Name:
Power Series
I. The series
∞
P
xn = 1+x+x2 +x3 +· · · is geometric with a = 1 and r = x. Thus, it converges
n=0
1
to 1−x
when |x| < 1. Thus, R = 1 and the interval of convergence is (−1, 1). Since we did
not use ratio test, you do not need to test endpoints. Use facts about geometric series to
determine a power series representation and interval of convergence for the functions
1
(a) f (x) = 3−x
(b) f (x) =
2
3−x
(c) f (x) =
x
9+x2
II. For more complicated power series, we often have to differentiate or integrate. This makes
1
the function look similar to 1−x
and allows us to write a power series representation. For
example,
1
d
=
(1 + x)2
dx
=−
−1
1+x
d
=−
dx
1
1 − (−x)
∞
d X
=−
(−x)n
dx n=0
∞
∞
∞
X
X
d X
d
(−1)n (−x)n = −
(−1)n (x)n = −
(−1)n nxn−1
dx n=0
dx
n=0
n=0
∞
X
=
(−1)n+1 nxn−1
n=0
This series is no longer geometric, but its radius of convergence is the same as the series we
differentiated. Thus, R = 1. Use a similar technique to find a power series representation
and radius of convergence for the given functions.
(a) f (x) = ln (1 − x)
(b) f (x) =
x
(1+4x)2
2
III. Knowing how to represent functions as power series allows us to integrate functions we could
not integrate by other methods. For instance,
∞
X
1
1
=
(−x7 )n ⇒
=
1 + x7
1 − (−x7 ) n=0
Z X
Z
∞
∞
7n+1
X
1
7 n
n x
(−x
)
dx
=
C
+
(−1)
dx
=
1 + x7
7n + 1
n=0
n=0
We can evaluate this at x = 0 to get
the final answer is
∞
P
7n+1
1 + (−1)n x7n+1 .
n=0
1
1+07
=C+
∞
P
7n+1
(−1)n 07n+1 ⇒ 1 = C. Consequently,
n=0
This series converges when | − x7 | < 1. Thus, |x| < 1.
Use a similar technique to evaluate the integral as a power series. Also, determine its radius
of convergence.
R x
(a) 1−x
3 dx
(b)
R
ln 1−x
x
dx