Math 131A: Real Analysismjandr/Math131A/lectures.pdf · Math 131A: Real Analysis ... every real number is either bigger than or equal to 0, ... by the end of this class, you will

Math 131A: Real Analysis

Michael AndrewsUCLA Mathematics Department

March 19, 2018

Contents

1 Introduction 4

2 Truth Tables 42.1 NOT(P ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 P and Q, P or Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 P =⇒ Q (read this as “P arrow Q,” not as “P implies Q”) . . . . . . . . . . . . . . 5

3 Some familiar sets 6

4 Quantifiers 74.1 Unspecified variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74.2 ∃ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84.3 ∀ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84.4 A well-formed sentence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94.5 Verifying a sentence involving quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . 94.6 Negating a sentence involving quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . 114.7 Parentheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.8 Negating a sentence with parentheses carefully . . . . . . . . . . . . . . . . . . . . . 144.9 Math 95 notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

5 Absolute value 155.1 Definition and basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.2 The point of the absolute value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 A little more about sets 17

7 Sequences and the definition of convergence 197.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197.2 Convergence of sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207.3 Divergence of sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217.4 How to spend your time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227.5 Constructing the proof of theorem 7.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . 23

1

7.6 Motivating the definition of convergence . . . . . . . . . . . . . . . . . . . . . . . . . 257.7 Another sequence that converges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267.8 Constructing the proof of theorem 7.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . 287.9 The algebra of limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317.10 The game and the computer program, finishing the proof of 7.9.1 . . . . . . . . . . . 327.11 Another algebra of limits proof, more game/computer program . . . . . . . . . . . . 33

8 Continuity 368.1 The definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368.2 Motivating the definition of continuity . . . . . . . . . . . . . . . . . . . . . . . . . . 368.3 Examples of continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

9 Uniform Continuity 409.1 The definition and a comparison with the definition of continuity . . . . . . . . . . . 409.2 Examples of uniformly continuous and non-uniformly continuous functions . . . . . . 40

10 What are the real numbers? The completeness axiom 4310.1 What are the real numbers? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4310.2 Boundedness, maximum and minimum elements . . . . . . . . . . . . . . . . . . . . 4510.3 The completeness axiom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4710.4 Suprema and Infima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4810.5 Contrapositive and contradiction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5010.6 An almost-proof of corollary 10.3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

11 Some more on sequences (not lectured) 5111.1 Uniqueness of limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5111.2 Bounded sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5311.3 Algebra of limits (continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

12 Sequences and the completeness of R 5512.1 Cauchy sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5512.2 Monotone Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5612.3 Subsequences and the Bolzano-Weierstrass theorem . . . . . . . . . . . . . . . . . . . 5812.4 Completeness of R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6012.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

13 More on continuous functions 6213.1 The sequence definition of continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . 6213.2 New continuous functions from old ones (not lectured) . . . . . . . . . . . . . . . . . 6413.3 The Intermediate Value Theorem and the Extreme Value Theorem . . . . . . . . . . 6513.4 A theorem that should be included somewhere . . . . . . . . . . . . . . . . . . . . . 67

14 Calculus 6814.1 Limits along the domain of a function . . . . . . . . . . . . . . . . . . . . . . . . . . 6814.2 The derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7014.3 Rules for differentiating (not lectured or examined explicitly) . . . . . . . . . . . . . 7114.4 Fermat and Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

2

14.5 The Mean Value Theorem and important results from calculus . . . . . . . . . . . . 74

15 Series 7715.1 From sequences to series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7715.2 Tests for convergence and divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 7915.3 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

16 More functions: cos, sin, exp, log 8516.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8516.2 cos, sin and π. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8516.3 exp, log, exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8616.4 De Moivre’s theorem (off syllabus) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

17 Taylor’s theorem 8917.1 Taylor series and Taylor polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 8917.2 Taylor series of polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9017.3 A rubbish Taylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9017.4 An even more rubbish Taylor series (definitely off syllabus) . . . . . . . . . . . . . . 9117.5 Taylor’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

18 Miscellaneous 9518.1 Denseness of the rationals and irrationals in R . . . . . . . . . . . . . . . . . . . . . 9518.2 Irrationality of

√2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

18.3 Countability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

19 Integration 10019.1 Goals for our theory of integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10019.2 Deficiencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10119.3 What is integration anyway? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10119.4 The definition of integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10119.5 Basic examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10319.6 Integration: two auxilary theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10419.7 Properties of the integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10519.8 Preliminary examples to the fundamental theorems of calculus . . . . . . . . . . . . 10919.9 The fundamental theorems of calculus . . . . . . . . . . . . . . . . . . . . . . . . . . 10919.10One more theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11119.11Picard Iterates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

20 Final exam expectations 11320.1 Sections you can safely ignore . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11320.2 Sections which will be most heavily examined? . . . . . . . . . . . . . . . . . . . . . 11320.3 Other comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

3

1 Introduction

Many people find mathematics satisfying because a (well-constructed) statement in math is eithertrue or false. For example, 0 = 0 is true and 0 6= 0 is false. Another example of a true statement is:

if f : R −→ R is a differentiable function, then f is continuous.

The first two examples are very different from the last example. The first examples are ridiculouslyeasy, whereas the latter is more complicated. The latter is more complicated because of the wordswhich appear. It is unlikely that the word “continuous” has ever been defined carefully enough foryou so that whatever function I give you, you can answer with certainty, either “it is continuous”or “it is not continuous”, and you never answer “it is and it’s not.” Even if such a definition wasgiven to you, it is unlikely that you were given the time to master verifying such definitions to thestandard of a pure mathematician.

The purpose of this class is to give you precise definitions of concepts such as continuity anddifferentiability, and to teach you how to verify such definitions (or their negation). We will learnhow concepts like these relate to one another; the sentence above gives an example of how continuityand differentiability relate to one another. We will learn how to prove such relationships rigorously.

2 Truth Tables

Mathematics is concerned with statements which are true or false. We gave two simple examplesand a more complicated example in the introduction. Just like when we speak English, mathematicsis built up by putting together lots of simpler pieces. In this section, we learn a few simple ways ofconstructing statements which are true or false from existing statements which are true or false.

2.1 NOT(P )

Suppose P is a statement which is true or false. We normally call such a thing a proposition. Wehave a new proposition NOT(P ). The truth of NOT(P ) is completely determined by the truth ofP . This is demonstrated in the following truth table.

P NOT(P )

T F

F T

For example, the statement “I am your lecturer for math 131A” is true, and the statement “Iam NOT your lecturer for math 131A” is false. “I am a cat” is false; “I am NOT a cat” is true.

You can see that NOT(NOT(P )) always has the same truth value as P by completing thefollowing truth table.

P NOT(P ) NOT(NOT(P ))

T F

F T

“I am NOT NOT your lecturer for math 131A” is true.

4

2.2 P and Q, P or Q

Here is the truth table for (P and Q), and for (P or Q). It is important to note that mathematically“or” is taken to be inclusive unless we explicitly say otherwise. If you consider such rules for non-mathematical propositions, you might think that this seems like a strange convention, but I hopeyou will find that in mathematics such a convention makes a lot of sense. The following sentenceseems fine to me: every real number is either bigger than or equal to 0, or less than or equal to 0.

P Q P and Q P or Q

T T T T

T F F T

F T F T

F F F F

De Morgan’s Laws state the following.

1. NOT(P or Q) has the same truth values as (NOT(P ) and NOT(Q));

2. NOT(P and Q) has the same truth values as (NOT(P ) or NOT(Q)).

If you’ve never seen these laws before you should think up examples that convince you they work,and check them with truth tables.

2.3 P =⇒ Q (read this as “P arrow Q,” not as “P implies Q”)

Here is the truth table for (P =⇒ Q). This is the most bizarre truth table. Whenever P is false,(P =⇒ Q) is true. This normally upsets people at first. We’ll see why this is a sensible conventiononce we introduce quantifiers.

P Q P =⇒ Q NOT(P =⇒ Q)

T T T F

T F F T

F T T F

F F T F

1. You can use a truth table to see that

(P =⇒ Q) has the same truth values as (NOT(P ) or Q).

2. De Morgan’s Laws tell us that

NOT(P =⇒ Q) has the same truth values as (P and NOT(Q)).

5

3 Some familiar sets

To make our discussion of quantifiers easier I need to introduce some concepts and some notation.

Definition 3.1. A set is any collection of individuals. The members of a set are often called itselements. Two sets are equal if and only if they have the same elements. We write x ∈ X to meanthat “x is an element of a set X.”

Notation 3.2. Curly brackets (braces) are often used to show sets. The set whose elements area1, a2, a3, . . . , an is written {a1, a2, a3, . . . , an}. Similarly, the set whose members are those of aninfinite sequence a1, a2, a3, . . . of objects is denoted {a1, a2, a3, . . .}.

Definition 3.3. A common way to define a set is as “those elements with a certain property.”This can be problematic (Russell’s paradox), but usually it is okay. We write

{x : x has property P}

for the set consisting of elements x with a property P , or

{x ∈ X : x has property P}

for the elements of a previously defined set X with a property P .The colons should be read as “such that.”

Definition 3.4. If X and Y are sets then we write X ∪ Y and say “X union Y ” for the set

{a : a is an element of X or Y },

and we write X ∩ Y and say “X intersect Y ” for the set

{a : a is an element of X and Y }.

We write X \ Y and say “X take away Y ” for

{x ∈ X : x /∈ Y }.

Here x /∈ Y is read “x is not an element of Y .”

Notation 3.5. Here is some notation for familiar sets.

1. the natural numbers N = {1, 2, 3, . . .},

2. the integers Z = {0} ∪ {−n : n ∈ N} ∪ N = {0,−1, 1,−2, 2,−3, 3, . . .},

3. the rationals Q = {mn : m ∈ Z, n ∈ N},

4. the reals R (the stars of the show and yet you can’t even tell me what they are),

5. the irrationals R \Q,

6. the complex numbers C = {x+ iy : x, y ∈ R}.

6

4 Quantifiers

Quantifiers are the key to making precise, strong definitions. If you do not understand quanitifiersby the end of this class, you will fail the class. They are tricky, and they trip lots of people up, butthey are worth investing lots of time into understanding properly. This class is difficult for multiplereasons. My hope is for us to conquer quantifiers early on so that we can focus on other difficultaspects later without quantifiers getting in our way.

4.1 Unspecified variables

In section 2 we talked about statements which are true or false, and some ways that we can formnew such statements from existing ones. Here is a formula.

a2 + b2 = c2

Is it true or false? What about the following formula? Is that true or false?

x =−b±

√b2 − 4ac

2a

I just fed you nonsense. I wrote down familiar formulae, and gave you no context for them atall. It is impossible to say whether they are true or false, because the truth of the first depends onwhat a, b, and c are, and the truth of the second depends on what a, b, c and x are. If you knowwhat these familiar looking formula relate to, you will also know that the significance of a, b, andc in each equation is completely different, even though we have used the same letters.

Here’s are some even simpler examples of formulae.

1. x = 0.

Is it true or false? Again, it depends on what x is. However, there are two observations thatI can make.

(a) It is not true for all possible choices for x.

(If you find this sentence ambiguous, the simplicity of the equation x = 0 should resolvethat ambiguity. Quantifiers will remove the ambiguity from more complicated sentences.Maybe you prefer, “it is not true for all possible choices for x.”)

(b) There is at least one choice of x which makes it true.

2. (x+ 1)2 = x2 + 2x+ 1.

A priori, the truth of this formula might depend on x, but you have seen this one a milliontimes in your life. You know this is always true.

(a) It is true for all possible choices for x.

(b) As a consequence, there is at least one choice of x which makes it true.

3. x = x+ 1.

(a) It is not true for all possible choices for x.

(b) In fact, it is never true. There is not at least one choice of x which makes it true.

7

4.2 ∃

We now introduce our first quantifier ∃ which is read “there exists.” Oddly enough, giving a precisedefinition of this symbol in terms of the truth values of propositions is beyond the scope of thisclass. It is not too difficult, but it is for another time, when you take a class in mathematical logic.We demonstrate how it is used with the examples of the previous section.

The following sentence is true.∃x ∈ R : x = 0.

This is read “there exists an x in R such that x = 0” or “there exists a real number x such thatx = 0.” The following sentence is also true.

∃x ∈ R : (x+ 1)2 = x2 + 2x+ 1.

You can see these sentences are true because, in both cases, taking x to be 0 makes the subsequentformula true.

The following sentence is false. ∃x ∈ R : x = x+ 1.It is possible to use more than one ∃. For instance,

∃a ∈ R : ∃b ∈ R : ∃c ∈ R : a2 + b2 = c2

is true because when a, b, and c are equal to 0, the formula is true. Don’t dwell on this last examplejust yet because we will talk about how to deal with multiple quantifiers at a time shortly.

4.3 ∀

Our next quantifier is ∀ which is read “for all.”The following sentence is false.

∀x ∈ R, x = 0.

This is read “for all x in R, x = 0” or “for all real numbers x, x = 0.” It is false because 1 is a realnumber which is not equal to 0.

The following sentence is true.

∀x ∈ R, (x+ 1)2 = x2 + 2x+ 1.

That’s because you know how to multiply (x + 1)2 = (x + 1)(x + 1) using distributivity of multi-plication.

The following sentence is false. ∀x ∈ R, x = x+ 1.The following sentence is also false.

∀a ∈ R, ∀b ∈ R, ∀c ∈ R, a2 + b2 = c2

That’s because taking a = 0, b = 0 and c = 1, makes the formula false.

Example 4.3.1. I think the following example is useful because it shows why the truth table for(P =⇒ Q) is what it is. We’ll be coming back to this one.

∀n ∈ Z, (n > 10 =⇒ n > −10).

It expresses the sentence “if an integer is bigger than 10, then it is bigger than −10” and so it is atrue sentence. It is not crazy to believe that this is what the symbols encode, but I need to giveyou a way of verifying the truth of quantified sentences for you to really see why this is the case.

8

4.4 A well-formed sentence

In mathematical logic, the concept of a “sentence” is defined. To be a well-formed sentence, thevariables appearing in the formulae of the sentence must be “quantified over.” This means that foreach variable there must be a corresponding quantifier. This is to guarantee that the sentence hasa well-defined truth value. In this class, you should be suspicious if you see a variable that has notbeen previously declared or quantified over.

4.5 Verifying a sentence involving quantifiers

Most sentences we will encounter in this course will be of the form

Q1Q2 . . . Qn ϕ

where each individual Qi is either of the form ∃xi ∈ Xi or ∀xi ∈ Xi, and ϕ is something constructedfrom a load of formulae and the operations of “and,” “or,” and “ =⇒ .” (It can be more complicatedthan this because ϕ could also involve quanitfiers and look, for example, like (∃x ∈ X : F1) =⇒ F2.The second example below is more complicated in a similar way.)

• A proof of the truth of such a sentence should consist of (n+ 1) steps.

• At the i-th step, to deal with ∃xi ∈ Xi you should say “let xi = BLAH” where BLAH needsto be an element of Xi written only in terms of variables which have already been specified.You should always check that the element xi that you declare is an element of the set Xi.

• To deal with ∀xi ∈ Xi you simply say “let xi ∈ Xi.”

• ϕ will be some proposition involving the variables x1, . . . , xn. At the (n+1)-st step, you mustverify the truth of ϕ based on your previous declared variables. For clarity, you should breakthis step up in to further substeps (a), (b), (c), . . . if necessary.

Example 4.5.1. The following sentence is true.

∀n ∈ Z, (n > 10 =⇒ n > −10).

Proof. We write the proof in the format just described.

1. Let n ∈ Z.

2. We must verify the truth of (n > 10 =⇒ n > −10).

(a) Case 1: n ≤ 10.

n > 10 is false, so the truth table for “ =⇒ ” says (n > 10 =⇒ n > −10) is true.

(b) Case 2: n > 10.

n > 10 is true. For (n > 10 =⇒ n > −10) to be true we must verify that n > −10.This is true because n > 10 > −10.

I hope this example illustrates why the truth table for P =⇒ Q is what it is. The sentenceis supposed to say “if an integer is bigger than 10, then it is bigger than −10.” Thinking ofthis statement as one about every integer, we see that we don’t really care about integers lessthan or equal to 10 and we want (n > 10 =⇒ n > −10) to be true in this case.

9


∀x ∈ R, ∀y ∈ R,(x ≤ y =⇒

(∃z ∈ R : x+ z2 = y

)).

Proof.

1. Let x ∈ R.

2. Let y ∈ R.

3. We must verify the truth of

x ≤ y =⇒(∃z ∈ R : x+ z2 = y

).

(a) Case 1: x > y.

x ≤ y is false, so the truth table for “ =⇒ ” says the relevant sentence is true.

(b) Case 2: x ≤ y.

x ≤ y is true and we must verify the truth of

∃z ∈ R : x+ z2 = y.

Notice that because the original sentence is not of the nice form I mentioned at the startof the section, we’ve hit a point where we have to verify another proposition involvinga quantifier.

i. Let z =√y − x. z is a well-defined real number: we are working under the assump-

tion that x ≤ y, and so y − x ≥ 0.

ii. We must verify that x+ z2 = y. This is trivial algebra:

x+ z2 = x+ (√y − x)2 = x+ (y − x) = y.


∃x ∈ R : ∀y ∈ R, ∃z ∈ R : x2 + y2 = z2.

Proof.

1. Let x = 0.

2. Let y ∈ R.

3. Let z = y. Notice z ∈ R.

4. We see that x2 + y2 = 02 + z2 = z2.

There’s something important to note about this proof: we let z = y. This is fine because we madethis declaration in step 3, and y was declared in step 2. Letting x = y in step 1 would have been anillegal declaration. We CANNOT define a variable in terms of a variable that’s declared later.

10

4.6 Negating a sentence involving quantifiers

Example 4.6.1. Let’s try to show the following sentence to be true.

∃x ∈ R : ∀y ∈ R, ∀z ∈ R, x2 + y2 = z2.

Our proof would read as follows.

1. Let x = BLAH. Notice x ∈ R.

2. Let y ∈ R.

3. Let z ∈ R.

4. We need to check the truth of x2 + y2 = z2.

We need to think up what BLAH should be. Suppose we tried x = 0 in step 1. Then, in step 4, wewould have to show that y2 = z2. Checking that this is true is impossible since we know absolutelynothing about y and z (other than the fact that they are real numbers). y could be 1, and z couldbe 0, and in this case the equation is false. Similarly, if we tried x = 1 in step 1, then, in step 4,we would have to show that 1 + y2 = z2. Checking that this is true is impossible too.

Hmm, it seems the sentence is false. How do we verify this? First, let’s note some things.

• The reason it is false is not because one choice of x failed us; it is because all choices of xfail us. Saying it another way, for all x ∈ R, the following proposition is false.

∀y ∈ R, ∀z ∈ R, x2 + y2 = z2.

• Why is the proposition above false? When x = 0, we argued it was false by thinking aboutthe case when y = 1 and z = 0. In the case when x = 0, this actually gives a proof that

∃y ∈ R : ∃z ∈ R : x2 + y2 6= z2

is true.

• In the end, the reason that the orginal sentence is false is that the following sentence is true.

∀x ∈ R, ∃y ∈ R : ∃z ∈ R : x2 + y2 6= z2.


∀x ∈ R, ∃y ∈ R : ∃z ∈ R : x2 + y2 6= z2.

Proof. 1. Let x ∈ R.

2. Let y = 1. Notice y ∈ R.

3. Let z = 0. Notice z ∈ R.

4. We see that x2 + y2 = x2 + 1 ≥ 1 > 0 = z2. In particular, x2 + y2 6= z2.

11

Remark 4.6.3. In our attempt to verify the original sentence, the declaration “let x =√z2 − y2”

would be wrong for two reasons. First, y and z are declared after x. Second, z2 − y2 could be lessthan zero, and so it is not possible to argue that x ∈ R.

Remark 4.6.4. Note the way that I used “in particular” in the previous proof. “In particular” isused in the following way.

[Something holding less often]. In particular, [something holding more often].

Some examples in plain English are as follows.

I am a man. In particular, I am a human.

I am a human. In particular, I am an animal.

I love all burgers. In particular, I love In ‘N’ Out burgers.

Some math examples.

x = y. In particular, x ≤ y.x > y. In particular, x 6= y.

n ∈ Z is divisible by 4. In particular, n is even.

Just for the hell of it, let’s note another proof of the previous example.

Proof.

1. Let x ∈ R.

2. Let y = x+ 2. Notice y ∈ R.

3. Let z =√

2(x+ 1). Notice z ∈ R.

4. We see that

x2 + y2 = x2 + (x+ 2)2 = 2x2 + 4x+ 4 = 2(x+ 1)2 + 2 6= 2(x+ 1)2 = z2.

Notice that the sentences

∃x ∈ R : ∀y ∈ R, ∀z ∈ R, x2 + y2 = z2

∀x ∈ R, ∃y ∈ R : ∃z ∈ R : x2 + y2 6= z2

are related to each other by replacing ∃ with ∀, ∀ with ∃, and = with 6=.In general, there is a procedure called negation which is always guaranteed to change a true

sentence to a false one, and a false sentence to a true one.

12

Definition 4.6.5. The negation of a sentence S is NOT(S ); NOT(S ) can be obtained from Svia the following rules.

• Being careful with parentheses when there is a complicated arrangement of clauses;

• NOT(∃x ∈ X : ϕ) is (∀x ∈ X, NOT(ϕ));

• NOT(∀x ∈ X, ϕ) is (∃x ∈ X : NOT(ϕ));

• NOT(NOT(ϕ)) is ϕ;

• NOT(ϕ or ψ) is (NOT(ϕ) and NOT(ψ));

• NOT(ϕ and ψ) is (NOT(ϕ) or NOT(ψ));

• NOT(ϕ =⇒ ψ) is (ϕ and NOT(ψ)).

I’m not a fan of “rules.” “Rules” exist because they are true for a reason. Thinking carefully aboutwhat your quantified sentences say will prevent you from negating incorrectly, and you should neverapply the rules mindlessly.

Remark 4.6.6. The procedure for checking a sentence is false is now as follows.

• Negate the sentence.

• Verify that the negated sentence is true.

Example 4.6.7. The following sentence is false.

∀n ∈ Z, (n > −10 =⇒ n > 10).

Proof. The negation of the sentence is

∃n ∈ Z : (n > −10 and n ≤ 10).

This is true, because we can verify it as follows.

1. Let n = 0. Notice n ∈ Z.

2. Because −10 < 0 ≤ 10, we see that (n > −10 and n ≤ 10) is true.

Remark 4.6.8. Notice that (n > −10 =⇒ n > 10) is true when n = −20 and when n = 20.However, what matters is that it is not true when n = 0. The quantifier “∀n ∈ N” ensures that thesentence in the example has one truth value.

13

4.7 Parentheses

In this section I have used more parentheses than necessary so that you do not get confused. Thereis a convention that a quantifier automatically opens up parentheses. So

∀x ∈ R, ∃y ∈ R : x = y =⇒ 0 = 1

means

∀x ∈ R,(∃y ∈ R :

(x = y =⇒ 0 = 1

)).

(This pointless sentence is true; can you prove it?) It does not mean(∀x ∈ R,

(∃y ∈ R : x = y

))=⇒ 0 = 1.

(This pointless sentence is false; can you prove it?)Because of this convention, the parentheses in

∀a ∈ R \ {0}, ∀b ∈ R, ∀c ∈ R,(∃x ∈ R : ax2 + bx+ c = 0

)=⇒ b2 − 4ac ≥ 0

are necessary to distinguish it from

∀a ∈ R \ {0}, ∀b ∈ R, ∀c ∈ R, ∃x ∈ R : ax2 + bx+ c = 0 =⇒ b2 − 4ac ≥ 0.

If you find yourself confused by some parentheses (or lack of them), ask for clarification.

4.8 Negating a sentence with parentheses carefully

Example 4.8.1. The negation of the sentence

∀a ∈ R \ {0}, ∀b ∈ R, ∀c ∈ R,(∃x ∈ R : ax2 + bx+ c = 0

)=⇒ b2 − 4ac ≥ 0

is given by

∃a ∈ R \ {0} : ∃b ∈ R : ∃c ∈ R :

(∃x ∈ R : ax2 + bx+ c = 0

)and b2 − 4ac < 0.

Example 4.8.2. The negation of the sentence

∀a ∈ R \ {0}, ∀b ∈ R, ∀c ∈ R, b2 − 4ac ≥ 0 =⇒(∃x ∈ R : ax2 + bx+ c = 0

)is given by

∃a ∈ R \ {0} : ∃b ∈ R : ∃c ∈ R : b2 − 4ac ≥ 0 and

(∀x ∈ R, ax2 + bx+ c 6= 0

).

4.9 Math 95 notes

The first 27 pages of http://math.ucla.edu/~mjandr/Math95/lectures.pdf cover the previousmaterial even more thoroughly.

14

http://math.ucla.edu/~mjandr/Math95/lectures.pdf

5 Absolute value

5.1 Definition and basic properties

Definition 5.1.1. The absolute value function | · | : R −→ R is defined piecewise by

|x| =

{x if x ≥ 0

−x if x < 0.

Lemma 5.1.2. ∀x ∈ R, (|x| ≥ 0 and |x| = | − x|).

Proof. Left to you as preparation for the first quiz.

Lemma 5.1.3. ∀x ∈ R, x ≤ |x|.

Proof.

1. Let x ∈ R.

2. (a) Case 1: x ≥ 0. In this case, x = |x|. In particular, x ≤ |x|.(b) Case 2: x < 0. In this case, x < 0 < −x = |x|.

Corollary 5.1.4. ∀x ∈ R, −|x| ≤ x ≤ |x|.

Proof.

1. Let x ∈ R.

2. The previous lemma says x ≤ |x|. Since −x ∈ R the previous lemma also says that −x ≤ |−x|.The first lemma says that | − x| = |x|, so this gives −x ≤ |x| and thus, −|x| ≤ x.

Remark 5.1.5. P ⇐⇒ Q is shorthand for ((P =⇒ Q) and (Q =⇒ P )).

P Q P =⇒ Q Q =⇒ P P ⇐⇒ Q

T T T T T

T F F T F

F T T F F

F F T T T

Lemma 5.1.6. ∀x ∈ R, ∀y ∈ R, |x| ≤ |y| ⇐⇒ −|y| ≤ x ≤ |y|.

Proof.

1. Let x ∈ R.

2. Let y ∈ R.

15

3. From the truth table for P ⇐⇒ Q, we see that we must show that it is not possible forexactly one of |x| ≤ |y|, −|y| ≤ x ≤ |y| to be true. This time, we will do this by showing thatwhenever one of them is true, the other is also true.

(a) Suppose |x| ≤ |y| is true. Then −|y| ≤ −|x| is true too. By using the previous corollary,we see that −|y| ≤ −|x| ≤ x ≤ |x| ≤ |y| is true; in particular, −|y| ≤ x ≤ |y| is true.

(b) Suppose that −|y| ≤ x ≤ |y| is true. Then x ≤ |y| and −x ≤ |y| are true.

i. Case 1: x ≥ 0. We see that |x| = x ≤ |y| is true.

ii. Case 2: x < 0. We see that |x| = −x ≤ |y| is true.

The following is the most important inequality in analysis.

Theorem 5.1.7. (The triangle inequality) ∀x ∈ R, ∀y ∈ R, |x+ y| ≤ |x|+ |y|.

Proof.

1. Let x ∈ R.

2. Let y ∈ R.

3. • By the previous lemma we have −|x| ≤ x ≤ |x|.• By the previous lemma we have −|y| ≤ y ≤ |y|.• Because of how inequalities work - you should think about such things more than I do

since simple inequality mistakes on quizzes and finals make me furious - this gives

−|x| − |y| ≤ x+ y ≤ |x|+ |y|,

i.e. −(|x|+ |y|) ≤ x+ y ≤ |x|+ |y|. By the previous lemma we have |x+ y| ≤ |x|+ |y|.

Corollary 5.1.8. (The triangle inequality’s cousin) ∀x ∈ R, ∀y ∈ R, ||x| − |y|| ≤ |x− y|.

Proof.

1. Let x ∈ R.

2. Let y ∈ R.

3. A dirty trick and the triangle inequality show that |x| = |(x− y) + y| ≤ |x− y|+ |y|. Thus,|x|− |y| ≤ |x− y|. Similarly, −(|x|− |y|) = |y|− |x| ≤ |y−x| = |x− y|. The last lemma showsthat ||x| − |y|| ≤ |x− y|.

The trick used in this proof is very common. Remember it!

Remark 5.1.9. Inequalities involving the absolute values of expressions should only come fromthe application of one of the previous results or the following (trivial) result.

∀x ∈ R \ {0}, ∀y ∈ R \ {0}, |x| ≤ |y| ⇐⇒ 1

|y|≤ 1

|x|.

Anything else will look suspicious and will annoy me even if it happens to be true.

16

5.2 The point of the absolute value

When you first learn about absolute values, you are told that they “ignore the sign.” This is true:for example, |227 | =

227 and | − e| = e. But what is the point of an absolute value? In analysis, they

appear so frequently because they allow us to measure the distance between two numbers. What’sthe distance between a and b? Taking the difference of the numbers, a− b, is a good start, but wedo not care about the sign.

Theorem 5.2.1. The distance fuction defined by d(a, b) = |a− b| has the following properties.

1. (positivity) ∀a ∈ R, ∀b ∈ R,(d(a, b) ≥ 0 and

(d(a, b) = 0 ⇐⇒ a = b

)).

2. (symmetry) ∀a ∈ R, ∀b ∈ R, d(a, b) = d(b, a).

3. (triangle) ∀a ∈ R, ∀b ∈ R, ∀c ∈ R, d(a, c) ≤ d(a, b) + d(b, c).

a c

b

Proof. You can prove it if you want. We’ll never need it explicitly. I just wanted to give you somecontext.

6 A little more about sets

Definition 6.1. The empty set, written ∅, is the set with no elements.

Definition 6.2. Given two sets X and Y , we say that X is a subset of Y and write X ⊆ Y iff

∀z, z ∈ X =⇒ z ∈ Y

Remark 6.3.

1. In this definition “iff” stands for “if and only if.” This means that saying “X is a subset ofY ” is exactly the same as saying “the quantified sentence is true.”

Often, in definitions, mathematicians say “if” even though the meaning is “iff.” I normallydo this, and I feel a little silly for writing “iff,” but I decided that it’s the least confusingthing I can do. To make this “iff” feel different from a non-definitional “iff” I have used bold.

2. The quantified sentence is a little strange since we don’t say where z lives; we just have “∀z.”Things become funny once you start talking about sets. Basically everything in math is a setin disguise, so this means “for every mathematical thing z.”

Example 6.4. Suppose X is a set. Then ∅ ⊆ X.

17

Proof. Let X be a set. We must check that the following sentence is true.

∀x, x ∈ ∅ =⇒ x ∈ X.

1. Let x be some mathematical thing.

2. We must check the truth of (x ∈ ∅ =⇒ x ∈ X). Since ∅ has no elements, x ∈ ∅ is false. Thetruth table for “ =⇒ ” tells us that (x ∈ ∅ =⇒ x ∈ X) is true.

Example 6.5. For any sets X and Y we have the following inclusions.

1. X ⊆ X ∪ Y , Y ⊆ X ∪ Y .

2. X ∩ Y ⊆ X, X ∩ Y ⊆ Y .

3. X ∩ Y ⊆ X ∪ Y .

Now we can improve our definition of two sets being equal.

Definition 6.6. Suppose X and Y are sets. We say that X = Y iff X ⊆ Y and Y ⊆ X, i.e. iffthe following sentence is true

∀z, z ∈ X ⇐⇒ z ∈ Y.

Example 6.7.

1. {0, 1} = {1, 0}.

2. {2, 2} = {2}.

3. {∅} 6= ∅.

Finally, we record some commonly occuring sets.

Definition 6.8 (Real intervals). Let a, b ∈ R. The following are called intervals.

1. (a, b) := {x ∈ R : a < x < b} [open interval];

2. [a, b] := {x ∈ R : a ≤ x ≤ b} [closed interval];

3. (a, b] := {x ∈ R : a < x ≤ b} [half open interval];

4. [a, b) := {x ∈ R : a ≤ x < b} [half open interval];

5. (a,∞) := {x ∈ R : a < x};

6. [a,∞) := {x ∈ R : a ≤ x};

7. (−∞, b) := {x ∈ R : x < b};

8. (−∞, b] := {x ∈ R : x ≤ b};

9. (−∞,∞) := R.

18

Remark 6.9.

1. Read the symbol := as “is defined to be.” It is quite different from =, “is equal to,” whichindicates that two previously defined entities are the same.

2. If you see any of the first four sets above, you can assume, unless otherwise stated, that a < b;this is because if a ≥ b, then these sets are either empty or consist of one element and so itis unlikely we’ll want to consider these cases.

3. Although ∞ is used for notation in the above sets, it is not something that we will treat asa number. You can not plug it into an equation. ∞ will always require careful considerationand there will be careful definitions concerning it. It is often used as a notational device.

7 Sequences and the definition of convergence

“Black (1)then (1)white are (2)all I see (3)in my in-fan-cy, (5)red and yel-low then came to be (8). . . ”

Tool, Los Angeles, 2001.

7.1 Sequences

A sequence is what you think it is.

Definition 7.1.1 (Informal). A sequence is an infinite list of numbers.

Here are some examples:

1, 2, 3, 4, 5, . . . , n, . . .

1,1

2,

1

3,

1

4,

1

5, . . . ,

1

n, . . .

−1, 1, −1, 1, −1, . . . , (−1)n, . . .

2,(3

2

)2,(4

3

)3,(5

4

)4,(6

5

)5, . . . ,

(1 +

1

n

)n, . . .

1, 1, 2, 3, 5, . . . , Fn = Fn−1 + Fn−2, . . .

1

1,

2

1,

3

2,

5

3,

8

5, . . . ,

Fn+1

Fn, . . .

19

We frequently write sn for the n-th term in a sequence, and so we could express the first foursequences above, concisely, as

sn = n, sn =1

n, sn = (−1)n, sn =

(1 +

1

n

)n.

(Is there a formula for the n-th Fibonacci number which is not iterative? Can you find evidence ofthe Fibonacci sequence in the Tool song “Lateralus,” other than that highlighted in the quote atthe start of the lecture?)

Notation 7.1.2. To indicate an abstract sequence we use parentheses: (sn). Sometimes we maywant to emphasize that the list is indexed by natural numbers and we write (sn)n∈N. We might alsowrite (sn)∞n=1. It is convenient to be able to start a sequence with sm, where m might be greaterthan 1. In this case we write (sn)∞n=m. So, for example,(

1

n

)∞n=8

=1

8,

1

9,

1

10,

1

11,

1

12, . . . .

7.2 Convergence of sequences

Here’s our first major definition of the class.

Definition 7.2.1. Suppose (sn)∞n=1 is a sequence.We say that (sn) converges iff the following sentence is true.

∃L ∈ R : ∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒ |sn − L| < ε.

Remark 7.2.2. The quanitifier “∀ε > 0” is shorthand for “∀ε ∈ (0,∞).”

First, let me show you a sequence which converges and a sequence which does not converge,and proofs of those facts without any explanation of where the proofs come from.

Theorem 7.2.3. The sequence ( 1n)∞n=1 converges.

Proof. By definition, we must verify that the following sentence is true.

∃L ∈ R : ∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒∣∣∣∣ 1n − L

∣∣∣∣ < ε.

1. Let L = 0.

2. Let ε > 0.

3. Let N = d1ε e+ 1.

Here d−e denotes the ceiling function. It rounds up to the nearest integer. For example,⌈5

2

⌉= 3 = d3e.

This is useful frequently, so remember it.

4. Let n ∈ N.

20

5. We must verify the truth of (n ≥ N =⇒ | 1n − L| < ε).

(a) Case 1: n < N .

n ≥ N is false, so the truth table for “ =⇒ ” says (n ≥ N =⇒ | 1n − L| < ε) is true.

(b) Case 2: n ≥ N .

We must show that | 1n − L| < ε.

We have N = d1ε e+ 1 > 1ε and also∣∣∣∣ 1n − L

∣∣∣∣ =

∣∣∣∣ 1n − 0

∣∣∣∣ =1

n≤ 1

N< ε.

7.3 Divergence of sequences

Definition 7.3.1. Suppose (sn)∞n=1 is a sequence.We say that (sn) diverges iff (sn) does not converge, i.e. iff the following sentence is true.

∀L ∈ R, ∃ε > 0 : ∀N ∈ N, ∃n ∈ N : n ≥ N and |sn − L| ≥ ε.

Theorem 7.3.2. The sequence ((−1)n)∞n=1 diverges.

Proof. By definition, we must verify that the following sentence is true.

∀L ∈ R, ∃ε > 0 : ∀N ∈ N, ∃n ∈ N : n ≥ N and |(−1)n − L| ≥ ε.

1. Let L ∈ R.

2. Let ε = 1.

3. Let N ∈ N.

4. (a) Case 1: L ≥ 0. Let n = 2N + 1.

(b) Case 2: L < 0. Let n = 2N .

5. We must verify that (n ≥ N and |(−1)n − L| ≥ ε) is true.

(a) Case 1: L ≥ 0.

n = 2N + 1 = N + (N + 1) ≥ N . Also,

|(−1)n − L| = |(−1)2N+1 − L| = | − 1− L| = 1 + L ≥ 1 = ε.

(b) Case 2: L < 0.

n = 2N = N +N ≥ N . Also,

|(−1)n − L| = |(−1)2N − L| = |1− L| = 1− L ≥ 1 = ε.

21

7.4 How to spend your time

In terms of your learning, your current goals should be as follows.

• Make sure you are happy with all of the quantifier verifications of the previous sections.

• Make sure you are happy with the correctness of the last two proofs given, even if you haveno clue where the idea for them came from.

• – Look at the previous two proofs.

– Figure out which parts of the proof are mindless, and will be the same for any sequence.

– Figure out which bits require thought. Think about these parts and see if you can figureout what thinking I did to come up with them.

You should spend a long time on this step before going on to the next one, either until youthink that you have figured it out, or until two hours has passed.

• Try and understand what the definition of convergence is saying. It is a wonderful and cleverdefinition. It encodes everything you could ever want.

You should spend a long time on this step before going on to the next one, either until youthink that you have figured it out, or until two hours has passed.

• Read further and understand my description of where the proofs came from, and what thedefinition of convergence is saying.

22

7.5 Constructing the proof of theorem 7.2.3

Let’s revisit theorem 7.2.3, the fact that ( 1n)∞n=1 converges.

We must verify that the following sentence is true.

∃L ∈ R : ∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒∣∣∣∣ 1n − L

∣∣∣∣ < ε.

A lot of the proof is fairly mindless and it is likely that writing what follows is a step in the rightdirection.

1. Let L = BLAH.

2. Let ε > 0.

3. Let N = BLAH.

4. Let n ∈ N.

5. We must verify the truth of (n ≥ N =⇒ | 1n − L| < ε).

(a) Case 1: n < N .

n ≥ N is false, so the truth table for “ =⇒ ” says (n ≥ N =⇒ | 1n − L| < ε) is true.

(b) Case 2: n ≥ N .

We must show that | 1n − L| < ε. We have∣∣∣∣ 1n − L∣∣∣∣ = . . . .

BLAH BLAH BLAH.

We have to fill in the BLAHs. First, we must figure out what L is. What’s the significance of L? Ihope that after thinking about the definition of convergence and the proofs of theorem 7.2.3 longenough you will have come to the conclusion that L is supposed to be the limit of the sequence.

Even if you have not parsed all the quantifiers just yet, you know that the condition |sn−L| < εsays “sn is within ε of L.” The limit of a sequence is precisely the real number that the terms in asequence get close to, so L has got to be the limit, right?! Let’s take a minute to make a definition.

Definition 7.5.1. Suppose (sn)∞n=1 is a sequence and L ∈ R.We say that (sn) converges to L iff the following sentence is true.

∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒ |sn − L| < ε.

In this case L is said to be the limit of the sequence (sn) and we will often express this by writinglimn→∞ sn = L or, if it is not confusing, lim sn = L.

Remark 7.5.2.

1. If you look at the quantified sentence in the previous definition, you will see that L has notbeen quantified over. This is okay because L was declared in the preamble. By the time westate the quantified sentence, L is a constant real number.

23

2. (a) If (sn) converges, then there is some L ∈ R such that (sn) converges to L.

(b) If (sn) converges to some L ∈ R, then (sn) converges.

These statements are trivial from the definitions, but in this early stage of analysis, you shouldcheck that you understand them 100%.

I hope that while taking math 31B or a similar class you were told that limn→∞1n = 0. We are

now proving this fact. This is the reason that I took L to be 0 in my proof.Now we can improve upon the beginning of the proof written above a little bit.

1. Let L = 0.

5. (b) Case 2: n ≥ N .

We must show that | 1n − L| < ε. We have∣∣∣∣ 1n − L∣∣∣∣ =

∣∣∣∣ 1n − 0

∣∣∣∣ =1

n. . .

Now to N .

• The purpose of N is to specify how far along in the sequence we have to go beforeterms in the sequence are within ε of the limit L.

• N will almost always depend on ε.

• If ε is small, N will almost always be large. Formulae like N = ε are stupid.

In the case that we’re considering we see that beyond N , 1n is within 1

N of 0, and so we just need1N < ε, i.e. N > 1

ε . I’ve decided to go back to my Oxford routes and demand that N be a naturalnumber. The textbook says otherwise. Sorry about this. It’ll be an exercise later to show that thischoice doesn’t matter. d1ε e+ 1 is the simplest way to write down an explicit natural number biggerthan 1

ε .We’ve now completed the proof.

3. Let N = d1ε e+ 1.

5. (a) Case 2: n ≥ N .

We must show that | 1n − L| < ε.

We have N = d1ε e+ 1 > 1ε and also∣∣∣∣ 1n − L

∣∣∣∣ =

∣∣∣∣ 1n − 0

∣∣∣∣ =1

n≤ 1

N< ε.

24

7.6 Motivating the definition of convergence

A picture is worth a thousand words.

sN−5

sN−4

sN−3

N

sN+1

L

L+ ε

L− ε

A sequence of real numbers (sn) converges to a real number L, the limit, if

• whenever a small positive quantity ε is specified

• by going far enough along in the sequence

• we ensure that the terms of the sequence differ from the limit by less than the small quantityspecified.

The first bullet point is said mathematically as ∀ε > 0.The last two are said mathematically as follows.

• ∃N ∈ N (there exists a number telling us how far we need to go)

• : ∀n ∈ N, n ≥ N =⇒ (if we go this far along in the sequence)

• |sn − L| < ε (then the terms of the sequence are within ε of L).

25

7.7 Another sequence that converges

Theorem 7.7.1. The sequence ( nn2−101)∞n=1 converges.

Before writing the proof let’s think what it must look like. I hope that you can see that thelimit of this sequence will be 0. Thus, our proof is going to look as follows.

We must verify that the following sentence is true.

∃L ∈ R : ∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒∣∣∣∣ n

n2 − 101− L

∣∣∣∣ < ε.

1. Let L = 0.

2. Let ε > 0.

3. Let N = BLAH.

4. Let n ∈ N.

5. We must verify the truth of (n ≥ N =⇒ | nn2−101 − L| < ε).

(a) Case 1: n < N .

n ≥ N is false, so the truth table for “ =⇒ ” says (n ≥ N =⇒ | nn2−101 −L| < ε) is true.

(b) Case 2: n ≥ N .

We must show that | nn2−101 − L| < ε. We have∣∣∣∣ n

n2 − 101− L

∣∣∣∣ =

∣∣∣∣ n

n2 − 101

∣∣∣∣ = . . . .

BLAH BLAH BLAH.

Now I get to say more about N . The clause

∃N ∈ N : ∀n ∈ N, n ≥ N =⇒∣∣∣∣ n

n2 − 101

∣∣∣∣ < ε

says that | nn2−101 | < ε is true as long as n is sufficiently large, and N is a natural number quantifying

exactly how large n needs to be.When considering the expression | n

n2−101 | my first thought is “is nn2−101 positive?” The answer

is, “not necessarily, but as long as n is sufficiently large.” We can improve on this and say that “itis positive as long as n >

√101.”

∀n ∈ N, n >√

101 =⇒ n

n2 − 101> 0.

We can now pretty much work under the assumption that nn2−101 positive. But by saying “pretty

much,” I mean that we have to remember the conditions under which it actually holds: n >√

101.

26

My next thought is that nn2−101 should behave somewhat like 1

n . A useful inequality towards aproof would be

∀n ∈ N,n

n2 − 101≤ 1

n

but this is UTTER NONSENSE!! In fact, the following sentence is true

∀n ∈ N, n >√

101 =⇒ n

n2 − 101>

1

n.

The inequality goes the wrong way. Instead, we note that

∀n ∈ N, n ≥√

202 =⇒ 0 ≤ n

n2 − 101≤ 2

n(7.7.2)

This is because the condition 0 ≤ nn2−101 ≤

2n is equivalent to 202 ≤ n2 by basic algebra. The idea

I had here was “ 1n doesn’t give a correct inequality, but maybe 2

n does.” In the end, the 2 made allthe difference.

We finally note that the following sentence is true.

∀n ∈ N, n >2

ε=⇒ 2

n< ε. (7.7.3)

Equations (7.7.2) and (7.7.3) suggest taking N = dmax{√

202, 2ε}e+ 1.

Proof of theorem 7.7.1. We must verify that the following sentence is true.

∃L ∈ R : ∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒∣∣∣∣ n

n2 − 101− L

∣∣∣∣ < ε.

1. Let L = 0.

2. Let ε > 0.

3. Let N = dmax{√

202, 2ε}e+ 1.

4. Let n ∈ N.

5. We must verify the truth of (n ≥ N =⇒ | nn2−101 − L| < ε).

(a) Case 1: n < N .

n ≥ N is false, so the truth table for “ =⇒ ” says (n ≥ N =⇒ | nn2−101 −L| < ε) is true.

(b) Case 2: n ≥ N .

Since N = dmax{√

202, 2ε}e+ 1 >√

101, we have n >√

101 and n2 − 101 > 0.

Since N = dmax{√

202, 2ε}e+ 1 >√

202, we have n ≥√

202 and nn2−101 ≤

2n .

Since N = dmax{√

202, 2ε}e+ 1 > 2ε , we have n > 2

ε .

Thus, ∣∣∣∣ n

n2 − 101− L

∣∣∣∣ =

∣∣∣∣ n

n2 − 101

∣∣∣∣ =n

n2 − 101≤ 2

n< ε.

27

7.8 Constructing the proof of theorem 7.3.2

Let’s revisit theorem 7.3.2, the fact that ((−1)n)∞n=1 diverges.We must verify that the following sentence is true.

∀L ∈ R, ∃ε > 0 : ∀N ∈ N, ∃n ∈ N : n ≥ N and |(−1)n − L| ≥ ε.

A lot of the proof is fairly mindless and it is likely that writing what follows is a step in the rightdirection.

1. Let L ∈ R.

2. Let ε = BLAH.

3. Let N ∈ N.

4. Let n = BLAH.


BLAH.

This looks tricky because ε is allowed to depend on L, and n is allowed to depend on L, ε, andN . That’s a lot to juggle.

It turns out that whenever (sn) is a sequence,

“∀L ∈ R, ∃ε > 0 : ∀N ∈ N, ∃n ∈ N : n ≥ N and |sn − L| ≥ ε” is true

if and only if “∃ε > 0 : ∀L ∈ R, ∀N ∈ N, ∃n ∈ N : n ≥ N and |sn − L| ≥ ε” is true.

This is quite a subtle fact, one which we will only be able to prove later on. For now, this simplifiesthings a little. It tells you that ε does not need to depend on L; that is the point in swapping thefirst two quantifiers.

In the case of ((−1)n)∞n=1, my intuition for choosing ε = 1 is the following.

• The clause∀N ∈ N, ∃n ∈ N : n ≥ N and |(−1)n − L| ≥ ε

says that however far along in the sequence we go, we can always find a term in the sequencewhich is ε or further away from L.

• We don’t know what L is but we can think about how difficult it would be to prove that thesentence above is true for different Ls.

• Suppose L is 1. Then the even terms of the sequence are 0 away from L. However, the oddterms are 2 away from L. This suggests taking ε = 2.

• Suppose L is −1. Then the odd terms of the sequence are 0 away from L. However, the eventerms are 2 away from L. This suggests taking ε = 2.

• Suppose L is 500. Then the odd terms of the sequence are 501 away from L and the eventerms are 499 away from L. This continues to suggest that ε = 2 would be a fine choice.

28

• Suppose L is 0. Then all of the terms are 1 away from L. This suggests that ε = 1 would bea better choice.

• For the first three choices of L that we spoke about, we considered the even and odd termsseparately. We did not have to do this for the choice L = 0. All of this hinted to me thatconsidering the cases L ≥ 0 and L < 0 separately might be useful; there is some symmetrygoing on.

• Suppose L ≥ 0. Then the odd terms of the sequence are at least 1 away from L and thiscontinues to suggest that ε = 1 would be a fine choice.

• Suppose L < 0. Then the even terms of the sequence are at least 1 away from L and thiscontinues to suggest that ε = 1 would be a fine choice.

All of the above reasoning suggests taking ε = 1. It also suggests that when specifying n we shouldbreak up into two cases, when L ≥ 0 and when L < 0.

Our incomplete proof now becomes.

1. Let L ∈ R.

2. Let ε = 1.

3. Let N ∈ N.

4. (a) Case 1: L ≥ 0. Let n = BLAH.

(b) Case 2: L < 0. Let n = BLAH.


(a) Case 1: L ≥ 0.

BLAH.

(b) Case 2: L < 0.

BLAH.

Let’s consider how to choose n in the case that L ≥ 0. When L ≥ 0, the odd terms of thesequence are at least 1 away from L. This tells us we should choose n to be an odd number. Wealso want n to be bigger than or equal to N . 2N + 1 is an odd number bigger than or equal to N .

When L < 0, the even terms of the sequence are at least 1 away from L. This tells us we shouldchoose n to be an even number. We also want n to be bigger than or equal to N . 2N is an evennumber bigger than or equal to N .

This allows us to fill in.

4. (a) Case 1: L ≥ 0. Let n = 2N + 1.

(b) Case 2: L < 0. Let n = 2N .

We are just left to complete step 5. By this point this is just a matter of writing down everythingcorrectly. However, this step is important and I’ll be annoyed when I see nonsense inequalities, orincorrect absolute values, so make sure you get things correct!

29

Remark 7.8.1. At this point I may as well mention that my order of preference for a quiz or finalsolution is as follows.

• A complete and correct solution.

• An incomplete solution where nothing that is written is incorrect. In addition, separate ideasare given for completing the proof with difficulties highlighted.

• An incomplete solution where nothing that is written is incorrect. No idea on how to proceed.

• A solution which uses inequalities that are nonsense in the attempt of making the solutionappear complete. This type of solution will be dealt with harshly.

30

7.9 The algebra of limits

It is a little tedious to verify the definition of a convergent sequence every time we want to showthat a sequence converges. In mathematics, we often prove theorems to save ourselves from havingto do similar work over and over again. That is the purpose of the algebra of limits. It says thingslike, “the sum of two convergent sequences is a convergent sequence.”

The first result in the algebra of limits which we will prove is the following.

Theorem 7.9.1. Suppose (sn)∞n=1 is a sequence and s, c ∈ R. If (sn) converges to s, then (csn)∞n=1

converges to cs. We could also state this theorem by saying that the following sentence is true.

∀sequences (sn)∞n=1, ∀s ∈ R, ∀c ∈ R, limn→∞

sn = s =⇒ limn→∞

csn = cs.

Proof attempt.

1. Let (sn) be a sequence.

2. Let s ∈ R.

3. Let c ∈ R.

4. We must verify the truth of (limn→∞ sn = s =⇒ limn→∞ csn = cs).

(a) Case 1: limn→∞ sn = s is false. Trivial.

(b) Case 2: limn→∞ sn = s is true. We must show that limn→∞ csn = cs is true. Both ofthese expressions have a definition associated with them. So we can expand on this andsay the following.

• The following sentence is true.

∀ε′ > 0, ∃N ′ ∈ N : ∀n ∈ N, n ≥ N ′ =⇒ |sn − s| < ε′.

(The use of ε′ and N ′ will become clear in the next subsection.)

• We must show that the following sentence is true.

∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒ |csn − cs| < ε.

We verify the truth of the last statement.

i. Let ε > 0

ii. Let N = BLAH.

iii. Let n ∈ N.

iv. We must verify the truth of (n ≥ N =⇒ |csn − cs| < ε).

A. Case 1: n < N . Trivial.

B. Case 2: n ≥ N . In this case

|csn − cs| = |c||sn − s| . . .BLAH.

We have to fill in the BLAHs. How can we possibly write down an N if we have no idea what(csn) looks like?

31

7.10 The game and the computer program, finishing the proof of 7.9.1

In earlier subsections we verified that some sequences converge. After figuring out the limit L of asequence, this verification comes down to playing an “ε-N game.” The goal of the ε-N game is tofigure out N in terms of ε, and it must be an N ∈ N making the relevant statement

∀n ∈ N, n ≥ N =⇒ |sn − s| < ε (7.10.1)

true. When we wrote our proofs, we wrote them in the style of a computer program.The information that a sequence converges to a limit can be viewed as giving us such a computer

program. If we give the computer program a postive real number ε′, it will return a natural numberN ′ making the relevant statement (7.10.1)′ true.

The key to completing the proof of theorem 7.9.1 is to use such a “computer program” to specifyan N .

Completing the proof of theorem 7.9.1.

4. (b) i. Let ε > 0.

ii. This is where the bulk of the argument happens.

A. Let ε′ = ε1+|c| . Then ε′ > 0.

B. Since∀ε′ > 0, ∃N ′ ∈ N : ∀n ∈ N, n ≥ N ′ =⇒ |sn − s| < ε′

is true, we obtain an N ′ ∈ N such that the following sentence is true.

∀n ∈ N, n ≥ N ′ =⇒ |sn − s| < ε′.

C. Let N = N ′.

iii. Let n ∈ N.

iv. We must verify the truth of (n ≥ N =⇒ |csn − cs| < ε).


B. Case 2: n ≥ N .Since N = N ′, we have n ≥ N ′, and so |sn − s| < ε′. Thus,

|csn − cs| = |c||sn − s| ≤ |c|ε′ = |c|ε

1 + |c|< ε.

Notice how algorithmically this proof is written.

• We say what ε′ is in terms of c and ε which were already declared in 2. and 4.(b)i., respectively.

• We obtain N ′ from the “computer program” and ε′ which were already declared in 4.(b)• and4.(b)ii.A., respectively.

• We say what N is in terms of N ′ which was declared in the step before.

• Finally, we check that everything works as it is supposed to.

32

7.11 Another algebra of limits proof, more game/computer program

Let’s see another similar example.

Theorem 7.11.1. Suppose (sn)∞n=1 and (tn)∞n=1 are sequences, and that s, t ∈ R. If (sn) convergesto s and (tn) converges to t, then (sn + tn)∞n=1 converges to s+ t.

We could also state this theorem by saying that the following sentence is true.

∀sequences (sn)∞n=1, ∀sequences (tn)∞n=1, ∀s ∈ R, ∀t ∈ R,(limn→∞

sn = s and limn→∞

tn = t

)=⇒ lim

n→∞(sn + tn) = s+ t.

Proof.

1. Let (sn) be a sequence.

2. Let (tn) be a sequence.

3. Let s ∈ R.

4. Let t ∈ R.

5. We must verify the truth of(limn→∞

sn = s and limn→∞

tn = t

)=⇒ lim

n→∞(sn + tn) = s+ t.

(a) Case 1: (limn→∞ sn = s and limn→∞ tn = t) is false. Trivial.

(b) Case 2: (limn→∞ sn = s and limn→∞ tn = t) is true.

We must show that limn→∞(sn + tn) = s+ t is true. Since all of these expressions havea definition associated with them, we can expand on this and say the following.


∀ε1 > 0, ∃N1 ∈ N : ∀n ∈ N, n ≥ N1 =⇒ |sn − s| < ε1.


∀ε2 > 0, ∃N2 ∈ N : ∀n ∈ N, n ≥ N2 =⇒ |tn − t| < ε2.

• We must show that the following sentence is true.

∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒ |(sn + tn)− (s+ t)| < ε.

We verify the truth of the last statement.

i. Let ε > 0.

ii. This is where the bulk of the argument happens.

A. Let ε1 = ε2 . Then ε1 > 0.

B. Let ε2 = ε2 . Then ε2 > 0.

33

C. Since∀ε1 > 0, ∃N1 ∈ N : ∀n ∈ N, n ≥ N1 =⇒ |sn − s| < ε1.

is true, we obtain an N1 ∈ N such that the following sentence is true.

∀n ∈ N, n ≥ N1 =⇒ |sn − s| < ε1.

D. Since∀ε2 > 0, ∃N2 ∈ N : ∀n ∈ N, n ≥ N2 =⇒ |tn − t| < ε2.

is true, we obtain an N2 ∈ N such that the following sentence is true.

∀n ∈ N, n ≥ N2 =⇒ |tn − t| < ε2.

E. Let N = max{N1, N2}.iii. Let n ∈ N.

iv. We must verify the truth of (n ≥ N =⇒ |(sn + tn)− (s+ t)| < ε).


B. Case 2: n ≥ N .Since N = max{N1, N2} ≥ N1, we have n ≥ N1, and so |sn − s| < ε1.Since N = max{N1, N2} ≥ N2, we have n ≥ N2, and so |tn − t| < ε2.Thus,

|(sn + tn)− (s+ t)| = |(sn − s) + (tn − t)| ≤ |sn − s|+ |tn − t| < ε1 + ε2 = ε.

Again notice how algorithmically this proof is written. Here is how you would see most math-ematicians write such a proof.

The proof of theorem 7.11.1 written differently. Suppose (sn) and (tn) are sequences, that s, t ∈ R,that limn→∞ sn = s and limn→∞ tn = t.

We wish to show that limn→∞(sn + tn) = s+ t, so let ε > 0.Since (sn) converges to s, there exists an N1 ∈ N such that

∀n ∈ N, n ≥ N1 =⇒ |sn − s| <ε

2.

Since (tn) converges to t, there exists an N2 ∈ N such that

∀n ∈ N, n ≥ N2 =⇒ |tn − t| <ε

2.

Let N = max{N1, N2}, and suppose that n ∈ N and n ≥ N . Then

|(sn + tn)− (s+ t)| = |(sn − s) + (tn − t)| ≤ |sn − s|+ |tn − t| <ε

2+ε

2= ε.

34

You can see that this proof does exactly what the previous proof does. However, it misses outcertain trivialities. It also misses out explaining exactly what quantified sentence is being checkedat what time, and is less explicit about how known-to-be-true quantified sentences are being used.The order of declarations is still exactly the same.

Eventually, I’ll let you write proofs like the one just written. The reason I won’t just yet is thatthe way I’ve been doing it forces you to declare variables in the correct order. Thus, mistakes youmake should be elsewhere, and we can discuss them as opposed to quantifier errors.

Remark 7.11.2. Let me call your attention to something about the way I wrote the previousalgebra of limits proofs. In the first proof, I wrote. . .

Since∀ε′ > 0, ∃N ′ ∈ N : ∀n ∈ N, n ≥ N ′ =⇒ |sn − s| < ε′

is true, we obtain an N ′ ∈ N such that the following sentence is true.

∀n ∈ N, n ≥ N ′ =⇒ |sn − s| < ε′.

When I write∀ε′ > 0, ∃N ′ ∈ N : ∀n ∈ N, n ≥ N ′ =⇒ |sn − s| < ε′,

I write it as a formula, and remark that the whole statement is true. Then I go ahead and actuallyuse the truth of this statement (ε′ was already specified). I write “we obtain an N ′ ∈ N” in the bulkof the text, in words as opposed to quantifiers, to emphasize that this is an N ′ that I am actuallygoing to use. I follow this up by stating the property that this N ′ has (as a formula):

∀n ∈ N, n ≥ N ′ =⇒ |sn − s| < ε′.

These are subtle uses of text/formulae. They are probably “me-teaching-131A-things” but thereis method in the madness, and it does emphasize the different things which going on. If you writethis way, you will be clearer than most mathematicians.

35

8 Continuity

8.1 The definition

Here’s another important definition.

Definition 8.1.1. Suppose D is a subset of R, f : D −→ R is a function, and a ∈ D.(Here D stands for “domain.”)We say that f is continuous at a iff the following sentence is true.

∀ε > 0, ∃δ > 0 : ∀x ∈ D, |x− a| < δ =⇒ |f(x)− f(a)| < ε.

Remark 8.1.2. If you look at the quantified sentence in the previous definition, you will see thatf and a have not been quantified over. This is okay because they were declared in the preamble.By the time we state the quantified sentence, f and a are “constants.”

Definition 8.1.3. Suppose D is a subset of R and f : D −→ R is a function.We say that f is continuous iff for all a ∈ D, f is continuous at a, i.e. iff the following sentence

is true.∀a ∈ D, ∀ε > 0, ∃δ > 0 : ∀x ∈ D, |x− a| < δ =⇒ |f(x)− f(a)| < ε.

8.2 Motivating the definition of continuity

A picture is worth a thousand words and I will draw at least one in class; just sketch any continuousfunction, and label a on the x-axis and f(a) on the y-axis.

Suppose f : D −→ R is a function and a ∈ D.f is continuous at a if

• whenever a small positive quantity ε is specified

• by keeping the “x-values” close enough to a

• we can ensure that the “y-values” differ from f(a) by less than the small quantity specified.

The first bullet point is said mathematically as ∀ε > 0.The last two are said mathematically as follows.

• ∃δ > 0 (there exists a number telling us how close we need to stay to a)

• : ∀x ∈ D, |x− a| < δ =⇒ (for “x-values” which are this close to a)

• |f(x)− f(a)| < ε (the “y-values” are within ε of f(a)).

8.3 Examples of continuous functions

Theorem 8.3.1. The function f : R −→ R defined by f(x) = x2 is continuous at 2.

Proof. We must verify that the following sentence is true.

∀ε > 0, ∃δ > 0 : ∀x ∈ R, |x− 2| < δ =⇒ |x2 − 4| < ε.

36

1. Let ε > 0.

2. Let δ = min{1, ε5}. We have δ > 0.

3. Let x ∈ R.

4. Suppose that |x− 2| < δ. We must show that |x2 − 4| < ε.

Using the fact that δ = min{1, ε5} ≤ 1, we see that

|x+ 2| = |(x− 2) + 4| ≤ |x− 2|+ 4 < δ + 4 ≤ 5.

Using this together with the fact that δ = min{1, ε5} ≤ε5 , we see that

|x2 − 4| = |x+ 2||x− 2| ≤ 5|x− 2| < 5δ ≤ ε.

Remark 8.3.2. The thinking here was as follows.We have control over |x − 2|: in the part of the proof that requires reasoning, |x − 2| is taken

to be less than δ, and we get to choose δ earlier on in the proof.We want to use our control over |x− 2| to force |x2 − 4| to be less than ε. First, we apply high

school techniques (factoring) to write |x2− 4| as |x+ 2||x− 2|. In this expression we have completecontrol over |x− 2|, but |x+ 2| is a little more confusing.

Thinking about the graph of |x + 2| will convince you that this term can be arbitrarily large(this makes things difficult for us), but if x is close to 2, then it remains smaller: |x+ 2| is less thanor equal to 5 as long as x is within 1 of 2. We can arrange for x to be within 1 of 2, by taking δ tobe less than or equal to 1. We use the triangle inequality to prove all of this without appealing tothe graph.

The other ideas which go into the proof are similar to ones which we have already seen.

By being careful we can amend the argument above so that it works for all points a ∈ R.

Theorem 8.3.3. The function f : R −→ R defined by f(x) = x2 is continuous.


∀a ∈ R, ∀ε > 0, ∃δ > 0 : ∀x ∈ R, |x− a| < δ =⇒ |x2 − a2| < ε.

1. Let a ∈ R.

2. Let ε > 0.

3. Let δ = min{1, ε1+2|a|}. δ is well-defined since 1 + 2|a| 6= 0, and δ > 0.

4. Let x ∈ R.

5. Suppose that |x− a| < δ. We must show that |x2 − a2| < ε.

Using the fact that δ = min{1, ε1+2|a|} ≤ 1, we see that

|x+ a| = |(x− a) + 2a| ≤ |x− a|+ 2|a| < δ + 2|a| ≤ 1 + 2|a|.

Using this together with the fact that δ = min{1, ε1+2|a|} ≤

ε1+2|a| , we see that

|x2 − a2| = |x+ a||x− a| ≤ (1 + 2|a|) · |x− a| < (1 + 2|a|) · δ ≤ ε.

37

Theorem 8.3.4. The function f : R \ {0} −→ R defined by f(x) = 1x is continuous at 1.


∀ε > 0, ∃δ > 0 : ∀x ∈ R \ {0}, |x− 1| < δ =⇒∣∣∣∣1x − 1

∣∣∣∣ < ε.

1. Let ε > 0.

2. Let δ = min{1,ε}2 . We have δ > 0.

3. Let x ∈ R \ {0}.

4. Suppose that |x− 1| < δ. We must show that | 1x − 1| < ε.

Since δ = min{1,ε}2 ≤ 1

2 , we have

1− x ≤ |x− 1| < δ ≤ 1

2.

Thus, x ≥ 12 and |x| = x ≥ 1

2 .

Using the last fact together with δ = min{1,ε}2 ≤ ε

2 , we see that∣∣∣∣1x − 1

∣∣∣∣ =|x− 1||x|

≤ |x− 1|12

= 2|x− 1| < 2δ ≤ ε.

Remark 8.3.5. The thinking here was as follows.We have control over |x− 1|. We want to use our control over |x− 1| to force | 1x − 1| to be less

than ε. First, we apply high school techniques (common denominator) to write | 1x−1| as |x−1|/|x|.In this expression we have complete control over |x− 1|, but our control over |x| is less obvious.

The issue is that |x| being close to 0 makes |x − 1|/|x| large. We can keep |x| away from 0 bykeeping x within 1

2 of 1 and we arrange this by taking δ to be less than or equal to 12 .

The other ideas which go into the proof are similar to ones which we have already seen.

38

By being careful we can amend the argument above so that it works for all points a ∈ R \ {0}.

Theorem 8.3.6. The function f : R \ {0} −→ R defined by f(x) = 1x is continuous.


∀a ∈ R \ {0}, ∀ε > 0, ∃δ > 0 : ∀x ∈ R \ {0}, |x− a| < δ =⇒∣∣∣∣1x − 1

a

∣∣∣∣ < ε.

1. Let a ∈ R \ {0}.

2. Let ε > 0.

3. Let δ = min{|a|,a2ε}2 . We have δ > 0 because a and ε are non-zero.

4. Let x ∈ R \ {0}.

5. Suppose that |x− a| < δ. We must show that | 1x −1a | < ε.

Since δ = min{|a|,a2ε}2 ≤ |a|2 , we have

|a| − |x| ≤∣∣∣∣|x| − |a|∣∣∣∣ ≤ |x− a| < δ ≤ |a|

2.

Thus, |x| ≥ |a|2 .

Using the last fact together with δ = min{|a|,a2ε}2 ≤ a2ε

2 , we see that∣∣∣∣1x − 1

a

∣∣∣∣ =|x− a||x||a|

≤ |x− a||a|2 |a|

=2

a2· |x− a| < 2

a2· δ ≤ ε.

39

9 Uniform Continuity

9.1 The definition and a comparison with the definition of continuity

Uniform continuity is a strange concept. It is not clear to me why it is on the 131A syllabus but itis useful to check a student’s understanding of quantifiers.

Recall definition 8.1.3. It says the following.

Definition. Suppose D is a subset of R and f : D −→ R is a function.We say that f is continuous iff the following sentence is true.

∀a ∈ D, ∀ε > 0, ∃δ > 0 : ∀x ∈ D, |x− a| < δ =⇒ |f(x)− f(a)| < ε.

Compare this with the following definition.

Definition 9.1.1. Suppose D is a subset of R and f : D −→ R is a function.We say that f is uniformly continuous iff the following sentence is true.

∀ε > 0, ∃δ > 0 : ∀x ∈ D, ∀a ∈ D, |x− a| < δ =⇒ |f(x)− f(a)| < ε.

Remark 9.1.2. The only difference in the definitions is that the quantifier ∀a ∈ D has moved frombefore ∃δ > 0 to after ∃δ > 0. Recall from subsection 4.5, the second bullet point that to deal with∃δ > 0 you should say “let δ = BLAH” where BLAH needs to be a positive real number writtenonly in terms of variables which have already been specified. In the definition of continuity thismeans that δ can depend on ε and a. In the definition of uniform continuity this means that δ canonly depend on ε; it can not depend on a.

9.2 Examples of uniformly continuous and non-uniformly continuous functions

Theorem 9.2.1. The function f : [−88, 888] −→ R defined by f(x) = x2 is uniformly continuous.


∀ε > 0, ∃δ > 0 : ∀x ∈ [−88, 888], ∀a ∈ [−88, 888], |x− a| < δ =⇒ |x2 − a2| < ε.

1. Let ε > 0.

2. Let δ = ε2·888 .

3. Let x ∈ [−88, 888].

4. Let a ∈ [−88, 888].

5. Suppose that |x− a| < δ. We must show that |x2 − a2| < ε.

This is given by factoring, applying the triangle inequality, and the fact that |x|, |a| ≤ 888:

|x2 − a2| = |x+ a||x− a| ≤ (|x|+ |a|)|x− a| ≤ 2 · 888 · |x− a| < ε.

40

Remark 9.2.2. I hope that the previous proof did not feel too difficult. We factored |x2 − a2| sothat we could see |x− a|. The domain of the function ensures that |x| and |a| do not get too large,and so applying the triangle inequality to |x+ a| is useful.

Remark 9.2.3. In theorem 8.3.3 we showed that the function f : R −→ R defined by f(x) = x2 iscontinuous. We could try to rearrange this proof into one showing that f is uniformly continuous.It would read as follows.

1. Let ε > 0.

2. Let δ = min{1, ε1+2|a|}. δ is well-defined since 1 + 2|a| 6= 0, and δ > 0.

3. Let x ∈ R.

4. Let a ∈ R.

5. Suppose that |x− a| < δ . . .

This makes NO SENSE AT ALL. The declaration in 2. uses a, but a is not declared until 4. Infact, f is not uniformly continuous.

After doing a few continuity proofs, I hope that you will start to see that we need a smaller δ(relative to ε) when the graph of the function under consideration is steeper. When the domain off(x) = x2 is [−88, 888], f is steepest at 888. The issue when we take the domain to be all of Ris that there is no upper bound on steepness: f becomes steeper and steeper as x is made larger.You can see in the following proof that when δ is small, x and a are large, and so the proof makesuse of this fact.

Theorem 9.2.4. The function f : R −→ R defined by f(x) = x2 is not uniformly continuous.


∃ε > 0 : ∀δ > 0, ∃x ∈ R : ∃a ∈ R : |x− a| < δ and |x2 − a2| ≥ ε.

1. Let ε = 1.

2. Let δ > 0.

3. Let x = 1δ .

4. Let a = x+ δ2 .

5. We have |x− a| = δ2 < δ. We also have

|x2 − a2| =(x+

δ

2

)2

− x2 = δx+δ2

4≥ δx = 1 = ε.

41

Theorem 9.2.5. The function f : [1,∞) −→ R defined by f(x) = 1x is uniformly continuous.


∀ε > 0, ∃δ > 0 : ∀x ∈ [1,∞), ∀a ∈ [1,∞), |x− a| < δ =⇒∣∣∣∣1x − 1

a

∣∣∣∣ < ε.

1. Let ε > 0.

2. Let δ = ε.

3. Let x ∈ [1,∞).

4. Let a ∈ [1,∞).

5. Suppose that |x− a| < δ. We must show that | 1x −1a | < ε.

Since a, x ∈ [1,∞) we have ∣∣∣∣1x − 1

a

∣∣∣∣ =|x− a||x||a|

≤ |x− a| < δ = ε.

Remark 9.2.6. As with f(x) = x2, the domain makes a huge difference to the uniform continuityof f(x) = 1

x . The graph of f(x) = 1x becomes steeper and steeper as x approaches 0. You can see

in the following proof that when δ is small, x and a are small, and so the proof makes use of thisfact.

Theorem 9.2.7. The function f : (0,∞) −→ R defined by f(x) = 1x is not uniformly continuous.


∃ε > 0 : ∀δ > 0, ∃x ∈ (0,∞) : ∃a ∈ (0,∞) : |x− a| < δ and

∣∣∣∣1x − 1

a

∣∣∣∣ ≥ ε.1. Let ε = 1.

2. Let δ > 0.

3. Let x = δδ+1 . We could also write this as x = 1

1+ 1δ

.

4. Let a = δ.

5. We have 0 < δδ+1 < δ. Thus, x, a ∈ (0, δ] and so |x− a| < δ. Moreover,∣∣∣∣1x − 1

a

∣∣∣∣ =

∣∣∣∣(1 +1

δ

)− 1

δ

∣∣∣∣ = 1 ≥ ε.

42

10 What are the real numbers? The completeness axiom

10.1 What are the real numbers?

We have just spent three weeks getting used to using quantifiers and some important definitions.We are now in a position to do some serious mathematics.

The main difference between lower division “mathematics” and (upper division) mathematicsis that we address the following questions.

• All these rules that we keep using. . . where do they come from, why are they true, and howdo we prove them?

• These mathematical objects that we’ve been talking about. . . what are they?

We’ve already started addressing the first questions. This section starts to address the questionof the second bullet point.

What are the real numbers?

This is a surprisingly difficult question. In this class, we will take the answer to be:

• a set together with

• operations of addition + and multiplication ·,

• an ordering ≤,

• and these obey various axioms.

The approach taken might be unsatisfying for you in a couple of ways.

• We will not be explicit about all of the axioms. Most of them hold for Q, and you know howaddition, subtraction, multiplication, division, and ≤ work for Q. For this reason, I chose notto talk about them.

• We will not prove that there actually exists something satisfying the axioms. I would love totell you about a construction of the real numbers, but there is not enough time. If you wantto convince yourself of the existence of the real numbers, you can look up Dedekind cuts.

We will focus on the axiom that distinguishes R from Q. First, let’s recap how you learned aboutQ. The natural numbers N are the simplest numbers we can think of, the numbers we use to count:

1, 2, 3, 4, . . .

They seem like enough until we try and solve the equation x+ 2 = 1. We realize that to solve thisequation we need the number −1 and are led to consider the integers Z:

0,±1,±2,±3, . . .

These seem pretty great until we try and solve an equation like 2x = 1. To solve this equation weneed the number 1

2 . We are lead to consider the rationals Q:{m

n: m ∈ Z, n ∈ N

}.

Just in case you are philosophically minded (otherwise you can ignore this). . .

43

• N∪{0} can be constructed by starting with the empty set and iterating the same constructionover and over again.

0 = ∅, 1 = {∅}, 2 = {∅, {∅}}, 3 = {∅, {∅}, {∅, {∅}}}, . . .

• The integers can be constructed by imposing an equivalence relation on the set N × N (thisset contains pairs of natural numbers).

• The rational numbers can be constructed by imposing an equivalence relation on the set Z×N(this set contains pairs constisting of one integer and one natural number).

• You would know all of this after taking classes in abstract algebra and set theory.

What deficiency do the rationals suffer from that lead us to consider the real numbers? We moti-vated all our previous considerations by the inability to solve an equation. We cannot solve x2 = −1in Q, but this is not a reason to consider the reals, for the solution to this equation is not a realnumber; it is i ∈ C. (The complex numbers C actually have the magic property that any polynomialhas a root, that is, they are algebraically closed.)

N ⊆ Z ⊆ Q ⊆ R ⊆ C.

When did you first learn about real numbers which are not rational? My guess is that you wouldprobably say, “when I was told about

√2, π, and e,” but I think you knew about them even before

that. After you were told about fractions, you were probably taught to write them in their decimalexpansion and, soon enough, you became happy with the idea that if you specify a non-negativeinteger a0 and a sequence a1, a2, a3, a4, . . . with each an ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, then the notation

a0 . a1 a2 a3 a4 · · ·

determines a non-negative real number. What does this decimal expansion mean? It means

a0 +

∞∑k=1

10−kak.

There’s an infinity sign here, which we haven’t explained yet. Define a sequence by

s1 = a0 . a1,

s2 = a0 . a1 a2,

s3 = a0 . a1 a2 a3,

...

sn = a0 . a1 a2 a3 · · · an = a0 +

n∑k=1

10−kak,

We expect this sequence to converge and a0 . a1 a2 a3 a4 · · · represents the limit of the sequence.The reason we expect the sequence to converge is that we do not think of the real number line as

44

having “holes” in it. The problem with the rationals is that there are “holes” and such a sequencedoes not necessarily converge. As an example, consider the sequence

s1 = 1.4, s2 = 1.41, s3 = 1.414, s4 = 1.4142, s5 = 1.41421, . . . , sn =b10n√

2c10n

, . . .

Each number in the sequence is rational but the limit is√

2, which is irrational. The reals are thesmallest set containing the rationals which don’t have holes. This is expressed by the completenessaxiom, which is the topic of the next section.

N − want to be able to count

Z − want to be able to solve x+ 2 = 1

Q − want to be able to solve 2x = 1

R − want there to be no holes, i.e. for the completeness axiom to hold

C − want to be able to solve x2 = −1

10.2 Boundedness, maximum and minimum elements

The completeness axiom distinguishes the reals from the rationals. Before being able to state thecompleteness axiom, we have to make some important definitions.

Definition 10.2.1. Suppose S is a nonempty subset of R.

1. Suppose M ∈ R. M is said to be an upper bound for S iff ∀s ∈ S, s ≤M .

2. Suppose m ∈ R. m is said to be an lower bound for S iff ∀s ∈ S, s ≥ m.

Definition 10.2.2. Suppose S is a nonempty subset of R and s0 ∈ S.

1. s0 is said to be the maximum or greatest element of S iff s0 is an upper bound for S, i.e. iff

∀s ∈ S, s ≤ s0.

In this case, we write maxS = s0.

2. s0 is said to be the minimum or least element of S iff s0 is a lower bound for S, i.e. iff

∀s ∈ S, s ≥ s0.

In this case, we write minS = s0.

Remark 10.2.3. Suppose S is a nonempty subset of R. The difference “between being an upperbound for S” and “being the maximum element of S” is “containment in S”: an upper bound doesnot necessarily have to be in S; the maximum element does have to be in S.

45

Example 10.2.4.

• 1 is the maximum element of [0, 1].

• 1 is an upper bound for [0, 1].

• 2 is not the maximum element of [0, 1].

• 2 is an upper bound for [0, 1].

Example 10.2.5.

• 1 is not the maximum element of [0, 1).

• 1 is an upper bound for [0, 1).

• 2 is not the maximum element of [0, 1).

• 2 is an upper bound for [0, 1).

Example 10.2.6. [0, 1) does not have a maximum element.

Proof. The statement that [0, 1) has a maximum element is

∃s0 ∈ [0, 1) : ∀s ∈ [0, 1), s ≤ s0.

The negation is (∀s0 ∈ [0, 1), ∃s ∈ [0, 1) : s > s0). To verify this we let s0 ∈ [0, 1) and s = s0+12 .

Example 10.2.7. N does not have an upper bound.

Proof. The statement that N has an upper bound is

∃M ∈ R : ∀n ∈ N, n ≤M.

The negation is (∀M ∈ R, ∃n ∈ N : n > M). To verify this we let M ∈ R and n = d|M |e+ 1.

Definition 10.2.8. Suppose S is a nonempty subset of R.

1. S is said to be bounded above iff there is an upper bound for S, i.e. iff

∃M ∈ R : ∀s ∈ S, s ≤M.

2. S is said to be bounded below iff there is a lower bound for S, i.e. iff

∃m ∈ R : ∀s ∈ S, s ≥ m.

3. S is said to be bounded iff S is bounded above and bounded below.

Remark 10.2.9. If a nonempty subset of R has a maximum element, it is bounded above.

46

Example 10.2.10.

1. Let X = {1, 4, 5, 6, 8}.

• X is bounded above because 1742 is an upper bound for X.

• X is bounded below because −π is a lower bound for X.

• X is bounded.

• minX = 1 and maxX = 8. Both are guarenteed to exist because X is finite.

2. Let Y = {q ∈ Q : 0 ≤ q ≤√

3}.

• Y is bounded above because 10 is an upper bound.

• Y is bounded below because −1 is a lower bound.

• Y is bounded.

• minY = 0. However, maxY does not exist (proving this is a good exercise).

3. N, Z, Q and R are not bounded above.

4. N is bounded below because −227 is a lower bound for N.

5. Z, Q and R are not bounded below.

10.3 The completeness axiom

Definition 10.3.1. Here is what we mean by the completeness axiom.

• Suppose S is a nonempty subset of R which is bounded above. Then the set

{M ∈ R : M is an upper bound for S}

has a minimum element.

Remark 10.3.2.

1. Stating the completeness axiom required three definitions: bounded above, upper bound, andminimum.

2. Another way to say the completeness axiom which rolls off the tongue a little more easily:

A nonempty subset of R which is bounded above has a least upper bound.

3. We do not state the analogous version of the completeness axiom for bounded below sets asan axiom because it is a corollary of the completeness axiom just stated. . .

Corollary 10.3.3. Suppose S is a nonempty subset of R which is bounded below. Then the set

{m ∈ R : m is a lower bound for S}

has a maximum element.

Proof. This is a little too involved for a first example involving these concepts. It is a good exercisefor you later. Here’s the idea you should try and implement. Define −S := {−s : s ∈ S}. Since Sis bounded below, −S is bounded above; thus, −S has a least upper bound M ; m := −M is thegreatest lower bound for S.

47

10.4 Suprema and Infima

The completeness axiom and its corollary allow us to make the following definition.

Definition 10.4.1. Let S be a nonempty subset of R.

1. If S is bounded above, then the supremum of S is the least upper bound of S. We write supSfor the supremum of S.

2. If S is bounded below, then the infimum of S is the greatest lower bound of S. We use inf Sto denote the infimum of S.

This definition reads nice and concisely but how do we use it? Suppose S is a nonempty subsetof R which is bounded above and M ∈ R. In order to show that supS = M we have to show twothings (and we should definitely do part 1 before part 2).

1. M is an upper bound for S, i.e. ∀s ∈ S, s ≤M .

2. M is the least upper bound for S. The most useful way of expressing this is using the followingsentence.

∀M ′ ∈ R, M ′ < M =⇒ (∃s ∈ S : s > M ′).

(Why is this the correct sentence? Well, its contrapositive is given by

∀M ′ ∈ R, (∀s ∈ S : s ≤M ′) =⇒ M ≤M ′

and this says “if M ′ is an upper bound for S, then M ≤M ′.”)

Let’s see an example.

Theorem 10.4.2. Suppose that b ∈ R. Then sup(−∞, b) = b.

Proof. We proceed as just discussed. Let S = (−∞, b).

1. By definition of S, for all s ∈ S, we have s < b. In particular, for all s ∈ S, s ≤ b, and thisshows that b is an upper bound for S.

2. Suppose b′ < b. Letting s = b+b′

2 , we have b′ < s < b. The definition of S and the fact thats < b tells us that s ∈ S, and so s > b′ demonstrates that b′ is not an upper bound for S. Weconclude that b is the least upper bound for S.

By checking these two points we have confirmed that supS = b.

Remark 10.4.3. In the above proof, we let s = b+b′

2 because it is nice to give an explicit exampleof an existence statement when it is possible to do so. However, it will not always be possible togive such a simple example. Also, the fact that this s is an element of S is a consequence of the Sin question, (−∞, b); such an averaging technique will not usually give an element of a set S.

Remark 10.4.4. Part 1 of checking supS = M actually shows that supS ≤M .What we are saying is that if we show M to be an upper bound of S, then supS ≤ M , and

this is because supS is the least upper bound. This strategy for showing such an inequality is usedfrequently. Note it: if S is a nonempty bounded above subset of R, then the following sentence istrue.

∀M ∈ R, (∀s ∈ S, s ≤M) =⇒ supS ≤M.

This sentence expresses, again, “supS is less than or equal to all upper bounds of S.”

48

You should check that you can prove the results recorded in the following example.

Example 10.4.5.

1. Let X = {1, 4, 5, 6, 8}. Then inf X = 1 and supX = 8.

2. Let Y = {q ∈ Q : 0 ≤ q ≤√

3}. Then inf Y = 0 and supY =√

3.

3. inf N = 1.

4. inf(a, b) = inf[a, b] = inf[a, b) = inf(a, b] = a.

5. sup(a, b) = sup[a, b] = sup[a, b) = sup(a, b] = b.

The completeness axiom is supposed to fill in the gaps of Q. In particular, the existence of√

2should follow from the completeness axiom and this is the result of the following theorem, whichgives another example of using the definitions carefully. It is a little tricky because of the messyinequalities. The proof is the first one that came to mind when I tried to prove it. Maybe you canimprove it and make it a little nicer? It is a good exercise to check you understand the applicationof the main definitions even if the inequalities concern you.

Theorem 10.4.6. There is a positive real number x such that x2 = 2, i.e.√

2 exists.

Proof. Consider the setS := {y ∈ R : y2 < 2}.

Since 1 ∈ S, S is nonempty.If y > 2, then y2 > 4 ≥ 2, so that y /∈ S. The contrapositive says that if y ∈ S, then y ≤ 2, and

so we see that 2 is an upper bound for S. In particular, S is bounded above.Let x = supS. We wish to show x is positive and x2 = 2.Because 1 ∈ Y and x is an upper bound for Y , we have 1 ≤ x.Because 2 is an upper bound for Y and x is the least upper bound for Y , we have x ≤ 2.Summarizing, we have 1 ≤ x ≤ 2, and we are left to show that x2 = 2.First, since this will be useful later, note that 8x ≥ 8 > 2 = 22 − 2 ≥ x2 − 2.Suppose for contradiction that x2 < 2. Let δ = 2 − x2. Then δ > 0 and we can choose ε with

0 < ε < min{1, δ5}. Since we know x ≤ 2, we have

(x+ ε)2 = x2 + 2εx+ ε2 ≤ x2 + 4ε+ ε2 = x2 + ε(4 + ε) < x2 + 5ε < x2 + δ = 2,

and so x+ ε ∈ S. Since x+ ε > x, this contradicts x being an upper bound for S. We have checkedthat we cannot have x2 < 2.

Suppose for contradiction that x2 > 2. Let ε = x2−22 . Since x ≤ 2 < x2, we have

2 =2 + 2

2<x2 + 2

2= x2 − x2 − 2

2= x2 − ε ≤ x2 − ε · x

2< x2 − ε · x

2+ε2

16=

(x− ε

4

)2

.

We see that ε > 0, and so x− ε4 < x. Since x is the least upper bound of S, x− ε

4 is not an upperbound for S, and this means there is a y ∈ S with y > x− ε

4 . It is possible to check that x− ε4 > 0

(using the fact that 8x > x2 − 2), and so(x− ε

4

)2

< y2 < 2.

We obtain 2 < 2, a contradiction. Thus, we have checked that we cannot have x2 > 2 either.We conclude that x2 = 2, as required.

49

10.5 Contrapositive and contradiction

To verify the truth of a statement (P =⇒ Q) there are three standard ways to proceed.

1. Verifying it directly: we suppose that P is true and demonstrate that Q must be true.

2. Verifying the contrapositive, the statement that (NOT(Q) =⇒ NOT(P )).

3. We use a proof by contradiction: we suppose NOT(P =⇒ Q) is true, i.e. (P and NOT(Q))is true, and demonstrate that nonsense follows, e.g. 0 6= 0.

A proof which makes use of the contrapositive can always be turned into a proof by contradiction:if we can prove (NOT(Q) =⇒ NOT(P )) is true and we suppose that (P and NOT(Q)) is true,then we can prove (P and NOT(P )) is true, and this is nonsense.

Consider the following proposition, which was used in the last theorem.

Proposition 10.5.1. Let S := {y ∈ R : y2 < 2}. Then the following sentence is true.

∀y ∈ R, y ∈ S =⇒ y ≤ 2.

Proof 1. We prove the contrapositive.Suppose that y > 2. Then y2 > 4 and so y2 ≥ 2. By definition of S, this means that y /∈ S.

Proof 2. We use a proof by contradiction.Suppose that y ∈ S. Suppose for contradiction that y > 2. Then y2 > 4 and so y2 ≥ 2. By

definition of S, this means that y /∈ S. This is a contradiction to the fact that y ∈ S.

Proof 3. We use a proof by contradiction.Suppose that y ∈ S. Suppose for contradiction that y > 2. Then y2 > 4 and so y2 ≥ 2. Since

y ∈ S, we have y2 < 2. This is a contradiction to the fact that y2 ≥ 2.

Proof 1 verified the contrapositive of the statement. Turning this into a proof by contradiction,in the way decribed above, results in proof 2. Proof 3 gives a different proof by contradiction.

Hopefully proofs 2 and 3 demonstrate the following fact. When proving something by contradic-tion, there will be lots of potential contradictions and it is important to highlight which statementsare contradicting one another.

10.6 An almost-proof of corollary 10.3.3

Theorem 10.6.1. Suppose S is a nonempty subset of R which is bounded below. Define

−S := {−s : s ∈ S}.

Then −S is bounded above and inf S = − sup(−S).

Proof. Suppose S is a nonempty subset of R which is bounded below. Let −S be as in the theoremstatement. First, we show that −S is bounded above. Since inf S is a lower bound for S we have

∀s ∈ S, s ≥ inf S.

We claim that (∀t ∈ −S, t ≤ − inf S). Let t ∈ −S. Then, by definition of −S, for some s ∈ S, wehave t = −s. Since s ≥ inf S, we have t = −s ≤ − inf S. Thus, −S is bounded above by − inf S.

50

Since −S is bounded above, sup(−S) exists. We wish to show that − sup(−S) is the greatestlower bound of S. First, we show that it is a lower bound of S, i.e. that

∀s ∈ S, s ≥ − sup(−S).

Let s ∈ S. Then −s ∈ −S and, since sup(−S) is an upper bound for −S, we have −s ≤ sup(−S).Thus, s ≥ − sup(−S).

Next, we must show that − sup(−S) is the greatest lower bound of S, i.e.

∀m′ ∈ R, m′ > − sup(−S) =⇒ (∃s ∈ S : s < m′).

Suppose that m′ ∈ R and that m′ > − sup(−S). Then −m′ < sup(−S). Because sup(−S) is theleast upper bound of −S, −m′ is not an upper bound for −S. This means that there is a t ∈ −Ssuch that t > −m′. By definition of −S, for some s ∈ S, we have t = −s. Thus, s = −t < m′. Thisshows that m′ is not a lower bound for S, so we have demonstrated that − sup(−S) is the greatestlower bound of S. By definition, inf S = − sup(−S).

Remark 10.6.2. The proof just given was longer than necessary.After showing that (∀t ∈ −S, t ≤ − inf S), we can conclude that sup(−S) ≤ − inf S.After showing that (∀s ∈ S, s ≥ − sup(−S)), we can conclude that inf(S) ≥ − sup(−S).This allows us to conclude that − sup(−S) ≥ inf S ≥ − sup(−S), which gives equality.However, the proof given has the benefit that with small alterations it proves corollary 10.3.3.

It’s a good exercise for you to see what alterations need to be made.

11 Some more on sequences (not lectured)

I will not lecture this section in much detail. You should read it yourself. It fills in some gaps.

11.1 Uniqueness of limits

Recall the definition of a sequence converging to a particular limit, and a sequence converging.

Definition.

• Suppose (sn)∞n=1 is a sequence and L ∈ R.

We say that (sn) converges to L iff the following sentence is true.

∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒ |sn − L| < ε.

• Suppose (sn)∞n=1 is a sequence.

We say that (sn) converges iff the following sentence is true.

∃L ∈ R : ∀ε > 0, ∃N ∈ N : ∀n ∈ N, n ≥ N =⇒ |sn − L| < ε.

The definition of convergence makes it sounds like there could be more than one limit; there isnot. Before proving this result, we make note of a useful fact which is used frequently in analysis.

51

Lemma 11.1.1. ∀x ∈ R, (∀ε > 0, |x| < ε) =⇒ x = 0.

Proof. The contrapositive is

∀x ∈ R, x 6= 0 =⇒ (∃ε > 0 : |x| ≥ ε).

The proof of this is: suppose x ∈ R and x 6= 0; then let ε = |x|.

Theorem 11.1.2. A convergent sequence has a unique limit, i.e. the following sentence is true.

∀sequences (sn)∞n=1, ∀L1 ∈ R, ∀L2 ∈ R,(

limn→∞

sn = L1 and limn→∞

sn = L2

)=⇒ L1 = L2.

Proof. Suppose that (sn)∞n=1 is a sequence, that L1, L2 ∈ R, limn→∞ sn = L1, and limn→∞ sn = L2.We must show that L1 = L2.

Remark 11.1.3. If one thinks of limn→∞ sn as a number, then this looks like it should be trivial.The problem is that we have been using “limn→∞ sn = L” as an abbreviation for “(sn) convergesto L;” it has not meant “limn→∞ sn is a number and it equals L.” The point of proving theorem11.1.2 is to show that “limn→∞ sn = L” does also serve as an abbreviation for the latter statement.

To demonstrate L1 = L2 it is enough, by the previous lemma, to verify that

∀ε > 0, |L1 − L2| < ε.

Let ε > 0.Let ε1 = ε

2 . Since limn→∞ sn = L1, there is an N1 ∈ N such that

∀n ∈ N, n ≥ N1 =⇒ |sn − L1| < ε1.

Let ε2 = ε2 . Since limn→∞ sn = L2, there is an N2 ∈ N such that

∀n ∈ N, n ≥ N2 =⇒ |sn − L2| < ε2.

Let N = max{N1, N2}.Since N ≥ N1, we have |sN − L1| < ε1. Since N ≥ N2, we have |sN − L2| < ε2. Thus,

|L1 − L2| ≤ |sN − L1|+ |sN − L2| < ε1 + ε2 = ε.

Remark 11.1.4. To reiterate what the previous remark said. . . Now that we have proved theorem11.1.2, limn→∞ sn = L means “(sn) converges, limn→∞ sn denotes the unique limit of (sn), and thisnumber is equal to L.” Provided that (sn) converges, limn→∞ sn is now meaningful by itself.

52

11.2 Bounded sequences

Definition 11.2.1.

1. We say that a sequence (sn)∞n=1 is bounded above iff the set {sn : n ∈ N} is bounded above.

2. We say that a sequence (sn)∞n=1 is bounded below iff the set {sn : n ∈ N} is bounded below.

3. We say that a sequence (sn)∞n=1 is bounded iff the set {sn : n ∈ N} is bounded.

Proposition 11.2.2. Suppose that (sn)∞n=1 is a sequence. (sn) is bounded if and only if the followingsentence is true.

∃M > 0 : ∀n ∈ N, |sn| ≤M.

Proof. A good exercise.

Theorem 11.2.3. Suppose that (sn)∞n=1 is a sequence. If (sn) converges, then (sn) is bounded.

Proof. Suppose (sn)∞n=1 is a sequence, and that it converges to L ∈ R. We must show that (sn) isbounded, and we will do this by showing that the following sentence is true.

∃M > 0 : ∀n ∈ N, |sn| ≤M.

By taking ε = 1, the definition of convergence tells us that there exists an N ∈ N such that

∀n ∈ N, n ≥ N =⇒ |sn − L| < 1.

(Here, it is like we have used a “computer program” to get N ; upon giving it 1, it returned N .)Thus, when n ∈ N and n ≥ N , we have

|sn| = |sn − L+ L| ≤ |sn − L|+ |L| < 1 + |L|.

Let M be the maximum element in the set

{|sn| : n ∈ N, n < N} ∪ {1 + |L|}.

Then M ≥ 1 + |L| > 0, and for all n ∈ N, |sn| ≤M . This demonstrates that (sn) is bounded.

Remark 11.2.4. In the previous proof we used the definition of convergence to bound the termsin the sequence appearing after the N -th term. The terms before the N -th term are easily boundedsince there are finitely many of them: we can take the maximum of them.

Theorem 11.2.5. Suppose (sn)∞n=1 and (tn)∞n=1 are sequences. If (sn) converges to 0 and (tn) isbounded, then (sntn)∞n=1 converges to 0, i.e. the following sentence is true.

∀sequences (sn)∞n=1, ∀sequences (tn)∞n=1,(limn→∞

sn = 0 and (tn) is bounded

)=⇒ lim

n→∞sntn = 0.

Proof. Left to you as preparation for the second quiz.

53

11.3 Algebra of limits (continued)

Theorem 11.3.1. Suppose that (sn)∞n=1 and (tn)∞n=1 are convergent sequences, and that c ∈ R.

1. limn→∞ csn = c · limn→∞ sn.

2. limn→∞(sn + tn) = limn→∞ sn + limn→∞ tn.

3. limn→∞ sntn = (limn→∞ sn)(limn→∞ tn).

4. If limn→∞ tn 6= 0, then there exists an N ∈ N such that ( sntn )∞n=N makes sense, and

limn→∞

sntn

=limn→∞ snlimn→∞ tn

.

Proof. Suppose that (sn) and (tn) are convergent sequences, and that c ∈ R.

1. This is theorem 7.9.1.

2. This is theorem 7.11.1.

3. Let s = limn→∞ sn and t = limn→∞ tn.

It is easy to prove that limn→∞ s = s, limn→∞ t = t, and limn→∞ st = st.

The previous part of the theorem tells us that limn→∞(sn − s) = 0 and limn→∞(tn − t) = 0.

Theorems 11.2.3 and 11.2.5 tells us that limn→∞(sn − s)t = 0 and limn→∞ sn(tn − t) = 0.

Since sntn = (sn − s)t+ sn(tn − t) + st, the previous part of the theorem tells us that

limn→∞

sntn = 0 + 0 + st = ( limn→∞

sn)( limn→∞

tn).

4. Suppose that limn→∞ tn 6= 0.

The issue with ( sntn ) making sense is that we cannot divide by zero.

Let t = limn→∞ tn. Then |t| > 0. Since limn→∞ tn = t, there is an N ∈ N such that

∀n ∈ N, n ≥ N =⇒ |tn − t| < |t|.

When n ∈ N and n ≥ N we have

|t| ≤ |t− tn|+ |tn| < |t|+ |tn|

and so when n ∈ N and n ≥ N , we have |tn| > 0. This means that ( sntn )∞n=N makes sense.

For the rest, since sntn

= sn · 1tn

, it is enough, by the previous part of the theorem, to show

that limn→∞1tn

= 1t . We use a cheap trick for this. . .

Let f : R \ {0} −→ R be defined by f(x) = 1x . Recall from theorem 8.3.6 that f is continuous

at t. Because ( 1tn

)∞n=N is a sequence converging to t with each term in R \ {0}, the sequence

definition of continuity (definition 13.1.1) gives us limn→∞ f(tn) = f(t), i.e. limn→∞1tn

= 1t .

54

12 Sequences and the completeness of R

We have said already that the completeness axiom distinguishes the reals from the rationals. Inthis section, we will prove results about sequences which depend on the completeness axiom. Thesetheorems are not true for the rationals; we are finally doing real analysis.

12.1 Cauchy sequences

At least four of you have asked me,

“how do I show a sequence converges if I cannot figure out its limit L?”

((1 + 1n)n) is an example of a sequence for which this question is relevant. This sequence converges

to the number e. But what is e?!? Some people define e to be the limit of this sequence. Thissounds circular! We would like to be able to assert that this sequence converges without referenceto its limit. That is the purpose of Cauchy sequences.

A sequence is convergent if the terms in the sequence become arbitrarily close to a specified limitL. In this case, the triangle inequality tells us that the terms in the sequence become arbitrarilyclose to one another, too (see theorem 12.1.3 below). This is the motivation for the definition of aCauchy sequence.

Definition 12.1.1. Suppose (sn)∞n=1 is a sequence.We say that (sn) is Cauchy iff the following sentence is true.

∀ε > 0, ∃N ∈ N : ∀n ∈ N, ∀m ∈ N, n,m ≥ N =⇒ |sn − sm| < ε.

Theorem 12.1.2. Suppose that

• a0 ∈ N ∪ {0},

• (an)∞n=1 is a sequence,

• ∀n ∈ N, an ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

Then sequence (sn)∞n=1 defined by sn = a0 . a1 a2 a3 · · · an = a0 +∑n

k=1 10−kak is Cauchy.

Proof. Let ε > 0, N = d1ε e+ 1, n,m ∈ N, and suppose that n,m ≥ N . Without loss of generality,suppose that n > m (the n = m case is easy and |sn − sm| = |sm − sn|). Then

|sn − sm| =n∑

k=m+1

10−kak ≤ 10−m ≤ 1

m≤ 1

N< ε.

(The first inequality says 0.0 · · · 00am+1 · · · an ≤ 0.0 · · · 01, where the “underline” is used to indicatethat these are the same zeroes.)

One property that the reals have, which the rationals do not, is that Cauchy sequences converge.This is stated in the next theorem. We will have to wait until (12.4) before we can prove this result.

Theorem (Completeness of R). Suppose that (sn)∞n=1 is a sequence. If (sn) is Cauchy, then (sn)converges.

55

The opposite implication is far easier to prove.

Theorem 12.1.3. Suppose that (sn)∞n=1 is a sequence. If (sn) converges, then (sn) is Cauchy.

Proof. Suppose (sn)∞n=1 is a sequence which converges to a real number L. We must show that (sn)is Cauchy.

Let ε > 0. Since ε2 > 0 and (sn) converges to L, there exists an N ∈ N such that

∀n ∈ N, n ≥ N =⇒ |sn − L| <ε

2.

Thus, when n,m ∈ N and n,m ≥ N , we have

|sn − sm| ≤ |sn − L|+ |sm − L| <ε

2+ε

2= ε.

Thus, we have demonstrated that (sn) is Cauchy.

Our first step towards proving that Cauchy sequences converge is the following theorem.

Theorem 12.1.4. Suppose that (sn)∞n=1 is a sequence. If (sn) is Cauchy, then (sn) is bounded.

Proof. This is almost identical to the proof of theorem 11.2.3.Suppose (sn)∞n=1 is a Cauchy sequence. We must show that (sn) is bounded.By taking ε = 1, the definition of a Cauchy sequence tells us that there exists an N ∈ N such

that∀n ∈ N, ∀m ∈ N, n,m ≥ N =⇒ |sn − sm| < 1.

Thus, when n ∈ N and n ≥ N , we have

|sn| = |sn − sN + sN | ≤ |sn − sN |+ |sN | < 1 + |sN |.

Let M be the maximum element in the set

{|sn| : n ∈ N, n < N} ∪ {1 + |sN |}.

Then M ≥ 1 + |sN | > 0, and for all n ∈ N, |sn| ≤M . This demonstrates that (sn) is bounded.

12.2 Monotone Sequences

Once we prove that Cauchy sequences converge we are in good shape: we can then prove a sequenceconverges without having to know its limit. Let’s deal with another situation where we can do thesame.

Definition 12.2.1. Let (sn)∞n=1 be a sequence.

1. (sn) is said to be increasing iff for all n ∈ N, sn+1 ≥ sn.

2. (sn) is said to be decreasing iff for all n ∈ N, sn+1 ≤ sn.

3. We say a sequence is monotone iff it is increasing or decreasing.

56

Example 12.2.2.

1. The sequence (1− 1n)∞n=1 is increasing.

2. The sequence (0)∞n=1 is increasing and decreasing.

3. The sequence ((−1)n)∞n=1 is not monotone.

Theorem 12.2.3. Suppose that (sn)∞n=1 is a sequence. If (sn)∞n=1 is increasing and bounded above,then (sn) is convergent.

Proof. Suppose (sn) is an increasing sequence which is bounded above. To say that (sn) is boundedabove means, by definition, that the set {sn : n ∈ N} is bounded above. The completeness axiomsays that this set has a supremum. Let

S = sup{sn : n ∈ N}.

We’ll show that (sn) converges to S.Let ε > 0. Then S− ε < S, which means S− ε is not an upper bound for {sn : n ∈ N}. So there

is an N ∈ N with sN > S − ε. Since (sn) is increasing, when n ∈ N and n ≥ N we have sn ≥ sN .For all n ∈ N, we also have sn ≤ S. Thus, when n ∈ N and n ≥ N , we have

0 ≤ S − sn ≤ S − sN < ε and |sn − S| < ε.

This demonstrates that (sn) converges to S which is what we wanted to show.

Remark 12.2.4. Notice how the completeness axiom provided us with the limit for the sequencein the previous proof. Without the completeness axiom, we would have been completely stumped.

Corollary 12.2.5. Suppose that (sn)∞n=1 is a sequence. If (sn)∞n=1 is decreasing and bounded below,then (sn) is convergent.

Proof. Suppose (sn)∞n=1 is an decreasing sequence which is bounded below. Then (−sn)∞n=1 is anincreasing sequence which is bounded above. The previous theorem tells us that (−sn) converges,and then the algebra of limits (11.3.1) tells us that (sn) converges.

Putting the last two results together gives the following theorem.

Theorem 12.2.6. Suppose that (sn)∞n=1 is a sequence. If (sn)∞n=1 is monotone and bounded, then(sn) is convergent.

Example 12.2.7. Let s1 = 1 and, for n ∈ N, let sn+1 = [1− 14n2 ]sn.

• Claim 1: each sn ≥ 0.

• Claim 2: (sn) is decreasing.

• Thus (sn) converges. Its limit is, in fact, 2π . How could you calculate this?! This result says

2

π=

1 · 32 · 2

· 3 · 54 · 4

· 5 · 76 · 6

· 7 · 98 · 8

· 9 · 11

10 · 10· 11 · 13

12 · 12· · · ·

57

12.3 Subsequences and the Bolzano-Weierstrass theorem

Definition 12.3.1. Suppose that (sn)∞n=1 is a sequence and that (nk)∞k=1 is sequence such that

∀k ∈ N, nk ∈ N and nk+1 > nk.

Then (snk)∞k=1 = sn1 , sn2 , sn3 , . . . is a sequence. (snk)∞k=1 is said to be a subsequence of (sn)∞n=1.

Example 12.3.2. Suppose that we have:

• (sn) = 8, 4, 6, 4, 7, 4, 5, 5, 5, 34, 35, 35, 7, 3, 35, 24, 6, 456, 7, 5, 7, 2, 5, 4, 25, 5, 6, . . .

• (nk) = 1, 2, 4, 7, 8, 9, 12, 15, 16, 17, 24, 25, . . .

What is (snk)? Well,

(sn) = 8, 4, 6, 4, 7, 4, 5, 5, 5, 34, 35, 35, 7, 3, 35, 24, 6, 456, 7, 5, 7, 2, 5, 4, 25, 5, 6, . . . ;

and (snk) = 8, 4, 4, 5, 5, 5, 35, 35, 24, 6, 4, 25, . . . .

Notice that a different choice of (nk)∞k=1 can give the same (snk)∞k=1 . . .

Suppose that we have:

• (sn) as before;

• (nk) = 1, 2, 6, 7, 8, 9, 11, 15, 16, 17, 24, 25, . . . .

What is (snk)∞k=1? Well,

(sn) = 8, 4, 6, 4, 7, 4, 5, 5, 5, 34, 35, 35, 7, 3, 35, 24, 6, 456, 7, 5, 7, 2, 5, 4, 25, 5, 6, . . .

and (snk) = 8, 4, 4, 5, 5, 5, 35, 35, 24, 6, 4, 25, . . .

Remark 12.3.3. What’s going on?(nk)

∞k=1 is a sequence of natural numbers. It tells us which terms of the sequence (sn)∞n=1 appear

in our subsequence (snk)∞k=1. The sequence (nk)∞k=1 is strictly increasing. This says that the terms

of (snk)∞k=1 appear in the same order that they appear in (sn)∞n=1, and that they cannot repeat.A useful observation, which follows from the fact that (nk)

∞k=1 is strictly increasing: we have

∀k ∈ N, nk ≥ k.

Theorem 12.3.4. Suppose (sn)∞n=1 is a sequence and L ∈ R. If (sn) converges to L and (snk)∞k=1

is a subsequence of (sn), then (snk) converges to L.

Proof. Suppose that (sn) is a sequence, L ∈ R, limn→∞ sn = L, and that (snk) is a subsequence of(sn). We must show that limk→∞ snk = L.

Let ε > 0. Since (sn) converges to L, we can find an N ∈ N so that

∀n ∈ N, n ≥ N =⇒ |sn − L| < ε.

Let k ∈ N, and suppose that k ≥ N . Then nk ≥ k ≥ N and so |snk − L| < ε.

58

Example 12.3.5. The sequence (sn)∞n=1 where sn = (−1)n has the subsequence consisting of theeven terms (s2k)

∞k=1 = 1, 1, 1, . . . and the subsequence consisting of the odd terms

(s2k−1)∞k=1 = −1,−1,−1, . . . .

Since these sequences converge to 1 and −1, respectively, (sn) must diverge. . .Suppose for contradiction that (sn) converges to a number L. Then the previous theorem says

that (s2k)∞k=1 and (s2k−1)

∞k=1 converge to L. Thus, L = 1 and L = −1, and so 1 = −1. This is a

contradiction. Thus, (sn) diverges.

We saw in theorem 11.2.3 that every convergent sequence is bounded. The previous exampleserves as a reminder that the converse is false: there are bounded sequences which are not conver-gent. However, the Bolzano-Weierstrass theorem uses subsequences to get as close as possible to aconverse result.

Theorem 12.3.6 (Bolzano-Weierstrass Theorem). Suppose (sn)∞n=1 is a sequence. If (sn)∞n=1 isbounded, then there exists a subsequence (snk)∞k=1 of (sn)∞n=1 which converges.

Proof. Suppose (sn)∞n=1 is a bounded sequence. We wish to find a convergent subsequence (snk)∞k=1.Because a subsequence of a bounded sequence is bounded, and because a bounded monotone

subsequence is convergent (12.2.6), it is enough to find a monotone subsequence (snk) of (sn). Thisis the content of the proposition which follows.

Proposition 12.3.7. Suppose (sn)∞n=1 is a sequence. Then there exists a subsequence (snk)∞k=1 of(sn)∞n=1 which is monotone.

Proof. Let (sn) be a sequence.

Definition. We’ll say the n-th term of (sn) is dominant iff it is strictly greater than every termwhich follows it. That is, (∀m ∈ N, n < m =⇒ sn > sm).

There are two cases:

1. There are infinitely many dominant terms.

Let (snk) be any subsequence consisting solely of dominant terms. Then (snk) is a (strictly)decreasing subsequence of (sn).

2. There are finitely many dominant terms.

We give a procedure for constructing a subsequence which is increasing. The procedure firstsays how to find the first term of the subsequence. It then describes how to find the nextterm in the subsequence supposing that we already have k terms of the subsequence.

Choose n1 ∈ N so that all the dominant terms sn have n < n1, i.e. sn1 is beyond all dominantterms. Now suppose that we have n1 < n2 < . . . < nk so that sn1 ≤ sn2 ≤ . . . ≤ snk . Sincenk ≥ n1, snk is not dominant and we can find an nk+1 > nk with snk+1

≥ snk . In this waywe obtain an increasing subsequence (snk) of (sn).

59

12.4 Completeness of R

We are almost ready to prove that Cauchy sequences converge in R. We need one more theorem.

Theorem 12.4.1. Suppose (sn) is a Cauchy sequence. If there is a subsequence (snk) of (sn) whichconverges, then (sn) converges.

Proof. Suppose (sn) is a Cauchy sequence and that (snk) is a subsequence of (sn) which converges.Let L = limk→∞ snk . We will show that limn→∞ sn = L.

Let ε > 0. Since (sn) is Cauchy, there exists an N ∈ N so that

∀n ∈ N, ∀m ∈ N, n,m ≥ N =⇒ |sn − sm| <ε

2.

We wish to show that

∀n ∈ N, n ≥ N =⇒ |sn − L| < ε. (12.4.2)

First, note that, since limk→∞ snk = L, there exists an N ′ ∈ N so that

∀k ∈ N, k ≥ N ′ =⇒ |snk − L| <ε

2.

Let K = max{N,N ′} and M = nK .We are now ready to demonstrate (12.4.2). Let n ∈ N, and suppose that n ≥ N . We have

|sn − L| = |sn − sM + snK − L| ≤ |sn − sM |+ |snK − L|

Since n ≥ N and M = nK ≥ K ≥ N , we have |sn − sM | < ε2 .

Since K ≥ N ′, we have |snK − L| < ε2 .

Thus, |sn − L| < ε.

We are finally ready to prove the most important theorem of the course.

Theorem 12.4.3 (Completeness of R). Suppose that (sn)∞n=1 is a sequence. If (sn)∞n=1 is Cauchy,then (sn)∞n=1 converges.

Proof. Suppose (sn) is a Cauchy sequence. Then, by 12.1.4, (sn) is bounded. Thus, by the Bolzano-Weierstrass theorem, (sn) has a convergent subsequence (snk). We are now done, by the previoustheorem, since (sn) is a Cauchy sequence with a convergent subsequence.

An analysis of the proof reveals that it hinges on the result that a sequence which is increasingand bounded above is convergent; the proof of this made explicit use of the completeness axiom.

12.5 Examples

In this section we have studied the relationship between different kinds of sequences. For example,in subsection 12.1 we prove the following inclusions.

{(sn)∞n=1 : (sn) is convergent} ⊂ {(sn)∞n=1 : (sn) is Cauchy} ⊂ {(sn)∞n=1 : (sn) is bounded}

Summarizing the results of this section as I have done in the equation above may be useful for you.We will use the facts that bounded monotone sequences and Cauchy sequences converge repeat-

edly. We will often do this in an abstract context in the course of a proof; it is useful to see someexamples of applying these concepts to explicit sequences as well.

60

Example 12.5.1. ((1 + 1n)n)∞n=1 converges. On quiz preparation 3, I guide you towards showing

that the sequence is increasing and bounded above. The limit of this sequence is called e.

Example 12.5.2. Let F0 = 0, F1 = 1, and for n ∈ N, let Fn+1 = Fn + Fn−1. (Fn)∞n=1 is called theFibonacci sequence. We have

(Fn)∞n=1 = 1, 1, 2, 3, 5, 8, 13, . . .

For n ∈ N, let Rn = Fn+1

Fn, so that (Rn)∞n=1 is the ratio of successive Fibonacci numbers.

We wish to show that (Rn)∞n=1 converges. To demonstrate the power of the results from thissubsection, we will show that (Rn)∞n=1 converges by proving that it is Cauchy. To do this, first, wemake note of some properties of (Fn) and (Rn).

Lemma 12.5.3. For all n,m ∈ N,

• Fn−1 ≤ Fn;

• Fn+1 ≤ 2Fn;

• 3Fn+1 ≤ 2Fn+2;

• 1 ≤ Rn ≤ 2;

• 32 ≤ Rn+1;

• Rn+1 = 1 + 1Rn

;

• Rn+1 −Rm+1 = Rm−RnRmRn

.

• |Rn+1 −Rm+1| ≤ 23 · |Rn −Rm|.

Proof. Left as an exercise except for the last claim.Suppose n,m ∈ N. If n = m, then |Rn+1 − Rm+1| = 0 = 2|Rn−Rm|

3 . If n 6= m, either n ≥ 2 sothat Rn ≥ 3

2 , or m ≥ 2 so that Rm ≥ 32 . The inequality then follows from the previous claim and

the fact that Rn, Rm ≥ 1.

Lemma 12.5.4. For all k ∈ N, (23)k ≤ 2k+1 .

Proof. Let k ∈ N. Then the binomial theorem gives(2

3

)k=

2k

(1 + 2)k≤ 2k

k2k−1 + 2k=

2

k + 2≤ 2

k + 1.

We can now show that (Rn)∞n=1 is Cauchy.Let ε > 0. Let N = d2ε e+ 1. Suppose that n,m ∈ N and that n,m ≥ N . Then

|Rn −Rm| ≤(

2

3

)N−1∣∣∣∣Rn−N+1 −Rm−N+1

∣∣∣∣ ≤ (2

3

)N−1≤ 2

N< ε.

The first inequality follows from the last bullet point of the first lemma applied N − 1 times. Thesecond inequality follows from the fact that Rn−N+1, Rm−N+1 ∈ [1, 2]. The third inequality followsfrom the second lemma with k = N − 1. The last inequality follows from the choice of N .

Thus, limn→∞Rn exists, call it ϕ. Since, for all n ∈ N, 1 ≤ Rn ≤ 2, we have 1 ≤ ϕ ≤ 2. Since

limn→∞Rn+1 = limn→∞Rn = ϕ 6= 0 and Rn+1 = 1 + 1Rn

, we have ϕ = 1 + 1ϕ . Thus, ϕ = 1+

√5

2 .

61

13 More on continuous functions

13.1 The sequence definition of continuity

Here’s another way that mathematicians define continuity at a point a.

Definition 13.1.1. Suppose D is a subset of R, f : D −→ R is a function, and a ∈ D.We say that f is continuous at a iff the following sentence is true.

∀sequences (xn),

[[∀n ∈ N, xn ∈ D

]and

[limn→∞

xn = a

]]=⇒ lim

n→∞f(xn) = f(a).

That is, whenever (xn) is a sequence converging to a with all its terms in D, (f(xn))∞n=1 convergesto f(a).

Remark 13.1.2. Suppose f : R −→ R is continuous and (xn) is a convergent sequence. Since f iscontinuous, f is continuous at a := limn→∞ xn (definition 8.1.3). By the previous definition,

limn→∞

f(xn) = f

(limn→∞

xn

).

So continuity tells us that we can interchange certain limits. This is not something we can always do.If you’re about to interchange limits, you shoud think “is the function I’m dealing with continuous?”

Is it okay for a concept to have two definitions? Not really. In math, we should only mean onething by the statement “f : D −→ R is continuous at a.” However, it turns out to be okay becausethe following theorem tells us that the two definitions are equivalent: whichever definition someoneuses, they will say the same functions are continuous at a.

Theorem 13.1.3. Suppose D is a subset of R, f : D −→ R is a function, and a ∈ D. The followingsentence is true.[

∀ε > 0, ∃δ > 0 : ∀x ∈ D, |x− a| < δ =⇒ |f(x)− f(a)| < ε

]⇐⇒

[∀sequences (xn),

[[∀n ∈ N, xn ∈ D

]and

[limn→∞

xn = a

]]=⇒ lim

n→∞f(xn) = f(a)

].

That is, definitions (8.1.1) and (13.1.1) are equivalent.

Proof. Suppose D is a subset of R, f : D −→ R is a function, and a ∈ D. We will prove:

1.

[∀ε > 0, ∃δ > 0 : ∀x ∈ D, |x− a| < δ =⇒ |f(x)− f(a)| < ε

]=⇒[

∀sequences (xn),

[[∀n ∈ N, xn ∈ D

]and

[limn→∞ xn = a

]]=⇒ limn→∞ f(xn) = f(a)

];

2. (the contrapositive of the arrow in the other direction)[∃ε > 0 : ∀δ > 0, ∃x ∈ D : |x− a| < δ and |f(x)− f(a)| ≥ ε

]=⇒[

∃sequence (xn) :

[∀n ∈ N, xn ∈ D

],

[limn→∞ xn = a

],

and (f(xn))∞n=1 does not converge to f(a)

].

62

1. First, suppose that definition (8.1.1) is satisfied. We must verify

∀sequences (xn),

[[∀n ∈ N, xn ∈ D

]and

[limn→∞

xn = a

]]=⇒ lim

n→∞f(xn) = f(a).

Let (xn)∞n=1 be a sequence. Suppose that for all n ∈ N, xn ∈ D, and that (xn) converges toa. We need to show that (f(xn))∞n=1 converges to f(a), so let ε > 0. By our assumption thatdefinition (8.1.1) is satisfied, we can find a δ > 0 such that

∀x ∈ D, |x− a| < δ =⇒ |f(x)− f(a)| < ε.

By our assumption that (xn)∞n=1 converges to a, we can find an N ∈ N such that

∀n ∈ N, n ≥ N =⇒ |xn − a| < δ.

Suppose that n ∈ N and n ≥ N . By the property of N , we have |xn − a| < δ. Since xn ∈ D,by the property of δ, we have |f(xn)− f(a)| < ε. Thus, (f(xn))∞n=1 converges to f(a).

2. Suppose that the negation of definition (8.1.1) is satisfied. Then there exists an ε > 0 suchthat

∀δ > 0, ∃x ∈ D : |x− a| < δ and |f(x)− f(a)| ≥ ε.

We wish to construct a sequence (xn)∞n=1 which shows that the negation of definition (13.1.1)is satisfied. Let n ∈ N. To get the n-th term, we note that 1

n > 0, and so, by the statementabove, there exists an xn ∈ D such that

|xn − a| <1

nand |f(xn)− f(a)| ≥ ε.

Doing this for each n ∈ N gives a sequence (xn). It has

• ∀n ∈ N, xn ∈ D.

• ∀n ∈ N, a− 1n < xn < a+ 1

n .

• ∀n ∈ N, |f(xn)− f(a)| ≥ ε.

The first condition is one of the properties we want of (xn). The second condition togetherwith the squeeze lemma gives limn→∞ xn = a. The last condition shows that (f(xn))∞n=1 doesnot converge to f(a).

Recall definition 8.1.3. Because of the previous theorem we could also have defined continuity(not mentioning a particular point a) as follows.

Definition 13.1.4. Suppose D is a subset of R and f : D −→ R is a function.We say that f is continuous iff for all a ∈ D, f is continuous at a, i.e. iff the following sentence

is true.

∀a ∈ D, ∀sequences (xn),

[[∀n ∈ N, xn ∈ D

]and

[limn→∞

xn = a

]]=⇒ lim

n→∞f(xn) = f(a).

63

We can use the sequence definition to show that functions defined by polynomials are continuous.

Theorem 13.1.5. Suppose that d ∈ N and c0, c1, . . . , cd−1, cd ∈ R. Then the function p : R −→ Rdefined by p(x) = cdx

d + cd−1xd−1 + . . .+ c1x+ c0 is continuous.

Proof. Suppose a ∈ R. We must show p is continuous at a, that is, we must verify

∀sequences (xn), limn→∞

xn = a =⇒ limn→∞

p(xn) = p(a).

(We omitted writing the part that is trivially always true because D = R.)Suppose that (xn) is a sequence and that limn→∞ xn = a. Then

limn→∞

p(xn) = limn→∞

(cdx

dn + cd−1x

d−1n + . . .+ c1xn + c0

)= cda

d + cd−1ad−1 + . . .+ c1a+ c0 = p(a),

where the first and third equality follows from the definition of p and the second equality followsfrom the algebra of limits (11.3.1).

Example 13.1.6. Let f : R −→ R be defined by

f(x) =

{x if x ∈ Q0 if x ∈ R \Q.

f is discontinuous at 1 because we can verify the negation of definition 13.1.1:

∃sequence (xn) : limn→∞

xn = 1 and (f(xn))∞n=1 does not converge to f(1).

(We omitted writing the part that is trivially always true because D = R.)

We let xn = 1 +√2n . Then for all n ∈ N, xn ∈ R \Q, f(xn) = 0, and we have

limn→∞

f(xn) = 0 6= 1 = f(1).

13.2 New continuous functions from old ones (not lectured)

Theorem 13.2.1. Suppose D is a subset of R, a ∈ D, f : D −→ R and g : D −→ R are functionswhich are continuous at a, and k ∈ R. Then:

1. kf , defined by (kf)(x) = k · f(x), is continuous at a;

2. f + g, defined by (f + g)(x) = f(x) + g(x), is continuous at a;

3. fg, defined by (fg)(x) = f(x)g(x), is continuous at a;

4. The domain of fg is {x ∈ D : g(x) 6= 0}. If g(a) 6= 0 then f

g is defined and continuous at a.

Proof. You use the algebra of limits (11.3.1) just like in the proof that polynomials define continuousfunctions.

For example, to demonstrate part 1, suppose that (xn) is a sequence converging to a whoseterms are in D. Since f is continuous at a, we have limn→∞ f(xn) = f(a). Thus,

limn→∞

(kf)(xn) = limn→∞

k(f(xn)) = k limn→∞

f(xn) = k(f(a)) = (kf)(a).

The first and last equality are by definition of kf ; the second uses the algebra of limits.

64

Theorem 13.2.2. Suppose f : Df −→ R and g : Dg −→ R are functions. Let

Dh = {x ∈ Df : f(x) ∈ Dg}

and define h : Dh −→ R by h(x) = g(f(x)). Suppose that a ∈ Dh, that f is continuous at a, and gis continuous at f(a). Then h is continuous at a.

Proof. Let (xn) be a sequence converging to a whose terms are in Dh. Since Dh ⊂ Df and f iscontinuous at a, limn→∞ f(xn) = f(a). Moreover, (f(xn)) is a sequence whose terms are in Dg andso, since g is continuous at f(a), limn→∞ g(f(xn)) = g(f(a)), i.e. limn→∞ h(xn) = h(a).

Theorem 13.2.3. The function | · | : R −→ R is (uniformly) continuous.

Proof. For all x, a ∈ R, we have ||x| − |a|| ≤ |x− a|. Thus, given ε, we can take δ = ε.

Corollary 13.2.4. Suppose D is a subset of R, f : D −→ R is a function, and a ∈ D. If f iscontinuous at a, then the function |f | : D −→ R defined by |f |(x) = |f(x)| is continuous at a.

Proof. |f | is the composite of the function f and the function | · |. We have just shown that thelatter is continuous and so we can use theorem 13.2.2.

13.3 The Intermediate Value Theorem and the Extreme Value Theorem

Theorem 13.3.1 (Intermediate value theorem). Suppose a, b, c ∈ R, a < b, and that f : [a, b] −→ Ris a function satisfying f(a) < c < f(b). If f is continuous, then there is an x0 ∈ (a, b) such thatf(x0) = c.

Concise proof. Suppose a, b, c ∈ R, a < b, that f : [a, b] −→ R is a continuous function, and thatf(a) < c < f(b). We will find x0 using the “bisection method” which you might have encounteredbefore.

Let a0 = a and b0 = b. Then f(a0) < c < f(b0). Suppose that we have

a0 ≤ a1 ≤ . . . ≤ an ≤ bn ≤ . . . ≤ b1 ≤ b0,

bn − an = b−a2n , and that f(an) < c < f(bn). If f(an+bn2 ) = c, we can take x0 = an+bn

2 and finish. If

f(an+bn2 ) < c, then let an+1 = an+bn2 and bn+1 = bn. Otherwise, let an+1 = an and bn+1 = an+bn

2 .Either the procedure terminates and we have x0, or we obtain an increasing sequence (an) and a

decreasing sequence (bn). Both are bounded and so they converge. Let x0 = lim an and y0 = lim bn.Since f is continuous and (−∞, c] is closed, f(x0) = lim f(an) ≤ c. Similarly, f(y0) = lim f(bn) ≥ c.

Now y0 − x0 = lim bn − lim an = lim(bn − an) = lim b−a2n = 0. Thus x0 = y0 and f(x0) = c.

Remark 13.3.2. I’ll expand on this proof a lot more in lecture. The write up above is very terse.After I present the expanded proof in class, it is a good exercise to go back to this one, make sure

that you understand the procedure used in the proof, and mention the theorems and/or definitionswhich I am using in the latter two paragraphs more explicitly.

Remark 13.3.3. We could also state the intermediate value theorem as follows. Suppose a, b ∈ R,a < b, and that f : [a, b] −→ R is a continuous function. If c ∈ R and f(a) < c < f(b), then thereexists an x0 ∈ (a, b) such that f(x0) = c. This has the same meaning, however, the way it is writtenmeans that the contrapositive looks a little different.

65

Corollary 13.3.4. Suppose a, b, d ∈ R, a < b, and that g : [a, b] −→ R is a function satisfyingg(a) > d > g(b). If g is continuous, then there is an x0 ∈ (a, b) such that g(x0) = d.

Proof. Suppose a, b, d ∈ R, a < b, g : [a, b] −→ R is a continuous function, and that g(a) > d > g(b).Let f = −g. Then f : [a, b] −→ R is continuous. Let c = −d. Then

f(a) = −g(a) < −d = c = −d < −g(b) = f(b).

The intermediate value theorem gives an x0 ∈ (a, b) such that f(x0) = c. Thus, g(x0) = −f(x0) =−c = d.

Example 13.3.5. Suppose f : [0, 1]→ [0, 1] is a function. If f is continuous, then there exists anx0 ∈ [0, 1] such that f(x0) = x0.

Proof. Let g(x) = x− f(x). g is continuous since it is the difference of continuous functions. Also,g(0) = 0 − f(0) ≤ 0 and g(1) = 1 − f(1) ≥ 0. Either g(0) = 0, or g(1) = 0, or we can apply theintermediate value theorem to find an x0 ∈ (0, 1) with g(x0) = 0. In any case, we have an x0 ∈ [0, 1]with g(x0) = 0, i.e. f(x0) = x0.

Definition 13.3.6. Suppose D is a subset of R and f : D −→ R is a function. We say that f isbounded iff the set {f(x) : x ∈ D} is bounded. Equivalently, we say f is bounded iff if there existsan M > 0 such that for all x ∈ D, |f(x)| ≤M , that is (∃M > 0 : ∀x ∈ D, |f(x)| ≤M).

Theorem 13.3.7 (Extreme Value Theorem). Suppose a, b ∈ R, a < b, and that f : [a, b]→ R is afunction. If f continuous, then f is bounded and f obtains its maximum and minimum values, i.e.there exist x0, X0 ∈ [a, b] with the property that for all x ∈ [a, b], f(x0) ≤ f(x) ≤ f(X0).

Proof. Suppose a, b ∈ R, a < b, and that f : [a, b] −→ R is a function.First, we show that if f continuous, then f is bounded. To do this we verify the contrapositive,

that is, that if f : [a, b] −→ R is not bounded, then f is not continuous.Suppose that f is not bounded.Let n ∈ N. Because f is not bounded, there exists an xn ∈ [a, b] with |f(xn)| > n. Doing this

for each n ∈ N, we obtain a sequence (xn)∞n=1. Since [a, b] is bounded, (xn) is bounded, and so theBolzano-Weierstrass theorem tells us that (xn) has a convergent subsequence (xnk); let c = limxnk .We showed on quiz preparation 3 that [a, b] is closed. This implies that c ∈ [a, b].

We will show that f is not continuous at c. For all k ∈ N, we have |f(xnk)| > nk ≥ k. Thus,(f(xnk))∞k=1 is not bounded, and so it is not convergent. However, (xnk)∞k=1 converges to c, and sothis demonstrates that f is not continuous at c.

We now move to the second claim which we prove directly. Suppose that f is continuous. Thenf is bounded. Let R = {f(x) : x ∈ [a, b]}, m = infR, and M = supR. It is enough to show thatthere exist x0, X0 ∈ [a, b] such that f(x0) = m and f(X0) = M .

Again, we use sequences. Let n ∈ N. m is the greatest lower bound for R and m < m+ 1n , so

m+ 1n is not a lower bound for R. Thus, there exists an xn ∈ [a, b] with f(xn) < m+ 1

n . Moreover,since m is a lower bound for R, we have m ≤ f(xn). By doing this for every n ∈ N, we obtain asequence (xn) such that for all n ∈ N, m ≤ f(xn) < m+ 1

n .Since [a, b] is bounded, the Bolzano-Weierstrass gives us a convergent subsequence (xnk) of (xn);

let x0 = limxnk . Because [a, b] is closed, x0 ∈ [a, b]. Since f is continuous at x0, f(x0) = lim f(xnk),and the squeeze lemma tells us that lim f(xnk) = m.

Similarly, we can find an X0 ∈ [a, b] such that f(X0) = M .

66

Remark 13.3.8. Notice that the properties of [a, b] that we used were closed and bounded. Thetheorem would hold for any function f : D −→ R such that D is a closed and bounded subset of R.

13.4 A theorem that should be included somewhere

Theorem 13.4.1. Suppose a, b ∈ R, a < b and that f : [a, b] −→ R is a function. If f is continuous,then f is uniformly continuous.

Proof. Suppose a, b ∈ R, a < b, and that f : [a, b] −→ R is a continuous function, and suppose forcontradiction that f is not uniformly continuous. This means that there exists an ε > 0 such thatthe following sentence is true.

∀δ > 0, ∃x ∈ [a, b], ∃y ∈ [a, b] : |x− y| < δ and |f(x)− f(y)| ≥ ε.

Let n ∈ N. Then 1n > 0. By the sentence above, there exist an xn ∈ [a, b] and yn ∈ [a, b] such

that |xn−yn| < 1n and |f(xn)−f(yn)| ≥ ε. Doing this for each n ∈ N, we obtain sequences (xn)∞n=1,

(yn)∞n=1.Since [a, b] is bounded, using Bolzano-Weierstrass, we can find a convergent subsequence (xnk)∞k=1

of (xn)∞n=1. This determines a subsequence (ynk)∞k=1 of (yn)∞n=1. For all k ∈ N, we have

|xnk − ynk | <1

nk≤ 1

k.

Thus, for all k ∈ N, xnk − 1k < ynk < xnk + 1

k , and the squeeze lemma tells us that (ynk) converges,and moreover, that limxnk = lim ynk . Since [a, b] is closed this number is an element of [a, b], andbecause f is continuous, we deduce that

lim f(xnk) = f(limxnk) = f(lim ynk) = lim f(ynk).

Thus, lim |f(xnk)− f(ynk)| = 0. However, for all k ∈ N, we have |f(xnk)− f(ynk)| ≥ ε. This is therequired contradiction.

Remark 13.4.2. Once again, the properties of [a, b] that we used were closed and bounded. Thetheorem would hold for any function f : D −→ R such that D is a closed and bounded subset of R.

67

14 Calculus

14.1 Limits along the domain of a function

In math 31A limits are discussed, but not the limits which we have been talking about. We havebeen discussing limits of sequences; the limits discussed in math 31A are limits along the domainof a function. As in math 31A, the most important reason for talking about this concept is so thatwe can define differentiability.

The limit limx→a f(x) tells you what value you expect f(x) to take at a based on the values off(x) nearby but not equal to a. Let’s see an example before even giving the definition.

Example 14.1.1 (From Math 31A).

1 2 3 4 5 6 7 8 9

1

2

3

4

1. limx→1 f(x) = f(1) = 1 (continuous at 1).

2. limx→2− f(x) = 2 6= 4 = f(2) = limx→2+ f(x) (jump discontinuity, right continuous at 2).

3. limx→4− f(x) = f(4) = 2 6= 1 = limx→2+ f(x) (jump discontinuity, left continuous at 4).

4. limx→6 f(x) = 3 6= 4 = f(6) (removable discontinuity at 6).

5. limx→2−(f(x)f(x+ 2)) = limx→2− f(x) limx→2− f(x+ 2) = 2 · 2 = 4.

6. limx→2+(f(x)f(x+ 2)) = limx→2+ f(x) limx→2+ f(x+ 2) = 4 · 1 = 4.

7. limx→2(f(x)f(x+2)) = 4 (if the left and right-hand limits agree, the (both-sided) limit existsand takes on the same value).

The definitions of limx→a f(x) = L, limx→a+ f(x) = L, limx→a− f(x) = L are all similar innature to the definitions of continuity. Just like with continuity, they each have equivalent ε-δdefinitions and sequence definitions, and depending on context, one may be more useful than theother.

The simplest difference between the definitions which follow and that of continuity is that f(a)has been replaced by L. The more subtle difference is how the (P =⇒ Q) statements depend onx. Because limx→a f(x) should not depend on the value of f(a):

68

• |x− a| < δ has become 0 < |x− a| < δ;

• and xn ∈ D has become xn ∈ D \ {a}.

• The definitions of right and left-hand limits have other appropriate changes.

Definition 14.1.2. Suppose D is a subset of R, f : D −→ R is a function, and a, L ∈ R.

1. We say that the limit of f(x) as x tends to a is L, and write limx→a f(x) = L, iff the followingsentence is true.

∀ε > 0, ∃δ > 0 : ∀x ∈ D, 0 < |x− a| < δ =⇒ |f(x)− L| < ε.

2. We say that the right-hand limit of f(x) as x tends to a is L, and write limx→a+ f(x) = L,iff the following sentence is true.

∀ε > 0, ∃δ > 0 : ∀x ∈ D, a < x < a+ δ =⇒ |f(x)− L| < ε.

3. We say that the left-hand limit of f(x) as x tends to a is L, and write limx→a− f(x) = L, iffthe following sentence is true.

∀ε > 0, ∃δ > 0 : ∀x ∈ D, a− δ < x < a =⇒ |f(x)− L| < ε.

Definition 14.1.3. Suppose D is a subset of R, f : D −→ R is a function, and a, L ∈ R.

1. We say that the limit of f(x) as x tends to a is L, and write limx→a f(x) = L, iff the followingsentence is true.

∀sequences (xn),

[[∀n ∈ N, xn ∈ D \ {a}

]and

[limn→∞

xn = a

]]=⇒ lim

n→∞f(xn) = L.

2. We say that the right-hand limit of f(x) as x tends to a is L, and write limx→a+ f(x) = L,iff the following sentence is true.

∀sequences (xn),

[[∀n ∈ N, xn ∈ D ∩ (a,∞)

]and

[limn→∞

xn = a

]]=⇒ lim

n→∞f(xn) = L.

3. We say that the left-hand limit of f(x) as x tends to a is L, and write limx→a− f(x) = L, iffthe following sentence is true.

∀sequences (xn),

[[∀n ∈ N, xn ∈ D ∩ (−∞, a)

]and

[limn→∞

xn = a

]]=⇒ lim

n→∞f(xn) = L.

Remark 14.1.4. Notice that a does not have to be an element of D.

1. For the function f : (1,∞) −→ R defined by f(x) = x+ 1, we have limx→1 f(x) = 2.

2. The closure of a set E is defined as the smallest closed set containing E :{a ∈ R : there exists a sequence (xn) such that

[∀n ∈ N, xn ∈ E

]and

[limn→∞

xn = a

] }.

3. As long as a is in the closure of D\{a} (respectively, D∩(a,∞), D∩(−∞, a)), limx→a f(x) = L(respectively, limx→a+ f(x) = L, limx→a− f(x) = L) can only be true for one value of L. Thisis because the limit of a sequence is unique (11.1.2).

4. If a is not in the relevant closure, then limx→a f(x) = L is trivially true regardless of what Lis. Choosing such an a is silly, and I will rarely do it.

69

Here’s a theorem that should be familiar from 31A.

Theorem 14.1.5. Suppose D is a subset of R, f : D −→ R is a function, and a, L ∈ R.Then limx→a f(x) = L if and only if

limx→a+

f(x) = L and limx→a−

f(x) = L.

Proof. We prove the first implication using the sequence definition and the second implication usingthe ε-δ definition.

Suppose that limx→a f(x) = L. We wish to show limx→a+ f(x) = L. So suppose (xn) is a se-quence converging to a with all of its terms in D∩(a,∞). In particular, (xn) is a sequence convergingto a with all of its terms in D \{a}. Thus, since limx→a f(x) = L, we have limn→∞ f(xn) = L, andthis completes the proof that limx→a+ f(x) = L. Similarly, we can prove that limx→a− f(x) = L.

Conversely, suppose that limx→a+ f(x) = L and that limx→a− f(x) = L. We wish to show thatlimx→a f(x) = L so let ε > 0. Since limx→a+ f(x) = L, there exists a δ+ > 0 so that

∀x ∈ D, a < x < a+ δ+ =⇒ |f(x)− L| < ε.

Since limx→a− f(x) = L, there exists a δ− > 0 so that

∀x ∈ D, a− δ− < x < a =⇒ |f(x)− L| < ε.

Let δ = min{δ+, δ−}. Then δ > 0. Moreover, if x ∈ D and 0 < |x−a| < δ, then either a−δ < x < aor a < x < a+ δ, so that either a− δ− < x < a or a < x < a+ δ+, and we have |f(x)−L| < ε.

Proposition 14.1.6. Suppose D is a subset of R, f : D −→ R is a function, and a ∈ D.f is continuous at a if and only if limx→a f(x) = f(a).

Proof. A good exercise for you.

14.2 The derivative

The derivative is a concept which will be very familiar from previous calculus courses.

• It describes the instantaneous rate of change of a function.

• It is the limit of the slopes of secant lines.

Definition 14.2.1. Suppose D is a subset of R, f : D −→ R is a function, a ∈ D. We say that fis differentiable at a, or f has derivative at a iff

∃L ∈ R : limx→a

f(x)− f(a)

x− a= L.

In this case, we shall write f ′(a) for such an L.

Example 14.2.2. Let n ∈ N∪{0} and let f : R −→ R be defined by f(x) = xn+1. Then, for x 6= a

f(x)− f(a)

x− a=xn+1 − an+1

x− a=

n∑k=0

xn−kak.

70

The function g : R −→ R, g(x) =∑n

k=0 xn−kak is continuous because it is defined by a polynomial.

Thus,

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

x→ag(x) = g(a) =

n∑k=0

an−kak = (n+ 1)an.

Theorem 14.2.3. Suppose D is a subset of R, f : D −→ R is a function, a ∈ R.If f is differentiable at a point a, then f is continuous at a.

Proof. Suppose f is differentiable at a. For x 6= a, we have

f(x) = f(a) +f(x)− f(a)

x− a· (x− a)

Thus, using the algebra of limits, we see that limx→a f(x) = f(a) + f ′(a) · 0 = f(a), which showsf is continuous at a (14.1.6).

Definition 14.2.4. Suppose D is a subset of R and f : D −→ R is a function. Suppose D′ is asubset of D. We say that f is differentiable on D′ iff for all a ∈ D′, f is differentiable at a.

In this case, we obtain a new function f ′ : D′ −→ R.

Example 14.2.5. The function g : [0,∞) −→ R defined by g(x) =√x is differentiable on (0,∞).

For a > 0, we have (using continuity of 1√x+√a

at a)

g′(a) = limx→a

√x−√a

x− a= lim

x→a

1√x+√a

=1

2√a.

We have g′ : (0,∞) −→ R, g′(x) = 12√x.

14.3 Rules for differentiating (not lectured or examined explicitly)

Theorem 14.3.1. Suppose D is a subset of R, a ∈ D, f : D −→ R and g : D −→ R are functionswhich are differentiable at a, and k ∈ R. Then:

1. For k ∈ R, kf is differentiable at a and (kf)′(a) = k · f ′(a).

2. f + g is differentiable at a and (f + g)′(a) = f ′(a) + g′(a).

3. fg is differentiable at a and (fg)′(a) = f(a)g′(a) + f ′(a)g(a) (product rule).

4. The domain of fg is {x ∈ D : g(x) 6= 0}. If g(a) 6= 0 then f

g is differentiable at a and(f

g

)′(a) =

g(a)f ′(a)− f(a)g′(a)

g(a)2(quotient rule).

Proof. See the textbook if you care.

Example 14.3.2 (Power rule for integers). Let n ∈ Z. If h : R\{0} −→ R is defined by h(x) = xn,then h is differentiable (on R \ {0}) and h′(x) = nxn−1.

We have already shown the result when n ∈ N. It is easy when n = 0. For n < 0, we use thequotient rule, taking f(x) = 1 and g(x) = x−n. Then

h′(x) =

(f

g

)′(x) =

g(x)f ′(x)− f(x)g′(x)

g(x)2=−g′(x)

g(x)2=nx−n−1

x−2n= nxn−1.

71

Theorem 14.3.3 (The chain rule). Suppose f : Df −→ R and g : Dg −→ R are functions. Let

Dh = {x ∈ Df : f(x) ∈ Dg}

and define h : Dh −→ R by h(x) = g(f(x)). Suppose that a ∈ Dh, that f is differentiable at a, andg is differentiable at f(a). Then h is differentiable at a, and

h′(a) = g′(f(a)) · f ′(a).

Proof. The idea of the proof is that

g(f(x))− g(f(a))

x− a=g(f(x))− g(f(a))

f(x)− f(a)· f(x)− f(a)

x− a

=

[(g(f(x))− g(f(a))

f(x)− f(a)− g′(f(a))

)+ g′(f(a))

]· f(x)− f(a)

x− a.

This equation makes sense only if x 6= a and f(x) 6= f(a). To account the fact that f(x) might beequal to f(a) even though x 6= a, we need to define an auxilary function v. Write b for f(a).

Define a function v : Dg −→ R by

v(y) =

{g(y)−g(b)y−b − g′(b) if y 6= b

0 if y = b.

Since f is differentiable at a, f is continuous at a. Since g is differentiable at b, v is continuous atb. So Dh −→ R, x 7−→ v(f(x)) is continuous at a and

limx→a

v(f(x)) = v(f(a)) = v(b) = 0.

Note that for all y ∈ Dg we have[v(y) + g′(b)

]· (y − b) = g(y)− g(b).

In particular, letting x ∈ Dh and y = f(x), we have[v(f(x)) + g′(f(a))

]· (f(x)− f(a)) = g(f(x))− g(f(a))

so that for all x ∈ Dh \ {a}[v(f(x)) + g′(f(a))

]· f(x)− f(a)

x− a=g(f(x))− g(f(a))

x− a.

Applying limx→a and using the algebra of limits gives the result.

72

14.4 Fermat and Rolle’s Theorem

In math 31A the derivative is used to find local maxima and minima. The result students of math31A are making use of is the following.

Theorem 14.4.1 (Fermat). Suppose D is a subset of R, f : D −→ R is a function, a ∈ R, δ > 0,(a− δ, a+ δ) ⊂ D, and that f is differentiable at a.

If min{f(x) : x ∈ (a− δ, a+ δ)} exists and is equal to f(a), then f ′(a) = 0.Similarly, if max{f(x) : x ∈ (a− δ, a+ δ)} exists and is equal to f(a), then f ′(a) = 0.

Proof. Suppose D is a subset of R, f : D −→ R is a function, a ∈ R, δ > 0, (a− δ, a+ δ) ⊂ D, thatf is differentiable at a, and that min{f(x) : x ∈ (a− δ, a+ δ)} = f(a).

For all x ∈ (a − δ, a + δ), f(x) − f(a) ≥ 0. So, for x ∈ (a, a + δ), we have f(x)−f(a)x−a ≥ 0 which

gives (it’s a good exercise to prove this)

f ′(a) = limx→a

f(x)− f(a)

x− a≥ 0.

For x ∈ (a− δ, a), we have f(x)−f(a)x−a ≤ 0 which gives f ′(a) = limx→a

f(x)−f(a)x−a ≤ 0.

Thus, f ′(a) = 0.

Remark 14.4.2. The hypothesis that (a− δ, a+ δ) ⊂ D is to prevent a from being an endpoint ofD. The function f : [0, 1] −→ R defined by f(x) = x has a minimum at 0 but f ′(0) = 1 6= 0.

The hypothesis that f is differentiable at a is necessary. f : [−1, 1] −→ R defined by f(x) = |x|has a minimum at 0, but f is not differentiable at 0, and so we do not have f ′(0) = 0.

A consequence of Fermat’s theorem is Rolle’s theorem.

Theorem 14.4.3 (Rolle’s Theorem). Suppose a, b ∈ R, a < b, and that f : [a, b]→ R is a function.If f is continuous, differentiable on (a, b), and f(a) = f(b), then there exists a c ∈ (a, b) such thatf ′(c) = 0.

Proof. Suppose a, b ∈ R, a < b, and that f : [a, b]→ R is a function. Suppose, in addition, that fis continuous, differentiable on (a, b), and f(a) = f(b).

Since f is continuous and defined on a closed and bounded set, the extreme value theorem saysthat f is bounded and obtains its bounds, i.e. there exist x0, x1 ∈ [a, b] with the property that forall x ∈ [a, b], f(x0) ≤ f(x) ≤ f(x1).

There are now two cases.

1. At least one of x0 or x1 is in the open interval (a, b).

Suppose that x0 ∈ (a, b). We can apply Fermat’s theorem to f with δ = min{b− x0, x0 − a}to see that f ′(x0) = 0, and take c = x0. Similarly, if x1 ∈ (a, b), we have f ′(x1) = 0 and cantake c = x1.

2. x0, x1 ∈ {a, b}, i.e. x0 and x1 are endpoints of the interval. Then

f(x0) = f(x1) = f(a) = f(b).

Thus, the fact that for all x ∈ (a, b), we have f(x0) ≤ f(x) ≤ f(x1), tells us that f is constant.So the derivative is zero everywhere, and we can take c = a+b

2 .

73

14.5 The Mean Value Theorem and important results from calculus

In math 31A students use the derivative to say things about the graph of a function. To make suchdeductions one needs consequences of mean value theorem and other related results.

Theorem 14.5.1 (Mean Value Theorem). Suppose a, b ∈ R, a < b, and that f : [a, b] −→ R is afunction. If f is continuous, and differentiable on (a, b), then there exists a c ∈ (a, b) such that

f ′(c) =f(b)− f(a)

b− a.

Proof. Suppose a, b ∈ R, a < b, and that f : [a, b] → R is a continuous function which is differen-tiable on (a, b).

Let g : [a, b] −→ R be defined by

g(x) = f(x)−(f(b)− f(a)

b− a

)(x− a).

Then g(a) = f(a) = g(b). Moreover, since f and (x−a) are continuous, and differentiable on (a, b),g is continuous, and differentiable on (a, b). So Rolle’s Theorem says that there is a c ∈ (a, b) withg′(c) = 0. For x ∈ (a, b) we have

g′(x) = f ′(x)−(f(b)− f(a)

b− a

),

and so this means that f ′(c) = f(b)−f(a)b−a .

The following result is so so so so so important and depends entirely on the mean value theorem.

Theorem 14.5.2. Suppose a, b ∈ R, a < b, and f : (a, b) −→ R is a function. If f is differentiableon (a, b) and for all x ∈ (a, b), f ′(x) = 0, then f is a constant function.

Proof. Suppose a, b ∈ R, a < b, and f : (a, b)→ R is a function, that f is differentiable on (a, b) andthat for all x ∈ (a, b), f ′(x) = 0. To show that f is constant, we will show that for all x0, x1 ∈ (a, b),if x0 < x1, then f(x0) = f(x1).

Suppose x0, x1 are in (a, b) and that x0 < x1. f is continuous on [x0, x1] and differentiable on(x0, x1). So the mean value theorem says that there exists a c ∈ (x0, x1) such that

f ′(c) =f(x0)− f(x1)

x0 − x1.

Since f ′(x) = 0 for all x ∈ (x0, x1), this shows that f(x0) = f(x1).

Theorem 14.5.3. Suppose a, b ∈ R, a < b, and that f : (a, b) −→ R, g : (a, b) −→ R are functionswhich are differentiable on (a, b). If for all x ∈ (a, b), f ′(x) = g′(x), then there is a constant c ∈ Rsuch that f(x) = g(x) + c for all x ∈ (a, b).

Proof. Suppose a, b ∈ R, a < b, f : (a, b)→ R, g : (a, b)→ R are functions which are differentiableon (a, b), and that for all x ∈ (a, b), f ′(x) = g′(x). Define h : (a, b) −→ R by h(x) = f(x) − g(x).Then h′(x) = 0 for all x ∈ (a, b). The previous corollary says that h is a constant function on (a, b),i.e. there is a constant c ∈ R such that h(x) = c for all x ∈ (a, b). Thus, f(x) = g(x) + c for allx ∈ (a, b).

74

Remark 14.5.4. The domain is important in the previous theorem. What are the anti-derivativesof the function f : (−1, 0) ∪ (0, 1) −→ R defined by f(x) = 1

x2?

Definition 14.5.5. Suppose D is a subset of R, and f : D −→ R is a function. We say that f is

1. increasing iff

[∀x0 ∈ D, ∀x1 ∈ D, x0 < x1 =⇒ f(x0) ≤ f(x1)

];

2. decreasing iff

[∀x0 ∈ D, ∀x1 ∈ D, x0 < x1 =⇒ f(x0) ≥ f(x1)

];

3. strictly increasing iff

[∀x0 ∈ D, ∀x1 ∈ D, x0 < x1 =⇒ f(x0) < f(x1)

];

4. strictly decreasing iff

[∀x0 ∈ D, ∀x1 ∈ D, x0 < x1 =⇒ f(x0) > f(x1)

].

When graphing a function one uses the following result.

Theorem 14.5.6. Suppose a, b ∈ R, a < b, and f : (a, b) −→ R is a function which is differentiableon (a, b). Then

1.

[∀x ∈ (a, b), f ′(x) ≥ 0

]=⇒ f is increasing;

2.

[∀x ∈ (a, b), f ′(x) ≤ 0

]=⇒ f is decreasing;

3.

[∀x ∈ (a, b), f ′(x) > 0

]=⇒ f is strictly increasing;

4.

[∀x ∈ (a, b), f ′(x) < 0

]=⇒ f is strictly decreasing.

Proof. Suppose a, b ∈ R, a < b, and f : (a, b) −→ R is a function which is differentiable on (a, b).Suppose x0, x1 are in (a, b) and that x0 < x1. f is continuous on [x0, x1] and differentiable on

(x0, x1). So the mean value theorem says that there exists a c ∈ (x0, x1) such that

f ′(c) =f(x0)− f(x1)

x0 − x1.

1. Suppose that for all x ∈ (a, b), f ′(x) ≥ 0. Then, in particular, f ′(c) ≥ 0. The equation abovesays that f(x0)− f(x1) and x0 − x1 have the same sign. Thus, f(x0) ≤ f(x1).

2. Suppose that for all x ∈ (a, b), f ′(x) ≤ 0. Then, in particular, f ′(c) ≤ 0. The equation abovesays that f(x0)− f(x1) and x0 − x1 have opposite signs. Thus, f(x0) ≥ f(x1).

3. Suppose that for all x ∈ (a, b), f ′(x) > 0. Then, in particular, f ′(c) > 0. The equation abovesays that f(x0) − f(x1) and x0 − x1 have the same sign and that f(x0) − f(x1) 6= 0. Thus,f(x0) < f(x1).

75

4. Suppose that for all x ∈ (a, b), f ′(x) < 0. Then, in particular, f ′(c) > 0. The equation abovesays that f(x0)−f(x1) and x0−x1 have the opposite signs and that f(x0)−f(x1) 6= 0. Thus,f(x0) > f(x1).

Remark 14.5.7. Suppose a, b ∈ R, a < b, and f : (a, b) −→ R is a function which is differentiableon (a, b).

• The reverse implications to 1 and 2 are true.

• The reverse implications to 3 and 4 are not necessarily true. For example, 3 fails in the casethat f : (−1, 1) −→ R is defined by f(x) = x3, and 4 fails for g = −f .

Students of math 31A are taught the “first derivative test.” When plotting a graph, one findsthe critical points (where the derivative vanishes or does not exist), and then calculates the sign ofthe derivative inbetween to decide whether each critical point is a local maximum, local minimum,or neither. The following theorem says it is enough to look at just one value of f ′ between eachcritical point in order to determine the sign of the derivative on the entire interval.

Theorem 14.5.8 (Intermediate value theorem for derivatives). Suppose a, b ∈ R, a < b, and thatf : [a, b] −→ R is a function which is differentiable on [a, b]. If f ′(a) < 0 < f ′(b), then there existsan x0 ∈ (a, b) such that f ′(x0) = 0.

Proof. Suppose a, b ∈ R, a < b, f : [a, b]→ R is a function which is differentiable on [a, b], and thatf ′(a) < 0 < f ′(b).

Since f is differentiable, f is continuous, and so f obtains its minimum at a point x0 ∈ [a, b]. Ifwe can show that x0 ∈ (a, b), then Fermat’s theorem will give f ′(x0) = 0 which is what we want.

We have f ′(a) < 0, which tells us that

limx→a

f(x)− f(a)

x− a< 0.

This means that there is an x1 ∈ (a, b) with f(x1)−f(a)x1−a < 0; f(x1) < f(a) and a is not a miniumum

for f . Similarly, b is not a miniumum for f because

limx→b

f(x)− f(b)

x− b> 0.

We conclude that x0 ∈ (a, b). Fermat’s theorem applied to f with δ = min{b−x0, x0−a} completesthe proof.

Remark 14.5.9. Later on, once we have sin : R −→ R, we can define a function f : [− 12π ,

1π ] −→ R

by f(0) = 0 and f(x) = x2 sin( 1x) when x 6= 0. This function satisfies the hypotheses of the previoustheorem, but f ′ is discontinuous at 0. This shows that applying the intermediate value theorem tof ′ is an invalid proof.

Corollary 14.5.10 (Intermediate value theorem for derivatives). Suppose a, b ∈ R, a < b, andthat g : [a, b] −→ R is a function which is differentiable on [a, b]. If g′(a) > 0 > g′(b), then thereexists an x0 ∈ (a, b) such that g′(x0) = 0.

76

15 Series

15.1 From sequences to series

Definition 15.1.1 (Series notation). If n,m ∈ Z, n ≥ m, and am, am+1, . . . , an−1, an ∈ R, we write∑nk=m ak for

am + am+1 + . . .+ an−1 + an.

We wish to consider infinite summations∑∞

k=m ak; infinite summations are referred to as series.

Definition 15.1.2. Given a series∑∞

k=m ak, the sequence (sn)∞n=m with sn =∑n

k=m ak is calledthe sequence of partial sums. If the sequence of partial sums (sn) converges then we denote its limit(as well as the series we started with) by

∞∑k=m

ak;

we say the series∑∞

k=m ak converges. Otherwise, we say that∑∞

k=m ak diverges.

Notice that when the series∑∞

k=m ak converges we have, by definition,

∞∑k=m

ak = limn→∞

n∑k=m

ak,

since∑n

k=m ak is the n-th term of the sequence of partial sums.

Theorem 15.1.3. Suppose (ak)∞k=0 is a sequence and that m ∈ N.

∑∞k=0 ak converges if and only

if∑∞

k=m ak converges and in this case

∞∑k=0

ak =m−1∑k=0

ak +∞∑k=m

ak.

Proof. Suppose (ak)∞k=0 is a sequence and that m ∈ N.

We have to use the sequences of partial sums. So define sequences (sn)∞n=m and (tn)∞n=m by

sn =n∑k=0

ak and tn =n∑

k=m

ak.

Then sn =∑m−1

k=0 ak + tn. Notice that∑m−1

k=0 ak is just some fixed number, so the algebra of limitstells us that (sn)∞n=m converges if and only if (tn)∞n=m converges, i.e.

∑∞k=0 ak converges if and only

if∑∞

k=m ak converges.The algebra of limits also tell us that, in this case, limn→∞ sn =

∑m−1k=0 ak+limn→∞ tn, i.e. that

∞∑k=0

ak =

m−1∑k=0

ak +

∞∑k=m

ak.

Notation 15.1.4. The previous theorem demonstrates that if we only care about the convergenceor divergence of a series, where we start summing does not matter. We can then be lazy and write∑ak for the series instead of

∑∞k=m ak.

77

Proposition 15.1.5. Suppose that∑∞

k=m ak and∑∞

k=m bk are convergent series and that c ∈ R.Then

1.∑∞

k=m cak is convergent; it converges to c∑∞

k=m ak.

2.∑∞

k=m(ak + bk) is convergent; it converges to∑∞

k=m ak +∑∞

k=m bk.

Proof. Part 1 and 2 follow from theorem 11.3.1 part 1 and 2, respectively.

Definition 15.1.6.∑ak is Cauchy iff the following sentence is true.

∀ε > 0, ∃N ∈ N : ∀n ∈ N, ∀m ∈ N, n ≥ m ≥ N =⇒∣∣∣∣ n∑k=m

ak

∣∣∣∣ < ε.

The point of this definition is that it is equivalent to saying that the sequence of partial sumsis Cauchy.

Proposition 15.1.7. A series is Cauchy if and only if its sequence of partial terms is Cauchy.

Proof. For n,m ∈ N with n ≥ m ≥ 2, we have∑n

k=m ak = sn − sm−1.It is left as an exercise to finish the proof.

Since a sequence is convergent if and only if it is Cauchy we obtain the following corollary.

Corollary 15.1.8. A series converges if and only if it is Cauchy.

Corollary 15.1.9. If∑ak converges, then limk→∞ ak = 0.

Proof. Suppose∑ak converges. We wish to show limk→∞ ak = 0.

Let ε > 0. Since∑ak converges, it is Cauchy, and so we can find N ∈ N such that

∀n ∈ N, ∀m ∈ N, n ≥ m ≥ N =⇒∣∣∣∣ n∑k=m

ak

∣∣∣∣ < ε.

Taking k = n = m we see that

∀k ∈ N, k ≥ N =⇒ |ak| < ε.

The converse of this result is not true; this is demonstrated by the second example below.

Example 15.1.10.

1. Suppose a, r ∈ R.∑∞

k=0 ark = a+ ar + ar2 + ar3 + . . . is called a geometric series.

In order to be interesting, suppose a 6= 0.

For r 6= 1, we have∑n

k=0 ark = a1−rn+1

1−r , since

(1− r)n∑k=0

ark = a(1− r)(1 + r + . . .+ rn) = a((1 + r + . . .+ rn)− (r + . . .+ rn + rn+1))

= a(1− rn+1).

When |r| < 1, limn→∞ rn+1 = 0 (can you prove this?). In this case, the algebra of limits gives∑∞

k=0 ark = a

1−r . When |r| ≥ 1 the series diverges since the terms do not converge to 0 (weare using the contrapositive of the corollary just proved).

78

2.∑∞

k=11k is called the harmonic series. Here are some in the terms of the sequence of partial

sums.

s1 = 1 ≥ 2

2

s2 = 1 +1

2≥ 3

2

s4 = 1 +1

2+

(1

3+

1

4

)≥ 4

2

s8 = 1 +1

2+

(1

3+

1

4

)+

(1

5+

1

6+

1

7+

1

8

)≥ 5

2

s2n ≥n+ 2

2

We see that (sn) is not bounded above and so it diverges. By definition,∑∞

k=11k diverges.

15.2 Tests for convergence and divergence

Calculating limits of series is harder than for sequences. How can we show a series converges? Well,we have a load of tests we can apply.

The first result shows that if the series in question has minus signs involved, it can be useful toconsider the series obtained by changing them all to pluses.

Definition 15.2.1. A series∑ak converges absolutely iff

∑|ak| converges.

Theorem 15.2.2. If a series∑ak is absolutely convergent, then it is convergent.

Proof. Suppose∑ak is absolutely convergent. We wish to show that

∑ak is convergent; we will

show it is Cauchy.Let ε > 0. By definition of what it means to convergence absolutely,

∑|ak| converges, and so∑

|ak| is Cauchy. This means that there exists an N ∈ N such that

∀n ∈ N, ∀m ∈ N, n ≥ m ≥ N =⇒∣∣∣∣ n∑k=m

|ak|∣∣∣∣ < ε.

Since∣∣∑n

k=m ak∣∣ ≤∑n

k=m |ak| =∣∣∑n

k=m |ak|∣∣ we also have

∀n ∈ N, ∀m ∈ N, n ≥ m ≥ N =⇒∣∣∣∣ n∑k=m

ak

∣∣∣∣ < ε.

Note that the converse of this theorem is false. . .

Definition 15.2.3. A series∑ak is conditionally convergent iff it is convergent and not absolutely

convergent.

Example 15.2.4. The series∑∞

k=1(−1)k−1

k is conditionally convergent.

Let ak = (−1)k−1

k . We have seen that∑|ak| diverges; it is the harmonic series. At the end of

this section we’ll prove the alternating series test which will tell us that∑ak converges.

79

Theorem 15.2.5 (Comparison Test). Suppose that∑ak and

∑bk are series and that ak ≥ bk ≥ 0

for all k. If∑ak converges, then

∑bk converges.

Proof. Suppose∑ak and

∑bk are series, that ak ≥ bk ≥ 0 for all k, and that

∑ak converges. Let

(sn)∞n=m and (tn)∞n=m be the sequences of partial sums corresponding to∑ak and

∑bk, respectively,

so that

sn =n∑

k=m

ak and tn =n∑

k=m

bk.

Since∑ak converges, (sn) converges. Moreover, because ak ≥ bk ≥ 0 for all k, the sequences (sn)

and (tn) are increasing and we have for all n,

tn ≤ sn ≤ limn→∞

sn,

Thus, (tn) is increasing and bounded above, and so (tn) converges, i.e.∑bk converges.

The contrapositive of this result is also known as a comparison test.

Corollary 15.2.6 (Comparison Test). Suppose that∑ak and

∑bk are series and that ak ≥ bk ≥ 0

for all k. If∑bk diverges, then

∑ak dinverges.

We also have a corollary to the first comparison test and the fact that absolutely convergentseries are convergent.

Corollary 15.2.7. Suppose∑ak and

∑bk are series and ak ≥ |bk| for all k. If

∑ak converges,

then∑bk converges.

Proof. Suppose that∑ak and

∑bk are series, ak ≥ |bk| for all k, and

∑ak converges. The first

comparison test tells us that∑|bk| converges, i.e. that

∑bk is absolutely convergent. Thus,

∑bk

converges.

Let’s summarize everything in one statement.

Theorem 15.2.8 (Comparison Tests). Let∑ak and

∑bk be series.

1. Suppose that ak ≥ |bk| for all k, and that∑ak converges. Then

∑bk converges.

2. Suppose that ak ≥ bk ≥ 0 for all k, and that∑bk diverges. Then

∑ak diverges.

Example 15.2.9.

1.∑∞

k=0(−1)k3k+5

converges since 13k≥∣∣∣ (−1)k3k+5

∣∣∣ for each k ∈ N ∪ {0}.

2.∑∞

k=15

k− 12

diverges since 5k− 1

2

≥ 1k ≥ 0 for each k ∈ N.

Example 15.2.10. The following is a very wrong statement of the comparison tests.Let

∑ak and

∑bk be series.

1. Suppose that ak ≥ bk for all k, and that∑ak converges. Then

∑bk converges.

2. Suppose that |ak| ≥ bk ≥ 0 for all k, and that∑bk diverges. Then

∑ak diverges.

80

Counter-examples are as follows:

1. Take ak = 0 and bk = −1.

2. Take ak = (−1)k−1

k and bk = 1k .

I’ll leave it to you to figure out the best way of making sure you remember what is true; hopefully,you’ll decide that it is to understand the sequence of arguments that I just made.

Theorem 15.2.11 (Ratio test). Let (ak) be a sequence of non-zero terms such that

α = limk→∞

∣∣∣∣ak+1

ak

∣∣∣∣exists.

1.∑ak converges if α < 1;

2.∑ak diverges if α > 1;

(α = +∞ is allowed here, although you need to look up the definition of “diverge to +∞”).

Proof. Let (ak) be a sequence of non-zero terms such that α = limk→∞ |ak+1

ak| exists.

First, suppose that α < 1. Since (|ak+1

ak|)∞k=1 converges to α, there is an N ∈ N with the property

that

∀k ∈ N, k ≥ N =⇒∣∣∣∣∣∣∣∣ak+1

ak

∣∣∣∣− α∣∣∣∣ < 1− α2

.

Let r = 1+α2 . Then 0 < r < 1 and when k ∈ N and k ≥ N we have∣∣∣∣ak+1

ak

∣∣∣∣ ≤ ∣∣∣∣∣∣∣∣ak+1

ak

∣∣∣∣− α∣∣∣∣+ |α| < 1− α2

+ α = r

which tells us that |ak+1| < r|ak|. Define a sequence (bk) by

bk =

{|ak| when k < N

rk−N |aN | when k ≥ N.

|ak| ≤ bk for each k, and∑bk converges since eventually it just looks like a geometric series with

ratio r and |r| < 1. The comparison test tells us that∑ak converges.

To clarify this proof you might want to go through it in the case that ak = 10k

k! . Then α = 0,r = 1

2 and you can take N = 20.Now suppose that α > 1. Similarly to above, we can find an N ∈ N with the property that

∀k ∈ N, k ≥ N =⇒ |ak+1| > |ak|

Thus, whenever k ∈ N and k ≥ N , we have |ak| ≥ |aN |. This means that the sequence (ak) cannotconverge to 0 and so

∑ak diverges.

Remark 15.2.12. When α = 1 nothing can be said: the series∑∞

k=11k diverges; the series

∑∞k=1

1k2

converges. (You can prove the latter claim by showing that∑n

k=11k2≤ 2 − 1

n ≤ 2; an increasingsequence which is bounded above converges.)

81

Theorem 15.2.13 (Root test). Let (ak) be a sequence such that α = limk→∞ |ak|1k exists.

1.∑ak converges if α < 1.

2.∑ak diverges α > 1;

(α = +∞ is allowed here, although you need to look up the definition of “diverge to +∞”).

Proof. Very similar to the previous proof. In fact, it is a little bit easier. When α < 1, we can findan N ∈ N so that

∀k ∈ N, k ≥ N =⇒ |ak|1k < r

where r = 1+α2 which is less than 1. Thus, when k ∈ N and k ≥ N , we have |ak| < rk . . .

Remark 15.2.14. Just as for the ratio test, when α = 1, nothing can be said.

Remark 15.2.15. If one of the ratio test or root test is inconclusive because the relevant limit is1, then the other will also be inconclusive.

Remark 15.2.16. There are more general versions of the ratio test and root test which make useof lim sup and lim inf. They are not much harder to prove, but I don’t want to discuss lim sup andlim inf.

Example 15.2.17.

1.∑∞

k=0 3(−1)k−k converges.

The ratio test does not tell us anything (even the more general one). But the root test appliessince

limk→∞

|3(−1)k−k|1k = lim

k→∞3

(−1)k

k−1 = 3−1 < 1.

Here we have used continuity of exponentiation; we have not proved continuity of exponenti-ation yet but it will follow from definition 16.3.9.

2.∑∞

k=1k!3k

diverges sinceak+1

ak= k+1

3 and the sequence (k+13 )∞k=1 diverges to +∞.

Theorem 15.2.18 (Alternating series). Suppose (ak)∞k=1 is a decreasing sequence converging to 0

(with each ak ≥ 0). Then∑

(−1)k−1ak converges.

Proof. Suppose (ak)∞k=1 is a decreasing sequence which converges to 0.

Let sn =∑n

k=1(−1)kak. Then we have the following facts.

• For all n ∈ N, s2n+1 ≤ s2n (since −a2n+1 ≤ 0)

• (s2n+1)∞n=1 is increasing (since a2n − a2n+1 ≥ 0)

• (s2n)∞n=1 is decreasing (since −a2n−1 + a2n ≤ 0)

• (s2n+1)∞n=1 is bounded above by 0 (since s2n+1 ≤ s2n ≤ s2n−2 ≤ s2n−4 ≤ . . . ≤ s4 ≤ s2 ≤ 0)

• (s2n)∞n=1 is bounded below by −a1 (since s2n ≥ s2n+1 ≥ s2n−1 ≥ . . . ≥ s3 ≥ s1 = −a1.)

82

Since bounded monotone sequences converge, we deduce that there exist s, t ∈ R such that (s2n)converges to s and (s2n+1) converges to some t. We have (s2n− s2n+1) = (a2n+1). Taking the limitgives s− t = 0. So limn→∞ s2n = limn→∞ s2n+1 = s.

One can now show that limn→∞ sn = s using an N = max{N1, N2} style argument; if you writethis out you’ll probably end up taking N = max{2N1, 2N2 + 1}.

Thus, (−sn) converges, as required.

Example 15.2.19.∑∞

k=1(−1)k−1

k = 1− 12 + 1

3 −14 + . . . converges.

15.3 Power series

Definition 15.3.1. Let (an)∞n=0 be a sequence of reals and c ∈ R. A power series (centered at c)is a series of the form

∞∑n=0

an(x− c)n = a0 + a1(x− c) + a2(x− c)2 + a3(x− c)3 + . . .

It defines a function f with domain D = {x ∈ R :∑∞

n=0 an(x− c)n converges}.

Example 15.3.2.

1. f(x) =∑∞

n=0 xn has domain (−1, 1). On this set it agrees with the function 1

1−x .

2. f(x) =∑∞

n=0(x−3)nn! has domain R (by the ratio test, since limn→∞

|x−3|n+1 = 0). In fact, it

defines the function exp(x− 3).

3. f(x) =∑∞

n=0 n!(x+2)n has domain {−2} (by the ratio test, since limn→∞(n+1)|x+2| = +∞when x 6= −2). At this point it takes the value 0! · 00 = 1.

Theorem 15.3.3. Let∑∞

n=0 an(x− c)n be a power series such that β = limn→∞ |an|1n exists. Let

R = 1β [with the convention that 1

0 = +∞ and 1+∞ = 0].

1. The power series converges absolutely on the set {x ∈ R : |x− c| < R}.

2. The power series diverges on the set {x ∈ R : |x− c| > R}.

Thus, the domain D of the function f defined by∑∞

n=0 an(x− c)n satisfies

(c−R, c+R) ⊆ D ⊆ [c−R, c+R].

Proof. Follows from the root test.

Definition 15.3.4. The R in the theorem is called the radius of convergence of∑∞

n=0 an(x− c)n.

Remark 15.3.5. The theorem says nothing about convergence or divergence at the end points ofthe interval, i.e. when x = c−R or c+R.

Remark 15.3.6. It is unsatisfying that this theorem is not applicable for all power series, namelythose for which limn→∞ |an|

1n does not exist. A more powerful version of the theorem uses the fact

that lim supn→∞ |an|1n always exists.

83

Remark 15.3.7. If lim |an+1

an| is easier to calculate than lim |an|

1n , we can make use of the following

lemma.

Lemma 15.3.8. Suppose (an) is a sequence consisting of non-zero terms and lim |an+1

an| exists

(either it converges, or it diverges to +∞). Then lim |an|1n exists and equals lim |an+1

an|.

Proof. Omitted.

Example 15.3.9. f(x) =∑∞

n=1xn

n has domain [−1, 1).

1. β = lim nn+1 = 1, so the radius of convergence R is 1

1 = 1.

This tells us the series converges on (−1, 1) and diverges outside of [−1, 1].

2. The series diverges when x = 1, since this gives the harmonic series.

3. The series converges when x = −1 since this gives the alternating series series.

We will see, shortly, that the function f agrees with − log(1− x) on [−1, 1).

Example 15.3.10. f(x) =∑∞

n=01

(n+1)22n(x − 2)n has domain [0, 4], where the power series con-

verges absolutely.

1. β = lim (n+1)22n

(n+2)22n+1 = 12 , so the radius of convergence R is 2.

So the power series converges absolutely on (0, 4).

2. The power series converges absolutely for x = 0 and x = 4 because∑∞

n=01

(n+1)2converges

(15.2.12).

The reason power series are so useful is that they allow one to define many continuous anddifferentiable functions. Here is the theorem one needs.

Theorem (Not on syllabus). Suppose the power series∑∞

n=0 an(x− c)n has radius of convergenceR and let f(x) =

∑∞n=0 an(x− c)n. Then

1. f(x) is continuous on its domain;

2. f(x) is differentiable on (c−R, c+R) and there we have f ′(x) =∑∞

n=1 nan(x− c)n−1.

Proof. Omitted. Continuity at the endpoints is actually quite tricky (Abel’s theorem).

Up until now we didn’t have too many continuous or differentiable functions: pretty much justpolynomials. We used infinity to our advantage again, realizing that infinite polynomials, i.e. powerseries, are a manageable concept.

The following corollary motivates attempting to write an infinitely differentiable function usinga power series, the Taylor series.

Corollary. Suppose the power series∑∞

n=0 an(x− c)n has a positive radius of convergence R > 0.Let f(x) =

∑∞n=0 an(x− c)n. Then f is infinitely differentiable on (c−R, c+R) and

f (n)(c) = n!an, so that f(x) =

∞∑n=0

f (n)(c)

n!(x− c)n.

84

16 More functions: cos, sin, exp, log

16.1 Definitions

We have proved a lot of theorems about continuous functions and differentiable functions. Beforethe last section we only really had polynomials as examples. Now we have power series and theseenable us to talk about our favorite functions - cosine, sine, and the exponential function.

Definition 16.1.1. exp, cos, sin : R −→ R are defined by

expx =∞∑n=0

xn

n!= 1 + x+

x2

2!+x3

3!+ . . .

cosx =

∞∑n=0

(−1)nx2n

(2n)!= 1− x2

2!+x4

4!− x6

6!+ . . .

sinx =

∞∑n=0

(−1)nx2n+1

(2n+ 1)!= x− x3

3!+x5

5!− x7

7!+ . . . .

It is straightforward to check that these power series all have infinite radius of convergence usingthe ratio test.

16.2 cos, sin and π.

Theorem 16.2.1. cos′ = − sin, sin′ = cos.

Proof. This follows from the off-syllabus theorem, which tells us we can differentiate term-by-term,and polynomial differentiation.

Theorem 16.2.2. For all x ∈ R, cos2 x+ sin2 x = 1.

Proof. By inspection cos(0) = 1 and sin(0) = 0, and so the result is true when x = 0.Let f : R −→ R be defined by f(x) = cos2 x+ sin2 x. Then by the chain rule

f ′(x) = −2 cosx sinx+ 2 sinx cosx = 0,

so that f(x) = 1 for all x ∈ R.

Corollary 16.2.3. For all x ∈ R, | cosx| ≤ 1 and | sinx| ≤ 1.

Theorem 16.2.4 (The defining properties of cos and sin).Let p : R −→ R2 be defined by p(t) = (cos t, sin t). Then

1. ∀t ∈ R, p(t) · p(t) = p′(t) · p′(t) = 1;

2. p(0) = (1, 0), p′(0) = (0, 1);

3. ∀t ∈ R, (p(t), 0)× (p′(t), 0) = (0, 0, 1).

Proof. Easy.

85

The first property is that p(t) is a path on the unit circle which is parametrized by arc-length.The second property is that p(t) starts at (1, 0) and travels upwards. The third property is that p(t)travels counter-clockwise. So p(t) moves around the unit circle, counterclockwise, at unit speed,starting at (1, 0). This is how cos and sin should have been defined to you in high school.

What is a sensible definition of π? π should be the length of the path p(t) from (1, 0) to (−1, 0)along the unit circle. Using the theorem just proved this suggest the following.

Definition 16.2.5. π is the first nonzero solution of sinx = 0.

Lemma 16.2.6. For all x ∈ (0, 1], sinx > 0.

Proof. Postponed until we have Taylor’s theorem. See lemma 17.5.8.

Theorem 16.2.7. There is an x0 ∈ (1, 4) with sinx0 = 0.

Proof. Postponed until we have Taylor’s theorem. See theorem 17.5.9.

Theorem 16.2.8. limx→0sinxx = 1.

Proof. Let g(x) =∑∞

n=0(−1)nx2n(2n+1)! .

Then g is continuous at 0 and g(0) = 1. Moreover, xg(x) = sinx.Also, this follows from the fact that sin′ 0 = cos 0 = 1.

16.3 exp, log, exponents

Theorem 16.3.1. exp′ = exp.

Proof. Differentiate the power series term-by-term.

Lemma 16.3.2. exp(0) = 1 and for x > 0, expx > 1 + x.

Proof. exp(x) = 1 + x+ x2[∑∞

n=2xn−2

n!

]and for x ≥ 0,

∑∞n=2

xn−2

n! ≥12 .

Theorem 16.3.3. exp(x) > 0 for all x ∈ R.

Proof. Let f : R→ R be defined by f(x) = exp(x) exp(−x). The product rule and chain rule allowus to show that f ′(x) = 0 for all x ∈ R. Moreover, f(0) = 1 so that

exp(x) exp(−x) = 1 for all x ∈ R.

For x ≥ 0, we have exp(x) ≥ 1 > 0. The formula above shows that exp(x) > 0 for all x ∈ R.

Theorem 16.3.4. exp(x+ y) = exp(x) exp(y) for all x, y ∈ R.

Proof. Fixing y ∈ R, we can define f : R −→ R by

f(x) =exp(x) exp(y)

exp(x+ y).

f is continuous and differentiable. The quotient rule and chain rule allow us to show that f ′(x) = 0.Moreover, f(0) = 1.

86

Theorem 16.3.5. exp is a strictly increasing function with image (0,∞).

Proof. For all x ∈ R, exp′ x = expx > 0, so that exp is strictly increasing.Given c ∈ (1,∞), we have exp(0) = 1 < c < exp(c − 1) using lemma 16.3.2. The intermediate

value theorem says there is an x0 ∈ (0,∞) with exp(x0) = c. Given c ∈ (0, 1), 1c ∈ (1,∞), so there

is an x0 with exp(x0) = 1c and exp(−x0) = c. Finally, exp(0) = 1.

Theorem. There is a function log : (0,∞) −→ R with

1. log(exp(x)) = x for all x ∈ R;

2. exp(log(y)) = y for all y ∈ (0,∞).

Proof. First, we need to define log, so let y ∈ (0,∞). The theorem just proved gives us a x ∈ Rwith exp(x) = y. Moreover, since exp is strictly increasing, there is only one such x. Let log(y) = x;then the second property is immediate. For the other property, let x ∈ R. Then exp(log(exp(x))) =exp(x). Since exp is strictly increasing, exp is injective so that log(exp(x)) = x.

Theorem 16.3.6. log is continuous.

Proof. Let y0 ∈ (0,∞) and (yn) be a sequence in (0,∞) converging to y0. For each n ∈ N ∪ {0},let xn = log(yn). We need to show that limn→∞ xn = x0.

Suppose for contradiction that (xn) does not converge to x0. Then there exists a δ > 0 so thatwhatever N ∈ N we pick, we can find an n ∈ N such that n ≥ N and |xn − x0| ≥ δ.

Let ε = exp(x0)− exp(x0 − δ). Then ε > 0 and | exp(x)− exp(x0)| ≥ ε whenever |x− x0| ≥ δ.(Proving this uses the fact that exp and its derivative are increasing and the mean value theorem.)

Whatever N ∈ N we pick we can find an n ∈ N with n ≥ N and |xn − x0| ≥ δ. For such an n,|yn − y0| = | exp(xn)− exp(x0)| ≥ ε. This contradicts (yn) converging to y0.

Theorem 16.3.7. log is differentiable with log′(x) = 1x .

Proof. Let y0 ∈ (0,∞) and (yn) be a sequence in (0,∞) \ {y0} converging to y0. We need to showthat

limn→∞

log(yn)− log(y0)

yn − y0=

1

y0.

For n ∈ N ∪ {0}, let xn = log(yn). Since log is continuous we have

limn→∞

xn = limn→∞

log(yn) = log(y0) = x0.

Moreover, since log is injective and for all n ∈ N, yn 6= y0, we have for all n ∈ N, xn 6= x0. So (xn)is a sequence in R \ {x0} converging to x0. This means that

limn→∞

log(yn)− log(y0)

yn − y0= lim

n→∞

xn − x0exp(xn)− exp(x0)

= limx→x0

x− x0exp(x)− exp(x0)

=1

exp′(x0)=

1

exp(x0)=

1

y0.

87

Theorem 16.3.8. The power series∞∑n=1

(−1)n−1

nxn

converges to log(1 + x) for x ∈ (−1, 1].

Proof. Let f(x) =∑∞

n=0(−1)n−1

n xn. Then f is defined on (−1, 1]. The off-syllabus theorem fromlast section says that f is differentiable on (−1, 1) and f ′(x) is given by

∞∑n=1

(−1)n−1xn−1 =∞∑n=0

(−x)n =1

1 + x.

Let g : (−1, 1] −→ R be defined by g(x) = log(1 + x). The chain rule shows that f ′(x) = g′(x)on (−1, 1). Moreover, f(0) = g(0) = 0, so that f(x) = g(x) on (−1, 1). The off-syllabus theoremsays that f is continuous at 1; so is g. Thus f(1) = g(1), too.

The exponential function and the logarithmic function allow us to make sense of all exponentswith one definition. Before now you may not have known what 2

√2 means.

Definition 16.3.9 (Exponents). If x > 0 and y ∈ R, then define

xy := exp(y log x).

The properties that we proved about exp and log allow one to prove the following lemma.

Lemma 16.3.10. For x > 0 and y, z ∈ R, xy+z = xy · xz.Corollary 16.3.11. For x > 0 and y ∈ Q, xy means what you think it does.

Theorem 16.3.12. Suppose that y ∈ R. Then f : (0,∞) −→ R defined by f(x) = xy is differen-tiable and f ′(x) = yxy−1.

Proof. Using the definition of xy, the chain rule and some previously proved results give

f ′(x) = exp′(y log x) · (y log′ x) = exp(y log x) · yx

= xy · yx

= yxy−1.

16.4 De Moivre’s theorem (off syllabus)

By plugging in complex numbers into the power series defining exp, cos, sin one obtains De Moivre’stheorem:

exp(iθ) = cos(θ) + i sin(θ).

One reason this is interesting is because it tells us that

∀n ∈ Z, exp(2πin) = 1.

There are many choices for log 1. This creates a lot for mathematicians to play with, for instance,xy becomes multivalued. A simple example of this is that x

12 now gives the positive and negative

square root.Also, exp(πi) = −1, so log(−1) = πi + 2πin. This means we can define xy for negative x. We

can even define it for x ∈ C \ 0 and y ∈ C.

ii = exp(i log i) = exp

[i · i(π

2+ 2πn

)]= exp

[−(π

2+ 2πn

)]∈ R.

88

17 Taylor’s theorem

17.1 Taylor series and Taylor polynomials

In the section on power series, we saw the following development of ideas.

• Starting with a sequence (an)∞n=0 of real numbers and c ∈ R,

• we get a power series∑∞

n=0 an(x− c)n centered at c.

• Define a function by f(x) =∑∞

n=0 an(x− c)n.

Its domain contains (c−R, c+R) where R = 1β and β = lim sup |an|

1n .

• If R > 0, then f is infinitely differentiable on (c−R, c+R), and the sequence (f (n)(c))∞n=0 isdetermined by (an)∞n=0 in a simple manner: for all n ∈ N ∪ {0}, we have f (n)(c) = n!an.

The consequence of this chain of ideas is that if we want to define a function f , with f (n)(c) = kn,then we can try the function f(x) =

∑∞n=0

knn! (x − c)

n; as long as the power series has a positiveradius of convergence, we succeed.

Power series are convenient, and so we might start with a function f which is infinitely differ-entiable at a point c, and try to write it in terms of a power series centered at c. The above chainof ideas shows that there is only one power series we should bother trying. It is called the Taylorseries of f about c.

Definition 17.1.1. Suppose f : (a, b) −→ R is infinitely differentiable at c ∈ (a, b). Then

T (x) =

∞∑n=0

f (n)(c)

n!(x− c)n = f(c) + f ′(c)(x− c) +

f ′′(c)

2!(x− c)2 +

f ′′′(c)

3!(x− c)3 + . . .

is called the Taylor series of f about c.

The following lemma is essentially what motivated the definition of the Taylor series.

Lemma 17.1.2. Suppose f : (a, b) −→ R is infinitely differentiable at c ∈ (a, b) and T (x) is theTaylor series of f about c.

If T (x) has a positive radius of convergence then, for all n ∈ N ∪ {0}, we have

T (n)(c) = f (n)(c).

An interesting question is as follows.

Question. When does T (x) converge to f(x)?Answer. Sometimes, but not always. We will see this shortly.

In the context of complex analysis Taylor’s theorem says that “everything differentiable functionis locally defined by a power series.” Things are a little more tricky in real analysis. We still havesomething called Taylor’s theorem but it is important because it allows one to estimate the error thatoccurs when one approximates a function by polynomials, a useful method in applied mathematics.

There are numerous ways one could try to approximate a function by a polynomial dependingon what one thinks of as a “good” approximation. The Taylor polynomials work in the same wayas Taylor series: they ensure that some of the derivatives are correct at the point about which weapproximate.

89

Definition 17.1.3. Suppose f : (a, b) −→ R is n times differentiable at c ∈ (a, b). Then

Tn(x) =n∑k=0

f (k)(c)

k!(x− c)k = f(c) + f ′(c)(x− c) +

f ′′(c)

2!(x− c)2 + . . .+

f (n)(c)

n!(x− c)n

is called the n-th Taylor polynomial of f about c.

Lemma 17.1.4. Suppose f : (a, b)→ R is n times differentiable at c ∈ (a, b) and Tn(x) is the n-thTaylor polynomial of f about c. Then, for all k ∈ N with k ≤ n, we have

T (k)n (c) = f (k)(c).

For k ∈ N with k > n we have T(k)n (x) = 0.

Tn(x) is the unique polynomial for which the lemma just stated is true.

17.2 Taylor series of polynomials

One should note that polynomials are simple instances of power series; they’re power series in whichonly finitely many of the coefficients are nonzero. As a result, finding the Taylor series and Taylorpolynomials of polynomials is very easy.

Example 17.2.1. Let f(x) = x3 + x+ 1.The Taylor series of f(x) about 0 is T (x) = 1 + x+ x3, i.e. f(x) itself.The Taylor polynomials about 0 are as follows: T0(x) = 1, T1(x) = 1 + x = T2(x), and for all

n ∈ N with n ≥ 3, we have Tn(x) = 1 + x+ x3.

Example 17.2.2. Let f(x) = x3 + x+ 1.The Taylor series of f(x) about 1 is

3 + 4(x− 1) + 3(x− 1)2 + (x− 1)3,

i.e. f(x) itself but with some algebra done to write it in terms of powers of (x− 1) instead of x.The Taylor polynomials about 1 are as follows:

T0(x) = 3, T1(x) = 3 + 4(x− 1), T2(x) = 3 + 4(x− 1) + 3(x− 1)2,

and for all n ∈ N with n ≥ 3, we have Tn(x) = 3 + 4(x− 1) + 3(x− 1)2 + (x− 1)3.

17.3 A rubbish Taylor series

Consider the function g : R −→ R defined by g(x) = |x|. g is infinitely differentiable at 1 and itsTaylor series about 1 is given by T (x) = 1 + (x− 1). Although the Taylor series has infinite radiusof convergence, it only agrees with g on [0,∞). This is not so bad because at least there is an openinterval, centered around 1, on which it agrees with g, namely (0, 2). The reason things go wrong,in this example, is because g is not even differentiable at 0. You might be led to speculate thatfor an infinitely differentiable function f : R −→ R, that the Taylor series at a point c agrees withthe function on its domain of convergence, or at least on a smaller open interval centered aroundc. This is not the case.

90

Example 17.3.1. Define f : R −→ R by

f(x) =

{exp

(−1x

)for x > 0

0 for x ≤ 0.

When x > 0, f(x) 6= 0, but the next two lemmas show that the Taylor series of f at 0 is zero.

Lemma 17.3.2. With f as above, there are polynomials {pn(x) : n ∈ N ∪ {0}} such that

∀n ∈ N, ∀x > 0, f (n)(x) =pn(x)

x2nf(x).

Proof. We can take p0(x) = 1. Suppose f (n)(x) = pn(x)x2n

f(x). Then

f (n+1)(x) =x2np′n(x)− 2nx2n−1pn(x)

x4nf(x) +

pn(x)

x2nf ′(x).

=x2p′n(x)− 2nxpn(x) + pn(x)

x2(n+1)· f(x)

so we can take pn+1(x) = x2p′n(x)− 2nxpn(x) + pn(x).

Lemma 17.3.3. With f as above, f is infinitely differentiable at 0 and for all n ∈ N, f (n)(0) = 0.

Proof. Suppose we have shown that f is n times differentiable and f (n)(0) = 0 starting with n = 0.Then

limx→0+

f (n)(x)− f (n)(0)

x− 0= pn(0) · lim

x→0+

f(x)

x2n+1= pn(0) · lim

y→∞+

y2n+1

exp(y)= 0.

It is easy to see limx→0−f (n)(x)−f (n)(0)

x−0 = 0, so f (n+1)(0) = 0.

Thus, when f is as above, its Taylor series about 0 is zero, but when x > 0, f(x) > 0; in thisexample, the Taylor series does not converge to the original function, even at points close by to 0.

17.4 An even more rubbish Taylor series (definitely off syllabus)

Consider the function f : R −→ R defined by

f(x) =

∞∑m=1

exp(−m) cos(m2x).

Supposing we can differentiate term-by-term we have, for n ∈ N,

f (4n)(0) =

∞∑m=1

[exp(−m) · (m2)4n

]≥ exp(−4n) · ((4n)2)4n =

[(4n)2

exp(1)

]4n≥ 0.

Thus,

β = lim sup

∣∣∣∣f (n)(0)

n!

∣∣∣∣ 1n ≥ lim sup

[f (4n)(0)

(4n)!

] 14n

≥ lim sup

[f (4n)(0)

(4n)4n

] 14n

≥ lim sup

[4n

exp(1)

]= +∞,

and so the radius of convergence of the Taylor series of f about 0 is equal to 1β = 0.

We can check that differentiation term-by-term is valid. This is a special case of a more generalresult that you can find in an analysis textbook.

91

Lemma. Let f : R −→ R be as above. Then f is infinitely differentiable and

∀x ∈ R, f (n)(x) =∞∑m=1

[exp(−m) · (m2)n · cos(n)(m2x)

].

Proof. We proceed by induction. The result is true when n = 0 so suppose that n ∈ N ∪ {0} andthat f (n)(x) has the description in the statement of the lemma. For m ∈ N, let gm(x) = cos(m2x)and fix x ∈ R. Using the triangle inequality for series, we have, for y 6= x,∣∣∣∣f (n)(y)− f (n)(x)

y − x−∞∑m=1

[exp(−m) ·g(n+1)

m (x)

]∣∣∣∣ ≤ ∞∑m=1

[exp(−m) ·

∣∣∣∣g(n)m (y)− g(n)m (x)

y − x−g(n+1)

m (x)

∣∣∣∣].Using the mean value theorem, the triangle inequality, and the fact that all derivatives of cos arebounded by 1, we find that, for all m ∈ N,∣∣∣∣g(n)m (y)− g(n)m (x)

y − x− g(n+1)

m (x)

∣∣∣∣ ≤ 2(m2)n+1.

Suppose ε > 0. Since∑∞

m=1

[exp(−m) · 2(m2)n+1

]converges we can find an M ∈ N so that

∞∑m=M+1

[exp(−m) · 2(m2)n+1

]<ε

2.

We can find a δ > 0 so that

∀y ∈ R, 0 < |y − x| < δ =⇒M∑m=1

[exp(−m) ·

∣∣∣∣g(n)m (y)− g(n)m (x)

y − x− g(n+1)

m (x)

∣∣∣∣] < ε

2.

Then

∀y ∈ R, 0 < |y − x| < δ =⇒∣∣∣∣f (n)(y)− f (n)(x)

y − x−∞∑m=1

[exp(−m) · g(n+1)

m (x)

]∣∣∣∣ < ε.

Thus, f (n+1)(x) =∑∞

m=1

[exp(−m) · g(n+1)

m (x)

]which completes the inductive step.

17.5 Taylor’s theorem

The examples of the last two sections show that when given an infinitely differentiable function f ,the Taylor series of f may not converge to f .

In real analysis (as opposed to complex analysis), Taylor’s theorem is not concerned with Taylorseries, but with Taylor polynomials. It allows one to estimate the difference between the value thatthe actual function f takes and the value which a Taylor polynomial Tn−1(x) takes.

Theorem 17.5.1 (Taylor’s theorem). Suppose a, b ∈ R, a < b, and f : (a, b) −→ R is a function.Suppose n ∈ N, f is n times differentiable on (a, b), that c ∈ (a, b) and x ∈ (a, b) \ {c}. Then thereexists a y between c and x such that

f(x) =n−1∑k=0

f (k)(c)

k!(x− c)k +

f (n)(y)

n!(x− c)n.

92

Remark 17.5.2. Let f be as in the statement of Taylor’s theorem and let Tn−1(x) be the (n−1)-stTaylor polynomial of f at c ∈ (a, b). Then the above equation can be rewritten as follows.

f(x) = Tn−1(x) +f (n)(y)

n!(x− c)n

Remark 17.5.3. Reading Taylor’s theorem in the case when n = 1, assuming c < x, writing d forx, and x0 for y, gives the following statement.

Suppose f : (a, b) −→ R is differentiable. Suppose that c, d ∈ R and that a < c < d < b. Thenthere is an x0 ∈ (c, d) so that

f(d) = f(c) + f ′(x0)(d− c), i.e. f ′(x0) =f(d)− f(c)

d− c.

This follows from the mean value theorem since f is continuous on [c, d] and differentiable on (c, d).

Proof of Taylor’s theorem. Suppose a, b ∈ R, a < b, and f : (a, b) −→ R is a function. Supposen ∈ N, f is n times differentiable on (a, b), that c ∈ (a, b) and x ∈ (a, b) \ {c}.

Let M ∈ R be the unique solution to

f(x) =

n−1∑k=0

f (k)(c)

k!(x− c)k +

M

n!(x− c)n.

We need to show that there is a y between c and x with f (n)(y) = M .Let g : (a, b) −→ R be defined by

g(t) =n−1∑k=0

f (k)(c)

k!(t− c)k +

M

n!(t− c)n − f(t).

One checks that g(k)(c) = 0 for k < n. Let x0 = x. By choice of M , we have g(x0) = 0. Rolle’stheorem provides x1 between c and x0 with g′(x1) = 0.

Suppose k ∈ N, k < n, and that we have x1, . . . , xk such that

xj is between c and xj−1 for 1 ≤ j ≤ k, and g(j)(xj) = 0 for j ≤ k.

Rolle’s theorem applied to g(k) provides xk+1 between c and xk such that g(k+1)(xk+1) = 0. So napplications of Rolle’s theorem provides xn between c and x such that g(n)(xn) = 0. But

g(n)(t) = M − f (n)(t),

so this says f (n)(xn) = M and we can take y = xn.

In math 31B, questions on Taylor’s theorem look like the following examples.

Example 17.5.4. Let Tn(x) denote the n-th Taylor polynomial of sinx about 0. Find an n ∈ Nso that | sin(1)− Tn(1)| < 1

5000 .We apply Taylor’s theorem. We take f = sin, a = −1, b = 2, use n+ 1 instead of n, c = 0 and

x = 1. Taylor’s theorem says that there is a y ∈ (0, 1) with

sin(1)− Tn(1) =sin(n+1)(y)

(n+ 1)!.

Since | sin(n+1)(y)| ≤ 1, | sin(1)−Tn(1)| ≤ 1(n+1)! and so we see that n = 6 works ( 1

7! = 15040 <

15000 .)

93

Example 17.5.5. Let Tn(x) denote the n-th Taylor polynomial of sinx about 0. Find an n ∈ Nso that | sin(2)− Tn(2)| < 1

5000 .We apply Taylor’s theorem. We take f = sin, a = −1, b = 100, use n + 1 instead of n, c = 0

and x = 2. Taylor’s theorem says that there is a y ∈ (0, 2) with

sin(2)− Tn(2) =sin(n+1)(y)

(n+ 1)!2n+1.

Since | sin(n+1)(y)| ≤ 1, | sin(2)− Tn(2)| ≤ 2n+1

(n+1)! and so we see that n = 10 works:

211

11!=

2

1· 2

2· 2

3· 2

4· 2

5· 2

6· 2

7· 2

8· 2

9· 2

10· 2

11

< 2 · 1 · 1 · 1

2· 1

2· 1

3· 1

3· 1

4· 1

4· 1

5· 1

5

=2

(5!)2<

2

1002=

1

5000.

Example 17.5.6. Let Tn(x) denote the n-th Taylor polynomial of log x about 1. Find an n ∈ Nso that | log(1310)− Tn(1310)| < 1

600 .We apply Taylor’s theorem. We take f = log, a = 0, b = 2, use n + 1 instead of n, c = 1 and

x = 1310 . Taylor’s theorem says that there is a y ∈ (1, 1310) with

log

(13

10

)− Tn

(13

10

)=

log(n+1)(y)

(n+ 1)!

(13

10− 1

)n+1

=log(n+1)(y)

(n+ 1)!

(3

10

)n+1

.

One checks that | log(n+1)(y)| = n!yn+1 < n! so that∣∣∣∣ log

(13

10

)− Tn

(13

10

)∣∣∣∣ < 1

n+ 1

and we see that n = 600 will do.

We can also use Taylor’s theorem to fill in some gaps from the cos and sin section.

Lemma 17.5.7. For x ∈ (0, 1), cosx ≥ 12 .

Proof. We apply Taylor’s theorem. We take f = cos, a = −1, b = 1, n = 2, c = 0, and x ∈ (0, 1).Taylor’s theorem says that there is a y ∈ (0, x) with

cosx = 1− cos(y)

2x2.

Since 0 ≤ x2 ≤ 1 and cos y ≤ 1, we have

cosx = 1− cos(y)

2x2 ≥ 1− x2

2≥ 1− 1

2=

1

2.

94

Lemma 17.5.8. For all x ∈ (0, 1], sinx > 0.

Proof. Suppose for contradiction that there is an x ∈ (0, 1] such that sinx ≤ 0. Since sin 0 = 0, themean value theorem tells us that there is a x0 ∈ (0, 1) with sin′ x0 ≤ 0, i.e. cosx0 ≤ 0. The lemmajust proven says that cosx0 ≥ 1

2 , so this is the required contradiction.

Theorem 17.5.9. There is an x0 ∈ (1, 4) with sinx0 = 0.

Proof. We have just seen that sin 1 > 0. Use Taylor’s theorem with f = sin, n = 9, c = 0 and x = 4to deduce that sin 4 < 0. Use the intermediate value theorem to deduce that there is an x0 ∈ (1, 4)with sinx0 = 0.

18 Miscellaneous

18.1 Denseness of the rationals and irrationals in R

Theorem 18.1.1. Suppose a, b ∈ R and that a < b. Then there exists a q ∈ Q such that a < q < b.

Proof. Suppose a, b ∈ R and that a < b. Let n = d 2b−ae and

q =

⌊10n(a+b2 )

⌋10n

.

We claim that q ∈ Q and a < q < b.Checking that q is rational isn’t too bad. Because a < b, we have b− a > 0. Thus, 2

b−a > 0 and

n = d 2b−ae ∈ N. Since b10n(a+b2 )c ∈ Z and n ∈ N, we have q ∈ Q.

Before verifying that a < q < b, let us address where our formula for q came from. First, wespotted that a < a+b

2 < b. Then we obtained q by truncating the decimal expansion for a+b2 at the

n-th decimal place. Truncating a decimal expansion results in a smaller number and so it shouldbe true that q ≤ a+b

2 < b. We also need to show a < q. The point is that, by choosing n to be large

enough, we ensured q approximates a+b2 well enough so that it is larger than a.

We verify that a < q < b carefully (making no reference to decimal expansions).

• b10n(a+b2 )c ≤ 10n(a+b2 ), so q ≤ a+b2 .

• a+b2 < b+b

2 = b.

• Putting the last two bullet points together gives q < b.

• 10n(a+b2 )− b10n(a+b2 )c < 1, so a+b2 − q <

110n .

• 10n ≥ n, so 110n ≤

1n .

• n = d 2b−ae ≥

2b−a > 0, so 1

n ≤b−a2 .

• Putting the last three bullet points together gives a+b2 − q <

b−a2 , so q > a+b

2 −b−a2 = a.

Easier proof. Suppose a, b ∈ R and that a < b. Let n = d 1b−ae+ 1, m = bna+ 1c, and q = m

n .

q ∈ Q. a < q < b is equivalent to a < mn < b which is equivalent to na < m < nb. n > 1

b−a , sona+ 1 < nb, thus na < m < nb is implied by na < m ≤ na+ 1, and m has this property.

95

Theorem 18.1.2. Suppose a, b ∈ R and that a < b. There exists an r ∈ R \Q such that a < r < b.

Proof. Suppose a, b ∈ R and that a < b. By the previous theorem, there exists a q ∈ Q such thata+√

2 < q < b+√

2. Then a < q −√

2 < b. Let r = q −√

2.We just need to show r ∈ R \Q. Suppose for contradiction that r ∈ Q. Then

√2 = q − r ∈ Q,

the required contradiction.

If you look up the proof of the first theorem in the textbook, you’ll find that they make use ofthe Archimedean property and that to prove this they make use of the completeness axiom. Whyis it so complicated? This comes down to what we can or cannot assume. At the start of this class,I assumed that you were familiar with using the axioms of an ordered field, even if you were notused to addressing such axioms explicitly. For this reason, I used them without mention. I assumedthe completeness axiom and we spoke about this a lot. Finally, I assumed the existence of the floorand ceiling functions. We can prove the existence of such functions from the completeness axiom.

Theorem 18.1.3. We can prove that N is not bounded above using only the axioms of an orderedfield and the completeness axiom.

Proof. Suppose for contradiction that N is bounded above. The completeness axiom tells us N hasa supremum, supN. Because supN− 1 < supN, supN− 1 is not an upper bound for N and thereexists an n ∈ N such that supN− 1 < n. We obtain supN < n+ 1 and n+ 1 ∈ N. This contradictsthe fact that supN is an upper bound for N.

Theorem 18.1.4. The following sentence can be proved using only the axioms of an ordered fieldand the completeness axiom.

∀x ∈ R, ∃m ∈ Z : m− 1 < x ≤ m.

Proof. Let x ∈ R and let S = {n ∈ Z : n ≥ x}.First, we show that S is nonempty. Suppose for contradiction that S = ∅. Then for all n ∈ Z,

we have n < x. In particular, for all n ∈ N, we have n < x. This says that x is an upper bound forN, and this contradicts the previous theorem.

S is nonempty and bounded below by x. The completeness axiom tells us that S has an infimuminf S. Let m = inf S. Over the next two paragraphs, we show that m ∈ S.

Since m + 1 > m, and m is the greatest lower bound of S, m + 1 is not a lower bound for S.This means that there exists an n ∈ S such that n < m + 1. Since m is a lower bound for S andn ∈ S, we have m ≤ n. Summarizing, we have n ∈ S and m ≤ n < m+ 1. We claim that m = n.

Suppose for contradiction that m 6= n. Since n ≥ m, we have n > m. Thus, n is not a lowerbound for S and there exists an n′ ∈ S such that n′ < n. Since m is a lower bound for S, we havem ≤ n′. Summarizing, n′, n ∈ S and m ≤ n′ < n < m+ 1. Let k = n− n′. Since S is a subset ofZ, we have n′, n ∈ Z, and so k ∈ Z. The last noted inequality gives 0 < k < 1. This is the requiredcontradiction. Thus, m = n, and since n ∈ S, we have shown that m ∈ S.

We conclude that m ∈ Z, m ≥ x, and m = minS. Suppose for contradiction that m − 1 ≥ x.Since m− 1 ∈ Z, this gives m− 1 ∈ S, and this contradicts m = minS. Thus, m− 1 < x.

96


∀x ∈ R, ∃! m ∈ Z : m− 1 < x ≤ m.

Proof. The exclamation mark means “unique.” In general “∃! a : ϕ(a)” means[∃a : ϕ(a)

]and

[∀a1, ∀a2,

(ϕ(a1) and ϕ(a2)

)=⇒ a1 = a2

].

We have already shown existence; we just have to deal with uniqueness.Suppose x ∈ R, and that m1,m2 ∈ Z satisfy m1 − 1 < x ≤ m1 and m2 − 1 < x ≤ m2. Then

m1 − 1 < x ≤ m2 and m2 − 1 < x ≤ m1,

so m1 −m2 < 1 and m2 −m1 < 1. Thus, |m1 −m2| < 1. This gives m1 = m2.

Definition 18.1.6. The ceiling function d−e : R −→ R is defined by the following property.

∀x ∈ R,[dxe ∈ Z and dxe − 1 < x ≤ dxe

].

Remark 18.1.7. By the previous theorem, assuming only the axioms of an ordered field and thecompleteness axiom, the ceiling function is well-defined.

Similarly, we have the following theorem and definition.


∀x ∈ R, ∃! m ∈ Z : m ≤ x < m+ 1.

Definition 18.1.9. The floor function b−c : R −→ R is defined by the following property.

∀x ∈ R,[bxc ∈ Z and bxc ≤ x < bxc+ 1

].

18.2 Irrationality of√2

We have used this result all throughtout the class. I include the proof for completeness.

Theorem 18.2.1.√

2 ∈ R \Q.

Proof. Suppose for contradiction that√

2 ∈ Q. Then there exist m,n ∈ N such that m and n haveno common factor and m

n =√

2.

We obtain m2

n2 = 2, so m2 = 2n2. Thus, m2 is even, and this means m is even (the square ofan odd number is odd). Let k = m

2 . Then k ∈ N and we have (2k)2 = 2n2, i.e. 4k2 = 2n2. Thus,2k2 = n2, n2 is even, and n is even.

We have show m and n are even and this contradicts the fact that m and n have no commonfactor.

97

18.3 Countability

Studying the size of sets requires making the following definition.

Definition 18.3.1. Suppose X and Y are sets and that f : X −→ Y is a function.

1. We say f is injective iff

[∀x1 ∈ X, ∀x2 ∈ X, f(x1) = f(x2) =⇒ x1 = x2

].

2. We say f is surjective iff

[∀y ∈ Y, ∃x ∈ X : f(x) = y

].

3. We say f is a bijection iff f is injective and surjective.

Definition 18.3.2. Suppose X is a nonempty set.

1. We say X is finite iff there exists an N ∈ N and a bijection f : {1, . . . , N} −→ X.

2. We say X is countably infinite iff there exists a bijection f : N −→ X.

3. We say X is countable iff it is finite or countably infinite.

Theorem 18.3.3. Suppose that X is a nonempty set. X is countable if and only if there exists asurjection g : N −→ X.

Proof. Suppose X is finite. Then there exists an N ∈ N and a bijection f : {1, . . . , N} → X. Defineg : N −→ X by

g(n) =

{f(n) if n ≤ Nf(N) if n > N.

Then g is a surjection.Suppose X is countably infinite. Then there exists a bijection f : N −→ X. Let g = f . Then g

is a surjection.Conversely, suppose g : N −→ X is a surjection. Define h : X −→ N by

h(x) = min{n ∈ N : g(n) = x}.

Note that for all x ∈ X, g(h(x)) = x. We claim g|imh is a bijection. If g|imh(h(x1)) = g|imh(h(x2)),then x1 = x2, so h(x1) = h(x2). Thus, g|imh is injective. If x ∈ X, then g|imh(h(x)) = x, so g|imh

is surjective.If {h(x) : x ∈ X} is unbounded, define j : N −→ imh by j(n) = min(imh\{j(1), . . . , j(n−1)}).

Then g ◦ j : N −→ X is a bijection, so X is countably infinite.If {h(x) : x ∈ X} is bounded we can find an N ∈ N, and a bijection j′ : {1, . . . , N} −→ imh

defined by the same formula as j was. Then g ◦ j′ is a bijection, so X is finite.

Theorem 18.3.4. N is countable.

Proof. id : N −→ N is a bijection.

98

Theorem 18.3.5. Z is countable.

Proof. Define f : N −→ Z by f(1) = 0, f(2) = −1, f(3) = 1, f(4) = −2, f(5) = 2, . . ., that is

f(n) =

{n−12 if n is odd,

−n2 if n is even.

f is a bijection.

Theorem 18.3.6. N× N is countable.

Proof. Define g : N −→ N by

g(N) = max

{m ∈ N :

N

2m−1∈ N

}.

Define h : N −→ N× N by h(N) = (g(N), N2g(N) + 1

2).

Define j : N× N −→ N by j(m,n) = 2m−1(2n− 1). h and j are inverse bijections.

Theorem 18.3.7. Z× N is countable.

Proof. Define k : Z×N −→ N by k(m,n) = j(f−1(m), n), where f and j are as in the previous twoproofs. k is a bijection.

Theorem 18.3.8. Q is countable.

Proof. Define p : Z × N −→ Q by p(m,n) = mn . Define q : N −→ Q by q(n) = p(k−1(n)), where k

is as in the previous proof. Since p is a surjection, so is q.

Definition 18.3.9. Suppose X is a set. The powerset of X is defined to be the set of subsets of X

P(X) = {A : A ⊆ X}.

Theorem 18.3.10. P(N) is uncountable.

Proof. This will follow from the following result.

Proposition 18.3.11. Suppose ϕ : N −→ P(N) is a function. Define

E := {n ∈ N : n /∈ ϕ(n)}.

Then, E ∈ P(N), and for all n ∈ N, ϕ(n) 6= E.

Proof. A good exercise for you.

Theorem 18.3.12. [0, 1] is uncountable.

Proof. The “same” proof as above, but using decimals. See “Cantor’s diagonal proof.”Alternatively, make use of the injection i : P(N) −→ [0, 1] defined by i(A) =

∑n∈A 3−n.

Corollary 18.3.13. R is uncountable.

Corollary 18.3.14. R \Q is uncountable.

99

19 Integration

19.1 Goals for our theory of integration

Given a, b ∈ R with a < b, we wish to define a notion of integration∫ b

a: L̃[a, b] −→ R,

where L̃[a, b] is a subset of the set of all real-valued functions with domain [a, b], the set of integrablefunctions on [a, b].

We would like the following properties:

1. Every continuous function on [a, b] is integrable on [a, b], i.e. C[a, b] ⊆ L̃[a, b].

2.∫ ba is linear, by which we mean the following.

Suppose f and g are integrable on [a, b] and c ∈ R, then

(a) cf is integrable on [a, b] and∫ ba cf = c

∫ ba f ;

(b) f + g is integrable on [a, b] and∫ ba (f + g) =

∫ ba f +

∫ ba g.

3. (a) If f is integrable on [a, b] and f(x) ≥ 0 for all x ∈ [a, b], then∫ ba f ≥ 0 (positivity).

(b) If f is continuous on [a, b] and f(x) ≥ 0 for all x ∈ [a, b], then∫ ba f = 0 if and only if

f = 0 (that is,∫ ba seperates continuous functions).

4. If f is integrable on [a, b], then |f | is integrable on [a, b] and |∫ ba f | ≤

∫ ba |f |.

5. If a < c < b, f : [a, b] −→ R, and f is integrable on [a, c] and [c, b], then f is integrable on[a, b] and ∫ b

af =

∫ c

af +

∫ b

cf.

6. The fundamental theorems of calculus:

(a) Suppose g is a continuous function on [a, b] that is differentiable on (a, b) and that g′ isintegrable on [a, b], then ∫ b

ag′ = g(b)− g(a).

(b) Suppose f is an integrable function on [a, b]. Define F : [a, b] −→ R by

F (x) =

∫ x

af(t) dt.

Then F is continuous on [a, b].

Moreover, when f is continuous at x0 ∈ (a, b), F is differentiable at x0, and in this caseF ′(x0) = f(x0).

7. We have integration by part and change of variables theorems.

8. Every piecewise monotonic function on [a, b] is integrable.

100

19.2 Deficiencies

Our theory will have deficiencies. Here are some.

1. The function f : [0, 1]→ R, given by f(0) = 0 and f(x) = 1√x

for x 6= 0 will not be integrable

on [0, 1] according to our theory. For any ε ∈ (0, 1), f will be integrable on [ε, 1] and we willfind that

limε−→0+

∫ 1

ε

dx√x

= 2,

as expected. A better theory of integration (the Lebesgue theory) would include the functionf as integrable on [0, 1].

2. The function g : [0, 1] → R with g(x) = 1 when x ∈ Q and g(x) = 0 when x /∈ Q will not beintegrable on [0, 1] according to our theory. In the Lebesgue theory this function is integrable,and

∫ 10 g = 0. This is because “countable sets are measure zero.”

However, all theories of integration have their limitations.

Example. This example is definitely non-examinable, but too interesting to omit. I might give aquestion on the last quiz preparation to explain this example.

For x ∈ R, let x + Q = {x + r : r ∈ Q}. Suppose there is a set E ⊆ [−1, 1] with the followingtwo properties.

1. ∀e0 ∈ E, ∀e1 ∈ E, e0 + Q = e1 + Q =⇒ e0 = e1.

2.⋃e∈E(e+ Q) = R.

Let f : [−1, 1] −→ R be defined by

f(x) =

{1 if x ∈ E,0 if x /∈ E.

f is bounded and yet it is not integrable on [−1, 1] regardless of what theory we use.To construct such an E requires the axiom of choice. The point is that this allows us to construct

a set so crazy that its length does not make sense.

19.3 What is integration anyway?

Integration is often introduced as the “opposite of differentiation.” This is very wrong! Maybe, youwere taught better. If so, integration will have been described as “the signed area under a curve.”

Our way of defining the integral will be intuitive. We will approximate the area “under thecurve” in two ways. One will underestimate it, the other will overestimate it. We let our approxi-mations improve and, if they tend towards the same answer, we’ll say our function is integrable.

19.4 The definition of integrability

We will only talk about integrating bounded functions on closed bounded intervals.

101

Definition 19.4.1. Let f be a bounded function on a closed interval [a, b]. If S is any subset of[a, b], we let

M(f, S) = sup{f(x) : x ∈ S}, m(f, S) = inf{f(x) : x ∈ S}.

To keep things in perspective, notice that given a bounded function f : [a, b] −→ R,

M(f, [a, b]) · (b− a), m(f, [a, b]) · (b− a)

give first approximations to∫ ba f . The first number is too big, the second is too small. How do we

get better approximations? We slice up the interval.

Definition 19.4.2. A partition of [a, b] is a finite ordered subset of [a, b]:

P = {a = t0 < t1 < . . . < tn−1 < tn = b}.

Definition 19.4.3. Let P be a partition of [a, b] as in the previous definition.The upper Darboux sum U(f, P ) with respect to P is

U(f, P ) =n∑k=1

M(f, [tk−1, tk]) · (tk − tk−1)

and the lower Darboux sum L(f, P ) with respect to P is

L(f, P ) =n∑k=1

m(f, [tk−1, tk]) · (tk − tk−1).

These give better approximations to∫ ba f . Indeed, the following lemma shows that these num-

bers are closer.

Lemma 19.4.4. Suppose f : [a, b] −→ R is a bounded function and P is a partition of [a, b], then

m(f, [a, b]) · (b− a) ≤ L(f, P ) ≤ U(f, P ) ≤M(f, [a, b]) · (b− a).

Proof. Let P = {a = t0 < . . . < tn = b}. For each k ∈ {1, . . . , n}, M(f, [tk−1, tk]) ≤M(f, [a, b]), sothat

U(f, P ) ≤n∑k=1

M(f, [a, b]) · (tk − tk−1) = M(f, [a, b]) · (b− a).

Similarly, m(f, [a, b]) ≤ m(f, [tk−1, tk]) ≤M(f, [tk−1, tk]) implies the other inequalities.

This lemma also shows that for a bounded function f : [a, b] −→ R the sets

{U(f, P ) : P is a partition of [a, b]}, {L(f, P ) : P is a partition of [a, b]}

are bounded and so we can make the following definition.

Definition 19.4.5. Let f : [a, b] −→ R be a bounded function.The upper Darboux integral is

U(f) = inf{U(f, P ) : P is a partition of [a, b]}.

The lower Darboux integral is

L(f) = sup{L(f, P ) : P is a partition of [a, b]}.

102

You might expect that L(f) ≤ U(f) and, indeed, we will see that this is true a little later on. If

L(f) < U(f), then something funny is happening and we cannot assign a value to∫ ba f . Otherwise

we can integrate.

Definition 19.4.6. A bounded function f : [a, b] −→ R is integrable on [a, b] iff L(f) = U(f) andin this case we write ∫ b

af =

∫ b

af(x) dx

for L(f) = U(f).

19.5 Basic examples

Example 19.5.1. Let b > 0 and f(x) = x3 on [0, b].Consider the partition

Pn = {t0 < t1 < . . . < tn−1 < tn = b}

where tk = bkn .

U(f, Pn) =

n∑k=1

t3k · (tk − tk−1) =b4

n4·n∑k=1

k3 =b4

n4·(n(n+ 1)

2

)2

=b4

4·(

1 +2

n+

1

n2

)

L(f, Pn) =

n∑k=1

t3k−1 · (tk − tk−1) =b4

n4·n∑k=1

(k − 1)3 =b4

n4·(

(n− 1)n

2

)2

=b4

4·(

1− 2

n+

1

n2

)So we have

U(f) ≤ limn→∞

U(f, Pn) =b4

4= lim

n→∞L(f, Pn) ≤ L(f).

As long as we can show that L(f) ≤ U(f), we will see that x3 is integrable on [0, b] and that∫ b

0x3 dx =

b4

4.

The inequality L(f) ≤ U(f) will be a theorem of the next subsection.

Example 19.5.2. Let a < b,

g(x) =

{1 if x ∈ Q ∩ [a, b]

0 if x ∈ [a, b] \Q

and let P = {a = t0 < . . . < tn = b} be a partition of [a, b]. Then

U(f, P ) =n∑k=1

M(f, [tk−1, tk]) · (tk − tk−1) =n∑k=1

1 · (tk − tk−1) = b− a

L(f, P ) =n∑k=1

m(f, [tk−1, tk]) · (tk − tk−1) =n∑k=1

0 · (tk − tk−1) = 0.

So L(f) = 0 < b− a = U(f), so that g is not integrable with respect to our theory, even though itis in the Lebesgue theory.

103

Example 19.5.3. For each rational number r ∈ (0, 1), write r = pq where p, q are natural numbers

with no common factors. Then consider the following function on [0, 1].

h(x) =

{1q if x = p

q is rational, and not equal to 0 or 1

0 if x is irrational, or equal to 0 or 1.

Then L(h) = 0 as in the previous example.Let Pn be the partition {0 = t0 < t1 < . . . < tn−1 < tn = 1} where tk = k

n . Then (exercise)

inf{U(h, Pn) : n ∈ N} = 0 so that U(h) = 0. This means h is integrable on [0, 1] and∫ 10 h = 0.

19.6 Integration: two auxilary theorems

To make calculations using the definition directly the following theorem is useful.

Theorem 19.6.1. Suppose f : [a, b] −→ R is a bounded function, then L(f) ≤ U(f).

Proof. Suppose that given any two partitions P and Q, we can show that L(f, P ) ≤ U(f,Q). We’llcome back and prove this.

Let Q be any partition of [a, b]. Then U(f,Q) is an upper bound for the set

{L(f, P ) : P is a partition of [a, b]}

so that L(f) ≤ U(f,Q). Since Q was an arbitrary partition of [a, b], this shows that L(f) is a lowerbound for the set

{U(f,Q) : Q is a partition of [a, b]}

so that L(f) ≤ U(f).Let P and Q be two partitions. We need to show that L(f, P ) ≤ U(f,Q). We have a partition

P ∪Q. Of course, it is enough to show that

L(f, P ) ≤ L(f, P ∪Q) ≤ U(f, P ∪Q) ≤ U(f,Q).

We already know the middle inequality. Because P,Q ⊆ P ∪ Q, it is enough to show that givenpartitions P ′ and Q′ with P ′ ⊆ Q′ we have

L(f, P ′) ≤ L(f,Q′), U(f,Q′) ≤ U(f, P ′).

[We apply the left inequality with P ′ = P and Q′ = P ∪ Q and the right inequality with P ′ = Qand Q′ = P ∪Q.] This is the next lemma.

Lemma 19.6.2. Given partitions P and Q with P ⊆ Q we have

L(f, P ) ≤ L(f,Q), U(f,Q) ≤ U(f, P ).

Proof. We can prove this by induction on the size of Q \ P . For the inductive step, we have to dothe case when Q has one more element than P . So suppose

P = {a = t0 < t1 < . . . < tn−1 < tn = b}

andQ = {a = t0 < t1 < . . . < tk−1 < u < tk < . . . < tn−1 < tn = b}.

104

We have M(f, [tk−1, u]),M(f, [u, tk]) ≤M(tk−1, tk) so that

M(f, [tk−1, u]) · (u− tk−1) +M(f, [u, tk]) · (tk − u)

≤ M(f, [tk−1, tk]) · (u− tk−1) +M(f, [tk−1, tk]) · (tk − u)

= M(f, [tk−1, tk]) · (tk − tk−1).

By adding in the other terms in this tells us that U(f,Q) ≤ U(f, P ).Similarly, L(f, [tk−1, u]), L(f, [u, tk]) ≥ L(f, [tk−1, tk]) implies L(f, P ) ≤ L(f,Q).

If we want to prove a function is integrable without knowing the value of the integral (oftenuseful in proofs) then the following criterion is useful.

Theorem 19.6.3 (A criterion for integration). A bounded function f : [a, b] −→ R is integrable ifand only if for all ε > 0, there exists a partition P of [a, b] such that

U(f, P )− L(f, P ) < ε.

Proof. Suppose f is integrable and let ε > 0. By definition of inf, U(f) + ε2 is not a lower bound

for{U(f, P ) : Q is a partition of [a, b]},

so there is a partition Q of [a, b] with U(f,Q) < U(f) + ε2 . Similarly, there is a partition P of [a, b]

with L(f, P ) > L(f)− ε2 . Since f is integrable, L(f) = U(f) and we write this as

∫ ba f . Moreover,

using the previous lemma as in the proof before the lemma, we have(∫ b

af

)− ε

2< L(f, P ) ≤ L(f, P ∪Q) ≤ U(f, P ∪Q) ≤ U(f,Q) <

(∫ b

af

)+ε

2

so that U(f, P ∪Q)− L(f, P ∪Q) < ε.Conversely, for each partition P , we have

L(f, P ) ≤ L(f) ≤ U(f) ≤ U(f, P )

so that 0 ≤ U(f) − L(f) ≤ U(f, P ) − L(f, P ). We deduce that if, for each ε > 0, we can find apartition P with U(f, P )− L(f, P ) < ε, then L(f) = U(f), so that f is integrable.

19.7 Properties of the integral

Theorem 19.7.1. Every continuous function on [a, b] is integrable on [a, b].

Proof. Let f : [a, b] −→ R be a continuous function and let ε > 0. Since [a, b] is closed and bounded,f is uniformly continuous on [a, b] (theorem 13.4.1). Thus, there is a δ > 0 so that

∀x ∈ [a, b], ∀y ∈ [a, b], |x− y| < δ =⇒ |f(x)− f(y)| < ε

b− a.

Choose n ∈ N so that b−an < δ and let

P = {a = t0 < t1 < . . . < tn−1 < tn = b}

105

be given by tk = a+ (b−a)kn .

Fix k ∈ {1, . . . , n}. Since f is continuous on [tk−1, tk], it assumes its maximum and minimumvalues, i.e. there are x, y ∈ [tk−1, tk] with f(x) = M(f, [tk−1, tk]) and f(y) = m(f, [tk−1, tk]). Since

x, y ∈ [a, b] and |x− y| ≤ tk − tk−1 =b− an

< δ,

we haveM(f, [tk−1, tk])−m(f, [tk−1, tk]) = |f(x)− f(y)| < ε

b− a.

This holds for each k ∈ {1, . . . , n} and so

U(f, P )− L(f, P ) =n∑k=1

(M(f, [tk−1, tk])−m(f, [tk−1, tk])) · (tk − tk−1)

<n∑k=1

ε

b− a· (tk − tk−1) = ε.

By the criterion for integrability, f is integrable on [a, b].

Theorem 19.7.2 (Linearity). Suppose f and g are integrable on [a, b] and c ∈ R, then

1. cf is integrable on [a, b] and∫ ba cf = c

∫ ba f ;

2. f + g is integrable on [a, b] and∫ ba (f + g) =

∫ ba f +

∫ ba g.

Proof.

1. The result is easy to check for c = 0.

Notice that M(−f, S) = −m(f, S) and m(−f, S) = −M(f, S) for any S ⊆ [a, b]. Thus,U(−f, P ) = −L(f, P ) and L(−f, P ) = −U(f, P ) for any partition P , and so

U(−f) = −L(f), L(−f) = −U(f).

This means U(f) = L(f) if and only if U(−f) = L(−f), so that f is integrable if and only if−f is, and in this case ∫ b

a−f = U(−f) = −L(f) = −

∫ b

af,

i.e. the result holds for c = −1.

We are just left to show the result for c > 0. Similarly to the c = −1 case, this follows fromM(cf, S) = cM(f, S) and m(cf, S) = cm(f, S).

2. Notice that M(f + g, S) ≤ M(f, S) + M(g, S) so that U(f + g, P ) ≤ U(f, P ) + U(g, P ) forany partition P .

106

Let ε > 0. Since f is integrable there is a partition P with U(f, P )− L(f, P ) < ε2 . Since g is

integrable there is a partition Q with U(g,Q)− L(g,Q) < ε2 . Now

U(f + g) ≤ U(f + g, P ∪Q) ≤ U(f, P ∪Q) + U(g, P ∪Q)

≤ U(f, P ) + U(g,Q)

< L(f, P ) + L(g,Q) + ε ≤ L(f) + L(g) + ε =

∫ b

af +

∫ b

ag + ε.

Similarly, L(f + g) >∫ ba f +

∫ ba g − ε. Summarizing,∫ b

af +

∫ b

ag − ε < L(f + g) ≤ U(f + g) <

∫ b

af +

∫ b

ag + ε.

Since ε > 0 was arbitrary this shows

L(f + g) = U(f + g) =

∫ b

af +

∫ b

ag,

which completes the proof.

Theorem 19.7.3.

1. If f is integrable on [a, b] and f(x) ≥ 0 for all x ∈ [a, b], then∫ ba f ≥ 0.

Thus, if f and g are integrable and f(x) ≥ g(x) for all x ∈ [a, b], then∫ ba f ≥

∫ ba g.

2. If f is continuous on [a, b] and f(x) ≥ 0 for all x ∈ [a, b], then∫ ba f = 0 if and only if f = 0.

Proof.

1. Suppose f(x) ≥ 0 for all x ∈ [a, b] and let P = {a = t0 < t1 = b}. Then

0 ≤ m(f, [a, b]) · (b− a) = L(f, P ) ≤ L(f) =

∫ b

af.

The second part follows, by applying the first part to f − g and using linearity.

2. Suppose f is continuous on [a, b], f(x) ≥ 0 for all x ∈ [a, b], and that there is an x0 ∈ [a, b]with f(x0) > 0. Since f is continuous at x0, there is a δ > 0 so that

∀x ∈ [a, b], |x− x0| < δ =⇒ |f(x)− f(x0)| <f(x0)

2.

Then for x ∈ [a, b] ∩ (x0 − δ, x0 + δ), we have f(x) > f(x0)2 .

Let P be a partition {a = t0 < . . . < tn = b} with some tk−1, tk ∈ (x0 − δ, x0 + δ). Then∫ b

af = L(f) ≥ L(f, P ) ≥ m(f, [tk−1, tk]) · (tk − tk−1) ≥

f(x0)

2· (tk − tk−1) > 0.

107

By the contrapositive, if f is continuous on [a, b], f(x) ≥ 0 for all x ∈ [a, b], and∫ ba f = 0, we

must have f = 0. The other implication is easy.

In lecture, I decided to torture myself and you by writing down a partition with some tk−1, tk ∈(x0 − δ, x0 + δ). Here is one:

P = {a, x0 −min{δ, x0 − a}

2, x0 +

min{δ, b− x0}2

, b}.

It has three points if x0 ∈ {a, b}; otherwise, it has four points.

Theorem 19.7.4. Suppose f is integrable on [a, b]. Then |f | is integrable on [a, b] and∣∣∣∣ ∫ b

af

∣∣∣∣ ≤ ∫ b

a|f |.

Proof. We just need to show |f | is integrable, since then −|f | ≤ f ≤ |f |, together with linearity,will imply that

−∫ b

a|f | ≤

∫ b

af ≤

∫ b

a|f |,

giving |∫ ba f | ≤

∫ ba |f |.

It is enough to show that for any S ⊆ [a, b], M(|f |, S)−m(|f |, S) ≤M(f, S)−m(f, S), for thisimplies that for any partition P of [a, b]

U(|f |, P )− L(|f |, P ) ≤ U(f, P )− L(f, P ),

and so the criterion for integrability for |f | will follow from that for f .To check M(|f |, S)−m(|f |, S) ≤M(f, S)−m(f, S) there are three cases

1. f ≥ 0 on S, in which case |f | = f on S and we have equality;

2. f ≤ 0 on S, in which case |f | = −f on S and we have equality again;

3. m(f, S) < 0 < M(f, S), in which case the result follows from the facts that

M(|f |, S)−m(|f |, S) ≤M(|f |, S)

and that either

(a) M(|f |, S) = M(f, S) < M(f, S)−m(f, S) or

(b) M(|f |, S) = −m(f, S) < M(f, S)−m(f, S).

Theorem 19.7.5. If a < c < b, f : [a, b] −→ R, and f is integrable on [a, c] and [c, b], then f isintegrable on [a, b] and ∫ b

af =

∫ c

af +

∫ b

cf.

108

Proof. f is bounded on [a, b] since it is on [a, c] and [c, b]. Using self-explanatory notation, this resultis true because of the following fact: given a partion P of [a, c], and a partition Q of [c, b], P ∪ Qis a partition of [a, b], U ba(f, P ∪Q) = U ca(f, P ) +U bc (f,Q) and Lba(f, P ∪Q) = Lca(f, P ) +Lbc(f,Q).

Given ε > 0, there is a partition P of [a, c] with U ca(f, P )−Lca(f, P ) < ε2 , and there is a partition

Q of [c, b] with U bc (f,Q)−Lbc(f,Q) < ε2 . Thus, U ba(f, P ∪Q)−Lba(f, P ∪Q) < ε and f is integrable

on [a, b].∫ b

af ≤ U ba(f, P ∪Q) = U ca(f, P ) + U bc (f,Q) < Lca(f, P ) + Lbc(f,Q) + ε ≤

∫ c

af +

∫ b

cf + ε,

and similarly,∫ ba f >

∫ ca f +

∫ bc f − ε.

19.8 Preliminary examples to the fundamental theorems of calculus

Example 19.8.1. The function defined by f(0) = 0 and f(x) = sin( 1x) for x 6= 0 is integrable on[0, 1]; in fact, on any interval [a, b]. Note that it is not continuous at 0.

Proof. Let ε ∈ (0, 1) and δ = ε3 . Then f is continuous on [δ, 1] and so integrable on [δ, 1]. Thus, there

is a partition P of [δ, 1] with U1δ (f, P )−L1

δ(f, P ) < δ. On the other hand, M(f, [0, δ])−m(f, [0, δ]) ≤2, so that letting Q = {0} ∪ P we have

U10 (f,Q)− L1

0(f,Q) < 2δ + δ = ε.

We conclude f is integrable on [0, 1]. The argument for an arbitrary closed interval is similar.

Example 19.8.2. Let f(0) = 0 and f(x) = 2x cos(1/x) + sin(1/x) for x 6= 0.Notice that f is not continuous at 0, but it is integrable on any interval [a, b], for it is the sum

of two integrable functions (the first is continuous and we have just checked the second one). Thefollowing theorem allows us to deduce that∫ b

af(t) dt = g(b)− g(a)

where g(0) = 0 and g(x) = x2 cos( 1x) for x 6= 0.

19.9 The fundamental theorems of calculus

Theorem 19.9.1 (Fundamental theorem of calculus part I). Suppose g is a continuous functionon [a, b] that is differentiable on (a, b) and that g′ is integrable on [a, b], then∫ b

ag′ = g(b)− g(a).

Proof. g′ is function on (a, b). To say it is integrable means that any extension to a function on[a, b] is integrable. By an exercise on quiz preparation 5, the extension is irrelevant, so let g′ denoteany function on [a, b] extending the derivative of g.

Let P = {a = t0 < . . . < tn = b} be a partition. The mean value theorem says that for eachk ∈ {1, . . . , n} there is an xk ∈ (tk−1, tk) with

g′(xk) · (tk − tk−1) = g(tk)− g(tk−1).

109

By definition of m(g′, S) and M(g′, S) we have

L(g′, P ) ≤n∑k=1

g′(xk) · (tk − tk−1) ≤ U(g′, P ).

Moreover,n∑k=1

g′(xk) · (tk − tk−1) =n∑k=1

(g(tk)− g(tk−1)) = g(b)− g(a),

so that L(g′, P ) ≤ g(b)− g(a) ≤ U(g′, P ).Since this holds for all partitions P , we have L(g′) ≤ g(b)− g(a) ≤ U(g′). Moreover, since g′ is

integrable, we have L(g′) = U(g′), so that∫ b

ag′ = g(b)− g(a),

as required.

Theorem 19.9.2 (Fundamental theorem of calculus part II). Suppose f is an integrable functionon [a, b]. Define F : [a, b] −→ R by

F (x) =

∫ x

af(t) dt.

Then F is continuous on [a, b]. Moreover, if f is continuous at x0 ∈ (a, b), then F is differentiableat x0 and F ′(x0) = f(x0).

Proof. Suppose f is integrable on [a, b]. Let x, y ∈ [a, b] with x < y. Then

|F (y)− F (x)| =∣∣∣∣ ∫ y

xf(t) dt

∣∣∣∣ ≤ ∫ y

x|f(t)| dt ≤ |y − x| ·M(|f |, [a, b]).

This means that given ε > 0 we can choose δ = εM(|f |,[a,b])+1 to show F is uniformly continuous on

[a, b].Suppose f is continuous at x0 ∈ (a, b) and let ε > 0. Then there is a δ > 0 so that

∀t ∈ [a, b], |t− x0| < δ =⇒ |f(t)− f(x0)| <ε

2.

For x ∈ [a, b] with x ∈ (x0, x0 + δ) we have∣∣∣∣F (x)− F (x0)

x− x0− f(x0)

∣∣∣∣ =

∣∣∣∣ 1

x− x0

∫ x

x0

[f(t)− f(x0)

]dt

∣∣∣∣ ≤ 1

|x− x0|

∫ x

x0

∣∣∣∣f(t)− f(x0)

∣∣∣∣dt < ε.

Similarly, for x ∈ [a, b] with x ∈ (x0 − δ, x0) we have∣∣F (x)−F (x0)

x−x0 − f(x0)∣∣ < ε. Thus,

limx→x0

F (x)− F (x0)

x− x0= f(x0),

i.e. F is differentiable at x0 and F ′(x0) = f(x0).

110

19.10 One more theorem

Theorem. Every monotonic function on [a, b] is integrable on [a, b].

Proof. Suppose f : [a, b] −→ R is increasing and let ε > 0. Choose n ∈ N such that

n > (f(b)− f(a)) · (b− a)

ε.

Let P be the partition P = {a = t0 < t1 < . . . < tn−1 < tn = b} given by tk = a+ (b−a)kn . Then

U(f, P )− L(f, P ) =

n∑k=1

(M(f, [tk−1, tk])−m(f, [tk−1, tk])) · (tk − tk−1)

=

n∑k=1

(f(tk)− f(tk−1)) ·b− an

= (f(b)− f(a)) · b− an

< ε.

19.11 Picard Iterates

Example 19.11.1. Suppose we wanted to find a function y : R −→ R, which is differentiable, hasy(0) = 1 and y′(x) = y(x) for all x ∈ R, and we did not know about exp, how would we proceed?

If y is such a function, the fundamental theorem of calculus (part I) tells that

y(x) = 1 +

∫ x

0y(t)dt.

So we might hope to obtain y(x) as a limit of functions yn(x) defined inductively by y0(x) = 1, andfor n ≥ 1

yn(x) = 1 +

∫ x

0yn−1(t)dt.

Let’s see if this works.

1. y0(x) = 1;

2. y1(x) = 1 +∫ x0 y0(t)dt = 1 +

∫ x0 dt = 1 + x;

3. y2(x) = 1 +∫ x0 y1(t)dt = 1 +

∫ x0 (1 + t)dt = 1 + x+ x2

2 ;

4. y3(x) = 1 +∫ x0 y2(t)dt = 1 +

∫ x0 (1 + t+ t2

2 )dt = 1 + x+ x2

2! + x3

3! ;

5. . . .

6. Suppose inductiely that yn−1(x) =∑n−1

k=0xk

k! ;

7. Then

yn(x) = 1 +

∫ x

0yn−1(t)dt = 1 +

∫ x

0

n−1∑k=0

tk

k!dt = 1 +

n−1∑k=0

∫ x

0

tk

k!dt = 1 +

n−1∑k=0

xk+1

(k + 1)!

=n∑k=0

xk

k!.

111

Indeed, the sequence of function (yn) converges (uniformly) to

exp(x) =∞∑k=0

xk

k!.

This is how one should discover the exponential function.

Definition 19.11.2. Suppose f(x, y) is a function of two variables. Then the equation

y′(x) = f(x, y(x))

is called an ordinary differential equation (ODE). If x0, y0 ∈ R, then the condition that

y(x0) = y0

is called an initial condition.

Definition 19.11.3. Suppose

y′(x) = f(x, y(x)), y(x0) = y0

is an ordinary differential equation with initial condition. Then the associated Picard iterates arethe functions yn(x), where for n ≥ 1,

yn(x) = y0 +

∫ x

x0

f(t, yn−1(t))dt.

Theorem 19.11.4 (Picard’s Existence theorem for ODE’s). Suppose

y′(x) = f(x, y(x)), y(x0) = y0

is an ordinary differential equation with initial condition.Then, provided f is nice enough, the Picard iterates (yn) converge to a solution of the differential

equation close to x0.

To prove such a theorem, one needs uniform continuity of f , in particular, a Lipschitz condition.One shows that the Picard iterates converge uniformly. One uses this to interchange a limit with anintegral in order to check that the relevant integral equation is satisfied. One uses the fundamentaltheorem of calculus to verify that this implies the differential equation is satisfied.

All of this will seem like too much to take in, but I wanted to emphasize that everything we havelearned this quarter has a legitimate application: it tells us solutions to ODEs exist and gives us amethod to approximate their solutions. Doing more analysis, we can even learn how quickly theseapproximations converge. Often one cannot write down analytic solutions to differential equationseven though we know a solution exists, and so this information is particularly useful in areas ofapplied math, physics, engineering, biology, chemistry: anywhere where solutions to differentialequations need to be approximated for simulations.

112

20 Final exam expectations

20.1 Sections you can safely ignore

Sections 15, 16, 17, 18, 19.9.2, 19.10, 19.11 will not be examined at all.The proof of 12.3.7 will not be examined, though you should know the result.The proof of 13.4.1 will not be examined, though you should know the result.Definitions 14.1.2 and 14.1.3 come in three parts; you only need to know the first of each. 14.1.5

can be ignored. While proving 14.1.6 was a good exercise, I will not ask it on the exam - it’s tooboring.

I’ll never ask for the proof of 14.3.1 or 14.3.3. You can use them.You do not need to know 14.5.8, 14.5.9, or 14.5.10.You do not need to be able to write the proofs of 19.4.4, 19.6.1, or 19.6.2, but you should know

why they’re true (the picture). The proofs of 19.7.2, 19.7.3, and 19.7.4 can be left out as well.

20.2 Sections which will be most heavily examined?

Section 4 will be examined implicitly. It should be very clear by now: if you don’t understand thissection, then you’re doomed.

There will definitely be one question examining section 7. If you have struggled with the class,this is a good question to focus on. Aim to get this one perfect to avoid failing!

There will definitely be one question examining section 8 / 13.1. If you have struggled with theclass, this is another good question to focus on. Aim to get this one perfect to avoid failing!

It seems fairly likely I’ll have a questions on sup’s and/or inf’s, section 10.The proofs in 12 probably felt quite difficult when we first did them. I hope when you go back

to them now, they feel easier. This is my favorite part of the course. I could ask for any of theseproofs (except 12.3.7). The likelihood of this is hard to say. It depends on how the rest of the examis looking.

I feel things get more difficult after this. If you’ve struggled, I would suggest focusing far moreon what I have mentioned before this point.

I think the proofs of the IVT and EVT are the most involved in the class. I would never askfor all of the proof of the EVT. However, it comes in two parts, and one part could be okay as oneof the longest questions on the exam.

Limit along domains are only likely to be examined via checking a function is differentiable.I’m also more likely to examine a thorough understanding of the exact hypotheses and why they’renecessary in Fermat’s Theorem, Rolle’s Theorem, and the MVT, than to ask you to prove thesetheorems. You might focus on such things for the IVT and EVT.

20.3 Other comments

Although I didn’t lecture section 11, I think you should know it. However, my proofs of 11.3.1 3and 4 are quick and concise and maybe not the best to read.

You could try and pove 3 using the inequality |sntn− st| ≤ |sn||tn− t|+ |sn− s||t| together withthe fact that (sn) is bounded when it is convergent.

You could try and prove the second part of 4 using the fact that∣∣∣∣ 1

tn− 1

t

∣∣∣∣ =|tn − t||tn||t|

,

113

and that when (tn) converges to t 6= 0, there exists an N ∈ N such that

∀n ∈ N, n ≥ N, |tn − t| <|t|2,

which gives

∀n ∈ N, n ≥ N, |tn| >|t|2.

114

Documents

Math 131A: Real Analysismjandr/Math131A/lectures.pdf · Math 131A: Real Analysis ... every real number is either bigger than or equal to 0, ... by the end of this class, you will