MATH1151 Calculus Notes

7/25/2019 MATH1151 Calculus Notes

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MATH1151 – Maths for Actuarial Studies & Finance

Calculus Lecture 01

Real Numbers and Least Upper Bounds

Dr Jonathan Kress

Slides by A/Prof Thanh Tran

School of Mathematics and StatisticsUniversity of New South Wales

Dr Jonathan Kress (UNSW Maths & Stats) MATH1151 1 / 20

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1 Sets of numbers

2 An historical problem

3 The least upper bound axiom

4 Bernouilli’s question – Revisited


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Sets and elements


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Sets and elements

A set is a collection of distinct objects.


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Sets and elements


The objects in a set are called the elements or members of the

set.


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Sets and elements



set.

Example 1


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Sets and elements



set.

Example 1

MATH1151 is an element of the set of all mathematical courses

offered at UNSW.


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Sets and elements



set.

Example 1

MATH1151 is an element of the set of all mathematical courses

offered at UNSW.

UCLA is not a member of the Australian universities.


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Some notation

N = {0, 1, 2, 3, . . .} (set of all natural numbers)


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Some notation


Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .} (set of all integers)


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Some notation



Z+ =

{1, 2, 3, . . .

} (set of all positive integers)


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Some notation



Z+ = {1, 2, 3, . . .} (set of all positive integers)

Z− = {. . . , −3, −2, −1} (set of all negative integers)


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Some notation





Q = {p q : p ,q ∈ Z,q = 0} (set of all rational numbers)


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Some notation






R (set of all real numbers).


S i

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Some notation






R (set of all real numbers).

C (set of all complex numbers)


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Example 2

We write2 ∈ N and read 2 is an element of N

and

−4

3 /∈ Z and read −4/3 is not an element of Z


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Example 2


and

−4


A remark


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Example 2


and

−4


A remark

The equation x 2 = 2 does not have a solution in Q.


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Example 2


and

−4


A remark


Intuitively, R is the number line or decimal expansions.


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Example 2


and

−4


A remark


Intuitively, R is the number line or decimal expansions.

A real number which is not a rational is called an irrationalnumber. E.g.,

√ 2, π, e .


The empty set ∅

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The empty set ∅

The empty set or null set, denoted by ∅, is a set which has no

members at all.


The empty set ∅

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The empty set ∅


members at all.

Example 3


The empty set ∅

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The empty set ∅


members at all.

Example 3The set of all students doing MATH1151 in 2012 whose ages arebelow 10 is an empty set.


The empty set ∅

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The empty set ∅


members at all.

Example 3The set of all students doing MATH1151 in 2012 whose ages arebelow 10 is an empty set.

The set of all real numbers x satisfying x 2 + 1 = 0 is an empty set.


Two ways to define a set

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List all members inside a pair of curly brackets (or braces).



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Give a rule to determine which members of some previouslydefined set are to be members of the set being defined.



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Example 4



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Example 4

The set of all natural numbers less than 3 can be represented byone of the following ways



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o ays to de e a set

List all members inside a pair of curly brackets (or braces).Give a rule to determine which members of some previouslydefined set are to be members of the set being defined.

Example 4


{0, 1, 2} or {2, 1, 0} or {n ∈ N : n < 3}.



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y

List all members inside a pair of curly brackets (or braces).Give a rule to determine which members of some previouslydefined set are to be members of the set being defined.

Example 4


{0, 1, 2} or {2, 1, 0} or {n ∈ N : n < 3}.

The equation x 2 = 1 has two real solutions 1 and −1. The set ofall solutions of this equation can be written as



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y



Example 4

The set of all natural numbers less than 3 can be represented by

one of the following ways

{0, 1, 2} or {2, 1, 0} or {n ∈ N : n < 3}.

The equation x 2 = 1 has two real solutions 1 and

−1. The set of

all solutions of this equation can be written as

{−1, 1} or {x ∈ R : x 2 = 1}.


Some useful jargon

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j g

Definition 5 (Subset)


Some useful jargon

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j g

Definition 5 (Subset)Suppose that A and B are two sets. If x ∈ A implies that x ∈ B ,then A is called a subset of B . We write A ⊂ B .


Some useful jargon

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If A is a subset of B we can also say that B contains A, and write

B ⊃ A.


Some useful jargon

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If A is a subset of B we can also say that B contains A, and write

B ⊃ A.

Example 6

N⊂Z (N is a subset of Z)

R ⊃ Q (R contains Q).


Intervals (open and closed)

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Suppose that a and b are two real numbers satisfying a < b . Then



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(a ,b ) = {x ∈ R : a < x < b }



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(a ,b ) = {x ∈ R : a < x < b }

[a ,b ] = {x ∈ R : a ≤ x ≤ b }



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(a ,b ) = {x ∈ R : a < x < b }

[a ,b ] = {x ∈ R : a ≤ x ≤ b }

[a ,b ) = {x ∈ R : a ≤ x < b }



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(a ,b ) = {x ∈ R : a < x < b }

[a ,b ] = {x ∈ R : a ≤ x ≤ b }

[a ,b ) = {x ∈ R : a ≤ x < b }

(a ,b ] = {x ∈ R : a < x ≤ b }



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(a ,b ) = {x ∈ R : a < x < b }

[a ,b ] = {x ∈ R : a ≤ x ≤ b }

[a ,b ) = {x ∈ R : a ≤ x < b }

(a ,b ] = {x ∈ R : a < x ≤ b }

(a ,∞

) = {x ∈

R : a < x }



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(a ,b ) = {x ∈ R : a < x < b }

[a ,b ] = {x ∈ R : a ≤ x ≤ b }

[a ,b ) = {x ∈ R : a ≤ x < b }

(a ,b ] = {x ∈ R : a < x ≤ b }

(a ,∞

) = {x ∈

R : a < x }

(−∞,b ] = {x ∈ R : x ≤ b }



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(a ,b ) = {x ∈ R : a < x < b }

[a ,b ] = {x ∈ R : a ≤ x ≤ b }

[a ,b ) = {x ∈ R : a ≤ x < b }

(a ,b ] = {x ∈ R : a < x ≤ b }

(a ,∞

) = {x ∈

R : a < x }

(−∞,b ] = {x ∈ R : x ≤ b }

a b

a b

a b

a b

a

b


An historical problem

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Assume that $A is invested at an interest rate of 100r % per annum.

If we compound annually then after 1 year the total amount is



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If we compound annually then after 1 year the total amount isIf we compound every 6 months then



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If we compound annually then after 1 year the total amount isIf we compound every 6 months then

◮ after 6 months the total amount is



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If we compound every 6 months then◮ after 6 months the total amount is

◮ after 1 year the total amount is



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If we compound every 4 months then



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◮

after 1 year the total amount is



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◮

after 1 year the total amount is

If we compound every 1n

th year (i.e., n times per year) then at the

end of the year we have


Question (Bernouilli, circa 1680)

Does $A(1 + r n)n grows without bound as n becomes arbitrarily large?

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$ ( + n ) g y g

[In other words, does our investment grow without bound if wecompound the interest over arbitrarily short time period?]

Let’s look at the case when r = 1 and let x n =

1 + 1n

n . Then using

the binomial theorem

x n =n

k =0

n

k 1

n k = 1 +

n

k =1

n (n − 1)(n − 2) · · · (n − k + 1)

k ! ×

1

n k

= 1 +n

k =1

1

1 − 1n

1 − 2

n

· · · 1 − k −1n

k !

< 1 +

n k =1

1

k ! < 1 +

n k =1

1

2k −1 (since 2k −1

< k ! for k ≥ 3)

= 1 + 1 −

12

n 1 − 1

2

< 1 + 1

1 − 12

= 3. Hence A

1 +

1

n

n

< 3A.


So effectively we have answered Bernoulli’s question: we can’t getinfinitely rich via continuous compounding of interest. However, more

can be said as every investor knows

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can be said, as every investor knows

1 +

1

n n

<

1 +

1

n + 1n +1

Proof: Invoke the binomial theorem again. For k = 1, . . . , n − 1, the(k + 1)st term in the expansion of

1 + 1

n

n is

n k

× 1n k

= 1

1 − 1n

1 − 2n · · · 1 −

k −2n

1 − k −1n

k !

whilst the corresponding term in the expansion of

1 + 1n +1

n +1is

n + 1

k

× 1

(n + 1)k =

1

1 − 1n +1

1 − 2n +1 · · ·1 − k −1n +1

k !

Therefore the first expansion is smaller and the result follows because

the extra term in the second expansion is positive.Dr Jonathan Kress (UNSW Maths & Stats) MATH1151 12 / 20

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We have proved

x 1 < x 2 < x 3 < · · · < x n < x n +1 < · · · < 3.

In other words,

The sequence {x n }∞n =1 is increasing;


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We have proved

x 1 < x 2 < x 3 < · · · < x n < x n +1 < · · · < 3.

In other words,


The sequence is bounded above (by, say, 3).


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We have proved

x 1 < x 2 < x 3 < · · · < x n < x n +1 < · · · < 3.

In other words,


The sequence is bounded above (by, say, 3).

Question:

Does it get arbitrarily close to a real number?


An axiom (SH10 – Section 11.1)

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Definition 7A set S of real numbers is said to be bounded above if there existsM ∈ R (M may not be in S ) such that

x

≤ M for all x

∈ S .

M is called an upper bound of S .



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Definition 7A set S of real numbers is said to be bounded above if there existsM ∈ R (M may not be in S ) such that

x

≤ M for all x

∈ S .


The least upper bound axiom

Every non-empty set of real numbers that has an upper bound has aleast upper bound.



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Definition 7

A set S of real numbers is said to be bounded above if there existsM ∈ R (M may not be in S ) such that

x ≤ M for all x ∈ S .


The least upper bound axiom

Every non-empty set of real numbers that has an upper bound has a

least upper bound.The least upper bound of a set S , if it exists, is denoted by lub S orsupS (supremum).


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Example 8

Let S = [0, 1] = {x ∈ R : 0 ≤ x ≤ 1}. Find supS .


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Example 8

Let S = [0, 1] = {x ∈ R : 0 ≤ x ≤ 1}. Find supS .

Example 9

Let S = [0, 1) = {x ∈ R : 0 ≤ x < 1}. Find supS .


Characterizing the least upper bound

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If M = sup S then any number less than M is not an upper bound

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of S .




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of S .

In other words, given any ǫ > 0, there exists x ∈ S such thatM − ǫ < x .




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of S .


In notation: ∀ǫ > 0, ∃x ∈ S such that M − ǫ < x .




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of S .



∀ reads “for all” or “for any” and ∃ reads “there exists”.




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If M sup S then any number less than M is not an upper bound

of S .




To show that L = supS




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If M sup S then any number less than M is not an upper bound

of S .





Show that L is an upper bound of S , i.e., show that

x

≤ L for all x

∈ S ,




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p y pp

of S .





Show that L is an upper bound of S , i.e., show that

x

≤ L for all x

∈ S ,

Show that for any ǫ > 0 there exists x 0 ∈ S such that

L − ǫ < x 0.


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Example 10

Use the method shown above to show that supS = 1 where S = [0, 1]and S = [0, 1), respectively; see Example 8 and Example 9.


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Example 11

Let S = {4 − 1n : n = 1, 2, 3, . . .}. Find supS if it exists.


Bernouilli’s question – Revisited

We proved, for x n =

1 +

1

n

n

,

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p

n




1 +

1

n

n

,

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n

◮ x n is increasing;




1 +

1

n

n

,

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n

◮ x n is increasing;◮ x n is bounded above.




1 +

1

n

n

,

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n


◮ x n is bounded above.Let

S =

1 +

1

n

n

: n = 1, 2, 3, . . .

.

Then S has a least upper bound by the least upper bound axiom.




1 +

1

n

n

,

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n



S =

1 +

1

n

n

: n = 1, 2, 3, . . .

.


Let L = supS . Then for any ǫ > 0, L − ǫ is not an upper bound ofS .




1 +

1

n

n

,

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n



S =

1 +

1

n

n

: n = 1, 2, 3, . . .

.



Hence there exists n 0 ∈ Z+ such that x n 0 > L− ǫ.




1 +

1

n

n

,

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n



S =

1 +

1

n

n

: n = 1, 2, 3, . . .

.



Hence there exists n 0 ∈ Z+ such that x n 0 > L− ǫ.Since {x n } is increasing we have

L − ǫ < x n ≤ L for all n ≥ n 0.




1 +

1

n

n

,

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n



S =

1 +

1

n

n

: n = 1, 2, 3, . . .

.



Hence there exists n 0 ∈ Z+ such that x n 0 > L− ǫ.Since {x n } is increasing we have

L − ǫ < x n ≤ L for all n ≥ n 0.Hence

∀ǫ > 0, ∃n 0 ∈ Z+ : ∀n ≥ n 0, |x n − L| < ǫ. (1)


We say that L is the limit of the sequence {x n }∞n =1 and write

limn →∞

x n = L or x n → L as n → ∞meaning x approaches (or converges to) L as n tends to infinity

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meaning x n approaches (or converges to) L as n tends to infinity

(or increases arbitrarily).



limn →∞

x n = L or x n → L as n → ∞meaning xn approaches (or converges to) L as n tends to infinity

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The order in (1) is important: ǫ is given first and n 0 depends on ǫ.

If ǫ = 10−6 then we would generally have to choose a larger n 0than that when ǫ = 10−2.



limn →∞


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The convergence of {x n }∞n =1 is slow.

x 103 = 2.7169239322 · · ·x 106 = 2.718280469 · · ·

L = 2.7182818284590 · · ·



limn →∞


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x 103 = 2.7169239322 · · ·x 106 = 2.718280469 · · ·

L = 2.7182818284590 · · ·L is denoted (by Euler) by e .



limn →∞


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x 103 = 2.7169239322 · · ·x 106 = 2.718280469 · · ·

L = 2.7182818284590 · · ·L is denoted (by Euler) by e .

Later in this course we will learn that

limn →∞

1 +

x

n

n = e x for all x ∈ R.


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MATH1151 Calculus Notes

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