Taylor and Maclaurin Series
If a function f has derivatives of all orders at x = a (i.e. if f (k) (a) exists for all k = 0, 1, 2, 3, . . .)
then we can construct the following power series in (x − a):
f (a) +
1 ′
1 ′′
1
f (a)(x − a) + f (a)(x − a)2 + . . . + f (n) (a)(x − a)n + . . .
1!
2!
n!
∞
∑
1 (r)
=
f (a)(x − a)r .
r!
r=0
This series is called the Taylor series of f about x = a. If a = 0 then we usually call this the
Maclaurin series of f .
We write
∞
∑
1 (r)
f (x) =
f (a)(x − a)r
r!
r=0
for
|x − a| < R
to mean that for all x = c, such that |c − a| < R, the sum
∞
∑
1 (r)
f (a)(c − a)r
r!
r=0
is finite and equals f (c). We call R the radius of convergence. If R = ∞ then the Taylor series
equals f (x) for all x ∈ R.
1. Standard Maclaurin Series
• For α ∈ R, (1 + x)α = 1 + αx +
•
•
•
•
•
•
α(α − 1) 2 α(α − 1)(α − 2) 3
x +
x + ...
2!
3!
x2 x3 x4
ln(1 + x) = x −
+
−
+ . . . for −1 < x ≤ 1
2
3
4
x2 x 3 x4
ex = 1 + x +
+
+
+ . . . for x ∈ R
2!
3!
4!
x3 x5 x 7
+
−
+ . . . for x ∈ R
sin(x) = x −
3!
5!
7!
x 2 x4 x 6
+
−
+ . . . for x ∈ R
cos(x) = 1 −
2!
4!
6!
x3 x5 x7
+
+
+ . . . for x ∈ R
sinh(x) = x +
3!
5!
7!
x 2 x4 x6
+
+
+ . . . for x ∈ R
cosh(x) = 1 +
2!
4!
6!
1
for |x| < 1
Example 1
Using the standard series, find the Maclaurin series for:
(i) f (x) = sin(3x),
(ii) f (x) = cos(x) + cosh(x),
(iii) f (x) = e2x ln(1 − x).
Include all terms up to the fifth power.
(i) Use the series for sin(x) with x replaced by 3x:
(3x)3 (3x)5
+
− ...
3!
5!
9x3 81x5
= 3x −
+
− ...
2
40
sin(3x) = 3x −
The above series is valid for 3x ∈ R, i.e. for x ∈ R.
(ii) The series for cos(x) and cosh(x) can be added term by term to obtain the series for their
sum:
(
) (
)
x2 x4 x6
x2 x4 x6
cos(x) + cosh(x) =
1−
+
−
+ ... + 1 +
+
+
+ ...
2!
4!
6!
2!
4!
6!
2x4
= 2+
+ ...
4!
x4
= 2+
+ ...
12
The above series is valid for x ∈ R.
(iii) The series for this product is obtained by multiplying the series for e2x and ln(1 − x),
respectively:
=
=
=
=
e2x ln(1 − x)
(
)(
)
(2x)2 (2x)3 (3x)4
(−x)2 (−x)3 (−x)4 (−x)5
1 + (2x) +
+
+
+ ...
(−x) −
+
−
+
− ...
2!
3!
4!
2
3
4
5
(
)(
)
4x3 27x4
x 2 x3 x4 x5
2
1 + 2x + 2x +
+
+ ...
−x −
−
−
−
− ...
3
8
2
3
4
5
x2 x 3 x4 x5
2x4 x5
2x5 4x4 2x5 27x5
−x −
−
−
−
− 2x2 − x3 −
−
− 2x3 − x4 −
−
−
−
− ...
2
3
4
5
3
2
3
3
3
8
5x2 10x3 13x4 649x5
−
−
−
− ...
−x −
2
3
4
120
The above series is valid for −1 < (−x) ≤ 1, i.e. for −1 ≤ x < 1. Note that at each stage
of the above calculations we kept only enough terms to ensure that we could get all terms
with powers up to and including x5 .
2
Example 2
Find the first three nonzero terms in the Taylor Series of f about x = a if
π
8
(ii) f (x) = (x + 1)3 , a = 2
(i) f (x) = cos(2x), a =
(iii) f (x) = x ln(x), a = 1
(i) f ′ (x) = −2 sin(2x), f ′′ (x) = −4 cos(2x)
(π )
(π ) (
π ) 1 ′′ ( π ) (
π )2
+ f′
x−
+ f
x−
f
+ ...
8
8
8
2
8
8
(π ) (
( π )) (
( π )) (
π) 1 (
π )2
= cos
+ −2 sin
x−
+
−4 cos
x−
+ ...
4
4
8
2
4
8
√ (
1
π) √ (
π )2
= √ − 2 x−
− 2 x−
+ ...
8
8
2
(ii) f ′ (x) = 3(x + 1)2 , f ′′ (x) = 6(x + 1)
1
f (2) + f ′ (2)(x − 2) + f ′′ (2)(x − 2)2 + . . .
2
1
= 33 + 3 × 32 (x − 2) + × 6 × 3(x − 2)2
2
= 27 + 27(x − 2) + 9(x − 2)2 + . . .
1 ′′′
1
, f (x) = − 2
x
x
1
1
f (1) + f ′ (1)(x − 1) + f ′′ (1)(x − 1)2 + f ′′′ (1)(x − 1)3 + . . .
2
6
1 1
1 −1
= 1 ln(1) + (ln(1) + 1)(x − 1) + × (x − 1)2 + ×
(x − 1)3 + . . .
2 1
6
1
1
1
= (x − 1) + (x − 1)2 − (x − 1)3 + . . .
2
6
(iii) f ′ (x) = ln(x) + 1, f ′′ (x) =
2. Error Estimation
The Taylor series of f about the point x = a can be truncated in order to provide an approximating polynomial for the function. We also obtain an expression describing the error between
this polynomial and f .
f (x) = f (a) + f ′ (a)(x − a) +
f (n) (a)
f ′′ (a)
(x − a)2 + . . . +
(x − a)n + Rn
2!
n!
= Pn (x) + Rn
• Pn (x) is the approximating (or Taylor) polynomial of degree n
• Rn is the truncating error
• The truncating error can be written in various ways; we shall use Lagrange’s form
Rn =
f (n+1) (c)
(x − a)n+1 ,
(n + 1)!
where c is some value that lies between a and x. We don’t know the value of c, but we
can obtain upper and/or lower bounds for the error.
3
π
Example 3 Find the Taylor polynomial of degree 3 for sin(x) about x = . Hence estimate
3
sin 64◦ and find the maximum error in this approximation.
(
◦
First note that 64 =
)
π
π
+
radians.
3 45
( π ) √3
f (x) = sin(x) ⇒ f
=
3
2
(π ) 1
f ′ (x) = cos(x) ⇒ f ′
=
3
2
√
(π )
3
(2)
(2)
f (x) = − sin(x) ⇒ f
=−
3
2
(π )
1
f (3) (x) = − cos(x) ⇒ f (3)
=−
3
2
60 + 4
180
π radians =
f (4) (x) = sin(x) ⇒ f (4) (c) = sin(c)
Therefore
and so
√ (
3 1(
π)
3
π )2
π )3
1 (
x−
−
x−
x−
sin(x) ≃
+
−
2
2
3
4
3
12
3
√
(π
π)
sin 64◦ = sin
+
√ 3 45 √ ( )
3 1π
3 π 2
1 ( π )3
=
−
+
−
2 45 √ 4 45
12 45
√2
3
π
3π 2
π3
=
+
+
+
2
90
8100
1093500
The truncating error is
R3 =
f (4) (c) (
π )4 sin(c) ( π )4
x−
=
,
4!
3
4!
45
. In order to estimate the maximum error we are required to find the
where π3 < c < 64π
180
maximum value of sin(c) (as all other terms are known). The best we can say is that sin(c) ≤ 1
and so
1 ( π )4
π4
R3 ≤
=
.
4! 45
24 × 454
4