Math 140A Elementary Analysis Homework Questions 6
2.15
Alternating Series and Integral Tests
2 Determine which of the following series converge. Justify your answers.
∑
(a)
h
sin(
nπ in
)
6
(b)
∑
h
sin(
nπ in
)
7
4 Repeat question 2 for the following.
∞
(a)*
∑
√
n =2
1
n ln n
∞
(b)
ln n
n =2 n
∑
∞
(c)*
1
n(ln n)(ln ln n)
n =4
∑
∞
(d)
ln n
2
n =2 n
∑
6 * (a) Give an example of a divergent series ∑ an for which ∑ a2n converges.
(b) Prove that if ∑ an is a convergent series of nonnegative terms, then ∑ a2n also converges.
3
Continuity
3.17
Continuous Functions
(
2 Let f ( x ) =
4
0
x≥0
and g( x ) = x2 for all x. Thus dom( f ) = dom( g) = R.
x<0
(a) Determine the following functions: f + g, f g, f ◦ g, g ◦ f . Be sure to specify their domains.
(b) Which of the functions f , g, f + g, f g, f ◦ g, g ◦ f is continuous?
√
4 * Prove that the function x is continuous on its domain [0, ∞).
8 Let f and g be real-valued functions.
(a) Show that min( f , g) = 12 ( f + g) − 21 | f − g|.
(b) Show that min( f , g) = − max(− f , − g).
(c) Use (a) or (b) to prove that if f and g are continuous at x0 in R, then min( f , g) is continuous
at x0 .
10 Prove that the following functions are discontinuous at the indicated point x0 . You may use
either the Definition, or the e − δ property.
(
1 x>0
(a) f ( x ) =
, x0 = 0;
0 x≤0
(
sin( 1x ) x 6= 0
, x0 = 0;
(b) * g( x ) =
0
x=0


x>0
1
(c) sgn( x ) = 0
x = 0 , x0 = 0;


−1 x < 0
1
(d) P( x ) = 15 for 0 ≤ x < 1 and P( x ) = 15 + 13n for n ≤ x < n + 1, for x0 a positive integer.
12 * (a) Let f be a continuous real-valued function with domain ( a, b). Show that if f (r ) = 0 for
each rational number r ∈ ( a, b), then f ( x ) = 0 for all x ∈ ( a, b).
(b) Let f and g be continuous real-valued functions on ( a, b) such that f (r ) = g(r ) for each
rational number r ∈ ( a, b). Prove that f ( x ) = g( x ) for all x ∈ ( a, b).
14 (Hard but important: try this, and take the time to understand the answer when it’s released!)
p
For each rational number x, write x = q where p, q are integers with no common factors and
q > 0, and then define f ( x ) = 1q . Also define f ( x ) = 0 for all x ∈ R \ Q. Then f ( x ) = 1 for each
integer, and f ( 12 ) = f (− 12 ) = f ( 32 ) = · · · = 21 , etc. Show that f is continuous at each point of
R \ Q and discontinuous at each point of Q.
2
Math 140A Elementary Analysis Homework Answers 6
2.15
2
Alternating Series and Integral Tests
n
(a) If an = sin( nπ
)
, then a3+12m = 1 for all m, so there exists a subsequence of ( an ) which
6
n
diverges.
does not converge to zero. Thus an → 0, whence ∑ sin( nπ
6 )
(b) Let an = sin( nπ
7 ), then | an | = |sin( nπ/7)| ≤ sin(3π/7) ≤ 0.975 (draw the graph of Sine!)
for all n. Therefore | ann | ≤ 0.975n , and, since ∑ 0.975n is a convergent geometric series, ∑ an
converges by the comparison test.
4
(a) For n ≥ 4, ln n <
√
√
n =⇒
√
n ln n < n. Thus, by the comparison test,
∞
∞
1
1
1
1
1
1
< =⇒ ∑ √
=√
+√
+∑ √
n
n ln n
n ln n
3 ln 3 n=4 n ln n
2 ln 2
n =2
∞
=⇒
∑
n =2
√
∞
1
1
1
1
>√
+√
+∑
n ln n
3 ln 3 n=4 n
2 ln 2
which diverges to +∞.
(b) Setting u = ln x, we obtain
Z ∞
ln x
x
2
thus ∑∞
n =2
(c) Since
d
dx
ln n
n
dx =
4
ln 2
u dx = +∞
diverges. (Or use comparison test with ∑ n1 )
ln ln ln x =
Z ∞
Z ∞
1
,
x (ln x )(ln ln x )
whence
1
dx = ln ln ln x |4∞ = +∞
x (ln x )(ln ln x )
Therefore ∑∞
n =4
1
n(ln n)(ln ln n)
diverges to +∞.
(d) Substituting α = ln x and applying integration by parts, we obtain
Z ∞
ln x
2
thus ∑∞
n =2
6
(a) If an =
x2
ln n
n2
1
,
n2/3
dx =
Z ∞
ln 2
αe−α dα =
1
(ln 2 + 1) < +∞
2
converges.
1
then ∑ an diverges, but ∑ a2n = ∑ n4/3
converges.
(b) If ∑ an converges, then an → 0. Let e = 1, then ∃ N s.t.
n > N =⇒ an < 1 =⇒ a2n < an
Thus ∑ a2n = ∑n< N a2n + ∑n≥ N a2n < ∑n< N a2n + ∑n≥ N an which converges.
3
3.17
Continuous Functions
2 Let f ( x ) = 4 for x ≥ 0, f ( x ) = 0 for x < 0 and g( x ) = x2 for all x. Thus dom( f ) = dom( g) = R.
(
(
(
4 + x2 x ≥ 0
4x2 x ≥ 0
16 x ≥ 0
(a) ( f + g)( x ) =
, ( f g)( x ) =
, f ◦ g( x ) = 4, g ◦ f =
.
2
x
x<0
0
x<0
0
x<0
All domains are R.
(b) Only g, f g, and f ◦ g are continuous.
√
√
4 Fix a ≥ 0 and let e > 0 be given. Let δ = e a. Note that dom( x ) = R≥0 . Thus
√
√
√ √
e a
| x − a|
√ <√
√ ≤e
x− a = √
x ∈ dom( x ) and | x − a| < δ =⇒
x+ a
x+ a
√
whence x is continuous.
8
(a) Suppose that f ≤ g. Then
1
1
1
1
( f + g) − | f − g| = ( f + g) − ( g − f ) = f = min( f , g)
2
2
2
2
Conversely, if g ≤ f , then
1
1
1
1
( f + g) − | f − g| = ( f + g) − ( f − g) = g = min( f , g)
2
2
2
2
Hence result.
(b) Recall that max( f , g) = 12 ( f + g) + 12 | f − g|. Then
1
1
1
1
− max(− f , − g) = − (− f − g) − |− f + g| = ( f + g) − | f − g| = min( f , g)
2
2
2
2
(c) If f and g are continuous at x0 , then, by the Theorem from lectures, f + g and | f − g| are
coninuous, whence min( f , g) = 12 ( f + g) − 12 | f − g| is continuous at x0 .
10
(a) Let xn = 1/n. Then xn → 0, but f ( xn ) = 1 → 1 6= f (0), whence f is discontinuous at
x0 = 0.
Alternatively, let e = 1. For any δ > 0, there exist x ∈ (−δ, δ) for which f ( x ) = 1 6< e (i.e.
any x ∈ (0, δ)).
(b) Let xn = (2n+11/2)π . Then xn → 0, but f ( xn ) = 1 → 1 6= f (0), whence f is discontinuous at
x0 = 0.
Alternatively, let e = 1. For any δ > 0, there exists n ∈ N with (2n+11/2)π < δ, and thus
some x ∈ (−δ, δ) for which f ( x ) = 1 6< e.
(c) Proof identical to that for (a).
(d) Let xk = n + 1 − k+1 1 . Then xk → n + 1, but f ( xk ) = 15 + 13n → 15 + 13n 6= f (n + 1),
whence f is discontinuous at x0 = n + 1.
Alternatively, let x0 = n + 1 and e = 1 (or anything else ≤ 13). For any δ > 0, there exists
k ∈ N with k+1 1 < δ. Then x = n + 1 − k+1 1 ∈ ( x0 − δ, x0 + δ), but
| f ( x ) − f (n + 1)| = |15 + 13n − (15 + 13(n + 1))| = 13 6< e
4
12
(a) Let c ∈ ( a, b) and let e > 0 be given. Since f is continuous, there exists δ > 0 for which
x ∈ ( a, b) and | x − c| < δ =⇒ | f ( x ) − f (c)| < e
Since the rational numbers are dense in any interval, there certainly exist rational numbers
x satisfying the above hypothesis, whence f ( x ) = 0 and so | f (c)| < e for any e > 0. This
forces f (c) = 0 for all c ∈ ( a, b).
(b) If f and g are continuous on ( a, b), then f − g is continuous. But ( f − g)(r ) = 0 for each
rational number r ∈ ( a, b) by hypothesis, so part (a) immediately gives ( f − g)( x ) = 0
∀ x ∈ ( a, b). Thus f ( x ) = g( x ).
∈ Q where qp is written in lowest terms and q > 0. Suppose that f is continuous at r,
and let e = 1q . Then ∃δ > 0 such that
14 Let r =
p
q
| x − r | < δ =⇒ | f ( x ) − f (r )| < e =⇒ f ( x ) −
1 1
<
q
q
However, the irrational numbers are dense in the interval (r − δ, r + δ), so there are irrational
numbers x satisfying the above. Selecting such an x we conclude that 1q < 1q : a contradiction.
Thus f is discontinuous at r ∈ Q.
Now let i ∈ R \ Q and let e > 0 be given. Let q = de−1 e be the least integer greater than 1/e
and consider the set A := {r ∈ Q : f (r ) ≥ 1q }. Successive elements of A are separated by at
least q12 . Let δ = min{|i − r | : r ∈ A} be the least distance from i to an element of A. Certainly
δ > 0. Now, if | x − i | < δ we know that either x ∈ R \ Q, or that x ∈ Q \ A: in either case,
| f ( x ) − f (i )| = f ( x ) <
1
≤e
q
Thus f is continuous at i.
5