elementary analysis: the theory of calculususers.math.msu.edu/users/albert/math320.pdf ·...

Elementary Analysis: The Theory of Calculus

Albert Cohen

C336 Wells HallDepartment of MathematicsDepartment of StatisticsMichigan State University

East Lansing MI48823

[email protected]

Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 1 / 124

Course Information

Syllabus to be posted on class page in first week of classes

Homework assignments will posted there as well

Page can be found at https://mathdata.msu.edu/CP/RW/C320.html

Many examples within these slides are found in textbook - Springerowns the copyright and I make no claim on the material used here.

Text: Elementary Analysis: The Theory of Calculus by Kenneth A.Ross (Springer Undergraduate Texts in Mathematics)

Can be found in MSU bookstores now

This book will be our reference, and questions for assignments will bechosen from it. Copyright for all questions used from this bookbelongs to Springer

What set of numbers do you prefer?

The set N of Natural Numbers

Properties (Peano Axioms)

N1: 1 ∈ NN2: If n ∈ N then its successor n + 1 ∈ NN3: 1 is not the successor of any element of NN4: If n and m have the same successor, then n = mN5: If A ⊂ N, with 1 ∈ A and n + 1 ∈ A whenever n ∈ A, then A = N

Can we prove N5?

For sake of contradiction, assume N5 is false.

Consider the smallest element n0 ∈ {x ∈ N | x ∈ Ac}.If n0 6= 1, then n0 is the successor of some natural number.

Then n0 − 1 ∈ A and so its successor n0 ∈ A, a contradiction.

Is this a complete argument?

Can we prove N5?

Mathematical Induction

Argument above made too many assumptions -but introducesMathematical Induction

To prove list of statements P1,P2, ... is true, must show

I 1 P1 is trueI 2 Pn+1 is true whenever Pn is true

I 1 is known as Basis for Induction and I 2 is known as Induction Step

Examples

Theorem

1 + 2 + ...+ n = n(n+1)2

Proof.

I 1 1 = 1(1+1)2 , so we have Basis for Induction.

I 2 Check Pn implies Pn+1. Assume Pn holds. Then

1 + 2 + ...+ n + (n + 1) =n(n + 1)

2+ (n + 1)

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

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

2(n + 1)(n + 2)

Examples

Theorem

For all n ∈ N, 7n − 2n is divisible by 5.

Proof.

I 1 5 = 71 − 21 is divisible by 5, so we have Basis for Induction.

I 2 Check Pn implies Pn+1. Assume Pn holds. Then there exists aninteger m = 7n−2n

7n+1 − 2n+1 = 7n+1 − 7 · 2n + 7 · 2n − 2n+1

= 7 · (7n − 2n) + 5 · 2n

= 5 · (7m + 2n)

Examples

Theorem

| sin(nx) |≤ n | sin(x) | for all n ∈ N, x ∈ R.

Proof.

I 1 | sin(1 · x) |≤ 1· | sin(x) | , so we have Basis for Induction.

I 2 Check Pn implies Pn+1. Assume Pn holds. Then

| sin((n + 1)x) | =| sin(nx + x) |=| sin(nx) cos(x) + cos(nx) sin(x) |≤| sin(nx) cos(x) | + | cos(nx) sin(x) |≤| sin(nx) | · | cos(x) | + | cos(nx) | · | sin(x) |≤| sin(nx) | + | sin(x) |≤ n | sin(x) | + | sin(x) |= (n + 1) | sin(x) |

HW Questions

1.1, 1.3

1.6, 1.7

1.8a, b

1.12a, b, c

Integer, Rationa, and Algebraicl Numbers

Definition

Define the set Z of Integers as {0, 1,−1, 2,−2, ...}Define the set Q of Rationals as

{x = m

n | m, n ∈ Z, n 6= 0}

Q contains all terminating decimalsWhat happens if we close this set to include all non-terminatingdecimals ?Can we include numbers like

√2 ? Define the set A of Algebraic

numbers as..

A ≡ {x | a0 + a1x + ...+ anxn = 0, n ≥ 1, a0, a1, , .., an ∈ Z}

Examples 1a.)− d .)

Theorem

a.) 417 , b.)

√3, c.)17

13 , d .)

√2 + 5

13 are algebraic numbers.

Proof.

For a.) through c .), the numbers are solutions of the correspondingpolynomial equations:

a.) 17x − 4 = 0,

b.) x2 − 3 = 0

c.) x3 − 17 = 0

For d .), if we set x =

√2 + 5

13 then x2 − 2 = 5

13 and so

(x2 − 2

which implies

√2 + 5

13 is a solution of x6 − 6x4 + 12x2 − 13 = 0

Rational Zeros Theorem (Theorem 2.2)

Theorem

Suppose that a0, a1, , .., an ∈ Z , r ∈ Q , n ≥ 1 , and an, a0 6= 0.Suppose also that r satisfies

anrn + an−1rn−1 + ...+ a1r1 + a0 = 0 (4)

with r = pq : p, q ∈ Z.

Finally, suppose that p, q have no common factors and q 6= 0.Then q divides an and p divides a0.

Rational Zeros Theorem

Proof: By given assumptions,

+ an−1

)n−1+ ...+ a1

+ a0 = 0 (5)

Multiplying through by qn, this equation can be rewritten

n∑k=0

an−kpn−kqk = 0 (6)

Rational Zeros Theorem

Proof (continued)It follows that

anpn = −q ·n∑

an−kpn−kqk−1 (7)

Therefore q divides anpn. Since p, q have no common factors, q dividesan. Similarly,

a0qn = −p ·n−1∑k=0

an−kpn−k−1qk (8)

Since p, q have no common factors, p divides a0. QED

Examples 2 and 4

Theorem√

2 /∈ Q, 3√

6 /∈ Q.

Proof.

By Theorem 2.2, the only rational numbers that could solve x2− 2 = 0 are{1,−1, 2,−2} and the only rational numbers that could solve x3 − 6 = 0are {1,−1, 2,−2, 3,−3, 6,−6}.

Examples 2 and 4

Theorem√

2 /∈ Q, 3√

6 /∈ Q.

Proof.

By Theorem 2.2, the only rational numbers that could solve x2− 2 = 0 are{1,−1, 2,−2} and the only rational numbers that could solve x3 − 6 = 0are {1,−1, 2,−2, 3,−3, 6,−6}.

HW Questions

2.1, 2.3

2.5, 2.6

Axioms and Properties of Q

Assume a, b, c ∈ QProperties

A1: a + (b + c) = (a + b) + cA2: a + b = b + aA3: a + 0 = aA4: ∀a, ∃ − a : a + (−a) = 0M1: a(bc) = (ab)cM2: ab = baM3: a · 1 = aM4: ∀a 6= 0, ∃a−1 : aa−1 = 1DL: a(b + c) = ab + ac

A system that contains more than one element and satisfies these 9properties is known as a Field

Axioms and Properties of Q

Assume a, b, c ∈ QOrder Structure ≤

O1: Either a ≤ b or b ≤ aO2: If a ≤ b and b ≤ a then a = bO3: If a ≤ b and b ≤ c then a ≤ cO4: If a ≤ b then a + c ≤ b + cO5: If a ≤ b and 0 ≤ c then ac ≤ bc

A field satisfying O1− O5 is called an Ordered Field.

Theorem 3.1

Theorem

Assume a, b, c ∈ RThe following are consequences of the field properties:

(i): a + c = b + c ⇒ a = b(ii): a · 0 = 0(iii): (−a)b = −ab(iv): (−a)(−b) = ab(v): ab = 0⇒ either a = 0 or b = 0

Theorem 3.2

Theorem

Assume a, b, c ∈ R. Note that a < b means a ≤ b and a 6= b

The following are consequences of an ordered field:

(i): If a ≤ b then −b ≤ −a(ii): If a ≤ b and c ≤ 0 then bc ≤ ac(iii): If 0 ≤ a and 0 ≤ b then 0 ≤ ab(iv): 0 ≤ a2

(v): 0 < 1(vi): If 0 < a then 0 < a−1

(vii): If 0 < a < b then 0 < b−1 < a−1

Definitions and Theorem 3.5

Define | a | as

| a |= a when a ≥ 0,

| a |= −a when a ≤ 0.

dist(a, b) ≡| a− b |Theorem 3.5:

Theorem

Assume a, b ∈ R(i): | a |≥ 0

(ii): | ab |=| a || b |

(iii): | a + b |≤| a | + | b |

Corollary 3.6 and the Triangle Inequality

Corollary 3.6:

Corollary

Assume a, b, c ∈ R. Then

dist(a, c) ≤ dist(a, b) + dist(b, c) (9)

Triangle Inequality: Assume a, b, c ∈ R. Then

| a + b |≤| a | + | b | (10)

The Completeness Axiom

Let S ⊂ R, S 6= φ.

If S contains a largest element s0 then it is called the maximum of S :s0 = max S

If S contains a smallest element s1 then it is called the minimum ofS : s0 = min S

Examples:

[a, b), (a, b){r ∈ Q : 0 ≤ r ≤

√2}{

n−1n

: n ∈ N}

Let S ⊂ R, S 6= φ.

If a real number M satisfies s ≤ M for all s ∈ S , then M is called anupper bound of S and S is said to be bounded above.

If a real number m satisfies s ≥ m for all s ∈ S , then m is called alower bound of S and S is said to be bounded below.

S is said to be bounded if it is bounded above and below; i.e. ifS ⊂ [m,M].

Examples

[a, b), (a, b){r ∈ Q : 0 ≤ r ≤

√2}{

n−1n

: n ∈ N}

Let S ⊂ R, S 6= φ.

If S is bounded above and has a least upper bound x , then x ≡ sup S .

If S is bounded below and has a greatest lower bound y , theny ≡ inf S .

Examples

[a, b), (a, b){r ∈ Q : 0 ≤ r ≤

√2}{

n−1n

: n ∈ N}

The Completeness Axiom and other properties of R

The Completeness Axiom: Assume S ⊂ R, S 6= φ is bounded above.Then it has a least upper bound.

Corollary 4.6: Assume S ⊂ R, S 6= φ is bounded below. Then it has agreatest lower bound.

Archimedian Property: If a, b > 0, then ∃n ∈ N such that na > b.

Denseness of Q: If a, b ∈ R, and a < b then ∃r ∈ Q such thata < r < b.

HW Questions

4.1a, i , j , k , r , u,w

4.6a, b

4.7a, b

4.10, 4.15, 4.16

The symbols ∞ and −∞

These symbols are concepts and not real numbers. We extend orderings toR ∪ {−∞,∞}, but not algebraic structure.

R = (−∞,∞)

[a,∞) = {x ∈ R : a ≤ x}If S ⊂ R, S 6= φ is not bounded above, then sup S =∞If S ⊂ R, S 6= φ is not bounded below, then inf S = −∞

HW Questions

5.1a, b, c , d

Limits of Sequences

A sequence s will be taken to be a map from the set N∪{0} into the reals:

s : N ∪ {0} → R (11)

We often denote such a function by its elements: {sn}∞n=0 or (s0, s1, ...)

Examples

{sk}∞k=1 : sn =1

{ak}∞k=0 : an = (−1)n

{ak}∞k=1 : an = cos(nπ

){ak}∞k=0 : an = (n)

{bk}∞k=0 : bn =

Definition of Convergence

Definition

A sequence sn is said to converge to its limit s if for each ε > 0 thereexists an N such that n ≥ N implies | sn − s |< ε.

A sequence that does not converge to any limit diverges.

Take sn = 3n+17n−4 . Then

n > 0⇒| 3n+17n−4 −

37 |< 1

n > 4⇒| 3n+17n−4 −

37 |< 0.1

n > 39⇒| 3n+17n−4 −

37 |< 0.01

n > 388⇒| 3n+17n−4 −

37 |< 0.001

n > 387, 755⇒| 3n+17n−4 −

37 |< 0.000001

Is there a general rule? How about the previous sequences (examples ?) Ifa sequence converges to a limit, can there be multiple limits?

Take sn = 3n+17n−4 . Then

n > 0⇒| 3n+17n−4 −

37 |< 1

n > 4⇒| 3n+17n−4 −

37 |< 0.1

n > 39⇒| 3n+17n−4 −

37 |< 0.01

n > 388⇒| 3n+17n−4 −

37 |< 0.001

n > 387, 755⇒| 3n+17n−4 −

37 |< 0.000001

Is there a general rule? How about the previous sequences (examples ?) Ifa sequence converges to a limit, can there be multiple limits?

HW Questions

7.1, 7.3a− l

7.4, 7.5

Scratchwork versus Complete Proof

Theorem: limn→∞1n2

Discussion: If we want | 1n2− 0 |< ε then we need n > 1√

Formal Proof: Let ε > 0. Let N = 1√ε. Then n > N ⇒ n2 > 1

ε and so1n2< ε. This implies | 1

n2− 0 |< ε. �

Theorem: limn→∞3n+17n−4 = 3

Discussion: If we want | 3n+17n−4 −

37 |=|

197(7n−4) |< ε then we need

n > 47 + 19

49ε .

Formal Proof: Let ε > 0. Let N = 47 + 19

49ε . Then n > N implies7n − 4 > 19

7ε and so 197(7n−4) < ε. This implies | 3n+1

7n−4 −37 |< ε. �

37 |=|

n > 47 + 19

49ε .

7n−4 −37 |< ε. �

37 |=|

n > 47 + 19

49ε .

7n−4 −37 |< ε. �

Theorem

limn→∞(−1)n does not exist

Proof.

For sake of contradiction, assume there is an a = limn→∞(−1)n. Letε = 1

2 . By the definition of a limit, there exists an N such that n > Nimplies | (−1)n − a |< ε. This means both | 1− a |< ε and | −1− a |< ε,which implies

2 =| 1− (−1) |=| 1− a + a− (−1) |≤| 1− a | + | −1− a |

which is a contradiction.

Theorem

limn→∞(−1)n does not exist

Proof.

For sake of contradiction, assume there is an a = limn→∞(−1)n. Letε = 1

2 . By the definition of a limit, there exists an N such that n > Nimplies | (−1)n − a |< ε. This means both | 1− a |< ε and | −1− a |< ε,which implies

2 =| 1− (−1) |=| 1− a + a− (−1) |≤| 1− a | + | −1− a |

which is a contradiction.

Square Roots and Limits: Theorem 8.5

Theorem

Let sn be a sequence of non-negative real numbers and supposes = limn→∞sn exists. If s > 0, then

√s = limn→∞

Proof.

Let ε > 0. Since s = limn→∞sn, there exists N such that n > N implies| sn − s |<

√sε. It follows that

sn −√

s | =| sn − s |√

sn +√

≤ | sn − s |√s

√sε√s

This argument can of course be extended to s = 0, and is left to you!

Square Roots and Limits: Theorem 8.5

Theorem

Let sn be a sequence of non-negative real numbers and supposes = limn→∞sn exists. If s > 0, then

√s = limn→∞

Proof.

Let ε > 0. Since s = limn→∞sn, there exists N such that n > N implies| sn − s |<

√sε. It follows that

sn −√

s | =| sn − s |√

sn +√

≤ | sn − s |√s

√sε√s

This argument can of course be extended to s = 0, and is left to you!Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 38 / 124

Example 8.6

Theorem

Let sn be a convergent sequence of real numbers and suppose

sn 6= 0 for all n ∈ Nlimn→∞sn = s 6= 0.

Then inf {| sn |: n ∈ N} > 0.

Proof.

Let ε = |s|2 . Since s = limn→∞sn, there exists N such that n > N implies

| sn− s |< |s|2 . It follows from the triangle inequality that | sn |≥ |s|2 . Define

m = min

{| s |

2, s1, s2, ..., sN

}. (14)

Then clearly m > 0 and | sn |≥ m for all n ∈ N. It follows thatinf {| sn |: n ∈ N} ≥ m > 0.

Example 8.6

Theorem

Let sn be a convergent sequence of real numbers and suppose

sn 6= 0 for all n ∈ Nlimn→∞sn = s 6= 0.

Then inf {| sn |: n ∈ N} > 0.

Proof.

Let ε = |s|2 . Since s = limn→∞sn, there exists N such that n > N implies

| sn− s |< |s|2 . It follows from the triangle inequality that | sn |≥ |s|2 . Define

m = min

{| s |

2, s1, s2, ..., sN

}. (14)

Then clearly m > 0 and | sn |≥ m for all n ∈ N. It follows thatinf {| sn |: n ∈ N} ≥ m > 0.

HW Questions

8.1a, c, 8.2a, c , d

8.4, 8.5, 8.6

8.6, 8.7

8.9a, 8.10

Theorem 9.1

Theorem

Convergent sequences are bounded

Proof.

Let sn be a convergent sequence, and limn→∞sn = s. Let ε = 1, and sothere exists N such that n > N implies | sn − s |< 1. It follows from thetriangle inequality that | sn |<| s | +1. Define

M = max {| s | +1, | s1 |, | s2 |, ..., | sN |} . (15)

Then clearly | sn |≤ M for all n ∈ N and it follows that {sn} is a boundedsequence.

Theorem 9.1

Theorem

Convergent sequences are bounded

Proof.

Let sn be a convergent sequence, and limn→∞sn = s. Let ε = 1, and sothere exists N such that n > N implies | sn − s |< 1. It follows from thetriangle inequality that | sn |<| s | +1. Define

M = max {| s | +1, | s1 |, | s2 |, ..., | sN |} . (15)

Then clearly | sn |≤ M for all n ∈ N and it follows that {sn} is a boundedsequence.

Some Useful Limit Theorems

Theorem 9.2:

Theorem

If {sn} converges to s and k ∈ R, then {ksn} converges to ks

Proof.

Assume k 6= 0, otherwise this is trivial. Let ε > 0, and so there exists Nsuch that n > N implies | sn − s |< ε

|k| . Then clearly | ksn − ks |≤ ε.

Theorem 9.2:

Theorem

If {sn} converges to s and k ∈ R, then {ksn} converges to ks

Proof.

Assume k 6= 0, otherwise this is trivial. Let ε > 0, and so there exists Nsuch that n > N implies | sn − s |< ε

|k| . Then clearly | ksn − ks |≤ ε.

Theorems 9.3, 9.4:

Theorem

If sn → s and tn → t then

sn + tn → s + t

sntn → st

We prove the product rule. Recall that convergence of a sequenceguarantees its boundedness:

Proof.

Define M := sup {| sn |} . For any ε > 0, there exists N1,N2 such that

n > N1 ⇒| sn − s |< ε

2 | t |

n > N2 ⇒| tn − t |< ε

It follows that for n > N := max {N1,N2},

| sntn − st | =| sntn − tsn + tsn − st |≤| sntn − tsn | + | tsn − st |≤| sn || tn − t | + | t || sn − s |

< M · ε

2M+ | t | · ε

2 | t |= ε.

Lemma 9.5:

If {sn} converges to s and sn, s 6= 0 for all n, then 1sn

converges to 1s .

Proof.

By Example 8.6, we have that X ≡ inf {| sn |: n ∈ N} > 0. Then thereexists N(ε | X | s) such that n > N(ε· | X | s) implies∣∣∣ 1

sn− 1

∣∣∣ =∣∣∣sn − s

∣∣∣ ≤ ∣∣∣sn − s

∣∣∣ < ε | X | s

| X | s= ε. (18)

Lemma 9.5:

If {sn} converges to s and sn, s 6= 0 for all n, then 1sn

converges to 1s .

Proof.

By Example 8.6, we have that X ≡ inf {| sn |: n ∈ N} > 0. Then thereexists N(ε | X | s) such that n > N(ε· | X | s) implies∣∣∣ 1

sn− 1

∣∣∣ =∣∣∣sn − s

∣∣∣ ≤ ∣∣∣sn − s

∣∣∣ < ε | X | s

| X | s= ε. (18)

Theorem 9.6:

Theorem

If sn → s and tn → t, where tn, t 6= 0 for all n then sntn→ s

Proof.

Try now!

Theorem 9.6:

Theorem

If sn → s and tn → t, where tn, t 6= 0 for all n then sntn→ s

Proof.

Try now!

More Examples

Example 1 : n3+6n2+74n3+3n−4 →

Example 2 : n−5n2+7

Prove these now!

More Examples

Definition

limn→∞ sn =∞ if ∀M > 0 there exists an N : n > N ⇒ sn > M.

Example 1 :√

n + 7→∞Example 2 : n2+3

n+1 diverges

Prove these now!

More Examples

Definition

limn→∞ sn =∞ if ∀M > 0 there exists an N : n > N ⇒ sn > M.

Example 1 :√

n + 7→∞Example 2 : n2+3

n+1 diverges

Prove these now!

Theorem 9.9

Theorem

If {sn} , {tn} are such that sn →∞ and limn→∞ tn = t > 0, thenlimn→∞ sntn =∞.

Proof.

Let M > 0. Choose 0 < m < t. By definition, ∃N1 such thatn > N1 ⇒ tn > m. Since limn→∞ tn, ∃N2 such that n > N2 ⇒ sn >

Take N = max {N1,N2}. Then n > N ⇒ sntn >Mm m = M.

Theorem 9.9

Theorem

If {sn} , {tn} are such that sn →∞ and limn→∞ tn = t > 0, thenlimn→∞ sntn =∞.

Proof.

Let M > 0. Choose 0 < m < t. By definition, ∃N1 such thatn > N1 ⇒ tn > m. Since limn→∞ tn, ∃N2 such that n > N2 ⇒ sn >

Take N = max {N1,N2}. Then n > N ⇒ sntn >Mm m = M.

Theorem 9.10:

Theorem

If {sn} ⊂ R+ then sn →∞ if and only if 1sn→ 0

Proof.

Try now!

Theorem 9.10:

Theorem

If {sn} ⊂ R+ then sn →∞ if and only if 1sn→ 0

Proof.

Try now!

HW Questions

9.1a, c, 9.2a

9.4, 9.5, 9.6

9.9a, 9.10, 9.18

Monotone sequences

A sequence sn where sn ≤ sn+1 is said to non-decreasing. If sn ≥ sn+1 thenit is said to be is said to non-increasing. A non-increasing ornon-decreasing sequence is said to be monotonic.

A Very Useful Limit Theorem: Theorem 10.2

Theorem

All bounded monotone sequences converge.

Proof.

Wlog, let {sn} be a bounded non-decreasing sequence. DefineS = {sn : n ∈ N} , and define u = sup S . Since S is bounded, u ∈ R. Itremains to show that sn → u. To do so, choose ε > 0. Since u − ε is notan upper bound of S , there exists N such that sN > u − ε. But, since sn isnon-decreasing, we have sN ≤ sn for all n ≥ N. Of course, sn ≤ u for all n,so so n > N implies u − ε < sn ≤ u, which implies | sn − u |< ε.

A Very Useful Limit Theorem: Theorem 10.2

Theorem

All bounded monotone sequences converge.

Proof.

Wlog, let {sn} be a bounded non-decreasing sequence. DefineS = {sn : n ∈ N} , and define u = sup S . Since S is bounded, u ∈ R. Itremains to show that sn → u. To do so, choose ε > 0. Since u − ε is notan upper bound of S , there exists N such that sN > u − ε. But, since sn isnon-decreasing, we have sN ≤ sn for all n ≥ N. Of course, sn ≤ u for all n,so so n > N implies u − ε < sn ≤ u, which implies | sn − u |< ε.

More Monotonic Limit Theorems

Theorem 10.3: All unbounded monotone increasing sequences divergeto ∞.

Theorem 10.4: All unbounded monotone decreasing sequencesdiverge to −∞.

Corollary 10.5: All monotone sequences converge or diverge to ∞ or−∞.

lim inf and lim sup

Definition

A sequence sn in R has

lim sup sn = limN→∞

sup {sn : n > N}

lim inf sn = limN→∞

inf {sn : n > N}(19)

A sequence sn in R is called a Cauchy sequence if ∀ε > 0 there exists anumber N such that m, n > N implies | sn − sm |< ε

Theorem 10.7

Theorem

Let sn be a sequence in R. Then

If limn→∞ sn = s is defined as a real number or ∞,−∞, thenlim inf sn = s = lim sup sn

If lim inf sn = lim sup sn then s ∈ R and lim inf sn = s = lim sup sn

Lemma 10.9

Convergent sequences are Cauchy sequences.

Proof.

Suppose that s = limn→∞ sn exists. Then for all n,m > N(ε2

)it follows

| sn − sm | =| sn − s + s − sm |≤| sn − s | + | sm − s |

Thus, sn is a Cauchy sequence.

Lemma 10.9

Convergent sequences are Cauchy sequences.

Proof.

Suppose that s = limn→∞ sn exists. Then for all n,m > N(ε2

)it follows

| sn − sm | =| sn − s + s − sm |≤| sn − s | + | sm − s |

Thus, sn is a Cauchy sequence.

Lemma 10.10

Cauchy sequences are bounded.

Proof.

For all n,m > N (1) we have | sn − sm |< 1. In particular,| sn − sN(1)+1 |< 1 for all n > N(1). Define

M = max{| sN(1)+1 | +1, | s1 |, | s2 |, ...., | sN(1) |

Then | sn |≤ M for all n.

Lemma 10.10

Cauchy sequences are bounded.

Proof.

For all n,m > N (1) we have | sn − sm |< 1. In particular,| sn − sN(1)+1 |< 1 for all n > N(1). Define

M = max{| sN(1)+1 | +1, | s1 |, | s2 |, ...., | sN(1) |

Then | sn |≤ M for all n.

Theorem 10.11

Theorem

A sequence is convergent if and only if it is a Cauchy sequence.

Theorem 10.11

Proof.

The previous Lemma has shown that all convergent sequences are Cauchy.It remains to show the other implication holds. Namely, that all Cauchysequences are convergent.

Hence, assume sn is Cauchy. Then by Lemma10.9, we know sn is bounded.

We need only show now that lim inf sn ≥ lim sup sn.

But, this can be seen by taking m, n > N(ε) which implies sn < sm + ε.

It follows that

sup {sn : n > N} ≤ sm + ε

⇒ sup {sn : n > N} ≤ inf {sm : m > N}+ ε(22)

Taking limits as ε→ 0, and so N →∞, gives us our result.

Theorem 10.11

Proof.

The previous Lemma has shown that all convergent sequences are Cauchy.It remains to show the other implication holds. Namely, that all Cauchysequences are convergent.Hence, assume sn is Cauchy. Then by Lemma10.9, we know sn is bounded.

It follows that

Theorem 10.11

Proof.

It follows that

Theorem 10.11

Proof.

It follows that

Theorem 10.11

Proof.

It follows that

Theorem 10.11

Proof.

It follows that

Babylonian Method of Computing Square Roots

Consider a positive real number A > 1. Now, define

sn+1 = f (sn)

s0 = A

f (x) =1

) (23)

Since f (A) = A+12 < A and ∀x > 0,

dx(f (x)− x) = −1

)< 0, (24)

we know that {sn}∞n=0 is a bounded (by 0) monotonic decreasing sequence.

Furthermore, f (√

A) =√

Have we proven convergence of this recursively defined sequence?

sn+1 = f (sn)

s0 = A

f (x) =1

) (23)

Since f (A) = A+12 < A and ∀x > 0,

dx(f (x)− x) = −1

)< 0, (24)

Furthermore, f (√

A) =√

sn+1 = f (sn)

s0 = A

f (x) =1

) (23)

Since f (A) = A+12 < A and ∀x > 0,

dx(f (x)− x) = −1

)< 0, (24)

Furthermore, f (√

A) =√

sn+1 = f (sn)

s0 = A

f (x) =1

) (23)

Since f (A) = A+12 < A and ∀x > 0,

dx(f (x)− x) = −1

)< 0, (24)

Furthermore, f (√

A) =√

sn+1 = f (sn)

s0 = A

f (x) =1

) (23)

Since f (A) = A+12 < A and ∀x > 0,

dx(f (x)− x) = −1

)< 0, (24)

Furthermore, f (√

A) =√

Markov Chain Model of Employment

The US government has studied models of employment and has come upwith the following observation:

P[Unemployed finds job by end of the year] = pf ∈ (0, 1)

P[Employed loses job by end of the year] = pl ∈ (0, 1)

Define Wk to be the probability a worker is employed at the beginning ofyear k , Nk the probability she is not working at the beginning of year k .

Then (Wk+1

[1− pl pf

pl 1− pf

Finally, assume that Wk + Nk = 1.

Question: Does Wk →W for some W ∈ (0, 1)?

The US government has studied models of employment and has come upwith the following observation:

P[Unemployed finds job by end of the year] = pf ∈ (0, 1)

P[Employed loses job by end of the year] = pl ∈ (0, 1)

Define Wk to be the probability a worker is employed at the beginning ofyear k , Nk the probability she is not working at the beginning of year k .

Then (Wk+1

[1− pl pf

pl 1− pf

Finally, assume that Wk + Nk = 1.

Question: Does Wk →W for some W ∈ (0, 1)?

It follows that the matrix can be diagonalized as

(1− pl pf

pl 1− pf

(pfpl−1

)(1 00 1− pf − pl

)( plpf +pl

plpf +pl

− plpf +pl

pfpf +pl

Define

pf +pl

plpf +pl

− plpf +pl

pfpf +pl

It follows that the matrix can be diagonalized as

(1− pl pf

pl 1− pf

(pfpl−1

)(1 00 1− pf − pl

)( plpf +pl

plpf +pl

− plpf +pl

pfpf +pl

Define

pf +pl

plpf +pl

− plpf +pl

pfpf +pl

Hence,

~qk+1 =

(1 00 1− pf − pl

)~qk ⇒ ~qk =

B(1− pf − pl)k

Returning to our original notation,(Wk

(pfpl−1

)(1 00 1− pf − pl

(pfpl−1

)(1 00 1− pf − pl

B(1− pf − pl)k

(pfpl−1

B(1− pf − pl)k+1

pl− B(1− pf − pl)

A + B(1− pf − pl)k+1

Hence,

~qk+1 =

(1 00 1− pf − pl

)~qk ⇒ ~qk =

B(1− pf − pl)k

Returning to our original notation,(Wk

(pfpl−1

)(1 00 1− pf − pl

(pfpl−1

)(1 00 1− pf − pl

B(1− pf − pl)k

(pfpl−1

B(1− pf − pl)k+1

A + B(1− pf − pl)k+1

Solving for our parameters A,B, we see that(W0

A + B(1− pf − pl)

pfplpf

plN0−W0

1−pf−pl

1+pfpl

pl+pfpl

) (30)

Solving for our parameters A,B, we see that(W0

A + B(1− pf − pl)

pfplpf

plN0−W0

1−pf−pl

1+pfpl

pl+pfpl

) (30)

Nonlinear Predator-Prey Model

Consider a simplified Lotka-Volterra model of predator-prey dynamics.Define

Rn the number of Rabbits at time n

Fn the number of Foxes at time n

a is the natural growth rate of rabbits in the absence of foxencounters,

c is the natural death rate of foxes in the absence of food (rabbits),

b is the death rate of rabbits per encounter with foxes,

e is the efficiency of converting the energy from eaten rabbits intomore foxes.

and the dynamical system

Rn+1 = Rn + aRn − bRnFn

Fn+1 = Fn + ebRnFn − cFn(31)

Question: Does (Rn,Fn)→ (R,F ) ? If so, how do we characterize thispair in terms of the given parameters?

Nonlinear Predator-Prey Model

Consider a simplified Lotka-Volterra model of predator-prey dynamics.Define

Rn the number of Rabbits at time n

Fn the number of Foxes at time n

a is the natural growth rate of rabbits in the absence of foxencounters,

c is the natural death rate of foxes in the absence of food (rabbits),

b is the death rate of rabbits per encounter with foxes,

e is the efficiency of converting the energy from eaten rabbits intomore foxes.

and the dynamical system

Rn+1 = Rn + aRn − bRnFn

Fn+1 = Fn + ebRnFn − cFn(31)

Question: Does (Rn,Fn)→ (R,F ) ? If so, how do we characterize thispair in terms of the given parameters?

HW Questions

10.1, 10.2

10.8, 10.9

10.10, 10.11

Subsequences

Suppose {sn}n∈N is a sequence. Then a subsequence of this sequence isanother sequence of the form

tk = snkn1 < n2 < ... < nk < nk+1 < ...

As an example, consider

sn = n2(−1)n

nk = 2k

tk = snk

= (2k)2(−1)2k

Check out more examples in book!

Subsequences

tk = snkn1 < n2 < ... < nk < nk+1 < ...

sn = n2(−1)n

nk = 2k

tk = snk

= (2k)2(−1)2k

Subsequences

tk = snkn1 < n2 < ... < nk < nk+1 < ...

sn = n2(−1)n

nk = 2k

tk = snk

= (2k)2(−1)2k

Theorem 11.2

Theorem

If a sequence {sn}n∈N converges to a limit s, then every subsequenceconverges to the same limit.

Proof.

For sake of contradiction, assume not. Then there exists a subsequence{snk}k∈N that does not converge to s. Then there exists an ε > 0 and afurther subsequence kl such that | snkl − s |> ε, with kl →∞monotonically. This contradicts the assumption that sn → s.

Question: Could we have come to the same conclusion if we hadattempted to prove this theorem using the concepts of lim sup sn, lim inf sn?

Theorem 11.2

Theorem

Proof.

Theorem 11.2

Theorem

Proof.

Theorem 11.3 and friends

Theorem 11.3:

Theorem

Every sequence {sn}n∈Nhas a monotonic subsequence.

Proof: In book, but let’s try to draw a picture why this works!

Corollary 11.4:

Corollary

Let {sn}n∈N be any sequence. Then there exists a monotonic subsequencewhose limit is lim sup sn and another monotonic subsequence whose limit islim inf sn.

Proof: Again, in book, but let’s try to analyze and come up with examples!

Theorem 11.3:

Theorem

Corollary 11.4:

Corollary

Theorem 11.3:

Theorem

Corollary 11.4:

Corollary

Theorem 11.3:

Theorem

Corollary 11.4:

Corollary

Bolzano-Weierstrass Theorem

Theorem

Every bounded sequence has a convergent subsequence.

Proof.

By Theorem 11.3, every bounded sequence has a monotonic subsequence.But, by Theorem 10.2, this subsequence is convergent.

Bolzano-Weierstrass Theorem

Theorem

Every bounded sequence has a convergent subsequence.

Proof.

By Theorem 11.3, every bounded sequence has a monotonic subsequence.But, by Theorem 10.2, this subsequence is convergent.

Subsequential Limits

Definition

A subsequential limit of a sequence {sn}n∈N is any real number or thesymbols ∞,−∞ that is the limit of some subsequence of sn.

Can we think of any examples?

cos (nπ).

(−1)n.

(−1)nn2.

(1 + (−1)n) n.

Subsequential Limits

Definition

A subsequential limit of a sequence {sn}n∈N is any real number or thesymbols ∞,−∞ that is the limit of some subsequence of sn.

Can we think of any examples?

cos (nπ).

(−1)n.

(−1)nn2.

(1 + (−1)n) n.

Theorem 11.7

Theorem

Let {sn}n∈N be a sequence in R. Define S as the set of subsequentiallimits of {sn}n∈N. Then

S 6= φ.

sup S = lim sup sn and inf S = lim inf sn.

limn→∞ sn exists if and only if S has exactly one element, which islimn→∞ sn.

Theorem 11.8

Theorem

Let S denote the set of all subsequential limits of a sequence {sn}n∈N.Suppose {tn}n∈N ⊂ S

⋂R and that t = lim tn. Then t ∈ S .

HW Questions

11.1, 11.2

Theorem 12.1

Theorem

If {sn}n∈N has a positive real limit s, and {tn}n∈N is any sequence of realnumbers, then lim sup sntn = s · lim sup tn.

Theorem 12.1: Proof (part 1)

Proof.

We consider only the case β ≡ lim sup tn <∞. By Corollary 11.4, thereexists a subsequence {tnk}k∈N ⊂ {tn}n∈N such that limk→∞ tnk = β. ByTheorem 11.2, we also have limk→∞ snk = s, and so limk→∞ snk tnk = sβ.Since lim sup sntn is bigger than any subsequential limit of sntn, we haveshown

lim sup sntn ≥ s · lim sup tn. (34)

Theorem 12.1: Proof (part 2)

Proof.

For the reverse inequality, as s > 0, the elements sn are eventually all > 0for all n > N for some N. It follows that 1

sn→ 1

s . Hence

lim sup tn = lim sup1

snsntn

slim sup sntn

and so

lim sup sntn ≤ s · lim sup tn. (36)

Combining the inequalities, we obtain lim sup sntn = s · lim sup tn.

Theorem 12.2

Theorem

If {sn}n∈N⋂{0} = φ, then we have

lim inf | sn+1

sn| ≤ lim inf | sn |

≤ lim sup | sn |1n

≤ lim sup | sn+1

Proof.

See book and exercises!

Corollary 12.3

Corollary

If lim | sn+1

sn| = L ∈ R then we have lim | sn |

1n = L.

Proof.

If lim | sn+1

sn| = L, then all four values in Theorem 12.2 must equal L.

Hence by Theorem 10.7, we have lim | sn |1n = L.

Corollary 12.3

Corollary

If lim | sn+1

sn| = L ∈ R then we have lim | sn |

1n = L.

Proof.

If lim | sn+1

sn| = L, then all four values in Theorem 12.2 must equal L.

Hence by Theorem 10.7, we have lim | sn |1n = L.

Some Definitions

Definition

A series is a sum of elements of a sequence:

(Partial Sum)n∑

ak ≡ am + am+1 + ...+ an

(Infinite Sum)∞∑

ak ≡ limn→∞

n∑k=m

(Absolutely Convergent Sum)∞∑

| ak | ∈ R

Question: Does Convergence ⇒ Absolute Convergence? Or doesAbsolute Convergence ⇒ Convergence? Any counterexamples?

Some Definitions

Definition

(Partial Sum)n∑

ak ≡ am + am+1 + ...+ an

ak ≡ limn→∞

n∑k=m

| ak | ∈ R

Some Definitions

Definition

(Partial Sum)n∑

ak ≡ am + am+1 + ...+ an

ak ≡ limn→∞

n∑k=m

| ak | ∈ R

Example 1

(Partial Sum)n∑

a · rk = a · 1− rn+1

1− r

(Infinite Sum)∞∑k=0

a · rk ≡ limn→∞

a · 1− rn+1

1− r

= a · 1

1− rif | r |< 1

Example 1

(Partial Sum)n∑

a · rk = a · 1− rn+1

1− r

a · 1− rn+1

1− r

= a · 1

1− rif | r |< 1

Example 1

(Partial Sum)n∑

a · rk = a · 1− rn+1

1− r

a · 1− rn+1

1− r

= a · 1

1− rif | r |< 1

Example 2

Define f (p) ≡∑∞

k=11kp . Then

(Condition for convergence) f (p) <∞ if and only if p > 1

(p = 2)∞∑k=1

k2=π2

(p = 4)∞∑k=1

k4=π4

How about for general p?

Example 2

k=11kp . Then

(p = 2)∞∑k=1

k2=π2

(p = 4)∞∑k=1

k4=π4

Example 2

k=11kp . Then

(p = 2)∞∑k=1

k2=π2

(p = 4)∞∑k=1

k4=π4

Cauchy Criterion

A series is said to satisfy the Cauchy Criterion if

∀ε > 0,∃N(ε) such that m, n > ε implies |∑n

k=j ak −∑m

k=j ak |< ε.

This is of course equivalent to |∑n

k=m+1 ak |< ε.

Theorem 14.4

Theorem

A series converges if and only if it satisfies the Cauchy criterion.

Corollary 14.5

Corollary

If a series∑∞

k=0 ak converges then lim an = 0.

Proof.

Follows directly from Cauchy critierion.

Cauchy Criterion

k=j ak −∑m

k=j ak |< ε.

k=m+1 ak |< ε.

Theorem 14.4

Theorem

Corollary 14.5

Corollary

If a series∑∞

Proof.

Cauchy Criterion

k=j ak −∑m

k=j ak |< ε.

k=m+1 ak |< ε.

Theorem 14.4

Theorem

Corollary 14.5

Corollary

If a series∑∞

Proof.

Cauchy Criterion

k=j ak −∑m

k=j ak |< ε.

k=m+1 ak |< ε.

Theorem 14.4

Theorem

Corollary 14.5

Corollary

If a series∑∞

Proof.

Comparison Test: Theorem 14.6, Corollary 14.7

Theorem

Let∑

an be a series where an ≥ 0 for all n. Then

an converges, and | bn |≤ an for all n, then∑

bn converges.

an =∞, and bn ≥ an for all n, then∑

bn =∞ .

Proof.

Now on chalkboard!

Corollary

Absolutely convergent series are convergent.

Proof.

Now on chalkboard!

Theorem

Let∑

bn converges.

bn =∞ .

Proof.

Now on chalkboard!

Corollary

Proof.

Now on chalkboard!

Theorem

Let∑

bn converges.

bn =∞ .

Proof.

Now on chalkboard!

Corollary

Proof.

Now on chalkboard!

Theorem

Let∑

bn converges.

bn =∞ .

Proof.

Now on chalkboard!

Corollary

Proof.

Now on chalkboard!

Theorem

Let∑

bn converges.

bn =∞ .

Proof.

Now on chalkboard!

Corollary

Proof.

Now on chalkboard!

Root Test: Theorem 14.9

Theorem

Let∑

an be a series and let α = lim sup | an |1n . Then

∑an converges absolutely if α < 1.∑an diverges if α > 1.

The test gives no information if α = 1.

Theorem

Let∑

an be a series and let α = lim sup | an |1n . Then∑

an converges absolutely if α < 1.

∑an diverges if α > 1.

Theorem

Let∑

an converges absolutely if α < 1.∑an diverges if α > 1.

Theorem

Let∑

an converges absolutely if α < 1.∑an diverges if α > 1.

Root Test

Proof.

If α < 1, select ε > 0 somall enough so that α + ε < 1. Then, byDefinition 10.6 there is an N ∈ N such that

α− ε < sup{| an |

1n : n > N

}< α + ε (41)

In particular, we have | an |1n< α + ε so

| an |< (α + ε)n (42)

By Comparison, and the fact that α+ ε < 1, we have∑

an converges

If α > 1, then a subsequence of | an |1n has limit α > 1, and so

an 6→ 0. By Corollary 14.5, our series cannot converge.

Ratio Test: Theorem 14.8

Theorem

Let∑

an be a series where an 6= 0 for all n. Then

∑an converges absolutely if lim sup | an+1

an| < 1.∑

an diverges if lim inf | an+1

an| > 1.

If lim inf | an+1

an| ≤ 1 ≤ lim sup | an+1

an| then the test gives no

information.

Theorem

Let∑

an be a series where an 6= 0 for all n. Then∑an converges absolutely if lim sup | an+1

an| < 1.

∑an diverges if lim inf | an+1

an| > 1.

If lim inf | an+1

information.

Theorem

Let∑

an| < 1.∑

an| > 1.

If lim inf | an+1

information.

Theorem

Let∑

an| < 1.∑

an| > 1.

If lim inf | an+1

information.

Ratio Test

Proof.

Let α = lim sup | an |1n . Then, by Theorem 12.2 we have

lim inf | an+1

an| ≤ α ≤ lim sup | an+1

an| (43)

If lim sup | an+1

an| < 1, then α < 1 and the series converges by the

Root Test.

If lim inf | an+1

an| > 1, then α > 1 and the series diverges by the Root

Ratio Test

Proof.

lim inf | an+1

an| (43)

If lim sup | an+1

Root Test.

If lim inf | an+1

Ratio Test

Proof.

lim inf | an+1

an| (43)

If lim sup | an+1

Root Test.

If lim inf | an+1

Examples

∑∞n=0

nn2+3∑∞

n2+1∑∞n=0

n3n∑∞

−3−(−1)n

)n∑∞

n=0 2(−1)n−n∑∞

n=1(−1)n√

Examples

∑∞n=0

)n∑∞n=0

∑∞n=0

1n2+1∑∞

n=0n3n∑∞

−3−(−1)n

)n∑∞

n=0 2(−1)n−n∑∞

n=1(−1)n√

Examples

∑∞n=0

)n∑∞n=0

nn2+3∑∞

∑∞n=0

n3n∑∞

−3−(−1)n

)n∑∞

n=0 2(−1)n−n∑∞

n=1(−1)n√

Examples

∑∞n=0

)n∑∞n=0

nn2+3∑∞

n2+1∑∞n=0

∑∞n=0

−3−(−1)n

)n∑∞

n=0 2(−1)n−n∑∞

n=1(−1)n√

Examples

∑∞n=0

)n∑∞n=0

nn2+3∑∞

n2+1∑∞n=0

n3n∑∞

−3−(−1)n

∑∞n=0 2(−1)

n−n∑∞n=1

(−1)n√n

Examples

∑∞n=0

)n∑∞n=0

nn2+3∑∞

n2+1∑∞n=0

n3n∑∞

−3−(−1)n

)n∑∞

n=0 2(−1)n−n

∑∞n=1

(−1)n√n

Examples

∑∞n=0

)n∑∞n=0

nn2+3∑∞

n2+1∑∞n=0

n3n∑∞

−3−(−1)n

)n∑∞

n=0 2(−1)n−n∑∞

n=1(−1)n√

Alternating Series and Integral Test

Recall the Comparison test. One can extend this thinking to comparisonwith an appropriate integral and Riemann sum. Consider

∫∞1

1xp , for

example.

Also, periodically changing signs may help us in terms of achievingconvergence.

Alternating Series and Integral Test

Recall the Comparison test. One can extend this thinking to comparisonwith an appropriate integral and Riemann sum. Consider

∫∞1

1xp , for

example.

Also, periodically changing signs may help us in terms of achievingconvergence.

Theorem 15.3

Theorem

If a1 ≥ a2 ≥ ... ≥ an ≥ .. ≥ 0 and lim an = 0 then∑

(−1)nan <∞.

Proof: In book.

Theorem 15.3

Theorem

If a1 ≥ a2 ≥ ... ≥ an ≥ .. ≥ 0 and lim an = 0 then∑

(−1)nan <∞.

Proof: In book.

HW Questions

12.1, 12.2, 12.3

12.4, 12.6, 12.8, 12.10, 12.13, 12.14

14.1, 14.2, 14.3, 14.6, 14.7, 14.9

15.1, 15.2, 15.5, 15.7

Definitions

Definition

The domain of a function f is the set on which f is defined, and iswritten dom(f ).

We work with real valued functions. This means f (x) ∈ R for allx ∈ dom(f ).

The natural domain of a function f is the largest subset of R onwhich f is defined.

A function f is continuous at x∗ ∈ dom(f ) if for every sequence{xn} ⊆ dom(f ) with lim xn = x∗, we have lim f (xn) = f (x∗)

If f is continuous at every point x ∈ S ⊆ dom(f ), then f is said to becontinuous on S.

If f is continuous at every point x ∈ dom(f ), then f is said to becontinuous.

Definitions

Definition

Definitions

Definition

Definitions

Definition

Definitions

Definition

Definitions

Definition

Theorem 17.2

Theorem

Let f be a real valued function whose domain is a subset of R. Then f iscontinuous at x∗ ∈ dom(f ) if and only if ∀ε > 0 there exists a δ > 0 suchthat x ∈ dom(f ) and | x − x∗ |< δ imply | f (x)− f (x∗) |< ε

Examples

Assume f (0) = 0 and

f (x) = x2 sin ( 1x ). Prove f is continuous at 0.

f (x) = 1x sin ( 1

x2). Prove f is not continuous at 0.

Examples

Assume f (0) = 0 and

f (x) = x2 sin ( 1x ). Prove f is continuous at 0.

f (x) = 1x sin ( 1

x2). Prove f is not continuous at 0.

Theorem 17.3

Theorem

Let f be a real valued function whose domain is a subset of R. If f iscontinuous at x∗ ∈ dom(f ), then | f | and k · f , k ∈ R, are continuous atx∗.

Theorem 17.4

Theorem

Let f , g be real valued functions that are continuous at x∗ ∈ R. Then

f + g is continuous at x∗.

f · g is continuous at x∗.fg is continuous at x∗.

As an example, max(f , g) = 12(f + g) + 1

2 | f − g | is continuous.

Theorem 17.4

Theorem

f · g is continuous at x∗.

fg is continuous at x∗.

Theorem 17.4

Theorem

Theorem 17.4

Theorem

Theorem 17.5

Theorem

If f is continuous at x∗ ∈ R, and g is continuous at f (x∗) then g(f (x)) iscontinuous at x∗.

Proof.

By definition of continuity, g(yn)→ g(f (x∗)) for all sequences {yn}where yn → f (x∗).

Since f (x) is continuous at x∗, for all sequences {xn} where xn → x∗,it follows that f (xn)→ f (x∗).

Hence, g(f (xn))→ g(f (x∗)) for all sequences {xn} where xn → x∗.

Theorem 17.5

Theorem

If f is continuous at x∗ ∈ R, and g is continuous at f (x∗) then g(f (x)) iscontinuous at x∗.

Proof.

By definition of continuity, g(yn)→ g(f (x∗)) for all sequences {yn}where yn → f (x∗).

Since f (x) is continuous at x∗, for all sequences {xn} where xn → x∗,it follows that f (xn)→ f (x∗).

Hence, g(f (xn))→ g(f (x∗)) for all sequences {xn} where xn → x∗.

Theorem 18.1

Theorem

If f is continuous on an closed interval [a, b], then it is a bounded function.Moreover, f assumes its maximum and minimum values on [a, b].

Proof.

Assume that f is unbounded on [a, b]. Then ∀n ∈ N, ∃xn ∈ [a, b] suchthat | f (xn) |> n

But, by Bolzano-Weirstrass, there exists a subsequence {xnk}∞k=1 such that

xnk → x0 ∈ [a, b]. Since f is continuous at x0, we have by definitionf (xnk )→ f (x0). This contradicts the assumption | f (xn) |> n, and so f isbounded. Proof of attainment of extrema in book!

Theorem 18.1

Theorem

Proof.

Theorem 18.1

Theorem

Proof.

xnk → x0 ∈ [a, b]. Since f is continuous at x0, we have by definitionf (xnk )→ f (x0). This contradicts the assumption | f (xn) |> n, and so f isbounded.

Proof of attainment of extrema in book!

Theorem 18.1

Theorem

Proof.

Theorem 18.2 : Intermediate Value Theorem

Theorem

If f is continuous and real-valued on an interval I , then whenevera, b ∈ I , a < b, and f (a) < y < f (b) then there exists at least onex ∈ (a, b) such that f (x) = y.

Proof.

Without loss of generality, assume f (a) < y < f (b). DefineS = {x ∈ [a, b] : f (x) < y}. Since a ∈ S , S is non-empty.,so x0 = sup Srepresents a number in [a, b]. For each n ∈ N, x0 − 1

n is not an upperbound for S . This implies the existence of a sequence {sn} ⊆ S such thatx0 − 1

n < sn ≤ x0. Thus, lim sn = x0 and since f (sn) < y for all n, we havef (x0) = lim f (sn) ≤ y .

Let tn = min{

b, x0 + 1n

}. Since x0 ≤ tn ≤ x0 + 1

n we have lim tn = x0.Each tn belongs to [a, b] , but not to S , so f (tn) ≥ y for all n. Therefore,f (x0) = lim f (tn) ≥ y . This implies f (x0) = y .

Theorem

Proof.

Let tn = min{

b, x0 + 1n

}. Since x0 ≤ tn ≤ x0 + 1

Theorem

Proof.

Let tn = min{

b, x0 + 1n

}. Since x0 ≤ tn ≤ x0 + 1

Corollary 18.3, Theorem 18.4, and some Examples

Corollary

If f is continuous, real-valued on an interval I , then the setf (I ) = {f (x) : x ∈ I} is also an interval or a single point.

Examples

Let f be a continuous function mapping [0, 1] into [0, 1]. Then f hasa fixed point, i.e. a point x0 ∈ [0, 1] such that f (x0) = x0

If y > 0 and m ∈ N, then y has a positive mth root.

Theorem

Let f be a continuous, strictly increasing function on some interval I .Then f (I ) is an interval J by Corollary 18.3, and f −1 represents a functionwith domain J. The function f −1 is a continuous strictly increasingfunction on J.

Corollary

Examples

Theorem

Corollary

Examples

Theorem

Corollary

Examples

Theorem

Theorems 18.5, 18.6

Theorem

Let g be a strictly increasing function on an interval J such that g(J) isan interval I . Then g is continuous on J.

Theorem

Let f be a one-to-one continuous function on an interval I . Then f isstrictly increasing or strictly decreasing.

Theorems 18.5, 18.6

Theorem

Let g be a strictly increasing function on an interval J such that g(J) isan interval I . Then g is continuous on J.

Theorem

Let f be a one-to-one continuous function on an interval I . Then f isstrictly increasing or strictly decreasing.

Uniform Continuity

Definition 19.1:

Definition

Let f be a real valued function defined on a set S ⊆ R.

Then f is uniformly continuous on S if ∀ε > 0 there exists a δ > 0 suchthat x , y ∈ S and | x − y |< δ imply | f (x)− f (y) |< ε.

We say f is uniformly continuous if it is uniformly continuous on dom(f ).

Example 2: Let f (x) = 1x2

and fix a > 0. Then f is uniformly cts on[a,∞).

Example 4: Let f (x) = x2 and fix a > 0. Then f is uniformly cts on[−a, a].

Uniform Continuity

Definition 19.1:

Definition

Uniform Continuity

Definition 19.1:

Definition

Uniform Continuity

Definition 19.1:

Definition

Uniform Continuity

Definition 19.1:

Definition

Uniform Continuity: Theorem 19.2

Theorem

Let f be continuous function on a closed interval [a, b]. Then f isuniformly continuous on [a, b]

Proof.

For sake of contradiction, assume that f is not uniformly continuous.

n and | f (xn)− f (yn) |≥ ε. By theBolzano- Weirstrass Theorem, a subsequence {xnk} converges to a limitx∗ ∈ [a, b], and ynk → x∗ as well. But, f is cts at x∗ and so

limk→∞

f (xnk ) = f (x∗) = limk→∞

f (ynk )

limk→∞

[f (xnk )− f (ynk )] = 0(44)

This contradicts the assumption above that | f (xnk )− f (ynk ) |≥ ε for allk .

Proof.

For sake of contradiction, assume that f is not uniformly continuous.Then, ∀δ > 0, there exists x , y ∈ [a, b] such that | x − y |< δ and| f (x)− f (y) |≥ ε.

It follows that for each n ∈ N, there existsxn, yn ∈ [a, b] such that | xn − yn |< 1

limk→∞

f (xnk ) = f (x∗) = limk→∞

f (ynk )

limk→∞

[f (xnk )− f (ynk )] = 0(44)

Proof.

n and | f (xn)− f (yn) |≥ ε. By theBolzano- Weirstrass Theorem, a subsequence {xnk} converges to a limitx∗ ∈ [a, b], and ynk → x∗ as well.

But, f is cts at x∗ and so

limk→∞

f (xnk ) = f (x∗) = limk→∞

f (ynk )

limk→∞

[f (xnk )− f (ynk )] = 0(44)

Proof.

limk→∞

f (xnk ) = f (x∗) = limk→∞

f (ynk )

limk→∞

[f (xnk )− f (ynk )] = 0(44)

Theorem 19.4

Theorem

Let f be uniformly continuous on a set S and {sn} a Cauchy sequence inS. Then {f (sn)} is a Cauchy sequence.

Proof.

Let {sn} a Cauchy sequence in S and let ε > 0. Since f is uniformlycontinuous on S , there exists δ > 0 where x , y ∈ S and | x − y |< δ imply| f (x)− f (y) |< ε.

Since {sn} is a Cauchy sequence, there exists N suchthat m, n > N implies | sn − sm |< δ. But by definition of UniCts, we havem, n > N implies | f (sn)− f (sm) |< ε. Hence, f (sn) is Cauchy.

Theorem 19.4

Theorem

Let f be uniformly continuous on a set S and {sn} a Cauchy sequence inS. Then {f (sn)} is a Cauchy sequence.

Proof.

Let {sn} a Cauchy sequence in S and let ε > 0. Since f is uniformlycontinuous on S , there exists δ > 0 where x , y ∈ S and | x − y |< δ imply| f (x)− f (y) |< ε. Since {sn} is a Cauchy sequence, there exists N suchthat m, n > N implies | sn − sm |< δ. But by definition of UniCts, we havem, n > N implies | f (sn)− f (sm) |< ε. Hence, f (sn) is Cauchy.

Theorem 19.4: Example

Example 6: f (x) = 1x2

is not uniformly continuous.

To show this, take the Cauchy sequence

{sn}∞n=1 :={

}∞n=1⊂ (0, 1).

Then f (sn) = (n + 1)2 and is not Cauchy.

Hence, f is not uniformly continuous on (0, 1)

Function Extensions

If f is a real valued function and f̃ is defined such that

dom(f ) ⊂ dom(f̃ )

f̃ (x) = f (x) on dom(f )

then f̃ is an extension of f .

Example 7: Extend f (x) = x sin ( 1x )

Example 8: Extend g(x) = sin ( 1x )

Function Extensions

If f is a real valued function and f̃ is defined such that

dom(f ) ⊂ dom(f̃ )

f̃ (x) = f (x) on dom(f )

then f̃ is an extension of f .

Example 7: Extend f (x) = x sin ( 1x )

Example 8: Extend g(x) = sin ( 1x )

Theorems 19.5, 19.6

Theorem

A real-valued function f on (a, b) is uniformly continuous on (a, b) if andonly if it can be extended to a continuous function f̃ on [a, b].

Theorem

Let f be a real-valued continuous function on an interval I . Let I o be theinterval obtained by removing from I any endpoints that happen to be inI . If f is differentiable on I o and if f ′ is bounded on I o , then f is uniformlycontinuous on I .

Theorems 19.5, 19.6

Theorem

A real-valued function f on (a, b) is uniformly continuous on (a, b) if andonly if it can be extended to a continuous function f̃ on [a, b].

Theorem

Let f be a real-valued continuous function on an interval I . Let I o be theinterval obtained by removing from I any endpoints that happen to be inI . If f is differentiable on I o and if f ′ is bounded on I o , then f is uniformlycontinuous on I .

Power Series

Given a sequence of real numbers {an}∞n=0, the series∑∞

n=0 anxn is calleda Power Series. It turns out that one of the following holds for the powerseries:

the power series converges for all x ∈ Rthe power series converges only for x = 0

the power series converges only for values of x ∈ I , where I is abounded interval centered at 0.

Power Series

the power series converges for all x ∈ R

the power series converges only for x = 0

Power Series

Theorem 23.1

Theorem

For the power series∑∞

n=0 anxn, let

β = lim sup | an |1n

R = 0 if β =∞R =∞ if β = 0

the power series converges for | x |< R

the power series diverges for | x |> R

Theorem 23.1

Theorem

n=0 anxn, let

R = 0 if β =∞R =∞ if β = 0

Theorem 23.1

Theorem

n=0 anxn, let

R = 0 if β =∞R =∞ if β = 0

Examples

Consider:

Example 1 :∑∞

n=01n!x

Example 2 :∑∞

n=0 xn

Example 3 :∑∞

n=01nxn

Example 4 :∑∞

n=01n2

Example 5 :∑∞

n=0 n!xn

Example 6 :∑∞

n=0 2−nx3n

Example 7 :∑∞

n=0(−1)n+1

n (x − 1)n

Examples

Consider:

Example 1 :∑∞

n=01n!x

Example 2 :∑∞

n=0 xn

Example 3 :∑∞

n=01nxn

Example 4 :∑∞

n=01n2

Example 5 :∑∞

n=0 n!xn

Example 6 :∑∞

n=0 2−nx3n

Example 7 :∑∞

n=0(−1)n+1

n (x − 1)n

Examples

Consider:

Example 1 :∑∞

n=01n!x

Example 2 :∑∞

n=0 xn

Example 3 :∑∞

n=01nxn

Example 4 :∑∞

n=01n2

Example 5 :∑∞

n=0 n!xn

Example 6 :∑∞

n=0 2−nx3n

Example 7 :∑∞

n=0(−1)n+1

n (x − 1)n

Examples

Consider:

Example 1 :∑∞

n=01n!x

Example 2 :∑∞

n=0 xn

Example 3 :∑∞

n=01nxn

Example 4 :∑∞

n=01n2

Example 5 :∑∞

n=0 n!xn

Example 6 :∑∞

n=0 2−nx3n

Example 7 :∑∞

n=0(−1)n+1

n (x − 1)n

Examples

Consider:

Example 1 :∑∞

n=01n!x

Example 2 :∑∞

n=0 xn

Example 3 :∑∞

n=01nxn

Example 4 :∑∞

n=01n2

Example 5 :∑∞

n=0 n!xn

Example 6 :∑∞

n=0 2−nx3n

Example 7 :∑∞

n=0(−1)n+1

n (x − 1)n

Examples

Consider:

Example 1 :∑∞

n=01n!x

Example 2 :∑∞

n=0 xn

Example 3 :∑∞

n=01nxn

Example 4 :∑∞

n=01n2

Example 5 :∑∞

n=0 n!xn

Example 6 :∑∞

n=0 2−nx3n

Example 7 :∑∞

n=0(−1)n+1

n (x − 1)n

Examples

Consider:

Example 1 :∑∞

n=01n!x

Example 2 :∑∞

n=0 xn

Example 3 :∑∞

n=01nxn

Example 4 :∑∞

n=01n2

Example 5 :∑∞

n=0 n!xn

Example 6 :∑∞

n=0 2−nx3n

Example 7 :∑∞

n=0(−1)n+1

n (x − 1)n

Pointwise Convergence

Definition 24.1: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.

Then fn converges pointwise on S to a function f if lim fn(x) = f (x) for allx ∈ S .

Example 2: Let fn(x) = xn for x ∈ [0, 1]. Then fn(x)→ f (x) ≡ 1{x=1}.

Example 4: Let fn(x) = 1n sin (nx) for x ∈ R. Then fn(x)→ 0 pointwise

Uniform Convergence

Then fn converges uniformly on S to a function f if ∀ε > 0 there exists anumber N such that | fn(x)− f (x) |< ε for all x ∈ S and all n > N.

Example 3: Let fn(x) = (1− | x |)n for x ∈ (−1, 1). Thenfn(x)→ f (x) ≡ 1{x=0} pointwise, but not uniformly.

Example 4: Let fn(x) = 1n sin (nx) for x ∈ R. Then fn(x)→ 0 uniformly

Example 5: Let fn(x) = nxn for x ∈ [0, 1). Then fn(x)→ 0 pointwise on[0, 1), but again not uniformly.

Uniform Convergence

Uniform Convergence: Theorem 24.3

Theorem

The uniform limit of a sequence of continuous function is continuous.

Proof.

(Sketch)By the Triangle Inequality,

| f (x)− f (x∗) | ≤| f (x)− fn(x) | + | fn(x)− fn(x∗) | + | fn(x∗)− f (x∗) |(46)

Uniform continuity forces the first term ahead of the inequality to be < ε3

for all n ≥ N(ε3

)+ 1. And, for all x such that | x − x∗ |< δ

), we have

the second term < ε3 . The third term follows from pointwise convergence.

Uniform Convergence: Theorem 24.3

Theorem

The uniform limit of a sequence of continuous function is continuous.

Proof.

(Sketch)By the Triangle Inequality,

| f (x)− f (x∗) | ≤| f (x)− fn(x) | + | fn(x)− fn(x∗) | + | fn(x∗)− f (x∗) |(46)

Uniform continuity forces the first term ahead of the inequality to be < ε3

for all n ≥ N(ε3

)+ 1. And, for all x such that | x − x∗ |< δ

), we have

the second term < ε3 . The third term follows from pointwise convergence.

Uniform Convergence

Remark 24.4: fn(x) converges uniformly on S if and only iflimn→∞[sup {| fn(x)− f (x) |: x ∈ S}] = 0

Example 7: Let fn(x) = x1+nx2

for x ∈ R. Use the above remark to showif the sequence converges uniformly on R or not.

Proof.

Notice that f ′n(x) = 1−nx2(1+nx2)

. It follows that fn attains its max at xn = 1√n

with value fn(xn) = 12√n

.It follows that

limn→∞[sup {| fn(x)− f (x) |: x ∈ S}] = limn→∞[ 12√n

] = 0.

Example 8: Let fn(x) = xn(1− x) for x ∈ [0, 1]. Use the above remark toshow if the sequence converges uniformly on [0, 1] or not.

Proof.

fn attains its max at xn = 1n+1 with value fn(xn) = n

(n+1)n . It follows that

limn→∞[sup {| fn(x)− f (x) |: x ∈ S}] = limn→∞n

(n+1)n+1 = 0.

Uniform Convergence

Remark 24.4: fn(x) converges uniformly on S if and only iflimn→∞[sup {| fn(x)− f (x) |: x ∈ S}] = 0

Example 7: Let fn(x) = x1+nx2

for x ∈ R. Use the above remark to showif the sequence converges uniformly on R or not.

Proof.

Notice that f ′n(x) = 1−nx2(1+nx2)

. It follows that fn attains its max at xn = 1√n

with value fn(xn) = 12√n

.It follows that

limn→∞[sup {| fn(x)− f (x) |: x ∈ S}] = limn→∞[ 12√n

] = 0.

Example 8: Let fn(x) = xn(1− x) for x ∈ [0, 1]. Use the above remark toshow if the sequence converges uniformly on [0, 1] or not.

Proof.

fn attains its max at xn = 1n+1 with value fn(xn) = n

(n+1)n . It follows that

limn→∞[sup {| fn(x)− f (x) |: x ∈ S}] = limn→∞n

(n+1)n+1 = 0.

Uniform Convergence

Then fn is uniformly Cauchy on S if ∀ε > 0 there exists a number N suchthat | fn(x)− fm(x) |< ε for all x ∈ S and all m, n > N.

Theorem 25.4: Let fn be a sequence of uniformly Cauchy functions on aset S ⊆ R. Then there exists a function f on S such that fn → funiformly on S

Theorem 25.5: Let∑∞

k=0 gk be a series of functions on a set S ⊆ R.Suppose that each gk is continuous on S and that the series convergesuniformly on S . Then the series is a continuous function on S .

k=0 gk be a series of functions that satisfies theCauchy criterion uniformly on a set S ⊆ R. Then the series convergesuniformly on S .

Uniform Convergence

Weirstrass M-test

Theorem 25.4: Let Mk be a sequence of nonnegative real numbers where∑∞k=0 Mk <∞. If | gk |≤ Mk for all x ∈ S , then

∑∞k=0 gk converges

uniformly on S .

Proof To verify the Cauchy criterion on S , let ε > 0. Since the series∑Mk converges, it satisfies the Cauchy criterion.So there exists a number

N such that n ≥ m > N implies |∑n

k=m Mk |< ε. Hence, if n ≥ m > Nand x ∈ S , then

gk(x) |≤n∑

| gk(x) |≤n∑

Mk < ε. (47)

It follows that the series∑

gk satisfies the Cauchy criterion uniformly onS and the Theorem 25.6 shows that it converges uniformly on S .

Weirstrass M-test

uniformly on S .

Proof To verify the Cauchy criterion on S , let ε > 0. Since the series∑Mk converges, it satisfies the Cauchy criterion.

So there exists a numberN such that n ≥ m > N implies |

∑nk=m Mk |< ε. Hence, if n ≥ m > N

and x ∈ S , then

gk(x) |≤n∑

| gk(x) |≤n∑

Mk < ε. (47)

Weirstrass M-test

uniformly on S .

k=m Mk |< ε.

Hence, if n ≥ m > Nand x ∈ S , then

gk(x) |≤n∑

| gk(x) |≤n∑

Mk < ε. (47)

Weirstrass M-test

uniformly on S .

gk(x) |≤n∑

| gk(x) |≤n∑

Mk < ε. (47)

Weirstrass M-test

uniformly on S .

gk(x) |≤n∑

| gk(x) |≤n∑

Mk < ε. (47)

Differentiation and Integration of Power Series

The following are nice applications of the Weirstrass M-Test:

n=0 anxn be a power series with radius ofconvergence R. If 0 < R1 < R, then the power series converges uniformlyon [−R1,R1] to a continuous function.

Corollary 26.2: The power series∑∞

n=0 anxn converges to a continuousfunction on the open interval (−R,R).

The following are nice applications of the Weirstrass M-Test:

n=0 anxn be a power series with radius ofconvergence R. If 0 < R1 < R, then the power series converges uniformlyon [−R1,R1] to a continuous function.

Corollary 26.2: The power series∑∞

n=0 anxn converges to a continuousfunction on the open interval (−R,R).

Lemma 26.3: If the power series∑∞

n=0 anxn has radius of convergence R,then the related power series

∑∞n=0 nanxn−1 and

∑∞n=0

ann+1xn+1 also have

radius of convergence R.

Proof First observe that∑∞

n=0 nanxn−1 and∑∞

n=0 nanxn have the sameradius of convergence, as do the pair

∑∞n=0

ann+1xn+1 and

∑∞n=0

ann+1xn.

Next recall that R = 1β and β = lim sup | an |

1n . For the series

∑∞n=0 nanxn

we consider lim sup n | an |1n = lim sup n

1n | an |

1n . Since lim n

1n = 1, it

follows that lim sup n | an |1n = β. Hence the series

∑nanxn has a radius

of convergence R. Similar work goes into proving the result for∑∞n=0

ann+1xn+1

∑∞n=0

ann+1xn+1 also have

∑∞n=0

ann+1xn+1 and

∑∞n=0

ann+1xn.

1n . For the series

∑∞n=0 nanxn

1n | an |

1n . Since lim n

1n = 1, it

ann+1xn+1

∑∞n=0

ann+1xn+1 also have

∑∞n=0

ann+1xn+1 and

∑∞n=0

ann+1xn.

1n . For the series

∑∞n=0 nanxn

1n | an |

1n . Since lim n

1n = 1, it

ann+1xn+1

∑∞n=0

ann+1xn+1 also have

∑∞n=0

ann+1xn+1 and

∑∞n=0

ann+1xn.

1n . For the series

∑∞n=0 nanxn

1n | an |

1n . Since lim n

1n = 1, it

ann+1xn+1

Theorem 25.2: Let fn be a sequence of continuous functions on [a, b],and suppose that fn → f uniformly on [a, b]. Then

limn→∞

afn(x)dx =

af (x)dx (48)

A useful application of this theorem for power series is

Theorem 26.4: Suppose that f (x) =∑∞

0 anxn has a radius ofconvergence R > 0. Then, for | x |< R we have∫ x

0f (t)dt =

∞∑n=0

ann + 1

xn+1 (49)

limn→∞

afn(x)dx =

af (x)dx (48)

0f (t)dt =

∞∑n=0

ann + 1

xn+1 (49)

limn→∞

afn(x)dx =

af (x)dx (48)

0f (t)dt =

∞∑n=0

ann + 1

xn+1 (49)

0 anxn has a radius ofconvergence R > 0. Then, for | x |< R we have f is differentiable and

f ′(x) =∞∑n=1

nanxn−1 (50)

ProofBegin with g(x) =

∑∞n=1 nanxn−1, which Lemma 26.3 guarantees is

convergent for | x |< R. Theorem 26.4 shows that we can integrate g termby term ∫ x

0g(t)dt =

∞∑n=1

anxn = f (x)− a0 for | x |< R. (51)

f ′(x) =∞∑n=1

nanxn−1 (50)

convergent for | x |< R. Theorem 26.4 shows that we can integrate g termby term

0g(t)dt =

∞∑n=1

anxn = f (x)− a0 for | x |< R. (51)

f ′(x) =∞∑n=1

nanxn−1 (50)

convergent for | x |< R. Theorem 26.4 shows that we can integrate g termby term ∫ x

0g(t)dt =

∞∑n=1

anxn = f (x)− a0 for | x |< R. (51)

So, if 0 < R1 < R then

f (x) =

g(t)dt + k for | x |< R1. (52)

where k is a constant. By the continuity of g and the FTC, we know thatf is differentiable and that f ′(x) = g(x)QED

Example 1 Recall∑∞

n=0 xn = 11−x for | x |< 1. Differentiating, and

integrating, term by term, we obtain for | x |< 1,

∞∑n=1

nxn−1 =1

(1− x)2

∞∑n=1

n + 1xn+1 =

1− tdt = − ln (1− x)

f (x) =

g(t)dt + k for | x |< R1. (52)

where k is a constant. By the continuity of g and the FTC, we know thatf is differentiable and that f ′(x) = g(x)

∞∑n=1

nxn−1 =1

(1− x)2

∞∑n=1

n + 1xn+1 =

1− tdt = − ln (1− x)

f (x) =

g(t)dt + k for | x |< R1. (52)

∞∑n=1

nxn−1 =1

(1− x)2

∞∑n=1

n + 1xn+1 =

1− tdt = − ln (1− x)

f (x) =

g(t)dt + k for | x |< R1. (52)

∞∑n=1

nxn−1 =1

(1− x)2

∞∑n=1

n + 1xn+1 =

1− tdt = − ln (1− x)

elementary analysis: the theory of calculususers.math.msu.edu/users/albert/math320.pdf ·...

Documents