This book starts with a brief introduction of mathematical induction and provides 101 practice problems of mathematical

*391*
*72*
*900KB*

*English*
*Pages 250*

- Author / Uploaded
- Duc Tran

Mathematical Induction 101 With 101 Practice Problems Duc Van Khanh Tran

Illustrations by Vy Nguyen Tong

Table of Contents About the Author

1

Preface

2

Prerequisites: Mathematical Induction

4

Problems 1-10

7

Problems 11-20

28

Problems 21-30

49

Problems 31-40

71

Problems 41-50

93

Problems 51-60

118

Problems 61-70

140

Problems 71-80

164

Problems 81-90

190

Problems 91-101

216

Acknowledgements

245

Reviews

246

About the Author Hi! My name is Duc Van Khanh Tran, and I am a Vietnamese undergraduate at the University of Texas at Austin majoring in Mathematics. When I was in elementary school, I studied at Morinosato Elementary School in Kanazawa, Ishikawa province, Japan for about two years. For middle school education, I studied at Le Quy Don Middle School in Ho Chi Minh City, Vietnam. When I was in 9th grade, I came to the USA and started my high school study at Brentwood Christian School in Austin, Texas. The classes in my high school were more relaxing and less stressful than the classes in my middle school, so I began understanding and enjoying the things I studied at school. After a while, I realized that I enjoyed math the most out of all the subjects. Also, there was a math team in my high school, and participating in the math team made me love mathematics even more. That is how I “fell in love” with mathematics. When I was in 11th grade, I came across @daily math , a very popular math account on Instagram. Inspired by the @daily math page, I also created a math account on Instagram called @dvkt math with about 33,000 followers currently. Other than sharing mathematics knowledge by the means of a math page on Instagram, I decided to also write math books. Before writing this book, I also wrote three other books: • An Introduction to Calculus: With Hyperbolic Functions, Limits, Derivatives, and More (May 2021); • Integrals and Sums Fiesta: An Integral Part of a Math Enthusiast’s Life (November 2021); • Basic Linear Algebra: An Introduction with an Intuitive Approach (May 2022).

1

Preface Mathematical induction is one of the common methods of proof in mathematics. It is a method to prove a statement about natural numbers n by figuring out the pattern from some particular cases, such as n = 1, n = 2, n = 3, etc. The method of mathematical induction is explained further in the “Prerequisites” chapter at the beginning of the book. After the “Prerequisites” chapter, there are 101 problems for the readers to practice with mathematical induction. This book aims to introduce the method of mathematical induction to those who are unfamiliar with it while also providing some interesting mathematical induction problems to those who are already familiar with it. So, this book would be suitable for students who want to learn about mathematical induction or who are looking for more problems to practice with mathematical induction. Most of these problems are either created by myself or collected from online sources such as Math Stack Exchange, Instagram, Twitter, etc. A few problems are taken from or inspired by problems in the book The Cauchy-Schwarz Master Class by J. Michael Steele and past preliminary exams of Singaporean junior colleges. (All of the solutions are written by myself and not taken from others.) While the practice problems cover a wide variety of topics (as you can see at the top of each problem), most problems in this book only require the prerequisite knowledge of high school Algebra and introductory Calculus. A few problems also require some familiarity with the basics of Linear Algebra and Set Theory. There might be some unfamiliar concepts to those who have only taken high school Algebra and introductory Calculus, but some important definitions required to understand the problems are provided in each problem. Lastly, please note that since we are only proving using mathematical induction in this book and the solutions are written by myself alone, some problems in this book could have other ways to prove, either with or without using mathematical induction. The readers are encouraged to find other possible methods of proof to improve their proof skills. I hope you enjoy the variety of mathematical induction problems awaiting ahead! Duc Van Khanh Tran Texas, USA, 2023 2

Prerequisites Mathematical Induction Mathematical induction is a proof method that can be used to prove a statement about natural numbers n ∈ N. To clarify the notations, in this book N does not include 0, and the letters k, m, and n always represent a natural number. Before discussing the mathematical induction method itself, let us discuss why it is called “mathematical induction.” There are mainly two types of proof: deductive proof and inductive proof. A deductive proof is where we prove a general statement directly by using axioms, definitions, proven theorems, etc. An inductive proof is where we prove a general statement by proving some specific cases first and then somehow relying on those specific cases to show that the statement is true for all cases. As you will see, mathematical induction relies on some specific cases called base cases. In this book we will focus on two most common types of mathematical induction called weak induction and strong induction.

Weak Induction To prove a statement S(n) for all natural numbers n, we prove the following. 1. S(1) is true; 2. If S(k) is true for some k ≥ 1, then S(k + 1) is also true.

Now let us discuss how this method can prove S(n) for all n ∈ N. First, we prove that the base case S(1) is true. Then, we prove the induction step, which is S(k +1) is true if S(k) is true. By applying the induction step to S(1), we can conclude that S(2) is true. Then, we can also conclude that S(3) is true because S(2) is true, S(4) is true because S(3) is true, S(5) is true because S(4) is true, and so on. Thus, this method can prove that S(n) is true for all n ∈ N.

4

This method of weak induction can be modified slightly to prove a statement S(n) for all n ≥ a for some natural number a > 1. Weak Induction (Modified) To prove a statement S(n) for all n ≥ a, we prove the following. 1. S(a) is true; 2. If S(k) is true for some k ≥ a, then S(k + 1) is also true.

Since we are proving a statement S(n) is true for all n ≥ a, the base case is S(a) instead of S(1). Then, applying the induction step, we can conclude that S(a + 1), S(a + 2), S(a + 3), etc. are true. Thus, we can prove that S(n) is true for all n ≥ a.

Strong Induction To prove a statement S(n) for all natural numbers n, we prove the following. 1. S(1), S(2), · · · , S(`) are true for some natural number `; 2. If S(m) is true for all 1 ≤ m ≤ k for some k ≥ `, then S(k + 1) is also true.

Now how does this method prove that S(n) is true for all n ∈ N? First, we prove the base cases S(1), S(2), · · · , S(`). Then, we prove the induction step, which is S(k + 1) is true if S(m) is true for all 1 ≤ m ≤ k. By applying the induction step to S(1), S(2), · · · , S(`), we can conclude that S(` + 1) is true. Then, we can also conclude that S(` + 2) is true because S(1), S(2), · · · , S(` + 1) are true, S(` + 3) is true because S(1), S(2), · · · , S(` + 2) are true, and so on. Thus, this method can prove that S(n) is true for all n ∈ N. Similarly to weak induction, strong induction can also be modified slightly to prove a statement S(n) for all n ≥ a for some natural number a > 1. Strong Induction (Modified) To prove a statement S(n) for all n ≥ a, we prove the following. 1. S(a), S(a + 1), · · · , S(a + `) are true for some natural number `; 2. If S(m) is true for all a ≤ m ≤ k for some k ≥ a + `, then S(k + 1) is also true.

Since we are proving a statement S(n) is true for all n ≥ a, the base cases would start with S(a) instead of S(1). So, the base cases would be S(a), S(a+1), · · · , S(a+ 5

`). Then, applying the induction step, we can conclude that S(a+`+1), S(a+`+2), S(a + ` + 3), etc. are true. Thus, we can prove that S(n) is true for all n ≥ a.

Now let us discuss the difference between weak induction and strong induction. Note that in weak induction, we only need one previous case S(k) to prove S(k +1). On the other hand, in strong induction, we need to use at least one previous case other than S(k) to prove S(k + 1). Usually, we use two or more previous cases in strong induction. However, like in problem 16, we do have some problems where we only use one previous case in strong induction. In problem 16, we use S(k − 1), NOT S(k), to prove S(k + 1), so it is strong induction instead of weak induction. Thus, the difference between weak induction and strong induction mainly lies in the induction step. That is why you will not be sure about whether to use weak or strong induction and how many base cases you need to prove until you prove the induction step. When we prove the induction step, we need to assume that the previous case(s) is/are true and then prove that S(k + 1) is true by using that assumption. The induction step could be quite challenging because sometimes you would need to be creative in connecting the previous cases to S(k + 1). To be more familiar with mathematical induction and to be better at deciding between weak and strong inductions, the best way is to practice as much as possible, and this book provides plenty of such practice problems. For each of the problems in this book, I suggest that you try it on your own first and then look at the solution after you figure it out or get stuck for a long time.

6

Problem 1 Topic: Series

Problem: Prove

n X

j=

j=1

n(n + 1) . 2

Solution Base Cases We need to prove that it is true for n = 1. We can check that 1 X

j=1=

j=1

1(1 + 1) , 2

so it is true for n = 1.

Induction Step Assume k X j=1

j=

k(k + 1) 2

for some k ≥ 1. To prove the induction step, we need to prove k+1 X j=1

j=

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

(1.1)

Using assumption (1.1), k+1 X

j=

j=1

k X

j + (k + 1)

j=1

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

=

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n X j=1

j=

n(n + 1) 2

for all n ∈ N.

8

Problem 2 Topic: Series

Problem: Prove

n X

j2 =

j=1

n(n + 1)(2n + 1) . 6

Solution Base Cases We need to prove that it is true for n = 1. We can check that 1 X

j 2 = 12 = 1 =

j=1

1(1 + 1)(2 · 1 + 1) , 6

so it is true for n = 1.

Induction Step Assume k X j=1

j2 =

k(k + 1)(2k + 1) 6

for some k ≥ 1. To prove the induction step, we need to prove k+1 X j=1

j2 =

(k + 1)(k + 1 + 1)(2(k + 1) + 1) (k + 1)(k + 2)(2k + 3) = . 6 6 9

(2.1)

Using assumption (2.1), k+1 X

j2 =

j=1

k X

j 2 + (k + 1)2

j=1

= = = = =

k(k + 1)(2k + 1) + (k + 1)2 6 k(k + 1)(2k + 1) (k + 1)(6k + 6) + 6 6 (k + 1) [k(2k + 1) + (6k + 6)] 6 (k + 1) 2k 2 + 7k + 6 6 (k + 1)(k + 2)(2k + 3) . 6

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n X j=1

j2 =

n(n + 1)(2n + 1) 6

for all n ∈ N.

10

Problem 3 Topic: Series

Problem: Prove

n X

n(n + 1) j = 2 j=1 3

2 .

Solution Base Cases We need to prove that it is true for n = 1. We can check that 1 X

j 3 = 13 = 1 =

j=1

1(1 + 1) 2

2 ,

so it is true for n = 1.

Induction Step Assume k X

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

2 (3.1)

for some k ≥ 1. To prove the induction step, we need to prove k+1 X

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

11

2

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

2 .

Using assumption (3.1), k+1 X j=1

j3 =

k X

j 3 + (k + 1)3

j=1

2 k(k + 1) + (k + 1)3 = 2 k 2 (k + 1)2 (4k + 4)(k + 1)2 + = 4 4 (k + 1)2 k 2 + 4k + 4 = 4 2 (k + 1)2 (k + 2) = 4 2 (k + 1)(k + 2) = . 2

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n X

n(n + 1) j = 2 j=1 3

for all n ∈ N.

12

2

Problem 4 Topic: Harmonic Number, Inequality

Problem: Prove that Hn ≤ n for all n ∈ N where Hn denotes the n-th Harmonic number.

Definition - Harmonic Number: Hn =

n X 1 j=1

j

.

Solution Base Cases We need to prove that it is true for n = 1. We can check that H1 =

1 X 1 j=1

j

=

1 = 1 ≤ 1, 1

so it is true for n = 1.

Induction Step Assume Hk ≤ k for some k ≥ 1. To prove the induction step, we need to prove Hk+1 ≤ k + 1. 13

(4.1)

Using assumption (4.1), Hk+1 =

k+1 X j=1

=

1 j

k X 1 j=1

j

+

1 k+1

1 k+1 1 ≤k+ . k+1

= Hk +

Since k + 1 ≥ 1, 1 ≤ 1, k+1 and we obtain Hk+1 ≤ k + 1. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Hn ≤ n for all n ∈ N.

14

Problem 5 Topic: Harmonic Number, Inequality

Problem: Prove

2n n+1 for all n ∈ N where Hn denotes the n-th Harmonic number. Hn ≥

Definition - Harmonic Number: Hn =

n X 1 j=1

j

.

Solution Base Cases We need to prove that it is true for n = 1. We can check that H1 =

1 X 1 j=1

j

=

1 2·1 =1≥ , 1 1+1

so it is true for n = 1.

Induction Step Assume Hk ≥

2k k+1

15

(5.1)

for some k ≥ 1. To prove the induction step, we need to prove Hk+1 ≥

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

Using assumption (5.1), Hk+1 =

k+1 X j=1

=

1 j

k X 1 j=1

j

+

1 k+1

1 k+1 1 2k + ≥ k+1 k+1 2k + 1 = . k+1 = Hk +

Note that

2k + 1 2k + 2 k − = ≥0 k+1 k+2 (k + 1)(k + 2)

because k ≥ 0, k + 1 > 0, and k + 2 > 0, so 2k + 1 2k + 2 ≥ . k+1 k+2 Therefore, 2k + 2 . k+2 Thus, we proved that it is true for n = k + 1 if it is true for n = k. Hk+1 ≥

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Hn ≥

2n n+1

for all n ∈ N.

16

Problem 6 Topic: Inequality

Problem: Prove en > n + 1 for all n ∈ N where e is the base of natural logarithm, e ≈ 2.7183.

Solution Base Cases We need to prove that it is true for n = 1. We can check that e1 = e ≈ 2.7183 > 1 + 1, so it is true for n = 1.

Induction Step Assume ek > k + 1 for some k ≥ 1. To prove the induction step, we need to prove ek+1 > (k + 1) + 1 = k + 2. Multiplying e to both sides of the inequality in assumption (6.1), ek+1 > ek + e. Since e > 2 > 1, ek + e > k + 2. Therefore, ek+1 > k + 2. Thus, we proved that it is true for n = k + 1 if it is true for n = k. 17

(6.1)

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, en > n + 1 for all n ∈ N.

18

Problem 7 Topic: Inequality

Problem: Prove en−1 > 2n for n ≥ 4 where e is the base of natural logarithm, e ≈ 2.7183.

Solution Base Cases We need to prove that it is true for n = 4. We can check that e4−1 = e3 ≈ (2.7183)3 > (2.7)3 = 19.683 > 24 , so it is true for n = 4.

Induction Step Assume ek−1 > 2k for some k ≥ 4. To prove the induction step, we need to prove e(k+1)−1 = ek > 2k+1 . Multiplying e to both sides of the inequality in assumption (7.1), ek > e · 2k . Since e > 2, e · 2k > 2 · 2k = 2k+1 . Therefore, ek > 2k+1 . Thus, we proved that it is true for n = k + 1 if it is true for n = k. 19

(7.1)

Conclusion We proved that it is true for n = 4. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, en−1 > 2n for n ≥ 4.

20

Problem 8 Topic: Inequality

Problem: Prove n! > 2n + 2n for n ≥ 5.

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 5. We can check that 5! = 120 > 42 = 2 · 5 + 25 , so it is true for n = 5.

Induction Step Assume k! > 2k + 2k for some k ≥ 5. To prove the induction step, we need to prove (k + 1)! > 2(k + 1) + 2k+1 . Multiplying both sides of assumption (8.1) by k + 1, (k + 1)! > 2k(k + 1) + 2k (k + 1). 21

(8.1)

Since k ≥ 5 > 1 and k + 1 ≥ 6 > 2, 2k(k + 1) > 2(k + 1) and 2k (k + 1) > 2k · 2 = 2k+1 . So, (k + 1)! > 2(k + 1) + 2k+1 . Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 5. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n! > 2n + 2n for n ≥ 5.

22

Problem 9 Topic: Inequality

Problem: Prove nn > 3n2 + 6 for n ≥ 4.

Solution Base Cases We need to prove that it is true for n = 4. We can check that 44 > 3 · 42 + 6, so it is true for n = 4.

Induction Step Assume k k > 3k 2 + 6 for some k ≥ 4. To prove the induction step, we need to prove (k + 1)k+1 > 3(k + 1)2 + 6 = 3k 2 + 6k + 9. Since k + 1 > k, (k + 1)k+1 = (k + 1) · (k + 1)k > (k + 1) · k k . Using assumption (9.1), (k + 1) · k k > (k + 1) 3k 2 + 6 = 3k 3 + 3k 2 + 6k + 6. Since k ≥ 4, 3k 3 + 6 ≥ 3 · 43 + 6 > 9, 23

(9.1)

so 3k 3 + 3k 2 + 6k + 6 > 3k 2 + 6k + 9. Therefore, (k + 1)k+1 > 3k 2 + 6k + 9 = 3(k + 1)2 + 6. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 4. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, nn > 3n2 + 6 for n ≥ 4.

24

Problem 10 Topic: Inequality

Problem: Prove Bernoulli’s inequality, which states that (1 + x)n ≥ 1 + nx for x > −1 and n ∈ N.

Solution Base Cases We need to prove that it is true for n = 1. Since (1 + x)1 = 1 + x and 1 + 1 · x = 1 + x, it is indeed true that (1 + x)1 ≥ 1 + 1 · x.

Induction Step Assume (1 + x)k ≥ 1 + kx for some k ≥ 1. To prove the induction step, we need to prove (1 + x)k+1 ≥ 1 + (k + 1)x. 25

(10.1)

Since x > −1, 1 + x > 0. So, we can multiply 1 + x to both sides of the inequality in assumption (10.1) without changing the sign: (1 + x)k+1 ≥ (1 + x)(1 + kx) = 1 + (k + 1)x + kx2 . Since k > 0 and x2 ≥ 0, kx2 ≥ 0, and thus (1 + x)k+1 ≥ 1 + (k + 1)x. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, (1 + x)n ≥ 1 + nx if x > −1 for all n ∈ N.

26

Problem 11 Topic: Series

Problem: For a 6= 1, prove n−1 X

aj =

j=0

an − 1 . a−1

Solution Base Cases We need to prove that it is true for n = 1. We can check that 1−1 X

aj = a0 = 1 =

j=0

a1 − 1 , a−1

so it is true for n = 1.

Induction Step Assume k−1 X

ak − 1 a−1

aj =

j=0

for some k ≥ 1. To prove the induction step, we need to prove (k+1)−1

X j=0

j

a =

k X

aj =

j=0

28

ak+1 − 1 . a−1

(11.1)

Using assumption (11.1), k X

aj =

j=0

k−1 X

aj + ak

j=0

ak − 1 + ak a−1 ak − 1 ak (a − 1) = + a−1 a−1 ak − 1 ak+1 − ak = + a−1 a−1 k+1 a −1 . = a−1 =

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n−1 X

aj =

j=0

if a 6= 1 for all n ∈ N.

29

an − 1 a−1

Problem 12 Topic: Series, Fibonacci Number

Problem: Prove

n X

Fj = Fn+2 − 1

j=1

where Fn denotes the n-th Fibonacci number.

Definition - Fibonacci Numbers: The Fibonacci numbers are a sequence of numbers defined as F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.

Solution Base Cases We need to prove that it is true for n = 1. When n = 1, 1 X

Fj = F1 = 1

j=1

and F1+2 − 1 = F3 − 1 = F2 + F1 − 1 = F1 + F0 + F1 − 1 = 1 + 0 + 1 − 1 = 1, so it is indeed true that 1 X

Fj = F1+2 − 1.

j=1

30

Induction Step Assume k X

Fj = Fk+2 − 1

(12.1)

j=1

for some k ≥ 1. To prove the induction step, we need to prove k+1 X

Fj = F(k+1)+2 − 1 = Fk+3 − 1.

j=1

Using assumption (12.1), k+1 X

Fj =

j=1

k X

Fj + Fk+1

j=1

= Fk+2 − 1 + Fk+1 . Using the recursive relation of Fn , Fn = Fn−1 + Fn−2 , we have Fk+2 + Fk+1 = Fk+3 . Therefore, k+1 X

Fj = Fk+3 − 1.

j=1

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n X

Fj = Fn+2 − 1

j=1

for all n ∈ N.

31

Problem 13 Topic: Series, Fibonacci Number

Problem: Prove

n X

F2j+1 = F2n+2

j=0

where Fn denotes the n-th Fibonacci number.

Definition - Fibonacci Numbers: The Fibonacci numbers are a sequence of numbers defined as F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.

Solution Base Cases We need to prove that it is true for n = 1. When n = 1, 1 X

F2j+1 = F1 + F3

j=0

and F2·1+2 = F4 = F2 + F3 . Since F2 = F1 + F0 = F1 + 0 = F1 , it is indeed true that 1 X

F2j+1 = F2·1+2 .

j=0

32

Induction Step Assume k X

F2j+1 = F2k+2

(13.1)

j=0

for some k ≥ 1. To prove the induction step, we need to prove k+1 X

F2j+1 = F2(k+1)+2 = F2k+4 .

j=0

Using assumption (13.1), k+1 X

F2j+1 =

j=0

k X

F2j+1 + F2k+3

j=0

= F2k+2 + F2k+3 . Using the recursive relation of Fn , Fn = Fn−1 + Fn−2 , we have F2k+2 + F2k+3 = F2k+4 . Therefore, k+1 X

F2j+1 = F2k+4 .

j=0

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n X

F2j+1 = F2n+2

j=0

for all n ∈ N.

33

Problem 14 Topic: Series, Fibonacci Number

Problem: Prove

n X

F2j = F2n+1 − 1

j=1

where Fn denotes the n-th Fibonacci number.

Definition - Fibonacci Numbers: The Fibonacci numbers are a sequence of numbers defined as F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.

Solution Base Cases We need to prove that it is true for n = 1. When n = 1, 1 X

F2j = F2 = F1 + F0 = 1 + 0 = 1

j=1

and F2·1+1 − 1 = F3 − 1 = F2 + F1 − 1 = F1 + F0 + F1 − 1 = 1 + 0 + 1 − 1 = 1, so it is indeed true that 1 X

F2j = F2·1+1 − 1.

j=1

34

Induction Step Assume k X

F2j = F2k+1 − 1

(14.1)

j=1

for some k ≥ 1. To prove the induction step, we need to prove k+1 X

F2j = F2(k+1)+1 − 1 = F2k+3 − 1.

j=1

Using assumption (14.1), k+1 X

F2j =

j=1

k X

F2j + F2k+2

j=1

= F2k+1 − 1 + F2k+2 . Using the recursive relation of Fn , Fn = Fn−1 + Fn−2 , we have F2k+1 + F2k+2 = F2k+3 . Therefore, k+1 X

F2j = F2k+3 − 1.

j=1

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n X

F2j = F2n+1 − 1

j=1

for all n ∈ N.

35

Problem 15 Topic: Series, Fibonacci Number

Problem: Prove

n X

Fj2 = Fn Fn+1

j=1

where Fn denotes the n-th Fibonacci number.

Definition - Fibonacci Numbers: The Fibonacci numbers are a sequence of numbers defined as F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.

Solution Base Cases We need to prove that it is true for n = 1. When n = 1, 1 X

Fj2 = F12 = 1

j=1

and F1 F1+1 = F1 F2 = F1 (F1 + F0 ) = 1, so it is indeed true that 1 X

Fj2 = F1 F1+1 .

j=1

36

Induction Step Assume k X

Fj2 = Fk Fk+1

(15.1)

j=1

for some k ≥ 1. To prove the induction step, we need to prove k+1 X

Fj2 = Fk+1 F(k+1)+1 = Fk+1 Fk+2 .

j=1

Using assumption (15.1), k+1 X

Fj2 =

j=1

k X

2 Fj2 + Fk+1

j=1 2 = Fk Fk+1 + Fk+1

= Fk+1 (Fk + Fk+1 ). Using the recursive relation of Fn , Fn = Fn−1 + Fn−2 , we have Fk + Fk+1 = Fk+2 . Therefore, k+1 X

Fj2 = Fk+1 Fk+2 .

j=1

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n X

Fj2 = Fn Fn+1

j=1

for all n ∈ N.

37

Problem 16 Topic: Series

Problem: Prove

n X j=0

1 n+1 = n+1 . 2n + 4j 2

Solution Base Cases We need to prove that it is true for n = 1 and n = 2. We can check that 1 X j=0

and

2 X j=0

1 1 1 1+1 1 + = = 1+1 = 21 + 4 j 2+1 2+4 2 2

1 1 1 3 2+1 1 + + = = 2+1 , = 22 + 4j 4 + 1 4 + 4 4 + 16 8 2

so it is true for n = 1 and n = 2.

Induction Step Assume

m X j=0

2m

m+1 1 = m+1 j +4 2

(16.1)

for all 1 ≤ m ≤ k for some k ≥ 2. To prove the induction step, we need to prove k+1 X j=0

1 (k + 1) + 1 k+2 = (k+1)+1 = k+2 . 2k+1 + 4j 2 2 38

Note that we can write the sum when n = k + 1 as k+1 X j=0

k

X 1 1 1 1 = + + 2k+1 + 4j 2k+1 + 1 2k+1 + 4k+1 j=1 2k+1 + 4j k

1

1X 1 1 = k+1 + . + k+1 2 +1 2 + 4k+1 4 j=1 2k−1 + 4j−1 Substituting j − 1 → j, k X j=1

k−1

X 1 1 = . k−1 j−1 k−1 + 4j 2 +4 2 j=0

Then, by assumption (16.1) when m = k − 1, k−1 X j=0

1 (k − 1) + 1 k = (k−1)+1 = k . 2k−1 + 4j 2 2

So, k+1 X j=0

1 1 1 k 1 = k+1 + + · k 2k+1 + 4j 2 + 1 2k+1 + 4k+1 4 2 1 k + k+2 k+1 +1 (1 + 2 ) 2 k+1 2 +1 k + = k+1 k+1 2 (2 + 1) 2k+2 k 1 = k+1 + k+2 2 2 k+2 = k+2 . 2 =

1

2k+1

+

2k+1

Thus, we proved that it is true for n = k +1 if it is true for n = m for all 1 ≤ m ≤ k.

Conclusion We proved that it is true for n = 1 and n = 2. Then, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k. Therefore, by strong induction, n X j=0

n+1 1 = n+1 2n + 4 j 2

for all n ∈ N.

39

Problem 17 Topic: Divisibility

Problem: Prove 3 | n3 − n

for all n ∈ N.

Definition - Divisibility: For a ∈ Z \ {0} and b ∈ Z, a divides b, denoted as a | b, when a is a factor of b. Formally, a | b ⇔ b = a` for some ` ∈ Z.

Solution Base Cases We need to prove that it is true for n = 1. Since 13 − 1 = 0 = 3 · 0 where 0 is an integer, 3 | 13 − 1 by definition of divisibility.

Induction Step Assume 3 | k3 − k 40

(17.1)

for some k ≥ 1. To prove the induction step, we need to prove 3 | (k + 1)3 − (k + 1) . By assumption (17.1) and the definition of divisibility, k 3 − k = 3` for some ` ∈ Z. Then, (k + 1)3 − (k + 1) = k 3 + 3k 2 + 3k + 1 − k − 1 = 3` + 3k 2 + 3k = 3 ` + k2 + k . Since ` + k 2 + k is an integer, 3 | (k + 1)3 − (k + 1)

by definition of divisibility. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, 3 | n3 − n for all n ∈ N.

41

Problem 18 Topic: Divisibility

Problem: Prove 5 | n5 − n

for all n ∈ N.

Definition - Divisibility: For a ∈ Z \ {0} and b ∈ Z, a divides b, denoted as a | b, when a is a factor of b. Formally, a | b ⇔ b = a` for some ` ∈ Z.

Solution Base Cases We need to prove that it is true for n = 1. Since 15 − 1 = 0 = 5 · 0 where 0 is an integer, 5 | 15 − 1 by definition of divisibility.

Induction Step Assume 5 | k5 − k 42

(18.1)

for some k ≥ 1. To prove the induction step, we need to prove 5 | (k + 1)5 − (k + 1) . By assumption (18.1) and the definition of divisibility, k 5 − k = 5` for some ` ∈ Z. Then, (k + 1)5 − (k + 1) = k 5 + 5k 4 + 10k 3 + 10k 2 + 5k + 1 − k − 1 = 5` + 5k 4 + 10k 3 + 10k 2 + 5k = 5 ` + k 4 + 2k 3 + 2k 2 + k . Since ` + k 4 + 2k 3 + 2k 2 + k is an integer, 5 | (k + 1)5 − (k + 1)

by definition of divisibility. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, 5 | n5 − n for all n ∈ N.

43

Problem 19 Topic: Divisibility

Problem: Prove 7 | (9n − 2n ) for all n ∈ N.

Definition - Divisibility: For a ∈ Z \ {0} and b ∈ Z, a divides b, denoted as a | b, when a is a factor of b. Formally, a | b ⇔ b = a` for some ` ∈ Z.

Solution Base Cases We need to prove that it is true for n = 1. Since 91 − 21 = 7 = 7 · 1 where 1 is an integer, 7 | 91 − 21

7 | 9k − 2k

by definition of divisibility.

Induction Step Assume

44

(19.1)

for some k ≥ 1. To prove the induction step, we need to prove 7 | 9k+1 − 2k+1 . By assumption (19.1) and by definition of divisibility, 9k − 2k = 7` for some ` ∈ Z. Then, 9k+1 − 2k+1 = 9 · 9k − 2 · 2k = (7 + 2) · 9k − 2 · 2k = 7 · 9k + 2 · 9k − 2k

= 7 · 9k + 2 · 7` = 7 9k + 2` . Since 9k + 2` is an integer, 7 | 9k+1 − 2k+1

by definition of divisibility. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, 7 | (9n − 2n ) for all n ∈ N.

45

Problem 20 Topic: Divisibility

Problem: Prove 9 | 82n − 3 · 7n + 2

for all n ∈ N.

Definition - Divisibility: For a ∈ Z \ {0} and b ∈ Z, a divides b, denoted as a | b, when a is a factor of b. Formally, a | b ⇔ b = a` for some ` ∈ Z.

Solution Base Cases We need to prove that it is true for n = 1. Since 82·1 − 3 · 71 + 2 = 45 = 9 · 5 where 5 is an integer, 9 | 82·1 − 3 · 71 + 2

by definition of divisibility.

Induction Step Assume 9 | 82k − 3 · 7k + 2 46

(20.1)

for some k ≥ 1. To prove the induction step, we need to prove 9 | 82(k+1) − 3 · 7k+1 + 2 . By assumption (20.1) and by definition of divisibility, 82k − 3 · 7k + 2 = 9` for some ` ∈ Z. Then, 82(k+1) − 3 · 7k+1 + 2 = 64 · 82k − 7 · 3 · 7k + 2 = 63 · 82k − 6 · 3 · 7k + 82k − 3 · 7k + 2 = 63 · 82k − 6 · 3 · 7k + 9` = 9 7 · 82k − 2 · 7k + ` . Since 7 · 82k − 2 · 7k + ` is an integer, 9 | 82(k+1) − 3 · 7k+1 + 2 by definition of divisibility. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, 9 | 82n − 3 · 7n + 2 for all n ∈ N.

47

Problem 21 Topic: Inequality

Problem: For x, y ≥ 0, prove n x+y xn + y n ≤ 2 2 for all n ∈ N.

Solution Base Cases We need to prove that it is true for n = 1. We can check that

x+y 2

1 =

x+y x1 + y 1 ≤ , 2 2

so it is true for n = 1.

Induction Step Assume

x+y 2

k ≤

xk + y k 2

for some k ≥ 1. To prove the induction step, we need to prove

x+y 2

k+1 ≤ 49

xk+1 + y k+1 . 2

(21.1)

Multiplying both sides of the inequality in assumption (21.1) by

x+y 2

k+1

≤

x+y 2

xk + y k 2

x+y 2 ,

xk+1 + y k+1 + xy k + yxk 4 xk+1 + y k+1 xk+1 + y k+1 xy k + yxk = − + 2 4 4 xk+1 + y k+1 xk+1 + y k+1 − xy k − yxk = − 2 4 xk − y k (x − y) xk+1 + y k+1 − . = 2 4 =

For any real numbers x and y, it is always either x ≥ y or y ≥ x. If x ≥ y, then x − y ≥ 0 and xk − y k ≥ 0, so xk − y k (x − y) ≥ 0. If y ≥ x, then x − y ≤ 0 and xk − y k ≤ 0, so xk − y k (x − y) ≥ 0. In both cases, xk − y k (x − y) ≥ 0. So, xk − y k (x − y) ≤ 0, − 4 and thus we have xk − y k (x − y) xk+1 + y k+1 xk+1 + y k+1 − ≤ . 2 4 2 Therefore,

x+y 2

k+1 ≤

xk+1 + y k+1 . 2

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n x+y xn + y n ≤ 2 2 for x, y ≥ 0 and for all n ∈ N.

50

Problem 22 Topic: Inequality

Problem: If a > b > 0, prove (2a − 2b)n ≥ 2an − (2b)n for all n ∈ N.

Solution Base Cases We need to prove that it is true for n = 1. We can check that (2a − 2b)1 = 2a − 2b ≥ 2a1 − (2b)1 , so it is true for n = 1.

Induction Step Assume (2a − 2b)k ≥ 2ak − (2b)k

(22.1)

for some k ≥ 1. To prove the induction step, we need to prove (2a − 2b)k+1 ≥ 2ak+1 − (2b)k+1 . Since a > b, 2a − 2b > 0, so we can multiply 2a − 2b to both sides of the inequality in assumption (22.1) without changing the sign: (2a − 2b)k+1 ≥ (2a − 2b) 2ak − (2b)k = 4ak+1 + (2b)k+1 − 4bak − (2a)(2b)k = 2ak+1 + 2ak+1 − (2b)k+1 + 2(2b)k+1 − 4bak − (2a)(2b)k = 2ak+1 − (2b)k+1 + 2(a − 2b) ak − (2b)k . 51

For any real numbers a and 2b, it is always either a ≥ 2b or 2b ≥ a. If a ≥ 2b, then a − 2b ≥ 0 and ak − (2b)k ≥ 0, so (a − 2b) ak − (2b)k ≥ 0. If 2b ≥ a, then a − 2b ≤ 0 and ak − (2b)k ≤ 0, so (a − 2b) ak − (2b)k ≥ 0. In both cases, (a − 2b) ak − (2b)k ≥ 0. So, 2ak+1 − (2b)k+1 + 2(a − 2b) ak − (2b)k ≥ 2ak+1 − (2b)k+1 . Therefore, (2a − 2b)k+1 ≥ 2ak+1 − (2b)k+1 . Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, (2a − 2b)n ≥ 2an − (2b)n if a > b > 0 for all n ∈ N.

52

Problem 23 Topic: Inequality

Problem: For y ≥ 0, prove ny − y n ≤ n − 1 for all n ∈ N.

Solution Base Cases We need to prove that it is true for n = 1 and n = 2. We can check that 1 · y − y1 = 0 ≤ 1 − 1 and 2 · y − y 2 = −(y − 1)2 + 1 ≤ 1 = 2 − 1, so it is true for n = 1 and n = 2.

Induction Step Assume my − y m ≤ m − 1

(23.1)

for all 1 ≤ m ≤ k for some k ≥ 2. To prove the induction step, we need to prove (k + 1)y − y k+1 ≤ (k + 1) − 1 = k. By assumption (23.1) when m = k, ky − y k ≤ k − 1 ⇔ −y k ≤ k − 1 − ky. 53

Since y ≥ 0, we can multiply y to both sides of the inequality without changing the sign: −y k+1 ≤ y(k − 1 − ky). So, (k + 1)y − y k+1 ≤ (k + 1)y + y(k − 1 − ky) = 2ky − ky 2 = k 2y − y 2 . By assumption (23.1) when m = 2, 2y − y 2 ≤ 2 − 1 = 1. Therefore, (k + 1)y − y k+1 ≤ k. Thus, we proved that it is true for n = k +1 if it is true for n = m for all 1 ≤ m ≤ k.

Conclusion We proved that it is true for n = 1 and n = 2. Then, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k. Therefore, by strong induction, ny − y n ≤ n − 1 if y ≥ 0 for all n ∈ N.

54

Problem 24 Topic: Matrices

Problem: For a, b ∈ R, prove a 0

0 b

n

n a = 0

0 bn

for all n ∈ N.

Definition - Matrix Multiplication: For those who are unfamiliar with matrices, for our purpose here, we only need to know that a11 a12 b11 b12 a11 b11 + a12 b21 a11 b12 + a12 b22 = . a21 a22 b21 b22 a21 b11 + a22 b21 a21 b12 + a22 b22

Solution Base Cases We need to prove that it is true for n = 1. We can check that 1 1 a 0 a 0 a 0 = = , 0 b 0 b 0 b1 so it is true for n = 1.

Induction Step Assume a 0

k k 0 a = b 0 55

0 bk

(24.1)

for some k ≥ 1. To prove the induction step, we need to prove

a 0 0 b

k+1 =

k+1 a 0

0 bk+1

.

Using assumption (24.1), a 0

k k+1 a 0 a 0 0 = 0 b 0 b b k a 0 a 0 = 0 bk 0 b k a · a + 0 · 0 ak · 0 + 0 · b = 0 · a + bk · 0 0 · 0 + bk · b k+1 a 0 = . 0 bk+1

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n n a 0 a 0 = 0 b 0 bn for a, b ∈ R and for all n ∈ N.

56

Problem 25 Topic: Matrices

Problem: For a ∈ R, prove 1 0

n a 1 = 1 0

na 1

for all n ∈ N.

Definition - Matrix Multiplication: For those who are unfamiliar with matrices, for our purpose here, we only need to know that a11 a12 b11 b12 a11 b11 + a12 b21 a11 b12 + a12 b22 = . a21 a22 b21 b22 a21 b11 + a22 b21 a21 b12 + a22 b22

Solution Base Cases We need to prove that it is true for n = 1. We can check that 1 1 a 1 1·a = , 0 1 0 1 so it is true for n = 1.

Induction Step Assume 1 0

k a 1 = 1 0 57

ka 1

(25.1)

for some k ≥ 1. To prove the induction step, we need to prove 1 0

k+1 a 1 = 1 0

(k + 1)a . 1

Using assumption (25.1), 1 0

k+1 k a 1 a 1 a = 1 0 1 0 1 1 ka 1 a = 0 1 0 1 1 · 1 + ka · 0 1 · a + ka · 1 = 0·1+1·0 0·a+1·1 1 (k + 1)a . = 0 1

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n 1 a 1 na = 0 1 0 1 for all n ∈ N.

58

Problem 26 Topic: Matrices, Fibonacci Number

Problem: Prove

1 1

n 1 F = n+1 0 Fn

Fn Fn−1

for all n ∈ N where Fn denotes the n-th Fibonacci number.

Definition - Matrix Multiplication: For those who are unfamiliar with matrices, for our purpose here, we only need to know that a11 a12 b11 b12 a11 b11 + a12 b21 a11 b12 + a12 b22 = . a21 a22 b21 b22 a21 b11 + a22 b21 a21 b12 + a22 b22

Definition - Fibonacci Numbers: The Fibonacci numbers are a sequence of numbers defined as F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.

Solution Base Cases We need to prove that it is true for n = 1. When n = 1, 1 1

1 0

1

1 = 1

1 F = 2 0 F1

F1 F0

because F0 = 0, F1 = 1, and F2 = F1 + F0 = 1 + 0 = 1, so it is true for n = 1. 59

Induction Step Assume 1 1

1 0

k =

Fk+1 Fk

Fk Fk−1

(26.1)

for some k ≥ 1. To prove the induction step, we need to prove 1 1

1 0

k+1 =

F(k+1)+1 Fk+1

Fk+1

=

F(k+1)−1

Fk+2 Fk+1

Fk+1 . Fk

Using assumption (26.1), 1 1

k+1 k 1 1 1 1 1 = 0 1 0 1 0 Fk+1 Fk 1 1 = Fk Fk−1 1 0 Fk+1 · 1 + Fk · 1 Fk+1 · 1 + Fk · 0 = Fk · 1 + Fk−1 · 1 Fk · 1 + Fk−1 · 0 Fk + Fk+1 Fk+1 = . Fk−1 + Fk Fk

Using the recursive relation Fn = Fn−1 + Fn−2 of Fn , Fk + Fk+1 = Fk+2 and Fk−1 + Fk = Fk+1 . Therefore, 1 1

k+1 1 F = k+2 0 Fk+1

Fk+1 . Fk

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n 1 1 Fn+1 Fn = 1 0 Fn Fn−1 for all n ∈ N.

60

Problem 27 Topic: Floor Function, Inequality

Problem: Prove

0≤

j k n+1 n ≤1 − 2 2

for all n ∈ N where b·c denotes the floor function.

Definition - Floor Function: For x ∈ R, the floor function bxc is the largest integer not exceeding x. For example, b2c = 2, b3.6c = 3, b−1.2c = −2, etc. Formally, for any ` ∈ Z, bxc = ` ⇔ ` ≤ x < ` + 1. With this definition of floor function, it should be easy to show that bxc has the property bx + ac = bxc + a for any a ∈ Z.

Solution Base Cases We need to prove that it is true for n = 1. When n = 1, 1+1 1 − = 1 − 0 = 1, 2 2 so it is true that

1+1 1 0≤ − ≤ 1. 2 2 61

Induction Step Assume

0≤

k+1 k − ≤1 2 2

(27.1)

for some k ≥ 1. To prove the induction step, we need to prove (k + 1) + 1 k+1 0≤ − ≤ 1. 2 2 Note that

(k + 1) + 1 k+1 k+2 k+1 − = − 2 2 2 2 k+1 k +1 − = 2 2 k k+1 = +1− 2 2 k+1 k =1− − . 2 2

By assumption (27.1),

k+1 k 0≤ − ≤ 1, 2 2 so

0≤1−

k k+1 − ≤ 1. 2 2

Therefore, k+1 (k + 1) + 1 − ≤ 1. 0≤ 2 2

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, j k n n+1 − ≤1 0≤ 2 2 for all n ∈ N.

62

Problem 28 Topic: Floor Function, Inequality

Problem: Prove

n+1 n ≥ 2 2

for all n ∈ N where b·c denotes the floor function.

Definition - Floor Function: For x ∈ R, the floor function bxc is the largest integer not exceeding x. For example, b2c = 2, b3.6c = 3, b−1.2c = −2, etc. Formally, for any ` ∈ Z, bxc = ` ⇔ ` ≤ x < ` + 1. With this definition of floor function, it should be easy to show that bxc has the property bx + ac = bxc + a for any a ∈ Z.

Solution Base Cases We need to prove that it is true for n = 1 and n = 2. We can check that 1+1 1 =1≥ 2 2 and

2+1 2 =1≥ , 2 2

so it is true for n = 1 and n = 2. 63

Induction Step Assume

m+1 m ≥ 2 2

(28.1)

for all 1 ≤ m ≤ k for some k ≥ 2. To prove the induction step, we need to prove (k + 1) + 1 k+1 ≥ . 2 2 Note that

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

By assumption (28.1) when m = k − 1, (k − 1) + 1 k k−1 . = ≥ 2 2 2 Therefore,

(k + 1) + 1 k−1 k+1 ≥ +1= . 2 2 2

Thus, we proved that it is true for n = k +1 if it is true for n = m for all 1 ≤ m ≤ k.

Conclusion We proved that it is true for n = 1 and n = 2. Then, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k. Therefore, by strong induction, n+1 n ≥ 2 2 for all n ∈ N.

64

Problem 29 Topic: Floor Function, Inequality

Problem: Prove

√ n n ≤ 2 for n ≥ 2 where b·c denotes the floor function.

Definition - Floor Function: For x ∈ R, the floor function bxc is the largest integer not exceeding x. For example, b2c = 2, b3.6c = 3, b−1.2c = −2, etc. Formally, for any ` ∈ Z, bxc = ` ⇔ ` ≤ x < ` + 1. With this definition of floor function, it should be easy to show that bxc has the property bx + ac = bxc + a for any a ∈ Z.

Solution Base Cases We need to prove that it is true for n = 2 and n = 3. When n = 2, √ √ √ √ 1 < 2 < 4 ⇔ 1 < 2 < 2, so When n = 3,

j√ k 2 2 =1≤ . 2 √

1

b1 . Let an =

ban−1 c 1 + 3 2

and

1 bn−1 + 3 6 for n ≥ 2. Prove that an > bn for all n ∈ N. bn =

Definition - Floor Function: For x ∈ R, the floor function bxc is the largest integer not exceeding x. For example, b2c = 2, b3.6c = 3, b−1.2c = −2, etc. Formally, for any ` ∈ Z, bxc = ` ⇔ ` ≤ x < ` + 1. With this definition of floor function, it should be easy to show that bxc has the property bx + ac = bxc + a for any a ∈ Z.

Solution Base Cases We need to prove that it is true for n = 1. Since we are given that a1 > b1 , it is true for n = 1. 90

Induction Step Assume ak > bk

(40.1)

for some k ≥ 1. To prove the induction step, we need to prove ak+1 > bk+1 . By the recursive formula of an , bak c 1 ak+1 = + . 3 2 From the definition of floor function, we know that ak < bak c + 1 ⇔ bak c > ak − 1. So,

ak 1 ak − 1 1 + = + . 3 2 3 6 Using assumption (40.1), since ak > bk , ak+1 >

ak+1 >

bk 1 + . 3 6

By the recursive formula of bn , 1 bk + = bk+1 . 3 6 Hence, ak+1 > bk+1 . Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, an > bn for all n ∈ N.

91

Problem 41 Topic: Harmonic Number, Inequality

Problem: Prove Hn Hn(2) + 2 ≥ Hn + Hn(2) + Hn(3) (m)

for all n ∈ N where Hn denotes the n-th Harmonic number and Hn the n-th generalized Harmonic number of order m.

denotes

Definition - Harmonic Number: Hn =

n X 1 j=1

j

.

Definition - Generalized Harmonic Number: Hn(m) =

n X 1 . m j j=1

Solution Base Cases We need to prove that it is true for n = 1. We can check that (2)

(2)

(3)

H1 H1 + 2 = 1 · 1 + 2 = 3 ≥ 3 = 1 + 1 + 1 = H1 + H1 + H1 , so it is true for n = 1. 93

Induction Step Assume (2)

(2)

(3)

Hk Hk + 2 ≥ Hk + Hk + Hk

(41.1)

for some k ≥ 1. To prove the induction step, we need to prove (2)

(2)

(3)

Hk+1 Hk+1 + 2 ≥ Hk+1 + Hk+1 + Hk+1 . Note that k+1 X

Hk+1 =

j=1

(2)

Hk+1 =

k+1 X j=1

k

1 1 1 X1 = + = Hk + , j j k+1 k+1 j=1 k

X 1 1 1 1 (2) = + = Hk + , 2 2 2 j j (k + 1) (k + 1)2 j=1

and (3)

Hk+1 =

k+1 X j=1

k

X 1 1 1 1 (3) = + = Hk + . 3 3 3 j j (k + 1) (k + 1)3 j=1

Using assumption (41.1), (2) Hk+1 Hk+1

+2=

1 Hk + k+1

(2) Hk +

1 (k + 1)2

+2

(2)

(2)

= Hk Hk +

Hk Hk 1 + + +2 k + 1 (k + 1)2 (k + 1)3 (2)

(2)

(3)

≥ Hk + Hk + Hk + +

Hk 1 Hk + + k + 1 (k + 1)2 (k + 1)3

(2)

= Hk +

Hk Hk (3) (2) + Hk + + Hk+1 . k+1 (k + 1)2

For k ≥ 1, Hk = 1 +

1 1 + · · · + ≥ 1, 2 k

and (2)

Hk

=1+

1 1 + · · · + 2 ≥ 1. 22 k

So, (2)

Hk+1 Hk+1 + 2 ≥ Hk +

1 1 (2) (3) + Hk + + Hk+1 k+1 (k + 1)2 (2)

(3)

= Hk+1 + Hk+1 + Hk+1 . Thus, we proved that it is true for n = k + 1 if it is true for n = k. 94

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Hn Hn(2) + 2 ≥ Hn + Hn(2) + Hn(3) for all n ∈ N.

95

Problem 42 Topic: Harmonic Number, Inequality

Problem: Prove Hn Hn(2)

p

Solution Base Cases We need to prove that it is true for n = 1. We can check that 1 X 1 1−j j 1 1 0 1 0 1 x y = x y + x y = x + y = (x + y)1 , j 0 1 j=0 so it is true for n = 1. 170

Induction Step Assume (x + y)k =

k X k k−j j x y j j=0

for some k ≥ 1. To prove the induction step, we need to prove (x + y)

k+1

=

k+1 X

k + 1 k+1−j j x y . j

j=0

Multiplying both sides of (73.1) by x + y, (x + y)k+1 =

k X k

j

j=0

= xk+1 +

xk+1−j y j +

k X k

j

j=0

xk−j y j+1

k X

k−1 k k+1−j j X k k−j j+1 x y + x y + y k+1 . j j j=0

j=1

Let j + 1 → j, k−1 X j=0

k k X k k−j j+1 X k k x y = xk−(j−1) y j = xk+1−j y j . j j − 1 j − 1 j=1 j=1

So, (x + y)k+1 = xk+1 +

k X k

j

j=1

k xk+1−j y j + y k+1 . j−1

+

Since k! k k k! + + = j!(k − j)! (j − 1)!(k − j + 1)! j j−1 j · k! (k − j + 1) · k! + = j!(k − j + 1)! j!(k − j + 1)! (k + 1)! k+1 = = , j!(k + 1 − j)! j which is known as Pascal’s identity, we have (x + y)

k+1

=x

k+1

+

k X k+1 j=1

Note that xk+1 = and y

k+1

j

xk+1−j y j + y k+1 .

k + 1 k+1−0 0 x y 0

k + 1 k+1−(k+1) k+1 = x y . k+1 171

(73.1)

Hence, (x + y)k+1 =

k+1 X j=0

k + 1 k+1−j j x y . j

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, (x + y)n =

n X n n−j j x y j j=0

for all n ∈ N.

172

Problem 74 Topic: Binomial Coefficient

Problem: For n, r ∈ N, prove the Hockey-Stick identity, n X j j=r

r

=

n+1 , r+1

for n ≥ r.

Definition - Binomial Coefficient: For non-negative integers p and r, ( p! if r ≤ p p . = r!(p−r)! r 0 if r > p

Solution Base Cases We need to prove that it is true for n = r. We can check that r X j j=r

r

=

r r+1 =1= , r r+1

so it is true for n = r. 173

Induction Step Assume k X j j=r

r

=

k+1 r+1

(74.1)

for some k ≥ r. To prove the induction step, we need to prove k+1 X j=r

j r

=

k+2 . r+1

Using assumption (74.1), k+1 X j=r

j r

k X j

k+1 r r j=r k+1 k+1 = + . r+1 r

=

+

In problem 73, we have calculated that k k k+1 + = . j j−1 j Let k → k + 1 and j → r + 1, k+1 k+1 k+2 + = . r+1 r r+1 So, k+1 X j=r

j r

=

k+2 . r+1

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = r. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n X j j=r

r

=

for n ≥ r.

174

n+1 r+1

Problem 75 Topic: Binomial Coefficient, Inequality

Problem: Prove

2n n

≤ 22n−1

for all n ∈ N.

Definition - Binomial Coefficient: For non-negative integers p and r, ( p! if r ≤ p p . = r!(p−r)! r 0 if r > p

Solution Base Cases We need to prove that it is true for n = 1. We can check that 2 2! = = 2 ≤ 2 = 22·1−1 , 1 1! 1! so it is true for n = 1.

Induction Step Assume 2k ≤ 22k−1 k 175

(75.1)

for some k ≥ 1. To prove the induction step, we need to prove 2k + 2 ≤ 22(k+1)−1 = 22k+1 . k+1 Note that

2k + 2 k+1

(2k + 2)! ((k + 1)!)2 (2k + 2)(2k + 1) (2k)! = · (k + 1)2 (k!)2 (2k + 2)(2k + 1) 2k . = (k + 1)2 k =

Since 2k + 1 ≤ 2k + 2, using assumption (75.1), 2k + 2 (2k + 2)2 2k ≤ (k + 1)2 k k+1 2 2k =2 k ≤ 22 · 22k−1 = 22k+1 . Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, 2n ≤ 22n−1 n for all n ∈ N.

176

Problem 76 Topic: Binomial Coefficient, Fibonacci Number

Problem: For any non-negative integer `, prove n X n j=0

j

F`+j = F`+2n

for all n ∈ N where Fn denotes the n-th Fibonacci number.

Definition - Binomial Coefficient: For non-negative integers p and r, ( p! if r ≤ p p = r!(p−r)! . r 0 if r > p

Definition - Fibonacci Numbers: The Fibonacci numbers are a sequence of numbers defined as F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.

Solution Base Cases We need to prove that it is true for n = 1. Since 1 X 1 1 1 F`+j = F`+0 + F`+1 = F` + F`+1 = F`+2 , j 0 1 j=0 it is true for n = 1. 177

Induction Step Assume k X k j=0

j

F`+j = F`+2k

(76.1)

for some k ≥ 1. To prove the induction step, we need to prove k+1 X k + 1 F`+j = F`+2(k+1) = F`+2k+2 . j j=0 First, note that k k+1 X X k + 1 k+1 k+1 k+1 F`+j = F`+0 + F`+j + F`+k+1 0 j k+1 j j=1 j=0 = F` +

k X k+1 j=1

j

F`+j + F`+k+1 .

In problem 73, we have calculated that k+1 k k = + . j j j−1 So, k X k k k+1 + F`+j + F`+k+1 F`+j = F` + j−1 j j j=1

k+1 X j=0

= F` +

k X k j=1

j

F`+j +

k X k F`+j + F`+k+1 . j−1 j=1

Substituting j − 1 → j, k k−1 X X k k F`+j = F`+j+1 . j−1 j j=1 j=0 So, k+1 X j=0

k k−1 X X k k+1 k F`+j = F` + F`+j + F`+1+j + F`+k+1 . j j j j=1 j=0

Note that F` +

k X k j=1

j

F`+j

k k X X k k k = F`+0 + F`+j = F`+j 0 j j j=1 j=0

and k−1 X j=0

k−1 k X k X k k k F`+1+j + F`+k+1 = F`+1+j + F`+1+k = F`+1+j . j j k j j=0 j=0 178

Hence, k+1 X j=0

k k X X k+1 k k F`+j = F`+j + F`+1+j . j j j j=0 j=0

By assumption (76.1), k X k j=0

j

F`+j = F`+2k

and k X k j=0

j

F`+1+j = F`+1+2k ,

so k+1 X j=0

k+1 F`+j = F`+2k + F`+1+2k = F`+2k+2 j

by using the recursive relation of Fn . Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n X n j=0

j

F`+j = F`+2n

for all n ∈ N.

179

Problem 77 Topic: Binomial Coefficient, Fibonacci Number

Problem: Prove

n X n−j j=0

j

= Fn+1

for all n ∈ N where Fn denotes the n-th Fibonacci number.

Definition - Binomial Coefficient: For non-negative integers p and r, ( p! if r ≤ p p . = r!(p−r)! r 0 if r > p

Definition - Fibonacci Numbers: The Fibonacci numbers are a sequence of numbers defined as F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.

Solution Base Cases We need to prove that it is true for n = 1 and n = 2. When n = 1, 1 X 1−j j=0

j

1 0 = + = 1 + 0 = 1 = F2 0 1 180

because F2 = F1 + F0 = 1 + 0 = 1. When n = 2, 2 X 2−j j

j=0

2 1 0 = + + = 1 + 1 + 0 = 2 = F3 0 1 2

because F3 = F2 + F1 = 1 + 1 = 2. So, it is true for n = 1 and n = 2.

Induction Step Assume

m X m−j j

j=0

= Fm+1

(77.1)

for all 1 ≤ m ≤ k for some k ≥ 2. To prove the induction step, we need to prove k+1 X j=0

k+1−j j

= Fk+2 .

Note that k+1 X j=0

k+1−j j

=

X k k+1 k+1−j 0 + + 0 j k+1 j=1

=1+

k X k+1−j j

j=1

.

In problem 73, we have calculated that k+1 k k = + . j j j−1 Let k → k − j,

k+1−j j

=

k−j j

k−j + . j−1

So, k+1 X j=0

k+1−j j

=1+

k X k−j j

j=1

k X k−j + . j−1 j=1

Note that 1+

k X k−j j=1

j

X X k k k−0 k−j k−j = + = . 0 j j j=1 j=0

Substituting j − 1 → j, k X k−j j=1

j−1

=

k−1 X j=0

181

k−1−j . j

Hence, k+1 X j=0

k+1−j j

=

k X k−j j

j=0

+

k−1 X j=0

k−1−j . j

By assumption (77.1) when m = k, k X k−j j

j=0

= Fk+1 .

By assumption (77.1) when m = k − 1, k−1 X j=0

k−1−j j

= Fk .

So, k+1 X j=0

k+1−j j

= Fk+1 + Fk = Fk+2

by using the recursive relation of Fn . Thus, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k.

Conclusion We proved that it is true for n = 1 and n = 2. Then, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k. Therefore, by strong induction, n X n−j j=0

j

for all n ∈ N.

182

= Fn+1

Problem 78 Topic: Derivative

Problem: Prove

dn n x = n! dxn

for all n ∈ N.

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. We can check that d x = 1 = 1!, dx so it is true for n = 1.

Induction Step Assume dk k x = k! dxk 183

(78.1)

for some k ≥ 1. To prove the induction step, we need to prove dk+1 k+1 x = (k + 1)!. dxk+1 Using assumption (78.1), dk+1 k+1 dk x = k k+1 dx dx

d k+1 x dx

dk (k + 1)xk dxk dk = (k + 1) k xk dx = (k + 1) · k! =

= (k + 1)!. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, dn n x = n! dxn for all n ∈ N.

184

Problem 79 Topic: Derivative

Problem: If f (x) is an even function, then prove that f (2n) (x) is even for all n ∈ N where f (n) (x) denotes the n-th derivative of f (x).

Solution Base Cases We need to prove that it is true for n = 1. If f (x) is even, then f (x) = f (−x). Differentiating both sides with respect to x twice, f (1) (x) = −f (1) (−x) f (2) (x) = − −f (2) (−x) = f (2) (−x). So, f (2) (x) is even.

Induction Step Assume that f (2k) (x) is even if f (x) is even for some k ≥ 1. To prove the induction step, we need to prove that f (2k+2) (x) is even. By the assumption that f (2k) (x) is even, f (2k) (x) = f (2k) (−x). Differentiating both sides with respect to x twice, f (2k+1) (x) = −f (2k+1) (−x) f (2k+2) (x) = − −f (2k+2) (−x) = f (2k+2) (−x). 185

Hence, f (2k+2) (x) is even. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, if f (x) is an even function, then f (2n) (x) is even for all n ∈ N.

186

Problem 80 Topic: Derivative, Trigonometry

Problem: Prove

dn nπ sin x = sin x + dxn 2

for all n ∈ N.

Solution Base Cases We need to prove that it is true for n = 1. Since d sin x = cos x dx and

π π π sin x + = sin x cos + cos x sin = cos x, 2 2 2

it is true that

d π sin x = sin x + . dx 2

Induction Step Assume

dk kπ sin x = sin x + dxk 2

for some k ≥ 1. To prove the induction step, we need to prove dk+1 (k + 1)π sin x = sin x + . dxk+1 2 187

(80.1)

Differentiating both sides of (80.1) with respect to x using the chain rule, d kπ kπ dk+1 sin x = sin x + = cos x + . dxk+1 dx 2 2 By using the addition formula of sine, π π kπ kπ kπ = sin x + cos + cos x + sin cos x + 2 2 2 2 2 (k + 1)π = sin x + . 2 Hence, (k + 1)π dk+1 sin x = sin x + . dxk+1 2 Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, dn nπ sin x = sin x + dxn 2 for all n ∈ N.

188

Problem 81 Topic: Derivative, Trigonometry

Problem: Prove

dn nπ cos x = cos x + dxn 2

for all n ∈ N.

Solution Base Cases We need to prove that it is true for n = 1. Since d cos x = − sin x dx and

π π π cos x + = cos x cos − sin x sin = − sin x, 2 2 2

it is true that

d π cos x = cos x + . dx 2

Induction Step Assume

dk kπ cos x = cos x + dxk 2

for some k ≥ 1. To prove the induction step, we need to prove dk+1 (k + 1)π cos x = cos x + . dxk+1 2 190

(81.1)

Differentiating both sides of (81.1) with respect to x using the chain rule, d kπ kπ dk+1 cos x = cos x + = − sin x + . dxk+1 dx 2 2 By using the addition formula of cosine, π π kπ kπ kπ = cos x + cos − sin x + sin − sin x + 2 2 2 2 2 (k + 1)π = cos x + . 2 Hence, (k + 1)π dk+1 cos x = cos x + . dxk+1 2 Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, dn nπ cos x = cos x + dxn 2 for all n ∈ N.

191

Problem 82 Topic: Derivative

Problem: Prove

dn (−1)n−1 (n − 1)! ln x = n dx xn

for all n ∈ N.

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. We can check that d 1 (−1)1−1 (1 − 1)! ln x = = , dx x x1 so it is true for n = 1.

Induction Step Assume (−1)k−1 (k − 1)! dk ln x = k dx xk 192

(82.1)

for some k ≥ 1. To prove the induction step, we need to prove (−1)k k! dk+1 ln x = . dxk+1 xk+1 Differentiating both sides of (82.1) with respect to x, dk+1 d (−1)k−1 (k − 1)! ln x = k+1 dx dx xk d 1 dx xk −k = (−1)k−1 (k − 1)! × k+1 x (−1)k k! = . xk+1 = (−1)k−1 (k − 1)! ×

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, dn (−1)n−1 (n − 1)! ln x = dxn xn for all n ∈ N.

193

Problem 83 Topic: Derivative

Problem: For a, b, c, d ∈ R and b, c 6= 0, prove dn ax + b n!(−c)n−1 (ad − bc) = n dx cx + d (cx + d)n+1 for all n ∈ N.

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. By the quotient rule, a(cx + d) − c(ax + b) d ax + b ad − bc 1!(−c)1−1 (ad − bc) = = = , 2 2 dx cx + d (cx + d) (cx + d) (cx + d)1+1 so it is true for n = 1.

Induction Step Assume

k!(−c)k−1 (ad − bc) dk ax + b = dxk cx + d (cx + d)k+1 194

(83.1)

for some k ≥ 1. To prove the induction step, we need to prove (k + 1)!(−c)k (ad − bc) dk+1 ax + b . = dxk+1 cx + d (cx + d)k+2 Differentiating both sides of (83.1) with respect to x, dk+1 ax + b d k!(−c)k−1 (ad − bc) = k+1 dx cx + d dx (cx + d)k+1 d 1 dx (cx + d)k+1 c(−k − 1) = k!(−c)k−1 (ad − bc) × (cx + d)k+2 (k + 1)!(−c)k (ad − bc) . = (cx + d)k+2 = k!(−c)k−1 (ad − bc) ×

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, dn ax + b n!(−c)n−1 (ad − bc) = n dx cx + d (cx + d)n+1 for all n ∈ N.

195

Problem 84 Topic: Derivative

Problem: Prove

1 (2n)! 1 dn √ = n · n dx 4 n! (1 − x)n+ 12 1−x

for all n ∈ N.

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. We can check that 1 1 d 1 2! √ · = = 1 , dx 1 − x 4 · 1! (1 − x)1+ 21 2(1 − x)3/2 so it is true for n = 1.

Induction Step Assume dk 1 (2k)! 1 √ = k · dxk 1 − x 4 k! (1 − x)k+ 21 196

(84.1)

for some k ≥ 1. To prove the induction step, we need to prove 1 (2k + 2)! 1 dk+1 √ = k+1 · . k+1 dx 4 (k + 1)! (1 − x)k+1+ 21 1−x Differentiating both sides of (84.1) with respect to x, 1 d (2k)! 1 dk+1 √ = · dxk+1 1 − x dx 4k k! (1 − x)k+ 12 d 1 (2k)! = k × 4 k! dx (1 − x)k+ 12 k + 12 (2k)! × 1 k 4 k! (1 − x)k+ 2 +1 1 (2k)! (2k + 1)(2k + 2) × = k × 1 4 k! 4(k + 1) (1 − x)k+1+ 2 1 (2k + 2)! · . = k+1 4 (k + 1)! (1 − x)k+1+ 12 =

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, dn 1 1 (2n)! √ = n · dxn 1 − x 4 n! (1 − x)n+ 12 for all n ∈ N.

197

Problem 85 Topic: Derivative, Trigonometry

Problem: Prove dn (−1)n−1 (n − 1)! −1 sin n cot−1 x tan x = n/2 2 dxn (1 + x ) for all n ∈ N.

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. Since sin2 cot−1 x + cos2 cot−1 x = 1 and cos cot−1 x = cot cot−1 x = x, −1 sin cot x we have 1 sin cot−1 x = √ . 1 + x2 198

So, d 1 tan−1 x = dx 1 + x2 1 1 ·√ =√ 1 + x2 1 + x2 (−1)1−1 (1 − 1)! = sin cot−1 x . 1/2 (1 + x2 )

Induction Step Assume

dk (−1)k−1 (k − 1)! −1 sin k cot−1 x tan x = k/2 2 dxk (1 + x )

for some k ≥ 1. To prove the induction step, we need to prove (−1)k k! dk+1 tan−1 x = sin (k + 1) cot−1 x . k+1 (k+1)/2 dx (1 + x2 ) Differentiating both sides of (85.1) with respect to x using the product rule, dk+1 d (−1)k−1 (k − 1)! −1 sin k cot−1 x tan x = k+1 k/2 dx dx (1 + x2 ) d 1 =(−1)k−1 (k − 1)! sin k cot−1 x × dx (1 + x2 )k/2 +

(−1)k−1 (k − 1)! (1 +

k/2 x2 )

×

d sin k cot−1 x . dx

Using the chain rule, d 1 −kx = k/2 (k+2)/2 dx (1 + x2 ) (1 + x2 ) and

d −k sin k cot−1 x = cos k cot−1 x · . dx 1 + x2

So, dk+1 −kx tan−1 x =(−1)k−1 (k − 1)! sin k cot−1 x × k+1 (k+2)/2 dx (1 + x2 ) (−1)k−1 (k − 1)! −k cos k cot−1 x × + k/2 1 + x2 (1 + x2 ) (−1)k k! x √ = sin k cot−1 x (k+1)/2 2 2 1+x (1 + x ) 1 +√ cos k cot−1 x . 1 + x2 199

(85.1)

Since sin2 cot−1 x + cos2 cot−1 x = 1 and

cos cot−1 x = cot cot−1 x = x, −1 sin cot x

we have

1 sin cot−1 x = √ 1 + x2

and

x cos cot−1 x = √ . 1 + x2

Hence, (−1)k k! dk+1 −1 tan x = cos cot−1 x sin k cot−1 x k+1 (k+1)/2 dx (1 + x2 ) + sin cot−1 x cos k cot−1 x (−1)k k! = sin (k + 1) cot−1 x . (k+1)/2 2 (1 + x ) Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, dn (−1)n−1 (n − 1)! −1 tan x = sin n cot−1 x n/2 2 dxn (1 + x ) for all n ∈ N.

200

Problem 86 Topic: Derivative, Hyperbolic Functions

Problem: Prove d2n (2n − 1)! −1 x tanh−1 x = n sinh 2n tanh 2n 2 dx (1 − x ) for all n ∈ N.

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. The second derivative of tanh−1 x is d2 d 1 2x tanh−1 x = = 2. dx2 dx 1 − x2 (1 − x2 ) The hyperbolic cosine cosh x and the hyperbolic sine sinh x satisfy the identity cosh2 x − sinh2 x = 1, which can be proven using the definitions cosh x =

ex + e−x 2

201

and sinh x =

ex − e−x . 2

Since cosh2 tanh−1 x − sinh2 tanh−1 x = 1 and

sinh tanh−1 x −1

cosh tanh we have

x

= tanh tanh−1 x = x,

x sinh tanh−1 x = √ 1 − x2

and

1 cosh tanh−1 x = √ . 1 − x2

Using the identity sinh (2x) = 2 sinh x cosh x, sinh 2 tanh−1 x = 2 sinh tanh−1 x cosh tanh−1 x 1 2x x ·√ = . =2· √ 2 2 1 − x2 1−x 1−x So, (2 · 1 − 1)! (1 −

1 x2 )

sinh 2 tanh−1 x =

1 2x 2x · = 2. 2 2 1−x 1−x (1 − x2 )

Hence, it is true that d2 (2 · 1 − 1)! −1 tanh−1 x = x . 1 sinh 2 tanh 2 dx2 (1 − x )

Induction Step Assume

d2k (2k − 1)! tanh−1 x = sinh 2k tanh−1 x k 2k dx (1 − x2 )

for some k ≥ 1. To prove the induction step, we need to prove d2k+2 (2k + 1)! sinh (2k + 2) tanh−1 x . tanh−1 x = k+1 2k+2 2 dx (1 − x ) Differentiating both sides of (86.1) with respect to x using the product rule, d 1 d2k+1 tanh−1 x =(2k − 1)! sinh 2k tanh−1 x × dx2k+1 dx (1 − x2 )k (2k − 1)! d + × sinh 2k tanh−1 x . k 2 dx (1 − x ) 202

(86.1)

Using the chain rule, 2kx d 1 = k k+1 2 dx (1 − x ) (1 − x2 ) and d 2k . sinh 2k tanh−1 x = cosh 2k tanh−1 x · dx 1 − x2 Hence, 2kx d2k+1 tanh−1 x =(2k − 1)! sinh 2k tanh−1 x × k+1 2k+1 dx (1 − x2 ) 2k cosh 2k tanh−1 x (2k − 1)! × + k 1 − x2 (1 − x2 ) (2k)! x √ = sinh 2k tanh−1 x 1 k+ 2 1−x (1 − x2 ) 2 1 cosh 2k tanh−1 x +√ 1 − x2 (2k)! −1 −1 = sinh tanh x sinh 2k tanh x 1 k+ (1 − x2 ) 2 + cosh tanh−1 x cosh 2k tanh−1 x . Using the identity cosh(x + y) = cosh x cosh y + sinh x sinh y, we get d2k+1 (2k)! −1 tanh−1 x = x . 1 cosh (2k + 1) tanh 2k+1 k+ dx (1 − x2 ) 2 Differentiating both sides with respect to x again using the product rule, 1 d2k+2 d tanh−1 x =(2k)! cosh (2k + 1) tanh−1 x × 2k+2 dx dx (1 − x2 )k+ 21 (2k)! d + cosh (2k + 1) tanh−1 x . 1 × k+ dx (1 − x2 ) 2 Using the chain rule, d 1 (2k + 1)x = k+ 3 dx (1 − x2 )k+ 12 (1 − x2 ) 2 and 2k + 1 d cosh (2k + 1) tanh−1 x = sinh (2k + 1) tanh−1 x · . dx 1 − x2 203

Hence, (2k + 1)x d2k+2 tanh−1 x =(2k)! cosh (2k + 1) tanh−1 x × 2k+2 k+ 3 dx (1 − x2 ) 2 +

(2k)! k+ 1

(2k + 1) sinh (2k + 1) tanh−1 x × 1 − x2

(1 − x2 ) 2 x (2k + 1)! √ cosh (2k + 1) tanh−1 x = k+1 2 2 1−x (1 − x ) 1 +√ sinh (2k + 1) tanh−1 x 1 − x2 (2k + 1)! = sinh tanh−1 x cosh (2k + 1) tanh−1 x k+1 (1 − x2 ) + cosh tanh−1 x sinh (2k + 1) tanh−1 x . Using the identity sinh(x + y) = sinh x cosh y + cosh x sinh y, we get (2k + 1)! d2k+2 tanh−1 x = sinh (2k + 2) tanh−1 x . k+1 2k+2 dx (1 − x2 ) Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, d2n (2n − 1)! −1 x tanh−1 x = n sinh 2n tanh 2n 2 dx (1 − x ) for all n ∈ N.

204

Problem 87 Topic: Derivative, Binomial Coefficient

Problem: Prove the Leibniz rule, n X dn n (j) (f (x)g(x)) = f (x)g (n−j) (x), dxn j j=0

for all n ∈ N where f (n) (x) denotes the n-th derivative of f (x).

Definition - Binomial Coefficient: For non-negative integers p and r, ( p! if r ≤ p p = r!(p−r)! . r 0 if r > p

Solution Base Cases We need to prove that it is true for n = 1. By definition of derivative, f 0 (x) = lim

h→0

f (x + h) − f (x) . h

So, d f (x + h)g(x + h) − f (x)g(x) (f (x)g(x)) = lim . h→0 dx h 205

Adding and subtracting f (x)g(x + h) in the numerator, d (f (x)g(x)) dx f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x) = lim h→0 h f (x + h)g(x + h) − f (x)g(x + h) f (x)g(x + h) − f (x)g(x) = lim + lim h→0 h→0 h h f (x + h) − f (x) g(x + h) − g(x) = lim · g(x) + f (x) · lim h→0 h→0 h h =f 0 (x)g(x) + f (x)g 0 (x). For the right-hand side, 1 X 1 j=0

j

f

(j)

(x)g

(1−j)

1 (0) 1 (1) (1) (x) = f (x)g (x) + f (x)g (0) (x) 0 1 = f 0 (x)g(x) + f (x)g 0 (x).

So, it is true that 1 X 1 (j) d f (x)g (1−j) (x). (f (x)g(x)) = j dx j=0

Induction Step Assume

m X dm m (j) (f (x)g(x)) = f (x)g (m−j) (x) m j dx j=0

(87.1)

for all 1 ≤ m ≤ k for some k ≥ 1. To prove the induction step, we need to prove k+1 X k + 1 dk+1 (f (x)g(x)) = f (j) (x)g (k+1−j) (x). dxk+1 j j=0

By assumption (87.1) when m = k, k X dk k (j) (f (x)g(x)) = f (x)g (k−j) (x). k dx j j=0

Differentiating both sides with respect to x, k dk+1 d X k (j) (f (x)g(x)) = f (x)g (k−j) (x) dxk+1 dx j=0 j

=

k X k d (j) f (x)g (k−j) (x) . j dx j=0

206

By assumption (87.1) when m = 1, 1 X d 1 (j) (f (x)g(x)) = f (x)g (1−j) (x) = f 0 (x)g(x) + f (x)g 0 (x). dx j j=0

So, d (j) f (x)g (k−j) (x) = f (j+1) (x)g (k−j) (x) + f (j) (x)g (k+1−j) (x), dx and we have k h i X dk+1 k (f (x)g(x)) = f (j+1) (x)g (k−j) (x) + f (j) (x)g (k+1−j) (x) k+1 dx j j=0

=

k X k j=0

j

f

(j+1)

(x)g

(k−j)

(x) +

k X k j=0

j

f (j) (x)g (k+1−j) (x).

Substituting j + 1 → j, k X k j=0

j

f

(j+1)

(x)g

(k−j)

(x) =

k+1 X j=1

k f (j) (x)g (k+1−j) (x). j−1

Hence, dk+1 (f (x)g(x)) dxk+1 k k+1 X X k k (j) (j) (k+1−j) = f (x)g (x) + f (x)g (k+1−j) (x) j − 1 j j=1 j=0 k X k (k+1) k (k+1−k−1) = f (x)g (x) + f (j) (x)g (k+1−j) (x) k j − 1 j=1 +

k X k j=1

j

f (j) (x)g (k+1−j) (x) +

=f (k+1) (x)g(x) +

k (0) f (x)g (k+1−0) (x) 0

k X k k + f (j) (x)g (k+1−j) (x) + f (x)g (k+1) (x). j − 1 j j=1

In problem 73, we have calculated that

k k k+1 + = . j−1 j j 207

So, dk+1 (f (x)g(x)) dxk+1 k X k + 1 (j) =f (k+1) (x)g(x) + f (x)g (k+1−j) (x) + f (x)g (k+1) (x) j j=1 k X k + 1 (k+1) k + 1 (j) (x)g (k+1−k−1) (x) + f f (x)g (k+1−j) (x) k+1 j j=1 k + 1 (0) + f (x)g (k+1−0) (x) 0 k+1 X k + 1 = f (j) (x)g (k+1−j) (x). j j=0 =

Thus, we proved that it is true for n = k +1 if it is true for n = m for all 1 ≤ m ≤ k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k. Therefore, by strong induction, n X n (j) dn (f (x)g(x)) = f (x)g (n−j) (x) j dxn j=0

for all n ∈ N.

208

Problem 88 Topic: Derivative

Problem: Prove n n n Y X fj0 (x) d Y fj (x) = fj (x) · dx j=1 f (x) j=1 j=1 j for all n ∈ N.

Solution Base Cases We need to prove that it is true for n = 1 and n = 2. When n = 1, 1 1 1 0 0 Y Y X fj (x) d f (x) . fj (x) = f10 (x) = f1 (x) · 1 = fj (x) · dx j=1 f1 (x) j=1 f (x) j=1 j When n = 2, by the product rule of derivative, 2 d Y d fj (x) = (f1 (x)f2 (x)) = f10 (x)f2 (x) + f1 (x)f20 (x) dx j=1 dx

= f1 (x)f2 (x) ·

f10 (x) f20 (x) + f1 (x) f2 (x)

So, it is true for n = 1 and n = 2. 209

=

2 Y j=1

fj (x) ·

2 X fj0 (x) j=1

fj (x)

.

Induction Step Assume m m m 0 Y Y X f (x) d j fj (x) = fj (x) · dx j=1 f (x) j=1 j=1 j

(88.1)

for all 1 ≤ m ≤ k for some k ≥ 2. To prove the induction step, we need to prove k+1 k+1 k+1 Y X fj0 (x) d Y . fj (x) · fj (x) = dx j=1 f (x) j j=1 j=1 By assumption (88.1) when m = 2, d (f1 (x)f2 (x)) = f1 (x)f2 (x) · dx Let f1 (x) →

k Y

f10 (x) f20 (x) + f1 (x) f2 (x)

.

fj (x) and f2 (x) → fk+1 (x),

j=1

−1 k k 0 Y Y f (x) d d fj (x) · fj (x) = fj (x) · fj (x) + k+1 . dx j=1 dx f k+1 (x) j=1 j=1 j=1 k+1 Y

k+1 Y

By assumption (88.1) when m = k, k k k Y X fj0 (x) d Y . fj (x) = fj (x) · dx j=1 f (x) j j=1 j=1 So, k+1 d Y fj (x) dx j=1 −1 k+1 k k k 0 0 Y Y Y X fj (x) f (x) + k+1 fj (x) · = fj (x) · fj (x) · f (x) f k+1 (x) j=1 j=1 j j=1 j=1

k 0 X fj0 (x) fk+1 (x) = + fj (x) · f (x) fk+1 (x) j=1 j=1 j k+1 k+1 Y X fj0 (x) . = fj (x) · f (x) j=1 j=1 j k+1 Y

Thus, we proved that it is true for n = k +1 if it is true for n = m for all 1 ≤ m ≤ k. 210

Conclusion We proved that it is true for n = 1 and n = 2. Then, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k. Therefore, by strong induction, n n n Y X fj0 (x) d Y fj (x) = fj (x) · dx j=1 f (x) j j=1 j=1 for all n ∈ N.

211

Problem 89 Topic: Matrices

Problem: If A is an idempotent matrix, then prove that An = A for all n ∈ N.

Definition - Idempotent Matrix: An idempotent matrix is a matrix A such that A2 = A.

Solution Base Cases We need to prove that it is true for n = 1. The base case n = 1 is trivial since it is always true that A1 = A.

Induction Step Assume that Ak = A if A is an idempotent matrix for some k ≥ 1. To prove the induction step, we need to prove that Ak+1 = A if A is an idempotent matrix. By the assumption that Ak = A, we have Ak+1 = AAk = AA = A2 . If A is an idempotent matrix, then by definition of idempotent matrix, A2 = A. So, Ak+1 = A2 = A if A is an idempotent matrix. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, if A is an idempotent matrix, then An = A for all n ∈ N. 212

Problem 90 Topic: Matrices

n Problem: Let A be an invertible matrix. Prove that the inverse of A is equal −1 n −1 −1 n to the n-th power of A , i.e. (A ) = A , for all n ∈ N.

Definition - Inverse of Matrix: For an invertible matrix A, the inverse of A is the matrix A−1 such that AA−1 = A−1 A = I, where I is the identity matrix. In this problem, we will use the property (AB)−1 = B −1 A−1 . We can check that this is true by checking that AB B −1 A−1 = B −1 A−1 AB = I. However, note that this is NOT a rigorous proof of the property.

Solution Base Cases We need to prove that it is true for n = 1. Since A1

−1

= A−1

it is true for n = 1. 213

1

= A−1 ,

Induction Step Assume Ak

−1

= A−1

k

(90.1)

for some k ≥ 1. To prove the induction step, we need to prove Ak+1

−1

= A−1

k+1

.

By using the assumption (90.1) and the property (AB)−1 = B −1 A−1 , Ak+1

−1

−1 = AAk −1 −1 = Ak A k = A−1 A−1 k+1 = A−1 .

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, n −1 (An ) = A−1 for all n ∈ N.

214

Problem 91 Topic: Matrices

Problem: Let A be a diagonalizable matrix, i.e. A = P DP −1 where D is a diagonal matrix. Then, prove that An = P Dn P −1 for all n ∈ N.

Definition - Inverse of Matrix: For an invertible matrix A, the inverse of A is the matrix A−1 such that AA−1 = A−1 A = I, where I is the identity matrix.

Solution Base Cases We need to prove that it is true for n = 1. Since A1 = A = P DP −1 = P D1 P −1 , it is true for n = 1.

Induction Step Assume Ak = P Dk P −1 for some k ≥ 1. To prove the induction step, we need to prove Ak+1 = P Dk+1 P −1 . 216

(91.1)

By assumption (91.1), Ak+1 = Ak A = P Dk P −1 = P Dk

P DP −1 P −1 P DP −1

= P Dk IDP −1 = P Dk DP −1 = P Dk+1 P −1 . Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, An = P Dn P −1 for all n ∈ N.

217

Problem 92 Topic: Set Theory

Problem: If Sn is a set with n elements, then prove that Sn has 2n subsets.

Definition - Subset: For two sets A and B, A is a subset of B, denoted as A ⊆ B, when all elements in A are in B. For example, {1, 3, 5} ⊆ {1, 2, 3, 5, 6}. Formally, A ⊆ B ⇔ [∀x, x ∈ A ⇒ x ∈ B]. Also, the empty set ∅ is a subset of any set, but this needs to be proven using the formal definition of subset. However, we will not prove this here.

Definition - Set Union: For two sets A and B, the union of A and B, denoted as A ∪ B, is the set containing all elements in A and all elements in B. For example, {1, 2, 3} ∪ {1, 3, 5} = {1, 2, 3, 5}. Formally, x ∈ A ∪ B ⇔ [x ∈ A or x ∈ B].

Solution Base Cases We need to prove that it is true for n = 1. Let S1 be a set with 1 element, S1 = {a1 }. Then, it has 2 subsets, ∅ and {a1 }. So, it is true that a set with 1 element S1 has 21 subsets. 218

Induction Step Assume that a k-element set Sk , Sk = {a1 , a2 , · · · , ak }, has 2k subsets. To prove the induction step, we need to prove that a k + 1-element set Sk+1 , Sk+1 = {a1 , a2 , · · · , ak , ak+1 }, has 2k+1 subsets. Note that Sk+1 = {a1 , a2 , · · · , ak } ∪ {ak+1 } = Sk ∪ {ak+1 }. A subset of Sk+1 either contains or does not contain ak+1 . Subsets of Sk+1 that do not contain ak+1 are subsets of Sk . By the assumption that Sk has 2k subsets, Sk+1 has 2k subsets that do not contain ak+1 . Subsets of Sk+1 that contain ak+1 are the unions of a subset of Sk with {ak+1 }. Since there are 2k subsets of Sk by assumption, there are 2k sets formed by taking the union of a subset of Sk with {ak+1 }. So, Sk+1 has 2k subsets that contain ak+1 . Hence, Sk+1 has 2k + 2k = 2k+1 subsets. Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Sn has 2n subsets if it has n elements for all n ∈ N.

219

Problem 93 Topic: Set Theory

Problem: Let S1 , S2 , · · · , Sn be sets such that Sn ⊆ Sn−1 ⊆ · · · ⊆ S2 ⊆ S1 , then prove that n \ Si = Sn i=1

for all n ∈ N.

Definition - Set Intersection: For two sets A and B, the intersection of A and B, denoted as A ∩ B, is the set containing all elements that are in both A and B. For example, {1, 2, 3} ∩ {1, 3, 5} = {1, 3}. Formally, x ∈ A ∩ B ⇔ [x ∈ A and x ∈ B]. For more than 2 sets, we have n \

Ai = A1 ∩ A2 ∩ · · · ∩ An .

i=1

Definition - Subset: For two sets A and B, A is a subset of B, denoted as A ⊆ B, when all elements in A are in B. For example, {1, 3, 5} ⊆ {1, 2, 3, 5, 6}. Formally, A ⊆ B ⇔ [∀x, x ∈ A ⇒ x ∈ B].

Definition - Equal Sets: For two sets A and B, A is equal to B, denoted as A = B, when all elements in A are in B and all elements in B are in A. For

220

example, {1, 3, 5} = {1, 1, 3, 5, 5}. Formally, A = B ⇔ [∀x, x ∈ A ⇔ x ∈ B], or equivalently A = B ⇔ [A ⊆ B and B ⊆ A].

Solution Base Cases We need to prove that it is true for n = 1 and n = 2. The case when n = 1 is trivial since it is always true that 1 \

Si = S1 .

i=1

For n = 2, we need to prove that 2 \

Si = S1 ∩ S2 = S2

i=1

if S2 ⊆ S1 . First, we need to prove S1 ∩ S2 ⊆ S2 . Let x ∈ S1 ∩ S2 . By definition of set intersection, x ∈ S1 ∩ S2 ⇒ [x ∈ S1 and x ∈ S2 ]. Since the statement “x ∈ S1 and x ∈ S2 ” is true only when “x ∈ S1 ” and “x ∈ S2 ” are both true, [x ∈ S1 and x ∈ S2 ] ⇒ x ∈ S2 . Hence, ∀x, x ∈ S1 ∩ S2 ⇒ x ∈ S2 , and S1 ∩ S2 ⊆ S2 by definition of subset. Next, we need to prove S2 ⊆ S1 ∩ S2 . Let x ∈ S2 . If S2 ⊆ S1 , then by definition of subset, x ∈ S2 ⇒ x ∈ S1 . So, x ∈ S2 ⇒ [x ∈ S1 and x ∈ S2 ]. By definition of set intersection, [x ∈ S1 and x ∈ S2 ] ⇒ x ∈ S1 ∩ S2 . 221

Hence, ∀x, x ∈ S2 ⇒ x ∈ S1 ∩ S2 , and S2 ⊆ S1 ∩ S2 by definition of subset. Thus, S1 ∩ S2 ⊆ S2 and S2 ⊆ S1 ∩ S2 . By definition of equal sets, S1 ∩ S2 = S2 if S2 ⊆ S1 .

Induction Step Assume

m \

Si = Sm

(93.1)

i=1

if Sm ⊆ Sm−1 ⊆ · · · ⊆ S2 ⊆ S1 for all 1 ≤ m ≤ k for some k ≥ 2. To prove the induction step, we need to prove k+1 \

Si = Sk+1

i=1

if Sk+1 ⊆ Sk ⊆ · · · ⊆ S2 ⊆ S1 . By the assumption (93.1) when m = k, k \

Si = Sk

i=1

if Sk ⊆ Sk−1 ⊆ · · · ⊆ S2 ⊆ S1 . So, k+1 \ i=1

Si =

k \

Si ∩ Sk+1 = Sk ∩ Sk+1 .

i=1

By the assumption (93.1) when m = 2, S1 ∩ S2 = S2 if S2 ⊆ S1 . Let S1 → Sk and S2 → Sk+1 , Sk ∩ Sk+1 = Sk+1 if Sk+1 ⊆ Sk . Hence, k+1 \

Si = Sk+1

i=1

if Sk+1 ⊆ Sk ⊆ · · · ⊆ S2 ⊆ S1 . Thus, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k.

Conclusion We proved that it is true for n = 1 and n = 2. Then, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k. Therefore, by strong induction, n \

Si = Sn

i=1

if Sn ⊆ Sn−1 ⊆ · · · ⊆ S2 ⊆ S1 for all n ∈ N. 222

Problem 94 Topic: Set Theory

Problem: Let S1 , S2 , · · · , Sn be sets such that Sn ⊆ Sn−1 ⊆ · · · ⊆ S2 ⊆ S1 , then prove that n [ Si = S1 i=1

for all n ∈ N.

Definition - Set Union: For two sets A and B, the union of A and B, denoted as A ∪ B, is the set containing all elements in A and all elements in B. For example, {1, 2, 3} ∪ {1, 3, 5} = {1, 2, 3, 5}. Formally, x ∈ A ∪ B ⇔ [x ∈ A or x ∈ B]. For more than 2 sets, we have n [

Ai = A1 ∪ A2 ∪ · · · ∪ An .

i=1

Definition - Subset: For two sets A and B, A is a subset of B, denoted as A ⊆ B, when all elements in A are in B. For example, {1, 3, 5} ⊆ {1, 2, 3, 5, 6}. Formally, A ⊆ B ⇔ [∀x, x ∈ A ⇒ x ∈ B].

Definition - Equal Sets: For two sets A and B, A is equal to B, denoted as A = B, when all elements in A are in B and all elements in B are in A. For

223

example, {1, 3, 5} = {1, 1, 3, 5, 5}. Formally, A = B ⇔ [∀x, x ∈ A ⇔ x ∈ B], or equivalently A = B ⇔ [A ⊆ B and B ⊆ A].

Solution Base Cases We need to prove that it is true for n = 1 and n = 2. The case when n = 1 is trivial since it is always true that 1 [

Si = S1 .

i=1

For n = 2, we need to prove that 2 [

Si = S1 ∪ S2 = S1

i=1

if S2 ⊆ S1 . First, we need to prove S1 ∪ S2 ⊆ S1 . Let x ∈ S1 ∪ S2 . By definition of set union, x ∈ S1 ∪ S2 ⇒ [x ∈ S1 or x ∈ S2 ]. If S2 ⊆ S1 , then by definition of subset, x ∈ S2 ⇒ x ∈ S1 . So, [x ∈ S1 or x ∈ S2 ] ⇒ [x ∈ S1 or x ∈ S1 ] ⇒ x ∈ S1 . Hence, ∀x, x ∈ S1 ∪ S2 ⇒ x ∈ S1 , and S1 ∪ S2 ⊆ S1 by definition of subset. Next, we need to prove S1 ⊆ S1 ∪ S2 . Let x ∈ S1 . Since the statement “x ∈ S1 or x ∈ S2 ” is true when either “x ∈ S1 ” or “x ∈ S2 ” is true, x ∈ S1 ⇒ [x ∈ S1 or x ∈ S2 ]. 224

By definition of set union, [x ∈ S1 or x ∈ S2 ] ⇒ x ∈ S1 ∪ S2 . Hence, ∀x, x ∈ S1 ⇒ x ∈ S1 ∪ S2 , and S1 ⊆ S1 ∪ S2 by definition of subset. Thus, S1 ∪ S2 ⊆ S1 and S1 ⊆ S1 ∪ S2 . By definition of equal sets, S1 ∪ S2 = S1 if S2 ⊆ S1 .

Induction Step Assume

m [

Si = S1

(94.1)

i=1

if Sm ⊆ Sm−1 ⊆ · · · ⊆ S2 ⊆ S1 for all 1 ≤ m ≤ k for some k ≥ 2. To prove the induction step, we need to prove k+1 [

Si = S1

i=1

if Sk+1 ⊆ Sk ⊆ · · · ⊆ S2 ⊆ S1 . By the assumption (94.1) when m = k, k [

Si = S1

i=1

if Sk ⊆ Sk−1 ⊆ · · · ⊆ S2 ⊆ S1 . So, k+1 [ i=1

Si =

k [

Si ∪ Sk+1 = S1 ∪ Sk+1 .

i=1

By the assumption (94.1) when m = 2, S1 ∪ S2 = S1 if S2 ⊆ S1 . Let S2 → Sk+1 , S1 ∪ Sk+1 = S1 if Sk+1 ⊆ S1 . Hence, k+1 [

Si = S1

i=1

if Sk+1 ⊆ Sk ⊆ · · · ⊆ S2 ⊆ S1 . Thus, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k.

Conclusion We proved that it is true for n = 1 and n = 2. Then, we proved that it is true for n = k + 1 if it is true for n = m for all 1 ≤ m ≤ k. Therefore, by strong induction, n [

Si = S1

i=1

if Sn ⊆ Sn−1 ⊆ · · · ⊆ S2 ⊆ S1 for all n ∈ N. 225

Problem 95 Topic: Integrals

Problem: Prove

∞

Z

xn e−x dx = n!.

0

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. By integration by parts, Z ∞ Z ∞ −x −x ∞ xe dx = −xe + e−x dx 0 0 0 Z ∞ = e−x dx 0 ∞ = −e−x 0 = 1 = 1!, so it is true for n = 1.

Induction Step Assume

Z

∞

xk e−x dx = k!

0

226

(95.1)

for some k ≥ 1. To prove the induction step, we need to prove Z ∞ xk+1 e−x dx = (k + 1)!. 0

By integration by parts, Z ∞ Z ∞ ∞ xk+1 e−x dx = −xk+1 e−x 0 + (k + 1) xk e−x dx 0 0 Z ∞ = (k + 1) xk e−x dx. 0

Using the assumption (95.1), Z ∞ xk+1 e−x dx = (k + 1) · k! = (k + 1)!. 0

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Z ∞ xn e−x dx = n! 0

for all n ∈ N.

227

Problem 96 Topic: Integrals, Trigonometry

Problem: Prove Z π4

tan2n+1 x dx =

0

n (−1)n ln 2 (−1)n X (−1)j + . 2 2 j=1 j

Solution Base Cases We need to prove that it is true for n = 1. By using the identity 1+tan2 x = sec2 x, Z

π 4

tan3 x dx =

Z

0

π 4

tan x sec2 x − 1 dx

0

Z =

π 4

tan x sec2 x dx −

0

Z

π 4

tan x dx 0

π4 tan2 x = + ln (cos x) 2 0 1 ln 2 = − 2 2 1 (−1)1 ln 2 (−1)1 X (−1)j = + , 2 2 j=1 j

so it is true for n = 1. 228

Induction Step Assume π 4

Z

tan2k+1 x dx =

0

k (−1)k ln 2 (−1)k X (−1)j + 2 2 j=1 j

(96.1)

for some k ≥ 1. To prove the induction step, we need to prove Z

π 4

tan2k+3 x dx =

0

k+1 (−1)k+1 ln 2 (−1)k+1 X (−1)j + . 2 2 j j=1

By using the identity 1 + tan2 x = sec2 x, Z

π 4

2k+3

tan

Z

π 4

x dx =

0

tan2k+1 x sec2 x − 1 dx

0

Z =

π 4

tan2k+1 x sec2 x dx −

0

Z

π 4

tan2k+1 x dx

0

π Z π 4 tan2k+2 x 4 − tan2k+1 x dx = 2k + 2 0 0 Z π4 1 = − tan2k+1 x dx. 2k + 2 0

By the assumption (96.1), Z

π 4

tan2k+3 x dx =

0

k 1 (−1)k ln 2 (−1)k X (−1)j − − 2k + 2 2 2 j=1 j

=

k (−1)k+1 (−1)k+1 (−1)k+1 ln 2 (−1)k+1 X (−1)j + + 2(k + 1) 2 2 j j=1

=

k+1 (−1)k+1 ln 2 (−1)k+1 X (−1)j + . 2 2 j j=1

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Z 0

π 4

tan2n+1 x dx =

n (−1)n ln 2 (−1)n X (−1)j + 2 2 j=1 j

for all n ∈ N.

229

Problem 97 Topic: Integrals, Trigonometry

Problem: Prove

π 2

Z

sin2n x dx =

0

(2n)! π · . 4n (n!)2 2

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. By using the identity sin2 x = Z 0

π 2

π 2

1 − cos (2x) dx 2 0 π x sin (2x) 2 = − 2 4 0 π 2! π = = 1 · , 4 4 (1!)2 2

sin2 x dx =

Z

so it is true for n = 1. 230

1−cos (2x) , 2

Induction Step Assume Z

π 2

sin2k x dx =

0

(2k)! π · 4k (k!)2 2

(97.1)

for some k ≥ 1. To prove the induction step, we need to prove Z

π 2

sin2k+2 x dx =

0

(2k + 2)! π · . 2 + 1)!) 2

4k+1 ((k

By using the identity sin2 x + cos2 x = 1, π 2

Z

2k+2

sin

π 2

Z x dx =

0

sin2k x 1 − cos2 x dx

0 π 2

Z =

Z

2k

x dx −

sin 0

π 2

sin2k x cos2 x dx.

0

By integration by parts, π 2

Z

# π2 Z π2 1 sin2k+1 x x cos x dx = · cos x + sin2k+2 x dx 2k + 1 2k + 1 0 0 Z π2 1 = sin2k+2 x dx. 2k + 1 0 "

2k

sin 0

2

Substituting this into (97.2), Z

π 2

2k+2

sin

Z x dx =

0

2k + 2 2k + 1

Z

2k

sin 0

π 2

2k+2

sin

Z x dx =

0

Z

π 2

π 2

1 x dx − 2k + 1

Z

π 2

sin2k+2 x dx

0

sin2k x dx

0 π 2

sin2k+2 x dx =

0

2k + 1 2k + 2

Z

π 2

sin2k x dx.

0

By the assumption (97.1), Z 0

π 2

2k + 1 (2k)! π · · 2k + 2 4k (k!)2 2 2k + 2 2k + 1 (2k)! π = · · · 2k + 2 2k + 2 4k (k!)2 2 (2k + 2)! π = k+1 · . 4 ((k + 1)!)2 2

sin2k+2 x dx =

Thus, we proved that it is true for n = k + 1 if it is true for n = k. 231

(97.2)

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Z

π 2

sin2n x dx =

0

for all n ∈ N.

232

(2n)! π · 4n (n!)2 2

Problem 98 Topic: Integrals, Trigonometry

Problem: Prove Z

π 4

2n

sin 0

n (2n)! π X 2j (j!)2 x dx = n − . 4 (n!)2 4 j=1 2j(2j)!

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. By using the identity sin2 x = Z π4 Z π4 1 − cos (2x) 2 sin x dx = dx 2 0 0 π x sin (2x) 4 = − 2 4 0 π 1 = − 8 4 1

=

2! π X 2j (j!)2 − , 41 (1!)2 4 j=1 2j(2j)! 233

1−cos (2x) , 2

so it is true for n = 1.

Induction Step Assume k j 2 X π (2k)! 2 (j!) − sin2k x dx = k 4 (k!)2 4 j=1 2j(2j)!

π 4

Z 0

(98.1)

for some k ≥ 1. To prove the induction step, we need to prove π 4

Z

sin2k+2 x dx =

0

k+1 X

j

2

(2k + 2)! 2 (j!) π − . 4k+1 ((k + 1)!)2 4 j=1 2j(2j)!

By using the identity sin2 x + cos2 x = 1, Z

π 4

sin2k+2 x dx =

0

Z

π 4

sin2k x 1 − cos2 x dx

0

Z

π 4

=

sin2k x dx −

π 4

Z

0

sin2k x cos2 x dx.

(98.2)

0

By integration by parts, Z

π 4

"

2k

sin 0

sin2k+1 x x cos x dx = · cos x 2k + 1

# π4

2

=

+ 0

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

Z π4 1 sin2k+2 x dx 2k + 1 0 Z π4 sin2k+2 x dx. 0

Substituting this into (98.2), Z

π 4

sin2k+2 x dx =

0

Z

π 4

sin2k x dx −

0

1 1 − 2k+1 (2k + 1) 2k + 1

Z

π 4

sin2k+2 x dx.

0

So, 2k + 2 2k + 1

Z

π 4

sin2k+2 x dx =

0

Z 0

Z

π 4

sin2k x dx −

0 π 4

sin2k+2 x dx =

2k + 1 2k + 2 234

Z 0

π 4

1 2k+1 (2k

sin2k x dx −

+ 1)

1 . 2k+2 (k + 1)

By the assumption (98.1), Z 0

π 4

k j 2 X 1 2k + 1 (2k)! π 2 (j!) − k+2 sin2k+2 x dx = · − 2k + 2 4k (k!)2 4 j=1 2j(2j)! 2 (k + 1) k j 2 X 1 (2k + 2)! 2 (j!) π − − k+2 = k+1 4 ((k + 1)!)2 4 j=1 2j(2j)! 2 (k + 1) k+1 j 2 X 2 (j!) (2k + 2)! π − = k+1 . 4 ((k + 1)!)2 4 j=1 2j(2j)!

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Z π4 n j 2 X (2n)! π 2 (j!) − sin2n x dx = n 4 (n!)2 4 j=1 2j(2j)! 0 for all n ∈ N.

235

Problem 99 Topic: Integrals, Trigonometry

Problem: Prove π 4

Z

sec2n x dx =

0

n−1 22n−1 (n!)2 X (2j)! . n (2n)! j=0 2j (j!)2

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. Since Z

π 4

π

sec2 x dx = [tan x]04 = 1 =

0

1−1 22·1−1 (1!)2 X (2j)! , 1 · 2! 2j (j!)2 j=0

it is true for n = 1.

Induction Step Assume Z

π 4

sec 0

2k

k−1 22k−1 (k!)2 X (2j)! x dx = k (2k)! j=0 2j (j!)2

236

(99.1)

for some k ≥ 1. To prove the induction step, we need to prove Z

π 4

sec2k+2 x dx =

0

k 22k+1 ((k + 1)!)2 X (2j)! . (k + 1)(2k + 2)! j=0 2j (j!)2

By integration by parts and using the identity 1 + tan2 x = sec2 x, π 4

Z

sec2k+2 x dx =

0

Z

π 4

sec2k x sec2 x dx

0

π = sec2k x tan x 04 − 2k π 4

Z

k

= 2 − 2k

π 4

Z

sec2k x tan2 x dx

0

sec2k x sec2 x − 1 dx

0

= 2k − 2k

π 4

Z

sec2k+2 x dx + 2k

Z

0

So, Z

π 4

sec2k+2 x dx =

0

π 4

sec2k x dx.

0

2k 2k + 2k + 1 2k + 1

Z

π 4

sec2k x dx.

0

By the assumption (99.1), Z

π 4

sec2k+2 x dx =

0

k−1 2k 2k 22k−1 (k!)2 X (2j)! + · 2k + 1 2k + 1 k (2k)! j=0 2j (j!)2

=

k−1 22k+1 ((k + 1)!)2 X (2j)! 2k + 2k + 1 (k + 1)(2k + 2)! j=0 2j (j!)2

=

k 22k+1 ((k + 1)!)2 X (2j)! . (k + 1)(2k + 2)! j=0 2j (j!)2

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Z 0

π 4

sec2n x dx =

n−1 22n−1 (n!)2 X (2j)! n (2n)! j=0 2j (j!)2

for all n ∈ N.

237

Problem 100 Topic: Integrals, Trigonometry

Problem: Prove Z

π 2

x2n cos x dx =

0

n X (−1)n (2n)! π 2j

(−1)j (2j)!

j=0

2

.

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. By integration by parts twice, Z π2 Z π2 2 π2 2 x cos x dx = x sin x 0 − 2 x sin x dx 0

0

= = =

π 2 2 π 2 2 π 2 2

Z

π 2

−2

x sin x dx 0 π

π 2

Z

+ [2x cos x]02 − 2

cos x dx 0

Z −2

cos x dx = 0

238

π 2

π 2 2

− 2.

Since 1 X (−1)1 2! π 2j π 2 = − 2, (−1)j (2j)! 2 2 j=0

it is true that Z

π 2

x2 cos x dx =

0

1 X (−1)1 2! π 2j . (−1)j (2j)! 2 j=0

Induction Step Assume Z

π 2

x2k cos x dx =

0

k X (−1)k (2k)! π 2j j=0

(−1)j (2j)!

(100.1)

2

for some k ≥ 1. To prove the induction step, we need to prove Z

π 2

x

2k+2

k+1 X

cos x dx =

0

j=0

(−1)k+1 (2k + 2)! π 2j . (−1)j (2j)! 2

By integration by parts twice, π 2

Z

x2k+2 cos x dx

0

π = x2k+2 sin x 02 − (2k + 2) = = =

π 2k+2 2 π 2k+2 2 π 2k+2 2

Z − (2k + 2)

π 2

Z

π 2

x2k+1 sin x dx

0

x2k+1 sin x dx

0

π + (2k + 2) x2k+1 cos x 02 − (2k + 2)(2k + 1) Z − (2k + 2)(2k + 1)

π 2

Z

π 2

x2k cos x dx

0

x2k cos x dx.

0

By the assumption (100.1), Z

π 2

x2k+2 cos x dx =

0

=

=

π 2k+2 2

k+1 X j=0

k X (−1)k (2k)! π 2j j=0

π 2k+2 2

− (2k + 2)(2k + 1)

+

(−1)j (2j)!

k X (−1)k+1 (2k + 2)! π 2j j=0

(−1)j (2j)!

2

(−1)k+1 (2k + 2)! π 2j . (−1)j (2j)! 2

Thus, we proved that it is true for n = k + 1 if it is true for n = k. 239

2

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Z 0

π 2

x2n cos x dx =

n X (−1)n (2n)! π 2j j=0

for all n ∈ N.

240

(−1)j (2j)!

2

Problem 101 Topic: Integrals, Trigonometry

Problem: Prove Z

π 2

x cos

2n

0

n−1 (2n)! π 2 X 4j (j!)2 x dx = n − . 4 (n!)2 8 (2j + 2)! j=0

Definition - Factorial: n! = n · (n − 1) · (n − 2) · · · 2 · 1. 0! = 1.

Solution Base Cases We need to prove that it is true for n = 1. By using the identity cos2 x = Z π2 Z π2 x + x cos (2x) dx x cos2 x dx = 2 0 0 2 π x x sin (2x) cos (2x) 2 = + + 4 4 8 0 π2 1 − 16 4 1−1 2! π 2 X 4j (j!)2 = 1 − , 4 (1!)2 8 (2j + 2)! j=0

=

241

1+cos (2x) , 2

so it is true for n = 1.

Induction Step Assume Z

π 2

x cos2k x dx =

0

k−1 X

2

j

2

(2k)! π 4 (j!) − 4k (k!)2 8 (2j + 2)! j=0

(101.1)

for some k ≥ 1. To prove the induction step, we need to prove π 2

Z 0

k j 2 X 4 (j!) (2k + 2)! π − . x cos2k+2 x dx = k+1 4 ((k + 1)!)2 8 (2j + 2)! j=0 2

By using the identity sin2 x + cos2 x = 1, π 2

Z

x cos

2k+2

Z

π 2

x dx =

0

x cos2k x 1 − sin2 x dx

0

Z =

π 2

x cos

2k

Z x dx −

0

π 2

x cos2k x sin2 x dx.

0

By integration by parts, π 2

Z −

x cos2k x sin2 x dx

0 π 2

Z =

(x sin x) − cos2k x sin x dx

0

π Z π2 cos2k+1 x 2 1 = x sin x · − (sin x + x cos x) cos2k+1 x dx 2k + 1 0 2k + 1 0 Z π2 Z π2 1 1 2k+1 =− sin x cos x dx − x cos2k+2 x dx 2k + 1 0 2k + 1 0 Z π2 1 1 =− − x cos2k+2 x dx. (2k + 2)(2k + 1) 2k + 1 0 Substituting this into (101.2) and simplifying algebraically gives Z 0

π 2

x cos2k+2 x dx =

2k + 1 2k + 2

Z 0

242

π 2

x cos2k x dx −

1 . 4(k + 1)2

(101.2)

By the assumption (101.1), Z

π 2

k−1 X

j

2

1 2k + 1 (2k)! π 4 (j!) − · − 2k + 2 4k (k!)2 8 (2j + 2)! 4(k + 1)2 j=0 k−1 2 j 2 X 1 (2k + 2)! 4 (j!) π − − = k+1 4 ((k + 1)!)2 8 (2j + 2)! 4(k + 1)2 j=0 k j 2 2 X 4 (j!) (2k + 2)! π − = k+1 . 4 ((k + 1)!)2 8 (2j + 2)! j=0

x cos2k+2 x dx =

0

2

Thus, we proved that it is true for n = k + 1 if it is true for n = k.

Conclusion We proved that it is true for n = 1. Then, we proved that it is true for n = k + 1 if it is true for n = k. Therefore, by weak induction, Z π2 n−1 2 X 4j (j!)2 (2n)! π x cos2n x dx = n − 4 (n!)2 8 (2j + 2)! 0 j=0 for all n ∈ N.

243

Acknowledgements In this section, I would like to mention some people whom I would like to thank for their help during the process of producing this book. On Instagram, I have established an edit team consisting of a few members with various math backgrounds. I would like to thank those members for their helpful feedback and suggestions for the book. The following is the list of the members of the edit team in no particular order: • Thang Pang Ern • Keyvon Rashidi • Mislav Plavac I also appreciate Andrzej Kukla, the cover designer of this book, and Vy Nguyen Tong, the illustrator of this book, for their artistic contribution to make this book more than just a boring math book with a simple cover and a long list of problems.

245

Reviews “In the ideal situation, every college student should learn some mathematics, with the depth and difficulty largely varying. As in the practice of most refined universities, mathematical education should never be standardized. Meanwhile, math induction is one of the key tools that students need to know. Based on my teaching experiences to many Chinese and international students, I found Tran’s book fairly interesting and accessible. Mathematical Induction 101 contains 101 carefully chosen exercise problems. I believe that Tran’s book will prove to be useful for college students in the broad range of arts and sciences. Hopefully many will appreciate the beauty and power of mathematical reasoning with mastery of math induction after using this book.” Dr. Chunwei Song Professor of Mathematics, Peking University

“This book provides an interesting collection of identities and inequalities that can be proved by mathematical induction. These 101 practices will be useful to learn standard techniques. Some of the topics and their solutions would also be found stimulating even for advanced students of mathematics.” Dr. Hiroaki Nakamura Professor of Mathematics, Osaka University

“This book provides almost all types of problems in high school and collegiate mathematics which can be solved by mathematical induction. To each problem, a self-contained and detailed solution is given. Difficulty ranges from quite elementary to somewhat complex. Among them there are many interesting ones. I am particularly interested in problems related to the Fibonacci sequence, floor function, and iterated functions.” Dr. Young-One Kim Professor Emeritus of Mathematics, Seoul National University

246

“As a mathematics student, this is a book I would have loved in my introductory proof class. Duc Van Khanh Tran has written up a fantastic way to both learn and practice one of the most essential tools in a mathematician’s toolkit: Induction! It’s difficult to find such a large, comprehensive array of well-written and demonstrative practice problems, which makes this all the more valuable a resource. Duc Van Khanh Tran carefully pens each solution to be as intuitive and explanatory as possible, so the risk of confusion at a solution is minimal. The vast amount of problems is complemented by the vast array of topics the book covers. Inequalities, sequences, trigonometry, power series, derivatives, and integrals are only some of the concepts found, guaranteeing that anyone interested in math will get something new out of this book. I’ve tried a few of these for fun! I really like the integral problems myself. All in all, if you’d like to learn about induction, hone your induction skills, or just want a bunch of fun induction problems, purchase away! You won’t regret it!” Said Kaili Undergraduate Student of Mathematics, University of Virginia

“Despite minimal prior proof background, I can grasp the main scopes of the book without much difficulty. Indeed, Duc Tran’s book provides clear instructions, comprehensive content, and helpful problems that ease students into learning mathematical induction.” Ky Minh Vinh Nguyen Undergraduate Student of Informatics, University of Texas at Austin

247