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Transcript
Math 500 – Intermediate Analysis
Homework 8 – Solutions
Fall 2014
23.1 For each of the following power series, find the radius of convergence and determine the exact interval of convergence.
P∞ n3 n
(d)
n=1 3n x .
Solution: Setting an =
n3
,
3n
we have that
1/n
|an |
n1/n
=
3
3
and hence limn→∞ |an |1/n = 13 . It follows that the radius of convergence is
R = 3. To determine the exact interval of convergence, we must check
P the3
convergence of the series at x = ±3. When x = 3, the series is ∞
n=1 n
th
3
which diverges by the n term test (limn→∞ n 6= 0). Similarly, at x = −3,
the series again diverges by the nth term test, and hence the interval of
convergence is precisely (−3, 3).
P∞ (−1)n n
x .
(h)
n=1
n2 4n
Solution: Set an =
(−1)n
n2 4n
and notice that
|an |1/n =
1
2
(n1/n ) 4
and hence limn→∞ |an |1/n = 41 . Hence, the radius of convergence is R = 4.
Moreover, at the endpoints we have convergence at x = 4 (by the integral
test) and convergence at x = −4 (by the alternating series test).
n n
.
23.4 For n = 0, 1, 2, . . ., let an = 4+2(−1)
5
an+1 (a) Find lim sup |an |1/n , lim inf |an |1/n , lim sup an+1
,
and
lim
inf
an .
an Solution: Notice that
6 n
, if n is even
5 n
an =
2
, if n is odd
5
1
In particular, the subsequential limits of the sequence |an |1/n are 65 and 25 .
Thus, lim sup |an |1/n = 56 and lim inf |an |1/n = 52 . Moreover, we have
1 n
an+1 , if n is even
3
an = 3n ,
if n is odd
an+1 an+1 and hence lim inf an = 0 and lim sup an = +∞.
P
P∞
n
(b) Do the series ∞
n=0 an and
n=0 (−1) an converge? Explain.
Solution: The easiest way to do this is to note that the sequence (an ) does
n
not converge to zero as n → ∞. Indeed, when n is even we get an = 56
which clearly does not converge to zero as n → ∞. Thus, neither of the
two series listed converges.
P
n
(c) Now, consider the power series ∞
n=0 an x with the coefficients an as above.
Find the radius of convergence and determine the exact interval of convergence of the series.
Solution: From part (a), the radius of convergence of the given power
series is R = 56 . To determine the exact interval of convergence, we must
now test the endpoints. At x = 56 , we have the series
n
∞ X
2 + (−1)n
3
n=0
which diverges by the nth term test (CHECK!). Similarly, at x = − 56 , we
have
n
∞
X
2 + (−1)n
n
(−1)
3
n=0
which again diverges by the nth term test (CHECK!).
Thus, the interval
of convergence for this power series is − 65 , 56 .
24.2 For x ∈ [0, ∞), let fn (x) = nx .
(a) Find f (x) = limn→∞ fn (x).
Solution: For a given x0 ∈ [0, ∞), the numerical sequence fn (x0 ) = xn0
clearly converges to zero as n → ∞. Thus, the sequence of functions (fn )
converges pointwise to the function f (x) = 0 on [0, ∞).
(b) Determine whether fn → f uniformly on [0, 1].
Solution: Let ε > 0 be given and notice that for any x ∈ [0, 1], we have
|fn (x) − f (x)| = |f (x)| =
2
|x|
1
≤ .
n
n
Thus choosing N = 1ε (which, notice, does NOT depend on x), it follows
for all n > N we have
|fn (x) − f (x)| ≤
1
< ε.
n
Since ε > 0 was arbitrary, we conclude that fn → f uniformly on [0, 1].
(c) Determine whether fn → f uniformly on [0, ∞).
Solution: The convergence here is not uniform. If it were, then there
would exist and N ∈ N such that |fn (x)| < 1 for all x ∈ [0, 1] and all
n > N . However, this is clearly can’t be the case since
|fN +1 (x)| > 1
for all x > N + 1. Thus, the we conclude that (fn ) does not converge to
the function f uniformly on the set [0, ∞].
1
24.3 Repeat exercise 24.2 for fn (x) = 1+x
n.
Solution: In class, we argued that xn → 0 pointwise on [0, 1), and that xn → 1
at x = 1. Moreover, for x > 1 we have xn → +∞. Thus, we see that the
sequence (fn ) converges pointwise to the function f : [0, ∞) → R defined by

 1, if x ∈ [0, 1)
1
, if x = 1
f (x) =
 2
0, if x > 1.
In particular, f is not continuous on either of the sets [0, 1] nor [0, ∞) and hence
the convergence can not be uniform on either of these sets by Theorem 24.3.
24.7 Let fn (x) = x − xn for x ∈ [0, 1].
(a) Does the sequence (fn ) converge pointwise on the set [0, 1]? If so, give the
limit function.
Solution: Given a fixed x0 ∈ [0, 1), we know from class that xn0 → 0 and
that xn0 → 1 for x0 = 1. Thus, the sequence of functions (fn ) converges
pointwise to the function
x,
if x ∈ [0, 1)
f (x) =
x − 1, if x = 1
on the set [0, 1].
3
(b) Does (fn ) converge uniformly on [0, 1]? Prove your assertion.
Solution: No, the convergence is not uniform. Indeed, notice that although each function fn is continuous on [0, 1], the pointwise limit f is
not continuous on [0, 1]. Thus, the convergence can not be uniform by
Theorem 24.3.
24.12 Let (fn ) be a sequence of functions defined on a set S ⊂ R. Then (fn ) converges
uniformly to a function f : S → R if and only if
lim sup |f (x) − fn (x)| = 0.
n→∞ x∈S
Solution: First, suppose that limn→∞ supx∈S |f (x) − fn (x)| = 0 and let ε > 0
be given. Then by definition there exists an N ∈ R such that
0 ≤ sup |f (x) − fn (x)| < ε
x∈S
for all n > N . By the definition of the supremum then, it follows that we have
|f (x) − fn (x)| < ε
for all n > N and for all x ∈ S. That is, the sequence (fn ) converges to f
uniformly on S.
Conversely, suppose now that fn → f uniformly on a set S and let ε > 0 be
given. Then by definition, given any ε > 0 there exists a N ∈ R such that
|f (x) − fn (x)| < ε
for all n > N and for all x ∈ S. Thus, for all n > N we have
sup |f (x) − fn (x)| ≤ ε
x∈S
and hence the numerical sequence (supx∈S |f (x) − fn (x)|)∞
n=0 must converge to
zero as n → +∞, i.e. we must have
lim sup |f (x) − fn (x)| = 0
n→∞ x∈S
as claimed.
24.15 Let fn (x) =
nx
1+nx
for x ∈ [0, ∞).
4
(a) Find f (x) = limn→∞ f (x).
Solution: First, notice that fn (0) = 0 for all n ∈ N, and hence fn (0) → 0
as n → ∞. Moreover, for x > 0 we have by the limit theorems that
x
nx
lim
= lim 1
=1
n→∞ 1 + nx
n→∞
+x
n
and hence the sequence of functions (fn ) converges pointwise to the function
0, if x = 0
f (x) =
1, if x > 0
on the set [0, ∞).
(b) Does fn → f uniformly on [0, 1]? Explain.
Solution: Notice that although each function fn is continuous on [0, 1],
the pointwise limit f is not continuous on [0, 1]. Thus, the convergence
can not be uniform by Theorem 24.3.
(c) does fn → f uniformly on [1, ∞)? Explain.
Solution: Notice that on [1, ∞) the pointwise limit f is continuous, so the
argument above does not hold. Instead, we use the result from exercise
24.12. To this end, we use calculus to find the global extrema of the
1
functions fn . First, notice that fn0 (x) = (1+nx)
2 > 0 for all x ∈ [1, ∞) and
hence the function fn is strictly increasing on the set [1, ∞). Moreover,
n
fn (1) = 1+n
and limx→+∞ fn (x) = 1. Thus, given any n ∈ N and x ∈ [1, ∞)
we have
n
.
|fn (x) − f (x)| = |fn (x) − 1| ≤ 1 −
n+1
n
Since limn→∞ n+1
= 1 by the limit theorems, it follows that given any
n
< ε provided that
ε > 0 there exists an N >∈ R such that 0 < 1 − 1+n
n > N and hence
|fn (x) − f (x)| < ε
for all x ∈ [1, ∞). This directly verifies that the convergence is indeed
uniform on [1, ∞).
P
P∞
n
25.6 (a) Show that if ∞
n=1 |an | < ∞, then the series
n=1 an x converges uniformly on [−1, 1] to a continuous function.
n
Solution: Notice
P∞that given any x ∈ [−1, 1], we have |an x | < |an |.
Since the series n=1 |an | converges by hypothesis,
it follows by the WeierP∞
strass M-Test that the series of functions n=1 an xn converges uniformly
P
on [−1, 1]. Since each partial sum kn=1 an xn is aP
continuous on [−1, 1], it
n
follows by Theorem 24.3 that the uniform limit ∞
n=1 an x must also be
continuous on [−1, 1] as claimed.
5
P
xn
(b) Does ∞
n=1 n2 represent a continuous
P∞ 1 function on [−1, 1]?
Solution: Since the
series
P∞ xn n=1 n2 is convergent by the integral test, the
series of functions n=1 n2 converges uniformly to a continuous function
on [−1, 1] by part (a).
P
xn
25.8 Show that ∞
n=1 n2 2n has radius of convergence 2, and that the series converges
uniformly to a continuous function on [−2, 2].
Solution: First, notice that if an = n212n , then
|an |1/n =
1
2
(n1/n ) 2
and hence limn→∞ |an |1/n = 21 . It follows therefore that the radius of convergence
of this series is R = 2. Secondly, notice that the series is convergent at the
endpoints x = ±2 and hence the interval of convergence is [−2, 2]. Now, to
see this represents a continuous P
function on [−2, 2] notice that |an xn | ≤ n12 for
1
all x ∈ [−2, 2]. Since the series ∞
is convergent by the integral test, the
n=1P
n2
n
Weierstrass M-Test implies the series ∞
n=1 an x converges uniformly on [−2, 2]
and hence the infinite series must represent a continuous function on [−2, 2] by
Theorem 24.3.
6