Jim Lambers
MAT 169
Fall Semester 2009-10
Lecture 16 Notes
These notes correspond to Section 8.7 in the text.
Taylor and Maclaurin Series, cont’d
Previously, we have learned how to compute Taylor series for certain functions by computing the
coefficients directly and recognizing the pattern that they fit. However, for many functions, this
approach is not feasible, because any pattern in the coefficients is too complex to recognize. We
now illustrate how other techniques for obtaining power series can be used, sometimes in tandem,
to obtain Taylor series for certain well-known functions.
Example Recall the sum of a geometric series,
∞
∑
𝑥𝑛 =
𝑛=0
1
1−𝑥
provided ∣𝑥∣ < 1. That is, the radius of convergence is 𝑅 = 1. Because the power series representation of a function at a given center is unique, it follows that the Maclaurin series for 1/(1 − 𝑥) is
the above geometric series.
Substituting −𝑥2 for 𝑥 yields
∞
∑
(−𝑥2 )𝑛 =
𝑛=0
∞
∑
(−1)𝑛 𝑥2𝑛 =
𝑛=0
1
1
=
.
2
1 − (−𝑥 )
1 + 𝑥2
∣𝑥2 ∣
Because
< 1 implies that ∣𝑥∣ < 1, this series has the same radius of convergence, 𝑅 = 1.
Furthermore, the above series is the Maclaurin series for 1/(1 + 𝑥2 ).
By anti-differentiating this series term-by-term, we can obtain a Maclaurin series for tan−1 𝑥,
for ∣𝑥∣ < 1, which is an antiderivative of 1/(1 + 𝑥2 ):
∫ 𝑥
1
tan−1 𝑥 =
𝑑𝑠 + 𝐶
2
0 1+𝑠
∫ 𝑥∑
∞
=
(−1)𝑛 𝑠2𝑛 𝑑𝑠 + 𝐶
0 𝑛=0
=
=
∞
∑
(−1)𝑛
∫
𝑥
𝑠2𝑛 𝑑𝑠 + 𝐶
0
𝑛=0
∞
∑
(−1)𝑛
𝑛=0
1
𝑥2𝑛+1
+ 𝐶.
2𝑛 + 1
The unknown constant 𝐶 is included because we have computed a specific antiderivative of 1/(1 +
𝑥2 ), and all antiderivatives of a given function differ from one another by a constant. To compute
𝐶, we substitute 𝑥 = 0, and because tan−1 0 = 0, we obtain the equation 0 = 0 + 𝐶, so we conclude
that 𝐶 = 0 and the Maclaurin series for tan−1 𝑥 is
−1
tan
𝑥=
∞
∑
(−1)𝑛
𝑛=0
𝑥2𝑛+1
,
2𝑛 + 1
∣𝑥∣ < 1.
□
Two Taylor series with the same center can trivially be added or subtracted in order to obtain
a Taylor series for the sum or difference of the corresponding functions. The radius of convergence
of the sum or difference is the minimum of the radii of convergence of the two series that are being
added or subtracted.
It is also possible to multiply or divide Taylor series to obtain new Taylor series, as the following
examples show.
Example Recall the Maclaurin series for 𝑒𝑥 and cos 𝑥:
𝑒𝑥 =
∞
∑
𝑥𝑛
𝑛=0
cos 𝑥 =
∞
∑
𝑛!
=1+𝑥+
(−1)𝑛
𝑛=0
𝑥2 𝑥3 𝑥4
+
+
+ ⋅⋅⋅,
2!
3!
4!
𝑥2𝑛
𝑥2 𝑥4 𝑥6
=1−
+
−
+ ⋅⋅⋅.
(2𝑛)!
2!
4!
6!
𝑒𝑥 cos 𝑥
The Taylor series for
can be obtained by multiplying these series, in the same way that
polynomials are multiplied. This is because for each non-negative integer 𝑛, the number of terms
from each series that are needed to compute the coefficient 𝑐𝑛 in the product is finite.
We have
(∞
)( ∞
)
2𝑚
∑
∑ 𝑥𝑛
𝑥
𝑚 𝑥
𝑒 cos 𝑥 =
(−1)
𝑛!
(2𝑚)!
𝑛=0
=
∞ ∑
∞
∑
𝑚=0
2𝑚+𝑛
𝑚 𝑥
(−1)
𝑛=0 𝑚=0
=
∞ ⌊ℓ/2⌋
∑
∑
(−1)𝑚
ℓ=0 𝑚=0
𝑛!(2𝑚)!
𝑥ℓ
(ℓ − 2𝑚)!(2𝑚)!
1
1
1
= 1 + 𝑥 − 𝑥3 − 𝑥4 − 𝑥5 + ⋅ ⋅ ⋅ ,
3
6
30
where ℓ = 2𝑚 + 𝑛, and, for any real number 𝑥, ⌊𝑥⌋ is the “floor” of 𝑥, which is the greatest integer
that is less than or equal to 𝑥. This series converges for all 𝑥, like the series for 𝑒𝑥 and cos 𝑥. □
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Example Recall the Maclaurin series for sin 𝑥 and cos 𝑥:
sin 𝑥 =
∞
∑
(−1)𝑛
𝑛=0
cos 𝑥 =
∞
∑
𝑥2𝑛+1
𝑥3 𝑥5 𝑥7
=𝑥−
+
−
+ ⋅⋅⋅,
(2𝑛 + 1)!
3!
5!
7!
(−1)𝑛
𝑛=0
𝑥2 𝑥4 𝑥6
𝑥2𝑛
=1−
+
−
+ ⋅⋅⋅.
(2𝑛)!
2!
4!
6!
To compute the Maclaurin series for tan 𝑥 = sin 𝑥/ cos 𝑥, we can use polynomial division to divide the Maclaurin series for sin 𝑥 by that of cos 𝑥, even though these series are not polynomials
themselves, but sums of infinitely many monomials.
First, we divide the leading terms of the two series. The first term of the series for sin 𝑥 is 𝑥,
while the first term of the series for cos 𝑥 is 1. Therefore, the first term of the quotient of the series,
and therefore the series for tan 𝑥, is 𝑥/1 = 𝑥.
Next, we subtract 𝑥 times the series for cos 𝑥 from that of sin 𝑥 to obtain the remainder:
∞
∑
(−1)𝑛
𝑛=0
∞
∞
𝑛=0
𝑛=0
∑
𝑥2𝑛+1
𝑥2𝑛+1 ∑
2𝑛
−
(−1)𝑛
=
(−1)𝑛+1
𝑥2𝑛+1 .
(2𝑛 + 1)!
(2𝑛)!
(2𝑛 + 1)!
Then, we divide the leading term of this series, which is 𝑥3 /3, by the leading term of the series for
cos 𝑥, which is 1, to obtain the second term of the quotient, which is 𝑥3 /3.
Continuing this process yields the series
1
2
17 7
62 9
tan 𝑥 = 𝑥 + 𝑥3 + 𝑥5 +
𝑥 +
𝑥 + ⋅⋅⋅
3
15
315
2835
The coefficients arise from the following pattern:
1
3
2
15
17
315
62
2835
1
1
+ ,
3! 2!
1
1
11
=
− +
,
5! 4! 3 2!
1
1
11
2 1
= − + −
+
,
7! 6! 3 4! 15 2!
1
1
11
2 1
17 1
=
− +
−
+
.
9! 8! 3 6! 15 4! 315 2!
= −
Note that each coefficient is expressed in terms of the previous coefficients. □
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Summary
∙ Given a Maclaurin series for a function 𝑓 (𝑥), a Maclaurin series for 𝑓 (𝑥𝑝 ), where 𝑝 is a positive
integer, can be obtained by substitution of 𝑥𝑝 for 𝑥 in the series for 𝑓 (𝑥).
∙ A Taylor series for 𝑓 (𝑥) can be differentiated term-by-term to obtain a Taylor series for 𝑓 ′ (𝑥).
The center and radius of convergence of the series for 𝑓 ′ (𝑥) are the same as that of 𝑓 (𝑥).
∙ ∫A Taylor series for 𝑓 (𝑥) can be anti-differentiated term-by-term to∫ obtain a Taylor series for
𝑥
𝑥
𝑓 (𝑠) 𝑑𝑠. The center and radius of convergence of the series for
𝑓 (𝑠) 𝑑𝑠 are the same as
that of 𝑓 (𝑥).
∙ A Taylor series for 𝑓 (𝑥) can be integrated term-by-term from 𝑎 to 𝑏 to compute the integral
from 𝑎 to 𝑏 of 𝑓 (𝑥), provided that (𝑎, 𝑏) lies with in the interval of convergence of the series.
∙ Two Taylor series, for functions 𝑓 (𝑥) and 𝑔(𝑥), with the same center 𝑥0 and radius of convergence 𝑅, can be multiplied to obtain a Taylor series for 𝑓 (𝑥)𝑔(𝑥) whose radius of convergence
is also 𝑅.
∙ Two Taylor series, for functions 𝑓 (𝑥) and 𝑔(𝑥), with the same center 𝑥0 , can be divided using
polynomial long division to obtain a Taylor series for 𝑓 (𝑥)/𝑔(𝑥), provided that the leading
coefficient 𝑐0 of the series for 𝑔(𝑥) is nonzero.
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