Lecture 2icsrsss/teaching/ma2730/lec/print8lec...Brunel University September 21, 2015 Shaw bicom ,...

Overview (MA2730,2812,2815) lecture 2

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

September 21, 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 2

MA2730: topics for Lecture 2

Lecture 2

Finishing our example

Properties of Taylor polynomials

Theorems on existence and uniqueness

Examples

The introductory material covered in this lecture can be found inThe Handbook in Section 1.2




Lecture 2

Summary

We can derive the Maclaurin series expansions:

sin(x) = x− x3

3!+

x5

5!− x

7

7!+ · · · =

∞∑

n=0

(−1)nx2n+1(2n + 1)!

cos(x) = 1− x2

2!+

x4

4!− x

6

6!+ · · · =

∞∑

n=0

(−1)nx2n(2n)!

,

ex = 1 + x+x2

2!+

x3

3!+

x4

4!+ · · · =

∞∑

n=0

xn

n!.

Do you recall eiθ = cos(θ) + i sin(θ) from Level 1?




Lecture 2

Summary. We deduced the general form:

The Taylor and Maclaurin expansion (series)

The infinitely differentiable function f : R → R has the formalexpansion about a ∈ R given by:

f(x) =

∞∑

n=0

(x− a)n f(n)(a)

n!

If a = 0 this is called the Maclaurin expansion or series, otherwiseit is called the Taylor expansion or series.

And we used Taylor’s series to approximate a function about x 6= 0.This idea is useful. See also (1.1) in Section 1.1 of The Handbookwhere it is shown that,

1 + x+ x2 + x3 = 40 + 34(x − 3) + 10(x− 3)2 + (x− 3)3.Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 2

Exercise

Examples

Expand sin(x) about x = π/4 up to third powers. Use it toapproximate sin(47◦) = sin(47π/180) and determine the error.

You have 4 minutes.




Lecture 2

Solution: boardwork.




Lecture 2

Summarising

We found that:

sin(x) ≈ 1√2

(1 +

(x− π

4

)− 1

2!

(x− π

4

)2− 1

3!

(x− π

4

)3).

You can check this at http://www.wolframalpha.comJust type in: taylor series sin(x) about pi/4.

So sin(47◦) = 0.731353701 . . . and with x = 47π/180:

sin(x) ≈ 1√2

(1 +

(47π

180− π

4

)− 1

2!

(47π

180− π

4

)2− 1

3!

(47π

180− π

4

)3)

we get sin(47◦) ≈ 0.731353657.




Lecture 2

Maclaurin and Taylor polynomials

We have seen that,

The Taylor and Maclaurin expansion (series)

The infinitely differentiable function f : R → R has the formalexpansion about a ∈ R given by:

f(x) =∞∑

n=0

(x− a)n f(n)(a)

n!

If a = 0 this is called the Maclaurin expansion or series, otherwiseit is called the Taylor expansion or series.

But how can we deal with the infinity in the sum?One way is to chop the sum at a finite number of terms. . .This leaves us with a polynomial approximation to f .




Lecture 2

Maclaurin and Taylor polynomials

Definition

Let I ⊆ R be an open interval, f : I → R be an n-timesdifferentiable function at a ∈ I. The Taylor polynomial T anf ofdegree n at a point x = a of the function f is the polynomial

T anf(x) = f(a)+f′(a)(x−a)+f

′′(a)2!

(x−a)2+· · ·+f(n)(a)

n!(x−a)n.

If a = 0 we write the Maclaurin polynomial as Tnf = T0nf for

simplicity.

See Equation (1.2) and Definitions 1.4 and 1.7 in The Handbook.Notice that we can take n as large as we please. . .But there is no infinity.




Lecture 2

Properties of the Taylor polynomial

Theorem

If f : I → R is continuous at a point a ∈ I and if its first nderivatives are also continuous at a ∈ I then the Taylor polynomialp = T anf exists and is the only polynomial of degree n thatsatisfies,

p(a) = f(a) anddmp

dxm(a) =

dmf

dxm(a) for 1 6 m 6 n.

See Theorem 1.5 in The Handbook.

We’re going to spend a bit of time on the statement of thistheorem, and on its proof.

Stating and proving theorems is of central importance inmathematics.




Lecture 2

Proof — the plan

It is important to be able to read a theorem. Usually they are quitebrief, but every word means something.

Theorems have to be proven. . .

. . . and proofs have to be planned.

First parse the theorem and list all of the claims it containsand all of the assumptions it relies upon.

Some assumptions are tacit rather than stated.

Then decide whether or not one claim is contingent onanother. . .

. . . then decide on the order in which to tackle the proofs.

Let’s look again at our theorem and plan its proof.




Lecture 2

What are this theorem’s claims?

Theorem


p(a) = f(a) anddmp

dxm(a) =

dmf


See Theorem 1.5 in The Handbook.First it is saying that the Taylor polynomial exists. This isimportant — it’s easy to talk about things that don’t exist. . .For example: consider the number t > 0 such that

∫ t0 x

2 dx = −1.Second, the theorem is saying that T anf is unique.Which claim should we tackle first? Existence. Why?




Lecture 2

Proof — Part 1

Recall that

T anf(x) = a0 + a1(x− a) + a2(x− a)2 + · · ·+ an(x− a)n.

where each an = f(n)(a)/n! and is well defined as a real number

by the hypotheses of the theorem.Expanding the brackets and collecting terms gives a polynomial ofdegree n. Examples:

n = 1 gives p(x) = a1x+ (a0 − aa1).n = 2 gives p(x) = a2(x

2 − 2ax+ a2) + a1x+ (a0 − aa1). . .. . . so: p(x) = a2x

2 + (a1 − 2aa2)x+ (a2a2 + a0 − aa1)

Hence the Taylor polynomial exists. But is it unique?




Lecture 2

Proof — Part 2

A common technique for proving uniqueness is to assumenon-uniqueness and show that a contradiction results.So, let’s assume there are two Taylor polynomials

p(x) = a0 + a1(x− a) + a2(x− a)2 + · · ·+ an(x− a)n,q(x) = b0 + b1(x− a) + b2(x− a)2 + · · ·+ bn(x− a)n,

each satisfying p(m)(a) = q(m)(a) = f (m)(a) for 0 6 m 6 n.Hence:

p(a) = q(a) implies that a0 = b0.

p′(a) = q′(a) implies that a1 = b1.and so on until. . .

p(n) = q(n) implies an = bn.

Therefore p = q because all of the coefficients agree.Therefore there cannot be two distinct Taylor polynomials.




Lecture 2

The proof is now complete

Theorem


p(a) = f(a) anddmp

dxm(a) =

dmf


We showed that the theorem’s hypotheses were enough toguarantee that T anf exists. And that there was no otherpolynomial with the same properties. So T anf exists and is unique.You should work through the proof yourself. We made a meal of it,but in later proofs we will go faster. . . See Theorem 1.5 in TheHandbook for a different proof.




Lecture 2

A remark about proofs

Every step in a proof has to be justifiable, and a proof is usuallyexpected to be complete.That means that it contains all the detail required for a reader tounderstand why the theorem is true.This can cause some confusion. What does all the detail actuallymean? Do we have to go back to proving everything that camebefore?Surely not! That would mean starting with 1 + 1 = 2 each time. . .The way around this is as follows: a proof is a demonstration thatis accessible to other mathematicians who work in the field withwhich the proof is concerned.If auxilliary results are needed then they are written in a series ofsupporting lemmas. These have their own proofs and are presentedeither before the main theorem or in an appendix.




Lecture 2

A lemma

Lemma

Given an arbitrary a ∈ R, every polynomial of degree n,

A(x) = A0 +A1x+A2x2 +A3x

3 + · · · +Anxn,

can be written in the alternative form

A(x) = B0+B1(x−a)+B2(x−a)2+B3(x−a)3+· · ·+Bn(x−a)n.

This is important because our uniqueness proof above was basedon two polynomials each written in the ‘(x− a)’ form.But what if another Taylor polynomial existed that couldn’t bewritten in that form?This lemma shows that no such alternative can exist.But we need a proof. . .




Lecture 2

Proof

Introduce a new variable y = x− a and substitute x = y + a into

A(x) = A0 +A1x+A2x2 +A3x

3 + · · · +Anxn.

to get,

A(x) = A0+A1(y+a)+A2(y+a)2+A3(y+a)

3+· · ·+An(y+a)n.

Now multiply out the brackets and collect up all terms withcommon powers of y. This results in coefficients B0, B1, . . . , Bnsuch that:

A(x) = B0 +B1y +B2y2 +B3y

3 + · · · +Bnyn.

Substitute (x− a) for y and the lemma is proven.Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 2

This idea is useful. See also the first example in Section 1.1 onThe Handbook where in (1.1) it is shown that,

1 + x+ x2 + x3 = 40 + 34(x − 3) + 10(x− 3)2 + (x− 3)3.

For example, we can quickly see that if x is very close to 3 then(x− 3)2 and (x− 3)3 are negligible in comparison with the otherterms and

1 + x+ x2 + x3 ≈ 40 + 34(x − 3)is a useful local approximation.




Lecture 2

The approximation in pictures

2 2.5 3 3.5 40

10

20

30

40

50

60

70

801 + x+ x2 + x3

40 + 34(x− 3)

Figure : Illustration of f(x) = 1 + x+ x2 + x3 ≈ 40 + 34(x− 3)

Since f(3) = 40 and f ′(3) = 1 + 2(3) + 3(3)2 = 34 we see that40 + 34(x − 3) is T 31 f , where f(x) = 1 + x+ x2 + x3.




Lecture 2

Let’s zoom in. . .

2.7 2.8 2.9 3 3.1 3.2 3.3

30

35

40

45

501 + x+ x2 + x3

40 + 34(x− 3)

Figure : Zoomed illustration of f(x) = 1 + x+ x2 + x3 ≈ 40 + 34(x− 3)

This is the first degree Taylor polynomial of f . It agrees with boththe value of f and the first derivative of f at x = 3.




Lecture 2

Let’s see this again for a more interesting example

We found at the beginning of this lecture that,

sin(x) ≈ 1√2

(1 +

(x− π

4

)− 1

2!

(x− π

4

)2− 1

3!

(x− π

4

)3).

Let’s plot these polynomials up to degree 3 on the right and seewhat they look like.

What are we expecting to see?




Lecture 2

Constant - zero-th degree approximation: POOR!

0 0.5 1 1.5 2 2.5 3 3.5

−0.20

0.2

0.4

0.6

0.8

1 sin(x)1√2

Figure : f(x) = sin(x) and Tπ/40 f near π/4 = 0.785398 . . .

sin(x) ≈ 1√2

(1 +

(x− π

4

)− 1

2!

(x− π

4

)2− 1

3!

(x− π

4

)3).




Lecture 2

Linear - first degree approximation: BETTER!

0 0.5 1 1.5 2 2.5 3 3.5

−0.20

0.2

0.4

0.6

0.8

1

1.2sin(x)

1√2

(1 +

(x− π4

))


sin(x) ≈ 1√2

(1 +

(x− π

4

)− 1

2!

(x− π

4

)2− 1

3!

(x− π

4

)3).




Lecture 2

Quadratic - second degree approximation: GOOD!

0 0.5 1 1.5 2 2.5 3 3.5

−0.20

0.2

0.4

0.6

0.8

1sin(x)

1√2

(1 +

(x− π4

)− 12!

(x− π4

)2)


sin(x) ≈ 1√2

(1 +

(x− π

4

)− 1

2!

(x− π

4

)2− 1

3!

(x− π

4

)3).




Lecture 2

Cubic - third degree approximation: EXCELLENT!

0 0.5 1 1.5 2 2.5 3 3.5

−0.20

0.2

0.4

0.6

0.8

1 sin(x)

1√2

(1 +

(x− π4

)− 12!

(x− π4

)2 − 13!(x− π4

)3)


sin(x) ≈ 1√2

(1 +

(x− π

4

)− 1

2!

(x− π

4

)2− 1

3!

(x− π

4

)3).




Lecture 2

Any Questions?




Lecture 2

Summary

We know how to derive T anf from a smooth function f .

We know that T anf agrees with f at x = a.

We know that, at x = a, all derivatives of T anf up to order nagree with the same order derivative of f at a.

We know that theorems have to be read carefully. . .

. . . and that proofs have to be planned.

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




Lecture 2

Summary

We used http://www.wolframalpha.com to check our work.But understood that it doesn’t replace our own working.Be careful with the internet — it can be a bit of a junk yard.




End of Lecture

Computational andαpplie∂ Mathematics

This lecture was based onSection 1.2, in Chapter 1 of TheHandbook.Homework: attempt Question 3on Exercise Sheet 1a.



Lecture 2icsrsss/teaching/ma2730/lec/print8lec...Brunel University September 21, 2015 Shaw bicom ,...

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