Lecture 3 - Brunel University Londonpeople.brunel.ac.uk/~icsrsss/teaching/ma2730/lec/print8...Shaw bicom , mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 Overview

Overview (MA2730,2812,2815) lecture 3

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 6, 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 3

MA2730: topics for Lecture 3

Lecture 3

Shortcuts

Collected IMPORTANT results

Error bounds

Examples and Exercises

The material covered in this lecture can be found in Section 1.2 ofThe Handbook.




Lecture 3

Some remarks on the lectures

Now we are underway you will have noticed several things

I summarise The Handbook, but I don’t teach its detail

I make mistakes.

Don’t be intimidated — if you spot a mistake then tell me.

I ask for questions

Don’t be shy — if you have a question then ask me.

Everyone will be grateful

Mark Twain:

He who asks is a fool for five minutes, but he who does not askremains a fool forever




Lecture 3

Shortcuts

Sometimes the development of a Maclaurin or Taylorpolynomial can be greatly simplified by spotting patterns.

Pattern spotting plays a big role in maths research.

New results can sometimes be guessed by seeing a pattern inexisting results.

Of course, the guess is not enough. Its correctness has to beproven.

We are going to see how pattern spotting can be used toderive some famous and important Maclaurin polynomials.




Lecture 3

Consider the Maclaurin polynomial of cos(x)

Tnf(x) =n∑

m=0

xm

m!f (m)(0) =

n∑

m=0

amxm with am =1

m!f (m)(0)

and with f(x) = cos(x) we have,

f ′(x) = − sin(x), f ′′(x) = − cos(x), f ′′′(x) = sin(x), f (4)(x) = cos(x)

Then f (5)(x) = − sin(x) = f (1)(x).

Can you see the pattern? Since f(x) = f (0)(x), at x = 0 we have,

f (0)(0) = f (4)(0) = f (8)(0) = · · · = 1

f (1)(0) = f (5)(0) = f (9)(0) = · · · = 0

f (2)(0) = f (6)(0) = f (10)(0) = · · · = −1

f (3)(0) = f (7)(0) = f (11)(0) = · · · = 0




Lecture 3

derivatives of cos(x)

If f(x) = cos(x), at x = 0 we have,

f (0)(0) = f (4)(0) = f (8)(0) = · · · = 1

f (1)(0) = f (5)(0) = f (9)(0) = · · · = 0

f (2)(0) = f (6)(0) = f (10)(0) = · · · = −1

f (3)(0) = f (7)(0) = f (11)(0) = · · · = 0

Therefore, since am = f (m)(0)/m! we have m! am = f (m)(0) and,

0! a0 = 4! a4 = 8! a8 = · · · = 11! a1 = 5! a5 = 9! a9 = · · · = 02! a2 = 6! a6 = 10! a10 = · · · = −13! a3 = 7! a7 = 11! a11 = · · · = 0




Lecture 3

Maclaurin expansion of cos(x)

Using am = f (m)(0)/m! and knowing that,

0! a0 = 4! a4 = 8! a8 = · · · = 11! a1 = 5! a5 = 9! a9 = · · · = 02! a2 = 6! a6 = 10! a10 = · · · = −13! a3 = 7! a7 = 11! a11 = · · · = 0

The general Maclaurin expansion of f(x) = cos(x),

Tnf(x) =

n∑

m=0

xm

m!f (m)(0) = a0 + a1x+ a2x

2 + a3x3 + · · ·

becomes

cos(x) = 1− x2

2!+

x4

4!− x6

6!+ · · · .




Lecture 3

Maclaurin polynomials of f(x) = cos(x)

Using pattern spotting to make life easy we found that,

Tnf(x) =

n∑

m=0

xm

m!f (m)(0) = a0 + a1x+ a2x

2 + a3x3 + · · ·

for f(x) = cos(x) becomes

cos(x) = 1− x2

2!+

x4

4!− x6

6!+ · · · .

Example

Write down T10 cos(x) and T11 cos(x).

Comment on these results.




Lecture 3

Pause to summarise

We have found that:

Pattern spotting can ease our derivations.

Two Maclaurin (or Taylor) polynomials can be the same.

Another example of these observations is given inSubsection 1.2.2 of The Handbook for f(x) = sin(x).




Lecture 3

Extra homework: (1± x)−1 = a0 + a1x + a2x2 + · · ·

Use pattern spotting to determine the Maclaurin expansions of

(1 + x)−1 and (1− x)−1.

First: derive the following pattern and then take x = 0:

dm

dxm(1± x)−1 = (∓1)mm! (1± x)−(m+1).

Then: deduce that for (1± x)−1 we get am = (∓1)m. Hence,

(1± x)−1 = 1∓ x+ x2 ∓ x3 + x4 ∓ x5 + x6 ∓ x7 + · · ·

(1 + x)−1 = 1− x+ x2 − x3 + x4 − x5 + x6 − x7 + · · ·(1− x)−1 = 1 + x+ x2 + x3 + x4 + x5 + x6 + x7 + · · ·




Lecture 3

An important remark

We just saw that


Take x = 1 in the first of these and we get,

1

2= 1− 1 + 1− 1 + 1− 1 + 1− 1 + 1− 1 + · · ·

Do you believe this?Discussion and boardwork on allowable values of x.We’ll return to these issues in later lectures.




Lecture 3

Build on what you know

We just saw that


Now determine the Maclaurin expansions of

ln(1 + x) and ln(1− x).

Boardwork. . .




Lecture 3

Collected IMPORTANT results

You should know, remember for ever, and be able to derive:

sin(x) = x− x3

3!+

x5

5!− x7

7!+ · · · =

∞∑

n=0

(−1)nx2n+1

(2n+ 1)!

cos(x) = 1− x2

2!+

x4

4!− x6

6!+ · · · =

∞∑

n=0

(−1)nx2n

(2n)!,

ex = 1 + x+x2

2!+

x3

3!+

x4

4!+ · · · =

∞∑

n=0

xn

n!.

(1± x)−1 = 1∓ x+ x2 ∓ x3 + x4 ∓ x5 + x6 ∓ x7 + · · ·

± ln(1± x) = x∓ x2

2+

x3

3∓ x4

4+

x5

5∓ x6

6+ · · ·

Remember that x may only take certain values! More later. . .




Lecture 3

Extra Homework

Recall that

ln

(1 + x

1− x

)= ln(1 + x)− ln(1− x)

and use what you already know to show that

1

2ln

(1 + x

1− x

)= x+

x3

3+

x5

5+

x7

7+

x9

9+ · · ·




Lecture 3

Some more shortcuts

We have seen how pattern spotting can ease our pain ,We are now going to see how to use the idea of composition offunctions for another short cut.This material is based on the example in Subsection 1.2.3 of TheHandbook where it is shown that

sin(x2) = x2 − x6

3!+

x10

5!− x14

7!+ · · ·

Since sin(x2) = sin(u) with u = x2 we just use the Maclaurinseries for sin(u):

sin(u) = u− u3

3!+

u5

5!− u7

7!+ · · ·

and replace u with x2. Clever eh?Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 3

You try: derive the Maclaurin expansions of

Example

1 e2x.

21

1 + 4x.

You have 120 seconds:

Solution to follow: boardwork. . .Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 3

And now a hint at the hard stuff. . .




Lecture 3

Back to our first example

Remember that earlier we saw that by chopping off

sin(x) = x− x3

3!+

x5

5!− x7

7!+ · · · =

∞∑

n=0

(−1)nx2n+1

(2n + 1)!

at the x5 term gave a good approximation to sin(x) near to x = 0.

x− x3

3!+

x5

5!looks like:

Key Point: if x is close to zerothen x2, x3, x4, . . . all becomesmaller and more negligible asx → 0.

−10 −5 0 5 10−1.5

−1

−0.5

0

0.5

1

1.5

sin(x)polynomial




Lecture 3

A discussion of error

So, chopping

sin(x) = x− x3

3!+

x5

5!− x7

7!+ · · ·

at the x5 term gives a good approximation to sin(x) near to x = 0:

sin(x) ≈ T5 sin(x) with T5 sin(x) = x− x3

3!+

x5

5!.

Can we quantify the error? That is, can we answer this question:

sin(x)− T5 sin(x) =?

To do this reqires an error estimate: we will look at the Peanoform as described in Subsection 1.2.4 of The Handbook.




Lecture 3

The Peano form of the remainder

In general, the Taylor polynomial of degree n, about x = a, of thefunction f is denoted as T a

nf(x) and introduces an approximation:

f(x) ≈ T anf(x).

The error is introduced by chopping the series off at a finitenumber of terms.The part that is ‘chopped off’ is called the remainder.We denote it by

Ranf(x) = f(x)− T a

nf(x)

to emphasise that its size depends on n, a and x as well, of course,as the actual function f .




Lecture 3

Formality and precision

Definition 1.8 in The Handbook

Let I ⊆ R be an open interval, a ∈ I and f : I → R be an n-timesdifferentiable function, then Ra

nf(x) = f(x)− T anf(x) is called the

n-th order remainder (or error) term of the Taylor polynomial T anf

of f . We write R0nf simply as Rnf for Maclaurin polynomials.

Proposition 1.9 in The Handbook

Let I ⊆ R be an open interval, a ∈ I and f : I → R be an n-timesdifferentiable function, then there exists a neighbourhood ofa ∈ J ⊆ I and a function hn : J → R such that the error termRa

nf satisfies Ranf(x) = hn(x) (x− a)n with limx→a hn(x) = 0.

Explanation — Boardwork.




Lecture 3

That’s pretty technical!Don’t worry, we’ll come back to it in later lectures.For now it is important to understand that

f(x)− T anf(x) = Ra

nf(x) with Ranf(x) = (x− a)nhn(x).

Here hn is a function that gets smaller as x gets closer to a:

limx→a

hn(x) = 0

and (x− a)n → 0 as n → ∞ if |x− a| < 1.In general terms this is pretty hard to understand. We’ll just givean example of it for now and come back to it in later lectures.




Lecture 3

Example

The Peano form of the remainder is,

f(x)− T anf(x) = Ra

nf(x) with Ranf(x) = (x− a)nhn(x),

where hn is a function that gets smaller as x gets closer to a:

limx→a

hn(x) = 0.

To illustrate this accept, for now, the fact that with a = 0,

ex = 1 + x+x2

2︸︷︷︸this is T2ex

+Kx3︸︷︷︸R2ex

for an existing but unknown K(x).

Hence h2(x) = K(x)x so that R2ex = x2h2(x) as required.




Lecture 3

Any questions or comments?




Lecture 3

Summary

We can:

exploit pattern spotting to speed our work.

use function composition Maclaurin expansions.

write a remainder term for Taylor polynomials.

We saw:

that the remainder term requires more study.

that the Taylor or Maclaurin polynomials might give dubiousvalues.




End of Lecture

Computational andαpplie∂ Mathematics

He who asks is a fool for five minutes, but he who does not askremains a fool forever Mark Twain

The material covered in this lecture can be found in Section 1.2 ofThe Handbook.Homework: Questions 5, 6, 9, 10 on Exercise Sheet 1a.



Documents

Lecture 3 - Brunel University Londonpeople.brunel.ac.uk/~icsrsss/teaching/ma2730/lec/print8...Shaw bicom , mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 Overview