LECTURE 3: REVIEW OF POWER SERIES
MINGFENG ZHAO
July 09, 2015
Power series
Definition 1. A power series has the general form
∞
X
ck (x − a)k , where a and ck ’s are real numbers, and x is a variable.
k=0
The ck ’s are the coefficients of the power series, and a is the center of the power series.
Remark 1. For any non-negative integer m, then
∞
X
ck (x − a)k =
∞
X
cn+m (x − a)n+m .
n=0
k=m
Theorem 1. For the power series
∞
X
ck (x − a)k := c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + · · · , define the set
k=0
S = {x ∈ R :
∞
X
ck (x − a)k converges}.
k=0
Then S is an interval centered at a with radius R which can be computed in the following way: let
r = lim
k→∞
|ck+1 |
,
|ck |
then
R=
1
.
r
Moreover, we have
• If R = 0, then the power series
∞
X
ck (x − a)k converges only at x = a, and the power series
k=0
∞
X
ck (x − a)k
k=0
diverges for all x 6= a.
• If 0 < R < ∞, then the power series
∞
X
∞
X
ck (x − a)k absolutely converges for all |x − a| < R, and the power series
k=0
ck (x − a)k diverges for all |x − a| > R.
k=0
1
2
MINGFENG ZHAO
• If R = ∞, then the power series
∞
X
ck (x − a)k absolutely converges for all x.
k=0
In particular, the power series
∞
X
k
ck (x−a) absolutely converges for all |x−a| < R, and the power series
k=0
∞
X
ck (x−a)k
k=0
diverges for all |x − a| > R.
Remark 2. The set S in Theorem 1, is called the interval of convergence of the power series
radius R in Theorem 1 is called the radius of convergene of the power series
∞
X
∞
X
ck (x − a)k . The
k=0
k
ck (x − a) .
k=0
Remark 3. For a series
say that the series
∞
X
∞
X
ak , we say that the series
k=1
∞
X
ak absolutely converges if the series
k=1
ak conditionally coverages if the series
k=1
∞
X
∞
X
|ak | converges. We
k=1
ak converges, but the series
k=1
∞
X
|ak | diverges.
k=1
Example 1. Find the radius of convergence and the interval of convergence of the power series
∞
X
(x − 2)k
√
.
k
k=1
Notice that
r
√ 1 k+1 k+1
= 1.
r = lim = lim
k→∞
k→∞
k
√1k ∞
X
(x − 2)k
1
√
is R = = 1, which implies that the series absolutely
r
k
k=1
∞
X
(−1)k
√
converges for all x such that |x − 2| < 1, that is, 1 < x < 3. On the other hand, when x = 1, we get
which is
k
k=1
∞
X
1
√ which is divergent by p-series. Hence the interval of convergence
conditionally convergent; when x = 3, we get
k
k=1
∞
k
X
(x − 2)
of the power series
is [1, 3).
k
Then the radius of convergence of the power series
k=1
Taylor series and Maclaurin series
Definition 2. Suppose a function f (x) has derivatives of all orders on an interval centered at point a, then the Taylor
series for f (x) centered at a is:
∞
X
f (k) (a)
k=0
k!
· (x − a)k := f (a) + f 0 (a)(x − a) +
A Taylor series centered at 0 is called a Maclaurin series.
f 00 (a)
f (3) (a)
(x − a)2 +
(x − a)3 + · · · .
2!
3!
LECTURE 3: REVIEW OF POWER SERIES
Remark 4. Let f (x) =
∞
X
3
ck (x − a)k , then
k=0
c0 = f (a),
and ck =
f (k) (a)
,
k!
∀k ≥ 1.
Two important power series
Geometric series:
∞
X
1
= 1 + x + x2 + x3 + · · · =
xk ,
1−x
for −1 < x < 1.
k=0
Exponential function:
∞
ex = 1 + x +
X xk
x3
x2
+
+ ··· =
,
2!
3!
k!
for −∞ < x < ∞.
k=0
Example 2. Find the radius of convergence of the series
∞
X
(−1)k 2k+1 x2k+1 and evaluate the sum of this series.
k=0
Notice that
∞
X
(−1)k 2k+1 x2k−1
=
k=0
∞
X
2(−2)k · x · (x2 )k
k=0
=
2x ·
∞
X
(−2)k (x2 )k .
k=0
Let y = x2 , then
∞
X
(−1)k 2k+1 x2k−1 = 2x ·
k=0
∞
X
(−2)k y k .
k=0
k
Let ck = (−2) for all k ≥ 0, then
r
=
=
k→∞
|ck+1 |
|(−2)k+1 |
= lim
k→∞
|ck |
(−2)k
2.
∞
X
1
1
= , which implies that
(−1)k (−2)k y k converges
r
2
k=0
k=0
∞
∞
X
X
for |y| < 21 , and
(−1)k (−2)k y k diverges for |y| > 12 , Since y = x2 , then
(−1)k (−2)k x2k converges for |x2 | < 21 ,
So the radius of convergence of the
and
∞
X
∞
X
lim
(−1)k (−2)k y k is R =
k=0
k
k 2k
(−1) (−2) x
k=0
2
diverges for |x | >
1
2,
which implies that the radius of convergence of the
k=0
√
∞
X
k=0
2
R=
. Moreover, we have
2
∞
X
(−1)k 2k+1 x2k+1
k=0
=
2x ·
∞
X
k=0
(−2)k (x2 )k
(−1)k 2k+1 x2k+1 is
4
MINGFENG ZHAO
=
2x ·
∞
X
(−2x2 )k
k=0
=
=
2x
1 − (−2x2 )
2x
.
1 + 2x2
Since | − 2x2 | < 1
Differential and integral of power series
Theorem 2. Suppose the power series
∞
X
ck (x − a)k converges to f (x) for |x − a| < R, then
k=0
I. Then f (x) is differentiable for |x − a| < R, and
0
f (x) =
∞
X
kck (x − a)k−1 ,
for all |x − a| < R.
k=0
II. Then
Z
f (x) dx =
∞
X
ck
k=0
(x − a)k+1
+ C,
k+1
Example 3. Identify the function represented by the power series
Approach I: Let f (x) =
∞
X
xk
k=1
k
for all |x − a| < R.
∞
X
xk
k=1
k
.
, then
d
f (x) =
dx
0
∞
X
xk
k=1
!
k
∞
X
xk−1 =
k=1
1
,
1−x
if |x| < 1.
So we have
Z
f (x) =
1
dx = − ln |1 − x| + C.
1−x
Since f (0) = 0, then C = 0. Hence we get
∞
X
xk
k=1
k
= − ln |1 − x|.
Approach II: In fact, we have
∞
X
xk
k=1
k
=
∞ Z
X
xk−1 dx =
k=1
Z
=
=
1
dx
1−x
Z
∞
X
k=1
if |x| < 1
− ln |1 − x| + C.
!
xk−1
Z
dx =
∞
X
k=0
!
xk
dx
LECTURE 3: REVIEW OF POWER SERIES
5
Plug x = 0 into the above identities, then C = 0. So we get
∞
X
xk
k
k=1
= − ln |1 − x|.
Example 4. Identify the function represented by the power series
∞
X
(−1)k k
k=1
4k
x2k .
In fact, we have
∞
X
(−1)k k
k=1
4k
x
2k
2 k
∞
X
x
k −
4
=
k=1
∞
X
=
ky k
Let y = −
k=1
∞
X
x2
4
!
∞
∞
X
d k
d X k
y · ky
=y
(y ) = y
y
dy
dy
k=1
k=1
k=1
2
x d
y
y
if |y| < 1, then − < 1, that is, |x| < 2
dy 1 − y
4
=
=
y·
=
k−1
1−y+y
y
=
(1 − y)2
(1 − y)2
2
− x4
=
1+
Since y = −
4
x2
4
4x2
− 2
.
(x + 4)2
=
Remark 5. The function f (x) = −
x2 2
∞
X
(−1)k k 2k
4x2
is
defined
on
−∞
<
x
<
∞,
but
its
power
series
x is only
(x2 + 4)2
4k
k=1
defined on (−2, 2).
Recurrence relation of sequence
∞
Definition 3. Let m ∈ N and {an }∞
n=1 be a sequence, a recurrence relation of the sequence {an }n=1 is a sequence of
functions {fn : Rn → R}∞
n=1 such that
an+1 = fn (a1 , a2 , · · · , an ),
Example 5. Let {an }∞
n=0 satisfy that a1 = 1 and an =
n−1
X
k=1
Since an =
n−1
X
kak for all n ≥ 2, then
k=1
a2
=
1 · a1 = 1 · 1
∀n ≥ 1.
kak for all n ≥ 2, find a2 , a3 , a4 .
6
MINGFENG ZHAO
=
a3
1
2
X
=
kak = 1 · a1 + 2 · a2
k=1
a4
=
1·1+2·1
=
3
3
X
=
kak = 1 · a1 + 2 · a2 + 3 · a3
k=1
=
1·1+2·1+3·3
=
12.
Products of two power series with the same center
Theorem 3. The product of two power series
∞
X
bk (x − a)k and
k=0
"
∞
X
# "
bk (x − a)
k=0
k
·
∞
X
ck (x − a)k is given by:
k=0
#
k
ck (x − a)
k=0
∞
X
=
∞
n
X
X
n=0
!
bk cn−k
(x − a)n .
k=0
Proof. For simplicity, we can take a = 0, then
"∞
# "∞
#
X
X
bk xk ·
ck xk
= b0 + b1 x + b2 x2 + b3 x3 + · · · · c0 + c1 x + c2 x2 + c3 x3 + · · ·
k=0
k=0
=
b0 c0 + b0 c1 x + b0 c2 x2 + b0 c3 x3 + · · ·
+b1 c0 x + b1 c1 x2 + b1 c2 x3 + b1 c3 x4 + · · ·
+b2 c0 x2 + b2 c1 x3 + b2 c2 x4 + b2 c3 x5 + · · ·
+b3 c0 x3 + b3 c1 x4 + b3 c2 x5 + b3 c3 x6 + · · ·
+···
= b0 c0
+(b0 c1 + b1 c0 )x
+(b0 c2 + b1 c1 + b2 c0 )x2
+(b0 c3 + b1 c2 + b2 c1 + b3 c0 )x3
+(b0 c4 + b1 c3 + b2 c2 + b3 c1 + b4 c0 )x4
LECTURE 3: REVIEW OF POWER SERIES
7
+···
=
∞
n
X
X
n=0
!
bk cn−k
xn .
k=0
Series solution to a constant coefficient ODE
Example 6. Find the general solution to y 00 + y = 0 in powers of x.
∞
X
Let y =
ck xk be a power series solution to y 00 + y = 0, then
k=0
y0
=
∞
X
kck xk−1
k=0
=
∞
X
kck xk−1
k=1
=
∞
X
(n + 1)cn+1 xn
Let k = n + 1
n=0
=
∞
X
(k + 1)ck+1 xk
Let k = n
k=0
y 00
=
∞
X
k(k + 1)ck+1 xk−1
k=0
=
∞
X
k(k + 1)ck+1 xk−1
k=1
=
∞
X
(n + 1)(n + 2)cn+2 xn
Let k = n + 1
n=0
=
∞
X
(k + 1)(k + 2)ck+2 xk
Let k = n.
k=0
Since y 00 + y = 0, then
0=
∞
X
(k + 1)(k + 2)ck+2 xk +
k=0
∞
X
ck xk =
k=0
∞
X
[(k + 1)(k + 2)ck+2 + ck ] xk .
k=0
So we get (k + 1)(k + 2)ck+2 + ck = 0 for all k ≥ 0, that is,
ck+2 = −
Then
1
ck ,
(k + 1)(k + 2)
for all k ≥ 0.
8
MINGFENG ZHAO
(1) If we start from c0 , then c0 → c2 → c4 → · · · → c2n → · · · , that is,
c2n
=
=
=
=
−1
2n(2n − 1)
−1
2n(2n − 1)
−1
2n(2n − 1)
· c2n−2
−1
· c2n−4
(2n − 2)(2n − 3)
−1
1
·
· ··· ·
· c0
(2n − 2)(2n − 3)
2·1
·
(−1)n
· c0 ,
(2n)!
∀n ≥ 0.
.
(2) If we start from c1 , then c1 → c3 → c5 → · · · → c2n+1 → · · · , that is,
c2n+1
=
=
=
=
−1
(2n + 1)(2n)
−1
(2n + 1)(2n)
−1
(2n + 1)(2n)
· c2n−1
−1
· c2n−3
(2n − 1)(2n − 2)
−1
−1
·
· ··· ·
· c1
(2n − 1)(2n − 2)
3·2
·
(−1)n
· c1 ,
(2n + 1)!
∀n ≥ 0.
.
So we get
y
=
∞
X
k
ck x =
k=0
∞
X
n=0
c2n x
2n
+
∞
X
c2n+1 x2n+1
n=0
∞
∞
X
X
(−1)n 2n
(−1)n 2n+1
= c0
x + c1
x
.
(2n)!
(2n + 1)!
n=0
n=0
Remark 6. The general solution to y 00 + y = 0 is given by y = C1 cos(x) + C1 sin(x). Notice that
cos(x) =
∞
X
(−1)n 2n
x ,
(2n)!
n=0
and
sin(x) =
∞
X
(−1)n 2n+1
x
.
(2n + 1)!
n=0
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca