adv math[unit 4]

Advanced Engineering Mathematics

Power Series and Power Series Solutions of Differential Equations

POWER SERIES AND POWER SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS

4.1. Sequences and Series

A sequence is a list of numbers

a�, a�, a�, ⋯ , a�

in a given order. Each of a�, a�, a�and so on represents a number. These are the terms of the sequence.

An infinite sequence of numbers is a function whose domain is the set of positive integers. For example,

the function associated to the sequence

2, 4, 6, 8, 10, 12,⋯ , 2n,⋯

sends 1 to a� = 2, 2 to a� = 4 and so on. The general behavior of this sequence is described by the

formula

a� = 2n

Sequences can be described by writing rules that specify their terms or by listing terms.

Convergence and Divergence of Sequences. Sometimes, the numbers in a sequence approach a single

value as the index n increases. This happens in the sequence

�1, 12 , 13 , 14 ,⋯ , 1n ,⋯ � whose terms approach zero as n gets large, and in the sequence

�0, 12 , 23 , 34 , 45 ,⋯ ,1 −1n ,⋯ � whose terms approach one. Such sequences are said to converge to that value. On the other hand,

sequences like

�1, √2, √3,⋯ , √n,⋯ � have terms that get larger than any number as n increases; such sequences are said to diverge. Still there

are sequences like

�1, −1, 1,−1,⋯ , �−1��, ⋯ � that terms oscillate between two values and does not converge to a single value. The sequence �a�� converges to the number L if to every positive number ϵ there corresponds an integer N such that for all n, n > N and |a� − L| < ϵ , where L is the limit of the sequence �a�� . If no such L exists, then the

sequence is said to diverge.



A series (or for this matter, an infinite series) is the sum of an infinite sequence of numbers

a� + a� + a� +⋯+ a� +⋯

The sum of the first n terms of the series

s� = a� + a� + a� +⋯+ a�

is an ordinary finite sum and is called the nth partial sum. As n gets larger, we expect the partial sums to

get closer and closer to a limiting value in the same sense that the terms of a sequence approach a limit.

For example, the series

�1 + 12 + 14 + 18 + 116 +⋯� we form a sequence of partial sums as follows

s� = 1

s� = 1 + 12 = 32

s� = 1 + 12 + 14 = 74

s' = 1+ 12 + 14 + 18 = 158

⋮ s� = 1+ 12 + 14 + 18 + ⋯+ 12�)� = 2 − 12�)�

The partial sums form a sequence whose nth term is

s� = 2 − 12�)�

This sequence of partial sums converges to 2, because

lim�→. 12�)� = 0

Thus the sum of the sequence /1 + ��+ �

'+ �0+ �

�1 +⋯2 is 2.In general, the series

a� + a� + a� +⋯+ a� +⋯



has partial sums that form the terms of the sequence�s��. If the latter sequence converges to a limit L,

then L is the sum of the series. If the sequence of the partial sums of the series does not converge, then

the series diverges.

Example 4.1

1. Find the sum of the geometric series a + ar + ar� + ar� +⋯+ ar�)� +⋯ = ∑ ar�)�.�5� . Determine the necessary conditions for this series to converge and the value to which this series converges. 2. Using the results above, determine the sum of a geometric series whose first term is 1/9 and a common

ratio 1/3.

3. Using again the results of the first problem, determine the sum ∑ �)��89'8.�5: .

4. Find the sum of the series ∑ ��.�5� and ∑ 1

��)��.�5�

Answers:

1. The nth partial sum is s� = ;��)<8��)< , r ≠ 1. The series converges when |r| < 1 and this converges to

s� = ;�)<, otherwise, the series diverges.

2. s� = �1

3. s� = 4

4. ∑ ��.�5� = 1, ∑ 1

��)��.�5� = 3

The >th term test for divergence. The series

?n�.

�5�= 1 + 4 + 9 + ⋯+ n� +⋯

diverges because the partial sums grow beyond every number L. Thus we can observe that if the partial

sum s� = ∑ a�.�5� converges, then a� → 0. Thus, we can test if the series diverges based on the above

observation: that the partial sum s� = ∑ a�.�5� diverges if lim�→. a� fails to exist or is different from

zero.



Example 4.2

Determine the following series if it diverges using the nth term test for divergence.

1. ∑ n�.�5�

2. ∑ ��.�5�

3. ∑ �−1��.�5� Answers:

1. The series diverges because n� → ∞.

2. The series diverges because �� → 1

3. The series diverges because lim�→.�−1�� does not exist.

If the series ∑a� = A and ∑b� = B are convergent series, then

?�a� + b�� =?a� +?b� = A + B (4.1a)

?�a� − b�� =?a� −?b� = A − B (4.1b)

?ka� = k?a� = kA (4.1c)

The integral test. The harmonic series

?1n.

�5�= 1 + 12 + 13 + ⋯+ 1n +⋯

is divergent, even though �� → 0. Thus the nth term test for divergence fails in this case. The reason why it

diverges is because there is no upper bound for its partial sums.



If we let the sequence as �a��. Suppose that a� = f�n�, where f is a continuous, positive, decreasing

function of x for all x ≥ N (N is a positive integer. Then the series ∑ a�.�5H and the integral I f�x�.H dx

both converge or both diverge.

Example 4.3 Test the following series for convergence using the integral test.

1. ∑ ��.�5�

2. ∑ ��K.�5�

Answers: 1. The series diverges. 2. The series converges.

The ratio test. Let ∑a� be a series with positive terms and suppose that

lim�→. a��a� = p (4.2)

then,

(a) the series converges if p < 1,

(b) the series diverges if p > 1 or p is infinite,

(c) the test is inconclusive if p = 1.

Example 4.4 Investigate the convergence of the following series, and if the series is convergent, find its sum.

1. ∑ �8�9�8.�5:

2. ∑ ��!�!�!.�5�

3. ∑ '8�!�!��!.�5�

Answers:

1. The series converges. s� = �� .

2. The series diverges. 3. The series diverges.



The root test. Let ∑a� be a series with a� ≥ 0 for n ≥ N and suppose that

lim�→. Na�8 = p (4.3)

then,

(a) the series converges if p < 1,

(b) the series diverges if p > 1 or p is infinite,

(c) the test is inconclusive if p = 1.

Example 4.5 Investigate the convergence of the following series using root test.

1. ∑ �K�8.�5�

2. ∑ �8�K.�5�

3. ∑ O ��P�.�5�

Answers: 1. The series converges. 2. The series diverges. 3. The series converges.

Alternating series, absolute and conditional convergence. A series in which the terms are alternately

positive and negative is an alternating series. For example, the alternating harmonic series

1 − 12 + 13 − 14 + 15 − ⋯+�−1��n + ⋯

is convergent, as well as the alternating geometric series

−2 + 1 − 12 + 14 − 18 + ⋯+�−1��42� +⋯

with a common ratio r = −1/2. The series

1 − 2 + 3 − 4 + 5 − 6 +⋯+ �−1��n +⋯

however, is divergent because the nth term does not approach zero. In general, the alternating series



?�−1��u�.

�5�= u� − u� + u� − u' +⋯ (4.3)

converges if all three of the following conditions are satisfied:

1. The u�’s are all positive;

2. u� ≥ u�� for all n ≥ N, for some integer N; and

3. u� → 0.

A series ∑a� converges absolutely (is absolutely convergent) if the corresponding series of absolute

values, ∑|a�|, converges. In effect, a series that converge but does not converge absolutely converges

conditionally.

The absolute convergence test. If ∑|a�| converges, then ∑a� converges, although the converse of this

statement is generally not true (take a look at the harmonic series).

Example 4.6 Investigate the convergence of the following series.

1. ∑ �−1�� K.�5�

2. ∑ �)��8RS�T.�5� for p > 0, p > 1, and 0 < p ≤ 1.

Answers: 1. The series converges absolutely.

2. For p > 0, the series converges. For p > 1, the series converges absolutely. For 0 < p ≤ 1, the series converges conditionally.

Drill Problems 4.1 A. Which of the following sequences converges? Find the limit of each convergent sequence. 1. a� = 2 + �0.1��

2. a� = �)��

3. a� = O1 + W�P�

B. Find the sum of the following series.

4. ∑ �'8.�5�



5. ∑ '�'�)��'��.�5�

C. Investigate the convergence or divergence of the following series.

6. ∑ O �√�P�.�5:

7. ∑ X� ��.�5�

8. ∑ ��K��.�5�

9. ∑ O1 − ��P�.�5�

10. ∑ O��− ��KP�.�5�

11. ∑ �! X� ��!.�5�

12. ∑ �!�.�5�

13. ∑ �−1�� O ��:P�.�5�

14. ∑ �−1�� ln O1 + ��P.�5�

15. ∑ n�e)�.�5�

4.2. Power Series

A power series about x = 0 is a series of the form

?c�x�.

�5:= c: + c�x + c�x� +⋯+ c�x� +⋯ (4.4)

A power series about x = a is a series of the form

?c��x − a��.

�5:= c: + c��x − a� + c��x − a�� +⋯+ c��x − a�� +⋯ (4.5)

in which the center a and the coefficients c:, c�, c�,⋯ , c�, ⋯ are constants. For 4.4, taking all the

coefficients to be 1 gives the geometric power series



?x�.

�5:= 1 + x + x� +⋯+ x� +⋯

This series converges to ��)[ for |x| < 1. We express this fact by writing

11 − x = 1 + x + x� +⋯+ x� +⋯

(4.6)

for −1 < x < 1. Up to now, we have used the equation 4.6 as a formula for the sum of the series on the

right. At this point, we wish to emphasize the fact that the polynomial on the right P��x� approximates the

function on the left.

Figure 4.1. Approximation of the function 1/(1-x) by the polynomial

The power series

1 − 12 �x − 2� + 14 �x − 2�� +⋯+ ]−12^� �x − 2�� +⋯

matches equation 4.5 with a = 2 , c: = 1 , c� = −1/2 , c� = 1/4 , ⋯ , c� = �−1/2�� . This is a

geometric series with first term 1 and ratio r = − [)�� . The series converges for 0 < x < 4. The sum is



2x = 1 − 12 �x − 2� + 14 �x − 2�� +⋯+ ]−12^� �x − 2�� +⋯

(4.7)

Figure 4.2. Approximation of the function 2/x by a polynomial

Convergence of power series. If the power series

?a�x�.

�5:= a: + a�x + a�x� +⋯+ a�x� +⋯ (4.8)

converges for x = c ≠ 0, then it converges absolutely for all x with |x| < c. If the series diverges for x = d, then it diverges for all x with |x| > d.

Radius of convergence of power series. The convergence of the series ∑ c��x − a�� is described by

one of the following three possibilities:

1. There is a positive number R such that the series diverges for x with |x − a| > R but converges

absolutely for x with |x − a| < R. The series may or may not converge at either of the endpoints x = a −R and x = a + R.

2. The series converges absolutely for every x (R = ∞).

3. The series converges at x = a and diverges elsewhere (R = 0).

The number R is called the radius of convergence of the power series and the interval of radius R centered

at x = a is called the interval of convergence. The interval of convergence may be open, closed, or half-

open, depending on the particular series.



How to test a power series for convergence.

1. Use the ratio test (or the nth root test) to find the interval where the series converges absolutely.

Ordinarily, this is an open interval |x − a| < R or a − R < x < a + R.

2. If the interval of absolute convergence is finite, test for convergence at each end point. Use comparison

test, integral test, or the alternating series test.

3. If the interval of absolute convergence is a − R < x < a + R, the series diverges for |x − a| > R (it

does not even converge conditionally), because the nth term does not approach zero for those values of x.

Example 4.7 For what values of x do the following power series converge?

1. ∑ �−1��)� [8�.�5�

2. ∑ �−1��)� [K8RS��)�.�5�

3. ∑ [8�!.�5:

4. ∑ n! x�.�5:

Answers: 1. −1 < x ≤ 1 2. −1 ≤ x ≤ 1 3. All x 4. All x except x = 0

Term-by-term differentiation. If ∑c��x − a�� converges for a − R < x < a + R for some R > 0, it

defines a function f f�x� = ?c��x − a��

.

�5:

(4.9a)

Such a function f has derivatives of all orders inside the interval of convergence. We can obtain the

derivatives by differentiating the original series term-by-term.

f′�x� = ?nc��x − a��)�.

�5:

(4.9b)



f′′�x� = ?n�n − 1�c��x − a��)�.

�5:

(4.9c)

and so on. Each of these derived series converges at every interior point of the interval of convergence of

the original series.

Example 4.8 Obtain the first and second derivative of the function

f�x� = 11 − x = 1 + x + x� + x� +⋯+ x� +⋯ =?x�.

�5:

for −1 < x < 1. Answers:

f a�x� = 1�1 − x�� = 1 + 2x + 3x� + 4x� +⋯+ nx�)� +⋯ =?nx�)�.

�5�

f aa�x� = 2�1 − x�� = 2+ 6x + 12x� +⋯+ n�n − 1�x�)� +⋯ =?n�n − 1�x�)�.

�5�

all for −1 < x < 1.

Term-by-term integration. Suppose that

f�x� = ?c��x − a��.

�5:

(4.9a)

converges for a − R < x < a + R (R > 0), Then

bf�x�dx = ?c� �x − a��n + 1.

�5:

(4.10)

with

?c� �x − a��n + 1.

�5:

converging also for a − R < x < a + R (R > 0).



Example 4.9 Investigate the following series 1.

f�x� = x − x�5 + x95 − ⋯

for −1 ≤ x ≤ 1. 2. f�t� = 1 − t + t� − t� +⋯ for −1 < t < 1. Answer: 1. Using term-by-term differentiation and integration of the resulting series, we find that

f�x� = x − x�5 + x95 −⋯ = tan)� x

for −1 < x < 1. 2. Using term-by-term integration, we can generate the series for ln�1 + x� as

ln�1 + x� = x − x�2 + x�3 − x

'4 + ⋯

for −1 < x < 1.

Multiplication of power series. If A�x� = ∑ a�x�.�5: and B�x� = ∑ b�x�.�5: converges absolutely for |x| < R and

c� = a:b� + a�b�)� + a�b�)� +⋯+ a�)�b� + a�b: =?adb�)d�

d5:

then the series ∑ c�x�.�5: converges absolutely to A�x�B�x� for |x| < R:

e?a�x�.

�5:fe?b�x�

.

�5:f = ?c�x�

.

�5: (4.11)



Drill Problems 4.2 A. For the following series, find the series’ radius and interval of convergence. 1. ∑ �x + 5��.�5:

2. ∑ �−1��4x + 1��.�5:

3. ∑ ��[)��8�.�5�

4. ∑ �[8��.�5:

5. ∑ �)��8�[��8�.�5�

B. For the following series, find the series’ interval of convergence and within this interval, the sum of the series as a function of x.

6. ∑ �[)��K8'�.�5�

7. ∑ �[��K8g8.�5:

8. ∑ O√[� − 1P�.�5:

9. ∑ �lnx��.�5:

10. ∑ O[K�� P�.�5:

4.3. Taylor and Maclaurin Series

This section shows how functions that are infinitely differentiable generate power series called Taylor

series. This is when we think that a function f�x� has derivatives of all orders on an interval I can be

expressed as a power series on that interval.

If we have

f�x� = ?c��x − a��.

�5:= c: + c��x − a� + c��x − a�� +⋯+ c��x − a�� +⋯

(4.9)

Term-by-term differentiation gives

f a�x� = c� + 2c��x − a� + 3c��x − a�� +⋯+ nc��x − a��)� +⋯



f aa�x� = 1 ∙ 2c� + 2 ∙ 3c��x − a� + 3 ∙ 4c��x − a�� +⋯+ n ∙ �n − 1�c��x − a��)� +⋯

f aaa�x� = 1 ∙ 2 ∙ 3c� + 2 ∙ 3 ∙ 4c'�x − a� + 3 ∙ 4 ∙ 5c9�x − a�� +⋯+ n ∙ �n − 1� ∙ �n − 2�c��x − a��)� +⋯

with the nth derivative of, for all n being

f ��x� = n! c� + asumoftermswith�x − a�asfactor Since these equations all hold at x = a, we have

f a�a� = c�

f aa�a� = 1 ∙ 2c�

f aaa�a� = 1 ∙ 2 ∙ 3c�

⋮ f ��a� = n! c�

With this, a general formula for the nth term of the coefficients of the power series can be deduced, that is

c� = f ��a�n! (4.12)

If the function f�x� has a series representation, then it must be

f�x� = ? f �d��a�k! �x − a�d.

d5:= f�a� + f a�a��x − a� + f aa�a�2! �x − a�� +⋯+ f ��a�n! �x − a�� +⋯ (4.13)

which is the Taylor series expansion of f�x� about x = a. When a = 0, the series becomes

f�x� =?f �d��0�k! xd.

d5:= f�0� + f a�0�x + f aa�0�2! x� +⋯+ f ��0�n! x� +⋯ (4.14)

which is the Maclaurin series expansion of f�x�. The right hand side of equations 4.13 and 4.14 are called the Taylor (Maclaurin) polynomials of order n of

the function f�x�. P��x� = f�a� + f a�a��x − a� + f aa�a�2! �x − a�� +⋯+ f ��a�n! �x − a�� (4.15)



P��x� = f�0� + f a�0�x + f aa�0�2! x� +⋯+ f ��0�n! x� (4.16)

These polynomials provided a linear approximation of the function f�x� in the neighborhood of a (or 0)

Example 4.10

1. Find the Taylor series generated by f�x� = �[ at a = 2. Where, if anywhere, does the series converge to

�[. 2. Find the Taylor series expansion, and the Taylor polynomials generated by f�x� = e[ at x = 0. 3. Find the Taylor series and Taylor polynomials generated by f�x� = cos x at x = 0. Answer:

1. f�x� = ��− �[)��

�K + �[)��K�m −⋯+ �−1�� [)��8�8nS +⋯. The series converges for 0 < x < 4.

2. The Taylor series expansion of f�x� is f�x� = ∑ [od!.d5: . The Taylor polynomials generated by the

function is

P��x� = 1 + x + x�2 +⋯x�n!

Figure 3. Plot of approximation for exp(x)



3. The Taylor series generated by function f at x = 0 is ∑ �)��o[Ko��d�!.d5: . The 2nth order (as well as the

�2n + 1�th order) polynomial of cos x is then

P��x� = P��x� = 1 − x�2! + x'4! −⋯+ �−1�� x��2n�!

Figure 4. Taylor polynomial approximation of cos(x)

Drill Problems 4.3 A. Find the Maclaurin series (Taylor series at x = 0) of the following functions

1. f�x� = e)[

2. f�x� = e[/�

3. f�x� = ��[

4. f�x� = ��)[

5. sin 3x

6. cosh x

7. sinh x

8. x' − 2x� − 5x + 4

9. �x + 1��

10. sin [�



B. Find the Taylor series generated by the following functions at x = a

11. f�x� = x� − 2x + 4, a = 2

12. f�x� = 2x� + 2x� + 3x − 8, a = 1

13. f�x� = x' + x� + 1, a = −2

14. f�x� = 3x9 − x' + 2x� + x� − 2, a = −1

15. f�x� = �[K, a = 1

16. f�x� = [�)[, a = 0

17. f�x� = e[, a = 2

18. f�x� = 2[, a = 1

C. Find the first and second order Taylor polynomial approximation of the following functions at x = 0. 19. f�x� = ln�cos x�

20. f�x� = epq� [

4.4 Power Series Solution of Differential Equations: Solutions Near an Ordinary Point

For a linear differential equation

b:�x�y�� + b��x�y��)�� +⋯+ b��x�y = R�x� (4.17)

the point x = x: is called an ordinary point of 4.17 if b:�x:� ≠ 0. A singular point of the linear equation is

any point x = x� for which b:�x�� = 0. Any point that is not a singular point is an ordinary point.

The differential equation

�1 − x��yaa − 6xya − 4y = 0

has x = 1 and x = −1 as its only singular points in the finite complex plane. The equation

yaa + 2xya + y = 0

has no singular points in the finite plane. The equation

xyaa + ya + xy = 0

has the origin, x = 0, as the only singular point in the finite plane.



The solution of linear equations with constant coefficients can be accomplished by methods developed

earlier. For linear, ordinary differential equations with variable coefficients, and of order greater than one,

probably the most generally effective method of attack is that based upon the use of power series.

In solving differential equations near an ordinary point, we assume a solution of a power series of the form

y = ?c�x�.

�5: (4.18)

and substitute this to the given differential equation. The task then is to find the general term of the series c�. The method described is demonstrated in the following examples.

Example 4.11

1. Solve the equation yaa + 4y = 0 near the ordinary point x = 0.

2. Solve the equation �1 − x��yaa − 6xya − 4y = 0 near the ordinary point x = 0. Answers:

1. y = c: cos2x + �� c� sin 2x.

2. y = st��)[K�K + sSu�[)[mv��)[K�K

Drill Problems 4.4 Find the general solution of the following differential equations near the origin. 1. yaa + 3xya + 3y = 0

2. �1 − 4x��yaa + 8y = 0

3. �1 + x��yaa + 10xya + 20y = 0

4. �x� − 9�yaa + 3xya − 3y = 0

5. �x� + 4�yaa + 6xya + 4y = 0

6. yaa + x�y = 0

7. �1 + 2x��yaa + 3xya − 3y = 0

8. yaa + xya + 3y = x�

9. yaa + 3xya + 7y = 0



10. �x� + 4�yaa + xya − 9y = 0