Math1024 Answer to Homework 9
Exercise 4.4.3 (2)
1
√ = 1. The series converges for |x − 1| < 1, diverges for |x − 1| > 1. The radius of
lim n np
converges is 1.
Exercise 4.4.3 (3)
1
√ = 1. The series converges for |2x − 1| < 1, diverges for |2x − 1| > 1. The radius
lim n np
1
of converges is .
2
Exercise
(9)
4.4.3
q
√
an lim an+1 = (n+1)!
= n + 1 → ∞.
n!
R = +∞.
Exercise 4.4.3 (10)
((n + 1)!)2
(n + 1)2
1
(2n + 2)!
lim
=
lim
= . The radius of convergence is 4.
2
(n!)
(2n + 2)(2n + 1)
4
(2n)!
Exercise 4.4.3 (14)
2
an+1
2n+1 |x|(n+1) −1
=
= 2|x|2n+1
an
2n |x|n2 −1
Only when |x| < 1, the series absolutely converges, so the radius of convergence is 1.
Exercise s
4.4.4 (2)
n2
n
n+1
1
n
By lim
= lim 1 +
= e, the radius of convergence is e−1 .
n
n
Exercise 4.4.5
an+1
=
an
|x|2n+3
(n+1)!(n+2)!22n+3
|x|2n+1
(n)!(n+1)!22n+1
=
|x|
→ 0(n → ∞)
4(n + 1)(n + 2)
So the radius of convergence is ∞.
Exercise 4.4.6
an+1
=
an
|x|3n+3
(3n+2)!!!(n+2)!(3n+3)!!!
|x|3n
(3n−1)!!!(3n)!!!
=
|x|3
→ 0(n → ∞)
(3n + 2)(3n + 3)
So the radius of convergence is ∞.
Exercise 4.4.7
1
P
n
0
P(an + bn )xn converges for |x| ≤ min{R, R0}.
P(an −n bn )xn converges for |x| ≤ min{R,
PR }. n
an x .
P(−1) an x nhas the same radius R as
an (2x − 1) converges for |2x − 1| < R and diverges for |2x − 1| > R. The radius of
R
convergence is .
2
√
P
anP
x2n converges for |x2 | < R and diverges for |x2 | > R. The radius of convergence is R.
For
an xn+2 ,
|an+1 ||x|n+3
an+1
|x|
=
|an ||x|n+2
an
then
P then radius of convergence is R.
a2n x absolutely converges when
lim
p
n
|a2n ||x| < 1
Wand
> 1, the series diverges. so the P
radius of convergence is R2 .
P when nthe limit
P
an+2
x2 an xn has the same radiusp
R as
an x n .
p
Px =
n
n
n2
For
an x , if R 6= 0, +∞, when |x| < 1, |an ||x| → 0, and when |x| > 1, n |an ||x|n →
∞, so the
of convergence is 1.
P radius
n
For
an2 x , also by compare test, the radius of convergence is
lim p
n
1
|an2 |
So whenPR > 1, the radius is ∞; when R = 1, the radius is 1; when R < 1, the radius is 0.
For
a2n x2n ,
|a2n+2 | |a2n+1 |
|x|2
|a2n+2 ||x|2n+2 |
2
=
·
·
|x|
→
|a2n ||x|2n
|a2n+1 | |a2n |
R2
so the radius
of convergence is R.
P
2
For
an2 xn , by the root test
p
1
n
|an2 ||x|n = |an2 | n2 ·n |x|n
You can check when R−1 |x| < 1, the limit above is 0; when R−1 |x| = 1, the limit above is 1;
when R−1 |x| > 1, the limit above is ∞. So the radius of convergence is R.
Exercise 4.4.9 (3)
Substituting x by x2 in the Taylor expansion of sin x, we get
sin x2 = x2 −
x6 x10
x2n+1
+
− · · · + (−1)n
+ ··· .
3!
5!
(2n + 1)!
The radius of convergence is +∞.
2
Exercise 4.4.9 (6)
π
π
sin 2x = sin 2 x −
+ π = − sin 2 x −
2
2
2n+1
π 3 25 π 5
π 2n+1
π 23 n+1 2
+
x−
−
x−
+ · · · + (−1)
x−
+ ··· .
=− x−
2
3!
2
5!
2
(2n + 1)!
2
The radius of convergence is +∞.
Exercise 4.4.12
The Airy function in Exercise 4.4.6 is
A(x) = 1 +
∞
X
n=1
Q
∞
X
x3n nk=1 (3k − 2)
x3n
=1+
(3n)!
(3n)!
n=1
Qn
k=1 (3k − 2)
Then it’s easy to find that
Qn
∞
3n−2
X
x
k=1 (3k − 2)
A00 (x) =
(3n − 2)!
n=1
for all x.
and
A00 (x) = xA(x).
Exercise 4.4.14(2)
By taking derivative terms by term and multiplying by x with respect to the following series,
we get for |x| < 1,
+∞
X
n=0
+∞
X
nxn−1
n=0
+∞
X
nx
nx
n=0
x(3 − x)
,
(1 − x)3
0
x(3 − x)
3 + 7x − 2x2
=
=
,
(1 − x)3
(1 − x)4
n2 xn =
n=0
+∞
X
3 n−1
n=0
+∞
X
x
,
(1 − x)2
0
x
3−x
=
=
,
2
(1 − x)
(1 − x)3
nxn =
n=0
+∞
X
2 n−1
n=0
+∞
X
1
,
1−x
0
1
1
=
=
,
1−x
(1 − x)2
xn =
n 3 xn =
3x + 7x2 − 2x3
(1 − x)4
3
Exercise
(5) 4.4.14 2n+1
0
P
P∞ x
1
2n
= ∞
=
By
for |x| < 1, we get
n=0 x
n=0
2n + 1
1 − x2
Z x
∞
X
dt
1+x
x2n+1
1
=
= log
2
2n + 1
2
1−x
0 1−t
n=0
for|x| < 1.
Exercise 4.4.15 (2)
2n+1
P∞
x2n
1 P∞
1
n
n x
(−1)
=
(−1)
= (sin x − x).
n=1
n=1
(2n + 1)!
x
(2n + 1)!
x
for x ∈ (−∞, ∞)
Exercise 4.4.16 (2)
The power series
sin x P∞ (−1)n x2n
= n=0
converges for all x. Integrating term by term, we
x
(2n + 1)!
get
Z
x
0
∞
X
(−1)n x2n+1
sin t
dt =
t
(2n + 1)(2n + 1)!
n=0
for all x.
Exercise 4.4.16 (5)
The derivative (log(x +
√
√
1 + x2 ))0 = √
1
. For |x| < 1, we have
1 + x2
1
1 1 −1 4 1 1 −1 −3 6
1 + x2 = 1 + x 2 +
x +
x + ···
2
2! 2 2
3! 2 2 2
∞
∞
X
X
(2n)!
n 1 · 3 · 5 · · · (2n − 3) 2n
(−1)
(−1)n n
=1−
x =1−
x2n .
n
2
n!2
4
(2n
−
1)(n!)
n=1
n=1
Integrating term by term, we get
∞
X
√
2
log(x + 1 + x ) = x −
n=1
(−1)n (2n)!
x2n+1 ,
4n (4n2 − 1)(n!)2
The equality actually holds at ±1.
4
|x| < 1.