Solution Set #8
Section 9.1
1.
Yes. Take N = 11. Then for n $ N, |
2.
a.
For 0 # x < 1, we know that xn
f n (x)
sin n x
n
n
n
N
10
v 0 as n v 4. Hence, for 0 # x < 1, we have
v 0. For x = 1, we have x = 1; so f
n
pointwise on [0,1] to the function f(x)
3.
& 0 | ' | sin n x | # 1 # 1 < 1 .
'
(1) = ½. Hence, { f n } converges
0 0#x<1
.
1
x'1
2
n
b.
No. For any n, we have f n is continuous at x = 1. Hence, for any n there exists a
neighborhood of x = 1 on which | fn (x) - ½ | < ¼. But, then on that
neighborhood, for x ú 1, we cannot have | fn (x) - 0 | < ¼.
a.
Let x 0 [0,1]. If x = 0, then f n (0) = 0 for all n. Hence, f n (0)
0 as n
4.
If x > 0, then there exists (by the Archimedian principle) an N such that for n $ N
we have 1/n < x. Hence, for n $ N we have f n (x) = 0 and, hence, f n (0)
0 as
v
n
v 4.
b.
No. For any n, we have that fn (x) = n > ½ for 0 < x < 1/n.
c.
For each n,
m
1
0
d.
fn
'
m
1/n
0
n
' 1 . Hence, lim
n
We have from a. and c. that lim
n
v4
m
1
0
fn
ú
v4
m
1
m
1
0
fn
v
v
'1
lim fn
0 n
v4
Section 9.2
1.
Let {fn} converge uniformly on E to f and let {gn} converge uniformly on E to g. We
claim that {fn + gn } converges uniformly on E to f + g. Let g > 0 there exists N1 such that
for n > N1 we have | fn (x) - f (x) | < g/2 for all x 0 E and there exists N2 such that for n >
N2 we have | gn (x) - g (x) | < g/2 for all x 0 E. Let N = max(N1 , N2). Then, for n > N2 we
have | (fn + gn) (x) - (f + g) (x) | # | fn (x) - f (x) | + | gn (x) - g (x) | < g/2 + g/2 = g for all
x 0 E. Hence, {fn + gn } converges uniformly on E to f + g.
2.
On [0, 4) we have that | gn (x) | # gn (0) = 1/n since gn is a non-negative decreasing
function of x. The sequence {1/n} converges monotonically to 0. Let g > 0 there exists N
such that for n > N we have 1/n < g. Hence, for n > N we have | gn (x) - 0 | = | gn (x) | # gn
(0) = 1/n < g for all x in [0, 4). Thus, { gn } converges uniformly to 0 on [0,4).
4.
a.
On [0,½] we have that | fn (x) | # fn (½) =
1
2n % 1
since fn is a non-negative
1
increasing function of x on [0,½]. Since the sequence {
} converges
2 %1
converges monotonically to 0, we have by the same argument as employed in 2.
above that { fn } converges uniformly to 0 on [0,½].
b.
n
The conclusion of problem 2b. in Section 9.1 is that for g = ¼ there does not exist
a N such that for n $ N we can have | fn (x) - 0 | < g on [0,1). Hence, we cannot
have uniform convergence on [0,1].
Section 9.3
1.
If the sequence { fn } were to be uniformly convergent on [0,4), then since each fn is
continuous on [0,4) we would have to have the limit function continuous on [0,4). But,
1 0 # x <1
the limit function is f (x)
'
1
2
x ' 1 which is not continuous at x = 1. Hence, we
0 x>1
cannot have uniform convergence on [0,4).
Section 9.4
1.
a.
Apply the Weierstrass M-Test with Mn = 1/n2. Then, on [0,4) we have
4
1
1
1
# 2 ' Mn . Since the series
converges, then by the
2
2
2
n'1 n
n %x
n
j
4
Weierstrass M-Test we have
3.
j
n'1
1
n2 % x2
converges uniformly on [0,4).
Apply the Weierstrass M-Test with Mn = | an |. Then, on [0,1] we have | an xn | # | an | =
4
Mn. Since by hypothesis series
4
have
jax
n'1
n
n
j | a | converges, then by the Weierstrass M-Test we
n'1
n
converges uniformly on [0,1].
4
4.
If the series
ja
n'1
n
converges, then the power series f (x)
'
4
jax
n'1
n
n
converges for x =
x0 = 1. By Theorem 9.4F the power series converges uniformly on [-x1 , x1] for all x1
such that 0 < x1 < x0 = 1. Hence, on each such interval [-x1 , x1] with 0 < x1 < x0 = 1 the
power series f is a continuous function. But every x in (-1,1) belongs to an interval of the
form [-x1 , x1] with 0 < x1 < x0 = 1, namely the interval [-x,x].
Section 9.5
4
1.
Hence, by Theorem 9.5A
2.
j a x converges uniformly on [0,1].
a
.
dx ' j
a x dx ' j
m
n%1
By the above problem 9.4.3, the power series
4
mj
1
0 n '0
Since the power series sin x
an x n
n
n
n'1
4
n'0
1
0
n
n
4
n
n'0
3
5
' x & x % x & ã has radius of convergence of infinity,
3!
5!
i.e., since the power series converges on all intervals (-S , S) for all S > 0, then by
Theorem 9.5C the sin x is differentiable on all intervals (-S , S) for all S > 0, and
4
2
4
x2
cos x ' (sin x)) ' 1 & 3
% 5 x & ã ' 1 & x % x & ã on (-4, 4).
3!
5!
2!
4!