Adv math[unit 4]

Advanced Engineering Mathematics
Power Series and Power Series Solutions of Differential Equations Page 1
POWER SERIES AND POWER SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
4.1. Sequences and Series
A sequence is a list of numbers
aଵ, aଶ, aଷ, ⋯ , a୬
in a given order. Each of aଵ, aଶ, aଷand 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 aଵ = 2, 2 to aଶ = 4 and so on. The general behavior of this sequence is described by the
formula
a୬ = 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
൜1,
1
2
,
1
3
,
1
4
, ⋯ ,
1
n
, ⋯ ൠ
whose terms approach zero as n gets large, and in the sequence
൜0,
1
2
,
2
3
,
3
4
,
4
5
, ⋯ ,1 −
1
n
, ⋯ ൠ
whose terms approach one. Such sequences are said to converge to that value. On the other hand,
sequences like
൛1, √2, √3, ⋯ , √n, ⋯ ൟ
have terms that get larger than any number as n increases; such sequences are said to diverge. Still there
are sequences like
ሼ1, −1, 1, −1, ⋯ , ሺ−1ሻ୬ାଵ
, ⋯ ሽ
that terms oscillate between two values and does not converge to a single value. The sequence ሼa୬ሽ
converges to the number L if to every positive number ϵ there corresponds an integer N such that for all n,
n > N and |a୬ − L| < ϵ, where L is the limit of the sequence ሼa୬ሽ. 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
aଵ + aଶ + aଷ + ⋯ + a୬ + ⋯
The sum of the first n terms of the series
s୬ = aଵ + aଶ + aଷ + ⋯ + a୬
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
൜1 +
1
2
+
1
4
+
1
8
+
1
16
+ ⋯ ൠ
we form a sequence of partial sums as follows
sଵ = 1
sଶ = 1 +
1
2
=
3
2
sଷ = 1 +
1
2
+
1
4
=
7
4
sସ = 1 +
1
2
+
1
4
+
1
8
=
15
8
⋮
s୬ = 1 +
1
2
+
1
4
+
1
8
+ ⋯ +
1
2୬ିଵ
= 2 −
1
2୬ିଵ
The partial sums form a sequence whose nth term is
s୬ = 2 −
1
2୬ିଵ
This sequence of partial sums converges to 2, because
lim
୬→ஶ
1
2୬ିଵ
= 0
Thus the sum of the sequence ቄ1 +
ଵ
ଶ
+
ଵ
ସ
+
ଵ
଼
+
ଵ
ଵ଺
+ ⋯ ቅ is 2.In general, the series
aଵ + aଶ + aଷ + ⋯ + a୬ + ⋯

has partial sums that form the terms of the sequenceሼs୬ሽ. 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 + arଶ
+ arଷ
+ ⋯ + ar୬ିଵ
+ ⋯ = ∑ ar୬ିଵஶ
୬ୀଵ .
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 ∑
ሺିଵሻ౤ହ
ସ౤
ஶ
୬ୀ଴ .
4. Find the sum of the series ∑
ଵ
୬ሺ୬ାଵሻ
ஶ
୬ୀଵ and ∑
଺
ሺଶ୬ିଵሻሺଶ୬ାଵሻ
ஶ
୬ୀଵ
Answers:
1. The nth partial sum is s୬ =
ୟሺଵି୰౤ሻ
ଵି୰
, r ≠ 1. The series converges when |r| < 1 and this converges to
s୬ =
ୟ
ଵି୰
, otherwise, the series diverges.
2. s୬ =
ଵ
଺
3. s୬ = 4
4. ∑
ଵ
୬ሺ୬ାଵሻ
ஶ
୬ୀଵ = 1, ∑
଺
ሺଶ୬ିଵሻሺଶ୬ାଵሻ
ஶ
୬ୀଵ = 3
The ‫ܖ‬th term test for divergence. The series
෍ nଶ
ஶ
୬ୀଵ
= 1 + 4 + 9 + ⋯ + nଶ
+ ⋯
diverges because the partial sums grow beyond every number L. Thus we can observe that if the partial
sum s୬ = ∑ a୬
ஶ
୬ୀଵ converges, then a୬ → 0. Thus, we can test if the series diverges based on the above
observation: that the partial sum s୬ = ∑ a୬
ஶ
୬ୀଵ diverges if lim୬→ஶ a୬ 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. ∑ nଶஶ
୬ୀଵ
2. ∑
୬ାଵ
୬
ஶ
୬ୀଵ
3. ∑ ሺ−1ሻ୬ାଵஶ
୬ୀଵ
Answers:
1. The series diverges because nଶ
→ ∞.
2. The series diverges because
୬ାଵ
୬
→ 1
3. The series diverges because lim୬→ஶሺ−1ሻ୬ାଵ
does not exist.
If the series ∑ a୬ = A and ∑ b୬ = B are convergent series, then
෍ሺa୬ + b୬ሻ = ෍ a୬ + ෍ b୬ = A + B (4.1a)
෍ሺa୬ − b୬ሻ = ෍ a୬ − ෍ b୬ = A − B (4.1b)
෍ ka୬ = k ෍ a୬ = kA (4.1c)
The integral test. The harmonic series
෍
1
n
ஶ
୬ୀଵ
= 1 +
1
2
+
1
3
+ ⋯ +
1
n
+ ⋯
is divergent, even though
ଵ
୬
→ 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 ሼa୬ሽ. Suppose that a୬ = fሺnሻ, where f is a continuous, positive, decreasing
function of x for all x ≥ N (N is a positive integer. Then the series ∑ a୬
ஶ
୬ୀ୒ and the integral ‫׬‬ fሺxሻ
ஶ
୒
dx
both converge or both diverge.
Example 4.3
Test the following series for convergence using the integral test.
1. ∑
ଵ
୬
ஶ
୬ୀଵ
2. ∑
ଵ
୬మ
ஶ
୬ୀଵ
Answers:
1. The series diverges.
2. The series converges.
The ratio test. Let ∑ a୬ be a series with positive terms and suppose that
lim
୬→ஶ
a୬ାଵ
a୬
= 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. ∑
ଶ౤ାହ
ଷ౤
ஶ
୬ୀ଴
2. ∑
ሺଶ୬ሻ!
୬!୬!
ஶ
୬ୀଵ
3. ∑
ସ౤୬!୬!
ሺଶ୬ሻ!
ஶ
୬ୀଵ
Answers:
1. The series converges. s୬ =
ଶଵ
ଶ
.

The root test. Let ∑ a୬ be a series with a୬ ≥ 0 for n ≥ N and suppose that
lim
୬→ஶ
ඥa୬
౤
= 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. ∑
୬మ
ଶ౤
ஶ
୬ୀଵ
2. ∑
ଶ౤
୬మ
ஶ
୬ୀଵ
3. ∑ ቀ
ଵ
ଵା୬
ቁ
୬
ஶ
୬ୀଵ
Answers:
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 −
1
2
+
1
3
−
1
4
+
1
5
− ⋯ +
ሺ−1ሻ୬ାଵ
n
+ ⋯
is convergent, as well as the alternating geometric series
−2 + 1 −
1
2
+
1
4
−
1
8
+ ⋯ +
ሺ−1ሻ୬
4
2୬
+ ⋯
with a common ratio r = −1/2. The series
1 − 2 + 3 − 4 + 5 − 6 + ⋯ + ሺ−1ሻ୬ାଵ
n + ⋯
however, is divergent because the nth term does not approach zero. In general, the alternating series

෍ሺ−1ሻ୬
u୬
ஶ
୬ୀଵ
= uଵ − uଶ + uଷ − uସ + ⋯ (4.3)
converges if all three of the following conditions are satisfied:
1. The u୬’s are all positive;
2. u୬ ≥ u୬ାଵ for all n ≥ N, for some integer N; and
3. u୬ → 0.
A series ∑ a୬ converges absolutely (is absolutely convergent) if the corresponding series of absolute
values, ∑|a୬|, converges. In effect, a series that converge but does not converge absolutely converges
conditionally.
The absolute convergence test. If ∑|a୬| converges, then ∑ a୬ 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. ∑ ሺ−1ሻ୬ାଵ ଵ
୬మ
ஶ
୬ୀଵ
2. ∑
ሺିଵሻ౤షభ
୬౦
ஶ
୬ୀଵ 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. a୬ = 2 + ሺ0.1ሻ୬
2. a୬ =
ଵିଶ୬
ଵାଶ୬
3. a୬ = ቀ1 +
଻
୬
ቁ
୬
B. Find the sum of the following series.
4. ∑
ଵ
ସ౤
ஶ
୬ୀଶ

5. ∑
ସ
ሺସ୬ିଷሻሺସ୬ାଵሻ
ஶ
୬ୀଵ
C. Investigate the convergence or divergence of the following series.
6. ∑ ቀ
ଵ
√ଶ
ቁ
୬
ஶ
୬ୀ଴
7. ∑
୪୬ ୬
୬
ஶ
୬ୀଶ
8. ∑
୬
୬మାଵ
ஶ
୬ୀଵ
9. ∑ ቀ1 −
ଵ
ଷ୬
ቁ
୬
ஶ
୬ୀଵ
10. ∑ ቀ
ଵ
୬
−
ଵ
୬మቁ
୬
ஶ
୬ୀଵ
11. ∑
୬! ୪୬ ୬
୬ሺ୬ାଶሻ!
ஶ
୬ୀଵ
12. ∑
୬!
୬
ஶ
୬ୀଵ
13. ∑ ሺ−1ሻ୬ାଵ
ቀ
୬
ଵ଴
ቁ
୬
ஶ
୬ୀଵ
14. ∑ ሺ−1ሻ୬ାଵ
ln ቀ1 +
ଵ
୬
ቁஶ
୬ୀଵ
15. ∑ nଶ
eି୬ஶ
୬ୀଵ
4.2. Power Series
A power series about x = 0 is a series of the form
෍ c୬x୬
ஶ
୬ୀ଴
= c଴ + cଵx + cଶxଶ
+ ⋯ + c୬x୬
+ ⋯ (4.4)
A power series about x = a is a series of the form
෍ c୬ሺx − aሻ୬
ஶ
୬ୀ଴
= c଴ + cଵሺx − aሻ + cଶሺx − aሻଶ
+ ⋯ + c୬ሺx − aሻ୬
+ ⋯ (4.5)
in which the center a and the coefficients c଴, cଵ, cଶ, ⋯ , c୬, ⋯ are constants. For 4.4, taking all the
coefficients to be 1 gives the geometric power series

෍ x୬
ஶ
୬ୀ଴
= 1 + x + xଶ
+ ⋯ + x୬
+ ⋯
This series converges to
ଵ
ଵି୶
for |x| < 1. We express this fact by writing
1
1 − x
= 1 + x + xଶ
+ ⋯ + x୬
+ ⋯ (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 P୬ሺxሻ approximates the
function on the left.
Figure 4.1. Approximation of the function 1/(1-x) by the polynomial
The power series
1 −
1
2
ሺx − 2ሻ +
1
4
ሺx − 2ሻଶ
+ ⋯ + ൬−
1
2
൰
୬
ሺx − 2ሻ୬
+ ⋯
matches equation 4.5 with a = 2, c଴ = 1, cଵ = −1/2, cଶ = 1/4, ⋯, c୬ = ሺ−1/2ሻ୬
. This is a
geometric series with first term 1 and ratio r = −
୶ିଶ
ଶ
. The series converges for 0 < x < 4. The sum is

2
x
= 1 −
1
2
ሺx − 2ሻ +
1
4
ሺx − 2ሻଶ
+ ⋯ + ൬−
1
2
൰
୬
ሺx − 2ሻ୬
+ ⋯ (4.7)
Figure 4.2. Approximation of the function 2/x by a polynomial
Convergence of power series. If the power series
෍ a୬x୬
ஶ
୬ୀ଴
= a଴ + aଵx + aଶxଶ
+ ⋯ + a୬x୬
+ ⋯ (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 ∑ c୬ሺx − aሻ୬
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. ∑ ሺ−1ሻ୬ିଵ ୶౤
୬
ஶ
୬ୀଵ
2. ∑ ሺ−1ሻ୬ିଵ ୶మ౤షభ
ଶ୬ିଵ
ஶ
୬ୀଵ
3. ∑
୶౤
୬!
ஶ
୬ୀ଴
4. ∑ n! x୬ஶ
୬ୀ଴
Answers:
1. −1 < x ≤ 1
2. −1 ≤ x ≤ 1
3. All x
4. All x except x = 0
Term-by-term differentiation. If ∑ c୬ሺx − aሻ୬
converges for a − R < x < a + R for some R > 0, it
defines a function f
fሺxሻ = ෍ c୬ሺx − aሻ୬
ஶ
୬ୀ଴
(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′ሺxሻ = ෍ nc୬ሺx − aሻ୬ିଵ
ஶ
୬ୀ଴
(4.9b)

f′′ሺxሻ = ෍ nሺn − 1ሻc୬ሺx − aሻ୬ିଶ
ஶ
୬ୀ଴
(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
fሺxሻ =
1
1 − x
= 1 + x + xଶ
+ xଷ
+ ⋯ + x୬
+ ⋯ = ෍ x୬
ஶ
୬ୀ଴
for −1 < x < 1.
Answers:
fᇱሺxሻ =
1
ሺ1 − xሻଶ
= 1 + 2x + 3xଶ
+ 4xଷ
+ ⋯ + nx୬ିଵ
+ ⋯ = ෍ nx୬ିଵ
ஶ
୬ୀଵ
fᇱᇱሺxሻ =
2
ሺ1 − xሻଷ
= 2 + 6x + 12xଶ
+ ⋯ + nሺn − 1ሻx୬ିଶ
+ ⋯ = ෍ nሺn − 1ሻx୬ିଶ
ஶ
୬ୀଶ
all for −1 < x < 1.
Term-by-term integration. Suppose that
ஶ
୬ୀ଴
(4.9a)
converges for a − R < x < a + R (R > 0), Then
න fሺxሻdx = ෍ c୬
ሺx − aሻ୬ାଵ
n + 1
ஶ
୬ୀ଴
(4.10)
with
෍ c୬
ሺx − aሻ୬ାଵ
n + 1
ஶ
୬ୀ଴
converging also for a − R < x < a + R (R > 0).

Example 4.9
Investigate the following series
1.
fሺxሻ = x −
xଷ
5
+
xହ
5
− ⋯
for −1 ≤ x ≤ 1.
2.
fሺtሻ = 1 − t + tଶ
− tଷ
+ ⋯
for −1 < t < 1.
Answer:
1. Using term-by-term differentiation and integration of the resulting series, we find that
fሺxሻ = x −
xଷ
5
+
xହ
5
− ⋯ = tanିଵ
x
for −1 < x < 1.
2. Using term-by-term integration, we can generate the series for lnሺ1 + xሻ as
lnሺ1 + xሻ = x −
xଶ
2
+
xଷ
3
−
xସ
4
+ ⋯
for −1 < x < 1.
Multiplication of power series. If Aሺxሻ = ∑ a୬x୬ஶ
୬ୀ଴ and Bሺxሻ = ∑ b୬x୬ஶ
୬ୀ଴ converges absolutely for
|x| < R and
c୬ = a଴b୬ + aଵb୬ିଵ + aଶb୬ିଶ + ⋯ + a୬ିଵbଵ + a୬b଴ = ෍ a୩b୬ି୩
୬
୩ୀ଴
then the series ∑ c୬x୬ஶ
୬ୀ଴ converges absolutely to AሺxሻBሺxሻ for |x| < R:
൭෍ a୬x୬
ஶ
୬ୀ଴
൱ ൭෍ b୬x୬
ஶ
୬ୀ଴
൱ = ෍ c୬x୬
ஶ
୬ୀ଴
(4.11)

Drill Problems 4.2
A. For the following series, find the series’ radius and interval of convergence.
1. ∑ ሺx + 5ሻ୬ஶ
୬ୀ଴
2. ∑ ሺ−1ሻ୬ሺ4x + 1ሻ୬ஶ
୬ୀ଴
3. ∑
ሺଷ୶ିଶሻ౤
୬
ஶ
୬ୀଵ
4. ∑
୬୶౤
୬ାଶ
ஶ
୬ୀ଴
5. ∑
ሺିଵሻ౤ሺ୶ାଶሻ౤
୬
ஶ
୬ୀଵ
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. ∑
ሺ୶ିଵሻమ౤
ସ୬
ஶ
୬ୀଵ
7. ∑
ሺ୶ାଵሻమ౤
ଽ౤
ஶ
୬ୀ଴
8. ∑ ቀ
√୶
ଶ
− 1ቁ
୬
ஶ
୬ୀ଴
9. ∑ ሺlnxሻ୬ஶ
୬ୀ଴
10. ∑ ቀ
୶మାଵ
ଷ
ቁ
୬
ஶ
୬ୀ଴
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 fሺxሻ has derivatives of all orders on an interval I can be
expressed as a power series on that interval.
If we have
ஶ
୬ୀ଴
= c଴ + cଵሺx − aሻ + cଶሺx − aሻଶ
+ ⋯ + c୬ሺx − aሻ୬
+ ⋯
(4.9)
Term-by-term differentiation gives
fᇱሺxሻ = cଵ + 2cଶሺx − aሻ + 3cଶሺx − aሻଶ
+ ⋯ + nc୬ሺx − aሻ୬ିଵ
+ ⋯

fᇱᇱሺxሻ = 1 ∙ 2cଶ + 2 ∙ 3cଷሺx − aሻ + 3 ∙ 4cଶሺx − aሻଶ
+ ⋯ + n ∙ ሺn − 1ሻc୬ሺx − aሻ୬ିଶ
+ ⋯
fᇱᇱᇱሺxሻ = 1 ∙ 2 ∙ 3cଷ + 2 ∙ 3 ∙ 4cସሺx − aሻ + 3 ∙ 4 ∙ 5cହሺx − aሻଶ
+ ⋯ + n ∙ ሺn − 1ሻ ∙ ሺn − 2ሻc୬ሺx − aሻ୬ିଷ
+ ⋯
with the nth derivative of, for all n being
fሺ୬ሻሺxሻ = n! c୬ + a sum of terms with ሺx − aሻ as factor
Since these equations all hold at x = a, we have
fᇱሺaሻ = cଵ
fᇱᇱሺaሻ = 1 ∙ 2cଶ
fᇱᇱᇱሺaሻ = 1 ∙ 2 ∙ 3cଷ
⋮
fሺ୬ሻሺaሻ = n! c୬
With this, a general formula for the nth term of the coefficients of the power series can be deduced, that is
c୬ =
fሺ୬ሻሺaሻ
n!
(4.12)
If the function fሺxሻ has a series representation, then it must be
fሺxሻ = ෍
fሺ୩ሻ
ሺaሻ
k!
ሺx − aሻ୩
ஶ
୩ୀ଴
= fሺaሻ + fᇱሺaሻሺx − aሻ +
fᇱᇱሺaሻ
2!
ሺx − aሻଶ
+ ⋯ +
fሺ୬ሻሺaሻ
n!
ሺx − aሻ୬
+ ⋯ (4.13)
which is the Taylor series expansion of fሺxሻ about x = a. When a = 0, the series becomes
fሺxሻ = ෍
fሺ୩ሻ
ሺ0ሻ
k!
x୩
ஶ
୩ୀ଴
= fሺ0ሻ + fᇱሺ0ሻx +
fᇱᇱሺ0ሻ
2!
xଶ
+ ⋯ +
fሺ୬ሻሺ0ሻ
n!
x୬
+ ⋯ (4.14)
which is the Maclaurin series expansion of fሺxሻ.
The right hand side of equations 4.13 and 4.14 are called the Taylor (Maclaurin) polynomials of order n of
the function fሺxሻ.
P୬ሺxሻ = fሺaሻ + fᇱሺaሻሺx − aሻ +
fᇱᇱሺaሻ
2!
ሺx − aሻଶ
+ ⋯ +
fሺ୬ሻሺaሻ
n!
ሺx − aሻ୬ (4.15)

P୬ሺxሻ = fሺ0ሻ + fᇱሺ0ሻx +
fᇱᇱሺ0ሻ
2!
xଶ
+ ⋯ +
fሺ୬ሻሺ0ሻ
n!
x୬ (4.16)
These polynomials provided a linear approximation of the function fሺxሻ in the neighborhood of a (or 0)
Example 4.10
1. Find the Taylor series generated by fሺxሻ =
ଵ
୶
at a = 2. Where, if anywhere, does the series converge to
ଵ
୶
.
2. Find the Taylor series expansion, and the Taylor polynomials generated by fሺxሻ = e୶
at x = 0.
3. Find the Taylor series and Taylor polynomials generated by fሺxሻ = cos x at x = 0.
Answer:
1. fሺxሻ =
ଵ
ଶ
−
ሺ୶ିଶሻ
ଶమ +
ሺ୶ିଶሻమ
ଶయ − ⋯ + ሺ−1ሻ୬ ሺ୶ିଶሻ౤
ଶ౤శభ + ⋯. The series converges for 0 < x < 4.
2. The Taylor series expansion of fሺxሻ is fሺxሻ = ∑
୶ౡ
୩!
ஶ
୩ୀ଴ . The Taylor polynomials generated by the
function is
P୬ሺxሻ = 1 + x +
xଶ
2
+ ⋯
x୬
n!
Figure 3. Plot of approximation for exp(x)

3. The Taylor series generated by function f at x = 0 is ∑
ሺିଵሻౡ୶మౡ
ሺଶ୩ሻ!
ஶ
୩ୀ଴ . The 2nth order (as well as the
ሺ2n + 1ሻth order) polynomial of cos x is then
Pଶ୬ሺxሻ = Pଶ୬ାଵሺxሻ = 1 −
xଶ
2!
+
xସ
4!
− ⋯ + ሺ−1ሻ୬
xଶ୬
ሺ2nሻ!
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. fሺxሻ = eି୶
2. fሺxሻ = e୶/ଶ
3. fሺxሻ =
ଵ
ଵା୶
4. fሺxሻ =
ଵ
ଵି୶
5. sin 3x
6. cosh x
7. sinh x
8. xସ
− 2xଷ
− 5x + 4
9. ሺx + 1ሻଶ
10. sin
୶
ଶ

B. Find the Taylor series generated by the following functions at x = a
11. fሺxሻ = xଷ
− 2x + 4, a = 2
12. fሺxሻ = 2xଷ
+ 2xଶ
+ 3x − 8, a = 1
13. fሺxሻ = xସ
+ xଶ
+ 1, a = −2
14. fሺxሻ = 3xହ
− xସ
+ 2xଷ
+ xଶ
− 2, a = −1
15. fሺxሻ =
ଵ
୶మ, a = 1
16. fሺxሻ =
୶
ଵି୶
, a = 0
17. fሺxሻ = e୶
, a = 2
18. fሺxሻ = 2୶
, a = 1
C. Find the first and second order Taylor polynomial approximation of the following functions at x = 0.
19. fሺxሻ = lnሺcos xሻ
20. fሺxሻ = eୱ୧୬ ୶
4.4 Power Series Solution of Differential Equations: Solutions Near an Ordinary Point
For a linear differential equation
b଴ሺxሻyሺ୬ሻ
+ bଵሺxሻyሺ୬ିଵሻ
+ ⋯ + b୬ሺxሻy = Rሺxሻ (4.17)
the point x = x଴ is called an ordinary point of 4.17 if b଴ሺx଴ሻ ≠ 0. A singular point of the linear equation is
any point x = xଵ for which b଴ሺxଵሻ = 0. Any point that is not a singular point is an ordinary point.
The differential equation
ሺ1 − xଶሻyᇱᇱ
− 6xyᇱ
− 4y = 0
has x = 1 and x = −1 as its only singular points in the finite complex plane. The equation
yᇱᇱ
+ 2xyᇱ
+ y = 0
has no singular points in the finite plane. The equation
xyᇱᇱ
+ yᇱ
+ 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 = ෍ c୬x୬
ஶ
୬ୀ଴
(4.18)
and substitute this to the given differential equation. The task then is to find the general term of the series
c୬. The method described is demonstrated in the following examples.
Example 4.11
1. Solve the equation yᇱᇱ
+ 4y = 0 near the ordinary point x = 0.
2. Solve the equation ሺ1 − xଶሻyᇱᇱ
− 6xyᇱ
− 4y = 0 near the ordinary point x = 0.
Answers:
1. y = c଴ cos 2x +
ଵ
ଶ
cଵ sin 2x.
2. y =
ୡబ
ሺଵି୶మሻమ +
ୡభ൫ଷ୶ି୶య൯
ଷሺଵି୶మሻమ
Drill Problems 4.4
Find the general solution of the following differential equations near the origin.
1. yᇱᇱ
+ 3xyᇱ
+ 3y = 0
2. ሺ1 − 4xଶሻyᇱᇱ
+ 8y = 0
3. ሺ1 + xଶሻyᇱᇱ
+ 10xyᇱ
+ 20y = 0
4. ሺxଶ
− 9ሻyᇱᇱ
+ 3xyᇱ
− 3y = 0
5. ሺxଶ
+ 4ሻyᇱᇱ
+ 6xyᇱ
+ 4y = 0
6. yᇱᇱ
+ xଶ
y = 0
7. ሺ1 + 2xଶሻyᇱᇱ
+ 3xyᇱ
− 3y = 0
8. yᇱᇱ
+ xyᇱ
+ 3y = xଶ
9. yᇱᇱ
+ 3xyᇱ
+ 7y = 0

10. ሺxଶ
+ 4ሻyᇱᇱ
+ xyᇱ
− 9y = 0

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