Analysis

Analysis Lecture – with brief lecture notes Shirleen Stibbe

www.shirleenstibbe.co.uk

OPEN UNIVERSITY

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M203 Pure Mathematics Summerschool

Introduction Mathematical Analysis is one of those topics that appear so divorced from 'real' life, that it's quite hard to get enthusiastic about it.

It seems mostly to boil down to 'proving' stuff that looks so immediately obvious, that you begin to doubt the sanity of mathematicians.

But actually, things are not always as they appear.

I'm going to tell you a story about my friend Nigel, and how a rigorous use of analysis could have prevented his catastrophic demise.

But first, a bit of background.

2

Analysis The art of proving the blindingly obvious?

But things may not always be as they seem ‌..

3

The whole trouble started, when I began the unit on Analysis. If I'd gone for Sociology instead of Maths, Nigel would still be alive today!

Anyhow, the first thing I looked at were Sequences – easy stuff, obvious (!). A list of numbers, with a pretty clear pattern – piece of cake.

Easy to tell the next number in the sequence, and the general term – isn't it?

And by the way: c) also satisfied by an = 2 + (n – 1)/n d) your first example of things not being as they seem!

Null sequences – the basic weapons you'll need for analysing all sequences.

Just one point about the examples on the slide: We've taken N to be [1/ ε], i.e. the largest integer less than or equal to 1/ ε. This N is the smallest we could have chosen – but any N > [1/ ε] would have done as well.

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A Sequence {an} is an unending list of real numbers a1, a2, a3, …, with nth term an.

Examples:

/ , ..1/4 ✔

1/ , 1 2 3

a)

1,

b)

1, 2, 3, .. 4

c)

2, 2½, 22/3,

d)

3, 3, 3, .. 3

-597

✔

an = 1/n an =

✔

n ✔ an = 3 - 1/n

.. 23/4✔

✔

✘ an = 3 ✘ 3-100(n-1)(n-2)(n-3)

A sequence {an} is Null if, for every positive number ε, there is an integer N such that |an| < ε for all n > N Example: an = 1/n Particular

General

|an| = 1/n < 0.3

|an| = 1/n < ε

⇔ n > 1/0.3

⇔ n > 1/ ε

⇔ N = [10/3]

Sequence Diagram

ε ⇔ N = [1/ ε] -ε

N

n

= 3 5

Anyway, I'd just got to this point in the unit, when I had a visit from my friend, Nigel.

Although we'd been close at one time, I hadn't seen him for quite a while. So I was surprised when he turned up one day seeking my help with a problem.

Nigel is a carpet seller, He's not just any old carpet seller, though. He happens to be the best carpet seller in the entire universe!

There's nothing he doesn't know about carpets. And nothing he doesn't know about selling.

The trouble is, there's quite a lot a he doesn't know about mathematics. And that's why he called on me â&#x20AC;&#x201C; big mistake, as it turned out!

He was negotiating the sale of his most expensive carpet to a new client, a Mr Sierpinski, and was having difficulty working out the implications of the contract.

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My friend Nigel

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The carpet was stunningly beautiful, and Mr Sierpinski was blown away by it. The problem was he also loved his polished wooden floor with a deep and abiding passion.

Mr Sierpinski, being a mathematician, had come up with a really smart solution to his dilemma.

Suppose you divide the carpet (which is square) into nine equal squares, and remove the middle square, so the beautiful floor shows through.

Then divide each of the remaining squares into nine equal squares, and remove the middle one â&#x20AC;&#x201C; and so on.

You should be able to do it at least a hundred times, since the room is so big.

Nigel had agreed to give it a go, and that Mr Sierpinski would pay only for the actual amount of carpet finally laid. Seemed OK to him â&#x20AC;&#x201C; it was an eye-wateringly expensive carpet.

He asked me to calculate how much carpet he would have to supply.

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So I hauled out my maths kit (course notes, stack of paper, five pencils, three erasers, hot towels, bucket of coffee) and plunged in.

Apologies for the feet and inches, by the way, but it was a long time ago, when we were still living in black and white.

It took only four or so hours for me to spot a pattern â&#x20AC;&#x201C; I was dealing with a sequence!

So naturally, I returned to the course notes to find out what to do next.

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Nigel's Carpet Problem 30 ft

Total area = 30 × 30 = 900 sq feet = 1 unit an = units remaining after nth step

a0 = 1 a1 = 8/9 × a0 = 8/9

…

a2 = 8/9 × a1 = (8/9) 2 an = 8/9 × an–1 =

(8/9) n

{ an } = { (8/9) n } 11

The OU was really helpful. It obligingly gave me a list of basic null sequences.

Notice in the examples that the denominator is the dominant term – it grows faster than the numerator.

This is why they're all null sequences – converge to 0.

I didn't take me long to decide that the pattern of Nigel's sequence matched sequence b), since 8/9 < 1.

Then I did some calculations – and wished I hadn't!

One square inch? Not a lot of profit in that, no matter how expensive the carpet.

I dreaded breaking the news to Nigel.

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Basic Null Sequences

Examples

a) { 1/n p }

p>0

{ 1/n 2 }

b) { c n }

|c| < 1

{ (2/3) n }

c) { n p c n }

p > 0, |c| < 1

{ n4 /3n }

d) { c n/n! }

câ&#x2C6;&#x2C6;R

{ 10n/n! }

e) { n p/n! }

p>0

{ n 5/n! }

Nigel's Carpet Problem: { an } = { (8/9)n } is a null sequence! n

an (units)

area

20

0.095

85 sq feet

50

2.8 x 10-3

2.5 sq feet

100

7.6 x 10-6

1 sq inch 13

It was as bad as I expected.

Of course, Nigel went ballistic when he saw the bottom line.

But fortunately, when he's angry he lapses into an obscure Quenyan dialect, so my ears weren't offended by his comments.

It was all so upsetting, that when he left, a deeply unhappy man, I immediately turned back to the analysis units for solace.

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Nigel's comments

✯❈ ❂◆❇❇❅❒✁

✷❈❁▼ ❁ ❃❏❍❐●❅▼❅

❂❁▲▼❁❒❄✁ ★❅ ❋■❅◗

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Now we're onto more interesting sequences – a bit more robust than those null wimps which just fizzle out to nothingness.

Get to know the combination rules – they're extremely useful.

They allow you to decompose complex sequences into simpler ones whose limits may be much easier to find and prove.

Important Note Always show where you're using the rules in assignments and in the exam, and say that you're using them. You get marks for justifying your assertions (i.e. adding in the twiddly bits)– and lose them when you don't!

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The sequence {an} is convergent with limit k if {an – k} is a null sequence.

Notation: an → k as n → ∞ Example:

lim an = k

or

n→∞

an = 3 − 1 n k+ε

an − 3 = − 1 n

k-ε

Since {–1/n} is null, n

an → 3 as n → ∞

Combination Rules: if

an → k and bn → m , then

Sum rule:

an + bn → k + m

Multiple rule:

λan → λk

Product rule:

an bn → k m

Quotient rule:

if

m

Reciprocal rule:

if if

k ≠ 0 1 / an → 1 / k k = 0 { 1/an } diverges

≠0

λ∈R

an / bn → k / m

Squeeze Rule: if bn>an, {an} diverges

⇒ {bn} diverges 17

The strategy for finding limits that they suggest is almost a 'turn the handle' operation â&#x20AC;&#x201C; the only things you have to remember are the dominance order, and the basic null sequences â&#x20AC;&#x201C; both of which are in the Handbook.

Maybe you could save time and write out the template of the strategy before the exam starts, then just fill in the numbers when you see the question?

But take care with the details. This sort of question usually attracts 5 marks: The empty template is probably worth about 2 marks (for method), 1.5 marks for accuracy, and 1/2 a mark for each twiddly bit ("basic null sequences", "from the Combination Rules")

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Strategy: to find the limit of a convergent sequence 1 Identify the dominant term:

cn (c > 1) dominates n p (p > 0) ] 2 Divide numerator and denominator of an by the [ n! dominates

dominant term.

3 Identify any null sequences and apply the Combination Rules.

Example: limit of

an =

3n + 5n 2 − 2 n

4 + 3n + 1

as n → ∞

Template: fill in the missing bits 1

The dominant term is

4n .

2

Dividing through by

4n

gives ** Twiddly bits

3

As

{(3/4)n,}, {n2/4n}, {1/4n}, {n/4n} are basic null

sequences**, it follows from the Combination Rules** that

as n → ∞ .

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Here's a useful weapon for checking the convergence of sequences that may not be amenable to the 'template' strategy.

You may have spotted immediately that this (very simple) example is a null sequence. The bad news: If the question asks you to prove it's convergent, the smart-arse answer (Yes – basic null sequence – see unit p11) won't cut any ice!

And by the way, it's essential to show that the sequence satisfies both criteria of the theorem (i.e. decreasing and bounded below). And also to state which theorem you're appealing to ("by the MCT").

Back to Nigel's story I'd just finished this section, when Nigel turned up again.

Apparently, he'd gone to see Mr Sierpinski to turn down the contract – but Mr S had come up with another idea!

He worried that the cut edges of the carpet might look somewhat ragged and unfinished. So his idea was to sew a beautiful gold braid around all the edges. He invited Nigel to undertake the work.

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Monotone Convergence Theorem If the sequence {an} is Either increasing and bounded above Or

decreasing and bounded below

then {an} is convergent.

Example Prove that the sequence {a n} = convergent.

is

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So it was back to the maths kit for me, to work out how much gold braid Nigel would need.

This was much harder than before – I even had to dredge up the identity: 1 + x + x2 + … + xn-1 = (xn – 1) / (x – 1). Remember that?

It took three whole days and several buckets of coffee to come up with the formula.

All those Sequence rules convinced me that this was a diverging sequence, so I knew there might be a problem for Nigel, even before I crunched the numbers.

And I was right!

9 x 1018 miles ≈ the distance to Alpha Centauri and back 190,000 times.

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Nigel's Carpet Opportunity 10 yards

Perimeter = 4 × 10 = 40 yards = 1 unit bn = units of braid needed at the nth step b1 =

1/3

b2 =

b1 + b1 /3 × 8 = (1/3)[1 + 8/3]

b3 =

b2 + b1 /9 × 64 = (1/3)[1 + 8/3 + (8/3) 2]

… bn = (1/5) [(8/3) n– 1] > 2n–3 for n ≥ 1 { 1/

2n–3} is null , so {2n–3} diverges (Reciprocal

Rule), and {bn} also diverges (Squeeze Rule) n

b (units)

length of braid

10 20

3,636 66,130,931

82 miles 1,502,975 miles

50

n

3 ! 10

20

9 ! 10

18

miles

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I thought Nigel would explode when I told him the bad news.

He was absolutely delighted!

"But Nigel," I said, "don't you see? It's absolutely impossible. There just isn't that much gold braid in the entire universe!"

His reply alarmed me.

"Ah â&#x20AC;&#x201C; but Mr Sierpinski doesn't know that!

"This is a job for life. I'll never have to sell another carpet and I'll still be rich. That gold braid doesn't come cheap, you know!"

He was a truly happy man. He had visions of a golden future as a wealthy and well-connected socialite, being invited to the Queen's garden parties â&#x20AC;&#x201C; the works!

I didn't have the heart to spoil his dreams with mere reality, so he left, still euphoric.

And I turned to the next topic in the course, Continuity.

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Nigel's vision

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Here we discover that sequences are useful for more than just making carpet merchants miserable or ecstatic.

Just try testing continuity of functions without them!

And why do we care whether functions are continuous or not?

Well, if they're not they could kill someone!

The example is one of those instances when the answer is so patently obvious (once you've drawn the picture!), that it seems absurd to have to prove it.

But life doesn't normally provide such innocent functions. This example just gives you some practice in using the techniques that might be absolutely vital for more complex functions.

NB: when you draw the picture, make sure you show clearly which branch f(1) lies on (filled blob).

If you're sliding down the right branch from the right, you'll fall down a hole when you hit x =1. That's what you use to prove discontinuity at x = 1.

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A function f : A → R is continuous at a ∈ A if, for every sequence { xn } in A,

if xn → a then f ( xn ) → f (a ) Example: ⎧ x 2 if x ≤ 1 f ( x ) = ⎨ ⎩ x − 1 if x > 1

Is f continuous

a) at x = 0?

b) at x = 1?

First draw the picture.

x



x2 x



x− 1

⎧ x 2 if x ≤ 1 f ( x ) = ⎨ ⎩ x − 1 if x > 1 27

a) The same definition applies for x = 0 ± ε for every ε ≤ 1. So you can use the properties of a general null sequence to show that the condition is satisfied for every sequence converging to 0. b) Here you can see that the definition depends on whether x = 1 + ε or x = 1 – ε, where ε > 0.

You need only one rogue sequence - counterexample – to show that the proposition (i.e. that f is continuous at 1) is false.

So choose a sequence {an} that: i) is always > 1, (so that f(an) stays on the right hand branch and falls down the hole ii) converges to 1.

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⎧ x 2 if x ≤ 1 f ( x ) = ⎨ ⎩ x − 1 if x > 1

Is f continuous a) at x = 0? b) at x = 1?

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Note that the OU has looked after your interests again.

It gives you a list of basic continuous functions, and a set of rules you can use to avoid having to do the 'sequence test' for a wide range of functions.

There's something missing though, from the list of basic continuous functions.

It may not seem important to you that there's no mention of the domains of the BCFs. But, as you'll see, this apparently minor omission brought about an event of such catastrophic devastation, that it has cast a blight on my life ever since.

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Basic Continuous Functions a) polynomials and rational functions b) modulus function c)

f (x ) = x

square root function

f (x ) = x

d) trig functions sin, cos, tan e) exp/log functions

ex, xa,

Combination Rules If f and g are continuous at

loge(x)

a, then so are:

Sum Rule:

f + g

Multiple Rule:

λ f, λ ∈ R fg f / g if g(a) ≠ 0

Product Rule: Quotient Rule:

Composition Rule If f is continuous at a, and g is continuous at f(a), then g ( f ) is continuous at a.

Other useful rules:

g and h are continuous at a, then so is f if Local f = g on an interval around a Squeeze g ≤ f ≤ h (x ≠ a), g = f = h at a Glue f = g (x < a), f = h (x > a), g = f = h at a If

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The Intermediate Value Theorem can sometimes be quite useful.

This fairly innocuous example might have arisen if you were trying to work out whether there is a value x, such that x3 + 2x2 = ex. Note that it doesn't give you the actual value of x, just shows that the answer to question is yes!

The example shows you how to use the rules, and, in particular, which twiddly bits are necessary to write down to justify your arguments.

By the way, pick any numbers that are easy to work out. You don't have to do anything clever. I chose 1 and -10, so I didn't need a calculator.

By this stage, you're probably wondering where Nigel had got to in all this time â&#x20AC;Ś

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Intermediate Value Theorem If and then

f is continuous on [ a, b ], k lies between f(a) and f(b), there exists a number c in ] a, b [ such that f(c) = k.

Example Prove that f(x) = x3 + 2x2 - ex has at least one zero on R. Solution xn and ex are basic continuous functions.

So f is continuous on [–10, 0] (by the Combination Rules). f (1) = 1 + 2 – e > 0

f (–10) = – 1000 + 200 - e–10 < – 800 < 0 By the IVT, there exists c in ]–10, 0[ such that f (c) = 0.

Therefore f has a zero on R.

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The next time I heard from Nigel, he had - in a sense achieved his goal.

He was in fact a guest of Her Majesty – but not quite in the way he'd expected. HMP Broadmoor is a pretty poor substitute for Buckingham Palace.

So what happened to land Nigel in jail? Well, it appears that he started work on Mr Sierpinski's carpet, and at the end of the first month he asked for his month's pay.

But Mr S was a cunning old mathematician. "Oh no," he said. "You'll only be paid when the job is complete."

Now Nigel isn't one to hold back on his emotions. But he really lost it this time! The pictures of Mr Sierpinski from the police files, before and after Nigel had finished with him, tell the whole story.

His letter said that he was planning a breakout – couldn't stand the food in Broadmoor – and needed a formula to program a flightpath for his carpet. He included a picture of a peak he'd have to fly over.

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Crime Scene Report

Case number: 347/26A!

Waclaw Sierpinski 1882 - 1969

4.7 km

5 km

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It was a nightmare trying to find a mathematical function which would generate a shape like that â&#x20AC;&#x201C; but I'm not one to give up easily, and eventually (8 days later) I had it!

The function was pretty damned complicated, but Its graph fitted Nigel's picture almost exactly!

I checked it carefully to make sure it was kosher: ex, tan, log, xn are all basic continuous functions.

Apply the combination and composition rules and bob's your uncle.

I was reassured that it would do the trick, no problem.

I sent the formula to Nigel straight away, along with a suitably filled cake â&#x20AC;&#x201C; anything to help an old mate!

AND I NEVER HEARD FROM HIM AGAIN!

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The complicated function f(x) = 3e-3x (tan (loge(x + 1) ) + 10x6)

The suitably filled cake

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I did get a call from the police, however.

They explained that Nigel had broken out of prison.

They'd been tracking his progress from a helicopter, but he'd suddenly disappeared completely off the radar screen. They were absolutely baffled. Could I help?

I told them I knew nothing about it, of course. Well, you have to protect your friends, haven't you?

But it did start me wondering whether my formula might have had something to do with it. It looked like a nice, well-behaved, continuous path but â&#x20AC;Ś.

I'd initially plotted 50 points. Since Excel merely joins the plotted points, could the path misbehave inbetween?

Just to be sure, I plotted 1000 points. A blip appeared between 3.7 and 3.9. Hmmmm â&#x20AC;Ś So I plotted 10,000 points just in that interval, and OMG! The function is catastrophically discontinuous!

The path splits into two parts. The left branch suddenly turns upward, and goes up forever, the other goes downward, and down for ever!

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The catastrophic discontinuity f(x) = 3e-3x (tan (loge(x + 1) ) + 10x6)

50 points

1,000 points

Discontinuity at x â&#x2030;&#x2C6; 3.8 3.7

3.9

10,000 points

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Of course, I blamed Bill Gates to start with. Bloody Excel misled me!

Then I looked a bit more closely at the function.

You saw how carefully I'd checked the functions and rules. But what I didn't check was the domains!

The tan function is continuous – but only in the open intervals ] (2n-1)π/2, (2n+1)π/2 [ In particular, tan(x) is discontinuous at x = π/2.

So for f(x) = 3e-3x (tan (loge(x + 1) ) + 10x6), everything goes swimmingly in (0, 5), except: for x = eπ/2 – 1, loge (x + 1) = loge (eπ/2) = π/2 and then we're in dead lumber!

And by the way, there's no such word as discontinuitant – but there should be!

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The catastrophic discontinuitant g(x) = tan x

-π/2

π/2

3π/2

Discontinuous at x = π/2. (and lots of other places)

Therefore

f(x) = 3e-3x (tan (loge(x + 1) ) + 10x6) Discontinuous at x = eπ/2 – 1

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And that's how a discontinuous function can kill your friends.

And you know the worst part? It's not knowing which direction he was travelling in.

Upward path? He's headed towards UDFy-38135539 (look it up). On the other? Don't even think about it.

And just in case you were wondering what happened to Mr Sierpinsky – well that's not a pretty story either.

When the medics finally got the gold braid off his mouth, the guy started calling urgently for ... a carpenter!

He'd had this idea for a coffee table, you see.

You take a large cube of wood, divide it into 27 equal cubes, and remove the centre cube. Then divide each of the 26 remaining cubes into 27 equal cubes .......

The men in white coats were able to secure a place for him in a very pleasant facility. And now he spends his days happily building a gasket by dividing a triangle into 4 equal triangles and removing ……..

THE END

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Which way did he go?

Mr S's coffee table

Mr S's gasket