Download Infinite Sequences and Series

Chapter 3
Infinite Sequences and Series
3.1
Sequences
Definition 3.1.1 An infinite sequence of numbers is a function from the
set Z of integers into a set R.
The set R can be any set, but in our course it is usually the set R of
real numbers. Thus a sequence is usually denoted by writing down all the
numbers in the range with numbers in the domain as the indices:
{a1 , . . . , an , . . .} = {an ∈ R | n ∈ Z+ } = {an }.
For example, an = (−1)n+1 n1 means
½
¾
1 1
1
n+1 1
1, − , , − , . . . , (−1)
, ... .
2 3
4
n
Definition 3.1.2 A sequence {an } converges to a number L if for every
positive number ε there is an integer N such that for all n > N ,
|an − L| < ε.
In this case, we write
lim an = L, or an → L,
n→∞
and call L the limit of the sequence.
If no such a number L exists, we say {an } diverges.
67
68
Chapter 3.
Infinite Sequences and Series
Definition 3.1.3 A sequence {an } diverges to infinity if for every number M there is an integer N such that for all n > N ,
an > M.
In this case, we write
lim an = ∞, or an → ∞.
n→∞
Similarly, if for every number m there is an integer N such that for all
n > N,
an < m,
then we say {an } diverges to negative infinity and write
lim an = −∞, or an → −∞.
n→∞
Theorem 3.1.1 Let {an } and {bn } be convergent sequences to A and B,
respectively.
(1)
(2)
(3)
(4)
limn→∞ (an ± bn ) = A ± B.
limn→∞ (an · bn ) = A · B.
limn→∞ (k · bn ) = k · B.
A
limn→∞ abnn = B
, provided B 6= 0.
Example 3.1.1 (1) Show limn→∞ n1 = 0.
For a given any ε > 0, take N such that N > 1ε . Then for n > N ,
|
(2) limn→∞
n−1
n
1
1
− 0| <
< ε.
n
N
= limn→∞ (1 − n1 ) = 1 − 0 = 1.
¤
Theorem 3.1.2 (Sandwich Theorem) Let {an }, {bn } and {cn } be sequences of real numbers. If an ≤ bn ≤ cn for all n except for a finite
number, and if limn→∞ an = limn→∞ cn = L, then limn→∞ bn = L.
Example 3.1.2 (1) limn→∞ cosn n = 0, since − n1 ≤
(2) limn→∞ 21n = 0, since 0 ≤ 21n ≤ n1 .
cos n
n
≤ n1 .
¤
Theorem 3.1.3 Let {an } be a sequence of real numbers. If an → L and if
f is a function that is continuous at L and defined at all an , then f (an ) →
f (L).
3.1.
Sequences
69
p
Example 3.1.3 (1) (n + 1)/n → 1, since
p
√
and L = 1, (n + 1)/n → 1 = 1.
(2) limn→∞ 21/n = 20 = 1: since limn→∞
and L = 1.
n+1
n
1
n
→ 1 by taking f (x) =
√
x
= 0, take an = n1 , f (x) = 2x
¤
Theorem 3.1.4 Let {an } be a sequence of real numbers and f (x) be a function defined for all x ≥ n0 such that f (n) = an for n ≥ n0 . Then
lim f (x) = L ⇒ lim an = L.
x→∞
n→∞
Proof: If limx→∞ f (x) = L, then for any ε > 0 there is M such that for
any x > M |f (x) − L| < ε. Choose an integer N > max{M, n0 }, so that
∀ n > N |an − L| = |f (n) − L| < ε.
¤
1
Example 3.1.4 (1) limx→∞ lnnn = limx→∞ n1 = 0.
n
2n
2
(2) limn→∞ 5n
= limn→∞ 2 ·ln
= ∞.
5
³
´n
³
´
n+1
(3) For an = n−1 , take the natural logarithm: ln an = n ln n+1
n−1
and then
µ
¶
n+1
lim ln an = lim n ln
n→∞
n→∞
n−1
³
´
ln n+1
n−1
= lim
n→∞
1/n
=
lim
−2
n2 −1
n→∞ −1/n2
Thus an = eln an → e2 .
Theorem 3.1.5 (1) limn→∞
√
(2) limn→∞ n n = 1.
= lim
2n2
= 2.
−1
n→∞ n2
¤
ln n
n
= 0.
(3) limn→∞ x1/n = 1, for x > 0.
(4) limn→∞ xn = 0, for |x| < 1.
(5) limn→∞ (1 + nx )n = ex , for any x.
(6)) limn→∞
xn
n!
= 0, for any x.
Sometimes, a sequence may be given by a recursive formula: an is determined by the values of preceding terms with values of initial terms.
70
Chapter 3.
Infinite Sequences and Series
Example 3.1.5 (1) an = an−1 + 1 with an initial value a1 = 1 defines the
natural numbers an = n.
(2) an = ran−1 defines a geometric sequence an = rn with an initial
value a0 = 1.
(3) an = nan−1 with a1 = 1 defines the factorial an = n!.
(4) an+1 = an + an−1 with a1 = 1 and a2 = 1 defines the Fibonacci
numbers.
¤
Definition 3.1.4 A sequence {an } with the property that an ≤ an+1 for
all n is called a nondecreasing sequence.
Definition 3.1.5 A sequence {an } is bounded from above if there is a
number M such that an ≤ M for all n. The number M is called an upper
bound for {an }. If no number less than M is an upper bound for {an },
then M is called the least upper bound for {an }.
The completeness of the real numbers means that any set bounded above
has a least upper bound. Thus, a nondecreasing sequence bounded above
has a least upper bound L. In fact, (i) an ≤ L for all n, (ii) for any ε > 0
L − ε can not be an upper bound of the sequence since L is the least upper
bound. Thus, ∃ n0 such that an0 > L − ε. Since it is nondecreasing, for all
n > n0 ,
L ≥ an ≥ an0 > L − ε.
This means L is the limit of the sequence: i.e, an → L.
Theorem 3.1.6 A nondecreasing sequence of real numbers converges if and
only if it is bounded from above. If a nondecreasing sequence converges, it
converges to its least upper bound.
+ · · · + n1 − ln n, show that the limit
µ
¶
1 1
1
C = lim Cn = lim
+ + · · · + − ln n
n→∞
n→∞ 1
2
n
Example 3.1.6 For Cn =
1
1
+
1
2
exists.
Solution: Let Dn = Cn − n1 . Then Dn < Cn , and from the inequality:
1
1
1
< ln(1 + ) <
n+1
n
n
3.2. INFINITE SERIES
71
obtained in Example 2.3.1, we get
1
1
− ln(1 + ) < 0,
n+1
n
1
1
1
1
+ = − ln(1 + ) > 0.
= Cn+1 − Cn −
n+1 n
n
n
Cn+1 − Cn =
Dn+1 − Dn
That is, Cn is decreasing, while Dn is increasing. Hence, Cn converges, since
0 = D1 < Dn < Cn . The limit C is called the Euler’s constant. For an
application of this number, see the last part of Section 11.1.2.
¤
Example 3.1.7 (Logarithmic Limits) Find the limit of
¶µ
¶ µ
¶
µ
1
2
p
an = 1 +
1+
··· 1 +
,
np
np
np
where p is a positive integer.
Solution: Note that 1 ≤ an . By taking the logarithm of an , we get
¶
µ
¶
µ
¶
µ
2
p
1
+ ln 1 +
+ · · · + ln 1 +
0 ≤ ln an = ln 1 +
np
np
np
" µ
¶np
¶ np
¶ np #
µ
µ
1
1
2 2
p p
=
ln 1 +
+ 2 ln 1 +
+ · · · + p ln 1 +
np
np
np
np
≤
=
1
(ln e + 2 ln e + · · · + p ln e)
np
1
p+1
(1 + 2 + · · · + p) =
→ 0, as n → ∞,
np
2n
where the third inequality follows from Example 2.4.1. Therefore,
lim an = lim eln an = elimn→∞ ln an = e0 = 1.
n→∞
3.2
n→∞
Infinite Series
An infinite series is the sum of an infinite sequence of numbers:
a1 + a2 + · · · + an + · · ·
¤
72
Chapter 3.
Infinite Sequences and Series
How can we find this sum of infinite numbers in a finite life time? For
this, we look at the sequence of sums of finite number of terms, called the
sequence of partial sums:
S1 = a1 , S2 = a1 + a2 , S3 = a1 + a2 + a3 , . . . , Sn =
n
X
ak , . . . .
k=1
Definition 3.2.1 If the sequence {Sn } converges to a number L, we say
that the series converges to the sum L, written as
a1 + a2 + · · · + an + · · · =
∞
X
an = L.
n=1
If the sequence of partial sums does not converge, we say that the series
diverges.
Example 3.2.1 [Geometric Series] A geometric series is of the form:
a + ar + ar2 + · · · + arn−1 + · · · =
∞
X
arn−1
n=0
in which a 6= 0 and r are fixed real numbers. As it is well known, the n-th
partial sum is
a(1 − rn )
, r 6= 1.
1−r
Note that, if r = 1, Sn = na, and if r = −1, then Sn is alternating between
0 and 1. Thus the series diverges if |r| = 1. Since rn ≤ |rn | →
½
0
if |r| < 1,
as n → ∞,
∞ if |r| > 1
½ a
∞
X
if |r| < 1,
n−1
1−r
ar
=
¤
∞ (diverges)
if |r| ≥ 1.
Sn =
n=1
Example 3.2.2 [Telescoping Series] (1) For a series of the form
∞
X
an =
P∞
n=1
1
n=1 n(n+1)
Sn
an =
∞
X
n=1
1
,
n(n + 1)
1
− n+1
and so
µ
¶ µ
¶ µ
¶
µ
¶
1 1
1 1
1 1
1
1
=
−
+
−
+
−
+ ···
−
1 2
2 3
3 4
n n+1
1
= 1−
→ 1, as n → ∞.
n+1
=
1
n
¤
3.2.
Infinite series
73
Consider a convergent series
P∞
n=1 an
= S. Then
an = Sn − Sn−1 = (S − Sn ) − (S − Sn−1 ) → 0
as n → ∞. This proves that ”If
P∞
n=1 an
converges, then an → 0.”
Theorem 3.2.1 (n-th
P term test for divergence) If limn→∞ an 6= 0 or
fails to exist, then ∞
n=1 an diverges.
P
n+1
n+1
Example 3.2.3 (1) ∞
n=1 n diverges since n → 1 6= 0.
P∞
(2) n=1 (−1)n+1 diverges since limn→∞ (−1)n+1 does not exist.
(3) The Harmonic series
∞
X
1
1 1 1
1
= 1 + + + + · + + ···
n
2 3 4
n
n=1
diverges even if an → 0. In fact,
1 1
1
1
1
1
1
+ ( + ) + ( + ··· + ) + ( + ··· + ) + ···
2
3 4
5
8
9
16
1
1
1
1
> 1 + + ( ) + ( ) + ( ) + ···
2
2
2
2
→ ∞.
1+
Theorem 3.2.2 Suppose
series. Then
(1)
(2)
P∞
n=1 an
= A and
P∞
n=1 bn
¤
= B are convergent
P∞
n=1 (an
± bn ) = A + B.
P∞
n=1 kan = k
n=1 an = kA.
P∞
P
If ∞
n=1 an diverges, (1) and (2) in Theorem 3.2.2 also diverges. The
sum of two divergent series can be convergent: For example, if an = 1 and
bn = −1 for all n, then an + bn = 0 for all n.
Adding or deleting a finite number of terms does not alter the convergence or the divergence of the series.
74
Chapter 3.
3.3
Infinite Sequences and Series
Tests for Convergence of Series
P
We first assume that the terms an of a series ∞
n=1 an are all positive. Then
Sn+1 = Sn +an means the sequence of the partial sums form a nondecreasing
sequence. Thus for the convergence of those series all we need to find out is
that the partial sums are bounded from above.
P
Corollary 3.3.1 A series ∞
n=1 an of nonnegative terms converges if and
only if its partial sums are bounded from above.
3.3.1
The integral test
Example 3.3.1 For the series:
∞
X
1
1
1
1
1
= 1 + 2 + 2 + 2 + · + 2 + ··· ,
n2
2
3
4
n
n=1
R∞
R∞
consider the integral 1 f (x)dx = 1 x12 dx. The series may be considered
as the area of the column rectangles with base 1 and altitude n12 , while the
integral is the area under the graph of f from 0 to ∞ as the following figure
shows:
y
6
f (x) =
1
x2
1
12
1
22
1
n2
µ
0
1 2 3
n−1 n
- x
Then, for any n,
1
1
1
1
+ 2 + 2 +·+ 2
2
4
· n¸n
Z2 n 3
1
1
1
≤ 1+
dx = 1 + −
= 2 − < 2.
2
x 1
n
1 x
Sn = 1 +
3.3.
Tests for convergence
75
Thus that series converges.
¤
P
Theorem 3.3.2 (The integral test) Let ∞
n=1 an be a series of positive
terms. Suppose that an = f (n) for P
some continuous positive Rdecreasing
∞
function f of x > N . Then the series ∞
n=1 an and the integral N f (x)dx
both converge or both diverge.
Proof: Since f is decreasing and f (n) = an , from the figure below,
y
y = f (x)
6
a1
a2
a1
a2
9
an
an
1 2
-
x
3 4 n−1 n n+1
we obtain the following relations:
Z n+1
Z
f (x)dx ≤ a1 + a2 + a3 + · · · + an ≤ a1 +
1
n
f (x)dx.
1
These inequalities hold for all n. This shows the series and the integral both
finite or infinite.
¤
Example 3.3.2 [The p-Series test] For a real constant p, the series
∞
X
1
1
1
1
1
= 1 + p + p + p + · + p + ···
np
2
3
4
n
n=1
is called a p-series. Let f (x) = x1p . If p > 1, f (x) > 0 is decreasing function
of x, and
Z ∞
Z ∞
x−p+1 b
1
−p
dx
=
x
dx
=
lim
|1
b→∞ −p + 1
xp
1
1
1
1
=
lim ( p−1 − 1)
1 − p b→∞ b
1
1
=
(0 − 1) =
.
1−p
p−1
76
Chapter 3.
Infinite Sequences and Series
Thus the series converges.
If p < 1, then 1 − p > 0 and
Z ∞
1
1
1
dx =
lim ( p−1 − 1) = ∞.
p
b→∞
x
1
−
p
b
1
Thus the series diverges by the integral test.
If p = 1, we have the divergent harmonic series. Hence, the p-series
converges for p > 1, and diverges for every other value of p.
¤
3.3.2
The comparison test
P
Theorem 3.3.3 (The Comparison Test) Let ∞
n=1 an be a series of positive terms.
P∞
P∞
(1)
n=1 cn with an ≤ cn
n=1 an converges if there is a convergent series
for n > N .
P∞
P∞
(2)
n=1 dn of nonnegative
n=1 an diverges if there is a divergent series
terms with an ≥ dn for n > N .
Proof: (1) The partial sums of
P∞
n=1 an
are bounded from above by
M = a1 + · · · + aN +
∞
X
cn ,
n=N +1
so that they form a nondecreasing
sequence with a limit L ≤ M .
P
a
(2) The partial sums ofP ∞
n=1 n can not be bounded from above, oth∞
erwise the divergent series n=1 dn will be bounded from above, which will
be convergent by (1).
¤
P
5
5
Example 3.3.3 (1) ∞
n=1 5n−1 diverges, since 5n−1 =
P∞ 1
1
≤ 21n for n > 1.
(2) n=0 n! converges since n!
1
n− 15
> n1 .
¤
Theorem 3.3.4 (Limit Comparison Test) Suppose that an > 0 and bn >
0 for all n > N .
P
P∞
(1) If limn→∞ abnn = c > 0, then ∞
n=1 an and
n=1 bn both converge or
both diverge.
3.3.
Tests for convergence
(2) If limn→∞
(3) If limn→∞
Proof: (1) For
an
bn
an
bn
c
2
77
P
P∞
= 0 and ∞
n=1 bn converges, then
n=1 an converges.
P∞
P∞
= ∞ and n=1 bn diverges, then n=1 an diverges.
> 0, there exists N such that ∀ n > N ,
¯
¯
¯ an
¯
¯ − c¯ < c .
¯ bn
¯ 2
Thus
c
<
2
an
bn
3c
,
2
3c
< bn ,
2
<
c
bn < an
2
and so Theorem 3.3.3(1) follows.
(2) For ε > 0, there exists N such that ∀ n > N ,
an
bn
< ε. Thus
an < εbn ,
and so Theorem 3.3.3(2) follows.
(3) For M > 0, there exists N such that ∀ n > N ,
an
bn
> M. Thus
M bn < an ,
and so Theorem 3.3.3(3) follows.
¤
P∞
Example 3.3.4 (1) For an = n22n+1
,
n=1 an diverges: Compare it with
+2n+1
bn = n1 . Then
2n2 + n
an
= lim 2
= 2.
lim
n→∞ n + 2n + 1
n→∞ bn
ln n P∞
(2) For an = 1+n
, n=1 an diverges: Since an ≈ nnln2 n = lnnn > n1 , we
n2 +5
can take bn = n1 . Then
an
n + n2 ln n
= lim
= ∞.
n→∞
n→∞ bn
n2 + 5
lim
(3) If an =
and then
ln n
,
n3/2
then
ln n
n3/2
<
n1/4
n3/2
=
1
n5/4
for large n. Thus take bn =
1
,
n5/4
1
an
ln n
4
n
= lim 1/4 = lim
= lim 1/4 = 0
n→∞ bn
n→∞ n
n→∞ (1/4)n−3/4
n→∞ n
lim
shows
P∞
n=1 an
converges.
¤
78
Chapter 3.
3.3.3
Infinite Sequences and Series
The ratio and root tests
P∞
Theorem 3.3.5 (The Ratio Test) Let
n=1 an be a series of positive
terms. Suppose that
an+1
lim
= ρ.
n→∞ an
(1) the series converges if ρ < 1,
(2) the series diverges if ρ > 1,
(3) the test fails if ρ = 1.
Proof: (1) Choose an r such that ρ < r < 1, so that ε = r − ρ > 0.
an+1
an → ρ means ∃ N such that ∀ n ≥ N
an+1
< ρ + ε = r.
an
Thus aN +1 < raN ,
aN +2 < raN +1 < r2 aN ,
..
.
aN +m < raN +m−1 < rm aN .
Therefore,
∞
X
an =
n=1
<
N
−1
X
n=1
N
−1
X
n=1
(2)
an+1
an
an +
∞
X
an
n=N
an + aN
∞
X
n=0
n
r =
N
−1
X
an + aN
n=1
1
.
1−r
→ ρ > 1 means ∃ N such that ∀ n ≥ N
an+1
> 1, or aN < aN +1 < aN +2 < · · · 9 0.
an
n
and bn = n12 . In both cases, an+1
an = n+1 → 1, and
P
P
∞
n 2
= ( n+1
) → 1. But ∞
¤
n=1 an diverges, while
n=1 bn converges.
(3) Consider an =
bn+1
bn
1
n
3.3.
Tests for convergence
Example 3.3.5 (1)
P∞
n=1
79
2n +5
3n
an+1
1
=
an
3
(2)
P∞
(2n)!
n=1 n!n!
converges, since
Ã
2+
1+
5
2n
5
2n
!
→
2
< 1.
3
diverges, since
n!n!(2n + 2)(2n + 1)(2n)!
(2n + 2)(2n + 1)
4n + 2
an+1
=
=
=
→ 4 > 1.
an
(n + 1)!(n + 1)!(2n)!
(n + 1)(n + 1)
n+1
(3) For
P∞
4n n!n!
n=1 (2n)! ,
an+1
4(n + 1)2
2(n + 1)
=
=
→ 1.
an
(2n + 2)(2n + 1)
2n + 1
an+1
an
Thus the test fails. However,
shows the series diverges.
P∞
=
2n+2
2n+1
> 1 means a1 = 2 ≤ an 9 0
¤
½
n
2n ,
1
2n ,
for n odd
is not
for n even
a geometric series. Since lim an = 0, the test for divergence fails. The
integral test does not look promising, nor does the ratio test, since an+1
an =
½ 1
,
for
n
odd
2n
, which is alternating.
¤
n+1
for n even
2 ,
Example 3.3.6 A series
n=1 an
with an =
P
Theorem 3.3.6 (The Root Test) Let ∞
n=1 an be a series of positive terms.
Suppose that
√
lim n an = ρ.
n→∞
(1) the series converges if ρ < 1,
(2) the series diverges if ρ > 1,
(3) the test fails if ρ = 1.
Proof: (1) Choose an r such that ρ < r < 1, so that ε = r − ρ > 0.
√
n a → ρ means ∃ N such that ∀ n ≥ N
n
√
n
an < ρ + ε = r, or an < rn .
80
Chapter 3.
Infinite Sequences and Series
Then,
∞
X
N
−1
X
an =
n=1
n=1
N
−1
X
<
an +
∞
X
an
n=N
an + r N
n=1
∞
X
n=0
rn =
N
−1
X
an +
n=1
rN
.
1−r
Thus, by the comparison test, the series converges.
√
(2) n an → ρ > 1 means ∃ N such that ∀ n ≥ N
√
n
an > 1, or 1 < an 9 0.
√
√
1
1
n
n a
(3)
Consider
a
=
and
b
=
=
n → 1, and
n
n
n
2 . In both cases,
n
n
√
P
P
√
∞
∞
n
2
n
bn = ( n) → 1. But n=1 an diverges, while n=1 an converges.
¤
√
Example 3.3.7 (1) For the Example 3.3.6, n an =
1
2
½ √
n
n/2,
1
,
2
shows the series converges.
q
P
1 n
1 n
1
n
(
(2) ∞
)
converges,
since
( 1+n
) = 1+n
→ 0 < 1.
n=0 1+n
q
P
2n
2
n 2n
(3) ∞
= (√
n n)2 → 2 > 1.
n=1 n2 diverges, since
n2
3.3.4
for n odd
→
for n even
¤
The alternating series test
A series in which the terms are alternatively positive and negative is called
an alternating series.
Theorem 3.3.7 (The Alternating Series Test) An alternating series
∞
X
(−1)n+1 an = a1 − a2 + a3 − · · ·
n=1
converges if
(1) an > 0 for all n,
(2) an ≥ an+1 for all n ≥ N ,
(3) an → 0.
3.3.
Tests for convergence
81
Proof: If n = 2k is an even number, then
S2k = (a1 − a2 ) + (a3 − a4 ) + · · · + (an−1 − an )
= a1 − (a2 − a3 ) − (a4 − a5 ) − · · · − (an−2 − an−1 ) − an .
Since ak − ak+1 ≥ 0, the first equality shows S2k+2 ≥ S2k , which means S2k
is nondecreasing. The second equality shows S2k ≤ a1 , which means S2k is
bounded from above. Thus it has a limit: limk→∞ S2k = L.
If n = 2k + 1 is an odd number, then
S2k+1 = S2k + a2k+1 → L + 0 = L,
as k → ∞, since limk→∞ a2k+1 = 0. Therefore, limn→∞ Sn = L.
¤
Example 3.3.8 The alternating harmonic series
∞
X
(−1)n+1
n=1
1
1 1 1
= 1 − + − + ···
n
2 3 4
converges.
¤
Theorem 3.3.8 (The Alternating Series Estimation) If an alternating series
∞
X
(−1)n+1 an = a1 − a2 + a3 − · · ·
n=1
satisfies the three conditions in Theorem 3.3.7, then for n ≥ N
|L − Sn | < an+1 ,
where L is the limit of the sum.
P
1
2
n 1
Example 3.3.9 The series ∞
n=0 (−1) 2n converges to 1−(−1/2) = 3 . If we
P∞
1
1
take the sum upto 8-th term, the error is n=8 (−1)n n12 = 256
− 512
+ ···.
Thus
1
¤
|L − S7 | <
.
256
P
P∞
Definition 3.3.1 (1) An series ∞
n=1 an converges absolutely if
n=1 |an |
converges.
(2) A series that converges but does not converge absolutely converges
conditionally.
82
Chapter 3.
Infinite Sequences and Series
P∞
n+1 1 converges absolutely since
Example
3.3.10
(1)
The
series
n=1 (−1)
n2
P∞
P
∞
1
n 1
n=1 |(−1) n2 | =
n=1 n2 converges. P
n+1 1 converges condition(2) The alternating harmonic series ∞
n=1 (−1)
n
ally.
¤
Theorem 3.3.9
P (The Absolutely Convergence Test) If
verges, then ∞
n=1 an converges.
P∞
n=1 |an |
con-
Proof: Note that for any n,
−|an | ≤ an ≤ |an |, or 0 ≤ an + |an | ≤ 2|an |.
P
P
Thus if |an | converges, so does (an + |an |). Since an = (an + |an |) − |an |,
P
∞
n=1 an is the difference of two convergent series:
∞
X
an =
n=1
∞
X
(an + |an |) −
n=1
∞
X
|an |.
n=1
Thus the series converges.
¤
Example 3.3.11 (1) The series
P∞
sin n
n=1 n2
converges absolutely, since
∞
∞
X
X
sin n
1
| 2 |≤
.
n
n2
n=1
n=1
Thus it also converges.
P
(−1)n−1
(2) For p > 0, the alternating p-series ∞
converges, since for
n=1
np
p > 1 the series converges absolutely, and for 0 < p ≤ 1 the series converges
¤
conditionally by the alternating series test.
P
Note that the convergence of a series ∞
n=1 an does not usually imply
the convergence of a series with its term rearranged.
P
Theorem 3.3.10 (Rearranged Series) If ∞
n=1 an converges absolutely,
then any rearrangement of the terms gives rise absolutely convergent series.
3.4. POWER SERIES
83
P
n+1 1 converges
Example 3.3.12 The alternating harmonic series ∞
n=1 (−1)
n
conditionally. This series can be rearranged to diverge or to reach any preassigned sum:
X
1
(−1)n+1
n
=
X
=
3.4
X 1
1
−
→ ∞ − ∞.
2n − 1
2n
1 1 1 1 1 1 1 1
1
1
− + + − + + − +
+
−
1 2 3 5 4 7 9 6 11 13
1
1
1
1
1
1
1
1
1
+
+
−
+
+
−
+
+
−
8 15 17 10 19 21 12 23 25
1
1
1
+
−
+ ··· → 1
¤
14 27 16
Power Series
In the previous sections, we have studied the convergence of series with
terms of fixed numbers. Now the terms in a series can be some functions in
a variable, say x, varying in some interval.
Definition 3.4.1 A power series about x = 0 is a series of the form
∞
X
an xn = a0 + a1 x + a2 x2 + · · · + an xn + · · · .
n=0
A power series about x = a is a series of the form
∞
X
an (x − a)n = a0 + a1 (x − a) + a2 (x − a)2 + · · · + an (x − a)n + · · · .
n=0
We are concerned about the values of x for which the series converges.
Example 3.4.1 [Geometric Series] A series
∞
X
xn = 1 + x + x2 + · · · + xn + · · ·
n=0
1
for those x with |x| < 1.
is called a geometric series. This converges to 1−x
The partial sums of the original series are the polynomials that approximate
the limit.
84
Chapter 3.
Infinite Sequences and Series
The power series
1
1
1
1 − (x − 2) + (x − 2)2 + · · · + (− )n (x − 2)n + · · ·
2
4
2
is also a geometric series with the ratio − x−2
2 , which converges for x with
x−2
1
2
| − 2 | < 1 or 0 < x < 4 to 1+ x−2 = x . Thus,
2
2
1
1
1
= 1 − (x − 2) + (x − 2)2 + · · · + (− )n (x − 2)n + · · · , 0 < x < 4.
x
2
4
2
This series can be approximated by polynomials for values of x near 2:
P0 (x) = 1,
1
x
P1 (x) = 1 − (x − 2) = 2 − ,
2
2
1
3x 3x2
1
+
.
P2 (x) = 1 − (x − 2) + (x − 2)2 = 3 −
2
4
2
4
¤
Example 3.4.2
Find all the values of x for which the series converges:
P
n−1 xn = x − x2 + · · · + (−1)n−1 xn + · · · , since | an+1 | =
(−1)
(1) For ∞
n=1
n
2
n
an
n
n+1 |x| → |x|, by the ratio test, the series converges absolutely for |x| < 1
and diverges for |x| > 1. For x = 1, the series is the alternating harmonic
series which converges. For x = −1, the series is negative of the harmonic
series which diverges. Thus the interval in which the series converges is
(−1, 1].
P
n−1 x2n−1 = x − x3 + x5 + · · · + (−1)n−1 x2n−1 + · · · ,
(2) For ∞
n=1 (−1)
2n−1
3
5
2n−1
2n−1 2
2
since | an+1
an | = 2n+1 x → x , by the ratio test, the series converges absolutely
for x2 < 1 and diverges for x2 > 1. For x = 1, the series is an convergent
alternating series. For x = −1, the series is again negative of an alternating
series which converges. Thus the interval in which the series converges is
[−1, 1].
P
|x|
an+1
x2
xn
xn
(3) For ∞
n=0 n! = 1 + x + 2! + · · · + n! + · · · , since | an | = n+1 → 0, by
the ratio test, the series converges absolutely for all x ∈ R Thus the interval
in which thePseries converges is R.
an+1
∞
n
2
n
(4) For
n=0 n!x = 1 + x + x 2! + · · · + x n! + · · · , since | an | =
(n + 1)|x| → ∞, by the ratio test, the series diverges for all x ∈ R except
x = 0.
¤
Theorem 3.4.1 If a power series
∞
X
n=0
an xn = a0 + a1 x + a2 x2 + · · · + an xn + · · ·
3.4.
Power Series
85
converges for x = c 6= 0, then it converges absolutely for all x with |x| < |c|.
If it diverges for x = d, then it diverges for all x with |x| > |d|.
P
n
n
Proof: Suppose that ∞
n=0 an c converges. Then limn→∞ an c = 0. Thus
1
n
∃ N such that, for all n ≥ N , |an c | < 1 or |an | < |c|n . Now for any x with
|x| < |c|,
∞
X
n
|an x | =
n=0
≤
=
N
−1
X
n=0
N
−1
X
n=0
N
−1
X
n=0
n
|an x | +
|an xn | +
∞
X
n=N
∞
X
n=N
|an xn |
x
| |n
c
1
x
.
|an xn | + | |N
c 1 − | xc |
If it diverges for x = d and converges at c with |c| > |d|, then, by the
first part, the series converges for all x with |x| < |c|, especially at x = d,
which is a contradiction.
¤
P
n
0
For a series of P
the form ∞
n=0 an (x − a) , replace x − a by x and apply
∞
0
n
Theorem 3.4.1 to n=0 an (x ) .
Corollary 3.4.2 A power series
following three possible ways:
P∞
n=0 an (x
− a)n converges in one of the
(1) There is a positive number R such that the series diverges for x with
|x − a| > R and converges absolutely for x with |x − a| < R. At either
of the end points x = a − R and x = a + R, the series may or may not
converge (0 < R < ∞).
(2) The series converges absolutely for every x ∈ R (R = ∞).
(3) The series converges at x = a and diverges elsewhere (R = 0).
Proof: Cases (2) and (3) are trivial. Assuming a = 0, supposePthat it is
n
not in the case of (2) or (3). Thus there
a d 6= 0 such that ∞
n=0 an d
Pis
n
diverges, also there is a p 6= 0 such that ∞
n=0 an p converges. Let
S = {x ∈ R |
∞
X
n=0
an xn converges.}.
86
Chapter 3.
Infinite Sequences and Series
Then ∀ x ∈ S |x| ≤ d so that S is bounded from above. Thus
P∞ it has
a least upper bound R. If |x| > R > p, then x 6∈ S and so n=0 an xn
diverges. If |x| < R, |x| is
bound forPS, so that there is b ∈ S
Pnot an upper
∞
n converges, so
n
such that |x| < b. Since ∞
a
b
n=0 n
n=0 an |x| converges by
Theorem 3.4.1.
If a 6= 0, set x0 = x − a.
¤
R is called the radius of convergence of the series, and the interval
(a − R, a + R) is called the interval of convergence.
3.4.1
Term-by-term differentiation and integration
P
n
Theorem 3.4.3 If a power series ∞
n=0 an (x−a) converges on an interval
of convergence:
I = (a − R, a + R) for some R > 0, it defines a function
P
a
(x
− a)n on I, which is differentiable infinitely many times
f (x) = ∞
n=0 n
on the interval I and
f 0 (x) =
f 00 (x) =
∞
X
n=1
∞
X
nan (x − a)n−1
n(n − 1)an (x − a)n−2 ,
n=2
and so on. Each of these derived series converges on I.
P
n
2
Example 3.4.3 For f (x) = ∞
n=0 x = 1 + x + x + · · · =
0
f (x) =
f 00 (x) =
∞
X
n=1
∞
X
nxn−1 = 1 + 2x + 3x2 + 4x3 + · · · =
on (−1, 1),
1
(1 − x)2
n(n − 1)xn−2 = 2! + 6x + 12x2 + · · · =
n=2
1
1−x
2
.
(1 − x)3
¤
P
sin(n!x)
Remark: For the series f (x) = ∞
, which converges on R, its
n=0
n2
term by term derivative
∞
X
n! cos(n!x)
n2
n=1
diverges for all x: This is not a power series since it is not a sum of positive
integer powers of x.
3.4.
Power Series
87
P
n
Theorem 3.4.4 Suppose that a power series f (x) = ∞
n=0 an (x − a) converges on an interval of convergence: I = (a − R, a + R) for some R > 0.
Then its term by term integral
Z
∞
X
(x − a)n+1
f (x)dx =
+C
an
n+1
n=1
converges on I.
Example 3.4.4 For f (x) =
f 0 (x) =
∞
X
n=0 (−1)
n x2n+1
2n+1
3
5
= x− x3 + x5 +· · · on [−1, 1],
(−1)n−1 (x2 )n−1 = 1 − x2 + x4 + x6 + · · · =
n=1
Z
f (x) =
P∞
f 0 (x)dx =
Z
1
, −1 < x < 1.
1 + x2
1
dx = tan−1 x + C.
1 + x2
Since f (0) = 0, C = 0. Thus
f (x) = x −
x3 x5
+
+ · · · = tan−1 x, −1 < x < 1.
3
5
The series, in fact, also converges to tan−1 x at x = ±1.
P
n n
2
3
Example 3.4.5 For f (t) = ∞
n=0 (−1) t = 1 − t + t − t + · · · =
convergent on (−1, 1),
¯x
Z x
¯
t2 t3 t4
1
dx = t − + − + · · · ¯¯
ln(1 + x) =
2
3
4
0 1+t
0
= x−
1
1+t ,
x2 x3 x4
+
−
+ · · · , −1 < x < 1.
2
3
4
The series, in fact, also converges to ln 2 at x = 1.
3.4.2
¤
¤
Multiplication of power series
P
P∞
n
n
Theorem 3.4.5 Suppose that f (x) = ∞
n=0 an x and g(x) =
n=0 bn x
converge absolutely for |x| < R. Their multiplication is defined as
f (x) · g(x) = (
∞
X
n=0
an xn ) · (
∞
X
n=0
bn xn ) =
∞
X
n=0
cn xn ,
88
Chapter 3.
Infinite Sequences and Series
where
cn = a0 bn + a1 bn−1 + · · · + an b0 =
Then the series
P∞
n
X
ak bn−k .
k=0
n
n=0 cn x
converges to h(x) = f (x) · g(x) for |x| < R.
Example 3.4.6 For the series
∞
X
xn = 1 + x + x2 + · · · + xn + · · · =
n=0
f (x) · g(x) =
∞
∞
n=0
n=0
1
= f (x) = g(x), for |x| < 1,
1−x
X
X
1
n
=
c
x
=
(n + 1)xn , for |x| < 1,
n
(1 − x)2
since cn = 1 + 1 + · · · + 1 = n + 1. Note that
1
1
d
(
)=
.
dx 1 − x
(1 − x)2
3.5
¤
Taylor and Maclaurin Series
In the previous section, we have seen that some power series can be expressed
in terms of elementary functions on the interval of convergence by using
term by term differentiation and integration. The sum of a power series
is a continuous function with derivatives of all orders within its interval of
convergence.
Just like the number of functions with explicit antiderivatives that can
be expressed in terms of elementary functions is very small compared to
the number of integrable functions, the number of power series that can be
expressed in terms of elementary functions is small compared to the number
of power series that converges on some interval.
In this section, we will see that a function with derivatives of all orders
on an interval I can be expressed by a convergent power series on I, so that
its functional value can be approximated from the series by summing a finite
terms.
Suppose that the function is the sum of a power series
f (x) =
∞
X
an (x − a)n
n=0
= a0 + a1 (x − a) + a2 (x − a)2 + · · · + an (x − a)n + · · ·
3.5.
Taylor and Maclaurin Series
89
with positive radius of convergence R > 0. By repeated term by term
differentiation on I:
f 0 (x) = a1 + 2a2 (x − a) + 3a3 (x − a)2 + · · · + nan (x − a)n−1 + · · ·
f 00 (x) = 1 · 2a2 + 2 · 3a3 (x − a) + 3 · 4a4 (x − a)2 + · · ·
+(n − 1) · nan (x − a)n−1 + · · ·
f 000 (x) = 1 · 2 · 3a3 + 2 · 3 · 4a4 (x − a) + 3 · 4 · 5a5 (x − a)2 + · · · +
..
.
f (n) (x) = n!an + a sum of terms with (x − a) as a factor.
Thus
f (a) = a0 ,
f 0 (a) = 1a1 ,
f 00 (a)
,
2!
f 000 (a)
f 000 (a) = 3!a3 , or a3 =
,
3!
..
.
f 00 (a) = 2!a2 , or a2 =
f (n) (a) = n!an , , or an =
..
.
f (n) (a)
,
n!
.
Thus, for a function with
P∞ derivatives nof all orders, if it has a power series
representation f (x) = n=0 an (x − a) that converges to the value f (x) for
any x in an interval I centered at a, then there is only one such series with
(n)
its n-th coefficients: an = f n!(a) , so that it must be
f (x) =
∞
X
f (n) (a)
n=0
n!
(x − a)n
f 00 (a)
f (n) (a)
(x − a)2 + · · · +
(x − a)n + · · · ,
2!
n!
In general, if f (x) is a function with derivatives of all orders on some
interval I containing x = a, then one can make a power series as
= f (a) + f 0 (a)(x − a) +
P (x, a) =
∞
X
f (n) (a)
n=0
n!
(x − a)n = f (a) + f 0 (a)(x − a) +
+
f 00 (a)
(x − a)2 + · · ·
2!
f (n) (a)
(x − a)n + · · · ,
n!
90
Chapter 3.
Infinite Sequences and Series
which is called the Taylor series of f at x = a. When a = 0, the series is
called the Maclaurin series of f :
P (x) =
∞
X
f (n) (a)
n=0
n!
xn = f (a) + f 0 (a)x +
f 00 (a) 2
f (n) (a) n
x + ··· +
x + ··· .
2!
n!
Now, a question is whether this Taylor series converges to f (x) at each
point x in I or not. For some functions, it will do, but some others it will
not.
Example 3.5.1 Find the Taylor series of f (x) =
interval of convergence.
1
x
at a = 2, and the
Solution: Since
1
,
x
1
f 0 (x) = − 2 ,
x
2!
f 00 (x) = 3 ,
x
1
f (2) = ,
2
f (x) =
f (n) (x) = (−1)n
1
,
22
00
1
f (2)
= 3,
2!
2
f 0 (2) = −
..
.
(−1)n
f (n) (2)
= n+1 ,
n!
2
n!
,
xn+1
the Taylor series is
f (n) (2)
f 00 (2)
(x − 2)2 + · · · +
(x − 2)n + · · ·
2!
n!
n
1 (x − 2) (x − 2)2
n (x − 2)
−
+
−
·
·
·
+
(−1)
+ ··· .
2
22
23
2n+1
f (2) + f 0 (2)(x − 2) +
=
Since this is a geometric series with a0 =
absolutely to
1+
1
2
(x−2)
2
=
1
2
and ratio r = − x−2
2 , it converges
1
1
= ,
2 + (x − 2)
x
for |x − 2| < 2. Thus the Taylor series converges to f (x) =
|x − 2| < 2 or 0 < x < 4.
1
x
at a = 2 for
¤
3.5.
Taylor and Maclaurin Series
91
Example 3.5.2 Consider the following function:
½
0,
if x = 0,
f (x) =
2
−1/x
e
, if x 6= 0.
One can easily show that this function has derivatives of all orders at x = 0,
which are f (n) (0) = 0 for all n. This means that the Taylor series of f at
x = 0 is
f 00 (0) 2
f (n) (0) n
x + ··· +
x + ···
2!
n!
= 0 + 0 · x + 0 · x2 + · · · + 0 · xn = 0.
f (0) + f 0 (0)x +
The series converges to 0 6= f (x) for x 6= 0.
¤
Theorem 3.5.1 (Taylor’s Theorem) Suppose that f and its first n derivatives f 0 , f 00 , . . ., f (n) are continuous on a closed interval between a and b
and f (n) is differentiable on the open interval between a and b. Then there
is a number c ∈ (a, b) such that
f (b) = f (a) + f 0 (a)(b − a) +
+
f (n) (a)
f 00 (a)
(b − a)2 + · · · +
(b − a)n
2!
n!
f (n+1) (c)
(b − a)n+1 .
(n + 1)!
Proof: Assume a < b.
Z
f (b) − f (a) =
b
f 0 (t)dt.
a
v(t) = t − b so that du = f 00 (t)dt and dv = dt. Then
Z b
Z b
=
udv = uv|ba −
vdu, by integration by parts
a
a
Z b
£
¤b
= − (b − t)f 0 (t) a +
(b − t)f 00 (t)dt,
a
Z b
= f 0 (a)(b − a) +
f 00 (t)(b − t)dt,
Set u(t) =
f 0 (t),
a
(b − t)2
Set u = f 00 (t), dv = (b − t)dt, so v = −
2!
¸b Z b
·
2
(b − t)2 000
(b − t) 00
f (t) +
f (t)dt
= f 0 (a)(b − a) −
2!
2!
a
a
92
Chapter 3.
Infinite Sequences and Series
f 00 (a)
= f (a)(b − a) +
(b − a)2 +
2!
Z
b
0
a
f 000 (t)
(b − t)2
dt.
2!
By induction,
f 00 (a)
f (n) (a)
= f 0 (a)(b − a) +
(b − a)2 + · · · +
(b − a)n
2!
n!
Z b
(b − t)n
+
f (n+1) (t)
dt.
n!
a
Since f (n+1) (t) is bounded on [a, b], it has maximum and minimum at some
u, v ∈ [a, b] such that f (n+1) (v) ≤ f (n+1) (t) ≤ f (n+1) (u) for all t ∈ [a, b].
Thus,
Z b
Z b
Z b
(b − t)n
(b − t)n
(b − t)n
(n+1)
(n+1)
(n+1)
dt ≤
f
(t)
dt ≤ f
(u)
dt.
f
(v)
n!
n!
n!
a
a
a
However,
Z
a
b
·
¸b
(b − t)n
(b − t)n+1
(b − a)n+1
dt = −
=
,
n!
(n + 1)! a
(n + 1)!
and so
Rb
f
(n+1)
(v) ≤
a
n
f (n+1) (t) (b−t)
n! dt
(b−a)n+1
(n+1)!
≤ f (n+1) (u).
By the intermediate value Theorem 1.4.3, there is a number c ∈ [a, b] such
that
Z b
(b − t)n
(b − a)n+1
f (n+1) (t)
dt = f (n+1) (c)
= Rn (b, a).
n!
(n + 1)!
a
When a > b, almost the same proof holds.
¤
For a fixed a, b can now be treated as an independent variable, and
replaced by x. Hence, if f has derivatives of all orders in an interval I
containing a, then for each positive integer n and x ∈ I,
f (x) = f (a) + f 0 (a)(x − a) +
f 00 (a)
f (n) (a)
(x − a)2 + · · · +
(x − a)n +
2!
n!
f (n+1) (c)
(x − a)n+1 ,
(n + 1)!
where
Pn (x) = f (a) + f 0 (a)(x − a) +
f (n) (a)
f 00 (a)
(x − a)2 + · · · +
(x − a)n
2!
n!
3.5.
Taylor and Maclaurin Series
93
is called the Taylor polynomial of order n of f at x = a, and
f (x) − Pn (x) = Rn (x, a) =
f (n+1) (c)
(x − a)n+1
(n + 1)!
is called the remainder of order n or the error term.
P∞ f (n) (a)
n
Theorem 3.5.2 The Taylor series P (x, a) =
n=0
n! (x − a) of f
converges to f (x) on I, if Rn (x, a) → 0 as n → ∞ for all x ∈ I: we write
f (x) =
∞
X
f (n) (a)
n=0
n!
(x − a)n .
Since we do not know what c is in the remainder term, we can not
compute the precise value of the error term. However, we can estimate its
possible maximum on I: i.e., if there is a positive number M such that
|f (n+1) (t)| ≤ M for all t between a and x, then
|Rn (x, a)| ≤ M
|x − a|n+1
.
(n + 1)!
Thus, the Taylor polynomial Pn (x) approximates f (x) within this much
of error, and the Taylor series converges to f (x) on I if this condition on
the remainder term is satisfied for every n.
Example 3.5.3 Find the Taylor formula of each of the following functions:
(1) f (x) = ex at x = 0.
Since f (n) (x) = ex for all n, f (n) (0) = 1 for all n. Thus
x2
xn
xn+1
+ ··· +
+ ec
2!
n!
(n + 1)!
= Pn (x, 0) + Rn (x, 0),
f (x) = 1 + x +
n+1
|x|
for some c between 0 and x. Since |Rn (x, 0)| ≤ e|x| (n+1)!
→ 0 as n → ∞ for
any fixed x ∈ R, the Taylor series converges to f (x):
∞
ex = 1 + x +
X xn
xn
x2
+ ··· +
+ ··· =
= lim Pn (x, 0).
n→∞
2!
n!
n!
n=0
(2) f (x) = sin x at x = 0.
94
Chapter 3.
Infinite Sequences and Series
The derivatives are:
f 0 (x) = cos x,
f (x) = sin x
f 00 (x) = − sin x
f 000 (x) = − cos x,
..
.
f (2n) (x) = (−1)n sin x
f (2n+1) (x) = (−1)n cos x.
Thus f (2n) (0) = 0 and f (2n+1) (0) = (−1)n .
x3 x5
(−1)n x2n+1 (−1)n+1 x2n+3 cos c
+
− ··· +
+
3!
5!
(2n + 1)!
(2n + 3)!
= Pn (x) + Rn (x),
sin x = x −
2n+2
|x|
→ 0 as n → ∞
for some c between 0 and x. Note that |Rn (x, 0)| ≤ 1 (2n+2)!
for any fixed x ∈ R, the Taylor series converges to f (x):
∞
sin x = x −
X (−1)n x2n+1
x3 x5
(−1)n x2n+1
+
− ··· +
+ ··· =
.
3!
5!
(2n + 1)!
(2n + 1)!
n=0
(3) f (x) = cos x at x = 0.
The derivatives are:
f 0 (x) = − sin x,
f (x) = cos x
f 00 (x) = − cos x
f 000 (x) = sin x,
..
.
f (2n) (x) = (−1)n cos x
f (2n+1) (x) = (−1)n+1 sin x.
Thus f (2n) (0) = (−1)n and f (2n+1) (0) = 0.
x2 x4
(−1)n x2n (−1)n x2n+1 cos c
+
− ··· +
+
2!
4!
(2n)!
(2n + 1)!
= Pn (x) + Rn (x),
cos x = 1 −
2n+1
|x|
for some c between 0 and x. Note that |Rn (x, 0)| ≤ 1 (2n+1)!
→ 0 as n → ∞
for any fixed x ∈ R, the Taylor series converges to f (x):
∞
cos x = 1 −
X (−1)n x2n
(−1)n x2n
x2 x4
+
− ··· +
+ ··· =
.
2!
4!
(2n)!
(2n)!
n=0
(4) f (x) = cos 2x at x = 0.
3.5.
Taylor and Maclaurin Series
95
∞
cos 2x = 1 −
X (−1)n (2x)2n
(2x)2 (2x)4
(−1)n (2x)2n
+
− ··· +
+ ··· =
,
2!
4!
(2n)!
(2n)!
n=0
which converges for all x with −∞ < 2x < ∞. By the uniqueness of the
Taylor series, this must be the Taylor series of cos 2x.
(5) f (x) = x sin x at x = 0.
∞
x sin x = x(x −
X (−1)n x2n+2
x3 x5
(−1)n x2n+1
+
− ··· +
+ ···) =
,
3!
5!
(2n + 1)!
(2n + 1)!
n=0
which converges for all x with −∞ < x < ∞. By the uniqueness of the
Taylor series, this must be the Taylor series of cos 2x.
¤
Indeed, the uniqueness of the Taylor series says that it doesn’t matter
how we obtain the expression of f by a power series. As long as the function
is expressed by a convergent power series, it is the Taylor series by the
uniqueness.
Most of the elementary functions can be expressed by the Taylor series
and computing the series is the most practical method of evaluating the
functional values f (x) for given input number x.
However, the series has infinitely many terms and we can not take the
sum of infinitely many terms, all we can do is taking sum of only a finite
number of terms. Thus the Taylor polynomial is only way that can approximate the sum. Than question is how many terms do we need to take to
have the error sufficiently small?
Example 3.5.4 (1) Calculate e with an error of less than 10−6 .
1
For x = 1 in the Taylor formula of ex ,we get e = 1 + 1 + 2!1 + · · · + n!
+
Rn (1), with
1
Rn (1) = ec
.
(n + 1)!
1
Knowing e < 3, (n+1)!
< Rn (1) <
n + 1 = 10, or n = 9. Thus
e≈1+1+
3
(n+1)! .
Since
3
10!
<
1
106
1
1
+ · · · + ≈ 2.718282.
2!
9!
(2) Calculate sin 1 with an error of less than 3 · 10−4 .
<
1
9! ,
we choose
96
Chapter 3.
Infinite Sequences and Series
The Taylor formula is
x3
+ 0 · x4 +
3!
x3
= 0 + x − 0 · x2 −
+ 0 · x4 +
3!
sin x = 0 + x − 0 · x2 −
x5
− R5 (x)
5!
x5
− 0 · x6 − R6 (x).
5!
2n+1
x
Thus we take smaller error R2n (x) = sin c (2n+1)!
with 2n − 1 terms.
1
For x = 1, R2n (1) = sin c (2n+1)!
with c ∈ (0, 1). Thus
|R2n (1)| ≤
for 2n + 1 = 7, since
1
7!
<
2
104
<
1
3
< 4
(2n + 1)!
10
3
.
104
Hence, we get
sin 1 ≈ 1 −
with
|R6 (1)| ≤
3.6
1
1
+ ,
3! 5!
1
3
< 4.
7!
10
¤
Applications of Power Series
For an integer m ≥ 0, the binomial expansion is the well known expansion
of the function:
µ ¶
µ
¶
m 2
m
m
f (x) = (1 + x) = 1 + mx +
x + ··· +
xm−1 + xm .
2
m−1
For a general number m, one can still have a similar expansion like this.
Let f (x) = (1 + x)m . Then
f 0 (x) = m(1 + x)m−1
f 00 (x) = (m − 1)m(1 + x)m−2
f 000 (x) = (m − 2)(m − 1)m(1 + x)m−3
..
.
f (k) (x) = (m − k + 1) · · · (m − 1)m(1 + x)m−k .
3.6.
Applications of Power Series
97
Thus the Taylor series of f at x = 0 is:
(m − 1)m 2 (m − 2)(m − 1)m 3
x +
x + ··· +
2!
3!
(m − k + 1) · · · (m − 1)m k
x + ··· ,
k!
which is called the binomial series. If m is an integer ≥ 0, the series stops
after (m + 1) terms because the coefficients from k = m + 1 on are zero. If
m is not an integer or 0, then the series is infinite and converges for |x| < 1:
Indeed, by the ratio test,
¯
¯ ¯
¯
¯ ak+1 ¯ ¯ m − k ¯
¯
¯=¯
¯
¯ ak ¯ ¯ k + 1 x¯ → |x|,
P (x) = 1 + mx +
as k → ∞. One can also easily show that the remainder term approaches 0
as k → ∞, so that this Taylor series converges to f (x) = (1 + x)m : With a
notational convention, let
µ ¶
µ ¶
µ ¶
m
(m − k + 1) · · · (m − 1)m
m
m
(m − 1)m
,
=
.
= m,
=
2!
k
k!
1
2
Then
(1 + x)m = 1 +
∞ µ ¶
X
m k
x .
k
k=1
¡−1¢
¡ ¢ −1(−2)
Example 3.6.1 If m = −1, then 1 = −1, −1
= 1, and
2!
2 =
µ ¶
k!
−1
−1(−2)(−3) · · · (−1 − k + 1)
= (−1)k = (−1)k .
=
k!
k!
k
Thus
(1 + x)−1 = 1 − x + x2 − x3 + · · · + (−1)k xk + · · · .
¡1¢
√
Example 3.6.2 If m = 21 , then f (x) = 1 + x, and 12 =
1 1
·( −1)
2 2
2!
¤
1
2,
¡1¢
2
2
=
= − 18 , and
µ ¶
µ ¶
( 12 )(− 12 )(− 32 )
( 1 )(− 12 )(− 23 )(− 25 )
−1
1
−1
5
=
= ,
= 2
=
.
3
3!
16
4
4!
128
Hence,
√
x x2 x3 5x4
1+x = 1+ −
+
−
+ · · · for |x| small
2Ã 8
16 128
!
∞
X
1
(2n − 3)!
n−1
n
= 1+
x+
(−1)
x .
2
22n−3 (n − 2)!n!
n=2
¤
98
Chapter 3.
Example 3.6.3 Estimate
Note that
R1
0
Infinite Sequences and Series
sin x2 dx with in an error of less than 0.001.
(−1)n x4n+2
x6 x10
+
− ··· +
+ ··· ,
3!
5!
(2n + 1)!
Z
x3
x7
x11
(−1)n x4n+3
sin x2 dx = C +
−
+
− ··· +
+ ··· .
3
7 · 3! 11 · 5!
(4n + 3)(2n + 1)!
sin x2 = x2 −
Thus,
Z
1
sin x2 dx =
0
Since
1
11·5!
1
1
1
(−1)n
−
+
− ··· +
+ ··· .
3 7 · 3! 11 · 5!
(4n + 3)(2n + 1)!
≈ 0.00076 < 0.001,
Z
1
sin x2 dx =
0
1
1
−
≈ 0.310
3 7 · 3!
within the error required.
¤
Example 3.6.4 Recall that, in Example 3.4.4, from
d
1
tan−1 x =
= 1 − x2 + x4 − x6 + · · · ,
dx
1 + x2
by term by term integration we get
Z
1
x3 x5 x7
tan−1 x =
+
−
+ ··· .
dx
=
x
−
1 + x2
3
5
7
Since the term by term integration theorem was not proven, we derive
this series expression of tan−1 from an integral of a finite sum:
1
(−1)n+1 t2n+2
2
4
6
n 2n
=
1
−
t
+
t
−
t
+
·
·
·
+
(−1)
t
+
.
1 + t2
1 + t2
By integrating both sides from t = 0 to t = x, we get
tan−1 x = x −
x2n+1
x3 x5 x7
+
−
+ · · · + (−1)n
+ Rn (x),
3
5
7
2n + 1
where
Z
Rn (x) =
0
x
(−1)n+1 t2n+2
dt.
1 + t2
3.6.
Applications of Power Series
99
Since the denominator of the integrand is greater than or equal to 1,
Z |x|
|x|2n+3
|Rn (x)| ≤
t2n+2 dt =
→ 0,
2n + 3
0
as n → ∞ for |x| ≤ 1. Thus, we get
tan−1 x =
∞
X
(−1)n
n=0
x3 x5 x7
x2n+1
=x−
+
−
+ · · · , |x| ≤ 1.
2n + 1
3
5
7
For x = 1,
π
1 1 1
(−1)n
= 1 − + − + ··· +
+ ··· .
4
3 5 7
2n + 1
This series converges very slowly. However, for x close to 0, the series
converges rapidly, and so one can use various trigonometric formulas: Let
α = tan−1 21 and β = tan−1 31 . Then
tan(α + β) =
1
+1
π
tan α + tan β
= 2 13 = tan .
1 − tan α tan β
4
1− 6
¤
Example 3.6.5 Series can be used to evaluate the limit of functions. Find
the limit:
¶
µ
1
1
−
.
lim
x→0 sin x
x
1
1
−
sin x x
3
=
=
=
Hence,
µ
lim
x→0
5
x − (x − x3! + x5! − · · · )
x − sin x
=
3
5
x sin x
x(x − x3! + x5! − · · · )
x3 ( 3!1 −
x2
5!
+ ···)
2
4
− x3! + x5! − · · · )
2
x( 3!1 − x5! + · · · )
.
2
4
(1 − x3! + x5! − · · · )
x2 (1
1
1
−
sin x x
¶
= lim
x→0
x2
5! + · · · )
x2
x4
3! + 5! − · · · )
x( 3!1 −
(1 −
= 0.
On the other hand, for |x| small,
1
1
1
x
1 x
− ≈ x = , or csc x = + .
sin x x
3!
6
x 6
¤
100
3.7
Chapter 3.
Infinite Sequences and Series
Fourier Series
The Taylor polynomial gives a good approximation for functional values
for x near a particular point x = a where the functional value is given.
However, the error in the approximation can be large at points that are far
away. There is another method, called the Fourier series, that often gives
good approximation on wide intervals, and often works with discontinuous
functions for which Taylor polynomial fails. The Fourier series approximate
functions with sums of sine and cosine functions. It is well suited for analyzing periodic functions, such as radio signals and alternating currents, for
solving heat transfer problems, and for many other problems in science and
engineering.
Let f (x) be a function on the interval [0, 2π]. We wish to approximate
f by a function of the form:
fn (x) = a0 + (a1 cos x + b1 sin x) + (a2 cos 2x + b2 sin 2x) + · · ·
+(an cos nx + bn sin nx).
n
X
(ak cos kx + bk sin kx).
= a0 +
k=1
To choose ak and bk that make fn (x) a best possible approximation to f (x)
in the following sense:
(1) fn (x) and f (x) give the same value when integrated from 0 to 2π:
Z 2π
Z 2π
fn (x)dx =
f (x)dx.
0
0
(2) fn (x) cos kx and f (x) cos kx give the same value when integrated from
0 to 2π, for k = 1, 2, 3,. . ., n:
Z 2π
Z 2π
fn (x) cos kxdx =
f (x) cos kxdx.
0
0
(2) fn (x) sin kx and f (x) sin kx give the same value when integrated from
0 to 2π, for k = 1, 2, 3,. . ., n:
Z 2π
Z 2π
fn (x) sin kxdx =
f (x) sin kxdx.
0
0
3.7.
Fourier Series
101
From the expression of fn (x) defined above, the left side integrals are:
Z
2π
fn (x)dx = 2πa0
0
Z
2π
fn (x) cos kxdx = πak
0
Z
2π
0
fn (x) sin kxdx = πbk .
These follow from the fact that
Z
½
2π
cos px cos qxdx =
0
Z
½
2π
sin px sin qxdx =
Z
0
2π
π
0
if p = q,
if p =
6 q.
π
0
if p = q,
if p =
6 q.
sin px cos qxdx = 0.
0
Therefore, we choose fn (x) so that the integral on the left remain the same
when fn (x) is replaced by f (x):
a0 =
ak =
bk =
Z 2π
1
f (x)dx,
2π 0
Z
1 2π
f (x) cos kxdx,
π 0
Z
1 2π
f (x) sin kxdx,
π 0
k = 1, . . . , n,
k = 1, . . . , n.
The Fourier series for f (x) is obtained, as n → ∞,
a0 +
∞
X
(ak cos kx + bk sin kx).
k=1
½
Example 3.7.1 Let f (x) =
1
2
if 0 ≤ x ≤ π,
if π < x ≤ 2π.
102
Chapter 3.
Infinite Sequences and Series
The coefficients of the Fourier series of f are:
a0 =
=
ak =
=
=
bk =
=
=
=
Z 2π
1
f (x)dx,
2π 0
µZ π
¶
Z 2π
1
3
1dx +
2dx = .
2π
2
π
0
Z 2π
1
f (x) cos kxdx,
π 0
µZ π
¶
Z 2π
1
cos kxdx +
2 cos kxdx ,
π
0
π
Ã
¯
¯ !
1 sin kx ¯¯π
2 sin kx ¯¯2π
+
= 0, k ≥ 1.
π
k ¯0
k ¯π
Z
1 2π
f (x) sin kxdx,
π 0
µZ π
¶
Z 2π
1
sin kxdx +
2 sin kxdx ,
π
0
π
Ã
¯
¯ !
cos kx ¯¯π
2 cos kx ¯¯2π
1
−
+−
,
¯
π
k ¯0
k
π
cos kπ − 1
(−1)k − 1
=
,
kπ
kπ
k ≥ 1.
Thus, a0 = 32 , a1 = a2 = · · · = 0,
2
2
2
b1 = − , b2 = 0, b3 = − , b4 = 0, b5 = − , b6 = 0, · · · ,
π
3π
5π
and the Fourier series is
3 2
+
2 π
3.7.1
µ
¶
sin 3x sin 5x
sin x +
+
+ ··· .
3
5
¤
Convergence of Fourier series
Taylor series are computed from the value of a function and derivatives at a
single point x = a, and can not reflect the behavior of discontinuous function
such as f in Example 3.7.1 past a discontinuity. The reason that a Fourier
3.7.
Fourier Series
103
series can be used to represent such functions is that the Fourier series of
a function depends on the existence of certain integrals, whereas the Taylor
series depends on derivatives of a function near a single point. A function
can be fairly rough even discontinuous, and still be integrable.
The coefficients used to construct Fourier series are precisely those one
should choose to minimize the integral of the square of the distance between
f and fn :
Z
2π
0
(f (x) − fn (x))2 dx
is minimized by choosing a0 , a1 , . . ., an and b1 , b2 , . . ., bn as we did. While
Taylor series are useful to approximate a function and its derivatives near a
point, Fourier series minimizes an error which is distributed over an interval.
Theorem 3.7.1 Let f (x) be a function such that f and f 0 are piecewise
continuous on the interval [0, 2π]. Then f is equal to its Fourier series at
all points where f is continuous. At a point c where f has a discontinuity,
the Fourier series converges to
f (c+ ) + f (c− )
2
where f (c+ ) and f (c− ) are the right and left limits of f at c.