LectureslidesforMA2730AnalysisIicsrsss/teaching/ma2730/lec/lec18.pdf · 2015. 11. 10. · Overview(MA2730,2812,2815) lecture18 Lecture18. MA2730: topicsforLecture18. Lecture18. Taylor’s

Overview (MA2730,2812,2815) lecture 18

Lecture slides for MA2730 Analysis I

Simon Shawpeople.brunel.ac.uk/~icsrsss

[email protected]

College of Engineering, Design and Physical Sciences

bicom & Materials and Manufacturing Research Institute

Brunel University

November 10, 2015

Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel

MA2730, Analysis I, 2015-16

http://people.brunel.ac.uk/~icsrsss

mailto:[email protected]


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



http://people.brunel.ac.uk/~icsrsss/teaching/ma2730


Lecture 18

MA2730: topics for Lecture 18

Lecture 18

Taylor’s series with integral remainder

Lagrange’s form of the remainder

Convergence of Taylor’s series

Examples and Exercises

Reference: The Handbook, Chapter 6, Section 6.1.Homework: Q10, Sheet 5bSeminar: Ad Hoc. Student driven.




Lecture 18

Time Management — Tip 4

Begin with the end in mind.

Seek first to understand, then to be understood.

Stephen Covey, The Seven Habits of Highly Effective People




Lecture 18

Time Management — Tip 4




Think about this in the context of your assignments.

Don’t rush in. Plan. Gather. Execute.




Lecture 18

Reference: Stewart, Chapter 12.10

We now return to Taylor and Maclaurin series.

It’s time to understand when they can be used.

The sums go to infinity. . .

They are infinite series. . .

At the start of term we didn’t know how to deal with them.

Now we do. We’ve come a long way.

Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.




Lecture 18












Lecture 18












Lecture 18












Lecture 18












Lecture 18












Lecture 18












Lecture 18












Lecture 18












Lecture 18

In Lecture 1 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)nf (n)(a)

n!

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

Note the infinite sum — this is now our focus.

Recall the collected IMPORTANT results from Lecture 3. . .




Lecture 18

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+ · · ·

In Lecture 3 We said:“Remember that x may only take certain values! More later. . . ”




Lecture 18

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+ · · ·

In Lecture 3 We said:“Remember that x may only take certain values! More later. . . ”

This is “later”.Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 18

We didn’t know how to deal with the infinite sum so we chopped itand formed Taylor or Maclaurin polynomials.

This meant that the part we chopped off had to be carried along ina remainder. . .

We first saw this in Lecture 3, and today will study the remainderterm in more depth.




Lecture 18

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 a

nf(x).

The error is introduced by chopping the series off at a finitenumber of terms.




Lecture 18

The remainder



f(x) ≈ T a

nf(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.




Lecture 18

The remainder



f(x) ≈ T a

nf(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

Ra

nf(x) = f(x)− T a

nf(x) Taylor form, a ∈ R,

Rnf(x) = f(x)− Tnf(x) Maclaurin form, a = 0,

The size of the remainder depends on n, a and x as well, ofcourse, as the actual function f .




Lecture 18

Yet more from 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.

The point is this. . .

Since Ranf(x) = f(x)− T a

nf(x),

if limn→∞

Ra

nf(x) = 0 then f(x) = limn→∞

T a

nf(x)

and we are guaranteed that f is given by its Taylor series.




Lecture 18

We now have tools to prove that these are in fact valid

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+ · · ·

We’ll start with (1− x)−1 because it is quite special.




Lecture 18

We will deal only with Maclaurin series, where a = 0.

We have Rnf(x) = f(x)− Tnf(x) but, in general, we know onlythat Rnf(x) exists. Putting a specific form to it is much harder.

Except in the case f(x) = (1− x)−1. Here for x 6= 1, theMaclaurin polynomial of degree n is given by,

Tnf(x) = 1 + x+ x2 + x3 + x4 + · · · + xn =1− xn+1

1− x

=1

1− x−

xn+1

1− x

= f(x)−xn+1

1− x

Therefore Rnf(x) = f(x)− Tnf(x) =xn+1

1− x.




Lecture 18

So, for

f(x) =1

1− xwe have Rnf(x) =

xn+1

1− x

for x 6= 1.

Now for theBIG question:

For what values of x does Tnf(x) converge to f(x)?

We answer this question by analysis.

Convergence will occur if and only if

limn→∞

(

f(x)− Tnf(x))

= limn→∞

Rnf(x) = limn→∞

xn+1

1− x= 0.

The set of values for which this happens is the goal of the analysis.

We already know that x = 1 is not allowed.




Lecture 18

Theorem

1

1− x=

∞∑

n=0

xn if and only if |x| < 1.

Proof

For x ∈ R \ {1} we have 11−x

= K ∈ R, some real numberindependent of n. Therefore,

Rnf(x) =xn+1

1− x= Kxn+1.

Furthermore we know that xn+1 → 0 as n → ∞ providing that−1 < x < 1, and that xn+1 6→ 0 if x 6 −1 or x > 1.

Hence Rnf(x) = Kxn+1 → 0 as n → ∞ if and only if |x| < 1.




Lecture 18

Quite a lot happened there. Let’s look again at what we achieved.

Theorem

1

1− x=

∞∑

n=0

xn if and only if |x| < 1.

Proof — the structure

It is always true, by definition, that f(x)− Tnf(x) = Rnf(x).

Here we found that Rnf(x) = Kxn+1 for K independent of n.

Then, if limn→∞

(

f(x)− Tnf(x)))

= limn→∞

Rnf(x) = 0

we’ll have f(x) = T∞f(x) — the Maclaurin series. We proved that

limn→∞

Rnf(x) = limn→∞

Kxn+1 = 0 if and only if |x| < 1.




Lecture 18

We now have our first rigorously proven Maclaurin series:

1

1− x= 1+ x+ x2 + x3 + x4 + · · · =

∞∑

n=0

xn for − 1 < x < 1

where we can be certain that the ∞ is justified by our analysis.

We were lucky though, we knew Rnf(x) exactly.

Because f(x) = (1− x)−1 we could use the geometric series.

It isn’t usually that easy though. . .

Let’s look at another example and see where the difficulties lie.




Lecture 18


1

1− x= 1+ x+ x2 + x3 + x4 + · · · =

∞∑

n=0

xn for − 1 < x < 1









Lecture 18


1

1− x= 1+ x+ x2 + x3 + x4 + · · · =

∞∑

n=0

xn for − 1 < x < 1









Lecture 18


1

1− x= 1+ x+ x2 + x3 + x4 + · · · =

∞∑

n=0

xn for − 1 < x < 1









Lecture 18


1

1− x= 1+ x+ x2 + x3 + x4 + · · · =

∞∑

n=0

xn for − 1 < x < 1









Lecture 18

Here’s another familiar result: with f(x) = cos(x) we have,

Tnf(x) =

n∑

m=0

(−1)mx2m

(2m)!= 1−

x2

2!+

x4

4!−

x6

6!+ · · · +

(−1)nx2n

(2n)!

We would like to proceed as before and say that:

limn→∞

Rnf(x) = 0 so Tnf(x) → cos(x) as n → ∞.

But we don’t know what Rnf(x) is for f(x) = cos(x).

All we know so far about the remainder is in Definition 1.8 andProposition 1.9 from Lecture 5. . .




Lecture 18

From Lecture 5. . .

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.

Let’s look closer at this proposition. . .




Lecture 18




So, in general, for a Maclaurin polynomial Tnf(x) (where a = 0),

Rnf(x) = f(x)− Tnf(x) and Rnf(x) → 0 as x → 0.

But. . . This tells us nothing about convergence: limn→∞

Rnf(x) =?

We need to know much more about the remainder, Rnf(x), todeal with the convergence of Maclaurin series.Here’s a very general form. . .




Lecture 18







Rnf(x) =?





Lecture 18







Rnf(x) =?





Lecture 18







Rnf(x) =?





Lecture 18







Rnf(x) =?





Lecture 18

Taylor’s theorem, Theorem 6.6 in The Handbook

Denote the mth derivative of a function f : R → R by f (m). Letf (n+1) be integrable on R, for some n ∈ N, then

f(x) =n∑

m=0

(x− a)m

m!f (m)(a) +Ra

nf(x),

for all x, a ∈ R, and where

Ra

nf(x) =1

n!

∫

x

a

(x− ξ)nf (n+1)(ξ) dξ

is the remainder.




Lecture 18

Proof

We use induction. Assume the theorem is true forn = k ∈ {0, 1, 2, . . .} and use the definitions:

u = f (k+1)(ξ) =⇒ du = f (k+2)(ξ) dξ,

dv = (x− ξ)k dξ =⇒ v = −(x− ξ)k+1

k + 1,

to integrate Ra

kf by parts. This gives:

Ra

kf =(x− a)k+1

(k + 1)!f (k+1)(y) +Ra

k+1f,

and demonstrates that if the theorem holds for n = k, then it alsoholds for n = k + 1. For n = 0 Taylor’s theorem simply asserts thefundamental theorem of calculus and this completes the proof.




Lecture 18

So, we have from this very general result that,

f(x) = T a

nf(x) = Ra

nf =1

n!

∫

x

a

(x− ξ)nf (n+1)(ξ) dξ,

but we can do better than this. . .

Recall that if G takes non-negative values then

∫

b

a

G(ξ) dξ =

the area between thegraph of G and the x axisbetween x = a and x = b.

So:

∫

b

a

G(ξ) dξ 6 M(b−a) where M = max{G(x) : a 6 x 6 b}




Lecture 18

If M = max{G(x) : a 6 x 6 b} then, because M is a constant, bycomparing areas we have,

∫

b

a

G(ξ) dξ 6

∫

b

a

M dξ = M(b− a)

0 1 2 3 4 5 6 7 8 9 10 11 12

0

20

40

60

0 1 2 3 4 5 6 7 8 9 10 11 12

0

20

40

60




Lecture 18

If M = max{G(x) : a 6 x 6 b} then, because M is a constant, bycomparing areas we have,

∫

b

a

G(ξ) dξ 6

∫

b

a

M dξ = M(b− a)

0 1 2 3 4 5 6 7 8 9 10 11 12

0

20

40

60

0 1 2 3 4 5 6 7 8 9 10 11 12

0

20

40

60




Lecture 18

Furthermore, because graphs below the x axis create cancellationthrough negative areas, we get for any function that,. . .∫

b

a

G(ξ) dξ 6

∣

∣

∣

∣

∫

b

a

G(ξ) dξ

∣

∣

∣

∣

6

∫

b

a

|G(ξ)| dξ 6

∫

b

a

M dξ = M(b−a)

where, this time, M = max{|G(x)| : a 6 x 6 b}

0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5

Applying this to our remainder gives (if a 6 x),

|Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ . . .




Lecture 18


b

a

G(ξ) dξ 6

∣

∣

∣

∣

∫

b

a

G(ξ) dξ

∣

∣

∣

∣

6

∫

b

a

|G(ξ)| dξ 6

∫

b

a

M dξ = M(b−a)


0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5


|Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ . . .




Lecture 18


b

a

G(ξ) dξ 6

∣

∣

∣

∣

∫

b

a

G(ξ) dξ

∣

∣

∣

∣

6

∫

b

a

|G(ξ)| dξ 6

∫

b

a

M dξ = M(b−a)


0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5


|Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ . . .




Lecture 18


b

a

G(ξ) dξ 6

∣

∣

∣

∣

∫

b

a

G(ξ) dξ

∣

∣

∣

∣

6

∫

b

a

|G(ξ)| dξ 6

∫

b

a

M dξ = M(b−a)


0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5


|Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ . . .




Lecture 18


b

a

G(ξ) dξ 6

∣

∣

∣

∣

∫

b

a

G(ξ) dξ

∣

∣

∣

∣

6

∫

b

a

|G(ξ)| dξ 6

∫

b

a

M dξ = M(b−a)


0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5


|Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ . . .




Lecture 18


b

a

G(ξ) dξ 6

∣

∣

∣

∣

∫

b

a

G(ξ) dξ

∣

∣

∣

∣

6

∫

b

a

|G(ξ)| dξ 6

∫

b

a

M dξ = M(b−a)


0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12

−0.5

0

0.5

1

1.5


|Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ . . .




Lecture 18

Assuming that a 6 x we have,

|Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ,

and defining Mn = max{|f (n)(ξ)| : a 6 ξ 6 x},

gives: |Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ,

6Mn+1

n!

∫

x

a

(x− ξ)n dξ,

=Mn+1

n!

[

(x− ξ)n+1

n+ 1

]a

x

,

=Mn+1|x− a|n+1

(n+ 1)!.

(Challenge: what happens if a > x?)This is the Lagrange bound of Ra

nf(x). We now have a theorem. . .Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 18


|Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ,


gives: |Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ,

6Mn+1

n!

∫

x

a

(x− ξ)n dξ,

=Mn+1

n!

[

(x− ξ)n+1

n+ 1

]a

x

,

=Mn+1|x− a|n+1

(n+ 1)!.





Lecture 18


|Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ,


gives: |Ra

nf | 61

n!

∫

x

a

(x− ξ)n|f (n+1)(ξ)| dξ,

6Mn+1

n!

∫

x

a

(x− ξ)n dξ,

=Mn+1

n!

[

(x− ξ)n+1

n+ 1

]a

x

,

=Mn+1|x− a|n+1

(n+ 1)!.





Lecture 18

Taylor’s theorem (Lagrange bound of the remainder)Theorem 6.6 in The Handbook

Denote the mth derivative of a function f : R → R by f (m). Letf (n+1) be bounded on R for some n ∈ N then

f(x) =n∑

m=0

(x− a)m

m!f (m)(a) +Ra

nf(x),

for all x, a ∈ R, and where,

|Ra

nf(x)| 6Mn+1|x− a|n+1

(n+ 1)!

with Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between a and x}.




Lecture 18

That was all pretty tough.

But it is real progress.

The next step is to put this result to work for us.

The central question remains:

Given a function, f , and its Maclaurin polynomial, Tnf(x), when isit true that

f(x) = limn→∞

Tnf(x)?

Taking a = 0, this is what we now know for x ∈ R:

Rnf(x) = f(x)− Tnf(x) and |Rnf(x)| 6Mn+1|x|

n+1

(n+ 1)!.

where Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between 0 and x}.




Lecture 18






f(x) = limn→∞

Tnf(x)?



n+1

(n+ 1)!.





Lecture 18






f(x) = limn→∞

Tnf(x)?



n+1

(n+ 1)!.





Lecture 18






f(x) = limn→∞

Tnf(x)?



n+1

(n+ 1)!.





Lecture 18






f(x) = limn→∞

Tnf(x)?



n+1

(n+ 1)!.





Lecture 18






f(x) = limn→∞

Tnf(x)?



n+1

(n+ 1)!.





Lecture 18






f(x) = limn→∞

Tnf(x)?



n+1

(n+ 1)!.





Lecture 18



n+1

(n+ 1)!.


So, we’ll have,f(x) = lim

n→∞

Tnf(x)

and we can rigorously write f(x) = T∞f(x), if and only if

limn→∞

|Rnf(x)| 6 limn→∞

Mn+1|x|n+1

(n+ 1)!= 0.

In the next lecture we will use this for some familiar functions.Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel



Lecture 18

Summary

We can:

Give a complete analysis of the Maclaurin series for (1− x)−1.

Prove Taylor’s theorem with integral remainder.

Derive the Lagrange form of the remainder.




End of Lecture

Computational andαpplie∂ Mathematics




End of Lecture








End of Lecture





Reference: The Handbook, Chapter 6, Section 6.1.Homework: Q10, Sheet 5bSeminar: Ad Hoc. Student driven.



Documents

LectureslidesforMA2730AnalysisIicsrsss/teaching/ma2730/lec/lec18.pdf · 2015. 11. 10. · Overview(MA2730,2812,2815) lecture18 Lecture18. MA2730: topicsforLecture18. Lecture18. Taylor’s