Lecture 4 - Brunel University Londonpeople.brunel.ac.uk/~icsrsss/teaching/ma2730/lec/print8...Shaw...

Overview (MA2730,2812,2815) lecture 4

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 4

MA2730: topics for Lecture 4

Lecture 4

Calculus of Taylor polynomials

Examples and Exercises

The introductory material covered in this lecture can be found inThe Handbook, Chapter 1, Section 1.3: Calculus of TaylorPolynomials.




Lecture 4

We saw some useful short cuts in the last lecture.

They were examples of a general calculus for these polynomialexpansions.

In this lecture we are going to look at the detail of this calculus. Itis based on the content of Section 1.1, Calculus of Taylorpolynomials, in The Handbook

Any questions or comments before we begin?




Lecture 4

Can you feel the change in tempo? It’s getting harder and faster.

Today we are going to take a glimpse of the technicalities thatunderlie Taylor expansions.

The good news is that for most applications, most of thetime, we don’t need all these details.

BUT: some are very important - hence we study them.

Mathematics is a part of physics. Physics is an experimentalscience, a part of natural science. Mathematics is the part of

physics where experiments are cheap. V.I. Arnold




Lecture 4

Addition of Taylor polynomials

Part 1 of Theorem 1.11 in The Handbook says that:

Theorem 1.11, Part 1

If h is the sum or difference of two other functions, f and g,

h = f ± g

Then the Taylor polynomial of h is the sum of the Taylorpolynomials of f and g. Therefore,

T anh = T a

nf ± T ang

Boardwork: based on the example of ex + sin(x) underTheorem 1.11 in The Handbook.

This is an easy theorem to prove. Try It!




Lecture 4

Advanced concepts

Convince yourself that If α and β are arbitrary numbers then

T an (αf + βg) = αT a

nf + βT ang

This means that T an is a linear operator.

Challenge homework

Take x = iθ in the Maclaurin expansion for exp(x).

Simplify by using i2 = −1, i3 = −i etc.

Separate the expansion into two parts: one containing i andthe other not.

Notice that these two parts give sin(θ) and cos(θ).

Conclude that eiθ = cos(θ) + i sin(θ).




Lecture 4

Degree of combinations of polynomials

Let p be a polynomial of degree n and q be a polynomial of degreem. For example,

p(x) = 3x3 + 4x and q(x) = x9 + 3.

We can add them to get p+ q; subtract to get either p− q or q− p;multiply to get pq; or, compose them to get p ◦ q (i.e. p(q(x))).It is of interest to know the degree of the resulting polynomial.What is the degree of each of these?

p+ q

p− q

q − p

pq

p ◦ qBoardwork.




Lecture 4

Products of polynomials

Once again (needed for Part 2 of Theorem 1.11). . .Let p be a polynomial of degree n and q be a polynomial of degreem. For example,

p(x) = 3x3 + 4x and q(x) = x9 + 3.

It is of interest to know the degree of the resulting polynomial.what is the degree of each of these?

p+ q max{m,n}p− q max{m,n}q − p max{m,n}pq m+ n

p ◦ q mn

This is the first part of Lemma 1.10 in The Handbook.What does ‘Lemma’ mean?




Lecture 4

Chopping Taylor polynomials early

The second part of Lemma 1.10 in The Handbook says:

Suppose that we already have the Taylor polynomial, T anf , of

degree n for the function f .

Given k < n, how can we easily obtain T ak f?

Lemma 1.10

If 0 6 k < n then T ak f is obtained from T a

nf by ignoring all theterms in T a

nf containing (x− a)ℓ for ℓ = k + 1, k + 2, . . . , n.

Example: boardwork




Lecture 4

Taylor series of a Taylor series

This ought to be obvious but only after a bit of thought.

Mathematicians often use the word obvious to mean this is reallysubtle, and you have to think long and hard about it. . . ,Set g(x) = T a

nf(x) for some function f .

What is T ang(x)?

Have a think about it. Boardwork (and Theorem 1.5 in TheHandbook)

If k 6 n what is T ak g(x)?

Have a think about it. Boardwork

We can now move on and study Part 2 of Theorem 1.11.




Lecture 4

The product of Taylor polynomials

Part 2 of Theorem 1.11

If f and g have the Taylor polynomials T anf and T a

ng of the samedegree n, and about the same point x = a, then the product,h = fg, has the Taylor polynomial of degree 2n about x = a givenby the product of the Taylor polynomials of f and g:

T a2nh = T a

nf · T ang.

Moreover: T anh = T a

n (Tanf · T a

ng).

You need to know how to use this result BUT the proof istechnically difficult, not given and not examinable.There is, however, a prize for the first high quality hand in!Boardwork: based on the example of e2x(1 + 3x)−1 underTheorem 1.11 in The Handbook.




Lecture 4

Trig. identity: 2 sin(x) cos(x) = sin(2x)

Advanced homework

The trig indentity above is well known. Does it also hold for theMaclaurin expansions of sin(x), cos(x) and sin(2x)?

Hints: show first that,

sin(x) cos(x) =

(x− x3

3!+

x5

5!− x7

7!+ · · ·

)(1− x2

2!+

x4

4!− x6

6!+ · · ·

),

= x− 22x3

3!+

24x5

5!− 26x7

7!− · · ·

Multiply this by 2; compare to the Maclaurin expansion of sin(2x)What do you conclude? Is this a proof?




Lecture 4

Composition of Taylor polynomials

We have seen how to deal with the Taylor expansions of the sum,difference and product of two functions.

Now we look at the compostition of two functions.

f ◦ g — ‘f(g(x)), or f of g of x’

Recall that given two functions f : R → R and g : R → R theircomposition is denoted by f ◦ g and is defined by the rule(f ◦ g)(x) = f(g(x)), which we can read as ‘f of g of x’.

If f(x) = x2 + sin(x) and g(x) = 3x then (f ◦ g)(x) =?

If f(x) = (5 + 7x2)−2 and g(x) = xex then (f ◦ g)(x) =?

Now we can move on to Theorem 1.12Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 4

Theorem 1.12 in The Handbook

If h = f ◦ g thenT amnh = T g(a)

n f(T amg).

In particular:T anh = T a

n

(T g(a)n f(T a

ng)).

You need to know how to use this result. BUT: this result istechnically difficult to prove. The proof is not given and notexaminable.

This looks H O R R I B L E!Let’s take it apart and see what’s going on.




Lecture 4



n f(T amg).


n

(T g(a)n f(T a

ng)).

For the first part, with b = g(a):

set p = T bnf , a polynomial of degree n in (x− b),

set q = T amg, a polynomial of degree m in (x− a).

Then T amnh = T

g(a)n f(T a

mg) = p ◦ qAnd we recall from earlier that p ◦ q is of degreen mn.




Lecture 4



n f(T amg).


n

(T g(a)n f(T a

ng)).

For the second part, with b = g(a):

set p = T bnf , a polynomial of degree n in (x− b),

set q = T amg, a polynomial of degree m in (x− a).

Then T a2nh = T

g(a)n f(T a

ng) = p ◦ qAnd we recall from earlier that T a

nh is obtained from T a2nh by

chopping off the higher degree terms:

T anh = T a

n (Ta2nh) = T a

n

(T g(a)n f(T a

ng))




Lecture 4

Remark 1.13 on Theorem 1.12 in The Handbook

If h = f ◦ g and g(0) = 0 then

Tmnh = Tnf(Tmg).

In particular:Tnh = Tn

(Tnf(Tng)

).

This is from Remark 1.13 in The Handbook.This is the Maclaurin form of Theorem 1.12 and is the form that ismost useful in practice.

Notice the condition g(0) = 0. Compare to the Taylor form:

T anh = T a

n

(T g(a)n f(T a

ng)).

We’ll return to this a bit later. . .Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 4

It’s time for an example

From The Handbook: Determine T12f where f(x) =1

1 + x2.

Don’t try this:

T12f(x) = f(0)+xf ′(0)+x2

2!f ′′(0)+

x3

3!f (3)(0)+ · · ·+ x12

12!f (12)(0)

Instead: put g(x) = x2 (note: g(0) = 0) so that f(x) =1

1 + g(x).

Recall also that

1

1 + g= 1− g + g2 − g3 + g4 − · · ·

Hence: you finish it for homework

1

1 + x2=




Lecture 4

Summary

We can determine the degree of the sum, difference, product andcomposition of two polynomials.

We know how to derive Taylor polynomials for:

sums and differences of functions.

products of functions.

compositons of functions.




End of Lecture

Computational andαpplie∂ Mathematics

Mathematics is a part of physics. Physics is an experimentalscience, a part of natural science. Mathematics is the part of

physics where experiments are cheap. V.I. Arnold

The introductory material covered in this lecture can be found inThe Handbook in Chapter 1 of The Handbook, in Section 1.3:Calculus of Taylor Polynomials.Homework: derive equation (1.16) in The Handbook, and attemptQuestions 5, 6, 9, 10 on Exercise Sheet 1a.



Lecture 4 - Brunel University Londonpeople.brunel.ac.uk/~icsrsss/teaching/ma2730/lec/print8...Shaw...

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