Analysis Problem Book
Amol Sasane
Contents
Chapter 1. The real number system and preliminaries
§1.1. Sum of squares
§1.2. The number line, field and order axioms of R
§1.3. Rational and irrational numbers
§1.4. Least upper bound property of R
§1.5. Intervals
1
1
1
2
2
4
§1.6. Functions
4
§1.7. Cardinality
5
Chapter 2. Sequences
7
§2.1. Convergence of sequences
7
§2.2. Bounded and monotone sequences
7
§2.4. Sandwich Theorem
9
§2.3. Algebra of limits
§2.5. Subsequences, Bolzano-Weierstrass Theorem, Cauchy sequences
§2.6. Pointwise versus uniform convergence
Chapter 3. Continuity
§3.1. Definition of continuity
§3.2. Continuous functions at a point preserve convergence of sequences there
8
10
11
13
13
13
§3.3. Intermediate and Extreme Value Theorems
15
§3.5. Limits
16
§3.4. Uniform continuity
Chapter 4. Differentiation
§4.1. Definition of the derivative
§4.2. The Product, Quotient and Chain Rule
16
19
19
19
§4.3. Tangents and normals to curves, Newton-Raphson Method
20
§4.5. Optimisation, convexity
0
form of l’Hôpital’s Rule
§4.6.
0
23
§4.4. Mean Value Theorem, Rolle’s Theorem, monotonicity, Taylor’s Formula
21
25
iii
iv
Contents
Chapter 5. Integration
27
§5.1. Definition and properties of the Riemann integral
27
§5.3. Integration by Parts and by Substitution
28
§5.5. Improper integrals
29
§5.7. Applications: length, area, volume
37
§5.2. Fundamental Theorem of Integral Calculus
27
§5.4. Riemann sums
29
§5.6. Transcendental functions
32
Chapter 6. Series
41
§6.1. Convergence/divergence of series
41
§6.3. Comparison, Ratio, Root
42
§6.2. Absolute convergence and the Leibniz’s Alternating Series Test
§6.4. Power series
Solutions
42
45
47
Solutions to the exercises from Chapter 1
47
Solutions to the exercises from Chapter 2
68
Solutions to the exercises from Chapter 3
82
Solutions to the exercises from Chapter 4
93
Solutions to the exercises from Chapter 5
117
Solutions to the exercises from Chapter 6
153
Chapter 1
The real number system
and preliminaries
1.1. Sum of squares
1.1.1. The aim of this exercise is to offer a simple justification for the sum of squares formula.
n(n + 1)(2n + 1)
for all n ∈ N.
6
(2) Here is a pictorial “proof without words” of the fact that for n ∈ N,
(1) Show using induction that 12 + 22 + 32 + · · · + n2 =
1 + 2 + 3 + ··· + n =

•



•
n
•



•
•
• •
• •
+
◦ ◦
◦
•
n(n + 1)
.
2
◦ ◦
•
◦ ◦
•
=
•
◦ ◦
•
◦
|
◦
•
•
•
◦
◦
•
•
{z
n+1
◦
◦
◦
•
◦
◦
◦
◦
}
Using this, give a justification of the sum of squares formula by considering the following
picture, in which the triangles on the right are 60◦ -rotated versions of each other:
1
n
n
2 2
··· n
n ···
3 3 3
4 ··· n
n ··· 4
+
+
=
4 4 4 4
3 4 ··· n
n ··· 4 3
···
2 3 4 ··· n
n ··· 4 3 2
n n n ··· n
1 2 3 4 ··· n
n ··· 4 3 2 1
2n+1
2n+1 2n+1
···
2n+1 2n+1
···
2n+1
1.2. The number line, field and order axioms of
R
1.2.1. Depict −11/6 and
√
3 on the number line.
1.2.2. Using the field axioms of R, prove the following:
(1) Additive inverses are unique.
(2) For all a ∈ R, 0 · a = 0.
(3) For all a ∈ R, (−1) · a = −a.
1
2
1. The real number system and preliminaries
1.2.3. Using the order axioms for R, show the following:
(1) For all a ∈ R, a2 ≥ 0.
(2) 1 > 0.
(3) There is no real number x such that x2 + 1 = 0.
1.3. Rational and irrational numbers
1.3.1. We offer a different proof of the irrationality of
the irrationality of “surds 1”.
√
2, and en route learn a technique to prove
(1) Prove the Rational Zeros Theorem: Let c0 , c1 , · · · , cd be d ≥ 1 integers such that c0 and
cd are not zero. Let r = p/q where p, q are integers having no common factor and such
that q > 0. Suppose that r is a zero of the polynomial c0 + c1 x + · · · + cd xd . Then q
divides cd and p divides c0 .
√
(2) Show that 2 is irrational.
√
(3) Show that 3 6 is irrational.
1.4. Least upper bound property of R
1.4.1. Provide the following information about the set S
An
upper
bound
A
lower
bound
Is S
bounded?
sup S
inf S
If sup S exists,
then is sup S
in S?
If inf S exists,
then is inf S
in S?
max S
min S
where S is given by:
(1) (0, 1]
(2) [0, 1]
(3) (0, 1)
1
: n ∈ Z \ {0}
(4)
n
1
(5) − : n ∈ N
n
n
:n∈N
(6)
n+1
(7) {x ∈ R : x2 ≤ 2}
(8) {0, 2, 1, 3, 2013}
1
:n∈N
(9) (−1)n 1 +
n
(10) {x2 : x ∈ R}
x2
:
x
∈
R
.
(11)
1 + x2
1Surds refer to irrational numbers which arise as the nth root of a natural number. The mathematician al-Khwarizmi
(around 820 AD) called irrational numbers “inaudible”, which was later translated to the Latin surdus for “mute”.
1.4. Least upper bound property of R
3
1.4.2. For any nonempty bounded set S, prove that inf S ≤ sup S, and that the equality holds if
and only if S is a singleton set (that is a set with cardinality 1).
1.4.3. Let A and B be nonempty subsets of R that are bounded above and such that A ⊂ B.
Prove that sup A ≤ sup B.
1.4.4. Let A and B be nonempty subsets of R that are bounded above. Prove that sup(A ∪ B)
exists and that sup(A ∪ B) = max{sup A, sup B}.
1.4.5. Determine whether the following statements are TRUE or FALSE.
(1) If u is an upper bound of S (⊂ R), and u′ < u, then u′ is not an upper bound for S.
(2) If u∗ is the supremum of S (⊂ R), and ǫ > 0, then u∗ − ǫ is not an upper bound of S.
(3) Every subset of R has a maximum.
(4) Every subset of R has a supremum.
(5) Every bounded subset of R has a maximum.
(6) Every bounded subset of R has a supremum.
(7) Every bounded nonempty subset of R has a supremum.
(8) Every set that has a supremum is bounded above.
(9) For every set that has a maximum, the maximum belongs to the set.
(10) For every set that has a supremum, the supremum belongs to the set.
(11) For every set S that is bounded above, |S| defined by {|x| : x ∈ S} is bounded.
(12) For every set S that is bounded, |S| defined by {|x| : x ∈ S} is bounded.
(13) For every bounded set S, if inf S < x < sup S, then x ∈ S.
1.4.6. Let A and B be nonempty subsets of R that are bounded above and define
A + B = {x + y : x ∈ A and y ∈ B}.
Prove that sup(A + B) exists and that sup(A + B) ≤ sup A + sup B.
1.4.7. Let S be a nonempty subset of real numbers which is bounded below. Let −S denote the
set of all real numbers −x, where x belongs to S. Prove that inf S exists and inf S = − sup(−S).
1.4.8. Let S be a nonempty set of positive real numbers, and define
1
−1
S =
:x∈S .
x
Show that S −1 is bounded above if and only if inf S > 0. Furthermore, in case inf S > 0, show
that
1
.
sup S −1 =
inf S
1.4.9. Show that a subset S of R is bounded if and only if there exists a M ∈ R such that for all
x ∈ S, |x| ≤ M .
1.4.10. Show that if a, b ∈ R and a < b, then there exists an irrational number between a and b.
1
1
1
1
1−
2+
2−
1+
1
n
n
n
n
= = sup
.
1.4.11. Show that inf
n∈N
6
3 n∈N
6
4
1. The real number system and preliminaries
1.5. Intervals
1.5.1. Let An , n ∈ N, be a collection of sets. Then
and we use
[
n∈N
\
n∈N
\
An denotes their intersection, that is,
n∈N
An = {x : ∀n ∈ N, x ∈ An },
An to denote the union of the sets An , n ∈ N, that is,
[
n∈N
Prove that
An = {x : ∃n ∈ N such that x ∈ An }.
\ 1
.
0,
n
n∈N
\ 1
(2) {0} =
0,
.
n
n∈N
[ 1
1
(3) (0, 1) =
.
,1 −
n+2
n+2
n∈N
\ 1
1
.
(4) [0, 1] =
− ,1 +
n
n
(1) ∅ =
n∈N
1.5.2. Let x0 ∈ R and δ > 0. Prove that (x0 − δ, x0 + δ) = {x ∈ R : |x − x0 | < δ}.
1.5.3. Prove that if x, y are real numbers, then ||x| − |y|| ≤ |x − y|.
1.5.4. Show that the generalized triangle inequality: if n ∈ N and a1 , · · · , an are real numbers,
then |a1 + · · · + an | ≤ |a1 | + · · · + |an |. We say that the numbers a1 , · · · , an have the same sign if
either of the the following two cases is true:
1◦ a1 ≥ 0, · · · , an ≥ 0.
2◦ a1 ≤ 0, · · · , an ≤ 0.
In other words, the numbers have the same sign if on the number lie either they all lie on the
right of 0 including 0, or they all lie on the left of 0 including 0. Show that equality holds in the
generalized triangle inequality if and only if the numbers have the same sign. Hint: Consider the
n = 2 case first.
1.6. Functions
1.6.1. Let f, g : R → R be given by
x ∈ R. Compute the following:
(1) f (2) + g(2).
(2) f (2) − 2 · g(2).
(3) f (2) · g(2).
(4) (f (2))/(g(2)).
(5) f (g(2)).
f (x) =
1 + x,
g(x) =
1 − x,
5
1.7. Cardinality
(6) For a ∈ R, f (a) + g(−a).
(7) For t ∈ R, f (t) · g(−t).
1.6.2. Suppose that f : R → R has a graph as shown below in Figure 1: Sketch the graphs of the
0
Figure 1. The graph of f .
functions g1 , g2 , g3 , g4 , g5 : R → R, defined by
g1 (x)
:=
f (x + 1),
g2 (x)
:=
g3 (x)
:=
f (x − 1),
f (2x),
g4 (x)
:=
f (x/2),
g5 (x)
:=
f (−x),
for x ∈ R.
1.6.3. Sketch the graphs of x 7→ x3 and of x 7→ 2x2 + 4x + 1, x ∈ R.
1.6.4. The aim of this exercise is to show some elementary properties of polynomials. Let p be a
polynomial of degree d ≥ 1.
(1) If p(0) = 0, then p(x) = xq(x), where q is a polynomial of degree d − 1. (The degree of
the zero polynomial is defined as 0.)
(2) For each real a, pa defined by pa (x) = p(x + a), x ∈ R, is also a polynomial of degree d.
(3) If p(a) = 0 for some real a, then p(x) = (x − a)q(x), where q is a polynomial of degree
d − 1.
(4) If p(x) = 0 for d + 1 distinct real values of x, then p(x) = 0 for all real x.
(5) Let pe be polynomial of degree d ≥ de such that p(x) = pe(x) for d + 1 distinct values of x.
Then p(x) = pe(x) for all real x.
1.7. Cardinality
1.7.1. A set S is said to be countable if it is finite or if there is a bijective map from N onto S. A
set which is not countable is called uncountable.
(1) Show that Z is countable.
(2) Prove that every subset of a countable set is countable.
(3) If A, B are countable, then show that A × B is also countable.
(4) Prove that Q is countable.
6
1. The real number system and preliminaries
Remark 1.1. However, R is uncountable. Here is a proof of the fact that [0, 1] ⊂ R is uncountable.
Suppose, on the contrary, that [0, 1] is countable. Let x1 , x2 , x3 , · · · be an enumeration of [0, 1].
For each n ∈ N, construct a subinterval [an , bn ] of [0, 1] that does not contain xn inductively as
follows:
Initially, a0 := 0, b0 := 1.
Suppose for k ≥ 0, ak , bk have been chosen. Choose ak+1 , bk+1 like this:
If xk+1 ≤ ak or xk+1 ≥ bk , then
ak+1
:=
bk+1
:=
b k − ak
,
3
b k − ak
.
ak + 2 ·
3
ak +
If ak < xk+1 < bk , then
ak+1
bk+1
bk − xk+1
,
3
bk − xk+1
.
:= xk+1 + 2 ·
3
:= xk+1 +
See Figure 2
0
1
a0
b0
xk+1
ak
ak+1
ak
ak+1
xk+1
bk
bk+1
bk+1
xk+1
bk
Figure 2. Construction of [an , bn ], n ≥ 0.
Then for all n ∈ N, [an , bn ] 6= ∅ and xn 6∈ [an , bn ]. Moreover,
0 < a1 < a2 < a3 < · · · < an < · · · < bn < bn−1 < · · · < b2 < b1 < 1.
Let
a := sup an and b := inf bn .
n∈N
n∈N
Then a ≤ b, and so [a, b] 6= ∅. Also, for all n ∈ N, [a, b] ⊂ [an , bn ] and xn 6∈ [an , bn ]. So for all n ∈ N,
xn 6∈ [a, b]. So the points in [a, b] (⊂ [0, 1]) are missing from the enumeration, a contradiction!
Chapter 2
Sequences
2.1. Convergence of sequences
2.1.1. (Convergence/divergence)
(1) Prove that the constant sequence (1)n∈N is convergent.
(2) Can the limit of a convergent sequence be one of the terms of the sequence?
(3) If none of the terms of a convergent sequence equal its limit, then prove that the terms
of the sequence cannot consist of a finite number of distinct values.
(4) Prove that the sequence ((−1)n )n∈N is divergent.
2.1.2. Prove that lim
n→∞
1
6= 1.
n
2.1.3. In each of the cases listed below, give an example of a divergent sequence (an )n∈N that
satisfies the given conditions. Suppose that L = 1.
(1) For every ǫ > 0, there exists an N such that for infinitely many n > N , |an − L| < ǫ.
(2) There exists an ǫ > 0 and a N ∈ N such that for all n > N , |an − L| < ǫ.
2.1.4. Let S be a nonempty subset of R that is bounded above. Show that there exists a sequence
(an )n∈N contained in S (that is, an ∈ S for all n ∈ N) and which is convergent with limit equal to
sup S.
2.1.5. Let (an )n∈N be a sequence such that for all n ∈ N, an ≥ 0. Prove that if (an )n∈N is
convergent with limit L, then L ≥ 0. .
2.2. Bounded and monotone sequences
2.2.1. Let (an )n∈N be a sequence defined by
a1 = 1 and an =
2n + 1
an−1 for n ≥ 2.
3n
Prove that (an )n∈N is convergent.
2.2.2. If (bn )n∈N is a bounded sequence, then prove that
bn
n
is convergent with limit 0.
n∈N
7
8
2. Sequences
2.2.3. (∗) If (an )n∈N is a convergent sequence with limit L, then prove that the sequence (sn )n∈N ,
where
a1 + · · · + an
, n ∈ N,
sn =
n
is also convergent with limit L. Give an example of a sequence (an )n∈N such that (sn )n∈N is
convergent but (an )n∈N is divergent.
2.2.4. Given a bounded sequence (an )n∈N , define
ℓk = inf{an : n ≥ k} and uk = sup{an : n ≥ k},
k ∈ N.
Show that the sequences (ℓn )n∈N , (un )n∈N are bounded and monotone, and conclude that they
are convergent. (Their respective limits are denoted by lim inf an and lim sup an .)
n→∞
n→∞
2.2.5. Fill in the blanks in the following proof of the fact that every bounded decreasing sequence
of real numbers converges.
Let (an )n∈N be a bounded decreasing sequence of real numbers. Let ℓ∗ be the
lower
of the set of real numbers.
bound of {an : n ∈ N}. The existence of ℓ∗ is guaranteed by the
of (an )n∈N . Taking ǫ > 0, we must show that there exists a positive
We show that ℓ∗ is the
integer N such that
for all n > N . Since ℓ∗ + ǫ > ℓ∗ , ℓ∗ + ǫ is not
of {an : n ∈ N}.
≤ aN <
. Since (an )n∈N is
, we have for all
Therefore there exists N with
n ≥ N that ℓ∗ − ǫ < ℓ∗ ≤
≤ aN < ℓ∗ + ǫ, and so |an − ℓ∗ | < ǫ.
2.2.6. Circle the convergent sequences in the following list, and find the limit in case the sequence
converges:
(1) (cos(πn))n∈N .
(2) (1 + n2 )n∈N .
(3)
sin n
n
.
n∈N
2
3n
(4) 1 −
n+1
.
n∈N
(5) 0.9, 0.99, 0.999, · · · .
2.3. Algebra of limits
2.3.1. Recall the convergent sequence (an )n∈N from Exercise 2.2.1 on page 7 defined by
a1 = 1 and an =
2n + 1
an−1 for n ≥ 2.
3n
What is its limit?
Hint: If (an )n∈N is a convergent sequence with limit L, then (an+1 )n∈N is also a convergent
sequence with limit L.
2.3.2. Suppose that the sequence (an )n∈N is convergent, and assume that the sequence (bn )n∈N
is bounded. Prove that the sequence (cn )n∈N defined by
cn =
is convergent, and find its limit.
an bn + 5n
a2n + n
9
2.4. Sandwich Theorem
2.3.3. Let (an )n∈N be a convergent sequence with limit L and suppose
that an ≥ 0 for all n ∈ N.
√
√
Prove that the sequence ( an )n∈N is also convergent, with limit L.
Hint: First show that L ≥ 0. Let ǫ > 0. If L = 0, then choose N ∈ N large enough so that
for n > √
N , |an − L|√= an < ǫ2 . If L >
√ 0, then choose N ∈ N large enough so that for n > N ,
√
√
| an − L|| an + L| = |an − L| < ǫ L.
p
2.3.4. Show that ( n2 + n − n)n∈N is a convergent sequence and find its limit.
√
Hint: ‘Rationalize the numerator’ by using n2 + n + n.
2.3.5. Prove that if (an )n∈N and (bn )n∈N are convergent sequences such that for all n ∈ N, an ≤ bn ,
then
lim an ≤ lim bn .
n→∞
n→∞
Hint: Use Exercise 2.1.5 on page 7.
2.4. Sandwich Theorem
n!
n!
is convergent and lim n = 0.
2.4.1. Prove that the sequence
n
n→∞
n n∈N
n
n!
1 2
n
1
1
Hint: Observe that 0 ≤ n = · · · · · · ≤ · 1 · · · · · 1 ≤ .
n
n n
n
n
n
2.4.2. Prove that for all k ∈ N, the sequence
1 k + 2 k + 3 k + · · · + nk
nk+2
is convergent and
n∈N
1 k + 2 k + 3 k + · · · + nk
= 0.
n→∞
nk+2
lim
1
2.4.3. ( lim n n = 1.)
n→∞
(1) Using induction, prove that if x ≥ −1 and n ∈ N, then (1 + x)n ≥ 1 + nx.
2
√ 2
1
1
(2) Show that for all n ∈ N, 1 ≤ n n < (1 + n) n ≤ 1 + √
.
n
1
Hint: Take x = √ in the inequality above.
n
1
(3) Prove that (n n )n∈N is convergent and find its limit.
2.4.4. Let (an )n∈N be a sequence contained in the interval (a, b) (that is, for all n ∈ N, a < an < b).
If (an )n∈N is convergent with limit L, then prove that L ∈ [a, b].
Hint: Use Exercise 2.1.5 on page 7.
Give an example to show that L needn’t belong to (a, b).
2.4.5. Let (an )n∈N be a convergent sequence, and let (bn )n∈N satisfy |bn − an | <
Show that (bn )n∈N is also convergent. What is its limit?
1
1
Hint: Observe that − + an < bn < an + for all n ∈ N.
n
n
1
for all n ∈ N.
n
10
2. Sequences
2.5. Subsequences, Bolzano-Weierstrass
Theorem, Cauchy sequences
2.5.1. Determine if the following statements are true or false.
(1) Every subsequence of a convergent real sequence is convergent.
(2) Every subsequence of a divergent real sequence is divergent.
(3) Every subsequence of a bounded real sequence is bounded.
(4) Every subsequence of an unbounded real sequence is unbounded.
(5) Every subsequence of a monotone real sequence is monotone.
(6) Every subsequence of a nonmonotone real sequence is nonmonotone.
(7) If every subsequence of a real sequence converges, the sequence itself converges.
(8) If for a real sequence (an )n∈N , the sequences (a2n )n∈N and (a2n+1 )n∈N both converge,
then (an )n∈N converges.
(9) If for a real sequence (an )n∈N , the sequences (a2n )n∈N and (a2n+1 )n∈N both converge to
the same limit, then (an )n∈N converges.
2.5.2. (∗) Beginning with 2 and 7, the sequence 2, 7, 1, 4, 7, 4, 2, 8, 2, 8, . . . is constructed by multiplying successive pairs of its terms and adjoining the result as the next one or two members of
the sequence depending on whether the product is a one- or two-digit number. Thus we start with
2 and 7, giving the product 14, and so the next two terms are 1, 4. Proceeding in this manner, we
get subsequent terms as follows:
2, 7
2, 7, 1, 4
2, 7, 1, 4
2, 7, 1, 4, 7
2, 7, 1, 4, 7
2, 7, 1, 4, 7, 4
2, 7, 1, 4, 7, 4
2, 7, 1, 4, 7, 4, 2, 8
2, 7, 1, 4, 7, 4, 2, 8
2, 7, 1, 4, 7, 4, 2, 8, 2, 8
..
.
Prove that this sequence has the constant subsequence 6, 6, 6, . . . .
Hint: Show that 6 appears an infinite number of times as follows. Since the terms 2, 8, 2, 8 are
adjacent, they give rise to the adjacent terms 1, 6, 1, 6 at some point, which in turn give rise to
the adjacent terms 6, 6, 6 eventually, and so on. Proceeding in this way, find out if you get a loop
containing the term 6.
2.5.3. Show that if (an )n∈N is a sequence that does not converge to L, then there exists an ǫ > 0
and there exists a subsequence (ank )k∈N of (an )n∈N such that for all k ∈ N, |ank − L| ≥ ǫ.
2.5.4. (∗) Consider the bounded divergent sequence ((−1)n )n∈N . Note that there exist two subsequences (−1, −1, −1, . . . and 1, 1, 1, . . . ) which have distinct limits (−1 6= 1). In this exercise
we show that this is a general phenomenon. Show that if (an )n∈N is bounded and divergent, then
11
2.6. Pointwise versus uniform convergence
it has two subsequences which converge to distinct limits. Hint: Use the Bolzano-Weierstrass
theorem twice.
2.5.5. Show that if (an )n∈N is a Cauchy sequence, then (an+1 − an )n∈N converges to 0.
2.6. Pointwise versus uniform convergence
2.6.1. Suppose that I is an interval and fn : I → R (n ∈ N) be a sequence which is pointwise
convergent to f : I → R. Let the numbers an := sup{|fn (x) − f (x)| : x ∈ I} (n ∈ N) all exist.
Prove that (fn )n∈N converges uniformly to f if and only if lim an = 0.
n→∞
Let fn : (0, ∞) → R be given by fn (x) = xe−nx for x ∈ (0, ∞) and n ∈ N. Show that the
sequence (fn )n∈N converges uniformly on (0, ∞).
2.6.2. Let fn : [0, 1] → R be defined by
fn (x) =
x
1 + nx
(x ∈ [0, 1]).
Does (fn )n∈N converge uniformly on [0, 1]?
2.6.3. Let fn : R → R be defined by
fn (x) = 1 −
1
(1 + x2 )n
(x ∈ R, n ∈ N).
Show that the sequence (fn )n∈N of continuous functions converges pointwise to the function
1 if x 6= 0,
f (x) =
0 if x = 0,
which is discontinuous at 0.
2.6.4. (To be done after the relevant topics have been covered.) Uniform convergence often
implies that the limit function inherits properties possessed by the terms of the sequence. For
instance, we will see later on that if a sequence of continuous functions converges uniformly to a
function f , then f is also continuous. Morally, the reason nice things can happen with uniform
convergence is that we can exchange two limiting processes, which is not allowed always when
one just has pointwise convergence. The following exercises demonstrate the precariousness of
exchanging limiting processes arbitrarily.
m
.
(1) For n ∈ N and m ∈ N, set am,n =
m+n
Show that for each fixed n, lim am,n = 1, while for each fixed m, lim am,n = 0.
m→∞
n→∞
Is lim lim am,n = lim lim am,n ?
m→∞ n→∞
n→∞ m→∞
(2) Let fn : R → R be defined by
fn (x) =
sin(nx)
√
n
(x ∈ R, n ∈ N).
Show that (fn )n∈N converges pointwise to the zero function f . However, show that
(fn′ )n∈N does not converge pointwise to (the zero function) f ′ .
(3) Let fn : [0, 1] → R (n ∈ N) be defined by fn (x) = nx(1 − x2 )n (x ∈ [0, 1]). Show that
(fn )n∈N converges pointwise to the zero function f . However, show that
Z 1
Z 1
1
lim f (x)dx.
fn (x)dx = 6= 0 =
lim
n→∞ 0
2
0 n→∞
12
2. Sequences
Remark 2.1. Besides the preservation of continuity under uniform convergence, one also has the
following results associated with uniform convergence, which we won’t establish in this course (but
instead we refer the interested student to Rudin’s book Principles of Mathematical Analysis for
details).
Proposition 2.2. If fn : [a, b] → R (n ∈ N) is a sequence of Riemann-integrable functions on
[a, b] which converges uniformly to f : [a, b] → R, then f is also Riemann-integrable on [a, b], and
moreover
Z b
Z b
f (x)dx = lim
fn (x)dx.
a
n→∞
a
Proposition 2.3. Let fn : (a, b) → R (n ∈ N) be a sequence of differentiable functions on (a, b),
such that there exists a point c ∈ (a, b) for which (fn (c))n∈N converges. If the sequence (fn′ )n∈N
converges uniformly to g on (a, b), then (fn )n∈N converges uniformly to a differentiable function
f on (a, b), and moreover, f ′ (x) = g(x) for all x ∈ (a, b).
Chapter 3
Continuity
3.1. Definition of continuity
3.1.1. Given ǫ > 0, find some δ > 0 such that if x ∈ (0, ∞) satisfies |x − 2| < δ, then
1 1
− < ǫ.
x 2
Conclude that x 7→
1
: (0, ∞) → R is continuous at 2.
x
3.1.2. Let f : R → R be a function that satisfies f (x + y) = f (x) + f (y) for all x, y ∈ R.
(1) Suppose that f is continuous at some real number c. Prove that f is continuous on R.
Hint: Since f is continuous at c, given ǫ > 0, ∃δ > 0 such that for all x ∈ R satisfying
|x − c| < δ, |f (x) − f (c)| < ǫ. Show that given any other point c′ ∈ R, the function f is
continuous at c′ by showing that the same δ works (for this ǫ).
(2) Give an example of such a function.
3.1.3. Suppose that f : R → R and there exists a M > 0 such that for all x ∈ R, |f (x)| ≤ M |x|.
Prove that f is continuous at 0. Hint: Find f (0).
3.1.4. Let f : R → R be defined by
f (x) =
0
1
if x is rational,
if x is irrational.
Prove that for every c ∈ R, f is not continuous at c.
Hint: Use the fact that there are irrational numbers arbitrarily close to any rational number and
rational numbers arbitrarily close to any irrational number.
3.1.5. Let f : R → R be a continuous function. Prove that if for some c ∈ R, f (c) > 0, then there
exists a δ > 0 such that for all x ∈ (c − δ, c + δ), f (x) > 0.
3.2. Continuous functions at a point preserve
convergence of sequences there
3.2.1. Let c ∈ R, δ > 0 and f : (c − δ, c] → R be continuous and strictly increasing on (c − δ, c).
Show that f is strictly increasing on (c − δ, c].
13
14
3. Continuity
3.2.2. Prove that if f : R → R is continuous and f (x) = 0 if x is rational, then f (x) = 0 for all
x ∈ R. Revisit Exercise 3.1.4.
Hint: Given any real number c, there exists a sequence of rational numbers (rn )n∈N that converges
to c.
3.2.3. Let f : R → R be a function that preserves divergent sequences, that is, for every divergent
sequence (xn )n∈N , (f (xn ))n∈N is divergent as well. Prove that f is one-to-one.
Hint: Let x1 , x2 be distinct real numbers, and consider the sequence x1 , x2 , x1 , x2 , . . . .
3.2.4. Let I be an interval, c ∈ I, and f : I → R. Show that the following are equivalent:
(1) f is continuous at c.
(2) For every sequence (xn )n∈N contained in I such that (xn )n∈N converges to c, the sequence
(f (xn ))n∈N converges.
3.2.5. Consider the function f : R → R defined by
x if x is rational,
f (x) =
−x if x is irrational.
Prove that f is continuous only at 0. Hint: For every rational number, there is a sequence of
irrational numbers that converges to it, and for every irrational number, there is a sequence of
rational numbers that converges to it.
3.2.6. (∗) Every nonzero rational number x can be uniquely written as x = n/d, where n, d denote
integers without any common divisors and d > 0. When r = 0, we take d = 1 and n = 0. Consider
the function f : R → R defined by


if x is irrational,
 0
f (x) =
n
1

is rational.
if x =

d
d
Prove that f is discontinuous at every rational number, and continuous at every irrational number.
Hint: For an irrational number x, given any ǫ > 0, and any interval (N, N + 1) containing x, show
that there are just finitely many rational numbers r in (N, N + 1) for which f (r) ≥ ǫ. Use this to
show the continuity at irrationals.
3.2.7. Let f : R → R be a continuous function such that for all x, y ∈ R, f (x + y) = f (x) + f (y).
Show that there exists a real number a such that for all x ∈ R, f (x) = ax. Hint: Show first that
for natural numbers n, f (n) = nf (1). Extend this to integers n, and then to rational numbers
n/d. Finally use the density of Q in R to prove the claim.
3.2.8. Find all continuous functions f : R → R such that for all x ∈ R, f (x) + f (2x) = 0.
x
x
x
=f
= −f
= ···.
Hint: Show that f (x) = −f
2
4
8
3.2.9. Give an example of
(1) an interval I,
(2) a continuous function f : I → R, and
(3) a Cauchy sequence (xn )n∈N
for which (f (xn ))n∈N is not a Cauchy sequence in R. Compare with Exercise 3.4.4.
3.3. Intermediate and Extreme Value Theorems
15
3.3. Intermediate and Extreme Value Theorems
3.3.1. Let f : [0, ∞) → R be the function defined by
f (x) =
1
,
1 + x2
x ∈ [0, ∞).
Show that f is strictly decreasing and that f ([0, ∞)) = (0, 1]. Find an expression for the inverse
function f −1 : (0, 1] → [0, ∞) and explain why f −1 is continuous on (0, 1]. Sktech the graphs of f
and f −1 .
3.3.2. Consider a flat pancake of arbitrary shape. Show that there is a straight line cut that
divides the pancake into two parts having equal areas. Can the direction of the straight line cut
be chosen arbitrarily?
3.3.3. In each case, give an example of a continuous function f : S → R, such that f (S) = T or
else explain why there can be no such f .
(1) S = (0, 1), T = (0, 1].
(2) S = (0, 1), T = (0, 1) ∪ (1, 2).
(3) S = R, T = Q.
3.3.4. A function f : R → R is called periodic if there exists a T > 0 such that for all x ∈ R,
f (x + T ) = f (x). If f : R → R is continuous and periodic, then prove that f is bounded, that is,
the set S = {f (x) | x ∈ R} is bounded.
3.3.5. Let f : [a, b] → R be continuous on [a, b], and define f∗ as follows:
f (a)
if x = a,
f∗ (x) =
max{f (y) : y ∈ [a, x]} if x ∈ (a, b].
(1) Show that f∗ is a well-defined function.
(2) If f : [−1, 1] → R is given by f (x) = x − x2 , then find f∗ .
3.3.6. Suppose that f : [0, 1] → R is a continuous function such that for all x ∈ [0, 1], 0 ≤ f (x) ≤ 1.
Prove that there exists at least one c ∈ [0, 1] such that f (c) = c.
Hint: Consider the continuous function g(x) = f (x) − x, and use the intermediate value theorem.
16
3. Continuity
3.3.7. At 8:00 a.m. on Saturday, a hiker begins walking up the side of a mountain to his weekend
campsite. On Sunday morning at 8:00 a.m., he walks back down the mountain along the same
trail. It takes him one hour to walk up, but only half an hour to walk down. At some point on his
way down, he realizes that he was at the same spot at exactly the same time on Saturday. Prove
that he is right.
Hint: Let u(t) and d(t) be the position functions for the walks up and down, and apply the
intermediate value theorem to f (t) = u(t) − d(t).
3.3.8. Show that the polynomial function p(x) = 2x3 − 5x2 − 10x+ 5 has a real root in the interval
[−1, 2].
3.3.9. Let f : R → R be continuous. If S := {f (x) : x ∈ R} is neither bounded above nor bounded
below, prove that S = R.
Hint: If y ∈ R, then since S is neither bounded above nor bounded below, there exist x0 , x1 ∈ R
such that f (x0 ) < y < f (x1 ).
3.3.10. (∗) Show that given any continuous function f : R → R, there exists a x0 ∈ [0, 1] and a
m ∈ Z \ {0} such that f (x0 ) = mx0 . In other words, the graph of f intersects some nonhorizontal
line y = mx at some point x0 in [0, 1].
Hint: If f (0) = 0, take x0 = 0 and any m ∈ Z \ {0}. If f (0) > 0, then choose N ∈ N satisfying
N > f (1), and apply the intermediate value theorem to the continuous function g(x) = f (x) − N x
on the interval [0, 1]. If f (0) < 0, then first choose a N ∈ N such that N > −f (1), and consider
the function g(x) = f (x) + N x, and proceed in a similar manner.
3.3.11. (∗) Prove that there does not exist a continuous function f : R → R such that assumes
rational values at irrational numbers, and irrational values at rational numbers, that is,
f (Q) ⊂ R \ Q and f (R \ Q) ⊂ Q.
Hint: Note that for every m ∈ Z \ {0}, there does not exist a x0 ∈ R such that f (x0 ) = mx0 .
3.4. Uniform continuity
3.4.1. Show that f : R → R given by f (x) = x2 (x ∈ R) is continuous, but not uniformly
continuous. Hint: Consider x = n and y = n + n1 for large n.
3.4.2. Prove that the function f : R → R defined by f (x) = |x| (x ∈ R) is uniformly continuous.
3.4.3. A function f : R → R is said to be (globally) Lipschitz if there exists a number L > 0 such
that for all x, y ∈ R, |f (x) − f (y)| ≤ L|x − y|. Prove that every Lipschitz function is uniformly
continuous.
3.4.4. Let I be an interval and let f : I → R be uniformly continuous. Show that if (xn )n∈N is a
Cauchy sequence, then (f (xn ))n∈N is a Cauchy sequence. Compare this with Exercise 3.2.9.
3.5. Limits
3.5.1. Let f (a, c) ∪ (c, b) → R be such that lim f (x) and lim f (x) exist. Show that lim f (x)
x→c+
exists if and only if lim f (x) = lim f (x).
x→c+
x→c−
x→c−
x→c
17
3.5. Limits
3.5.2. In each of the following cases, calculate the limit if it exists.
(1) lim
x→0
|x|
.
x+1
(2) lim (⌊x⌋ − x).
x→1
(3) lim x⌊x⌋.
x→0
1
(4) lim sin .
x→0
x
3.5.3. Let the polynomials A, B of degrees α, β ≥ 1 be given by
A(x)
B(x)
= a0 + a1 x + · · · + aα xα ,
= b0 + b1 x + · · · + bβ xβ ,
where aα and bβ are nonzero. Show that


0






aα



A(x)  bβ
lim
=
x→∞ B(x)


+∞








 −∞
if α < β,
if α = β,
aα
> 0,
bβ
aα
if α > β and
< 0.
bβ
if α > β and
Chapter 4
Differentiation
4.1. Definition of the derivative
4.1.1. Use the definition of the derivative to find f ′ (x), where f (x) :=
√
x2 + 1, x ∈ R.
4.1.2. If f : (a, b) → R is differentiable at c ∈ (a, b), then show that
f (c + h) − f (c − h)
hց0
2h
lim
exists and equals f ′ (c). Is the converse true?
4.1.3. Let f : (0, ∞) → R be a function and let c > 0. Show that the following are equivalent:
(1) f is differentiable at c.
(2) lim
k→1
f (kc) − f (c)
exists.
k−1
Moreover, show that if either (1) or (2) hold, then f ′ (c) =
1
f (kc) − f (c)
lim
.
c k→1
k−1
4.1.4. Let f : (0, ∞) → R be such that for all x, y ∈ (0, ∞), f (xy) = f (x) + f (y). If f is
differentiable at 1, then show that f is differentiable at every c ∈ (0, ∞) and that f ′ (c) = f ′ (1)/c.
Conclude that f is infinitely differentiable. If f ′ (1) = 2, then find f (n) (3), n ∈ N.
4.1.5. (Differentiable but not continuously differentiable example.) Consider f : R → R defined
by f (0) = 0 and for x 6= 0,
1
f (x) = x2 sin .
x
Prove that f is differentiable, but f ′ is not continuous at 0.
4.2. The Product, Quotient and Chain Rule
4.2.1. Prove that if f, g are infinitely differentiable on an open interval I, then
n X
n (k)
(f g)(n) (x) =
f (x)g (n−k) (x), x ∈ I.
k
k=0
For a rational x and n a nonnegative integer, define x[n] := x(x − 1) · · · (x − n + 1). Show that if
x, y ∈ Q, then
n X
n [k] [n−k]
(x + y)[n] =
x y
.
k
k=0
19
20
4. Differentiation
Hint: Differentiate tx+y n times with respect to t ∈ I := (0, ∞).
(In fact after we have defined logarithms, and how to take real exponents, that is, the map
t 7→ tx : (0, ∞) → R, where x ∈ R, one can see that the same result holds even when x, y ∈ R.)
4.2.2. Let f : (0, ∞) → R be the strictly decreasing function given by
f (x) =
1
,
1 + x2
x ∈ (0, ∞).
From Exercise 3.3.1, it follows that f ((0, ∞)) = (0, 1). Show that the inverse f −1 : (0, 1) → (0, ∞)
is differentiable, and find the derivative (f −1 )′ .
4.2.3. Find f ′ (x) for each of the following functions:
(1) f (x) = sin(x sin x) + sin(sin(x2 )).
(2) f (x) = sin(6 cos(6 sin(6 cos x))).
3
2
(3) f (x) = sin (sin(sin x))
.
(You may use the fact that sin′ = cos and cos′ = − sin.)
4.2.4. By using the binomial expansion (1 + x)n =
n X
n k
x , find expressions for
k
k=0
(1)
(2)
(3)
(4)
n
X
k n
.
3
k
k=1
n
X
2 n
.
k
k
k=1
n
X
1
n
.
k+1 k
k=1
n
X
n
(2k + 1)
.
k
k=1
4.2.5. Evaluate the sum
n
X
k2
k=1
2k
by considering the function S given by S(x) =
n
X
k=1
4.3. Tangents and normals to curves,
Newton-Raphson Method
4.3.1. Find the values of the constants a, b, c for which the graphs of f, g given by
f (x)
:= x2 + ax + b,
g(x)
:= x3 − c,
x ∈ R, intersect at the point (1, 2) and have the same tangent there.
xk .
4.4. Mean Value Theorem, Rolle’s Theorem, monotonicity, Taylor’s Formula
4.3.2. Find the equation of the tangent at
x(t)
y(t)
for t ∈ (0, ∞).
1
21
, 4 to the curve t 7→ (x(t), y(t)), where
4
1
:= 2 ,
t
p
t2 + 12,
:=
4.3.3. Find the points on the curve given implicitly by x2 + xy + y 2 = 7 at which
(1) the tangent is parallel to the x-axis
(2) the tangent is parallel to the y-axis.
4.3.4. Find the tangents to the implicitly defined curve x2 y + xy 2 = 6 at the points for which
d2 y
at these points.
x = 1. Also compute
dx2
4.3.5. Consider the function f given by f (x) = x2 −2. Suppose that the Newton-Raphson
√ Method
converges with the initial guess x0 := 1. Generate a few rational approximations to 2.
4.3.6. Show that x4 − x3 − 75 = 0 for some x between 3 and 4. Use the Newton-Raphson Method
to find an approximate value of this x, starting with an initial guess of x0 := (3 + 4)/2 = 3.5.
4.4. Mean Value Theorem, Rolle’s Theorem,
monotonicity, Taylor’s Formula
4.4.1. Show that for every real a, b ∈ R, | cos a − cos b| ≤ |a − b|.
4.4.2. Suppose that f : R → R is differentiable, |f ′ (x)| ≤ 1 for all x ∈ R, and that there exists an
a > 0 such that f (−a) = −a, f (a) = a. Find f (0). Give an example of such a function f .
4.4.3. Suppose that f : R → R is differentiable and that there are L, L′ ∈ R such that
lim f (x) =
L,
lim f ′ (x)
L′ .
x→∞
x→∞
=
Prove that L′ = 0.
4.4.4. Let c ∈ (a, b), and let f : (a, b) → R be such that f is
(1) differentiable on (a, b) \ {c},
(2) continuous on (a, b), and
(3) lim f ′ (x) exists.
x→c
Then f is differentiable at c, and f ′ (c) = lim f ′ (x).
x→c
4.4.5. Let f : R → R be such that for all x, y ∈ R, |f (x) − f (y)| ≤ (x − y)2 . Prove that f is
constant.
4.4.6. Let f : (a, b) → R be differentiable on (a, b) and suppose that there is number M such that
for all x ∈ (a, b), |f ′ (x)| ≤ M . Show that f is uniformly continuous on (a, b).
22
4. Differentiation
4.4.7. Find all functions f : R → R such that f is differentiable on R and for all x ∈ R and all
n ∈ N,
f (x + n) − f (x)
.
f ′ (x) =
n
Hint: Conclude that f must be twice differentiable and calculate f ′′ (x).
4.4.8. Prove that if c0 , · · · , cd are any real numbers satisfying
c1
cd
c0
+
+ ··· +
= 0,
1
2
d+1
then the polynomial c0 + c1 x + · · · + cd xd has a zero in (0, 1).
4.4.9. Show that if f : (a, b) → R has a local minimum at c ∈ (a, b), and if f is twice differentiable
at c, then f ′′ (c) ≥ 0.
Suppose that F satisfies the differential equation
F ′′ (x) + F ′ (x)g(x) − F (x) = 0
for some function g. Prove that if F is 0 at two points, then F is 0 on the interval between them.
4.4.10. Suppose that f is n times differentiable and that f (x) = 0 for n + 1 distinct x. Prove
that f (n) (x) = 0 for some x.
4.4.11. Show that there are exactly two real values of x such that
and that they lie in
π π
.
− ,
2 2
x2 = x sin x + cos x
4.4.12. (On y ′′ + y = 0.)
(1) If y = y(x) is a solution of the differential equation
y ′′ + y = 0,
(4.1)
then show that y 2 + (y ′ )2 is constant.
(2) Use part (1) to show that every solution to (4.1) has the form
y(x) = A cos x + B sin x
for suitable constants A, B. Proceed as follows: It is easy to show that all functions of
the above form satisfy (4.1). Let y be a solution. For it to have the form A cos x+B sin x,
it is necessary that A = y(0) and B = y ′ (0). Now consider
f (x) := y(x) − y(0) cos x − y ′ (0) sin x,
and apply (1) to f , making use of the fact that f (0) = f ′ (0) = 0.
(3) Use part (2) to prove the trigonometric addition formulae
sin(a + b) =
(sin a)(cos b) + (cos a)(sin b),
cos(a + b) =
(cos a)(cos b) − (sin a)(sin b).
4.4.13. Find the shortest distance from a given point (0, b) on the y-axis with b > 0, to the
parabola y = x2 .
4.4.14. Does f given by f (x) = (sin x − cos x)2 , x ∈ R have a maximum value? If so, find it.
1
sin x − x
=− .
3
x→0
x
6
4.4.15. Use Taylor’s Formula to show that lim
23
4.5. Optimisation, convexity
4.4.16. (o-notation) A special notation introduced by Landau in 1909 is particularly suited to
Taylor’s formula. Given functions f, g in an open interval I containing a such that g is nonzero in
I, we write
“f (x) = o(g(x)) as x → a”
f (x)
= 0.
if lim
x→a g(x)
The symbol f (x) = o(g(x)) is read “f (x) is little-oh of g(x)” or “f (x) is of smaller order than
g(x)”, and is intended to convey the idea that for x near a, f (x) is small as compared to g(x).
For example,
f (x) = o(1) as x → a means that lim f (x) = 0,
x→a
and
f (x)
= 0.
x
Also, if h is a function on I, then we write “f (x) = g(x) + o(h(x)) as x → a” to mean that
“f (x) − g(x) = o(h(x)) as x → a”.
f (x) = o(x) as x → a means that lim
x→a
(1) If f has a continuous (n + 1)st derivative in some open interval containing the compact
interval [a − ǫ, a + ǫ], and if
M := max{|f n+1 (x)| : |x − a| ≤ ǫ},
then it follows from Taylor’s Formula that
n
X
f (k) (a)
(x − a)k + o((x − a)n ) as x → a.
f (x) =
k!
k=0
x3
+ o(x3 ) as x → 0.
3
tan x − x
(3) Show that lim
= 1.
x→0 sin x − x cos x
(2) Show that tan x = x +
4.4.17. Let f : R → R. We call x ∈ R a fixed point of f if f (x) = x.
(1) If f is differentiable, and for all x ∈ R, f ′ (x) 6= 1, then prove that f has at most one
fixed point.
(2) (∗) Show that if there is an M < 1 such that for all x ∈ R, |f ′ (x)| ≤ M , then there is
a fixed point x∗ of f , and that x∗ = lim xn , where x1 is arbitrary and xn+1 = f (xn )
n→∞
(n ∈ N).
(3) Visualize the process described in the part (2) above via the zig-zag path
(x1 , x2 ) −→ (x2 , x2 ) −→ (x2 , x3 ) −→ (x3 , x3 ) −→ (x3 , x4 ) −→ . . . .
(4) Prove that the function f : R → R defined by
1
(x ∈ R)
f (x) = x +
1 + ex
has no fixed point, although 0 < f ′ (x) < 1 for all x ∈ R. Is this a contradiction to the
result in part (2) above? Explain.
4.5. Optimisation, convexity
4.5.1. Sketch the curve y = 2x3 +2x2 −2x−1 after locating intervals of increase/decrease, intervals
of convexity/concavity1, points of maxima/minima, and points of inflection (places where f ′′ = 0).
How many times, and approximately where does the curve cross the x-axis?
1A function is concave if −f is convex.
24
4. Differentiation
4.5.2. If a1 < · · · < an , then find the minimum value of
f (x) :=
n
X
k=1
(x − ak )2 ,
x ∈ R.
Next find the minimum value of
g(x) :=
n
X
k=1
|x − ak |,
x ∈ R.
This is a problem where Calculus won’t help: on the intervals between the ak , the function is
linear, and so the minimum must occur at one of the ak , and these are points where g is not
differentiable. However, the answer is easy to find if one considers how f changes as one passes
from one such interval to another.
2+a
, where h is given by
Let a > 0. Show that the maximum value of h is
1+a
h(x) =
1
1
+
,
1 + |x| 1 + |x − a|
x ∈ R.
Hint: Consider (−∞, 0), (0, a), (a, ∞) separately, and find the derivative there.
4.5.3. The picture in Figure 1 below shows the graph of the derivative f ′ of f . Find all local
maximizers and local minimizers of f .
f′
2
0
1
3
4
Figure 1. Graph of f ′ .
4.5.4. A cannon ball is shot from the ground with velocity v in the vertical plane at an angle
α from the horizontal, so that it has an initial vertical component of velocity equal to v sin α,
and a horizontal component v cos α. Its horizontal velocity remains constant at v cos α, while the
vertical component changes under gravitational force. The height h(t) at time t above the ground
is given by h(t) = −5t2 + v(sin α)t for t ≥ 0. Find the angle α that maximizes the span (that is
the horizontal distance reached) of the cannon ball.
4.5.5. Show that x 7→ |x| : R → R is convex.
4.5.6. Let f : I → R be a function on an interval I ⊂ R. We define the epigraph of f by
[
U (f ) :=
{(x, y) : y ≥ f (x)} ⊂ I × R.
x∈I
In other words, U (f ) is the “region above the graph of f ”. A subset C ⊂ R2 is called a convex set
if for all v1 , v2 ∈ C and for all t ∈ (0, 1), (1 − t) · v1 + t · v2 ∈ C. Show that f is a convex function
if and only if U (f ) is a convex set.
0
form of l’Hôpital’s Rule
0
4.5.7. (The Cauchy-Schwarz Inequality.)
25
4.6.
(1) Let a > 0, b, c ∈ R and consider the function f : R → R given by
f (t) = at2 + bt + c,
t ∈ R.
b2 − 4ac
.
4a
(2) Show that for all n ∈ N and all a1 , · · · , an , b1 , · · · , bn ∈ R, there holds that
Show that f has a minimizer and that the minimum value of f is −
(a21 + · · · + a2n )(b21 + · · · + b2n ) ≥ (a1 b1 + · · · + an bn )2 .
Hint: Consider
n
X
(tak − bk )2 .
k=1
4.5.8. (The Arithmetic Mean-Geometric Mean Inequality.)
(1) Suppose that f : I → R is a convex function on an interval I ⊂ R. If n ∈ N, and
x1 , · · · , xn ∈ I, then show that
x1 + · · · + xn
f (x1 ) + · · · + f (xn )
f
≤
.
n
n
(2) Show that − log : (0, ∞) → R is convex.
(3) Prove the Arithmetic Mean-Geometric Mean Inequality: for nonnegative real numbers
a1 , · · · , an , there holds that
√
a1 + · · · + an
≥ n a1 · · · an .
n
(The left hand side above is called the arithmetic mean of a1 , · · · , an , while the right
hand side is called their geometric mean.)
4.6.
0
form of l’Hôpital’s Rule
0
4.6.1. What is wrong with the following application of l’Hôpital’s Rule?
3x2 + 1
6x
x3 + x − 2
= lim
= lim
= 3.
2
x→1 2x − 3
x→1 2
x→1 x − 3x + 2
Show that the limit above is actually equal to −4.
lim
4.6.2. Revisit Exercise 4.4.15, but now use l’Hôpital’s Rule to show that lim
x→0
sin x − x
1
=− .
x3
6
Chapter 5
Integration
5.1. Definition and properties of the Riemann
integral
5.1.1. Is the following function f Riemann integrable on [0, 1]?
0 if x ∈ [0, 1] \ Q,
f (x) =
x if x ∈ [0, 1] ∩ Q.
Hint: For any partition P = {x0 = 0 < x1 < · · · < xn−1 < xn = 1}, xk+1 ≥
k = 0, · · · , n − 1. Use this to find a positive lower bound on upper sums.
xk+1 + xk
,
2
5.1.2. Let f, g ∈ R[a, b]. Show that max{f, g} and min{f, g} also belong to R[a, b], where
max{f, g} :=
max{f (x), g(x)},
min{f, g} :=
min{f (x), g(x)},
for all x ∈ [a, b]. Hint: max{a, b} =
a + b + |a − b|
for a, b ∈ R.
2
5.1.3. Give examples of bounded functions f, g : [0, 1] → R that are not Riemann integrable, but
|f |, f + g, f g are all Riemann integrable.
5.1.4. (An integral mean value result.) Let f ∈ C[a, b], ϕ ∈ R[a, b], and let ϕ be pointwise
nonnegative. Use the Intermediate Value Theorem for f to show that there is a c ∈ [a, b] such
that
Z b
Z b
f (x)ϕ(x)dx = f (c)
ϕ(x)dx.
a
a
Give examples to show that neither the continuity of f nor the nonnegativity of ϕ can be omitted.
5.2. Fundamental Theorem of Integral Calculus
1
x→0+ x3
5.2.1. Evaluate lim
Z
0
x
t2
dt.
t4 + 1
x
5.2.2. If f is continuous on R, then evaluate lim 2
x→x0 x − x2
0
and x0 = 0 separately.
Z
x
x0
f (t)dt. Consider the cases x0 6= 0
27
28
5. Integration
5.2.3. Let T > 0, and f : R → R be a continuous function which is T -periodic, that is, f (x + T ) =
f (x) for all x ∈ R. Show that the integral
Z a+T
f (x)dx
a
has the same value for all a ∈ R.
5.2.4. (Leibniz’s Rule for Integrals.) If f ∈ C[a, b] and u, v are differentiable on [c, d] and
u([c, d]) ⊂ [a, b], v([c, d]) ⊂ [a, b], then
Z v(x)
d
f (t)dt = f (v(x)) · v ′ (x) − f (u(x) · u′ (x), x ∈ [c, d].
dx u(x)
5.2.5. For x ∈ R, define
F (x)
:=
Z
2x
1
dt,
1 + t2
x2
1
p dt.
1 + |t|
1
G(x)
:=
Z
0
Find F ′ and G′ .
5.2.6. Let f : [0, ∞) → R be continuous. Find f (2) if for all x ≥ 0,
Z x
f (t)dt = x2 (1 + x).
(1)
0
(2)
Z
f (x)
t2 dt = x2 (1 + x).
0
(3)
Z
x2
f (t)dt = x2 (1 + x).
0
(4)
Z
x2 (1+x)
f (t)dt = x.
0
5.2.7. Let f be a continuous function on R and λ 6= 0. Consider
Z
1 x
y(x) =
f (t) sin(λ(x − t))dt for x ∈ R.
λ 0
Show that y is a solution to the inhomogeneous differential equation y ′′ (x) + λ2 y(x) = f (x) for all
x ∈ R and with the initial conditions y(0) = 0 and y ′ (0) = 0.
5.3. Integration by Parts and by Substitution
5.3.1. Let a > 0 and f ∈ C[−a, a] be an odd function, that is, f (x) = −f (−x) for all x ∈ [−a, a].
Prove that
Z a
f (x)dx = 0.
−a
5.3.2. Let m, n be nonnegative integers. Show that
Z
0
1
xm (1 − x)n dx =
Hint: If I(m, n) denotes the integral, then show that I(m, n) =
m!n!
.
(m + n + 1)!
n
I(m + 1, n − 1) for n ∈ N.
m+1
29
5.5. Improper integrals
5.3.3. For f ∈ C[a, b], show that for all x ∈ [a, b],
Z x
Z x Z u
(x − u)f (u)du.
f (t)dt du =
a
a
0
Remark 5.1. More generally, one can show, using induction, that
Z x
Z x Z un Z u1
f (u) · (x − u)n
du.
···
f (t)dt du1 · · · dun =
n!
a
a
a
a
5.3.4. (Taylor’s Formula with Integral Remainder.) Let n be a nonnegative integer and f ∈
C n+1 [a, b]. Show that
f (n) (a)
1
f (b) = f (a) + f (a)(b − a) + · · · +
(b − a)n +
n!
n!
′
Z
b
a
(b − t)n f (n+1) (t)dt.
(Note that as opposed to the Taylor’s Formula we have met before, where the error term contained
an undetermined c, now the “integral remainder” does not involve such an undetermined number
c.)
5.3.5. Find lim
n→∞
5.3.6. Evaluate
Z
1
0
Z
1
4
0
5.3.7. Evaluate
Z
8
1
nxn−1
dx.
1+x
x
√
dx.
1 − 4x2
√
3
q√
3
x·
x4 − 1 dx.
5.4. Riemann sums
5.4.1. Find lim
n→∞
1
1
1
1
√
.
+√
+√
+ ··· + √
1 2 + n2
2 2 + n2
3 2 + n2
n2 + n2
5.5. Improper integrals
5.5.1. Determine whether or not the following improper integrals exist:
Z ∞
1
√
dx.
(1)
1 + x3
0
Z ∞
x
√ dx.
(2)
1 + x3
0
5.5.2. (Properties of the Gamma function Γ.)
(1) Show that Γ(1) = 1.
(2) For s > 0, Γ(s + 1) = s · Γ(s).
(3) Show that for all n ∈ N, Γ(n + 1) = n!.
30
5. Integration
5.5.3. Suppose that f : [0, ∞) → [0, ∞) is such that
Z ∞
f (x)dx
(5.1)
0
exists. Intuitively, we expect that the area under the graph of the nonnegative f in intervals [x, ∞)
to become smaller and smaller as x becomes larger and larger, and so one is tempted to conclude
that f itself must have limit 0 as x → ∞. Show with an example that this needn’t be the case.
Suppose now that we know that f : [0, ∞) → [0, ∞) is such that, besides having that the
improper integral (5.1) exists, also f is differentiable and that
Z ∞
f ′ (x)dx
0
exists. Show that lim f (x) = 0.
x→∞
5.5.4. (Convolution) For f, g : R → R, which are both zero outside some compact interval, we
define the convolution f ∗ g : R → R by
Z ∞
f (τ )g(t − τ )dτ, t ∈ R,
(f ∗ g)(t) =
−∞
assuming that the integral exists for each t.
(1) Note that the graph of g(−·) is obtained by reflecting the graph of g about the y-axis,
and for a fixed t, the graph of g(t − ·) is a shifted version of the graph of g(−·). So in
order to find out the value (f ∗ g)(t), one may proceed as follows.
(a) Draw the graph of f and g.
(b) Reflect the graph of g about the y-axis.
(c) Translate the graph of g(−·) by |t| units to the left if t < 0, and to the right if t ≥ 0.
(d) Multiply the functions f and g(t − ·) pointwise, and find the area under the graph
of this pointwise product.
Use this procedure to graphically determine the convolution 1[0,1] ∗ 1[0,1] , where 1[0,1] is
the indicator function of the interval [0, 1]:
1 if x ∈ [0, 1],
1[0,1] (x) =
0 if x ∈ R \ [0, 1].
(2) Show that f ∗ g = g ∗ f .
5.5.5. (Differentiation under the integral sign.1) While we don’t learn the theory behind this, let
us try to use this tool formally, that is, without paying attention to rigour. To find
Z ∞
sin x
dx,
x
0
we consider the more general integral
I(α) =
Z
0
∞
e−αx
sin x
dx,
x
1Differentiation under the integral sign is mentioned in Feynman’s memoir Surely You’re Joking, Mr. Feynman!
in the chapter “A Different Box of Tools”, where he mentions learning it from a Calculus book by Woods while in high
school. “But I caught on how to use that method, and I used that one damn tool again and again. So because I was
self-taught using that book, I had peculiar methods of doing integrals. The result was, when guys at MIT or Princeton
had trouble doing a certain integral, it was because they couldn’t do it with the standard methods they had learned in
school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found
it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for
doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it
before giving the problem to me.”
31
5.5. Improper integrals
by introducing the “parameter” α. (So our integral of interest corresponds to α = 0.) Then by
differentiating with respect to α we obtain
Z ∞
Z ∞
∞
sin x
1
(−x)e−αx
I ′ (α) =
−e−αx sin xdx = − 2
dx =
e−αx ((−α) sin x − cos x)
x
α +1
0
0
0
1
.
= − 2
α +1
Thus
π
0 − I(α) = I(∞) − I(α) = − + tan−1 α,
2
Z ∞
sin x
π
and so
dx = I(0) = . Try your hand at formally finding the values of the following
x
2
0
integrals.
Z ∞
Z ∞
(sin(αx))2
(sin x)2
dx
by
considering
I(α)
=
dx.
(1)
x2
x2
Z0 ∞
Z0 ∞
sin x
sin(αx)
(2)
e−x
dx by considering I(α) =
e−x
dx.
x
x
0
0
Z ∞ α
Z 1
x −1
x−1
dx by considering I(α) =
dx.
(3)
log x
0
0 log x
Z ∞
Z ∞
tan−1 (αx) − tan−1 x
tan−1 (πx) − tan−1 x
dx by considering I(α) =
dx.
(4)
x
x
0
0
5.5.6. (Gravitational potential energy and escape velocity.)
(1) The gravitational potential energy (due to the gravitational pull of the Earth) at distance
a R from the center of the earth can be thought to be the amount of work done to bring
an object from separation R to far away (“at infinity R = ∞”). By Newton’s Law of
Gravitation, the force experienced by a mass m at a separation r from the center of the
Earth is given by
GM m
F =
,
r2
where G is the universal gravitation constant and M is the mass of the Earth. The work
done to move a mass m at r through a small distance dr is given by
GM m
Work done = (Force) · (Displacement) =
dr,
r2
where we have assumed that the force is constant over [r, r + dr] for small dr. So the
potential energy V (R) at R is given by the (improper) integral
Z ∞
GM m
dr.
V (R) := − =
r2
R
Find an explicit expression for V (R).
(2) The escape velocity ve (R) at a separation R from the center of the earth is defined as to
be the one which imparts enough kinetic energy to the object in order to overcome the
gravitation potential energy V (R). Show that
r
2GM
.
ve =
R
Using the following values, determine the escape velocity of a rocket on the surface of
the earth:
Radius of the Earth
=
6, 371 km,
Mass of the Earth
=
5.97219 × 1024 kg,
Universal gravitational constant G =
6.67384 × 10−11 m3 kg−1 s−2 .
32
5. Integration
5.6. Transcendental functions
5.6.1. (Euler’s constant γ.) Another important number named after Euler, is the Euler’s constant
γ, which plays, among other things, a role in Number Theory.
(1) Prove that
1
1
≥ log(n + 1) − log n ≥
for all n ∈ N.
n
n+1
(2) If
1 1
1
+ + · · · + − log n,
2 3
n
is decreasing and that an ≥ 0 for all n ∈ N. Conclude that there
an := 1 +
then show that (an )n∈N
is a number
1 1
1
γ := lim 1 + + + · · · + − log n .
n→∞
2 3
n
(Approximately, γ = 0.5772156649 · · ·, but it is not known whether γ is rational or
irrational. The reason behind the choice of the symbol is that γ is closely related to the
Γ function. For example, γ = −Γ′ (1).)
5.6.2. Show that for all x ≥ 0, x −
x2
x2
x3
≤ log(1 + x) ≤ x −
+ .
2
2
3
5.6.3. In “hyperbolic geometry” of the unit disc D := {(x, y) ∈ R2 : x2 + y 2 < 1} in the plane,
straight lines in D are circular arcs that are orthogonal to the bounding circle T. See Figure 1.
The distance between two points A, B is then taken as
D
T
P
B
A
C
Q
Figure 1. A straight line passing through A, B, C in the picture on the left. In Escher’s “Circle
Limit I”, the backbones of the fish lie along hyperbolic lines, and all of the white fish are
“hyperbolically congruent” to each other, as are all of the black fish.
⌢
d(A, B) := log
AP
⌢
BQ
⌢ · ⌢
AQ BP
!
,
⌢
where AP denotes the circular Euclidean arc length, etc. Show that for three points A, B, C lying
on such a line, d(A, B) + d(B, C) = d(A, C).
5.6.4. Show that
Z b
(1)
log xdx = b log b − a log a + a − b for all a, b ∈ (0, ∞).
a
(2)
Z
a
b
ex dx = eb − ea for all a, b ∈ R.
33
5.6. Transcendental functions
5.6.5. (Stirling’s Formula.)
(1) Compute the improper integral
Z
1
log xdx.
0
(2) Based on the result in the previous part, try giving a formal nonrigorous explanation of
Stirling’s Formula: for large n, log n! ≈ n log n − n.
5.6.6. If a, b > 0 and b 6= 1, then we define “the logarithm of a to the base b”, by logb a :=
log a
.
log b
(1) Show that blogb a = a.
(2) Prove that log2 3 is irrational.
(3) Show that there are irrational numbers a, b such that ab is rational.
(4) Find the flaw in the following argument given for the claim that 1 > 2.
“We know that 4 > 2. As the logarithm is a strictly increasing function, taking
log1/2 of both sides, we obtain −2 = log1/2 4 > log1/2 2 = −1, and so 1 > 2.”
(5) Sketch the graphs of x 7→ log x, log1/2 x, log10 x in the same picture.
5.6.7. Show that for all x ∈ R,
Z x
1
√
dt
1
+ t2
0
Z xp
1 + t2 dt
0
p
= log(x + 1 + x2 ),
√
√
x 1 + x2 + log(x + 1 + x2 )
=
.
2
5.6.8. Let a, xi ∈ R, and consider the “Initial Value Problem”
′
x (t) = ax′ (t), t ≥ 0,
x(0) = xi ,
Show that this Initial Value Problem has a unique solution, given by x(t) = eta xi , t ≥ 0.
Hint: To show uniqueness, consider the derivative of f , where f (t) := e−ta x(t).
5.6.9. (Newton’s Law of Cooling.) Newton’s Law of Cooling states that an object cools at a rate
proportional to the difference of its temperature and the temperature of the surrounding medium.
Find the temperature Θ(t) of an object at time t, in terms of the its temperature Θ0 at time
0, assuming that the temperature of the surrounding medium is kept at a constant, M . What
happens as t → ∞?
Hint: To solve the differential equation, note that (ekt (Θ − M ))′ = ekt (Θ′ + k(Θ − M )) for a
constant k.
5.6.10. (Radioactive decay.) A radioactive substance diminishes at a rate proportional to the
amount present (since all atoms have equal probability of disintegrating, the total disintegration
is proportional to the number of atoms remaining). If A(t) is the amount at time t, this means that
A′ (t) = −cA(t) for some c > 0 (which represents the probability that an atom will disintegrate).
(1) Find A(t) in terms of the amount A(0) = A0 present at time 0.
(2) Show that there is a number τ (the half-life of the radioactive element) with the property
that A(t + τ ) = A(t)/2 for all t.
34
5. Integration
5.6.11. (Compound interest.) An amount of P SEK is deposited in a bank which pays an interest
at a rate r per year, compounded m times a year. Thus the total amount (of principal plus interest)
at the end of n years is
r mn
.
A=P 1+
m
If r, n are kept fixed, show that this amount approaches the limit P ern as m → ∞. This motivates
the following definition. We say that the money grows at an annual rate r when compounded
continuously if the amount A(t) after t years is P ert for all t ≥ 0. Approximately how long does
it take for a bank account to double in value if it receives interest at an annual rate of 6% if
(1) compounded continuously?
(2) it is just a simple interest?
5.6.12. (The hyperbolic functions.) The hyperbolic sin and hyperbolic cos functions sinh, cosh :
R → R are defined as
cosh x :=
sinh x :=
ex + e−x
,
2
ex − e−x
,
2
for x ∈ R.
(1) Show that for any t ∈ R, the point (cosh t, sinh t) is on the hyperbola x2 − y 2 = 1. Sketch
the portion of the hyperbola obtained.
(2) Show also that
sinh 0 = 0, cosh 0 = 1,
sinh′ x = cosh x, cosh′ x = sinh x, x ∈ R,
cosh(x + y) = (cosh x)(cosh y) + (sinh x)(sinh y), x, y ∈ R.
(3) Sketch the graphs of sinh and cosh.
5.6.13. Show that lim xx = 1. Conclude that lim n1/n = 1.
n→∞
x→0+
5.6.14. Show that the improper integral
Z
0
Note that for x > 0,
1
1
dx exists.
xx
∞
X
1
(−x log x)n
−x log x
=
e
=
.
x
x
n!
n=0
Based on this, try giving a formal argument to justify the curious identity
Z 1
∞
X
1
1
dx
=
.
x
n
x
n
0
n=1
1
1
2
n n
5.6.15. Find lim
1+
.
1+
··· 1 +
n→∞
n
n
n
5.6.16. Order the following functions from the fastest growing to the slowest growing:
2x , ex , xx , (log x)x , ex/2 , x1/2 , log2 x, log(log x), (log x)2 , xe , x2 , log x, (2x)x , x2x .
35
5.6. Transcendental functions
5.6.17. Evaluate the following limits:
log(log x)
.
x→∞
log x
3sin x − 1
lim
.
x→0
x
sin x − x + x3 /6
.
lim
x→0
x3
cos x − 1 + x2 /2
.
lim
x→0
x4 1
1
lim
.
−
x→0 x
sin x
(1) lim
(2)
(3)
(4)
(5)
5.6.18. Evaluate
Z
1
2
− 12
(cos x) · log
1−x
1+x
dx.
5.6.19. Find the derivative of f : (0, ∞) → R, where
log x
, x > 0.
x
Which is bigger: eπ or π e ? (You may use the estimates e < 3 < π.)
f (x) =
5.6.20. Prove that if m, n ∈ N, then
Z π
0
(sin(mx))(sin(nx))dx =
π
−π
Z π
(cos(mx))(sin(nx))dx = 0.
if m 6= n,
if m = n
=
Z
π
(cos(mx))(cos(nx))dx,
−π
−π
5.6.21. (Fourier Series.) Let n ∈ N and a0 , a1 , · · · , an and b1 , · · · , bn be real numbers. Consider
the function f : R → R defined by
f (x) := a0 +
n
X
(ak cos(kx) + bk sin(kx)) ,
k=1
x ∈ R.
(1) Show that f is 2π-periodic, that is, f (x + 2π) = f (x) for all x ∈ R.
(2) Prove that
a0 =
and for k = 1, · · · , n,
ak
=
bk
=
1
2π
Z
π
f (x)dx,
−π
Z
1 π
f (x) cos(kx)dx,
π −π
Z
1 π
f (x) sin(kx)dx.
π −π
(3) Let ak := 0 for 0 ≤ k ≤ n, and


 4
kπ
bk :=

 0
if k is even,
otherwise.
Take n = 3, n = 33 and n = 333, successively, and in each case, plot the resulting f
using a package such as Maple, Mathematica or Matlab on the computer. What do you
observe?
36
5. Integration
5.6.22. (Fixed points of sin and cos.)
(1) Show that 0 is the only fixed point of sin : R → R.
(2) Prove that cos : R → R has a unique fixed point c∗ .
(3) Determine experimentally the value of c∗ as follows. In your scientific calculator, key in
any number, and press the “cos key” repeatedly. After a while, the display stabilizes.
Explain by means of a picture, why this displayed value must be c∗ (approximately).
5.6.23. (Addition formula for tan.)
(1) Show that for real x, y such that x, y, x + y are not in πZ +
tan(x + y) =
π
, there holds that
2
tan x + tan y
.
1 − (tan x)(tan y)
(2) Prove that tan 1◦ 6∈ Q. (Here one degree, 1◦ , is the the angle measuring
5.6.24. Show that
n
lim lim (cos(2πm!x)) =
m→∞ n→∞
Using the fact that e =
π
radians.)
180
1 if x ∈ Q,
0 if x 6∈ Q.
∞
X
1
and the above result, show that e 6∈ Q.
n!
n=0
5.6.25. (Visual differentiation of tan.) We have learnt that
tan′ x =
1
= 1 + (tan x)2 .
(cos x)2
(5.2)
Here we offer a formal geometric explanation. Consider the triangle shown in Figure 2. If we
ℓdx
d tan x
ℓ
dx
tan x
x
1
Figure 2. Visual differentiation of tan.
increase x by a small amount dx, then tan x increases by the amount d tan x shown in the figure.
Show that (5.2) holds by imagining that in the limiting case when dx diminishes to 0, the blue
triangle is ultimately similar to the red one.
5.6.26. Find
Z
0
1
1
dt.
1 + t2
37
5.7. Applications: length, area, volume
5.6.27. Show that
Z
0
π
2
1
√
1 + (tan x)
2
dx =
π
.
4
(The integrand is taken as 1 when x = 0, and 0 when x =
Hint: Consider the symmetry in the graph of
1
√
1 + (tan x)
π
.)
2
2
about
π 1
.
,
4 2
5.6.28. For x ∈ (−1, 1), the inverse sin−1 : (−1, 1) → (−π/2, π/2) of the strictly increasing
continuous function sin : (−π/2, π/2) → (−1, 1) exists. Prove that
For y ∈ (−1, 1), show that
Z
1
,
(sin−1 )′ (y) = p
1 − y2
y
1
dt
1 − t2
0
Z yp
1 − t2 dt
√
0
=
=
y ∈ (−1, 1).
sin−1 y,
p
y 1 − y 2 + sin−1 y
.
2
5.6.29. Find the polar coordinates of the following points given in Cartesian/rectangular coordinates:
(1) (1, 1).
(2) (1, 0).
(3) (0, 1).
(4) (−1, 0).
(5) (−1, −1).
(6) (0, −1).
5.7. Applications: length, area, volume
5.7.1. One winter morning in Lund, it started snowing at a heavy and constant rate. A snowplough started out at 8 : 00 a.m. At 9 : 00 a.m. it had gone 2 swedish miles. By 10 : 00 a.m., it
had gone 3 swedish miles. Assuming that the snowplough removes a constant volume of snow per
hour, determine the time at which it started snowing.
5.7.2. Find the area enclosed by an ellipse described by
y2
x2
+
= 1,
a2
b2
where a, b > 0. What happens when a = b?
5.7.3. The horizontal line y = c intersects the curve y = 2x − 3x3 in the first quadrant as shown
in the figure below. Find c so that the areas of the two shaded regions are equal.
5.7.4. Calculate the area enclosed by the two “petals” of the lemniscate given in polar coordinates
by r2 = 2 cos(2θ).
5.7.5. Calculate the volume of a doughnut, with the radius of the greater circle equal to R (that
is, of the central circle lying midway in the annular region obtained by taking a horizontal cross
section of the doughnut), and that of the two little circles, obtained by taking a vertical cross
section of the doughnut, equal to r.
38
5. Integration
y=c
2x3 − 3x2
Figure 3. The two shaded areas.
5.7.6. Calculate the volume of an ellipsoid, namely the solid of revolution obtained by revolving
the region enclosed by the ellipse
x2
y2
+
= 1,
a2
b2
where a, b > 0, in the upper half-plane and the x-axis. What happens when a = b?
5.7.7. A round hole of radius
the volume cut out.
√
3 is bored through the center of a solid ball of radius 2 cm. Find
5.7.8. The position of a particle in the plane R2 at time t ≥ 0 is given by
x(t)
=
y(t) =
1
(2t + 3)1/3 ,
3
t2
+ t,
2
for t ≥ 0. Find the distance it travels between t = 0 and t = 3. What is its average speed?
5.7.9. Calculate the arclength of the cardioid given in polar coordinates by r = 2(1 + cos θ).
5.7.10. (The Elliptic Integral.) Show that the perimeter of an ellipse given by
x2
y2
+
= 1,
a2
b2
where b ≥ a > 0, is given by
b
Z
0
2π
p
1 − k 2 (sin θ)2 dθ,
for a suitable constant k. (This integral is called an elliptic integral, and cannot be expressed in
terms of elementary functions when b 6= a.) What happens when b = a?
5.7.11. Calculate the surface area of a doughnut, with the radius of the greater circle equal to R
(that is, of the central circle lying midway in the annular region obtained by taking a horizontal
cross section of the doughnut), and that of the two little circles, obtained by taking a vertical cross
section of the doughnut, equal to r.
5.7.12. How accurately should we measure the radius of a ball in order to calculate its surface
area within 3% of its exact value?
39
5.7. Applications: length, area, volume
5.7.13. Let a ∈ R with a > 0. An arc of the “catenary” given by
x
y = a cos
a
whose end points have x-coordinates 0 and a is revolved about the x-axis. Show that the surface
area A and the volume V of the solid thus generated are related by the formula
2V
.
A=
a
5.7.14. (Design of a clepsydra.) A clepsydra (literally meaning “water thief” in Greek) or a water
clock is designed by revolving the graph of x 7→ Cxm : [0, r] → R, where C, m > 0, as shown in
Figure 4.
Cxm
r
Figure 4. Clepsydra.
What should m be if the level of water is to decrease linearly as time passes? You may use Toricelli’s
Law
√ stating that the speed of water flowing out when the height of water is h is proportional to
h.
Chapter 6
Series
6.1. Convergence/divergence of series
6.1.1. (Tantalising tan−1 .) Show that
∞
X
tan−1
n=1
π
1
= .
2n2
4
1
(2n + 1) − (2n − 1)
tan a − tan b
Hint: Write
=
and use tan(a − b) =
.
2
2n
1 + (2n + 1)(2n − 1)
1 + tan a tan b
6.1.2. Show that for every real number x > 1, the series
4
2n
2
1
+
+
·
·
·
+
+
+ ...
1 + x 1 + x2
1 + x4
1 + x2n
converges. Hint: Add
1
.
1−x
6.1.3. Consider the Fibonnaci sequence (Fn )n∈N with F0 = F1 = 1 and Fn+1 = Fn + Fn−1 for
n ∈ N. Show that
∞
X
1
= 1.
F
F
n=2 n−1 n+1
6.1.4. Does the series
∞
X
n=1
cos
1
converge?
n
6.1.5. Prove that if a1 ≥ a2 ≥ a3 . . . is a sequence of nonnegative numbers, and if
∞
X
an < +∞,
n=1
then an approaches 0 faster than 1/n, that is, lim nan = 0.
n→∞
Hint: Consider the inequalities an+1 + · · · + a2n ≥ n · a2n and an+1 + · · · + a2n+1 ≥ n · a2n+1 .
Show that the assumption a1 ≥ a2 ≥ a3 . . . above cannot be dropped by considering the
lacunary series whose n2 th term is 1/n2 and all other terms are zero.
6.1.6. Consider the Arithmetic-Geometric progression
1, 2r, 3r2 , 4r3 , · · ·
41
42
6. Series
where r ∈ R. Note that 1, 2, 3, 4, · · · form an arithmetic progression, while 1, r, r2 , r3 , · · · form a
geometric progression. Show that if |r| < 1, then
1 + 2r + 3r2 + 4r3 + · · · =
1
.
(1 − r)2
6.2. Absolute convergence and the Leibniz’s
Alternating Series Test
6.2.1. Does the series
∞
X
sin n
converge?
n2
n=1
6.2.2. Let s > 0. Show that
∞
X
(−1)n
converges.
ns
n=1
√
n
converges.
6.2.3. Prove that
(−1)
n
+
1
n=1
∞
X
6.2.4. Prove that
∞
X
n
(−1)n sin
n=1
1
converges.
n
6.3. Comparison, Ratio, Root
6.3.1. Determine if the following series are convergent or not.
(1)
∞
X
n2
.
2n
n=1
∞
X
(n!)2
.
(2n)!
n=1
∞ n
X
4
(3)
n5 .
5
n=1
(2)
6.3.2. Prove that
∞
X
n
converges. (∗) Can you also find its value?
4 + n2 + 1
n
n=1
Hint: Write the denominator as (n2 + 1)2 − n2 .
6.3.3. Let (an )n∈N be a real sequence such that
∞
X
n=1
Hint: First conclude that for large n, |an | < 1.
a4n
converges. Show that
∞
X
a5n converges.
n=1
6.3.4. (∗) We have seen that the series
∞
X
1
np
n=1
converges if and only if p > 1. Suppose that we sum the series for all n that can be written without
using the numeral 9. (Imagine that the key for 9 on the keyboard is broken.)
X′
. Prove that the series
Call the resulting summation
X′ 1
np
43
6.3. Comparison, Ratio, Root
converges if and only if p > log10 9. Hint: First show that there are 8 · 9k−1 numbers without 9
between 10k−1 and 10k .
6.3.5. Determine if the following statements are true or false.
(1) If
(2) If
∞
X
n=1
∞
X
|an | is convergent, then so is
an is convergent, then so is
n=1
n→∞
∞
X
n=1
a2n .
an converges.
n→∞
log
n=1
∞
X
n=1
(4) If lim (a1 + · · · + an ) = 0, then
(5)
a2n .
n=1
(3) If lim an = 0, then
∞
X
∞
X
n+1
converges.
n
∞
X
an converges.
n=1
(6) If an > 0 (n ∈ N) and the partial sums of (an )n∈N are bounded above, then
converges.
(7) If an > 0 (n ∈ N) and
∞
X
an converges, then
n=1
∞
X
an
n=1
∞
X
1
diverges.
a
n=1 n
6.3.6. (Fourier series.) In order to understand a complicated situation, it is natural to try to
break it up into simpler things. For example, from Calculus we learn that an analytic function
can be expanded into a Taylor series, where we break it down into the simplest possible analytic
functions, namely monomials 1, x, x2 , . . . as follows:
f (x) = f (0) + f ′ (0)x +
f ′′ (0) 2
x + ....
2!
The idea behind the Fourier series is similar. In order to understand a complicated periodic
function, we break it down into the simplest periodic functions, namely sines and cosines. Thus
if T ≥ 0 and f : R → R is T -periodic, that is, f (x) = f (x + T ) (x ∈ R), then one tries to find
coefficients a0 , a1 , a2 , . . . and b1 , b2 , b3 , . . . such that
∞ X
2πn
2πn
an cos
f (x) = a0 +
x + bn sin
x .
(6.1)
T
T
n=1
Suppose that the Fourier series given in (6.1) converges pointwise to f on R. Show that if
∞
X
n=1
(|an | + |bn |) < ∞,
then in fact the series converges uniformly.
The aim of this part of the exercise is to give experimental evidence for two things. Firstly,
the plausibility of the Fourier expansion, and secondly, that the uniform convergence might fail if
the condition in the previous part of this exercise does not hold. To this end, let us consider the
square wave f : R → R given by
1 if x ∈ [n, n + 1) for n even,
f (x) =
−1 if x ∈ [n, n + 1) for n odd.
44
6. Series
Then f is a 2-periodic signal. From the theory of Fourier Series, which we will not discuss in this
course, the coefficients can be calculated, and they happen to be 0 = a0 = a1 = a2 = a3 = . . . and


 4 if n is odd,
nπ
bn =

 0
if n is even.
Write a Maple program to plot the graphs of the partial sums of the series in (6.1) with, say, 3,
33, 333 terms. Discuss your observations.
Figure 1. Partial sums of the Fourier series for the square wave.
6.3.7. Show that if
∞
X
an converges, where each an is nonnegative, then so does
n=1
∞
X
√
an an+1 .
n=1
6.3.8. Let an ≥ 0 for all n ∈ N. Prove that
∞
X
an converges if and only if
n=1
6.3.9. Let ℓ1 , ℓ2 be defined by
ℓ1
ℓ
Show that ℓ1 ⊂ ℓ2 . Is ℓ1 = ℓ2 ?
2
:=
:=
(
(
(an )n∈N :
(an )n∈N :
∞
X
n=1
∞
X
n=1
)
|an | < ∞ ,
2
)
|an | < ∞ .
∞
X
an
converges.
1
+
an
n=1
45
6.4. Power series
6.3.10. As the harmonic series
∞
X
1
diverges, the reciprocal 1/sn of the nth partial sum
n
n=1
1 1
1
+ + ···+
2 3
n
approaches 0 as n → ∞. So the necessary condition for the convergence of the series
∞
X
1
sn := 1 +
n=1
sn
is satisfied. But we don’t know yet whether or not it actually converges. It is clear that the harmonic series diverges very slowly, which means that 1/sn decreases very slowly, and this prompts
the guess that this series diverges. Show that in fact our guess is correct. Hint: sn < n.
6.3.11. Show that the series
∞
X
1
converges.
n
n
n=1
6.3.12. Consider the Fibonnaci sequence (Fn )n∈N with F0 = F1 = 1 and Fn+1 = Fn + Fn−1 for
n ∈ N. Show that
∞
X
1
< +∞.
F
n=0 n
Hint: Fn+1 = Fn + Fn−1 > Fn−1 + Fn−1 = 2Fn−1 . Using this, show that both
and
1
1
1
+
+
+ . . . converge.
F1
F3
F5
6.3.13. Show that
∞
X
k=1
1
1
1
+
+
+...
F0 F2 F4
p
sin π k 4 + 1 converges absolutely.
6.4. Power series
6.4.1. Show that for all x ∈ R,
∞
X
x3
x5
x2n−1
=x−
+
− + · · · , and
sin x =
(−1)n−1
(2n − 1)!
3!
5!
n=1
cos x
=
∞
X
n=0
(−1)n
x2
x4
x2n
=1−
+
− +··· .
(2n)!
2!
4!
6.4.2. (A C ∞ function which is not analytic.) Let f : R → R be given by
(
2
e−1/x if x 6= 0,
f (x) =
0
if x = 0.
(1) Sketch the graph of f .
f (x)
= 0.
x→0 xn
(3) Show that for each n ∈ N, there exists a polynomial pn such that for all nonzero x,
1
2
.
f (n) (x) = e−1/x p
x
(4) Prove that f (n) (0) = 0 for all n ≥ 1 (showing that the Taylor series for f at 0 is identically
zero, clearly not equal to f ).
(2) Prove that for every n ∈ N, lim
Solutions
Solutions to the exercises from Chapter 1
Solution to Exercise 1.1.1.
(1) 12 = 1 and
n(n + 1)(2n + 1) (n=1) 1 · 2 · 3
= 1.
=
6
6
So the claim is true for n = 1. If for some n ∈ N there holds that
n(n + 1)(2n + 1)
,
1 2 + 2 2 + 3 3 + · · · + n2 =
6
then
n(n + 1)(2n + 1)
12 + · · · + n2 + (n + 1)2 =
+ (n + 1)2
6
(n + 1) n(2n + 1) + 6(n + 1)
=
6
(n + 1)
(n + 1)
(2n2 + 7n + 6) =
(n + 2)(2n + 3)
=
6
6
(n + 1)((n + 1) + 1)(2(n + 1) + 1)
=
.
6
By induction, the claim is true for all n ∈ N.
(2) Let us first find the entry in the jth place from the left of the ith row for each of the
triangles.
In the first triangle on the left hand side, the ith row has all entries equal to i. In
particular, also the jth entry from the left in this ith row is equal to i.
In the second triangle on the left hand side, the first entry in the ith row is n − i + 1.
The entries along the row increase from left to right. Thus the second entry is n−i+1+1
and the jth entry will be n − i + 1 + (j − 1).
In the third triangle on the left hand side, the first entry in the ith row is n. The
entries along the row decrease from left to right. Hence the second entry is n − 1, and
the jth entry is n − (j − 1).
So if we add the triangular arrays on the left entry-wise, then the jth entry in the
ith row of the resulting triangular array is
i + (n − i + 1 + (j − 1)) + (n − (j − 1)) = 2n + 1.
But the sum of the numbers in each triangle on the left is
1 + 2 · 2 + 3 · 3 + · · · + n · n = 12 + 22 + 33 + · · · + n2 =: Sn .
47
48
Solutions
Hence
3Sn
=
sum of the entries of the trianular array on the right hand side
n(n + 1)
,
= (2n + 1) · (1 + 2 + 3 + · · · + n) = (2n + 1) ·
2
n(n + 1)(2n + 1)
and so Sn =
.
6
Solution to Exercise 1.2.1. See Figure 2 for the construction of −11/6 on the real number line.
O≡0
P ≡ −11/6
A≡1
P ′ ≡ 11/6
B1
B2
B3
B4
B5
B6
B7
B8
B9
B10
B11
Figure 2. Construction of −11/6 on the real number line.
The steps of the construction are as follows:
(1) Draw a ray OA and take |OA| = 1.
(2) Draw a ray OB1 not parallel to OA.
(3) Cut equal lengths |OB1 | = |B1 B2 | = |B2 B3 | = · · · = |B10 B11 | along the ray OB1 .
(4) Join A to B6 , and construct B1 1P ′ parallel to AB6 , meeting OA (extended) at P ′ . Then
P ′ ≡ 11/6.
(5) With O as center and radius |OP ′ |, draw a circle which meets the line passing through
O and A at the point P 6≡ P ′ . Then P ≡ −11/6.
√
See Figure 3 for the construction of 3 on the real number line.
The steps of the construction are as follows:
(1) Draw a ray OA and take |OA| = 1.
(2) Construct AB perpendicular to OA such that |AB| = 1. Then |OB| =
√
√
12 + 12 = 2.
49
Solutions to the exercises from Chapter 1
O≡0
B
B′
A≡1
A′ ≡
Figure 3. Construction of
√
√
2
A′′ ≡
√
3
3 on the real number line.
(3) With √
center O and radius |OB| draw a circular arc meeting OA extended at A′ . Then
′
A ≡ 2.
(4) Construct A′ B ′ perpendicular to OA′ such that |A′ B ′ | = 1. Then
q√
√
√
′
|OB | = ( 2)2 + 12 = 2 + 1 = 3.
(5) With center
O and radius |OB ′ | draw a circular arc meeting OA′ extended at A′′ . Then
√
′′
A ≡ 3.
Solution to Exercise 1.2.2.
(1) Let a ∈ R and suppose that b1 , b2 ∈ R are such that
a + b1 = 0 = b1 + a and a + b2 = 0 = b2 + a.
Then b1 = b1 + 0 = b1 + (a + b2 ) = (b1 + a) + b2 = 0 + b2 = b2 . So b1 = b2 .
(2) Let a ∈ R. We have 0 · a + 0 · a = (0 + 0) · a = 0 · a. Adding −(0 · a) on both sides, we
have
(0 · a + 0 · a) + (−(0 · a)) = 0 · a + (−(0 · a)) = 0,
that is, 0 · a + (·a + (−(0 · a))) = 0. So 0 · a + 0 = 0, that is, 0 · a = 0.
(3) Let a ∈ R. Then a + (−1) · a = 1 · a + (−1) · a = (1 + (−1)) · a = 0 · a = 0. By the
uniqueness of inverses, −a = (−1) · a.
Solution to Exercise 1.2.3.
(1) We have the following cases:
1◦ a = 0. Then a2 = 0 · 0 = 0. So a2 ≥ 0.
2◦ a ∈ P. Then a, a ∈ P gives a2 = a · a ∈ P. So a2 ≥ 0.
3◦ −a ∈ P. Then −a, −a ∈ P gives a2 = (−a) · (−a) ∈ P. So a2 ≥ 0.
(2) We have that 1 6= 0 and 1 = 1 · 1 = 12 ∈ P. Hence 1 > 0.
(3) Clearly with x = 0, x2 + 1 = 02 + 1 = 0 + 1 = 1 6= 0. So x 6= 0. But then
x2 = x · x = (−x) · (−x) ∈ P.
Also 1 ∈ P. Hence x2 + 1 ∈ P. But x2 + 1 = 0 6∈ P, a contradiction.
50
Solutions
Solution to Exercise 1.3.1.
(1) We have
pd
p
+ · · · + cd d = 0.
q
q
d
Multiplying throughout by q , we obtain
cd pd = − c0 q d + c1 pq d−1 + · · · + cd−1 pd−1 q .
c0 + c1
(6.2)
As q divides the right hand side, q divides c0 pd . But q has no common factors with p,
and this implies that q divides cd . (This is because if we decompose p, q as products
of powers of primes, then there can be no common prime occurring in their respective
decompositions.) Also by rearranging (6.2), we obtain
c0 q d = − c1 pq d−1 + · · · + cd−1 pd−1 q + cd pd ,
and since p divides the right hand side, p must divide c0 q d . But p and q have no common
factor. So p must divide c0 .
√
(2) Suppose that 2 is rational, and let
√
p
2= ,
q
where p, q ∈ Z, q > 0, and p, q have no common factor. Then p/q is a rational zero of
the polynomial x2 − 2. By the Rational Zeros Theorem, p divides −2 and q divides 1.
So p ∈ {2, −2, 1, −1} and q = 1. But then
p
∈ {2, −2, 1, −1}.
q
√
2 is not equal to any of the values 2, −2, 1, −1. This contradiction shows
But clearly
√
that 2 6∈ Q.
√
(3) Suppose that 3 6 is rational, and let
√
p
3
6= ,
q
where p, q ∈ Z, q > 0, and p, q have no common factor. Then p/q is a rational zero of
the polynomial x3 − 6. By the Rational Zeros Theorem, p divides −6 and q divides 1.
So p ∈ {6, −6, 3, −3, 2, −2, 1, −1} and q = 1. But then
p
∈ {6, −6, 3, −3, 2, −2, 1, −1}.
q
√
But none of these values satisfy x3 − 6 = 0. This contradiction shows that 3 6 6∈ Q.
Solution to Exercise 1.4.1.
(1) S = (0, 1].
An upper bound of S. 1 is an upper bound, since for all x ∈ S = (0, 1], we have x ≤ 1.
In fact any real number u ≥ 1 is an upper bound.
A lower bound of S. 0 is a lower bound, since for all x ∈ S = (0, 1], we have 0 < x. In
fact any real number ℓ ≤ 0 is a lower bound.
Is S bounded? Yes. S is bounded above, since 1 is an upper bound of S. S is also
bounded below, since 0 is a lower bound. Since S is bounded above as well as bounded
below, it is bounded.
Solutions to the exercises from Chapter 1
51
Supremum of S. sup S = 1. Indeed, 1 is an upper bound, and moreover, if u is also an
upper bound, then 1 ≤ u (since 1 ∈ S).
Infimum of S. inf S = 0. First of all, 0 is a lower bound. Let ℓ be a lower bound of S. We
prove that ℓ ≤ 0. (We do this by supposing that ℓ > 0, and arriving at a contradiction.
The contradiction is obtained as follows: if ℓ > 0, then we will see that the average of
0 and ℓ, namely 2ℓ , is an element in S that is less than the lower bound ℓ, which is a
contradiction to the definition of a lower bound!) If ℓ > 0, then 0 < 2ℓ . Moreover, since
ℓ ≤ 1 (ℓ is a lower bound of S and 1 ∈ S) it follows that 2ℓ ≤ 12 ≤ 1. Thus 2ℓ ∈ S. But
since ℓ > 0, it follows that 2ℓ < ℓ, a contradiction. Hence ℓ ≤ 0.
If sup S exists, then is sup S ∈ S? Yes, sup S = 1 ∈ (0, 1] = S.
If inf S exists, then is inf S ∈ S? No, inf S = 0 6∈ (0, 1] = S.
Maximum of S. max S = 1, since sup S = 1 ∈ S.
Minimum of S. min S does not exist since inf S = 0 6∈ S.
(2) S = [0, 1].
An upper bound of S. 1 is an upper bound, since for all x ∈ S = [0, 1], we have x ≤ 1.
A lower bound of S. 0 is a lower bound, since for all x ∈ S = (0, 1], we have 0 ≤ x.
Is S bounded? Yes, since S is bounded above (1 is an upper bound) and it is bounded
below (0 is a lower bound).
Supremum of S. sup S = 1. Indeed, 1 is an upper bound, and moreover, if u is also an
upper bound, then 1 ≤ u (since 1 ∈ S).
Infimum of S. inf S = 0. Indeed, 0 is an lower bound, and moreover, if ℓ is also an lower
bound, then ℓ ≤ 0 (since 0 ∈ S).
If sup S exists, then is sup S ∈ S? Yes, sup S = 1 ∈ [0, 1] = S.
If inf S exists, then is inf S ∈ S? Yes, inf S = 0 ∈ [0, 1] = S.
Maximum of S. max S = 1, since sup S = 1 ∈ S.
Minimum of S. min S = 0, since inf S = 0 ∈ S.
(3) S = (0, 1).
An upper bound of S. 1 is an upper bound, since for all x ∈ S = (0, 1), x < 1.
A lower bound of S. 0 is a lower bound, since for all x ∈ S = (0, 1), 0 < x.
52
Solutions
Is S bounded? Yes, since S is bounded above (1 is an upper bound) and it is bounded
below (0 is a lower bound).
Supremum of S. sup S = 1. First of all, 1 is a upper bound. Let u be an upper bound
of S. We prove that 1 ≤ u. (We do this by supposing that u < 1 and arriving at a
contradiction. The contradiction is obtained as follows: if u < 1, then we will see that
the average of u and 1, namely u+1
2 , is an element in S that is larger than the upper
bound u, which is a contradiction to the definition of an upper bound!) Since u is an
upper bound and since 12 ∈ S, it follows that 0 < u (since 0 < 12 ≤ u). So if u < 1, then
u+1
1+1
u+1
u+1
0 < u = u+u
2 < 2 < 2 = 1. Hence 2 ∈ S. But u < 2 contradicts the fact that
u is an upper bound of S.
Infimum of S. inf S = 0. First of all, 0 is a lower bound. Let ℓ be a lower bound of S.
We prove that ℓ ≤ 0. If ℓ > 0, then 0 < 2ℓ . Moreover, since 21 ∈ S and ℓ is a lower bound
of S, it follows that ℓ ≤ 12 . Thus we have 0 < 2ℓ < l ≤ 12 < 1, and so 2ℓ ∈ S. But 2ℓ < l
contradicts the fact that ℓ is a lower bound of S.
If sup S exists, then is sup S ∈ S? No, sup S = 1 6∈ (0, 1) = S.
If inf S exists, then is inf S ∈ S? No, inf S = 0 6∈ (0, 1) = S.
Maximum of S. max S does not exist, since sup S = 1 6∈ S.
Minimum of S. min S does not exist, since inf S = 0 6∈ S.
(4) S =
1
n
| n ∈ Z \ {0} = n1 | n ∈ N ∪ − n1 | n ∈ N .
An upper bound of S. 1 is an upper bound, since for all n ∈ N,
1
n
≤ 1 and − n1 ≤ 0 ≤ 1.
A lower bound of S. −1 is a lower bound, since for all n ∈ N, −1 ≤ 0 ≤
1
n
and −1 ≤ − n1 .
Is S bounded? Yes, since S is bounded above (1 is an upper bound) and it is bounded
below (−1 is a lower bound).
Supremum of S. sup S = 1. 1 is an upper bound. Moreover, if u is also an upper bound,
then since 1 = 11 ∈ S, 1 ≤ u.
Infimum of S. inf S = −1. −1 is a lower bound. Moreover, if ℓ is also a lower bound,
1
∈ S, it follows that ℓ ≤ −1.
then since −1 = −1
If sup S exists, then is sup S ∈ S? Yes, sup S = 1 ∈ S.
If inf S exists, then is inf S ∈ S? Yes, inf S = −1 ∈ S.
Maximum of S. max S = 1, since sup S = 1 ∈ S.
Minimum of S. min S = −1, since inf S = −1 ∈ S.
(5) S = − n1 | n ∈ N .
53
Solutions to the exercises from Chapter 1
An upper bound of S. 0 is an upper bound, since for all n ∈ N, − n1 < 0.
A lower bound of S. −1 is a lower bound, since for all n ∈ N, −1 ≤ − n1 .
Is S bounded? Yes, since S is bounded above (0 is an upper bound) and it is bounded
below (−1 is a lower bound).
Supremum of S. sup S = 0. 0 is an upper bound. Moreover, if u is an upper bound and
1
1
if u < 0, then let N ∈ N be such that −u
< N (Archimedean principle with y = −u
and
1
x = 1!), and so we obtain u < − N ∈ S, a contradiction to the fact that u is an upper
bound of S. Thus if u is an upper bound, then 0 ≤ u.
Infimum of S. inf S = −1. −1 is a lower bound, and if ℓ is another lower bound, then
since −1 = − 11 ∈ S, it follows that ℓ ≤ −1.
If sup S exists, then is sup S ∈ S? No, since sup S = 0 6∈ S.
If inf S exists, then is inf S ∈ S? Yes, inf S = −1 ∈ S.
Maximum of S. max S does not exist since sup S = 0 6∈ S.
Minimum of S. min S = −1 since inf S = −1 ∈ S.
(6) S =
n
n
n+1
o
|n∈N .
An upper bound of S. 1 is an upper bound, since for all n ∈ N,
A lower bound of S.
that is, 1 ≤ n).
1
2
is a lower bound because for all n ∈ N,
1
2
n
n+1
≤
< 1.
n
n+1
(since n + 1 ≤ 2n,
Is S bounded? Yes, since S is bounded above (1 is an upper bound) and it is bounded
below ( 21 is a lower bound).
Supremum of S. sup S = 1. 1 is an upper bound of S. If u < 1 is an upper bound, then
u
< N (Archimedean property). Then u < NN+1 , contradicting
let N ∈ N be such that u−1
the fact that u is an upper bound.
Infimum of S. inf S = 12 . 21 is a lower bound, and if ℓ is a lower bound, then since
1
1
1
2 = 1+1 ∈ S, it follows that ℓ ≤ 2 .
If sup S exists, then is sup S ∈ S? No, since sup S = 1 6∈ S.
If inf S exists, then is inf S ∈ S? Yes, since inf S =
1
2
∈ S.
Maximum of S. max S does not exist since sup S = 1 6∈ S.
Minimum of S. min S exists since inf S =
1
2
∈ S.
54
Solutions
(7) S = {x ∈ R | x2 ≤ 2}.
√
√
An upper bound of S. 2 is an √
upper bound.√ (x > 2 implies x2 > 2, and so x 6∈ S. In
other words, if x ∈ S, then x ≤ 2, that is, 2 is an upper bound of S.)
√
√
√
A lower bound of S. − 2 is a lower bound. (x < − 2 implies that −x > 2 > 1 > 0.
2
So using
>
√ fact
√ that if a > b and c > 0, then ac > bc, we get x = (−x)(−x)
√
√ the
(−x)
2
>
2
2
=
2,
and
so,
x
∈
6
S.
In
other
words,
if
x
∈
S,
then
x
≥
−
2,
and
so
√
− 2 is a lower bound.)
√
Is S bounded?
Yes, since S is bounded above ( 2 is an upper bound) and it is bounded
√
below (− 2 is a lower bound).
√
√
Supremum of S. sup S = 2. First of all,
2 is an upper bound of S. Let u be an
√
√
u+ 2
upper bound such that u < 2. Then 2 ∈ S. (As 0 ∈ S and u is an upper bound
√
√
√
√
√
of S, 0 ≤ u. As u < 2, u < u+2 2 < 2. Hence we have 0 < u+2 2 < 2, and
√ 2
√ √ √ √
√
√ √
u+ 2
u+ 2
so u+2 2
2 < 2 2 = 2. Thus u+2 2 ∈ S.) But
= u+2 2
<
2
2
√
u < u+2 2 contradicts the fact that u is an upper bound. So if u is an upper bound of
√
S, then 2 ≤ u.
√
√
Infimum
of S. inf S = − 2. −√ 2 is a lower bound. Let
√ ℓ be a lower bound such that
√
− 2 < ℓ. Then we have −ℓ < 2, and since sup S = 2, it follows that −ℓ cannot be
an upper bound of S. So there exists an x ∈ S such that −ℓ < x. But since x ∈ S, we
have (−x)2 = x2 ≤ 2. So it follows that −x ∈ S. As ℓ is a lower bound, ℓ ≤ −x, that
is, x ≤ −ℓ. From −ℓ < x and x ≤ −ℓ, we arrive
√ at the contradiction that −ℓ < −ℓ.
Consequently, if ℓ is a lower bound, then ℓ ≤ − 2.
If sup S exists, then is sup S ∈ S? Yes, sup S =
√
2 ∈ S.
√
If inf S exists, then is inf S ∈ S? Yes, inf S = − 2 ∈ S.
Maximum of S. max S =
√
2.
√
Minimum of S. min S = − 2.
(8) S = {0, 2, 1, 3, 2013}.
An upper bound of S. 2013 is an upper bound, since 0 ≤ 2013, 2 ≤ 2013,, 1 ≤ 2013,
3 ≤ 2013 and 2013 ≤ 2013.
A lower bound of S. 0 is a lower bound, since 0 ≤ 0, 0 ≤ 2, 0 ≤ 2, 0 ≤ 1, 0 ≤ 3 and
0 ≤ 2013.
Is S bounded? Yes, since S is bounded above (2013 is an upper bound) and it is bounded
below (0 is a lower bound).
Supremum of S. sup S = 2013. 2013 is an upper bound, and if u is also an upper bound,
then since 2013 ∈ S, it follows that 2013 ≤ u.
55
Solutions to the exercises from Chapter 1
Infimum of S. inf S = 0. 0 is an lower bound, and if ℓ is also an lower bound, then since
0 ∈ S, it follows that ℓ ≤ 0.
If sup S exists, then is sup S ∈ S? Yes, sup S = 2013 ∈ {0, 2, 1, 3, 2013} = S.
If inf S exists, then is inf S ∈ S? Yes, inf S = 0 ∈ {0, 2, 1, 3, 2013} = S.
Maximum of S. max S = 2013 since sup S = 2013 ∈ S.
Minimum of S. min S = 0 since inf S = 0 ∈ S.
(9) S = (−1)n 1 + n1 | n ∈ N .
(This set has the elements − 12 , 32 , − 43 , 45 , . . . .)
An upper bound of S. 23 is an upper bound. If n ∈ N and n is even, then (−1)n 1 + n1 = 1 + n1 ≤ 1 + 21 = 23 .
If n ∈ N and n is odd, then (−1)n 1 + n1 = −1 − n1 < 0 < 23 .
A lower bound of S. −2 is a lower bound.
If n ∈ N and n is even, then (−1)n 1 + n1 = 1 + n1 > 0 > −2.
If n ∈ N and n is odd, then (−1)n 1 + n1 = −1 − n1 ≥ −1 − 1 = −2.
Is S bounded? Yes, since S is bounded above ( 23 is an upper bound) and it is bounded
below (−2 is a lower bound).
Supremum of S. sup S = 23 . 23 is an upper bound, and if u is also an upper bound, then
since 32 = (−1)2 1 + 21 ∈ S, it follows that 32 ≤ u.
Infimum of S. inf S = −2.
−2 is a lower bound, and if ℓ is also a lower bound, then
1
1
since −2 = (−1) 1 + 1 ∈ S, it follows that ℓ ≤ −2.
If sup S exists, then is sup S ∈ S? Yes, sup S =
3
2
∈ S.
Maximum of S. max S = 32 .
Minimum of S. min S = −2.
(10) S = {x2 | x ∈ R}.
An upper bound of S. There does not exist an upper bound of S. Let u ∈ R be an upper
2
bound of S, then u + 12 ∈ R, and so u2 + u + 41 = u + 12 ∈ S. But since u2 + 14 > 0,
2
it follows that u < u2 + u + 41 = u + 21 ∈ S, contradicting the fact that u is an upper
bound of S. Thus S does not have an upper bound.
A lower bound of S. 0 is a lower bound of S, since for all x ∈ R, 0 ≤ x2 .
Is S bounded? No, since the set is not bounded above.
56
Solutions
Supremum of S. sup S does not exist, since the set does not have an upper bound, and
so it cannot have a least upper bound. (Recall that every least upper bound is an upper
bound).
Infimum of S. inf S = 0. 0 is a lower bound, and if ℓ is also a lower bound, then since
0 = 02 ∈ S, it follows that ℓ ≤ 0.
If sup S exists, then is sup S ∈ S? sup S does not exist.
If inf S exists, then is inf S ∈ S? Yes, inf S = 0 ∈ S.
Maximum of S. max S does not exist, since sup S does not exist.
Minimum of S. min S = 0, since inf S = 0 ∈ S.
(11) S =
n
x2
1+x2
o
|x∈R .
An upper bound of S. 1 is an upper bound of S, since for all x ∈ R,
x2
1+x2
A lower bound of S. 0 is a lower bound of S, since for all x ∈ R, 0 ≤
< 1.
x2
1+x2 .
Is S bounded? Yes, since S is bounded above (1 is an upper bound) and it is bounded
below (0 is a lower bound).
Supremum of S. sup S = 1. 1 is an upper bound. Let u be
q an upper bound and let
02
u
< N . Then we have
u < 1. Since 0 = 1+02 ∈ S, 0 ≤ u. Let N ∈ N be such that 1−u
2
N
u < 1+N
2 ∈ S, a contradiction to the fact that u is an upper bound of S. Hence if u is
an upper bound of S, then 1 ≤ u.
Infimum of S. inf S = 0. 0 is a lower bound, and if ℓ is also a lower bound, then since
02
0 = 1+0
2 ∈ S, it follows that ℓ ≤ 0.
If sup S exists, then is sup S ∈ S? No, sup S = 1 6∈ S.
If inf S exists, then is inf S ∈ S? Yes, inf S = 0 ∈ S.
Maximum of S. max S does not exist, since sup S 6∈ S.
Minimum of S. min S = 0, since inf S = 0 ∈ S.
Solution to Exercise 1.4.2. Since S is bounded, in particular, it is bounded above, and furthermore, since it is nonempty, sup S exists, by the least upper bound property of R.
Since S is bounded, in particular it is bounded below, and furthermore, since it is nonempty,
inf S exists, by the greatest lower bound property of R.
Let x ∈ S. Since inf S is a lower bound of S,
inf S ≤ x.
(6.3)
Solutions to the exercises from Chapter 1
57
Moreover, since sup S is an upper bound of S,
x ≤ sup S.
(6.4)
From (6.3) and (6.4), we obtain inf S ≤ sup S.
Let inf S = sup S. If x ∈ S, then we have
inf S ≤ x ≤ sup S,
(6.5)
and so inf S = x(= sup S) (for if inf S < x, then from (6.5), inf S < sup S, a contradiction). Thus
S is a singleton set. Conversely, if S = {x}, then clearly x is an upper bound. If u < x is an
upper bound, then x ≤ u < x gives x < x, a contradiction. So sup S = x. Clearly x is also a
lower bound. If ℓ > x is also a lower bound, then x > ℓ ≥ x gives x > x, a contradiction. So
inf S = x = sup S.
Solution to Exercise 1.4.3. Since sup B is an upper bound of B, we have x ≤ sup B for all
x ∈ B. Since A ⊂ B, in particular we obtain x ≤ sup B for all x ∈ A. So sup B is an upper bound
of A, and so by the definition of the least upper bound of A, we obtain sup A ≤ sup B.
Solution to Exercise 1.4.4. If a ∈ A, then a ≤ sup A ≤ max{sup A, sup B}. Also if b ∈ B,
then b ≤ sup B ≤ max{sup A, sup B}. So if x ∈ A ∪ B, then either x ∈ A or x ∈ B, and from the
above, x ≤ max{sup A, sup B}. Hence A ∪ B is bounded above by max{sup A, sup B}. Also, as
A is nonempty, A ∪ B is nonempty. Hence by the least upper bound property of R, sup(A ∪ B)
exists. Also, it follows from the above that
sup(A ∪ B) ≤ max{sup A, sup B}.
(6.6)
Since A ⊂ A ∪ B, we also have that sup A ≤ sup(A ∪ B). Similarly, as B ⊂ A ∪ B, sup B ≤
sup(A ∪ B). From here we obtain
max{sup A, sup B} ≤ sup(A ∪ B).
(6.7)
From (6.6) and (6.7), it follows that sup(A ∪ B) = max{sup A, sup B}.
Solution to Exercise 1.4.5.
(1) FALSE (if S = {1}, then u = 3 is an upper bound of S, and although u′ = 2 < 3 = u,
u′ (= 2) is an upper bound of {1} = S).
(2) TRUE (if ǫ > 0, then u∗ − ǫ < u∗ , and so u∗ − ǫ cannot be an upper bound of S).
(3) FALSE (N has no maximum).
(4) FALSE (N has no supremum).
(5) FALSE ([0, 1) is bounded, but it does not have a maximum).
(6) FALSE (∅ is bounded, but it does not have a supremum).
(7) TRUE (least upper bound property of R).
(8) TRUE (the supremum itself is an upper bound).
(9) TRUE (definition of maximum).
(10) FALSE (the set [0, 1) has supremum 1, but 1 6∈ [0, 1)).
(11) FALSE (−N is bounded above since 0 is an upper bound, but | − N| = N is not bounded).
(12) TRUE (ℓ ≤ x ≤ u implies x ≤ u and −x ≤ −ℓ, and so we have
x ≤ u ≤ max{−ℓ, u} and − x ≤ −ℓ ≤ max{−ℓ, u}.
Thus |x| ≤ max{−ℓ, u} and so max{−ℓ, u} is an upper bound of |S|. Moreover, for every
y ∈ |S|, we have y = |x| for some x ∈ S, and so y = |x| ≥ 0. Thus 0 is a lower bound of
|S|.)
58
Solutions
(13) FALSE (if S = {0, 1}, then inf S = 0 <
1
2
< 1 = sup S, but
1
2
6∈ S).
Solution to Exercise 1.4.6. Since A is bounded above, there exists MA ∈ R such that for all
a ∈ A, a ≤ MA . Also, since B is bounded above, there exists MB ∈ R such that for all b ∈ B,
b ≤ MB . Consequently if a ∈ A and b ∈ B, a+ b ≤ MA + MB . So for all x ∈ A+ B, x ≤ MA + MB ,
and so MA + MB is an upper bound for A + B. So A + B is bounded above.
Clearly A + B is nonempty. Indeed, since A is nonempty, there exists some element a ∈ A,
and as B is nonempty, there exists an element b ∈ B, and so a + b ∈ A + B, that is, A + B is not
empty.
Since A + B is bounded above and it is not empty, by the least upper bound property of R,
it follows that sup(A + B) exists.
Since sup A is an upper bound of A, we have that for all a ∈ A, a ≤ sup A. Also, as sup B
is an upper bound of B, we have that for all b ∈ B, b ≤ sup B. So for all a ∈ A and b ∈ B,
a + b ≤ sup A + sup B.
Since every x ∈ A + B is such that x = a + b for some a ∈ A and some b ∈ B, it follows
that for all x ∈ A + B, x ≤ sup A + sup B. So sup A + sup B is an upper bound of A + B, and
consequently, by the definition of the least upper bound of A + B, sup(A + B) ≤ sup A + sup B.
Solution to Exercise 1.4.7. Let ℓ be a lower bound of S: for all x ∈ S, ℓ ≤ x. So for all x ∈ S,
−x ≤ −ℓ, that is, for all y ∈ −S, y ≤ −ℓ. Thus −S is bounded above because −ℓ is an upper
bound of −S.
Since S is nonempty, it follows that there exists an element x ∈ S, and so we obtain that
−x ∈ −S. Hence −S is nonempty.
As −S is nonempty and bounded above, it follows that sup(−S) exists, by the least upper
bound property of R.
Since sup(−S) is an upper bound of −S, we have that for all y ∈ −S, y ≤ sup(−S), that is,
for all x ∈ S, −x ≤ sup(−S). Hence for all x ∈ S, − sup(−S) ≤ x. So − sup(−S) is a lower bound
of S.
Next we prove that − sup(−S) is the greatest lower bound of S. Suppose that ℓ′ is a lower
bound of S such that − sup(−S) < ℓ′ . Then we have that for all x ∈ S, − sup(−S) < ℓ′ ≤ x, that
is, for all x ∈ S, −x ≤ −ℓ′ < sup(−S). Hence for all y ∈ −S, y ≤ −ℓ′ < sup(−S). So −ℓ′ is
an upper bound of −S, and −ℓ′ < sup(−S), which contradicts the fact that sup(−S) is the least
upper bound of −S. Hence ℓ′ ≤ − sup(−S).
Consequently, inf S exists and inf S = − sup(−S).
Solution to Exercise 1.4.8. (S is nonempty and bounded below (0 is a lower bound), and so
by the greatest lower bound property of R, inf S exists.)
“If” part: Let inf S > 0. Since inf S is a lower bound, it follows that for all x ∈ S, inf S ≤ x, that
1
1
1
1
. Hence for all y ∈ S −1 , y ≤
. So
is an upper bound of S −1 .
is, for all x ∈ S, ≤
x
inf S
inf S
inf S
Thus S −1 is bounded above.
“Only if” part: Suppose S −1 is bounded above (with an upper bound u, say). Then for all
y ∈ S −1 , y ≤ u, that is,
for all x ∈ S,
1
≤ u.
x
(6.8)
Solutions to the exercises from Chapter 1
59
1
1
≤ u. But x∗ ∈ S implies that x∗ > 0, and so
> 0.
x∗
x∗
1
1
Consequently u > 0. Hence from (6.8), we have that for all x ∈ S,
≤ x. Thus
is a lower
u
u
1
1
1
> 0.
bound of S, and so ≤ inf S. But u > 0 implies > 0, and consequently inf S ≥
u
u
u
Since S is not empty, ∃x∗ ∈ S and so
1
is an upper bound of S −1 and so we obtain
inf S
1
sup S −1 ≤
.
(6.9)
inf S
1
1
Furthermore, since u := sup S −1 is an upper bound of S −1 , as in the “Only if” part, =
u
sup S −1
1
is a lower bound of S, and so
≤ inf S, that is,
sup S −1
1
≤ sup S −1 .
(6.10)
inf S
1
From (6.9) and (6.10), it follows that sup S −1 =
.
inf S
If inf S > 0, then as in the “If” part, we see that
Solution to Exercise 1.4.9. If S is bounded, then it is bounded above and it is bounded below.
Thus S has an upper bound, say u, and a lower bound, say ℓ. So for all x ∈ S, ℓ ≤ x ≤ u, that is,
x ≤ u and −x ≤ −ℓ, and so we have
x ≤ u ≤ max{−ℓ, u} and − x ≤ −ℓ ≤ max{−ℓ, u}.
Thus |x| ≤ max{−ℓ, u} =: M .
Conversely, if there exists a M such that for all x ∈ S, |x| ≤ M , we have −M ≤ x ≤ M . So
−M is a lower bound of S and M is an upper bound of S. Thus S is bounded.
Solution to Exercise 1.4.10. By the density of Q in R, there exists a r ∈ Q such that
√
√
a + b 2 < r < b + 2,
√
√
that is, a < r − 2 < b, and clearly r − 2 ∈ R \ Q.
Solution to Exercise 1.4.11. For all n ∈ N,
1
1
1
1
1−
1+
2+
2−
(1 + 0)(2 + 0)
1
(1 − 0)(2 − 0)
1
n
n
n
n
≥
=
and
≤
= .
6
6
3
6
6
3
Now suppose that ℓ is a lower bound for


1
1




2+
 1+

n
n
:n∈N


6




and that ℓ > 1/3. Thus for all n ∈ N,
1
1
2+
≥ 6ℓ > 2, that is,
1+
n
n
2+
and so
3
1
≥ 6ℓ > 2,
+
n n2
1
3
≥ 6ℓ − 2 > 0. Hence for all n ∈ N,
+
n n2
3
1
3
1
4
= + ≥ + 2 ≥ 6ℓ − 2 > 0.
n
n n
n n
60
Solutions
4
, which is absurd. So ℓ ≤ 1/3 and
6ℓ − 2
1
1
1+
2+
1
n
n
= .
inf
n∈N
6
3
The above gives us that for all n ∈ N, n ≤
Suppose that u <
1
and that u is an upper bound of
3


1
1




2
−
1
−


n
n
:n∈N .


6




Thus for all n ∈ N,
1
1
1−
2−
≤ 6u < 2, that is,
n
n
2−
1
3
+ 2 ≤ 6u < 2,
n n
3
1
≤ 6u − 2 < 0. Hence for all n ∈ N,
+
n n2
3
1
3
> − ≥ 2 − 6u > 0.
n
n n
3
, which is absurd. So u ≥ 1/3 and so
The above gives us that for all n ∈ N, n ≤
2 − 6u
1
1
1−
2−
1
n
n
= .
sup
6
3
n∈N
and so −
Solution to Exercise 1.5.1.
\ 1
(1) If x ∈
0,
, then
n
n∈N
for all n ∈ N, 0 < x <
Let N ∈ N be such that
(6.11). So
1
.
n
(6.11)
1
1
< N (Archimedean property). Thus
< x, which contradicts
x
N
!
\ 1
¬ ∃x ∈
0,
n
n∈N
\ 1
that is,
= ∅.
0,
n
n∈N
1
(2) Clearly {0} ⊂ 0,
for all n ∈ N and so
n
\ 1
{0} ⊂
0,
.
(6.12)
n
n∈N
\ 1
. Then x ∈ [0, 1] and so x ≥ 0. If x > 0, then let N ∈ N be such
Let x ∈
0,
n
n∈N
1
1
1
that
, and hence
< N (Archimedean property), that is,
< x. So x 6∈ 0,
x
N
N
61
Solutions to the exercises from Chapter 1
\
\ 1
1
. Consequently, if x ∈
, then x = 0, that is,
0,
n
n
n∈N
n∈N
\ 1
0,
⊂ {0}.
(6.13)
n
n∈N
\ 1
= {0}.
From (6.12) and (6.13),
0,
n
n∈N
1
1
1
1
(3) Let n ∈ N. If x ∈
, then 0 <
,1 −
≤ x ≤ 1−
< 1, and so
n+2
n+2
n+2
n+2
x ∈ (0, 1). Hence
[ 1
1
⊂ (0, 1).
(6.14)
,1 −
n+2
n+2
x 6∈
0,
n∈N
1
− 2 < N1 (Archimedean
If x ∈ (0, 1), then 0 < x < 1. Let N1 ∈ N be such that
x
1
1
property), that is,
< x. Let N2 ∈ N be such that
− 2 < N2 (Archimedean
N1 + 2
1−x
1
property), that is, x < 1 −
. Thus with N := max{N1 , N2 }, we have
N2 + 2
1
1
1
1
≤
<x<1−
≤1−
,
N +2
N1 + 2
N2 + 2
N +2
[
1
1
1
1
that is, x ∈
⊂
. So we have
,1 −
,1−
N +2
N +2
n+2
n+2
n∈N
[ 1
1
.
(6.15)
,1 −
(0, 1) ⊂
n+2
n+2
n∈N
[ 1
1
From (6.14) and (6.15), we obtain (0, 1) =
.
,1 −
n+2
n+2
n∈N
(4) If x ∈ [0, 1], then for any n ∈ N,
1
1
− <0≤x≤1<1+ ,
n
n
1
1
and so x ∈ − , 1 +
. Hence
n
n
\ 1
1
[0, 1] ⊂
− ,1+
.
n
n
n∈N
\ 1
1
. Then
Let x ∈
− ,1 +
n
n
(6.16)
n∈N
−
1
1
≤ x ≤ 1 + for all n ∈ N.
n
n
(6.17)
1
We prove that this gives 0 ≤ x ≤ 1. For if x < 0, then let N1 ∈ N be such that − < N1 ,
x
1
that is, x < − , a contradiction to (6.17). Similarly, if x > 1, then let N2 ∈ N be
N1
1
1
, a contradiction to (6.16). Hence we see that
< N2 , that is, x > 1 +
such that
x−1
N2
neither x < 0 nor x > 1 are possible, and hence x ∈ [0, 1]. Thus
\ 1
1
⊂ [0, 1].
(6.18)
− ,1 +
n
n
n∈N
62
Solutions
(6.16) and (6.18) imply
\ 1
1
− ,1+
= [0, 1].
n
n
n∈N
Solution to Exercise 1.5.2. If x ∈ (x0 − δ, x0 + δ), then x0 − δ < x < x0 + δ. Adding −x0 ,
we obtain −δ < x − x0 < δ. So x − x0 < δ, and −(x − x0 ) < δ. Thus |x − x0 | < δ, and
x ∈ {x ∈ R : |x − x0 | < δ}. Consequently (x0 − δ, x0 + δ) ⊂ {x ∈ R : |x − x0 | < δ}.
If x ∈ {x ∈ R : |x − x0 | < δ}, then |x − x0 | < δ. Since for any real number r, |r| ≥ r
and |r| ≥ −r, we obtain that x − x0 ≤ |x − x0 | < δ and −(x − x0 ) ≤ |x − x0 | < δ. Hence
−δ < x − x0 < δ. Adding x0 , this yields x0 − δ < x < x0 + δ, that is, x ∈ (x0 − δ, x0 + δ). So
{x ∈ R : |x − x0 | < δ} ⊂ (x0 − δ, x0 + δ).
Thus (x0 − δ, x0 + δ) = {x ∈ R : |x − x0 | < δ}.
Solution to Exercise 1.5.3. From the inequality
for all a, b ∈ R,
|a + b| ≤ |a| + |b|,
we have that if x, y ∈ R, then |x| = |x − y + y| ≤ |x − y| + |y|, that is,
|x| − |y| ≤ |x − y|.
(6.19)
Interchanging x and y in (6.19), we obtain
|y| − |x| ≤ |y − x| = | − (x − y)| = | − 1||x − y| = 1 · |x − y| = |x − y|,
and so,
−(|x| − |y|) ≤ |x − y|
(6.20)
for all x, y ∈ R. From (6.19) and (6.20), we obtain ||x| − |y|| ≤ |x − y| for all x, y ∈ R.
Solution to Exercise 1.5.4. This follows easily from induction on n. If n = 1, then clearly
equality holds (|a1 | = |a1 |). If the claim is true for some n ∈ N, and if a1 , · · · , an , an+1 ∈ R, then
we have using the triangle inequality and the induction hypothesis that
|a1 + · · · + an + an+1 | ≤ |a1 + · · · + an | + |an+1 | ≤ |a1 | + · · · + |an | + |an+1 |,
and so the result follows by induction.
Suppose that a, b ∈ R are such that |a| + |b| = |a + b|. Suppose that a, b don’t have the same
sign. Then one must be positive and the other must be negative. Without loss of generality, we
may assume that a < 0 < b. Then we have two cases:
1◦ If a + b ≥ 0, then −a + b = |a| + |b| = |a + b| = a + b gives a = 0, a contradiction.
2◦ If a + b < 0, then −a + b = |a| + |b| = |a + b| = −(a + b) gives b = 0, a contradiction.
So a, b must have the same sign.
For the general equality case, we use induction on n again. Suppose that the result holds for
some n and that
|a1 | + · · · + |an | + |an+1 | = |a1 + · · · + an + an+1 |.
By the triangle inequality, we have
|a1 | + · · · + |an | + |an+1 | = |a1 + · · · + an + an+1 | ≤ |a1 + · · ·+ an | + |an+1 | ≤ |a1 | + · · · + |an | + |an+1 |,
and so all the inequalities above must be equalities. In particular, |a1 + · · · + an | + |an+1 | =
|a1 | + · · · + |an | + |an+1 | gives |a1 + · · · + an | = |a1 | + · · · + |an |. By the induction hypothesis,
a1 , · · · , an must have the same sign. Also |a1 + · · · + an + an+1 | = |a1 + · · · + an | + |an+1 | shows
that an+1 must have the same sign as a1 + · · · + an . But as a1 , · · · , an all have the same sign,
a1 + · · · + an has the same sign as a1 , · · · , an . Consequently a1 , · · · , an , an+1 have the same sign.
63
Solutions to the exercises from Chapter 1
Solution to Exercise 1.6.1. We have f (2) = 1 + 2 = 3 and g(2) = 1 − 2 = −1, and so
(1) f (2) + g(2) = 3 + (−1) = 2,
(2) f (2) − 2 · g(2) = 3 − 2 · (−1) = 3 + 2 = 5,
(3) f (2) · g(2) = 3 · (−1) = −3,
(4) (f (2))/(g(2)) = 3/(−1) = −3,
(5) f (g(2)) = f (−1) = 1 + (−1) = 0,
(6) for a ∈ R, f (a) + g(−a) = 1 + a + (1 − (−a)) = 2 + 2a,
(7) for t ∈ R, f (t) · g(−t) = (1 + t)(1 − (−t)) = (1 + t)2 .
Solution to Exercise 1.6.2.
See Figures 4 and 5.
f
g1
g2
0
Figure 4. The graphs of f, g1 , g2 .
g3
g5
f
g4
0
Figure 5. The graphs of f, g3 , g4 , g5 .
Solution to Exercise 1.6.3. See Figure 6. We note that
2x2 + 4x + 1 = 2(x2 + 2x + 1) − 1 = 2(x + 1)2 − 1.
One can also use Maple to plot these graphs using for example the following commands:
> f := xˆ3
> plot(f(x), x = -2.5..1);
> f := 2·xˆ2+4·x+1;
> plot(g(x), x = -3..3);
Solution to Exercise 1.6.4.
(1) Let p(x) = c0 + c1 x + c2 x2 + c3 x3 + · · · + cd xd , x ∈ R with cd 6= 0. Then p(0) = c0 = 0.
Thus
p(x) =
=
0 + c1 x + c2 x2 + c3 x3 + · · · + cd xd
x(c1 + c2 x + c3 x2 + · · · + cd xd−1 ) = xq(x),
64
Solutions
Figure 6. The graphs of x3 and 2x2 + 4x + 1.
where q is the polynomial given by q(x) = c1 + c2 x + c3 x2 + · · · + cd xd−1 , x ∈ R, and
clearly q has degree d − 1 since cd 6= 0.
(2) We have for x ∈ R that
pa (x)
=
=
n X
n k n−k
p(x + 1) =
cn (x + a) =
cn
x a
k
n=0
n=0
k=0
!
d
d
d X
n X
X
X
n
n n−k
xk =
an−k xk .
a
cn
cn
k
k
n=0
d
X
n
d
X
k=0
k=0
n=k
d
The coefficient of x is
d
X
n=d
d d−d
n n−d
a
= cd 6= 0.
a
= cd
cn
d
d
So pa is a polynomial of degree d.
(3) If p(a) = 0, then pa (0) = p(0 + a) = p(a) = 0. By the result in the first part, it follows
that there must exist a polynomial qe of degree d − 1 such that
pa (x) = xe
q (x),
x ∈ R,
that is, p(x + a) = xe
q (x), x ∈ R. Hence
p(x) = p(x − a + a) = (x − a)e
q (x − a) = (x − a)e
q−a (x) = (x − a)q(x),
x ∈ R,
where q := qe−a , and from the previous part, we know that (q =) qe−a is a polynomial
with degree equal to the degree of qe,that is, d − 1.
(4) Suppose that α1 , · · · , αd+1 are distinct real numbers such that
p(αk ) = 0 for k = 1, · · · , d + 1.
Then we can factorize p as p(x) = (x − α1 )q1 (x), where q1 is a polynomial of degree d − 1.
But since p(α2 ) = · · · = p(αd+1 ) = 0, and as α2 − α1 , · · · , αd+1 − α1 are all nonzero, it
follows that q(α2 ) = · · · = q(αd+1 ) = 0. So we can repeat the above with q1 . Eventually,
we obtain
p(x) = (x − α1 )(x − α2 ) · · · (x − αd )qd (x),
65
Solutions to the exercises from Chapter 1
for a polynomial qd of degree 0, that is, qd is a constant. As
p(αd+1 ) = 0 = (αd+1 − α1 ) · · · (αd+1 − αd ) qd (αd+1 ),
{z
}
|
6=0
it follows that qd ≡ (qd (αd+1 ) =) 0. But then p(x) = (x−α1 )(x−α2 ) · · · (x−αd )qd (x) = 0
for all x.
(5) Consider the polynomial p − pe of degree d. Since p − pe is 0 for d + 1 distinct real values,
it follows that p − pe ≡ 0, that is, p ≡ pe.
Solution to Exercise 1.7.1.
(1) This is best seen by means of a picture, as shown in Figure 7.
−4
−3 −2 −1
0
1
2
3
4
Figure 7. Countability of Z.
Clearly the resulting map from N to Z is injective (since each integer is crossed by the
blue path only once ever) and surjective (since every integer will be crossed by the blue
path sometime).
(2) First let us show that any infinite subset C of N is countable. Let a1 := min C. If
a1 , · · · , ak have been constructed, then define
ak+1 := min C \ {a1 , · · · , ak } .
Then a1 < a2 < a3 < · · · . Define ϕ : N → C by ϕ(n) = an , n ∈ N. Then ϕ is injective
(because if n < m, then ϕ(n) < ϕ(m)). Also, ϕ is surjective. Indeed, for each element
m ∈ C, there are only finitely many natural numbers, and much less elements of C,
which are smaller than m. If the number of such elements of C that are smaller than m
is nm , then it is clear that ϕ(nm + 1) = m.
Now let S be countable and let T ⊂ S. Let ϕ : S → N be a bijection. If T is finite,
then we are done. If T is infinite, then so is S. There is a bijection from T to the range of
ϕ|T . But the range of ϕ|T is a subset of N, and so it is countable. Hence T is countable
too.
(3) Since A and B are countable, we can list their elements:
A =
B
=
{a1 , a2 , a3 , · · · },
{b1 , b2 , b3 , · · · }.
Arrange the elements of A × B in an array and list them by following the path as shown
in Figure 8.
The resulting map from N to A × B is clearly surjective (since every element (an , bm ) is
hit by the blue path sometime), and it is also injective (since the path never hits a point
after having crossed it because it moves on to a parallel antidiagonal below).
66
Solutions
(a1 , b1 )
(a1 , b2 )
(a1 , b3 ) · · ·
(a2 , b1 )
(a2 , b2 )
(a2 , b3 ) · · ·
(a3 , b1 )
(a3 , b2 )
..
.
(a3 , b3 ) · · ·
..
..
.
.
..
.
Figure 8. Countability of A × B when A, B are infinite countable sets.
(4) Each q ∈ Q can be written uniquely as
n
,
d
where n, d ∈ Z, d > 0 and the greatest common divisor gcd(n, d) of n, d is equal to 1
(that is, n and d have no common factor besides 1, or n, d are coprime/relatively prime).
Hence we can consider Q as a subset of Z × Z. But Z is countable, and so by the previous
part, Z × Z is countable. Consequently, Q is countable.
q=
Solutions to the exercises from Chapter 1
67
68
Solutions
Solutions to the exercises from Chapter 2
Solution to Exercise 2.1.1.
(1) We prove that (1)n∈N is convergent sequence with limit 1. Given ǫ > 0, let N ∈ N, say
N = 1. Then for all n > N = 1, we have |an − L| = |1 − 1| = |0| < ǫ.
(2) Yes. For instance, the constant sequence (1)n∈N converges to 1.
(3) Suppose that the terms of the convergent sequence (an )n∈N (with limit, say, L) lie in the
finite set {v1 , . . . , vm }. If L 6∈ {v1 , . . . , vm }, then with
ǫ := min{|v1 − L|, . . . , |vm − L|} > 0,
let N ∈ N be such that for all n > N , |an −L| < ǫ. In particular, with n = N +1 > N , we
have |aN +1 − L| < ǫ. But aN +1 ∈ {v1 , . . . , vm }. Let aN +1 = vk for some k ∈ {1, . . . , m}.
Then we have
|vk − L| = |aN +1 − L| < ǫ = min{|v1 − L|, . . . , |vm − L|} ≤ |vk − L|,
a contradiction. So L ∈ {v1 , . . . , vm }, that is, L must be one of the terms. Thus we have
shown that
terms of the sequence
take finitely many values
=⇒
L must be one
of the terms
,
that is,
L is not equal to
any of the terms
=⇒
terms of the sequence cannot
consist of finitely many values
.
(4) Suppose that ((−1)n )n∈N is a convergent sequence with limit L. Then from part (3)
above, it follows that L = 1 or L = −1: indeed the terms of the sequence take finitely
many values, namely 1 and −1, and so L must be one of these terms. So we have the
following two cases:
1◦ If the limit is 1, then given ǫ = 1 > 0, let N ∈ N be such that for all n > N ,
|(−1)n − 1| < ǫ = 1. Let n be any odd number > N . Then we have that for such n,
2 = | − 2| = | − 1 − 1| = |(−1)n − 1| < ǫ = 1, a contradiction.
◦
2 If the limit is −1, then given ǫ = 1 > 0, let N ∈ N be such that for all n > N ,
|(−1)n − (−1)| < ǫ = 1. Let n be any even number > N . Then we have that for
such n, 2 = |2| = |1 + 1| = |(−1)n − (−1)| < ǫ = 1, a contradiction.
So ((−1)n )n∈N is divergent.
Solution to Exercise 2.1.2. If
1
= 1,
n
then given any ǫ > 0, there exists an N ∈ N such that for all n ∈ N such that n > N ,
1
− 1 < ǫ.
n
lim
n→∞
Let ǫ = 21 . (This choice is motivated by Figure 9, from which we see that, for instance ǫ = 1 won’t
give us a contradiction, but ǫ = 21 would.)
Then there exists a N∗ ∈ N such that for all n > N∗ ,
1
− 1 < ǫ = 1 ,
n
2
69
Solutions to the exercises from Chapter 2
1
Figure 9.
that is,
Consequently, for all n > N∗ ,
1
.
n n∈N
1
1
1
1 − = − 1 < .
n
n
2
1
1
< ,
n
2
that is, n < 2. In particular, since n = N∗ + 1 > N∗ , we obtain N∗ + 1 < 2, that is, N∗ < 1. But
there does not exist any natural number N∗ that is less than 1. Hence we arrive at a contradiction,
1
6= 1.
and so lim
n→∞ n
1−
Solution to Exercise 2.1.3.
(1) We have seen that the sequence ((−1)n )n∈N is divergent. Let ǫ > 0, and let N ∈ N.
Then for all even n > N (there are obviously infinitely many such n), we have
|an − L| = |(−1)n − 1| = |1 − 1| = 0 < ǫ.
(2) Again, for the divergent sequence ((−1)n )n∈N , with ǫ = 3 > 0, for all N ∈ N and all
n > N , we have
|an − L| ≤ |an | + |L| ≤ 1 + 1 = 2 < 3 = ǫ.
Solution to Exercise 2.1.4. S is nonempty and bounded above, and so by the least upper
bound property of R, it follows that sup S exists.
Given n ∈ N, we have n1 > 0, and so sup S − n1 < sup S. Thus S − n1 is not an upper bound
of S. Hence there must exist an element in S, which we denote by an , such that
1
¬ an ≤ sup S −
,
n
that is, an > sup S − n1 . In this way we construct the sequence (an )n∈N .
As sup S is an upper bound of S, we also have an ≤ sup S for all n ∈ N. Consequently,
1
1
for all n ∈ N, sup S − < an ≤ sup S < sup S + ,
n
n
that is,
1
1
for all n ∈ N, − < an − sup S < ,
n
n
that is,
1
for all n ∈ N, |an − sup S| < .
n
1
Given ǫ > 0, let N ∈ N be such that N > . Then for all n > N , we have
ǫ
1
1
|an − sup S| < <
< ǫ.
n
N
Hence (an )n∈N is convergent with limit equal to sup S.
70
Solutions
Solution to Exercise 2.1.5. Suppose L < 0. Then ǫ := −L/2 > 0, and so there exists a N ∈ N
such that for all n > N , |an − L| < ǫ = −L/2. Hence with n = N + 1 (> N ), we obtain
L
aN +1 − L ≤ |aN +1 − L| < − ,
2
that is, aN +1 <
L
< 0, a contradiction.
2
Solution to Exercise 2.2.1. We prove that the sequence is monotone and bounded, and hence
it must be convergent.
We prove that |an | ≤ 1 for all n ∈ N. We prove this using induction. We have |a1 | = |1| = 1.
If k ∈ N is such that |ak | ≤ 1, then
2k + 3 2k + 3 2k + 3
|ak | ≤ 1 · 1 = 1,
ak =
|ak | =
|ak+1 | = 3k + 3 3k + 3 3k + 3
and so the claim follows from induction. So the sequence is bounded.
Since n ≥ 1, it follows that 2n + 1 ≤ 3n, and so
2n + 1
≤1
3n
for all n ∈ N. Furthermore, note that for all n ∈ N, an ≥ 0 (induction!). Hence for all n ≥ 2,
an =
2n + 1
an−1 ≤ 1 · an−1 = an−1 .
3n
So (an )n∈N is decreasing.
As the sequence is bounded and monotone, it is convergent.
Solution to Exercise 2.2.2. Let M > 0 be such that for all n ∈ N, |bn | ≤ M . Given ǫ > 0, let
N ∈ N be such that M/ǫ < N , that is, 1/N < ǫ/M . Then for all n > N ,
bn
− 0 = |bn | ≤ M < M < M · ǫ = ǫ.
n
n
n
N
M
bn
is convergent with limit 0.
Hence
n n∈N
Solution to Exercise 2.2.3.
(1) Given ǫ > 0, there exists an N1 ∈ N such that for all n > N1 , |an − L| < ǫ/2. Since
(an )n∈N is convergent, it is bounded: there exists an M > 0 such that for all n ∈ N,
|an | ≤ M . Choose1 N ∈ N such that
N1 (M + |L|)
< N,
max N1 ,
ǫ/2
and so,
N > N1 and
N1 (M + |L|)
ǫ
< .
N
2
1This is arrived at by working backwards; we wish to make a1 +···+an − L less than ǫ for all n > N , so we manipulate
n
this (as shown in the chain of inequalities that follow) to see if can indeed achieve this by choosing the N large enough.
71
Solutions to the exercises from Chapter 2
Then for n > N , we have:
a1 + · · · + aN1 + aN1 +1 + · · · + an
− L
n
a1 + · · · + aN1 + aN1 +1 + · · · + an − nL = n
|a1 + · · · + aN1 + aN1 +1 + · · · + an − nL|
=
n
|a1 − L| + · · · + |aN1 − L| + |aN1 +1 − L| + · · · + |an − L|
≤
n
(|a1 | + |L| + · · · + |aN1 | + |L|) + ǫ/2 + · · · + ǫ/2
≤
n
N1 (M + |L|) + (n − N1 )(ǫ/2)
≤
n
N1 (M + |L|)
ǫ
N1
≤
·
+ 1−
n
n
2
ǫ
N1 (M + |L|)
+1·
<
N
2
ǫ
ǫ
+ = ǫ.
<
2 2
a1 + · · · + an
So
is a convergent sequence with limit L.
n
n∈N
(2) If an = (−1)n , n ∈ N, then (an )n∈N is divergent, but the sequence with nth term



1
2
n
0 if n is even
(−1) + (−1) + · · · + (−1)
a1 + · · · + an
,
=
=
1

n
n
 −
if n is odd
n
is convergent with limit equal to 0. Indeed, given ǫ > 0, let N ∈ N be such that
Then for n > N , we have




 0
if n is even  1
1
< ǫ.
|an − 0| = |an | =
≤ <
1


n
N

if n is odd 
n
a1 + · · · + an
is convergent with limit 0.
So
n
n∈N
1
ǫ
< N.
Solution to Exercise 2.2.4. Since (an )n∈N is bounded, it follows that there exists a M such
that for all n ∈ N, |an | ≤ M , that is, −M ≤ an ≤ M . If k ∈ N, then in particular, for all n ≥ k,
−M ≤ an ≤ M , and so the set {an : n ≥ k} is bounded. By the least upper bound property of R,
it then follows that inf{an : n ≥ k} and sup{an : n ≥ k} exist, that is, ℓk and uk are well-defined.
Furthermore, for each k, −M ≤ inf{an : n ≥ k} ≤ sup{an : n ≥ k} ≤ M, and so we see that the
sequences (ℓk )k∈N and (uk )k∈N are bounded.
Clearly {an : n ≥ k + 1} ⊂ {an : n ≥ k}, and so
uk+1 = sup{an : n ≥ k + 1} ≤ sup{an : n ≥ k} = uk ,
and so (uk )k∈N is a decreasing sequence.
Similarly, {−an : n ≥ k + 1} ⊂ {−an : n ≥ k}, and so we have
sup{−an : n ≥ k + 1} ≤ sup{−an : n ≥ k}.
72
Solutions
Hence
inf{an : n ≥ k + 1} = − sup{−an : n ≥ k + 1} ≥ − sup{−an : n ≥ k} = inf{an : n ≥ k},
that is, ℓk+1 ≥ ℓk . Consequently, (ℓk )k∈N is a increasing sequence.
As the sequences (uk )k∈N , (ℓk )k∈N are both bounded and monotone, it follows that they are
convergent.
Solution to Exercise 2.2.5. Let (an )n∈N be a bounded decreasing sequence of real numbers.
Let ℓ∗ be the greatest lower bound of {an : n ∈ N}. The existence of ℓ∗ is guaranteed by the
greatest lower bound property of the set of real numbers. We show that ℓ∗ is the limit of (an )n∈N .
Taking ǫ > 0, we must show that there exists a positive integer N such that |an − ℓ∗ | < ǫ for
all n > N . Since ℓ∗ + ǫ > ℓ∗ , ℓ∗ + ǫ is not a lower bound of {an : n ∈ N}. Therefore there
exists N with ℓ∗ ≤ aN < ℓ∗ + ǫ. Since (an )n∈N is decreasing, we have for all n ≥ N that
ℓ∗ − ǫ < ℓ∗ ≤ an ≤ aN < ℓ∗ + ǫ, and so |an − ℓ∗ | < ǫ.
Solution to Exercise 2.2.6.
(1) (cos(πn))n∈N = ((−1)n )n∈N diverges.
(2) (1 + n2 )n∈N is not bounded, and so it can’t be convergent.
(3) Let ǫ > 0, and let N ∈ N be such that N > 1/ǫ. Then for n > N , we have
| sin n|
sin n
1
1
=
−
0
≤ <
< ǫ.
n
n
n
N
sin n
is convergent with limit 0.
Thus
n
n∈N
3n2
(4) 1 −
is unbounded, seen as follows. Let M > 0 be such that for all n ∈ N,
n + 1 n∈N
2 1 − 3n ≤ M.
n + 1
Then for n ∈ N,
3n2
n
3n2
3n2 −1≥
−n= ,
≥
M ≥ 1 −
n+1
n+1
n+n
2
and so for all n ∈ N,
n ≤ 2M,
3n2
is unbounded, and hence divergent.
a contradiction. So 1 −
n + 1 n∈N
(5) The sequence is increasing and bounded above by 1. So it is convergent, with limit
9
9
9
+
+ ···+ n
lim 0. 9| ·{z
· · 9} = lim
n→∞ 10
n→∞
100
10
n nines
10 − 1 10 − 1
10 − 1
= lim
+
+ ··· +
n→∞
10
100
10n
1
= lim 1 − n = 1 − 0 = 1.
n→∞
10
73
Solutions to the exercises from Chapter 2
Solution to Exercise 2.3.1. First we prove the
Claim: (an )n∈N is convergent with limit L, then (an+1 )n∈N is also convergent with limit L.
Proof. Let ǫ > 0. Then ∃N ∈ N such that for all n > N , |an − L| < ǫ. Thus for all n > N , we
have n + 1 > N + 1 > N , and so |an+1 − L| < ǫ. Hence (an+1 )n∈N is convergent with limit L.
Alternately, observe that (an+1 )n∈N is a subsequence of (an )n∈N , and so must also be convergent with the same limit L.
We now apply this result to our sequence (an )n∈N , which satisfies:
an+1
=
3
2+
2(n + 1) + 1
na ,
an =
3 n
3(n + 1)
3+
n
1
for all n ∈ N. Since
is convergent with limit 0, by the theorem on algebra of limits,
n n∈N
3
n = 2 + 3 · 0 = 2.
lim
3
n→∞
3+3·0
3
3+
n
Again applying the theorem on algebra of limits, we obtain



3
2+
2+


na 
L = lim an+1 = lim 
n  = lim 
3
n→∞
n→∞
n→∞
3+
3+
n
1
Hence L = 0, that is, L = 0. So lim an = 0.
n→∞
3
2+

3
2
n
lim a = L.
3  n→∞ n
3
n
Solution to Exercise 2.3.2. We have
an bn + 5n
cn =
=
a2n + n
bn
+5
n
for all n ∈ N.
1
an · an · + 1
n
an ·
bn
N. The sequence an ·
+5
is convergent.
n
n∈N
The sequence (an )n∈N is convergent with limit, say L. Since (bn )n∈N is bounded,
bn
bn
is convergent with limit 0. Hence an ·
is convergent
the sequence
n n∈N
n n∈N
with limit L · 0 = 0. The sequence (5)n∈N is convergent with limit 5. So the sequence
bn
an ·
+5
is convergent with limit 0 + 5 = 5.
n
n∈N
2
an
+1
D. The sequence
has nonzero terms for all n ∈ N and it is convergent with
n
n∈N
the nonzero limit 1.
a2
a2n
+ 1 ≥ 1 for all n ∈ N, and so n + 1 6= 0 for all n ∈ N.
n
n
Since the sequence (an )n∈N is convergent with limit L, it follows that the sequence
1
(a2n )n∈N is convergent with limit L2 . Since
is convergent with limit 0, it follows
n n∈N
We have
74
Solutions
1
that a2n ·
is convergent with limit L2 ·0 = 0. Finally, as (1)n∈N is convergent with
n n∈N
2
an
+1
is convergent with limit 0+1 = 1(6= 0).
limit 1, it follows that the sequence
n
n∈N
From N and D above and the algebra of limits, (cn )n∈N is convergent with limit
5
= 5.
1
Solution to Exercise 2.3.3. We begin by showing that L ≥ 0. If L < 0, then ǫ := −L/2 > 0.
Let N ∈ N be such that for all n > N , |an − L| < ǫ = −L/2. Then we have
L
an − L ≤ |an − L| < − ,
2
and so an < L/2 < 0 for all n > N , a contradiction to the fact that
an ≥ 0 for all n ∈ N.
So L ≥ 0.
√
√
Now we show that ( an )n∈N is convergent with limit L. Let ǫ >). We consider the only
two possible cases, namely L = 0 and L > 0:
1◦ If L = 0, then let N ∈ N be such that for all n > N ,
|an − L| = |an − 0| = |an | = an < ǫ2 .
√
Then for n > N , we have an < ǫ, that is,
√
√
√
√
√
√
| an − L| = | an − 0| = | an | = an < ǫ.
√
√
So ( an )n∈N is convergent with limit L.
√
2◦ If L > 0, then let N ∈ N be such that for n > N , |an − L| < ǫ L. Then for all n > N ,
we obtain
ǫ
>
=
=
=
and so
√
Hence ( an )n∈N
|an − L|
√ √
√
√
|( an − L)( an + L)|
√ √
√
√
| an − L|| an + L)|
√ √
√
√
| an − L|( an + L)
√
√
√
√
ǫ L
ǫ L
√ ≤ √ = ǫ.
| an − L| < √
an + L
L
√
is convergent with limit L.
Solution to Exercise 2.3.4. For all n ∈ N, we have
√
p
p
n2 + n − n2
n
n2 + n + n
2
2
=√
= √
n + n − n = ( n + n − n) · √
2
2
2
n +n+n
n +n+n
n +n+n
n(1)
1
1
p
=r
.
=
=r
1
1
n2 + n
n
n2 + n + 1
1+ +1
+1
n
n
n2
1
As (1)n∈N is convergent with limit 1 and
is convergent with limit 0, it follows that
n n∈N
1
1
is convergent with limit 1. Furthermore 1+ ≥ 0 for all n ∈ N, and so by the previous
1+
n n∈N
n
!
!
r
r
√
1
1
is convergent with limit 1 = 1. Hence
part, it follows that
is
1+
1+ +1
n
n
n∈N
n∈N
75
Solutions to the exercises from Chapter 2
convergent with limit 1 + 1 = 2(6= 0). Also

r
1+
1
+ 1 > 1 > 0. Thus by the theorem on algebra
n



1
1

of limits, we conclude that the sequence 
is convergent with limit , that is,
r

2
1
1+ +1
n
n∈N
√
1
( n2 + n − n)n∈N is convergent with limit .
2
Solution to Exercise 2.3.5. Consider the sequence (bn − an )n∈N . As an ≤ bn , it follows that
bn − an ≥ 0 for all n ∈ N. From the theorem on algebra of limits, it follows that the sequence
(bn − an )n∈N is convergent (being the sum of the convergent sequence (bn )n∈N and the convergent
sequence (−an )n∈N ). Moreover, its limit is lim bn − lim an . From Exercise 2.1.5 on page 7, we
n→∞
n→∞
obtain lim bn − lim an ≥ 0, that is lim bn ≥ lim an .
n→∞
n→∞
n→∞
n→∞
Solution to Exercise 2.4.1. For all n ∈ N, we have
1 2
n−1 n
1
1
n!
· ≤ · 1 ····· 1 ·1 = .
0 ≤ n = · · ····
n
n n
n
n
n
n
1
Since (0)n∈N and
are both convergent with the same limit 0, from the Sandwich theorem,
n
n∈N
n!
it follows that
is convergent with the limit 0.
nn n∈N
Solution to Exercise 2.4.2. Let k ∈ N. For all n ∈ N, we have
0≤
nk + nk + nk + · · · + nk
n · nk
1
1 k + 2 k + 3 k + · · · + nk
≤
≤
= .
nk+2
nk+2
nk+2
n
Thus
for all n ∈ N. As (0)n∈N
1
1 k + 2 k + 3 k + · · · + nk
≤
0≤
k+2
n
n
1
and
are convergent with limit 0, from the Sandwich theorem,
n n∈N
1 k + 2 k + 3 k + · · · + nk
= 0.
n→∞
nk+2
lim
Solution to Exercise 2.4.3.
(1) We prove the claim using induction. Let x ≥ −1. Clearly (1 + x)1 = 1 + x = 1 + 1 · x. If
for some k ∈ N, (1 + x)k ≥ 1 + kx, then we have
(1 + x)k+1
=
(1 + x)k (1 + x)
≥
(1 + kx)(1 + x) (induction hypothesis and as 1 + x ≥ 0)
=
1 + (k + 1)x + x2
≥
1 + (k + 1)x.
=
1 + kx + x + x2
Hence by induction, the result follows.
(2) For all n ∈ N,
(for if n
1
n
1
1
n
n
nn ≥ 1
n
(6.21)
< 1, then n = (n ) < 1 = 1, a contradiction!). Clearly for all n ∈ N,
√ 2 1
√ 2
√ 2
1
(6.22)
n n = ( n ) n = n n < (1 + n) n .
76
Solutions
Finally,
n
√
1
1
√
≥ 1 + n · √ = 1 + n,
1+
n
n
and so
2 n n2
√ 2
1
1
=
1+ √
1+ √
≥ (1 + n) n .
n
n
(6.23)
Combining (6.21), (6.22), and (6.23), we obtain
2
√ 2
1
1
1 ≤ n n < (1 + n) n ≤ 1 + √
n
(6.24)
for all n ∈ N.
1
1
= 0, it follows that lim √ = 0. Hence
(3) Since lim
n→∞
n→∞ n
n
2
1
lim 1 + √
= (1 + 0)2 = 1 = lim 1.
n→∞
n→∞
n
1
Using (6.24), it follows from the Sandwich theorem that (n n )n∈N is convergent and
1
lim n n = 1.
n→∞
Remark 6.1. Here are two alternative proofs.
1
1
(1) For n ≥ 2, we have n n > 1 n = 1. Set
1
bn := n n − 1 > 0 for n > 2.
1
Then n n = 1 + bn , and so
n = (1 + bn )n = 1 + nbn +
Thus
also
n(n − 1) 2
n(n − 1) 2
bn + · · · + nbnn−1 + bnn >
bn .
2
2
2
> b2n > 0 for n > 2, and so by the Sandwich Theorem, lim b2n = 0. Hence
n→∞
n−1
lim bn = 0,
n→∞
and so
1
lim n n = lim (1 + bn ) = 1 + 0 = 1.
n→∞
n→∞
(2) Yet another proof can be given using the Arithmetic Mean-Geometric Mean Inequality,
which says that for nonnegative real numbers a1 , . . . , an , there holds that
√
a1 + · · · + an
≥ n a1 . . . an =: their geometric mean.
n
√ √
Applying this to the n numbers 1, . . . , 1 , n, n gives
| {z }
their arithmetic mean :=
(n−2)times
and so,
√
√
q
√
1
(n − 2) · 1 + n + n
n √
≥
n · n = nn ,
n
1−
1
2
2
+ √ ≥ n n (≥ 1).
n
n
1
By the Sandwich Theorem, it follows that lim n n = 1.
n→∞
77
Solutions to the exercises from Chapter 2
Solution to Exercise 2.4.4. Consider the sequence (an − a)n∈N . As an ∈ (a, b) for all n ∈ N, we
have an − a ≥ 0. From the theorem on algebra of limits, it follows that the sequence (an − a)n∈N
is convergent (being the sum of the convergent sequence (an )n∈N and the convergent sequence
(−a)n∈N ). Moreover, its limit is L − a. From Exercise 2.1.5 on page 7, we obtain L − a ≥ 0, that
is a ≤ L.
Next consider the sequence (b − an )n∈N . As an ∈ (a, b) for all n ∈ N, we have b − an ≥ 0. From
the theorem on algebra of limits, it follows that the sequence (b − an )n∈N is convergent (being
the sum of the convergent sequence (−an )n∈N and the convergent sequence (b)n∈N ). Moreover, its
limit is −L + b. From Exercise 2.1.5 on page 7, we obtain −L + b ≥ 0, that is L ≤ b.
Consequently a ≤ L ≤ b, that is, L ∈ [a, b].
1
contained in (0, 1). It is convergent with limit 0 6∈ (0, 1).
Consider
n + 1 n∈N
Solution to Exercise 2.4.5. For all n ∈ N, we have
−
1
1
< b n − an < ,
n
n
and so by adding an , we have
1
1
+ an < b n < + an .
n
n
By the theorem on algebra of limits, we know that
1
1
lim − + an = lim − + lim an = 0 + lim an
n→∞
n→∞
n→∞
n
n n→∞
1
1
+ an = lim
+ lim an = 0 + lim an
lim
n→∞ n
n→∞
n→∞
n→∞ n
−
=
=
lim an
n→∞
lim an .
n→∞
So by the sandwich theorem, it follows that (bn )n∈N is convergent with the limit lim an .
n→∞
Solution to Exercise 2.5.1.
(1) True.
Let (an )n∈N be convergent with limit L, and let (ank )k∈N be a subsequence. If ǫ > 0,
then there exists an N ∈ N such that for all n > N , |an − L| < ǫ. Choose K ∈ N such
that for all k > K, nk > N . Then for all k > K, nk > N and so |ank − L| < ǫ. Thus
(ank )k∈N is convergent with limit L.
(2) False.
The sequence ((−1)n )n∈N is divergent, but it has as a subsequence ((−1)2n )n∈N = (1)n∈N ,
which is convergent with limit 1.
(3) True.
Let (an )n∈N be bounded, and let M > 0 be a number such that for all n ∈ N, |an | ≤ M .
If (ank )k∈N is a subsequence, then also |ank | ≤ M for all k ∈ N, and so it is bounded as
well.
(4) False.
The sequence 1, 0, 2, 0, 3, 0, . . . is unbounded, but has as a subsequence 0, 0, 0, . . . , which
is bounded.
(5) True.
Let (an )n∈N be an increasing sequence. (The argument is similar for a decreasing sequence.) Suppose that (ank )k∈N is a subsequence. For each k ∈ N, nk+1 > nk and so we
have that ank+1 ≥ ank+1 −1 ≥ ank+1 −2 ≥ · · · ≥ ank . Thus (ank )k∈N is increasing as well.
78
Solutions
(6) False.
The sequence 1, 0, 2, 0, 3, 0, . . . is not monotone, but it has as a subsequence 1, 2, 3, . . . ,
which is monotone.
(7) True.
The sequence itself is a subsequence of itself, and so it must be convergent too.
(8) False.
The sequence (an )n∈N = ((−1)n )n∈N is divergent, but we have that both (a2n )n∈N =
(1)n∈N and (a2n+1 )n∈N = (−1)n∈N are convergent (with limits 1 and −1, respectively).
(9) True.
Let L be the common limit, and let ǫ > 0. There exists an N1 ∈ N such that for
all n > N1 , |a2n − L| < ǫ. Also, there exists an N2 ∈ N such that for all n > N2 ,
|a2n+1 − L| < ǫ. Choose N ∈ N such that N > max{2N1 , 2N2 + 1}. Let n > N . If
n is even, say n = 2k for some k ∈ N, then 2k > N ≥ 2N1 gives k > N1 , and so
|a2k − L| = |an − L| < ǫ. If, on the other hand, n is odd, say n = 2k + 1 for some k ∈ N,
then 2k + 1 = n > N ≥ 2N2 + 1 gives k > N2 , and so |a2k+1 − L| = |an − L| < ǫ. Hence
for all n > N , |an − L| < ǫ. Consequently, (an )n∈N is convergent with limit L.
Solution to Exercise 2.5.2. Observe that
the terms 2, 8, 2, 8 appear adjacently
and so the terms 1, 6, 1, 6 appear adjacently
and so the terms 6, 6, 6 appear adjacently
and so the terms 3, 6, 3, 6 appear adjacently
and so the terms 1, 8, 1, 8 appear adjacently
and
and
and
and
so
so
so
so
the
the
the
the
terms
terms
terms
terms
8, 8, 8 appear adjacently
6, 4, 6, 4 appear adjacently
2, 4, 2, 4 appear adjacently
8, 8, 8 appear adjacently
..
.
Hence we get the loop
. . . , 8, 8, 8, · · · → . . . , 6, 4, 6, 4, · · · → . . . , 2, 4, 2, 4, · · · → . . . , 8, 8, 8, · · · →,
which contains 6, and so 6 appears infinite number of times. Thus we can choose indices
n1 < n2 < n3 < . . .
such that for all k ∈ N, ank = 6. So (6)k∈N is a subsequence of the given sequence.
Solution to Exercise 2.5.3. We know that
¬ ∀ǫ > 0, ∃N ∈ N such that ∀n > N, |an − L| < ǫ ,
that is,
∃ǫ > 0 such that ∀N ∈ N, ∃n > N such that |an − L| ≥ ǫ.
(6.25)
Take N = 1 in (6.25). Then there exists an n1 > N = 1 such that |an1 − L| ≥ ǫ. We construct
the subsequence terms inductively as follows. Suppose that an1 , · · · , ank have been constructed.
Take N = nk in (6.25). Then there exists an nk+1 > N = nk such that |ank+1 − L| ≥ ǫ. Thus we
obtain the subsequence (ank )k∈N which satisfies |ank − L| ≥ ǫ for all k ∈ N.
79
Solutions to the exercises from Chapter 2
Solution to Exercise 2.5.4. By the Bolzano-Weierstrass Theorem the bounded sequence (an )n∈N
has some convergent subsequence (ank )k∈N with some limit, say L1 . But we know that the sequence
(an )n∈N is not convergent to L1 . So by Exercise 2.5.3, there exists an ǫ > 0 and a subsequence,
say (amk )k∈N such that for all k, |amk − L1 | ≥ ǫ. As (ank )k∈N converges to L1 , we know that
there exists a K large enough so that for all k ≥ K, |ank − L1 | < ǫ. Then clearly the subsequence
am1 , am2 , am3 , · · · has terms which are all distinct from the subsequence anK , anK+1 , anK+2 , · · ·
(since for k ≥ K, |ank − L1 | < ǫ, while for all k, |amk − L1 | ≥ ǫ). Now by the Bolzano-Weiertrass
Theorem, there exists a subsequence of the bounded sequence am1 , am2 , am3 , · · · which converges,
say, to L2 . But from the inequalities |amk − L1 | ≥ ǫ, k ≥ 1, we see that |L2 − L1 | ≥ ǫ > 0, and so
L1 6= L2 .
Solution to Exercise 2.5.5. Let ǫ > 0. Since (an )n∈N is a Cauchy sequence, there exists an
N ∈ N such that for all n, m > N , |an − am | < ǫ. In particular for n > N , we have m := n + 1 > N
and so |an − an+1 | < ǫ, that is, |(an+1 − an ) − 0| < ǫ. So (an+1 − an )n∈N is convergent with limit
0.
Solution to Exercise 2.6.1. (“If” part) Suppose that (an )n∈N converges to 0. Let ǫ > 0. Then
there exists an N ∈ N such that whenever n > N , |an − 0| < ǫ. As an ≥ 0, we have that an < ǫ
for all n > N . But by the definition of an , this means that for any x ∈ I,
|fn (x) − f (x)| ≤ an < ǫ.
Hence (fn )n∈N is uniformly convergent to f .
(“Only if” part) Now suppose that (fn )n∈N is uniformly convergent to f . Let ǫ > 0. Then there
exists an N ∈ N such that whenever n > N , we have for all x ∈ I, |fn (x) − f (x)| < ǫ. But this says
that ǫ is an upper bound for the set {|fn (x) − f (x)| : x ∈ I}. By the definition of the least upper
bound, it now follows that an ≤ ǫ. Since an is nonnegative, it follows that |an − 0| = |an | = an ≤ ǫ
for all n > N . In other words, the sequence (an )n∈N converges to 0.
For a fixed x ∈ (0, ∞), we have
lim fn (x) = lim xe−nx = x · lim e−nx = x · 0 = 0.
n→∞
n→∞
n→∞
So the pointwise limit of (fn )n∈N is the function f which is identically zero on (0, ∞). We now
show that the convergence is uniform by using the result proved above. For x ∈ (0, ∞) and n ∈ N,
we have nx > 0, and so
enx =
∞
X
1
1
(nx)k ≥ nx.
k!
1!
k=0
Hence 0 < xe−nx ≤
1
n.
Consequently
0 ≤ an := sup{|fn (x) − f (x)| : x ∈ (0, ∞)} =
sup xe−nx ≤
x∈(0,∞)
1
,
n
and so by the Sandwich Theorem,
lim an = 0.
n→∞
Hence (fn )n∈N converges uniformly to the function which is identically zero on (0, ∞).
80
Solutions
Solution to Exercise 2.6.2. We will show that (fn )n∈N converges uniformly to the function
which is identically 0 on [0, 1]. We have for all x ∈ [0, 1] that
|fn (x) − 0| =
1
nx
1
x
= ·
≤ · 1.
1 + nx
n 1 + nx
n
Let ǫ > 0. Choose an N ∈ N such that N > 1/ǫ. Then whenever n > N , we have for all x ∈ [0, 1]
that
1
1
< ǫ.
|fn (x) − 0| ≤ <
n
N
Consequently, (fn )n∈N converges uniformly to the function which is identically zero on [0, 1].
Solution to Exercise 2.6.3. If x 6= 0, then
0<
1
< 1,
1 + x2
and so
lim fn (x) = 1 − lim
n→∞
On the other hand,
n→∞
1
1 + x2
n
= 1 − 0 = 1.
1
= 1 − 1 = 0.
n→∞
n→∞
(1 + 02 )n
So the sequence (fn )n∈N converges pointwise to f . Clearly f is discontinuous at 0: for example, the
sequence ( n1 )n∈N converges to 0, but the sequence (f ( n1 ))n∈N = (1)n∈N converges to 1 6= 0 = f (0).
lim fn (0) = lim 1 −
Solution to Exercise 2.6.4.
(1) We have
lim am,n = lim
m→∞
m→∞
m
1
1
= lim
= 1.
n =
m + n m→∞ 1 + m
1+0
On the other hand,
m
= lim
n→∞
n→∞ m + n
lim am,n = lim
n→∞
m
n
m
n
+1
=
0
= 0.
0+1
Hence
lim lim am,n = lim 0 = 0 6= 1 = lim 1 = lim lim am,n .
m→∞ n→∞
m→∞
(2) We have for x ∈ R that
|fn (x) − f (x)| = |fn (x)| =
n→∞
n→∞ m→∞
| sin(nx)|
1
√
≤ √ .
n
n
Let ǫ > 0. Choose an N ∈ N such that N > 1/ǫ2 . Then whenever n > N , we have for
all x ∈ R that
1
1
|fn (x) − f (x)| ≤ √ < √ < ǫ.
n
N
Consequently, (fn )n∈N converges pointwise (in fact even uniformly!) to the function f
which is identically zero on R.
We have
d
n cos(nx) √
√
fn′ (x) :=
fn (x) =
= n cos(nx).
dx
n
f ′ on the other hand is identically 0 on R. If we look at x = 0, we have (fn′ (x))n∈N =
√
( n)n∈N is clearly unbounded, and thus not convergent. Hence (fn′ )n∈N is not pointwise
convergent (to f ′ ).
Solutions to the exercises from Chapter 2
81
(3) If x = 0 or x = 1, fn (x) = 0, and so (fn (x))n∈N = (0)n∈N is convergent with limit 0
(= f (x)). Now suppose that x ∈ (0, 1). Then 0 < 1 − x2 < 1. So we can find a positive
number h such that 1 − x2 = 1/(1 + h). By the binomial expansion for (1 + h)n , we have
the inequality (1 + h)n > n(n − 1)h2 , which yields
nx
✚
.
0 < nx(1 − x2 )n <
n(n − 1)h2
✚
Thus by the Sandwich Theorem we obtain that (fn (x))n∈N converges to 0. Consequently
(fn )n∈N converges pointwise to the zero function f on [0, 1].
Using the variable substitution u = x2 , we see that
Z 1
Z 1
Z
1 1 n
1 n
nx(1 − x2 )n dx =
.
fn (x)dx =
nu du =
2 0
2n+1
0
0
Thus
Z 1
Z 1
Z 1
1
1 n
= 6= 0 =
f (x)dx =
lim fn (x)dx.
lim
fn (x)dx = lim
n→∞ 0
n→∞ 2 n + 1
2
0
0 n→∞
82
Solutions
Solutions to the exercises from Chapter 3
Solution to Exercise 3.1.1. We have for all x > 0 that
1 1 |x − 2|
− =
.
x 2
2|x|
(From here we see that we need to investigate the size of |x| for x satsifying |x − 2| < δ.) When
x ∈ R satisfies |x − 2| < δ, we have
2 − δ < x < x + 2,
and in particular, if δ ≤ 1, then
0 < 1 = 2 − 1 ≤ 2 − δ < x = |x|.
So whenever δ ≤ 1 and x ∈ R satisfies |x − 2| < δ, then
1 1 |x − 2|
|x − 2|
|x − 2|
− =
<
=
.
x 2
2|x|
2·1
2
(In order to ensure that this is < ǫ, we just need to make sure that δ/2 ≤ ǫ, and so δ ≤ 2ǫ.) So if
we set δ := min{1, 2ǫ} > 0, then for x ∈ R satisfying |x − 2| < δ, we have
1 1 |x − 2|
|x − 2|
|x − 2|
δ
2ǫ
− =
<
=
< ≤
= ǫ.
x 2
2|x|
2·1
2
2
2
By the definition of continuity of a function at a point, we immediately see from the above that
x 7→ 1/x : (0, ∞) → R is continuous at 2.
Solution to Exercise 3.1.2.
(1) Let c′ ∈ R and ǫ > 0. Since f is continuous at c, there exists a δ > 0 such that for all
x ∈ R satisfying |x − c| < δ, |f (x) − f (c)| < ǫ. Then for all x ∈ R satisfying |x − c′ | < δ,
we have2
|f (x) − f (c′ )| = |f (x) − f (c′ ) + f (c) − f (c)| = |f (x − c′ + c) − f (c)| < ǫ,
since |(x − c′ + c) − c| = |x − c′ | < δ. So f is continuous at c′ . Since the choice of c′ ∈ R
was arbitrary, it follows that f is continuous on R.
(2) Let α ∈ R, and let f : R → R be given by f (x) = αx, for all x ∈ R. Then
f (x + y) = α(x + y) = αx + αy = f (x) + f (y).
Solution to Exercise 3.1.3. We observe that as |f (0)| ≤ M |0| = M · 0 = 0, it follows that
|f (0)| = 0, that is, f (0) = 0. Given ǫ > 0, we define δ = ǫ/M . Then for all x ∈ R satisfying
|x| = |x − 0| < δ, we have
|f (x) − f (0)| = |f (x) − 0| = |f (x)| ≤ M |x| = M |x − 0| < M δ = M
ǫ
= ǫ.
M
Hence f is continuous at 0.
2Here we use the fact that f is ‘additive’. First of all, f (0) = f (0 + 0) = f (0) + f (0), and so f (0) = 0. Hence
it follows that −f (c′ ) = f (−c′ ), since 0 = f (0) = f (c′ − c′ ) = f (c′ ) + f (−c′ ). Finally we have f (x) − f (c′ ) + f (c) =
f (x) + f (−c′ ) + f (c) = f (x − c′ ) + f (c) = f (x − c′ + c).
83
Solutions to the exercises from Chapter 3
Solution to Exercise 3.1.4. Let c ∈ R. Suppose that f is continuous at c. Consider ǫ = 21 > 0.
Then there exists a δ > 0 such that for all x ∈ R satisfying |x − c| < δ, |f (x) − f (c)| < ǫ = 12 . We
have the following two cases:
1◦ c ∈ Q. Then there exists x ∈ R \ Q such that |x − c| < δ.
But |f (x) − f (c)| = |1 − 0| = |1| = 1 > 21 , a contradiction.
2◦ c ∈ R \ Q. Then there exists x ∈ Q such that |x − c| < δ.
But |f (x) − f (c)| = |0 − 1| = | − 1| = 1 > 21 , a contradiction.
Hence f is not continuous at c.
Solution to Exercise 3.1.5. Since
|x − c| < δ,
f (c)
2
> 0, there exists a δ > 0 such that for all x ∈ R satisfying
f (c)
.
2
Thus for all x ∈ R satisfying |x − c| < δ (that is, c − δ < x < c + δ), we have
|f (x) − f (c)| <
f (c) − f (x) ≤ |f (c) − f (x)| = |f (x) − f (c)| <
and so f (x) >
f (c)
2
f (c)
,
2
> 0.
Solution to Exercise 3.2.1. We only have to show that f (x) < f (c) for all x ∈ (c − δ, c). Let
(xn )n∈N be a sequence such that x < x1 < x2 < · · · and
lim xn = c.
n→∞
Then f (x) < f (x1 ) ≤ f (xn ) for all n ∈ N. So
f (x) < f (x1 ) ≤ lim f (xn ) = f
n→∞
lim xn = f (c),
n→∞
where we use the continuity of f in order to get the last but one equality. Thus f (x) < f (c).
Remark 6.2. From here it follows that a continuous function f : (c − δ, c + δ) → R which is
strictly increasing on (c − δ, c) and (c, c + δ) is strictly increasing on (c − δ, c + δ).
Solution to Exercise 3.2.2. Let x ∈ R. For each n ∈ N, pick a qn ∈ Q such that
x < qn < x + 1/n.
Then the sequence (qn )n∈N is convergent with limit x. (Let ǫ > 0. By the Archimedean principle,
there exists an N ∈ N such that 1/ǫ < N . Thus for n > N , |qn − x| = qn − x < 1/n < 1/N < ǫ.)
Since f is continuous at c, we have f (c) = lim f (qn ) = lim 0 = 0.
n→∞
n→∞
Solution to Exercise 3.2.3. If x1 6= x2 , then the sequence x1 , x2 , x1 , x2 , . . . is divergent. (Indeed, the subsequence x1 , x1 , x1 , . . . converges to x1 , while the subsequence x2 , x2 , x2 , . . . converges
to x2 , and so it follows that the sequence x1 , x2 , x1 , x2 , . . . is divergent.)
Thus the sequence f (x1 ), f (x2 ), f (x1 ), f (x2 ), . . . is divergent. Consequently f (x1 ) 6= f (x2 ):
for otherwise if f (x1 ) = f (x2 ), then the sequence
f (x1 ),
q
f (x1 ),
f (x2 ),
q
f (x1 ),
f (x1 ),
q
f (x1 ),
f (x2 ) . . .
q
f (x1 ) . . .
is a constant sequence, and so it is convergent with limit f (x1 ) (= f (x2 )).
So we have shown that if x1 6= x2 , then f (x1 ) 6= f (x2 ), that is, the function f is one-to-one.
84
Solutions
Solution to Exercise 3.2.4. That (1) implies (2′ ) is immediate.
Now suppose that (2) holds. Let (xn )n∈N be a sequence contained in I that converges to c ∈ I.
Then the new sequence x1 , c, x2 , c, x3 , c, . . . also converges to c. By the hypothesis (2), it follows
that the image of this sequence under f , namely the sequence f (x1 ), f (c), f (x2 ), f (c), f (x3 ), f (c), . . .
must be convergent, say with limit L. But then each of its subsequences must also be convergent
with the same limit L. By looking at the even indexed terms of this sequence, we obtain that the
subsequence f (c), f (c), f (c), . . . converges to L, and so L = f (c). On the other hand, by looking
at the odd indexed terms, we conclude that the sequence f (x1 ), f (x2 ), f (x3 ), . . . must be also
convergent with limit L (=f (c)). Consequently, f is continuous at c.
Solution to Exercise 3.2.5. Let x 6= 0. We consider two possible cases.
1◦ Let x be rational. Then we can find a sequence (yn )n∈N of irrational numbers that
converges to x. If f was continuous at x, then (f (yn ))n∈N = (−yn )n∈N would have
to be convergent with limit f (x) = x. But then we obtain x = −x and so x = 0, a
contradiction.
2◦ Let x be irrational. Then we can find a sequence (rn )n∈N of rational numbers that
converges to x. If f was continuous at x, then (f (rn ))n∈N = (rn )n∈N would have to
be convergent with limit f (x) = −x. But then we obtain x = −x and so x = 0, a
contradiction.
So f is not continuous at x whenever x 6= 0. We now show that f is continuous at 0. Let ǫ > 0.
Let δ := ǫ > 0. Since f (x) = x if x is rational, and −x if x is irrational, it follows that in either
case, |f (x)| = |x|. Thus, whenever x ∈ R and |x − 0| = |x| < δ = ǫ, we have
|f (x) − f (0)| = |f (x) − 0| = |f (x)| = |x| < δ = ǫ.
So f is continuous at 0.
Solution to Exercise 3.2.6. Let x be a rational number. Then f (x) > 0. We can find a
sequence (yn )n∈N of irrational numbers that converges to x. If f was continuous at x, then
(f (yn ))n∈N = (0)n∈N would have to be convergent with limit f (x) > 0, which is clearly not true.
Thus f is not continuous at x.
Now suppose that x is irrational. Then x 6= 0, and so there exists an integer N such that
x ∈ (N, N + 1). Let ǫ > 0. Suppose that r is a rational number in (N, N + 1) for which f (r) ≥ ǫ.
Let r = n/d, where n, d are integers without any common divisors and d > 0. Then we have
f (r) = 1/d ≥ ǫ and so d ≤ 1/ǫ. So there are just finitely many possibilities for d. Moreover,
since N < n/d < N + 1, it follows that N d < n < (N + 1)d, and so there are only finitely many
possibilities for d as well. So there are just finitely many rational numbers r in (N, N + 1) for
which f (r) ≥ ǫ. Now among these finitely many rational numbers, let r∗ be one which is the
closest to x, and set δ := min{|x − r∗ |/2, |x − N |, |x − (N + 1)|} > 0. Then whenever z ∈ R and
|z − x| < δ, we are guaranteed that z ∈ (N, N + 1), and moreover if z is rational, then it can’t be
one of the rational r:s in (N, N + 1) for which f (r) ≥ ǫ, and so it must be the case that f (z) < ǫ.
If z is irrational, then f (z) = 0 by definition of f , and so then too we have f (z) = 0 < ǫ. So we
have that for all z ∈ R such that |z − x| < δ, |f (z) − f (x)| = |f (z) − 0| = |f (z)| = f (z) < ǫ. This
completes the proof that f is continuous at every irrational number.
Solution to Exercise 3.2.7. We show by induction that f (n) = nf (1) for all natural numbers n.
For n = 1, we have f (n) = f (1) = 1·f (1) = nf (1), and so the claim is true when n = 1. If for some
n ∈ N there holds f (n) = nf (1), then we have f (n+1) = f (n)+f (1) = nf (1)+f (1) = (n+1)f (1).
This completes the induction step, and so the result holds for all natural numbers.
85
Solutions to the exercises from Chapter 3
Also f (0) = f (0 + 0) = f (0) + f (0) shows that f (0) = 0 = 0 · f (1). So f (n) = nf (1) for all
nonnegative integers. Let m be a negative integer. Then
0 = f (0) = f (m + (−m)) = f (m) + f (−m) = f (m) + (−m)f (1),
and so f (m) = −(−m)f (1) = mf (1). So we have that f (n) = nf (1) for all integers n.
Now every rational number r can be expressed as r = n/d for some integers n, d with d > 0.
Then
nf (1) = f (n) = f (d · (n/d)) = f (n/d) + · · · + f (n/d) = d · f (n/d).
|
{z
}
d times
Consequently, f (n/d) = (nf (1))/d, that is, f (r) = rf (1).
Let x ∈ R. Then we can find a sequence (rn )n∈N of rational numbers which converges to x.
Using the continuity of f we obtain
f (x) = f lim rn = lim f (rn ) = lim rn f (1) = xf (1).
n→∞
n→∞
n→∞
Consequently, f (x) = xf (1) for all x ∈ R.
Solution to Exercise 3.2.8. Let x ∈ R. Clearly f (2x) = −f (x), and so
f (x) = −f (x/2) = (−1)2 f (x/4) = · · · = (−1)n f (x/2n )
for all x ∈ R. Since the sequence (x/2n )n∈N converges to 0, it follows that
f (0) = f lim x/2n = lim f (x/2n ) = (−1)n f (x).
n→∞
n→∞
But f (0) = f (2 · 0) = −f (0), and so f (0) = 0. It follows from the above that
f (x) = (−1)n f (0) = (−1)n 0 = 0.
So if f is continuous and it satisfies the given identity then it must be the constant function 0.
Conversely, the constant function 0 is indeed continuous and also f (2x) + f (x) = 0 + 0 = 0
for all x ∈ R.
Solution to Exercise 3.2.9. Let I = (0, ∞). Define f : (0, ∞) → R by f (x) = 1/x, x ∈ (0, ∞).
Then f is continuous. Take (xn )n∈N to be the Cauchy sequence n1 n∈N . Then f (xn ) = n, and
we have |f (xn ) − (xm )| = |n − m| ≥ 1 for all n 6= m, showing that the sequence (f (xn ))n∈N is not
Cauchy.
Solution to Exercise 3.3.1. That f is stricly decreasing: If x1 > x2 ≥ 0, then x21 = x1 · x1 >
x1 · x2 ≥ x2 · x2 = x22 ≥ 0, and so 1 + x21 > 1 + x22 , giving
f (x1 ) =
1
1
<
= f (x2 ).
2
1 + x1
1 + x22
So f is strictly decreasing.
That f ([0, ∞)) = (0, 1]: Let ǫ > 0. Choose n ∈ N such that n > 1/ǫ − 1. Then n2 > n > 1/ǫ − 1,
and so
1
1
f (n) =
<
= ǫ.
2
1+n
1 + 1/ǫ − 1
On the other hand, f (0) = 1. So by the Intermediate Value Theorem, (ǫ, 1] ⊂ f ([0, ∞)). As ǫ > 0
was arbitrary, it follows that (0, 1] ⊂ ([0, ∞)). Thereverse inclusion follows by observing that
0≤
1
1
≤
= 1,
2
1+x
1+0
for x ∈ [0, ∞) so that f ([0, ∞)) ⊂ [0, 1], and clearly f (x) = 1/(1 + x2 ) is never zero for x ∈ [0, ∞).
86
Solutions
Expression for f −1 : Let y ∈ (0, 1] and x ∈ [0, ∞). Then
So f −1
1
1
1
=y ⇔
= 1 + x2 ⇔ x2 = − 1 ⇔ x =
1 + x2
y
y
r
1
−1
: (0, 1] → [0, ∞) is given by f (y) =
− 1, y ∈ (0, 1].
y
r
1
− 1.
y
For any n > 0, f |[0,n] : [0, n] → R is strictly decreasing and continuous. Clearly f |[0,n] ([0, n]) =
1
1
So its inverse function (f |[0,n] )−1 : [ 1+n
f ([0, n]) = [f (n), f (0)] = [ 1+n
2 , 1].
2 , 1] → [0, n] is
1
continuous. But if y ∈ [ 1+n2 , 1] = f |[0,n] ([0, n]), then y = f (x) for some x ∈ [0, n], and so
(f |[0,n] )−1 (y) = x = f −1 (y). This shows that (f |[0,n] )−1 = (f −1 )|[ 1 2 ,1] . But as n was arbitrary,
1+n
it follows that f −1 is continuous on (0, 1].
See Figure 10 for the graphs of f, f −1 .
Figure 10. The graphs of f and f −1
Solution to Exercise 3.3.2. Imagine a directed line ℓ such that the pancake lies entirely to one
side of the line ℓ (say if we look along the direction of the line, the pancake appears to our right).
Now translate the line to the right parallel to itself till the whole pancake appears to the left of
the line. Suppose that the total distance by which the line is translated is d. At each intermediate
distance x ∈ [0, d], let A(x) denote the area of the part of the pancake that lies to the right of
our line. Thus if S is the total area of the pancake, then A(0) = A and A(d) = 0. As the map
A : [0, d] → R is continuous and f (0) = S ≥ S/2 ≥ 0 = f (d), it follows by the Intermediate Value
Theorem, that there is an y ∈ [0, d] such that A(y) = S/2, or in other words, at some position of
our line, the pancake is divided into two parts with equal areas.
As the direction of the line in the above process was inconsequential, given any direction, we
can choose a straight line cut having that direction which divides the pancake into two equal parts.
Solution to Exercise 3.3.3.
(1) Let
f (x) =
2x
if x ∈ (0, 21 ),
2 − 2x if x ∈ [ 21 , 1).
Then f is continuous on (0, 1) and f (0, 1) = (0, 1].
87
Solutions to the exercises from Chapter 3
(2) If there existed such a continuous f , then since 1/2, 3/2 ∈ T and since f (S) = T , there
would exist a, b ∈ S = (0, 1) such that f (a) = 1/2 and f (b) = 3/2. But then since we
have that f (a) = 1/2 < 1 < 3/2 = f (b), it follows by the Intermediate Value Theorem
that there is a c between a and b such that f (c) = 1. Thus 1 ∈ f (S) = T , a contradiction.
(3) We have 0, 1 ∈ Q = T = f (S),
√ and so there exist a, b ∈ S = R such that f (a) = 1 and
Value Theorem
f (b) = 2. Since f (a) = 1 < 2 < 2 = f (b), it follows by
√
√ the Intermediate
that there is a c between a and b such that f (c) = 2. Thus 2 ∈ f (S) = T = Q, a
contradiction.
(4) Set
f (x) =
0
1
if x ∈ [0, 1],
if x ∈ [2, 3].
S
Then f is continuous on [0, 1] [2, 3]. Indeed if (xn )n∈N is a convergent sequence in S
with limit L in S, then beyond a certain index N , the terms all lie in [0, 1], or they all
lie in [2, 3]. Thus the terms of the sequence (f (xn ))n∈N beyond this index N are all 0,
or they are all 1. In either case, the sequence (f (xn ))n∈N is convergent, and this shows
that f is continuous. Also it is clear that f (S) = T .
Solution to Exercise 3.3.4. Consider the function g : [0, T ] → R, given by
g(x) = f (x) for all x ∈ [0, T ].
Then g is continuous on [0, T ].
(Indeed, the continuity of g on [0, T ] is a trivial consequence of the continuity of f on R: If
c ∈ [0, T ], and ǫ > 0, then there exists a positive δ such that for all x ∈ R satisfying |x − c| < δ,
|f (x) − f (c)| < ǫ. Hence for all x ∈ [0, T ] satisfying |x − c| < δ, |g(x) − g(c)| = |f (x) − f (c)| < ǫ.
So g is continuous at c. But the choice of c was arbitrary, and so g is continuous on [0, T ].)
Applying the extreme value theorem to g, we conclude that there exist c, d ∈ [0, T ] such that
g(c) =
g(d) =
max{g(x) : x ∈ [0, T ]} and
min{g(x) : x ∈ [0, T ]}.
So for all x ∈ [0, T ], g(d) ≤ g(x) ≤ g(c), that is, f (d) ≤ f (x) ≤ f (c).
g = f |[0,T ]
...
...
d
−T
0 c
T
2T
3T
4T
Figure 11. A continuous periodic function is bounded.
So far we have proved the fact that f is bounded on [0, T ]. We now prove that f is bounded on
R by using the periodicity of f . See Figure 11.
Now if x is any real number, there exists a n ∈ Z such that x = nT + r, where r ∈ R is such
that r ∈ [0, T ). (Indeed, we have
jxk
x
+Θ
=
T
T
where Θ ∈ [0, 1). Consequently, we obtain x = nT +r, where n := Tx ∈ Z and r := T ·Θ ∈ [0, T ).)
Thus f (x) = f (nT + r) = f (r). As f (d) ≤ f (r) ≤ f (c), it follows that f (d) ≤ f (x) ≤ f (c). Since
the choice of x ∈ R was arbitrary, it follows that f (d) ≤ f (x) ≤ f (c) for all x ∈ R. So f (c) and
f (d) are upper and lower bounds, respectively, of the set {f (x) : x ∈ R}, and so it is bounded.
88
Solutions
Solution to Exercise 3.3.5.
(1) If x ∈ (a, b], then let f |[a,x] denote the restriction of f to [a, x], defined by f |[a,x] (y) = f (y)
for all y ∈ [a, x]. We note that f |[a,x] is a continuous function. Applying the Extreme
Value Theorem to f |[a,x] , we see that
max{f |[a,x](y) : y ∈ [a, x]} = max{f (y) : y ∈ [a, x]}
exists, and so f∗ is well-defined.
(2) f∗ is given by
f∗ (x) =
See Figure 12.
x − x2
1
4
f
1
4
0
1
2
1
f∗
1
4
0
if 0 ≤ x ≤ 21 ,
if 21 < x ≤ 1.
1
2
1
Figure 12. f and f∗ .
Solution to Exercise 3.3.6. Since the function f : [0, 1] → R and the function x 7→ x from
[0, 1] to R are both continuous on the interval [0, 1], it follows that also the function g : [0, 1] → R
defined by
g(x) = f (x) − x, for all x ∈ [0, 1]
is continuous on [0, 1]. Since 0 ≤ f (x) ≤ 1 for all x ∈ [0, 1], we have
g(0) = f (0) − 0 = f (0) ≥ 0, and
g(1) = f (1) − 1 ≤ 0.
So by the intermediate value theorem, there exists a c ∈ [0, 1] such that g(c) = 0, that is, f (c) = c.
Solution to Exercise 3.3.7. Let the weekend campsite
be at altitude H. Let u : [0, 1] → R be
the position function for the walk up, and d : 0, 21 → R be the position function for
walk
the
1
down. (We assume that these are continuous functions.) Consider the function f : 0, 2 → R
given by
1
f (t) = u(t) − d(t), t ∈ 0,
.
2
Then f is also continuous, and moreover
f (0) = u(0) − d(0) = 0 − H = −H < 0, while
1
1
1
1
1
=u
−d
=u
−0=u
> 0.
f
2
2
2
2
2
See Figure 13. Hence by the intermediate value theorem, it follows that there exists a c ∈ 0, 12
89
Solutions to the exercises from Chapter 3
H
u
d
u(c) = d(c)
0
c
1
2
1
Figure 13. The walks up and down.
such that f (c) = 0, that is, u(c) = d(c). So at time c past 8:00, the hiker was exactly at the same
spot on Saturday and Sunday.
Solution to Exercise 3.3.8. The polynomial function p : [−1, 2] → R is continuous on the
interval [−1, 2]. Moreover,
p(−1) = 2 · (−1)3 − 5 · (−1)2 − 10 · (−1) + 5 = −2 − 5 + 10 + 5 = 8 > 0, and
p(2) = 2 · (2)3 − 5 · (2)2 − 10 · (2) + 5 = 16 − 20 − 20 + 5 = −19 < 0.
Since p(−1) > 0 > p(2), from the intermediate value theorem applied to the continuous function
p on the interval [−1, 2], we conclude that there must exist a c ∈ [−1, 2] such that p(c) = 0. So p
has a real root in the interval [−1, 2].
Solution to Exercise 3.3.9. Obviously S ⊂ R. We now show the reverse inclusion. Let y ∈ R.
As S is not bounded above, y is not an upper bound of S, that is, there exists a x0 ∈ R such
that f (x0 ) < y.
Similarly, since S is not bounded below, y is not a lower bound of S, and so there exists a
x1 ∈ R such that f (x1 ) > y.
Now consider the restriction of f to the interval with endpoints x0 and x1 with the endpoints
included in the interval. Applying the intermediate value theorem to this continuous function, it
follows that there exists a real number c such that f (c) = y.
This shows that S = R.
Solution to Exercise 3.3.10. The following three cases are possible:
1◦ f (0) = 0. Then let x0 = 0 and let m ∈ Z \ {0}. Clearly f (x0 ) = f (0) = 0 = m0 = mx0 .
2◦ f (0) > 0. Choose N ∈ N satisfying N > f (1) (that such a N exists follows from the
Archimedean property). Consider g : [0, 1] → R defined by g(x) = f (x) − N x, x ∈ [0, 1].
As f and x 7→ N x are continuous, so is g. Note that g(0) = f (0) − N 0 = f (0) > 0, while
g(1) = f (1) − N < 0. Applying the intermediate value theorem to g (with y = 0), it
follows that there exists a x0 ∈ [0, 1] such that g(x0 ) = 0, that is, f (x0 ) = N x0 .
3◦ f (0) < 0. Choose a N ∈ N such that N > −f (1) (again the Archimedean property
guarantees the existence of such a N ), and consider the continuous function g : [0, 1] → R
defined by g(x) = f (x)+N x. We observe that g(0) = f (0) < 0, and g(1) = f (1)+N > 0,
and so by the intermediate value theorem, it follows that there exists a x0 ∈ [0, 1] such
that g(x0 ) = 0, that is, f (x0 ) = −N x0 .
This completes the proof.
90
Solutions
Solution to Exercise 3.3.11. Suppose that such a continuous function exists. From the part
above, it follows that there exists a x0 ∈ R and a m ∈ Z \ {0} such that f (x0 ) = mx0 . We have
the following two possible cases:
1◦ x0 ∈ Q. But then f (x0 ) is irrational, while mx0 is rational, a contradiction.
2◦ x0 6∈ Q. But then f (x0 ) is rational, whereas mx0 is irrational, a contradiction.
So f cannot be continuous.
(Alternately, one could observe that f (R \ Q), being a subset of Q, is countable. Also f (Q) is
clearly countable. Hence f (R) = f (R \ Q) ∪ f (Q) is countable. But then f (R) must be a singleton,
since if it contained two distinct points a < b, then it would also contain the whole interval [a, b] by
the intermediate value
√ theorem, and f (R) would then be uncountable. Suppose that f (R) = {x∗ }.
But f (0) = x∗ = f ( 2) gives a contradiction.)
Solution to Exercise 3.4.1. Suppose that f is uniformly continuous. Let ǫ = 1 > 0, and
let δ > 0 be such that for all x, y ∈ R satisfying |x − y| < δ, there holds |x2 − y 2 | < 1. For
n ∈ N, set xn = n and yn = n + n1 . Then |xn − yn | = n1 < δ for all n:s large enough. However,
|x2n − yn2 | = |n2 − (n + n1 )2 | = 2 + n12 > 2 > 1 = ǫ, a contradiction. Thus f is not uniformly
continuous.
Solution to Exercise 3.4.2. We have for x, y ∈ R that ||x| − |y|| ≤ |x − y|. If ǫ > 0, then with
δ := ǫ > 0, we have for all x, y ∈ R satisfying |x − y| < δ = ǫ that ||x| − |y|| ≤ |x − y| < δ = ǫ. So
the absolute value function is uniformly continuous.
Solution to Exercise 3.4.3. Let ǫ > 0. Set δ = ǫ/L > 0. Then whenever x, y ∈ R and
|x − y| < δ, we have
|f (x) − f (y)| ≤ L|x − y| < Lδ = ǫ.
Thus f is uniformly continuous.
Solution to Exercise 3.4.4. Suppose that (xn )n∈N is a Cauchy sequence in I. We want to show
that (f (xn ))n∈N is a Cauchy sequence in R. Let ǫ > 0. Since f is uniformly continuous, there
exists a δ > 0 such that whenever |x − y| < δ, we have |f (x) − f (y)| < ǫ. As (xn )n∈N is Cauchy,
there exists an index N ∈ N such that for all n, m > N , |xn − xm | < δ. But then for all n, m > N ,
we have |f (xn ) − f (yn )| < ǫ. Hence (f (xn ))n∈N is a Cauchy sequence in R.
Solution to Exercise 3.5.1. (If part): Suppose that
L := lim f (x) = lim f (x).
x→c+
x→c−
Let ǫ > 0. Then there exists a δ+ > 0 such that for all x satisfying c < x < c + δ+ , |f (x) − L| < ǫ.
Also, there exists a δ− > 0 such that for all x satisfying c − δ− < x < c, |f (x) − L| < ǫ. Let
δ := min{δ+ , δ− }. Then for all x such that 0 < |x − c| < δ, we have either c < x < c + δ ≤ c + δ+
or c − δ− ≤ c − δ < x < c, and in either case we have |f (x) − L| < ǫ. Hence
lim f (x) = L
= lim f (x) = lim f (x) .
x→c
x→c+
x→c−
(Only if part): Let
L := lim f (x).
x→c
Let ǫ > 0. Then there exists a δ > 0 such that for all x satisfying 0 < |x − c| < δ, we have
|f (x) − L| < ǫ. In particular, for all x satisfying c < x < c + δ, we have |f (x) − L| < ǫ, and so
lim f (x) = L.
x→c+
91
Solutions to the exercises from Chapter 3
Also, for all x satisfying c − δ < x < c, we have |f (x) − L| < ǫ, and so
lim f (x) = L.
x→c−
This implies that lim f (x) = L = lim f (x).
x→c+
x→c−
Solution to Exercise 3.5.2. (1) As x 7→ |x|, x + 1 are continuous, we have
lim |x| = |0| = 0, and lim (x + 1) = 0 + 1 = 1 6= 0.
x→0
x→0
lim |x|
0
|x|
= x→0
= = 0.
x→0 x + 1
lim (x + 1)
1
Thus by the Algebra of Limits, lim
x→0
(2) We have
lim (⌊x⌋ − x)
= 1 − 1 = 0, while
lim (⌊x⌋ − x)
= 0 − 1 = −1,
x→1+
x→1−
and so lim (⌊x⌋ − x) does not exist.
x→1
(3) We have
lim ⌊x⌋
= 0,
lim ⌊x⌋
= −1,
x→0+
x→0−
and so
lim x⌊x⌋
=
0 · 0 = 0,
lim x⌊x⌋
=
0 · (−1) = 0.
x→0+
x→0−
Hence lim x⌊x⌋ = 0.
x→0
(4) Suppose that the limit exists, and is some number L. Take any θ ∈ [−π/2, π/2] such that
sin θ 6= L. Set
1
xn :=
, n ∈ N.
θ + 2nπ
Then (xn )n∈N converges to 0. But
sin
and so
1
= sin(θ + 2nπ) = sin θ,
xn
1
1
converges to sin θ 6= L, a contradiction. So lim sin does not exist.
sin
x→0
n n∈N
x
Solution to Exercise 3.5.3. We have that B is zero for only finitely many real values, and so
for all sufficiently large enough x, B(x) 6= 0. We have
aα−1
a0
+ ··· +
+1
A(x)
aα α−β aα xα
a0 + · · · + aα−1 xα−1 + aα xα
aα α−β
aα x
=
·x
·
·x
·ϕ(x),
=
=
β−1
β
b
b
B(x)
b0 + b1 x + · · · + bβ−1 x
+ bβ x
bβ
bβ
0
β−1
+
·
·
·
+
+
1
bβ xβ
bβ x
where
a0
+ ···+
a xα
ϕ := α
b0
+ ···+
bβ xβ
aα−1
+1
aα x
.
bβ−1
+1
bβ x
92
Solutions
Clearly
a0
+ ···+
a xα
lim ϕ(x) = lim α
b0
x→∞
x→∞
+ ···+
bβ xβ
aα−1
+1
0 + ···+ 0 + 1
aα x
=
= 1.
bβ−1
0 + ···+ 0 + 1
+1
bβ x
Suppose that β > α. Then lim xα−β = 0, and so
x→∞
aα
aα α−β
A(x)
·x
· ϕ(x) =
· 0 · 1 = 0.
= lim
x→∞ B(x)
x→∞ bβ
bβ
lim
Next suppose that α = β. Then
A(x)
aα
aα
aα
· ϕ(x) =
·1=
.
= lim
x→∞ B(x)
x→∞ bβ
bβ
bβ
lim
Finally suppose that α > β. Consider first the case when
aα
> 0.
bβ
Then there exists an R1 > 0 such that for all x > R1 , 1 − ϕ(x) ≤ |ϕ(x) − 1| < 1/2, and so
ϕ(x) > 1/2. Let M > 0. Then taking any R2 such that
1/(α−β)
bβ
R2 > 2M ·
,
aα
aα α−β
·x
> 2M . Hence for all x > max{R1 , R2 }, we obtain
we have for x > R2 that
bβ
1
aα α−β
A(x)
·x
· ϕ(x) > 2M · = M.
=
B(x)
bβ
2
A(x)
= +∞.
B(x)
Now consider the case when
Thus lim
x→∞
aα
< 0.
bβ
Then there exists an R1 > 0 such that for all x > R1 , ±(ϕ(x) − 1) ≤ |ϕ(x) − 1| < 1/2, and so
0 < 1/2 < ϕ(x) < 3/2. Let M > 0. Then taking any R2′ such that
1/(α−β)
bβ
2
′
,
R2 > − M ·
3
aα
2
aα α−β
·x
< − M . Hence for all x > max{R1 , R2′ }, we obtain
we have for x > R2′ that
bβ
3
2
aα α−β
3
A(x)
·x
· ϕ(x) < − M · = −M.
=
B(x)
bβ
3
2
Thus lim
x→∞
A(x)
= −∞.
B(x)
93
Solutions to the exercises from Chapter 4
Solutions to the exercises from Chapter 4
Solution to Exercise 4.1.1. For x 6= x0 , we have
p
√
1 + x2 − 1 + x20
f (x) − f (x0 )
x2 + 1 − (x20 + 1)
p
√
=
=
x − x0
x − x0
(x − x0 )( 1 + x2 + 1 + x20 )
=
So
x2 − x20
x + x0
p
p
= √
.
√
2
2
(x − x0 )( 1 + x + 1 + x0 )
1 + x2 + 1 + x20
x + x0
f (x) − f (x0 )
x0 + x0
x0
p
p
= lim √
= p
=p
.
x→x0
x − x0
1 + x2 + 1 + x20
1 + x20 + 1 + x20
1 + x20
x
, x ∈ R.
Hence f is differentiable everywhere, and f ′ (x) = √
1 + x2
lim
x→x0
Solution to Exercise 4.1.2. For h > 0, small enough, we have
f (c + h) − f (c − h)
2h
=
=
f (c + h) − f (c) + f (c) − f (c − h)
2h
1 f (c + h) − f (c) 1 f (c) − f (c − h)
+
.
2
h
2
h
Since f is differentiable at c, we have
lim
h→0
h>0
f (c) − f (c − h)
f (c + (−h)) − f (c)
f (c + h) − f (c)
= f ′ (c) and lim
= lim
= f ′ (c).
h→0
h→0
h
h
−h
h>0
Consequently, lim
h→0
h>0
h>0
1
1
f (c + h) − f (c − h)
= · f ′ (c) + · f ′ (c) = f ′ (c).
2h
2
2
The converse is not true. For example, take f given by f (x) = |x|, x ∈ R, and c = 0. Then for
h > 0, we have
f (c + h) − f (c − h)
|0 + h| − |0 − h|
h−h
=
=
= 0.
2h
2h
2h
f (c + h) − f (c − h)
exists (and is = 0). But f is not differentiable at c = 0.
Hence lim
h→0
2h
h>0
Solution to Exercise 4.1.3. (1)⇒(2): Let ǫ > 0. Let δ > 0 be such that whenever 0 < |x−c| < δ,
we have
f (x) − f (c)
′
− f (c) < ǫ.
x−c
e we have 0 < |kc − c| < δ, so that
Now take δe := δ/c. Then δe > 0 and whenever 0 < |k − 1| < δ,
f (kc) − f (c)
− f ′ (c) < ǫ,
kc − c
f (kc) − f (c)
f (kc) − f (c)
′
that is, − cf (c) < ǫc. So lim
= cf ′ (c).
k→1
k−1
k−1
(2)⇒(1): Suppose that
f (kc) − f (c)
= L.
k−1
Let ǫ > 0. Let δ > 0 be such that whenever 0 < |k − 1| < δ, we have
f (kc) − f (c)
< ǫ.
−
L
k−1
lim
k→1
94
Solutions
e we have with k := x/c that
Now let δe := δc. Then δe > 0 and whenever 0 < |x − c| < δ,
x
0 < c − 1 < δc,
c
that is, 0 < |k − 1| < δ, so that
x
f ( c c) − f (c)
< ǫ.
−
L
x
−1
c
e then we have
So if 0 < |x − c| < δ,
Hence
f (x) − f (c) L ǫ
− < .
x−c
c
c
f (x) − f (c)
x−c
exists and equals L/c. So f is differentiable at c and
lim
x→c
f ′ (c) =
1
f (kc) − f (c)
L
= · lim
.
c
c k→1
k−1
Solution to Exercise 4.1.4. First we note that f (1) = 0 because
f (1) = f (1 · 1) = f (1) + f (1).
We have
f (kc) − f (c)
1 f (k) + f (c) − f (c)
= lim ·
k→1 c
kc − c
k−1
1 f (k)
1 f (k) − 0
= lim ·
= lim ·
k→1 c k − 1
k→1 c
k−1
f (k) − f (1)
f ′ (1)
1
lim
=
.
=
c k→1
k−1
c
So f ′ (x) = f ′ (1)/x, x ∈ (0, ∞). So f ′ is infinitely differentiable, and
f ′ (c)
=
lim
k→1
(−1)n−1 (n − 1)!f ′ (1)
, x ∈ (0, ∞).
xn
(−1)n−1 (n − 1)!2
.
In particular, if f ′ (1) = 2, then f (n) (3) =
3n
f (n) (x) =
Solution to Exercise 4.1.5. If x 6= 0, then f is differentiable at x and we have
1
1
1
1
1
′
2
− 2 = 2x sin − cos .
f (x) = 2x sin + x cos
x
x
x
x
x
On the other hand, we have for x 6= 0 that
f (x) − f (0)
1
= x sin ,
x−0
x
and so, given ǫ > 0, if we set δ = ǫ > 0, then we have for all x ∈ R satisfying 0 < |x − 0| = |x| <
δ = ǫ that
f (x) − f (0)
1 1 − 0 = x sin = |x| · sin ≤ |x| · 1 < δ = ǫ.
x−0
x
x
Consequently, f is differentiable at 0, and its derivative is f ′ (0) = 0.
1
)n∈N converges to 0, it would follows that
If f ′ were continuous, then since the sequence ( 2πn
′
′ 1
also (f ( 2πn ))n∈N converges to f (0) = 0. However, for all n ∈ N,
2
1
′
=
sin(2πn) − cos(2πn) = 0 − 1 = −1.
f
2πn
2πn
hence f ′ is not continuous. (So f is differentiable, but not “continuously differentiable”.)
95
Solutions to the exercises from Chapter 4
Solution to Exercise 4.2.1. When n = 1, we have for x ∈ I that
1 (0)
1 (1)
′
′
′
(1−0)
(f g) (x) = f (x)g(x) + f (x)g (x) =
f (x)g
(x) +
f (x)g (1−1) (x),
0
1
and so the claim is true when n = 1.
Suppose the claim is true for some n ∈ N. Then for x ∈ I,
(n+1)
(f g)
(x)
=
=
=
′
(f g)(n) (x) =
n X
n (k) (n−k)
f g
k
k=0
!′
(x)
n ′
X
n
f (k) g (n−k) (x)
k
k=0
n
X n
f (k+1) (x)g (n−k) (x) + f (k) (x)g (n−k+1) (x) .
k
k=0
Using the facts that
n
n+1
=1=
,
0
0
n
n+1
=1=
,
n
n+1
n
n
n+1
+
=
,
k−1
k
k
n−1
n X n
X
n
f (k+1) (x)g (n−k) (x) =
f (k) (x)g (n−k+1) (x),
k
k−1
k=0
k=1
we obtain
(f g)(n+1) (x)
=
=
=
=
n n
X
X
n (k+1)
f
(x)g (n−k) (x) +
f (k) (x)g (n−k+1) (x)
k
k=0
k=0
n−1
X n
n (n+1)
(k+1)
(n−k)
f
(x)g
(x) +
f
(x)g (0) (x)
k
n
k=0
n X
n (0)
n (k)
+
f (x)g (n+1) (x) +
f (x)g (n−k+1) (x)
0
k
k=1
n + 1 (0)
f (x)g (n+1) (x)
0
n−1
X n n
+
+
f (k) (x)g (n−k+1) (x)
k−1
k
k=1
n + 1 (n+1)
+
f
(x)g (0) (x)
n+1
n+1
X n + 1
f (k+1) (x)g (n+1−k) (x).
k
k=0
Consequently, the claim follows by induction.
Consider the map Θx : (0, ∞) → R given by Θx (t) = tx , t > 0. Then
x−n
Θ(n)
= x[n] tx−n ,
x (t) = x(x − 1) · · · (x − n + 1)t
96
Solutions
and so by the first part of this exercise,
(n)
Θx+y (t)
= (tx+y )(n) = (tx ty )(n) = (Θx Θy )(n) (t) =
n X
n
(n−k)
Θ(k)
(t),
x (t)Θy
k
k=0
that is, (x + y)[n] tx+y−n
n X
n [k] x−k [n−k] y−(n−k)
=
x t
y
t
. Setting t = 1 gives
k
k=0
(x + y)[n] =
n X
n [k] [n−k]
x y
.
k
k=0
Solution to Exercise 4.2.2. For x ∈ (0, ∞), we have
f ′ (x) = −
1
2x
· 2x = −
6= 0.
(1 + x2 )2
(1 + x2 )2
Hence by the Differentiable Inverse Theorem, f −1 is differentiable and for
r
1
1
∈ (0, 1)
(so that x =
− 1),
y = f (x) =
1 + x2
y
we have
(f −1 )′ (y)
1
(1 + 1/y − 1)2
(1 + x2 )2
=− p
=−
2x
2x
2 1/y − 1
−
(1 + x2 )2
1
1/y 2
=− p
.
= − p
2 1/y − 1
2y y(1 − y)
=
1
=
′
f (x)
Solution to Exercise 4.2.3.
(1) We have for x ∈ R that
f ′ (x) = cos(x sin x) · 1 · sin x + x · cos x + cos sin(x2 ) · cos(x2 ) · 2x.
(2) We have for x ∈ R that
f ′ (x) = cos 6 cos 6 sin(6 cos x) · (−6) sin 6 sin(6 cos x) · 6 cos(6 cos x) · (−6 sin x)
= 6 · 6 · 6 · cos 6 cos 6 sin(6 cos x) · sin 6 sin(6 cos x) · cos(6 cos x) · sin x.
(3) We have for x ∈ R that
2
· cos (sin(sin x))2 · 2 · sin(sin x) · cos(sin x) · cos x.
f ′ (x) = 3 sin (sin(sin x))2
Solution to Exercise 4.2.4.
n n X
X
n k
n k
3 , and so
3 = 4n − 1.
k
k
k=0
k=1
n
X n
(2) We have upon differentiating (1 + x)n =
xk that
k
(1) Putting x = 3, we get 4n = (1 + 3)n =
k=0
n−1
n(1 + x)
=
n X
k=1
n
kxk−1 .
k
97
Solutions to the exercises from Chapter 4
n−1
So upon multiplying by x, we get nx(1 + x)
n X
n
=
kxk , and differentiating we
k
k=1
get,
n−1
n(1 + x)
n−2
+ n(n − 1)(1 + x)
n X
n 2 k−1
=
k x
.
k
k=1
Finally, by setting x = 1, we obtain
n X
n 2
k = n2n−1 + n(n − 1)2n−2 = n2n−2 (2 + n − 1) = n(n + 1)2n−2 .
k
k=1
(3) We have using the Fundamental Theorem of Calculus that
1 Z 1
2n+1 − 1
1
=
(1 + x)n+1 =
(1 + x)n dx
n+1
n+1
0
0
n Z 1
n X
X
n
n
1
.
=
xk dx =
k
k k+1
0
k=0
k=0
n X
2n+1 − 2 − n
n
1
=
.
Thus
n+1
k k+1
k=1
n n X
X
n 2k
n 2k+1
(4) As (1 + x2 )n =
x , x(1 + x2 )n =
x
, and differentiating,
k
k
k=0
k=0
n X
n
2 n
2 n−1
1 · (1 + x ) + x · n(1 + x )
· 2x =
(2k + 1)x2k
k
k=0
n
X
n
Setting x = 1, we get
(2k + 1)
= 2n + n2n−1 · 2 = (n + 1)2n , and so
k
k=0
n
X
n
(2k + 1)
= (n + 1)2n − 1.
k
k=1
Solution to Exercise 4.2.5. For x 6= 1, we have
n
X
x − xn+1
(1 − x)(x + x2 + · · · + xn )
=
.
S(x) :=
xk =
1−x
1−x
k=1
Differentiating with respect to x, we obtain
n
X
(1 − (n + 1)xn )(1 − x) − (x − xn+1 )(−1)
kxk−1 =
(1 − x)2
k=1
=
=
1 − (n + 1)xn − x + (n + 1)xn+1 + x − xn+1
(1 − x)2
n
1 − (n + 1)x + nxn+1
.
(1 − x)2
Multiplying by x, and differentiating again we obtain
n
X
2
1
k 2 xk−1 =
(x −(n+1)xn+1 + nxn+2 ) +
(1 −(n+1)2 xn + n(n +2)xn+1 )
(1 −x)3
(1 −x)2
k=1
=
1 + x − (n + 1)2 xn + (2n2 + 2n − 1)xn+1 − n2 xn+2
.
(1 − x)3
Multiplying by x again and then substituting x = 1/2 yields
n
X
k2
k=1
2k
=6−
n2 + 4n + 6
.
2n
98
Solutions
Solution to Exercise 4.3.1. We must have f (1) = g(1) = 2, since the graphs meet at (1, 2). So
1 + a + b = 2 and 1 − c = 2,
that is, c = −1 and a + b = 1.
The tangent line being the same at (1, 2) implies that f ′ (1) = g ′ (1), and so
(2x + a)
= (3x3 )
,
x=1
x=1
that is, 2 + a = 3, and so a = 1. Using a + b = 1 (derived earlier), we obtain b = 0. So we have
a = 1, b = 0, c = −1. Then
f (x) =
x2 + x and
g(x) =
x3 + 1.
So f (1) = 2 = g(1). Also, f ′ (x) = 2x + 1 and g ′ (x) = 3x2 , so that f ′ (1) = 3 and g ′ (1) = 3. Hence
the tangent lines will be the same. See Figure 14.
Figure 14. The common tangent line at the point (1, 2).
Solution to Exercise 4.3.2. The point (1/4, 4) corresponds
to t =√2. (Indeed, 1/t2 = 1/4 gives
√
t = ±2, and as t > 0, t = 2. We check that then y(2) = 22 + 12 = 16 = 4.) We have
x′ (t) = −
2
,
t3
t
,
y ′ (t) = √
2
t + 12
2
1
= − , and
3
2
4
2
1
′
y (2) = √
= .
2
2
2 + 12
x′ (2) = −
So the equation of the tangent at the point (1/4, 4) to the curve is given by
1/2
y−4
y ′ (2)
1 = x′ (2) = −1/4 = −2,
x− 4
1
. See Figure 15.
that is, y − 4 = −2 x −
4
Solutions to the exercises from Chapter 4
99
Figure 15. The tangent line at (1/4, 4) to the curve.
Solution to Exercise 4.3.3.
(1) We want (x0 , y0 ) satisfying x20 + x0 y0 + y02 = 7 such that
dy = 0.
dx x0
We have 2x + 1 · y + x
dy
dy
+ 2y
= 0, and so
dx
dx
dy
−2x − y
=
.
dx
x + 2y
So 2x0 + y0 = 0, that is, y0 = −2x0 and substituting this in x20 + x0 y0 + y02 = 7 gives
x20 + x0 (−2x0 ) + (−2x0 )2 = 7, that is,
(r
r )
7
7
.
,−
x0 ∈
3
3
r
r
r !
r !
7
7
7
7
So the points where the tangent is horizontal are
and −
.
, −2
,2
3
3
3
3
r r !
r
r !
7
7
7
7
(2) A similar calculation or a symmetry argument gives −2
and 2
.
,
,−
3
3
3
3
See Figure 16.
Solution to Exercise 4.3.4. When x = 1, y + y 2 = 6 and so y 2 + y − 6 = 0, that is,
(y − 2)(y + 3) = 0,
and so y ∈ {2, −3}. So (1, 2), (1, −3) are points on x2 + xy 2 = 6 where x = 1. Using implicit
differentiation, we obtain
dy
dy
2xy + x2
+ 1 · y 2 + x · 2y ·
= 0,
(6.26)
dx
dx
dy
= 0, and so
that is, 2xy + y 2 + (x2 + 2xy)
dx
dy
2xy + y 2
=− 2
.
dx
x + 2xy
100
Solutions
Figure 16. The horizontal and vertical tangent lines.
2·1·2+4
8
dy
at (1, 2) is −
= − , and the equation of the tangent is
dx
1+2·1·2
5
y−2
8
=− ,
x−1
5
8
8
that is, y − 2 = − x + .
5
5
2 · 1 · (−3) + (−3)2
3
dy
at (1, −3) is −
= , and the equation of the tangent is
dx
1 + 2 · 1 · (−3)
5
3
y+3
= ,
x−1
5
3
3
that is, y + 3 = x − .
5
5
See Figure 17.
So
Figure 17. The tangent lines at (1, 2) and (1, −3).
Using implicit differentiation, we have from(6.26) that
2 · 1 · y + 2x
dy
dy
d2 y
dy
dy dy
d2 y
dy
+ 2x
+ x2 2 + 2y
+ 1 · 2y ·
+x·2
·
+ x · 2y · 2 = 0,
dx
dx
dx
dx
dx
dx dx
dx
Solutions to the exercises from Chapter 4
and so
So
1
d2 y
=− 2
dx2
x + 2xy
d2 y
at (1, 2) is
dx2
and
1
−
1+4
d2 y
at (1, −3) is
dx2
1
−
1−6
101
dy dy 2y +
.
· 4x + 4y + 2x
dx
dx
−8 8 252
,
4− · 4+8+2·
=
5
5
125
3 252
3
−6 + · 4 − 12 + 2 ·
=−
.
5
5
125
Solution to Exercise 4.3.5. With f (x) = x2 − 2, we have f ′ (x) = 2x, and so the update
equation in the Newton-Raphson Method is
xn+1 = xn −
xn
xn
x2 − 2
1
1
f (xn )
= xn −
=
.
= xn − n
+
+
′
f (xn )
2xn
2
xn
2
xn
Taking x0 = 1, we get
x1
=
x2
=
x3
=
1
3
+ 1 = = 1.5,
2
2
17
3 2
+ =
= 1.41666 · · · ,
4 3
12
17 12
577
+
=
≈ 1.4142157.
24 17
408
Solution to Exercise 4.3.6. We have with p := x4 − x3 − 75 that
p(3) =
p(4) =
34 − 33 − 75 = 81 − 27 − 75 = 6 − 27 < 0,
44 − 43 − 75 = 256 − 64 − 75 = 256 − 139 > 0.
By the Intermediate Value Property, there exists an x∗ ∈ [3, 4] such that p(x∗ ) = 0.
The update equation in the Newton-Raphson Method is
xn+1
=
=
f (xn )
f ′ (xn )
4x4 − 3x3n − x4n + x3n + 75
3x4 − 2x3 + 75
x4 − x3 − 75
= n 3 n 2 .
xn − n 3 n 2 = n
3
2
4xn − 3xn
4xn − 3xn
4xn − 3xn
xn −
The following table shows the values of xn obtained for successive values of n:
n
Approximate value of xn
0
1
2
3
4
3.5
3.2611
3.2291
3.2286
3.2286
(Using MATLAB, one can check that x∗ = 3.228577 · · · .)
Solution to Exercise 4.4.1. If a, b ∈ R and a < b, then applying the Mean Value Theorem to
the function cos ·, we obtain
cos a − cos b
= − sin c
a−b
for some c ∈ (a, b). But as | sin θ| ≤ 1 for all real θ, it follows that | cos a − cos b| ≤ |a − b|.
102
Solutions
Solution to Exercise 4.4.2. We will show that f (0) = 0. Let us suppose f (0) 6= 0. Consider
first the case that f (0) > 0. Then by the Mean Value Theorem applied to [−a, 0], we have
f (0) − f (−a)
= f ′ (c)
0 − (−a)
for some c between −a and 0. Hence
1=0+1<
f (0) − (−a)
f (0) − f (−a)
f (0)
+1=
=
= f ′ (c) < 1,
a
a
0 − (−a)
a contradiction.
Next suppose that f (0) < 0. Then by the Mean Value Theorem applied to [0, a], we have
for some d between 0 and a. Hence
1=1−0<1−
a contradiction.
f (a) − f (0)
= f ′ (d)
a−0
f (0)
a − f (0)
f (a) − f (0)
=
=
= f ′ (d) < 1,
a
a
a−0
An example of such a function f is f (x) := x, x ∈ R. Then |f ′ (x)| = 1 ≤ 1 for all x, and for
any a > 0, f (−a) = −a, f (a) = a. And we do have f (0) = 0.
Solution to Exercise 4.4.3. Given ǫ > 0, let R′ > 0 be such that for all x > R′ , |f ′ (x) − L′ | < ǫ.
Take any R > R′ such that for all x ∈ R with x > R, |f (x) − L| < ǫ/2. Then by the Mean Value
Theorem applied to f in [R + 1, R + 2],
f (R + 2) − f (R + 1)
= f ′ (ξ)
1
for some ξ ∈ (R + 1, R + 2). So
|f ′ (ξ)| = |f (R + 2) − f (R + 1)| = |f (R + 2) − L + L − f (R + 1)| ≤
ǫ
ǫ
+ = ǫ.
2 2
Hence
|L′ | = |L′ − f ′ (ξ) + f ′ (ξ)| ≤ |L′ − f ′ (ξ)| + |f ′ (ξ)| ≤ ǫ + ǫ = 2ǫ.
As ǫ > 0 was arbitrary, it follows that L′ = 0.
Solution to Exercise 4.4.4. Let x ∈ (a, b) \ {c}. By the Mean Value Theorem applied to the
compact interval with endpoints x and c, we have
f (x) − f (c)
= f ′ (cx ),
x−c
for some point cx lying between x and c. Note that if the distance of x to c is less than a certain
amount δ, then the distance of cx above to c will also be less than the amount δ.
Let
L := lim f ′ (x).
x→c
Let ǫ > 0. Then there exists a δ > 0 such that for all x ∈ (a, b) satisfying 0 < |x − c| < δ, we have
that |f ′ (x) − L| < ǫ. Hence for x ∈ (a, b) satisfying 0 < |x − c| < δ, we have
f (x) − f (c)
− L = |f ′ (cx ) − L| < ǫ.
x−c
So f ′ (c) exists and equals L.
Solutions to the exercises from Chapter 4
103
Solution to Exercise 4.4.5. We have |f (x) − f (y)| ≤ (x − y)2 = |x − y|2 , and so for x 6= y we
obtain
f (x) − f (y)
− 0 ≤ |x − y|.
x−y
Let ǫ > 0. Then with δ := ǫ > 0, we have that whenever y satisfies 0 < |x − y| < δ = ǫ, we have
f (x) − f (y)
− 0 ≤ |x − y| < δ = ǫ.
x−y
Thus f ′ (x) = 0 for all x ∈ R.
Now suppose that a, b ∈ R and that a < b. Then by the Mean Value Theorem applied to f
on the interval [a, b], we obtain
f (a) − f (b)
= f ′ (c)
a−b
for some c ∈ (a, b). But f ′ (c) = 0, and so f (a) = f (b). Thus f is constant.
Solution to Exercise 4.4.6. If x, y ∈ (a, b) and x < y, then we have by the Mean Value Theorem
applied to the interval [x, y] that
f (x) − f (y)
= f ′ (c)
x−y
for some c ∈ (a, b). Since |f ′ (c)| ≤ M , we obtain that
|f (x) − f (y)| ≤ M |x − y|.
Clearly this is also true if x = y, and (by interchanging the roles of x and y) if x > y. Hence for
all x, y ∈ (a, b), there holds |f (x) − f (y)| ≤ M |x − y|.
Let ǫ > 0. Set δ = ǫ/M . Then if x, y ∈ (a, b) and |x − y| < δ, we have
|f (x) − f (y)| ≤ M |x − y| < M δ = ǫ.
Hence f is uniformly continuous on (a, b).
Solution to Exercise 4.4.7. Since x 7→ f (x + n) is differentiable too, it follows that
f (x + n) − f (x)
= f ′ (x)
n
is differentiable. Also, for all m ∈ N, we have
x 7→
f ′ (x + m) =
=
=
f (x + m + n) − f (x + m)
f (x + m + n) − f (x) + f (x) − f (x + m)
=
n
n
n + m f (x + m + n) − f (x) m f (x + m) − f (x)
·
−
n
n+m
n
m
m ′
n+m ′
f (x) − f (x) = f ′ (x).
n
n
Thus
f ′ (x) − f ′ (x)
f ′ (x + n) − f ′ (x)
=
= 0.
n
n
By the Mean Value Theorem applied to f ′ , this gives that f ′ is a constant, that is, there is a c ∈ R
such that for all x ∈ R, f ′ (x) = c. Applying the Mean Value Theorem to f , we obtain for every
nonzero real x that
f (x) − f (0)
= f ′ (z)
x−0
for some z between 0 and x. But since f ′ is constant, it follows that f (x) = f (0) + cx for all x ∈ R.
f ′′ (x) =
104
Solutions
Solution to Exercise 4.4.8. Consider f given by
f (x) =
c1
cd d+1
c0
x + x2 + · · · +
x ,
1
2
d+1
x ∈ R.
Then f (0) = 0 and
f (1) =
c1
cd
c0
+
+ ··· +
= 0.
1
2
d+1
By Rolle’s Theorem, f ′ (c) = 0 for some c ∈ (0, 1). But
f ′ (x) =
c0
c1
cd
·1+
· 2x + · · · +
· (d + 1)xd = c0 + c1 x + · · · + cd xd ,
1
2
d+1
and so the claim follows.
Solution to Exercise 4.4.9. Suppose on the contrary that f ′′ (c) < 0. With ǫ := −f ′′ (c)/2 > 0,
there exists a δ > 0 such that whenever 0 < |x − c| < δ, we have
′
′
′
f (x) − 0
f (x)
f (x) − f ′ (c)
f ′′ (c)
′′
′′
′′
− f (c) = − f (c) = − f (c) < ǫ = −
.
x−c
x−c
x−c
2
In particular,
f ′′ (c)
f ′ (x)
− f ′′ (c) < −
, that is,
x−c
2
f ′ (x)
f ′′ (c)
>−
> 0.
c−x
2
So f ′ (x) > 0 in (c − δ, c) showing that f is strictly increasing there, and by the continuity of f , f
is strictly increasing in (c − δ, c]. So f (x) < f (c) for all x ∈ (c − δ, c), a contradiction to the local
minimality of f at c.
Suppose that F (a) = F (b) = 0 and that a < b. If F is not identically equal to 0 in [a, b], then
there exists a maximizer or a minimizer F |[a,b] on (a, b).
1◦ Let us first suppose that c ∈ (a, b) is a maximizer for F |[a,b] . Then F (c) > 0, and
F ′ (c) = 0. We have
0 = F ′′ (c) + F ′ (c)g(c) − F (c) = F ′′ (c) − F (c),
and so F ′′ (c) = F (c) > 0, a contradiction to the result established in the first part of
this exercise.
2◦ If c ∈ (a, b) is a minimizer for F |[a,b] , then F (c) < 0, and F ′ (c) = 0. Again
0 = F ′′ (c) + F ′ (c)g(c) − F (c) = F ′′ (c) − F (c),
and so F ′′ (c) = F (c) < 0, a contradiction to the result established in the first part of
this exercise.
(0)
(0)
(0)
Solution to Exercise 4.4.10. Let x1 < · · · < xn+1 be such that f (xk ) = 0 for k = 1, · · · , n+1.
(0)
(0)
(1)
By Rolle’s Theorem applied to each [xk , xk+1 ], k = 1, · · · , n, we get the existence of xk ∈
(0)
(0)
(1)
(1)
(xk , xk+1 ), k = 1, · · · , n such that f ′ (xk ) = 0, k = 1, · · · , n. Clearly x1
(1)
(1)
(1)
< · · · < xn . By
Rolle’s Theorem applied to f ′ on each [xk , xk+1 ], k = 1, · · · , n − 1, we get the existence of
(2)
(1)
(2)
(1)
xk ∈ (xk , xk+1 ), k = 1, · · · , n − 1 such that f ′′ (xk ) = 0, k = 1, · · · , n − 1. Proceeding in this
manner, we eventually get the existence of a
(n)
xn
(n)
such that f (n) (xn ) = 0. See Figure 18.
105
Solutions to the exercises from Chapter 4
f
(0)
x1
(0)
x2
(1)
x1
(0)
(0)
x3
(0)
xn+1
xn
(1)
x2
f′
(1)
xn
(1)
x3
f ′′
(2)
x1
(2)
xn−1
(2)
x2
..
.
f (n)
(n)
xn
Figure 18. Repeated application of Rolle’s Theorem.
Solution to Exercise 4.4.11. Consider f : R → R given by f (x) = x2 − x sin x − cos x, x ∈ R.
Then
π
π
π2
π
π2
f −
+ ·1−0 =
+ > 0,
=
2
4
2
4
2
f (0) = 0 − 0 − 1 = −1 < 0,
π π2
π
π(π − 2)
=
f
− ·1−0 =
> 0.
2
4
2
4
So by the Intermediate Value Theorem, there exists a c1 ∈ (−π/2, 0) and there exists a c2 ∈ (0, π/2)
such that f (c1 ) = f (c2 ) = 0.
Suppose that f has more than two distinct roots. Then by Rolle’s Theorem, f ′ would be zero
for at least two distinct real numbers. But
f ′ (x) = 2x − 1 · sin x − x · cos x − (− sin x) = x (2 − cos x),
| {z }
6=0
as | cos x|≤1
and so f ′ (x) = 0 if and only if x = 0. Consequently, f has exactly two zeros.
Solution to Exercise 4.4.12.
(1) Let g := y 2 + (y ′ )2 . Then g ′ = 2y · y ′ + 2y ′ · y ′′ = 2y ′ (y + y ′′ ) = 0.
| {z }
=0
(2) We have
(A cos x + B sin x)′
=
′′
=
(A cos x + B sin x)
−A sin x + B cos x, and
−A cos x − B sin x.
Thus (A cos x + B sin x)′′ + (A cos x + B sin x) = 0.
If y = A cos x + B sin x, then
y(0) = A · 1 + B · 0 = A, and
y ′ (0) = −A sin x + B cos x
= −A · 0 + B · 1 = B.
x=0
So if y = A cos x + B sin x, then A = y(0) and B = y ′ (0).
106
Solutions
Now let y be any solution to y ′′ + y = 0. Set f := y − y(0) cos x − y ′ (0) sin x. Then
f (0) =
′
f (0) =
y(0) − y(0) · 1 − 0 = 0 and
y ′ (0) + y(0) · 0 − y ′ (0) · 1 = 0.
Also, f ′′ + f = y ′′ + y + 0 = 0 + 0 = 0. So f is also a solution. Hence by the
previous part, f 2 + (f ′ )2 = (f (0))2 + (f ′ (0))2 = 0. Consequently, f = f ′ ≡ 0. So
y = y(0) cos x + y ′ (0) sin x. Done!
(3) Let y = sin(a + x). Then y ′ = (cos(a + x)) · 1 and y ′′ = (− sin(a + x)) · 1. So y ′′ + y =
− sin(a + x) + sin(a + x) = 0. Hence by the previous part,
sin(a + x) = y(0) cos x + y ′ (0) sin x = (sin a)(cos x) + (cos a)(sin x).
Setting x = b, we get sin(a + b) = (sin a)(cos b) + (cos a)(sin b).
Next consider y := cos(a + x). Then y ′ = − sin(a + x) and y ′′ = − cos(a + x). So
y ′′ + y = − cos(a + x) + cos(a + x) = 0. Hence by the previous part,
cos(a + x) = y(0) cos x + y ′ (0) sin x = (cos a)(cos x) + (− sin a)(sin x).
Setting x = b, we get cos(a + b) = (cos a)(cos b) − (sin a)(sin b).
Solution to Exercise 4.4.13. The distance between (0, b) and a point (x, x2 ) on the parabola
is given by
p
d(x) = (x − 0)2 + (x2 − b)2 , x ∈ R.
We have
lim d(x) = +∞,
x→±∞
and so there exists a R > 0 such that whenever |x| > R, d(x) > d(0) = b. On [−R, R], it
follows from the Extreme Value Theorem that the continuous function d has a minimizer, say at
x∗ ∈ [−R, R], and in particular d(x∗ ) ≤ d(0) = b. But for x ∈ R \ [−R, R], d(x) > b ≥ d(x∗ ). So
x∗ is a global minimizer. Hence d′ (x∗ ) = 0. But
1
d′ (x∗ ) = p
(2x∗ + 2(x2∗ − b)2x∗ ),
2
2
2
2 (x∗ − 0) + (x∗ − b)
We have that if b ∈((0, 1/2], then d′ (x∗ ) =)0 if and only if x∗ = 0, while if b > 1/2, then d′ (x∗ ) = 0
r
r
1
1
. We have if b > 1/2 that
if and only if x∗ ∈ 0, b − , − b −
2
2
d ±
r
1
b+
2
!
=
r
b−
1
< b,
4
2
2
where
p the last inequality follows from that fact that (b − 1/2) > 0, that is, b > b − 1/4 and so
b > b − 1/4. Thus the minimum value of the distance is
 r
1


b−
4

 b
1
,
2
1
if 0 < b ≤ .
2
if b ≥
107
Solutions to the exercises from Chapter 4
Solution to Exercise 4.4.14. We have
f (x + π) = (sin(x + π) − cos(x + π))2 = (− sin x − (− cos x))2 = (sin x − cos x)2 = f (x),
x ∈ R,
and so f is π-periodic. It is also continuous, and so by the Extreme Value Theorem, it possesses
a maximum value on [0, π], which serves as the maximum value on R. We have that
f ′ (x) = 2(sin x − cos x) · (cos x − (− sin x)) = 2((sin x)2 − (cos x)2 ) = −2 cos(2x),
x ∈ R.
So f ′ (x) = 0 if and only if
o
n
π
x ∈ (2n + 1) : n ∈ Z .
4
Hence these x are candidates for the maximizers. We look for maximizers in [0, π]. Then the
candidates for maximizers are at π/4 and 3π/4. But f (π/4) = 0, while
2
1 1
= 2.
f (3π/4) = √ − − √
2
2
Thus the maximum value of f is 2.
Solution to Exercise 4.4.15. Taylor’s Formula gives for x 6= 0 that there exists a cx between 0
and x, such that
sin x =
=
cos 0
sin 0 2 cos 0 3 sin 0 4 cos cx 5
x−
x −
x +
x +
x
1!
2!
3!
4!
5!
cos cx 5
x3
+
x .
x−
3!
5!
sin 0 +
So for x 6= 0,
sin x − x 1 cos cx 2 1
=
+
x ≤ |x|2 .
x3
6
5!
5!
Since lim |x|2 = 0, it follows that lim
x→0
x→0
sin x − x
1
=− .
x3
6
Solution to Exercise 4.4.15.
(1) From Taylor’s Formula, for each x such that |x − a| ≤ ǫ, there exists a cx ∈ (a − ǫ, a + ǫ)
such that
f (x) =
n
X
f (k) (a)
k=0
k!
(x − a)k +
f (n+1) (cx )
(x − a)n+1 .
(n + 1)!
f (n+1) (cx )
(x − a)n+1 . Then for 0 < |x − a| < ǫ,
(n + 1)!
En (x) |f (n+1) (cx )| (x − a)n+1 M
≤
0≤
=
|x − a|,
(x − a)n (n + 1)! (x − a)n (n + 1)!
For |x − a| ≤ ǫ, set En (x) :=
and so by the Sandwich Theorem, lim
x→a
En (x)
= 0. Hence the claim follows.
(x − a)n
(2) tan is infinitely differentiable on (−π/2, π/2), and in particular, tan(4) is continuous. We
have for x ∈ (−π/2, π/2) that
tan′ x
= 1 + (tan x)2 ,
tan′′ x
= 2(tan x)(1 + (tan x)2 ),
tan′′′ x
= 2(1 + (tan x)2 ) + 6(tan x)2 (1 + (tan x)2 ).
108
Solutions
Thus
tan x
tan′′ 0 2 tan′′′ 0 3
tan′ 0
x+
x +
x + o(x3 )
1!
2!
3!
x3
x3
+ o(x3 ) = x +
+ o(x3 ).
= 0+x+0+
3
3
= tan 0 +
(3) We have
x3
+ o(x3 ),
3
x3
= x−
+ o(x3 ),
6
x2
= 1−
+ o(x2 ),
2
tan x − x
=
sin x
cos x
as x → 0. Thus
tan x − x
sin x
cos x
x3
+ h1 (x),
3
x3
= x−
+ h2 (x),
6
x2
= 1−
+ h3 (x),
2
=
where
lim
x→0
h2 (x)
h3 (x)
h1 (x)
= lim
= lim
= 0.
3
3
x→0
x→0
x
x
x2
So
tan x − x
lim
x→0 sin x − x cos x
=
=
=
lim
x→0
lim
x→0
x3
3
x−
x3
6
− 61 +
+ h1 (x)
+ h2 (x) − x(1 −
x2
2
+ h3 (x))
h1 (x)
1
3 + x3
h2 (x)
h3 (x)
1
x3 + 2 + x2
1
3
− 61
+0
= 1.
+ 0 + 21 + 0
Solution to Exercise 4.4.17.
(1) Suppose that there exist a, b ∈ R such that a < b and f (a) = a, f (b) = b. Applying the
Mean Value Theorem to f on [a, b], we obtain
b−a
f (b) − f (a)
=
= 1 = f ′ (c)
b−a
b−a
for some c ∈ (a, b) ⊂ (0, 1). But as f ′ (c) 6= 1, we have arrived at a contradiction. Hence
if f has a fixed point, then it is unique.
(2) Let x1 ∈ R, and set xn+1 := f (xn ) for n ∈ N. We claim that
|f (xn+1 ) − f (xn )| ≤ M |xn+1 − xn | (n ∈ N).
(6.27)
This is clearly true if xn+1 = xn . If they are not equal, then we apply the Mean Value
Theorem to the interval with endpoints xn+1 and xn to obtain
f (xn+1 ) − f (xn )
= f ′ (c),
xn+1 − xn
for some c between xn+1 and xn . As |f ′ (c)| ≤ M , this yields the inequality (6.27).
109
Solutions to the exercises from Chapter 4
Thus we have by a repeated application of (6.27) that
|xn+1 − xn | = |f (xn ) − f (xn−1 )| ≤
≤
≤
Now consider the series
x1 +
∞
X
M |xn − xn−1 |
M · M |xn−1 − xn−2 |
..
.
M n−1 |x2 − x1 |.
(xn+1 − xn ).
n=1
The estimate |xn+1 − xn | ≤ M n−1 |x2 − x1 |, with M < 1, shows that this series converges
absolutely, using the Comparison Test. As the partial sums of the above series telescope,
that is,
x1 + (x2 − x1 ) + (x3 − x2 ) + · · · + (xn+1 − xn ) = xn+1 ,
it follows that
x∗ := lim xn
n→∞
exists. Since f is continuous, we also have that
f (x∗ ) = f lim xn = lim f (xn ) = lim xn+1 = x∗ .
n→∞
n→∞
n→∞
So x∗ is a fixed point.
(3) See Figure 19.
f
y=x
(x2 ,x3 )
x∗
(x3 ,x4 )
(x1 ,x2 )
x∗
x1
Figure 19. The zig-zag path (x1 , x2 ) −→ (x2 , x2 ) −→ (x2 , x3 ) −→ . . . .
(4) We have
ex
1 + e2x + ex
=
,
(1 + ex )2
1 + e2x + 2ex
and so 0 < f ′ (x) < 1 for all x ∈ R.
If f had a fixed point x∗ , then f (x∗ ) = x∗ would imply that
f ′ (x) = 1 −
f (x∗ ) = x∗ +
1
= x∗ ,
1 + ex∗
110
Solutions
and so we would have
1
= 0,
1 + ex∗
which is clearly impossible, since ex∗ > 0.
We observe that although |f ′ (x)| < 1 for all x ∈ R, it is not uniformly bounded away
from 1, that is, there does not exist an M < 1 such that for all x ∈ R, |f (x)| < M .
Suppose, on the contrary, that such an M exists. Then we would have for all x ∈ R that
1−
ex
< M.
(1 + ex )2
In particular, setting x = −n (n ∈ N), we obtain for all n ∈ N that
1−
e−n
< M.
(1 + e−n )2
Passing the limit as n → ∞ yields
1−
0
= 1 ≤ M,
(1 + 0)2
a contradiction to the fact that M < 1. Thus there is no contradiction to the result from
part (2).
Solution to Exercise 4.5.1. We have y(x) = 2x3 + 2x2 − 2x − 1, and so
1
′
2
(x + 1).
y (x) = 6x + 4x − 2 = 6 x −
3
Thus y ′ (x) > 0 if and only if [x > 1/3 or x < −1], and here y is increasing.
Also, y ′ (x) < 0 if and only if −1 < x < 1/3, and here y is decreasing.
As
1
y ′′ (x) = 12x + 4 = 12 x +
,
3
we see that y ′′ (x) > 0 if and only if x > −1/3, where y is convex.
Similarly y ′′ (x) < 0 if and only if x < −1/3, where y is concave.
y ′′ (x) = 0 if and only if x = −1/3, which is a point of inflection.
y ′ (x) = 0 if and only if [x = 1/3 or x = −1].
Around x = −1, y is concave, and so x = −1 is a local maximizer.
Around x = 1/3, y is convex, and so x = 1/3 is a local minimizer.
y(0) = −1, y(1) = 2 + 2 − 2 − 1 = 1, y(−1) = −2 + 2 + 2 − 1 = 1,
2
2 2
−2 + 4 + 18 − 27
7
1
=− + + −1=
=− ,
y −
3
27 9 3
27
27
and y(−2) = −16 + 8 + 4 − 1 < 0. By the Intermediate Value Theorem, the graph of y crosses
the x-axis between −2 and −1, between −1 and 0, and between 0 and 1. So there are at least 3
distinct zeros.
There can’t be any more than 3 distinct zeros, because otherwise by Rolle’s Theorem y ′ would
be 0 at least 3 times, but we have seen above that y ′ has only 2 zeros.
See Figure 20.
111
Solutions to the exercises from Chapter 4
Figure 20. The graph of 2x3 + 2x2 − 2x − 1.
Solution to Exercise 4.5.2. We have
n
n
X
X
f ′ (x) =
2(x − ak ) = 2nx − 2
ak ,
k=1
and so f ′ (x) = 0 if and only if x =
1
n
k=1
n
X
ak . Also,
k=1
f ′′ (x) =
n
X
x ∈ R.
2 = 2n > 0,
k=1
n
So f is convex. Hence x∗ :=
1X
ak is a minimizer of f and it is the only one.
n
k=1
The minimum value of f is
n
n
n
X
X
X
f (x∗ ) =
(x∗ − ak )2 = nx2∗ − 2x∗
ak +
a2k
k=1
k=1
k=1
n
n
n
n
1 X X X 2
1 X 2
ak
ak +
ak
ak − 2 ·
= n· 2
n
n
k=1
k=1
=
n
X
k=1
a2k −
n
1X
n
k=1
ak
2
k=1
k=1
.
Remark 6.3. Since f (x) ≥ 0 for all x ∈ R, in particular, f (x∗ ) ≥ 0, that is,
v
u n
n
u1 X
1X
Quadratic mean := t
a2k ≥ Arithmetic mean :=
ak .
n
n
k=1
k=1
As x → ±∞, g(x) → +∞, and g is linear in (−∞, a1 ), (an , ∞). So
g(x) ≥ g(a1 )
g(x) ≥ g(an )
for x ∈ (−∞, a1 ], and
for x ∈ [an , ∞).
112
Solutions
Thus the minimizer of g must lie in [a1 , an ]. For x ∈ [ak , ak+1 ), and by continuity of g, for
x ∈ [ak , ak+1 ], we have
g(x) − g(ak ) =
=
=
(x − a1 ) + · · · + (x − ak ) + (ak+1 − x) + · · · + (an − x)
− (ak − a1 ) + · · · + (ak − ak ) + (ak+1 − ak ) + · · · + (an − ak )
kx − (n − (k + 1))x − kak + (n − (k + 1))ak
(2k − n + 1) (x − ak ),
| {z }
≥0
and similarly
g(ak+1 ) − g(x) = (2k − n + 1) (ak+1 − x) .
| {z }
≥0
For 2k − n + 1 < 0, that is, for k = 1, · · · , ⌊(n − 1)/2⌋, it follows that g is decreasing on [ak , ak+1 ],
and so
g(x) ≥ g a⌊ n−1 ⌋+1 for x ∈ (−∞, a⌊ n−1 ⌋+1 ].
2
2
On the other hand, for 2k − n + 1 > 0, in particular, for k = ⌊(n − 1)/2⌋ + 1, · · · , n, it follows that
g is increasing on [ak , 2k+1 ], and so
g(x) ≥ g a⌊ n−1 ⌋+1 for x ∈ [a⌊ n−1 ⌋+1 , ∞).
2
2
The minimum value is thus
(
an + · · · + a n2 +1 − (a1 + · · · + a n2 )
g a⌊ n−1 ⌋+1 =
2
an + · · · + a n+1 − (a1 + · · · + a n−1 )
2
2
if n is even,
if n is odd.
For x ∈ (−∞, 0), we have x < 0 < a, that is, x − a < −a < 0, and so
h(x) =
h′ (x)
=
1
1
1
1
+
=
+
, and
1 − x 1 − (x − a)
1−x 1+a−x
1
1
+
> 0.
2
(1 − x)
(1 + a − x)2
So h is strictly increasing on (−∞, 0). Using this and the continuity of h, we see that
h(x) ≤ h(0) = 1 +
For x ∈ (0, a),
h(x) =
h′ (x)
=
1
2+a
=
for x ∈ (−∞, 0].
1+a
1+a
1
1
+
, and
1+x 1+a−x
1
1
+
.
−
(1 + x)2
(1 + a − x)2
So if x ∈ (0, a) and h(x) = 0, then 1 + a − x = 1 + x or 1 + a − x = −(1 + x). But the latter is
impossible since otherwise a = −a < 0. Hence we have that h′ (x) = 0 implies that 1+a−x = 1+x,
that is, x = a/2. Also,
h′′ (x) =
2
2
+
> 0 for x ∈ (0, a),
3
(1 + x)
(1 + a − x)3
and so x = a/2 is a minimizer of h|(0,a) . So h|[0,a] has no maximizer in (0, a). Hence
h(x) ≤ max{h(0), h(a)} for x ∈ [0, a].
We have
h(a) =
1
1
2+a
+
=
= h(0).
1+a 1+0
1+a
113
Solutions to the exercises from Chapter 4
So far we have got h(x) ≤ h(0) = h(a) for all x ∈ (−∞, a]. For x ∈ (a, ∞),
h(x)
=
h′ (x)
=
1
1
+
and
1+x 1+x−a
1
1
−
< 0,
−
2
(1 + x)
(1 + x − a)2
and so h is strictly decreasing in (a, ∞). Consequently, h(x) ≤ h(a) = h(0) for all x ∈ R. So the
maximum value of h is (2 + a)/(1 + a).
Solution to Exercise 4.5.3. Since f ′ is zero only at x = 1, 2, 3, 4, these are the only possibilities
for local maximizers/minimizers. We will argue that f has a local maximum at 1, a local minimum
at 3, while the points 2, 4 are neither local maximizers nor local minimizers.
f′
+
0
2
1
−
+
−
3
+
4
Figure 21. Solution to Exercise 4.5.3.
In the interval (1 − δ, 1), with δ small enough, f ′ > 0 and so f is increasing there. In the
interval (1, 1 + δ), with a small enough δ, f ′ < 0, and so f is decreasing there. So f has a local
maximum at 1.
In the intervals (2 − δ, 2) and (a, 2 + δ), with a small enough δ, f ′ < 0, and so f is strictly
decreasing in (2 − δ, 2 + δ). Thus 2 is neither a local maximizer nor a local minimizer.
In the interval (3 − δ, 3), with a small enough δ, f ′ < 0, and so f strictly decreasing there. In
the interval (3, 2 + δ), with a small enough δ, f ′ > 0, and so f is strictly increasing there. So 3 is
a local minimizer of f .
In the intervals (4 − δ, 4) and (4, 4 + δ), with a small enough δ, f ′ > 0, and so f is strictly
increasing in (4 − δ, 4 + δ). Thus 4 is neither a local maximizer nor a local minimizer.
Solution to Exercise 4.5.4. The time T > 0 when the cannon ball hits the ground is given by
−16T 2 + (v sin α)T = 0, that is,
16
.
T =
v sin α
The span is
speed
v cos α
v2
v2
s=
=
=
(cos α)(sin α) =
sin(2α).
time
16/(v sin α)
16
32
114
Solutions
We want to maximize
s(α) :=
We have s′ (α) =
v2
sin(2α),
32
v2
· 2 cos(2α), and so
32
s′ (α) = 0 ⇔ cos(2α) = 0
0 ≤ α ≤ 90◦ .
α = 45◦ (since α ≤ 90◦ ).
⇔
Also,
v2
· 2 · 2 · (− sin(2α)) ≤ 0,
32
Hence s is concave, and s is maximized at α = 45◦ .
s′′ (α) =
0 ≤ α ≤ 90◦ .
Solution to Exercise 4.5.5. For x1 , x2 ∈ R and α ∈ (0, 1) we have by the triangle inequality
that
|(1 − α)x1 + αx2 | ≤ |(1 − α)x1 | + |αx2 | = |1 − α||x1 | + |α||x2 | = (1 − α)|x1 | + α|x2 |.
Thus | · | is convex.
Solution to Exercise 4.5.6. (“If” part) Let x1 , x2 ∈ I and α ∈ (0, 1). Then (x1 , f (x1 )) ∈ U (f )
and (x2 , f (x2 )) ∈ U (f ). Since U (f ) is convex, we have that
(1 − α) · (x1 , f (x1 )) + α · (x2 , f (x2 )) = ((1 − α) · x1 + α · x2 , (1 − α)f (x1 ) + αf (x2 )) ∈ U (f ).
|
{z
} |
{z
}
=:y
=:x∈I
Consequently, (1 − α)f (x1 ) + αf (x2 ) = y ≥ f (x) = f ((1 − α) · x1 + α · x2 ). Hence f is convex.
(“Only if” part) Let (x1 , y1 ), (x2 , y2 ) ∈ U (f ) and α ∈ (0, 1). Then we know that y1 ≥ f (x1 ) and
y2 ≥ f (x2 ) and so we also have that
(1 − α)y1 + αy2 ≥ (1 − α)f (x1 ) + αf (x2 ) ≥ f ((1 − α) · x1 + α · x2 ).
Consequently
((1 − α) · x1 + α · x2 , (1 − α)y1 + αy2 ) ∈ U (f ),
that is, (1 − α) · (x1 , f (x1 )) + α · (x2 , f (x2 )) ∈ U (f ). So U (f ) is convex.
Solution to Exercise 4.5.7.
(1) We have f ′ (t) = 2at + b, and f ′′ (t) = 2a > 0. So f is convex. We have f ′ (t) = 0 if and
only if 2at + b = 0, that is, t = −b/(2a). So it follows that −b/(2a) is a minimizer of f ,
and it is the only one. We have
2
b
b2
b
b2
b2
b2 − 4ac
b
=a −
+c=
+b −
−
+c=− +c=−
.
f −
2a
2a
2a
4a 2a
4a
4a
(2) Let f (t) :=
n
X
k=1
(tak − bk )2 , t ∈ R. Then f (t) ≥ 0. Also,
f (t) =
=
n
X
(t2 a2k − 2ak bk t + b2k )
k=1
n
X
k=1
where a :=
n
X
k=1
!
a2k
!
2
t −2
a2k , b := −2
n
X
k=1
n
X
ak b k
k=1
ak b k
!
!
t+
n
X
b2k = at2 + bt + c,
k=1
and c :=
n
X
k=1
b2k .
If a = 0, then each ak = 0 for k = 1, · · · , n, and so both sides of the inequality are
0, and the so the claim holds trivially.
115
Solutions to the exercises from Chapter 4
So we may assume that a > 0. Then f has the minimum value −(b2 − 4ac)/(4a).
As f (t) ≥ 0 for all t ∈ R, we must have that in particular −(b2 − 4ac)/(4a) ≥ 0, that is,
b2 ≤ 4ac. Hence
!2
! n
!
n
n
X
X
X
≤4
−2
ak b k
a2k
b2k ,
k=1
and so
n
X
k=1
ak b k
!2
≤
k=1
n
X
k=1
a2k
!
n
X
k=1
k=1
!
b2k .
Solution to Exercise 4.5.8. (1) We prove this using induction on n. The result is trivially true
when n = 1, and in fact we have equality in this case. Suppose that the inequality has been
established for some n ∈ N. Now if x1 , · · · , xn , xn+1 ∈ I, then we have with
t :=
1
∈ (0, 1)
n+1
that
f
1
(x1 + · · · + xn + xn+1 )
=
n+1
=
=
≤
≤
=
=
1
1
n
· (x1 + · · · + xn ) +
· xn+1
n+1 n
n+1
1
1
1
xn+1
f
1−
· (x1 + · · · + xn ) +
n+1
n
n+1
1
f (1 − t) · (x1 + · · · + xn ) + t · xn+1
n
1
(1 − t) · f
(x1 + · · · + xn ) + t · f (xn+1 )
n
f (x1 ) + · · · + f (xn )
+ t · f (xn+1 )
(1 − t) ·
n
n
f (x1 ) + · · · + f (xn )
1
✚
·
· f (xn+1 )
+
n+1
n
+
1
n
✚
f (x1 ) + · · · + f (xn ) + f (xn+1 )
,
n+1
f
and so the claim follows for all n.
(2) Let f : (0, ∞) → R be defined by f (x) = − log x, x ∈ (0, ∞). Then
f ′ (x) = −
1
1
and f ′′ (x) = 2 for x ∈ (0, ∞).
x
x
As f ′′ (x) = 1/x2 > 0 for all x ∈ (0, ∞), it follows that f is convex.
(3) By the first two parts, it follows for a1 , · · · , an ∈ (0, ∞) that
− log a1 − · · · − log an
1
a1 + · · · + an
≤
= − log(a1 · · · an ),
− log
n
n
n
√
1
a1 + · · · + an
≥ log(a1 · · · an ) = log n a1 · · · an . As exp is increasing,
that is, log
n
n
√
√
a1 + · · · + an
a1 + · · · + an
≥ exp log n a1 · · · an = n a1 · · · an .
= exp log
n
n
116
Solutions
Solution to Exercise 4.6.1. We have
(x3 + x − 2)
= 1 + 1 − 2 = 0 and (x2 − 3x + 2)
x=1
x=1
= 1 − 3 + 2 = 0.
So by l’Hôpital’s Rule, we do have
x3 + x − 2
3x2 + 1
= lim
,
2
x→1 x − 3x + 2
x→1 2x − 3
provided that the latter exists. However,
(3x2 + 1)
= 2 · 1 + 1 = 4 6= 0,
lim
x=1
and so l’Hôpital’s Rule is not applicable a second time. This is the error. In fact,
3 · 12 + 1
4
3x2 + 1
x3 + x − 2
= lim
=
=
= −4.
2
x→1 2x − 3
x→1 x − 3x + 2
2·1−3
−1
lim
Solution to Exercise 4.6.2. Using l’Hôpital’s Rule, we have
sin x − x
cos x − 1
lim
= lim
x→0
x→0
x3
3x2
− sin x
= lim
x→0
6x
− cos x
= lim
x→0
6
−1
.
=
6
117
Solutions to the exercises from Chapter 5
Solutions to the exercises from Chapter 5
Solution to Exercise 5.1.1. Let P = {x0 = 0 < x1 < · · · < xn−1 < xn = 1} be any partition of
[0, 1]. Since every interval [xk , xk+1 ] contains an irrational number, it follows that the lower sum
S(P ) =
n−1
X
k=0
0 · (xk+1 − xk ) = 0.
On the other hand, for every interval [xk , xk+1 ], we have an increasing sequence of rational numbers
(qn )n∈N in [xk , xk+1 ] converging to xk+1 , and so
xk+1 = sup qn = sup f (qn ) ≤
n∈N
n∈N
sup
x∈[xk ,xk+1 ]
f (x) ≤ xk+1 .
Hence the upper sum
S(P ) =
n−1
X
k=0
xk+1 · (xk+1 − xk ) ≥
n−1
X
k=0
n−1
xk+1 + xk
1X 2
1
1
· (xk+1 − xk ) =
(xk+1 − x2k ) = (12 − 02 ) = .
2
2
2
2
k=0
So we have
S = inf S(P ) ≥
P ∈P
1
> 0 = sup S(P ) = S.
2
P ∈P
Thus f 6∈ R[0, 1].
Solution to Exercise 5.1.2. As f, g ∈ R[a, b], also f − g ∈ R[a, b] and so |f − g| ∈ R[a, b]. Hence
f + g + |f − g|
∈ R[a, b],
2
that is, x 7→
R[a, b].
f (x) + g(x) + |f (x) − g(x)|
= max{f (x), g(x)} is in R[a, b]. Thus max{f, g} ∈
2
Since
min{a, b} = a + b − max{a, b} for all a, b ∈ R,
it follows that
x 7→ min{f (x), g(x)} = f (x) + g(x) − max{f (x), g(x)} = (f + g − max{f, g})(x)
is in R[a, b]. Thus min{f, g} ∈ R[a, b].
Finally, we justify the claims in the Hint and (6.28).
If a > b then max{a, b} = a and min{a, b} = b = a + b − a = a + b − max{a, b}.
If a < b then max{a, b} = b and min{a, b} = a = a + b − b = a + b − max{a, b}.
If a = b then max{a, b} = a = b and min{a, b} = b = a = 2a − a = a + b − max{a, b}.
This proves (6.28).
a + b + |a − b|
a+b+a−b
=
= a = max{a, b}.
2
2
a+b+b−a
a + b + |a − b|
=
= b = max{a, b}.
If a < b, then max{a, b} = b and
2
2
a + b + |a − b|
a+b
If a = b, then |a − b| = 0 and
=
= a = b = max{a, b}.
2
2
If a > b, then max{a, b} = a and
(6.28)
118
Solutions
Solution to Exercise 5.1.3. Let
1 if x ∈ [0, 1] ∩ Q,
−1 if x ∈ [0, 1] ∩ Q,
f (x) =
and g(x) =
.
−1 if x ∈ [0, 1] \ Q
1 if x ∈ [0, 1] \ Q
Then f + g ≡ 0, f g ≡ −1, |f | ≡ 1, and so f + g, f g, |f | ∈ R[0, 1].
Let 1[0,1]∩Q be the function which is equal to 1 if its argument belongs to [0, 1] ∩ Q, and equal
to 0 otherwise. Then f = 2 · 1[0,1]∩Q − 1 6∈ R[0, 1], since otherwise
f +1
∈ R[0, 1],
2
1[0,1]∩Q =
a contradiction. Similarly g = −f 6∈ R[0, 1].
Solution to Exercise 5.1.4. By the Extreme Value Theorem, for all x ∈ [a, b],
min f (x) =: m ≤ f (x) ≤ M := max f (x).
x∈[a,b]
x∈[a,b]
As ϕ is nonnegative, mϕ(x) ≤ ϕ(x)f (x) ≤ M ϕ(x) for all x ∈ [a, b]. Hence
Z b
Z b
Z b
m
ϕ(x)dx ≤
ϕ(x)f (x)dx ≤ M
ϕ(x)dx.
a
If
Z
a
Z
b
ϕ(x)dx = 0, then (6.29) shows that
a
If
Z
b
a
b
a
ϕ(x)f (x)dx = 0, and so any c ∈ [a, b] will do.
ϕ(x)dx 6= 0, then by dividing (6.29) throughout by
min f (x) = m ≤
x∈[a,b]
Z
(6.29)
a
Z
b
ϕ(x)dx > 0 we obtain
a
b
ϕ(x)f (x)dx
a
Z
≤ M = max f (x).
b
x∈[a,b]
ϕ(x)dx
a
By the Intermediate Value Theorem applied to the continuous f : [a, b] → R, it follows that there
is a c ∈ [a, b] such that
Z b
ϕ(x)f (x)dx
a
,
f (c) = Z b
ϕ(x)dx
that is,
Z
b
f (x)ϕ(x)dx = f (c)
a
a
Z
b
ϕ(x)dx.
a
That the continuity of f cannot be omitted: Take [a, b] = [0, 1], ϕ ≡ 1, and


 0 if 0 ≤ x ≤ 1 ,
2
f (x) =
1

 1 if < x ≤ 1.
2
Then
Z
a
and
Z
b
f (x)ϕ(x)dx =
Z
1
0
b
ϕ(x)dx = 1. But there is no c such that f (c) =
a
1
,
2
f (x)dx =
1
=
2
Z
b
ϕ(x)f (x)dx
a
Z
a
.
b
ϕ(x)dx
119
Solutions to the exercises from Chapter 5
That the nonnegativity of ϕ cannot be omitted: Take [a, b] = [0, 1], f (x) = x, and
ϕ(x) = x −
1
− ǫ,
2
for x ∈ [0, 1], where we will specify ǫ later. Then
Z b
Z b 1
1 1 1
1
ǫ
ϕ(x)f (x)dx =
x x − − ǫ dx = −
+ǫ =
− ,
2
3
2
2
12
2
a
a
Z b
Z b
1
ϕ(x)dx =
x − − ǫ dx = −ǫ.
2
a
a
Z b
1
ϕ(x)f (x)dx
1
1
a
24
Take ǫ =
=
. Then Z b
= − < 0, while f (x) ≥ 0 for all x ∈ [0, 1].
1
12
2
−
ϕ(x)dx
12
a
Solution to Exercise 5.2.1. We will use l’Hôpital’s Rule. We have for x > 0 that
Z x
Z x
x3
t2
2
0≤
dt
≤
t
dt
=
,
4
3
0 t +1
0
Z x
t2
dt = 0. Also lim x3 = 0.
and so lim
4
x→0+
x→0+ 0 t + 1
Z x
x2
t2
d
dt = 4
. We have
By the Fundamental Theorem of Calculus,
4
dx 0 t + 1
x +1
Z x
1
1 1
x2
1
1 d
t2
dt
=
lim
= lim
= .
lim
x→0+ 3x2 x4 + 1
x→0+ 3 x4 + 1
x→0+ 3x2 dx 0 t4 + 1
3
Z x
1
1
t2
Hence by l’Hôpital’s Rule, lim 3
dt = .
4
x→0+ x
t
+
1
3
0
Solution to Exercise 5.2.2. By the Fundamental Theorem of Calculus,
Z x
d
f (t)dt = f (x).
dx x0
We have
Z x
d
f (t)dt
dx x0
f (x)
= lim
= f (x0 ).
lim
d
x→x0
x→x0
1
(x − x0 )
dx
Also, we have for x ∈ [x0 − 1, x0 + 1] that
Z x
Z x
0≤
|f (t)|dt ≤
max
|f (x)| |x − x0 |.
f (t)dt ≤
So lim
x→x0
Z
x∈[x0 −1,x0 +1]
x0
x0
x
x0
f (t)dt = 0. Also lim (x − x0 ) = 0. So by l’Hôpital’s Rule,
x→x0
lim
x→x0
Also, we have
Z
x
f (t)dt
x0
x − x0
x
1
=
x + x0
2
x
lim
=1
x→x0 x + x0
lim
x→x0
= f (x0 ).
if x0 6= 0, and
if x0 = 0.
120
Solutions
Thus
x
− x20
lim
x→x0 x2
Z
x
f (t)dt = lim
x→x0
x0
Z
x
·
x + x0
x
f (t)dt
x0
x − x0


 f (x0 )
2
=

 f (x0 )
if x0 6= 0,
if x0 = 0.
Solution to Exercise 5.2.3. We have for x ∈ R that by Domain Additivity
Z x+T
Z x+T
Z x
F (x) :=
f (t)dt =
f (t)dt −
f (t)dt.
x
0
0
But
Z
x+T
f (t)dt
=
0
=
Z
x
(using the subtitution u = t − T )
f (u + t)dt
−T
Z x
f (u)du.
−T
So
F (x) =
Z
x
f (t)dt −
−T
Z
x
f (t)dt,
0
and F ′ (x) = f (x) − f (x) = 0. Thus F is constant, and in particular, F (a) = F (0) for all a ∈ R,
that is,
Z a+T
Z T
f (t)dt =
f (t)dt
a
0
for all a ∈ R.
Solution to Exercise 5.2.4. Let F (y) :=
Z
y
a
G(x) = F (v(x)) =
Z
f (t)dt, y ∈ [a, b], and G := F ◦ v, that is,
v(x)
f (t)dt,
a
x ∈ [a, b].
By the Chain Rule and the Fundamental Theorem of Calculus,
Z v(x)
d
f (t)dt = G′ (x) = F ′ (v(x)) · v ′ (x) = f (v(x)) · v ′ (x), x ∈ [c, d].
dx a
Z u(x)
d
Similarly,
f (t)dt = f (u(x) · u′ (x), x ∈ [c, d]. Thus
dx a
!
Z v(x)
Z u(x)
Z v(x)
d
d
f (t)dt −
f (t)dt = f (v(x)) · v ′ (x) − f (u(x) · u′ (x), x ∈ [c, d].
f (t)dt =
dx u(x)
dx
a
a
Solution to Exercise 5.2.5. We will use Leibniz’s Rule for Integrals. We note that the functions
t 7→
1
1
p
,
1 + t2 1 + |t|
are continuous on R, and x 7→ 2x, x2 are differentiable. Thus we have for x ∈ R that
F ′ (x)
=
G′ (x)
=
2
1
·2 =
, and
1 + (2x)2
1 + 4x2
2x
1
p
· 2x =
.
2
1 + |x|
1 + |x |
121
Solutions to the exercises from Chapter 5
Solution to Exercise 5.2.6.
(1) By Leibniz’s Rule for Integrals, f (x) = 2x(1 + x) + x2 · 1 = 2x + 3x2 .
(2) Since
d t3
= t2 ,
dt 3
we have by the Fundamental Theorem of Calculus that
(f (x))3
− 0 = x2 (1 + x).
3
So (f (x))3 = 3x2 (1 + x) ≥ 0 for all x ≥ 0, that is, f (x) =
p
3
3x2 (1 + x), x ≥ 0.
(3) By Leibniz’s Rule for Integrals, f (x2 ) · 2x = 2x · (1 + x) + x2 · 1 = 2x + 3x2 . For x 6= 0,
3
f (x2 ) = 1 + x,
2
and as x > 0, it follows from here that
3√
f (x) = 1 +
x, x > 0.
2
Since f is continuous on [0, ∞),
3√
3√
x =1+
0.
f (0) = lim f (x) = lim 1 +
x→0+
x→0+
2
2
3√
x, x ≥ 0.
Thus f (x) = 1 +
2
(4) By Leibniz’s Rule for Integrals, f (x2 (1 + x)) · (2x · (1 + x) + x2 · 1) = 1, that is,
f (x2 (1 + x)) · (2x + 3x2 ) = 1,
x ≥ 0.
But if x = 0, then f (0) · 0 = 1, a contradiction. So no such continuous f exists.
Solution to Exercise 5.2.7. Using sin(λ(x − t)) = (sin(λx))(cos(λt)) − (cos(λx))(sin(λt)), we
obtain
Z
Z
sin(λx) x
cos(λx) x
y(x) =
f (t) cos(λt)dt −
f (t) sin(λt)dt.
λ
λ
0
0
Using the Product Rule and the Fundamental Theorem of Calculus, we obtain
Z x
✘
sin(λx) ✘✘✘✘
′
f (x) cos(λx)
f (t) cos(λt)dt +
✘
y (x) = (cos(λx))
✘
✘✘λ
0
Z x
✘
cos(λx) ✘✘✘✘
+(sin(λx))
f (t) sin(λt)dt −
✘f (x) sin(λx)
✘
✘✘λ
Z x
Z x0
f (t) sin(λt)dt.
f (t) cos(λt)dt + (sin(λx))
= (cos(λx))
0
0
Thus
y ′′ (x)
=
=
−λ(sin(λx))
Z
x
f (t) cos(λt)dt + (cos(λx))f (x) cos(λx)
0
Z x
+λ(cos(λx))
f (t) sin(λt)dt + (sin(λx))f (x) sin(λx)
0
Z x
f (t) sin(λ(x − t))dt + f (x)((cos(λx))2 + (sin(λx))2 )
−λ
0
=
−λ2 y(x) + f (x) · 1,
and so y ′′ (x) + λ2 y(x) = f (x) for all x ∈ R. Also from the expressions for y(x) and y ′ (x) above,
we see that y(0) = 0 and y ′ (0) = (cos 0) cot 0 + (sin 0) · 0 = 0.
122
Solutions
Solution to Exercise 5.3.1. Using the substitution u = −x, we have
Z 0
Z 0
Z a
Z a
Z
f (x)dx =
f (−u)(−1)du =
f (−u)du =
−f (u)du = −
−a
Consequently,
Z
a
a
f (x)dx =
−a
Z
0
f (x)dx +
−a
Z
0
0
a
f (x)dx = −
Z
0
a
f (u)du +
0
Z
a
f (u)du.
0
a
f (x)dx = 0.
0
Solution to Exercise 5.3.2. Let n ∈ N. Then using Integration by Parts, we obtain
Z 1
xm (1 − x)n dx
I(m, n) :=
0
=
(1 − x)n
=
n
m+1
Consequently, for n ∈ N,
I(m, n) =
=
=
=
Z
xm+1 1
−
m+1 0
0
1
Z
0
1
n(1 − x)n−1 (−1)
xm+1 (1 − x)n−1 dx =
xm+1
dx
m+1
n
I(m + 1, n − 1).
m+1
n
n
n−1
I(m + 1, n − 1) =
·
I(m + 2, n − 2) = · · ·
m+1
m+1 m+2
n(n − 1)(n − 1) · · · 1
I(m + n, 0)
(m + 1)(m + 2) · · · (m + n)
Z 1
n!
xm+n dx
(m + 1)(m + 2) · · · (m + n) 0
n!
1
n!m!
·
=
.
(m + 1)(m + 2) · · · (m + n) m + n + 1
(m + n + 1)!
Solution to Exercise 5.3.3. We use Integration by Parts and the Fundamental Theorem of
Calculus
Z x
Z u
Z u
x Z x
(x − u)f (u)du = (x − u)
f (t)dt −
(−1)
f (t)dtdu
a
a
a
a
a
Z x Z u
Z x Z u
f (t)dt du.
f (t)dt du =
= 0 − (x − a) · 0 +
a
a
a
a
Solution to Exercise 5.3.4. We use induction on n. If n = 0, then
Z
Z b
1 b
(b − t)0 f (0+1) (t)dt =
f ′ (t)dt = f (b) − f (a),
0! a
a
by the Fundamental Theorem of Calculus.
Suppose the formula has been established for some n. Let f : [a, b] → R be such that
f , · · · , f (n+2) exist and f (n+2) ∈ C[a, b]. Then
f (n) (a)
n
′
(b − a)
f (b) − f (a) + f (a)(b − a) + · · · +
n!
Z
1 b
=
(b − t)n f (n+1) (t)dt
n! a
!
b Z b
n+1
n+1
1
(b
−
t)
(−1)
(b
−
t)
(−1)
=
f (n+1) (t)
f (n+2) (t)
−
n!
n+1
n+1
a
a
!
Z b
(b − a)n+1
1
1
(n+1)
n+1 (n+2)
0+f
(a)
+
(b − t)
f
(t)dt
=
n!
n+1
n+1 a
Z b
1
f (n+1) (a)
(b − a)n+1 +
=
(b − t)n+1 f (n+2) (t)dt.
(n + 1)!
(n + 1)! a
′
Solutions to the exercises from Chapter 5
123
This completes the proof.
Solution to Exercise 5.3.5. We have using Integration by Parts that
Z 1
Z 1
1 Z 1
1
1
xn
1
nxn−1
n
n
x
dx
=
dx.
dx =
x −
+
−
2
1+x
(1 + x)2
2
0
0 (1 + x)
0
0 1+x
For x ∈ [0, 1], x ≥ x2 ≥ x3 ≥ · · · and so the sequence
Z 1
xn
dx
2
0 (1 + x)
n∈N
Z 1
xn
is decreasing. Moreover it is bounded below. So lim
dx exists.
n→∞ 0 (1 + x)2
Let ǫ > 0. Then
Z 1−ǫ
Z 1
Z 1
xn
xn
xn
dx
=
dx
+
dx
2
2
(1 + x)2
0
1−ǫ (1 + x)
0 (1 + x)
Z 1
Z 1−ǫ
1
(1 − ǫ)n
dx
+
dx = (1 − ǫ)n + ǫ.
≤
2
2
(1
+
0)
(1
+
0)
1−ǫ
0
Hence
Z
1
xn
dx ≤ lim (1 − ǫ)n + ǫ = 0 + ǫ = ǫ.
n→∞
n→∞ 0 (1 + x)2
As the choice of ǫ > 0 was arbitrary, it follows that
Z 1
xn
dx = 0.
lim
n→∞ 0 (1 + x)2
Z 1
nxn−1
1
From (6.30), it follows that lim
dx = .
n→∞ 0 1 + x
2
0 ≤ lim
Solution to Exercise 5.3.6. We use Integration by Substitution/Change of Variables:
u = 1 − 4x2 ,
du = −8xdx,
x = 0 ⇒ u = 1,
x = 1/4 ⇒ u = 3/4.
See Figure 22.
Figure 22. Change of variables u = 1 − 4x2 .
(6.30)
124
Solutions
We have
Z
1
4
0
x
√
dx =
1 − 4x2
=
Z
Z
1 1 1
−1/8
√ du
√ du =
4 43 2 u
u
1
r !
1
1 √ 1
1 √
3
=
u 3 =
1−
4
4
4
4
4
3
4
√ !
3
1−
.
2
Solution to Exercise 5.3.7. We use Integration by Substitution/Change of Variables:
u = x4/3 − 1,
du = (4/3)x1/3 dx,
x = 1 ⇒ u = 0,
x = 8 ⇒ u = 15.
See Figure 23.
Figure 23. Change of variables u = x4/3 − 1.
We have
Z
1
8
√
3
x·
q√
3
x4 − 1 dx
=
Z
15
0
=
3
4
=
15
2
3√
udu
4
15
1
3 2
1
u 2 +1 = · (153/2 − 0)
4 3
0
+1
1
2
3/2
.
Solution to Exercise 5.4.1. We have
n
n
n
X
X
1
k
k−1
1
k
1 X
√
q
,
=
=
f
−
n
n
n
k 2 + n2
( k )2 + 1 n k=1
k=1
k=1
n
where f (x) := √
1
, x ∈ [0, 1]. So
x2 + 1
n
X
1
√
= S(f, Pn ),
2 + n2
k
k=1
125
Solutions to the exercises from Chapter 5
where
As lim S(f, Pn ) =
n→∞
Z
Pn
:=
ξk
=
n
1 2 3
0, , , , · · · ,
,
n n n
n
k+1
, k = 0, 1, 2, 3, · · · , n − 1.
n
1
f (x)dx, we obtain
0
lim
n→∞
n
X
k=1
√
1
=
k 2 + n2
Z
1
0
1
√
dx.
x2 + 1
We use Integration by Substitution/Change of Variables:
x = sinh u,
dx = cosh udu,
x = 0 ⇒ u = 0,
x = 1 ⇒ u = sinh−1 1.
We have
Z 1
0
√
1
x2
+1
dx
=
=
Z
Z
sinh−1 1
0
sinh−1 1
0
1
Z
sinh−1 1
1
p
p
cosh udu =
cosh udu
2
1 + (sinh u)
(cosh u)2
0
Z sinh−1 1
1
1du = sinh−1 1.
cosh udu =
cosh u
0
Let x = sinh−1 1. Then sinh x = 1, that is,
ex − e−x
= 1,
2
√
√
x 2
and so (e√
) − 2ex − 1 = 0, that√is, ex = 1 + 2 or ex = 1 − 2. As ex > 0, it follows that
ex = 1 + 2, and so x = log(1 + 2).
√
1
1
1
1
Consequently, lim √
= log(1 + 2).
+√
+√
+ ··· + √
2
2
2
2
2
2
2
2
n→∞
1 +n
2 +n
3 +n
n +n
Solution to Exercise 5.5.1. (1) For n > 1,
Z n
Z 1
Z n
1
1
1
√
√
√
dx =
dx +
dx
3
3
1+x
1+x
1 + x3
0
0
1
Z
Z 1
n
1
1
√
√ dx
dx +
≤
3
1+x
x3
1
0
Z ∞
Z 1
1
1
√
dx.
dx +
≤
3/2
3
x
1+x
1
0
Z n
1
√
So
dx
is bounded, and it is clearly increasing. Thus
1 + x3
0
n∈N
Z n
Z ∞
1
1
√
√
dx = lim
dx
3
n→∞
1
+
x
1
+ x3
0
0
exists.
126
Solutions
(2) For n > 1, we have
Z n
0
x
dx
1 + x3/2
Z
=
Since
n
1
1
√ dx
x
n∈N
does not exist.
x
dx +
1 + x3/2
1
x
dx +
1 + x3/2
0
Z
≥
Z
1
0
Z
n
x
dx
1 + x3/2
n
x
dx
x3/2 + x3/2
1
Z
1
Z
x
1 n 1
√ dx.
≥
dx
+
3/2
2 1
x
0 1+x
Z n
x
dx
is unbounded. Hence
diverges, it follows that
1
+
x3/2
0
n∈N
Z ∞
x
dx
1
+
x3/2
0
Z
1
Solution to Exercise 5.5.2. (1) We have
Z ∞
Z
Γ(1) =
e−t t1−1 dt =
0+
−1
∞
e−t dt =
0+
−x
=
1
e−t dt + lim
x→∞
0+
e − e−1
− e0
+ lim
x→∞
−1
−1
1
1
− + 1 − 0 + = 1.
e
e
e
=
Z
Z
x
e−t dt
1
(2) We have
Γ(s + 1) =
lim
ǫց0
=
lim
ǫց0
Using Integration by Parts, we have
Z 1
1 Z
e−t ts dt = −ts e−t −
ǫ
ǫ
s −ǫ
As 0 ≤ ǫ e
s
Z
1
−t s+1−1
e t
dt + lim
x→∞
ǫ
Z
1
−t s
e t dt + lim
ǫ
1
ǫ
x→∞
Z
x
Z
x
e−t ts+1−1 dt
1
e−t ts dt
1
sts−1 (−e−t )dt = ǫs e−ǫ −
1
+s
e
Z
1
e−t ts−1 dt.
ǫ
ǫց0
≤ ǫ · 1 −→ 0 (as s > 0), we obtain
Z 1
Z 1
1
−t s
lim
e−t ts−1 dt.
e t dt = − + s · lim
ǫց0 ǫ
ǫց0 ǫ
e
(6.31)
Similarly, using Integration by Parts, we have
Z x
Z x
x Z x
1
s −x
−t s
s −t s−1
−t
e−t ts−1 dt.
e t dt = −t e −
st (−e )dt = − x e + s
e
1
1
1
1
Choose n ∈ N such that n > s. Then for x > 0,
ex = 1 +
and so e−x xs xn−s ≤ n!, which gives
0 ≤ e−x xs ≤
and so lim xs e−x = 0. Thus
x→∞
Z
lim
x→∞
1
x
e−t ts dt =
x
x2
xn
+
+ ··· ≥
,
1!
2!
n!
n!
xn−s
xր∞
−→ 0 (as n > s),
1
− 0 + s · lim
x→∞
e
Z
1
x
e−t ts−1 dt.
(6.32)
127
Solutions to the exercises from Chapter 5
From (6.31) and (6.32), it follows that
Γ(s + 1) =
=
=
Z x
Z 1
1
1
−t s−1
e−t ts−1 dt
e t dt + + s · lim
− + s · lim
x→∞ 1
ǫց0 ǫ
e
e
Z 1
Z x
−t s−1
−t s−1
s lim
e t dt + lim
e t dt
x→∞ 1
ǫց0 ǫ
Z ∞
s
e−t ts−1 dt = sΓ(s).
0+
(3) We have Γ(2) = Γ(1 + 1) = 1 · Γ(1) = 1 = 1!, and so the claim holds for n = 1. Suppose that
for some k ∈ N, Γ(k + 1) = k!. Then Γ((k + 1) + 1) = (k + 1) · Γ(k + 1) = (k + 1) · k! = (k + 1)!.
So the result follows by induction.
Solution to Exercise 5.5.3. Consider the function f given by

1


 n if x ∈ n, n + n3 , n ≥ 2,
f (x) =
[
1


n, n + 3 .
 0 if x ∈ R \
n
n≥2
Then f : [0, ∞) → [0, ∞),
Z
∞
f (x)dx =
0
X
n≥2
n·
X 1
1
=
< +∞.
3
n
n2
n≥2
But f (n) = n, n ≥ 2, and so clearly we do not have that lim f (x) = 0.
x→∞
By the Fundamental Theorem of Calculus,
f (x) − f (0) =
Z
lim f (x) = f (0) +
Z
and so
x→∞
x
f ′ (x)dx,
0
∞
f ′ (x)dx =: L.
0
As f (x) ≥ 0 for all x, we must have that L ≥ 0. Suppose that L > 0. Then there exists an R > 0
such that for all x > R, L − f (x) ≤ |f (x) − L| < L/2, and in particular, f (x) > L/2 for all x > R.
Hence for all x > R,
Z ∞
Z x
Z x
L
f (x)dx ≥
f (x)dx ≥
f (x)dx > (x − R) ,
2
0
0
R
which is absurd. Hence L = 0.
Solution to Exercise 5.5.4. See Figure 24. Thus

0



t
(1[0,1] ∗ 1[0,1] )(t) =
2−t



0
if
if
if
if
t ≤ 0,
0 ≤ t ≤ 1,
1 ≤ t ≤ 2,
t ≥ 2.
128
Solutions
f
g(t − · )
0
1
f ∗g
0
1
2
Figure 24. 1[0,1] ∗ 1[0,1] .
We have for all t ∈ R that
(g ∗ f )(t)
=
=
=
=
lim
T →∞
lim
T →∞
lim
Z ∞
T →∞
−∞
Z
T
g(τ )f (t − τ )dτ
−T
Z t−T
t+T
Z t+T
t−T
g(t − s)f (s)(−1)ds(using the substitution t − sτ = s)
f (s)g(t − s)ds
f (τ )g(t − τ )dτ = (f ∗ g)(t).
Solution to Exercise 5.5.5.
129
Solutions to the exercises from Chapter 5
(1) We have
I ′ (α)
=
=
Z
∞
Z0 ∞
0
=
Z ∞
sin(2αx)
2(sin(αx)) · cos(αx)
dx
=
dx
x2
x
0
sin u
du (using the substitution u = 2αx)
u
π
.
2
Thus I(α) = I(α) − 0 = I(α) − I(0) =
Z
∞
0
π
α, and so
2
π
π
(sin x)2
dx = I(1) = · 1 = .
2
x
2
2
(2) We have
I ′ (α)
=
Z
∞
e−x
0
(cos(αx)) · x
dx =
x
Z
∞
e−x cos(αx)dx
0
∞
1
−x
e
(−
cos(αx)
+
α
sin(αx))
1 + α2
0
1
.
1 + α2
=
=
Thus I(α) = I(α) − 0 = I(α) − I(0) = tan−1 α, and so
Z ∞
π
sin x
dx = I(1) = tan−1 1 = .
e−x
x
4
0
(3) We have
′
I (α)
Z
=
1
0
(log x)eα log x − 0
dx =
log x
Z
1
xα dx =
0
Thus I(α) = I(α) − 0 = I(α) − I(0) = log(α + 1), and so
(4) We have
′
I (α)
=
Z
∞
1
1+α2 x2
·x−0
x
0
dx =
Z
0
Thus I(α) = I(α) − 0 = I(α) − I(1) =
Z
0
∞
∞
Z
0
1
1
.
α+1
x−1
dx = I(1) = log 2.
log x
∞
1
1
1 π
−1
dx
=
tan
(αx)
= · .
1 + α2 x2
α
α 2
0
π
log α, and so
2
π
tan−1 (πx) − tan−1 x
dx = I(π) = log π.
x
2
Z x
GM m x
GM m
GM m GM m
dr = −
Solution to Exercise 5.5.6. (1) We have
+
, and
=−
2
r
r
x
R
R
R
so
Z ∞
Z x
GM m
GM m
GM m GM m
GM m
GM m
V (R) =
= 0+
dr
=
lim
dx
=
lim
−
+
=
.
2
2
x→∞
x→∞
r
r
x
R
R
R
R
R
(2) We must have
Kinetic energy =
and so ve (R) =
r
2GM
.
R
GM m
1
m(ve (R))2 =
= V (R) = Potential Energy
2
R
130
Solutions
With the given values, we have that the escape velocity on the surface of the Earth (so that
the separation R is just the radius of the Earth),
r
r
2GM
2 × (6.67384 × 10−11 ) × (5.97219 × 1024 )
=
ve (R) =
R
6.371 × 106
p
1.2512145 × 108 = 1.1185769 × 104 m s−1 ,
=
that is, approximately 11.2 km/s.
Solution to Exercise 5.6.1. (1) We have
Z
log(n + 1) − log n =
n+1
1
1
dt −
t
Clearly
Z
n+1
n
n+1
Z
n
1
dt
t
≤
1
dt
t
≥
Z
n+1
n
n+1
Z
n
Z
1
n
1
dt =
t
Z
n+1
n
1
dt.
t
1
1
dt = and
n
n
1
1
dt =
.
n+1
n+1
(2) We have
an
1 1
+ + ···+
2 3
1 1
= 1 + + + ···+
2 3
1
− log n
n
1
+ (log 1 − log 2 + log 2 − log 3 + · · · + log(n − 1) − log n)
n
1
1
1
= (1 + log 1 − log 2) +
+ log 2 − log 3 + · · · +
+ log(n − 1) − log n +
|
{z
}
2
n−1
n
{z
}
{z
}
|
|
≥0
= 1+
≥
≥0
1
≥ 0.
n
≥0
Also,
1 1
1
1 1
1
1
1 + + + · · · + − log n − 1 + + + · · · + +
− log(n + 1)
2 3
n
2 3
n n+1
1
≥ 0.
= log(n + 1) − log n −
n+1
Since (an )n∈N is monotone and bounded, it is convergent with a limit, which we call γ.
an − an+1
=
Solution to Exercise 5.6.2. Let f (x) = log(1 + x), x ≥ 0. Then we have f (0) = log 1 = 0, and
1
, f ′ (0) = 1,
f ′ (x) =
1+x
−1
f ′′ (x) =
, f ′′ (0) = −1,
(1 + x)2
2
.
f ′′′ (x) =
(1 + x)3
If x ∈ (0, ∞), then Taylor’s Formula gives
f (x) = f (0) + f ′ (0)x +
for some θ ∈ (0, x). Thus
log(1 + x) = 0 + x −
1 ′′
1
f (0)x2 + f ′′′ (θ)x3 ,
2!
3!
2
1
x2
x3 ,
+
2
3! (1 + θ)3
x ≥ 0.
131
Solutions to the exercises from Chapter 5
But clearly
2
2
1
x3 /3
x3
1
3
3
x
≥
0
for
x
≥
0
and
x
=
≤
for x ≥ 0. hence
3! (1 + θ)3
3! (1 + θ)3
(1 + θ)3
3
x−
x2
x3
x2
≤ log(1 + x) ≤ x −
+
2
2
3
for x ≥ 0.
Solution to Exercise 5.6.3. We have
⌢
d(A, B) + d(B, C)
=
log
⌢
AP
BQ
⌢ · ⌢
AQ BP
⌢
=
log
AP
⌢
AQ
=
log
⌢
AQ
⌢
+ log
⌢
·
⌢
AP
!
·
⌢
BP
⌢
CQ
⌢ · ⌢
BQ CP
⌢ !
CQ
!
BQ BP
⌢ · ⌢ · ⌢
BP BQ CP
⌢ !
CQ
= d(A, C).
⌢
CP
Solution to Exercise 5.6.4. (1) Using Integration by Parts, we have
Z b
b Z b 1
xdx = b log b − a log a − (b − a) = b(log b − 1) − a(log a − 1).
log xdx = (log x)x −
a
a
a x
(2) By the Fundamental Theorem of Calculus,
Z
a
b
x
e dx =
Z
a
b
b
d x
e dx = ex = eb − ea .
dx
a
Solution to Exercise 5.6.5. (1) For y > 0, using Integration by Parts (see Exercise 5.6.4.(1),
we have
Z 1
log xdx = −y log y + y − 1.
y
By l’Hôpital’s Rule, we have
lim y log y = lim
y→0
Thus
Z
Z
1
log xdx = lim
0
y→0
(2) We associate with
y
Z
y→0
1/y
log y
= lim (−y) = 0.
= lim
y→0
y→0 −1/y 2
1/y
1
log xdx = lim (−y log y + y − 1) = 0 + 0 − 1 = −1.
y→0
1
log xdx, a “Riemann type of sum”; see Figure 25.
0
We have
−1 =
≈
=
So log
Z
1
log xdx
0
1
1
2
1
n
1
log + log + · · · + log
n
n n
n
n
n
1
1 · 2 · 2 · 3···n
n!
1
= log n .
log
n
nn
n
n
n!
≈ −n, that is, log n! ≈ n log n − n.
nn
132
Solutions
0
1
n
2
n
3
n
n−1
n
1
log x
Figure 25. A Riemann type of sum.
Solution to Exercise 5.6.6. (1) We have blogb a = elogb a·log b = elog a = a.
(2) Suppose that
p
,
q
where p, q ∈ Z, and q ≥ 0. We may assume that p, q > 0 (because 3 > 1, 2 > 1 imply that
log 3, log 2 > 0, and so log2 3 > 0 too). Then 2p/q = 3, that is, 2p = 3q . But as p 6= 0, 2p is even.
However, 3q is odd, a contradiction.
√ 2 log2 3
√ 2
√
= ( 2 )log2 3 = 2log2 3 = 3 ∈ Q.
(3) 2 is irrational and 2 log2 3 is irrational, but 2
log2 3 =
(4) The flaw is that log1/2 is not an increasing function. Indeed, for x > 0,
log1/2 x =
log x
,
log(1/2)
and as log is strictly increasing and log(1/2) < 0, it follows that log1/2 is strictly decreasing.
(5) See Figure 26.
Figure 26. The graphs of log, log1/2 and log10 . The decreasing one is the graph of log1/2 .
Among the increasing ones, the graph of log10 is the one which takes value 1 at x = 10.
133
Solutions to the exercises from Chapter 5
Solution to Exercise 5.6.7. We have using the Fundamental Theorem of Calculus that
Z
p
1
d x
√
dt − log(x + 1 + x2 )
dx 0
1 + t2
1
1
1
√
−
· 1+ √
· 2x
= √
1 + x2
x + 1 + x2
2 1 + x2
√
1
1
x + 1 + x2
√
−
= 0.
= √
· √
2
2
1+x
x+ 1+x
1 + x2
So for all x ∈ R,
Z x
Z 0
p
p
1
1
√
√
dt − log(x + 1 + x2 ) =
dt − log(0 + 1 + 02 ) = 0.
1 + t2
1 + t2
0
0
Z x
p
1
√
dt = log(x + 1 + x2 ).
Hence
1 + t2
0
Also,
Z
d √
p
p
1
x 1 + x2
2
− log(x + 1 + x2 )
1 + t dt −
dx 0
2
2
√
p
1
x
1
x·x
1 · 1 + x2
2
√
1+x −
1+ √
−
− √
=
2
2 x + 1 + x2
2 1 + x2
1 + x2
√
p
1 + x2
1
x2
1 + x2 −
=
− √
− √
2
2 1 + x2
2 1 + x2
√
p
1 + x2
1 + x2
2
=
− √
1+x −
2
2 1 + x2
√
√
p
1 + x2
1 + x2
2
1+x −
=
−
= 0.
2
2
Hence for all x ∈ R,
√
Z xp
p
1
x 1 + x2
2
− log(x + 1 + x2 )
1 + t dt −
2
2
0
√
Z 0p
p
0 1 + 02
1
2
1 + t dt −
− log(0 + 1 + 02 ) = 0.
=
2
2
0
√
Z xp
p
2
1
x 1+x
+ log(x + 1 + x2 ).
1 + t2 dt =
Consequently,
2
2
0
x
Solution to Exercise 5.6.8. We have x′ (t) = aeta xi = ax(t), t ≥ 0, and x(0) = e0 xi = 1xi = xi .
Thus t 7→ eta xi is certainly a solution. Let us show that it is the only solution. Let f : [0, ∞) → R
defined by f (t) := e−ta x(t), t ≥ 0, where x is a solution to the given Initial Value Problem. Then
f ′ (t) = −ae−ta x(t) + e−ta x′ (t) = −ae−ta x(t) + e−ta ax(t) = 0,
t ≥ 0,
and so f is a constant. In particular, f (t) = e−ta x(t) = f (0) = xi , and so x(t) = eta xi , t ≥ 0.
Solution to Exercise 5.6.9. Newton’s Law of Cooling gives Θ′ (t) = −k(T − M ), where k > 0
is the constant of proportionality. Then
(ekt (Θ − M ))′ = ekt (Θ − M )′ + kekt (Θ − M ) = ekt (Θ′ + k(Θ − M )) = 0.
{z
}
|
=0
So ekt (Θ − M ) is constant, and so ekt (Θ(t) − M ) = ek·0 (Θ(0) − M ) = 1 · (Θ0 − M ). Rearranging,
we obtain
Θ(t) = e−kt (Θ0 − M ) + M = e−kt Θ0 + M (1 − e−kt ).
As t → ∞, Θ(t) → M , the ambient temperature, as expected.
134
Solutions
Solution to Exercise 5.6.10. (1) We have
(ect A(t))′ = ect A′ (t) + cect A(t) = ect (A′ (t) + cA(t)) = 0.
{z
}
|
=0
So ect A(t) is constant, and so ect A(t) = ec·0 A(0) = 1 · A0 . Hence A(t) = e−ct A0 for all t.
(2) (We want in particular that A(0 + τ ) = A(0)/2, that is, A(τ ) = A0 /2. So e−cτ A0 = A0 /2,
that is, e−cτ = 1/2. Thus −cτ = log(1/2) = − log 2. Thus we must have τ = (log 2)/c.) Indeed,
with
log 2
,
τ :=
c
1
1
A(t)
we have A(t + τ ) = e−c(t+τ ) A0 = e−ct e−cτ A0 = e−ct A0 = e−ct A0 =
.
2
2 | {z }
2
=A(t)
Solution to Exercise 5.6.11. We have
r·n
r·n
r mr
r mn
r mr
= P
1+
lim P 1 +
= P lim 1 +
= P er·n ,
m→∞
m→∞
m
m
m
where to get the last equality, we used the fact that
lim (1 + h)1/h = e,
h→0
which follows from lim
h→0
log(1 + h)
= 1 and the continuity of exp.
h
(1) We seek n such that
P e0.06n
= 2,
P
that is, e0.06n = 2. Hence 0.06n = log 2, that is, n =
log 2
0.6931
≈
≈ 11.6 years.
0.06
0.06
(2) We seek n such that
P (1 + nr)
= 2,
P
that is, 1 + 0 − 06n = 2, and so 0.06n = 1. Thus n =
1
≈ 16.7 years.
0.06
Solution to Exercise 5.6.12. (1) With (x, y) = (cosh t, sinh t), t ∈ R, we have
2 t
2
t
e − e−t
e + e−t
−
x2 − y 2 = (cosh t)2 − (sinh t)2 =
2
2
=
=
e2t − 2 + e−2t
e2t + 2 + e−2t
−
4
4
4
= 1.
4
See Figure 27.
(2) We have
sinh 0
=
cosh 0
=
1−1
e0 − e−0
=
= 0,
2
2
0
−0
e +e
1+1
=
= 1.
2
2
Solutions to the exercises from Chapter 5
135
Figure 27. The curve t 7→ (cosh t, sinh t). As t increases, the point on the curve moves upwards.
Also, for all x ∈ R we have
sinh′ x
=
cosh′ x
=
ex + e−x
d ex − e−x
=
= cosh x and
dx
2
2
ex − e−x
d ex + e−x
=
= sinh x.
dx
2
2
For x, y ∈ R, we have
(cosh x)(cosh y) +(sinh x)(sinh y)
=
=
=
ex + e−x ey + e−y
ex − e−x ey − e−y
·
+
·
2
2
2
2
x+y
x+y
x−y
−x+y
−(x+y)
e
− ex−y − e−x+y + e−(x+y)
e
+e
+e
+e
+
4
4
2ex+y + 2e−(x+y)
ex+y + e−(x+y)
=
= cosh(x + y).
4
2
(3) We have cosh′′ x = cosh ≥ 0 for all x and so cosh is convex. Also, limx→±∞ cosh x = +∞. See
Figure 28.
Figure 28. Graphs of ex , e−x on the left, and the graph of cosh on the right.
136
Solutions
We have sinh′′ x = sinh x > 0 for x > 0, and sinh′′ x < 0 for x < 0. Thus sinh is convex for x > 0,
and concave for x < 0. Also, limx→±∞ sinh x = ±∞. See Figure 28. See Figure 29.
Figure 29. Graphs of ex , −e−x on the left, and the graph of cosh on the right.
Solution to Exercise 5.6.13. Let f (x) = xx , x > 0. Then
h(x) := log f (x) = log(exp(x log x)) = x log x.
Thus
lim h(x) = lim x log x = lim
x→0+
x→0+
x→0+
By the continuity of exp, we obtain
1/x
log x
= lim −x = 0.
= lim
x→0+
1/x
−1/x2 x→0+
lim f (x) = lim exp h(x) = exp
x→0+
x→0+
lim h(x) = exp 0 = 1.
x→0+
1
= 1. So for every ǫ > 0, there exists a δ > 0 such that for all x satisfying
xx
0 < x < δ,
1
< ǫ.
−
1
xx
1
1
In particular for all n ∈ N such that n > , we have that x := satisfies 0 < x < δ and so
δ
n
1
1
n
|n − 1| = x − 1 < ǫ.
x
Hence also lim
x→0+
Consequently, lim n1/n = 1.
n→∞
Solution to Exercise 5.6.14. (1) Define
−x log x
e
g(x) :=
1
for x > 0,
for x = 0.
Then g is continuous. Indeed
lim x log x = lim
x→0+
x→0+
log x
1/x
= lim −x = 0.
= lim
x→0+
x→0+ −1/x2
1/x
137
Solutions to the exercises from Chapter 5
As exp is continuous, it follows that lim g(x) = lim e−x log x = e−0 = 1 = g(0). Thus
x→0+
lim
ǫց0
Z
1
ǫ
1
dx =
xx
lim
ǫց0
Z
=
1
Z
x→0+
1
g(x)dx = lim
ǫց0
ǫ
g(x)dx − lim
ǫց0
0
Z
ǫ
Z
0
1
g(x)dx −
g(x)dx =
0
Z
1
Z
ǫ
0
g(x)dx
g(x)dx,
0
where we have used
Z ǫ
Z ǫ
0 ≤ g(x)dx ≤
|g(x)|dx ≤ max |g(x)| ǫ
x∈[0,1]
0
0
in order to get the last equality.
(2) We have
Z
0
1
1
dx
xx
Z
=
1
e
−x log x
∞
X
(−1)n
n!
n=0
=
dx =
0
Z
0
Z
0
Z
∞
X
(−1)n 1 n
(−x log x)n
dx =
x (log x)n dx
n!
n!
0
n=0
n=0
∞
1X
e−nt (−t)n (−e−t )dt
∞
where we used the substitution x = e−t (and so dx = −e−t dt, when x = 0 we have t = ∞, and
when x = 1 we have t = 0) in order to get the last equality. Thus
Z
0
1
1
dx =
xx
=
Z
∞
X
1 ∞ −(n+1)t n
e
t dt
n! 0
n=0
Z
∞
X
1
1 ∞ −u un
du,
e
n
n! 0
(n + 1) n + 1
n=0
where we used the substitution u = (n + 1)t (and so du = (n + 1)dt, when t = 0 we have u = 0,
and when t = ∞ we have u = ∞) in order to get the last equality. Hence
Z
1
0
1
dx =
xx
=
=
Z ∞
∞
X
1
1
e−u u(n+1)−1 du
n+1
n!
(n
+
1)
0
n=0
∞
∞
X
X
1
1
1
1
Γ(n + 1) =
n!
n+1
n!
(n
+
1)
n!
(n
+
1)n+1
n=0
n=0
∞
X
1
1
=
.
n+1
(n + 1)
nn
n=1
n=0
∞
X
Solution to Exercise 5.6.15. With f defined by f (x) = log(1 + x), x ∈ [0, 1], we have that
S(f, Pn ) =
=
1
1
2
n 1
1
+ log 1 +
+ · · · + log 1 +
log 1 +
n
n
n
n
n
n
1
1
2
n n
log
1+
1+
··· 1 +
.
n
n
n
138
Solutions
Since lim S(f, Pn ) =
n→∞
Z
1
f (x)dx, we obtain
0
1
1
2
n n
lim log
1+
1+
··· 1 +
n→∞
n
n
n
=
Z
1
log(1 + x)dx
0
=
=
=
1 Z 1 1
xdx
(log(1 + x))x −
0
0 1+x
Z 1
1
1−
log 2 −
dx
1+x
0
1
log 2 − 1 + log(1 + x) = 2 log 2 − 1
0
4
= log .
e
1
1
4
4
2
n n
Since exp is continuous, it follows that lim log
1+
= elog e = .
1+
··· 1 +
n→∞
n
n
n
e
Solution to Exercise 5.6.16. In decreasing order of growth, we have
x2x , (2x)x , xx , (log x)x , ex , 2x , ex/2 , xe , x2 , x1/2 , (log x)2 , log2 x ≥ log x , log(log x).
Indeed,
e2x log x
x2x
= lim x x log x = lim ex(log x−log 2) = +∞;
x
x→∞ 2 e
x→∞
x→∞ (2x)
x
(2x)
= lim 2x = +∞;
(2) lim
x→∞
x→∞ xx
x
x
ex log x
(3) lim
= lim x log(log x) = lim ex log(x/ log x) = +∞;
x
x→∞ (log x)
x→∞ e
x→∞
(1) lim
(log x)x
ex log(log x)
=
lim
= lim ex(log(log x)−1) = +∞;
x→∞
x→∞
x→∞
ex
ex
ex
ex
lim
= lim x log 2 = lim ex log(e/2) = +∞ since e > 2;
x→∞ 2x
x→∞ e
x→∞
x
x log 2
2
e
lim
= lim
= lim ex(log 2−1/2) = +∞ since e < 4;
x→∞ ex/2
x→∞ ex/2
x→∞
x/2
e
x4
ex/2
x4−e x→∞
x/2
lim
=
+∞,
since
e
≥
,
and
so
≥
−→ ∞;
x→∞ xe
4!24
xe
4!24
e
x
lim
= lim xe−2 = +∞;
x→∞ x2
x→∞
2
x
= lim x3/2 = +∞;
lim
x→∞
x→∞ x1/2
√
1
√
1/2
x
1 x
2 x
= lim
= +∞;
lim
= lim
x→∞ 2(log x) 1
x→∞ (log x)2
x→∞ 4 log x
x
(4) lim
(5)
(6)
(7)
(8)
(9)
(10)
(log x)2
(log x)2
= lim
· log 2 = lim (log x) · log 2 = +∞;
x→∞ log2 x
x→∞ log x
x→∞
1
log2 x
=
> 1;
(12) lim
x→∞ log x
log 2
(11) lim
log x
= lim
x→∞
x→∞ log(log x)
(13) lim
1
x
1 1
log x x
= lim log x = +∞.
x→∞
139
Solutions to the exercises from Chapter 5
Solution to Exercise 5.6.17. (1) We use the
and g := log x, we have g, g ′ > 0 on (1, ∞) and
∞
form of l’Hôpital’s Rule. With f := log(log x)
∞
lim g(x) = lim log x = ∞,
x→∞
x→∞
and
1 1
1
f ′ (x)
log x x
= lim
= 0,
= lim
lim
1
x→∞
x→∞ log x
x→∞ g ′ (x)
x
f (x)
f ′ (x)
and so lim
= lim ′
= 0.
x→∞ g(x)
x→∞ g (x)
(2) We use the
0
form of l’Hôpital’s Rule. With
0
f (x) = 3sin x − 1,
g(x)
= x,
we have
lim f (x)
=
lim g(x)
=
x→0
x→0
lim (3sin x − 1) = 30 − 1 = 1 − 1 = 0,
x→0
lim x = 0,
x→0
and
e(log 3)(sin x) (log 3)(cos x)
f ′ (x)
= lim
= e(log 3)·0 (log 3)(cos 0) = log 3,
′
x→0
x→0 g (x)
1
f ′ (x)
f (x)
= lim ′
= log 3.
and so lim
x→0 g (x)
x→0 g(x)
lim
(3) We use the
and so
0
form of l’Hôpital’s Rule twice. We have
0
x3 = 0 and x3 = 0,
sin x − x +
6 x=0
x=0
sin x − x + x3 /6
cos x − 1 + x2 /2
= lim
3
x→0
x→0
x
3x2
lim
if the latter exists. But
and so
cos x − 1 + x2 /2 x=0
= 0 and 3x2 x=0
= 0,
sin x − x + x3 /6
1 1
cos x − 1 + x2 /2
− sin x + x
= lim
= lim
= − + = 0.
3
2
x→0
x→0
x→0
x
3x
6x
6 6
(4) We have cos x − 1 + x2 /2 = 0 and x4 = 0, and so
lim
x=0
x=0
− sin x + x
cos x − 1 + x2 /2
= lim
,
lim
4
x→0
x→0
x
4x3
if the latter exists. We have (− sin x + x)
= 0 and 4x3 = 0, and so
x=0
x=0
− cos x + 1
− sin x + x
= lim
,
lim
x→0
x→0
4x3
12x2
if the latter exists. We have (− cos x + 1)
= 0 and 12x2 = 0, and so
x=0
x=0
− cos x + 1
sin x
1
lim
= lim
=
.
2
x→0
x→0 24x
12x
24
140
Solutions
− sin x + x
− cos x + 1
sin x
1
cos x − 1 + x2 /2
= lim
= lim
= lim
=
.
x→0
x→0
x→0 24x
x→0
x4
4x3
12x2
24
(5) We have (sin x − x)
= 0 and x2 = 0, and so
Thus lim
x=0
x=0
cos x − 1
sin x − x
= lim
,
x→0
x2
2x
if the latter exists. We have (cos x − 1)
= 0 and 2x
= 0, and so
lim
x→0
x=0
x=0
cos x − 1
− sin x
lim
= lim
= 0.
x→0
x→0
2x
2
Thus
lim
x→0
1
1
−
x sin x
sin x − x
x
x
sin x − x
sin x − x
·
lim
= lim
= lim
= 0 · 1 = 0.
x→0 sin x
x→0
x sin x
x2
sin x x→0
x2
= lim
x→0
Solution to Exercise 5.6.18. We note that the integrand f is an odd function, that is,
f (x) = −f (−x) for all x ∈ [−1/2, 1/2].
Indeed,
f (−x) =
=
1 − (−x)
1+x
(cos(−x)) log
= (cos x) log
1 + (−x)
1−x
−1 !
1−x
1−x
= (cos x) − log
= −f (x),
(cos x) log
1+x
1+x
for all x ∈ [−1/2, 1/2]. Thus
Z 1/2
Z 0
Z
f (x)dx =
f (x)dx +
−1/2
−1/2
=
Z
0
f (−u)(−1)du +
Z
1/2
f (−u)du +
0
=
Z
1/2
0
=
−
f (x)dx
0
1/2
=
1/2
Z
−f (u)du +
1/2
f (x)dx +
0
Z
Z
Z
1/2
f (x)dx
0
(using the substitution u = −x
in the first integral)
1/2
f (x)dx
0
1/2
f (x)dx (as f is odd)
0
Z
1/2
f (x)dx = 0.
0
Solution to Exercise 5.6.19. We have
′
log x
1
1 1
1
= − 2 · log x + · = (1 − log x)
x
x
x x
x
for x ∈ (0, ∞).
We have
′
1
log x
= (1 − log x) ≤ 0
x
x
for x ∈ [e, ∞) because log x ≥ log e = 1 (as log is increasing). As π > 3 > e, we have that
log e
log π
<
,
π
e
that is, e log π < π log e. Hence π e < eπ , and so eπ is bigger.
141
Solutions to the exercises from Chapter 5
Solution to Exercise 5.6.20. We have for m, n ∈ N and x ∈ R that
cos((m + n)x)
=
cos((m − n)x)
=
(cos(mx))(cos(nx)) − (sin(mx))(sin(nx)),
(cos(mx))(cos(nx)) + (sin(mx))(sin(nx)),
and so 2(sin(mx))(sin(nx)) = cos((m − n)x) − cos((m + n)x).
If m 6= n, then
Z π
(sin(mx))(sin(nx))dx
=
−π
=
If m = n, then
Z
Z
cos((m − n)x) − cos((m + n)x) dx
−π
1 sin((m − n)x) sin((m + n)x) π
−
= 0.
2
m−n
m+n
−π
1
2
π
Z
1 − cos((m + n)x) dx
−π
sin((m + n)x) π
1
2π −
= π.
2
m+n
−π
π
(sin(mx))(sin(nx))dx
1
2
=
−π
=
π
Similarly
Z π
Z
1 π 0
(cos(mx))(cos(nx))dx =
cos((m − n)x) + cos((m + n)x) dx =
π
2
−π
−π
Z π
Also
(cos(mx)) (sin(nx)) dx = 0.
{z } | {z }
−π |
even
if m 6= n,
if m = n.
odd
Solution to Exercise 5.6.21. (1) For any integer k and for x ∈ R, we have
cos(k(x + 2π)) =
cos(kx + 2πk) = cos(kx) and
sin(k(x + 2π)) =
sin(kx + 2πk) = sin(kx).
Thus
f (x + 2π) =
a0 +
=
a0 +
n
X
k=1
n
X
(ak cos(k(x + 2π)) + bk sin(k(x + 2π)))
(ak cos(kx) + bk sin(kx)) = f (x).
k=1
(2) We have
1
2π
Z
π
f (x)dx
=
−π
=
=
1
2π
Z
1
2π
Z
π
−π
a0 +
n
X
k=1
(ak cos(kx) + bk sin(kx)) dx


Z π
Z π
n
1 X
a0 dx +
ak
cos(kx)dx + bk
sin(kx) dx
2π
−π
−π
−π | {z }
π
1
1
a0 · 2π +
2π
2π
k=1
n
X
k=1
ak
odd
sin(kx) π
−
+ 0 = a0 + 0 = a0 .
k
−π
142
Solutions
Using the result from Exercise 5.6.20, for ℓ = 1, · · · , n, we have
Z
Z
Z
n
X
1 π
1 π
1 π
f (x) cos(ℓx)dx =
a0 cos(ℓx)dx +
ak
(cos(kx))(cos(ℓx))dx
π −π
π −π
π −π
k=1
Z
n
X
1 π
+
bk
(sin(kx))(cos(ℓx))dx
π −π
k=1
− sin(ℓx) π
1
1
a0
=
+ aℓ π + 0 = aℓ ,
π
ℓ
π
−π
and
1
π
Z
π
f (x) sin(ℓx)dx
−π
=
1
π
+
Z
π
a0 sin(ℓx)dx +
−π
n
X
k=1
bk
1
π
Z
n
X
k=1
π
1
ak
π
Z
π
(cos(kx))(sin(ℓx))dx
−π
(sin(kx))(sin(ℓx))dx
−π
1
= 0 + 0 + bℓ π = bℓ .
π
(3) One can use the Maple commands
with(plots):
2
k
· (1 − (−1) ) · sin(k · x), k = 1..333 ;
f:=sum
(k · Pi)
plot(f,x=-10..10)
See Figure 30.
Figure 30. Partial sums of the Fourier series.
We observe that the graphs seem to look more and more like the graph of the “square wave”
function.
Solution to Exercise 5.6.22. (1) For x ∈ (0, π/2], we have by the Mean Value Theorem that
for some c ∈ (0, π/2), and so
sin x − sin 0
= cos c
x−0
sin x
= cos c < 1.
x
Thus sin x < x for x ∈ (0, π/2]. For x > π/2, x > 1 ≥ sin x. So sin x 6= x for all x ∈ (0, ∞). Hence
also sin(−x) = − sin x 6= −x for x ∈ (0, ∞). This sin x 6= x for all x ∈ R \ {0}. On the other hand,
sin 0 = 0. So 0 is the only fixed point of sin.
143
Solutions to the exercises from Chapter 5
(2) Consider f : R → R given by f (x) = cos x − x, x ∈ R. Then f (0) = cos 0 − 0 = 1 − 0 = 1 > 0,
while f (π/2) = cos(π/2)− π/2 = 0 − π/2 = −π/2 < 0. Hence by the Intermediate Value Theorem,
there exists a c∗ ∈ (0, π/2) such that f (c∗ ) = 0, that is, cos c∗ = c∗ . So there exists a fixed point.
We note that for x ≥ π/2, we have that f (x) = cos x − x ≤ 1 − π/2 < 0. For x ∈ (−π/2, 0],
f (x) = cos x − x > 0 − x > 0. For x ∈ (−∞, −π/2], f (x) = cos x − x ≥ −1 + π/2 > 0. So there
are no fixed points in (−∞, 0] ∪ [π/2, ∞). In (0, π/2), cos is strictly decreasing, and so cos can
have at most one fixed point there (for otherwise if 0 < c1 < c2 < π/2 are two fixed points, then
we get the contradiction that cos c1 = c1 < c2 = cos c2 ).
(3) Consider the sequence (xn )n≥0 defined by x0 = any real number, and xn+1 = cos xn for n > 0.
Then xn+1 −xn = cos xn −cos xn−1 . We note that x1 = cos x0 ∈ [−1, 1], and so x2 = cos x1 ∈ [0, 1],
and xn+1 = cos xn ∈ [0, 1] for all n ≥ 2. Thus there is a cn ∈ (0, 1) such that
|xn+1 −xn | = | cos xn −cos xn−1 | = (sin cn )|xn −xn−1 | ≤ (sin 1)|xn −xn−1 | = r|xn −xn−1 |, (6.33)
where r := sin 1 ∈ (0, 1). Using xn = x1 + (x2 − x1 ) + (x3 − x2 ) + · · · + (xn − xn−1 ) and (6.33),
we can show that (xn )n≥0 converges, to say x∗ . Then
f (x∗ ) = f lim xn = lim f (xn ) = lim xn+1 = x∗ .
n→∞
n→∞
n→∞
Thus x∗ = c∗ . See Figure 31.
Figure 31. Convergence to the fixed point of cos.
(Proof of the convergence of (xn )n≥0 : We have
|xn+k − xn | = x1 + (x2 − x1 ) + · · · + (xn − xn−1 ) + (xn+1 − xn ) + · · · + (xn+k − xn+k−1 )
− x1 + (x2 − x1 ) + · · · + (xn − xn−1 ) = (xn+1 − xn ) + · · · + (xn+k − xn+k−1 )
≤ |xn+1 − xn | + · · · + |xn+k − xn+k−1 |
≤ rn−1 |x2 − x1 | + · · · + rn+k−1 |x2 − x1 |
≤ (rn−1 + rn−2 + · · · )|x2 − x1 |
=
rn−1
n→∞
|x2 − x1 | −→ 0.
1−r
So the sequence (xn )n≥0 is Cauchy, and hence convergent.)
144
Solutions
Solution to Exercise 5.6.23. (1) We have cos(x+y) = (cos x)(cos y)−(sin x)(sin y) for x, y ∈ R.
Fix y and differentiate both sides with respect to x to get
− sin(x + y) = (− sin x)(cos y) − (cos x)(sin y),
that is, sin(x + y) = (sin x)(cos y) + (cos x)(sin y). Since x + y 6∈ πZ + π/2, dividing both sides by
cos(x + y) gives
tan(x + y) =
(sin x)(cos y) + (cos x)(sin y)
.
(cos x)(cos y) − (sin x)(sin y)
As x, y 6∈ πZ + π/2, we may divide the numerator and the denominator on the right hand side by
(cos x)(cos y) to obtain
(cos x)(sin y)
(sin x)(cos y)
+
tan x + tan y
(cos x)(cos y) (cos x)(cos y)
=
.
tan(x + y) =
(sin x)(sin y)
(cos x)(cos y)
1 − (tan x)(tan y)
−
(cos x)(cos y) (cos x)(cos y)
(2) Suppose that tan 1◦ ∈ Q. But whenever tan n◦ ∈ Q, for some natural number (< 89), it follows
from the addition formula
tan n◦ + tan 1◦
tan(n + 1)◦ =
1 − (tan n◦ )(tan 1◦ )
that also tan n◦ ∈ Q. Thus tan 2◦ , tan 3◦ , · · · , √
tan 60◦ all belong to Q. But from the picture shown
π
◦
in Figure 32, we have that tan 60 = tan 3 = 3 6∈ Q. This contradiction shows that tan 1◦ 6∈ Q.
Area =
1
π
6
√
3
2
60◦
1/2
Figure 32. tan 60◦ =
√
3
Remark 6.4.
π
(1) We remark that 1◦ is the angle 180
radians, and so 60◦ is the angle π/3 radians.
√
√
(2) That 3 is irrational
√ can be seen as follows. Let 3 be rational and suppose that for
integers p, q 6= 0, 3 = p/q, where the greatest common divisor of p and q is 1. Then
3q 2 = p2 and so p2 = 3q 2 . So 3 divides p2 and so 3 must also divide p. Let p = 3p′ .
Then 9p′2 = 3q 2 and so q 2 = 3p′2 . So 3 divides q 2 and so 3 must also divide q. But now
3 divides both p and q, contradicting the fact that p and q had no common factor.
145
Solutions to the exercises from Chapter 5
Solution to Exercise 5.6.24. Let
p
,
q
where p, q are integers and q > 0. Then for m > q,
p
p
m!x = m! = m · · · (q + 1) · q · = an integer.
q
q
x=
n
Thus m!x2π ∈ 2πZ, and so cos(2πm!x) = 1. Thus for all n ∈ N, and all m > q, (cos(2πm!x)) = 1,
so that
n
lim (cos(2πm!x)) = lim 1 = 1.
n→∞
n
n→∞
Hence lim lim (cos(2πm!x)) = lim 1 = 1.
m→∞ n→∞
m→∞
Now suppose that x 6∈ Q. Then for all m ∈ N, m!2πx 6∈ πZ. (Indeed if m!2πx = πk for some
k ∈ Z, then
k
∈ Q,
x=
2m!
a contradiction.) So cos(m!2πx) 6= ±1. Hence −1 < cos(m!2πx) < 1 for all m, and so for each m
we have that
lim (cos(m!2πx))n = 0.
n→∞
Consequently lim lim (cos(2πm!x))n = lim 0 = 0.
m→∞ n→∞
m→∞
Finally we show the irrationality of e. Fix m ∈ N. Then
X 1
X 1
m!e = m!
= an integer + m!
.
k!
k!
k≥0
So
k>m
X 1
cos(2πm!e) = cos 2π · (an integer) + 2πm!
k!
k>m
But
m!
1
1
1
+
+ ···+
(m + 1)! (m + 2)!
(m + k)!
=
≤
≤
So it follows that
m!
!
X 1
= cos 2πm!
k!
k>m
!
.
1
1
1
+
+ ···+
m + 1 (m + 1)(m + 2)
(m + 1) · · · (m + k)
1
1
1
+
+ ···+
m + 1 (m + 1)2
(m + 1)k
1
1/(m + 1)
1
+ ··· =
+
.
m + 1 (m + 1)2
1 − 1/(m + 1)
X 1
1/(m + 1) n→∞
−→ 0.
≤
k!
1 − 1/(m + 1)
k>m
We can choose a m0 large enough so that for all m > m0 ,
X 1
1
m!
< .
k!
2
k>m
Then for m > m0 ,
(0 <) 2πm!
X 1
1
< 2π = π.
k!
2
k>m
Then for all m ≥ m0 ,
X 1
cos(2πm!e) = cos 2π
k!
k>m
!
∈ (−1, 1),
n
and so for each m ≥ m0 , lim (cos(m!2πe))n = 0. Hence lim lim (cos(2πm!e)) = 0. From the
n→∞
m→∞ n→∞
first part of the exercise, we can conclude that e 6∈ Q.
146
Solutions
Solution to Exercise 5.6.25. By the similarity of the two triangles we have that
ℓ
d tan x
= ,
ℓdx
1
and so
d
tan x = ℓ2 = 1 + (tan x)2 .
dx
Solution to Exercise 5.6.26. We know that
1
d
tan−1 t =
for t ∈ R.
dt
1 + t2
Z 1
Z 1
1
1
π
d
π
−1
−1 Thus
dt
=
tan
tdt
=
tan
t
= −0= .
2
4
4
0
0 1+t
0 dt
Solution to Exercise 5.6.26. See Figure 33.
1
√ . There is visible symmetry about the
Figure 33. Symmetry in the graph of
1
+
(tan
x) 2
π 1
point
,
. If we cut out the rectangle and cut along the graph of the function, then we get
4 2
two pieces of paper which overlap perfectly, that is they are congruent.
Let f (x) :=
f
1
√
2
,
x ∈ [0, π/2). Let 0 < y <
π
. Then
4
1 + (tan x)
1
1
√ =
+y
=
√2
2
4
π
tan
y
+
1
1 + tan y +
1+
4
1 − tan y
π
√
2
=
(1 − tan y)
√
=1−
√
2
√
(1 + tan y) 2 + (1 − tan y) 2
+ (1 +
π
1
1
√ =1−
√ =1−f
= 1−
−
y
.
π
2
2
4
1 − tan y
1 + tan
−y
1+
4
1 + tan y
(1 − tan y)
2
√
tan y) 2
√
(1 + tan y)
This symmetry property implies that
Z π2
π π 1
1
√ dx =
· Area of the rectangle formed by (0, 0),
,0 ,
, 1 , (0, 1)
2
2
2
1 + (tan x) 2
0
1 π
π
=
· ·1 = .
2 2
4
147
Solutions to the exercises from Chapter 5
Solution to Exercise 5.6.28. By the Differentiable Inverse Theorem, we have that at y = sin x ∈
(−1, 1),
1
1
1
1
.
= p
=
(sin−1 )′ (y) =
=p
2
cos x
sin′ x
1 − (sin x)
1 − y2
In order to find the integrals, we use Integration by Substitution/Change of Variables:
t = sin u,
dt = cos udu,
t = 0 ⇒ u = 0,
t = y ⇒ u = sin−1 y.
See Figure 34.
Figure 34. The change of variable t = sin u.
Thus
Z
y
0
Also,
Z
0
y
1
√
dt =
1 − t2
p
1 − t2 dt
=
Z
sin−1 y
0
Z
0
=
=
=
=
1
· cos udu =
cos u
Z
sin−1 y
1du = sin−1 y.
0
sin−1 y
(cos u) · cos udu =
Z
0
sin−1 y
cos(2u) + 1
du
2
−1
sin(2u) sin y sin−1 y
+
4
2
0
sin(2 sin−1 y) sin−1 y
+
4
2
2(sin(sin−1 y))(cos(sin−1 y)) sin−1 y
+
4
2
p
−1
2
y 1−y
sin y
+
.
2
2
Solution to Exercise 5.6.29. See Figure 35.
√
(1) The point with Cartesian coordinates (1, 1) has polar coordinates ( 2, π4 ).
(2) The point with Cartesian coordinates (1, 0) has polar coordinates (1, 0).
(3) The point with Cartesian coordinates (0, 1) has polar coordinates (1, π2 ).
(4) The point with Cartesian coordinates (−1, 0) has polar coordinates (1, π).
148
Solutions
1
−1
0
1
−1
Figure 35. Exercise 5.6.29.
√
(5) The point with Cartesian coordinates (−1, −1) has polar coordinates ( 2, − 3π
4 ).
(6) The point with Cartesian coordinates (0, −1) has polar coordinates (1, − π2 ).
Solution to Exercise 5.7.1. Suppose that we choose the origin of time t to be the instant it
started snowing, and let T be the time at which the plough went out. Let x(t) be the distance the
snow plough has traversed up to time t. Thus x(T ) = 0. Let h(t) denote the height of the snow
at time t.
We have been given that there are constants c, k such that h′ (t) = c for all t ≥ 0, and that
x (t)h(t) = k for all t ≥ T . Hence h(t) = h(t) − 0 = h(t) − h(0) = ct, and so
′
x′ (t) =
k
,
ct
which gives
x(t) = x(t) − 0 = x(t) − x(T ) =
where α :=
k
t
t
log = α log ,
c
T
T
k
. But we know that
c
2 =
3 =
T +1
, and
T
T +2
α log
,
T
α log
T +2
log
3
(T + 1)3
(T + 2)2
T
so that =
giving
=
. Thus T 2 + T − 1 = 0, and as T ≥ 0, we have
3
T +1
2
T
T2
log
T
√
5−1
≈ 0.618 hours ≈ 37 minutes, 5 seconds.
T =
2
So it started snowing at about 7 : 22 : 55 a.m.
149
Solutions to the exercises from Chapter 5
Solution to Exercise 5.7.2. We have that the area enclosed is
Z π2 p
Z a r
x2
b 1 − (sin θ)2 a cos θdθ (using the substitution x = a sin θ)
4
b 1 − 2 dx = 4
a
0
0
Z π2
Z π2
1 + cos(2θ)
2
= 4ab
dθ
(cos θ) dθ = 4ab
2
0
0
π
1π
sin(2θ) π2
= 4ab 0 +
= πab.
= 4ab
+
2
22
4
0
When a = b, we recover the expression for the area of the disc (πa2 ).
Solution to Exercise 5.7.3. Let (b, c) be the rightmost intersection point, and (a, b) be the
leftmost intersection point. We wish to find c such that
Z a
Z b
(c − (2x − 3x3 ))dx =
(2x − 3x3 − c)dx,
0
and so
Z
0
b
a
(2x − 3x3 − c)dx = 0, that is,
3
b2 − b3 = bc.
4
(6.34)
2b − 3b3 = c.
(6.35)
But also as (b, c) is an intersection point,
2
9
Multiplying (6.35) by b and subtracting (6.34) gives b2 − b4 = 0, and so b = . Hence
4
3
8
4
2
c = 2b − 3b3 = 2 · − 3 ·
= .
3
27
9
To validate the solution, we check that (2/3, 4/9) is indeed the rightmost intersection point of
y = 4/9 and y √
= 2x − 3x3 : the zeros of 2x − 3x3 − 4/9 = (2/3 − x)(3x2 + 2x − 2/3) other than
2/3 are (−1 ± 3)/2, which are less than 2/3.
Solution to Exercise 5.7.4. We have that the area enclosed is
Z π4
Z π4
sin(2θ) π4
2 cos(2θ)
1
4
dθ = 4
cos(2θ)dθ = 4
= 4 · = 2.
2
2
2
0
0
0
Solution to Exercise 5.7.5.
√ of revolution of the planar region bounded
√ The doughnut is the solid
by the two curves x 7→ R + r2 − x2 and x 7→ R − r2 − x2 . Thus the volume is given by
Z r
p
p
π (R + r2 − x2 )2 − (R − r2 − x2 )2 dx
−r
Z r p
p
=
π r2 − x2 + R2 + 2R r2 − x2 − R2 − (r2 − x2 ) + 2R r2 − x2 dx
−r
Z r p
= 4πR
r2 − x2 dx.
Z
−r
p
πr2
.
r2 − x2 dx is the area of a semicircular disc with radius r, and so this equals
2
−r
So the volume of the doughnut is
Z r p
Z r p
p
πr2
r2 − x2 dx = 4πR ·
= 2π 2 Rr2 .
π (R + r2 − x2 )2 − (R − r2 − x2 )2 dx = 4πR
2
−r
−r
But
r
150
Solutions
Solution to Exercise 5.7.6. The volume is given by
!
r
Z a
Z a
x2
x2 2
2
2
πb 1 − 2 dx
π
b 1 − 2 − 0 dx =
a
a
−a
−a
Z a
x2
2
1 − 2 dx
= 2πb
a
0
3
2
4π 2
a
b a.
= 2πb2 a − 2 = 2πb2 a =
2a
3
3
If a = b, then we recover the expression for the volume of a sphere,
4 3
πa .
3
Solution to Exercise 5.7.7. See Figure 36, from which we see that the remaining
portion of the
√
√
2
ball is a solid of revolution of the planar region bounded by the curves x 7→ 4 − x and x 7→ 3
for x ∈ [−1, 1].
1
√
3
−1
2
1
Figure 36. Planar region.
Thus the volume of the remaining portion is
Z 1 p
Z 1
Z
√ 2
π ( 4 − x2 )2 − 3 dx =
π(4 − x2 − 3)dx = π2
−1
1 4π
=
.
(1 − x2 )dx = 2π 1 −
3
3
−1
−1
1
Thus the volume cut out is the volume of the sphere minus the volume of the remaining portion,
that is,
28π
4 3 4π
π2 −
=
.
3
3
7
Solution to Exercise 5.7.8. We have
x′ (t) =
y ′ (t) =
√
13
(2t + 3)1/2 2 = 2t + 3,
32
2t
+ 1 = t + 1,
2
and so (x′ (t))2 + (y ′ (t))2 = 2t + 3 + t2 + 2t + 1 = t2 + 4t + 4 = (t + 2)2 . Hence the distance travelled
is
2
Z 3
Z 3p
t
21
9
3
(t + 2)dt =
(x′ (t))2 + (y ′ (t))2 dt =
+ 2t = + 6 =
.
2
2
2
0
0
0
Thus the average speed is
distance travelled
21/2
7
=
= .
time taken
3
2
151
Solutions to the exercises from Chapter 5
Solution to Exercise 5.7.9. The curve t 7→ γ(t) = (x(t), y(t)), t ∈ [−π, π] is given by
x(t)
= r(t) cos t = 2(1 + cos t) cos t, and
y(t)
= r(t) sin t = 2(1 + cos t) sin t.
Thus
x′ (t)
′
y (t)
= 2(− sin t) cos t + 2(1 + cos t)(− sin t) = −2 sin t − 4 sin t cos t = −2 sin t − 2 sin(2t),
= 2(− sin t) sin t + 2(1 + cos t) cos t = 2 cos t + 2((cos t)2 − (sin t)2 ) = 2 cos t + 2 cos(2t).
Hence
(x′ (t))2 + (y ′ (t))2
= 4 (sin t)2 + (sin(2t))2 + 2(sin t)(sin(2t)) + (cos t)2 + (cos(2t))2 + 2(cos t)(cos(2t))
= 8 1 + (cos(2t)) cos t + 2(cos t) cos(2t)
t 2
.
= 8(1 + cos t) = 8 · 2 · cos
2
Z π p
Z π
t
t π
Thus the arc length is
(x′ (t))2 + (y ′ (t))2 dt =
4 cos dt = 8 sin = 16.
2
2 −π
−π
−π
Solution to Exercise 5.7.10. Consider the curve γ : [0, 2π] → R2 given by
t ∈ [0, 2π].
γ(t) = (a cos θ, b sin θ),
Then γ(t) goes round the perimeter of the ellipse once as t increases from 0 to 2π. We have that
the arclength of γ is
Z 2π p
Z 2π p
(−a sin θ)2 + (b cos θ)2 dθ =
b2 + (a2 − b2 )(sin θ)2 dθ
0
0
Z 2π s
a2
= b
1 − 1 − 2 (sin θ)2 dθ
b
0
Z 2π p
1 − k 2 (sin θ)2 dθ,
= b
where k :=
r
1−
0
a2
.
b2
Solution to Exercise 5.7.11. The surface area is given by
Z
Z 2π
p
2
2
2π(R + r sin t) (−r sin t) + (r cos t) dt =
2π
2π(R + r sin t)rdt
0
0
=
=
2π 2πr R · 2π + r(− cos t)
0
2πr(R · 2π + 0) = 4π 2 Rr.
Solution to Exercise 5.7.12. Suppose that the radius is changed from r to r + δr. Then we
want to know what the largest M is such that
A(r + δr) − A(r) |δr|
≤ 0.03,
≤ M ⇒ r
A(r)
where A(ρ) := 4πρ2 for ρ ≥ 0. We have
A(r + δr) − A(r)
δr δr 2
4π(r2 + 2rδr + (δr)2 − r2
= 2θ + θ2 ,
=2 +
=
2
A(r)
4πr
r
r
152
Solutions
where θ :=
δr
. But
r
√
√




2
−1 − 1.03 ≤ θ ≤ 1.03 − 1

 2θ + θ ≤ 0.03 

|2θ + θ2 | ≤ 0.03 ⇔
⇔
and
and
√
√




2θ + θ2 ≥ −0.03
θ ≤ −1 − 0.97 or − 1 + 0.97 ≤ θ.
0.03
−0.03
−2
So 1 +
0
√
√
√
√
0.97 ≤ θ 1.03 − 1 or −1 − 1.03 ≤ θ ≤ −1 − 0.97. Hence the largest value of M is
√
√
max{| − 1 + 0.97|, | 1.03 − 1|} ≈ max{0.0151, 0.0149} = 0.0151.
So we need to know the value of the radius to 1.51% of its exact value.
Solution to Exercise 5.7.13. We have that the surface area is
r
Z a
x
x 2
1 + sinh
A :=
2πa cosh
dx
a
a
0
Z a
Z a
x
x 2
x cosh
dx =
2πa cosh
dx,
=
2πa cosh
a
a
a
0
0
Z a
x 2
dx, and so
while the volume is given by V :=
πa2 cosh
a
0
Z a
x 2
dx
2πa cosh
A
2
a
0
= .
= Z a
2
x
V
a
dx
πa2 cosh
a
0
Solution to Exercise 5.7.14. The volume when the height is h is given by
Z h
V =
π(y 1/m )2 dy.
0
Thus by the Fundamental Theorem of Calculus and by the Chain Rule
dV dh
dh
dV
=
·
= π(h1/m )2 ·
.
dt
dh dt
dt
√
dV
But since
is proportional to h and as we require that
dt
dh
= constant,
dt
(since we want the height to decrease linearly with time), we must have that
2
1
= ,
m
2
that is, m = 4.
153
Solutions to the exercises from Chapter 6
Solutions to the exercises from Chapter 6
Solution to Exercise 6.1.1. We have
(2n + 1) − (2n − 1)
1
1
1
= tan−1
− tan−1
,
tan−1 2 = tan−1
2n
1 + (2n + 1)(2n − 1)
2n − 1
2n + 1
and so the partial sums telescope to give
∞
X
tan−1
n=1
1
1
π
π
1
= tan−1 − lim tan−1
= −0= .
2
n→∞
2n
1
2n + 1
4
4
Solution to Exercise 6.1.2. We have
1
4
2n
1
2
+
+
·
·
·
+
+
+
1 − x 1 + x 1 + x2
1 + x4
1 + x2n
2
2
4
2n
=
+
+
+ ···+
2
2
4
1−x
1+x
1+x
1 + x2n
n
4
2
4
+
+ ··· +
=
1 − x4
1 + x4
1 + x2n
..
.
2n
2n
2n+1
+
=
.
1 − x2n
1 + x2n
1 − x2n+1
Since x > 1, we can write x = 1 + h with h > 0, and so for integers k > 2 we have
k · (k − 1) 2
k 2
h .
xk = (1 + h)k >
h =
2
2
=
Thus
0<
and so by the Sandwich Theorem,
2
k
<
,
k
x
(k − 1) · h2
lim
k→∞
k
= 0.
xk
Hence for x > 1, we obtain
1
4
2n
1
2
+
+
·
·
·
+
+
+
+ ... =
1 − x 1 + x 1 + x2
1 + x4
1 + x2n
=
=
and so
1
2
4
2n
1
+
+
+
·
·
·
+
.
n + ··· =
2
4
2
1+x 1+x
1+x
1+x
x−1
2n+1
n+1
n→∞ 1 − x2
lim
lim
n→∞
2n+1
x2n+1
1
x2n+1
0
= 0,
0−1
−1
Solution to Exercise 6.1.3. We have
1
1
1
1
1
1
1
=
−
=
−
.
=
Fn−1 Fn+1
Fn−1 (Fn + Fn−1 )
Fn−1
Fn + Fn−1 Fn
Fn−1 Fn
Fn Fn+1
It is easy to see by induction that Fn ≥ n for all n ∈ N. Indeed, F1 = 1 ≥ 1, and if Fn ≥ n for
some n, then Fn+1 = Fn−1 + Fn ≥ 1 + Fn ≥ 1 + n. (Here the first inequality follows from the
obvious fact that n 7→ Fn is increasing, so that F1 = 1 ≤ F2 ≤ F3 ≤ · · · Fn−1 .) Thus
n
X
k=2
1
1
1
1
1
1
1
1
1
n→∞ 1
=
−
+
−
+ · · ·+
−
=
−
−→ .
Fk−1 Fk+1 F1 F2 F2 F3 F2 F3 F3 F4
Fn−1 Fn Fn Fn+1
F1 F2 Fn Fn+1
2
154
Solutions
Solution to Exercise 6.1.4. The sequence (1/n)n∈N is convergent with limit 0. The function
cos : R → R is continuous. Since continuous functions preserve convergence of sequences, it follows
that (cos(1/n))n∈N is convergent with limit cos 0 = 1. So we can conclude that since
1
= 1 6= 0,
lim cos
n→∞
n
1
does not converge.
the series
cos
n
n=1
∞
X
Solution to Exercise 6.1.5. Let sn := a1 + · · · + an denote the nth partial sum of the series for
n ∈ N. Since the series converges, (sn )n∈N converges to some limit L. So the sequence (s2n −sn )n∈N
converges to L − L = 0. We have
s2n − sn = an+1 + · · · + a2n ≥ a2n + · · · + a2n = n · a2n ≥ 0.
|
{z
}
n times
Thus by the Sandwich Theorem, we obtain lim n · a2n = 0. Hence also
n→∞
lim 2na2n = 0.
n→∞
(6.36)
Also, the sequence (s2n+1 − sn+1 )n∈N converges to L − L = 0. We have
s2n+1 − sn = an+2 + · · · + a2n+1 ≥ a2n+1 + · · · + a2n+1 = n · a2n+1 ≥ 0.
{z
}
|
n times
Thus by the Sandwich Theorem, we obtain lim n · a2n+1 = 0. As the sequence (an )n∈N converges
n→∞
to 0, we also have that (a2n+1 )n∈N converges to 0. Hence we obtain
lim (2n + 1)a2n+1 = 2 lim na2n+1 + lim a2n+1 = 2 · 0 + 0 = 0.
n→∞
n→∞
n→∞
(6.37)
It follows now from (6.36) and (6.37) that lim nan = 0.
n→∞
In order to show that the assumption a1 ≥ a2 ≥ a3 . . . cannot be dropped, we consider the
lacunary series whose n2 th term is 1/n2 and all other terms are zero:
a1 =
1
1
1
, a2 = a3 = 0, a22 = 2 , a22 +1 = · · · = a32 −1 = 0, a32 = 2 , . . . .
2
1
2
3
The sequence (sn )n∈N of partial sums is clearly increasing since an ≥ 0 for all n ∈ N. We have
sn2 =
1
1
1
1
+ 2 + 2 + ···+ 2,
12
2
3
n
which is the nth partial sum of the convergent series
∞
X
1
.
n2
n=1
It thus follows from the convergence of (sn2 )n∈N that (sn2 )n∈N is bounded, and so also (sn )n∈N
is bounded too (after all, given any m ∈ N, there is a perfect square N 2 exceeding it, so that
sm ≤ sN 2 ≤ M , where M is a bound for the sequence (sn2 )n∈N ). As the sequence (sn )n∈N is
monotone and bounded, it is convergent (by the Bolzano-Weierstrass Theorem). Note however
that the sequence (nan )n∈N has as a subsequence (n2 an2 )n∈N = (n2 n12 )n∈N = (1)n∈N , and so it
follows that (nan )n∈N does not converge to 0.
155
Solutions to the exercises from Chapter 6
Solution to Exercise 6.1.6. For n ∈ N, define sn := 1 + 2r + 3r2 + · · · + nrn−1 . Then
rsn = r + 2r2 + · · · + (n − 1)rn−1 + nrn ,
and so
(1 − r)sn
= sn − rsn = 1 + r + r2 + · · · + rn−1 − nrn
1 − rn
(1 − r)(1 + r + r2 + · · · + rn−1
− nrn =
− nrn .
1−r
1−r
=
Thus sn =
nrn
1 − rn
1
−
. Let h :=
− 1 > 0. Then for n ≥ 2,
2
(1 − r)
1−r
|r|
n
n 2
n n
n(n − 1) 2
n 2
(1 + h)n = 1 +
h+
h + ···+
h ≥
h =
h .
1
2
n
2
2
So for n ≥ 2,
0 ≤ n|r|n =
2
2
n
≤n
=
.
(1 + h)n
n(n − 1)h2
(n − 1)h2
Hence by the Sandwich Theorem, lim |nrn | = 0. So lim nrn = 0 as well, and thus
n→∞
lim sn = lim
n→∞
n→∞
n→∞
n
n
nr
1−r
−
(1 − r)2
1−r
=
1−0
0
1
−
.
=
(1 − r)2
1−r
(1 − r)2
Solution to Exercise 6.2.1. Since −1 ≤ sin n ≤ 1 for all n ∈ N, we have
sin n 1
n2 ≤ n2 (n ∈ N).
∞ X
sin n Thus the sequence (sn )n∈N of the partial sums of the series
n2 is bounded:
n=1
sn ≤
∞
X
1
1
1
1
+
+
·
·
·
+
≤
< +∞.
12
22
n2 n=1 n2
Moreover, it is an increasing sequence, and so by the Bolzano-Weierstrass Theorem (sn )n∈N is
convergent, that is,
∞ X
sin n n2 < +∞.
n=1
As the series converges absolutely, it is convergent, that is,
∞
X
sin n
converges.
n2
n=1
(When we learn the “Comparison Test” later, we can give a one line justification.)
Solution to Exercise 6.2.2. We have
1
(1) an := s ≥ 0 for all n ∈ N.
n
(2) As s > 0, (n + 1)s > ns and so
(3) lim
n→∞
1
1
>
for all n ∈ N.
s
n
(n + 1)s
1
= 0.
ns
∞
X
(−1)n
converges.
Thus by the Leibniz Alternating Series Theorem
ns
n=1
156
Solutions
Solution to Exercise 6.2.3. We have
√
n
≥ 0 for all n ∈ N.
(1) an :=
n+1
√
√
n
n+1
(2) (We want to know if for all n ∈ N, n+1
≥ n+2
. In order to see this, we square both
sides, and rearrange terms to arrive at the equivalent inequality n2 + n ≥ 1, which is
clearly true.)
We note that for all n ∈ N, n2 + n ≥ 1, and so
n(n + 2)2 = n3 + 42 + 4n ≥ n3 + 3n2 + 3n + 1 = (n + 1)3 .
Rearranging, we obtain
n
n+1
≥
,
2
(n + 1)
(n + 2)2
√
√
n
n+1
≥
for all n ∈ N.
and finally, taking square roots, we obtain
n+1
n+2
√
√1
n
0
n
(3) lim
= lim
= 0.
=
n→∞ n + 1
n→∞ 1 + 1
1
+
0
n
√
∞
X
n
n
converges.
Thus by the Leibniz Alternating Series Theorem
(−1)
n+1
n=1
Solution to Exercise 6.2.4. We have
(1) We know that sin x ≥ 0 for all x ∈ [0, π]. Since 0 ≤
that an := sin n1 ≥ 0 for all n ∈ N.
1
n
(2) The function x 7→ sin x : [0, π2 ] is increasing. Since 0 <
1
for all n ∈ N.
sin n1 ≤ sin n+1
≤ 1 < π for all n ∈ N, it follows
1
n
<
1
n+1
<1<
π
2,
it follows that
(3) Finally, the function x 7→ sin x : R → R is continuous, and the sequence ( n1 )n∈N is
convergent with limit 0, and so it follows that
1
1
= sin 0 = 0.
lim sin = sin lim
n→∞ n
n→∞
n
Thus by the Leibniz Alternating Series Theorem
∞
X
(−1)n sin
n=1
1
converges.
n
Solution to Exercise 6.3.1. We will use the Ratio Test in each part.
n2
(1) We have that an := n 6= 0 for each n ∈ N, and
2
2
an+1 (n + 1)2 2n
1
1
=
1
+
,
·
=
an 2n+1
n2
2
n
2
an+1 1
1
= lim 1 1 + 1
and so lim = (1 + 0)2 = < 1. Thus the series
n→∞
an n→∞ 2
n
2
2
∞
X
n2
2n
n=1
converges (absolutely).
(n!)2
6= 0 for each n ∈ N, and
(2) We have that an :=
(2n)!
an+1 ((n + 1)!)2 (2n)!
1 + n1
(n + 1)2
n+1
=
=
·
=
=
,
an (2n + 2)! (n!)2
(2n + 1)(2n + 2)
2(2n + 1)
2(2 + n1 )
157
Solutions to the exercises from Chapter 6
1
an+1 = lim 1 + n = 1 + 0 = 1 < 1. Thus the series
and so lim n→∞
an n→∞ 2(2 + n1 )
2(2 + 0)
4
∞
X
(n!)2
(2n)!
n=1
converges (absolutely).
n
4
(3) We have that an :=
n5 6= 0 for each n ∈ N, and
5
n+1
5
5
n
an+1 1
1
4
4 n+1
5
5
= 4
1
+
(n
+
1)
·
=
,
=
an 5
4
n5
5
n
5
n
5
an+1 1
4
4
4
1+
= lim
= (1 + 0)5 = < 1. Thus the series
and so lim n→∞
an n→∞ 5
n
5
5
∞
n
X
4
n5
5
n=1
converges (absolutely).
n
1
n
≤ 4 = 3 for all n ∈ N. Since
n4 + n2 + 1
n
n
∞
∞
X
X
n
1
< +∞, it follows from the Comparison Test that also
converges.
3
4 + n2 + 1
n
n
n=1
n=1
Solution to Exercise 6.3.2. We have that
We have
∞
X
n
4 + n2 + 1
n
n=1
=
=
=
=
∞
X
n
2 + 1)2 − n2
(n
n=1
∞
X
n
2 + 1 − n)(n2 + 1 + n)
(n
n=1
∞ 1X
1
1
−
2 n=1 n2 + 1 − n n2 + 1 + n
∞ 1X
1
1
.
−
2 n=1 (n − 1) · n + 1 n · (n + 1) + 1
But the partial sum of this last series is the “telescoping sum”
1
1 ✘✘
1 ✟
1 ✟
1 ✟
1
− ✟✟ + ✟✟ − ✟✟ + · · · + ✘✘✘✘
−
✘
0·1+1 ✟
1·2+1 ✟
1·2+1 ✟
2·3+1
(n − 1) · n + 1 n · (n + 1) + 1
1
.
=1−
n · (n + 1) + 1
∞
X
n
1
1
1
Consequently,
=
lim 1 −
= .
4
2
n→∞
n +n +1
2
n · (n + 1) + 1
2
n=1
Solution to Exercise 6.3.3. Since the series
∞
X
a4n converges, we have necessarily that
n=1
lim a4
n→∞ n
= 0.
But then lim |an |4 = 0 as well, and so it follows from here that
n→∞
lim |an | = 0.
n→∞
158
Solutions
Thus there is an N ∈ N such that for all n > N , |an | < 1. But now for n > N ,
|a5n | = a4n |an | < a4n ,
and so it follows from the Comparison Test that
∞
X
a5n converges too.
n=1
Solution to Exercise 6.3.4. Clearly numbers n such that 10k−1 ≤ n < 10k have k digits. If
we are to avoid the digit 9 in any of these places, then we see that the first digit can only be
1, 2, 3, . . . , 7, 8, while the rest can be any of the digits 0, 1, 2, 3, . . . , 7, 8. Thus the total number of
X′ 1
, there are
allowed numbers is 8 · 9k−1 . Thus in our series
np
1
1
1
8
terms with
< p ≤ p,
p
10
n
1
1
1
1
< p ≤ p,
8·9
terms with
102p
n
10
1
1
1
8 · 92 terms with
< p ≤ 2p ,
103p
n
10
..
.
1
1
1
k−1
< p ≤ (k−1)p ,
8·9
terms with
kp
10
n
10
..
.
Thus
8
1
9
92
+ 2p + 3p + . . .
p
10
10
10
We now have the following:
<
X′ 1
9
92
93
1
≤
8
+
+
+
+
.
.
.
.
np
1p
10p
102p
103p
(6.38)
9
< 1. Thus the geometric series on the right hand
10p
X′ 1
side of (6.38) converges. So by the Comparison Test,
converges as well.
np
9
2◦ p ≤ log10 9. Then 9 ≥ 10p and so p ≥ 1. Thus the geometric series on the left hand
10
X′ 1
side of (6.38) diverges. So by the Comparison Test,
diverges as well.
np
1◦ p > log10 9. Then 9 < 10p and so
Solution to Exercise 6.3.5.
(1) True.
Since the series
∞
X
n=1
|an | converges, we have necessarily that
lim |an | = 0.
n→∞
Thus there is an N ∈ N such that for all n > N , |an | < 1. But now for n > N ,
|a2n | = |an ||an | < |an |,
and so it follows from the Comparison Test that
∞
X
a2n converges too.
n=1
(2) False.
By the Leibniz Alternating Series Theorem, it is easy to check that
∞
X
1
(−1)n √
n
n=1
159
Solutions to the exercises from Chapter 6
2
∞
∞ X
X
1
n 1
is the harmonic series
, which is divergent.
(−1) √
converges. But
n
n
n=1
n=1
(3) False.
The harmonic series
(4) True.
∞
X
1
1
is divergent, but lim
= 0.
n→∞
n
n
n=1
The partial sums converge to 0, and so the series
∞
X
an converges to 0, by definition.
n=1
(5) False.
The partial sum
2
3
n+1
2 · 3 · · · · · (n + 1)
+ log + · · · + log
= log
= log(n + 1).
1
2
n
1 ·2 · ···· n
∞
X
n+1
Since (log(n + 1))n∈N diverges, it follows that
log
diverges.
n
n=1
sn = log
(6) True.
We know that increasing sequences that are bounded above are convergent. As the
sequence of partial sums of the given series is increasing and bounded above, the series
is convergent.
(7) True.
Since
∞
X
an converges, lim an = 0. This implies that
n→∞
n=1
hence can’t be convergent to 0. So
∞
X
1
diverges.
a
n=1 n
1
an
is unbounded, and
n∈N
Solution to Exercise 6.3.6. Let ǫ > 0. Since the series
∞
X
S :=
|an | + |bn |
n=1
is convergent, there exists an N ∈ N such that for all n > N ,
n
∞
X
X
(|ak | + |bk |) < ǫ.
(|ak | + |bk |) − S =
k=1
k=n+1
As
ak cos 2πk x + bk sin 2πk x ≤ |ak | cos 2πk x + |bk | sin 2πk x ≤ |ak | · 1 + |bk | · 1,
T
T
T
T
we have that
∞
∞
X
X
ak cos 2πk x + bk sin 2πk x ≤
(|ak | · 1 + |bk | · 1) < ǫ.
T
T
k=n+1
k=n+1
Thus if n > N , then for all x ∈ R, we have
n X
2πk
2πk
ak cos
x + bk sin
x
f (x) − a0 +
T
T
k=1
∞
X
2πk 2πk
x + bk sin
x = ak cos
T
T
k=n+1
∞
X
ak cos 2πk x + bk sin 2πk x < ǫ.
≤
T
T
k=n+1
160
Solutions
This shows that the Fourier series converges uniformly to f .
The plots are displayed on page 44. From the plots, we do see that the partial sum functions
look increasingly like the square wave, but we also notice the persistent overshoot at the points
−1, 0, 1, 2, 3, suggesting nonuniform convergence.
Solution to Exercise 6.3.7. It is clear that if the series
∞
X
an converges, then so does
∞
X
an+1 .
n=1
n=1
∞
X
an + an+1
Thus
converges too. By the Arithmetic Mean-Geometric Mean inequality
2
n=1
a+b √
≥ ab
2
√
√
for nonnegative a, b (this is just a rearrangement of the trivial observation that ( a − b)2 ≥ 0),
it follows that for all n ∈ N,
an + an+1
√
,
an an+1 ≤
2
and so the result follows from the Comparison Test.
Solution to Exercise 6.3.8. (“If” part) Suppose that
lim
n→∞
∞
X
an
converges. Then
1
+
an
n=1
1
an
= 1 − lim
= 0,
n→∞ 1 + an
1 + an
1
= 1. Thus there exists an N ∈ N such that for all n > N ,
1 + an
1
1
> .
1 + an
2
But this implies that for n > N ,
an
an
>
.
1 + an
2
∞
∞
X
X
an
By the Comparison Test, it follows that
converges, and so
an converges as well.
2
n=1
n=1
and so lim
n→∞
(“Only if” part) Suppose that
∞
X
an converges. Since the an s are all positive, we have
n=1
an
< an (n ∈ N),
1 + an
∞
X
an
and so by the Comparison Test, it follows that
converges.
1
+
an
n=1
Solution to Exercise 6.3.9. If the series
∞
X
n=1
|an | converges, we have lim |an | = 0. Thus there
n→∞
|a2n |
is an N ∈ N such that for all n > N , |an | < 1. But now for n > N ,
= |an ||an | < |an |, and so
∞
X
by the Comparison Test
a2n converges too. So every element of ℓ1 belongs to ℓ2 .
n=1
The sequence (1/n)n∈N belongs to ℓ2 since
∞
∞ 2
X
X
1
1
=
< +∞,
2
n
n
n=1
n=1
but it does not belong to ℓ1 (since the harmonic series diverges). So ℓ1 ( ℓ2 .
161
Solutions to the exercises from Chapter 6
Solution to Exercise 6.3.10. We have
1
1 1
sn := 1 + + + · · · + < 1 + 1 + 1 + · · · + 1 = n,
{z
}
2 3
n |
n times
and so
∞
X
1
1
1
for all n ∈ N. Hence by the Comparison Test, the series
diverges.
<
n
sn
s
n=1 n
Solution to Exercise 6.3.11. We have that
r
1
1
1
n
= ≤ =r<1
nn
n
2
for all n ≥ 2. So it follows from the Root Test that the series
∞
X
1
converges.
n
n
n=1
Solution to Exercise 6.3.12. We know that the Fibonacci sequence is strictly increasing, since
each term is clearly seen to be positive, and the recurrence gives Fn+1 = Fn + Fn−1 > Fn + 0 = Fn .
From here, we obtain
Fn+1 = Fn + Fn−1 > Fn−1 + Fn−1 = 2Fn−1 .
Thus
1/Fn+1
Fn−1
1
=
< =: r < 1.
1/Fn−1
Fn+1
2
1
1
1
1
1
1
+
+
+ . . . and
+
+
+ . . . converge. So if
So by the Ratio Test, both
F0
F2
F4
F1
F3
F5
sn :=
1
1
1
1
+
+
+ ···+
,
F0
F1
F2
Fn
n ∈ N,
then both (s2n )n∈N and (s2n+1 )n∈N converge to the same limit, namely
∞
∞
X
X
1
1
+
,
F
F
n=0 2n
n=0 2n+1
and so (sn )n∈N converges, that is,
∞
X
1
converges.
F
n=0 n
Solution to Exercise 6.3.13. As k 2 is an integer, we have
p
p
p
sin π k 4 + 1 = sin πk 2 1 + 1/k 4 = sin πk 2 1 + 1/k 4 − πk 2 !
p
p1 + 1/k 4 − 1 π
= sin πk 2
1 + 1/k 4 − 1 = sin
.
k2
1/k 4
But by the Mean Value Theorem, we have for x > 0 that there exists a cx ∈ (0, 1) such that
√
1
1+x−1
1
= √
< .
x
2
2 1 + cx
Consequently,
p
1 + 1/k 4 − 1
1
< .
4
1/k
2
Also, for each x 6= 0, we have that there exists a c between 0 and x such that
sin x − sin 0 = | cos c| ≤ 1,
x−0
162
Solutions
so that | sin x| ≤ |x| for all x ∈ R. Hence
! p
p
p
π 1
π 1 + 1/k 4 − 1 π 1 + 1/k 4 − 1 4
≤ 2 · .
≤ 2
sin π k + 1 = sin
2
4
4
k 2
k
k
1/k
1/k
Since
∞
X
1
converges, it follows by the Comparison Test that
k2
k=1
∞ p
X
sin π k 4 + 1 k=1
converges. Hence
∞
X
k=1
p
sin π k 4 + 1 converges absolutely.
Solution to Exercise 6.4.1. We have for x ∈ R and n ≥ 0 that
(−1)n/2 sin x
if n is even,
(n)
sin x =
(−1)(n−1)/2 cos x if n is odd
Hence | sin(n) x| ≤ 1 = 1n for all n ≥ 0 and all x ∈ R. Thus for all x ∈ R,
sin x =
∞
X
sin(n) 0 n
x3
x5
x =x−
+
− +··· .
n!
3!
5!
n=0
Similarly, for x ∈ R and n ≥ 0 that
cos(n) x =
(−1)n/2 cos x
(−1)(n+1)/2 sin x
if n is even,
if n is odd
Hence | cos(n) x| ≤ 1 = 1n for all n ≥ 0 and all x ∈ R. Thus for all x ∈ R,
cos x =
∞
X
cos(n) 0 n
x2
x4
x =1−
+
− +··· .
n!
2!
4!
n=0
Solution to Exercise 6.4.2.
2
(1) For x ≈ 0, −1/x2 is negative with a large absolute value, and so e−1/x ≈ 0. For
2
x → ±∞, −1/x2 → 0, and so e−1/x → 1. See Figure 37.
Figure 37. Graph of f .
163
Solutions to the exercises from Chapter 6
(2) Let m > 0. Choose n ∈ N such that n > m/2. For x 6= 0,
∞
X
1 1
1 1
≥
,
k! x2k
n! x2n
k=0
e−1/x2 n!|x|2n
x→0
≤ n!x2n . So m ≤
= n!|x|2n−m −→ 0.
x
|x|m
2
e1/x =
and so e−1/x
2
(3) We prove this by induction on n. For x 6= 0,
f ′ (x) = e−1/x
2
2 2
2
1
2x = e−1/x 3 = e−1/x P1
4
x
x
1
,
x
where P1 (t) := 2t3 . Suppose the claim holds for all natural numbers ≤ k. We now show
this for k + 1. We know
2
1
,
f (k) (x) = e−1/x Pk
x
where Pk is a polynomial. Hence
f (k+1) (x)
=
=
=
1
2
1
1
−1/x2 ′
+
e
P
−
P
k
k
x3
x
x
x2
2
1
1
2
1
+ Pk′
− 2
e−1/x
Pk
x3
x
x
x
2
1
,
e−1/x Pk+1
x
e−1/x
2
where Pk+1 (t) := 2t3 Pk (t) + Pk′ (t)(−t2 ), which is clearly a polynomial.
2
e−1/x − 0
= 0. If f (n) (0) = 0 for some n, then
x→0
x
f (n) (x) − f (n) (0)
f (n+1) (0) = lim
x→0
x
2
e−1/x Pn x1 − 0
= lim
x→0
x
2
e−1/x cd
c1
= lim
+
·
·
·
+
+
c
0
x→0
x
xd
x
(where Pn is given by Pn (t) = cd td + · · · + c1 t + c0 )
!
2
2
2
e−1/x
e−1/x
e−1/x
= lim
cd + · · · +
c1 +
c0 = 0.
x→0
xd+1
x2
x
(4) We have f ′ (0) = lim