Section 10.7: Taylor and Maclaurin Series
Taylor and Maclaurin series are power series representations of functions. Let
∞
X
f (x) =
cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + c4 (x − a)4 + · · · .
n=0
Then f (a) = c0 .
Taking the derivative gives
f 0 (x) = c1 + 2c2 (x − a) + 3c3 (x − a)2 + 4c4 (x − a)3 + · · · .
Then f 0 (a) = c1 .
Similarly,
f 00 (x) = 2c2 + 3 · 2c3 (x − a) + 4 · 3c4 (x − a)2 + · · · .
f 00 (a)
00
Then f (a) = 2c2 and so c2 =
.
2
Likewise,
f 000 (x) = 3 · 2c3 + 4 · 3 · 2(x − a) + · · · .
f 000 (a)
.
Then f 000 (a) = 3!c3 and so c3 =
3!
Continuing in this fashion, we find that in general
cn =
f (n) (a)
.
n!
Definition: The Taylor series for f (x) centered at x = a is
f (x) =
∞
X
f (n) (a)
n=0
where f
(n)
n!
(x − a)n ,
(a) is the nth derivative of f (x) at x = a.
Example: Find the Taylor series for f (x) = e3x centered at x = 2. What is the associated
radius of convergence?
The higher-order derivatives of f (x) are
f (x)
f 0 (x)
f 00 (x)
f 000 (x)
=
=
=
=
..
.
e3x
3e3x
9e3x
27e3x
f (n) (x) = 3n e3x
f (2)
f 0 (2)
f 00 (2)
f 000 (2)
=
=
=
=
..
.
e6
3e6
9e6
27e6
f (n) (2) = 3n e6 .
So the Taylor series is
∞
X
3n e6
e3x =
n=0
n!
(x − 2)n .
Using the Ratio Test,
n+1 6
3 e (x − 2)n+1
3(x − 2) n!
= 0 < 1.
lim
· n 6
= lim
n→∞ (n + 1)!
3 e (x − 2)n n→∞ n + 1 The radius of convergence is R = ∞.
1
centered at x = 3. What is the associated
Example: Find the Taylor series for f (x) =
x
radius of convergence?
The higher-order derivatives of f (x) are
1
x
1
0
f (x) = − 2
x
2
00
f (x) = 3
x
6
000
f (x) = − 4
x
..
.
(−1)n n!
f (n) (x) =
xn+1
1
3
1
0
f (3) = −
9
2
00
f (3) =
27
6
000
f (3) = −
81
..
.
(−1)n n!
f (n) (3) =
.
3n+1
f (x) =
So the Taylor series is
f (3) =
∞
1 X (−1)n
=
(x − 3)n .
n+1
x n=0 3
Using the Ratio Test,
(−1)n+1 (x − 3)n+1
−(x − 3) |x − 3|
3n+1
=
lim
·
= lim
< 1.
n→∞ 3n+2
(−1)n (x − 3)n n→∞ 3
3
The radius of convergence is R = 3.
Definition: The Maclaurin series for f (x) is the Taylor series centered at x = 0. That is,
f (x) =
∞
X
f (n) (0)
n=0
n!
where f (n) (0) is the nth derivative of f (x) at x = 0.
xn ,
Example: Find the Maclaurin series for f (x) = ex . What is the associated radius of convergence?
For n ≥ 0, the nth derivative of f (x) is
f (n) (x) = f (x) = ex .
Then f (n) (0) = 1 and the Maclaurin series is
x
e =
∞
X
xn
n=0
n!
.
Using the Ratio Test,
n+1
x
x n! = 0 < 1.
lim
·
= lim n→∞ (n + 1)! xn n→∞ n + 1 The radius of convergence is R = ∞.
Example: Find the Maclaurin series for f (x) = sin x. What is the associated radius of
convergence?
The higher-order derivatives of f (x) = sin x are
f (x)
f 0 (x)
f 00 (x)
f 000 (x)
f (4) (x)
=
=
=
=
=
..
.
sin x
cos x
− sin x
− cos x
sin x
f (0)
f 0 (0)
f 00 (0)
f 000 (0)
f (4) (0)
=
=
=
=
=
..
.
0
1
0
−1
0
The Maclaurin series for sin x is
∞
sin x = x −
X (−1)n x2n+1
x3 x5 x 7
+
−
+ −··· =
.
3!
5!
7!
(2n
+
1)!
n=0
Using the Ratio Test,
2
(−1)n+1 x2n+3 (2n + 1)! −x
= lim = 0 < 1.
lim ·
n→∞
(2n + 3)!
(−1)n x2n+1 n→∞ (2n + 3)(2n + 2) The radius of convergence is R = ∞.
Example: Find the Maclaurin series for f (x) = cos x. What is the associated radius of
convergence?
The higher-order derivatives of f (x) = cos x are
f (x)
f 0 (x)
f 00 (x)
f 000 (x)
f (4) (x)
=
=
=
=
=
..
.
cos x
− sin x
− cos x
sin x
cos x
f (0)
f 0 (0)
f 00 (0)
f 000 (0)
f (4) (0)
=
=
=
=
=
..
.
1
0
−1
0
1
The Maclaurin series for cos x is
∞
X (−1)n x2n
x2 x4 x6
+
−
+ −··· =
.
cos x = 1 −
2!
4!
6!
(2n)!
n=0
Using the Ratio Test,
2
(−1)n+1 x2n+2
−x
(2n)!
= lim = 0 < 1.
lim ·
n→∞
(2n + 2)!
(−1)n x2n n→∞ (2n + 2)(2n + 1) The radius of convergence is R = ∞.
3
Example: Find the Maclaurin series for f (x) = ex . What is the associated radius of convergence?
The Maclaurin series for eu is
u
e =
∞
X
un
n=0
n!
.
3
Setting u = x3 , the Maclaurin series for ex is
x3
e
=
∞
X
(x3 )n
n=0
n!
=
∞
X
x3n
n=0
n!
.
The radius of convergence is R = ∞.
Example: Find the Maclaurin series for f (x) = x cos(x3 ). What is the associated radius of
convergence?
The Maclaurin series for cos u is
cos u =
∞
X
(−1)n u2n
n=0
(2n)!
.
Setting u = x3 , the Maclaurin series for x cos(x3 ) is
x cos(x3 ) = x
∞
X
(−1)n (x3 )2n
(2n)!
n=0
∞
X
(−1)n x6n+1
=
n=0
(2n)!
.
The radius of convergence is R = ∞.
x
. What is the associated radius of
Example: Find the Maclaurin series for f (x) = x2 sin
2
convergence?
The Maclaurin series for sin u is
∞
X
(−1)n u2n+1
sin u =
n=0
Setting u =
(2n + 1)!
.
x
x
, the Maclaurin series for x2 sin
is
2
2
x sin
x
2
∞
∞
X
(−1)n x 2n+1 X (−1)n x2n+2
=x
=
.
2n+1 (2n + 1)!
(2n
+
1)!
2
2
n=0
n=0
The radius of convergence is R = ∞.
Z
Example: Consider the definite integral
1
cos(x2 )dx.
0
(a) Evaluate the integral as a series.
The Maclaurin series for cos(x2 ) is
2
cos(x ) =
∞
X
(−1)n (x2 )2n
n=0
Then
Z
0
1
(2n)!
=
∞
X
(−1)n x4n
n=0
(2n)!
.
1
∞
∞
n 4n+1 X
X
(−1)
x
(−1)n
2
cos(x )dx =
.
=
(4n
+
1)(2n)!
(4n
+
1)(2n)!
n=0
n=0
0
(b) Use the sum of the first three terms to approximate the integral. How accurate is this
approximation?
The second partial sum (sum of the first three terms) is
Z 2
1
1
cos(x2 )dx ≈ S2 = 2 −
+
.
10 216
0
By the Remainder Estimate for Alternating Series,
|R2 | ≤ a3 =
Z
Example: Approximate
1
1
1
=
≈ 0.00011.
13(6)!
9360
2
e−x dx with error less than 0.001.
0
2
The Maclaurin series for e−x is
e
−x2
=
∞
X
(−x2 )n
n=0
Then
Z
0
1
n!
=
∞
X
(−1)n x2n
n=0
n!
.
1
∞
∞
n 2n+1 X
X
(−1) x
(−1)n
−x2
e dx =
.
=
(2n + 1)(n)! (2n + 1)(n)!
n=0
n=0
0
By the Remainder Estimate for Alternating Series,
|Rn | ≤ an+1 =
1
.
(2n + 3)(n + 1)!
Let
1
1
<
(2n + 3)(n + 1)!
1000
(2n + 3)(n + 1)! > 1000.
If n = 4, then (2n + 3)(n + 1)! = (11)5! = 1320 > 1000. Then
Z 1
1
1
1
1
2
e−x dx ≈ S4 = 1 − +
−
+
≈ 0.7475.
3 10 42 216
0
The true value is approximately 0.7469.
Example: Evaluate limx→0
sin(x3 ) − x3
.
x9
The Maclaurin series for sin(x3 ) is
∞
X
(−1)n (x3 )2n+1
n=0
(2n + 1)!
=
∞
X
(−1)n x6n+3
n=0
= x3 −
(2n + 1)!
x9 x15
+
− +··· .
3!
5!
Then
9
15
− x3! + x5! − + · · ·
sin(x3 ) − x3
lim
=
lim
x→0
x→0
x9
x9
1
x6
= lim − +
− +···
x→0
3!
5!
1
= − .
6
Example: Find the sum of the given series.
(a)
∞
X
(−1)n 3n x2n
n=0
n!
The series can be rewritten as
∞
X
(−3x2 )n
n!
n=0
2
= e−3x .
∞
X
(−1)n π 2n+1
(b)
32n+1 (2n + 1)!
n=0
The series can be rewritten as
∞
π √3
X
(−1)n π 2n+1
=
.
= sin
(2n + 1)! 3
3
2
n=0
(c)
∞
X
(−4)n x2n
n=1
n!
The series can be rewritten as
∞
X
(−4x2 )n
n=1
n!
=
∞
X
(−4x2 )n
n=0
n!
2
− 1 = e−4x − 1.