LECTURE 33: POWER SERIES AND TAYLOR SERIES
MINGFENG ZHAO
March 30, 2015
∞
X
|ck+1 |
, then the radius of convergence of the power
|ck |
k=0
∞
∞
X
X
1
series
ck (x − a)k is R = . In particular, the power series
ck (x − a)k absolutely converges for all |x − a| < R,
r
k=0
k=0
∞
X
and the power series
ck (x − a)k diverges for all |x − a| > R.
Theorem 1. For the power series
ck (x − a)k , let r = lim
k→∞
k=0
Example 1. Find the radius of convergence of the series
∞
X
(−1)k 2k+1 x2k+1 and evaluate the sum of this series.
k=0
Notice that
∞
X
(−1)k 2k+1 x2k−1 =
k=0
∞
X
2(−2)k · x · (x2 )k = 2x ·
k=0
∞
X
(−2)k (x2 )k .
k=0
Let y = x2 , then
∞
X
(−1)k 2k+1 x2k−1 = 2x ·
∞
X
(−2)k y k .
k=0
k=0
k
Let ck = (−2) for all k ≥ 0, then
r
=
=
=
lim
|ck+1 |
|ck |
lim
|(−2)k+1 |
(−2)k
k→∞
k→∞
2.
√
1
2
1
1
2
So the radius of convergence of the
(−1) (−2) y is R = = , that is, |y| = |x | < . Then |x| <
, which
r
2
2
2
k=0
√
∞
X
2
implies that the radius of convergence of the
(−1)k 2k+1 x2k+1 is R =
. Moreover, we have
2
∞
X
k
k k
k=0
∞
X
(−1)k 2k+1 x2k+1
=
2x ·
k=0
∞
X
(−2)k (x2 )k
k=0
=
2x ·
∞
X
k=0
1
(−2x2 )k
2
MINGFENG ZHAO
=
=
Theorem 2. Suppose the power series
∞
X
2x
1 − (−2x2 )
2x
.
1 + 2x2
Since | − 2x2 | < 1
ck (x − a)k converges to f (x) for |x − a| < R, then
k=0
I. Then f (x) is differentiable for |x − a| < R, and
f 0 (x) =
∞
X
kck (x − a)k−1 ,
for all |x − a| < R.
k=0
II. Then
Z
f (x) dx =
∞
X
ck
k=0
Example 2. Find the power series of
(x − a)k+1
+ C,
k+1
for all |x − a| < R.
x
centered at 0.
1 + 4x
In fact, we have
x
1 + 4x
=
=
=
1
1 + 4x
1
x·
1 − (−4x)
x·
x·
∞
X
(−4x)k
if | − 4x| < 1, that is, |x| <
k=0
=
x·
X
1
4
(−4)k xk
k=0
=
∞
X
(−4)k xk+1
k=0
=
∞
X
(−4)k−1 xk .
k=1
Hence the power series of
∞
X
x
centered at 0 is
(−4)k−1 xk , and the interval of convergence is (−1/4, 1/4).
1 + 4x
k=1
Example 3. Let f (x) =
1
, then
1−x
∞
f (x) =
X
1
=
xk = 1 + x + x2 + x3 + · · · ,
1−x
for all |x| < 1.
k=0
By Theorem 2, we have
f 0 (x)
=
∞
∞
k=0
k=1
X
X
1
k−1
=
kx
=
kxk−1
(1 − x)2
LECTURE 33: POWER SERIES AND TAYLOR SERIES
∞
X
=
(k + 1)xk ,
3
for all |x| < 1.
k=0
Hence we have,
∞
X
1
=
(k + 1)xk ,
(1 − x)2
for all |x| < 1.
k=0
Example 4. Find the power series of tan−1 (x) centered at 0.
Notice that
Z
So let’s find the power series of
1
dx = tan−1 (x) + C.
1 + x2
1
centered at 0. In fact, we have
1 + x2
1
1 + x2
1
1 − (−x2 )
=
=
∞
X
(−x2 )k
if | − x2 | < 1, that is, |x| < 1
k=0
=
∞
X
(−1)k x2k .
k=0
Integrate on the both sides, by Theorem 2, we have
Z
tan−1 (x)
=
=
1
dx =
1 + x2
∞
X
(−1)k
Z
Z X
∞
(−1)k x2k dx
k=0
x2k dx
k=0
=
∞
X
(−1)k
· x2k+1 + C.
2k + 1
k=0
Since tan−1 (0) = 0, then C = 0. Hence the power series of tan−1 (x) centered at 0 is
∞
X
(−1)k
· x2k+1 .
2k + 1
k=0
Remark 1. By By Example 4, then
∞
X (−1)k
π
= tan−1 (1) =
.
4
2k + 1
k=0
Example 5. Find the power series of ln
1+x
1−x
centered at 0.
Notice that
ln
1+x
1−x
= ln(1 + x) − ln(1 − x).
4
MINGFENG ZHAO
For the Example 10 in Lecture 32 , we have
ln(1 − x) = −
∞
X
xk
k=1
k
,
for all |x| < 1.
Then we get
ln
1+x
1−x
ln(1 + x) − ln(1 − x)
=
= −
∞
X
(−x)k
k=1
k
k=1
∞
X
=
k=0
Hence the power series of ln
1+x
1−x
∞
X
xk
k=1
∞
X
(−1)k+1 + 1
=
k
+
centered at 0 is
∞
X
k=0
k
· xk
2
· x2k+1 .
2k + 1
2
· x2k+1 .
2k + 1
Taylor series
Definition 1. Suppose a function f (x) has derivatives of all orders on an interval centered at point a, then the Taylor
series for f (x) centered at a is:
∞
X
f (k) (a)
k=0
k!
· (x − a)k := f (a) + f 0 (a)(x − a) +
f 00 (a)
f (3) (a)
(x − a)2 +
(x − a)3 + · · · .
2!
3!
A Taylor series centered at 0 is called a Maclaurin series.
Remark 2. The Taylor series for f (x) centered at a is exactly the power series of f (x) centered at a, and the Maclaurin
series for f (x) centered at a is exactly the power series of f (x) centered at 0.
Example 6. Find the Maclaurin series of ex and find its interval of convergence.
Let f (x) = ex , then
f 0 (x) = ex ,
f 00 (x) = ex ,
and f (k) (x) = ex ,
for all k ≥ 0.
So we get f (k) (0) = e0 = 1 for all k ≥ 0. Hence the Maclaurin series of ex is:
∞
X
f (k) (0)
k=0
k!
· xk =
∞
X
1 k
x2
x3
x4
x =1+x+
+
+
+ ··· .
k!
2!
3!
4!
k=0
LECTURE 33: POWER SERIES AND TAYLOR SERIES
Notice that
r
=
lim
k→∞
1
(k+1)!
1
k!
k!
(k + 1)!
1
= lim
k→∞ k + 1
=
=
So the radius of convergence is R =
lim
k→∞
0.
1
= ∞, which implies that the interval of convergence is (−∞, ∞).
0
∞
∞
X
X
1
1
1
, since
=
is
xk , then the Maclaurin series of
xk .
1−x
1−x
1−x
k=0
k=0
1
, then
Also we can prove it by Definition 1, let f (x) =
1−x
Example 7. For
f (x)
=
(1 − x)−1
f 0 (x)
=
−1 · (1 − x)−2 · (−1) = (1 − x)−2
f 00 (x)
=
(−2) · (1 − x)−3 · (−1) = 2(1 − x)−3 = 2!(1 − x)−3
f (3) (x)
=
2 · (−3) · (1 − x)−4 · (−1) = 3!(1 − x)−4
f (4) (x)
=
3! · (−4) · (1 − x)−5 · (−1) = 4!(1 − x)−5
..
.
f (k) (x)
..
.
=
k!(1 − x)−(k+1) .
So we get f (k) (0) = k!. By Definition 1, then the Maclaurin series of
∞
X
f (k) (0)
k!
k=0
Example 8. Find the Maclaurin series of e−2x
2
+1
· xk =
∞
X
k!
k=0
k!
xk =
1
is
1−x
∞
X
xk .
k=0
.
By Example 6, we know that
ex =
∞
X
xk
k=0
k!
,
for all −∞ < x < ∞.
So we get
2
e−2x
+1
= e · e−2x
2
5
6
MINGFENG ZHAO
= e·
∞
X
(−2x2 )k
k=0
= e·
∞
X
(−2)k
k=0
=
k!
∞
X
k=0
k!
· x2k
e · (−2)k 2k
·x .
k!
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca