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Elementary Analysis: The Theory of Calculususers.math.msu.edu/users/albert/math320.pdf ·...
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Elementary Analysis: The Theory of Calculus
Albert Cohen
C336 Wells HallDepartment of MathematicsDepartment of StatisticsMichigan State University
East Lansing MI48823
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 1 / 124
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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
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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
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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?
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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?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 4 / 124
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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?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 4 / 124
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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
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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
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 5 / 124
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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
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 5 / 124
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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)
=1
2(n(n + 1) + 2(n + 1))
=1
2(n + 1)((n + 1) + 1)
=1
2(n + 1)(n + 2)
(1)
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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
5 and
7n+1 − 2n+1 = 7n+1 − 7 · 2n + 7 · 2n − 2n+1
= 7 · (7n − 2n) + 5 · 2n
= 5 · (7m + 2n)
(2)
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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) |
(3)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 8 / 124
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HW Questions
1.1, 1.3
1.6, 1.7
1.8a, b
1.12a, b, c
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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}
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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
)3= 5
which implies
√2 + 5
13 is a solution of x6 − 6x4 + 12x2 − 13 = 0
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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.
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Rational Zeros Theorem
Proof: By given assumptions,
an
(p
q
)n
+ an−1
(p
q
)n−1+ ...+ a1
(p
q
)1
+ a0 = 0 (5)
Multiplying through by qn, this equation can be rewritten
n∑k=0
an−kpn−kqk = 0 (6)
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Rational Zeros Theorem
Proof (continued)It follows that
anpn = −q ·n∑
k=1
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
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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}.
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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}.
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HW Questions
2.1, 2.3
2.5, 2.6
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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
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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.
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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
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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
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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 |
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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)
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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}
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The Completeness Axiom
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}
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The Completeness Axiom
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}
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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.
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HW Questions
4.1a, i , j , k , r , u,w
4.5
4.6a, b
4.7a, b
4.10, 4.15, 4.16
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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 = −∞
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HW Questions
5.1a, b, c , d
5.3
5.4
5.6
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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, ...)
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Examples
{sk}∞k=1 : sn =1
n2
{ak}∞k=0 : an = (−1)n
{ak}∞k=1 : an = cos(nπ
3
){ak}∞k=0 : an = (n)
1n
{bk}∞k=0 : bn =
(1 +
1
n
)n
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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.
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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?
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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?
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HW Questions
7.1, 7.3a− l
7.4, 7.5
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Scratchwork versus Complete Proof
Theorem: limn→∞1n2
= 0.
Discussion: If we want | 1n2− 0 |< ε then we need n > 1√
ε.
Formal Proof: Let ε > 0. Let N = 1√ε. Then n > N ⇒ n2 > 1
ε and so1n2< ε. This implies | 1
n2− 0 |< ε. �
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Scratchwork versus Complete Proof
Theorem: limn→∞1n2
= 0.
Discussion: If we want | 1n2− 0 |< ε then we need n > 1√
ε.
Formal Proof: Let ε > 0. Let N = 1√ε. Then n > N ⇒ n2 > 1
ε and so1n2< ε. This implies | 1
n2− 0 |< ε. �
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Scratchwork versus Complete Proof
Theorem: limn→∞1n2
= 0.
Discussion: If we want | 1n2− 0 |< ε then we need n > 1√
ε.
Formal Proof: Let ε > 0. Let N = 1√ε. Then n > N ⇒ n2 > 1
ε and so1n2< ε. This implies | 1
n2− 0 |< ε. �
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Scratchwork versus Complete Proof
Theorem: limn→∞3n+17n−4 = 3
7 .
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. �
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Scratchwork versus Complete Proof
Theorem: limn→∞3n+17n−4 = 3
7 .
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. �
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Scratchwork versus Complete Proof
Theorem: limn→∞3n+17n−4 = 3
7 .
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. �
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Scratchwork versus Complete Proof
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 |
<1
2+
1
2= 1
(12)
which is a contradiction.
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Scratchwork versus Complete Proof
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 |
<1
2+
1
2= 1
(12)
which is a contradiction.
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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→∞
√sn
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 +√
s
≤ | sn − s |√s
<
√sε√s
= ε.
(13)
This argument can of course be extended to s = 0, and is left to you!
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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→∞
√sn
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 +√
s
≤ | sn − s |√s
<
√sε√s
= ε.
(13)
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
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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.
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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.
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HW Questions
8.1a, c, 8.2a, c , d
8.3
8.4, 8.5, 8.6
8.6, 8.7
8.9a, 8.10
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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.
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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.
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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.
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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.
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Some Useful Limit Theorems
Theorems 9.3, 9.4:
Theorem
If sn → s and tn → t then
sn + tn → s + t
sntn → st
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Some Useful Limit Theorems
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
2M
(16)
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.
(17)
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Some Useful Limit Theorems
Lemma 9.5:
Lemma
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
s
∣∣∣ =∣∣∣sn − s
sns
∣∣∣ ≤ ∣∣∣sn − s
Xs
∣∣∣ < ε | X | s
| X | s= ε. (18)
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Some Useful Limit Theorems
Lemma 9.5:
Lemma
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
s
∣∣∣ =∣∣∣sn − s
sns
∣∣∣ ≤ ∣∣∣sn − s
Xs
∣∣∣ < ε | X | s
| X | s= ε. (18)
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Some Useful Limit Theorems
Theorem 9.6:
Theorem
If sn → s and tn → t, where tn, t 6= 0 for all n then sntn→ s
t
Proof.
Try now!
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Some Useful Limit Theorems
Theorem 9.6:
Theorem
If sn → s and tn → t, where tn, t 6= 0 for all n then sntn→ s
t
Proof.
Try now!
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More Examples
Example 1 : n3+6n2+74n3+3n−4 →
14
Example 2 : n−5n2+7
→ 0
Prove these now!
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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!
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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!
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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 >
Mm .
Take N = max {N1,N2}. Then n > N ⇒ sntn >Mm m = M.
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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 >
Mm .
Take N = max {N1,N2}. Then n > N ⇒ sntn >Mm m = M.
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Theorem 9.10:
Theorem
If {sn} ⊂ R+ then sn →∞ if and only if 1sn→ 0
Proof.
Try now!
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Theorem 9.10:
Theorem
If {sn} ⊂ R+ then sn →∞ if and only if 1sn→ 0
Proof.
Try now!
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HW Questions
9.1a, c, 9.2a
9.3
9.4, 9.5, 9.6
9.9a, 9.10, 9.18
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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.
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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.
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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.
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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−∞.
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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−∞.
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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−∞.
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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
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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
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Lemma 10.9
Lemma
Convergent sequences are Cauchy sequences.
Proof.
Suppose that s = limn→∞ sn exists. Then for all n,m > N(ε2
)it follows
that
| sn − sm | =| sn − s + s − sm |≤| sn − s | + | sm − s |
<ε
2+ε
2
(20)
Thus, sn is a Cauchy sequence.
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Lemma 10.9
Lemma
Convergent sequences are Cauchy sequences.
Proof.
Suppose that s = limn→∞ sn exists. Then for all n,m > N(ε2
)it follows
that
| sn − sm | =| sn − s + s − sm |≤| sn − s | + | sm − s |
<ε
2+ε
2
(20)
Thus, sn is a Cauchy sequence.
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Lemma 10.10
Lemma
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) |
}(21)
Then | sn |≤ M for all n.
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Lemma 10.10
Lemma
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) |
}(21)
Then | sn |≤ M for all n.
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Theorem 10.11
Theorem
A sequence is convergent if and only if it is a Cauchy sequence.
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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.
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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.
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 60 / 124
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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.
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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.
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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.
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Babylonian Method of Computing Square Roots
Consider a positive real number A > 1. Now, define
sn+1 = f (sn)
s0 = A
f (x) =1
2
(x +
A
x
) (23)
Since f (A) = A+12 < A and ∀x > 0,
d
dx(f (x)− x) = −1
2
(1 +
A
x2
)< 0, (24)
we know that {sn}∞n=0 is a bounded (by 0) monotonic decreasing sequence.
Furthermore, f (√
A) =√
A.
Have we proven convergence of this recursively defined sequence?
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Babylonian Method of Computing Square Roots
Consider a positive real number A > 1. Now, define
sn+1 = f (sn)
s0 = A
f (x) =1
2
(x +
A
x
) (23)
Since f (A) = A+12 < A and ∀x > 0,
d
dx(f (x)− x) = −1
2
(1 +
A
x2
)< 0, (24)
we know that {sn}∞n=0 is a bounded (by 0) monotonic decreasing sequence.
Furthermore, f (√
A) =√
A.
Have we proven convergence of this recursively defined sequence?
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Babylonian Method of Computing Square Roots
Consider a positive real number A > 1. Now, define
sn+1 = f (sn)
s0 = A
f (x) =1
2
(x +
A
x
) (23)
Since f (A) = A+12 < A and ∀x > 0,
d
dx(f (x)− x) = −1
2
(1 +
A
x2
)< 0, (24)
we know that {sn}∞n=0 is a bounded (by 0) monotonic decreasing sequence.
Furthermore, f (√
A) =√
A.
Have we proven convergence of this recursively defined sequence?
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Babylonian Method of Computing Square Roots
Consider a positive real number A > 1. Now, define
sn+1 = f (sn)
s0 = A
f (x) =1
2
(x +
A
x
) (23)
Since f (A) = A+12 < A and ∀x > 0,
d
dx(f (x)− x) = −1
2
(1 +
A
x2
)< 0, (24)
we know that {sn}∞n=0 is a bounded (by 0) monotonic decreasing sequence.
Furthermore, f (√
A) =√
A.
Have we proven convergence of this recursively defined sequence?
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Babylonian Method of Computing Square Roots
Consider a positive real number A > 1. Now, define
sn+1 = f (sn)
s0 = A
f (x) =1
2
(x +
A
x
) (23)
Since f (A) = A+12 < A and ∀x > 0,
d
dx(f (x)− x) = −1
2
(1 +
A
x2
)< 0, (24)
we know that {sn}∞n=0 is a bounded (by 0) monotonic decreasing sequence.
Furthermore, f (√
A) =√
A.
Have we proven convergence of this recursively defined sequence?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 61 / 124
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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
Nk+1
)=
[1− pl pf
pl 1− pf
](Wk
Nk
)(25)
Finally, assume that Wk + Nk = 1.
Question: Does Wk →W for some W ∈ (0, 1)?
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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
Nk+1
)=
[1− pl pf
pl 1− pf
](Wk
Nk
)(25)
Finally, assume that Wk + Nk = 1.
Question: Does Wk →W for some W ∈ (0, 1)?
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Markov Chain Model of Employment
It follows that the matrix can be diagonalized as
(1− pl pf
pl 1− pf
)=
(pfpl−1
1 1
)(1 00 1− pf − pl
)( plpf +pl
plpf +pl
− plpf +pl
pfpf +pl
)(26)
Define
~qk =
(pl
pf +pl
plpf +pl
− plpf +pl
pfpf +pl
)(Wk
Nk
)(27)
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Markov Chain Model of Employment
It follows that the matrix can be diagonalized as
(1− pl pf
pl 1− pf
)=
(pfpl−1
1 1
)(1 00 1− pf − pl
)( plpf +pl
plpf +pl
− plpf +pl
pfpf +pl
)(26)
Define
~qk =
(pl
pf +pl
plpf +pl
− plpf +pl
pfpf +pl
)(Wk
Nk
)(27)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 63 / 124
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Markov Chain Model of Employment
Hence,
~qk+1 =
(1 00 1− pf − pl
)~qk ⇒ ~qk =
(A
B(1− pf − pl)k
)(28)
Returning to our original notation,(Wk
Nk
)=
(pfpl−1
1 1
)(1 00 1− pf − pl
)~qk
=
(pfpl−1
1 1
)(1 00 1− pf − pl
)(A
B(1− pf − pl)k
)=
(pfpl−1
1 1
)(A
B(1− pf − pl)k+1
)=
(Apf
pl− B(1− pf − pl)
k+1
A + B(1− pf − pl)k+1
)(29)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 64 / 124
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Markov Chain Model of Employment
Hence,
~qk+1 =
(1 00 1− pf − pl
)~qk ⇒ ~qk =
(A
B(1− pf − pl)k
)(28)
Returning to our original notation,(Wk
Nk
)=
(pfpl−1
1 1
)(1 00 1− pf − pl
)~qk
=
(pfpl−1
1 1
)(1 00 1− pf − pl
)(A
B(1− pf − pl)k
)=
(pfpl−1
1 1
)(A
B(1− pf − pl)k+1
)=
(Apf
pl− B(1− pf − pl)
k+1
A + B(1− pf − pl)k+1
)(29)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 64 / 124
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Markov Chain Model of Employment
Solving for our parameters A,B, we see that(W0
N0
)=
(Apf
pl− B(1− pf − pl)
A + B(1− pf − pl)
)
⇒(
AB
)=
11+
pfplpf
plN0−W0
1−pf−pl
∴
(Wk
Nk
)→
pfpl
1+pfpl
11+
pfpl
=
(pf
pl+pfpl
pl+pf
) (30)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 65 / 124
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Markov Chain Model of Employment
Solving for our parameters A,B, we see that(W0
N0
)=
(Apf
pl− B(1− pf − pl)
A + B(1− pf − pl)
)
⇒(
AB
)=
11+
pfplpf
plN0−W0
1−pf−pl
∴
(Wk
Nk
)→
pfpl
1+pfpl
11+
pfpl
=
(pf
pl+pfpl
pl+pf
) (30)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 65 / 124
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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?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 66 / 124
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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?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 66 / 124
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HW Questions
10.1, 10.2
10.6
10.8, 10.9
10.10, 10.11
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 67 / 124
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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 < ...
(32)
As an example, consider
sn = n2(−1)n
nk = 2k
tk = snk
= (2k)2(−1)2k
= 4k2
(33)
Check out more examples in book!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 68 / 124
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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 < ...
(32)
As an example, consider
sn = n2(−1)n
nk = 2k
tk = snk
= (2k)2(−1)2k
= 4k2
(33)
Check out more examples in book!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 68 / 124
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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 < ...
(32)
As an example, consider
sn = n2(−1)n
nk = 2k
tk = snk
= (2k)2(−1)2k
= 4k2
(33)
Check out more examples in book!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 68 / 124
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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?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 69 / 124
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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?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 69 / 124
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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?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 69 / 124
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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!
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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!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 70 / 124
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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!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 70 / 124
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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!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 70 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 71 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 71 / 124
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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?
Try
cos (nπ).
(−1)n.
(−1)nn2.
(1 + (−1)n) n.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 72 / 124
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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?
Try
cos (nπ).
(−1)n.
(−1)nn2.
(1 + (−1)n) n.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 72 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 73 / 124
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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 .
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 74 / 124
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HW Questions
11.1, 11.2
11.4
11.10
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 75 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 76 / 124
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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)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 77 / 124
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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
≥ 1
slim sup sntn
(35)
and so
lim sup sntn ≤ s · lim sup tn. (36)
Combining the inequalities, we obtain lim sup sntn = s · lim sup tn.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 78 / 124
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Theorem 12.2
Theorem
If {sn}n∈N⋂{0} = φ, then we have
lim inf | sn+1
sn| ≤ lim inf | sn |
1n
≤ lim sup | sn |1n
≤ lim sup | sn+1
sn|
(37)
Proof.
See book and exercises!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 79 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 80 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 80 / 124
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Some Definitions
Definition
A series is a sum of elements of a sequence:
(Partial Sum)n∑
k=m
ak ≡ am + am+1 + ...+ an
(Infinite Sum)∞∑
k=m
ak ≡ limn→∞
n∑k=m
ak
(Absolutely Convergent Sum)∞∑
k=m
| ak | ∈ R
(38)
Question: Does Convergence ⇒ Absolute Convergence? Or doesAbsolute Convergence ⇒ Convergence? Any counterexamples?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 81 / 124
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Some Definitions
Definition
A series is a sum of elements of a sequence:
(Partial Sum)n∑
k=m
ak ≡ am + am+1 + ...+ an
(Infinite Sum)∞∑
k=m
ak ≡ limn→∞
n∑k=m
ak
(Absolutely Convergent Sum)∞∑
k=m
| ak | ∈ R
(38)
Question: Does Convergence ⇒ Absolute Convergence? Or doesAbsolute Convergence ⇒ Convergence? Any counterexamples?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 81 / 124
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Some Definitions
Definition
A series is a sum of elements of a sequence:
(Partial Sum)n∑
k=m
ak ≡ am + am+1 + ...+ an
(Infinite Sum)∞∑
k=m
ak ≡ limn→∞
n∑k=m
ak
(Absolutely Convergent Sum)∞∑
k=m
| ak | ∈ R
(38)
Question: Does Convergence ⇒ Absolute Convergence? Or doesAbsolute Convergence ⇒ Convergence? Any counterexamples?
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Example 1
(Partial Sum)n∑
k=0
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
(39)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 82 / 124
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Example 1
(Partial Sum)n∑
k=0
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
(39)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 82 / 124
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Example 1
(Partial Sum)n∑
k=0
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
(39)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 82 / 124
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Example 2
Define f (p) ≡∑∞
k=11kp . Then
(Condition for convergence) f (p) <∞ if and only if p > 1
(p = 2)∞∑k=1
1
k2=π2
6
(p = 4)∞∑k=1
1
k4=π4
90
(40)
How about for general p?
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Example 2
Define f (p) ≡∑∞
k=11kp . Then
(Condition for convergence) f (p) <∞ if and only if p > 1
(p = 2)∞∑k=1
1
k2=π2
6
(p = 4)∞∑k=1
1
k4=π4
90
(40)
How about for general p?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 83 / 124
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Example 2
Define f (p) ≡∑∞
k=11kp . Then
(Condition for convergence) f (p) <∞ if and only if p > 1
(p = 2)∞∑k=1
1
k2=π2
6
(p = 4)∞∑k=1
1
k4=π4
90
(40)
How about for general p?
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 83 / 124
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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.
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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.
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 84 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 84 / 124
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Comparison Test: Theorem 14.6, Corollary 14.7
Theorem
Let∑
an be a series where an ≥ 0 for all n. Then
If∑
an converges, and | bn |≤ an for all n, then∑
bn converges.
If∑
an =∞, and bn ≥ an for all n, then∑
bn =∞ .
Proof.
Now on chalkboard!
Corollary
Absolutely convergent series are convergent.
Proof.
Now on chalkboard!
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Comparison Test: Theorem 14.6, Corollary 14.7
Theorem
Let∑
an be a series where an ≥ 0 for all n. Then
If∑
an converges, and | bn |≤ an for all n, then∑
bn converges.
If∑
an =∞, and bn ≥ an for all n, then∑
bn =∞ .
Proof.
Now on chalkboard!
Corollary
Absolutely convergent series are convergent.
Proof.
Now on chalkboard!
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Comparison Test: Theorem 14.6, Corollary 14.7
Theorem
Let∑
an be a series where an ≥ 0 for all n. Then
If∑
an converges, and | bn |≤ an for all n, then∑
bn converges.
If∑
an =∞, and bn ≥ an for all n, then∑
bn =∞ .
Proof.
Now on chalkboard!
Corollary
Absolutely convergent series are convergent.
Proof.
Now on chalkboard!
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Comparison Test: Theorem 14.6, Corollary 14.7
Theorem
Let∑
an be a series where an ≥ 0 for all n. Then
If∑
an converges, and | bn |≤ an for all n, then∑
bn converges.
If∑
an =∞, and bn ≥ an for all n, then∑
bn =∞ .
Proof.
Now on chalkboard!
Corollary
Absolutely convergent series are convergent.
Proof.
Now on chalkboard!
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Comparison Test: Theorem 14.6, Corollary 14.7
Theorem
Let∑
an be a series where an ≥ 0 for all n. Then
If∑
an converges, and | bn |≤ an for all n, then∑
bn converges.
If∑
an =∞, and bn ≥ an for all n, then∑
bn =∞ .
Proof.
Now on chalkboard!
Corollary
Absolutely convergent series are convergent.
Proof.
Now on chalkboard!
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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.
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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.
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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.
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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.
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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.
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 88 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 88 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 88 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 88 / 124
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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
Test.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 89 / 124
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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
Test.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 89 / 124
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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
Test.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 89 / 124
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Examples
∑∞n=0
(−1
3
)n
∑∞n=0
nn2+3∑∞
n=01
n2+1∑∞n=0
n3n∑∞
n=0
(2
−3−(−1)n
)n∑∞
n=0 2(−1)n−n∑∞
n=1(−1)n√
n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 90 / 124
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Examples
∑∞n=0
(−1
3
)n∑∞n=0
nn2+3
∑∞n=0
1n2+1∑∞
n=0n3n∑∞
n=0
(2
−3−(−1)n
)n∑∞
n=0 2(−1)n−n∑∞
n=1(−1)n√
n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 90 / 124
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Examples
∑∞n=0
(−1
3
)n∑∞n=0
nn2+3∑∞
n=01
n2+1
∑∞n=0
n3n∑∞
n=0
(2
−3−(−1)n
)n∑∞
n=0 2(−1)n−n∑∞
n=1(−1)n√
n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 90 / 124
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Examples
∑∞n=0
(−1
3
)n∑∞n=0
nn2+3∑∞
n=01
n2+1∑∞n=0
n3n
∑∞n=0
(2
−3−(−1)n
)n∑∞
n=0 2(−1)n−n∑∞
n=1(−1)n√
n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 90 / 124
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Examples
∑∞n=0
(−1
3
)n∑∞n=0
nn2+3∑∞
n=01
n2+1∑∞n=0
n3n∑∞
n=0
(2
−3−(−1)n
)n
∑∞n=0 2(−1)
n−n∑∞n=1
(−1)n√n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 90 / 124
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Examples
∑∞n=0
(−1
3
)n∑∞n=0
nn2+3∑∞
n=01
n2+1∑∞n=0
n3n∑∞
n=0
(2
−3−(−1)n
)n∑∞
n=0 2(−1)n−n
∑∞n=1
(−1)n√n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 90 / 124
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Examples
∑∞n=0
(−1
3
)n∑∞n=0
nn2+3∑∞
n=01
n2+1∑∞n=0
n3n∑∞
n=0
(2
−3−(−1)n
)n∑∞
n=0 2(−1)n−n∑∞
n=1(−1)n√
n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 90 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 91 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 91 / 124
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Theorem 15.3
Theorem
If a1 ≥ a2 ≥ ... ≥ an ≥ .. ≥ 0 and lim an = 0 then∑
(−1)nan <∞.
Proof: In book.
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Theorem 15.3
Theorem
If a1 ≥ a2 ≥ ... ≥ an ≥ .. ≥ 0 and lim an = 0 then∑
(−1)nan <∞.
Proof: In book.
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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
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 94 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 94 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 94 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 94 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 94 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 94 / 124
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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
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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.
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 96 / 124
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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∗.
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 98 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 98 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 98 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 98 / 124
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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∗.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 99 / 124
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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∗.
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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!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 100 / 124
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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!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 100 / 124
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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!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 100 / 124
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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!
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 100 / 124
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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 .
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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 .
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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 .
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 102 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 102 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 102 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 102 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 103 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 103 / 124
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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].
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 104 / 124
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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].
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 104 / 124
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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].
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 104 / 124
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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].
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 104 / 124
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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].
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 104 / 124
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Uniform Continuity: Theorem 19.2
Theorem
Let f be continuous function on a closed interval [a, b]. Then f isuniformly continuous on [a, b]
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 105 / 124
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Uniform Continuity: Theorem 19.2
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
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 .
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 106 / 124
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Uniform Continuity: Theorem 19.2
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
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 .
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 106 / 124
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Uniform Continuity: Theorem 19.2
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
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 .
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 106 / 124
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Uniform Continuity: Theorem 19.2
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
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 .
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 106 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 107 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 107 / 124
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Theorem 19.4: Example
Example 6: f (x) = 1x2
is not uniformly continuous.
To show this, take the Cauchy sequence
{sn}∞n=1 :={
1n+1
}∞n=1⊂ (0, 1).
Then f (sn) = (n + 1)2 and is not Cauchy.
Hence, f is not uniformly continuous on (0, 1)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 108 / 124
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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 )
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 109 / 124
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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 )
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 109 / 124
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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 .
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 110 / 124
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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 .
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 110 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 111 / 124
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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 ∈ R
the 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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 111 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 111 / 124
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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.
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 111 / 124
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Theorem 23.1
Theorem
For the power series∑∞
n=0 anxn, let
β = lim sup | an |1n
R =1
β
R = 0 if β =∞R =∞ if β = 0
(45)
Then
the power series converges for | x |< R
the power series diverges for | x |> R
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 112 / 124
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Theorem 23.1
Theorem
For the power series∑∞
n=0 anxn, let
β = lim sup | an |1n
R =1
β
R = 0 if β =∞R =∞ if β = 0
(45)
Then
the power series converges for | x |< R
the power series diverges for | x |> R
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 112 / 124
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Theorem 23.1
Theorem
For the power series∑∞
n=0 anxn, let
β = lim sup | an |1n
R =1
β
R = 0 if β =∞R =∞ if β = 0
(45)
Then
the power series converges for | x |< R
the power series diverges for | x |> R
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 112 / 124
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Examples
Consider:
Example 1 :∑∞
n=01n!x
n
Example 2 :∑∞
n=0 xn
Example 3 :∑∞
n=01nxn
Example 4 :∑∞
n=01n2
xn
Example 5 :∑∞
n=0 n!xn
Example 6 :∑∞
n=0 2−nx3n
Example 7 :∑∞
n=0(−1)n+1
n (x − 1)n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 113 / 124
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Examples
Consider:
Example 1 :∑∞
n=01n!x
n
Example 2 :∑∞
n=0 xn
Example 3 :∑∞
n=01nxn
Example 4 :∑∞
n=01n2
xn
Example 5 :∑∞
n=0 n!xn
Example 6 :∑∞
n=0 2−nx3n
Example 7 :∑∞
n=0(−1)n+1
n (x − 1)n
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Examples
Consider:
Example 1 :∑∞
n=01n!x
n
Example 2 :∑∞
n=0 xn
Example 3 :∑∞
n=01nxn
Example 4 :∑∞
n=01n2
xn
Example 5 :∑∞
n=0 n!xn
Example 6 :∑∞
n=0 2−nx3n
Example 7 :∑∞
n=0(−1)n+1
n (x − 1)n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 113 / 124
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Examples
Consider:
Example 1 :∑∞
n=01n!x
n
Example 2 :∑∞
n=0 xn
Example 3 :∑∞
n=01nxn
Example 4 :∑∞
n=01n2
xn
Example 5 :∑∞
n=0 n!xn
Example 6 :∑∞
n=0 2−nx3n
Example 7 :∑∞
n=0(−1)n+1
n (x − 1)n
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Examples
Consider:
Example 1 :∑∞
n=01n!x
n
Example 2 :∑∞
n=0 xn
Example 3 :∑∞
n=01nxn
Example 4 :∑∞
n=01n2
xn
Example 5 :∑∞
n=0 n!xn
Example 6 :∑∞
n=0 2−nx3n
Example 7 :∑∞
n=0(−1)n+1
n (x − 1)n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 113 / 124
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Examples
Consider:
Example 1 :∑∞
n=01n!x
n
Example 2 :∑∞
n=0 xn
Example 3 :∑∞
n=01nxn
Example 4 :∑∞
n=01n2
xn
Example 5 :∑∞
n=0 n!xn
Example 6 :∑∞
n=0 2−nx3n
Example 7 :∑∞
n=0(−1)n+1
n (x − 1)n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 113 / 124
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Examples
Consider:
Example 1 :∑∞
n=01n!x
n
Example 2 :∑∞
n=0 xn
Example 3 :∑∞
n=01nxn
Example 4 :∑∞
n=01n2
xn
Example 5 :∑∞
n=0 n!xn
Example 6 :∑∞
n=0 2−nx3n
Example 7 :∑∞
n=0(−1)n+1
n (x − 1)n
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 113 / 124
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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
on R.
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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
on R.
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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
on R.
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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
on R.
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Uniform Convergence
Definition 24.2: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.
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
on R.
Example 5: Let fn(x) = nxn for x ∈ [0, 1). Then fn(x)→ 0 pointwise on[0, 1), but again not uniformly.
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Uniform Convergence
Definition 24.2: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.
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
on R.
Example 5: Let fn(x) = nxn for x ∈ [0, 1). Then fn(x)→ 0 pointwise on[0, 1), but again not uniformly.
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Uniform Convergence
Definition 24.2: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.
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
on R.
Example 5: Let fn(x) = nxn for x ∈ [0, 1). Then fn(x)→ 0 pointwise on[0, 1), but again not uniformly.
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Uniform Convergence
Definition 24.2: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.
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
on R.
Example 5: Let fn(x) = nxn for x ∈ [0, 1). Then fn(x)→ 0 pointwise on[0, 1), but again not uniformly.
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Uniform Convergence
Definition 24.2: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.
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
on R.
Example 5: Let fn(x) = nxn for x ∈ [0, 1). Then fn(x)→ 0 pointwise on[0, 1), but again not uniformly.
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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
(ε3
), we have
the second term < ε3 . The third term follows from pointwise convergence.
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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
(ε3
), we have
the second term < ε3 . The third term follows from pointwise convergence.
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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.
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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.
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Uniform Convergence
Definition 25.3: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.
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 .
Theorem 25.6: Let∑∞
k=0 gk be a series of functions that satisfies theCauchy criterion uniformly on a set S ⊆ R. Then the series convergesuniformly on S .
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Uniform Convergence
Definition 25.3: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.
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 .
Theorem 25.6: Let∑∞
k=0 gk be a series of functions that satisfies theCauchy criterion uniformly on a set S ⊆ R. Then the series convergesuniformly on S .
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Uniform Convergence
Definition 25.3: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.
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 .
Theorem 25.6: Let∑∞
k=0 gk be a series of functions that satisfies theCauchy criterion uniformly on a set S ⊆ R. Then the series convergesuniformly on S .
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Uniform Convergence
Definition 25.3: Let {fn} be a sequence of real valued functions definedon a set S ⊆ R.
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 .
Theorem 25.6: Let∑∞
k=0 gk be a series of functions that satisfies theCauchy criterion uniformly on a set S ⊆ R. Then the series convergesuniformly on S .
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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
|n∑
k=m
gk(x) |≤n∑
k=m
| gk(x) |≤n∑
k=m
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 .
QED
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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 numberN such that n ≥ m > N implies |
∑nk=m Mk |< ε. Hence, if n ≥ m > N
and x ∈ S , then
|n∑
k=m
gk(x) |≤n∑
k=m
| gk(x) |≤n∑
k=m
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 .
QED
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 119 / 124
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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
|n∑
k=m
gk(x) |≤n∑
k=m
| gk(x) |≤n∑
k=m
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 .
QED
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 119 / 124
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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
|n∑
k=m
gk(x) |≤n∑
k=m
| gk(x) |≤n∑
k=m
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 .
QED
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 119 / 124
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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
|n∑
k=m
gk(x) |≤n∑
k=m
| gk(x) |≤n∑
k=m
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 .
QED
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 119 / 124
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Differentiation and Integration of Power Series
The following are nice applications of the Weirstrass M-Test:
Theorem 26.1: Let∑∞
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).
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 120 / 124
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Differentiation and Integration of Power Series
The following are nice applications of the Weirstrass M-Test:
Theorem 26.1: Let∑∞
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).
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 120 / 124
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Differentiation and Integration of Power Series
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
QED
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 121 / 124
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Differentiation and Integration of Power Series
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
QED
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 121 / 124
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Differentiation and Integration of Power Series
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
QED
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 121 / 124
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Differentiation and Integration of Power Series
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
QED
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 121 / 124
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Differentiation and Integration of Power Series
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→∞
∫ b
afn(x)dx =
∫ b
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)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 122 / 124
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Differentiation and Integration of Power Series
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→∞
∫ b
afn(x)dx =
∫ b
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)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 122 / 124
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Differentiation and Integration of Power Series
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→∞
∫ b
afn(x)dx =
∫ b
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)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 122 / 124
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Differentiation and Integration of Power Series
Theorem 26.5: Suppose that f (x) =∑∞
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)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 123 / 124
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Differentiation and Integration of Power Series
Theorem 26.5: Suppose that f (x) =∑∞
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)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 123 / 124
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Differentiation and Integration of Power Series
Theorem 26.5: Suppose that f (x) =∑∞
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)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 123 / 124
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Differentiation and Integration of Power Series
So, if 0 < R1 < R then
f (x) =
∫ x
−R1
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
1
n + 1xn+1 =
∫ x
0
1
1− tdt = − ln (1− x)
(53)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 124 / 124
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Differentiation and Integration of Power Series
So, if 0 < R1 < R then
f (x) =
∫ x
−R1
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
1
n + 1xn+1 =
∫ x
0
1
1− tdt = − ln (1− x)
(53)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 124 / 124
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Differentiation and Integration of Power Series
So, if 0 < R1 < R then
f (x) =
∫ x
−R1
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
1
n + 1xn+1 =
∫ x
0
1
1− tdt = − ln (1− x)
(53)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 124 / 124
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Differentiation and Integration of Power Series
So, if 0 < R1 < R then
f (x) =
∫ x
−R1
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
1
n + 1xn+1 =
∫ x
0
1
1− tdt = − ln (1− x)
(53)
Albert Cohen (MSU) Lecture Notes: Math 320 MSU Spring Semester 2013 124 / 124