Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Partial differential equation wikipedia, lookup

Chain rule wikipedia, lookup

Sobolev space wikipedia, lookup

Multiple integral wikipedia, lookup

Lp space wikipedia, lookup

Limit of a function wikipedia, lookup

Uniform convergence wikipedia, lookup

Distribution (mathematics) wikipedia, lookup

Lebesgue integration wikipedia, lookup

Function of several real variables wikipedia, lookup

Fundamental theorem of calculus wikipedia, lookup

Sobolev spaces for planar domains wikipedia, lookup

Divergent series wikipedia, lookup

Neumann–Poincaré operator wikipedia, lookup

Series (mathematics) wikipedia, lookup

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
```
Related documents