2k − 1 (−1)k−1 k2 19. −72 4. −5 − 2x + x2 8. 25 12. 16. 20. 30 k=1 4 k=1 2k−1 1 2k (x − 5)k 1023 1024 23. 14652 24. −5396 27. 31. 513 256 3383 333 28. 17771050 59049 32. − 5809 990 33. $76,163.67 34. $20,981.40 35. $29,479.69 36. $38,882.12 37. 49,272.55 38. 60,754.80 39. For $100,000, the monthly payment is ≈ $202.95. 9.3 Mathematical Induction 673 9.3 Mathematical Induction The Chinese philosopher Confucius is credited with the saying, “A journey of a thousand miles begins with a single step.” In many ways, this is the central theme of this section. Here we introduce a method of proof, Mathematical Induction, which allows us to prove many of the formulas we have merely motivated in Sections
9.1 and 9.2 by starting with just a single step. A good example is the formula for arithmetic sequences we touted in Equation 9.1. Arithmetic sequences are deﬁned recursively, starting with a1 = a and then an+1 = an + d for n ≥ 1. This tells us that we start the sequence with a and we go from one term to the next by successively adding d. In symbols, a, a + d, a + 2d, a + 3d, a + 4d +... The pattern suggested here is that to reach the nth term, we start with a and add d to it exactly n − 1 times, which lead us to our formula an = a + (n − 1)d for n ≥ 1. But how do we prove this to be the case? We have the following. The Principle of Mathematical Induction (PMI): Suppose P (n) is a sentence involving the natural number n. IF 1. P (1) is true and 2. whenever P (k) is true, it follows that P (k + 1) is also true THEN the sentence P (n) is true for all natural numbers n. The Principle of Mathematical Induction, or PMI for short, is exactly that - a principle.1 It is a property of the natural numbers we either choose to accept or reject. In English, it says that if we want to prove that a formula works for all natural numbers n, we start by showing it is true for n = 1 (the ‘base step’) and then show that if it is true for a generic natural number k, it must be true for the next natural number, k + 1 (the ‘inductive step’). The notation P (n) acts just like function notation. For example, if P (n) is the sentence (formula) ‘n2 + 1 = 3’, then P (1) would be ‘12 + 1 = 3’, which is false. The construction P (k + 1) would be ‘(k + 1)2 + 1 = 3’. As usual, this new concept is best illustrated with an example. Returning to our quest to prove the formula for an arithmetic sequence, we ﬁrst identify P (n) as the formula an = a + (n − 1)d. To prove this formula
is valid for all natural numbers n, we need to do two things. First, we need to establish that P (1) is true. In other words, is it true that a1 = a + (1 − 1)d? The answer is yes, since this simpliﬁes to a1 = a, which is part of the deﬁnition of the arithmetic sequence. The second thing we need to show is that whenever P (k) is true, it follows that P (k + 1) is true. In other words, we assume P (k) is true (this is called the ‘induction hypothesis’) and deduce that P (k + 1) is also true. Assuming P (k) to be true seems to invite disaster - after all, isn’t this essentially what we’re trying to prove in the ﬁrst place? To help explain this step a little better, we show how this works for speciﬁc values of n. We’ve already established P (1) is true, and we now want to show that P (2) 1Another word for this you may have seen is ‘axiom.’ 674 Sequences and the Binomial Theorem is true. Thus we need to show that a2 = a + (2 − 1)d. Since P (1) is true, we have a1 = a, and by the deﬁnition of an arithmetic sequence, a2 = a1 +d = a+d = a+(2−1)d. So P (2) is true. We now use the fact that P (2) is true to show that P (3) is true. Using the fact that a2 = a + (2 − 1)d, we show a3 = a +(3 − 1)d. Since a3 = a2 + d, we get a3 = (a +(2 − 1)d)+ d = a +2d = a +(3 − 1)d, so we have shown P (3) is true. Similarly, we can use the fact that P (3) is true to show that P (4) is true, and so forth. In general, if P (k) is true (i.e., ak = a+(k −1)d) we set out to show that P (k +
1) is true (i.e., ak+1 = a + ((k + 1) − 1)d). Assuming ak = a + (k − 1)d, we have by the deﬁnition of an arithmetic sequence that ak+1 = ak + d so we get ak+1 = (a + (k − 1)d) + d = a + kd = a + ((k + 1) − 1)d. Hence, P (k + 1) is true. In essence, by showing that P (k + 1) must always be true when P (k) is true, we are showing that the formula P (1) can be used to get the formula P (2), which in turn can be used to derive the formula P (3), which in turn can be used to establish the formula P (4), and so on. Thus as long as P (k) is true for some natural number k, P (n) is true for all of the natural numbers n which follow k. Coupling this with the fact P (1) is true, we have established P (k) is true for all natural numbers which follow n = 1, in other words, all natural numbers n. One might liken Mathematical Induction to a repetitive process like climbing stairs.2 If you are sure that (1) you can get on the stairs (the base case) and (2) you can climb from any one step to the next step (the inductive step), then presumably you can climb the entire staircase.3 We get some more practice with induction in the following example. Example 9.3.1. Prove the following assertions using the Principle of Mathematical Induction. 1. The sum formula for arithmetic sequences: n (a + (j − 1)d) = j=1 2. For a complex number z, (z)n = zn for n ≥ 1. n 2 (2a + (n − 1)d). 3. 3n > 100n for n > 5. 4. Let A be an n × n matrix and let A be the matrix obtained by replacing a row R of A with cR for some real number c. Use the deﬁnition of determinant to show det(A) = c det(A). Solution. 1. We set P (n) to be the equation we are asked to prove. For n = 1, we
compare both sides of the equation given in P (n) 1 j=1 (a + (j − 1)d) a + (1 − 1)2a + (1 − 1)d) (2a) a = a 2Falling dominoes is the most widely used metaphor in the mainstream College Algebra books. 3This is how Carl climbed the stairs in the Cologne Cathedral. Well, that, and encouragement from Kai. 9.3 Mathematical Induction 675 This shows the base case P (1) is true. Next we assume P (k) is true, that is, we assume k (a + (j − 1)d) = j=1 k 2 (2a + (k − 1)d) and attempt to use this to show P (k + 1) is true. Namely, we must show k+1 (a + (j − 1)d) = j=1 k + 1 2 (2a + (k + 1 − 1)d) To see how we can use P (k) in this case to prove P (k + 1), we note that the sum in P (k + 1) is the sum of the ﬁrst k + 1 terms of the sequence ak = a + (k − 1)d for k ≥ 1 while the sum in P (k) is the sum of the ﬁrst k terms. We compare both side of the equation in P (k + 1). k+1 (a + (j − 1)d) j=1 summing the ﬁrst k + 1 terms k (a + (j − 1)d) + (a + (k + 1 − 1)d) j=1 summing the ﬁrst k terms adding the (k + 1)st term? = k + 1 2 (2a + (k + 1 − 1)d)? = k + 1 2 (2a + kd) k 2 (2a + (k − 1)d) Using P (k) +(a + kd) k(2a + (k − 1)d) + 2(a + kd) 2 2ka + 2a + k2d + kd 2? =? = = (k + 1)(2a + kd) 2 2ka + k2d + 2a + kd 2 2ka + 2a
+ k2d + kd 2 Since all of our steps on both sides of the string of equations are reversible, we conclude that the two sides of the equation are equivalent and hence, P (k + 1) is true. By the Principle of Mathematical Induction, we have that P (n) is true for all natural numbers n. 2. We let P (n) be the formula (z)n = zn. The base case P (1) is (z)1 = z1, which reduces to z = z which is true. We now assume P (k) is true, that is, we assume (z)k = zk and attempt to show that P (k + 1) is true. Since (z)k+1 = (z)k z, we can use the induction hypothesis and 676 Sequences and the Binomial Theorem write (z)k = zk. Hence, (z)k+1 = (z)k z = zk z. We now use the product rule for conjugates4 to write zk z = zkz = zk+1. This establishes (z)k+1 = zk+1, so that P (k + 1) is true. Hence, by the Principle of Mathematical Induction, (z)n = zn for all n ≥ 1. 3. The ﬁrst wrinkle we encounter in this problem is that we are asked to prove this formula for n > 5 instead of n ≥ 1. Since n is a natural number, this means our base step occurs at n = 6. We can still use the PMI in this case, but our conclusion will be that the formula is valid for all n ≥ 6. We let P (n) be the inequality 3n > 100n, and check that P (6) is true. Comparing 36 = 729 and 100(6) = 600, we see 36 > 100(6) as required. Next, we assume that P (k) is true, that is we assume 3k > 100k. We need to show that P (k + 1) is true, that is, we need to show 3k+1 > 100(k + 1). Since 3k+1 = 3 · 3k, the induction hypothesis gives 3k+1 = 3 · 3k > 3(100k) = 300k. We are done if we can show 300
k > 100(k + 1) for k ≥ 6. Solving 300k > 100(k + 1) we get k > 1 2. Since k ≥ 6, we know this is true. Putting all of this together, we have 3k+1 = 3 · 3k > 3(100k) = 300k > 100(k + 1), and hence P (k + 1) is true. By induction, 3n > 100n for all n ≥ 6. 4. To prove this determinant property, we use induction on n, where we take P (n) to be that the property we wish to prove is true for all n × n matrices. For the base case, we note that if A is a 1 × 1 matrix, then A = [a] so A = [ca]. By deﬁnition, det(A) = a and det(A) = ca so we have det(A) = c det(A) as required. Now suppose that the property we wish to prove is true for all k × k matrices. Let A be a (k + 1) × (k + 1) matrix. We have two cases, depending on whether or not the row R being replaced is the ﬁrst row of A. Case 1: The row R being replaced is the ﬁrst row of A. By deﬁnition, det(A) = n p=1 1pC a 1p 1p = (−1)(1+p) det A where the 1p cofactor of A is C 1p is the k × k matrix obtained by deleting the 1st row and pth column of A.5 Since the ﬁrst row of A is c times the ﬁrst 1p = c a1p. In addition, since the remaining rows of A are identical to row of A, we have a 1p = A1p. (To obtain these matrices, the ﬁrst row of A is removed.) Hence those of A, A det A 1p = det (A1p), so that C 1p = C1p. As a result, we get and A 1p det(A) = n p=1 1pC a 1p = n p=1 n c a1pC1p = c p=1 a1pC1p = c det(A), as
required. Hence, P (k + 1) is true in this case, which means the result is true in this case for all natural numbers n ≥ 1. (You’ll note that we did not use the induction hypothesis at all in this case. It is possible to restructure the proof so that induction is only used where 4See Exercise 54 in Section 3.4. 5See Section 8.5 for a review of this notation. 9.3 Mathematical Induction 677 it is needed. While mathematically more elegant, it is less intuitive, and we stand by our approach because of its pedagogical value.) Case 2: The row R being replaced is the not the ﬁrst row of A. By deﬁnition, det(A) = n p=1 1pC a 1p, where in this case, a 1p = a1p, since the ﬁrst rows of A and A are the same. The matrices 1p and A1p, on the other hand, are diﬀerent but in a very predictable way − the row in A A 1p which corresponds to the row cR in A is exactly c times the row in A1p which corresponds to the row R in A. In other words, A 1p and A1p are k × k matrices which satisfy the induction hypothesis. Hence, we know det A 1p = c det (A1p) and C 1p = c C1p. We get det(A) = n p=1 1pC a 1p = n p=1 n a1pc C1p = c p=1 a1pC1p = c det(A), which establishes P (k + 1) to be true. Hence by induction, we have shown that the result holds in this case for n ≥ 1 and we are done. While we have used the Principle of Mathematical Induction to prove some of the formulas we have merely motivated in the text, our main use of this result comes in Section 9.4 to prove the celebrated Binomial Theorem. The ardent Mathematics student will no doubt see the PMI in many courses yet to come. Sometimes it is explicitly stated and sometimes it remains hidden in the background. If ever you see a property stated as being true ‘for all natural numbers n’, it’s a solid bet that the formal proof requires the Principle of Mathematical
Induction. 678 Sequences and the Binomial Theorem 9.3.1 Exercises In Exercises 1 - 7, prove each assertion using the Principle of Mathematical Induction. 1. 2. n j=1 n j=1 j2 = n(n + 1)(2n + 1) 6 j3 = n2(n + 1)2 4 3. 2n > 500n for n > 12 4. 3n ≥ n3 for n ≥ 4 5. Use the Product Rule for Absolute Value to show \|xn\| = \|x\|n for all real numbers x and all natural numbers n ≥ 1 6. Use the Product Rule for Logarithms to show log (xn) = n log(x) for all real numbers x > 0 and all natural numbers n ≥ 1. an a 0 0 0 b 0 bn n = for n ≥ 1. 7. 8. Prove Equations 9.1 and 9.2 for the case of geometric sequences. That is: (a) For the sequence a1 = a, an+1 = ran, n ≥ 1, prove an = arn−1, n ≥ 1. (b) n j=1 arn−1 = a 1 − rn 1 − r, if r = 1, n j=1 arn−1 = na, if r = 1. 9. Prove that the determinant of a lower triangular matrix is the product of the entries on the main diagonal. (See Exercise 8.3.1 in Section 8.3.) Use this result to then show det (In) = 1 where In is the n × n identity matrix. 10. Discuss the classic ‘paradox’ All Horses are the Same Color problem with your classmates. 9.3 Mathematical Induction 679 9.3.2 Selected Answers 1. Let P (n) be the sentence n j=1 j2 = n(n + 1)(2n + 1) 6. For the base case, n = 1, we get 1 j=1 j2? = (1)(1 + 1)(2(1) + 1) 6 12 = 1 We now assume P (k) is true and use it to show P (k + 1) is true. We have k+1 j=1 j2? = (k + 1)((k + 1) + 1)(2(k + 1) + 1
) 6 k j=1 j2 + (k + 1)2 k(k + 1)(2k + 1) 6 Using P (k) +(k + 1)2 k(k + 1)(2k + 1) 6 + 6(k + 1)2 6 k(k + 1)(2k + 1) + 6(k + 1)2 6 (k + 1)(k(2k + 1) + 6(k + 1)) 6 (k + 1) 2k2 + 7k + 6 6 (k + 1)(k + 2)(2k + 3k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 By induction, n j=1 j2 = n(n + 1)(2n + 1) 6 is true for all natural numbers n ≥ 1. 4. Let P (n) be the sentence 3n > n3. Our base case is n = 4 and we check 34 = 81 and 43 = 64 so that 34 > 43 as required. We now assume P (k) is true, that is 3k > k3, and try to show P (k + 1) is true. We note that 3k+1 = 3 · 3k > 3k3 and so we are done if we can show 3k3 > (k + 1)3 for k ≥ 4. We can solve the inequality 3x3 > (x + 1)3 using the techniques of Section 5.3, and doing so gives us x > 1 ≈ 2.26. Hence, for k ≥ 4, 3√ 3k+1 = 3 · 3k > 3k3 > (k + 1)3 so that 3k+1 > (k + 1)3. By induction, 3n > n3 is true for all natural numbers n ≥ 4. 3−1 680 Sequences and the Binomial Theorem 6. Let P (n) be the sentence log (xn) = n log(x). For the duration
of this argument, we assume x > 0. The base case P (1) amounts checking that log x1 = 1 log(x) which is clearly true. Next we assume P (k) is true, that is log xk = k log(x) and try to show P (k + 1) is true. Using the Product Rule for Logarithms along with the induction hypothesis, we get xk+1 log = log xk · x = log xk + log(x) = k log(x) + log(x) = (k + 1) log(x) Hence, log xk+1 = (k + 1) log(x). By induction log (xn) = n log(x) is true for all x > 0 and all natural numbers n ≥ 1. 9. Let A be an n × n lower triangular matrix. We proceed to prove the det(A) is the product of the entries along the main diagonal by inducting on n. For n = 1, A = [a] and det(A) = a, so the result is (trivially) true. Next suppose the result is true for k × k lower triangular matrices. Let A be a (k + 1) × (k + 1) lower triangular matrix. Expanding det(A) along the ﬁrst row, we have det(A) = n p=1 a1pC1p Since a1p = 0 for 2 ≤ p ≤ k + 1, this simpliﬁes det(A) = a11C11. By deﬁnition, we know that C11 = (−1)1+1 det (A11) = det (A11) where A11 is k × k matrix obtained by deleting the ﬁrst row and ﬁrst column of A. Since A is lower triangular, so is A11 and, as such, the induction hypothesis applies to A11. In other words, det (A11) is the product of the entries along A11’s main diagonal. Now, the entries on the main diagonal of A11 are the entries a22, a33,..., a(k+1)(k+1) from the main diagonal of A. Hence, det(A) = a11 det (A11) = a11 a22a33 · · · a(k
+1)(k+1) = a11a22a33 · · · a(k+1)(k+1) We have det(A) is the product of the entries along its main diagonal. This shows P (k + 1) is true, and, hence, by induction, the result holds for all n × n upper triangular matrices. The n × n identity matrix In is a lower triangular matrix whose main diagonal consists of all 1’s. Hence, det (In) = 1, as required. 9.4 The Binomial Theorem 681 9.4 The Binomial Theorem In this section, we aim to prove the celebrated Binomial Theorem. Simply stated, the Binomial Theorem is a formula for the expansion of quantities (a + b)n for natural numbers n. In Elementary and Intermediate Algebra, you should have seen speciﬁc instances of the formula, namely (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3 If we wanted the expansion for (a + b)4 we would write (a + b)4 = (a + b)(a + b)3 and use the formula that we have for (a+b)3 to get (a+b)4 = (a+b) a3 + 3a2b + 3ab2 + b3 = a4+4a3b+6a2b2+4ab3+b4. Generalizing this a bit, we see that if we have a formula for (a + b)k, we can obtain a formula for (a + b)k+1 by rewriting the latter as (a + b)k+1 = (a + b)(a + b)k. Clearly this means Mathematical Induction plays a major role in the proof of the Binomial Theorem.1 Before we can state the theorem we need to revisit the sequence of factorials which were introduced in Example 9.1.1 number 6 in Section 9.1. Deﬁnition 9.4. Factorials: For a whole number n, n factorial, denoted n!, is the term fn of the sequence f0 = 1, fn = n · fn−1, n ≥ 1. Recall this means 0! = 1
and n! = n(n − 1)! for n ≥ 1. Using the recursive deﬁnition, we get: 1! = 1 · 0! = 1 · 1 = 1, 2! = 2 · 1! = 2 · 1 = 2, 3! = 3 · 2! = 3 · 2 · 1 = 6 and 4! = 4 · 3! = 4 · 3 · 2 · 1 = 24. Informally, n! = n · (n − 1) · (n − 2) · · · 2 · 1 with 0! = 1 as our ‘base case.’ Our ﬁrst example familiarizes us with some of the basic computations involving factorials. Example 9.4.1. 1. Simplify the following expressions. (a) 3! 2! 0! (b) 7! 5! (c) 1000! 998! 2! (d) (k + 2)! (k − 1)!, k ≥ 1 2. Prove n! > 3n for all n ≥ 7. Solution. 1. We keep in mind the mantra, “When in doubt, write it out!” as we simplify the following. (a) We have been programmed to react with alarm to the presence of a 0 in the denominator, but in this case 0! = 1, so the fraction is deﬁned after all. As for the numerator, 3! = 3 · 2 · 1 = 6 and 2! = 2 · 1 = 2, so we have 3! 2! 0! = (6)(2) 1 = 12. 1It’s pretty much the reason Section 9.3 is in the book. 682 Sequences and the Binomial Theorem (b) We have 7 = 5040 while 5 = 120. Dividing, we get 120 = 42. While this is correct, we note that we could have saved ourselves some 7! 5! = 5040 of time had we proceeded as follows 7! 5 = 42 In fact, should we want to fully exploit the recursive nature of the factorial, we can write 7! 5! = 7 · 6 · 5! 5! = 7 · 6 · 5! 5! = 42 (c) Keeping in mind the lesson we learned from the previous problem, we have 1000! 998! 2! = 1000 · 999 · 998! 998! · 2! = 1000 ·
999 · 998! 998! · 2! = 999000 2 = 499500 (d) This problem continues the theme which we have seen in the previous two problems. We ﬁrst note that since k + 2 is larger than k − 1, (k + 2)! contains all of the factors of (k − 1)! and as a result we can get the (k − 1)! to cancel from the denominator. To see this, we begin by writing out (k + 2)! starting with (k + 2) and multiplying it by the numbers which precede it until we reach (k − 1): (k + 2)! = (k + 2)(k + 1)(k)(k − 1)!. As a result, we have (k + 2)! (k − 1)! = (k + 2)(k + 1)(k)(k − 1)! (k − 1)! = (k + 2)(k + 1)(k) (k − 1)! (k − 1)! = k(k + 1)(k + 2) The stipulation k ≥ 1 is there to ensure that all of the factorials involved are deﬁned. 2. We proceed by induction and let P (n) be the inequality n! > 3n. The base case here is n = 7 and we see that 7! = 5040 is larger than 37 = 2187, so P (7) is true. Next, we assume that P (k) is true, that is, we assume k! > 3k and attempt to show P (k + 1) follows. Using the properties of the factorial, we have (k + 1)! = (k + 1)k! and since k! > 3k, we have (k + 1)! > (k + 1)3k. Since k ≥ 7, k + 1 ≥ 8, so (k + 1)3k ≥ 8 · 3k > 3 · 3k = 3k+1. Putting all of this together, we have (k + 1)! = (k + 1)k! > (k + 1)3k > 3k+1 which shows P (k + 1) is true. By the Principle of Mathematical Induction, we have n! > 3n for all n ≥ 7. Of all of the mathematical animals we have discussed in the text, factorials grow most quickly. In problem 2 of Example 9.
4.1, we proved that n! overtakes 3n at n = 7. ‘Overtakes’ may be too polite a word, since n! thoroughly trounces 3n for n ≥ 7, as any reasonable set of data will show. It can be shown that for any real number x > 0, not only does n! eventually overtake xn, but the ratio xn Applications of factorials in the wild often involve counting arrangements. For example, if you have ﬁfty songs on your mp3 player and wish arrange these songs in a playlist in which the order of the n! → 0 as n → ∞.2 2This fact is far more important than you could ever possibly imagine. 9.4 The Binomial Theorem 683 songs matters, it turns out that there are 50! diﬀerent possible playlists. If you wish to select only ten of the songs to create a playlist, then there are 50! 40! such playlists. If, on the other hand, you just want to select ten song ﬁles out of the ﬁfty to put on a ﬂash memory card so that now the order 50! 40!10! ways to achieve this.3 While some of these ideas are explored no longer matters, there are in the Exercises, the authors encourage you to take courses such as Finite Mathematics, Discrete Mathematics and Statistics. We introduce these concepts here because this is how the factorials make their way into the Binomial Theorem, as our next deﬁnition indicates. Deﬁnition 9.5. Binomial Coeﬃcients: Given two whole numbers n and j with n ≥ j, the binomial coeﬃcient n j (read, n choose j) is the whole number given by n j = n! j!(n − j)! The name ‘binomial coeﬃcient’ will be justiﬁed shortly. For now, we can physically interpret as the number of ways to select j items from n items where the order of the items selected is n j unimportant. For example, suppose you won two free tickets to a special screening of the latest Hollywood blockbuster and have ﬁve good friends each of whom would love to accompany you to ways to choose who goes with you. Applying Deﬁnition 9.5, we get the
movies. There are 5 2 5 2 = 5! 2!(5 − 2)! = 5! 2!3! = 5 · 4 2 = 10 So there are 10 diﬀerent ways to distribute those two tickets among ﬁve friends. (Some will see it as 10 ways to decide which three friends have to stay home.) The reader is encouraged to verify this by actually taking the time to list all of the possibilities. We now state anf prove a theorem which is crucial to the proof of the Binomial Theorem. Theorem 9.3. For natural numbers n and j with n ≥ j = The proof of Theorem 9.3 is purely computational and uses the deﬁnition of binomial coeﬃcients, the recursive property of factorials and common denominators. 3For reference, 50! 50! 40! 50! 40!10! = 30414093201713378043612608166064768844377641568960512000000000000, = 37276043023296000, and = 10272278170 684 Sequences and the Binomial Theorem! (j − 1)!(n − (j − 1))! + n! j!(n − j)! n! (j − 1)!(n − j + 1)! + n! j!(n − j)! n! (j − 1)!(n − j + 1)(n − j)! + n! j(j − 1)!(n − j)! n! j j(j − 1)!(n − j + 1)(n − j)! + n!(n − j + 1) j(j − 1)!(n − j + 1)(n − j)! n! j j!(n − j + 1)! + n!(n − j + 1) j!(n − j + 1)! n! j + n!(n − j + 1) j!(n − j + 1)! n! (j + (n − j + 1)) j!(n − j + 1)! (n + 1)n! j!(n + 1 − j))! (n + 1)! j!((n + 1) − j))! n + 1 j We are now in position to state and prove the Binomial Theorem where we see that binomial coeﬃcients are just that - coe
ﬃcients in the binomial expansion. Theorem 9.4. Binomial Theorem: For nonzero real numbers a and b, (a + b)n = an−jbj n j=0 n j for all natural numbers n. To get a feel of what this theorem is saying and how it really isn’t as hard to remember as it may ﬁrst appear, let’s consider the speciﬁc case of n = 4. According to the theorem, we have 9.4 The Binomial Theorem 685 (a + b)4 = 4 a4−jbj 4 j j=0 4 0 4 0 = = a4−0b0 + a4−1b1 + 4 1 a4−2b2 + 4 2 a4−3b3 + a4−4b4 4 4 a4 + a3b + 4 1 4 2 a2b2 + 4 3 ab3 + 4 3 4 4 b4 We forgo the simpliﬁcation of the coeﬃcients in order to note the pattern in the expansion. First note that in each term, the total of the exponents is 4 which matched the exponent of the binomial (a + b)4. The exponent on a begins at 4 and decreases by one as we move from one term to the next while the exponent on b starts at 0 and increases by one each time. Also note that the binomial coeﬃcients themselves have a pattern. The upper number, 4, matches the exponent on the binomial (a + b)4 whereas the lower number changes from term to term and matches the exponent of b in that term. This is no coincidence and corresponds to the kind of counting we discussed earlier. If we think of obtaining (a + b)4 by multiplying (a + b)(a + b)(a + b)(a + b), our answer is the sum of all possible products with exactly four factors - some a, some b. If we wish to count, for instance, the number of ways we obtain 1 factor of b out of a total of 4 possible factors, thereby forcing the remaining 3 factors to be a, the answer is 4 a3b is in the expansion. The. Hence, the term 4 1 1 other terms which appear cover the remaining cases. While this discussion gives an indication as to why the theorem is true, a formal proof requires Mathemat
ical Induction.4 To prove the Binomial Theorem, we let P (n) be the expansion formula given in the statement of the theorem and we note that P (1) is true since (a + b)1 a + b? =? = 1 a1−jbj 1 j j=0 1 0 a1−0b0 + a1−1b1 1 1 a + b = a + b Now we assume that P (k) is true. That is, we assume that we can expand (a + b)k using the formula given in Theorem 9.4 and attempt to show that P (k + 1) is true. 4and a fair amount of tenacity and attention to detail. 686 Sequences and the Binomial Theorem (a + b)k+1 = (a + b)(a + b)k k = (a + b) k j ak−jbj j=0 = a k j=0 k j ak−jbj + b = k j=0 k j ak+1−jbj + k j=0 k j=0 k j ak−jbj ak−jbj+1 k j Our goal is to combine as many of the terms as possible within the two summations. As the counter j in the ﬁrst summation runs from 0 through k, we get terms involving ak+1, akb, ak−1b2,..., abk. In the second summation, we get terms involving akb, ak−1b2,..., abk, bk+1. In other words, apart from the ﬁrst term in the ﬁrst summation and the last term in the second summation, we have terms common to both summations. Our next move is to ‘kick out’ the terms which we cannot combine and rewrite the summations so that we can combine them. To that end, we note k j=0 k j ak+1−jbj = ak+1 + k j=1 k j ak+1−jbj ak−jbj+1 = k j=0 k j k−1 j=0 k j ak−jbj+1 + bk+1 and so that (a + b)k+1 = ak+1 + ak+1−jbj + k j=1 k j k−1 j=0
k j ak−jbj+1 + bk+1 We now wish to write ak+1−jbj + k j=1 k j k−1 j=0 k j ak−jbj+1 as a single summation. The wrinkle is that the ﬁrst summation starts with j = 1, while the second starts with j = 0. Even though the sums produce terms with the same powers of a and b, they do so for diﬀerent values of j. To resolve this, we need to shift the index on the second summation so that the index j starts at j = 1 instead of j = 0 and we make use of Theorem 9.1 in the process. 9.4 The Binomial Theorem 687 k−1 j=0 k j ak−jbj+1 = k−1+1 k j − 1 j=0+1 ak−(j−1)b(j−1)+1 = k k j − 1 j=1 ak+1−jbj We can now combine our two sums using Theorem 9.1 and simplify using Theorem 9.3 k j=1 k j ak+1−jbj + ak−jbj+1 = k−1 j=0 k j = = k j=1 k j=1 k j=1 ak+1−jbj + k k ak+1−jbj k j j=1 j − 1 ak+1−jbj ak+1−jbj Using this and the fact that k+1 0 = 1 and k+1 k+1 = 1, we get (a + b)k+1 = ak+1 + k k + 1 j ak+1−jbj + bk+1 j=1 k + 1 0 ak+1b0 + k+1 j=0 k + 1 j a(k+1)−jbj = = k j=1 k + 1 j ak+1−jbj + k + 1 k + 1 a0bk+1 which shows that P (k + 1) is true. Hence, by induction, we have established that the Binomial Theorem holds for all natural numbers n. Example 9.4.2. Use the Binomial Theorem to ﬁnd the following. 1. (x − 2)4 2. 2.13 3. The term containing x3 in the expansion (2x
+ y)5 Solution. 1. Since (x − 2)4 = (x + (−2))4, we identify a = x, b = −2 and n = 4 and obtain 688 Sequences and the Binomial Theorem (x − 2)4 = 4 x4−j(−2)j 4 j j=0 4 0 = x4−0(−2)0 + x4−1(−2)1 + 4 1 4 2 x4−2(−2)2 + 4 3 x4−3(−2)3 + x4−4(−2)4 4 4 = x4 − 8x3 + 24x2 − 32x + 16 2. At ﬁrst this problem seem misplaced, but we can write 2.13 = (2 + 0.1)3. Identifying a = 2, b = 0.1 = 1 10 and n = 3, we get 3 23−j 3 j j 1 10 2 + 3 1 10 = = j=0 3 0 23−0 0 1 10 3 1 + 23−1 1 1 10 3 2 + 23−2 2 1 10 3 3 + 23−3 3 1 10 = 8 + 12 10 + 6 100 + 1 1000 = 8 + 1.2 + 0.06 + 0.001 = 9.261 3. Identifying a = 2x, b = y and n = 5, the Binomial Theorem gives (2x + y)5 = (2x)5−jyj 5 j=0 5 j Since we are concerned with only the term containing x3, there is no need to expand the entire sum. The exponents on each term must add to 5 and if the exponent on x is 3, the exponent on y must be 2. Plucking out the term j = 2, we get 5 2 (2x)5−2y2 = 10(2x)3y2 = 80x3y2 We close this section with Pascal’s Triangle, named in honor of the mathematician Blaise Pascal. Pascal’s Triangle is obtained by arranging the binomial coeﬃcients in the triangular fashion below. 9.4 The Binomial Theorem 689... for all whole numbers n, we get that each row of Pascal’s Triangle = 1 and n Since n n 0 begins and ends with 1. To generate the numbers in the middle of the
rows (from the third row onwards), we take advantage of the additive relationship expressed in Theorem 9.3. For instance, and so forth. This relationship is indicated by the arrows in the + array above. With these two facts in hand, we can quickly generate Pascal’s Triangle. We start with the ﬁrst two rows, 1 and 1 1. From that point on, each successive row begins and ends with 1 and the middle numbers are generated using Theorem 9.3. Below we attempt to demonstrate this building process to generate the ﬁrst ﬁve rows of Pascal’s Triangle −−−−−−→ −−−−−−→ −−−−−−→ 690 Sequences and the Binomial Theorem To see how we can use Pascal’s Triangle to expedite the Binomial Theorem, suppose we wish to for j = 0, 1, 2, 3, 4 and are the numbers which expand (3x − y)4. The coeﬃcients we need are 4 j form the ﬁfth row of Pascal’s Triangle. Since we know that the exponent of 3x in the ﬁrst term is 4 and then decreases by one as we go from left to right while the exponent of −y starts at 0 in the ﬁrst term and then increases by one as we move from left to right, we quickly obtain (3x − y)4 = (1)(3x)4 + (4)(3x)3(−y) + (6)(3x)2(−y)2 + 4(3x)(−y)3 + 1(−y)4 = 81x4 − 108x3y + 54x2y2 − 12xy3 + y4 We would like to stress that Pascal’s Triangle is a very quick method to expand an entire binomial. If only a term (or two or three) is required, then the Binomial Theorem is deﬁnitely the way to go. 9.4 The Binomial Theorem 691 9.4.1 Exercises In Exercises 1 - 9, simplify the given expression. 1. (3!)2 4. 7. 9! 4!3!2! 8 3 2. 5. 8. 10! 7! (n + 1)! n! 117 0, n ≥ 0. 3
. 6. 9. 7! 233! (k − 1)! (k + 2)! n n − 2, k ≥ 1., n ≥ 2 In Exercises 10 - 13, use Pascal’s Triangle to expand the given binomial. 10. (x + 2)5 11. (2x − 1)4 12. 1 3 x + y23 13. x − x−14 In Exercises 14 - 17, use Pascal’s Triangle to simplify the given power of a complex number. 14. (1 + 2i)4 √ 16. 3 i 3 2 + 1 2 15. −1 + i √ √ 33 √ 4 i 2 2 − 2 2 17. In Exercises 18 - 22, use the Binomial Theorem to ﬁnd the indicated term. 18. The term containing x3 in the expansion (2x − y)5 19. The term containing x117 in the expansion (x + 2)118 √ 7 2 in the expansion ( 20. The term containing x 21. The term containing x−7 in the expansion 2x − x−35 22. The constant term in the expansion x + x−18 23. Use the Prinicple of Mathematical Induction to prove n! > 2n for n ≥ 4. x − 3)8 24. Prove n j=0 n j = 2n for all natural numbers n. (HINT: Use the Binomial Theorem!) 25. With the help of your classmates, research Patterns and Properties of Pascal’s Triangle. 26. You’ve just won three tickets to see the new ﬁlm, ‘8.9.’ Five of your friends, Albert, Beth, Chuck, Dan, and Eugene, are interested in seeing it with you. With the help of your classmates, list all the possible ways to distribute your two extra tickets among your ﬁve friends. Now suppose you’ve come down with the ﬂu. List all the diﬀerent ways you can distribute the three tickets among these ﬁve friends. How does this compare with the ﬁrst list you made?? = 5 What does this have to do with the fact that 5 3 2 692 Sequences and the Binomial Theorem 9.4.2 Answers 1. 36 4. 1260 7. 56 2. 720 5. n + 1
8. 1 3. 105 6. 1 k(k+1)(k+2) 9. n(n−1) 2 10. (x + 2)5 = x5 + 10x4 + 40x3 + 80x2 + 80x + 32 11. (2x − 1)4 = 16x4 − 32x3 + 24x2 − 8x + 1 12. 1 3 x + y23 13. x − x−14 = 1 27 x3 + 1 3 x2y2 + xy4 + y6 = x4 − 4x2 + 6 − 4x−2 + x−4 14. −7 − 24i 15. 8 16. i 17. −1 18. 80x3y2 19. 236x117 20. −24x 7 2 21. −40x−7 22. 70 Index nth root of a complex number, 1000, 1001 principal, 397 nth Roots of Unity, 1006 u-substitution, 273 x-axis, 6 x-coordinate, 6 x-intercept, 25 y-axis, 6 y-coordinate, 6 y-intercept, 25 abscissa, 6 absolute value deﬁnition of, 173 inequality, 211 properties of, 173 acidity of a solution pH, 432 acute angle, 694 adjoint of a matrix, 622 alkalinity of a solution pH, 432 amplitude, 794, 881 angle acute, 694 between two vectors, 1035, 1036 central angle, 701 complementary, 696 coterminal, 698 decimal degrees, 695 deﬁnition, 693 degree, 694 DMS, 695 initial side, 698 measurement, 693 negative, 698 obtuse, 694 of declination, 761 of depression, 761 of elevation, 753 of inclination, 753 oriented, 697 positive, 698 quadrantal, 698 radian measure, 701 reference, 721 right, 694 standard position, 698 straight, 693 supplementary, 696 terminal side, 698 vertex, 693 angle side opposite pairs, 896 angular frequency, 708 annuity annuity-due, 667 ordinary deﬁnition of, 666 future value, 667 applied domain of a function, 60 arccosecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 10
69 Index 1070 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arccosine deﬁnition of, 820 graph of, 819 properties of, 820 arccotangent deﬁnition of, 824 graph of, 824 properties of, 824 arcsecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arcsine deﬁnition of, 820 graph of, 820 properties of, 820 arctangent deﬁnition of, 824 graph of, 823 properties of, 824 argument of a complex number deﬁnition of, 991 properties of, 995 of a function, 55 of a logarithm, 425 of a trigonometric function, 793 arithmetic sequence, 654 associative property for function composition, 366 matrix addition, 579 matrix multiplication, 585 scalar multiplication, 581 vector addition, 1015 scalar multiplication, 1018 asymptote horizontal formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 of a hyperbola, 531 slant determination of, 312 formal deﬁnition of, 311 slant (oblique), 311 vertical formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 augmented matrix, 568 average angular velocity, 707 average cost, 346 average cost function, 82 average rate of change, 160 average velocity, 706 axis of symmetry, 191 back substitution, 560 bearings, 905 binomial coeﬃcient, 683 Binomial Theorem, 684 Bisection Method, 277 BMI, body mass index, 355 Boyle’s Law, 350 buﬀer solution, 478 cardioid, 951 Cartesian coordinate plane, 6 Cartesian coordinates, 6 Cauchy’s Bound, 269 center of a circle, 498 of a hyperbola, 531 of an ellipse, 516 Index 1071 central angle, 701 change of base formulas, 442 characteristic polynomial, 626 Charles’s Law, 355 circle center of, 498 deﬁnition
of, 498 from slicing a cone, 495 radius of, 498 standard equation, 498 standard equation, alternate, 519 circular function, 744 cis(θ), 995 coeﬃcient of determination, 226 cofactor, 616 Cofunction Identities, 773 common base, 420 common logarithm, 422 commutative property function composition does not have, 366 matrix addition, 579 vector addition, 1015 dot product, 1034 complementary angles, 696 Complex Factorization Theorem, 290 complex number nth root, 1000, 1001 nth Roots of Unity, 1006 argument deﬁnition of, 991 properties of, 995 conjugate deﬁnition of, 288 properties of, 289 deﬁnition of, 2, 287, 991 imaginary part, 991 imaginary unit, i, 287 modulus deﬁnition of, 991 properties of, 993 polar form cis-notation, 995 principal argument, 991 real part, 991 rectangular form, 991 set of, 2 complex plane, 991 component form of a vector, 1013 composite function deﬁnition of, 360 properties of, 367 compound interest, 470 conic sections deﬁnition, 495 conjugate axis of a hyperbola, 532 conjugate of a complex number deﬁnition of, 288 properties of, 289 Conjugate Pairs Theorem, 291 consistent system, 553 constant function as a horizontal line, 156 formal deﬁnition of, 101 intuitive deﬁnition of, 100 constant of proportionality, 350 constant term of a polynomial, 236 continuous, 241 continuously compounded interest, 472 contradiction, 549 coordinates Cartesian, 6 polar, 919 rectangular, 919 correlation coeﬃcient, 226 cosecant graph of, 801 of an angle, 744, 752 properties of, 802 cosine graph of, 791 of an angle, 717, 730, 744 properties of, 791 1072 cost average, 82, 346 ﬁxed, start-up, 82 variable, 159 cost function, 82 cotangent graph of, 805 of an angle, 744, 752 properties of, 806 coterminal angle, 698 Coulomb’s Law, 355 C
ramer’s Rule, 619 curve orientated, 1048 cycloid, 1056 decibel, 431 decimal degrees, 695 decreasing function formal deﬁnition of, 101 intuitive deﬁnition of, 100 degree measure, 694 degree of a polynomial, 236 DeMoivre’s Theorem, 997 dependent system, 554 dependent variable, 55 depreciation, 420 Descartes’ Rule of Signs, 273 determinant of a matrix deﬁnition of, 614 properties of, 616 Diﬀerence Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 diﬀerence quotient, 79 dimension of a matrix, 567 direct variation, 350 directrix of a conic section in polar form, 981 of a parabola, 505 discriminant Index of a conic, 979 of a quadratic equation, 195 trichotomy, 195 distance deﬁnition, 10 distance formula, 11 distributive property matrix matrix multiplication, 585 scalar multiplication, 581 vector dot product, 1034 scalar multiplication, 1018 DMS, 695 domain applied, 60 deﬁnition of, 45 implied, 58 dot product commutative property of, 1034 deﬁnition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to orthogonality, 1037 relation to vector magnitude, 1034 work, 1042 Double Angle Identities, 776 earthquake Richter Scale, 431 eccentricity, 522, 981 eigenvalue, 626 eigenvector, 626 ellipse center, 516 deﬁnition of, 516 eccentricity, 522 foci, 516 from slicing a cone, 496 guide rectangle, 519 major axis, 516 minor axis, 516 Index 1073 reﬂective property, 523 standard equation, 519 vertices, 516 ellipsis (... ), 31, 651 empty set, 2 end behavior of f (x) = axn, n even, 240 of f (x) = axn, n odd, 240 of a function graph, 239 polynomial, 243 entry in a matrix, 567 equation contradiction, 549 graph of, 23 identity, 549 linear of n
variables, 554 linear of two variables, 549 even function, 95 Even/Odd Identities, 770 exponential function algebraic properties of, 437 change of base formula, 442 common base, 420 deﬁnition of, 418 graphical properties of, 419 inverse properties of, 437 natural base, 420 one-to-one properties of, 437 solving equations with, 448 extended interval notation, 756 Factor Theorem, 258 factorial, 654, 681 ﬁxed cost, 82 focal diameter of a parabola, 507 focal length of a parabola, 506 focus of a conic section in polar form, 981 focus (foci) of a hyperbola, 531 of a parabola, 505 of an ellipse, 516 free variable, 552 frequency angular, 708, 881 of a sinusoid, 795 ordinary, 708, 881 function (absolute) maximum, 101 (absolute, global) minimum, 101 absolute value, 173 algebraic, 399 argument, 55 arithmetic, 76 as a process, 55, 378 average cost, 82 circular, 744 composite deﬁnition of, 360 properties of, 367 constant, 100, 156 continuous, 241 cost, 82 decreasing, 100 deﬁnition as a relation, 43 dependent variable of, 55 diﬀerence, 76 diﬀerence quotient, 79 domain, 45 even, 95 exponential, 418 Fundamental Graphing Principle, 93 identity, 168 increasing, 100 independent variable of, 55 inverse deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 linear, 156 local (relative) maximum, 101 local (relative) minimum, 101 logarithmic, 422 1074 Index notation, 55 odd, 95 one-to-one, 381 periodic, 790 piecewise-deﬁned, 62 polynomial, 235 price-demand, 82 product, 76 proﬁt, 82 quadratic, 188 quotient, 76 range, 45 rational, 301 revenue, 82 smooth, 241 sum, 76 transformation of graphs, 120, 135 zero, 95 fundamental cycle of y = cos(x), 791 Fundamental Graphing Principle for equations, 23 for functions, 93 for polar equations, 938 Fundamental Theorem of Algebra, 290 Gauss-Jordan Elimination, 571 Gaussian Elimination, 5
57 geometric sequence, 654 geometric series, 669 graph hole in, 305 horizontal scaling, 132 horizontal shift, 123 of a function, 93 of a relation, 20 of an equation, 23 rational function, 321 reﬂection about an axis, 126 transformations, 135 vertical scaling, 130 vertical shift, 121 greatest integer function, 67 growth model limited, 475 logistic, 475 uninhibited, 472 guide rectangle for a hyperbola, 532 for an ellipse, 519 Half-Angle Formulas, 779 harmonic motion, 885 Henderson-Hasselbalch Equation, 446 Heron’s Formula, 914 hole in a graph, 305 location of, 306 Hooke’s Law, 350 horizontal asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 horizontal line, 23 Horizontal Line Test (HLT), 381 hyperbola asymptotes, 531 branch, 531 center, 531 conjugate axis, 532 deﬁnition of, 531 foci, 531 from slicing a cone, 496 guide rectangle, 532 standard equation horizontal, 534 vertical, 534 transverse axis, 531 vertices, 531 hyperbolic cosine, 1062 hyperbolic sine, 1062 hyperboloid, 542 identity function, 367 matrix, additive, 579 Index 1075 matrix, multiplicative, 585 statement which is always true, 549 imaginary axis, 991 imaginary part of a complex number, 991 imaginary unit, i, 287 implied domain of a function, 58 inconsistent system, 553 increasing function interval deﬁnition of, 3 notation for, 3 notation, extended, 756 inverse matrix, additive, 579, 581 matrix, multiplicative, 602 of a function formal deﬁnition of, 101 intuitive deﬁnition of, 100 independent system, 554 independent variable, 55 index of a root, 397 induction base step, 673 induction hypothesis, 673 inductive step, 673 inequality absolute value, 211 graphical interpretation, 209 non-linear, 643 quadratic, 215 sign diagram, 214 inﬂection point, 477 information entropy, 477 initial side of an angle, 698 instantaneous rate of change, 161, 472, 707 integer deﬁnition of
, 2 greatest integer function, 67 set of, 2 intercept deﬁnition of, 25 location of, 25 interest compound, 470 compounded continuously, 472 simple, 469 Intermediate Value Theorem polynomial zero version, 241 interrobang, 321 intersection of two sets, 4 deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 inverse variation, 350 invertibility function, 382 invertible function, 379 matrix, 602 irrational number deﬁnition of, 2 set of, 2 irreducible quadratic, 291 joint variation, 350 Kepler’s Third Law of Planetary Motion, 355 Kirchhoﬀ’s Voltage Law, 605 latus rectum of a parabola, 507 Law of Cosines, 910 Law of Sines, 897 leading coeﬃcient of a polynomial, 236 leading term of a polynomial, 236 Learning Curve Equation, 315 least squares regression line, 225 lemniscate, 950 lima¸con, 950 line horizontal, 23 least squares regression, 225 linear function, 156 of best ﬁt, 225 parallel, 166 1076 Index perpendicular, 167 point-slope form, 155 slope of, 151 slope-intercept form, 155 vertical, 23 linear equation n variables, 554 two variables, 549 linear function, 156 local maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 local minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 logarithm algebraic properties of, 438 change of base formula, 442 common, 422 general, “base b”, 422 graphical properties of, 423 inverse properties of, 437 natural, 422 one-to-one properties of, 437 solving equations with, 459 logarithmic scales, 431 logistic growth, 475 LORAN, 538 lower triangular matrix, 593 main diagonal, 585 major axis of an ellipse, 516 Markov Chain, 592 mathematical model, 60 matrix addition associative property, 579 commutative property, 579 deﬁnition of, 578 properties of, 579 additive identity, 579 additive inverse, 579 adjoint, 622 augmented, 568 characteristic polynomial, 626 cofactor, 6
16 deﬁnition, 567 determinant deﬁnition of, 614 properties of, 616 dimension, 567 entry, 567 equality, 578 invertible, 602 leading entry, 569 lower triangular, 593 main diagonal, 585 matrix multiplication associative property of, 585 deﬁnition of, 584 distributive property, 585 identity for, 585 properties of, 585 minor, 616 multiplicative inverse, 602 product of row and column, 584 reduced row echelon form, 570 rotation, 986 row echelon form, 569 row operations, 568 scalar multiplication associative property of, 581 deﬁnition of, 580 distributive properties, 581 identity for, 581 properties of, 581 zero product property, 581 size, 567 square matrix, 586 sum, 578 upper triangular, 593 maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 measure of an angle, 693 1077 Index midpoint deﬁnition of, 12 midpoint formula, 13 minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 minor, 616 minor axis of an ellipse, 516 model mathematical, 60 modulus of a complex number deﬁnition of, 991 properties of, 993 multiplicity eﬀect on the graph of a polynomial, 245, 249 of a zero, 244 natural base, 420 natural logarithm, 422 natural number deﬁnition of, 2 set of, 2 negative angle, 698 Newton’s Law of Cooling, 421, 474 Newton’s Law of Universal Gravitation, 351 oblique asymptote, 311 obtuse angle, 694 odd function, 95 Ohm’s Law, 350, 605 one-to-one function, 381 ordered pair, 6 ordinary frequency, 708 ordinate, 6 orientation, 1048 oriented angle, 697 oriented arc, 704 origin, 7 orthogonal projection, 1038 orthogonal vectors, 1037 overdetermined system, 554 parabola axis of symmetry, 191 deﬁnition of, 505 directrix, 505 focal diameter, 507 focal length, 506 focus, 505 from slicing a cone,
496 graph of a quadratic function, 188 latus rectum, 507 reﬂective property, 510 standard equation horizontal, 508 vertical, 506 vertex, 188, 505 vertex formulas, 194 paraboloid, 510 parallel vectors, 1030 parameter, 1048 parametric equations, 1048 parametric solution, 552 parametrization, 1048 partial fractions, 628 Pascal’s Triangle, 688 password strength, 477 period circular motion, 708 of a function, 790 of a sinusoid, 881 periodic function, 790 pH, 432 phase, 795, 881 phase shift, 795, 881 pi, π, 700 piecewise-deﬁned function, 62 point of diminishing returns, 477 point-slope form of a line, 155 polar coordinates conversion into rectangular, 924 deﬁnition of, 919 equivalent representations of, 923 polar axis, 919 pole, 919 1078 Index polar form of a complex number, 995 polar rose, 950 polynomial division dividend, 258 divisor, 258 factor, 258 quotient, 258 remainder, 258 synthetic division, 260 polynomial function completely factored over the complex numbers, 291 over the real numbers, 291 constant term, 236 deﬁnition of, 235 degree, 236 end behavior, 239 leading coeﬃcient, 236 leading term, 236 variations in sign, 273 zero lower bound, 274 multiplicity, 244 upper bound, 274 positive angle, 698 Power Reduction Formulas, 778 power rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 price-demand function, 82 principal, 469 principal nth root, 397 principal argument of a complex number, 991 principal unit vectors, ˆı, ˆ, 1024 Principle of Mathematical Induction, 673 product rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 Product to Sum Formulas, 780 proﬁt function, 82 projection x−axis, 45 y−axis, 46 orthogonal, 1038 Pythagorean Conjugates, 751 Pythagorean Identities
, 749 quadrantal angle, 698 quadrants, 8 quadratic formula, 194 quadratic function deﬁnition of, 188 general form, 190 inequality, 215 irreducible quadratic, 291 standard form, 190 quadratic regression, 228 Quotient Identities, 745 quotient rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 radian measure, 701 radical properties of, 398 radicand, 397 radioactive decay, 473 radius of a circle, 498 range deﬁnition of, 45 rate of change average, 160 Index 1079 instantaneous, 161, 472 slope of a line, 154 rational exponent, 398 rational functions, 301 rational number deﬁnition of, 2 set of, 2 Rational Zeros Theorem, 269 ray deﬁnition of, 693 initial point, 693 real axis, 991 Real Factorization Theorem, 292 real number deﬁnition of, 2 set of, 2 real part of a complex number, 991 Reciprocal Identities, 745 rectangular coordinates also known as Cartesian coordinates, 919 conversion into polar, 924 rectangular form of a complex number, 991 recursion equation, 654 reduced row echelon form, 570 reference angle, 721 Reference Angle Theorem for cosine and sine, 722 for the circular functions, 747 reﬂection of a function graph, 126 of a point, 10 regression coeﬃcient of determination, 226 correlation coeﬃcient, 226 least squares line, 225 quadratic, 228 total squared error, 225 relation algebraic description, 23 deﬁnition, 20 Fundamental Graphing Principle, 23 Remainder Theorem, 258 revenue function, 82 Richter Scale, 431 right angle, 694 root index, 397 radicand, 397 Roots of Unity, 1006 rotation matrix, 986 rotation of axes, 974 row echelon form, 569 row operations for a matrix, 568 scalar multiplication matrix associative property of, 581 deﬁnition of, 580 distributive properties of, 581 properties of, 581 vector associative property of, 1018 deﬁnition of, 1017 distributive
properties of, 1018 properties of, 1018 scalar projection, 1039 secant graph of, 800 of an angle, 744, 752 properties of, 802 secant line, 160 sequence nth term, 652 alternating, 652 arithmetic common diﬀerence, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 deﬁnition of, 652 geometric common ratio, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 1080 recursive, 654 series, 668 set deﬁnition of, 1 empty, 2 intersection, 4 roster method, 1 set-builder notation, 1 sets of numbers, 2 union, 4 verbal description, 1 set-builder notation, 1 Side-Angle-Side triangle, 910 Side-Side-Side triangle, 910 sign diagram algebraic function, 399 for quadratic inequality, 214 polynomial function, 242 rational function, 321 simple interest, 469 sine graph of, 792 of an angle, 717, 730, 744 properties of, 791 sinusoid amplitude, 794, 881 baseline, 881 frequency angular, 881 ordinary, 881 graph of, 795, 882 period, 881 phase, 881 phase shift, 795, 881 properties of, 881 vertical shift, 881 slant asymptote, 311 slant asymptote determination of, 312 formal deﬁnition of, 311 slope deﬁnition, 151 Index of a line, 151 rate of change, 154 slope-intercept form of a line, 155 smooth, 241 sound intensity level decibel, 431 square matrix, 586 standard position of a vector, 1019 standard position of an angle, 698 start-up cost, 82 steady state, 592 stochastic process, 592 straight angle, 693 Sum Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 Sum to Product Formulas, 781 summation notation deﬁnition of, 661 index of summation, 661 lower limit of summation, 661 properties of, 664 upper limit of summation, 661 supplementary angles, 696 symmetry about
the x-axis, 9 about the y-axis, 9 about the origin, 9 testing a function graph for, 95 testing an equation for, 26 synthetic division tableau, 260 system of equations back-substitution, 560 coeﬃcient matrix, 590 consistent, 553 constant matrix, 590 deﬁnition, 549 dependent, 554 free variable, 552 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 Index 1081 inconsistent, 553 independent, 554 leading variable, 556 linear n variables, 554 two variables, 550 linear in form, 646 non-linear, 637 overdetermined, 554 parametric solution, 552 triangular form, 556 underdetermined, 554 unknowns matrix, 590 tangent graph of, 804 of an angle, 744, 752 properties of, 806 terminal side of an angle, 698 Thurstone, Louis Leon, 315 total squared error, 225 transformation non-rigid, 129 rigid, 129 transformations of function graphs, 120, 135 transverse axis of a hyperbola, 531 Triangle Inequality, 183 triangular form, 556 underdetermined system, 554 uninhibited growth, 472 union of two sets, 4 Unit Circle deﬁnition of, 501 important points, 724 unit vector, 1023 Upper and Lower Bounds Theorem, 274 upper triangular matrix, 593 variable dependent, 55 independent, 55 variable cost, 159 variation constant of proportionality, 350 direct, 350 inverse, 350 joint, 350 variations in sign, 273 vector x-component, 1012 y-component, 1012 addition associative property, 1015 commutative property, 1015 deﬁnition of, 1014 properties of, 1015 additive identity, 1015 additive inverse, 1015, 1018 angle between two, 1035, 1036 component form, 1012 Decomposition Theorem Generalized, 1040 Principal, 1024 deﬁnition of, 1012 direction deﬁnition of, 1020 properties of, 1020 dot product commutative property of, 1034 deﬁnition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to magnitude, 1034 relation to orthogonality, 1037 work, 1042 head, 1012 initial point, 1012 magnitude deﬁn
ition of, 1020 properties of, 1020 relation to dot product, 1034 normalization, 1024 orthogonal projection, 1038 1082 Index multiplicity of, 244 of a function, 95 upper and lower bounds, 274 orthogonal vectors, 1037 parallel, 1030 principal unit vectors, ˆı, ˆ, 1024 resultant, 1013 scalar multiplication associative property of, 1018 deﬁnition of, 1017 distributive properties, 1018 identity for, 1018 properties of, 1018 zero product property, 1018 scalar product deﬁnition of, 1034 properties of, 1034 scalar projection, 1039 standard position, 1019 tail, 1012 terminal point, 1012 triangle inequality, 1044 unit vector, 1023 velocity average angular, 707 instantaneous, 707 instantaneous angular, 707 vertex of a hyperbola, 531 of a parabola, 188, 505 of an angle, 693 of an ellipse, 516 vertical asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 vertical line, 23 Vertical Line Test (VLT), 43 whole number deﬁnition of, 2 set of, 2 work, 1041 wrapping function, 704 zeroace the endpoint, we proceed as we did in Example 10.1.1, successively halving the angle measure until we ﬁnd 5π 8 ≈ 1.96 which tells us our arc extends just a bit beyond the quarter mark into Quadrant III. 706 Foundations of Trigonometry 4. Since 117 is positive, the arc corresponding to t = 117 begins at (1, 0) and proceeds counterclockwise. As 117 is much greater than 2π, we wrap around the Unit Circle several times before ﬁnally reaching our endpoint. We approximate 117 2π as 18.62 which tells us we complete 18 revolutions counter-clockwise with 0.62, or just shy of 5 8 of a revolution to spare. In other words, the terminal side of the angle which measures 117 radians in standard position is just short of being midway through Quadrant III2 t = 117 10.1.1 Applications of Radian Measure: Circular Motion Now that we have paired angles with real numbers via radian measure, a whole world of applications awaits us. Our ﬁrst
excursion into this realm comes by way of circular motion. Suppose an object is moving as pictured below along a circular path of radius r from the point P to the point Q in an amount of time t. Q θ r s P Here s represents a displacement so that s > 0 means the object is traveling in a counter-clockwise direction and s < 0 indicates movement in a clockwise direction. Note that with this convention the formula we used to deﬁne radian measure, namely θ =, still holds since a negative value of s incurred from a clockwise displacement matches the negative we assign to θ for a clockwise rotation. In Physics, the average velocity of the object, denoted v and read as ‘v-bar’, is deﬁned as the average rate of change of the position of the object with respect to time.16 As a result, we s r 16See Deﬁnition 2.3 in Section 2.1 for a review of this concept. 10.1 Angles and their Measure 707 s t = time. The quantity v has units of length have v = displacement time and conveys two ideas: the direction in which the object is moving and how fast the position of the object is changing. The contribution of direction in the quantity v is either to make it positive (in the case of counter-clockwise motion) or negative (in the case of clockwise motion), so that the quantity \|v\| quantiﬁes how fast the object is moving - it is the speed of the object. Measuring θ in radians we have θ = thus s = rθ and s r v = s t = rθ t = r · θ t θ t is called the average angular velocity of the object. It is denoted by ω and is The quantity read ‘omega-bar’. The quantity ω is the average rate of change of the angle θ with respect to time and thus has units radians time. If ω is constant throughout the duration of the motion, then it can be shown17 that the average velocities involved, namely v and ω, are the same as their instantaneous counterparts, v and ω, respectively. In this case, v is simply called the ‘velocity’ of the object and is the instantaneous rate of change of the position of the object with respect to time.18 Similarly,
ω is called the ‘angular velocity’ and is the instantaneous rate of change of the angle with respect to time. If the path of the object were ‘uncurled’ from a circle to form a line segment, then the velocity of the object on that line segment would be the same as the velocity on the circle. For this reason, the quantity v is often called the linear velocity of the object in order to distinguish it from the angular velocity, ω. Putting together the ideas of the previous paragraph, we get the following. Equation 10.2. Velocity for Circular Motion: For an object moving on a circular path of radius r with constant angular velocity ω, the (linear) velocity of the object is given by v = rω. time. Thus the left hand side of the equation v = rω has units length We need to talk about units here. The units of v are length time, the units of r are length only, and the units of ω are radians time, whereas time = length·radians the right hand side has units length · radians. The supposed contradiction in units is resolved by remembering that radians are a dimensionless quantity and angles in radian measure are identiﬁed with real numbers so that the units length·radians time. We are long overdue for an example. reduce to the units length time time Example 10.1.5. Assuming that the surface of the Earth is a sphere, any point on the Earth can be thought of as an object traveling on a circle which completes one revolution in (approximately) 24 hours. The path traced out by the point during this 24 hour period is the Latitude of that point. Lakeland Community College is at 41.628◦ north latitude, and it can be shown19 that the radius of the earth at this Latitude is approximately 2960 miles. Find the linear velocity, in miles per hour, of Lakeland Community College as the world turns. Solution. To use the formula v = rω, we ﬁrst need to compute the angular velocity ω. The earth π makes one revolution in 24 hours, and one revolution is 2π radians, so ω = 2π radians 12 hours, 24 hours = 17You guessed it, using Calculus... 18See the discussion on Page 161 for more details on the idea of an ‘instantaneous’ rate of change. 19We will discuss how we
arrived at this approximation in Example 10.2.6. 708 Foundations of Trigonometry where, once again, we are using the fact that radians are real numbers and are dimensionless. (For simplicity’s sake, we are also assuming that we are viewing the rotation of the earth as counterclockwise so ω > 0.) Hence, the linear velocity is v = 2960 miles · π 12 hours ≈ 775 miles hour It is worth noting that the quantity 1 revolution in Example 10.1.5 is called the ordinary frequency 24 hours of the motion and is usually denoted by the variable f. The ordinary frequency is a measure of how often an object makes a complete cycle of the motion. The fact that ω = 2πf suggests that ω is also a frequency. Indeed, it is called the angular frequency of the motion. On a related note, the quantity T = is called the period of the motion and is the amount of time it takes for the 1 f object to complete one cycle of the motion. In the scenario of Example 10.1.5, the period of the motion is 24 hours, or one day. The concepts of frequency and period help frame the equation v = rω in a new light. That is, if ω is ﬁxed, points which are farther from the center of rotation need to travel faster to maintain the same angular frequency since they have farther to travel to make one revolution in one period’s time. The distance of the object to the center of rotation is the radius of the circle, r, and is the ‘magniﬁcation factor’ which relates ω and v. We will have more to say about frequencies and periods in Section 11.1. While we have exhaustively discussed velocities associated with circular motion, we have yet to discuss a more natural question: if an object is moving on a circular path of radius r with a ﬁxed angular velocity (frequency) ω, what is the position of the object at time t? The answer to this question is the very heart of Trigonometry and is answered in the next section. 10.1 Angles and their Measure 709 10.1.2 Exercises In Exercises 1 - 4, convert the angles into the DMS system. Round each of your answers to the nearest second. 1. 63.75◦ 2. 200.325◦ 3. −
317.06◦ 4. 179.999◦ In Exercises 5 - 8, convert the angles into decimal degrees. Round each of your answers to three decimal places. 5. 125◦50 6. −32◦1012 7. 502◦35 8. 237◦5843 In Exercises 9 - 28, graph the oriented angle in standard position. Classify each angle according to where its terminal side lies and then give two coterminal angles, one of which is positive and the other negative. 9. 330◦ 10. −135◦ 13. −270◦ 17. 3π 4 21. − π 2 25. −2π 14. 5π 6 18. − π 3 22. 7π 6 26. − π 4 11. 120◦ 15. − 11π 3 19. 7π 2 23. − 5π 3 27. 15π 4 12. 405◦ 16. 20. 5π 4 π 4 24. 3π 28. − 13π 6 In Exercises 29 - 36, convert the angle from degree measure into radian measure, giving the exact value in terms of π. 29. 0◦ 33. −315◦ 30. 240◦ 34. 150◦ 31. 135◦ 35. 45◦ 32. −270◦ 36. −225◦ In Exercises 37 - 44, convert the angle from radian measure into degree measure. 37. π 41. π 3 38. − 2π 3 42. 5π 3 39. 7π 6 43. − π 6 40. 44. 11π 6 π 2 710 Foundations of Trigonometry In Exercises 45 - 49, sketch the oriented arc on the Unit Circle which corresponds to the given real number. 45. t = 5π 6 46. t = −π 47. t = 6 48. t = −2 49. t = 12 50. A yo-yo which is 2.25 inches in diameter spins at a rate of 4500 revolutions per minute. How fast is the edge of the yo-yo spinning in miles per hour? Round your answer to two decimal places. 51. How many revolutions per minute would the yo-yo in exercise 50 have to complete if the edge of the yo-yo is to be spinning at a rate of 42 miles per hour? Round your answer to two decimal places. 52. In the
yo-yo trick ‘Around the World,’ the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. If the yo-yo string is 28 inches long and the yo-yo takes 3 seconds to complete one revolution of the circle, compute the speed of the yo-yo in miles per hour. Round your answer to two decimal places. 53. A computer hard drive contains a circular disk with diameter 2.5 inches and spins at a rate of 7200 RPM (revolutions per minute). Find the linear speed of a point on the edge of the disk in miles per hour. 54. A rock got stuck in the tread of my tire and when I was driving 70 miles per hour, the rock came loose and hit the inside of the wheel well of the car. How fast, in miles per hour, was the rock traveling when it came out of the tread? (The tire has a diameter of 23 inches.) 55. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet. (Remember this from Exercise 17 in Section 7.2?) It completes two revolutions in 2 minutes and 7 seconds.20 Assuming the riders are at the edge of the circle, how fast are they traveling in miles per hour? 56. Consider the circle of radius r pictured below with central angle θ, measured in radians, and subtended arc of length s. Prove that the area of the shaded sector is A = 1 (Hint: Use the proportion s circumference of the circle.) A area of the circle = 2 r2θ. r s θ r 20Source: Cedar Point’s webpage. 10.1 Angles and their Measure 711 In Exercises 57 - 62, use the result of Exercise 56 to compute the areas of the circular sectors with the given central angles and radii. 57. θ = π 6, r = 12 60. θ = π, r = 1 58. θ = 5π 4, r = 100 61. θ = 240◦, r = 5 59. θ = 330◦, r = 9.3 62. θ = 1◦, r = 117 63. Imagine a rope tied around the Earth at the equator. Show that you need to add only 2π feet of length to the rope in order to lift it one foot above
the ground around the entire equator. (You do NOT need to know the radius of the Earth to show this.) 64. With the help of your classmates, look for a proof that π is indeed a constant. 712 Foundations of Trigonometry 10.1.3 Answers 1. 63◦45 5. 125.833◦ 2. 200◦1930 3. −317◦336 4. 179◦5956 6. −32.17◦ 7. 502.583◦ 8. 237.979◦ 9. 330◦ is a Quadrant IV angle coterminal with 690◦ and −30◦ 10. −135◦ is a Quadrant III angle coterminal with 225◦ and −4954 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 11. 120◦ is a Quadrant II angle coterminal with 480◦ and −240◦ 12. 405◦ is a Quadrant I angle coterminal with 45◦ and −3154 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 13. −270◦ lies on the positive y-axis 14. coterminal with 90◦ and −630◦ is a Quadrant II angle 5π 6 coterminal with 17π 6 and − 7π 4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 10.1 Angles and their Measure 713 15. − 11π 3 is a Quadrant I angle 5π π 3 3 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 17. 19. is a Quadrant II angle 3π 4 coterminal with 11π 4 and − 5π 4 −3 −2 −1 −1 −2 −3 −4 7π 2 coterminal with 3π 2 and − π 2 y 4 3 2 1 16. is a Quadrant III angle 5π 4 c
oterminal with 13π 4 and − 3π 4 −3 −2 −1 −1 −2 −3 −4 18. − π 3 is a Quadrant IV angle 7π 3 and − 5π 3 coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x π 4 coterminal with 9π 4 and − 7π 4 y 4 3 2 1 lies on the negative y-axis 20. is a Quadrant I angle −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 714 21. − π 2 lies on the negative y-axis 5π 3π 2 2 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x Foundations of Trigonometry 22. is a Quadrant III angle 7π 6 coterminal with 19π 6 and − 5π 4 −3 −2 −1 −1 −2 −3 −4 23. − 5π 3 is a Quadrant I angle 24. 3π lies on the negative x-axis coterminal with y 4 3 2 1 π 3 and − 11π 3 coterminal with π and −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 25. −2π lies on the positive x-axis 26. − π 4 is a Quadrant IV angle coterminal with 2π and −4π coterminal with 7π 4 and − 9π 4 −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 10.1 Angles and their Measure 715 27. 15π 4 is a Quadrant IV angle π 4 and − 7π 4 coterminal with y 28. − 13π 6 is a Quadrant IV angle π 11π 6 6 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −

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