Section 10.9: Applications of Taylor Polynomials
Suppose that f (x) can be expressed as a Taylor series centered at a:
f (x) =
∞
X
f (n) (a)
n=0
n!
(x − a)n .
The nth partial sum of the Taylor series is
Tn (x) =
n
X
f (j) (a)
j=0
j!
(x − a)j = f (a) +
f 0 (a)
f 00 (a)
f (n) (a)
(x − a) +
(x − a)2 + · · · +
(x − a)n .
1!
2!
n!
The nth partial sum Tn (x) is a polynomial called the nth degree Taylor polynomial for
f (x) centered at x = a.
Example: Find the first, second, and third degree Taylor polynomials for f (x) = ex centered
at x = 0.
The Maclaurin series for f (x) = ex is
ex =
∞
X
xn
n=0
n!
=1+x+
x2 x3
+
+ ··· .
2
3!
So the first, second, and third degree Taylor polynomials are
T1 (x) = 1 + x,
x2
,
2
x2 x3
T3 (x) = 1 + x +
+ .
2
6
T2 (x) = 1 + x +
Example: Find the first and second degree Taylor polynomials for f (x) =
x = 4.
The derivatives of f (x) are
√
x
1
f 0 (x) = √
2 x
1
f 00 (x) = − 3/2
4x
f (x) =
f (4) = 2
1
f 0 (4) =
4
f 00 (4) = −
1
.
32
√
x centered at
So the first and second degree Taylor polynomials are
1
T1 (x) = 2 + (x − 4),
4
1
1
T2 (x) = 2 + (x − 4) − (x − 4)2 .
4
64
Example: Find the first, second, and third degree Taylor polynomials for f (x) = sin x cenπ
tered at x = .
3
The derivatives of f (x) are
f (x) = sin x
f 0 (x) = cos x
f 00 (x) = − sin x
f 000 (x) = − cos x
√
3
2
3
1
0 π
f
=
3
2√
π
3
00
= −
f
2
3
1
000 π
f
= − .
3
2
f
π =
So the first, second, and third degree Taylor polynomials are
√
3 1
π
T1 (x) =
+
x−
,
2
3
√2
√ 3 1
π
3
π 2
T2 (x) =
+
x−
−
x−
,
2
3
4
3
√
√2
π
π 2
1 π 3
3 1
3
+
x−
−
x−
−
x−
.
T3 (x) =
2
2
3
4
3
12
3
Note: Since Taylor polynomials are the partial sums of a Taylor series, they can be used
to approximate f (x) near x = a. If the nth degree Taylor polynomial Tn (x) is used to
approximate f (x) near x = a, then the remainder of the approximation is
Rn (x) = f (x) − Tn (x).
How good is this approximation? How large should we take n to achieve a desired accuracy?
These questions can be answered using Taylor’s Inequality.
Theorem: (Taylor’s Inequality)
If |f (n+1) (x)| ≤ M for |x − a| < R, then the remainder Rn (x) of the Taylor series satisfies
|Rn (x)| ≤
M
|x − a|n+1 .
(n + 1)!
Example: Consider the function f (x) = ln x.
(a) Find the third degree Taylor polynomial for f (x) centered at x = 3.
The derivatives of f (x) are
f (x) = ln(x)
1
f 0 (x) =
x
1
00
f (x) = − 2
x
2
000
f (x) = 3
x
f (3) = ln(3)
1
f 0 (3) =
3
1
00
f (3) = −
9
2
000
.
f (3) =
27
So the third degree Taylor polynomial is
1
1
1
T3 (x) = ln(3) + (x − 3) − (x − 3)2 + (x − 3)3 .
3
18
81
(b) Use Taylor’s Inequality to estimate the accuracy of the approximation f (x) ≈ T3 (x)
for 2 ≤ x ≤ 4.
The fourth-order derivative of f (x) is
f (4) (x) = −
6
.
x4
On the interval 2 ≤ x ≤ 4,
|f (4) (x)| =
6
6
3
≤ 4 = .
4
x
2
8
By Taylor’s Inequality, on the interval 2 ≤ x ≤ 4,
3 1
1
1
|R3 (x)| ≤
|x − 3|4 = |x − 3|4 ≤ .
8 4!
64
64
Example: Use the Alternating Series Estimation Theorem to estimate the range of values of
x for which the approximation
x2 x4
cos x ≈ 1 −
+
2
24
is accurate to within 0.005.
The Maclaurin series for f (x) = cos x is the alternating series
cos x =
∞
X
(−1)n x2n
n=0
(2n)!
=1−
x2 x4 x6
+
+
− +··· .
2
4!
6!
By the Alternating Series Estimation Theorem, the error in approximating cos x by the first
three nonzero terms of its Maclaurin series is at most
6
6
x = x .
6! 720
To ensure the indicated accuracy, set
x6
≤ 0.005
720
x6 ≤ 3.6
|x| ≤ 1.2380.
The largest such interval is [−1.2380, 1.2380].
Example: Find the third degree Taylor polynomial for f (x) = sin x centered at x =
T3 (x) to estimate sin 35◦ correct to five decimal places.
The derivatives of f (x) are
f (x) = sin x
f 0 (x) = cos x
f 00 (x) = − sin x
f 000 (x) = − cos x
f
π 6
π
=
1
2
√
3
2
6
1
00 π
f
= −
6
2
√
π
3
= −
.
f 000
6
2
f0
=
So the third degree Taylor polynomial is
√ √ 1
3
π 1 π 2
3
π 3
T3 (x) = +
x−
−
x−
−
x−
.
2
2
6
4
6
12
6
To find sin 35◦ , convert to radian measure
35π
◦
sin 35 = sin
180
7π
= sin
36
√ 2 √ 3
1
3 7π π
1 7π π
3 7π π
≈
+
−
−
−
−
−
2
2
36
6
4 36
6
12 36
6
√ √ 1
3 π
1 π 2
3 π 3
=
−
+
−
2
2 36
4 36
12 36
≈ 0.57358.
π
. Use
6