lecture 16 - brunel university londonicsrsss/teaching/ma2730/lec/print8lec16.pdf · lecture 16...

Overview (MA2730,2812,2815) lecture 16

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 26, 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 16

MA2730: topics for Lecture 16

Lecture 16

The (ǫ,N) definition of convergence

Divergence

Uniqueness of the limit

Examples and Exercises

Reference: The Handbook, Chapter 5, Section 5.1.Homework: attempt Questions 1 to 4 on Exercise Sheet 4aSeminar: Q2a and Q3




Lecture 16

Reference: Stewart, calculus, sixth edition, Chapter 12.1.

We move on now to the material in Section 5.1.

This is a return to sequences: recall that we assumed that oursequences {an} were generated by a function an = f(n).

Much of our theory for limits and properties of sequences wasgenerated from the corresponding theory for functions.

While this is often adequate it is not general enough foradvanced applications.

So today we will meet the so-called (ǫ,N) definition forconvergence of a sequence.




Lecture 16

Time Management — Tip 2

Study delayed, reduces grade. SS

The pace is increasing once again.

Are you keeping up?

Here’s the centerpiece of today’s lecture. . .




Lecture 16

The (ǫ, N) definition of a limit of a sequence

Definition 5.1 in The Handbook

A sequence {an} has limit a if for every ǫ > 0, there is an integerN , usually depending on ǫ, such that, whenever n > N , we have|an − a| < ǫ.

If {an} has limit a we write limn→∞

an = a.

Equivalently, if {an} has limit a we write an → a as n → ∞.

If limn→∞

an exists, we say that the sequence {an} converges.

Otherwise we say it diverges.

The first part of this needs unpicking. This is the work for thislecture. . .




Lecture 16

The (ǫ, N) definition of a limit of a sequence



For example, consider the sequence {n−1}. Does this converge?To what?

We check this with the definition by taking an = 1n and a = 0.

Then,|an − a| = |n−1 − 0| < ǫ

for all n > N where N = ⌊ǫ−1⌋ and so n−1 → 0 as n → ∞.

BOARDWORK.




Lecture 16



There is a lot going on here.

We note that we must have a candidate, a, for the limit.

That candidate can come from intuition, calculation,suspicion, experience and even from an intelligent guess.

We choose ǫ > 0 (in maths ǫ often denotes an arbitrarily smallpositive number).

We need to be able to find an upper bound: |an − a| 6 h(n)where h is a simple function of n.

Then we decide if there is an N such that h(n) < ǫ for everyn > N . If so, the sequence converges to a.




Lecture 16



Here is that list again — just the main points.

Find a candidate, a, for the limit.

Choose an ǫ > 0.

Find a simple h : N → R such that |an − a| 6 h(n).

Is there an N such that h(n) < ǫ for every n > N?

If so, the sequence converges to a.




Lecture 16



Notice that N = N(ǫ); the value of N depends on our choice of ǫ.

The smaller we choose ǫ, the closer we force an to be to a.

This means that we have to go further into the infinite tail of thesequence. . .

. . . and therefore need N to get larger as ǫ gets smaller.

It is important to appreciate that convergence is determined bythis infinite tail — not by any finite number of the first terms. SeeRemarks 5.3 and 5.5 in The Handbook.




Lecture 16

What about divergence?


We say that limn→∞ an = +∞ if for every M ∈ R there exists aninteger N such that

an > M for all n > N.

We say that limn→∞ an = −∞ if for every M ∈ R there exists aninteger N such that

an < M for all n > N.

DISCUSSION




Lecture 16

Examples — Boardwork

Let’s apply our new definitions to these examples.{1

n2

}.

{2n}.{

1

n+ n2

}.

{cos(πn)}The last does not fit into any of our definitions.




Lecture 16

To deal with sequences such as {cos(πn)} in this new frameworkwe need to be able to determine non-convergence.

The next result is from Subsection 5.1.2 of The Handbook.

Proposition 5.7 in The Handbook

A sequence {an} is not convergent, and is therefore divergent, ifand only if for all a ∈ R, there exists an ǫ > 0, usually dependingon a, such that for all N ∈ N, there exists n > N such that

|an − a| > ǫ.

Proof: the proof given in The Handbook uses the notion of thecontraposition of the convergence criteria given in Definition 5.1.

We’ll give a different argument.




Lecture 16



|a− an| > ǫ.

Proof — the ‘if’ part. We choose a candidate for the limit,a ∈ R, and then choose the ǫ > 0 as given in the propositionabove. For these choices we want to find N such that,

|a− an| < ǫ for all n > N.

However, whatever value of N we select provides an n > N suchthat |a− an| > ǫ.

Therefore we cannot fulfil the conditions of Definition 5.1 and so{an} cannot be convergent.




Lecture 16



|a− an| > ǫ.

Proof — the ‘only if’ part. If {an} is divergent then for anychoice of a ∈ R we have limn→∞ |a− an| 6= 0. Hence for everyn > 1 large enough (i.e. far enough down in the infinite tail of thesequence) we can find an ǫ > 0 such that |a− an| > ǫ for alln > N . We can choose N = 1.




Lecture 16



|a− an| > ǫ.

For the sequence {cos(πn)} = {(−1)n} we can choose anyarbitrary number a ∈ R and then if a > 0,

|a− cos(πn)| = |a− (−1)| > 1 for n = 1, 3, 5, . . .

and if a < 0,

|a− cos(πn)| = |a− (+1)| > 1 for n = 2, 4, 6, . . .

So, whatever N ∈ N is chosen we can find n > N such that|a− cos(πn)| > 1. Therefore {cos(πn)} is not convergent.




Lecture 16

We now have techniques for the analysis of sequences {an} thatdo not rely on finding a function such that an = f(n):

1 we first determine a candidate for the limit,

2 and then show that |an − a| is as small as we please so longas we go far enough into the tail of the sequence.

3 That is: given an ǫ > 0 we find an N such that

|an − a| < ǫ for all n > N.

But without a function to give f(a), at the limit, how do we knowthat we guessed the correct candidate in step 1?

Fortunately, the limit is unique. So there is only one way to guessthis candidate correctly. That will be our last result for today.




Lecture 16

The limit is unique, Theorem 5.8 in The Handbook

If the sequence {an} converges then its limit is unique.

Proof: Boardwork.




Lecture 16

Summary

We can test a sequence for:

convergence using Definition 5.1.

divergence to ±∞ using Definition 5.4.

divergence (non-convergence) using Proposition 5.7.

In addition we can prove that the limit of a convergent sequence isunique.




End of Lecture

Computational andαpplie∂ Mathematics

Study delayed, reduces grade. SS

Reference: The Handbook, Chapter 5, Section 5.1.Homework: attempt Questions 1 to 4 on Exercise Sheet 4aSeminar: Q2a and Q3



lecture 16 - brunel university londonicsrsss/teaching/ma2730/lec/print8lec16.pdf · lecture 16...

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