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8.6. POWER SERIES
While we could use the methods of Section 8.5 and differentiate f (x) =
1
1 + x 2 several
times to look for patterns and find the Taylor series for f (x), we seek an alternate
approach because of how complicated the derivatives of f (x) quickly become.
(a) What is the Taylor series expansion for g(x) =
1
1−x ? What is the interval of
convergence of this series?
(b) How is g(−x 2 ) related to f (x)? Explain, and hence substitute −x 2 for x in the
power series expansion for g(x). Given the relationship between g(−x 2 ) and
f (x), how is the resulting series related to f (x)?
(c) For which values of x will this power series expansion for f (x) be valid? Why?
⊳
In a previous section we determined several important Maclaurin series and their
intervals of convergence. Here, we list these key functions and remind ourselves of their
corresponding expansions.
sin(x) =
∞
k=0
(−1) k x 2k+1
(2k + 1)!
for −∞ < x < ∞
cos(x) =
∞
k=0
(−1) k x 2k
(2k)!
for −∞ < x < ∞
e x =
∞
k=0
x k
k!
for −∞ < x < ∞
1
1 − x
=
∞
k=0
x
k
for −1 < x < 1
As we saw in Activity 8.29, we can use these known series to find other power series
expansions for related functions such as sin(x 2 ), e 5x 3 , and cos(x 5 ). Another important
way that we can manipulate power series is illustrated in the next activity.
Activity 8.30.
Let f be the function given by the power series expansion
f (x) =
∞
k=0
(−1)
k x 2k
(2k)!
.
(a) Assume that we can differentiate a power series term by term, just like we can
differentiate a (finite) polynomial. Use the fact that
f (x) = 1 −
x 2
2!
+
x 4
4!
−
x 6
6!
+ · · · + (−1)
k x 2k
(2k)!
+ · · ·