LECTURE 34: WORKING WITH TAYLOR SERIES
MINGFENG ZHAO
April 01, 2015
Definition 1. 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) +
f 00 (a)
f (3) (a)
(x − a)2 +
(x − a)3 + · · · .
2!
3!
A Taylor series centered at 0 is called a Maclaurin series.
Example 1. Find the Maclaurin series of cos(x) and sin(x).
Let f (x) = cos(x), then
f (0) (x) = f (x) = cos(x),
f (0) = 1
f 0 (x) = − sin(x),
f 0 (0) = 0
f 00 (x) = − cos(x),
f 00 (0) = −1
f (3) (x) = sin(x),
f (3) (0) = 0
f (4) (x) = cos(x),
f (4) (0) = 1.
So f (k) (0) = 0 if k is odd, and

 1,
if k is even
f (2k) (0) =
 −1, if k is odd.
Hence the Maclaurin series for f (x) is:
∞
X
(−1)k
k=0
(2k)!
x2k = 1 −
x2
x4
x6
+
−
+ ··· .
2!
4!
6!
Differentiate the above series, we get
− sin(x) =
∞
X
(−1)k 2k−1
x
.
(2k − 1)!
k=1
1
2
MINGFENG ZHAO
So we get
sin(x) = −
∞
∞
X
(−1)k 2k−1 X (−1)k+1 2k−1
x
=
x
.
(2k − 1)!
(2k − 1)!
k=1
k=1
Limits by Taylor series
x2 + 2 cos(x) − 2
.
x→0
3x4
By Example in 1, we have
Example 2. Evaluate lim
cos(x) =
∞
X
(−1)k
k=0
(2k)!
x2k = 1 −
x2
x4
x6
+
−
+ ··· .
2!
4!
6!
Then
x2 + 2 cos(x) − 2
x4
x6
x2
+
−
+ ···
x2 − 2 + 2 · 1 −
2!
4!
6!
6
4
x
x
2
2
+ 2 − + ···
= x −2+2−x +
12
6!
6
4
x
x
=
+ 2 − + ··· .
12
6!
=
So we get
2
x + 2 cos(x) − 2
=
3x4
x4
12
h 6
i
+ 2 − x6! + · · ·
3x4
2
1
x
1
=
+ 2 − + ··· →
,
36
6!
36
Hence we have
x2 + 2 cos(x) − 2
1
.
=
x→0
3x4
36
1
Example 3. Evaluate lim 6x5 sin
− 6x4 + x2 .
x→∞
x
1
Let t = , then t → 0 as x → ∞, which implies that
x
1
6 sin(t)
6
1
lim 6x5 sin
− 6x4 + x2
= lim
−
+
x→∞
t→0
x
t5
t4
t2
lim
=
lim
t→0
6 sin(t) − 6t + t3
.
t5
By Example 1, we have
sin(t) =
∞
X
(−1)k+1 2k−1
t3
t5
8t7
t
=t− +
−
− ··· .
(2k − 1)!
6
120
8!
k=1
as x → 0.
LECTURE 34: WORKING WITH TAYLOR SERIES
3
Then
6 sin(t) − 6t + t
3
t3
t5
t7
6 t− +
− − · · · − 6t + t3
6
120 7!
7
t5
t
6t − t3 +
−6
− · · · − 6t + t3
20
7!
7
5
t
t
−6
− ··· .
20
7!
=
=
=
So we get
3
6 sin(t) − 6t + t
=
t5
t5
20
−6
h
8t7
8!
− ···
i
t5
2
1
t
=
−6
− ··· ,
20
7!
as t → 0.
Hence we have
lim
x→∞
1
6 sin(t) − 6t + t3
1
=
.
− 6x4 + x2 = lim
6x5 sin
5
t→0
x
t
20
Remark 1. One can use l’Hospital’s Rule to check your answers for the above examples: Let f (x) and g(x) be
differential functions, and a be a number or ±∞, then
a.
f 0 (x)
0
type: Assume that lim f (x) = lim g(x) = 0 and lim 0
= L, then
x→a
x→a
x→a g (x)
0
lim
x→a
b.
f (x)
= L.
g(x)
∞
f 0 (x)
type: Assume that lim f (x) = ±∞, lim g(x) = ±∞ and lim 0
= L, then
x→a
x→a
x→a g (x)
∞
lim
x→a
f (x)
= L.
g(x)
sin(x)
.
x
By l’Hospital’s rule, we have
Example 4. Evaluate lim
x→0
lim
x→0
sin(x)
cos(x)
= lim
= cos(0) = 1.
x→0
x
1
Differentiating power series
Example 5. Find a power series solution of the differential equation y 0 (x) = y with initial condition y(0) = 1.
Let
y(x) =
∞
X
k=0
ck xk .
4
MINGFENG ZHAO
Since y(0) = 1, then c0 = 1. Notice that
y 0 (x)
=
∞
X
kck xk−1
k=0
=
∞
X
kck xk−1
k=1
=
∞
X
(k + 1)ck+1 xk
k=0
0
Since y (x) = y, then
∞
X
(k + 1)ck+1 xk =
k=0
∞
X
ck x k .
k=0
Then we have
(k + 1)ck+1 = ck ,
for all k ≥ 0.
1
· ck ,
k+1
for all k ≥ 0.
That is, we have
ck+1 =
Since c0 = 1, then
c1
=
c2
=
c3
=
c4
=
..
.
ck+1
So we get ck =
=
c0 = 1
1
1
c1 =
2
2
1
1
c2 =
3
3!
1 1
1
1
c3 = ·
=
4
4 3!
4!
..
.
1
1
1 1
1
· · · · · · c0 =
.
k+1 k
2 1
(k + 1)!
1
for all k ≥ 0. Hence we have
k!
y(x) =
∞
X
xk
k=0
k!
.
Hence we know that
y(x) =
∞
X
xk
k=0
k!
= ex .
LECTURE 34: WORKING WITH TAYLOR SERIES
Integrating power series
1
Z
Since ex =
∞
X
xk
k=0
k!
2
e−x dx.
Example 6. Evaluate the integral
0
, then
2
e−x =
∞
X
(−x2 )k
k!
k=0
=
∞
X
(−1)k
k!
k=0
· x2k .
So we get
1
Z
Z
∞
X
(−1)k
2
e−x dx =
0
k!
k=0
1
x2k dx =
0
∞
X
(−1)k
k=0
k!
·
1
2k + 1
Representing real numbers
Example 7. Identify the function represented by the power series
∞
X
(1 − 2x)k
k=1
k!
.
Let y = 1 − 2x, then
∞
X
(1 − 2x)k
k=1
k!
=
∞
X
yk
=
k!
k=1
∞
X
yk
k=0
k!
− 1 = ey − 1 = e1−2x − 1.
Example 8. 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
x2k
=
2 k
∞
X
x
k −
4
k=1
=
∞
X
ky k
Let y = −
k=1
=
=
=
∞
X
x2
4
!
∞
∞
X
d X k
d 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
=
2
(1 − y)
(1 − y)2
2
− x4
=
1+
=
−
x2 2
4
4x2
.
(x2 + 4)2
Since y = −
x2
4
5
6
MINGFENG ZHAO
Remark 2. The function f (x) = −
∞
X
4x2
(−1)k k 2k
is
defined
on
−∞
<
x
<
∞,
but
its
power
series
x is only
2
2
(x + 4)
4k
k=1
defined on (−2, 2).
Remark 3. For the problems like Example 8, the Step I is to write the series has the form of xm
X
g(k)y k by using
X
substitution, where g(k) is a rational function of k; the Step II is to identify the function for the series
g(k)y k by
using differentiation of integration; the Step II is to substitute y back to x.
Example 9. Identify the function represented by the power series
Approach I: Let f (x) =
∞
X
xk
k=1
k
∞
X
xk
k=1
k
.
, then
∞
X
xk
d
f 0 (x) =
dx
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
x
k−1
Z
dx =
k=1
Z
=
=
∞
X
!
x
k−1
k=1
1
dx
1−x
Z
dx =
∞
X
!
x
k
dx
k=0
if |x| < 1
− ln |1 − x| + C.
Plug x = 0 into the above identities, then C = 0. So we get
∞
X
xk
k=1
k
= − ln |1 − x|.
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