Math 3400
Homework 8
Section 5.1
Determine the radius of convergence.
∞
X
1.
(x − 3)n
n=0
Apply the ratio test. In this case
lim |
n→∞
(x − 3)n+1
| = |x − 3|.
(x − 3)n
The series converges if |x − 3| < 1, so the radius of convergence is r = 1.
2.
∞
X
n n
x
2n
n=0
Apply the ratio test. In this case
lim |
n→∞
(n + 1)2n xn+1
(n + 1) x
|
=
lim
|
| = |x/2|.
n→∞
n2n+1 xn
n 2
The series converges if |x/2| < 1, so the radius of convergence is r = 2.
3.
∞
X
x2n
n=0
n!
Apply the ratio test. In this case
lim |
n→∞
n!x2(n+1)
1
|
=
lim
|
x2 | = 0.
n→∞ n + 1
(n + 1)!x2n
The series converges for all x, so r = ∞.
4.
∞
X
2n xn
n=0
Apply the ratio test. In this case
lim |
n→∞
2n+1 xn+1
| = |2x|.
2n xn
The series converges when |2x| < 1, so r = 1/2.
For problems 9-12, determine the Taylor series
f (x) ∼
∞
X
f (n) (x0 )
n=0
n!
and determine the radius of convergence.
1
(x − x0 )n ,
9. f (x) = sin(x),
The formulas
d
sin(x) = cos(x),
dx
x0 = 0.
d
cos(x) = − sin(x),
dx
lead to
f (0) (0) = sin(0) = 0, f (1) (0) = cos(0) = 1, f (2) (0) = − sin(0) = 0, f (3) = − cos(0) = −1, . . . .
Thus
∞
X
x2n+1
x3 x5
sin(x) ∼ x −
+
+ ··· =
(−1)n
.
3!
5!
(2n
+
1)!
n=0
Apply the ratio test. In this case
lim |
n→∞
x2
x2n+3 (2n + 1)!
|
=
lim
|
| = 0.
n→∞ (2n + 3)(2n + 2)
x2n+1 (2n + 3)!
The series converges for all x, so r = ∞.
10. f (x) = ex ,
In this case since
d x
e
dx
x0 = 0.
= ex we have
f (n) (0) = e0 = 1
for all n. Thus the series is
∞
ex ∼ 1 + x +
X xn
x2 x3
+
+ ··· =
.
2!
3!
(n)!
n=0
Apply the ratio test. In this case
lim |
n→∞
xn+1 n!
x
| = lim |
| = 0.
n
n→∞ n + 1
x (n + 1)!
The series converges for all x, so r = ∞.
11. f (x) = x,
x0 = 1.
The Taylor series for a polynomial is simply that polynomial, written with the appropriate
center. Since the series has only finitely many terms, it converges for all 1. In this case the
series is
x = (x − 1) + 1 = 1 + (x − 1).
12. f (x) = x2 ,
x0 = −1.
This time the identity is
x2 = (x + 1)2 − 2(x + 1) + 1.
2
Rewrite as series in xn .
∞
X
21.
n(n − 1)an x
n−2
∞
X
=
(n + 2)(n + 1)an+2 xn .
n=2
n=0
∞
X
22.
an x
n+2
∞
X
=
n=0
23. x
∞
X
nan xn−1 +
n=1
∞
X
n=2
ak x k =
∞
X
nan xn +
n=1
k=0
an−2 xn .
∞
X
ak xk = a0 +
∞
X
(n + 1)an xn
n=1
k=0
Section 5.2
Find the recurrence relation, etc. For x0 = 0 the series are
y(x) =
∞
X
n
an x ,
0
y (x) =
∞
X
nan x
n−1
00
,
y (x) =
n(n − 1)an xn−2 .
n=2
n=1
n=0
∞
X
2. y 00 − xy 0 − y,
x0 = 0.
Plugging in the series gives
∞
X
n(n − 1)an x
n−2
−x
n=2
∞
X
nan x
n−1
−
n=1
∞
X
an xn = 0.
n=0
Rewriting in a consistent form gives
∞
X
n
(n + 2)(n + 1)an+2 x −
n=0
∞
X
n
nan x −
∞
X
n=1
an xn = 0.
n=0
Since the second series starts with n = 1,
2a2 − a0 = 0,
(n + 2)(n + 1)an+2 − nan − an = 0,
n = 1, 2, 3, . . . .
Solving the recurrence relation for the term of highest index gives
an+2 =
an
.
n+2
5. (1 − x)y 00 + y
x0 = 0.
Plugging in the series gives
(1 − x)
∞
X
n(n − 1)an xn−2 +
n=2
∞
X
n=0
3
an xn = 0.
Split the first term in two to get
∞
∞
∞
X
X
X
n(n − 1)an xn−2 −
n(n − 1)an xn−1 +
an xn = 0.
n=2
n=2
n=0
Rewriting in a consistent form gives
∞
∞
∞
X
X
X
n
n
(n + 2)(n + 1)an+2 x −
(n + 1)nan+1 x +
an xn = 0.
n=0
n=1
n=0
Since the second series starts with n = 1,
(n + 2)(n + 1)an+2 − (n + 1)nan+1 + an = 0,
2a2 + a0 = 0,
n = 1, 2, 3, . . . .
Solving the recurrence relation for the term of highest index gives
n(n + 1)an+1 − an
an+2 =
.
(n + 2)(n + 1)
8. xy 00 + y 0 + xy = 0,
x0 = 1,
expressed as a series with center x0 = 1. This means assuming
∞
∞
∞
X
X
X
k−1
00
k
0
k(k − 1)ck (x − 1)k−2 .
kck (x − 1) , y (x) =
ck (x − 1) , y (x) =
y(x) =
k=2
k=1
k=0
Express the coefficients as polynomials in x − 1, getting
[1 + (x − 1)]y 00 + y 0 + [1 + (x − 1)]y = 0.
Putting in the series gives
∞
∞
X
X
k−2
k(k − 1)ck (x − 1)k−1
k(k − 1)ck (x − 1)
+
k=2
k=2
+
∞
X
k−1
kck (x − 1)
+
k=1
∞
X
k
ck (x − 1) +
k=0
∞
X
ck (x − 1)k+1 = 0.
k=0
After the appropriate substitutions we get
∞
∞
X
X
(k + 2)(k + 1)ck+2 (x − 1)k +
k(k + 1)ck+1 (x − 1)k
k=0
+
∞
X
k=1
k
(k + 1)ck+1 (x − 1) +
k=0
∞
X
k
ck (x − 1) +
k=0
∞
X
ck−1 (x − 1)k = 0.
k=1
Since the series do not start with the same index, we first have
2c2 + c1 + c0 = 0,
and then
(k + 2)(k + 1)ck+2 + k(k + 1)ck+1 + (k + 1)ck+1 + ck + ck−1 = 0.
This simplifies to
ck+2 = −
(k + 1)2 ck+1 + ck + ck−1
.
(k + 2)(k + 1)
4