Lecture 17 - people.brunel.ac.ukpeople.brunel.ac.uk/~icsrsss/teaching/ma2730/lec/print4lec17.pdf · Overview (MA2730,2812,2815) lecture 17 Lecture slides for MA2730 Analysis I

Overview (MA2730,2812,2815) lecture 17

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

November 10, 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 17

MA2730: topics for Lecture 17

Lecture 17

Conditional and absolute convergence

The Leibniz test for alternating series

Rearrangements of conditionally convergent series: Jedi maths

Examples and Exercises

Reference: The Handbook, Chapter 5, Section 5.3.Homework: Finish Exercise Sheet 4aSeminar: Q5




Lecture 17

Time Management — Tip 3

Put First Things FirstStephen Covey, The Seven Habits of Highly Effective People

IMPORTANT NOT IMPORTANT

URGENT

Next LecturesTomorrow’s testBuy birthday present

Pay billsFood shopping

NOT URGENT

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Playstation




Lecture 17

Reference: Stewart, calculus, sixth edition, Chapter 12.5.

Recall that in Lecture 13 we ‘concluded’ that ln(2) = 2 ln(2).

It’s false, of course: we did it by taking x = 1 in theMaclaurin series for ln(1 + x) to get,

ln(2) = 1− 1

2+

1

3− 1

4+

1

5− 1

6+

1

7− 1

8+ · · ·

and then re-ordering the way in which the terms added up.

Today we study the material in Subsection 5.3 of TheHandbook. We’ll then see what went wrong. . .

Specifically, we will begin with Alternating Series and thenmove on to understand conditional and absolute convergence,as well as some striking implications for working withalternating series.




Lecture 17

We start with the concept of Alternating Series as described inSubsection 5.3.1 of The Handbook.

Definition 5.12 in The Handbook

The sequence {an} is alternating if the sequence {(−1)nan} is ofone sign, that is, every term is either non-positive or non-negative.Such sequences generate an alternating series

∑∞n=1 an.

Example

The sequence {an} for an =cos(πn)

n2generates the alternating

series,

∞∑

n=1

cos(πn)

n2= −1 +

1

22− 1

32+

1

42− 1

52+ · · ·

We see that (−1)nan is a sequence of positive terms.




Lecture 17

Until now a few of our convergence tests for series cannot beapplied to alternating series because they required the terms of theseries to be of one sign.

But we can work with geometric series,∑

ark, for which r < 0.

The ratio and root tests also allowed for alternating series.

And we have seen this — our first test:

Our first test. Lemma 4.14: absolute value convergence

If∑∞

k=1 |ak| converges, then∑∞

k=1 ak converges.

We’ll return to this result later.




Lecture 17

In a moment we shall state and prove the Leibniz test foralternating series.

Afterwards we’ll move to a discussion of the big issues that arisewhen dealing with alternating series.

Here’s the Leibniz test.




Lecture 17

Leibniz Test, Theorem 5.13 in The Handbook

Let {an} be an alternating sequence such that {|an|} is strictlymonotonically decreasing with limn→∞ an = 0.

Then the series∑∞

n=1 an converges.

Moreover, the rate of convergence of the partial sum sn to thelimit sum s can be estimated by |s− sn| 6 |an+1| for n > 1.

Proof: boardwork




Lecture 17

Examples

∞∑

n=1

(−1)n+1

n

∞∑

n=1

(−1)n+1

n2

∞∑

n=1

(−1)n

∞∑

n=0

(−1)n

(2n)!

Boardwork and discussion




Lecture 17

The first of those examples was the Macalurin expansion of ln(2):

ln(2) =

∞∑

n=1

(−1)n+1

n= 1− 1

2+

1

3− 1

4+

1

5− 1

6+

1

7− 1

8+ · · ·

It is in fact perfectly well behaved. But if we change the negativesigns to positive we get,

∞∑

n=1

1

n= 1 +

1

2+

1

3+

1

4+

1

5+

1

6+

1

7+

1

8+ · · ·

This is the Harmonic series. We’ve seen from the integral test thatit diverges to ∞.




Lecture 17

So, we have an alternating harmonic series that converges eventhough the harmonic series diverges:

1− 1

2+

1

3− 1

4+

1

5− 1

6+

1

7− 1

8+ · · · =

∞∑

n=1

(−1)n+1

n= ln(2)

1 +1

2+

1

3+

1

4+

1

5+

1

6+

1

7+

1

8+ · · · =

∞∑

n=1

1

n= ∞

What’s going on?

First let’s recall our first test, from Lecture 15. . .




Lecture 17

Our first test. Lemma 4.14: absolute value convergence

If∑∞


k=1 ak converges.

In particular, this says that

If∞∑

n=1

1

nconverges, then

∞∑

n=1

(−1)n+1

nconverges.

But this is of no use because the harmonic series on the left doesnot converge.

This, obviously, cannot be true in general:

If∑∞

k=1 ak converges, then∑∞

k=1 |ak| converges.




Lecture 17

So, to summarize,

If∑∞


k=1 ak converges.

If∑∞

k=1 ak converges, then∑∞

k=1 |ak| might converge. . .

. . . or it might not!

Examples:∑ 1n2 and

∑ (−1)n

n2 both converge.

∑ (−1)n

n converges to ln 2 BUT∑ 1

n diverges to ∞.

What a mess! Let’s tidy our thoughts up a bit.




Lecture 17

We are starting to uncover an important concept. Here it is.

Definition 5.11

The series∑∞

n=1 an is said to be absolutely convergent if the series∑∞n=1 |an| converges.

The series∑∞

n=1 an is said to be conditionally convergent if∑∞n=1 an converges but

∑∞n=1 |an| diverges.

So, for example:

∑ (−1)n

n2is absolutely convergent.

∑ (−1)n

nis conditionally convergent.




Lecture 17

What have found?

Let {an} be an alternating sequence.

We now see that an alternating series may converge whereasits one-signed counterpart might not.

Such a series,∑

an, is conditionally convergent.

Example:∑ (−1)n

n converges to ln 2 BUT∑ 1

n diverges to ∞and so

∑ (−1)n

n is conditionally convergent.

When both∑

an and∑ |an| converge the alternating series

is called absolutely convergent

Example:∑ (−1)n

n2is absolutely convergent.

Any ideas as to what is going on here?Any intuition? Any guesses?




Lecture 17

Some Intuition. . .

Let’s return to the harmonic series and its alternating counterpart:

1− 1

2+

1

3− 1

4+

1

5− 1

6+

1

7− 1

8+ · · · =

∞∑

n=1

(−1)n+1

n= ln(2)

1 +1

2+

1

3+

1

4+

1

5+

1

6+

1

7+

1

8+ · · · =

∞∑

n=1

1

n= ∞

The clue is in the proof technique used earlier for the Leibniz test.

In a divergent series of positive terms,∑ |an|, the terms do not

approach zero sufficiently fast for the sum to have a limit.

In an alternating series,∑

an, there is the possibility that thenegative terms offset the accumulation of the positive terms. Thesum may then have a limit, and the alternating series will converge.




Lecture 17

This raises an interesting question. . .

If the negative terms are cancelling out some of the accumulationof the positive terms then. . .

. . . if we move the terms around can we change this cancellationand get a different sum?

The answer is YES — and we’ve already seen an example of itback in Lecture 13.




Lecture 17

Alternating series are W E I R D!

Recall this from Lecture 13:

ln(2) = 1− 1

2+

1

3−1

4+

1

5− 1

6+

1

7−1

8+

1

9− 1

10+

1

11− 1

12+ · · ·

=

(1− 1

2

)−1

4+

(1

3− 1

6

)−1

8+

(1

5− 1

10

)− 1

12+ · · ·

=1

2− 1

4+

1

6− 1

8+

1

10− 1

12+ · · ·

=1

2

(1− 1

2+

1

3− 1

4+

1

5− 1

6+ · · ·

)

=1

2ln(2) So ln(2) =

1

2ln(2).

Discussion. . . by moving terms around we get a different sum forthe series. Which equals sign is WRONG?




Lecture 17

Rearrangements: Jedi maths!

To answer this we need to think about what a rearrangementactually is.

Let φ : N → N be a bijection (that is: φ is one-to-one and onto).

This means that as we feed each n ∈ N into φ our output willeventually produce the whole of N without duplication.

Two examples:

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, . . .} → {2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, . . .}{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, . . .} → {1, 3, 5, 7, 9, 2, 4, 6, 8, 10, 11, 13, . . .}

So, we can allow for any such rearrangement φ. . .




Lecture 17

Definition 5.14 in The Handbook

A series∑∞

m=1 bm is a re-arrangement of a series∑∞

n=1 an if thereexists a bijection φ : N → N such that bm = an for m = φ(n).

Rearrangements Theorem, Theorem 5.15

Every rearrangement of an absolutely convergent series isabsolutely convergent and all have the same sum.

Discussion: boardwork

Remark: Since ln(2) = −∑(−1)nn−1 is conditionally convergent

we cannot expect Theorem 5.15 to generalise to conditionallyconvergent series.




Lecture 17

Conditionally convergent series therefore have the potential todiverge — after some rearrangement.

This suggests that there must be some divergent behaviourembedded into each conditionally convergent series. This is thesubject of our next result.

Lemma 5.16

If∑∞

n=1 an is only conditionally convergent, then the series of itspositive terms, and of its negative terms, are both divergent.

Proof: we assume that none of the an are zero and set,

a+n =

{an if an > 00 if an < 0

and a−n =

{0 if an > 0an if an < 0

so that we can expect∞∑

n=1

an =

∞∑

n=1

a+n +

∞∑

n=1

a−n .




Lecture 17

We expect that∞∑

n=1

an =

∞∑

n=1

a+n +

∞∑

n=1

a−n .

There are now just three possibilities for∑

a+n and∑

a−n . . .

1) Both converge: P =∑

a+n and M =∑

a−n .

But this means that∑ |an| = P −M which is not possible

because∑

an is not absolutely convergent.

2) Only one converges: M =∑

a−n and ∞ =∑

a+n for example.

But now∑

an = ∞+M which is not possible because∑

an isgiven as conditionally convergent.

3) Both∑

a+n and∑

a−n diverge.

This completes the proof.




Lecture 17

Remarkably, a conditionally convergent series can be rearranged togive any value whatsoever, including ±∞

Riemann’s rearrangment theorem, Theorem 5.17

If a series∑∞

n=1 an converges conditionally, then for any numbers ∈ R there exists a re-arrangement, φ : N → N, such thats =

∑∞m=1 bm, where bm = an for m = φ(n).

Discussion: boardwork




Lecture 17

Summary

We can:

recognise an alternating series and find its absolute valuecounterpart.

apply and prove the Leibniz convergence test for alternatingseries.

test for absolute and conditional convergence.

apply the rearrangement theorem.

appreciate the result of Riemann’s rearrangement theorem.




End of Lecture

Computational andαpplie∂ Mathematics

Put First Things FirstStephen Covey, The Seven Habits of Highly Effective People

Reference: The Handbook, Chapter 5, Section 5.3.Homework: Finish Exercise Sheet 4aSeminar: Q5



Documents

Lecture 17 - people.brunel.ac.ukpeople.brunel.ac.uk/~icsrsss/teaching/ma2730/lec/print4lec17.pdf · Overview (MA2730,2812,2815) lecture 17 Lecture slides for MA2730 Analysis I