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 ...
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, ...
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 ...
 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 = ...
? 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 ask...
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...
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 dete...
= 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. 2...
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 co...
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 ...
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. O...
. 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 ...
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...
)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 fr...
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 quotie...
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...
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 ch...
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 + 21...
+ 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 −x...
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...
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...
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 throug...
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 nonl...
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 ...
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 fac...
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 conﬁrm...
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 ing...
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 exa...
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 inters...
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 outsi...
= 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 =...
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...
. 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 ...
. − 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 ...
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 thin...
. 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 ...
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 simp...
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 De...
! = 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 f...
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, − 2...
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...
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 commo...
. 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 g...
(−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 i...
’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...
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 1...
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...
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 ...
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, summat...
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 Not...
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 u...
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...
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). Assumi...
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 ac...
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 ...
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 =...
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. Geometri...
+ 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 ...
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...
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 s...
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 wheneve...
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 ar...
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...
+ 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 ...
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, wher...
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 S...
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...
) 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)(2...
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 ...
+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...
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 b...
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 beg...
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 i...
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 possi...
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 t...
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 us...
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 thi...
+ 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 ...
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....
. 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...
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...
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 deﬁniti...
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...
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 Sign...
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 equatio...
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...
, 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,...
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 mino...
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 ...
, 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...
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...
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 Eliminat...
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 pr...
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 ...
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 quanti...
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 velo...
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...
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 de...
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...
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...

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