ENGI 3424 Chapter 5ggeorge/3424/thebook/Week11.pdf · 2020. 10. 27. · ENGI 3424 5.01 –...

ENGI 3424 5 – Series Page 5-01

5. Series

Contents:

5.01 Sequences; general term, limits, convergence

5.02 Series; summation notation, convergence, divergence test

5.03 Series; telescoping series, geometric series, p-series

5.04 Tests for Convergence: comparison and limit comparison tests

5.05 Tests for Convergence: alternating series; absolute and conditional convergence

5.06 Tests for Convergence: ratio and root tests

5.07 Power Series, radius and interval of convergence

5.08 Taylor and Maclaurin Series, remainder term

5.09 Binomial Series

5.10 Introduction to Fourier Series

5.11 Introduction to Series Solutions of ODEs

Appendix:

5.A Integral Test [not examinable; for reference only]

ENGI 3424 5.01 – Sequences Page 5-02

5.01 Sequences; general term, limits, convergence

A sequence is a set of related objects that all follow the same rule. There is a logical

progression from current and previous elements to the next element.

Example 5.01.1

The alphabet: { a, b, c, ... , y, z }

The rule is: The nth

element na (the nth

letter of the alphabet)

Example 5.01.2

The set of all natural numbers = { 1, 2, 3, ... }

Two ways to express the rule are:

The nth

element 1

1n na a

(with 1 1a ) [a recurrence relationship]

or the direct definition

na n .

We are usually interested in two features:

An explicit form for the general term (often from a recurrence relationship);

The behaviour of the terms of a sequence as n .

Example 5.01.3

The terms of a sequence form an arithmetic progression if consecutive terms differ by a

common non-zero constant difference d :

1 1n nn na a d a a d

, with an initial term 1a a

Writing down the first few terms of the sequence,

, , 2 , 3 ,a a d a d a d

The general term is therefore 1n a da n

Clearly 1 sgnlim limnn n

a a d n d

and the sequence diverges.

Note the definition

1 0

sgn 0 0

1 0

x

x x

x


Example 5.01.4

The terms of a sequence form a geometric progression if consecutive terms change by a

common constant ratio r :

11

0, 1n

n nn

ar a a r r r

a

, with an initial term 1 0a a

Writing down the first few terms of the sequence,

2 3, , , ,a ar ar ar

The general term is therefore 1nna ar [or n

na ar if the count starts at n = 0]

The sequence alternates in sign if 0r

The sequence converges to a limit of zero if 1r

If 1r then lim nn

a

does not exist ( na alternates between +a and –a forever).

The sequence does not converge to a single value and therefore diverges, though not to

infinity.

1 nr a a n

so that the sequence

converges instantly to

a .

Otherwise

the sequence diverges

(to infinity).


The power sequence 1 , 2 , 3 , 4 ,r r r r rn

converges to 0 for r < 0,

converges to 1 for r = 0 and

diverges for r > 0.

ENGI 3424 5.01 – Sequences -05

Example 5.01.5

Find the general term of this sequence and determine whether it converges.

1 4 7 10 13, , , , ,

5 9 13 17 21

The numerators are increasing by 3 with each term

3n must be present in the numerator of the general term

The denominators are increasing by 4 with each term

4n must be present in the denominator of the general term

If n starts at 1, then 3 2

4 1n

na

n

Alternatives are possible:

If n starts at 0, then 3 1

4 5n

na

n

2

1

33 2 3lim lim lim

4 1 4 4

nnn n n

n

na

n

or, using l’Hôpital’s rule,

3 2 3 3lim lim lim

4 1 4 4

H

nn n n

na

n

Therefore the sequence converges to the limit 3

4.

Algebra of Sequences:

Given two convergent sequences na with limit value A and nb with limit value B ,

n npa qb is a convergent sequence with limit value (pA + qB) for all constants p, q;

n na b is a convergent sequence with limit value AB;

/n na b is a convergent sequence with limit value A/B (provided that 0B );

If f x is a continuous function with limx

f x L

and if na f n n [for all n]

then na converges with limit value L ;

If n n na c b n and if nc converges with limit value C then A C B .


Example 5.01.6

Find the limit of the sequence sin n

n

.

1 sin 11 sin 1

nn n

n n n

But 1

n

is a convergent sequence, with limit 1

lim 0n n

and 1

n

is a convergent sequence, with limit

1lim 0

n n

Therefore sin sin

0 lim 0 lim 0n n

n n

n n

Therefore sin n

n

is a convergent sequence, with limit 0.

[This solution contains an example of the “squeeze theorem”;

the general term of the sequence in question is trapped between the general terms of two

other sequences which both converge to the same limit.]

The “Race to Infinity”

There is a hierarchy of functional forms that all diverge to infinity with increasing n:

! 1 0 log 1n n cb

n n a a n c n b

for all sufficiently large n.

Example 5.01.7

!lim 0nn

n

n because “n

n goes to infinity faster than n ! ”

OR the sequence is positive and

1 2 3 2 1! 1 11n

n n nn n n n n nn

n n n n n n n n n n n n n n

Therefore all terms are bounded above by the terms of a sequence that converges to 0.


If every term of a sequence is no less than all of its predecessors 1 2 3 4a a a a ,

then the sequence is an increasing sequence.

If every term of a sequence is no greater than all of its predecessors 1 2 3a a a ,

then the sequence is a decreasing sequence.

A sequence is monotonic if and only if it is either increasing or decreasing.

If na U n for some constant value U, then the sequence na is upper-bounded.

If na L n for some constant value L, then the sequence na is lower-bounded.

A sequence is bounded if it is both upper and lower bounded.

Every upper-bounded increasing sequence is convergent.

Every lower-bounded decreasing sequence is convergent.

Every bounded monotonic sequence is convergent.

ENGI 3424 5.02 – Series Page 5-08

5.02 Series

The partial sum nS of a sequence is the sum of the first n terms of that sequence:

1 2 3

1

n

n n kk

S a a a a a

This partial sum is also a finite series.

The sum of an infinite series is the sum of all terms in the associated infinite sequence,

1 2 3

1

lim nk nk

S a a a a S

.

Example 5.02.1

Find the first three partial sums of the sequence 1

, 1, 2, 3,2n n

and find the limit of the sequence of partial sums.

1 1

1 11

2 2S a

2 1 2 1 2 2

1 1 3 11

2 4 4 2S a a S a

3 1 2 3 2 3 3

3 1 7 11

4 8 8 2S a a a S a

which suggests

1lim lim 1 1 0 1

2n nn n

S S

A more convincing proof is:

1 2 3 4

1 1 1 1

2 4 8 16S a a a a

1 1 1

2 4 8 16

S

1 1 1 1 1 1 1 1 11

2 2 4 8 16 4 8 16 2S

1

11 1

2nn

S

OR Take a metre-long rule. Split it in half. Split one piece in half again. Split one of the

shortest pieces in half again. Repeat forever. Then the one metre has been divided into

infinitely many pieces. Adding the lengths together: 1 1 1 1

12 4 8 16


Divergence Test

If the general term of a series does not converge to zero, then the series diverges.

1

lim 0n nnn

a a

diverges (the sum does not exist)

Example 5.02.2

Prove that all arithmetic series 1

1n

a n d

diverge, (where 0d ).

lim lim 1 lim 1 sgn 0nn n na a n d a d n d

Therefore, by the divergence test, the arithmetic series diverges.

OR

2 1 1 2 3 1nS a a d a d a n d na d n

Let 1 2 3 1x n

then 1 2 3 1x n n n

1

2 12

n nx n n n n n n x

Therefore the nth

partial sum of the arithmetic series is 1

2n

n ndS na

and clearly lim nnS

does not exist (unless 0a d ).

Therefore the arithmetic series diverges.

However, the converse of the divergence test is false.

Some divergent series do have general terms that converge to zero,

or equivalently: lim 0nna

does not guarantee that

1n

n

a

converges

(except for alternating series, section 5.05 below).


Example 5.02.3

Show that the harmonic series

1

1

nn

diverges.

1lim lim 0nn n

an

The divergence test therefore provides no information.

1 3

1 1 1 1 31, 1 1

2 3 4 4 2S S

7

1 1 1 1 1 1 1 1 1 1 1 1 41 1

2 3 4 5 6 7 4 4 8 8 8 8 2S

15 7

1 1 1 1 1 1 1 1 4 1 58

8 9 10 11 12 13 14 15 2 16 2S S

The pattern continues as 2 1

1

2k

kS

There is no upper bound to this sequence of partial sums lim nnS

Therefore the harmonic series diverges.

The divergence of the harmonic series is very slow.

4 terms are needed for the partial sum to exceed 2.



The sum of the first billion (109) terms is still well under 25. [approximately 21.3]

Yet the sum of the complete series does not exist (diverges to infinity)!


5.03 Series; telescoping series, geometric series, p-series

Example 5.03.1

Find the exact sum of the series 2

1

1

5 6n

Sn n

.

1

2 3 2 3n

A Ba

n n n n

By the cover-up rule,

1

2 2A

1

2 3

and

1

3 2 3 3B

1

1 1

2 3na

n n

1 1

3 4nS

1

4

1

5

1

5

1

6

1

1n

1

2n

1

2n

1

3n

The partial sum has telescoped to

1 1 1lim

3 3 3n nn

S S Sn

This is an example of a telescoping series.

The simplest such series have general terms similar to 1na f n f n .

Such series often involve the use of partial fractions.


Example 5.03.2

Find the exact sum of the series 3 2

2

8

3 3n

Sn n n

.

3 2 23 3 3 1 1 1 3n n n n n n n n

8

1 1 3 1 1 3n

A B Ca

n n n n n n

By the cover-up rule,

8

1 1A

81

2 41 1 1 3

8

1 1 1 1B

82

2 21 3

8

3 1 3 1 3 3C

81

4 2

1 2 1

1 1 3na

n n n

, n = 2, 3, 4, ...


Example 5.03.2 (continued)

nS

Collecting the surviving terms from this telescoping series, we find that

1 1 1 1 1 1 11

2 3 4 1 2 3nS

n n n n

1 1 1 12 6 4 3 11lim 1 0 0 0 0

2 3 4 12 12nn

S S

Note that this series starts at n = 2, not n = 1.

1a is undefined (division by zero).


Geometric Series

Each term in a geometric series is a constant multiple r (the common ratio) of the

immediately preceding term, except for the [non-zero] first term a . Quoting the general

term from the geometric progression on page 5.03, the geometric series is

1

1 0

orn n

n n

S ar ar

Applying the divergence test:

1 1

0 1

lim lim lim 1

otherwise

n nnn n n

r

a ar a r a r

DNE

[DNE = “does not exist”]

If r = +1 then the series becomes a + a + a + ... which diverges to infinity.

If r = –1 then the series becomes a – a + a – ... which oscillates forever between a

and 0 , so that the series diverges.

By the divergence test, the geometric series diverges whenever 1r .

The divergence test yields no information when 1r .

Examine the sequence of partial sums:

2 3 2 11

1

n nn

kn

k

S ar a ar ar ar ar ar

2 3 2 1n n nnrS ar ar ar ar ar ar

1 1n nnr S a ar a r

Therefore the nth partial sum of any geometric series (except r = 1) is

1

1

n

n

a rS

r

for 1 nr S na

1 0lim iff 1

1 1nn

a aS S r

r r


Example 5.03.3 (Example 5.02.1 revisited)

Evaluate

1

1

2nn

S

.

2 3 4

1 1 1 1

2 2 2 2S

This is a geometric series, with 1

2a r .

The series converges, because 1 1

12 2

r .

Immediately, 12

12

1 11

1 1 2 2

aS

r

Therefore S = 1.

Example 5.03.4

Find the sum of the power series

2 3 4f x x x x x

f x is a geometric series, with a x and r x .

This series converges iff 1 1x .

For these values of x,

1 1

a xf x S

r x

Therefore

1 11

undefined otherwise

xx

f x x


p-series

Another important family of standard series is the p-series

(also known as “hyperharmonic” series):

1

1p

nn

The divergence test establishes divergence for 0p .

It can be shown [section 5.A] that p-series

converge for all 1p and

diverge for all 1p .

The case 1p is the harmonic series, for which Example 5.02.3 established divergence.

In 1741 Leonhard Euler proved that 2

2

1

1

6n

n

.

The removal or insertion of a finite number of finite terms does not affect the

convergence of a series.

If

3n

n

S a

converges, then so does 1 2 1 2

1 3n n

n n

a a a a a a S

,

provided that 1a and 2a are finite.

ENGI 3424 5.04 – Comparison Tests Page 5-17

5.04 Tests for Convergence: comparison and limit comparison tests

If all terms na in a series are positive, then the sequence of partial sums nS is

necessarily increasing. 1 1n nn n

S S a S n

If another series

1n

n

b

is convergent with a sum B and if n na b n , then

B is an upper bound to nS

Recall that an increasing sequence that has an upper bound is convergent [page 5.07]

nS must converge to a value B

1n

n

a

converges to some value A B .

This is the basis of the comparison test.

Example 5.04.1

Is the series 2

1

1

5 6n n n

convergent?

[This is Example 5.03.1, for which the exact value of the sum was found by telescoping

the series.]

2

1

5 6na

n n

Compare with 2

1nb

n

For all positive n, 2 25 6 0 n nn n n a b and 1

nn

b

converges

[ p-series with 2 1p ]

Therefore yes, 2

1

1

5 6n n n

is convergent.

[Note that the comparison test may reveal convergence but cannot specify the exact value

of the sum.]


Example 5.04.2

Is the series

2

1

lnn n

convergent?

1

lnna

n

Compare with 1

nbn

0 ln 1n n n

2n na b n

But 2 2 1

1 11n

n n n

bn n

diverges

(harmonic series, except for finite first term)

Therefore no,

2 2

1

lnn

n n

an

diverges.

Put very briefly, the essence of the comparison test is:

A convergent ceiling forces convergence:

0 andn n na b n b converges na converges.

A divergent floor forces divergence:

0 andn n na b n b diverges na diverges.


Limit Comparison Test

All terms na and nb must be non-negative and L is some finite constant.

lim 0 andnnn

n

aL b

b converges na converges.

lim 0 andnnn

n

ab

b diverges na diverges.

The reference series

1n

n

b

is often the geometric series or a p-series.

Example 5.04.3

Is the series 3

3

ln

8n

n n

n

convergent?

[First note that the summation starts at n = 3, because 2a involves a division by zero,

which renders 3

1

ln

8n

n n

n

divergent instantly.]

Choose a p-series as the comparison series: 1

n pbn

3

3

2ln ln

881

p pn

n

n n na n n

b n

n

If 2p then lim n

nn

a

b

But 1p

n converges for 2p . We have no information if we choose 2p .

Limit comparison test, for 2p :

3

2 28lim lim ln lim 1 lim ln 1 0

p pn

n n n nn

an n n n

b n

1

2 1

lnlim lim

2

H

p pn n

n n

n p n

2

1 1lim 0

2 pnp n

- a finite limit

For 1 p the comparison series nb is convergent

and we require 2p . 1.5p is an example. We may select 1.5

1nb

n .

LCT: lim 0n

nn

a

b and nb is convergent YES, na is convergent.


Example 5.04.4

Is the series 2

3

1

4

9n

n

n n

convergent?

First note that 1 0a and 2 0a , while 0 3na n and all terms are finite.

However, deletion or addition of a finite number of finite terms to a series does not affect

the convergence of that series. Therefore the convergence status of 2

3

1

4

9n

n

n n

is the

same as that of 2

3

3

4

9n

n

n n

, to which the limit comparison test can be applied.

The overall order of n in the rational expression 2

3

4

9n

na

n n

is

2

3

1n

n n

Therefore choose 2

3

1n

nb

n n

Limit comparison test:

2 3 2

3 2

2

41

4 1 0lim lim lim 1

99 1 01

n

n n nn

a n n n

b n n n

n

lim 1 0n

nn

a

b and nb is divergent [the harmonic series]

Therefore no, na is divergent.

Note that the comparison test fails for this example - the inequality required to establish

divergence n na b is false!

ENGI 3424 5.05 – Alternating Series Test Page 5-21

5.05 Tests for Convergence: alternating series;

absolute and conditional convergence

Alternating Series Test

If consecutive terms of a series alternate in sign, then a simpler test for convergence is

available.

If 1 0n

n

an

a

and lim 0nna

then

1n

n

a

converges.

Example 5.05.1

Is the alternating harmonic series

1

1 1

1 1 1 11 1

2 3 4

nn

n n

an

convergent?

First note that the signature of an alternating series is a factor such as 1n

, which is +1

for all even n and –1 for all odd n.

This series is clearly an alternating series, for which 1 0n

n

an

a

.

1 1

lim lim 1 0n

nn na

n

Therefore, by the alternating series test, 1

1

11

n

nn

converges.

Note that if we take the absolute values of the terms in the alternating harmonic series,

then we obtain the harmonic series 1 1 1

12 3 4

, which diverges.


Absolute Convergence

If and only if the series

1n

n

a

converges,

then the series

1n

n

a

is absolutely convergent.

If and only if the series

1n

n

a

converges but the series

1n

n

a

diverges,

then the series

1n

n

a

is conditionally convergent.

The alternating harmonic series is an example of a conditionally convergent series.

If a series is absolutely convergent, then the alternating series test is a waste of time.

Another test to establish the convergence of

1n

n

a

automatically establishes that

1n

n

a

is absolutely convergent in one step.

If a series is conditionally convergent, then the alternating series test is often required to

establish convergence of

1n

n

a

, but another test is required on

1n

n

a

to complete the

proof of conditional convergence. Therefore test the convergence of 1

nn

a

first!

Absolute convergence allows the algebra of finite series to be extended to infinite series.

In particular, the order of an infinite number of terms may be changed without affecting

the sum of the series. The derivative of an absolutely convergent series is the sum of the

derivatives of the individual terms.

This is no longer necessarily true for conditionally convergent series. For example,

1 1 1 1 1 11 ln 2

2 3 4 5 6 7 but

1 1 1 1 1 31 ln 2

3 2 5 7 4 2 .


Example 5.05.2

Investigate the convergence of the alternating p-series

1

1 1

1 1 1 11 1

2 3 4

nn p p p p

n n

an

This is obviously an alternating series.

1n

n

a

is the p-series, which converges iff 1p .

1n

n

a

is absolutely convergent iff 1p .

0 0lim

undefined 0nn

pa

p

By the alternating series test, the series converges iff 0p .

Therefore the alternating p-series is

absolutely convergent for 1p

conditionally convergent for 0 1p

divergent for 0p

Example 5.05.3

Is

1

1

1 1 1 11

2 3 4

n

nn

absolutely convergent, conditionally

convergent or divergent? Note that 1/2

1 1

nn

This series is just the alternating p-series with 12

p .

12

0 1

Therefore the series is conditionally convergent.


Approximation Errors

For any convergent alternating series, the error in approximating the sum S of the entire

series caused by evaluating the partial sum Sn of just the first n terms is

1n nS S a

Example 5.05.4

Estimate

0

1

2 !

n

n

Sn

correct to four decimal places.

The signs of the terms of this series are clearly alternating and

1lim 0

2 !

n

n n

this series converges (by the alternating series test).

The error caused by using nS to estimate S is

1

1 1

2 2 !2 1 !n n

S S ann

No error in the xth

decimal place requires an error of less than 5 in the next decimal place.

We require

1

0.00005 2 2 ! 200002 2 !

nn

But 7 ! = 5 040 and 8 ! = 40 320

2 2 8 3n n

3

1 1 1 1 1 1 3891 1 0.54027

2! 4! 6! 2 24 720 720S

Therefore S = 0.540 3 (correct to 4 d.p.)

The exact value is S = cos 1 = 0.540 302 31...

[See Example 5.08.4 below]

ENGI 3424 5.06 – Ratio and Root Tests Page 5-25

5.06 Tests for Convergence: ratio and root tests

Among the most useful tests for series convergence is the ratio test:

If 1lim 1n

nn

a

a

then

1

n

n

a


If 1lim 1n

nn

a

a

then

1

n

n

a

is divergent.

If 1lim 1n

nn

a

a

then the test fails and there is no information on series convergence.

The ratio test is most useful when the general term includes an exponential or factorial

factor. The test fails when the general term is an algebraic function (terms of the form constantn in the numerator and/or denominator) only.

Example 5.06.1

Is the series

1 1!

n

n

n n

na

n

absolutely convergent, conditionally convergent or

divergent?

The general term suggests the use of the ratio test:

11 1 1 1! ! 1

11 ! 1 !

n n nn

n nn

a n n nn n

a n n n n n n

1lim 1 1

n

ne

n

[see the note on the next page]

Therefore

1

n

n

a

is divergent.

OR

Use the divergence test!

nn goes to infinity faster than !n

lim 0nna

series diverges.


Note:

Proof that 1

lim 1

n

ne

n→

+ =

:

11

n

xn

= +

1ln ln 1x n

n

= +

1ln 1

1

n

n

+

=

Recall that ( )( )( )

( )ln

f xdf x

dx f x

=

( )( )2

1

1

0ln 1

1

d nn

dn n

−−

−

− + =

+

( )1

1

ln 1lim ln lim

n n

nx

n

−

−→ →

+=

20lim

H

n

n−

→

−=

1 2

1

1 n n− −

+ −

( )

( )( )1

1lim

1 1nn

→

+=

+ +

11

1 0= = +

+

1limn

x e→

=1

lim 1

n

ne

n→

+ =

The next two sections will see extensive use of the ratio test.


Root Test:

If lim 1nnn

a

then

1

n

n

a


If lim 1nnn

a

then

1

n

n

a

is divergent.

If lim 1nnn

a

then the root test fails (provides no information).

Like the ratio test, the root test fails for purely algebraic functions.

The root test fails whenever the ratio test fails.

The root test works best for general terms of the form n

na f n .

Usually the ratio test can be applied more easily whenever the root test provides a result.

Example 5.06.2

Is the series

2 2 ln

n

n nn n

xa

n

absolutely convergent, conditionally convergent or

divergent?

The general term suggests the use of the root test:

1lim lim lim 0 1

ln lnn

nn n n

xa x x

n n

Therefore the series is absolutely convergent for all x.

ENGI 3424 5.07 – Power Series Page 5-28

5.07 Power series, radius and interval of convergence

A power series with centre x c is of the form

0 1 2 3

2 3

0

bn d d b b bn

n

f x a x c x c a a x c a x c a x c

where b and d are real constants and 0b .

A common choice of parameters for power series is 1b and 0d :

0 1 2 3

2 3

0

nn

n

f x a x c a a x c a x c a x c

Let nu now represent the general term bn d

na x c

and

let us apply the ratio test for convergence of a power series:

1 1 1

bn b d b

n n nbn d

n nn

u a x c a x c

u aa x c

1 1lim limbn n

n nn n

u ax c

u a

The series is absolutely convergent whenever

1 1

1

lim 1 1 lim limbn nn

n n nnn n

u a ax c

au a

1

lim nbn

n

ax c R

a

where R is the radius of convergence

The interval of convergence I includes

c R x c R and may include one or both

endpoints, which must be tested separately.

I is also the domain of the function f x .

Also note that 0f c a .

All power series are absolutely convergent at the centre, even if 0R .

If R , then the power series is absolutely convergent for all x.


Example 5.07.1

Determine the radius of convergence and interval of convergence for the power series

00

2

3

nn

n

n

n

a x cx

f xn

The centre of this series is at 2c .

4

31 1

11 1 4 1 0lim lim 1

3 4 3 1 1 0

n n n

n nnn n

a anR

a n n n a

Therefore the radius of convergence for this series is 1R

The boundaries of the interval of convergence are at 2 1 3x c R

and at 2 1 1x c R .

0 0

3 2 1 1 1 1 13

3 3 3 4 5 6

n n

n n

fn n

which is the alternating harmonic series (except for the first two terms).

The series is therefore conditionally convergent at 3x .

0 0

1 2 1 1 1 1 11

3 3 3 4 5 6

n

n n

fn n

which is the harmonic series (except for the first two terms).

The series is therefore divergent at 1x .

The interval of convergence is therefore 3 1x ,

with conditional convergence at x = –3.


Example 5.07.2

Determine the radius of convergence and interval of convergence for the power series

1

1

3

n

nn

xf x

n n

3/2

1

11 , where

3

nn n n

n

a x an

The centre of this series is at 1c .

3/21

3/21

3 1 3

3

n nn

nn

na

a n

13

3n

3/21n

n

3/2 3/21

1

lim 3 lim 1 3 1 0 3nnn n

n

aR

a

Therefore the radius of convergence for this series is 3R

The boundaries of the interval of convergence are at 1 3 2x c R

and at 1 3 4x c R .

3/2 3/21 1

3

2 1 12

n

n n

n nn n

f

which is the alternating p-series with 32

1p

The series is therefore absolutely convergent at 2x .

3/2 3/21 1

4 1 14

3

n

n nn

fn n

which is the p-series with 32

1p

The series is therefore absolutely convergent at 4x .

The interval of convergence is therefore 2, 4I ,

with absolute convergence at both ends.

For power series of the type 0

nn

n

a x c

with a finite interval of convergence, the

series usually (i) converges absolutely at both ends; or (ii) diverges at both ends; or

(iii) converges conditionally at one end and diverges at the other end.

ENGI 3424 5.08 – Taylor and Maclaurin series Page 5-31

5.08 Taylor and Maclaurin Series; Remainder Term

A function f x that is represented by a power series (using only non-negative integer

powers of x c ) is a Taylor series:

0

nn

n

f x a x c

0 1 2 3

1 2 3a a x c a x c a x c

0 00 0f c a a f c

Assuming that the series exists and is absolutely convergent, then

1

0

nn

n

f x a n x c

1 2 3

1 20 1 2 3a a x c a x c

1 10 1 0 01

f cf c a a

2

0

1n

nn

f x a n n x c

2 3

10 0 2 1 3 2a a x c

2 20 0 2 1 02 1

f cf c a a

3

0

1 2n

nn

f x a n n n x c

3 30 0 0 3 2 13 2 1

f cf c a a

and so on, so that the general term is

!

n

n

f ca

n and the Taylor series for f x is

0

!

nn

n

f cf x x c

n

The ratio test can determine the radius of convergence R .


Example 5.08.1

Find the Taylor series for lnf x x about 1x .

1c

lnf x x 0 1 ln1 0a f

11f x x

x

1

1 11

1! 1

fa

2

2 11f x x

x

2

1 1 1

2! 2 2

fa

3

3 2 11 2f x x

x

3

1 2 1

3! 3 2 3

fa

IV

4

4 3 2 12 3f x x

x

IV

4

1 3 2 1

4! 4 3 2 4

fa

The pattern continues as

1 1 !1

n nn

nf x

x

111 1 ! 1

1! !

n nn

n

f na

n n n

2 3 4

1

11 1 1 1 1

ln 11 2 3 4

n

nn x x x x

x xn

1

1 1 1 1lim lim lim lim 1 1 0 1

1

n

n n n nn

a nR

a n n n n

The endpoints of the interval of convergence are at 1 1 0, 2x c R .

1 1 1 1

01 2 3 4

f which is the negative of the harmonic series.

1 1 1 1

21 2 3 4

f which is the alternating harmonic series.

Therefore the Taylor series is valid on (0, 2]I

(with conditional convergence at x = 2, which shows that 1 1 1 1

ln 21 2 3 4 ).


Example 5.08.2

Prove L’Hôpital’s rule for the case of a (0/0) indeterminacy.

Let functions f (x) and g (x) be such that f (a) = g (a) = 0.

Then

f a

g a is a

0

0 type of indeterminacy.

Represent both functions by their Taylor series about x a , then

f af x

g x

212

x a f a x a f a

g a

21

2x a g a x a g a

Everywhere in the interval of convergence except at x = a,

1212

f a x a f af x

g x g a x a g a

0 0lim

0 0x a

f x f a f a

g x g a g a

(provided that 0g a ), which is L’Hôpital’s rule.

In the event that 0f a g a , then divide out by another factor of x a and

repeat as necessary.


Maclaurin Series

A Taylor series with a centre at 0x c is a Maclaurin series.

Example 5.08.3

Find the Maclaurin series for xf x e and find its interval of convergence.

0c

0

00 1xf x e a f e

1

001

1! 1

x f ef x e a

2

00 1

2! 2! 2!

x f ef x e a

The pattern continues as

00 1

! ! !

nn x

n

f ef x e a

n n n

Therefore the Maclaurin series for xf x e is

2 3 4

0

1! 2! 3! 4!

nx

n

x x x xe x

n

1

1 !1 !lim lim lim

!

n

n n nn

n nnaR

a n

!n lim 1

nn

Therefore the Maclaurin series is valid for all x : ,I

or, equivalently, I .


Example 5.08.4

Find the Maclaurin series for cosf x x and find its interval of convergence.

c = 0

0

cos0cos 1

0!f x x a

1

sin 0sin 0

1!f x x a

2

cos0 1cos

2! 2!f x x a

3

sin 0sin 0

3!f x x a

IV

4

cos0 1cos

4! 4!f x x a

Clearly the odd derivatives all lead to zero coefficients,

while the even derivatives alternate in sign.

The general even coefficient is

2

1

2 !

k

n ka a

k

The Maclaurin series for cosf x x is

2 4 62

0

1cos 1

2 ! 2! 4! 6!

kk

k

x x xx x

k

1

2 2 2 12 2 !lim lim

2 !lim

2 !

k

k k kk

k ka kR

k

k

a

2 !k

lim 2 2 2 1k

k k

Therefore the Maclaurin series is valid for all x : ,I

or, equivalently, I .


Example 5.08.4 (Alternative Solution)

Use the Euler expression for the cosine function in terms of the exponential function:

cos sinj

e j

and cos sinj

e j

2cos

j je e

1cos 1

2 1!

j

2 3

2! 3!

j j

1!

1j

2 3

2! 3!

j j

2 4 2 4 6

22cos 1 1 1

2 2! 4! 2! 4! 6!

j jj

24 2 2 1 1j j j

and because the Maclaurin series for both exponential functions converge absolutely for

all x, so does the Maclaurin series for the cosine function.

Using the summation notation,

2

0

cos 12 !

kk

k

xx

k

for all x .


Example 5.08.5

Use the Maclaurin series for the cosine function to find the Maclaurin series for the sine

function.

2 4

sin cos 12! 4!

x xx x dx dx

3 5

3! 5!

x xC x

But sin0 0 0 0 0 0 0C C

3 5 2 1

0

sin 13! 5! 2 1 !

kk

k

x x xx x

k

and this series converges absolutely for all x (because the series for cos x does).

Note that the Maclaurin series for any even function (such as cos x) can contain only even

powers of x, while the Maclaurin series for any odd function (such as sin x) can contain

only odd powers of x.

The Maclaurin series for sinf x x may also be found directly, through repeated

differentiations, or via the Euler equation and the Maclaurin series for the exponential

function (as in Example 5.08.4 for the cosine function) or by differentiation of –cos x .


Remainder Term

The practical use of Taylor (and Maclaurin) series arises when the partial sum of the first

few terms of the series nT x is used to approximate the value of a non-linear function.

The error caused by truncating the series at the nth

term is the nth

remainder term nR x :

11

1 !

nn

n n

fR x f x T x x c

n

where is some number between x and c .

The Taylor series is well-defined only if lim 0nnR x c R x c R

.

Example 5.08.6

The Maclaurin series for the exponential function xf x e is

2 3

0

1! 2! 3!

kx

k

x x xe x

k

11

10 0

1 ! 1 !

nn

nn

f e xR x x x

n n

Because factorial functions diverge to infinity faster than exponential functions,

1

lim lim 01 !

n

nn n

xR x e

n

and the Maclaurin series is therefore well-defined.

The error caused by approximating xf x e by the quadratic 2

2 12!

xT x x on

the interval 0 x a is at most 3

6

aa e.

For 0.1a , this maximum error is less than 2×10–4

, with a maximum relative error of

under 0.03%.

However, for 10a , the maximum error is substantial: more than 3×106. Many more

terms need to be taken to maintain accuracy, the further away from the centre one goes.

Note that 1y T x is the equation of the tangent line to y f x at x c .

ENGI 3424 5.09 – Binomial Series Page 5-39

5.09 Binomial Series

A special case of Maclaurin series arises for 1n

f x x .

01 0 1 1n

f x x f a

1

11 0

1!

n nf x n x f n a

2

2 11 1 0 1

2!

n n nf x n n x f n n a

and so on, leading to

1 2 11

!k

n n n n ka k

k

In the event that n is a positive integer,

!

! !k

na

k n k

(the binomial coefficients),

the series terminates at the term k = n and the series is therefore absolutely convergent

for all x when n is a positive integer.

For example, 4 2 3 41 1 4 6 4x x x x x

The series is absolutely convergent (trivially) for 0n , as 0

1 1x for all x .

For all other values of n the series does not terminate and the binomial expansion of

f x is

1

1 2 11 1

!

n k

k

n n n n kx x

k

Pascal’s triangle: 1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

⁞


The radius of convergence is

1

lim k

kk

aR

a

1 2 1 1 !lim

! 1 2 1k

n n n n k k

k n n n n k n k

111 1 0lim lim 1

1 1 0

knk kk

k

n k

Therefore the binomial expansion converges absolutely inside 1 1x .

The endpoints need to be checked separately [there is insufficient time in this course to

verify convergence at the endpoints].

Summary:

1

1 2 11 1

!

n k

k

n n n n kx x

k

Interval of convergence:

I for 0n or n [ the series terminates at k = n ]

1, 1I for 0n and not an integer, with absolute convergence at 1x .

1, 1I for 1 0n with conditional convergence at 1x .

1, 1I for 1n with divergence at 1x .


Example 5.09.1

Find the Maclaurin series for 1 2f x x and its interval of convergence.

In the binomial expansion, for "x" read "2x" and for "n" read "½".

1

1 2 11 1

!

n k

k

n n n n kX X

k

1/2

31 1 12 2 2 2

1

11 2 1 2

!

k

k

kf x x x

k

1 1 3 3 21

2k

k 2

!

k

k1

k

k

x

2 3 4 53 3 5 3 5 7

11! 2! 3! 4! 5!

x x x x x

2 3 4 55 7

1 2 12 2 8 8

x x x xx x

The series converges absolutely for 1 12 2

1 2 1x x .

0n the series converges absolutely at both endpoints.

Therefore the interval of convergence is 1 12 2

x .

Note that the general method for Maclaurin series could have been used instead, but the

sequence of differentiations would have been more tedious.

See the associated Excel file for an evaluation of the terms and partial sums of this series.

Problem Set 10 provides some more practice questions.

../demos/ex591.xls

ENGI 3424 5.10 – Fourier Series Page 5-42

5.10 Fourier Series

This section offers a very brief overview of [discrete] Fourier series. Some majors will

explore Fourier series and transforms in much greater detail in subsequent semesters.

The Fourier series of f x on the interval ,L L is

0

1

cos sin2

n nn

a n x n xf x a b

L L

where

1

cos , 0,1, 2, 3,L

Ln

n xa f x dx n

L L

and

1

sin , 1, 2, 3,L

nL

n xb f x dx n

L L

The ,n na b are the Fourier coefficients of f x .

Note that the cosine functions (and the function 1) are even, while the sine functions are

odd.

If f x is even ( f x f x for all x), then 0nb

for all n, leaving a Fourier cosine series (and perhaps a

constant term) only for f x .

If f x is odd ( f x f x for all x), then 0na

for all n, leaving a Fourier sine series only for f x .

Note that

0

0 if is odd

2 if is even

aa

a

g x

g x dxg x dx g x

with the multiplication table × | odd function even function

odd function | even odd

even function | odd even

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