engi 3424 chapter 5 gapped notes - memorial …ggeorge/3424/handout/h5series.pdfengi 3424 5.02 –...

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:

or the direct definition

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,

The general term is therefore na


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,

The general term is therefore na

The sequence alternates in sign if 0r

The sequence converges to a limit of zero if

If 1r then lim nn

a


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

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

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

.

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.


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 ).

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.

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.

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

.

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

.


Example 5.03.2 (continued)

nS

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


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:


Example 5.03.3 (Example 5.02.1 revisited)

Evaluate

1

1

2nn

S

.

Example 5.03.4

Find the sum of the power series

2 3 4f x x x 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.

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.]


Example 5.04.2

Is the series

2

1

lnn n

convergent?

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?


Example 5.04.4

Is the series 2

3

1

4

9n

n

n n

convergent?

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?

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.

Example 5.05.3

Is

1

1

1 1 1 11

2 3 4

n

nn

absolutely convergent, conditionally

convergent or divergent?


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

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?


Note:

Proof that 1

lim 1

n

ne

n

:


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?

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

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

0

2

3

n

n

xf x

n


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

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 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 10 1 0 01

f cf c a a

2

0

1n

nn

f x a n n 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 .


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


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.


Example 5.08.4

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


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


Example 5.08.5

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

function.

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 .


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".

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


Example 5.10.1

Expand

0 0

0

xf x

x x

in a Fourier series.


Example 5.10.1 (Additional Notes – also see

"www.engr.mun.ca/~ggeorge/3424/demos/")

The first few partial sums in the Fourier series

2

1

1 1 1cos sin

4n

n

f x nx nx xn n

are

04

S

1

2cos sin

4S x x

2

2 1cos sin sin 2

4 2S x x x

3

2 1 2 1cos sin sin 2 cos3 sin3

4 2 9 3S x x x x x

and so on.

The graphs of successive partial sums approach f (x) more closely, except in the vicinity

of any discontinuities, (where a systematic overshoot occurs, the Gibbs phenomenon).


Example 5.10.2

Find the Fourier series expansion for the standard square wave,

1 1 0

1 0 1

xf x

x

The graphs of the third and ninth partial sums (containing two and five non-zero terms

respectively) are displayed here, together with the exact form for f (x), with a periodic

extension beyond the interval (–1, +1) that is appropriate for the square wave.

3y S x



9y S x

Convergence

At all points ox x in (–L, L) where f x is continuous and is either differentiable or the

limits o

limx x

f x

and o

limx x

f x

both exist, the Fourier series converges to f x .

At finite discontinuities, (where the limits o

limx x

f x

and o

limx x

f x

both exist), the

Fourier series converges to o o

2

f x f x ,

(using the abbreviations o oo o

lim and limx x x x

f x f x f x f x

).

f (x) not continuous continuous but continuous and

at ox x not differentiable differentiable

In all cases, the Fourier series at ox x converges to o o

2

f x f x (the red dot).


Half-Range Fourier Series

A Fourier series for f x , valid on 0, L , may be constructed by extension of the

domain to ,L L .

An odd extension leads to a Fourier sine series:

1

sinn

n

n xf x b

L

where

0

2sin , 1, 2, 3,n

Ln x

b f x dx nL L

An even extension leads to a Fourier cosine series:

0

1

cos2

n

n

a n xf x a

L

where

0

2cos , 0,1, 2, 3,

L

nn x

a f x dx nL L

and there is automatic continuity of the Fourier cosine series at 0x and at x L .


Example 5.10.3

Find the Fourier sine series and the Fourier cosine series for on 0,1f x x .

f x x happens to be an odd function of x for any domain centred on 0x . The odd

extension of f x to the interval 1,1 is f x itself.

Evaluating the Fourier sine coefficients,

This function happens to be continuous and differentiable at 0x , but is clearly

discontinuous at the endpoints of the interval 1x .

Fifth order partial sum of the Fourier sine series for f x x on [0, 1]



The even extension of f x to the interval [–1, 1] is f x x .

Evaluating the Fourier cosine coefficients,



Third order partial sum of the Fourier cosine series for on 0,1f x x

Note how rapid the convergence is for the cosine series compared to the sine series.

3y S x for cosine series and 5y S x for sine series for on 0,1f x x

ENGI 3424 5.11 – Series Solutions of ODEs Page 5-51

5.11 Introduction to Series Solutions of ODEs

If the functions ,p x q x and r x in the ODE

2

2

d y dyp x q x y r x

dx dx

are all analytic in some interval o ox h x x h (and possess Taylor series

expansions around ox with radii of convergence of at least h), then a series solution to the

ODE around ox with a radius of convergence of at least h exists:

o

o

0

,!

n nn

nn y x

y x a x x an

Example 5.11.1

Find a series solution as far as the term in x3, to the initial value problem

2

24 ; 0 1, 0 4xd y dy

x e y y ydx dx

None of our previous methods apply to this problem.

The functions –x, ex and 4 are all analytic everywhere.

The solution of this ODE, expressed as a power series, is


Example 5.11.2

Find the general solution (as a power series about 0x ) to the ordinary differential

equation 2

2

20

d yx y

dx

Let the general solution be 0

nn

ny x a x

.

Then 0 1

1 10n nn n

n ny x n a x na x

and 0 2

2 21 0 0 1n nn n

n ny x n n a x n n a x

Substitute into the ODE:

2

2 0 2 0

2 2 21 0 1 0n n n nn n n n

n n n nn n a x x a x n n a x a x

We want the exponent of x to be the same in both summations, so that we can bring the

two summations together.

2 3 4 22 0

2 0 1 21 2 1 3 2 4 3 2 1n nn n

n nn n a x a x a x a x n n a x

and

0 1 2 20 2

2 2 3 4n n

n n

n na x a x a x a x a x

2 20 2

2 1 0n nn n

n nn n a x a x

Bring the two summations together for all terms from n = 2 onwards:

0 12 3 2 2

2

2 1 3 2 2 1 0n nn

na x a x n n a a x

But this equation must be true regardless of the choice of x.

Therefore the coefficient of each power of x must be zero.

2

2 3 20 and 2, 3, 4,2 1

nn

aa a a n

n n



and 0 0a y and 1 0a y are arbitrary.

Therefore the general solution is

0 10 1

0 1

2 3 4 5 6 7

8 9 10 11

0 0 0 012 20

0 0672 1440

a ay x a a x x x x x x x

a ax x x x

or

0 14 8 5 91 1 1 1

12 672 20 14401y x a x x a x x x

The arbitrary constants can be determined from initial conditions:

0 10 and 0a y a y

The Maple commands > with(DEtools):

> ode := diff(y(x),x,x) + x*x*y(x) = 0;

> Order := 15;

> dsolve(ode, y(x), series);

produce the output

4 5 81 1 10 0 0 0 0

12 20 672y x y D y x y x D y x y x

9 12 13 151 1 10 0 0 O

1440 88704 224640D y x y x D y x x

Follow the link for this example on the course web site, at

"http://www.engr.mun.ca/~ggeorge/3424/demos/index.html".


The two principal methods of series solution of ODEs of the form

2

2

d y dyp x q x y r x

dx dx

using a Maclaurin series

2 30 0

0 02! 3!

y yy x y y x x x

are

1) use the ODE to find 0 0 0 0 0 0y r p y q y

and differentiate y x r x p x y x q x y x successively

to find the higher derivatives

ny x and hence

0

!

n

n

ya

n

(as in Example 5.11.1)

and

2) Substitute 0n

nny x a x

into the ODE and match coefficients of nx to find

the values of the na (as in Example 5.11.2).

Method 2 is not advised for Example 5.11.1, as that would involve the product of two

infinite series,

0 0!

n n

nx n

nx

e y a xn

Method 1 could be used for Example 5.11.2, as shown on the next two pages.


Example 5.11.2 Alternative Solution to 2

2

20

d yx y

dx

1 2 30 0 0

01! 2! 3!

y y yy x y x x x

Let 0A y and 0B y

2 0 0y x x y x y

22 0 0y x xy x x y x y

2 2 22 2 2 2 4IVy x y x xy x xy x x y x y x xy x x x y x

4 2 4 0 2IV IVy x x y x xy x y A

3 44 2 4 4Vy x x y x x y x y x xy x

3 4 2 3 44 6 4 8 6 0 6Vx y x x y x x x y x x y x x y x y B

2 3 3 424 8 4 6VIy x x y x x y x x y x x y x

2 3 4 2 2 6 324 12 6 30 12x y x x y x x x y x x x y x x y x

0 0VIy

5 2 6 2 360 6 30 36 12VIIy x x x y x x x y x x y x x y x

5 2 6 3 2 5 2 660 6 66 12 60 18 66x x y x x x y x x x y x x x y x x x y x

0 0VIIy

4 5 5 2 660 90 60 18 132 6 66VIIIy x x y x x x y x x x y x x x y x

4 8 560 156 192 24 0 60VIIIx x y x x x y x y A

IXy x

3 7 4 8 4 5 2624 8 60 156 192 120 192 24x x y x x x y x x y x x x x y x

3 7 4 8816 32 252 276 0 252IXx x y x x x y x y B etc.

2 3 4 5 6 7 8 92 6 60 2520 0 0 0

4! 5! 8! 9!

A B A By x A Bx x x x x x x x x

4 8 5 91 1 1 11

12 672 20 1440y x A x x B x x x


Example 5.11.2 Alternative Solution (continued)

However, the above does not allow us to deduce a general pattern for na . We require a

generalized form of the product rule of differentiation for higher order derivatives.

The product rule of differentiation is

uv u v uv

where ,du dv

u vdx dx

Extending the product rule to higher orders of differentiation,

2

u v u v

uv uv u v uv u v uv

u v u v uv

2 2

2

3 3

u v u v

u v u vuv u v u v uv

u v uv

u v u v u v uv

The coefficients of n k k

n k k

d u d v

dx dx

are clearly the entries

0,1,2, ,k n in row n of Pascal’s triangle, which

are also the binomial coefficients

!

! !

nk

nC

k n k

It is straightforward to verify that

4 44

44 6 4

duv u v u v u v u v uv

dx

The general form for any natural number n is

0

n

k

n n k kn

n k n k k

d d u d vuv C

dx dx dx

which is also

1 2 2

1 2 2

3 3

3 3

1

2 1

1 2

3 2 1

n n n n

n n n n

n n

nn

n nd d u d u dv d u d vuv v n

dx dx dxdx dx dx

n n n d u d v d vu

dxdx dx


Example 5.11.2 Alternative Solution (continued)

Starting over:

2y x x y x

2 2

22 2 22

2 2 20

n k n kn

k k n kk

n nn

n n

d d d dy x y x x y x C x y x

dx dx dx dx

2 3 2 4

2 2 2 2 2 2

0 1 22 3 2 4

n n nn n n

n n n

d d d d dC x y x C x y x C x y x

dx dx dx dx dx

2 2 3 42 3

2 2 2 0 02

n n nn nx y x n x y x y x

Therefore, for 4n ,

2

2 3 42 2 2 3

n n nny x x y x n x y x n n y x

40 0 0 2 3 0

nny n n y

40 2 3

0! !

nn

n

y n na y

n n

But

4

4

0

4 !

n

n

ya

n

4 4

2 3 2 3 4 !4 !

! 1 2 3 4 !n n n

n n n n na n a a

n n n n n n

4

1

1n n

a an n

, which is the recurrence relation for 4n

54 0 1 6 2

1 1 1, ,

4 3 5 4 6 5a a a a a a

, etc.

We also have 0a A , 1a B , 2 3 0a a

The same conclusion follows,

4 8 12

5 9 13

1 1 11

4 3 8 7 4 3 12 11 8 7 4 3

1 1 1

5 4 9 8 5 4 13 12 9 8 5 4

y x A x x x

B x x x x


Example 5.11.3

Find the complete solution (as a power series about x = 0) to the initial value problem

2

2

21 5 3 0 , 0 0, 0 1

d y dyx x y y y

dx dx


nn

ny x a x

.

Then 1 2

1 2and 1n nn n

n ny x na x y x n n a x

.


2 2 1 0

2 2 2 1 11 1 5 3 0n n n nn n n n

n n n nn n a x n n a x na x a x

We prefer all summations to involve the same exponent nx in the general term.

Substitute m = n – 2 (and thus n = m + 2) into the first term:

22 0

21 2 1n mn m

n mn n a x m m a x

Start each summation from n = 2:

2 2 3 20 2

0 12 1 2 1 3 2 2 1m nm n

m nm m a x a x a x n n a x

,

1 11 2 2

15 5 1 5 5n n nn n n

n n nna x a x na x a x na x

and

0 10 1 0 1

0 2 2

3 3 3 3 3n n nn n n

n n na x a x a x a x a a x a x

and combine the summations for all n from 2 onwards:

22

2 1 1 5 3n n nnn

nn n a n n a na a x

2 3 1 0 12 6 5 3 3 0a a x a x a a x

2

2 0 3 122

2 1 5 3 2 3 6 8 0nnn

nn n a n n n a x a a a a x

2 0 3 1 22

2 3 6 8 2 1 3 1 0nnn

na a a a x n n a n n a x

This method can be difficult to follow. A slower alternative is on the next page.


From before,

2 2 1 0

21 1 5 3 0n n n nn n n n

n n n nn n a x n n a x na x a x

Write out enough terms from each series to spot a pattern:

0 1 2 3 4

2 3 4 5 62 1 3 2 4 3 5 4 6 5a x a x a x a x a x

– 2 3 4

2 3 42 1 3 2 4 3a x a x a x

– 1 2 3 4

1 2 3 45 1 2 3 4a x a x a x a x

– 0 1 2 3 4

0 1 2 3 43 0a x a x a x a x a x

Collecting the coefficients of 0x and 1x together, the series begins with

Clearly a general term can be constructed for n = 2 onwards.

The general term is of the form

The series can be re-written as

2 0 3 1 22

2 3 6 8 1 2 3 0nnn

na a a a x n n a n a x



Equating coefficients of xn :

x0:

x1:

All higher powers of x :

The Maple commands > with(DEtools):

> ode := (1-x^2)*diff(y(x),x,x) - 5*x*diff(y(x),x) - 3*y(x) = 0;

> Order := 11;

> dsolve(ode, y(x), series);

produce the output

2 3 4 53 4 15 80 0 0 0 0 0

2 3 8 5y x y D y x y x D y x y x D y x

6 7 8 9 10 1135 64 315 128 6930 0 0 0 0 O

16 35 128 63 256y x D y x y x D y x y x x

Follow the link for this example on the course web site, at

"http://www.engr.mun.ca/~ggeorge/3424/demos/index.html".


Example 5.11.4

Find the general solution (as a power series about 0x ) to the ordinary differential

equation 2

2

d yx y

dx

2 2

2 20

d y d yxy xy

dx dx


nn

ny x a x

.

Then 1 2

1 2and 1n nn n

n ny x na x y x n n a x

.


2 0

2 11 0n nn n

n nn n a x a x

Writing out the two series, term by term:

Equating coefficients of 0x :



For all 1n , equating coefficients of nx :

ENGI 3424 5.A – Integral Test Page 5-63

5.A Integral Test [for reference only – not examinable in this course]

If f x is a continuous function with positive values that are decreasing for all x k ,

, 1, 2,na f n n k k k and the improper integral limL

L x kf x dx

exists as a finite real number, then the series nn k

a

converges and

n kk kn k

f x dx a a f x dx

If the improper integral does not have a finite value, then the series diverges.

Proof:

Examine the area between 1x i and x i under the curve y f x :

Both rectangles have width one unit.

The smaller rectangle has height f i and area smaller than that under the curve.

The larger rectangle has height 1f i and area larger than that under the curve.

Therefore

1

1 1 1i

if i f x dx f i

This inequality is true for all areas from 1i k onwards.

Adding all of these areas together, we find

1 1

1i k i k

kf i f x dx f i

The left side can be re-arranged as

i k i k

k kf x dx f i f k f i f k f x dx

The right side can be re-arranged as

1

1i k i k i k

k kf i f i f x dx f x dx f i

ENGI 3424 5.A – Integral Test Page 5-64

Also note that the ith

term of the series is ia f i . It then follows that

n kk kn k

f x dx a a f x dx

The convergence or divergence of the series is therefore linked completely to the

convergence or divergence of the improper integral.

The double inequality also allows upper and lower bounds to be evaluated for the sum of

a convergent series whose terms form a sequence drawn from a positive, continuous and

decreasing function f x .

The convergence of the p-series can be established by a combination of the divergence

and integral tests:

01

lim 1 0

0 0

pn

p

pn

p

the p-series diverges for 0p .

When 0p the function 1pf x

x is positive, continuous and decreasing on 1,

so that we may use the integral test.

1

1

1

11lim 1 11 1

0 1lim ln 1

np

np

n

n

xp pp px dx

px p

Therefore the p-series

1

1p

nn

converges for 1p and diverges otherwise.

Another example of the integral test arises in showing that

2

1

lnp

nn n

converges if and only if 1p [– details omitted].

END OF CHAPTER 5

END OF ENGI 3424 !

engi 3424 chapter 5 gapped notes - memorial …ggeorge/3424/handout/h5series.pdfengi 3424 5.02 –...

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