math for success!jladouc2/files/textbook.pdfcontents 1 review 1 1.1 working with exponentials . . ....

Math for Success!

Jonathan LadouceurCarleton University

October 25, 2012

Contents

1 Review 11.1 Working with Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.3 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.4 Assignment I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Working with Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.1 Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.2 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.3 Assignment II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Polynomials 82.1 Modular Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.2 Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.3 Problem set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.4 Assignment III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 Synthetic Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2.1 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2.2 Assignment IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Greatest Common Divisor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3.1 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3.2 Assignment V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.4 Finding Roots of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4.1 Reduced Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4.2 Descartes Rule of Signs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4.3 Vieta’s Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4.4 Rational Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4.5 Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4.7 Assignment VI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.5 Symmetric Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.5.1 Vieta’s Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.5.2 Power Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.5.3 Assignment VII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

i

CONTENTS ii

3 Proofs 203.1 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.1.1 Summation Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.1.2 Rules for Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.1.3 Assignment VIII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2 Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2.1 Product Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2.2 Fermat’s Little Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2.3 Divisibility Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2.4 Assignment IX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.3 Logic Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.3.1 Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4 Trigonometry 254.1 Shifting Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2 De Moivre’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.3 Shifting Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.4 Intro to Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.5 Solve Trig Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.5.1 New Identites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.5.2 In class Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.5.3 Assignment X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.6 Working with Sum Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.6.1 New Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.6.2 In class Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.6.3 Assignment XI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.7 Working with Sum Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.7.1 New Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.7.2 In class Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.7.3 Assignment XII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.8 End of Trig Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.8.1 Assignment XIII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5 Combinatorics 295.1 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5.1.1 Addition Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.1.2 Multiplication Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.1.3 Complement Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.1.4 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.1.5 Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.1.6 Assignment XIV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5.2 Ordering, Replacement, Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.2.1 Formulas and Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.2.2 Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.2.3 Assignment XV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

CONTENTS iii

A Answers to Assignments 33A.1 Assignment I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33A.2 Assignment II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34A.3 Assignment III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35A.4 Assignment IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37A.5 Assignment V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38A.6 Assignment VI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42A.7 Assignment VII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43A.8 Assignment VII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43A.9 Midterm Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

A.9.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48A.9.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

A.10 Exam Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51A.10.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51A.10.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

B Handouts 55B.1 Trig Handout . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

Chapter 1

Review

1.1 Working with Exponentials

1.1.1 Rules

1. axay = ax+y

2. ax

ay= ax−y if a 6= 0

3. (ax)y = axy

4. ax · bx = (ab)x

5. ax

bx= (a

b)x if b 6= 0

6. a1/n = n√a

7. a0 = 1 if a 6= 0

8. a−x = 1ax

if a 6= 0

9. an − bn = (a− b)(an−1 + an−2b+ ...+ abn−2 + bn−1)

(difference of squares)

10. an + bn = (a+ b)(an−1 − an−2b+ ...− abn−2 + bn−1) (n odd)

(sum of squares)

WARNING: 00 is not defined and 0−x for x > 0 has no value.

1.1.2 Examples

1. What is 5 ·25 ·125 ·625 ·3125 ·15625 ·78125 ·390625 ·1953125 ·9765625 ·48828125 ·244140625 ·1220703125?

51 · 52 · · · 512 · 513

51+2+···+12+13

591

1

CHAPTER 1. REVIEW 2

2. Rationalise 2+√3

2−√3.

2 +√

3

2−√

3

2 +√

3

2−√

3

(2 +√

3

2 +√

3

)(2 +

√3)2

22 −√

32

4 + 4√

3 + 3

4− 3

7 + 4√

3

3. If n = m5 and m2 = 7, what is n3 ·m?

n3 ·m = (m5)3 ·m = m15 ·m = m16 = (m2)8 = 78

4. If x3y5 = 211 · 313 and xy2

= 127

, solve for x and y. Rearrange the second equation.

x

y2=

1

27

x =y2

27

Sub x = y2

27into the first equation.

x3y5 = 211 · 313

(y227

)3y5 = 211 · 313

(y2)3

273y5 = 211 · 313

(y6)

(33)3y5 = 211 · 313

(y6)

39y5 = 211 · 313

y11 = 39 · 211 · 313

y11 = 211 · 322

y11 = (2 · 32)11

y = 2 · 32 = 18

Sub y = 2 · 32 into the second equation.

x

y2=

1

27

CHAPTER 1. REVIEW 3

x =y2

27

x =(2 · 32)2

27

x =22 · 34

33

x = 22 · 34−3

x = 22 · 31 = 12

5. Solve for x: 28x = 162x

28x = 162x

2(23)x = (24)2x

223x = 24·2x

23x = 4 · 2x

23x = 22 · 2x

23x = 22+x

3x = 2 + x

2x = 2

x = 1

6. Simplify: (ab

+b

a+ 2)(a− b

2a+

b

a− b

)÷[(a+ 2b+ b2

a

)( a

a+ b+

b

a− b

)](a2 + b2 + 2ab

ab

)((a− b)(a− b) + 2ab

2a(a− b)

)÷[(a2 + 2ab+ b2

a

)(a(a− b) + b(b+ a)

(a+ b)(a− b)

)]((a+ b)2

ab

)( a2 + b2

2a(a− b)

)÷[((a+ b)2

a

)(a2 − ab+ b2 + ab)

(a+ b)(a− b)

)]((a+ b)2

ab

)( a2 + b2

2a(a− b)

)÷[((a+ b)2

a

)( a2 + b2

(a+ b)(a− b)

)]((a+ b)2(a2 + b2)

2a2b(a− b)

)÷[(a+ b)2(a2 + b2)

a(a+ b)(a− b)

]((a+ b)2(a2 + b2)

2a2b(a− b)

)·[ a(a+ b)(a− b)

(a+ b)2(a2 + b2)

]Cancel out similar terms.

a+ b

2ab

CHAPTER 1. REVIEW 4

7. Simplify:x− 1

x3/4 + x1/2· x

1/2 + x1/4

x1/2 + 1· x1/4 + 1

Basic method:x− 1

x3/4 + x1/2· x

1/2 + x1/4

x1/2 + 1· x1/4 + 1

x− 1

x1/2(x1/4 + 1)· x

1/4(x1/4 + 1)

x1/2 + 1· x1/4 + 1

Cancel out similar terms.x− 1

x1/2 + 1+ 1

Rationalise.(x− 1)(x1/2 − 1)

(x1/2 + 1)(x1/2 − 1)+ 1

(x− 1)(x1/2 − 1)

x− 1+ 1

(x1/2 − 1) + 1

x1/2

1.1.3 Problem Set

1. A beaker containing bacteria is left to sit in the lab. This bacteria divides once every minute,doubling the volume in the beaker. If the beaker was left out for 60 minutes, when was thebeaker 1/4 full? (58 minutes)

2. Solve 2√

2 = 8k. (k = 12)

3. Solve 230 = n10. (n = 8)

4. Solve 2−3

4n+5 = 82n−1 (n = −54)

5. Simplify:

6−a

2b

(2a3−b

3a

)2(2a−b3−2b−3a)

6. Sherri plans to retire in 5 years. She has $10,000 in a savings account with a 50% interestrate. If the interest compounds yearly, how much will she have? What about if it compoundsmonthly, weekly, daily, (or continuously)? (Assuming 1 year, 12 months, 52 weeks, 365 daysgives $75, 937.50, $115, 804.66, $120, 378.67, $121, 616.70, $121,824.94)

7. If m and k are integers, find all solutions to the equation 9(7k + 7k+2) = 5m+3 + 5m (m = 2,k = 1)

8. If 2x+3 + 2x = 3y+2 − 3y and x and y are integers, determine the values of x and y. (x = 3,y = 2)

CHAPTER 1. REVIEW 5

9. Rationalise 11+x1/2

(1−√x

1−x )

10. The function y = axr passes through the points (3,1) and (27,3). Calculate r and a. (r =12,a = 1√

3)

11. Simplify:x− 1

x3/4 + x1/2· x

1/2 + x1/4

x1/2 + 1· x1/4 + 1

(√x)

12. Simplify when z > 0, q > p: ((z2/p + z2/q)2 − 4z2/p+2/q

(z1/p − z1/q)2 + 4z1/p+1/q

)1/2(z1/p − z1/q)

1.1.4 Assignment I

1. A radioactive isotope has a half-life of 4 weeks. In 20 weeks, you have 16 grams of it. Howmuch did you start out with?

2. Solve 5√

125 = 5x+1

3. If x3y5 = 210 · 313 and x2

y= 1

4, solve for x and y.

4. Simplify(x−4)6y−3(w2)5

x−5(y7)−2w−8

5. Rationalise3xy√x−√y

6. If m and k are integers, find all solutions to the equation

8(5k−1 + 5k+1) = 2m+6 + 2m

7. The function y = axr passes through the points (2,9) and (3,4). Calculate r and a.

8. Replace b with -b in an − bn = (a− b)(an−1 + an−2b+ ...+ abn−2 + bn−1).

(a) What happens when n is odd?

(b) What happens when n is even?

9. Simplify(a2 + ab+ b2)(a3 + b3)(a2 − b2)

(a3 − b3)(a+ b)2

10. Simplify:x− 1

x3/4 + 1· x+ x1/4

x3/4 + x1/4· x1/2 − x

CHAPTER 1. REVIEW 6

1.2 Working with Logarithms

Definition: loga x = b or x = ab

1.2.1 Rules

1. loga (xy) = loga (x) + loga (y)

2. loga

(xy

)= loga(x)− loga (y)

3. loga (xy) = y loga (x)

4. loga (ax) = aloga (x) = x

5. loga (1) = 0

6. loga (x) = 1logx (a)

7. logy (a) · loga (x) = loga (x)loga (y)

= logy (x)

8. logab (xy) = yb

loga (x)

WARNING: In loga b, both a AND b must be positive! Also loga (1) = 0 implies that log1 (a) isundefined.NOTATION: I will often use lg instead of log10 to shorten questions.

1.2.2 Problem Set

1. Solve for x: lg (8) = x lg (4)

(32)

2. Solve for x (x > 1): log3 (√x− 1) + log3 (

√x+ 1) = 2

(x = 10)

3. What is the sum of the following series?

lg2

1+ lg

3

2+ lg

4

3+ lg

5

4+ ...+ lg

100

99

(2)

4. Find log3 10 and log3 1.2 in terms of x = log3 4 and y = log5 3.

(log3 10 = x2

+ 1y

and log3 1.2 = 1 + x2− 1

y)

5. What is the value of y in this equation?

log4 12

log4 2+

log2 4

lg 2= log2 y

(y = 1200)

CHAPTER 1. REVIEW 7

6. How many digits are there in 21000 if 0.3010 < lg 2 < 0.3011?

(302 digits)

7. Find all values of x so that log6 (x+ 3) + log6 (x− 2) = 2

(x = 6, x 6= −7)

8. What is x if log2 (27) + log4 (9) + log8 (3) = x log2 3?

(133

)

9. Solve for x in:

log2

x+ 5

x− 5+ log2 (x2 − 25) = 0

(x = −6, x 6= −4)

10. What is − log2 (log2 (√

4√

2)) + log3 (log3 ( 3√

3))?

(2)

1.2.3 Assignment II

1. I want to use my calculator to evaluate log2 3, but my calculator only does logarithms inbase 10. Should I go find a better calculator, or should I be able to find a way to make mycalculator tell me log2 3.

2. Evaluate the product (log2 3)(log3 4)(log4 5)(log5 6)(log6 7)(log7 8)

3. In how many points do the graphs of y = 2 lg x and y = lg(2x) intersect?

4. Find all x such that log6(x+ 2) + log6(x+ 3) = 1

5. If lg 36 = a and lg 125 = b, express log ( 112

) in terms of a and b.

6. Given that 0.3010 < lg 2 < 0.3011, how many digits are in 5200?

7. For all positive numbers x 6= 1, simplify:

1

log3 x+

1

log4 x+

1

log5 x

8. If log8 3 = P and log3 5 = Q, express log10 5 in terms of P and Q.

Chapter 2

Polynomials

2.1 Modular Arithmetic

Basically, think clock arithmetic: 5 hours + 9 hours = 2 hours (analog clock). The idea is to add,subtract, or multiply like usual, and then subtract a multiple of the base (m).

2.1.1 Examples

In base 12:

5 + 9 = 14 (mod 12)= 14− 12 (mod 12)= 2 (mod 12)

Or in base 10:

5× 9 = 45 (mod 10)= 45− 40 (mod 10)= 5 (mod 10)

This gives an addition or multiplication table from 0 to m-1.Multiplication Table for base 5 and base 6

1 2 3 41 1 2 3 42 2 4 1 33 3 1 4 24 4 3 2 1

1 2 3 4 51 1 2 3 4 52 2 4 0 2 43 3 0 3 0 34 4 2 0 4 25 5 4 3 2 1

Notice that for every number x in base 5, there is another number y in base 5 which gives xy = 1.We call y the inverse of x, and we call x a unit. Note that if y is the inverse of x, then x is theinverse of y.

2.1.2 Division

All that really gives trouble is division. If m is a prime, there isnt really anything to deal with. Wecan divide by any number other than zero. But if m isn’t a prime, some questions are impossibleor have multiple answers.

8

CHAPTER 2. POLYNOMIALS 9

1. 2x = 3 (mod 5)2x = 3 + 5 (mod 5)2x = 8 (mod 5)x = 4 (mod 5)

2. 2x = 4 (mod 6)x = 2 (mod 3)x = 2, 5 (mod 6)

Examples

When the we divide by a factor of the mod, we divide the mod as well.

3. 2x = 3 (mod 6) is impossible

4. 4x = 2 (mod 6)4x = 2 + 6 (mod 6)4x = 8 (mod 6)2x = 4 (mod 3)x = 2 (mod 3)x = 2, 5 (mod 6)

2.1.3 Problem set

Solve for x Some questions may have no answer:

1. 5x = 7 (mod 13) (4)

2. 13x = 8 (mod 31) (3)

3. 32x = 2 (mod 47) (3)

4. 40x = 25 (mod 43) (6)

5. 88x = 45 (mod 89) (44)

6. 6x = 8 (mod 15) (impossible)

7. 25x = 15 (mod 45) (6)

8. What are the units of base 8? (1,3,5,7)

1 2 3 4 5 6 71 1 2 3 4 5 6 72 2 4 6 0 2 4 63 3 6 1 4 7 2 54 4 0 4 0 4 0 45 5 2 7 4 1 6 36 6 4 2 0 6 4 27 7 6 5 4 3 2 1

Since 1,3,5,7 all have a

1 in their column, they are units.


9. Find all the pairs of inverses base 11. ((1,1),(2,6),(3,4),(5,9),(7,8),(10,10))

1 2 3 4 5 6 7 8 9 101 1 2 3 4 5 6 7 8 9 102 2 4 6 8 10 1 3 5 7 93 3 6 9 1 4 7 10 2 5 84 4 8 1 5 9 2 6 10 3 75 5 10 4 9 3 8 2 7 1 66 6 1 7 2 8 3 9 4 10 57 7 3 10 6 2 9 5 1 8 48 8 5 2 10 7 4 1 9 6 39 9 7 5 3 1 10 8 6 3 210 10 9 8 7 6 5 4 3 2 1

Find all the 1s in the table

1 · 1 = 1

2 · 6 = 1

3 · 4 = 1

5 · 9 = 1

7 · 8 = 1

10 · 10 = 1

This gives you a list of inverses.

10. Let’s say we have 8 people in a circle and we would like to pick one. If we exclude the 1001stperson we count from the group and then continue counting starting at the next person, whowill be chosen? (7th person) First label the people

ABCDEFGH

1001 mod 8 = 1

Take out the 1st person and list the people starting at the next person.

BCDEFGH

1001 mod 7 = 7

Take out the 7th person and list the people starting at the next person.

BCDEFG

1001 mod 6 = 5

Take out the 5th person and list the people starting at the next person.

GBCDE

1001 mod 5 = 1


BCDE

1001 mod 4 = 1


CDE


1001 mod 3 = 2

Take out the 2nd person and list the people starting at the next person.

EC

1001 mod 2 = 1

So the remaining person is C.

2.1.4 Assignment III

1. 7x = 5 (mod 13)

2. 13x = 8 (mod 31)

3. 32x = 8 (mod 47)

4. 35x = 30 (mod 43)

5. 2x = 45 (mod 89)

6. 2x = 9 (mod 15)

7. 20x = 30 (mod 45)

8. What are the units of base 10?

9. Find all the pairs of inverses base 13.

10. Do Problem Set question 10 with 9 people and the 999th is chosen.

2.2 Synthetic Division

We have a way to divide polynomials. With numbers, when we divide f by g, we get quotient qand remainder r so that: f = q · g + r. If we say that f , g, q, and r are polynomials instead ofnumbers, that is exactly what we get.Let’s do an example:

4x3 − 2x2 − 2

2x2 − 4x+ 2

Write the coefficients of the polynomial to be divided at the top.4 -2 0 -2

Negate the coefficients of the divisor except the first. Then write them on the left in reverse order.4 -2 0 -2

-242

Look at the first empty column. Add all the values above the line and divide by the value in thebottom left corner. Place this at the bottom of the column.


4 -2 0 -2-242 2

Multiply values on the left by the latest value and place them diagonally to the right from thelatest value.

4 -2 0 -2-2 -44 82 2

Repeat the previous two steps until the diagonal reaches the very last column.4 -2 0 -2

-2 -4 -64 8 122 2 3

Then place a line marking the quotient from the remainder.4 -2 0 -2

-2 -4 -64 8 122 2 3

Add up the remaining columns without dividing by the corner.4 -2 0 -2

-2 -4 -64 8 122 2 3 8 -8

The result of our division is:

4x3 − 2x2 − 2

2x2 − 4x+ 2= 2x+ 3 +

8x− 8

2x2 − 4x+ 2

2.2.1 Problem Set

1. Divide x2 − 1 by x− 1.(x+ 1 remainder 0)

2. Divide x4 + 2x3 + 2x2 − 10x− 20 by x− 2.(x3 + 4x2 + 10x− 10 remainder −40)

3. Divide 3x4 + 2x3 − x2 + 4x− 5 by x− 1. (x3 + 5x2 + 4x− 8 remainder 3)

4. Divide −x4 − 4x3 + 14x2 + 2x− 1 by x2 − 2x+ 1. (−x2 − 6x+ 3 remainder 14x− 4)

5. Divide −x5 + x4 − 2x3 + 2x2 + 4x+ 1 by x3 − 2x+ 2. (−x2 + x− 4 remainder 6x2 − 6x+ 9)

6. Divide 2x5 + x4 − x3 + 8x2 + x+ 1 by 2x2 − 1x+ 2. (x3 + x2 − 2x+ 2 remainder 7x− 3)

7. Divide x6 − 4x5 − 7x4 − 3x3 − 2x2 + 6x+ 9 by x− 1. (x5 − 3x4 − 10x3 − 13x2 − 15x− 9)

8. Divide 4x6 +x5−x4 + 3x3 + 5x2 + 3x+ 10 by x+ 1. (4x5− 3x4 + 2x3 +x2 + 4x− 1 remainder11)

9. Divide x6 + x5 + x4 + x3 + x2 + x− 1 by x2 + x− 1. (x4 + 2x2 − x+ 4 remainder −4x+ 3)

10. Divide 4x6 + 2x5 − 4x4 + 3x3 + 5x2 + 3x+ 10 by 2x3 + x− 1. (2x3 + x2 − 3x+ 2 remainder9x2 − x+ 11)


2.2.2 Assignment IV

1. Divide x3 + 3x2 + 3x+ 1 by x2 + 2x+ 1.

2. Divide 2x4 + 2x3 + 2x2 − 10x− 20 by x− 2.

3. Divide 2x4 + 2x3 − x2 + 4x− 5 by x− 1.

4. Divide −2x4 − 4x3 + 14x2 + 2x− 1 by x2 − 2x+ 1.

5. Divide −3x5 + x4 − 2x3 + 2x2 + 4x+ 1 by x3 − 2x+ 2.

6. Divide 2x6 − 4x5 − 7x4 − 3x3 − 2x2 + 6x+ 9 by x− 1.

7. Divide 3x6 + x5 − x4 + 3x3 + 5x2 + 3x+ 10 by x+ 1.

8. Divide 2x6 + x5 + x4 + x3 + x2 + x− 1 by x2 + x− 1.

9. Divide 3x6 + x5 − x4 + 3x3 + 5x2 + 3x+ 10 by x3 + x2 + x+ 1.

10. Divide −3x6 + 6x5 − 4x4 + 6x3 + 5x2 + 3x+ 5 by 3x3 + x− 1.

2.3 Greatest Common Divisor

This is one of the oldest and most efficient algorithms in existence.Algorithm steps:

1. Write out GCD(a, b) where a is bigger than b.

2. Divide a by b, getting remainder r (Find a (mod b)).

3. Now the problem is to find GCD(b, r).

4. Repeat steps 2 and 3 until r is a unit or 0.

5. If r is a unit, then 1 is the greatest common divisor, and if r is 0 than the previous remainderis the greatest common divisor. NOTE: For integers, 1 and -1 are the only units. Forpolynomials, any real number is a unit.

This algorithm works because the GCD is a multiple of both a and b, so the remainder must alsobe a multiple of the GCD.ExamplesExample 1: Find the greatest common divisor of 29753 and 8739.GCD (29754,8739)= (29754)-3(8739)=3537=GCD (8739,3537) (8739)-2(3537)=1665=GCD (3537,1665) (3537)-2(1665)=207=GCD (1665,207) (1665)-8(207)=9=GCD (207,9) (207)-23(9)=0

So the GCD of 29753 and 8739 is 9.Example 2: Find the greatest common divisor of −2x2 − 7x+ 6x+ 4x− 1 and x2 + 3x− 4

GCD(−2x2 − 7x+ 6x+ 4x− 1, x2 + 3x− 4)


-2 -7 6 4 -14 -8 -4 4

-3 6 3 -31 -2 -1 1 -3 3

GCD(x2 + 3x− 4,−3x+ 3)

GCD(x2 + 3x− 4, x− 1)

1 3 -41 1 41 1 4 0

So the GCD is x− 1.

2.3.1 Problem Set

1. Find GCD(255,128). (1)

2. Find GCD(1644,9348). (12)

3. Find GCD(8767,4436). (1)

4. Find GCD(12345,9876). (2469)

5. Find GCD(1234,56789). (1)

6. Find GCD(x4 + 3x+ 2, x2 − 1).(x+1)

7. Find GCD(x3 + 3x+ 5, x2 − 1). (1)

8. Find GCD(x4 + 5x3 + 5x2 − x+ 2, x2 − x− 6). (x+2)

9. Find GCD(x4 + 5x3 + 5x2 + 3x− 6, x2 + x− 6). (1)

10. Simplify this fraction completely.

8844x5 − 4422x4 − 30954x3 − 4422x2 + 22110x+ 8844

4576x4 − 4576x3 − 13728x2 + 4576x+ 9152

(201208

(2x+ 1))

2.3.2 Assignment V

1. Find GCD(511,256).

2. Find GCD(1644,9360).

3. Find GCD(8771,4436).

4. Find GCD(12345,6789).

5. Find GCD(1234,98765).

6. Find GCD(x4 + 4x+ 3, x2 − 1).


7. Find GCD(x3 + 4x+ 5, x2 − 1).

8. Find GCD(x4 − 5x3 + 5x2 − 3x+ 2, x2 + x− 2).

9. Find GCD(2x4 + 5x3 + 5x2 + 3x+ 6, x2 + x− 6).

10. Simplify this fraction completely:

255x5 − 255

102x3 − 102

2.4 Finding Roots of Polynomials

2.4.1 Reduced Form

For this section, we always reduce rational polynomials. To reduce, first multiply by the greatestcommon denominator. Then we divide by the GCD of all coefficients.

Example

15x2 +5

2x+ 20→ 6x2 + x+ 8 (2.1)

2.4.2 Descartes Rule of Signs

(NOTE (x−a)n means we say a is a root n times. This is called multiplicity.) The highest possiblenumber of positive roots is the number of times a polynomial switches signs. The actual numberis this number minus some multiple of two. Something similar happens for the number of possiblenegative roots when the polynomial doesn’t switch signs.

Example

So, f(x) = 3x2 + 2x+ 5− 2 has 1 positive root and has 2 or 0 negative roots.

2.4.3 Vieta’s Formulas

If you expand a polynomial (x − a1)(x − a2) · · · (x + an), you should notice that the first term isalways xn, the second term is always −(a1+a2+...+an)xn−1 and the last term is always a1a2 · · · an.Each coefficient of the polynomial can be written in terms of a1, ..an.

Example

f(x) = x3 + x2 + 3x+ 4 has the same roots as:−(a+ b+ c) = 1(ab+ ac+ bc) = 3−abc = 4


2.4.4 Rational Roots

If we have a polynomial in reduced form then only a few rational numbers can be roots. Let’s sayr = n

dis a root. Then n is a divisor of the last term of the polynomial and d is a divisor of the

first term of the polynomial.So if 2x2 + 11x+ 5 has rational roots, then the numerators are divisors of 5 and the denominatorsare divisors of 2.

2x2 + 11x+ 5 = (2x+ 1)(x+ 5)

r = −12,−5

2.4.5 Remainder Theorem

f(x) has a root a when f(x) = (x− a)g(x) or f(a) = 0.

Example

f(x) = 2x2 + 11x+ 52 11 5

-5 -10 -51 2 1 0

This means that -5 is a root of f .2 11 5

-1/2 -1 -51 2 10 0

This means -1/2 is a root of f .

2.4.6 Problem Set

Factor each of these polynomials:

1. f(x) = 2x2 + 3x− 2 = (2x− 1)(x+ 2)

2. f(x) = 6x2 + x− 2 = (2x− 1)(3x+ 2)

3. f(x) = 2x2 + 7x+ 3 = (x+ 3)(2x+ 1)

4. f(x) = −x2 + 4x− 3 = −(x− 1)(x− 3)

5. Annie is a kid of a very specific age. In 12 years, her age will be the same as her present agesquared. How old is she? (4)

6. f(x) = x3 − 6x2 + 11x− 6 = (x− 1)(x− 2)(x− 3)

7. f(x) = 3x3 + 454x2 + 51

4x+ 9

2= 3

4(x+ 1)(x+ 2)(4x+ 3)

8. f(x) = 4x4 − 293x3 − 13

6x2 + 19

2x+ 3 = 1

6(x− 2)(3x+ 1)(2x− 3)(4x+ 3)

9. f(x) = 2x4 − 73x3 − 24x2 + 13

3x+ 4 = 1

3(x+ 3)(x− 4)(2x− 1)(3x+ 1)

10. What is the sum and product of all the roots of x4 + x3 + x2 + x+ 1?


2.4.7 Assignment VI


1. 10x5 − 87x4 + 227x3 − 227x2 + 87x− 10

2. 3x5 − 4x4 − 17x3 + 18x2 + 20x− 8

3. 2x5 + 2x4 − 16x3 − 16x2 + 32x+ 32

4. −72x5 − 132x4 − 14x3 + 37x2 + 4x− 3

5. What is the sum of the squares of the roots of x3− x2 + x+ 1? (HINT: Expand (a+ b+ c)2.This gives an equation where everything is in terms of the coefficients except the sum of thesquares of the roots.) )

2.5 Symmetric Polynomials

Examples

x+ y + z (2.2)

x3 + y3 + z3 (2.3)

x2yz + xy2z + xyz2 (2.4)

Each of these polynomials stay the same if you switch x with y, x with z, or y with z.

2.5.1 Vieta’s Formulas

In particular, we can write any symmetric polynomial in terms of Vieta’s formulas:

Examples in three variables

S1 = x+ y + z, S2 = xy + xz + yz, S3 = xyz

x+ y + z = S1 (2.5)

x3 + y3 + z3 = S31 − 3S1S2 + S3 (2.6)

2.5.2 Power Sums

Or we can describe the same polynomials in terms sums of each variable to some power.

Examples in three variables

P1 = x+ y + z, P2 = x2 + y2 + z2, P3 = x3 + y3 + z3

x+ y + z = P1 (2.7)

x3 + y3 + z3 = P3 (2.8)


Method of Undetermined Coefficients

If we are given a symmetric polynomial, we break it into parts with the same degree and thenwrite each part in terms of possible basis polynomials.

x2 + y2 + z2 = aS21 + bS2 (2.9)

We then expand the right side.

x2 + y2 + z2 = a(x+ y + z)2 + b(xy + xz + yz) (2.10)

x2 + y2 + z2 = a(x+ y + z + 2xy + 2xz + 2yz) + b(xy + xz + yz) (2.11)

x2 + y2 + z2 = ax2 + ay2 + az2 + (2a+ b)xy + (2a+ b)xz + (2a+ b)yz) (2.12)

This gives two equations based on the coefficients.

1 = a (2.13)

0 = 2a+ b (2.14)

The solution to these two is a = 1, b = −2

2.5.3 Assignment VII

Determine if these are symmetric polynomials and if so, write them in terms of Si and Pi.

1. x2y + y2x (n=2)

2. x2y + y2z + z2x (n=3)

3. (x+ y)(x+ z)(y + z) (n=3)

4. (x− y)(y − z)(z − x) (n=3)

5. What is the sum of the squares of the roots of x3 − 2?

6. The discriminant of f(x) = ax2 + bx+x is D = b2−4ac. Write this out in terms of the rootsof f using S1 = −b/a and S2 = c/a.

TIPS for Assignment:

M3,0,0 = x3 + y3 + z3 = P3 (2.15)

M2,1,0 = x2y + x2z + y2x+ y2z + z2x+ z2y (2.16)

M1,1,1 = xyz = S3 (2.17)

aS31 + bS2S1 + cS3 (2.18)

= a(x+ y + z)3 + b(xy + xz + yz)(x+ y + z) + c(xyz) (2.19)

= a(M3,0,0 + 3M2,1,0 + 6M1,1,1) + b(M2,1,0 + 3M1,1,1) + cM1,1,1 (2.20)


= aM3,0,0 + (3a+ b)M2,1,0 + (6a+ 3b+ c)M1,1,1 (2.21)

aP 31 + bP2S1 + cP3 (2.22)

= a(x+ y + z)3 + b(x2 + y2 + z2)(x+ y + z) + c(x3 + y3 + z3) (2.23)

= a(M3,0,0 + 3M2,1,0 + 6M1,1,1) + b(M3,0,0 +M2,1,0) + cM3,0,0 (2.24)

= (a+ b+ c)M3,0,0 + (3a+ b)M2,1,0 + 6aM1,1,1 (2.25)

Chapter 3

Proofs

3.1 Induction

3.1.1 Summation Notation

Mathematical notation for adding a large number of terms is very simple.

15∑i=1

[i− 1] = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 (3.1)

Essentially substitute i = 1 into the function “i − 1”. Then increase i by one and repeat until ireaches the top number. Finally substitute this last term into the function. This shorthand savesspace whenever there is a rule to generate every term.

3.1.2 Rules for Induction

Induction is a way to prove something for every single natural number (1,2,3,...). Let’s go througha simple example.We want to prove this:

n∑i=1

i =n(n+ 1)

2(3.2)

The first thing to do is check that it is true for n = 1

1∑i=1

i = 1 (3.3)

1(1 + 1)

2= 1 (3.4)

Then assume that it is true for some n = k.

k∑i=1

i =k(k + 1)

2(3.5)

20

CHAPTER 3. PROOFS 21

Based on our assumption, can we prove that it is true for n = k + 1? (This is the next numberafter k)

k+1∑i=1

i =k∑i=1

[i] + (k + 1) =k(k + 1)

2+ k + 1 = (3.6)

=k2 + k + 2k + 2

2=k2 + 3k + 2

2=

(k + 1)(k + 2)

2=

(k + 1)((k + 1) + 1)

2(3.7)

This is exactly what we want to prove. Since our statement is true for k = 1, it is true for k = 2.Since our statement is true for k = 2, it is true for k = 3. We can repeat until we get to k=n. Sowe have proved the statement for every natural number.

3.1.3 Assignment VIII

Prove each statement by induction on natural numbers.

1.n∑i=1

(2i− 1) = n2

2.n∑i=1

i2 =n(n+ 1)(2n+ 1)

6

3.n∑i=1

i3 =n2(n+ 1)2

4

4.n∑i=1

(i+ 1)2i = n2n+1

5.n∑i=1

1

i(i+ 1)=

n

n+ 1

6.n∑i=0

ari =a(rn+1 − 1)

r − 1

(Hint start with n = 0 instead of n = 1)

7. The Fibonnacci sequence is formed by starting with F1 = 1,F2 = 1 and then each term isthe sum of the last two terms.

1, 1, 2, 3, 5, 8, 13, ...

Use induction and the rule (Fn = Fn−1 + Fn−2) to prove that the following is true.

1 + 1 + 2 + 3 + 5 + ...+ Fn =n∑i=1

Fn = Fn+2 − 1 (3.8)


3.2 Divisibility

3.2.1 Product Notation

Mathematical notation for multiplying a large number of terms is very simple.

15∏i=1

[i] = 1× 2× 3× 4× 5× 6× 7× 8× 9× 10× 11× 12× 13× 14× 15 (3.9)

Essentially substitute i = 1 into the function “i”. Then increase i by one and repeat until i reachesthe top number. Finally substitute this last term into the function. This shorthand saves spacewhenever there is a rule to generate every term.

3.2.2 Fermat’s Little Theorem

Here is a simple fact about mods.ap−1 = 1(mod p)Where p is a prime number and a is not divisible by p.Consider the p− 1 numbers: a, 2a, 3a, ..., (p− 1)a. We can show that no wo of these numbers havethe same remainder when divided by p. If ka = na(mod p) then k = n(mod p). This is impossibleif k and n are different and less than p. If we multiply all our numbers together, what do we get:

a · 2a · 3a · · · (p− 1)a = 1 · 2 · · · (p− 1)(mod p)

We get the right side by simply reordering all the remainders mod p.This means that ap−1 = 1(mod p) because we can divide both sides by numbers smaller than p.

3.2.3 Divisibility Tests

Let’s consider the usual tests for when a number divides another and create more.The simplest test is for multiples of 2 and 5. For 2n or 5n we only need to check the last n digits.This test works because a× 10n + b = b (mod 2 or 5).Example: 927893567904 is divisible by 4 because the last two digits are divisible by 4. It is notdivisible by 5 because the last digit is not divisible by 5.For any other prime p, we need to be more creative. Let’s look one test for 3 and 9. For thesenumbers, we add the digits up and check if this new number is a multiple of 9. The reason thistest works is because 10a+ b = a+ b (mod 3 or 9). This rule inspires tests for other numbers.Let’s consider 7. First, 10=3 (mod 7). This gives a test we use in the next example:

Example

We want to test 2,438,195,760 to see if it is divisible by 7. We can do this in two ways.Rule: Take the first digit. Multiply the digit by 3. Add it to the next digit. Repeat the last twosteps. (Do this mod 7) If we get 0, we have a multiple of 7.2× 3 + 4 = 10 = 33× 3 + 3 = 12 = 55× 3 + 8 = 23 = 2


2× 3 + 1 = 7 = 00× 3 + 9 = 9 = 22× 3 + 5 = 11 = 44× 3 + 7 = 19 = 55× 3 + 6 = 21 = 00× 3 + 0 = 0Therefore it is a multiple of 7.For 13, 100=9 (mod 11). The difference is that we take two digits at a time when we multiply andadd.

Example

We want to test 14118 to see if it is a multiple of 13.First break the number into pairs starting at the last digit. 1—41—18Then multiply the first group by 9 and add to the second group 50—18Mod the groups by 13 to get 11—5.Repeat to get 104 = 8× 13 = 0(mod 13).As long as we mod a power of 10 greater than the prime, we can get a divisibility test for a number.

3.2.4 Assignment IX

1. (2 marks) The number 4,876,391,520 uses every digit once and is divisible by every numberfrom 1 to 18. Prove it for 7 and 11.

2. Prove that 3003000 − 1 is divisible by 1001.

3. Prove that n5 + 4n is divisible by 5 for any integer n. Use induction on n for the next fewquestions.

4. 2n+3 does not divide 32n − 1.

5.n∏i=1

[i2]

is divisible by 2n−1 (break into cases where n is odd and even.

6. Use induction to prove that 133 divides 11n+1 + 122n−1 for any n ≥ 1.

(Tricky-need to add and subtract something to make induction assumption appear)

7. Use induction to prove that (2n− 1)2 − 1 is divisible by 8 for any n ≥ 1.

BONUS Use symmetric polynomials to prove that if a+ b+ c is divisible by 6, then a3 + b3 + c3 isdivisible by 6. (HINT one of a,b,c must be even)


3.3 Logic Problems

3.3.1 Practice Problems

For the next three problems you face a number of doors which each have statements which aretrue or false. Only one is the correct one to pass through.

Problem 1:A: Do not go through door CB: At least one of these signs is falseC: If door A is not the one to go through, then door B is.

Problem 2:A: Exactly two of these signs are trueB: Go this wayC: This is not the door to go throughD: Door B is not the way to go

Problem 3:A: This is not the door to go through unless the sign on the next door is true.B: Exactly two of these signs are falseC: This is the door to go through unless the sign on the previous door is false.

Problem 4 : My house has a number (integer):1. If my house number is a multiple of 3, then it is a number from 50 through 59.2. If my house number is not a multiple of 4, then it is a number from 60 through 69.3. If my house number is not a multiple of 6, then it is a number from 70 through 79.

Problem 5 : I want to play a game with the following rules:1. The game starts with 21 sticks.2. At each turn, alternating between 2 players, the player removes up 1, 2, or 3 sticks.3. The player who picks up the last stick loses.If I play second, what strategy will guarantee that I win?

Problem 6 : Philip tells the truth on one day of the week and lies every other day.1. One day he said: “I lie on Mondays and Tuesdays.”2. On the next day he said: “Today is either Thursday, Saturday, or Sunday.”3. On the next day he said: “I lie on Wednesdays and Fridays.”On which day of the week does Philip tell the truth?

Chapter 4

Trigonometry

4.1 Shifting Functions

4.2 De Moivre’s Formula

4.3 Shifting Functions

4.4 Intro to Problems

4.5 Solve Trig Identities

4.5.1 New Identites

tanx

2=

sinx

1 + cos x=

1− cosx

sinx(4.1)

sinx =2 tan x

2

1 + tan2 x2

(4.2)

cosx =1− tan2 x

2

1 + tan2 x2

(4.3)

4.5.2 In class Problems

cosα + cos 2α + cos 3α + ..+ cos kα =sin kα

2cos (k+1)α

2

sin α2

(4.4)

sinα + sin 2α + sin 3α + ..+ sin kα =sin kα

2sin (k+1)α

2

sin α2

(4.5)

What is A if tan 5α = cos(π9) cos(2π

9) cos(4π

9) where A equals:

A =sin 8α + sin 9α + sin 10α + sin 11α + sin 12α

cos 8α + cos 9α + cos 10α + cos 11α + cos 12α(4.6)

25

CHAPTER 4. TRIGONOMETRY 26

4.5.3 Assignment X

Answer at least 5 of the following:

1) (1 + sec 2α + tan 2α)(1− sec 2α = tan 2α) = 2 tan 2α (4.7)

2) (sec 2α + cot(5π

2+ 2α)) cot(

5π

4− α) = 1 (4.8)

3)cos(3π − 2α)

2 sin2(5π4

+ α)= tan(α− 5π

4) (4.9)

4)tan 2α + cot 3β

cot 2α + tan 3β=

tan 2α

tan 3β(4.10)

5) cosα + cos 2α + cos 6α + cos 7α = 4 cos(α/2) cos(5α/2) cos(4α) (4.11)

6) sin 9α + sin 10α + sin 11α + sin 12α = 4 cos(α/2) cosα sin(21α/2) (4.12)

7) cos 2α− cos 3α− cos 4α + cos 5α = −4 sin(α/2) sinα cos(7α/2) (4.13)

8) sin 4α− sin 5α− sin 6α + sin 7α = −4 sin(α/2) sinα sin(11α/2) (4.14)

BONUS

tanα

1− cotα+

1− cotα

tanα= 1 + secα cscα (4.15)

4.6 Working with Sum Identities

4.6.1 New Identity

tan 2x =2 tanx

1− tan2 x(4.16)


sinα+ sin β+ sin γ− sin(α+β) cos γ− cos(α+β) sin γ = 4 sinα + β

2sin

β + γ

2sin

γ + α

2(4.17)

(sinα− sin β)(sinα + sin β) = sin(α− β) sin(α + β) (4.18)

4.6.3 Assignment XI


1) sinα + sin

(α +

14

3π

)+ sin

(α− 8

3π

)= 0 (4.19)

2) cot2 α− cot2 β =cos2 α− cos2 β

sin2 α sin2 β(4.20)


3) (cosα− cos β)2 + (sinα− sin β)2 = 4 sin2 α− β2

(4.21)

4)(tanα + secα)(cosα− cotα)

(cosα + cotα)(tanα− secα)= 1 (4.22)

5)sin 4α

1 + cos 4α· cos 2α

1 + cos 2α= cot

(3π

2− α

)(4.23)

6) cos2(α− π

2

)+ cot2

(α− 3π

2

)=

1

sin2(α + π2)− cos2(α + π) (4.24)

7)1− tan

(π2

+ α)

1 + cot(2π − α)=

tan(π + α) + 1

cot(3π2− α

)− 1

(4.25)

8)tan 2α sec 2β − tan 2β sec 2α

sec 2α + sec 2β= tan(α− β) (4.26)

BONUS

tan

(α− 3π

4

)(1− sin 2α) = cos 2α (4.27)

4.7 Working with Sum Identities

4.7.1 New Identity

tanx =2 tan x

2

1− tan2 x2

(4.28)


cot2 2α− 1

2 cot 2α− cos 8α cot 4α = sin 8α (4.29)

sin 2α + sin 5α− sin 3α

cosα + 1− sin2 2α= 2 sinα (4.30)

4.7.3 Assignment XII


1)tan 3α

tan2 3α− 1· 1− cot2 3α

cot 3α= 1 (4.31)

2)1− cos 4α

sec2 2α− 1+

1 + cos 4α

csc2 2α− 1= 2 (4.32)

3)tanα− secα

cosα− cotα= tanα secα (4.33)

4)1− 2 sin2 α

1 + sin 2α=

1− tanα

1 + tanα(4.34)


5)cos 4α + 1

cotα− tanα=

1

2sin 4α (4.35)

6) cot(π

4+ 2α

)=

cos 4α

1 + sin 4α(4.36)

7)(sin2 α + tan2 α + 1)(cos2 α− cot2 α + 1)

(cos2 α + cot2 α + 1)(sin2 α + tan2 α− 1)= 1 (4.37)

8)

(√tanα +

√cotα

sinα + cosα

)2

=2

sin 2α(4.38)

BONUS

cos 4α tan 2α− sin 4α

cos 4α cot 2α + sin 4α= − tan2 2α (4.39)

4.8 End of Trig Identities

4.8.1 Assignment XIII


1) cot2 α + cot2 β − 2 cos(β − α)

sinα sin β+ 2 =

sin2(α− β)

sin2 α sin2 β(4.40)

2) sin 2α(2 cos 4α + 1) cot(π

6− 2α) cot(

π

6+ 2α) = sin 6α cot 2α tan 6α (4.41)

3) 4 cosα cos β cos(α− β)− 2 cos2(α− β)− cos 2β = cos 2α (4.42)

4) cotα− tanα− 2 tan 2α = 4 cot 4α (4.43)

5) cotα− tanα− 2 tan 2α− 4 tan 4α = 8 cot 8α (4.44)

6) sin2 β − cos2(α− β) + 2 cosα cos β cos(α− β) = cos2 α (4.45)

7) cos2 β + cos2(α− β)− 2 cosα cos β cos(α− β) = sin2 α (4.46)

8) tan 6α− tan 4α− tan 2α = tan 6α tan 4α tan 2α (4.47)

BONUS

cotα− tanα− 2 tan 2α− 4 tan 4α− ...− 2n tan 2nα = 2n+1 cot 2n+1α (4.48)

Chapter 5

Combinatorics

5.1 Counting

5.1.1 Addition Rule

If we have a choice between two possibilities, then we add the counts together. (OR, EACH)Example:How many ways can we pick one of three kinds of soda OR two kinds of juice?Solution:3+2=5

5.1.2 Multiplication Rule

If we have two choices which are both made one after the other, then we multiply the countstogether.(AND,Example:How many ways can we pick one of three kinds of soda AND one of two kinds of juice?Solution:3*2=6

5.1.3 Complement Rule

If it is easier to count the number of objects we don’t want, subtract from all possibilities.Example:How many two digit numbers have 1 as one of the two digits?Solution:9*10-8*9=18 (first digit can’t be zero)

5.1.4 Permutations

How many ways can we order n objects.Example: How many ways can you order these three letters: A,B,C.Solution:{ABC,ACB,BAC,BCA,CAB,CBA}In each position, we pick one of the remaining letters, so there are 3 in the first position, 2 remainingin the second position, and 1 remaining in the first position.There are 3 · 2 · 1 = 6 in total.By generalising to any number of objects, there are n! permutations (orderings) of n objects.

29

CHAPTER 5. COMBINATORICS 30

5.1.5 Practice Problems

1. There are 5 different teacups and 3 different tea saucers. How many ways are there to buya cup and a saucer?

2. There are 4 different teacups and 3 different tea saucers. How many ways are there to buya cup or a saucer?

3. There are three towns:A,B,C. There are 6 ways to travel from A to B and 4 ways to travelfrom B to C. How many ways are there to travel from A to C?

4. Each box in a 2× 2 table can be coloured white or black. How many ways can we colour thetable?

5. How many different ways can the letters in “VECTOR” be rearranged?

6. How many different ways can the letters in “TRUST” be rearranged?

7. How many six-digit numbers have all odd or all even digits.

8. How many six-digit numbers using 1,2,3,4,5,6 (any number of times) have at least two similardigits?

5.1.6 Assignment XIV

1. There are 5 teacups, 3 saucers, and 4 spoons. How many ways are there to buy two objectsof different types.

2. How many 1,2,3, or 4 letter words be made with the letters A,B,C?

3. How many different ways can the letters in “CARAVAN” be rearranged?

4. How many different ways can the letters in “CLOSENESS” be rearranged?

5. How many different ways can the letters in “MATHEMATICAL” be rearranged?

6. There are N power plugs and N power sockets. How many ways can these be plugged in?

7. There are three groups with 1, 2, and 4 people. How many ways can these groups be chosenfrom 7 people.

8. How man words can be made using 5 “A”s and at most 3 “B”s.

9. There are 2 apples, 3 pears, and 4 oranges. In how many different orderings can I eat them?

10. How many 6-digit numbers have at least one even digit. HINT:Count all odd digit 6-digitnumbers and subtract from all 6-digit numbers.


5.2 Ordering, Replacement, Similarity

5.2.1 Formulas and Derivation

Let’s pick k from n objects.

1. Order matters, with replacement:nk

2. Order doesn’t matter, with replacement:(n+k−1

k

)3. Order matters, without replacement:n!/k!

4. Order doesn’t matter, without replacement:(nk

)5.2.2 Practice Problems

1.∑n

i=1

(ni

)= 2n

2.(

nn−k

)=(nk

)3. How many ways are there to choose a team of 3 students from 30 students.

4. How many ways are there to divide 10 boys into 2 basketball teams of 5 each.

5. In how many ways can you choose 10 cards from a deck of 52 cards so that there is exactly1 ace?

6. In how many ways can you choose 10 cards from a deck of 52 cards so that there is at least1 ace?

7. How many six digit numbers have 3 even and 3 odd digits?

8. A person has 10 friends. Over several days he invites some of them to a dinner party so thatthe group chosen never repeats (possibly inviting no-one). How many days can he do thisfor?

9. How many ways are there to represent 1000000 as the product of three factors, if ordermatters?

5.2.3 Assignment XV

1. In how many ways are there to divide 15 people into three teams of 5?

2. How many ways are there to put 12 white and 12 black checkers on the black squares of achessboard?

3. How many ten-digit numbers have the sum of their digits equal to 4?

4. There are 10 types of postcards in a post office. How many ways are there to buy 12postcards?

5. A train with m passengers makes n stops. How many ways are there for passengers to getoff the train at the stops? (No passengers get on the train)


6. How many ways are there to arrange five red, five blue, and five green balls in a row so thatno two blue balls lie next to each other.

7. A theater group consists of 20 actors. How many ways are there to choose two groups of 6actors so that none of the actors are in both groups.

8. How many ways are there to divide a deck of 36 cards, including 4 aces, into halves so thateach half contains exactly 2 aces.

9. Find the sum of all 3-digit numbers that can be written using the digits 1,2,3,4 (with repe-tition allowed).

10. How many ways are there to put 3 one dollar bills and 10 quarters into 4 different boxes?

Appendix A

Answers to Assignments

A.1 Assignment I

1. A radioactive isotope has a half-life of 4 weeks. In 20 weeks, you have 16 grams of it. Howmuch did you start out with? (512 g)

2. Solve 5√

125 = 5x+1 (x = 32)

3. If x3y5 = 210 · 313 and x2

y= 1

4, solve for x and y.

Sub y = 4x2 into x3y5 = 210 · 313. (x = 3, y = 36)

4. Simplify(x−4)6y−3(w2)5

x−5(y7)−2w−8= x−19y11w18

5. Rationalise3xy√x−√y

=3xy(√x+√y)

x− y

6. If m and k are integers, find all solutions to the equation

8(5k−1 + 5k+1) = 2m+6 + 2m

23 · 5k−1(1 + 52) = 2m(26 + 1)

23 · 5k−1(1 + 25) = 2m(64 + 1)

23 · 5k−1 · 26 = 2m · 65

23 · 5k−1 · 2 · 131 = 2m · 51 · 131

23 · 5k−1 · 2 · 131 = 2m · 51 · 131

24 · 5k−1 = 2m · 51

So m = 4, k − 1 = 1

(m = 4, k = 2)

33

APPENDIX A. ANSWERS TO ASSIGNMENTS 34

7. The function y = axr passes through the points (2,9) and (3,4). Calculate r and a.

9 = a2r and 4 = a3r9

4=a2r

a3r

32

22=(2

3

)r(3

2

)2=(3

2

)−rSo r = −2

Then 9 = a2−2 so a = 36

8. Replace b with -b in an − bn = (a− b)(an−1 + an−2b+ ...+ abn−2 + bn−1).

(a) What happens when n is odd? We get a different formula: an + bn = (a + b)(an−1 −an−2b+ ...+ abn−2 + bn−1)

(b) What happens when n is even? We get a different way of writing an − bn:

an − bn = (a+ b)(an−1 − an−2b− ...+ abn−2 − bn−1)

9. Simplify(a2 + ab+ b2)(a3 + b3)(a2 − b2)

(a3 − b3)(a+ b)2

(a2 + ab+ b2)(a+ b)(a2 − ab+ b2)(a− b)(a+ b)

(a− b)(a2 + ab+ b2)(a+ b)2

a2 − ab+ b2

10. Simplify:x− 1

x3/4 + 1· x+ x1/4

x3/4 + x1/4· x1/2 − x

(x1/2 − 1)(x1/2 + 1)

x3/4 + 1· x

1/4(x3/4 + 1)

x1/4(x1/2 + 1)· x1/2 − x

(x1/2 − 1)x1/2 − xx− x1/2 − x−x1/2

A.2 Assignment II

1. I want to use my calculator to evaluate log2 3, but my calculator only does logarithms inbase 10. Should I go find a better calculator, or should I be able to find a way to make mycalculator tell me log2 3. (Take log10 3/ log10 2)

2. Evaluate the product (log2 3)(log3 4)(log4 5)(log5 6)(log6 7)(log7 8)

loga(b) logb(c) = loga(c)

This rule simplifies the product to log2 8 = 3


3. In how many points do the graphs of y = 2 lg x and y = lg(2x) intersect?

Simplify the first graph:lg(2x) = lg x+ lg(2). This only matches 2 lg x when x = 2

4. Find all x such that log6(x+ 2) + log6(x+ 3) = 1

log6(x+ 2)(x+ 3) = 1

(x+ 2)(x+ 3) = 6

x2 + 5x = 0

x = 0,−5

But only x = 0 is possible because logarithms of negatives have no value.

5. If lg 36 = a and lg 125 = b, express log ( 112

) in terms of a and b.

log ( 112

) = log ( 560

) = log(5)− log(6)− 1 = b3− a

2− 1

6. Given that 0.3010 < lg 2 < 0.3011, how many digits are in 5200?

1− lg 2 = lg 5

1− 0.3010 > 1− lg 2 > 1− 0.3011

200(1− 0.3010) > 200 lg 5 > 200(1− 0.3011)

139.8 > 200 lg 5 > 139.78

139 < 200 lg 5 < 140

10139 < 5200 < 10140

Then this number has 140 digits.

7. For all positive numbers x 6= 1, simplify:

1

log3 x+

1

log4 x+

1

log5 x

logx 3 + logx 4 + logx 5 = logx(3× 4× 5) = logx(60)

8. If log8 3 = P and log3 5 = Q, express log10 5 in terms of P and Q.

log10 5 = 1log5 10

log5 10 = log5 2 + 1

log2 5 = P3Q

Therefore log10 5 = 13

PQ+1

= PQ3+PQ

A.3 Assignment III

1. 7x = 5 (mod 13)

7x = 5− 26 (mod 13)

7x = −21 (mod 13)

x = −3 (mod 13)

x = 10 (mod 13)


2. 13x = 8 (mod 31)

13x = 8 + 31 (mod 31)

13x = 39 (mod 31)

x = 3 (mod 31)

3. 32x = 8 (mod 47)

4x = 1 (mod 47)

4x = 1 + 47 (mod 47)

4x = 48 (mod 47)

x = 12 (mod 47)

4. 35x = 30 (mod 43)

7x = 6 (mod 43)

7x = 6 + 43 (mod 43)

7x = 49 (mod 43)

x = 7 (mod 43)

5. 2x = 45 (mod 89)

2x = 45 + 89 (mod 89)

2x = 134 (mod 89)

x = 67

6. 2x = 9 (mod 15)

2x = 9 + 15 (mod 15)

2x = 24 (mod 15)

x = 12 (mod 15)

7. 20x = 30 (mod 45)

4x = 6 (mod 9)

2x = 3 (mod 9)

2x = 3 + 9 (mod 9)

2x = 12 (mod 9)

x = 6 (mod 9)

x = 6, 15, 24, 33, 42 (mod 45)

8. What are the units of base 10?

1,3,7,9

(GCD between 10 and these numbers is 1)

9. Find all the pairs of inverses base 13.

(1,1); (12,12); (2,7); (3,9); (4,10); (5,8); (6,11)


10. Do Problem Set question 10 with 9 people and the 999th is chosen.

Tricky without full solution, not on midterm. See example problem for method of solution.

A.4 Assignment IV

1. Divide x3 + 3x2 + 3x+ 1 by x2 + 2x+ 1.

1 3 3 1-1 -1 -1-2 -2 -21 1 1 0 0

x+ 1 remainder 0

2. Divide 2x4 + 2x3 + 2x2 − 10x− 20 by x− 2.

2 2 2 -10 -202 4 12 28 361 2 6 14 18 16

2x3 + 6x2 + 14x+ 18 remainder 16

3. Divide 2x4 + 2x3 − x2 + 4x− 5 by x− 1.

2 2 -1 4 -51 2 4 3 71 2 4 3 7 2

2x4 + 4x2 + 3x+ 7 remainder 2

4. Divide −2x4 − 4x3 + 14x2 + 2x− 1 by x2 − 2x+ 1.

-2 -4 14 2 -1-1 2 8 02 -4 -16 01 -2 -8 0 10 -1

−2x2 − 8x remainder 10x− 1

5. Divide −3x5 + x4 − 2x3 + 2x2 + 4x+ 1 by x3 − 2x+ 2.

-3 1 -2 2 4 1-2 6 -2 162 -6 2 -160 0 0 01 -3 1 -8 10 -14 17

−3x2 + x− 8 remainder 10x2 − 14x+ 17

6. Divide 2x6 − 4x5 − 7x4 − 3x3 − 2x2 + 6x+ 9 by x− 1.

2 -4 -7 -3 -2 6 91 2 -2 -9 -12 -14 -81 2 -2 -9 -12 -14 -8 1

2x5 − 2x4 − 9x3 − 12x2 − 14x− 8 remainder 1


7. Divide 3x6 + x5 − x4 + 3x3 + 5x2 + 3x+ 10 by x+ 1.

3 1 -1 3 5 3 10-1 -3 2 -1 -2 -3 01 3 -2 1 2 3 0 10

3x5 − 2x4 + x3 + 2x2 + 3x remainder 10

8. Divide 2x6 + x5 + x4 + x3 + x2 + x− 1 by x2 + x− 1.

2 1 1 1 1 1 -11 2 -1 4 -4 9

-1 -2 1 -4 4 -91 2 -1 4 -4 9 -12 8

2x4 − x3 + 4x2 − 4x+ 9 remainder −12x+ 8

9. Divide 3x6 + x5 − x4 + 3x3 + 5x2 + 3x+ 10 by x3 + x2 + x+ 1.

3 1 -1 3 5 3 10-1 -3 2 2 -4-1 -3 2 2 -4-1 -3 2 2 -41 3 -2 -2 4 5 1 6

3x3 − 2x2 − 2x+ 4 remainder 5x2 + x+ 6

10. Divide −3x6 + 6x5 − 4x4 + 6x3 + 5x2 + 3x+ 5 by 3x3 + x− 1.

-3 6 -4 6 5 3 51 -1 2 -1 1

-1 1 -2 1 -10 0 0 0 03 -1 2 -1 1 8 1 6

−2x3 + x2 − x+ 1 remainder 8x2 + x+ 6

A.5 Assignment V

1. Find GCD(511,256).

GCD(511, 256)

=GCD(256, 511− 2(256))

=GCD(256,−1)

=GCD(256, 1)

=1

2. Find GCD(1644,9360). (12)

GCD(9360, 1644)=

=GCD(1644, 9360− 6(1644))

=GCD(1644,−504)


=GCD(1644, 504)

=GCD(504, 1644− 3(504))

=GCD(504, 132)

=GCD(132, 504− 4(132))

=GCD(132,−24)

=GCD(132, 24)

=GCD(24, 132− 5(24))

=GCD(24, 12)

=GCD(12, 24− 2(12))

=GCD(12, 0)

=12

3. Find GCD(8771,4436).

GCD(8771, 4436)=

=GCD(4436, 8771− 2(4436))

=GCD(4436,−101)

=GCD(4436, 101)

=GCD(101, 4436− 44(101))

=GCD(101,−8)

=GCD(101, 8)

=GCD(8, 101− 13(8))

=GCD(8,−3)

=GCD(8, 3)

=GCD(3, 8− 3(3))

=GCD(3,−1)

=GCD(3, 1)

=1

4. Find GCD(12345,6789).

GCD(12345, 6789)=

=GCD(6789, 12345− 2(6789))

=GCD(6789,−1233)

=GCD(6789, 1233)

=GCD(1233, 6789− 6(1233))

=GCD(1233,−609)

=GCD(1233, 609)

=GCD(609, 1233− 2(609))


=GCD(609, 15)

=GCD(15, 609− 41(15))

=GCD(15,−6)

=GCD(15, 6)

=GCD(6, 15− 2(6))

=GCD(6, 3)

=GCD(3, 6− 2(3))

=GCD(3, 0)

=3

5. Find GCD(1234,98765).

GCD(98765, 1234)=

=GCD(1234, 98765− 80(1234))

=GCD(1234, 98765− 80(1234))

=GCD(1234, 45)

=GCD(45, 1234− 27(45))

=GCD(45, 19)

=GCD(19, 45− 2(19))

=GCD(19, 7)

=GCD(7, 19− 3(7))

=GCD(7,−2)

=GCD(7, 2)

=GCD(2, 7− 3(2))

=GCD(2,1)

=1

6. Find GCD(x4 + 4x+ 3, x2 − 1). (x+ 1)

1 0 0 4 31 1 0 10 0 0 01 1 0 1 4 4

GCD(x2 − 1, 4x+ 4)

GCD(x2 − 1, x+ 1)

1 0 -1-1 -1 11 1 -1 0

GCD(x+ 1, 0) = x+ 1


7. Find GCD(x3 + 4x+ 5, x2 − 1).

1 0 4 51 1 00 0 01 1 0 5 5

GCD(x2 − 1, 5x+ 5)

GCD(x2 − 1, x+ 1)

1 0 -1-1 -1 11 1 -1 0

GCD(x+ 1, 0) = x+ 1

8. Find GCD(x4 − 5x3 + 5x2 − 3x+ 2, x2 + x− 2).

1 -5 5 -3 22 2 -12 26

-1 -1 6 -131 1 -6 13 -28 28

GCD(x2 + x− 2,−28x+ 28)

GCD(x2 + x− 2, x− 1)

1 1 -21 1 21 1 2 0

GCD(x− 1, 0) = x+ 1

9. Find GCD(2x4 + 5x3 + 5x2 + 3x+ 6, x2 + x− 6).

1 -5 5 -3 22 2 -12 26

-1 -1 6 -131 1 -6 13 -28 28

GCD(x2 + x− 2,−28x+ 28)

GCD(x2 + x− 2, x− 1)

1 1 -21 1 21 1 2 0

GCD(x− 1, 0) = x+ 1

10. Simplify this fraction completely:

255x5 − 255

102x3 − 102

255(x5 − 1)

102(x3 − 1)


First find GCD of the constants. GCD(255, 102)=

=GCD(102, 255− 2(102))

=GCD(102, 51)

=GCD(51, 102− 2(51))

=GCD(51, 0)

=51

Now find GCD(x5 − 1, x3 − 1).

1 0 0 0 0 -11 1 00 0 0 00 0 0 01 1 0 0 1 0 -1

GCD(x3 − 1, x2 − 1)

1 0 0 -11 1 00 0 0 11 1 0 1 -1

GCD(x2 − 1, x− 1)

1 0 -11 1 11 1 1 0

GCD(x− 1, 0) = x− 1

5(51)(x− 1)(x4 + x3 + x2 + 1)

2(51)(x− 1)(x2 + x+ 1)

5(x4 + x3 + x2 + 1)

(x2 + x+ 1)

A.6 Assignment VI


1. 10x5 − 87x4 + 227x3 − 227x2 + 87x− 10

All positive roots by Newton’s rule.

Possibilities are 1,2,5,10,1/2,1/5,1/10,2/5,5/2.

(x− 1)(x− 2)(x− 5)(2x− 1)(5x− 1)

2. 3x5 − 4x4 − 17x3 + 18x2 + 20x− 8

2 positive and 3 negative roots by Newton’s rule.

Possibilities are ±1,±2,±4,±8,±1/3,±2/3,±4/3,±8/3.

(x− 2)2(x+ 1)(3x− 1)(x+ 2)(x− 2)


3. 2x5 + 2x4 − 16x3 − 16x2 + 32x+ 32


Possibilities are ±1,±2,±4,±8,±16,±32,±1/2.

2(x+ 1)(x+ 2)2(x− 2)2

4. −72x5 − 132x4 − 14x3 + 37x2 + 4x− 3


Possibilities are ±1,±3,±1/2,±1/3,±1/4,±1/6,±1/8,±1/9,±1/12,±1/18,±1/24,±36.

−3(2x+ 1)2(3x− 1)2(x+ 1)

5. What is the sum of the squares of the roots of x3− x2 + x+ 1? (HINT: Expand (a+ b+ c)2.This gives an equation where everything is in terms of the coefficients except the sum of thesquares of the roots.) )

This sum is -1.

A.7 Assignment VII

Determine if these are symmetric polynomials and if so, write them in terms of Si and Pi.

1. x2y + y2x (n=2)

Symmetric: S1S2 =P 31

2− P1P2

2

2. x2y + y2z + z2x (n=3)

Not symmetric when you switch x and y.

3. (x+ y)(x+ z)(y + z) (n=3) (S1S2 − S3 =P 31

3− P3

3)

4. (x− y)(y − z)(z − x) (n=3)

Not symmetric when you switch x and y.

5. What is the sum of the squares of the roots of x3 − 2? P2 = = S21 − 2S2 = 0− 2(0) = 0

6. The discriminant of f(x) = ax2 + bx+x is D = b2−4ac. Write this out in terms of the rootsof f using S1 = −b/a and S2 = c/a.

(D = a2(S21 − 4S2)

A.8 Assignment VII

Prove each statement by induction on natural numbers.

1.

n∑i=1

(2i− 1) = n2 (A.1)


Base case n = 1

1∑i=1

(2i− 1) = 12 (A.2)

2− 1 = 1 (A.3)

Assume for n = k

k∑i=1

(2i− 1) = k2 (A.4)

Proof for n = k + 1

k+1∑i=1

(2i− 1) =k∑i=1

(2i− 1) + 2(k + 1)− 1 (A.5)

= k2 + 2k + 1 (A.6)

= (k + 1)2 (A.7)

2.n∑i=1

i2 =n(n+ 1)(2n+ 1)

6(A.8)

Base case n = 1

1∑i=1

i2 =1(1 + 1)(2× 1 + 1)

6(A.9)

12 =2× 3

6(A.10)

Assume for n = k

k∑i=1

i2 =k(k + 1)(2k + 1)

6(A.11)

Proof for n = k + 1

k+1∑i=1

i2 =k∑i=1

i2 + (k + 1)2 (A.12)

=k(k + 1)(2k + 1)

6+ (k + 1)2 (A.13)

= (k + 1)(2k2 + k) + 6(k + 1)

6(A.14)

= (k + 1)2k2 + 7k + 6

6(A.15)

=(k + 1)(k + 2)(2k + 3)

6(A.16)

(A.17)


3.

n∑i=1

i3 =n2(n+ 1)2

4(A.18)

Base case n = 1

1∑i=1

i3 = 12 (A.19)

13 =12(1 + 1)2

4(A.20)

Assume for n = k

k∑i=1

i3 =k2(k + 1)2

4(A.21)

Proof for n = k + 1

k+1∑i=1

i3 =k∑i=1

i3 + (k + 1)3 (A.22)

=k2(k + 1)2

4+ (k + 1)3 (A.23)

= (k + 1)2(k2 + 4(k + 1))

4(A.24)

=(k + 1)2(k + 2)2

4(A.25)

(A.26)

4.

n∑i=1

(i+ 1)2i = n2n+1 (A.27)

Base case n = 1

1∑i=1

(i+ 1)2i = 1× 21+1 (A.28)

2× 2 = 22 (A.29)

Assume for n = k

k∑i=1

(i+ 1)2i = k2k+1 (A.30)


Proof for n = k + 1

k+1∑i=1

(i+ 1)2i =k∑i=1

(i+ 1)2i + (k + 1)2k+1 (A.31)

= k2k+1 + (k + 1)2k+1 (A.32)

= (2k + 2)2k+1 (A.33)

= (k + 1)2k+2 (A.34)

5.

n∑i=1

1

i(i+ 1)=

n

n+ 1(A.35)

Base case n = 1

1∑i=1

1

i(i+ 1)=

1

1 + 1(A.36)

1

1(1 + 1)=

1

2(A.37)

Assume for n = k

k∑i=1

1

i(i+ 1)=

k

k + 1(A.38)

Proof for n = k + 1

k+1∑i=1

1

i(i+ 1)=

k∑i=1

1

i(i+ 1)+

1

(k + 1)(k + 2)(A.39)

=k

k + 1+

1

(k + 1)(k + 2)(A.40)

=k(k + 2) + 1

(k + 1)(k + 2)(A.41)

=k2 + 2k + 1

(k + 1)(k + 2)(A.42)

=(k + 1)2

(k + 1)(k + 2)(A.43)

=k + 1

k + 2(A.44)

6.

n∑i=0

ari =a(rn+1 − 1)

r − 1(A.45)


(Hint start with n = 0 instead of n = 1) Base case n = 0

1∑i=0

ari =a(r0+1 − 1)

r − 1(A.46)

ar0 =a(r1 − 1)

r − 1(A.47)

Assume for n = k

k∑i=0

ari =a(rk+1 − 1)

r − 1(A.48)

Proof for n = k + 1

k+1∑i=0

ari =k∑i=0

ari + ark+1 (A.49)

=a(rk+1 − 1)

r − 1+ ark+1 (A.50)

= ark+1 − 1 + (r − 1)rk+1

r − 1(A.51)

= ark+2 − 1

r − 1(A.52)

(A.53)

7. The Fibonnacci sequence is formed by starting with F1 = 1,F2 = 1 and then each term isthe sum of the last two terms.

1, 1, 2, 3, 5, 8, 13, ...

Use induction and the rule (Fn = Fn−1 + Fn−2) to prove that the following is true.

1 + 1 + 2 + 3 + 5 + ...+ Fn =n∑i=1

Fn = Fn+2 − 1 (A.54)

Base case n = 1

1∑i=1

Fi = F1+2 − 1 (A.55)

1 = 2− 1 (A.56)

Assume for n = k

1∑i=k

Fi = Fk+2 − 1 (A.57)


Proof for n = k + 1

k+1∑i=1

Fi =k∑i=1

Fi + Fk+1 (A.58)

= Fk+2 − 1 + Fk+1 (A.59)

= Fk+3 − 1 (A.60)

(A.61)

A.9 Midterm Review

A.9.1 Questions

1. Solve for x:

44x = 168x

2. Rationalise and simplify 1(x+√3)(x−

√4)

.

3. For what value of x does log4 (23√x5) = log8 (4x3)?

4. How many digits are there in 5200?

5. Factor f(x) = 2x3 + 7x2 + 7x+ 2

6. A child’s age increased by 10 years gives a number which has a square root (whole number).Decreased by 10 years, the child’s age gives that square root. How old is the child?

7. Divide using Synthetic division with remainder: x6−x5 +x4 +x3 +x2−x−1 by x2 + 2x−2.

8. Find GCD(512,1424).

9. Find GCD(x4 + 3x3 − 3x2 − 11x− 6, 2x4 + 13x3 + 28x2 + 23x+ 6).

10. For any n ≥ 1, show that

n∑i=3

3i2 + 3i+ 1 = 1 + 7 + 19 + ...+ (3(n− 1)2 + 2(n− 1) + 1) + (3n2 + 3n+ 1) = n3

11. For any n ≥ 1, prove that 2n+2 does divide 32n − 1. (only one of (x-1),(x+1) can be divisibleby four at the same time, if one is even both are even)

12. Test to see if 538273 is divisible by 7.

13. What is 3x=9 (mod 6)?

14. What is 4x=9 (mod 7)?


A.9.2 Answers

1. Solve for x:

44x = 168x

22(4x) = 24(8x)

2(4x) = 4(8x)

21(22x) = 22(23x)

22x+1 = 23x+2

2x+ 1 = 3x+ 2

x = −1

2. Rationalise and simplify 1(x+√3)(x−

√4)

.

( (x−√3)(x+

√4)

x2−7x+12)

3. For what value of x does log4 (23√x5) = log8 (4x3)?

log4 (2x5/3 = log8 (4x3)

log4 2 + log4 x5/3 = log8 4 + log8 x

3

12

log2 2 + 12

log2 x5/3 = 1

3log2 4 + 1

3log2 x

3

12

+ 1253

log2 x = 23

log2 2 + 33

log2 x12

+ 56

log2 x = 23

+ log2 x12− 2

3= log2 x− 5

6log2 x

−16

= 16

log2 x

log2 x = −1

x = 2−1 = 12

4. How many digits are there in 5200?

1− lg 2 = lg 5

1− 0.3010 > 1− lg 2 > 1− 0.3011

200(1− 0.3010) > 200 lg 5 > 200(1− 0.3011)

139.8 > 200 lg 5 > 139.78

139 < 200 lg 5 < 140

10139 < 5200 < 10140

Then this number has 140 digits.


5. Factor f(x) = 2x3 + 7x2 + 7x+ 2

All negative roots.

Possibilities:−1/2, 1,−2

f(x) = (2x+ 1)(x+ 2)(x+ 1)

6. A child’s age increased by 10 years gives a number which has a square root (whole number).Decreased by 10 years, the child’s age gives that square root. How old is the child?

Let x be the child’s age.

x+ 10 = (x− 10)2

x2 − 21x+ 90 = 0

(x− 15)(x− 6) = 0

The only possibility is that the child is 15, since 6 would give a negative age ten years ago.

7. Divide using Synthetic division with remainder: x6−x5 +x4 +x3 +x2−x−1 by x2 + 2x−2.

1 -1 1 1 1 -1 -12 2 -6 18 -46 130

-2 -2 6 -18 46 -1301 1 -3 9 -23 65 -177 129

x4 − 3x3 + 9x2 − 23x+ 65 remainder −177x+ 129

8. Find GCD(512,1424). (4)

9. Find GCD(x4 + 3x3 − 3x2 − 11x− 6, 2x4 + 13x3 + 28x2 + 23x+ 6). (x2 + 4x+ 3)

GCD((x− 2)(x+ 1)2(x+ 3), (x+ 1)(x+ 2)(x+ 3)(2x+ 1))

= (x+ 1)(x+ 3)

10. For any n ≥ 1, show that

n∑i=3

3i2 + 3i+ 1 = 1 + 7 + 19 + ...+ (3n2 + 3n+ 1) = n3

11. For any n ≥ 1, prove that 2n+2 does divide 32n − 1. (only one of (x-1),(x+1) can be divisibleby four at the same time, if one is even both are even)

12. Test to see if 538273 is divisible by 7.

3(5) + 3 = 4 (mod 7)

3(4) + 8 = 6 (mod 7)

3(6) + 2 = 6 (mod 7)

3(6) + 7 = 4 (mod 7)

3(4) + 3 = 1 (mod 7)

Therefore 538273 = 1 (mod 7) and it is not divisible by 7.


13. What is 3x = 9 (mod 6)?

3x = 3 (mod 6)

x = 1 (mod 2)

14. What is 4x = 9 (mod 7)?

4x = 2 (mod 7)

2x = 1 (mod 7)

2x = 1 + 7 (mod 7)

2x = 8 (mod 7)

x = 4 (mod 7)

A.10 Exam Review

A.10.1 Questions

1. Find GCD(−x6 + 2x4 − 4x3 − x2 + x− 3, x5 − 3x4 + 4x3 − 3x2 + 2x− 1)

2. Find GCD(−x6 − x5 + 3x4 + 4x2 + 6x− 1, x5 − 3x4 + x3 + 4x2 − 4x− 3)

3. What is x in 5x = 17 (mod 11)?

4. What is x in 5x = 17 (mod 97)?

5. Find GCD(512,800).

6. What is x in 5x− 2 = 13 (mod 25)?

Prove the following

7.1

1− cosx+

1

1 + cos x= 2 csc2 x

8.(cot2 x+ 1)(sin2 x− 1) = − cot2 x

9.cscα + cotα = cot

α

2

10.

cos(α) + cos(2α) + cos(7α) + cos(8α) = 4 cosα

2cos 3α cos

9α

2

11.

(cos(α)− cos(β))2 + (sin(α)− sin(β))2 = 4 sin2(α− β

2

)12. Solve for (at least one) x:

tanx sin2 x = 2 tan x


13. Solve for (at least one) x:

4 cotx− sinx = 2 sin2 x

2

14. Five people get into an elevator. There are seven floors on which they may get off. Howmany ways can they leave the elevator if:

(a) Each person gets of on a different floor.

(b) We only care how many people get of on a floor, not which person.

(c) Each person gets of on any floor.

15. A box has 5 red apples and 4 green apples. How many ways are there to pick 2 red applesand 1 green apple (order matters)?

A.10.2 Answers

1. Find GCD(−x6 + 2x4 − 4x3 − x2 + x− 3, x5 − 3x4 + 4x3 − 3x2 + 2x− 1)

−1 0 2 −4 −1 1 −3

1 ↓ −1 −3−2 ↓ 2 6

3 ↓ −3 −9−4 ↓ 4 12

3 ↓ −3 9

1 −1 −3 −3 −19 10 6 −6

GCD(x5 − 3x4 + 4x3 − 3x2 + 2x− 1, 6x− 6)

GCD(x5 − 3x4 + 4x3 − 3x2 + 2x− 1, x− 1)

1 −3 4 −3 2 −1

1 ↓ 1 −2 2 −1 1

1 1 −2 2 −1 1 0

Therefore the GCD is x− 1.

2. Find GCD(−x6 − x5 + 3x4 + 4x2 + 6x− 1, x5 − 3x4 + x3 + 4x2 − 4x− 3)

−1 −1 3 0 4 6 −1

3 ↓ −3 −124 ↓ −4 −16−4 ↓ 4 16−1 ↓ 1 4

3 ↓ −3 −12

1 −1 −4 −8 8 16 −13 −13


GCD(x5 − 3x4 + x3 + 4x2 − 4x− 3,−13x− 13) GCD(x5 − 3x4 + x3 + 4x2 − 4x− 3, x+ 1)

1 −3 1 4 −4 −3

−1 ↓ −1 4 −5 1 3

1 1 −4 5 −1 −3 0

Therefore the GCD is x+ 1.

3. What is x in 5x = 17 (mod 11)?

5x = 17 (mod 11)

5x = 17 + 33 (mod 11)

5x = 50 (mod 11)

x = 10 (mod 11)

4. What is x in 5x = 17 (mod 97)?

5. Find GCD(512,800).

6. What is x in 5x− 2 = 13 (mod 25)?

Prove the following identities:

7.1

1− cosx+

1

1 + cos x= 2 csc2 x

8.(cot2 x+ 1)(sin2 x− 1) = − cot2 x

9.cscα + cotα = cot

α

2

10.


2cos 3α cos

9α

2


2cos 3α cos

9α

2

2 cos(4α) cos(3α) + 2 cos(5α) cos(3α) = 4 cosα

2cos 3α cos

9α

2

2 cos(3α)[cos(4α) + cos(5α)] = 4 cos 3α cosα

2cos

9α

2

cos(4α) + cos(5α) = 2 cosα

2cos

9α

2


11.

(cos(α)− cos(β))2 + (sin(α)− sin(β))2 = 4 sin2(α− β

2

)cos2 α + cos2 β − 2 cosα cos β + sin2 α + sin2 β − 2 sinα sin β = 4

1− cos(α− β)

2

cos2 α + sin2 α + cos2 β + sin2 β − 2 cosα cos β − 2 sinα sin β = 2− 2 cos(α− β)

1 + 1− 2(cosα cos β + sinα sin β) = 2− 2 cos(α− β)

12. Solve for (at least one) x:tanx sin2 x = 2 tan x

13. Solve for (at least one) x:

4 cotx− sinx = 2 sin2 x

2

14. Five people get into an elevator. There are seven floors on which they may get off. Howmany ways can they leave the elevator if:

(a) Each person gets of on a different floor.

(b) We only care how many people get of on a floor, not which person.

(c) Each person gets of on any floor.

15. A box has 5 red apples and 4 green apples. How many ways are there to pick 2 red applesand 1 green apple (order matters)?

Appendix B

Handouts

B.1 Trig Handout

WARNING: These formulas have specific values which make them undefined/infinite, especiallywhen they involve tangent.

Definitions

cotx =1

tanx(B.1)

secx =1

cosx(B.2)

cscx =1

sinx(B.3)

Pythagorean

cos2 x+ sin2 x = 1 (B.4)

A

1 + tan2 x = sec2 x (B.5)

cot2 x+ 1 = csc2 x (B.6)

Double Angle

sin 2x = 2 sin x cosx (B.7)

cos 2x = cos2 x− sin2 x (B.8)

sin 2x =2 tanx

1 + tan2 x(B.9)

cos 2x =1− tan2 x

1 + tan2 x(B.10)

tan 2x =2 tanx

1− tan2 x(B.11)

cot 2x =cot2 x− 1

2 cotx(B.12)

Table B.1: Trigonometric Substitution

New Function

Original sin2 x cos2 x tan2 x cot2 x

sin2 x sin2 x 1− cos2 xtan2 x

1 + tan2 x

1

1 + cot2 x

cos2 x 1− sin2 x cos2 x1

1 + tan2 x

cot2 x

1 + cot2 x

tan2 xsin2 x

1− sin2 x

1− cos2 x

cos2 xtan2 x

1

cot2 x

cot2 x1− sin2 x

sin2 x

cos2 x

1− cos2 x

1

tan2 xcot2 x

55

APPENDIX B. HANDOUTS 56

Table B.2: Trigonometric Shifts

Same Function Different Function

x π − α π + α −α π2− α 3π

2+ α 3π

2− α π

2+ α

sinx sinα − sinα − sinα cosα − cosα − cosα cosαcosx − cosα − cosα cosα sinα sinα − sinα − sinαtanx − tanα tanα − tanα cotα − cotα cotα − cotαcotx − cotα cotα − cotα tanα − tanα tanα − tanα

Addition

sin(x± y) = sin x cos y ± cosx sin y (B.13)

cos(x± y) = cos x cos y ∓ sinx sin y (B.14)

tan(x± y) =tanx± tan y

1∓ tanx tan y(B.15)

cot(x± y) =cotx cot y ∓ 1

cot y ± cotx(B.16)

Half-Angle

sin2 x

2=

1− cosx

2(B.17)

cos2x

2=

1 + cos x

2(B.18)

tanx

2=

sinx

1 + cos x=

1− cosx

sinx(B.19)

Sum to Product

sinx+ sin y = 2 sinx+ y

2cos

x− y2

(B.20)

sinx− sin y = 2 cosx+ y

2sin

x− y2

(B.21)

cosx+ cos y = 2 cosx+ y

2cos

x− y2

(B.22)

cosx− cos y = −2 sinx+ y

2sin

x− y2

(B.23)

Product to Sum

sinx sin y =1

2[cos(x−y)−cos(x+y)] (B.24)

cosx cos y =1

2[cos(x−y)+cos(x+y)] (B.25)

sinx cos y =1

2[sin(x−y)+cos(x+y)] (B.26)

math for success!jladouc2/files/textbook.pdfcontents 1 review 1 1.1 working with exponentials . . ....

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