Appendix A — Revision Guide

A.1 Recommended revision strategy

1. Make sure you are very comfortable with the Definitions, Statements, Proofs, and Examples contained in this Revision Guide.

2. Make sure you are comfortable with the Tutorial and Homework questions.

3. Make sure you are comfortable with the 2023/24 Exam Paper questions.

4. Make sure you are comfortable with the Checklist in Section A.2.

A.2 Checklist

Overall, you should be comfortable with the following topics/taks:

Preliminaries

• Prove that \(\sqrt{p} \notin \mathbb{Q}\) for \(p\) a prime number.

Complex Numbers

• Sum, multiplication and division of complex numbers
• Computing the complex conjugate
• Computing the inverse of a complex number
• Find modulus and argument of a complex number
• Compute Cartesian, Trigonometric and Exponential form of a complex number
• Complex exponential and its properties
• Computing powers of complex numbers
• Solving degree 2 polynomial equations in \(\mathbb{C}\)
• Long division of polynomials
• Solving higher degree polynomial equations in \(\mathbb{C}\)
• Finding the roots of unity
• Finding the n-th roots of a complex number

A.3 Preliminaries

Theorem 1

The number \(\sqrt{2}\) does not belong to \(\mathbb{Q}\).

Proof

Aassume by contradiction that \[ \sqrt{2} \in \mathbb{Q}\,. \tag{A.1}\]

1. Therefore, there exists \(q \in \mathbb{Q}\) such that \[ q = \sqrt{2} \, . \tag{A.2}\]

2. Since \(q \in \mathbb{Q}\), by definition we have \[ q = \frac{m}{n} \] for some \(m,n \in \mathbb{N}\) with \(n \neq 0\).

3. Recalling (A.2), we then have \[ \frac{m}{n} = \sqrt{2} \,. \]

4. We can square the above equation to get \[ \frac{m^2}{n^2} = 2 \,. \tag{A.3}\]

5. Withouth loss of generality, we can assume that \(m\) and \(n\) have no common factors.

6. Equation (A.3) implies \[ m^2 = 2 n^2 \,. \tag{A.4}\] Therefore the integer \(m^2\) is an even number.

7. Since \(m^2\) is an even number, it follows that also \(m\) is an even number. Then there exists \(p \in \mathbb{N}\) such that \[ m = 2p \,. \tag{A.5}\]

8. If we substitute (A.5) in (A.4) we get \[ m^2 = 2n^2 \, \implies \, (2p)^2 = 2n^2 \, \implies \, 4 p^2 = 2 n^2 \] Dividing both terms by \(2\), we obtain \[ n^2 = 2p^2\,. \tag{A.6}\]

9. We now make a series of observations:

   • Equation (A.6) says that \(n^2\) is even.
   • The same argument in Step 7 guarantees that also \(n\) is even.
   • We have already seen that \(m\) is even.
   • Therefore \(n\) and \(m\) are both even.
     
   • Hence \(n\) and \(m\) have \(2\) as common factor.
   • But Step 5 says that \(n\) and \(m\) have no common factors.
   • Contradiction

10. Our reasoning has run into a contradiction, stemming from assumption (A.1). Therefore (A.1) is FALSE, and so \[ \sqrt{2} \notin \mathbb{Q}\, \] ending the proof.

A.3.1 Set Theory

Definition 2

For two sets \(A\) and \(B\) we define their union as the set \[ A \cup B := \{ x\, \colon \, x \in A \, \text{ or } \, x \in B \} \,. \] The intersection of \(A\) and \(B\) is defined by \[ A \cap B := \{ x\, \colon \, x \in A \, \text{ and } \, x \in B \} \,. \] We denote the empty set by the symbol \(\emptyset\). Two sets are disjoint if \[ A \cap B = \emptyset \,. \] Given two sets \(A\) and \(B\), we say that \(A\) is contained in \(B\), in symbols \[ A \subseteq B \,, \] if all the elements of \(A\) are also contained in \(B\). Two sets \(A\) and \(B\) are equal, in symbols \[ A=B \,, \] if they contain the same elements.

Proposition 3

Let \(A\) and \(B\) be sets. Then \[ A=B \quad \iff \quad A \subseteq B \,\, \text{ and } \,\, B \subseteq A \,. \]

Definition 4

Let \(\Omega\) be a set, and \(A_n \subseteq \Omega\) a family of subsets, where \(n \in \mathbb{N}\).

1. The infinte union of the \(A_n\) is the set \[ \bigcup_{n \in \mathbb{N}} A_n := \{ x \in \Omega \, \colon \, x \in A_n \,\, \text{ for at least one } \,\, n \in \mathbb{N}\} \,. \]

2. The infinte intersection of the \(A_n\) is the set \[ \bigcap_{n \in \mathbb{N}} A_n := \{ x \in \Omega \, \colon \, x \in A_n \,\, \text{ for all } \,\, n \in \mathbb{N}\} \,. \]

Example 5

Question. Define \(\Omega:=\mathbb{N}\) and a family \(A_n\) by \[ A_n = \{ n, n+1, n+2, n+3, \ldots \} \,, \quad n \in \mathbb{N}\,. \]

1. Prove that \[ \bigcup_{n \in \mathbb{N}} A_n = \mathbb{N}\,. \tag{A.7}\]

2. Prove that \[ \bigcap_{n \in \mathbb{N}} A_n = \emptyset \,. \tag{A.8}\]

Solution.

1. Assume that \(m \in \cup_n A_n\). Then \(m \in A_n\) for at least one \(n \in \mathbb{N}\). Since \(A_n \subseteq \mathbb{N}\), we conclude that \(m \in \mathbb{N}\). This shows \[ \bigcup_{n \in \mathbb{N}} A_n \subseteq \mathbb{N}\,. \] Conversely, suppose that \(m \in \mathbb{N}\). By definition \(m \in A_m\). Hence there exists at least one index \(n\), \(n=m\) in this case, such that \(m \in A_n\). Then by definition \(m \in \cup_{n \in \mathbb{N}} A_n\), showing that \[ \mathbb{N}\subseteq \bigcup_{n \in \mathbb{N}} A_n \,. \] This proves (A.7).

2. Suppose that (A.8) is false, i.e., \[ \bigcap_{n \in \mathbb{N}} A_n \neq \emptyset \,. \] This means there exists some \(m \in \mathbb{N}\) such that \(m \in \cap_{n \in \mathbb{N}} A_n\). Hence, by definition, \(m \in A_n\) for all \(n \in \mathbb{N}\). However \(m \notin A_{m+1}\), yielding a contradiction. Thus (A.8) holds.

Definition 6

Let \(A,B \subseteq \Omega\). The complement of \(A\) with respect to \(B\) is the set of elements of \(B\) which do not belong to \(A\), that is \[ B \smallsetminus A := \{ x \in \Omega \, \colon \, x \in B \, \text{ and } \, x \notin A \} \,. \] In particular, the complement of \(A\) with respect to \(\Omega\) is denoted by \[ A^c := \Omega \smallsetminus A := \{ x \in \Omega \, \colon \, x \notin A \} \,. \]

Example 7

Question. Suppose \(A, B \subseteq \Omega\). Prove that \[ A \subseteq B \iff B^c \subseteq A^c \,. \]

Solution. Let us prove the above claim:

• First implication \(\implies\):
  Suppose that \(A \subseteq B\). We need to show that \(B^c \subseteq A^c\). Hence, assume \(x \in B^c\). By definition this means that \(x \notin B\). Now notice that we cannot have that \(x \in A\). Indeed, assume \(x \in A\). By assumption we have \(A \subseteq B\), hence \(x \in B\). But we had assumed \(x \in B\), contradiction. Therefore it must be that \(x \notin A\). Thus \(B^c \subseteq A^c\).

• Second implication \(\impliedby\): Note that, for any set, \[ (A^c)^c = A \,. \] Hence, by the first implication, \[ B^c \subseteq A^c \, \implies \, (A^c)^c \subseteq (B^c)^c \, \implies \, A \subseteq B \,. \]

Proposition 8: De Morgan’s Laws

Suppose \(A,B \subseteq \Omega\). Then \[ (A \cap B)^c = A^c \cup B^c \,, \qquad (A \cup B)^c = A^c \cap B^c \,. \]

Definition 9

Let \(\Omega\) be a set. The power set of \(\Omega\) is \[ \mathcal{P}(\Omega) := \{ A \, \colon \,A \subseteq \Omega \}\,. \]

Example 10

Question. Compute the power set of \[ \Omega = \{ x, y, z \}\,. \]

Solution. \(\mathcal{P}(\Omega)\) has \(2^3 = 8\), and \[\begin{align} \mathcal{P}(\Omega) = \{ \emptyset, & \, \{x\} , \, \{y\} , \, \{z\} , \{x,y\} \\ & \{x,z\}, \, \{y,z\} , \, \{x,y,z\} \} \,. \end{align}\]

Definition 11

Let \(A,B\) be sets. The product of \(A\) and \(B\) is the set of pairs \[ A \times B := \{ (a,b) \, \colon \, a \in A, \, b \in B \} \,. \]

A.3.2 Relations

Definition 12

Suppose \(A\) is a set. A binary relation \(R\) on \(A\) is a subset \[ R \subseteq A \times A \,. \]

Definition 13: Equivalence relation

A binary relation \(R\) is called an equivalence relation if it satisfies the following properties:

1. Reflexive: For each \(x \in A\) one has \[ (x,x) \in R \,, \]

2. Symmetric: We have \[ (x,y) \in R \implies (y,x) \in R \]

3. Transitive: We have \[ (x,y) \in R \,, \,\, (y,z) \in R \implies (x,z) \in R \]

If \((x,y) \in R\) we write \[ x \sim y \] and we say that \(x\) and \(y\) are equivalent.

Definition 14: Equivalence classes

Suppose \(R\) is an equivalence relation on \(A\). The equivalence class of an element \(x \in A\) is the set \[ [x]:=\{ y \in A \, \colon \, y \sim x \} \,. \] The set of equivalence classes of elements of \(A\) with respect to the equivalence relation \(R\) is denoted by \[ A / R \, := A / \!\sim \, := \,\{ [x] \, \colon \, x \in A \} \,. \]

Proposition 15

Let \(\sim\) be an equivalence relation on \(A\). Then

1. For each \(x \in A\) we have \[ [x] \neq \emptyset \]

2. For all \(x,y \in A\) it holds \[ x \sim y \quad \iff \quad [x] = [y] \,. \]

Example 16: Equality is an equivalence relation

Question. The equality defines a binary relation on \(\mathbb{Q}\times \mathbb{Q}\), via \[ R:=\{ (x,y) \in \mathbb{Q}\times \mathbb{Q}\, \colon \, x=y \} \,. \]

1. Prove that \(R\) is an equivalence relation.
2. Prove that \([x] = \{x\}\) and compute \(\mathbb{Q}/ R\).

Solution.

1. We need to check that \(R\) satisfies the 3 properties of an equivalence relation:

   • Reflexive: It holds, since \(x=x\) for all \(x \in \mathbb{Q}\),

   • Symmetric: Again \(x = y\) if and only if \(y = x\),

   • Transitive: If \(x = y\) and \(y = z\) then \(x = z\).

   Therefore, \(R\) is an equivalence relation.

2. The class of equivalence of \(x \in \mathbb{Q}\) is given by \[ [x] = \{ x \}\,, \] that is, this relation is quite trivial, given that each element of \(\mathbb{Q}\) can only be related to itself. The quotient space is then \[ \mathbb{Q}/ R = \{ [x] \, \colon \,x \in \mathbb{Q}\} = \{ \{ x\} \, \colon \,x \in \mathbb{Q}\} \,. \]

Example 17

Question. Let \(R\) be the binary relation on the set \(\mathbb{Q}\) of rational numbers defined by \[ x \sim y \iff x-y \in \mathbb{Z}\,. \]

1. Prove that \(R\) is an equivalence relation on \(\mathbb{Q}\).
2. Compute \([x]\) for each \(x \in \mathbb{Q}\).
3. Compute \(\mathbb{Q}/R\).

Solution.

1. We have:

   • Reflexive: Let \(x \in \mathbb{Q}\). Then \(x-x = 0\) and \(0 \in \mathbb{Z}\). Thus \(x \sim x\).

   • Symmetric: If \(x \sim y\) then \(x-y \in \mathbb{Z}\). But then also
     \[ -(x-y) = y - x \in \mathbb{Z} \] and so \(y \sim x\).

   • Transitive: Suppose \(x \sim y\) and \(y \sim z\). Then \[ x - y \in \mathbb{Z}\, \text{ and } \, y - z \in \mathbb{Z}\,. \] Thus, we have \[ x-z = (x-y) + (y-z) \in \mathbb{Z} \] showing that \(x \sim z\).

   Thus, we have shown that \(R\) is an equivalence relation on \(\mathbb{Q}\).

2. Note that \[ y -x \in \mathbb{Z} \] is equivalent to \[ \exists \, n \in \mathbb{Z}\, \text{ s.t. } \, y - x = n \] which again is equivalent to \[ \exists \, n \in \mathbb{Z}\, \text{ s.t. } \, y = x + n \,. \] In summary we have \[ x \sim y \quad \iff \quad \exists \, n \in \mathbb{Z}\, \text{ s.t. } \, y = x + n \,. \] Therefore the equivalence classes with respect to \(\sim\) are \[ [x] = \{ x + n \ \colon \ n \in \mathbb{Z}\} \,. \] Each equivalence class has exactly one element in \([0,1) \cap \mathbb{Q}\), meaning that: \[ \forall x \in \mathbb{Q}\,, \,\, \exists! \, q \in \mathbb{Q}\, \text{ s.t. } \, 0 \leq q < 1 \, \text{ and } \, q \in [x]\,. \tag{A.9}\] Indeed: take \(x \in \mathbb{Q}\) arbitrary. Then \(x \in [n,n+1)\) for some \(n \in \mathbb{Z}\). Setting \(q:=x-n\) we obtain that \[ x = q + n \,, \qquad q \in [0,1) \,, \] proving (A.9). In particular (A.9) implies that for each \(x \in \mathbb{Q}\) there exists \(q \in [0,1) \cap \mathbb{Q}\) such that \[ [x] = [q] \,. \]

3. From Point 2 we conclude that \[ \mathbb{Q}/ R = \{ [x] \, \colon \,x \in \mathbb{Q}\} = \{ q \in \mathbb{Q}\, \colon \,0 \leq q <1 \} \,. \]

Definition 18: Partial order

A binary relation \(R\) on \(A\) is called a partial order if it satisfies the following properties:

1. Reflexive: For each \(x \in A\) one has \[ (x,x) \in R \,, \]

2. Antisymmetric: We have \[ (x,y) \in R \, \text{ and } \, (y,x) \in R \implies x = y \]

3. Transitive: We have \[ (x,y) \in R \,, \,\, (y,z) \in R \implies (x,z) \in R \]

Definition 19: Total order

A binary relation \(R\) on \(A\) is called a total order relation if it satisfies the following properties:

1. Partial order: \(R\) is a partial order on \(A\).
2. Total: For each \(x,y \in A\) we have \[ (x, y) \in R \,\, \mbox{ or } \,\, (y,x) \in R \,. \]

Example 20: Set inclusion is a partial order but not total order

Question. Let \(\Omega\) be a non-empty set and consider its power set \[ \mathcal{P}(\Omega) = \{ A \, \colon \,A \subseteq \Omega \}\,. \] The inclusion defines binary relation on \(\mathcal{P}(\Omega) \times \mathcal{P}(\Omega)\), via \[ R:=\{ (A,B) \in \mathcal{P}(\Omega) \times \mathcal{P}(\Omega) \, \colon \, A \subseteq B \} \,. \]

1. Prove that \(R\) is an order relation.
2. Prove that \(R\) is not a total order.

Solution.

1. Check that \(R\) is a partial order relation on \(\mathcal{P}(\Omega)\):
   • Reflexive: It holds, since \(A \subseteq A\) for all \(A \in \mathcal{P}(\Omega)\).
   • Antisymmetric: If \(A \subseteq B\) and \(B \subseteq A\), then \(A=B\).
   • Transitive: If \(A \subseteq B\) and \(B \subseteq C\), then, by definition of inclusion, \(A \subseteq C\).
2. In general, \(R\) is not a total order. For example consider \[ \Omega = \{x, y\}\,. \] Thus \[ \mathcal{P}(\Omega) = \{ \emptyset , \, \{x\} , \, \{y\} , \, \{x,y\} \}\,. \] If we pick \(A=\{x\}\) and \(B=\{y\}\) then \(A \cap B = \emptyset\), meaning that \[ A \not\subseteq B \,, \quad B \not\subseteq A \,. \] This shows \(R\) is not a total order.

Example 21: Inequality is a total order

Question. Consider the binary relation \[ R:=\{ (x,y) \in \mathbb{Q}\times \mathbb{Q}\, \colon \, x \leq y \} \,. \] Prove that \(R\) is a total order relation.

Solution. We need to check that:

1. Reflexive: It holds, since \(x \leq x\) for all \(x \in \mathbb{Q}\),

2. Antisymmetric: If \(x \leq y\) and \(y \leq x\) then \(x = y\).

3. Transitive: If \(x \leq y\) and \(y \leq z\) then \(x \leq z\).

Finally, we halso have that \(R\) is a total order on \(\mathbb{Q}\), since for all \(x,y \in \mathbb{Q}\) we have \[ x \leq y \,\, \text{ or } \,\, y \leq x \,. \]

A.3.3 Absolute value

Definition 22: Absolute value

For \(x \in \mathbb{R}\) we define its absolute value as the quantity \[ |x| = \begin{cases} x & \text{ if } \, x \geq 0 \\ -x & \text{ if } \, x < 0 \end{cases} \]

Proposition 23

For all \(x \in \mathbb{R}\) they hold:

1. \(|x| \geq 0\).
2. \(|x| = 0\) if and only if \(x = 0\).
3. \(|x| = |-x|\).

Lemma 24

Let \(x,y \in \mathbb{R}\). Then \[ |x| \leq y \iff -y \leq x \leq y \,. \]

Corollary 25

Let \(x,y \in \mathbb{R}\). Then \[ |x| < y \iff -y < x < y \,. \]

Theorem 26: Triangle inequality

For every \(x, y \in \mathbb{R}\) we have \[ | |x| - |y| | \leq |x+y| \leq |x| + |y| \,. \tag{A.10}\]

Proposition 27

For any \(x,y \in \mathbb{R}\) it holds \[ ||x|-|y|| \leq |x-y| \leq |x|+|y|\,. \tag{A.11}\] Moreover for any \(x,y,z \in \mathbb{R}\) it holds \[ |x-y| \leq |x-z| + |z-y| \,. \]

A.3.4 Induction

Definition 28: Principle of Inducion

Let \(\alpha(n)\) be a statement which depends on \(n \in \mathbb{N}\). Suppose that

1. \(\alpha(1)\) is true, and
2. Whenever \(\alpha(n)\) is true, then \(\alpha(n+1)\) is true.

Then \(\alpha(n)\) is true for all \(n \in \mathbb{N}\).

Example 29: Formula for summing first \(n\) natural numbers

Question. Prove by induction that the following formula holds for all \(n \in \mathbb{N}\): \[ 1 + 2 + 3 + \ldots + (n-1) + n = \frac{n(n+1)}{2} \,. \tag{A.12}\]

Solution. Define \[ S(n) = 1 + 2 + \ldots + n \,. \] This way the formula at (A.12) is equivalent to \[ S(n) = \frac{n(n+1)}{2} \,, \quad \forall \, n \in \mathbb{N}\,. \]

1. It is immediate to check that (A.12) holds for \(n=1\).
2. Suppose (A.12) holds for \(n = k\). Then \[\begin{align} S(k+1) & = 1 + \ldots + k + (k+1) \\ & = S(k) + (k+1) \\ & = \frac{k(k+1)}{2} + (k+1) \\ & = \frac{ k(k+1) + 2(k+1) }{2} \\ & = \frac{(k+1)(k+2)}{2} \end{align}\] where in the first equality we used that (A.12) holds for \(n=k\). We have proven that \[ S(k+1) = \frac{(k+1)(k+2)}{2}\,. \] The RHS in the above expression is exactly the RHS of (A.12) computed at \(n = k+1\). Therefore, we have shown that formula (A.12) holds for \(n = k+1\).

By the Principle of Induction, we conclude that (A.12) holds for all \(n \in \mathbb{N}\).

Example 30: Bernoulli’s inequality

Question. Let \(x \in \mathbb{R}\) with \(x>-1\). Bernoulli’s inequality states that \[ (1+x)^{n} \geq 1+n x \,, \quad \forall \, n \in \mathbb{N}\,. \tag{A.13}\] Prove Bernoulli’s inequality by induction.

Solution. Let \(x \in \mathbb{R}, x>-1\). We prove the statement by induction:

• Base case: (A.33) holds with equality when \(n=1\).

• Induction hypothesis: Let \(k \in \mathbb{N}\) and suppose that (A.33) holds for \(n=k\), i.e., \[ (1+x)^{k} \geq 1+k x \,. \] Then \[\begin{align*} (1+x)^{k+1} & = (1+x)^{k}(1+x) \\ & \geq(1+k x)(1+x) \\ & =1+k x+x+k x^{2} \\ & \geq 1+(k+1) x \,, \end{align*}\] where we used that \(kx^2 \geq 0\). Then (A.33) holds for \(n=k+1\).

By induction we conclude (A.33).

A.4 Real Numbers

A.4.1 Fields

Definition 31: Binary operation

A binary operation on a set \(K\) is a function \[ \circ \ \colon K \times K \to K \] which maps the ordered pair \((x,y)\) into \(x \circ y\).

Definition 32

Let \(K\) be a set and \(\circ \ \colon K \times K \to K\) be a binary operation on \(K\). We say that:

1. \(\circ\) is commutative if \[ x \circ y = y \circ x \,, \quad \, \forall \, x,y \in K \]
2. \(\circ\) is associative if \[ (x \circ y) \circ z = x \circ (y \circ z) \,, \quad \, \forall \, x,y,z \in K \]
3. An element \(e \in K\) is called neutral element of \(\circ\) if \[ x \circ e = e \circ x = x \,, \quad \, \forall \, x \in K \]
4. Let \(e\) be a neutral element of \(\circ\) and let \(x \in K\). An element \(y \in K\) is called an inverse of \(x\) with respect to \(\circ\) if \[ x \circ y = y \circ x = e \,. \]

Example 33

Question. Let \(K=\{0,1\}\) be a set with binary operation \(\circ\) defined by the table \[ \begin{array}{c|cc} \circ & 0 & 1 \\ \hline 0 & 1 & 1 \\ 1 & 0 & 0 \\ \end{array} \]

1. Is \(\circ\) commutative? Justify your answer.

2. Is \(\circ\) associative? Justify your answer.

Solution.

1. We have \[ 0 \circ 1 = 1 \, , \quad 1 \circ 0 = 0 \] and therefore \[ 0 \circ 1 \neq 1 \circ 0 \,. \] showing that \(\circ\) is not commutative.

2. We have \[ (0 \circ 1) \circ 1 = 1 \circ 1 = 0 \,, \] while \[ 0 \circ (1 \circ 1) = 0 \circ 0 = 1 \,, \] so that \[ (0 \circ 1) \circ 1 \neq 0 \circ (1 \circ 1)\,. \] Thus, \(\circ\) is not associative.

Definition 34: Field

Let \(K\) be a set with binary operations of addition \[ +\ \colon K \times K \to K \,, \quad (x,y) \mapsto x + y \] and multiplication \[ \cdot\ \colon K \times K \to K \,, \quad (x,y) \mapsto x \cdot y = xy \,. \] We call the triple \((K, + , \cdot)\) a field if:

1. The addition \(+\) satisfies: \(\,\forall \, x,y,z \in K\)
   • (A1) Commutativity and Associativity: \[ x+y = y+x \] \[ (x+y)+z = x+(y+z) \]
   • (A2) Additive Identity: There exists a neutral element in \(K\) for \(+\), which we call \(0\). It holds: \[ x + 0 = 0 + x = x \]
   • (A3) Additive Inverse: There exists an inverse of \(x\) with respect to \(+\). We call this element the additive inverse of \(x\) and denote it by \(-x\). It holds \[ x + (-x) = (-x) + x = 0 \]
2. The multiplication \(\cdot\) satisifes: \(\,\forall \, x,y,z \in K\)
   • (M1) Commutativity and Associativity: \[ x \cdot y = y \cdot x \] \[ (x \cdot y) \cdot z = x \cdot (y \cdot z) \]
   • (M2) Multiplicative Identity: There exists a neutral element in \(K\) for \(\cdot\), which we call \(1\). It holds: \[ x \cdot 1 = 1 \cdot x = x \]
   • (M3) Multiplicative Inverse: If \(x \neq 0\) there exists an inverse of \(x\) with respect to \(\cdot\). We call this element the multiplicative inverse of \(x\) and denote it by \(x^{-1}\). It holds \[ x \cdot x^{-1} = x^{-1} \cdot x = 1 \]
3. The operations \(+\) and \(\cdot\) are related by
   • (AM) Distributive Property: \(\,\forall \, x,y,z \in K\) \[ x \cdot (y + z) = (x \cdot y) + (y \cdot z) \,. \]

Theorem 35

Consider the sets \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\) with the usual operations \(+\) and \(\cdot\). We have:

• \((\mathbb{N}, + , \cdot)\) is not a field.

• \((\mathbb{Z}, + , \cdot)\) is not a field.

• \((\mathbb{Q}, + , \cdot)\) is a field.

Theorem 36

Let \(K\) with \(+\) and \(\cdot\) defined by \[ \begin{array}{c|cc} + & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \qquad \qquad \begin{array}{c|cc} \cdot & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array} \] Then \((K,+,\cdot)\) is a field.

Proposition 37: Uniqueness of neutral elements and inverses

Let \((K,+,\cdot)\) be a field. Then

1. There is a unique element in \(K\) with the property of \(0\).
2. There is a unique element in \(K\) with the property of \(1\).
3. For all \(x \in K\) there is a unique additive inverse \(-x\).
4. For all \(x \in K\), \(x \neq 0\), there is a unique multiplicative inverse \(x^{-1}\).

Proof

1. Suppose that \(0 \in K\) and \(\widetilde{0} \in K\) are both neutral element of \(+\), that is, they both satisfy (A2). Then \[ 0 + \widetilde{0} = 0 \] since \(\widetilde{0}\) is a neutral element for \(+\). Moreover \[ \widetilde{0} + 0 = \widetilde{0} \] since \(0\) is a neutral element for \(+\). By commutativity of \(+\), see property (A1), we have \[ 0 = 0 + \widetilde{0} = \widetilde{0} + 0 = \widetilde{0} \,, \] showing that \(0 = \widetilde{0}\). Hence the neutral element for \(+\) is unique.
2. Exercise.
3. Let \(x \in K\) and suppose that \(y, \widetilde{y} \in K\) are both additive inverses of \(x\), that is, they both satisfy (A3). Therefore \[ x + y = 0 \] since \(y\) is an additive inverse of \(x\) and \[ x + \widetilde{y} = 0 \] since \(\widetilde{y}\) is an additive inverse of \(x\). Therefore we can use commutativity and associativity and of \(+\), see property (A1), and the fact that \(0\) is the neutral element of \(+\), to infer \[\begin{align*} y & = y + 0 = y + (x + \widetilde{y}) \\ & = (y + x) + \widetilde{y} = (x + y) + \widetilde{y} \\ & = 0 + \widetilde{y} = \widetilde{y} \,, \end{align*}\] concluding that \(y = \widetilde{y}\). Thus there is a unique additive inverse of \(x\), and \[ y = \widetilde{y} = -x \,, \] with \(-x\) the element from property (A3).
4. Exercise.

Definition 38

Let \(K\) be a set with binary operations \(+\) and \(\cdot\), and with an order relation \(\leq\). We call \((K,+,\cdot,\leq)\) an ordered field if:

1. \((K,+,\cdot)\) is a field

2. There \(\leq\) is of total order on \(K\): \(\, \forall \, x, y, z \in K\)

   • (O1) Reflexivity: \[ x \leq x \]
   • (O2) Antisymmetry: \[ x \leq y \, \mbox{ and } \, y \leq x \,\, \implies \,\, x = y \]
   • (O3) Transitivity: \[ x \leq y \,\, \mbox{ and } \,\, y \leq z \,\, \implies \,\, x = z \]
   • (O4) Total order:
     \[ x \leq y \,\, \mbox{ or } \,\, y \leq x \]

3. The operations \(+\) and \(\cdot\), and the total order \(\leq\), are related by the following properties: \(\, \forall x, y, z \in K\)

   • (AM) Distributive: Relates addition and multiplication via \[ x \cdot (y + z) = x \cdot y + x \cdot z \]
   • (AO) Relates addition and order with the requirement: \[ x \leq y \,\, \implies \,\, x + z \leq y + z \]
   • (MO) Relates multiplication and order with the requirement: \[ x \geq 0, \, y \geq 0 \,\, \implies \,\, x \cdot y \geq 0 \]

Theorem 39

\((\mathbb{Q},+,\cdot,\leq)\) is an ordered field.

A.4.2 Supremum and infimum

Definition 40: Upper bound, bounded above, supremum, maximum

Let \(A \subseteq K\):

1. We say that \(b \in K\) is an upper bound for \(A\) if \[ a \leq b \,, \quad \forall \, a \in A \,. \]

2. We say that \(A\) is bounded above if there exists and upper bound \(b \in K\) for \(A\).

3. We say that \(s \in K\) is the least upper bound or supremum of \(A\) if:

   • \(s\) is an upper bound for \(A\),
   • \(s\) is the smallest upper bound of \(A\), that is, \[ \mbox{If } \, b \in K \, \mbox{ is upper bound for } \, A \, \mbox{ then } \, s \leq b \,. \] If it exists, the supremum is denoted by \[ s = \sup \ A \,. \]

4. Let \(A \subseteq K\). We say that \(M \in K\) is the maximum of \(A\) if: \[ M \in A \,\, \mbox{ and } \,\, a \leq M \,, \, \forall a \in A \,. \] If it exists, we denote the maximum by \[ M = \max A \,. \]

Remark 41

Note that if a set \(A \subseteq K\) in NOT bounded above, then the supremum does not exist, as there are no upper bounds of \(A\).

Proposition 42: Relationship between Max and Sup

Let \(A \subseteq K\). If the maximum of \(A\) exists, then also the supremum exists, and \[ \sup A = \max A \,. \]

Definition 43: Upper bound, bounded below, infimum, minimum

Let \(A \subseteq K\):

1. We say that \(l \in K\) is a lower bound for \(A\) if \[ l \leq a \,, \quad \forall \, a \in A \,. \]

2. We say that \(A\) is bounded below if there exists a lower bound \(l \in K\) for \(A\).

3. We say that \(i \in K\) is the greatest lower bound or infimum of \(A\) if:

   • \(i\) is a lower bound for \(A\),
   • \(i\) is the largest lower bound of \(A\), that is, \[ \mbox{If } \, l \in K \, \mbox{ is a lower bound for } \, A \, \mbox{ then } \, l \leq i \,. \] If it exists, the infimum is denoted by \[ i = \inf A \,. \]

4. We say that \(m \in K\) is the minimum of \(A\) if: \[ m \in A \,\, \mbox{ and } \,\, m \leq a \,, \, \forall a \in A \,. \] If it exists, we denote the minimum by \[ m = \min A \,. \]

Proposition 44

Let \(A \subseteq K\). If the minimum of \(A\) exists, then also the infimum exists, and \[ \inf A = \min A \,. \]

Proposition 45

Let \(A \subseteq K\). If \(\inf A\) and \(\sup A\) exist, then \[ \inf A \leq a \leq \sup A \,, \quad \forall \, a \in A \,. \]

Proposition 46: Relationship between sup and inf

Let \(A \subseteq K\). Define \[ - A := \{ - a \, \colon \,a \in A \} \,. \] They hold

1. If \(\sup A\) exists, then \(\inf A\) exists and \[ \inf(-A) = - \sup A \,. \]
2. If \(\inf A\) exists, then \(\sup A\) exists and \[ \sup(-A) = - \inf A \,. \]

A.4.3 Axioms of Real Numbers

Definition 47: Completeness

Let \((K,+,\cdot,\leq)\) be an ordered field. We say that \(K\) is complete if it holds the property:

• (AC) For every \(A \subseteq K\) non-empty and bounded above \[ \sup A \in K \,. \]

Theorem 48

\(\mathbb{Q}\) is not complete. In particular, there exists a set \(A \subseteq \mathbb{Q}\) such that

• \(A\) is non-empty,
• \(A\) is bounded above,
• \(\sup A\) does not exist in \(\mathbb{Q}\).

One of such sets is, for example, \[ A = \{ q \in \mathbb{Q}\, \colon \,q \geq 0 \,, \,\, q^2 < 2 \} \,. \]

Proposition 49

Let \((K,+,\cdot,\leq)\) be a complete ordered field. Suppose that \(A \subseteq K\) is non-empty and bounded below. Then \[ \inf A \in K\,. \]

Definition 50: System of Real Numbers \(\mathbb{R}\)

A system of Real Numbers is a set \(\mathbb{R}\) with two operations \(+\) and \(\cdot\), and a total order relation \(\leq\), such that

• \((\mathbb{R},+,\cdot, \leq)\) is an ordered field

• \(\mathbb{R}\) sastisfies the Axiom of Completeness

A.4.4 Inductive sets

Definition 51: Inductive set

Let \(S \subseteq \mathbb{R}\). We say that \(S\) is an inductive set if they are satisfied:

• \(1 \in S\),
• If \(x \in S\), then \((x + 1) \in S\).

Example 52

Question. Prove the following:

1. \(\mathbb{R}\) is an inductive set.

2. The set \(A=\{0,1\}\) is not an inductive set.

Solution.

1. We have that \(1 \in \mathbb{R}\) by axiom (M2). Moreover \((x + 1) \in \mathbb{R}\) for every \(x \in \mathbb{R}\), by definition of sum \(+\).

2. We have \(1 \in A\), but \((1 + 1) \notin A\), since \(1 + 1 \neq 0\).

Proposition 53

Let \(\mathcal{M}\) be a collection of inductive subsets of \(\mathbb{R}\). Then \[ S := \bigcap_{M \in \mathcal{M}} \, M \] is an inductive subset of \(\mathbb{R}\).

Definition 54: Set of Natural Numbers

Let \(\mathcal{M}\) be the collection of all inductive subsets of \(\mathbb{R}\). We define the set of natural numbers in \(\mathbb{R}\) as \[ \mathbb{N}:= \bigcap_{M \in \mathcal{M}} \, M \,. \]

Proposition 55: \({\mathbb{N}}_{\mathbb{R}}\) is the smallest inductive subset of \(\mathbb{R}\)

Let \(C \subseteq \mathbb{R}\) be an inductive subset. Then \[ \mathbb{N}\subseteq C \,. \] In other words, \(\mathbb{N}\) is the smallest inductive set in \(\mathbb{R}\).

Theorem 56

Let \(x \in \mathbb{N}\). Then \[ x \geq 1 \,. \]

A.5 Properties of \(\mathbb{R}\)

Theorem 57: Archimedean Property

Let \(x \in \mathbb{R}\) be given. Then:

1. There exists \(n \in \mathbb{N}\) such that \[ n > x \,. \]

2. Suppose in addition that \(x>0\). There exists \(n \in \mathbb{N}\) such that \[ \frac1n < x \,. \]

Theorem 58: Archimedean Property (Alternative formulation)

Let \(x,y \in \mathbb{R}\), with \(0<x<y\). There exists \(n \in \mathbb{N}\) such that \[ nx > y \,. \]

Theorem 59: Nested Interval Property

For each \(n \in \mathbb{N}\) assume given a closed interval \[ I_n := [a_n,b_n] = \{ x \in \mathbb{R}\, \colon \,a_n \leq x \leq b_n \} \,. \] Suppose that the intervals are nested, that is, \[ I_n \supset I_{n+1}\,, \quad \forall \, n \in \mathbb{N}\,. \] Then \[ \bigcap_{n=1}^\infty I_n \neq \emptyset \,. \tag{A.14}\]

Example 60

Question. Consider the open intervals \[ I_n := \left( 0, \frac1n \right) \,. \] These are clearly nested \[ I_n \supset I_{n+1}\,, \quad \forall \, n \in \mathbb{N}\,. \] Prove that \[ \bigcap_{n=1}^\infty I_n = \emptyset \,. \tag{A.15}\]

Solution. Suppose by contradiction that the intersection is non-empty. Then there exists \(x \in \mathbb{N}\) such that \[ x \in I_n \,, \quad \forall \, n \in \mathbb{N}\,. \] By definition of \(I_n\) the above reads \[ 0 < x < \frac1n \,, \quad \forall \, n \in \mathbb{N}\,. \tag{A.16}\] Since \(x>0\), by the Archimedean Property in Theorem 57 Point 2, there exists \(n_0 \in \mathbb{N}\) such that \[ 0 < \frac{1}{n_0} < x \,. \] The above contradicts (A.16). Therefore (A.15) holds.

A.5.1 Revisiting Sup and Inf

Proposition 61: Characterization of Supremum

Let \(A \subseteq \mathbb{R}\) be a non-empty set. Suppose that \(s \in \mathbb{R}\) is an upper bound for \(A\). They are equivalent:

1. \(s = \sup A\)
2. For every \(\varepsilon> 0\) there exists \(x \in A\) such that \[ s - \varepsilon< x \,. \]

Proposition 62: Characterization of Infimum

Let \(A \subseteq \mathbb{R}\) be a non-empty set. Suppose that \(i \in \mathbb{R}\) is a lower bound for \(A\). They are equivalent:

1. \(i = \inf A\)
2. For every \(\varepsilon\in \mathbb{R}\), with \(\varepsilon> 0\), there exists \(x \in A\) such that \[ x < i + \varepsilon\,. \]

Proposition 63

Let \(a, b \in \mathbb{R}\) with \(a<b\). Let \[ A:= (a,b) = \{ x \in \mathbb{R}\, \colon \,a < x < b \}\,. \] Then \[ \inf A = a \,, \quad \sup A = b \,. \]

Corollary 64

Let \(a, b \in \mathbb{R}\) with \(a<b\). Let \[ A:= (a,b) = \{ x \in \mathbb{R}\, \colon \,a < x < b \}\,. \] Then \(\min A\) and \(\max A\) do not exist.

Corollary 65

Let \(a, b \in \mathbb{R}\) with \(a<b\). Let \[ A:= [a,b) = \{ x \in \mathbb{R}\, \colon \,a \leq x < b \}\,. \] Then \[ \min A = \inf A = a \,, \quad \sup A = b \,, \] \(\max A\) does not exist.

Proposition 66

Define the set \[ A := \left\{ \frac1n \, \colon \,n \in \mathbb{N}\right\} \,. \] Then \[ \inf A = 0 \,, \quad \sup A = \max A = 1 \,. \]

Proof

Part 1. We have \[ \frac1n \leq 1 \,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore \(1\) is an upper bound for \(A\). Since \(1 \in A\), by definition of maximum we conclude that \[ \max A = 1 \,. \] Since the maximum exists, we conclude that also the supremum exists, and \[ \sup A = \max A = 1 \,. \]

Part 2. We have \[ \frac1n > 0 \,, \quad \forall \, n \in \mathbb{N}\,, \] showing that \(0\) is a lower bound for \(A\). Suppose by contradiction that \(0\) is not the infimum. Therefore \(0\) is not the largest lower bound. Then there exists \(\varepsilon\in \mathbb{R}\) such that:

• \(\varepsilon\) is a lower bound for \(A\), that is, \[ \varepsilon\leq \frac1n \,, \quad \forall \, n \in \mathbb{N}\,, \tag{A.17}\]

• \(\varepsilon\) is larger than \(0\): \[ 0 < \varepsilon\,. \]

As \(\varepsilon>0\), by the Archimedean Property there exists \(n_0 \in \mathbb{N}\) such that \[ 0 < \frac{1}{n_0} < \varepsilon\,. \] This contradicts (A.17). Thus \(0\) is the largest lower bound of \(A\), that is, \(0 = \inf A\).

Part 3. We have that \(\min A\) does not exist. Indeed suppose by contradiction that \(\min A\) exists. Then \[ \min A = \inf A \,. \] As \(\inf A = 0\) by Part 2, we conclude \(\min A = 0\). As \(\min A \in A\), we obtain \(0 \in A\), which is a contradiction.

A.5.2 Cardinality

Definition 67: Cardinality, Finite, Countable, Uncountable

Let \(X\) be a set. The cardinality of \(X\) is the number of elements in \(X\). We denote the cardinality of \(X\) by \[ |X| := \# \, \mbox{ of elements in } \, X \,. \] Further, we say that:

1. \(X\) is finite if there exists a natural number \(n \in \mathbb{N}\) and a bijection \[ f \colon \{1,2,\ldots, n \} \to X \,. \] In particular \[ |X| = n \,. \]

2. \(X\) is countable if there exists a bijection \[ f \colon \mathbb{N}\to X \,. \] In this case we denote the cardinality of \(X\) by \[ |X| = |\mathbb{N}| \,. \]

3. \(X\) is uncountable if \(X\) is neither finite, nor countable.

Proposition 68

Let \(X\) be a countable set and \(A \subseteq X\). Then either \(A\) is finite or countable.

Example 69

Question. Prove that \(X = \{a,b,c\}\) is finite.

Solution. Set \(Y=\{ 1,2,3\}\). The function \(f \colon X \to Y\) defined by \[ f(1)=a \,, \quad f(2) = b \,, \quad f(3) = c \,, \] is bijective. Therefore \(X\) is finite, with \(|X| = 3\).

Example 70

Question. Prove that the set of natural numbers \(\mathbb{N}\) is countable.

Solution. The function \(f \colon X \to \mathbb{N}\) defined by \[ f(n):=n \,, \] is bijective. Therefore \(X = \mathbb{N}\) is countable.

Example 71

Question. Let \(X\) be the set of even numbers \[ X = \{ 2n \, \colon \,n \in \mathbb{N}\} \,. \] Prove that \(X\) is countable.

Solution. Define the map \(f \colon \mathbb{N}\to X\) by \[ f(n):= 2n \,. \] We have that:

1. \(f\) is injective, because \[ f(m)=f(k) \, \implies \, 2m=2k \, \quad \, m=k \,. \]

2. \(f\) is surjective: Suppose that \(m \in X\). By definition of \(X\), there exists \(n \in \mathbb{N}\) such that \(m = 2n\). Therefore, \(f(n) = m\).

We have shown that \(f\) is bijective. Thus, \(X\) is countable.

Example 72

Question. Prove that the set of integers \(\mathbb{Z}\) is countable.

Solution. Define \(f \colon \mathbb{N}\to \mathbb{Z}\) by \[ f(n):= \begin{cases} \dfrac{n}{2} & \,\, \mbox{ if } \, n \, \mbox{ even } \\ -\dfrac{n+1}{2} & \,\, \mbox{ if } \, n \, \mbox{ odd } \end{cases} \] For example \[\begin{align*} f(0) & = 0 \,, \quad f(1) = -1 \,, \quad f(2) = 1 \,, \quad f(3) = -2 \,, \\ f(4) & = 2 \,, \quad f(5) = -3 \,, \quad f(6) = 3 \,, \quad f(7) = -4 \,. \end{align*}\] We have:

1. \(f\) is injective: Indeed, suppose that \(m \neq n\). If \(n\) and \(m\) are both even or both odd we have, respectively \[\begin{align*} f(m) = \frac{m}{2} & \neq \frac{n}{2} = f(n) \\ f(m) = -\frac{m+1}{2} & \neq - \frac{n+1}{2} = f(n) \,. \end{align*}\] If instead \(m\) is even and \(n\) is odd, we get \[ f(m) = \frac{m}{2} \neq - \frac{n+1}{2} = f(n) \,, \] since the LHS is positive and the RHS is negative. The case when \(m\) is odd and \(n\) even is similar.

2. \(f\) is surjective: Let \(z \in \mathbb{Z}\). If \(z \geq 0\), then \(m:=2z\) belongs to \(\mathbb{N}\), is even, and \[ f(m) = f(2z) = z \,. \] If instead \(z < 0\), then \(m := -2z -1\) belongs to \(\mathbb{N}\), is odd, and \[ f(m) = f(-2z-1) = z \,. \]

Therefore \(f\) is bijective, showing that \(\mathbb{Z}\) is countable.

Proposition 73

Let the set \(A_n\) be countable for all \(n \in \mathbb{N}\). Define \[ A = \bigcup_{n \in \mathbb{N}} \, A_n \, . \] Then \(A\) is countable.

Theorem 74: \(\mathbb{Q}\) is countable

The set of rational numbers \(\mathbb{Q}\) is countable.

Theorem 75: \(\mathbb{R}\) is uncountable

The set of Real Numbers \(\mathbb{R}\) is uncountable.

Theorem 76

The set of irrational numbers \[ \mathcal{I}:=\mathbb{R}\smallsetminus \mathbb{Q} \] is uncountable.

Proof

We know that \(\mathbb{R}\) in uncountable and \(\mathbb{Q}\) is countable. Suppose by contradiction that \(\mathcal{I}\) is countable. Then \[ \mathbb{Q}\cup \mathcal{I} \] is countable by Proposition 73, being union of countable sets. Since by definition \[ \mathbb{R}= \mathbb{Q}\cup \mathcal{I} \,, \] we conclude that \(\mathbb{R}\) is countable. Contradiction.

A.6 Complex Numbers

Definition 77: Complex Numbers

The set of complex numbers \(\mathbb{C}\) is defined as

\[ \mathbb{C}:= \mathbb{R}+ i \mathbb{R}:= \{x + i y \, \colon \,x, y \in \mathbb{R}\} \,. \] For a complex number \[ z = x + i y \in \mathbb{C} \] we say that

• \(x\) is the real part of \(z\), and denote it by \[ x = \operatorname{Re}(z) \]
• \(y\) is the imaginary part of \(z\), and denote it by \[ y = \operatorname{Im}(z) \]

We say that

• If \(\operatorname{Re}z = 0\) then \(z\) is a purely imaginary number.
• If \(\operatorname{Im}z = 0\) then \(z\) is a real number.

Definition 78: Addition and multiplication in \(\mathbb{C}\)

Let \(z_1,z_2 \in \mathbb{C}\), so that \[ z_1 = x_1 + i y_1 \,\,, \quad z_2 = x_2 + i y_2 \,\, , \] for some \(x_1,x_2,y_1,y_2 \in \mathbb{R}\):

1. The sum of \(z_1\) and \(z_2\) is \[ z_1 + z_2 :=\left(x_{1}+x_{2}\right) + i \left(y_{1}+y_{2}\right) \,. \]

2. The multiplication of \(z_1\) and \(z_2\) is \[\begin{align*} z_1 \cdot z_2 := \left(x_{1} \cdot x_{2} - y_{1} \cdot y_{2}\right) + i \left(x_{1} \cdot y_{2} + x_{2} \cdot y_{1} \right) \,, \end{align*}\]

Example 79

Question. Compute \(zw\), where \[ z = -2+3 i \,, \quad w = 1- i \,. \]

Solution. Using the definition we compute \[\begin{align*} z \cdot w & = (-2+3 i) \cdot (1 - i) \\ & = (-2-(-3))+(2+3) i \\ & = 1 + 5 i \, . \end{align*}\] Alternatively, we can proceed formally: We just need to recall that \(i^2\) has to be replaced with \(-1\): \[\begin{align*} z \cdot w & = (-2+3 i) \cdot (1 - i) \\ & = - 2 + 2i + 3i - 3 i^2 \\ & = (-2 + 3 ) + ( 2 + 3 ) i \\ & = 1 + 5 i \, . \end{align*}\]

Proposition 80: Additive inverse in \(\mathbb{C}\)

The neutral element of addition in \(\mathbb{C}\) is the number \[ 0 := 0 + 0 i \,. \] For any \(z = x + i y \in \mathbb{C}\), the unique additive inverse is given by \[ - z := -x - i y \,. \]

Proposition 81: Multiplicative inverse in \(\mathbb{C}\)

The neutral element of multiplication in \(\mathbb{C}\) is the number \[ 1 := 1 + 0 i \,. \] For any \(z = x + i y \in \mathbb{C}\), the unique multiplicative inverse is given by \[ z^{-1} := \frac{x}{x^{2}+y^{2}}+ i \, \frac{-y}{x^{2}+y^{2}} \,. \]

Proof

It is immediate to check that \(1\) is the neutral element of multiplication in \(\mathbb{C}\). For the remaining part of the statement, set \[ w := \frac{x}{x^{2}+y^{2}}+ i \, \frac{-y}{x^{2}+y^{2}} \,. \] We need to check that \(z \cdot w = 1\) \[\begin{align*} z \cdot w & = (x+i y) \cdot \left(\frac{x}{x^{2}+y^{2}}+ i \, \frac{-y}{x^{2}+y^{2}} \right) \\ & = \left(\frac{x^{2}}{x^{2}+y^{2}} - \frac{y \cdot(-y)}{x^{2}+y^{2}}\right) + i \, \left(\frac{x \cdot(-y)}{x^{2}+y^{2}}+\frac{x y}{x^{2}+y^{2}}\right) \\ & =1 \,, \end{align*}\] so indeed \(z^{-1}=w\).

Example 82

Question. Let \(z = 3 + 2 i\). Compute \(z^{-1}\).

Solution. By the formula in Propostion 81 we immediately get \[ z^{-1} = \frac{3}{3^{2}+2^{2}} + \, \frac{-2}{3^{2}+2^{2}} \, i = \frac{3}{13}-\frac{2}{13} \, i \,. \] Alternatively, we can proceed formally: \[\begin{align*} (3+2 i)^{-1} & = \frac{1}{3+2 i} \\ & = \frac{1}{3+2 i} \, \frac{3-2 i}{3-2 i} \\ & = \frac{3-2 i}{3^2+2^2} \\ & = \frac{3}{13}-\frac{2}{13} i \,, \end{align*}\] and obtain the same result.

Theorem 83

\((\mathbb{C},+, \cdot)\) is a field.

Example 84

Question. Let \(w=1+i\) and \(z=3-i\). Compute \(\frac{w}{z}\).

Solution. We compute \(w/z\) using the two options we have:

1. Using the formula for the inverse from Proposition 81 we compute \[\begin{align*} z^{-1} & = \frac{x}{x^{2}+y^{2}} + i \, \frac{-y}{x^{2}+y^{2}} \\ & = \frac{3}{3^2 + 1^2} - i \, \frac{-1}{3^2 + 1^2} \\ & = \frac{3}{10} + \frac{1}{10} \, i \end{align*}\] and therefore \[\begin{align*} \frac{w}{z} & = w \cdot z^{-1} \\ & = (1 + i) \, \left( \frac{3}{10} + \frac{1}{10} \, i \right) \\ & = \left(\frac{3}{10}-\frac{1}{10}\right)+\left(\frac{1}{10}+\frac{3}{10}\right) i \\ & = \frac{2}{10}+\frac{4}{10} i \\ & = \frac{1}{5}+\frac{2}{5} i \end{align*}\]

2. We proceed formally, using the multiplication by \(1\) trick. We have \[\begin{align*} \frac{w}{z} & = \frac{1+i}{3-i} \\ & = \frac{1+i}{3-i} \frac{3+i}{3+i} \\ & = \frac{3-1+(3+1) i}{3^2+1^2} \\ & =\frac{2}{10}+\frac{4}{10} i \\ & = \frac{1}{5}+\frac{2}{5} i \end{align*}\]

Definition 85: Complex conjugate

Let \(z=x+iy\). We call the complex conjugate of \(z\), denoted by \(\bar{z}\), the complex number

\[ \bar{z}=x- i y \, . \]

Theorem 86

For all \(z_1, z_2 \in \mathbb{C}\) it holds:

• \(\overline{z_1 + z_2 }=\overline{z_1}+\overline{z_2}\)

• \(\overline{z_1 \cdot z_2}=\overline{z_1} \cdot \overline{z_2}\)

A.6.1 The complex plane

Definition 87: Modulus

The modulus of a complex number \(z=x+i y\) is defined by \[ |z|: = \sqrt{x^{2}+y^{2}} \,. \]

Definition 88: Distance in \(\mathbb{C}\)

Given \(z_1,z_2 \in \mathbb{C}\), we define the distance between \(z_1\) and \(z_2\) as the quantity \[ |z_1 - z_2| \,. \]

Theorem 89

Given \(z_1,z_2 \in \mathbb{C}\), we have \[ |z_1 - z_2| = \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } \,. \]

Example 90

Question. Compute the distance between \[ z = 2-4 i \,, \quad w = -5+i \,. \]

Solution. The distance is \[\begin{align*} |z- w| & = |(2-4 i)-(-5+i)| \\ & = |7-5 i| \\ & =\sqrt{7^{2}+(-5)^{2}} \\ & =\sqrt{74} \end{align*}\]

Theorem 91

Let \(z, z_{1}, z_{2} \in \mathbb{C}\). Then

1. \(\left|z_1 \cdot z_2\right|=\left|z_{1}\right|\left|z_{2}\right|\)

2. \(\left|z^{n}\right|=|z|^{n}\) for all \(n \in \mathbb{N}\)

3. \(z \cdot \bar{z}=|z|^{2}\)

Theorem 92: Triangle inequality in \(\mathbb{C}\)

For all \(x, y, z \in \mathbb{C}\),

1. \(|x+y| \leq|x|+|y|\)

2. \(|x-z| \leq|x-y|+|y-z|\)

A.6.2 Polar coordinates

Definition 93: Argument

Let \(z \in \mathbb{C}\). The angle \(\theta\) between the line connecting the origin and \(z\) and the positive real axis is called the argument of \(z\), and is denoted by \[ \theta := \arg (z) \,. \]

Example 94

We have the following arguments: \[\begin{align*} & \arg (1) = 0 & \quad & \arg (i) = \frac{\pi}{2} \\ & \arg (-1) = \pi & \quad & \arg (-i) = -\frac{\pi}{2} \\ & \arg (1+i)=\frac{1}{4} \pi & \quad & \arg (-1-i)=-\frac{3}{4} \pi \end{align*}\]

Theorem 95: Polar coordinates

Let \(z \in \mathbb{C}\) with \(z = x + i y\) and \(z \neq 0\). Then \[ x = \rho \cos(\theta) \,, \quad y = \rho \sin(\theta) \,, \] where \[ \rho := |z| = \sqrt{x^2 + y^2} \,, \quad \theta := \arg(z) \,. \]

Definition 96: Trigonometric form

Let \(z \in \mathbb{C}\). The trigonometric form of \(z\) is \[ z = |z| \left[ \cos (\theta)+i \sin (\theta) \right] \,, \] where \(\theta = \arg(z)\).

Example 97

Question. Suppose that \(z \in \mathbb{C}\) has polar coordinates \[ \rho = \sqrt{8}\,, \quad \theta = \frac{3}{4} \pi \,. \] Therefore, the trigonometric form of \(z\) is \[ z = \sqrt{8} \left[ \cos \left( \frac{3}{4} \pi \right) + i \sin \left( \frac{3}{4} \pi \right) \right] \,. \] Write \(z\) in cartesian form.

Solution. We have \[\begin{gather*} x = \rho \cos (\theta) = \sqrt{8} \cos \left( \frac{3}{4} \pi \right) = - \sqrt{8} \cdot \frac{\sqrt{2}}{2} = -2 \\ y = \rho \sin (\theta) = \sqrt{8} \sin \left( \frac{3}{4} \pi \right) = \sqrt{8} \cdot \frac{\sqrt{2}}{2} = 2 \,. \end{gather*}\] Therefore, the cartesian form of \(z\) is \[ z = x + i y = - 2 + 2 i \,. \]

Corollary 98: Computing \(\arg(z)\)

Let \(z \in \mathbb{C}\) with \(z = x + i y\) and \(z \neq 0\). Then \[ \arg(z) = \begin{cases} \arctan \left( \dfrac{y}{x} \right) & \,\, \mbox{ if } x>0 \\ \arctan \left( \dfrac{y}{x} \right) + \pi & \,\, \mbox{ if } x<0\,\, \mbox{and} \,\, y \geq 0 \\ \arctan \left( \dfrac{y}{x} \right) - \pi & \,\, \mbox{ if } x<0\,\, \mbox{and} \,\, y < 0 \\ \dfrac{\pi}{2} & \,\, \mbox{ if } x=0\,\, \mbox{and} \,\, y > 0 \\ -\dfrac{\pi}{2} & \,\, \mbox{ if } x=0\,\, \mbox{and} \,\, y < 0 \\ \end{cases} \] where \(\arctan\) is the inverse of \(\tan\).

Example 99

Question. Compute the arguments of the complex numbers \[ z = 3+4 i\,, \quad \bar{z} = 3-4 i \,, \quad - \bar{z} = -3+4 i \,, \quad - z= -3-4 i \,. \]

Solution. Using the formula for \(\arg\) in Corollary 98 we have \[\begin{align*} \arg (3+4 i) & =\arctan \left(\frac{4}{3}\right) \\ \arg (3-4 i) & =\arctan \left(-\frac{4}{3}\right) = -\arctan \left(\frac{4}{3}\right) \\ \arg (-3+4 i) & = \arctan \left(-\frac{4}{3}\right) + \pi = - \arctan \left(\frac{4}{3}\right) + \pi \\ \arg (-3-4 i) & = \arctan \left(\frac{4}{3}\right) - \pi \end{align*}\]

A.6.3 Exponential form

Theorem 100: Euler’s identity

For all \(\theta \in \mathbb{R}\) it holds \[ e^{i \theta}=\cos (\theta)+ i \sin (\theta) \,. \]

Theorem 101

For all \(\theta \in \mathbb{R}\) it holds \[ \left|e^{i \theta}\right|=1 \,. \]

Theorem 102

Let \(z \in \mathbb{C}\) with \(z = x+iy\) and \(z \neq 0\). Then \[ z = \rho e^{i\theta}\,, \] where \[ \rho := |z| = \sqrt{x^2 + y^2} \,, \qquad \theta := \arg(z) \,. \]

Definition 103: Exponential form

The exponential form of a complex number \(z \in \mathbb{C}\) is \[ z= \rho e^{i\theta} = |z| \, e^{i \arg(z)} \,. \]

Example 104

Question. Write the number \[ z=-2+2 i \] in exponential form.

Solution. From Example 97 we know that \(z = -2+2i\) can be written in trigonometric form as \[ z = \sqrt{8} \left[ \cos \left( \frac{3}{4} \pi \right) + i \sin \left( \frac{3}{4} \pi \right) \right] \,. \] By Euler’s identity we hence obtain the exponential form \[ z=\sqrt{8} e^{i \frac{3}{4} \pi} \,. \]

Remark 105: Periodicity of exponential

For all \(k \in \mathbb{Z}\) we have \[ e^{i \theta}=e^{i(\theta+2 \pi k)} \,, \tag{A.18}\] meaning that the complex exponential is \(2\pi\)-periodic.

Proposition 106

Let \(z, z_1,z_2 \in \mathbb{C}\) and suppose that \[ z= \rho e^{i \theta} \,, \quad z_1 = \rho_1 e^{i \theta_1}\,, \quad z_2 = \rho_2 e^{i \theta_2}\,. \] We have \[ z_1 \cdot z_2 = \rho_1 \rho_2 e^{i (\theta_1 + \theta_2)} \,, \quad z^n = \rho^n e^{i n \theta} \,, \] for all \(n \in \mathbb{N}\).

Example 107

Question. Compute \((-2+2 i)^{4}\).

Solution. We have two possibilities:

1. Use the binomial theorem: \[\begin{align*} (-2+2 i)^{4} & =(-2)^{4}+\left(\begin{array}{l} 4 \\ 1 \end{array}\right)(-2)^{3} \cdot 2 i+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)(-2)^{2} \cdot(2 i)^{2} \\ & \quad +\left(\begin{array}{l} 4 \\ 3 \end{array}\right)(-2) \cdot(2 i)^{3}+(2 i)^{4} \\ & =16-4 \cdot 8 \cdot 2 i-6 \cdot 4 \cdot 4+4 \cdot 2 \cdot 8 i+16 \\ & =16-64 i-96+64 i+16=-64 \,. \end{align*}\]

2. A much simpler calculation is possible by using the exponential form: We know that \[ -2+2 i = \sqrt{8} e^{i \frac{3}{4} \pi} \] by Example 104. Hence \[ (-2+2 i)^{4}=\left(\sqrt{8} e^{i \frac{3}{4} \pi}\right)^{4}=8^{2} e^{i 3 \pi}=-64 \,, \] where we used that \[ e^{i 3 \pi} = e^{i \pi} = \cos(\pi) + i \sin(\pi) = - 1 \] by \(2\pi\) periodicity of \(e^{i\theta}\) and Euler’s identity.

Definition 108: Complex exponential

The complex exponential of \(z \in \mathbb{C}\) is defined as \[ e^z = |z| e^{i\theta} \,, \quad \theta = \arg(z) \,. \]

Theorem 109

Let \(z,w \in \mathbb{C}\). Then \[ e^{z+w} = e^z e^w \,, \quad (e^z)^{w} = e^{zw} \,. \tag{A.19}\]

Example 110

Question. Compute \(i^{i}\).

Solution. We know that \[ |i| = 1 \,, \quad \arg(i) = \frac{\pi}{2} \,. \] Hence we can write \(i\) in exponential form \[ i= |i| e^{i\arg(i)} = e^{i \frac{\pi}{2}} \,. \] Therefore \[ i^{i}=\left(e^{i \frac{\pi}{2}}\right)^{i}=e^{i^2\frac{\pi}{2}}=e^{-\frac{\pi}{2}} \,. \]

A.6.4 Fundamental Theorem of Algebra

Theorem 111: Fundamental theorem of algebra

Let \(p_{n}(z)\) be a polynomial of degree \(n\) with complex coefficients, i.e., \[ p_{n}(z)=a_{n} z^{n}+a_{n-1} z^{n-1}+\ldots+a_{1} z+a_{0}, \] for some coefficients \(a_{n}, \ldots, a_{0} \in \mathbb{C}\) with \(a_{n} \neq 0\). There exist \[ z_{1}, \ldots, z_{n} \in \mathbb{C} \] solutions to the polynomial equation \[ p_n(z) = a_{n} z^{n}+a_{n-1} z^{n-1}+\ldots+a_{1} z+a_{0}=0 \,. \tag{A.20}\] In particular, \(p_n\) factorizes as \[ p_{n}(z)=a_{n}\left(z-z_{1}\right)\left(z-z_{2}\right) \cdots\left(z-z_{n}\right) \,. \tag{A.21}\]

Example 112

Question. Find all the complex solutions to \[ z^{2}=-1 \tag{A.22}\]

Solution. The equation \(z^2 = -1\) is equivalent to \[ p(z) = 0 \,, \quad p(z):=z^{2}+1 \,. \] Since \(p\) has degree \(n=2\), the Fundamental Theorem of Algebra tells us that there are two solutions to (A.22). We have already seen that these two solutions are \(z=i\) and \(z=-i\). Then \(p\) factorizes as \[ p(z) = z^{2} + 1 = (z-i)(z+i) \,. \]

Example 113

Question. Find all the complex solutions to \[ z^{4}-1=0 \,. \tag{A.23}\]

Solution The associated polynomial equation is \[ p(z) = 0 \,, \quad p(z) := z^4 - 1 \,. \] Since \(p\) has degree \(n=4\), the Fundamental Theorem of Algebra tells us that there are \(4\) solutions to (A.23). Let us find such solutions. We use the well known identity \[ a^2-b^2 = (a+b)(a-b) \,, \quad \forall \, a,b \in \mathbb{R}\,, \] to factorize \(p\). We get: \[ p(z) = (z^4-1) = (z^2+1)(z^2-1) \,. \] We know that \[ z^2 + 1 = 0 \] has solutions \(z = \pm i\). Instead
\[ z^2 - 1 = 0 \] has solutions \(x = \pm 1\). Hence, the four solutions of (A.23) are given by \[ z=1,-1, i,-i \,, \] and \(p\) factorizes as \[ p(z) = z^4 - 1 = (z-1)(z+1)(z-i)(z+i) \,. \]

Definition 114

Suppose that the polynomial \(p_n\) factorizes as \[ p_n(z) = a_n (z-z_1)^{k_1} (z-z_2)^{k_2} \cdots (z-z_m)^{k_m} \] with \(a_n \neq 0\), \(z_1,\ldots, z_m \in \mathbb{C}\) and \(k_1, \ldots , k_m \in \mathbb{N}\), \(k_i \geq 1\). In this case \(p_n\) has degree \[ n = k_1 + \ldots + k_m = \sum_{i=1}^m k_i \,. \] Note that \(z_i\) is solves the equation \[ p_n(z) = 0 \] exactly \(k_i\) times. We call \(k_i\) the multiplicity of the solution \(z_i\).

Example 115

The equation \[ (z-1)(z-2)^{2}(z+i)^{3}=0 \] has 6 solutions:

• \(z=1\) with multiplicity \(1\)
• \(z=2\) with multiplicity \(2\)
• \(z=-i\) with multiplicity \(3\)

A.6.5 Solving polynomial equations

Proposition 116: Quadratic formula

Let \(a, b, c \in \mathbb{R}, a \neq 0\) and consider the equation \[ ax^2 + bx + c = 0 \,. \tag{A.24}\] Define \[ \Delta := b^2 - 4 ac \in \mathbb{R}\,. \] The following hold:

1. If \(\Delta > 0\) then (A.24) has two distinct real solutions \(z_1,z_2 \in \mathbb{R}\) given by \[ z_1 = \frac{-b - \sqrt{\Delta}}{2 a} \,, \quad z_2 = \frac{-b + \sqrt{\Delta}}{2 a} \,. \]
2. If \(\Delta = 0\) then (A.24) has one real solution \(z \in \mathbb{R}\) with multiplicity \(2\). Such solution is given by \[ z = z_1 = z_2 = \frac{-b}{2 a} \,. \]
3. If \(\Delta < 0\) then (A.24) has two distinct complex solutions \(z_1,z_2 \in \mathbb{C}\) given by \[ z_1 = \frac{-b - i\sqrt{-\Delta}}{2 a} \,, \quad z_2 = \frac{-b + i\sqrt{-\Delta}}{2 a} \,, \] where \(\sqrt{-\Delta} \in \mathbb{R}\), since \(-\Delta>0\).

In all cases, the polynomial at (A.24) factorizes as \[ a z^{2}+b z+c = a (z-z_1)(z-z_2) \,. \]

Example 117

Question. Solve the following equations:

1. \(3 z^{2}-6 z+2 = 0\)
2. \(4 z^{2}-8 z+4=0\)
3. \(z^{2}+2 z+3=0\)

Solution.

1. We have that \[ \Delta = (-6)^{2}-4 \cdot 3 \cdot 2 = 12 > 0 \] Therefore the equation has two distinct real solutions, given by \[ z=\frac{-(-6) \pm \sqrt{12}}{2 \cdot 3}=\frac{6 \pm \sqrt{12}}{6}=1 \pm \frac{\sqrt{3}}{3} \] In particular we have the factorization \[ 3 z^{2}-6 z+2 = 3 \left( z - 1 - \frac{\sqrt{3}}{3} \right) \left( z - 1 + \frac{\sqrt{3}}{3} \right) \,. \]

2. We have that \[ \Delta = (-8)^{2}-4 \cdot 4 \cdot 4 = 0 \,. \] Therefore there exists one solution with multiplicity \(2\). This is given by \[ z=\frac{-(-8)}{2 \cdot 4} = 1 \,. \] In particular we have the factorization \[ 4 z^{2}-8 x+4 = 4 (z-1)^2 \,. \]

3. We have \[ \Delta = 2^{2}-4 \cdot 1 \cdot 3 = - 8 < 0 \,. \] Therefore there are two complex solutions given by \[ z=\frac{-2 \pm i \sqrt{8}}{2 \cdot 1} = -1 \pm i \sqrt{2} \,. \] In particular we have the factorization \[ z^{2}+2 z+3 = (z + 1 - i \sqrt{2}) (z + 1 + i\sqrt{2}) \,. \]

Proposition 118: Quadratic formula with complex coefficients

Let \(a, b, c \in \mathbb{C}, a \neq 0\). The two solutions to \[ a z^{2}+b z + c=0 \] are given by \[ z_1 = \frac{-b + S_1}{2 a} \,, \quad z_2 = \frac{-b + S_2}{2 a} \,, \] where \(S_1\) and \(S_2\) are the two solutions to \[ z^2 = \Delta \,, \quad \Delta := b^2 - 4ac \,. \]

Example 119

Question Find all the solutions to \[ \frac{1}{2} z^{2}-(3+i) z+(4-i)=0 \,. \tag{A.25}\]

Solution. We have \[\begin{align*} \Delta & = (-(3+i))^{2}-4 \cdot \frac{1}{2} \cdot(4-i) \\ & = 8+6 i-8+2 i \\ & =8 i \,. \end{align*}\] Therefore \(\Delta \in \mathbb{C}\). We have to find solutions \(S_1\) and \(S_2\) to the equation \[ z^2 = \Delta = 8i \,. \tag{A.26}\] We look for solutions of the form \(z=a+ i b\). Then we must have that \[ z^{2}=(a+ ib)^{2}=a^{2}-b^{2}+2 a b i = 8 i \,. \] Thus \[ a^{2}-b^{2}=0 \,, \quad 2 a b = 8 \,. \] From the first equation we conclude that \(|a|=|b|\). From the second equation we have that \(ab=4\), and therefore \(a\) and \(b\) must have the same sign. Hence \(a=b\), and \[ 2 a b = 8 \quad \implies \quad a = b = \pm 2 \,. \] From this we conclude that the solutions to (A.26) are \[ S_{1} = 2+2 i \,, \quad S_{2}=-2-2 i \,. \] Hence the solutions to (A.25) are \[\begin{align*} z_1 & = \frac{3+i+S_{1}}{2 \cdot \frac{1}{2}} = 3 + i + S_{1} \\ & = 3 + i + 2 + 2i = 5 + 3i \,, \end{align*}\] and \[\begin{align*} z_2 & = \frac{3+i+S_{2}}{2 \cdot \frac{1}{2}} = 3+i+S_{2} \\ & = 3+i -2 - 2i = 1 - i \,. \end{align*}\] In particular, the given polynomial factorizes as \[\begin{align*} \frac{1}{2} z^{2}-(3+i) z+(4-i) & = \frac12 (z - z_1)(z-z_2) \\ & = \frac12 (z - 5 - 3i)(z - 1 + i) \,. \end{align*}\]

Example 120

Question. Consider the equation \[ z^{3}-7 z^{2}+6 z=0 \,. \]

1. Check whether \(z=0,1,-1\) are solutions.
2. Using your answer from Point 1, and polynomial division, find all the solutions.

Solution.

1. By direct inspection we see that \(z=0\) and \(z=1\) are solutions.

2. Since \(z = 0\) is a solution, we can factorize \[ z^{3}-7 z^{2}+6 z=z\left(z^{2}-7 z+6\right) \,. \] We could now use the quadratic formula on the term \(z^{2}-7 z+6\) to find the remaining two roots. However, we have already observed that \(z=1\) is a solution. Therefore \(z-1\) divides \(z^{2}-7 z+6\). Using polynomial long division, we find that \[ \frac{z^{2}-7 z+6}{z-1}=z-6 \,. \] Therefore the last solution is \(z=6\), and \[ z^{3}-7 z^{2}+6 z=z(z-1)(z-6) \,. \]

Example 121

Question. Find all the complex solutions to \[ z^{3}-7 z+6=0 \,. \]

Solution. It is easy to see \(z=1\) is a solution. This means that \(z-1\) divides \(z^{3}-7 z+6\). By using polynomial long division, we compute that \[ \frac{z^{3}-7 z+6}{z-1}=z^{2}+z-6 \,. \] We are now left to solve \[ z^{2}+z-6 = 0\,. \] Using the quadratic formula, we see that the above is solved by \(z=2\) and \(z=-3\). Therefore the given polynomial factorizes as \[ z^{3}-7 z+6 = (z-1)(z-2)(z+3) \,. \]

A.6.6 Roots of unity

Theorem 122

Let \(n \in \mathbb{N}\) and consider the equation \[ z^{n}=1 \,. \tag{A.27}\] All the \(n\) solutions to (A.27) are given by \[ z_k = \exp \left(i \frac{2 \pi k}{n}\right) \,, \quad k = 0, \ldots, n-1 \,, \] where \(\exp(x)\) denotes \(e^x\).

Definition 123

The \(n\) solutions to \[ z^{n}=1 \] are called the roots of unity.

Example 124

Question. Find all the solutions to \[ z^{4}=1 \,. \]

Solution. The \(4\) solutions are given by \[ z_k = \exp \left(i \frac{2 \pi k}{4} \right) = \exp \left(i \frac{\pi k}{2} \right) \,, \] for \(k=0,1,2,3\). We compute: \[\begin{align*} z_0 & = e^{i 0} = 1 \,, & \quad & z_1 = e^{i \frac{\pi}{2}}=i \,, \\ z_2 & = e^{i \pi}=-1 \,, & \quad & z_3 = e^{i \frac{3 \pi}{2}}=-i \, . \end{align*}\] Note that for \(k=4\) we would again get the solution \(z=e^{i 2 \pi}=1\).

Example 125

Question. Find all the solutions to \[ z^{3}=1 \,. \]

Solution. The \(3\) solutions are given by \[ z_k = \exp \left( i \frac{2 \pi k}{3} \right) \,, \] for \(k=0,1,2\). We compute: \[ z_0=e^{i 0}=1, \quad z_1=e^{i \frac{2 \pi}{3}}, \quad z_2=e^{i \frac{4 \pi}{3}} . \] We can write \(z_1\) and \(z_2\) in cartesian form: \[ z_1 = e^{i \frac{2 \pi}{3}}=\cos \left(\frac{2 \pi}{3}\right)+i \sin \left(\frac{2 \pi}{3}\right)=-\frac{1}{2}+\frac{\sqrt{3}}{2} i \] and \[ z_2 = e^{i \frac{4 \pi}{3}}=\cos \left(\frac{4 \pi}{3}\right)+i \sin \left(\frac{4 \pi}{3}\right)=-\frac{1}{2}-\frac{\sqrt{3}}{2} i \,. \]

A.6.7 Roots in \(\mathbb{C}\)

Theorem 126

Let \(n \in \mathbb{N}\), \(c \in \mathbb{C}\) and consider the equation \[ z^{n}=c \,. \tag{A.28}\] All the \(n\) solutions to (A.28) are given by \[ z_k = \sqrt[n]{|c|} \, \exp \left(i \, \frac{\theta + 2 \pi k}{n}\right) \,, \quad k = 0, \ldots, n-1 \,, \] where \(\sqrt[n]{|c|}\) is the \(n\)-th root of the real number \(|c|\), and \(\theta = \arg(c)\).

Example 127

Question. Find all the \(z \in \mathbb{C}\) such that \[ z^{5}=-32 \,. \]

Solution. Let \(c = -32\). We have \[ |c| = |-32|=32=2^{5}\,, \quad \theta = \arg (-32)=\pi \,. \] The \(5\) solutions are given by \[ z_k = \left(2^{5}\right)^{\frac{1}{5}} \exp \left(i \pi \, \frac{1+2 k}{5} \right) \,, \quad k \in \mathbb{Z}\,, \] for \(k=0,1,2,3,4\). We get \[\begin{align*} z_0 & = 2 e^{i \frac{\pi}{5}} \, & \quad & z_1 = 2 e^{i \frac{3 \pi}{5}} \\ z_2 & = 2 e^{i \pi}=-2 \, & \quad & z_3=2 e^{i \frac{7 \pi}{5}} \\ z_4 &= 2 e^{i \frac{9 \pi}{5}} & \quad & \end{align*}\]

Example 128

Question. Find all the \(z \in \mathbb{C}\) such that \[ z^{4}=9\left(\cos \left(\frac{\pi}{3}\right)+i \sin \left(\frac{\pi}{3}\right)\right) \,. \]

Solution. Set \[ c:=9\left(\cos \left(\frac{\pi}{3}\right)+i \sin \left(\frac{\pi}{3}\right)\right) \,. \] The complex number \(c\) is already in the trigonometric form, so that we can immediately obtain \[ |c| = 9 \,, \quad \theta = \arg(c) = \frac{\pi}{3} \,. \] The \(4\) solutions are given by \[\begin{align*} z_k & = \sqrt[4]{9} \, \exp \left( i \, \frac{ \pi/3 + 2 \pi k}{4} \right) \\ & = \sqrt{3} \exp \left( i \pi \, \frac{1+6 k}{12} \right) \end{align*}\] for \(k=0,1,2,3\). We compute \[\begin{align*} z_0 & = \sqrt{3} e^{i \pi \frac{1}{12}} & \quad & z_1 = \sqrt{3} e^{i \pi \frac{7}{12}} \\ z_2 & = \sqrt{3} e^{i \pi \frac{13}{12}} & \quad & z_3 = \sqrt{3} e^{i \pi \frac{19}{12}} \end{align*}\]

A.7 Sequences in \(\mathbb{R}\)

Definition 129: Convergent sequence

The real sequence \((a_n)\) converges to \(a\), or equivalently has limit \(a\), denoted by \[ \lim_{n \rightarrow \infty} a_{n}=a \,, \] if for all \(\varepsilon\in \mathbb{R}, \varepsilon>0\), there exists \(N \in \mathbb{N}\) such that for all \(n \in \mathbb{N}, n \geq N\) it holds that \[ \left|a_{n}-a\right| < \varepsilon\,. \] Using quantifiers, we can write this as \[ \forall \, \varepsilon>0 , \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \,\left|a_{n}-a\right| < \varepsilon\,. \] The sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) is convergent if it admits limit.

Theorem 130

Let \(p>0\). Then \[ \lim _{n \rightarrow \infty} \frac{1}{n^{p}}=0 \,. \]

Proof

Let \(p>0\). We have to show that \[ \forall \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|\frac{1}{n^{p}}-0\right| < \varepsilon\,. \] Let \(\varepsilon>0\). Choose \(N \in \mathbb{N}\) such that \[ N > \frac{1}{\varepsilon^{1 / p}} \,. \tag{A.29}\] Let \(n \geq N\). Since \(p>0\), we have \(n^{p} \geq N^{p}\), which implies \[ \frac{1}{n^p} \leq \frac{1}{N^p} \,. \] By (A.29) we deduce \[ \frac{1}{N^p} < \varepsilon\,. \] Then \[ \left|\frac{1}{n^{p}}-0\right|=\frac{1}{n^{p}} \leq \frac{1}{N^{p}} < \varepsilon\,. \]

Example 131

Question. Using the definition of convergence, prove that \[ \lim_{n \to \infty} \frac{n}{2n+3} = \frac{1}{2} \,. \]

Solution.

1. Rough Work: Let \(\varepsilon>0\). We want to find \(N \in \mathbb{N}\) such that \[ \left|\frac{n}{2n+3}-\frac{1}{2}\right| < \varepsilon\,, \quad \forall \, n \geq N \,. \] To this end, we compute: \[\begin{align*} \left|\frac{n}{2 n+3}-\frac{1}{2}\right| & =\left|\frac{2n -(2n + 3)}{2(2n +3)}\right| \\ & =\left|\frac{- 3}{4n + 6}\right| \\ & = \frac{3}{4n + 6} \,. \end{align*}\] Therefore \[\begin{align*} \left|\frac{n}{2n+3}-\frac{1}{2}\right| < \varepsilon \quad \iff & \quad \frac{3}{4n + 6} < \varepsilon\\ \quad \iff & \quad \frac{4n + 6}{3} > \frac{1}{\varepsilon} \\ \quad \iff & \quad 4n + 6 > \frac{3}{\varepsilon} \\ \quad \iff & \quad 4n > \frac{3}{\varepsilon} - 6 \\ \quad \iff & \quad n > \frac{3}{4\varepsilon} - \frac{6}{4} \,. \end{align*}\] Looking at the above equivalences, it is clear that \(N \in \mathbb{N}\) has to be chosen so that \[ N > \frac{3}{4\varepsilon} - \frac{6}{4} \,. \]

2. Formal Proof: We have to show that \[ \forall \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|\frac{n}{2n+3}-\frac{1}{2}\right| < \varepsilon\,. \] Let \(\varepsilon>0\). Choose \(N \in \mathbb{N}\) such that \[ N > \frac{3}{4\varepsilon} - \frac{6}{4} \,. \tag{A.30}\] By the rough work shown above, inequality (A.30) is equivalent to \[ \frac{3}{4N + 6} < \varepsilon\,. \] Let \(n \geq N\). Then \[\begin{align*} \left|\frac{n}{2n+3}-\frac{1}{2}\right| & = \frac{3}{4 n+ 6 } \\ & \leq \frac{3}{4 N+ 6 } \\ & < \varepsilon\,, \end{align*}\] where in the third line we used that \(n \geq N\).

Definition 132: Divergent sequence

We say that a sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) in \(\mathbb{R}\) is divergent if it is not convergent.

Theorem 133

Let \(\left(a_{n}\right)\) be the sequence defined by \[ a_{n}=(-1)^{n} \,. \] Then \(\left(a_{n}\right)\) does not converge.

Proof

Suppose by contradiction that \(a_n \to a\) for some \(a \in \mathbb{R}\). Let \[ \varepsilon:=\frac12 \,. \] Since \(a_n \to a\), there exists \(N \in \mathbb{N}\) such that \[ |a_n - a| < \varepsilon= \frac13 \, \quad \forall \, n \geq N \,. \] If we take \(n = 2N\), then \(n \geq N\) and \[ |a_{2N} - a| = | 1 - a | < \frac12 \,. \] If we take \(n = 2N + 1\), then \(n \geq N\) and \[ |a_{2N+1} - a| = | -1 - a | < \frac12 \,. \] Therefore \[\begin{align*} 2 & = | (1 - a) - (-1 - a) | \\ & \leq |1 - a| + |-1 - a| \\ & < \frac12 + \frac12 = 1 \,, \end{align*}\] which is a contradiction. Hence \((a_n)\) is divergent.

Theorem 134: Uniqueness of limit

Let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) be a sequence. Suppose that \[ \lim_{n \rightarrow \infty} a_{n} = a \,, \quad \lim_{n \rightarrow \infty} a_{n} = b \,. \] Then \(a=b\).

Definition 135: Bounded sequence

A sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) is called bounded if there exists a constant \(M \in \mathbb{R}\), with \(M>0\), such that \[ \left|a_{n}\right| \leq M \,, \quad \forall \, n \in \mathbb{N}\,. \]

Theorem 136

Every convergent sequence is bounded.

Example 137

The sequence \[ a_n = (-1)^n \] is bounded but not convergent.

Corollary 138

If a sequence is not bounded, then the sequence does not converge.

Remark 139

For a sequence \((a_n)\) to be unbounded, it means that \[ \forall \, M > 0 \,, \,\, \exists \, n \in \mathbb{N}\, \text{ s.t. } \, \left|a_{n}\right| > M \,. \]

Theorem 140

For all \(p>0\), the sequence \[ a_n = n^p \] does not converge.

Theorem 141

The sequence \[ a_n = \log n \] does not converge.

Theorem 142: Algebra of limits

Let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) and \(\left(b_{n}\right)_{n \in \mathbb{N}}\) be sequences in \(\mathbb{R}\). Suppose that \[ \lim_{n \rightarrow \infty} a_{n}=a \,, \quad \lim_{n \rightarrow \infty} b_{n}=b \,, \] for some \(a,b \in \mathbb{R}\). Then,

1. Limit of sum is the sum of limits: \[ \lim_{n \rightarrow \infty}\left(a_{n} \pm b_{n}\right)=a \pm b \]

2. Limit of product is the product of limits: \[ \lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right) = a b \]

3. If \(b_{n} \neq 0\) for all \(n \in \mathbb{N}\) and \(b \neq 0\), then \[ \lim_{n \rightarrow \infty} \left(\frac{a_{n}}{b_{n}}\right)=\frac{a}{b} \]

Example 143

Question. Prove that \[ \lim_{n \to \infty} \, \frac{3 n}{7 n+4} = \frac{3}{7} \,. \]

Solution. We can rewrite \[ \frac{3 n}{7 n+4}=\frac{3}{7+\dfrac{4}{n}} \] From Theorem 130, we know that \[ \frac{1}{n} \rightarrow 0 \,. \] Hence, it follows from Theorem 143 Point 2 that \[ \frac{4}{n} = 4 \cdot \frac1n \rightarrow 4 \cdot 0 = 0 \,. \] By Theorem 143 Point 1 we have \[ 7 + \frac{4}{n} \rightarrow 7 + 0 = 7 \,. \] Finally we can use Theorem 143 Point 3 to infer \[ \frac{3 n}{7 n+4}=\frac{3}{7+\dfrac{4}{n}} \rightarrow \frac{3}{7} \,. \]

Example 144

Question. Prove that \[ \lim_{n \rightarrow \infty} \frac{n^{2}-1}{2 n^{2}-3} = \frac{1}{2} \,. \]

Solution. Factor \(n^2\) to obtain \[ \frac{n^{2}-1}{2 n^{2}-3} = \frac{1-\dfrac{1}{n^{2}}}{2-\dfrac{3}{n^{2}}} \,. \] By Theorem 130 we have \[ \frac{1}{n^2} \to 0 \,. \] We can then use the Algebra of Limits Theorem 143 Point 2 to infer \[ \frac{3}{n^2} \to 3 \cdot 0 = 0 \] and Theorem 143 Point 1 to get \[ 1 - \frac{1}{n^2} \to 1 - 0 = 1 \,, \quad 2 - \frac{3}{n^2} \to 2 - 0 = 2 \,. \] Finally we use Theorem 143 Point 3 and conclude \[ \frac{1-\dfrac{1}{n^{2}}}{2-\dfrac{3}{n^{2}}} \to \frac{1}{2} \,. \] Therefore \[ \lim_{n \to \infty } \, \frac{n^{2}-1}{2 n^{2}-3} = \lim_{n \to \infty} \, \frac{1-\dfrac{1}{n^{2}}}{2-\dfrac{3}{n^{2}}} = \frac{1}{2} \,. \]

Example 145

Question. Prove that the sequence \[ a_n = \frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1} \] does not converge.

Solution. To show that the sequence \(\left(a_n\right)\) does not converge, we divide by the largest power in the denominator, which in this case is \(n^2\) \[\begin{align*} a_n & = \frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1} \\ & =\frac{4 n+ \dfrac{8}{n} + \dfrac{1}{n^{2}} }{7+ \dfrac{2}{n} + \dfrac{1}{n^{2}} } \\ & = \frac{b_n}{c_n} \end{align*}\] where we set \[ b_n := 4 n+ \dfrac{8}{n} + \dfrac{1}{n^{2}}\,, \quad c_n := 7 + \dfrac{2}{n} + \dfrac{1}{n^{2}} \,. \] Using the Algebra of Limits Theorem 143 we see that \[ c_n = 7+ \dfrac{2}{n} + \dfrac{1}{n^{2}} \to 7 \,. \] Suppose by contradiction that \[ a_n \to a \] for some \(a \in \mathbb{R}\). Then, by the Algebra of Limits, we would infer \[ b_n = c_n \cdot a_n \to 7 a \,, \] concluding that \(b_n\) is convergent to \(7a\). We have that \[ b_n = 4n + d_n \,, \quad d_n := \dfrac{8}{n} + \dfrac{1}{n^{2}} \,. \] Again by the Algebra of Limits Theorem 143 we get that \[ d_n = \dfrac{8}{n} + \dfrac{1}{n^{2}} \to 0 \,, \] and hence \[ 4n = b_n - d_n \to 7a - 0 = 7a \,. \] This is a contradiction, since the sequence \((4n)\) is unbounded, and hence cannot be convergent. Hence \((a_n)\) is not convergent.

Example 146

Question. Define the sequence \[ a_n := \frac{2 n^{3}+7 n+1}{5 n+9} \cdot \frac{8 n+9}{6 n^{3}+8 n^{2}+3} \,. \] Prove that \[ \lim_{n \to \infty} a_n = \frac{8}{15} \,. \]

Solution. The first fraction in \((a_n)\) does not converge, as it is unbounded. Therefore we cannot use Point 2 in Theorem 143 directly. However, we note that \[\begin{align*} a_{n} & =\frac{2 n^{3}+7 n+1}{5 n+9} \cdot \frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\ & = \frac{8 n+9}{5 n+9} \cdot \frac{2 n^{3}+7 n+1}{6 n^{3}+8 n^{2}+3} \,. \end{align*}\] Factoring out \(n\) and \(n^3\), respectively, and using the Algebra of Limits, we see that \[ \frac{8 n+9}{5 n+9}=\frac{8+9 / n}{5+9 / n} \to \frac{8+0}{5+0}=\frac{8}{5} \] and \[ \frac{2+7 / n^{2}+1 / n^{3}}{6+8 / n+3 / n^{3}} \to \frac{2+0+0}{6+0+0}=\frac{1}{3} \] Therefore Theorem 143 Point 2 ensures that \[ a_{n} \to \frac{8}{5} \cdot \frac{1}{3}=\frac{8}{15} \,. \]

Example 147

Question. Prove that \[ a_n = \frac{n^{7 / 3}+2 \sqrt{n}+7}{4 n^{3 / 2}+5 n} \] does not converge.

Solution. The largest power of \(n\) in the denominator is \(n^{3/2}\). Hence we factor out \(n^{3/2}\) \[\begin{align*} a_n & = \frac{n^{7 / 3}+2 \sqrt{n}+7}{4 n^{3 / 2}+5 n} \\ & = \frac{n^{7 / 3-3 / 2}+2 n^{1/2 - 3/2} + 7 n^{-3 / 2}}{4 + 5 n^{-3/2} } \\ & = \frac{n^{5/6}+2 n^{- 1} + 7 n^{-3 / 2}}{4 + 5 n^{-3/2} } \\ & = \frac{b_n}{c_n} \end{align*}\] where we set \[ b_n := n^{5/6}+2 n^{- 1} + 7 n^{-3 / 2} \,, \quad c_n := 4 + 5 n^{-3/2} \,. \] We see that \(b_n\) is unbounded while \(c_n \to 4\). By the Algebra of Limits (and usual contradiction argument) we conclude that \((a_n)\) is divergent.

Theorem 148

Let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) be a sequence in \(\mathbb{R}\) such that \[ \lim_{n \to \infty} \, a_n = a \,, \] for some \(a \in \mathbb{R}\). If \(a_{n} \geq 0\) for all \(n \in \mathbb{N}\) and \(a \geq 0\), then \[ \lim _{n \rightarrow \infty} \sqrt{a_{n}}=\sqrt{a} \,. \]

Example 149

Question. Define the sequence \[ a_n = \sqrt{9 n^{2}+3 n+1}-3 n \,. \] Prove that \[ \lim_{n \to \infty} \, a_n = \frac12 \,. \]

Solution. We first rewrite \[\begin{align*} a_n & = \sqrt{9 n^{2}+3 n+1}-3 n \\ & = \frac{\left(\sqrt{9 n^{2}+3 n+1}-3 n\right)\left(\sqrt{9 n^{2}+3 n+1}+3 n\right)}{\sqrt{9 n^{2}+3 n+1}+3 n} \\ & =\frac{9 n^{2}+3 n+1-(3 n)^{2}}{\sqrt{9 n^{2}+3 n+1}+3 n} \\ & =\frac{3 n+1}{\sqrt{9 n^{2}+3 n+1}+3 n} \, . \end{align*}\] The biggest power of \(n\) in the denominator is \(n\). Therefore we factor out \(n\): \[\begin{align*} a_n & = \sqrt{9 n^{2}+3 n+1}-3 n \\ & =\frac{3 n+1}{\sqrt{9 n^{2}+3 n+1}+3 n} \\ & =\frac{3+ \dfrac{1}{n} }{\sqrt{9+ \dfrac{3}{n} + \dfrac{1}{n^{2}}} + 3 } \,. \end{align*}\] By the Algebra of Limits we have \[ 9+ \frac{3}{n} + \frac{1}{n^{2}} \to 9 + 0 + 0 = 9 \,. \] Therefore we can use Theorem 148 to infer \[ \sqrt{ 9 + \frac{3}{n} + \frac{1}{n^{2}} } \to \sqrt{9} \,. \] By the Algebra of Limits we conclude: \[ a_n = \frac{3+ \dfrac{1}{n} }{\sqrt{9+ \dfrac{3}{n} + \dfrac{1}{n^{2}} }+ 3 } \to \frac{ 3 + 0 }{ \sqrt{9} + 3 } = \frac12 \,. \]

Example 150

Question. Prove that the sequence \[ a_n = \sqrt{9 n^{2}+3 n+1}-2 n \] does not converge.

Solution. We rewrite \(a_n\) as \[\begin{align*} a_n & = \sqrt{9 n^{2}+3 n+1}-2 n \\ & =\frac{ (\sqrt{9 n^{2}+3 n+1} - 2 n) (\sqrt{9 n^{2}+3 n+1}+2 n) }{\sqrt{9 n^{2}+3 n+1}+2 n} \\ & =\frac{9 n^{2}+3 n+1-(2 n)^{2}}{\sqrt{9 n^{2}+3 n+1}+2 n} \\ & =\frac{5 n^{2}+3 n+1}{\sqrt{9 n^{2}+3 n+1}+2 n} \\ & =\frac{5 n + 3 + \dfrac{1}{n} }{\sqrt{9+ \dfrac{3}{n} + \dfrac{1}{ n^2 } } + 2 } \\ & = \frac{b_n}{c_n} \,, \end{align*}\] where we factored \(n\), being it the largest power of \(n\) in the denominator, and defined \[ b_n : = 5 n + 3 + \dfrac{1}{n}\,, \quad c_n := \sqrt{9+ \dfrac{3}{n} + \dfrac{1}{ n^2 } } + 2 \,. \] Note that \[ 9+ \dfrac{3}{n} + \dfrac{1}{ n^2 } \to 9 \] by the Algebra of Limits. Therefore \[ \sqrt{9+ \dfrac{3}{n} + \dfrac{1}{ n^2 }} \to \sqrt{9} = 3 \] by Theorem 148. Hence \[ c_n = \sqrt{9+ \dfrac{3}{n} + \dfrac{1}{ n^2 }} + 2 \to 3 + 2 = 5 \,. \] The numerator \[ b_n = 5 n + 3 + \dfrac{1}{n} \] is instead unbounded. Therefore \((a_n)\) is not convergent, by the Algebra of Limits and the usual contradiction argument.

A.8 Limit Tests

In this section we discuss a number of Tests to determine whether a sequence converges or not. These are known as Limit Tests.

A.8.1 Squeeze Theorem

When a sequence \((a_n)\) oscillates, it is difficult to compute the limit. Examples of terms which produce oscillations are \[ (-1)^n \,, \quad \sin(n) \,, \quad \cos(n) \,. \] In such instance it might be useful to compare \((a_n)\) with other sequences whose limit is known. If we can prove that \((a_n)\) is squeezed between two other sequences with the same limiting value, then we can show that also \((a_n)\) converges to this value.

Theorem 151: Squeeze theorem

Let \(\left(a_{n}\right), \left(b_{n}\right)\) and \(\left(c_{n}\right)\) be sequences in \(\mathbb{R}\). Suppose that \[ b_{n} \leq a_{n} \leq c_{n} \,, \quad \forall \, n \in \mathbb{N}\,, \] and that \[ \lim_{n \rightarrow \infty} b_{n} = \lim_{n \rightarrow \infty} c_{n} = L \, . \] Then \[ \lim_{n \rightarrow \infty} a_{n}= L \, . \]

Proof

Let \(\varepsilon>0\). Since \(b_{n} \to L\) and \(c_n \to L\) , there exist \(N_1, N_2 \in \mathbb{N}\) such that \[\begin{align*} -\varepsilon< b_n - L < \varepsilon\,, \,\, \forall \, n \geq N_1 \,, \\ - \varepsilon< c_n - L < \varepsilon\,, \,\, \forall \, n \geq N_2 \,. \end{align*}\] Set \[ N := \max\{N_1,N_2\} \,. \] Let \(n \geq N\). Using the assumption that \(b_n \leq a_{n} \leq c_{n}\), we get \[ b_n - L \leq a_{n} - L \leq c_{n} - L \,. \] In particular \[ - \varepsilon< b_n - L \leq a_n - L \leq b_n - L < \varepsilon\,. \] The above implies \[ - \varepsilon< a_n - L < \varepsilon\quad \implies \quad \left|a_{n}-L\right| < \varepsilon\,. \]

Example 152

Question. Prove that \[ \lim_{n \to \infty} \, \frac{(-1)^{n}}{n} = 0 \,. \]

Solution. For all \(n \in \mathbb{N}\) we can estimate \[ -1 \leq(-1)^{n} \leq 1 \,. \] Therefore \[ \frac{-1}{n} \leq \frac{(-1)^{n}}{n} \leq \frac{1}{n} \,, \quad \forall \, n \in \mathbb{N}\,. \] Moreover \[ \lim_{n \to \infty} \frac{-1}{n}= -1 \cdot 0=0 \,, \quad \lim_{n \to \infty} \frac{1}{n}=0 \,. \] By the Squeeze Theorem 151 we conclude \[ \lim_{n \to \infty} \frac{(-1)^{n}}{n}=0 \,. \]

Example 153

Question. Prove that \[ \lim_{n \to \infty} \frac{\cos (3 n) + 9 n^{2}}{11 n^{2}+15 \sin (17 n)} = \frac{9}{11} \,. \]

Solution. We know that \[ -1 \leq \cos(x) \leq 1 \,, \quad - 1 \leq \sin(x) \leq 1 \,, \quad \forall \, x \in \mathbb{R}\,. \] Therefore, for all \(n \in \mathbb{N}\) \[ - 1 \leq \cos(3n) \leq 1 \,, \quad -1 \leq \sin(17n) \leq 1 \,. \] We can use the above to estimate the numerator in the given sequence: \[ -1 + 9 n^{2} \leq \cos (3 n)+9 n^{2} \leq 1+ 9 n^{2} \,. \tag{A.31}\] Concerning the denominator, we have \[ 11 n^{2}-15 \leq 11 n^{2}+15 \sin (17 n) \leq 11 n^{2} + 15 \] and therefore \[ \frac{1}{11 n^{2} + 15} \leq \frac{1}{11 n^{2}+15 \sin (17 n)} \leq \frac{1}{11 n^{2}-15} \,. \tag{A.32}\] Putting together (A.31)-(A.32) we obtain \[ \frac{-1 + 9 n^{2}}{11 n^{2} + 15} \leq \frac{\cos (3 n)+9 n^{2}}{11 n^{2}+15 \sin (17 n)} \leq \frac{1+ 9 n^{2}}{11 n^{2}-15} \,. \] By the Algebra of Limits we infer \[ \frac{-1+9 n^{2}}{11 n^{2}+15}=\frac{-\dfrac{1}{n^{2}} + 9}{11 + \dfrac{15}{n^{2}}} \to \frac{0+9}{11+0}=\frac{9}{11} \] and \[ \frac{1+9 n^{2}}{11 n^{2} - 15}=\frac{ \dfrac{1}{n^{2}} + 9}{ 11 - \dfrac{15}{n^{2}}} \to \frac{0+9}{11+0}=\frac{9}{11} \,. \] Applying the Squeeze Theorem 151 we conclude \[ \lim_{n \to \infty} \frac{\cos (3 n)+ 9 n^{2}}{11 n^{2}+15 \sin (17 n)}=\frac{9}{11} \,. \]

Warning

Suppose that the sequences \((a_n), (b_n), (c_n)\) satisfy \[ b_n \leq a_n \leq c_n \,, \quad \forall n \in \mathbb{N}\,, \] and \[ b_n \to L_1 \,, \quad c_n \to L_2 \,, \quad L_1 \neq L_2 \,. \] In general, we cannot conclude that \(a_n\) converges.

Example 154

Consider the sequence \[ a_n = \left( 1 + \frac{1}{n} \right) (-1)^n \,. \] For all \(n \in \mathbb{N}\) we can bound \[ - 1 - \frac{1}{n} \leq \left( 1 + \frac{1}{n} \right) (-1)^n \leq 1 + \frac{1}{n} \,. \] However \[ - 1 - \frac{1}{n} \longrightarrow - 1 - 0 = -1 \] and \[ 1 + \frac{1}{n} \longrightarrow 1 + 0 = 1 \,. \] Since \[ - 1 \neq 1 \,, \] we cannot apply the Squeeze Theorem 151 to conclude convergence of \((a_n)\). Indeed, \((a_n)\) is a divergent sequence.

Proof. Suppose by contradiction that \(a_n \to a\). We have \[ a_n = (-1)^n + \frac{(-1)^n}{n} = b_n + c_n \] where \[ b_n := (-1)^n \,, \quad c_n := \frac{(-1)^n}{n} \,. \] We have seen in Example 153 that \(c_n \to 0\). Therefore, by the Algebra of Limits, we have \[ b_n = a_n - c_n \longrightarrow a - 0 = a \,. \] However, Theorem 133 says that the sequence \(b_n = (-1)^n\) diverges. Contradiction. Hence \((a_n)\) diverges.

A.8.2 Geometric sequences

Definition 155

A sequence \(\left(a_{n}\right)\) is called a geometric sequence if \[ a_{n}=x^{n} \,, \] for some \(x \in \mathbb{R}\).

The value of \(|x|\) determines whether or not a geometric sequence converges, as shown in the following theorem.

Theorem 156: Geometric Sequence Test

Let \(x \in \mathbb{R}\) and let \(\left(a_{n}\right)\) be the sequence defined by \[ a_{n}:=x^{n} \,. \] We have:

1. If \(|x|<1\), then \[ \lim_{n \to \infty} a_{n} = 0 \,. \]

2. If \(|x|>1\), then sequence \(\left(a_{n}\right)\) is unbounded, and hence divergent.

Warning

The Geometric Sequence Test in Theorem 156 does not address the case \[ |x|=1 \,. \] This is because, in this case, the sequence \[ a_n = x^n \] might converge or diverge, depending on the value of \(x\). Indeed, \[ |x| = 1 \quad \implies \quad x = \pm 1 \,. \] We therefore have two cases:

• \(x = 1\): Then \[ a_n = 1^n = 1 \] so that \(a_n \to 1\) and \((a_n)\) is convergent.

• \(x=-1\): Then \[ a_n = x^n = (-1)^n \] which is divergent by Theorem 133.

To prove Theorem 156 we need the following inequality, known as Bernoulli’s inequality.

Lemma 157: Bernoulli’s inequality

Let \(x \in \mathbb{R}\) with \(x>-1\). Then \[ (1+x)^{n} \geq 1+n x \,, \quad \forall \, n \in \mathbb{N}\,. \tag{A.33}\]

Proof

Let \(x \in \mathbb{R}, x>-1\). We prove the statement by induction:

• Base case: (A.33) holds with equality when \(n=1\).

• Induction hypothesis: Let \(k \in \mathbb{N}\) and suppose that (A.33) holds for \(n=k\), i.e., \[ (1+x)^{k} \geq 1+k x \,. \] Then \[\begin{align*} (1+x)^{k+1} & = (1+x)^{k}(1+x) \\ & \geq(1+k x)(1+x) \\ & =1+k x+x+k x^{2} \\ & \geq 1+(k+1) x \,, \end{align*}\] where we used that \(kx^2 \geq 0\). Then (A.33) holds for \(n=k+1\).

By induction we conclude (A.33).

We are ready to prove Theorem 156.

Proof: Proof of Theorem 156

Part 1. The case \(|x|<1\).

If \(x=0\), then \[ a_n = x^n = 0 \] so that \(a_n \to 0\). Hence assume \(x \neq 0\). We need to prove that \[ \forall \, \varepsilon> 0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \,\, |x^n - 0| < \varepsilon\,. \] Let \(\varepsilon>0\). We have \[ |x| < 1 \quad \implies \quad \frac{1}{|x|} > 1 \,. \] Therefore \[ |x|= \frac{1}{1+u} \,, \quad u:=\frac{1}{|x|} - 1 > 0 \,. \] Let \(N \in \mathbb{N}\) be such that \[ N > \frac{1}{\varepsilon u} \,, \] so that \[ \frac{1}{N u} < \varepsilon\,. \] Let \(n \geq N\). Then \[\begin{align*} \left|x^{n}-0\right| & =|x|^{n} \\ & = \left(\frac{1}{1+u}\right)^{n} \\ & =\frac{1}{(1+u)^{n}} \\ & \leq \frac{1}{1+n u} \\ & \leq \frac{1}{n u} \\ & \leq \frac{1}{N u} \\ & < \varepsilon\,, \end{align*}\] where we used Bernoulli’s inequality (A.33) in the first inequality.

Part 2. The case \(|x|>1\).

To prove that \((a_n)\) does not converge, we prove that it is unbounded. This means showing that \[ \forall \, M > 0 \,, \, \exists n \in \mathbb{N}\, \text{ s.t. } \, \left| a_{n}\right| >M \,. \] Let \(M > 0\). We have two cases:

• \(0 < M \leq 1\): Choose \(n=1\). Then \[ \left|a_{1}\right|=|x|>1 \geq M \,. \]

• \(M>1\): Choose \(n \in \mathbb{N}\) such that \[ n> \frac{\log M}{\log |x|} \,. \] Note that \(\log |x|>0\) since \(|x|>1\). Therefore \[\begin{align*} n>\frac{\log M}{\log |x|} & \iff n \log |x|>\log M \\ & \iff \log |x|^n>\log M \\ & \iff |x|^{n}>M \,. \end{align*}\] Then \[ \left|a_{n}\right|=\left|x^{n}\right|=|x|^{n} > M \,. \]

Hence \((a_n)\) is unbounded. By Corollary 138 we conclude that \((a_n)\) is divergent.

Example 158

We can apply Theorem 156 to prove convergence or divergence for the following sequences.

1. We have \[ \left(\frac{1}{2}\right)^{n} \longrightarrow 0 \] since \[ \left|\frac{1}{2}\right|=\frac{1}{2}<1 \,. \]

2. We have \[ \left(\frac{-1}{2}\right)^{n} \longrightarrow 0 \] since \[ \left|\frac{-1}{2}\right|=\frac{1}{2}<1 \,. \]

3. The sequence \[ a_n = \left(\frac{-3}{2}\right)^{n} \] does not converge, since \[ \left|\frac{-3}{2}\right|=\frac{3}{2}>1 \,. \]

4. As \(n \rightarrow \infty\), \[ \frac{3^{n}}{(-5)^{n}}=\left(-\frac{3}{5}\right)^{n} \longrightarrow 0 \] since \[ \left|-\frac{3}{5}\right|=\frac{3}{5}<1 \,. \]

5. The sequence \[ a_{n}=\frac{(-7)^{n}}{2^{2 n}} \] does not converge, since \[ \frac{(-7)^{n}}{2^{2 n}}=\frac{(-7)^{n}}{\left(2^{2}\right)^{n}}=\left(-\frac{7}{4}\right)^{n} \] and \[ \left|-\frac{7}{4}\right|=\frac{7}{4}>1 \,. \]

A.8.3 Ratio Test

Theorem 159: Ratio Test

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{R}\) such that \[ a_{n} \neq 0 \,, \quad \forall \, n \in \mathbb{N}\,. \]

1. Suppose that the following limit exists: \[ L:=\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \,. \] Then,

   • If \(L<1\) we have \[ \lim_{n \to\infty} a_{n}=0 \,. \]

   • If \(L>1\), the sequence \(\left(a_{n}\right)\) is unbounded, and hence does not converge.

2. Suppose that there exists \(N \in \mathbb{N}\) and \(L>1\) such that \[ \left|\frac{a_{n+1}}{a_{n}}\right| \geq L \,, \quad \forall \, n \geq N \,. \] Then the sequence \(\left(a_{n}\right)\) is unbounded, and hence does not converge.

Proof

Define the sequence \(b_{n}=\left|a_{n}\right|\). Then,

\[ \left|\frac{a_{n+1}}{a_{n}}\right|=\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}=\frac{b_{n+1}}{b_{n}} \]

Part 1. Suppose that there exists the limit \[ L:=\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \,. \] Therefore \[ \lim_{n \to \infty} \frac{b_{n+1}}{b_{n}} = L \,. \tag{A.34}\]

• \(L<1\): Choose \(r>0\) such that \[ L<r<1 \,. \] Set \[ \varepsilon:= r-L \] By the convergence at (A.34) there exists \(N \in \mathbb{N}\) such that \[ \left|\frac{b_{n+1}}{b_{n}}-L\right| < \varepsilon= r - L \,, \quad \forall \, n \geq N \,. \] In particular \[ \frac{b_{n+1}}{b_{n}}-L < r-L \,, \quad \forall \, n \geq N \,, \] which implies \[ b_{n+1} < r \,{b_{n}} \,, \quad \forall \, n \geq N \,. \tag{A.35}\] Let \(n \geq N\), we can use (A.35) recursively and obtain \[ 0 \leq b_{n} < \, r b_{n-1} < \ldots < r^{n-N} \, b_{N} = r^{n} \, \frac{b_{N}}{r^{N}} \,. \] In particular, we have proven that \[ 0 \leq b_{n} < r^{n} \, \frac{b_{N}}{r^{N}} \,, \quad \forall \, n \in \mathbb{N}\,. \tag{A.36}\] Since \(|r|<1\), by the Geometric Sequence Test Theorem 156 we infer \[ r^{n} \rightarrow 0 \,. \] The Algebra of Limits the yields \[ r^{n} \, \frac{b_{N}}{r^{N}} \rightarrow 0 \,. \] By the Squeeze Theorem 151 applied to (A.36), it follows that \[ b_{n} = |a_n| \to 0 \,. \] Since \[ -\left|a_{n}\right| \leq a_{n} \leq\left|a_{n}\right| \,, \] and \[ -|a_n| \to 0 \,, \quad |a_n| \to 0 \,, \] we can again apply the Squeeze Theorem 151 to infer \[ a_{n} \rightarrow 0 \,. \]

• \(L>1\): Choose \(r>0\) such that \[ 1<r<L \,. \] Define \[ \varepsilon:= L - r > 0 \,. \] By the convergence (A.34), there exists \(N \in \mathbb{N}\) such that \[ \left|\frac{b_{n+1}}{b_{n}}-L\right| < \varepsilon= L - r \,, \quad \forall \, n \geq N \,. \] In particular, \[ -(L-r) < \frac{b_{n+1}}{b_{n}}-L \,, \quad \forall \, n \geq N \,, \] which implies \[ b_{n+1} > r \, b_{n} \,, \quad \forall \, n \geq N \,. \tag{A.37}\] Let \(n \geq N\). Applying (A.37) recursively we get \[ b_{n} > r^{n-N} \, b_{N} = r^{n} \, \frac{b_{N}}{r^{N}} \,, \quad \forall \, n \geq N \,. \tag{A.38}\] Since \(|r|>1\), by the Geometric Sequence Test we have that the sequence \[ (r^n) \] is unbounded. Therefore also the right hand side of (A.38) is unbounded, proving that \((b_n)\) is unbounded. Since \[ b_n = |a_n|\,, \] we conclude that \((a_n)\) is unbounded. By Corollary 138 we conclude that \((a_n)\) does not converge.

Part 2. Suppose that there exists \(N \in \mathbb{N}\) and \(L>1\) such that \[ \left|\frac{a_{n+1}}{a_{n}}\right| \geq L \,, \quad \forall \, n \geq N \,. \] Since \(b_n = |a_n|\), we infer \[ b_{n+1} \geq L \, b_n \,, \quad \forall \, n \geq N \,. \] Arguing as above, we obtain \[ b_n \geq L^n \, \frac{b_N}{L^N} \,, \quad \forall \, n \geq N \,. \] Since \(L>1\), we have that the sequence \[ L^n \, \frac{b_N}{L^N} \] is unbounded, by the Geometric Sequence Test. Hence also \((b_n)\) is unbounded, from which we conclude that \((a_n)\) is unbounded. By Corollary 138 we conclude that \((a_n)\) does not converge.

Let us apply the Ratio Test to some concrete examples.

Example 160

Question. Let \[ a_{n}=\frac{3^{n}}{n !} \,, \] where we recall that \(n!\) (pronounced \(n\) factorial) is defined by \[ n! := n \cdot (n-1) \cdot (n-2) \cdot \ldots \cdot 3 \cdot 2 \cdot 1 \,. \] Prove that \[ \lim_{n \to \infty} a_n = 0 \,. \]

Solution. We have \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & = \dfrac{\left( \dfrac{3^{n+1}}{(n+1) !} \right) }{ \left( \dfrac{3^{n}}{n !} \right) } \\ & = \frac{3^{n+1}}{3^{n}} \, \frac{n !}{(n+1) !} \\ & = \frac{3 \cdot 3^n}{3^n} \, \frac{n!}{(n+1) n!} \\ & =\frac{3}{n+1} \longrightarrow L = 0 \,. \end{align*}\] Hence, \(L=0<1\) so \(a_{n} \to 0\) by the Ratio Test in Theorem 159.

Example 161

Question. Consider the sequence \[ a_{n}=\frac{n ! \cdot 3^{n}}{\sqrt{(2 n) !}} \,. \] Prove that \((a_n)\) is divergent.

Solution. We have \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & =\frac{(n+1) ! \cdot 3^{n+1}}{\sqrt{(2(n+1)) !}} \frac{\sqrt{(2 n) !}}{n ! \cdot 3^{n}} \\ & =\frac{(n+1) !}{n !} \cdot \frac{3^{n+1}}{3^{n}} \cdot \frac{\sqrt{(2 n) !}}{\sqrt{(2(n+1)) !}} \end{align*}\] For the first two fractions we have \[ \frac{(n+1) !}{n !} \cdot \frac{3^{n+1}}{3^{n}} = 3(n+1) \,, \] while for the third fraction \[\begin{align*} \frac{\sqrt{(2 n) !}}{\sqrt{(2(n+1)) !}} & =\sqrt{\frac{(2 n) !}{(2 n+2) !}} \\ & = \sqrt{\frac{ (2n)! }{ (2n+2) \cdot (2n+1) \cdot (2n)! }} \\ & = \frac{1}{\sqrt{(2 n+1)(2 n+2)}} \,. \end{align*}\] Therefore, using the Algebra of Limits, \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & = \frac{3(n+1)}{\sqrt{(2 n+1)(2 n+2)}}\\ & = \frac{3n \left(1+ \dfrac{1}{n} \right)}{\sqrt{n^2 \left( 2 + \dfrac{1}{n}\right) \left(2 + \dfrac{2}{n} \right)}} \\ & = \frac{3 \left(1+ \dfrac{1}{n} \right)}{\sqrt{ \left( 2 + \dfrac{1}{n}\right) \left(2 + \dfrac{2}{n} \right)}} \longrightarrow \frac{3}{\sqrt{4}} = \frac{3}{2} > 1 \,. \end{align*}\] By the Ratio Test we conclude that \((a_n)\) is divergent.

Example 162

Question. Prove that the following sequence is divergent \[ a_{n}=\frac{n !}{100^{n}} \,. \]

Solution. We have \[ \left|\frac{a_{n+1}}{a_{n}}\right| = \frac{100^{n}}{100^{n+1}} \frac{(n+1) !}{n !} =\frac{n+1}{100} \,. \] Choose \(N=101\). Then for all \(n \geq N\), \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & = \frac{n+1}{100} \\ & \geq \frac{N+1}{100} \\ & = \frac{101}{100} > 1 \,. \end{align*}\] Hence \(a_{n}\) is divergent by the Ratio Test.

Warning

The Ratio Test in Theorem 159 does not address the case \[ L=1 \,. \] This is because, in this case, the sequence \((a_n)\) might converge or diverge.

For example:

• Define the sequence \[ a_n = \frac1n \,. \] We have \[ \left|\frac{a_{n+1}}{a_{n}}\right| = \frac{n}{n+1} \rightarrow L = 1 \,. \] Hence we cannot apply the Ratio Test. However we know that \[ \lim_{n \to \infty} \, \frac1n = 0 \,. \]

• Consider the sequence \[ a_n = n \,. \] We have \[ \frac{|a_{n+1}|}{|a_n|} = \frac{n+1}{n} \rightarrow L = 1 \,. \] Hence we cannot apply the Ratio Test. However we know that \((a_n)\) is unbounded, and thus divergent.

If the sequence \((a_n)\) is geometric, the Ratio Test of Theorem 159 will give the same answer as the Geometric Sequence Test of Theorem 156. This is the content of the following remark.

Remark 163

Let \(x \in \mathbb{R}\) and define the geometric sequence \[ a_{n}=x^{n} \,. \] Then \[ \left|\frac{a_{n+1}}{a_{n}}\right| = \frac{\left|x^{n+1}\right|}{\left|x^{n}\right|} = \frac{|x|^{n+1}}{|x|^{n}} =|x| \rightarrow |x| \,. \] Hence:

• If \(|x|<1\), the sequence \((a_n)\) converges by the Ratio Test
• If \(|x|>1\), the sequence \((a_n)\) diverges by the Ratio Test.
• If \(|x|=1\), the sequence \((a_n)\) might be convergent or divergent.

These results are in agreement with the Geometric Sequence Test.

A.9 Monotone sequences

We have seen in Theorem 136 that convergent sequences are bounded. We noted that the converse statement is not true. For example the sequence \[ a_n = (-1)^n \] is bounded but not convergent, as shown in Theorem 133. On the other hand, if a bounded sequence is monotone, then it is convergent.

Definition 164: Monotone sequence

Let \((a_n)\) be a real sequence. We say that:

1. \((a_n)\) is increasing if \[ a_n \leq a_{n+1} \,, \quad \forall \, n \geq N \,. \]

2. \((a_n)\) is decreasing if \[ a_n \geq a_{n+1} \,, \quad \forall \, n \geq N \,. \]

3. \((a_n)\) is monotone if it is either increasing or decreasing.

Example 165

Question. Prove that the following sequence is increasing \[ a_n = \frac{n-1}{n} \,. \]

Solution. We have \[ a_{n+1} = \frac{n}{n+1} > \frac{n-1}{n} = a_n \,, \] where the inequality holds because \[\begin{align*} \frac{n}{n+1} > \frac{n-1}{n} \quad & \iff \quad n^2 > (n-1)(n+1) \quad \\ & \iff \quad n^2 > n^2 - 1 \\ & \iff \quad 0 > - 1 \end{align*}\]

Example 166

Question. Prove that the following sequence is decreasing \[ a_n = \frac1n \,. \]

Solution. We have \[ a_n = \frac1n > \frac{1}{n+1} = a_{n+1} \,, \] concluding.

The main result about monotone sequences is the Monotone Convergence Theorem.

Theorem 167: Monotone Convergence Theorem

Let \((a_n)\) be a sequence in \(\mathbb{R}\). Suppose that \((a_n)\) is bounded and monotone. Then \((a_n)\) converges.

Proof

Assume \((a_n)\) is bounded and monotone. Since \((a_n)\) is bounded, the set \[ A:=\{ a_n \, \colon \,n \in \mathbb{N}\} \subseteq \mathbb{R} \] is bounded below and above. By the Axiom of Completeness of \(\mathbb{R}\) there exist \(i,s \in \mathbb{R}\) such that \[ i = \inf A \,, \quad s = \sup A \,. \]

We have two cases:

1. \((a_n)\) is increasing: We are going to prove that \[ \lim_{n \to \infty} a_n = s \,. \] Equivalently, we need to prove that \[ \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \,\, |a_n - s| < \varepsilon\,. \tag{A.39}\] Let \(\varepsilon> 0\). Since \(s\) is the smallest upper bound for \(A\), this means \[ s - \varepsilon \] is not an upper bound. Therefore there exists \(N \in \mathbb{N}\) such that \[ s - \varepsilon< a_N \,. \tag{A.40}\] Let \(n \geq N\). Since \(a_n\) is increasing, we have \[ a_N \leq a_n \,, \quad \forall \, n \geq N \,. \tag{A.41}\] Moreover \(s\) is the supremum of \(A\), so that \[ a_n \leq s < s + \varepsilon\,, \quad \forall \, n \in \mathbb{N}\,. \tag{A.42}\] Putting together estimates (A.40)-(A.41)-(A.42) we get \[ s - \varepsilon< a_N \leq a_n \leq s < s + \varepsilon\,, \quad \forall \, n \geq N \,. \] The above implies \[ s - \varepsilon< a_n < s + \varepsilon\,, \quad \forall \, n \geq N \,, \] which is equivalent to (A.39).

2. \((a_n)\) is decreasing: With a similar proof, one can show that \[ \lim_{n \to \infty} a_n = i \,. \] This is left as an exercise.

A.9.1 Example: Euler’s Number

As an application of the Monotone Convergence Theorem we can give a formal definition for the Euler’s Number \[ e = 2.71828182845904523536 \dots \]

Theorem 168

Consider the sequence \[ a_n = \left( 1 + \frac{1}{n} \right)^n \,. \]

We have that:

1. \((a_n)\) is monotone increasing,
2. \((a_n)\) is bounded.

In particular \((a_n)\) is convergent.

Proof

Part 1. We prove that \((a_n)\) is increasing \[ a_{n} \geq a_{n-1} \,, \quad \forall \, n \in \mathbb{N}\,, \] which by definition is equivalent to \[ \left( 1 + \frac{1}{n} \right)^n \geq \left( 1 + \frac{1}{n - 1} \right)^{n-1} \,, \quad \forall \, n \in \mathbb{N}\,. \] Summing the fractions we get \[ \left( \frac{n+1}{n} \right)^n \geq \left( \frac{n}{n-1} \right)^{n-1} \,. \] Multiplying by \(((n-1)/n)^n\) we obtain \[ \left( \frac{n-1}{n} \right)^n \left( \frac{n+1}{n} \right)^n \geq \frac{n-1}{n} \,, \] which simplifies to \[ \left( 1 - \frac{1}{n^2} \right)^n \geq 1 - \frac1n \,, \quad \forall \, n \in \mathbb{N}\,. \tag{A.43}\] Therefore \((a_n)\) is increasing if and only if (A.43) holds. Recall Bernoulli’s inequality from Lemma 157: For \(x \in \mathbb{R}\), \(x>-1\), it holds \[ (1 + x )^n \geq 1 + nx \,, \quad \forall \, n \in \mathbb{N}\,. \] Appliying Bernoulli’s inequality with \[ x = -\frac{1}{n^2} \] yields \[ \left( 1 - \frac{1}{n^2} \right)^n \geq 1 + n \left( -\frac{1}{n^2} \right) = 1 - \frac{1}{n} \,, \] which is exactly (A.43). Then \((a_n)\) is increasing.

Part 2. We have to prove that \((a_n)\) is bounded, that is, that there exists \(M > 0\) such that \[ |a_n| \leq M \,, \quad \forall \, n \in \mathbb{N}\,. \] To this end, introduce the sequence \((b_n)\) by setting \[ b_n := \left( 1 + \frac1n \right)^{n+1} \,. \] The sequence \((b_n)\) is decreasing.

To prove \((b_n)\) is decreasing, we need to show that \[ b_{n-1} \geq b_n \,, \quad \forall \, n \in \mathbb{N}\,. \] By definition of \(b_n\), the above reads \[ \left( 1 + \frac{1}{n-1} \right)^{n} \geq \left( 1 + \frac{1}{n} \right)^{n+1} \,, \quad \forall \, n \in \mathbb{N}\,. \] Summing the terms inside the brackets, the above is equivalent to \[ \left( \frac{n}{n-1} \right)^{n} \geq \left( \frac{n+1}{n} \right)^{n} \left( \frac{n+1}{n} \right) \,. \] Multiplying by \((n/(n+1))^n\) we get \[ \left( \frac{n^2}{n^2 - 1} \right)^{n} \geq \left( \frac{n+1}{n} \right) \,. \] The above is equivalent to \[ \left( 1 + \frac{1}{n^2 - 1} \right)^{n} \geq \left( 1 + \frac{1}{n} \right) \,. \tag{A.44}\] Therefore \((b_n)\) is decreasing if and only if (A.44) holds for all \(n \in \mathbb{N}\). By choosing \[ x = \frac{1}{n^2 - 1} \] in Bernoulli’s inequality, we obtain \[\begin{align*} \left( 1 + \frac{1}{n^2 - 1} \right)^{n} & \geq 1 + n \left( \frac{1}{n^2 - 1} \right) \\ & = 1 + \frac{n}{n^2 - 1} \\ & \geq 1 + \frac1n \,, \end{align*}\] where in the last inequality we used that \[ \frac{n}{n^2 - 1} > \frac1n \,, \] which holds, being equivalent to \(n^2 > n^2 - 1\). We have therefore proven (A.44), and hence \((b_n)\) is decreasing.

We now observe that For all \(n \in \mathbb{N}\) \[\begin{align*} b_n & = \left( 1 + \frac1n \right)^{n+1} \\ & = \left( 1 + \frac1n \right)^n \left( 1 + \frac1n \right) \\ & = a_n \left( 1 + \frac1n \right) \\ & > a_n \,. \end{align*}\] Since \((a_n)\) is increasing and \((b_n)\) is decreasing, in particular \[ a_n \geq a_1 \,, \quad b_n \leq b_1 \,. \] Therefore \[ a_1 \leq a_n < b_n \leq b_1 \,, \quad \forall \, n \in \mathbb{N}\,. \] We compute \[ a_1 = 2 \,, \quad b_1 = 4 \,, \] from which we get \[ 2 \leq a_n \leq 4 \,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore \[ |a_n| \leq 4 \,, \quad \forall \, n \in \mathbb{N}\,, \] showing that \((a_n)\) is bounded.

Part 3. The sequence \((a_n)\) is increasing and bounded above. Therefore \((a_n)\) is convergent by the Monotone Convergence Theorem 167.

Thanks to Theorem 168 we can define the Euler’s Number \(e\).

Definition 169: Euler’s Number

The Euler’s number is defined as \[ e := \lim_{n \to \infty } \, \left( 1 + \frac{1}{n} \right)^n \,. \]

Setting \(n=1000\) in the formula for \((a_n)\), we get an approximation of \(e\): \[ e \approx a_{1000} = 2.7169 \,. \]

A.10 Some important limits

In this section we investigate limits of some sequences to which the Limit Tests do not apply.

Theorem 170

Let \(x \in \mathbb{R}\), with \(x > 0\). Then \[ \lim_{n \to \infty} \sqrt[n]{x} = 1 \,. \]

Proof

Step 1. Assume \(x \geq 1\). In this case \[ \sqrt[n]{x} \geq 1 \,. \] Define \[ b_n := \sqrt[n]{x} - 1 \,, \] so that \(b_n \geq 0\). By Bernoulli’s Inequality we have \[ x = (1 + b_n)^n \geq 1 + n b_n \,. \] Therefore \[ 0 \leq b_n \leq \frac{x-1}{n} \,. \] Since \[ \frac{x-1}{n} \longrightarrow 0 \,, \] by the Squeeze Theorem we infer \(b_n \to 0\), and hence \[ \sqrt[n]{x} = 1 + b_n \longrightarrow 1 + 0 = 1\,, \] by the Algebra of Limits.

Step 2. Assume \(0< x < 1\). In this case \[ \frac{1}{x} > 1 \,. \] Therefore \[ \lim_{n \to \infty} \, \sqrt[n]{ 1/x } = 1 \,. \] by Step 1. Therefore \[ \sqrt[n]{x} = \frac{1}{\sqrt[n]{ 1/x }} \longrightarrow \frac{1}{1} = 1\,, \] by the Algebra of Limits.

Theorem 171

Let \((a_n)\) be a sequence such that \(a_n \to 0\). Then \[ \sin(a_n) \to 0 \,, \quad \cos(a_n) \to 1 \,. \]

Proof

Assume that \(a_n \to 0\) and set \[ \varepsilon:= \frac{\pi}{2} \,. \] By the convergence \(a_m \to 0\) there exists \(N \in \mathbb{N}\) such that \[ |a_n| < \varepsilon= \frac{\pi}{2} \, \quad \forall \, n \geq N \,. \tag{A.45}\]

Step 1. We prove that \[ \sin(a_n) \to 0 \,. \] By elementary trigonometry we have \[ 0 \leq |\sin (x)| = \sin |x| \leq |x| \,, \quad \forall \, x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \,. \] Therefore, since (A.45) holds, we can substitute \(x=a_n\) in the above inequality to get \[ 0 \leq |\sin (a_n)| \leq |a_n| \,, \quad \forall \, n \geq \mathbb{N}\,. \] Since \(a_n \to 0\), we also have \(|a_n|\to 0\). Therefore \(|\sin (a_n)| \to 0\) by the Squeeze Theorem. This immediately implies \(\sin(a_n) \to 0\).

Step 2. We prove that \[ \cos(a_n) \to 1 \,. \] Inverting the relation \[ \cos^2(x) + \sin^2 (x) = 1 \,, \] we obtain \[ \cos(x) = \pm \sqrt{ 1 - \sin^2 (x) } \,. \] We have that \(\cos(x) \geq 0\) for \(-\pi/2 \leq x \leq \pi/2\). Thus \[ \cos(x) = \sqrt{ 1 - \sin^2 (x) } \,, \quad \forall \, x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \,. \] Since (A.45) holds, we can set \(x=a_n\) in the above inequality and obtain \[ \cos(a_n) = \sqrt{ 1 - \sin^2 (a_n) } \,, \quad \forall \, n \geq N \,. \] By Step 1 we know that \(\sin(a_n) \to 0\). Therefore, by the Algebra of Limits, \[ 1 - \sin^2 (a_n) \longrightarrow 1 - 0 \cdot 0 = 1 \,. \] Using Theorem 148 we have \[ \cos(a_n) = \sqrt{1 - \sin^2 (a_n)} \longrightarrow \sqrt{1} = 1 \,, \] concluding the proof.

Theorem 172

Suppose \((a_n)\) is such that \(a_n \to 0\) and \[ a_n \neq 0 \,, \quad \forall \, n \in \mathbb{N}\, . \] Then \[ \lim_{n \to \infty} \frac{\sin(a_n)}{a_n} = 1 \,. \]

Proof

The following elementary trigonometric inequality holds: \[ \sin (x) < x < \tan (x) \,, \quad \forall \, x \in \left[0 ,\frac{\pi}{2} \right] \,. \] Note that \(\sin x >0\) for \(0 < x < \pi/2\). Therefore we can divide the above inequality by \(\sin(x)\) and take the reciprocals to get \[ \cos(x) < \frac{\sin (x)}{x} < 1 \,, \quad \forall \, x \in \Big( 0 ,\frac{\pi}{2} \Big] \,. \] If \(-\pi/2<x<0\), we can apply the above inequality to \(-x\) to obtain \[ \cos(-x) < \frac{\sin(-x)}{-x} < 1 \,. \] Recalling that \(\cos(-x) = \cos(x)\) and \(\sin(-x)=-\sin(x)\), we get \[ \cos(x) < \frac{\sin(x)}{x} < 1 \,, \quad \forall \, x \in \Big( -\frac{\pi}{2}, 0 \Big] \,. \] Thus \[ \cos(x) < \frac{\sin(x)}{x} < 1 \,, \quad \forall \, x \in \left[-\frac{\pi}{2}, \frac{\pi}{2} \right] \smallsetminus \{0\} \,. \tag{A.46}\] Let \[ \varepsilon:= \frac{\pi}{2} \,. \] Since \(a_n \to 0\), there exists \(N \in \mathbb{N}\) such that \[ |a_n| < \varepsilon= \frac{\pi}{2} \,, \quad \forall \, n \geq N \,. \] Since \(a_n \neq 0\) by assumption, the above shows that \[ a_n \in \Big[ -\frac{\pi}{2}, \frac{\pi}{2} \Big] \smallsetminus \{0\} \,, \quad \forall \, n \geq \mathbb{N}\,. \] Therefore we can substitute \(x=a_n\) into (A.46) to get \[ \cos(a_n) < \frac{\sin(a_n)}{a_n} < 1 \,, \quad \forall \, n \geq N \,. \] We have \[ \cos(a_n) \to 1 \] by Theorem 171. By the Squeeze Theorem we conclude that \[ \lim_{n \to \infty} \frac{\sin(a_n)}{a_n} = 1 \,. \]

Warning

You might be tempted to apply L’Hôpital’s rule (which we did not cover in these Lecture Notes) to compute \[ \lim_{x \to 0} \frac{\sin(x)}{x} \,. \] This would yield the correct limit \[ \lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{(\sin(x))'}{(x)'} = \lim_{x \to 0} \cos(x) = 1 \,. \] However this is a circular argument, since the derivative of \(\sin(x)\) at \(x=0\) is defined as the limit \[ \lim_{x \to 0} \frac{\sin(x)}{x} \,. \]

Theorem 173

Suppose \((a_n)\) is such that \(a_n \to 0\) and \[ a_n \neq 0 \,, \quad \forall \, n \in \mathbb{N}\, . \] Then \[ \lim_{n \to \infty} \frac{1 - \cos(a_n)}{(a_n)^2} = \frac{1}{2} \,, \quad \lim_{n \to \infty} \frac{1 - \cos(a_n)}{a_n} = 0 \, . \]

Proof

Step 1. By Theorem 171 and Theorem 172, we have \[ \cos(a_n) \to 1 \,, \quad \frac{\sin(a_n)}{a_n} \to 1 \,. \] Therefore \[\begin{align*} \frac{1 - \cos(a_n)}{(a_n)^2} & = \frac{1 - \cos(a_n)}{(a_n)^2} \, \frac{1 + \cos(a_n)}{1 + \cos(a_n)} \\ & = \frac{1 - \cos^2(a_n)}{(a_n)^2} \, \frac{1}{1 + \cos(a_n)} \\ & = \left( \frac{\sin(a_n)}{a_n} \right)^2 \, \frac{1}{1 + \cos(a_n)} \longrightarrow 1 \cdot \frac{1}{1 + 1} = \frac12 \,, \end{align*}\] where in the last line we use the Algebra of Limits.

Step 2. We have \[ \frac{1 - \cos(a_n)}{a_n} = a_n \cdot \frac{1 - \cos(a_n)}{(a_n)^2} \longrightarrow 0 \cdot \frac{1}{2} = 0 \,, \] using Step 1 and the Algebra of Limits.

Example 174

Question. Prove that \[ \lim_{n \to \infty} \, n \sin \left( \frac{1}{n} \right) = 1 \,. \tag{A.47}\]

Solution. This is because \[ n \sin \left( \frac{1}{n} \right) = \frac{ \sin \left( \dfrac{1}{n} \right) }{ \dfrac{1}{n} } \longrightarrow 1 \,, \] by Theorem 172 with \(a_n = 1/n\).

Example 175

Question. Prove that \[ \lim_{n \to \infty} \, n^2 \left( 1 - \cos \left( \dfrac{1}{n} \right) \right) = \frac12 \,. \tag{A.48}\]

Solution. Indeed, \[ n^2 \left( 1 - \cos \left( \frac{1}{n} \right)\right) = \dfrac{1 - \cos \left( \dfrac{1}{n} \right)}{\dfrac{1}{n^2}} \longrightarrow \frac12 \,, \] by applying Theorem 173 with \(a_n = 1/n\).

Example 176

Question. Prove that \[ \lim_{n \to \infty} \, \frac{n \left( 1- \cos \left( \dfrac{1}{n} \right) \right) }{ \sin \left( \dfrac{1}{n} \right) } = \frac12 \,. \]

Solution. Using (A.48)-(A.47) and the Algebra of Limits \[\begin{align*} \frac{n \left( 1- \cos \left( \dfrac{1}{n} \right) \right) }{ \sin \left( \dfrac{1}{n} \right) } & = \frac{n^2 \left( 1- \cos \left( \dfrac{1}{n} \right) \right) }{ n \sin \left( \dfrac{1}{n} \right) } \\ & \longrightarrow \frac{1/2}{1} = \frac12 \,. \end{align*}\]

Example 177

Question. Prove that \[ \lim_{n \to \infty} \, n \cos \left( \frac{2}{n} \right) \sin \left( \frac{2}{n} \right) = 2 \,. \]

Solution. We have \[ \cos \left( \frac{2}{n} \right) \longrightarrow 1 \,, \] by Theorem 171 applied with \(a_n = 2/n\). Moreover \[ \frac{\sin \left( \dfrac{2}{n} \right)}{\dfrac{2}{n}} \longrightarrow 1 \,, \] by Theorem 172 applied with \(a_n = 2/n\). Therefore \[\begin{align*} n \cos \left( \frac{2}{n} \right) \sin \left( \frac{2}{n} \right) & = 2 \cdot \cos \left( \frac{2}{n} \right) \cdot \frac{\sin \left( \dfrac{2}{n} \right)}{\dfrac{2}{n}} \\ & \longrightarrow 2 \cdot 1 \cdot 1 = 2 \,, \end{align*}\] where we used the Algebra of Limits.

Example 178

Question. Prove that \[ \lim_{n \to \infty} \, \frac{n^2+1}{n+1} \sin \left( \dfrac{1}{n} \right) = 1 \,. \]

Solution. Note that \[\begin{align*} \frac{n^2+1}{n+1} \sin \left( \dfrac{1}{n} \right) & = \left( \frac{1+\dfrac{1}{n^2}}{1+ \dfrac{1}{n}} \right) \cdot \left( n \sin \left( \dfrac{1}{n} \right) \right) \\ & \longrightarrow \frac{1 + 0}{1 + 0} \cdot 1 = 1\,, \end{align*}\] where we used (A.47) and the Algebra of Limits.