Overview (MA2730,2812,2815) lecture 9 - Brunel

Overview (MA2730,2812,2815) lecture 9

Lecture slides for MA2730 Analysis I

Simon Shawpeople.brunel.ac.uk/~icsrsss

[email protected]

College of Engineering, Design and Physical Sciencesbicom & Materials and Manufacturing Research InstituteBrunel University

October 16, 2015

Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel

MA2730, Analysis I, 2015-16


Contents of the teaching and assessment blocks

MA2730: Analysis I

Analysis — taming infinity

Maclaurin and Taylor series.

Sequences.

Improper Integrals.

Series.

Convergence.

LATEX2ε assignment in December.

Question(s) in January class test.

Question(s) in end of year exam.

Web Page:http://people.brunel.ac.uk/~icsrsss/teaching/ma2730




Lecture 9

MA2730: topics for Lecture 9

Lecture 9

Bounded sequences

Convergence intervals for bounded sequences

Examples and Exercises

The material covered in this lecture can be found in TheHandbook, Chapter 2, Section 2.3.

Either write something worth reading or do something worthwriting.Benjamin Franklin




Lecture 9

Reference: Stewart, Chapter 12.

This week we will describe the concepts of a bounded sequenceand a monotone sequence.

We’ll see that, when taken together, these furnish an importanttool with which we can study convergence of sequences.

The material covered in this lecture can be found in TheHandbook, Chapter 2, Section 2.3.Homework: attempt all remaining Questions on Exercise Sheet 2a.Seminar: outline of 4, 5, 6c, 7 on Exercise Sheet 2a




Lecture 9

Where are we?

So far we have seen,

sequences as lists, {an}.sequences as functions an = f(n).

convergence: an → a as n → ∞.

divergence: an unbounded or ‘unsettled’ as n → ∞.

the algebra of limits

standard convergent sequences.

This week we take a glimpse at the very hard stuff.




Lecture 9

Abstractness

Given a sequence, {an}, there are two fundamental questions:

Does it converge to a limit?

If so, what is its limit?

If the answer to the first is NO then, obviously, the second cannotbe answered.

The first question is the most important for us.

The second can sometimes be very difficult to answer, and mayrequire special (and approximate) techniques.

So, often we are asking the abstract question of does the limitexist?, rather than the specific question of what is its value?.




Lecture 9

This isn’t so unusual

Consider∫ 10 e−x2

dx. This is impossible to calculate in closed form,and yet we know it exists because it represents a well-defined area.

−3 −2 −1 0 1 2 3

0

0.2

0.4

0.6

0.8

1




Lecture 9

Bounded sequences

We have seen that if a sequence has a limit then

an → a as n → ∞,

and this means that as n gets large, all the terms an must getcloser and closer to the limit a.

Since a must be a real number, it must be true that each ancannot be much smaller than a, or much bigger that a.

Boardwork for

{4 +

cos(πn)

n

}.

This leads us to a new concept.




Lecture 9

Bounded above, Bounded below, Bounded. . .

Definition 2.15, Subsection 2.3.1 in The Handboook

We say that:

a sequence {an} is bounded below if there is m ∈ R such thatan > m for all n > 1.

a sequence {an} is bounded above if there is M ∈ R suchthat an 6 M for all n > 1.

a sequence is bounded if it is bounded below and above. Thatis, if m 6 an 6 M for all n > 1.

Boardwork: examples from The Handbook,

{1− e−n cos(n)}, {arctan(n)}, {(−1)n}.

Which converge?




Lecture 9

It is tempting to think that a bounded sequence converges, but itneed not. For example, {(−1)n}.

However, a convergent sequence must be bounded.

Theorem 2.16, Subsection 2.3.1 in The Handboook

A convergent sequence is bounded.

That’s probably the shortest theorem we have yet seen!

Short theorems don’t always imply short proofs however. . .

Recall: proofs have to be planned. So let’s plan first.




Lecture 9

Why is planning important?




Lecture 9

Plan of the proof

First let’s collect up all the things we know.

We have a convergent sequence {an}, with limit a:

limn→∞

an = a.

We know that an gets as close as we please to a providing that wetake n large enough because limn→∞(an − a) = 0. (Why?)

We want to prove that there exist real numbers m and M suchthat m 6 an 6 M for all n > 1.

We’ll then have proven that {an} is bounded.

Let’s see how to prove the theorem. . .Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 9

Proof: a convergent sequence is bounded

There is an N such that |a− an| 6 1 for all n > N . Why?

Hence −1 6 a− an 6 1 for all n > N .

Hence a− 1 6 an 6 a+ 1 for all n > N .

Therefore

an > min{a− 1, aN , aN−1, aN−2, . . . , a2, a1} = m

and

an 6 max{a+ 1, aN , aN−1, aN−2, . . . , a2, a1} = M

Hence m 6 an 6 M for all n > 1 and {an} is bounded.




Lecture 9

The contrapositive

So, we have proven that a convergent sequence is bounded.

Applying the contrapositive to this tell us that.

an unbounded sequence is not convergent.

Recall: the contrapostive of ‘if A is true then B is true’.

is ‘if B is not true then A is not true’.




Lecture 9

Another useful theorem

Let’s move on now to The Handbook, Subsection 2.3.2:Convergent sequences and closed bounded intervals.

Theorem 2.17, non-negative sequences

If {an} is a sequence of real numbers such that an > 0 for alln ∈ N and an → a then a > 0.

Similarly, if an 6 0, for all n then a 6 0.

This says that if all the terms in the sequence are of one sign (orzero) then the limit cannot be of a different sign.

Example: if {(n+ | sin(n)|)−2} has a limit, it cannot be negative.

Proof: deferred until we see the (ǫ, N) definition of convergence.Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 9

An important corollary

What is a corollary?This one is on convergent sequences and closed bounded intervals.

Corollary 2.18

Verbal statement:If a closed bounded interval contains all terms of a convergentsequence then the limit of that sequence belongs to the interval.Equivalently:if xn → x ∈ R and each xn ∈ I = [a, b] ⊆ R, then x ∈ I.

It’s interesting to consider the many ways a statement can beformulated. Dont be afraid to experiment. . .




Lecture 9

Planning the proof

Corollary 2.18


Proof plan:

Show that x > a and x 6 b in separate steps.

Consider the sequence {yn} where yn = xn − a > 0 and useTheorem 2.17 to get lim(xn − a) = lim yn > 0.

This gives 0 6 x− a and proves that x > a. Let’s see the proof. . .Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 9

The proof — part 1

Corollary 2.18


Homework: you prove that x 6 b and conclude that x ∈ [a, b].




Lecture 9

Example

Consider the sequence with general term:

an = 4 + e−n cos(n) for n = 1, 2, 3, . . .

Since for n > 1,

0 6 e−n 6 1 and − 1 6 cos(n) 6 1,

we can see that4− 1 6 an 6 4 + 1

and so the limit, if it exists satisfies a ∈ [3, 5]. Comments?

In fact 3 < an < 5 because e−n < 1. Discussion.




Lecture 9

Summary

We can:

analyse a sequence and determine if it is bounded.

prove that a convergent sequence is bounded.

analyse non-negative (positive) sequences.

prove that the limit lies within the sequence’s bound(s).




End of Lecture

Computational andαpplie∂ Mathematics

Either write something worth reading or do something worthwriting.Benjamin Franklin

The material covered in this lecture can be found in TheHandbook, Chapter 2, Section 2.3.Homework: attempt all remaining Questions on Exercise Sheet 2a.Seminar: outline of 4, 5, 6c, 7 on Exercise Sheet 2a



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Overview (MA2730,2812,2815) lecture 9 - Brunel