Math 131A: Real Analysis

Transcription

1 Math 131A: Real Analysis Michael Andrews UCLA Mathematics Department February 6, 018 Contents 1 Introduction 4 Truth Tables 4.1 NOT(P ) P and Q, P or Q P = Q (read this as P arrow Q, not as P implies Q ) Some familiar sets 6 4 Quantifiers Unspecified variables A well-formed sentence Verifying a sentence involving quantifiers Negating a sentence involving quantifiers Parentheses Negating a sentence with parentheses carefully Math 95 notes Absolute value Definition and basic properties The point of the absolute value A little more about sets 17 7 Sequences and the definition of convergence Sequences Convergence of sequences Divergence of sequences How to spend your time Constructing the proof of theorem

2 7.6 Motivating the definition of convergence Another sequence that converges Constructing the proof of theorem The algebra of limits The game and the computer program, finishing the proof of Another algebra of limits proof, more game/computer program Continuity The definition Motivating the definition of continuity Examples of continuous functions Uniform Continuity The definition and a comparison with the definition of continuity Examples of uniformly continuous and non-uniformly continuous functions What are the real numbers? The completeness axiom What are the real numbers? Boundedness, maximum and minimum elements The completeness axiom Suprema and Infima Contrapositive and contradiction An almost-proof of corollary Some more on sequences (not lectured) Uniqueness of limits Bounded sequences Algebra of limits (continued) Sequences and the completeness of R Cauchy sequences Monotone Sequences Subsequences and the Bolzano-Weierstrass theorem Completeness of R Examples More on continuous functions The sequence definition of continuity New continuous functions from old ones (not lectured) The Intermediate Value Theorem and the Extreme Value Theorem A theorem that should be included somewhere Calculus Limits along the domain of a function The derivative Rules for differentiating (not lectured or examined explicitly) Fermat and Rolle s Theorem

3 14.5 The Mean Value Theorem and important results from calculus Series From sequences to series Tests for convergence and divergence Power series More functions: cos, sin, exp, log Definitions cos, sin and π exp, log, exponents De Moivre s theorem (off syllabus) Taylor s theorem Taylor series and Taylor polynomials Taylor series of polynomials A rubbish Taylor series An even more rubbish Taylor series (definitely off syllabus) Taylor s theorem Miscellaneous Denseness of the rationals and irrationals in R Irrationality of Countability Integration Goals for our theory of integration Deficiencies What is integration anyway? The definition of integrability Basic examples Integration: two auxilary theorems Properties of the integral Preliminary examples to the fundamental theorems of calculus The fundamental theorems of calculus One more theorem Picard Iterates

4 1 Introduction Many people find mathematics satisfying because a (well-constructed) statement in math is either true or false. For example, 0 = 0 is true and 0 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 ridiculously easy, whereas the latter is more complicated. The latter is more complicated because of the words which appear. It is unlikely that the word continuous has ever been defined carefully enough for you 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 was given to you, it is unlikely that you were given the time to master verifying such definitions to the standard of a pure mathematician. The purpose of this class is to give you precise definitions of concepts such as continuity and differentiability, and to teach you how to verify such definitions (or their negation). We will learn how concepts like these relate to one another; the sentence above gives an example of how continuity and differentiability relate to one another. We will learn how to prove such relationships rigorously. Truth Tables Mathematics is concerned with statements which are true or false. We gave two simple examples and a more complicated example in the introduction. Just like when we speak English, mathematics is built up by putting together lots of simpler pieces. In this section, we learn a few simple ways of constructing statements which are true or false from existing statements which are true or false..1 NOT(P ) Suppose P is a statement which is true or false. We normally call such a thing a proposition. We have a new proposition NOT(P ). The truth of NOT(P ) is completely determined by the truth of P. This is demonstrated in the following truth table. P NOT(P ) T F For example, the statement I am your lecturer for math 131A is true, and the statement I am 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 the following truth table. F T P NOT(P ) NOT(NOT(P )) I am NOT NOT your lecturer for math 131A is true. T F F T 4

5 . 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 nonmathematical propositions, you might think that this seems like a strange convention, but I hope you will find that in mathematics such a convention makes a lot of sense. The following sentence seems fine to me: every real number is either bigger than or equal to 0, or less than or equal to 0. De Morgan s Laws state the following. P Q P and Q P or Q T T T T T F F T F T F T F F F F 1. NOT(P or Q) has the same truth values as (NOT(P ) and NOT(Q));. 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..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 convention once 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).. De Morgan s Laws tell us that NOT(P = Q) has the same truth values as (P and NOT(Q)). 5

6 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 its elements. Two sets are equal if and only if they have the same elements. We write x X to mean that x is an element of a set X. Notation 3.. Curly brackets (braces) are often used to show sets. The set whose elements are a 1, a, a 3,..., a n is written {a 1, a, a 3,..., a n }. Similarly, the set whose members are those of an infinite sequence a 1, a, a 3,... of objects is denoted {a 1, a, a 3,...}. 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 We write X \ Y and say X take away Y for Here x / Y is read x is not an element of Y. {a : a is an element of X and Y }. {x X : x / Y }. Notation 3.5. Here is some notation for familiar sets. 1. the natural numbers N = {1,, 3,...},. the integers Z = {0} { n : n N} N = {0, 1, 1,,, 3, 3,...}, 3. the rationals Q = { m n : 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

7 4 Quantifiers Quantifiers are the key to making precise, strong definitions. If you do not understand quanitifiers by the end of this class, you will fail the class. They are tricky, and they trip lots of people up, but they are worth investing lots of time into understanding properly. This class is difficult for multiple reasons. My hope is for us to conquer quantifiers early on so that we can focus on other difficult aspects later without quantifiers getting in our way. 4.1 Unspecified variables In section we talked about statements which are true or false, and some ways we can form new such statements from existing ones. Here is a formula. a + b = c Is it true or false? What about the following formula? Is that true or false? x = b ± b 4ac a I just fed you nonsense. I wrote down familiar formulae, and gave you no context for them at all. It is impossible to say whether they are true or false, because the truth of the first depends on what a, b, and c are, and the truth of the second depends on what a, b, c and x are. If you know what these familiar looking formula relate to, you will also know that the significance of a, b, and c 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 that I 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 resolve that 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.. (x + 1) = x + x + 1. A priori, the truth of this formula might depend on x, but you have seen this one a million times 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

8 4. We now introduce our first quantifier which is read there exists. Oddly enough, giving a precise definition of this symbol in terms of the truth values of propositions is beyond the scope of this class. 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 that x = 0. The following sentence is also true. x R : (x + 1) = x + x + 1. You can see these sentences are true because, in both cases, taking x to be 0 makes the subsequent formula 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 : a + b = c is true because when a, b, and c are equal to 0, the formula is true. Don t dwell on this last example just 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 real number which is not equal to 0. The following sentence is true. x R, (x + 1) = x + x + 1. That s because you know how to multiply (x + 1) = (x + 1)(x + 1) using distributivity of multiplication. The following sentence is false. x R, x = x + 1. The following sentence is also false. a R, b R, c R, a + b = c That s because taking a = 0, b = 0 and c = 1, makes the formula false. Example 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 a true sentence. It is not crazy to believe that this is what the symbols encode, but I need to give you a way of verifying the truth of quantified sentences for you to really see why this is the case. 8

9 4.4 A well-formed sentence In mathematical logic, the concept of a sentence is defined. To be a well-formed sentence, the variables appearing in the formulae of the sentence must be quantified over. This means that for each variable there must be a corresponding quantifier. This is to guarantee that the sentence has a well-defined truth value. In this class, you should be suspicious if you see a variable that has not been 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 Q 1 Q... Q n ϕ where each individual Q i is either of the form x i X i or x i X i, and ϕ is something constructed from a load of formulae and the operations of and, or, and =. (It can be more complicated than this because ϕ could also involve quanitfiers and look, for example, like ( x X : F 1 ) = F. 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 x i X i you should say let x i = BLAH where BLAH needs to be an element of X i written only in terms of variables which have already been specified. You should always check that the element x i that you declare is an element of the set X i. To deal with x i X i you simply say let x i X i. ϕ will be some proposition involving the variables x 1,..., x n. At the (n+1)-st step, you must verify the truth of ϕ based on your previous declared variables. For clarity, you should break this step up in to further substeps (a), (b), (c),... if necessary. Example 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.. 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 : 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 sentence is supposed to say if an integer is bigger than 10, then it is bigger than 10. Thinking of this statement as one about every integer, we see that we don t really care about integers less than or equal to 10 and we want (n > 10 = n > 10) to be true in this case. 9

10 Example The following sentence is true. ( x R, y R, x y = ( )) z R : x + z = y. Proof. 1. Let x R.. Let y R. 3. We must verify the truth of x y = ( ) z R : x + z = y. (a) Case 1: x > y. x y is false, so the truth table for = says the relevant sentence is true. (b) Case : x y. x y is true and we must verify the truth of z R : x + z = y. Notice that because the original sentence is not of the nice form I mentioned at the start of the section, we ve hit a point where we have to verify another proposition involving a quantifier. i. Let z = y x. z is a well-defined real number: we are working under the assumption that x y, and so y x 0. ii. We must verify that x + z = y. This is trivial algebra: x + z = x + ( y x) = x + (y x) = y. Example The following sentence is true. Proof. 1. Let x = 0.. Let y R. 3. Let z = y. Notice z R. 4. We see that x + y = 0 + z = z. x R : y R, z R : x + y = z. There s something important to note about this proof: we let z = y. This is fine because we made this declaration in step 3, and y was declared in step. Letting x = y in step 1 would have been an illegal declaration. We CANNOT define a variable in terms of a variable that s declared later. 10

11 4.6 Negating a sentence involving quantifiers Example Let s try to show the following sentence to be true. Our proof would read as follows. 1. Let x = BLAH. Notice x R.. Let y R. 3. Let z R. x R : y R, z R, x + y = z. 4. We need to check the truth of x + y = z. We need to think up what BLAH should be. Suppose we tried x = 0 in step 1. Then, in step 4, we would have to show that y = z. Checking that this is true is impossible since we know absolutely nothing about y and z (other than the fact that they are real numbers). y could be 1, and z could be 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 + y = z. 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 x fail us. Saying it another way, for all x R, the following proposition is false. y R, z R, x + y = z. Why is the proposition above false? When x = 0, we argued it was false by thinking about the case when y = 1 and z = 0. In the case when x = 0, this actually gives a proof that is true. y R : z R : x + y z In the end, the reason that the orginal sentence is false is that the following sentence is true. Example The following sentence is true. Proof. 1. Let x R.. Let y = 1. Notice y R. 3. Let z = 0. Notice z R. x R, y R : z R : x + y z. x R, y R : z R : x + y z. 4. We see that x + y = x > 0 = z. In particular, x + y z. 11

12 Remark In our attempt to verify the original sentence, the declaration let x = z y would be wrong for two reasons. First, y and z are declared after x. Second, z y could be less than zero, and so it is not possible to argue that x R. Remark Note the way that I used in particular in the previous proof. In particular is used in the following way. [Something holding less often]. In particular, [something holding more often]. Some examples in plain English are as follows. Some math examples. 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. x = y. In particular, x y. x > y. In particular, x 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.. Let y = x +. Notice y R. 3. Let z = (x + 1). Notice z R. 4. We see that x + y = x + (x + ) = x + 4x + 4 = (x + 1) + (x + 1) = z. Notice that the sentences x R : y R, z R, x + y = z x R, y R : z R : x + y z are related to each other by replacing with, with, and = with. 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. 1

13 Definition The negation of a sentence S is NOT(S ); NOT(S ) can be obtained from S via 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 about what your quantified sentences say will prevent you from negating incorrectly, and you should never apply the rules mindlessly. Remark The procedure for checking a sentence is false is now as follows. Negate the sentence. Verify that the negated sentence is true. Example The following sentence is false. Proof. The negation of the sentence is This is true, because we can verify it as follows. 1. Let n = 0. Notice n Z. n Z, (n > 10 = n > 10). n Z : (n > 10 and n 10).. Because 10 < 0 10, we see that (n > 10 and n 10) is true. Remark Notice that (n > 10 = n > 10) is true when n = 0 and when n = 0. However, what matters is that it is not true when n = 0. The quantifier n N ensures that the sentence in the example has one truth value. 13

14 4.7 Parentheses In this section I have used more parentheses than necessary so that you do not get confused. There is 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 : ax + bx + c = 0 = b 4ac 0 are necessary to distinguish it from a R \ {0}, b R, c R, x R : ax + bx + c = 0 = b 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 The negation of the sentence ( ) a R \ {0}, b R, c R, x R : ax + bx + c = 0 = b 4ac 0 is given by a R \ {0} : b R : c R : ( ) x R : ax + bx + c = 0 and b 4ac < 0. Example The negation of the sentence a R \ {0}, b R, c R, b 4ac 0 = ( ) x R : ax + bx + c = 0 is given by a R \ {0} : b R : c R : b 4ac 0 and ( ) x R, ax + bx + c Math 95 notes The first 7 pages of cover the previous material even more thoroughly. 14

15 5 Absolute value 5.1 Definition and basic properties Definition The absolute value function : R R is defined piecewise by { x if x 0 x = x if x < 0. Lemma x R, ( x 0 and x = x ). Proof. Left to you as preparation for the first quiz. Lemma x R, x x. Proof. 1. Let x R.. (a) Case 1: x 0. In this case, x = x. In particular, x x. (b) Case : x < 0. In this case, x < 0 < x = x. Corollary x R, x x x. Proof. 1. Let x R.. 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 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 x R, y R, x y y x y. Proof. 1. Let x R.. Let y R. 15

16 3. From the truth table for P Q, we see that we must show that it is not possible for exactly one of x y, y x y to be true. This time, we will do this by showing that whenever 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 : x < 0. We see that x = x y is true. The following is the most important inequality in analysis. Theorem (The triangle inequality) x R, y R, x + y x + y. Proof. 1. Let x R.. 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 (The triangle inequality s cousin) x R, y R, x y x y. Proof. 1. Let x R.. 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 shows that x y x y. The trick used in this proof is very common. Remember it! Remark Inequalities involving the absolute values of expressions should only come from the 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

17 5. 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, 7 = 7 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 s the distance between a and b? Taking the difference of the numbers, a b, is a good start, but we do not care about the sign. Theorem 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.. (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). b a c Proof. You can prove it if you want. We ll never need it explicitly. I just wanted to give you some context. 6 A little more about sets Definition 6.1. The empty set, written, is the set with no elements. Definition 6.. Given two sets X and Y, we say that X is a subset of Y and write X Y iff Remark 6.3. z, z X = z Y 1. In this definition iff stands for if and only if. This means that saying X is a subset of Y is exactly the same as saying the quantified sentence is true. Often, in definitions, mathematicians say if even though the meaning is iff. I normally do this, and I feel a little silly for writing iff, but I decided that it s the least confusing thing I can do. To make this iff feel different from a non-definitional iff I have used bold.. 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 set in disguise, so this means for every mathematical thing z. Example 6.4. Suppose X is a set. Then X. 17

18 Proof. Let X be a set. We must check that the following sentence is true. 1. Let x be some mathematical thing. x, x = x X.. We must check the truth of (x = x X). Since has no elements, x is false. The truth 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.. 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. iff the following sentence is true z, z X z Y. Example {0, 1} = {1, 0}.. {, } = {}. 3. { }. 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];. [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

19 Remark Read the symbol := as is defined to be. It is quite different from =, is equal to, which indicates that two previously defined entities are the same.. 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 it is 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 as a number. You can not plug it into an equation. will always require careful consideration and 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 () all I see (3) in my in-fan-cy, (5) red and yel-low then came to be (8)... Tool, Los Angeles, Sequences A sequence is what you think it is. Definition (Informal). A sequence is an infinite list of numbers. Here are some examples: 1,, 3, 4, 5,..., n,... 1, 1, 1 3, 1 4, 1 5,..., 1 n,... 1, 1, 1, 1, 1,..., ( 1) n,..., ( 3 ), ( 4 3 ) 3, ( 5 4 ) 4, ( 6 5) 5,..., ( n) n,... 1, 1,, 3, 5,..., F n = F n 1 + F n, , 1, 3, 5 3, 8 5,..., F n+1 F n,... 19

20 We frequently write s n for the n-th term in a sequence, and so we could express the first four sequences above, concisely, as s n = n, s n = 1 ( n, s n = ( 1) n, s n = 1 + n) 1 n. (Is there a formula for the n-th Fibonacci number which is not iterative? Can you find evidence of the Fibonacci sequence in the Tool song Lateralus, other than that highlighted in the quote at the start of the lecture?) Notation To indicate an abstract sequence we use parentheses: (s n ). Sometimes we may want to emphasize that the list is indexed by natural numbers and we write (s n ) n N. We might also write (s n ) n=1. It is convenient to be able to start a sequence with s m, where m might be greater than 1. In this case we write (s n ) n=m. So, for example, ( 1 n ) n=8 7. Convergence of sequences Here s our first major definition of the class. = 1 8, 1 9, 1 10, 1 11, 1 1,.... Definition Suppose (s n ) n=1 is a sequence. We say that (s n ) converges iff the following sentence is true. L R : ɛ > 0, N N : n N, n N = s n L < ɛ. Remark 7... The quanitifier ɛ > 0 is shorthand for ɛ (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 The sequence ( 1 n ) n=1 converges. Proof. By definition, we must verify that the following sentence is true. L R : ɛ > 0, N N : n N, n N = 1 n L < ɛ. 1. Let L = 0.. Let ɛ > Let N = 1 ɛ + 1. Here denotes the ceiling function. It rounds up to the nearest integer. For example, 5 = 3 = 3. This is useful frequently, so remember it. 4. Let n N. 0

21 5. We must verify the truth of (n N = 1 n L < ɛ). (a) Case 1: n < N. n N is false, so the truth table for = says (n N = 1 n L < ɛ) is true. (b) Case : n N. We must show that 1 n L < ɛ. We have N = 1 ɛ + 1 > 1 ɛ and also 1 n L = 1 n 0 = 1 n 1 N < ɛ. 7.3 Divergence of sequences Definition Suppose (s n ) n=1 is a sequence. We say that (s n ) diverges iff (s n ) does not converge, i.e. iff the following sentence is true. L R, ɛ > 0 : N N, n N : n N and s n L ɛ. Theorem The sequence (( 1) n ) n=1 diverges. Proof. By definition, we must verify that the following sentence is true. 1. Let L R.. Let ɛ = Let N N. L R, ɛ > 0 : N N, n N : n N and ( 1) n L ɛ. 4. (a) Case 1: L 0. Let n = N + 1. (b) Case : L < 0. Let n = N. 5. We must verify that (n N and ( 1) n L ɛ) is true. (a) Case 1: L 0. n = N + 1 = N + (N + 1) N. Also, (b) Case : L < 0. n = N = N + N N. Also, ( 1) n L = ( 1) N+1 L = 1 L = 1 + L 1 = ɛ. ( 1) n L = ( 1) N L = 1 L = 1 L 1 = ɛ. 1

22 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 have no 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 figure out 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 you think 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 clever definition. 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 you think 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 the definition of convergence is saying.

23 7.5 Constructing the proof of theorem 7..3 Let s revisit theorem 7..3, the fact that ( 1 n ) n=1 converges. We must verify that the following sentence is true. L R : ɛ > 0, N N : n N, n N = 1 n L < ɛ. A lot of the proof is fairly mindless and it is likely that writing what follows is a step in the right direction. 1. Let L = BLAH.. Let ɛ > Let N = BLAH. 4. Let n N. 5. We must verify the truth of (n N = 1 n L < ɛ). (a) Case 1: n < N. n N is false, so the truth table for = says (n N = 1 n L < ɛ) is true. (b) Case : n N. We must show that 1 n L < ɛ. We have 1 n 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? I hope that after thinking about the definition of convergence and the proofs of theorem 7..3 long enough 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 s n L < ɛ says s n is within ɛ of L. The limit of a sequence is precisely the real number that the terms in a sequence get close to, so L has got to be the limit, right?! Let s take a minute to make a definition. Definition Suppose (s n ) n=1 is a sequence and L R. We say that (s n ) converges to L iff the following sentence is true. ɛ > 0, N N : n N, n N = s n L < ɛ. In this case L is said to be the limit of the sequence (s n ) and we will often express this by writing lim n s n = L or, if it is not confusing, lim s n = L. Remark If you look at the quantified sentence in the previous definition, you will see that L has not been quantified over. This is okay because L was declared in the preamble. By the time we state the quantified sentence, L is a constant real number. 3

24 . (a) If (s n ) converges, then there is some L R such that (s n ) converges to L. (b) If (s n ) converges to some L R, then (s n ) converges. These statements are trivial from the definitions, but in this early stage of analysis, you should check that you understand them 100%. 1 I hope that while taking math 31B or a similar class you were told that lim n n 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. = 0. We are 1. Let L = (b) Case : n N. We must show that 1 n L < ɛ. We have 1 n L = 1 n 0 = 1 n... Now to N. The purpose of N is to specify how far along in the sequence we have to go before terms in the sequence are within ɛ of the limit L. N will almost always depend on ɛ. If ɛ is small, N will almost always be large. Formulae like N = ɛ are stupid. In the case that we re considering we see that beyond N, 1 n is within 1 N of 0, and so we just need 1 N < ɛ, i.e. N > 1 ɛ. I ve decided to go back to my Oxford routes and demand that N be a natural number. The textbook says otherwise. Sorry about this. It ll be an exercise later to show that this choice doesn t matter. 1 ɛ + 1 is the simplest way to write down an explicit natural number bigger than 1 ɛ. We ve now completed the proof. 3. Let N = 1 ɛ (a) Case : n N. We must show that 1 n L < ɛ. We have N = 1 ɛ + 1 > 1 ɛ and also 1 n L = 1 n 0 = 1 n 1 N < ɛ. 4

25 7.6 Motivating the definition of convergence A picture is worth a thousand words. s N 5 s N 4 s N+1 L + ɛ L s N 3 L ɛ N A sequence of real numbers (s n ) converges to a real number L, the limit, if whenever a small positive quantity ɛ 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 quantity specified. The first bullet point is said mathematically as ɛ > 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) s n L < ɛ (then the terms of the sequence are within ɛ of L). 5

26 7.7 Another sequence that converges Theorem The sequence ( n 101 ) n=1 converges. n Before writing the proof let s think what it must look like. I hope that you can see that the limit 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 : ɛ > 0, N N : n N, n N = n n 101 L < ɛ. 1. Let L = 0.. Let ɛ > Let N = BLAH. 4. Let n N. n 5. We must verify the truth of (n N = L < ɛ). n 101 (a) Case 1: n < N. n N is false, so the truth table for = says (n N = (b) Case : n N. We must show that BLAH BLAH BLAH. n n 101 L < ɛ. We have n n 101 L = n n 101 =.... n n 101 L < ɛ) is true. Now I get to say more about N. The clause N N : n N, n N = n n 101 < ɛ n says that < ɛ is true as long as n is sufficiently large, and N is a natural number quantifying n 101 exactly how large n needs to be. n n When considering the expression my first thought is is positive? The answer n 101 n 101 is, not necessarily, but as long as n is sufficiently large. We can improve on this and say that it is positive as long as n > 101. n N, n > 101 = n n 101 > 0. n n 101 We can now pretty much work under the assumption that positive. But by saying pretty much, I mean that we have to remember the conditions under which it actually holds: n >

27 My next thought is that proof would be n n 101 should behave somewhat like 1 n. A useful inequality towards a n N, n n n but this is UTTER NONSENSE!! In fact, the following sentence is true n N, n > 101 = The inequality goes the wrong way. Instead, we note that n N, n 0 = 0 n n n 101 > 1 n. n n 101 n (7.7.) This is because the condition 0 n 101 n is equivalent to 0 n by basic algebra. The idea I had here was 1 n doesn t give a correct inequality, but maybe n does. In the end, the made all the difference. We finally note that the following sentence is true. n N, n > ɛ = n < ɛ. (7.7.3) Equations (7.7.) and (7.7.3) suggest taking N = max{ 0, ɛ } + 1. Proof of theorem We must verify that the following sentence is true. L R : ɛ > 0, N N : n N, n N = n n 101 L < ɛ. 1. Let L = 0.. Let ɛ > Let N = max{ 0, ɛ } Let n N. n 5. We must verify the truth of (n N = L < ɛ). n 101 (a) Case 1: n < N. n N is false, so the truth table for = says (n N = n n 101 (b) Case : n N. Since N = max{ 0, ɛ } + 1 > 101, we have n > 101 and n 101 > 0. Since N = max{ 0, ɛ } + 1 > 0, we have n n 0 and n 101 n. Since N = max{ 0, ɛ } + 1 > ɛ, we have n > ɛ. Thus, n n 101 L = n n 101 = n n 101 n < ɛ. L < ɛ) is true. 7

28 7.8 Constructing the proof of theorem 7.3. Let s revisit theorem 7.3., the fact that (( 1) n ) n=1 diverges. We must verify that the following sentence is true. L R, ɛ > 0 : N N, n N : n N and ( 1) n L ɛ. A lot of the proof is fairly mindless and it is likely that writing what follows is a step in the right direction. 1. Let L R.. Let ɛ = BLAH. 3. Let N N. 4. Let n = BLAH. 5. We must verify that (n N and ( 1) n L ɛ) is true. BLAH. This looks tricky because ɛ is allowed to depend on L, and n is allowed to depend on L, ɛ, and N. That s a lot to juggle. It turns out that whenever (s n ) is a sequence, if and only if L R, ɛ > 0 : N N, n N : n N and s n L ɛ is true ɛ > 0 : L R, N N, n N : n N and s n L ɛ is true. This is quite a subtle fact, one which we will only be able to prove later on. For now, this simplifies things a little. It tells you that ɛ does not need to depend on L; that is the point in swapping the first two quantifiers. In the case of (( 1) n ) n=1, my intuition for choosing ɛ = 1 is the following. The clause N N, n N : n N and ( 1) n L ɛ says that however far along in the sequence we go, we can always find a term in the sequence which is ɛ 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 the sentence above is true for different Ls. Suppose L is 1. Then the even terms of the sequence are 0 away from L. However, the odd terms are away from L. This suggests taking ɛ =. Suppose L is 1. Then the odd terms of the sequence are 0 away from L. However, the even terms are away from L. This suggests taking ɛ =. Suppose L is 500. Then the odd terms of the sequence are 501 away from L and the even terms are 499 away from L. This continues to suggest that ɛ = would be a fine choice. 8

29 Suppose L is 0. Then all of the terms are 1 away from L. This suggests that ɛ = 1 would be a better choice. For the first three choices of L that we spoke about, we considered the even and odd terms separately. We did not have to do this for the choice L = 0. All of this hinted to me that considering the cases L 0 and L < 0 separately might be useful; there is some symmetry going on. Suppose L 0. Then the odd terms of the sequence are at least 1 away from L and this continues to suggest that ɛ = 1 would be a fine choice. Suppose L < 0. Then the even terms of the sequence are at least 1 away from L and this continues to suggest that ɛ = 1 would be a fine choice. All of the above reasoning suggests taking ɛ = 1. It also suggests that when specifying n we should break up into two cases, when L 0 and when L < 0. Our incomplete proof now becomes. 1. Let L R.. Let ɛ = Let N N. 4. (a) Case 1: L 0. Let n = BLAH. (b) Case : L < 0. Let n = BLAH. 5. We must verify that (n N and ( 1) n L ɛ) is true. (a) Case 1: L 0. BLAH. (b) Case : L < 0. BLAH. Let s consider how to choose n in the case that L 0. When L 0, the odd terms of the sequence are at least 1 away from L. This tells us we should choose n to be an odd number. We also want n to be bigger than or equal to N. N + 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 should choose n to be an even number. We also want n to be bigger than or equal to N. N is an even number bigger than or equal to N. This allows us to fill in. 4. (a) Case 1: L 0. Let n = N + 1. (b) Case : L < 0. Let n = N. We are just left to complete step 5. By this point this is just a matter of writing down everything correctly. However, this step is important and I ll be annoyed when I see nonsense inequalities, or incorrect absolute values, so make sure you get things correct! 9

30 Remark At this point I may as well mention that my order of preference for a quiz or final solution is as follows. A complete and correct solution. An incomplete solution where nothing that is written is incorrect. In addition, separate ideas are 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 solution appear complete. This type of solution will be dealt with harshly. 30

31 7.9 The algebra of limits It is a little tedious to verify the definition of a convergent sequence every time we want to show that a sequence converges. In mathematics, we often prove theorems to save ourselves from having to do similar work over and over again. That is the purpose of the algebra of limits. It says things like, 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 Suppose (s n ) n=1 is a sequence and s, c R. If (s n) converges to s, then (cs n ) n=1 converges to cs. We could also state this theorem by saying that the following sentence is true. Proof attempt. sequences (s n ) n=1, s R, c R, 1. Let (s n ) be a sequence.. Let s R. 3. Let c R. lim s n = s = n 4. We must verify the truth of (lim n s n = s = lim n cs n = cs). (a) Case 1: lim n s n = s is false. Trivial. lim cs n = cs. n (b) Case : lim n s n = s is true. We must show that lim n cs n = cs is true. Both of these expressions have a definition associated with them. So we can expand on this and say the following. The following sentence is true. ɛ > 0, N N : n N, n N = s n s < ɛ. (The use of ɛ and N will become clear in the next subsection.) We must show that the following sentence is true. ɛ > 0, N N : n N, n N = cs n cs < ɛ. We verify the truth of the last statement. i. Let ɛ > 0 ii. Let N = BLAH. iii. Let n N. iv. We must verify the truth of (n N = cs n cs < ɛ). A. Case 1: n < N. Trivial. B. Case : n N. In this case cs n cs = c s n s... BLAH. We have to fill in the BLAHs. How can we possibly write down an N if we have no idea what (cs n ) looks like? 31

32 7.10 The game and the computer program, finishing the proof of In earlier subsections we verified that some sequences converge. After figuring out the limit L of a sequence, this verification comes down to playing an ɛ-n game. The goal of the ɛ-n game is to figure out N in terms of ɛ, and it must be an N N making the relevant statement n N, n N = s n s < ɛ (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 ɛ, it will return a natural number N making the relevant statement (7.10.1) true. The key to completing the proof of theorem is to use such a computer program to specify an N. Completing the proof of theorem (b) i. Let ɛ > 0. ii. This is where the bulk of the argument happens. A. Let ɛ = ɛ 1+ c. Then ɛ > 0. B. Since ɛ > 0, N N : n N, n N = s n s < ɛ is true, we obtain an N N such that the following sentence is true. n N, n N = s n s < ɛ. C. Let N = N. iii. Let n N. iv. We must verify the truth of (n N = cs n cs < ɛ). A. Case 1: n < N. Trivial. B. Case : n N. Since N = N, we have n N, and so s n s < ɛ. Thus, cs n cs = c s n s c ɛ ɛ = c 1 + c < ɛ. Notice how algorithmically this proof is written. We say what ɛ is in terms of c and ɛ which were already declared in. and 4.(b)i., respectively. We obtain N from the computer program and ɛ which were already declared in 4.(b) and 4.(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. 3

33 7.11 Another algebra of limits proof, more game/computer program Let s see another similar example. Theorem Suppose (s n ) n=1 and (t n) n=1 are sequences, and that s, t R. If (s n) converges to s and (t n ) converges to t, then (s n + t n ) n=1 converges to s + t. We could also state this theorem by saying that the following sentence is true. Proof. 1. Let (s n ) be a sequence.. Let (t n ) be a sequence. 3. Let s R. 4. Let t R. sequences (s n ) n=1, sequences (t n ) n=1, s R, t R, ( ) lim s n = s and n lim t n = t n = lim n (s n + t n ) = s + t. 5. We must verify the truth of ( lim n s n = s and ) lim t n = t n = lim n (s n + t n ) = s + t. (a) Case 1: (lim n s n = s and lim n t n = t) is false. Trivial. (b) Case : (lim n s n = s and lim n t n = t) is true. We must show that lim n (s n + t n ) = s + t is true. Since all of these expressions have a definition associated with them, we can expand on this and say the following. The following sentence is true. ɛ 1 > 0, N 1 N : n N, n N 1 = s n s < ɛ 1. The following sentence is true. ɛ > 0, N N : n N, n N = t n t < ɛ. We must show that the following sentence is true. ɛ > 0, N N : n N, n N = (s n + t n ) (s + t) < ɛ. We verify the truth of the last statement. i. Let ɛ > 0. ii. This is where the bulk of the argument happens. A. Let ɛ 1 = ɛ. Then ɛ 1 > 0. B. Let ɛ = ɛ. Then ɛ > 0. 33