Math 315 Homework 10 Solutions
§ 11.1 #8: Find the radius of convergence of the following series:
∞
X
2n n
x
n+1
n=0
Solution: There are two methods we can use. The first is to use the ratio test directly. The second
is to use the formula in your book.
Method 1: First, recall the ratio test:
Ratio Test: Given a series
∞
P
n=0
an , let L = lim aan+1
. If L < 1, then the series converges. If
n
n→∞
L > 1, the series diverges. If L = 1, the test is inconclusive.
Now, let an =
2n
n
n+1 x .
Note
an+1 L = lim n→∞
an n+1 n+1
2
x
n + 1 = lim · n n
n→∞
n+2
2 x
2(n + 1)
= |x| lim
n→∞ n + 2
= 2|x|
In order for the series to converge, by the ratio test we need L < 1. Thus, we need 2|x| < 1, or
|x| < 1/2. Thus, the radius of convergence is R = 1/2.
Method 2: Suppose we are given a power series
∞
X
an (x − x0 )n .
n=0
To find the radius of convergence, we can use the formula
|an |
n→∞ |an+1 |
R = radius of convergence = lim
which was derived from the ratio test and is given in (1.8) on p. 535 of your book. So for us, we
have the series
∞
X
2n n
x
n+1
n=0
and so in this case we let an =
2n
n+1 .
Then the radius of convergence is given by
|an |
n→∞ |an+1 |
2n
n+2
= lim
· n+1
n→∞ n + 1
2
1
n+1
=
lim
2 n→∞ n + 2
= 1/2.
R = lim
§ 11.1 #16: Find the Taylor series for f (x) = ln x about the point x0 = 1. Find the radius of
convergence.
Solution: Note f (1) = ln(1) = 0. Also,
f 0 (x) = x1
f 00 (x) = − x12
f 000 (x) = x23
f (4) (x) = − 2·3
x4
f (5) (x) = 2·3·4
5
x
f (6) (x) = − 2·3·4·5
x6
⇒
⇒
⇒
⇒
⇒
⇒
f 0 (1) = 1
f 00 (1) = −1
f 000 (1) = 2
f (4) (1) = −3!
f (5) (1) = 4!
f (6) (1) = −5!
In general,
(n − 1)!
xn
for n ≥ 1
f (n) (1) = (−1)n+1 (n − 1)!
for n ≥ 1.
f (n) (x) = (−1)n+1
and so
Hence, we get that
f 000 (1)
f (n) (1)
f 00 (1)
(x − 1)2 +
(x − 1)3 + · · · +
(x − 1)n + · · ·
2!
3!
n!
1
2
3!
(−1)n+1 (n − 1)!
= 0 + 1(x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 + · · · +
(x − 1)n + · · ·
2!
3!
4!
n!
1
1
1
(−1)n+1
= (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 + · · · +
(x − 1)n + · · ·
2
3
4
n
∞
X
(−1)n+1
=
(x − 1)n .
n
n=1
f (x) = f (1) + f 0 (1)(x − 1) +
Let an =
(−1)n+1
.
n
Then the radius of convergence is given by
(−1)n+1
n + 1 |an |
n+1
= lim ·
= 1.
R = lim
= lim
n→∞ |an+1 |
n→∞
n→∞
n
(−1)n+2 n
§ 11.1 #20: Find the power series
∞
P
an xn for the function
n=0
f (x) = (1 + x2 )
∞
X
xn
.
n
n=1
Solution: Note
f (x) = (1 + x2 )
=
∞
X
xn
n
n=1
∞
∞
X
xn X xn+2
+
.
n
n
n=1
n=1
Now we want an xn in the second sum. Note, shifting the index, we obtain
∞
∞
X
X
xn+2
xn
=
n
n−2
n=1
n=3
(If this is unclear, we quickly go through the details: Let m = n + 2, or equivalently n = m − 2. So
when n = 1, which is the original starting value of the sum, m = 3. And plugging in m − 2 for n
inside the sum, we get
∞
∞
X
X
xn+2
xm
=
.
n
m−2
n=1
m=3
We can now replace m with n to get the desired sum.)
Thus, we have
f (x) =
∞
∞
X
xn X xn
+
.
n
n−2
n=1
n=3
We pull out the first two terms in the first sum to get
∞
∞
x2 X xn X xn
+
+
2
n
n−2
n=3
n=3
∞
x2 X 1
1
+
+
xn .
=x+
2
n
n
−
2
n=3
f (x) = x +
§ 11.2 #20: Verify that x0 = 0 is an ordinary point of the differential equation
(1 + x2 )y 00 + 2xy 0 − 2y = 0.
Then find two linearly independent solutions to the differential equation valid near x0 = 0. Estimate the radius of convergence of the solutions.
Solution: First, let us verify that x0 = 0 is an ordinary point of the differential equation. We
rewrite the differential equation in the form
2x 0
2
y −
y = 0.
y 00 +
2
1+x
1 + x2
Thus, the coefficients of y 0 and y, respectively, are given by
2x
2
P (x) =
, Q(x) = −
1 + x2
1 + x2
which are both analytic at x0 = 0. To show this in detail, note
1
1
=
2
1+x
1 − (−x2 )
which is a geometric series with a = 1 and r = −x2 , so
∞
X
1
=
(−x2 )n
1 + x2
n=0
=
∞
X
(−1)n x2n
n=0
2
provided |x | < 1 or |x| < 1. So
P (x) = 2x
∞
X
(−1)n x2n
n=0
=2
∞
X
(−1)n x2n+1
n=0
and
∞
X
Q(x) = −2
(−1)n x2n
n=0
provided |x| < 1. Since the series for P (x) and Q(x) converge for −1 < x < 1 which contains the
point x0 = 0, these function are analytic at x0 = 0, by Definition 1.30. Thus, x0 = 0 is an ordinary
point.
Alternatively, we can use Theorem 1.34 on p. 542 to verify that x0 = 0 is an ordinary point.
(This is much easier and just requires us to verify that the denominator 1 + x2 of P (x) and Q(x)
is a polynomial which is nonzero at x0 = 0.)
To estimate the radius of convergence of the solutions, note since the radius of convergence
for P (x) and Q(x) is R = 1, by Theorem 2.29 the radius of convergence of any solution to the
differential equation about x0 = 0 will have radius of convergence R ≥ 1.
Now we find two linearly independent solutions to the differential equation valid near x0 = 0.
We look for a solution of the form
∞
X
y(x) =
an (x − x0 )n
=
n=0
∞
X
n=0
an xn .
The first two derivatives of y are given by
y 0 (x) =
∞
X
y 00 (x) =
nan xn−1 ,
n=1
∞
X
n(n − 1)an xn−2 .
n=2
We plug these into our differential equation:
0 = (1 + x2 )y 00 + 2xy 0 − 2y
∞
∞
∞
X
X
X
= (1 + x2 )
n(n − 1)an xn−2 + 2x
nan xn−1 − 2
an xn
n=2
=
∞
X
n=1
n(n − 1)an xn−2 +
n=2
∞
X
n=0
n(n − 1)an xn + 2
n=2
∞
X
nan xn − 2
n=1
∞
X
an xn .
n=0
n
Now we re-index the first sum so that it contains an x , and we note that we can start the second
and third sums at n = 0. Thus,
0=
=
=
=
∞
X
(n + 2)(n + 1)an+2 xn +
n=0
∞
X
n=0
∞
X
∞
X
n(n − 1)an xn + 2
∞
X
nan xn − 2
n=0
n=0
n=0
h
xn (n + 2)(n + 1)an+2 + n(n − 1)an + 2nan − 2an
i
h
i
xn (n + 2)(n + 1)an+2 + (n + 2)(n − 1)an
n=0
∞
X
h
i
(n + 2)xn (n + 1)an+2 + (n − 1)an .
n=0
It follows that
(n + 1)an+2 + (n − 1)an = 0,
n≥0
So
an+2 = −
(1 − n)an
(n − 1)an
=
,
n+1
n+1
n ≥ 0.
Plugging in values for n, we see that
a2
a4
a6
a8
= a10 = a0
= − a32 = − a30
= − 3a54 = a50
= − 5a76 = − a70
a3
a5
a7
a9
=0
= − 2a43 = 0
= − 4a65 = 0
= − 6a87 = 0.
In general,
n+1
a0
a2n = (−1)
,
2n−1
a2n+1 = 0,
n≥1
n ≥ 1.
∞
X
an xn
Thus,
y(x) =
∞
X
a2n x2n +
n=0
∞
X
a2n+1 x2n+1
n=0
= a0 + a1 x +
= a0 + a1 x +
∞
X
a2n x2n
n=1
∞
X
(−1)n+1 a0 2n
x .
2n − 1
n=1
Now, choose y1 (x) so that a0 = y1 (0) = 1 and a1 = y10 (0) = 0. Then
y1 (x) = 1 +
∞
X
(−1)n+1 2n
x .
2n − 1
n=1
Choose y2 (x) so that a0 = y2 (0) = 0 and a1 = y20 (0) = 1. Then
y2 (x) =
∞
X
n=0
a2n+1 x2n+1 = x +
∞
X
a2n+1 x2n+1 = x + 0 = x.
n=1
By Proposition 2.14, y1 (x) and y2 (x) are two linearly independent solutions.