SOLUTION OF HW11
MINGFENG ZHAO
December 06, 2012
1. [20 Points] a. Prove
b. Prove
∞
X
∞
X
k=1
sin
x
converges uniformly on any bounded interval.
k2
e−kx converges uniformly on any closed subinterval of (0, ∞).
k=0
Proof. a. Let I be any bounded interval, then there exists some constant M > 0 such that
|x| ≤ M,
∀x ∈ I
Then we know that
x sin 2 k
Since
∞
X
M
k=1
k2
≤
|x|
k2
≤
M
,
k2
∀x ∈ I, ∀k ≥ 1
converges, by M -test, we know that
∞
X
k=1
sin
x
converges uniformly on I.
k2
b. Let I = [a, b] be any closed subinterval of (0, ∞), then a > 0. Then we know that
e−kx
≤ e−ka
=
1
ea
k
,
∀x ∈ I, ∀k ≥ 1
k
∞ ∞
X
X
1
1
Since a > 0, then a < 1, so
converges,
by
M
-test,
we
know
that
e−kx converges
e
ea
k=1
k=1
uniformly on I.
2. [0 Points] Suppose the power series
∞
X
ak (x − x0 )n converges for x = x1 with x1 6= x0 . Then the
k=0
series converges uniformly for all x ∈ [x0 −h, x0 +h], where h is any positive number with h < |x1 −x0 |.
1
2
MINGFENG ZHAO
Proof. Let R be the convergence radius of the power series
∞
X
ak (x − x0 )n . Since
k=0
∞
X
ak (x − x0 )n
k=0
converges for x = x1 with x1 ≥ x0 , then we know that 0 < |x1 − x0 | ≤ R. By the definition of R, we
know that
R=
1
p
n
lim supn→∞
|an |
For 0 < h < |x1 − x0 | ≤ R, then there exists some > 0 such that
lim sup
p
n
|an | <
n→∞
1
1
<
h+
h
So there exists some N ∈ N such that for all n ≥ N , we have
p
n
|an | <
1
h+
So we know that
|ak (x − x0 )n |
1
· hn
(h + )n
n
h
=
, ∀n ≥ N, |x − x0 | ≤ h
h+
≤
n
∞ ∞
X
X
h
Since
converges, by the M -test, we know that
an (x − x0 )n uniformly converges
h+
n=N
n=N
∞
X
in |x − x0 | ≤ h. Hence we know that the series
an (x − x0 )n converges uniformly for all x ∈
n=1
[x0 − h, x0 + h]
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu