MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 2, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. Describe, in a simpler form, A = ∪n∈N[−1/n, 1− 1/n). Prove that the union is as you described it.
Model solution:
(20 marks in total)
We have A = [−1, 1) [Marks: 2]. Let us prove this. If x ∈ A [Marks: 1] then there is an n ∈ N such
that x ∈ [−1/n, 1− 1/n) [Marks: 2]. Since n ≥ 1, we have 1/n ≤ 1 and 1− 1/n < 1 so −1 ≤ −1/n ≤ x <
1− 1/n < 1 [Marks: 3], which proves that x ∈ [−1, 1) [Marks: 2].
Assume now that x ∈ [−1, 1) [Marks: 1]. If x < 0 then x ∈ [−1/n, 1 − 1/n) for n = 1 (since [−1/n, 1 −
1/n) = [−1, 0) in that case) [Marks: 2]. Consider now that x ≥ 0. Since x < 1, we can find n ∈ N such
that x < 1 − 1/n [Marks: 2]. Indeed, this relation amounts to 1/n < 1 − x, that is, since 1 − x > 0,
1/(1 − x) < n; thus, for n large enough, x < 1 − 1/n [Marks: 2]. Then −1/n ≤ 0 ≤ x < 1 − 1/n, which
shows that x ∈ [−1/n, 1− 1/n) [Marks: 1]. We have therefore proved that x ∈ ∪n∈N[−1/n, 1− 1/n) = A
[Marks: 2], which concludes the proof that A = [−1, 1).
2. The following identity for the sum of a finite geometric series is valid for all r 6= 1 :
1 + r + r2 + r3 + ... + rn =
1− rn+1
1− r
(a) Use mathematical induction to prove this identity.
Model solution:
Let S be the set of natural numbers n for which the identity 1 + r + r2 + r3 + ...+ rn = 1−r
n+1
1−r holds.
(i) Base step: 1 ∈ S because 1−r1+11−r = 1−r
2
1−r =
(1−r)(1+r)
1−r = 1 + r.
(ii) Inductive step: Assuming k ∈ S we have to show k + 1 ∈ S. Since k ∈ S we have
1 + r + r2 + r3 + ... + rk =
1− rk+1
1− r
and so
1 + r + r2 + r3 + ... + rk+1 =
1− rk+1
1− r + r
k+1 =
1− rk+1 + rk+1(1− r)
1− r =
1− rk+2
1− r .
This proves that k + 1 ∈ S.
By induction, S = N which means the identity is valid for all natural numbers n.
(b) Do you recall another derivation of this result, say from MTH1030?
Model solution:
Let T = 1 + r+ r2 + r3 + ...+ rn. Then rT = r+ r2 + r3 + ...+ rn+1. Subtracting these expressions we
get T − rT = 1− rn+1 and thus (1− r)T = 1− rn+1. The identity follows by dividing by 1− r 6= 0.
3. The purpose of this exercise is to show that a finite union of sets that are bounded above is also bounded above.
(a) Let a and b be two real numbers. Under which condition can we say that a + b is larger than both a and b?
Model solution:
If a and b are both positive, then adding b to the inequality a ≥ 0 gives a+ b ≥ b, and adding a to the
inequality b ≥ 0 gives a + b ≥ a. Hence, a + b is larger than both a and b.
(b) Let B = {b1, . . . , bn} be a finite set. Show that B is bounded above.
Solution:
Note: claiming that “max(B)” or “max(b1, . . . , bn)” is an upper bound of B is not a valid proof since
we have not yet established that a finite set always has a maximum...
If all the elements in B are negative, then B is bounded above by 0. Assume that B has some positive
elements, and let bi1 , . . . , bir be these positive elements. Set M = bi1 + · · ·+ bir . Following the ideas in
Part (a), for all j = 1, . . . , r,
M = bi1 + · · ·+ bij−1 + bij + bij+1 + · · ·+ bir ≥ bij
since bi1 , . . . , bir are all positive. This proves that M is larger than all the positive element in B. Since
M ≥ 0, it is also larger than all the negative elements in B, and thus M is an upper bound of all the
elements of B.
(c) Let A1, . . . , An be sets that are all bounded above. Show that ∪ni=1Ai is bounded above.
Model solution:
For all i = 1, . . . , n, denote by bi an upper bound of Ai. The set B = {b1, . . . , bn} is bounded above by
some real number M (see Part (b)). Let us show that M is an upper bound of A = ∪ni=1Ai. For all
x ∈ A there exists i = 1, . . . , n such that x ∈ Ai. By definition of bi we therefore have x ≤ bi and, since
M is an upper bound of B, bi ≤M . Hence x ≤M , which proves that M is an upper bound of A.
(d) Is the result of Part (c) valid if we consider an infinite union of sets, each of which being bounded above?
Model solution:
No. Consider An = {n} for all n ∈ N. Then An is bounded above (by n itself), but ∪n∈NAn = N is
not bounded above.
4. Prove by induction on n that any set of cardinal n ≥ 1 has a maximum.
Model solution:
Base step. Let n = 1. A set A of cardinal 1 is made of only one element, which is obviously greater than all
the elements (itself!) in A and contained in A. This element is therefore the maximum of A.
Inductive step. Take n ≥ 1, assume that any set of cardinal n has a maximum, and let A be a set of cardinal
n + 1. Let x ∈ A and B = A\{x}. Then B has cardinal n and, by the induction hypothesis, B has a
maximum m. The real number m therefore belongs to B, and thus to A, and m is greater than any element
in B.
If m ≥ x then m is greater than any element in B ∪ {x} = A and m is therefore the maximum of A.
If m < x then for any y ∈ A, either y = x ≤ x, or y ∈ B, in which case y ≤ m < x. In either case, x is
greater than any element in A and, since it belongs to A, x is the maximum of A.
A therefore has a maximum and the proof is complete.
5. Let A = {n/(m+n) : m,n ∈ N}. Find, if they exist, the supremum, infimum, maximum and minimum of A. Prove
that the values you give are indeed the supremum, infimum, maximum and minimum.
Solution:
Supremum. [Initial reflexions, does not have to be presented in a solution: To make the element in A large
(in order to assess what its supremum could be), we have to make n large and m small. With m = 1, the
smallest possible in N, nn+m =
n
n+1 , which gets close to 1 as n→∞ (this is an indeterminate limit, find it
by writing nn+1 =
1
1+1/n).]
Let us show that 1 = sup(A). For all x ∈ A we have x = nn+m for some n,m ∈ N. Since m ≥ 0 we infer
n + m ≥ n and thus, dividing by n + m, 1 ≥ nn+m = x. Hence, 1 is an upper bound of A.
Let b be another upper bound of A. Then b ≥ nn+1 for all n ∈ N since nn+1 ∈ A (it corresponds to m = 1).
This gives b ≥ 11+1/n and thus (1 + 1/n)b ≥ 1, or b/n ≥ 1 − b. If b were stricly lower than 1 then we can
find n ∈ N such that b/n < 1 − b (if b ≤ 0 this is obvious, and if b > 0 take n any natural number strictly
larger than b/(1 − b): n > b/(1 − b) and 1 − b > 0 implies n(1 − b) > b and thus 1 − b > b/n). This is a
contradiction, which shows that b ≥ 1. Hence, 1 is the lowest upper bound of A, that is to say its supremum.
Note: a faster conclusion consists in saying “letting n→∞ in b ≥ 1
1+1/n
we see that b ≥ 1”. However, this assumes
that we can indeed “take the limit of an inequality”, which is actually a Theorem we will have to prove (Order Limit
Theorem).
Infimum. [Initial reflexions, does not have to be presented in a solution: We now want to make elements in
A that are as small as possible. This suggests to take n small, and m large. The smallest n is 1, so we’re
looking at 11+m for m large. This quantity tends to 0 as m→∞...]
Let us show that 0 is the infimum of A. Any x ∈ A is written nn+m for n,m positive. x is therefore positive,
and 0 is a lower bound of A.
If c is another lower bound of A then c ≤ 11+m since this real number belongs to A. Using a similar argument
as above we would reach a contradiction if c > 0: taking m strictly larger than 1/c − 1 we would find that
m + 1 > 1/c and thus c > 11+m . Hence, c ≤ 0 and 0 is the greatest lower bound of A, its infimum.
Maximum and minimum. If A had a maximum of A is would also be its supremum, that is, 1. This
maximum would have to belong to A, so there would be n,m ∈ N such that 1 = nn+m , that is, n + m = n.
This would give m = 0, which is excluded. Hence, by contradiction, A does not have a maximum.
Any element in A is written nn+m for n,m stritcly positive. Hence, any element in A is strictly positive and
the infimum 0 of A does not belong to A. This shows that A does not have a minimum.
Supplementary exercises
A. (Abbott, Exercise 1.3.6) Find, without proofs, the suprema, infima, maxima and minima (whenever they exist) of
the following sets:
(a) A = {n ∈ N : n2 < 10}
(b) B = {n/(2n + 1) : n ∈ N}
(c) C = {n/m : m,n ∈ N with m + n ≤ 10}
Model solution:
Set sup inf max min
A 3 1 3 1
B 12
1
3 none
1
3
C 9 19 9
1
9
B. (a) Suppose p is a natural number. Prove that if p2 is divisible by 5 then p is also divisible by 5. (Hint: Write
p = 5s + r with appropriate conditions on s and r.)
Model solution:
Write p = 5s + r where s ∈ N ∪ {0} and r = 0, 1, 2, 3 or 4. Then p2 = 25s2 + 10sr + r2. So if p2 is
divisible by 5 then so is r2 since 25s2 + 10sr = 5(5s2 + 2sr) is divisble by 5. But r2 = 0, 1, 4, 9 or 16.
Thus the only possibility is r2 = 0. Hence, r = 0 and p is divisible by 5.
(b) Prove that
√
5 is irrational.
Model solution:
Suppose on the contrary that
√
5 = pq where p, q are natural numbers. We may assume they have no
common factors. Then p2 = 5q2 so p2 is divisible by 5. By part a.we conclude p is divisible by 5, say
p = 5s. From p2 = 5q2 we get q2 = 5s2 showing q2 is divisible by 5. By part a. again we conclude q
is divisible by 5. As both p and q are divisible by 5 we have a contradiction. So
√
5 is not a rational
number.
(c) Suppose you try the same argument to prove
√
4 is irrational. Of course the proof must fail (since
√
4 = 2 is
rational), but exactly where does it fail?
Solution:
Suppose we try the same argument to prove
√
4 is irrational. Given that p2 is divisible by 4 it does
not follow that p is divisible by 4, just that it is divisible by 2. (Indeed if we write p = 4s + r where
s ∈ N ∪ {0} and r = 0, 1, 2 or 3 then p2 = 16s2 + 8sr + r2. So if p2 is divisible by 4 then so is r2. But
r2 = 0, 1, 4 or 9. Thus r2 = 0 or 4 and r = 0 or 2 which gives p = 4s+ r is divisible by 4 or maybe just
2.)
C. [MTH3140] (Abbott, Exercise 1.2.2) Decide which of the following represent true statements about the nature of
sets. For those that are false, provide a specific example in which the statement in question does not hold (in other
words, provide a counter-example). For those that are true, give a proof.
(a) If A1 ⊇ A2 ⊇ A3 ⊇ A4 ⊇ ... are all sets containing an infinite number of elements, then the intersection
∩∞n=1An is infinite as well.
Model solution:
False. For example if An = (0,
1
n ) then A1 ⊇ A2 ⊇ A3 ⊇ A4 ⊇ ... are all sets containing an infinite
number of elements, but the intersection ∩∞n=1An is empty.
Note: we wouldn’t ask you, when providing such a simple counter-example, to formally prove the empti-
ness of this intersection, but could you do it?.
(b) If A1 ⊇ A2 ⊇ A3 ⊇ A4 ⊇ ... are all finite, nonempty sets of real numbers, then the intersection ∩∞n=1An is
finite, nonempty as well.
Solution:
True. If A1 ⊇ A2 ⊇ A3 ⊇ A4 ⊇ ... are all finite, nonempty sets of real numbers, then at some point the
sets cannot decrease further in size. We must have a natural number k such that Ak = Ak+1 = Ak+2 =
.... Thus the intersection ∩∞n=1An = Ak which is finite and nonempty.
This “proof” is not a formal one, rather an intuitive one. In particular, the part “at some point the
sets cannot decrease further in size” is not a very precise mathematical argument (although a good
intuitive one). To formalise this intuitive argument, we need to prove that there exists indeed k ∈ N
such that Ak = Ak+1 = . . .. The idea is to look at the cardinal of the sets Ak and to use the property
of N that any non-empty subset of N has a minimum element.
Let us prove that such a k exists. The set R = {Card(Ak) : k ∈ N} is a non-empty subset of N and
therefore has a minimum element m (property of N). There exists k ∈ N such that m = Card(Ak)
(definition of the minimum). Then for any p ≥ k, we have Card(Ap) ≤ Card(Ak) (because Ap ⊂ Ak)
and Card(Ak) = m ≤ Card(Ap) (because m is the minimum of R). Thus Card(Ak) = Card(Ap) for all
p ≥ k and, since Ap ⊂ Ak, this proves that Ap = Ak for all p ≥ k.
We therefore proved that Ak = Ak+1 = . . . and the rest is as written above: the intersection A =
∩∞n=1An is reduced to A1 ∩ A2 ∩ · · · ∩ Ak (precisely because Ak = Ak+1 = . . . and there is no need
repeating the same set in an intersection), and since Ak is contained in A1, A2, . . . , Ak−1, we have in
fact A1 ∩A2 ∩ · · · ∩Ak = Ak (a). Hence, A = Ak is indeed non-empty and finite.
This proof is a bit complex, we would not have required it from you without guidance. But you should
have been able to have some form of (educated) intuition on the result.
aCould you prove this in more length? Note that, in such a complicated proof, we we would accept “A1∩A2∩· · ·∩Ak =
Ak” without further argument.
♦ More exercises from Abbott’s textbook: 1.2.4, 1.2.5, 1.2.8, 1.2.9, 1.2.10, 1.3.5.
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 3, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. Let A be non-empty bounded above and x ∈ R such that x < sup(A). Do we have x ∈ A?
Solution:
Not in general. A = [0, 1] and x = −1 give a counter-example. This is not even true if we know that “x is
close to sup(A)”; consider for example A = {1}: any x < sup(A) = 1 does not belong to A.
This kind of assertion, “we have x < sup(A) and therefore x ∈ A” (or similar false assertions), is however
often found in assignments and exams. This exercise is here to remind you to be cautious when stating such
false properties.
2. Let A be bounded above. Prove that s = sup(A) if and only if
(a) s is an upper bound of A, and
(b) for any ε > 0 there exists a ∈ A such that s− ε < a ≤ s.
Model solution:
Direction ⇒: s is the supremum of A so, by definition, it is an upper bound of A. Let ε > 0. Then s− ε < s
so s− ε is not an upper bound of A. There is therefore a ∈ A such that a > s− ε. Since s is an upper bound
of A, we also have a ≤ s, and so s− ε < a ≤ s.
Direction ⇐: s is an upper bound of A and, for any ε > 0 there exists a ∈ A such that s− ε < a ≤ s. Since
we already know that s is an upper bound of A, we only have to prove that if x < s then x is not an upper
bound of A. For such an x, define ε = s−x > 0. Then by assumption there is a ∈ A such that a > s−ε = x,
and thus x is not an upper bound of A.
3. Let an =
2n+1
3n+7 . Prove directly from the definition that an → 23 as n → ∞ (in other words, given any ε > 0 find
Nε ∈ N such that |an − a| < ε whenever n ≥ Nε).
Model solution:
We write ∣∣∣∣an − 23
∣∣∣∣ = ∣∣∣∣2n+ 13n+ 7 − 23
∣∣∣∣ = ∣∣∣∣3 (2n+ 1)− 2(3n+ 7)(3n+ 7) 3
∣∣∣∣ = 119n+ 21 .
Since 9n+ 21 ≥ 9n, this yields ∣∣∣∣an − 23
∣∣∣∣ ≤ 119n.
Let ε > 0. We want the n ∈ N such that 119n ≤ ε, which means that n ≥ 119ε . Consider therefore Nε the first
natural number greater than 119ε . If n ≥ Nε then 119n ≤ ε and thus |an − 23 | ≤ ε.
4. For each pair of sequences (un)n∈N and (vn)n∈N defined below, indicate if (vn)n∈N is a subsequence of (un)n∈N or
not.
(a) un = 1/n and vn = 1/(2n), for all n ∈ N.
Model solution:
Yes, since vn = u2n for all n ∈ N.
(b) un = 2n and vn = 2n+ 1, for all n ∈ N.
Model solution:
No: v1 = 3 but 3 is not a value taken by the sequence (un)n∈N (which only takes even values).
(c) (un)n∈N = (0, 1, 2, 3, . . .) and (vn)n∈N = (1, 0, 2, 3, . . .).
Solution:
No.
Short justification: two values of (un)n∈N have been swapped in (vn)n∈N, which is not authorised when
constructed subsequences.
Long (and more correct) justification: if (vn)n∈N were a subsequence, we would have vn = uφ(n) for
a strictly increasing function φ : N → N. Then 1 = v1 = uφ(1) and since the only index k such that
uk = 1 is k = 2, we would have φ(1) = 2. Similarly, 0 = v2 = uφ(2) means, since 0 is only attained
in the sequence (un)n∈N for the index 1, that φ(2) = 1. Hence, φ(2) < φ(1) and φ cannot be strictly
increasing.
5. Let (un)n∈N be a sequence. Prove that un → ` as n→∞ if and only if |un − `| → 0 as n→∞.
Model solution:
The definition of “|un − `| → 0 as n → ∞” is: for any ε > 0, there is Nε ∈ N such that, for all n ≥ Nε,
| |un − `| − 0| ≤ ε. The latter inequality can be simplified into |un − `| ≤ ε. Hence, the definition of
“|un − `| → 0 as n→∞” is exactly the same as the definition of “un → ` as n→∞”.
Supplementary exercises
A. Let (an) be a sequence of reals. If (|an|) converges, can we infer that (an) converges? What about the converse?
(hint: the “second” triangle inequality ||x| − |y|| ≤ |x− y| might be helpful.)
Model solution:
No to the first question. Consider an = (−1)n. Then (|an|)n∈N = (1, 1, . . .) converges to 1 but (an)n∈N
diverges.
The converse is however true. Denote by a the limit of an. The second triangle inequality yields ||an|−|a|| ≤
|an − a|. Let ε > 0 and take Nε from the definition of an → a as n → ∞. Then, for all n ≥ Nε,
||an| − |a|| ≤ |an − a| ≤ ε. This shows that |an| → |a| as n→∞.
Note: another proof is possible by using, from ||an| − |a|| ≤ |an − a|, Exercise 5 and the squeeze theorem.
B. Let (un)n∈N be a sequence that is not bounded above. Show that (un)n∈N has a subsequence that diverges to +∞.
Model solution:
Since (un)n∈N is not bounded above, 1 is not an upper bound of (un)n∈N and there exists therefore n1 ∈ N
such that un1 > 1. The real number max(2, u1, . . . , un1) is not an upper bound of (un)n∈N either and there
exists therefore n2 ∈ N such that un2 > max(2, u1, . . . , un1). This shows in particular that un2 6= uj for all
j = 1, . . . , n1 and thus that n2 is not one of these j; so n2 > n1.
Continuing the reasoning we construct a sequence nk such that unk+1 > max(k + 1, u1, . . . , unk) for all
k ∈ N, which shows that unk > k and nk+1 > nk for all k ∈ N. The second relation shows that (nk)k∈N is
increasing, so (vk)k∈N = (unk)k∈N is a subsequence of (un)n∈N. The relation vk = unk ≥ k for all k ∈ N
shows that vk → +∞ as k →∞ (use the definition, or a squeeze theorem for diverging sequences).
C. [MTH3140] Let (un)n∈N be a sequence of real numbers and l ∈ R. Translate in terms of ε and N the following
two properties.
(a) (un) does not converge to l.
Solution:
This is the negation of “(un) converges to l”, that is of “for all ε > 0, there exists N ∈ N such that for
all n ≥ N we have |un − l| ≤ ε”.
Translating the contradiction of “for all ε the property Pε is true” amounts to saying that there is at
least one ε such that Pε does not hold. Here, Pε is “there exists N ∈ N such that for all n ≥ N we
have |un − l| ≤ ε”. Hence, our translation will look like “there exists ε > 0 such that Pε is false”.
Pε has the form “there exists N ∈ N such that QN is true”. The negation of this is “for any N ∈ N,
QN is false”. Here, QN is “for all n ≥ N we have |un − l| ≤ ε”.
Negating QN is done as in the beginning: it is stated as “there exists an n ≥ N such that ”|un− l| ≤ ε”
is false”, that is “there exists an n ≥ N such that |un − l| > ε”.
To summarise, the translation of “(un) does not converge to l” is:
there exists ε > 0 such that, for any N ∈ N,
there exists n ≥ N for which |un − l| > ε.
Notice the generic pattern here:
ˆ The negation of a sentence “for all X the property PX is true” is “there exists X such that PX
is false”.
ˆ The negation of a sentence “there exists Y such that the property QY is true” is “for all Y the
property QY is false”.
ˆ The negation of a complex sentence involving consecutive “for all” or “there exists” is done by
cutting this sentence in bits and dealing with each of these terms in sequence.
(b) (un) does not converge.
Model solution:
It is equivalent to “for any real number l, (un) does not converge to l”. From the previous part, a
translation is:
for any l ∈ R, there exists ε > 0 such that, for any N ∈ N,
there exists n ≥ N for which |un − l| > ε.
D. [MTH3140] (Abbott, Exercise 1.3.3)
(a) Let A ⊂ R be non-empty and bounded below, and define
B = {b ∈ R : b is a lower bound for A}.
Prove that supB = inf A (beware: part of the proof consists in proving that these sup and inf exist...).
Solution:
B is not empty (since A has at least one lower bound) and it is bounded from above (because A is not
empty and any element in A is an upper bound for all elements in B). Hence, supB exist.
We will now prove that inf A exists and is equal to supB. One way to prove that supB is the g.l.b.
of A is to prove that it is a lower bound of A and that it is greater than any other lower bound of A.
The second part is immediate: by definition, any lower bound of A belongs to B and, therefore, is less
than or equal to supB. Let us now prove that supB itself is a lower bound for A. If supB were not a
lower bound for A, we could find a ∈ A such that supB > a. By definition of supB, a is not an upper
bound of B and we can therefore find b ∈ B such that b > a. This gives a contradiction as b ∈ B is a
lower bound for A. Hence, supB is indeed a lower bound for A.
Another way to prove that supB is the infimum of A is to show (as we did) that it is a lower bound of
A, and that for any x > sup(B) there is a ∈ A such that a < x. This can be done by saying that since
x > supB we cannot have x ∈ B (as sup(B) is greater than any element in B) and therefore x is not a
lower bound of A. There is therefore a ∈ A such that a < x, as we wanted.
(b) Use (a) to explain why there is no need to assert that infima exist as part of the Axiom of Completeness.
Model solution:
An “infimum” version of the Axiom of Completeness would say that any non-empty set bounded from
below has an infimum. We precisely saw in (a), by using the standard AoC (stated in the lecture notes),
that if A is bounded by below, it has an infimum. Hence, this “infimum” AoC can actually be deduced
from the AoC in the lecture notes.
♦ More exercises from Abbott’s textbook: 1.3.2, 1.3.4, 1.3.8, 2.2.1, 2.2.2, 2.2.3, 2.2.4, 2.2.6.
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 4, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. Let bn = sin(
npi
2 ). Prove that (bn) diverges (hint: look for particular subsequences of (bn)...)
Model solution:
The subsequence (b1, b5, b9, b13, . . .) = (1, 1, 1, 1, . . .) converges to 1 whereas the subsequence
(b4, b8, b12, b16, . . .) = (0, 0, 0, 0, . . .) converges to 0. If (bn)n∈N were converging, by the subsequence limit
theorem all its subsequences should converge to the same limit, so (bn)n∈N diverges.
2. Let an =
n+1
2n−101 .
(a) Explain why (an)n∈N converges, and find its limit. Justify your answer.
Solution:
(10 marks in total)
The easiest to do is probably to try and use the ALT [Marks: 2].
After division by the dominating term in the numerator, we obtain a n = [1+1/n] /
[2-101/n] [Marks: 3]. As n tends to infinity, 1/n -> 0 so 1+1/n->1 (ALT) [Marks: 2].
Using again the ALT, 101/n-> 0 and thus, invoking again the ALT, 2-101/n-> 2
[Marks: 1]. An ultimate usage of the ALT then shows that a n -> 1/2 [Marks: 2].
Note: we pinpointed here each and every usage of the ALT, to show you the omnipresence of this
theorem. In practice, you don’t have to recall the ALT three times in a line...
(b) Prove from the definition that (an)n∈N converges to the limit you found.
Model solution:
(15 marks in total)
We write∣∣∣∣an − 12
∣∣∣∣ = ∣∣∣∣ n+ 12n− 101 − 12
∣∣∣∣ = ∣∣∣∣2(n+ 1)− (2n− 101)2(2n− 101)
∣∣∣∣ = ∣∣∣∣ 1032(2n− 101)
∣∣∣∣ = 1032|2n− 101| . [Marks : 4]
If 2n > 101, that is n > 101/2 (or n ≥ 51 since n ∈ N), we infer∣∣∣∣an − 12
∣∣∣∣ = 1032(2n− 101) . [Marks : 2] (1)
Let ε > 0 [Marks: 2]. We want n ≥ 51 and 1032(2n−101) ≤ ε [Marks: 2], which is equivalent to
n ≥ 51 and n ≥ 1
2
(
103
2ε
+ 101
)
. [Marks : 2]
Take therefore Nε any natural number greater than 51 and
103
4ε +
101
2 [Marks: 2]. For any n ≥ Nε,
since n ≥ 51 the estimate (1) is valid and therefore |an − 1/2| ≤ ε. [Marks: 1]
Note: in retrospective, it can be noticed that n ≥ 12
(
103
2ε + 101
)
actually implies n ≥ 51.
3. (Abbott Exercise 2.5.3) Give an example of each of the following or argue that such a request is impossible.
(a) A sequence that does not contain 0 or 1 as a term but contains subsequences converging to each of these values.
Model solution:
For example, ( 12 ,
3
2 ,
1
3 ,
4
3 ,
1
4 ,
5
4 , . . .).
(b) A sequence that contains subsequences converging to every point in the infinite set {1, 12 , 13 , 14 , ...}.
Model solution:
For example (1, 12 , 1,
1
2 ,
1
3 , 1,
1
2 ,
1
3 ,
1
4 , 1,
1
2 ,
1
3 ,
1
4 ,
1
5 , ...).
(c) An unbounded sequence with a convergent subsequence.
Model solution:
For example (0, 1, 0, 2, 0, 3, 0, 4, . . .).
(d) A sequence that has a subsequence that is bounded but contains no subsequence that converges.
Model solution:
Impossible. If a sequence has a subsequence that is bounded then by the Bolzano-Weierstrass theorem
that subsequence has a convergent subsequence. This convergent subsequence is also a convergent
subsequence of the original sequence.
4. Let (un)n∈N be a sequence that is bounded above by 1 and converges to `. Prove rigorously that ` ≤ 1.
Model solution:
We have un ≤ 1 for all n ∈ N. Let (vn)n be the constant sequence equal to 1; it clearly converges to 1 (for
any ε > 0, pick Nε = 1!). Then un ≤ vn for all n ∈ N and, since both sequences converge, we can apply the
OLT to see that their limit have the same order: ` ≤ 1.
5. Let (an)n∈N be a decreasing sequence that is not bounded below. Prove that limn→∞ an = −∞.
Solution:
“Bounded below” means that there exists a real M such that, for any n ∈ N, an ≥M . Saying that (an)n∈N
is not bounded below consist of saying that for any M ∈ R, there exists N ∈ N such that aN < M . But
since (an) is decreasing, we see that for any n ≥ N , an ≤ aN < M . Reading the boxed terms show that we
just proved limn→∞ an = −∞, by definition of this limit.
Supplementary exercises
A. Could you prove the squeeze theorem from the OLT?
Model solution:
Not directly... the squeeze theorem requires you to prove that the middle sequence converges (it is not part of
the assumptions), but the OLT does not directly state that a particular sequence converges (this is assumed
in the hypotheses of the OLT).
B. State and prove a squeeze theorem for infinite limits.
Model solution:
Theorem: let (an)n∈N and (bn)n∈N be sequences such that an ≤ bn for all n ∈ N. If an → +∞ as n → ∞,
then bn → +∞ as n→∞. If bn → −∞ as n→∞, then an → −∞ as n→∞.
Proof (only the first case, the second being similar). Let M ∈ R. By definition of an → +∞ as
n → ∞, there is NM ∈ N such that an ≥ M for all n ≥ NM . Then, for the same natural numbers
n, bn ≥ an ≥M . This concludes the proof that bn → +∞ as n→∞.
C. Let A be non-empty and bounded above. Show that there exists a sequence (an)n∈N in A and a sequence (bn)n∈N
outside A such that an ≤ sup(A) ≤ bn, limn→∞ an = sup(A) and limn→∞ bn = sup(A).
Model solution:
Let n ∈ N. Pick ε = 1/n > 0 in a characterisation of the supremum to see that there exists an ∈ A such
that sup(A)− 1/n ≤ an ≤ sup(A). By the squeeze theorem, an → sup(A) as n→∞.
Take bn = sup(A) + 1/n. Since bn > sup(A), bn cannot belong to A (as sup(A) would be greater than bn
otherwise). We obviously have bn → sup(A) as n→∞.
D. For each of the following sets, decide if they are compact, and explain why they are or are not.
(a) A = { 1p : p ∈ N}.
Solution:
Remember that a set is compact means that any sequence in it has a subsequence that converges in the
set. It is as important that there is a convergent subsequence than that the limit of this subsequence
is in the set. Assume you can find a sequence (an)n∈N in A that converges to some b 6∈ A. Then
by the subsequence limit theorem, any subsequence of (an)n∈N converges to b, and in particular no
subsequence will converge to an element of A, thus showing that A is not compact.
Here, finding such a subsequence is easy: an = 1/n does the trick, since 0 6∈ A. So A is not compact.
(b) [MTH3140] B = {0}∪{ 1p : p ∈ N} (for this one, do not necessarily try to be too rigourous, but try to develop
a good intuition).
Solution:
B “fills the gap” of A by including the limit of the sequence (1/n)n∈N previously considered. This just
means, at this stage, that the previous proof does not enable us to say that B is not compact. It does
not however prove that B is compact... indeed, we would need to consider any sequence in B, and there
are many other sequences than (1/n)n∈N. For example: (1, 1, 1, 1, . . .), or (1, 1/2, 1, 1/2, 1, 1/2, 1 . . .),
or (0, 1, 1/2, 1, 0, 1/6, 1/7, 1/101, 1/105, 1/2, 1, 1/4, . . .). It seems daunting to really try to consider all
sequences here.
We can however come up with a clean reasoning. It turns out that B is compact. Consider any sequence
(bn)n∈N in B. We want to show that there is a subsequence that converges to some element of B. We
first consider the case where this sequence takes the value 0 infinitely many times; then it has (0, 0, . . .)
as a subsequence that converges to 0 ∈ B, as required.
We now consider that (bn)n∈N only takes the value 0 finitely many times. By considering the subse-
quence of (bn)n∈N in which all such 0 are removed, we have a sequence of strictly positive numbers,
and two situations arise: either (i) there is η > 0 such that bn ≥ η for all n ∈ N, or (ii) for all η > 0 we
can find n ∈ N such that bn < η.
Case (i). Since the only values in B that are greater than η are 1, 1/2, . . . , 1/r with r the last integer
before 1/η, we see that (bn)n∈N can only take this finite number of values. In particular, one of these
values is taken an infinite number of times (since (bn)n∈N has infinitely many terms), and it is then easy
to construct a subsequence of (bn)n∈N that remains constant at that particular value – and therefore
converges to this element of B.
Case (ii). We construct a subsequence of (bn)n∈N that converges to 0 ∈ B. Take n1 = 1 and η =
min(1/2, b1) > 0. By assumption, we can find n2 ∈ N such that bn2 < η. In particular, n2 6= 1 and
thus n2 > n1 = 1. Now, take η = min(1/3, b1, . . . , bn2) > 0. There is n3 ∈ N such that bn3 < η, which
implies that n3 > n2 (as bn3 < bk for all k = 1, . . . , n2). Continuing the construction we find a sequence
(nk)k∈N such that nk+1 > nk and bnk+1 < min(1/(k+ 1), b1, . . . , bnk). Then (bnk)k∈N is a subsequence
of (bn)n∈N and, since 0 < bnk < 1/k for all k ∈ N, the squeeze theorem shows that bnk → 0 as k →∞.
E. [MTH3140]
(a) Prove that any compact set is bounded.
Model solution:
Proof by way of contradiction. Assume that a compact set K is not bounded. Then it is either not
bounded below, or not bounded above (or both). Let us assume it is not bounded above, the other
case being similar. Any n ∈ N is not an upper bound of K, and we we can thus find an ∈ K such that
an > n. The sequence (an)n∈N is made of elements of K, and it diverges to +∞ (from the definition,
or by a squeeze theorem for infinite limits). Any subsequence of (an)n∈N therefore also diverges to +∞
by the subsequence limit theorem for infinite limits. This is a contradiction with the fact that K is
compact, since (an)n∈N should have at least one subsequence that converges in K. The proof by way
of contradiction is therefore complete.
(b) Prove that any non-empty compact set has a maximum and a minimum.
Model solution:
We only prove the existence of a maximum, the case of the minimum being handled the same way.
Let K be compact and non-empty. By the preceding part, it is bounded so, by the AoC, it has a
supremum s. The conclusion follows if we can prove that s ∈ K. Let n ∈ N. Since 1/n > 0, by
Exercise 1 in Problem Set 2 we can find an ∈ K such that s − 1/n < an ≤ s. The squeeze theorem
shows that an → s as n → ∞. K being compact, we can find a subsequence (ank)k∈N that converges
to some real number r in K. Now, the subsequence limit theorem ensures that ank → s as k → ∞,
which shows that s = r. Hence, s ∈ K as required.
♦ More exercises from Abbott’s textbook: 2.3.1, 2.3.5, 2.3.7, 2.3.8, 2.3.11, 2.5.5
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 5, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
1. (Abbott Exercise 2.4.4) Show that the sequence
√
2,
√
2
√
2,
√
2
√
2
√
2, ... converges and find the limit. Proceed as
follows. Set a1 =
√
2 and for n ∈ N define an inductively by an+1 =
√
2an.
(a) Use mathematical induction to prove that an = 2
1− 12n (hint: Start by letting S be the set of all natural numbers
n such that an = 2
1− 12n ).
Model solution:
Let S be the set of natural numbers n such that an = 2
1− 12n . For the base step, we have a1 =
√
2 =
2
1
2 = 21−
1
2 = 21−
1
21 . Hence, 1 ∈ S.
For the inductive step, let us suppose that n ∈ S, that is, an = 21− 12n . Then, by definition of an+1 and
the algebraic rules on exponents,
an+1 =
√
2an = (2an)
1
2 = 2
1
2 a
1
2
n = 2
1
2 2
1
2 (1− 12n ) = 2
1
2 2
1
2− 12×2n = 2
1
2+
1
2− 12n+1 = 21−
1
2n+1 .
This shows that the formula holds for an+1 – that is, n + 1 ∈ S. Hence, by induction, S = N and the
formula holds for any n ∈ N.
(b) Can you deduce from the first question that (an)n∈N converges?
Solution:
We would need to know that 21−
1
2n converges as n → ∞. However, we cannot prove that from the
theorems we have established so far. Even if we know/prove that 2n → ∞, the ALT shows that
1/2n → 0 and thus (still by the ALT) 1− 1/2n → 1, but we do not know how to exploit that to show
that 21−
1
2n converges... if a → 1, can we say that 2a converges to something? This is something that
we will only be able to establish after the chapter on the continuity of functions.
(c) Prove by induction that an < an+1 for all n ∈ N.
Model solution:
Let S be the set of all natural numbers n such that an < an+1. For the base step we have 1 ∈ S because
(a1)
2
= 2 < (a2)
2
= 2
√
2. For the inductive step suppose n ∈ S. Thus an < an+1. So 2an < 2an+1
and
√
2an <
√
2an+1. This shows an+1 < an+2 which means n + 1 ∈ S. By induction S = N so the
inequality holds for all natural numbers n.
(d) Prove that an < 2 for all n ∈ N
Model solution:
We use induction. Let T be the set of all natural numbers n such that an < 2. For the base step we
have 1 ∈ S because a1 =
√
2 which is less than 2. For the inductive step suppose n ∈ T. Thus an < 2.
So 2an < 4 and
√
2an < 2. This shows an+1 < 2 which means n + 1 ∈ T. By induction T = N so the
inequality holds for all natural numbers n.
(e) Deduce that (an)n∈N converges (hint: apply a theorem from the lectures).
Model solution:
The sequence is increasing and bounded above. By the monotone convergence theorem (an) converges
to some limit a.
(f) Use the identity an+1 =
√
2an, the algebraic limit theorem and the subsequence limit theorem to find the limit.
Model solution:
The sequence (an+1)n∈N is a subsequence of (an)n∈N and therefore converges to a. Since an+1 =
√
2an,
we have a2n+1 = 2an and the ALT enables us to take the limit on both sides. This shows that a
2 = 2a,
and thus a = 0 or a = 2. We reject a = 0 because each an is greater than a1 =
√
2. So a = 2.
2. Let (an)n∈N be a bounded sequence and set bn = sup{ak : k ≥ n}.
(a) Explain why bn is well-defined.
Model solution:
Since (an)n∈N is bounded, the set {ak : k ≥ n} is bounded (it is contained in the set of all values of
the sequence). It is obviously not empty as it contains an, so by AoC it has a supremum.
(b) Show that (bn)n∈N converges (Hint: if A ⊂ B are bounded sets, sup(A) ≤ sup(B)).
Model solution:
We have {ak : k ≥ n + 1} ⊂ {ak : k ≥ n} so, by the hint, their suprema are ordered: bn+1 ≤ bn.
(bn)n∈N is therefore a decreasing sequence.
Let M be a lower bound of (an)n∈N. Then, for all n ∈ N, bn ≥ an (since bn is an upper bound of
{ak : k ≥ n}) and an ≥M (by definition of M). Hence bn ≥M for all n ∈ N.
(bn)n∈N is therefore decreasing and bounded below. By the MCT, it converges.
3. If
∑
an is a convergent series of positive terms, does
∑√
an converge?
Model solution:
Not necessarily: an = 1/n
2 provides a counter-example.
∑
an converges since it’s a p-series with p = 2, but∑√
an diverges since it’s a p-series with p = 1.
4. Determine whether the following series converge or diverge.
(a)
∑
1
n2+1 .
Model solution:
We use the comparison test (the terms are all positive): we have n2 +1 ≥ n2 so 1n2+1 ≤ 1n2 . Since
∑
1
n2
converges, as a p-series with p = 2,
∑
1
n2+1 also converges.
(b)
∑ cos(n2+1)
2n .
Model solution:
We have | cos(n2+1)2n | ≤ 12n = ( 12 )n. As a geometric series with ratio 12 ,
∑
( 12 )
n converges and thus, by
comparison test for series with positive terms,
∑ | cos(n2+1)2n | also converges. The absolute convergence
test then show that
∑ cos(n2+1)
2n converges.
(c)
∑
en
en+1 .
Model solution:
Dividing by en we have e
n
en+1 =
1
1+e−n . As n → ∞, e−n → 0 and thus, by the ALT for sequences,
1
1+e−n → 1. The n-th term of
∑
en
en+1 therefore does not converge to 0 and, by n-th term test, the
series does not converge.
5. Suppose that
∑
an is absolutely convergent. The purpose of this exercise is to show that
∑
a2n is then convergent.
Set sm = |a1|+ |a2|+ |a3|+ ... + |am| and tm = a21 + a22 + · · ·+ a2m.
(a) Why does the sequence (sm) converge?
Model solution:
The series
∑ |an| is convergent. Hence the sequence of partial sums (sm) converges to some limit s.
(b) Prove that tm ≤ s2m for each m ∈ N.
Model solution:
By developping,
s2m = (|a1|+ |a2|+ · · ·+ |am|)2
= a21 + a
2
2 + · · ·+ a2m + 2|a1| |a2|+ 2|a1| |a3|+ · · ·
= tm + positive terms.
Hence s2m ≥ tm for each m ∈ N.
(c) Prove that
∑
a2n converges.
Solution:
We have 0 ≤ tm ≤ s2m ≤ s2, where s = limm→∞ sm (this limit is greater than or equal to sm, for any m,
since the sequence (sm)m∈N is increasing). Hence, (tm)m∈N is an increasing sequence that is bounded
above and, by the MCT, it converges. This precisely states that
∑
a2n converges.
Alternatively, we could also directly apply, after having proved that the partial sums of
∑
a2n are
bounded, Theorem 3.2 (“behaviour of series with positive terms”) from the lecture notes.
Supplementary exercises
A. Provide a proof of the comparison test, for series of positive terms, based on the Cauchy criterion for series.
Model solution:
If 0 ≤ an ≤ bn then
|an + . . . + ap| = an + . . . + ap ≤ bn + . . . + bp ≤ |bn + . . . + bp|. (1)
Hence, if
∑
bn converges, by the Cauchy criterion we can find, for any ε > 0, an integer Nε such that, if
n ≥ p ≥ Nε, |bn + . . . + bp| ≤ ε. Using (1), we see that
∑
an satisfies the Cauchy criterion and therefore
converges.
Conversely, if
∑
an does not converge, it does not satisfy the Cauchy criterion. There exists therefore ε > 0
such that, for an integer N , we can find n ≥ p ≥ N such that the LHS of (1) is greater than ε. This shows
that the RHS of (1) is also greater that ε and therefore that
∑
bn does not converge (it does not satisfy the
Cauchy criterion).
B. [MTH13140]
(a) Prove directly from the definition (without using the theorem which states that any Cauchy sequence converges)
that if a Cauchy sequence has a converging subsequence, then it converges.
Model solution:
Let (an)n∈N be Cauchy and (ank)k be its convergent subsequence, say to a. Let ε > 0 and take Nε
such that
for any n, p ≥ Nε, |an − ap| ≤ ε (2)
(definition of a Cauchy sequence). Let K be such that for any k ≥ Kε we have |ank − a| ≤ ε (definition
of a converging sequence). Fix k = max(Kε, Nε), and recall that nk ≥ k. We then have nk ≥ Nε.
Apply therefore (2) with p = nk and the triangle inequality: for n ≥ Nε,
|an − a| = |an − ank + ank − a| ≤ |an − ank |+ |ank − a| ≤ ε + ε = 2ε.
This is true for any ε and any n ≥ Nε, and the proof is thus complete.
(b) Deduce from this and the Bolzano-Weierstraß theorem that any Cauchy sequence converges.
Model solution:
A Cauchy sequence is bounded and, by the BW theorem, has a converging subsequence. Part (a) then
shows that the whole Cauchy sequence converges.
C. [MTH3140] (Abbott, Exercise 2.5.4) Assume that (an)n∈N is a bounded sequence with the property that every
convergent subsequence of (an)n∈N converges to the same limit a ∈ R. Show that (an)n∈N must converge to a.
Solution:
We prove the result by contradiction. Negating the definition of the limit, we find ε > 0 such that, for any
N ∈ N, there exists n(N) ≥ N such that |an(N)−a| > ε. In other words, infinitely many (an)n∈N lie outside
[a − ε, a + ε]. We can then construct a subsequence of (an)n∈N which lie outside this interval. Let us do
that properly. Fix n1 = n(1) ≥ 1, so that |an1 − a| > ε. Take then n2 = n(n1 + 1) ≥ n1 + 1 > n1, so that
|an2−a| > ε. Continuing on with this construction, we build (nk)k∈N such that nk+1 > nk and |ank−a| > ε.
(ank)k∈N is thus a subsequence of (an)n∈N, since (nk)k∈N is (strictly) increasing, and (ank)k∈N obviously
does not converge to a since there is no rank k for which |ank − a| ≤ ε.
We would be finished if we could say that (ank)k∈N is a convergent sequence. Indeed, in that case, this would
provide a convergent subsequence of (an)n∈N which does not converge to a, a contradiction to our assumption.
But nothing ensures at this stage that (ank)k∈N is convergent. There is however one assumption we haven’t
used yet: (an)n∈N is bounded. This shows that (ank)k∈N is also bounded and, by the Bolzano-Weierstraß
theorem, that it has a convergent subsequence (ankl )l∈N. This subsequence of a subsequence of (an)n∈N is
a subsequence of (an)n∈N, which converges but does not converge to a since, for any l, |ankl − a| > ε by the
propery of (ank)k∈N. This concludes our proof by contradiction.
♦ More exercises from Abbott’s textbook: 2.4.6 (completes Ex. 2 above), 2.6.1, 2.6.3, 2.6.4, 2.6.5, 2.7.4, 2.7.10,
2.7.11.
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 6, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. Determine whether the following series converge or diverge.
(a)
∑ 1+(−1)n
n2
Model solution:
(Note: the ratio test cannot be applied since some of the coefficients vanish.)
Since 1 + (−1)n is equal to 0 or 2 depending on the parity of n, the coefficients an satisfy 0 ≤ an ≤
2/n2 = bn. The series
∑
bn = 2
∑
1/n2 converges by the ALT (because of the multiplication by 2)
and since
∑
1/n2 is a p-series with p = 2 > 1. Hence, by the comparison test (valid since each an is
positive),
∑
an converges.
(b)
∑
(−1)n+1 n2+2n2+1
Model solution:
(Note: the ratio test fails, as the limit of |an+1/an| is 1.)
Let cn = (−1)n+1 n2+2n2+1 be the generic term of this series. We have
|cn| = n
2 + 2
n2 + 1
=
1 + 2n2
1 + 1n2
→ 1 as n→∞
by the ALT. Hence, |cn| 6→ 0 as n→∞ and thus cn 6→ 0 as n→∞ (see Problem Set 2 if you doubt).
The n-th term test then shows that
∑
cn diverges.
(c)
∑ 2n cos(n)
3n
Model solution:
Let an =
2n cos(n)
3n . Since | cos | ≤ 1, we have |an| ≤ 2n/3n = (2/3)n. The series
∑
(2/3)n converges, as
a geometric series with ratio 2/3 < 1. Hence, by the comparison test,
∑ |an| converges. The absolute
convergence test then shows that
∑
an converges.
(d)
∑
rn
n2 for a fixed real number r ∈ (0,∞) (the convergence/divergence might depend on the value of r).
Model solution:
We use the ratio test with an =
rn
n2 . We have
|an+1|
|an| =
an+1
an
=
rn+1
(n+ 1)
2
n2
rn
=
rn2
(n+ 1)
2 =
r(
1 + 1n
)2 .
By the ALT, the limit of this last term exists and is r. The ratio test then shows that the series
converges if r < 1, and diverges if r > 1. For r = 1, the ratio test is inconclusive, but in that case the
series is
∑
1/n2, which is a p-series with p = 2 > 1 and therefore converges.
2. We proved in the previous problem set that if
∑
an is absolutely convergent, then
∑
a2n is convergent. Could you
prove this result by only assuming that
∑
an is conditionally convergent?
Model solution:
No, the result would be false in general. Consider an = (−1)n/
√
n. Then
∑
an is an alternating series and
therefore converges, but
∑
a2n is the p-series with p = 1, which diverges.
3. Prove directly from the definition that:
(a) limx→−2(x2 + x− 5) = −3.
Model solution:
We write
|x2 + x− 5− (−3)| = |x2 + x− 2| = |(x+ 2)(x− 1)| = |x+ 2| |x− 1|.
Assume that |x + 2| ≤ 1, that is −3 ≤ x ≤ −1. Then x is within maximal distance 4 of 1, that is
|x − 1| ≤ 4. Thus, |x2 + x − 5 − (−3)| ≤ 4|x + 2|. Let ε > 0 and take δ = min(1, ε/4). For all x such
that |x− (−2)| = |x+ 2| ≤ δ, we have |x+ 2| ≤ 1 and thus, by the previous estimate,
|x2 + x− 5− (−3)| ≤ 4|x+ 2| ≤ 4δ ≤ ε.
The proof is complete.
(b) limx→1
√
x2 + 1 =
√
2.
Solution:
We have to evaluate
√
x2 + 1 − √2 and we use the classical trick to get rid of √ : we multiply by√
x2 + 1 +
√
2. √
x2 + 1−
√
2 =
(
√
x2 + 1)2 − (√2)2√
x2 + 1 +
√
2
=
x2 + 1− 2√
x2 + 1 +
√
2
.
We then take the absolute values and use
√
x2 + 1 +
√
2 ≥ √2 to find
|
√
x2 + 1−
√
2| ≤ |x
2 − 1|√
2
=
|x+ 1| |x− 1|√
2
.
If |x− 1| ≤ 1 then |x+ 1| ≤ |x|+ 1 ≤ |x− 1|+ 1 + 1 ≤ 3 and therefore
|
√
x2 + 1−
√
2| ≤ 3√
2
|x− 1|.
The conclusion is now straightforward. For any ε > 0, let δ = min(
√
2ε/3, 1). If |x − 1| ≤ δ, then
|x− 1| ≤ 1 and the preceding inequality is therefore valid. It shows that |√x2 + 1−√2| ≤ 3√
2
δ ≤ ε, as
wanted.
4. Study the existence of the limit as x→ 0 of the following functions. Compute this limit if it exists.
(a) h(x) = cos(e1/x
2
) defined on R\{0}.
Model solution:
Let xn = 1/
√
ln(npi). Then e1/x
2
= npi and therefore h(xn) = cos(npi) = (−1)n. xn → 0 but
(h(xn))n∈N does not converge so limx→0 h(x) does not exist (sequential characterisation of functional
limits)
(b) g(x) =
√|x| cos(1/x), defined on R\{0}.
Model solution:
We have |g(x)| ≤√|x|. Let ε > 0 and take δ = ε2. Then if |x| = |x− 0| ≤ δ we have |g(x)| ≤ √δ = ε.
This proves that g(x)→ 0 as x→ 0.
5. Let f : A→ R and c ∈ L(A).
(a) Give a definition of “f(x)→ +∞ as x→ c”.
Solution:
The part “x→ c” is the same as for finite limits, so it will be evaluated in terms of |x−c| being less than
some δ. The part “f(x)→ +∞” cannot be evaluated, as for finite limits, by considering |f(x)− (+∞)|,
as this simply does not make sense (+∞ is not a number and cannot be used in inequalities, algebraic
operations, etc.). To state that f(x) is “close” to +∞, we’ll simply say that f(x) can be as large as we
want for the proper x, i.e. greater than any fixed number.
This leads to: f(x) → +∞ as x → c if for all M ∈ R, there is δ > 0 such that, for all x ∈ A with
0 < |x− c| ≤ δ, we have f(x) ≥M .
(b) Prove from the definition you have given that 1/x2 → +∞ as x→ 0.
Model solution:
We obviously consider f(x) = 1/x2 over its maximal domain A = R\{0}. Let M ∈ R. We want
f(x) ≥ M , that is 1/x2 ≥ M . We can put aside the case M ≥ 0 as the inequality is trivially
satisfied in that case. For M > 0, the relation 1/x2 ≥ M is true provided that 1/M ≥ x2, that is
|x| =
√
x2 ≤ 1/√M . Taking δ = 1/√M then shows that f(x) ≥M for all x ∈ [−δ, δ] ∩A.
6. State and prove a squeeze theorem for functional limits.
Model solution:
Theorem: let f, g, h be functions A → R, ` ∈ R and c ∈ L(A). Assume that f(x) ≤ g(x) ≤ h(x) for all
x ∈ A, and that f(x)→ ` and h(x)→ ` as x→ c. Then g(x)→ ` as x→ c.
Proof: we use the sequential characterisation of functional limits. Let (xn)n∈N in A such that xn 6= c for all
n ∈ N, and xn → c as n→∞. Then by sequential characterisation, f(xn) → ` and h(xn) → ` as n→∞.
Since f(xn) ≤ g(xn) ≤ h(xn), the squeeze theorem for limits of sequences show that g(xn) → ` as n→∞.
Using again the sequential characterisation of functional limits, but in the other direction, this proves that
limx→c g(x) = `.
Supplementary exercises
A. Let an =
n+1
n2+1 .
(a) Prove that (an)n∈N is a decreasing sequence.
Model solution:
The quantity
an − an+1 = n+ 1
n2 + 1
− (n+ 1) + 1
(n+ 1)
2
+ 1
=
(n+ 1)
(
n2 + 2n+ 2
)− (n+ 2) (n2 + 1)
(n2 + 2n+ 2) (n2 + 1)
=
n2 + 3n
(n2 + 2n+ 2) (n2 + 1)
is strictly positive. So an > an+1 for each n ∈ N. This shows that the sequence (an)n∈N is decreasing.
(b) Prove that
∑
(−1)n+1an is a convergent series.
Model solution:
We have limn→∞ an = limn→∞ n+1n2+1 = limn→∞
1
n+
1
n2
1+ 1
n2
= 0+01+0 = 0 by the ALT. Given than (an)n∈N is
decreasing and tends to 0, the AST shows that
∑
(−1)n+1an converges.
(c) Find a constant c > 0 such that an ≥ cn for all n ∈ N.
Model solution:
We want an ≥ cn , which is equivalent to n(n+ 1) ≥ cn2 + c, that is n2 +n ≥ cn2 + c. If we take c = 1/2
then, n2 ≥ cn2 and n ≥ c for all n ∈ N, which shows that n2 + n ≥ cn2 + c.
(d) Determine whether the series
∑
(−1)n+1an is absolutely or conditionally convergent.
Model solution:
We have
∣∣(−1)n+1an∣∣ = an ≥ cn ≥ 0. The series ∑ 1n diverges (p-series with p = 1). By the comparison
test,
∑∣∣(−1)n+1an∣∣ also diverges. Therefore the series ∑(−1)n+1an is only conditionally convergent.
B. (Abbott, Exercise 1.2.6) Given a function f and a subset A of its domain, let f(A) represent the range of f over
the set A; that is, f(A) = {f(x) : x ∈ A}.
We consider f(x) = x2. Find two sets A and B for which f(A ∩B) 6= f(A) ∩ f(B).
Model solution:
If A = [0, 2] and B = [−2, 0] then A ∩ B = {0}, f(A) = [0, 4] = f(B), f(A ∩ B) = {0} and so f(A ∩ B) 6=
f(A) ∩ f(B).
C. Assume that (ε, δ) is a valid pair in the definition of the continuity of some function at some point. Which of those
are always still valid pairs? Justify your answer.
(a) (ε, δ′) with δ′ > δ.
(b) (ε, δ′) with δ′ < δ.
(c) (ε′, δ) with ε′ > ε.
(d) (ε′, δ) with ε′ < ε.
Solution:
If (ε, δ) is a valid pair in the definition of “f is continuous at c” then for any x ∈ A such that |x− c| ≤ δ, we
have |f(x)− f(c)| ≤ ε.
If δ′ < δ and |x − c| ≤ δ′, then |x − c| ≤ δ and therefore |f(x) − f(c)| ≤ ε. Hence, (ε, δ′) is valid if δ′ < δ.
We can find some counter-examples (do it!) such that, for some δ′ > δ, (ε, δ′) is not valid.
If ε′ > ε then |f(x) − f(c)| ≤ ε implies |f(x) − f(c)| ≤ ε′ and therefore (ε′, δ) is valid. Here again, some
counter-examples show that this pair is not valid if ε′ < ε.
As a general intuitive rule, the part “|x− c|” in the definition is a condition we impose on x; so if we impose
a more restrictive condition (i.e. if δ is reduced), we are sure to still satisfy the estimate we want. This
estimate |f(x) − f(c)| ≤ ε is what we want to achieve; increasing ε makes it easier to achieve, under the
same condition |x− c| ≤ δ.
D. [MTH3140] Let f : A 7→ R and c ∈ L(A). Prove that limx→c f(x) exists if and only if:
For any sequence (xn)n∈N in A converging to c, and xn 6= c for all n,
(f(xn))n∈N has a limit in R.
(1)
In that case, all sequences (f(xn))n∈N in (1) converge to the same real number L, which is limx→c f(x).
(Hint: prove by contradiction that if (f(xn)) converges for any choice of the sequence (xn)n∈N converging to c then
all these sequences (f(xn))n∈N must have the same limit.
Assume therefore the existence of (xn)n∈N and (yn) converging to c, but never taking the value c, such that f(xn)→ `
and f(yn)→ `′ with ` 6= `′. Consider (zn)n∈N defined by z2n = xn, z2n+1 = yn for any n. (zn)n∈N converges to c,
zn 6= c for all n, but (f(zn))n∈N does not have any limit (why?)...).
Solution:
If limx→c f(x) = L then (1) follows from the sequential characterisation of the limit (and this characterisation
even tells us that f(xn)→ L, not only that the limit exists).
We now assume (1) and want to prove that limx→c f(x) exists. We will prove by contradiction, following the
hint in Remark 5.7 of the lecture notes, that all sequences (f(xn))n∈N in (1) have the same limit L. Then
the sequential characterisation of the limit can be applied and gives the existence of limx→c f(x) (and its
equality with L).
If we have two sequences (xn)n∈N and (yn)n∈N in A converging to c such that limn→∞ f(xn) = L and
limn→∞ f(yn) = L′ with L 6= L′, then we construct (zn) by z2n = xn and z2n+1 = yn for any n ∈ N. Then
(zn)n∈N converges to c. Indeed, for any ε > 0, there exists N1 such that n ≥ N1 implies |xn − c| ≤ ε and
n ≥ N2 implies |yn− c| ≤ ε. Taking N = 2 max(N1, N2) + 1 shows that, if k ≥ N , then k = 2n or k = 2n+ 1
with n ≥ max(N1, N2), and therefore zk = xn or zk = yn with, respectively, |xn − c| ≤ ε and |yn − c| ≤ ε.
In both cases, we see that for any k ≥ N , |zk − c| ≤ ε, as desired.
But (f(zn))n∈N does not converge. Indeed, if it converged to some S, then its subsequences (f(z2n))n∈N =
(f(xn))n∈N and (f(z2n+1))n∈N = (f(yn))n∈N would also converge to S. But these subsequences respectively
converge to L and L′, which gives L = S = L′ and thus the contradiction.
♦ More exercises from Abbott’s textbook: 2.7.3, 2.7.6, 2.7.12, 2.7.13, 2.7.14, 4.2.1, 4.2.3, 4.2.5, 4.2.8, 4.3.1, 4.3.3
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 7, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. Show that if f is continuous on some interval [a, b] and f(x) > 0 for all x ∈ [a, b], then ln(f) is bounded on [a, b].
Model solution:
[a, b] is a compact set (Corollary 2.30) and thus, f being continuous on this compact set, it has a maximum
and a minimum. That is, there are x1 and x2, such that f(x1) ≤ f(x) ≤ f(x2) for all x ∈ [a, b]. By
assumption, f(x1) > 0. Since ln is an increasing function and all three numbers are in the domain of it, we
infer ln(f(x1)) ≤ ln(f(x)) ≤ ln(f(x2)). The function ln(f) is therefore, on [a, b], bounded below by ln(f(x1))
and above by ln(f(x2)).
Alternative solution: by assumption, f : [a, b] → (0,∞) is continuous. We know that ln : (0,∞) → R is
continuous. Hence, by composition, ln(f) : [a, b] → R is continuous. Since [a, b] is compact, we infer that
ln(f) is bounded on [a, b].
2. (Abbott Exercise 4.4.6) Give an example of each of the following, or state that such a request is impossible. For
any that are impossible, supply a short explanantion (perhaps referencing the appropriate theorem(s)) for why this
is the case.
(a) A continuous function f : (0, 1)→ R and a Cauchy sequence (xn)n∈N in (0, 1) such that (f(xn))n∈N is not a
Cauchy sequence.
Model solution:
For example, f(x) = 1x and xn =
1
n . Then (xn)n∈N converges (though not to an element of the interval
(0, 1)) and is therefore Cauchy. However, f(xn) = n which is unbounded and therefore not Cauchy.
(b) A continuous function f : [0, 1] → R and a Cauchy sequence (xn)n∈N in [0, 1] such that (f(xn))n∈N is not a
Cauchy sequence.
Model solution:
This is impossible. If (xn)n∈N is a Cauchy sequence in [0, 1] then it converges to a limit x ∈ R. Since
0 ≤ xn ≤ 1, the OLT shows that 0 ≤ x ≤ 1, i.e. x ∈ [0, 1]. The function f is therefore continuous at x,
and thus f(xn)→ f(x) as n→∞. Hence, (f(xn))n∈N is a Cauchy sequence.
(c) A continuous function f : [0, 1]→ R which has a maximum but no minimum.
Model solution:
Impossible. [0, 1] is compact and f is continuous, so the EVT shows that f has a minimum (and a
maximum).
(d) A continuous bounded function f : (0, 1) → R that attains a maximum value but not a minimum value.
(Remember that a function is called bounded if its range is a bounded set.)
Model solution:
For example, f(x) = −(x− 12 )2 is continuous on (0, 1) and has range (− 14 , 0]. Thus f has a maximum
value of 0, but no minimum.
3. Prove Proposition 4.11 in the lecture notes, that is: if f : A 7→ R and c ∈ A ∩ L(A), “f is continuous at c” is
equivalent to “limx→c f(x) = f(c)”.
Solution:
(1) f is continuous at c means that: for any ε > 0, there exists δ > 0 such that, if x ∈ A and |x − c| ≤ δ
then |f(x)− f(c)| ≤ ε.
(2) “limx→c f(x) = f(c)” means that: for any ε > 0, there exists δ > 0 such that, if x ∈ A with 0 < |x−c| ≤ δ,
then |f(x)− f(c)| ≤ ε.
The only difference between the two definitions is the condition “0 < |x−c|” in (2). This consists in removing
the possibility x = c. However, if we include this point x = c, then the conclusion |f(x) − f(c)| ≤ ε also
clearly follows in (2). Hence, in this context (where the expected limit is f(c), not another real number), we
can indeed include x = c in the definition of lim and find back the definition of continuity.
4. Let f : R→ R such that there exists k > 0 satisfying |f(x)− f(y)| ≤ k|x− y| for all x, y ∈ R.
(a) Prove that f is continuous on R.
Model solution:
Let c ∈ R and ε > 0. Take δ = ε/k. Then, if |x − c| ≤ δ we have |f(x) − f(c)| ≤ k|x − c| ≤ ε. f is
therefore continuous at c and, since this is true for any c ∈ R, f is continuous on R.
(b) Let r > k. Prove that there exists a unique c ∈ R such that f(c) = rc. (Hint: consider g(x) = 1rf(x).)
Model solution:
The condition f(c) = rc translates into g(c) = c, that is, c is a fixed point of g. R is a closed interval
and, for all x, y,
|g(x)− g(y)| = 1
r
|f(x)− f(y)| ≤ k
r
|x− y|.
Hence, g is a contraction with contraction constant k/r < 1. By the contraction mapping theorem, g
has a unique fixed point.
5. (a) Prove the existence of the square root function, i.e. the function
√
: R+ → R+ such that, for any x ≥ 0,
(
√
x)2 = x.
Model solution:
For a given x ≥ 0, we look for the existence and uniqueness of y ≥ 0 such that y2 = x. The function
f(y) = y2 is continuous and satisfies f(0) = 0 and f(x+ 1) = x2 + 2x+ 1 > x (that is because 2x ≥ x;
beware, x2 is not necessarily greater than x). Hence, by the IVT, there exists y between 0 and x + 1
such that y2 = x.
To properly define the square root function, this y must be unique (square root must have and only one
image for each x). Assume thus that y2 = x = (y′)2 with y, y′ ≥ 0. Then y2 − (y′)2 = 0, which implies
(y + y′)(y − y′) = 0. Hence, either y + y′ = 0 or y = y′. In the first case, since y, y′ ≥ 0, we deduce
y = y′ = 0. Therefore, in any case, y = y′ and the uniqueness of the positive square root is proved.
(b) Prove that there is a number x such that
x4 + e1+sin
2(x) cos(x) cos(x5) = 10.
Model solution:
Set g(x) = x4 + e1+sin
2(x) cos(x) cos(x5). By the ACT and the composition of continuous functions,
g is continuouson R. Also g(0) = e < 10 and g(10pi) = 104pi4 + e cos(105pi5) ≥ 104pi4 − e (since
cos(105pi5) ≥ −1, so g(10pi) > 10. The IVT therefore shows that there is x, between 0 and 10, such
that g(x) = 0.
Supplementary exercises
A. [MTH3140] Give an example of each of the following, or state that such a request is impossible. For any that are
impossible, supply a short explanantion (perhaps referencing the appropriate theorem(s)) for why this is the case.
(a) A continuous function f : R 7→ R and a bounded set A such that f(A) is not bounded.
Model solution:
Impossible. If A is bounded, it is contained in a compact set [−M,M ] for M large enough, and
therefore f(A) ⊂ f([−M,M ]). We know that f([−M,M ]) is compact (preservation of compact sets)
and therefore bounded (Exercise 5, Problem Set 3). Hence, f(A) is bounded.
(b) A continuous function f : R 7→ R and a closed set A (i.e. if a sequence of elements in A converges to some
x ∈ R, then x ∈ A) such that f(A) is not closed.
Solution:
f(x) = arctan(x) and A = R. We have f(A) = (−pi/2, pi/2), which is clearly not closed as the sequence
yn = pi/2− 1/n ∈ f(A) shows (write the details!).
(c) A continuous function f : (0,∞) 7→ R and a bounded set A ⊂ (0,∞) such that f(A) is not bounded.
Model solution:
f(x) = 1/x and A = (0, 1). We have f(A) = (1,∞).
B. [MTH3140] We define f(x) = sin(1/x) if x 6= 0, f(0) = 0. Prove that f satisfies the IVT between any two points
of R (i.e. for any a < b ∈ R and any L between f(a) and f(b), there exists c ∈ (a, b) such that f(c) = L).
Solution:
If 0 < a < b or a < b < 0 then the conclusion follows from the IVT and the fact that f is continuous on
[a, b].
If a ≤ 0 ≤ b, then [a, b] contains an interval of the form [0, δ] or [−δ, 0], for some δ > 0. Let us assume
that [0, δ] ⊂ [a, b]. For n large enough, In = [ 12npi+pi/2 , 12npi−pi/2 ] is contained in [0, δ] (take n such that
1
2npi−pi/2 ≤ δ) and therefore f([a, b]) ⊃ f([0, δ]) ⊃ f(In). Letting g(x) = 1/x, we easily see (do it!) that
g(In) = [2npi − pi/2, 2npi + pi/2], and therefore f(In) = sin(g(In)) = [−1, 1] (why? prove it!). Hence,
[−1, 1] ⊂ f([a, b]) and, since f(a) and f(b) lie in [−1, 1] by definition of f , this shows that any real between
f(a) and f(b) lies in [−1, 1], and thus in f([a, b]), as required.
C. Two points on Earth are called antipodal if they are at exactly opposite points. For example, the North and South
poles are antipodal.
Assuming that the temperature at the surface of the Earth is continuous, prove that, at any given moment, there
are two antipodal points on the equator with exactly the same temperature. (Hint: Let T (θ) be the temperature
at the point on the equator with longitudinal angle θ radians where 0 ≤ θ ≤ 2pi. Assume that T is continuous and
consider T (θ + pi)− T (θ).)
Model solution:
Define S(θ) = T (θ+pi)−T (θ) which is a continuous function on the interval [0, pi]. Note that T (2pi) = T (0).
Thus, S(0) = T (pi)−T (0) and S(pi) = T (0)−T (pi). Hence S(0) and S(pi) have opposite signs. We can apply
the IVT to conclude that there is a θ ∈ [0, pi] with S(θ) = 0. In other words T (θ + pi) = T (θ) as required.
♦ More exercises from Abbott’s textbook: 4.3.5, 4.3.8, 4.4.4, 4.4.7, 4.5.3, 4.5.4
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 8, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. Prove that the function f defined below is continuous on R, differentiable on R, and that f ′ is discontinuous at 0.
f(x) = x2 sin
(
1
x
)
if x 6= 0 , f(0) = 0.
Model solution:
Differentiability outside 0 is obvious by the ADT. For x 6= 0 we have f ′(x) = 2x sin(1/x)− cos(1/x). At 0,
we use the definition of the derivative. Write
f(x)− f(0)
x
= x sin(1/x).
Since sin is bounded by 1, | f(x)−f(0)x | ≤ |x| and thus limx→0 f(x)−f(0)x = 0. This can be seen, e.g., by the
sequential characterisation of limit: if (xn)n∈N converges to 0 (but is never 0), then | f(xn)−f(0)xn | ≤ |xn| → 0
and by the squeeze theorem we infer that | f(xn)−f(0)xn | → 0, so the sequential characterisation of limit gives
limx→0
f(x)−f(0)
x = 0. Hence, f
′(0) exists and is equal to 0. f is therefore differentiable, and thus continuous,
on R.
However, f ′(1/(2npi)) = − cos(2npi) = −1 does not converge to f ′(0) = 0 as n→∞. Hence, f ′ is not
continuous at 0.
2. Let f and g be two differentiable functions on R, such that g does not vanish. Assume that c ∈ R is a extremum
of both f/g and fg.
(a) Prove that f ′(c) = 0 and either f(c) = 0 or g′(c) = 0.
Model solution:
(10 marks in total)
The functions f/g and fg are differentiable by the ADT [Marks: 1] and c is obviously in the interior
of R so we can apply the IET [Marks: 2]. We have (f/g)′(c) = 0 and (fg)′(c) = 0 which translates
into
f(c)g′(c)− f ′(c)g(c)
g(c)2
= 0 and f ′(c)g(c) + f(c)g′(c) = 0. [Marks : 2]
The first relation gives f(c)g′(c) = f ′(c)g(c) and the second f ′(c)g(c) = −f(c)g′(c). Hence, f ′(c)g(c) =
−f ′(c)g(c) so 2f ′(c)g(c) = 0 [Marks: 2] and, since g(c) 6= 0, f ′(c) = 0 [Marks: 1]. Coming back to
f(c)g′(c) = f ′(c)g(c) this shows that f(c)g′(c) = 0 and thus f(c) = 0 or g′(c) = 0 [Marks: 2].
(b) Find an example where f(c) = 0 and g′(c) 6= 0, and another example where f(c) 6= 0 and g′(c) = 0.
Solution:
As usual, when looking for example, we try simple functions. The first ones are constant or linear
functions. Since we do not want g to vanish anywhere, if we assume it to be linear it will actually be
constant. That will give g′(c) = 0. This leads to an example matching the second situation: taking for
example g = 1, we need to find a function f and a c such that c is an extremum of f/1 and f × 1, that
is f itself, and f(c) 6= 0. Here is an easy solution: f(x) = 1 + x2 has c = 0 as minimum but f(0) 6= 0.
For the other case, that is f(c) = 0 and g′(c) 6= 0, we do not want g constant. Still thinking about
polynomials, the first non-vanishing polynomial on R we can think of is probably 1+x2 (any linear one
will vanish). Since we want g′(c) 6= 0, the extremum c of f/g and fg cannot be c = 0. We also want
f(c) = f ′(c) = 0. Here too, thinking about a polynomial for f , this tells us that c will be a double root
of f . Let’s make c = 1 and f(x) = (x − 1)2. It remains to check if c = 1 is an extremum of f/g and
fg. At x = 1 both these functions vanish. f and g being both positive, f/g and fg are also positive.
Hence, c = 1 is indeed a minimum of f/g and fg.
[Marks: 3 for each case]
3. Define, for a > 0, ga(x) = a
x.
(a) What is the domain of ga? Is g continuous on its domain?
Model solution:
ax is defined, for a general x, as ex ln(a). This is defined for any x, so Dga = R. Moreover, ga(x) = e
x ln(a)
is continuous as a composition, and product by a constant, of continuous functions.
(b) On which domain is ga differentiable? Compute its derivative.
Model solution:
By the chain rule, x 7→ ex ln(a) is differentiable for any x ∈ R, and g′a(x) = ln(a)ex ln(a) = ln(a)ax.
4. Let g : R→ R be differentiable, and a < b be real numbers such that g′(a) < 0 < g′(b).
(a) Show that there exists x ∈ (a, b) such that g(a) > g(x), and y ∈ (a, b) such that g(y) < g(b).
Model solution:
By 1st order expansion, for h > 0 such that a+ h ∈ (a, b),
g(a+ h) = g(a) + hg′(a) + hr(h) = g(a) + h(g′(a) + r(h))
with r(h) → 0 as h → 0. Since ε = |g′(a)|/2 is strictly positive, for h > 0 small enough we have
|r(h)| ≤ |g′(a)|/2 and thus, since g′(a) < 0,
g′(a) + r(h) ≤ g′(a) + |g
′(a)|
2
≤ g′(a)− g
′(a)
2
=
g′(a)
2
< 0.
Hence, for this h, g(a+ h) = g(a) + h(g′(a) + r(h)) < g(a) and x = a+ h is suitable.
A similar reasoning show that, for h < 0 small enough, y = b+ h satisfies g(y) < g(b).
(b) Deduce the existence of c ∈ (a, b) such that g′(c) = 0.
Model solution:
Let c be a minimum of g on [a, b] (it exists by continuity of g on this compact set). Since g(a) > g(x) ≥
g(c) and g(b) > g(y) ≥ g(c), we cannot have c = a or c = b. Hence, c is an interior point and the IET
shows that g′(c) = 0.
Note: we have essentially proved here Darboux’s theorem.
Supplementary exercises
A. Let f : R\{c} → R, c ∈ R and ` ∈ R.
(a) Give a sound definition of the notion “f has limit ` at c from the right”, and an example of a function that
has limit 1 at 0 from the right, but that does not have a (standard) limit at 0.
Solution:
Obviously,“from the right” means that we only consider variables at the right of c. In other words,
we consider the restriction of f on (c,+∞), instead of f on its entire domain. As a consequence, a
sound definition follows by simply writing the continuity at c of this restriction. This amounts to: for
all ε > 0, there exists δ > 0 such that, for all x ∈ (c,+∞), if |x− c| ≤ δ then |f(x)− `| ≤ ε.
The function f(x) = 1 if x > 0 and f(x) = 0 if x < 0 provides an example of a function that is does
not have a limit at 0 (prove it!), but has a limit at 0 from the right since its restriction to (0,∞) is the
constant function equal to 1.
(b) Prove that if f has limit ` at c from the right and limit ` at c from the left (with an obvious adaptation of the
previous notion), then it has limit ` at c.
Model solution:
Let ε > 0 and take δr > 0 and δl > 0 from the definitions of “f has limit ` at c from the right” and
“f has limit ` at c from the left”. Fix δ = min(δr, δl) . For all x ∈ [c− δ, c+ δ]\{c} , we have either
x ∈ [c − δ, c) ⊂ [c − δl, c), in which case |f(x) − `| ≤ ε by choice of δl, or x ∈ (c, c + δ] ⊂ (c, c + δr],
in which case |f(x) − `| ≤ ε by choice of δr. Hence, in either case, |f(x)− `| ≤ ε and limx→c f(x)
therefore exists and is equal to `.
B. Let a ∈ R and define ha(x) =
{
xa if x > 0
0 if x ≤ 0.
(a) For what values of a is ha continuous at 0 ?
Solution:
For a general real number a and x > 0, xa is defined by xa = ex ln(x) = exp(a ln(x)). By the ACT and
the composition of continuous functions, x 7→ xa is continuous on (0,∞). The second part of ha, that
is the 0 function, is clearly continuous on (−∞, 0].
We then need to see for which a the function ha is continuous at 0. Based on the relation between
continuity and functional limits, and on Exercise A, we simply need to ensure that limx→0+ xa =
limx→0− 0 = 0. As x → 0+, ln(x) → −∞ so, if a > 0, a ln(x) → −∞ and, by behaviour of exp,
xa = exp(a ln(x)) → 0, so ha is continuous. If a = 0, ha is obviously discontinuous at 0 since it is
equal to 1 = exp(0 ln(x)) on (0,∞) and to 0 on (−∞, 0]. If a < 0, a ln(x) → +∞ as x → 0+ so
xa = exp(a ln(x))→ +∞ and ha is not continuous at 0.
(b) For what values of a is ha differentiable at 0 ?
Model solution:
We have
ha(x)− ha(0)
x− 0 =
{
xa−1 if x > 0
0 if x < 0.
By the proof in the previous item, limx→0
ha(x)−ha(0)
x−0 therefore exists if and only if a > 1. So these are
the values of a for which ha is differentiable at 0. In this case we also have h
′
a(0) = 0.
(c) For what values of a is the derivative function h′a continuous?
Model solution:
If a > 1 then
h′a(x) =
{
axa−1 if x > 0
0 if x ≤ 0.
This function h′a = aha−1 is continuous by Part (a), since a− 1 > 0.
C. (Abbott, exercise 5.3.6) Let g : [0, 1] 7→ R be twice differentiable with g′′(x) > 0 for all x ∈ (0, 1). Assume that
g(0) > 0 and g(1) = 1. The aim of this exercise is to show that g(d) = d for some d ∈ (0, 1) if and only if g′(1) > 1.
We set f(x) = g(x)− x.
(a) Assume that g(d) = d for some d ∈ (0, 1). Using Rolle’s theorem on f , deduce that f ′(c) = 0 for some c < 1.
Deduce that g′(1) > 1 (hint: g′ is strictly increasing (why?)).
Solution:
By choice of d, f(x) = g(x)−x satisfies f(d) = f(1) = 0 and therefore, by Rolle’s theorem, there exists
c ∈ (d, 1) such that f ′(c) = 0, that is g′(c) = 1. The assumption g′′ > 0 implies that g′ is strictly
increasing on (0, 1) (prove it, using the MVT), and therefore g′(1) > g′(c) = 1.
(b) Assume now that g′(1) > 1. Using a first order expansion, show that f(1 − z) < 0 if z > 0 is small enough.
Use then the IVT on f to obtain the existence of d ∈ (0, 1) such that g(d) = d.
Solution:
Since g′(1) > 1, we have f ′(1) > 0. Then, by the first order expansion, f(1 − z) = f(1) − zf ′(1) −
zr(−z) = −z(f ′(1) + r(−z)) where r(h) → 0 as h → 0. For z small enough, f ′(1) > −r(−z) (write
this “for z small enough” rigorously, by using the definition of limh→0 r(h) = 0 with ε = f ′(1)/2 > 0).
Hence, for these z, f(1 − z) < 0. Since f(0) > 0, the IVT provides d ∈ (0, 1 − z) such that f(d) = 0,
that is g(d) = d as requested.
D. [MTH3140] Let B : R2 × R2 → R be a bilinear form. Let γ, µ : (0, 1) → R2 be two differentiable functions
(that is, each of their components is a differentiable function (0, 1) → R). Show that the function t ∈ (0, 1) 7→
B(γ(t), µ(t)) ∈ R is differentiable, and find a formula for its derivative (Hint: (1) you already know the result for the
case of a bilinear form R×R→ R, (2) the bilinear form B is continuous, that is, it satisfies |B(a, b)| ≤ C‖a‖ ‖b‖
where C is a constant and ‖a‖ and ‖b‖ are the Euclidean norms of a and b).
Solution:
We have a similar result for bilinear functions R ×R → R: it’s the product rule! Indeed, B0 : (a, b) → ab
is bilinear R×R→ R and if f, g are functions (0, 1)→ R, B0(f, g) = fg.
To solve the exercise we simply reproduce the proof of the product rule. Let t ∈ (0, 1) and h small enough
so that t+ h ∈ (0, 1). We have, working component-by-component,
γ(t+ h) = γ(t) + hγ′(t) + hr(h) and µ(t+ h) = µ(t) + hµ′(t) + hq(h),
where r(h)→ (0, 0) and q(h)→ (0, 0) as h→ 0 (note that r(h) and q(h) are vectors in R2, whose components
are the remainders of the 1st order expansions of the components of γ and µ). Hence, using the bilinearity
of B,
B(γ(t+ h), µ(t+ h)) = B(γ(t) + hγ′(t) + hr(h), µ(t) + hµ′(t) + hq(h))
= B(γ(t), µ(t) + hµ′(t) + hq(h)) + hB(γ′(t), µ(t) + hµ′(t) + hq(h))
+ hB(r(h), µ(t) + hµ′(t) + hq(h))
= B(γ(t), µ(t)) + hB(γ(t), µ′(t)) + hB(γ(t), q(h)) + hB(γ′(t), µ(t))
+ h2B(γ′(t), µ′(t) + q(h)) + hB(r(h), µ(t) + hµ′(t) + hq(h))
= B(γ(t), µ(t)) + h[B(γ(t), µ′(t)) +B(γ′(t), µ(t))] + hR(h) (1)
where R(h) = B(γ(t), q(h)) + hB(γ′(t), µ′(t) + q(h)) + B(r(h), µ(t) + hµ′(t) + hq(h)). Using the continuity
of B and the triangle inequality,
|R(h)| ≤ C‖γ(t)‖ ‖q(h)‖+ Ch‖γ′(t)‖ ‖µ′(t) + q(h)‖+ C‖r(h)‖ ‖µ(t) + hµ′(t) + hq(h)‖
≤ C‖γ(t)‖ ‖q(h)‖+ Ch‖γ′(t)‖ (‖µ′(t)‖+ ‖q(h)‖) + C‖r(h)‖ (‖µ(t)‖+ h‖µ′(t)‖+ h‖q(h)‖).
Since r(h) → (0, 0) and q(h) → (0, 0) as h → 0, their norms tend to 0. By ALT and squeeze theorem,
the expression above shows that R(h) → 0 as h → 0, and (1) proves, by the 1st order expansion, that
t 7→ B(γ(t), µ(t)) is differentiable, with derivative B(γ(t), µ′(t)) +B(γ′(t), µ(t)).
♦ More exercises from Abbott’s textbook: 5.2.2, 5.2.3, 5.2.5
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 9, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. Let g : R→ R be given by g(x) = exp(−x2).
(a) Find g ′(x) and g ′′(x).
Model solution:
By the chain rule, g′ exists and g ′(x) = −2x exp(−x2). Using the chain rule and the ADT, g ′′(x) =
−2 exp(−x2) + 4x2 exp(−x2) = 4 exp(−x2) (x2 − 12).
(b) Find the maximum and minimum values of g′(x).
Solution:
We cannot directly apply the corollary of location of extrema because we look for the maximum and
minimum values of g′ on R, which is not a bounded closed interval.
We can however notice that g′(x) = −2x exp(−x2)→ 0 as x→ ±∞ (classical indeterminate limit) and
that g′(−1) > 0. Hence, there is an M ≥ 1 such that if |x| ≥ M then |g′(x)| < g′(−1). This tells
us that no maximum of g′ can exist outside [−M,M ], since the values of g′ in this interval cannot be
larger than g′(−1).
On [−M,M ], g′ is continuous so it has a maximum at c = ±M or c such that g′′(c) = 0, that is
c = ±1/√2 ∈ (−M,M) (we have M ≥ 1 > 1/√2). But x = ±M cannot be a maximum of g′ over
[−M,M ] since g′(±M) < g′(−1) by choice of M . Hence, the only possible maxima of g′ over [−M,M ]
are c = ±1/√2. The value +1/√2 is actually excluded since g′(1/√2) < 0 ≤ g′(−1). This shows that
the maximum of g on [−M,M ] is c = −1/√2. This is also the maximum over R since we know that
no maximum can exist outside [−M,M ].
Reasoning in a similar way, or using the fact that g′ is odd, we can see that 1/
√
2 is a global minimum
of g′ over R.
The maximum and minimum values of g′ over R are therefore
g′(
−1√
2
) =
√
2 exp(−1
2
) and g ′(
1√
2
) = −
√
2 exp(−1
2
).
(c) Explain why g is a contraction.
Model solution:
g : R → R is differentiable and, for any x, −√2 exp(−1/2) ≤ g′(x) ≤ √2 exp(−1/2), that is |g ′(x)| ≤√
2 exp(− 12 ) ≈ 0.857 76. By the Lipschitz estimate, we conclude that g is a contraction with contraction
constant γ =
√
2 exp(− 12 ).
(d) How many fixed points does g have?
Model solution:
By the contraction mapping theorem, g has exactly one fixed point. (Its value is approximately 0.65292).
2. (Abbott Exercise 5.3.7) Let
g(x) =
{
x
2 + x
2 sin(1/x) if x 6= 0,
0 if x = 0.
(a) Explain why g is differentiable at points x 6= 0 and find the derivative at these points.
Model solution:
By the algebraic differentiation theorem and the chain rule, g is differentiable at points x 6= 0 and
g ′(x) = 12 + 2x sin(1/x)− cos(1/x).
(b) Use the definition of derivative to prove that g is also differentiable at 0 and show g ′(0) > 0.
Model solution:
We compute
g(x)− g(0)
x− 0 =
x
2 + x
2 sin(1/x)− 0
x− 0 =
1
2
+ x sin(1/x).
Since sin remains bounded by 1, |x sin(1/x)| ≤ |x| and thus x sin(1/x) → 0 as x → 0. Thus g is
differentiable at 0 and g′(0) = 1/2 > 0.
(c) However, explain briefly why g is not increasing on any open interval containing 0.
Solution:
On any open interval containing 0, g ′ has strictly positive and strictly negative values (see figure below
– and try to rigorously prove it). Hence, g is not an increasing function on any such interval, otherwise
its derivative would be positive on that interval.
Plot of g′.
3. Let f : [a, b] → [a, b] be continuous, differentiable on (a, b) and such that f ′ is continuous on [a, b] (or, more
rigorously, can be extended in a continuous function on [a, b]). We assume that |f ′(x)| < 1 for all x ∈ [a, b]. Prove
that f is a contraction, and that it has a unique fixed point in [a, b].
Model solution:
f already sends [a, b] into [a, b]. The Lipschitz estimate gives, for x, y ∈ [a, b],
|f(x)− f(y)| ≤ |x− y| sup
z∈[x,y]
|f ′(z)| ≤ |x− y| sup
z∈[a,b]
|f ′(z)|.
The function |f ′| is continuous on [a, b], by composition of continuous fonctions (the absolute value, and
f ′). Hence it has a maximum on this interval (EVT). There is therefore z0 ∈ [a, b] such that |f ′(z0)| =
supz∈[a,b] |f ′(z)|. We therefore find
|f(x)− f(y)| ≤ |f ′(z0)| |x− y|
with |f ′(z0)| not depending on x, y, and strictly less than 1. This proves that f is a contraction. By the
contraction mapping theorem, f therefore has a unique fixed point in [a, b].
Note: the deduction that supz∈[a,b] |f ′(z)| < 1 from the assumption that each |f ′(z)| is strictly less than 1
requires invoking the EVT, and thus using the continuity of f ′. A pointwise strict estimate on a function
does not give a uniform strict estimate – it’s exactly as dealing with strict estimates in the OLT.
4. Let f : [−1, 1] 7→ R be continuous, twice differentiable on (−1, 1) and such that f(−1) = f(0) = f(1). Prove that
there exists c ∈ (−1, 1) such that f ′′(c) = 0.
Model solution:
(10 marks in total)
By Rolle’s theorem [Marks: 2] on [-1,0], there exists a in (-1,0) such that f’(a)=0
[Marks: 2].
By Rolle’s theorem on [0,1], there exists b in (0,1) such that f’(b)=0 [Marks: 2].
f’ is continuous on [a,b] since it is differentiable on (-1,1) which contains [a,b]
[Marks: 1], and differentiable on (a,b) [Marks: 1].
Since f’(a)=f’(b)=0 and a<0(a,b) such that f’’(c)=0 [Marks: 1].
5. (Abbott Exercise 5.3.5) Show that if f : R → R is differentiable and f ′(x) 6= 1 for all x ∈ R then f has at most
one fixed point.
Model solution:
(10 marks in total)
We prove the result by contradiction. [Marks: 2]
Assume that f has two fixed points ag is differentiable by ADT and g(a)=g(b)=0 by definition of a and b fixed points of f
[Marks: 2].
By Rolle’s theorem [Marks: 2], there is c in (a,b) such that g’(c)=0, which means that
f’(c)-1=0 [Marks: 2]. This is a contradiction with this assumption on f’.
Supplementary exercises
A. (Generalised MVT) Take a < b. Let f and g be two continuous functions on [a, b], which are differentiable on (a, b).
We assume that g′ is never zero on (a, b). Prove that g(b)− g(a) 6= 0 and that there exists c ∈ (a, b) satisfying
f ′(c)
g′(c)
=
f(b)− f(a)
g(b)− g(a) .
(Hint: consider h(x) = (f(b)− f(a))g(x)− (g(b)− g(a))f(x).)
Model solution:
By the MVT, g(b)− g(a) = g′(d)(b− a) for some d ∈ (a, b). Since g′(d) 6= 0, g(b)− g(a) 6= 0.
The function h is continuous on [a, b] by the ACT, and differentiable on (a, b) by the ADT. The MVT for h
states that there is c ∈ (a, b) satisfying
h′(c) =
h(b)− h(a)
b− a . (1)
We have h′(c) = (f(b)− f(a))g′(c)− (g(b)− g(a))f ′(c), and
h(b)− h(a) = (f(b)− f(a))g(b)− (g(b)− g(a))︸ ︷︷ ︸ f(b)− (f(b)− f(a))g(a) + (g(b)− g(a))︸ ︷︷ ︸ f(a)
= (f(b)− f(a))(g(b)− g(a))− (g(b)− g(a))︸ ︷︷ ︸(f(b)− f(a)) = 0.
Here, we have gathered similar common factors, denoted by underlines and underbraces. Hence, (1) yields
(f(b)− f(a))g′(c)− (g(b)− g(a))f ′(c) = 0, that is
(f(b)− f(a))g′(c) = (g(b)− g(a))f ′(c).
The result follows by dividing by g′(c) and g(b)− g(a), which are both non-zero.
B. [MTH3140]
(a) Let f : (a, b) 7→ R be differentiable and C ≥ 0. Prove that(
∀x, y ∈ (a, b) , |f(x)− f(y)| ≤ C|x− y|
)
⇔
(
C ≥ sup
z∈(a,b)
|f ′(z)|
)
.
Model solution:
Direction ⇒. By definition of the derivative, for any x ∈ (a, b) we have
f ′(x) = lim
y→x
f(y)− f(x)
y − x .
Hence, | · | being continuous,
|f ′(x)| = lim
y→x
∣∣∣∣f(y)− f(x)y − x
∣∣∣∣ .
We have | f(y)−f(x)y−x | ≤ C. Therefore, passing to the limit (using the OLT for limits of functions) we find
|f ′(x)| ≤ C. This is valid for any x ∈ (a, b) and proves that sup(a,b) |f ′| ≤ C.
Direction ⇐. By the Lipschitz estimate, for any x, y ∈ (a, b) we have
|f(x)− f(y)| ≤
(
sup
(x,y)
|f ′|
)
|x− y| ≤
(
sup
(a,b)
|f ′|
)
|x− y| ≤ C|x− y|.
(b) Let g(x) = x+ e−x if x ≥ 0 and g(x) = 1 if x < 0.
i. What is the best C from part (a) for g?
Solution:
g is differentiable on R. Outside 0 this is clear (and g′(x) = 0 if x < 0, g′(x) = 1− e−x if x > 0).
At 0 we write
g(x)− g(0)
x
=
{
1 + e
−x−e−0
x if x > 0,
0 if x < 0.
By differentiability of x 7→ exp(−x), e−x−e−0x → −e−0 = −1 as x → 0. Hence, as x → 0,
the previous expression shows that g(x)−g(0)x → 0 which proves that g is differentiable at 0 and
g′(0) = 0.
For any x ≥ 0, 0 ≤ 1 − e−x ≤ 1, and as x → ∞ we have 1 − e−x → 1. Hence, supR |g′| =
supx≥0 |1− e−x| = 1 (write details!) and the best possible C is C = 1.
ii. Prove that for any x 6= y, |g(x)− g(y)| < |x− y|.
Model solution:
The MVT gives g(x) − g(y) = g′(c)(x − y) for some c ∈ (x, y). For any c ∈ R, either g′(c) = 0 or
g′(c) = 1− e−c ∈ (0, 1). Hence, |g′(c)| < 1 and thus, since |x− y| 6= 0,
|g(x)− g(y)| = |g′(c)| |x− y| < |x− y|.
iii. Is g a contraction? Does it have a fixed point?
Solution:
The best possible C is C = 1 and g is therefore not a contraction. It does not have a fixed point,
as g(x) = 1 6= x if x ≤ 0 and g(x) = x− e−x < x if x ≥ 0.
Note that this reasoning is not an application of the contraction mapping theorem – this theorem
does not allow you to conclude “there is no fixed point”. The theorem states “contraction ⇒ fixed
point”, as usual you cannot deduce from that “not contraction ⇒ no fixed point”. Actually, if you
replace g with g − 1 then it has a fixed point, but it is still not a contraction.
This example also shows you that, to have a fixed point, g must be a “uniform” contraction, i.e.
|g(x)− g(y)| ≤ γ|x− y| for some γ < 1, not only |g(x)− g(y)| < |x− y|.
♦ More exercises from Abbott’s textbook: 5.3.3, 5.3.8
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 10, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. Using Taylor’s expansions, or otherwise, find the following limits:
(a) limx→0
exp(2x)−cos(x)
x ;
Model solution:
This limit is of indeterminate form 00 . Taylor’s expansions give
exp(2x)− cos(x)
x
=
1 + 2x+O(x2)− 1 +O(x2)
x
= 2 +O(x)
and therefore the limit as x→ 0 exists and is equal to 2.
(b) limx→2
x cos(pix)−1
x ;
Model solution:
This limit is not of indeterminate form. The function x cos(pix)−1x is continuous at x = 2 (ACT) so
limx→2
x cos(pix)−1
x =
2 cos(2pi)−1
2 =
1
2 .
(c) limx→0
exp(x)−1−x
x2 ;
Model solution:
This limit is of indeterminate form 00 . Taylor’s expansion gives
exp(x)− 1− x
x2
=
1 + x+ x2/2 +O(x3)− 1
x2
=
1
2
+O(x)
and therefore limx→0
exp(x)−1−x
x2 =
1
2 .
Note: if you know and want to use l’Hospital’s rule on this one, you need to do it twice in a row... the
final solution will probably be much longer than this one, if properly written!
2. Let f : R→ R be four times continuously differentiable. Show that, for all x ∈ I and h 6= 0 such that x+ h ∈ I,
f ′(x) =
f(x+ h)− f(x)
h
+O(h),
f ′(x) =
f(x+ h)− f(x− h)
2h
+O(h2),
f ′′(x) =
f(x+ h) + f(x− h)− 2f(x)
h2
+O(h2).
Note: Taylor expansions are extremely useful to compute approximations of high-order derivatives of f at a point
x using values of f around x. This is the foundation for developing many numerical methods for (ordinary and
partial) differential equations.
Solution:
An approach consists in developing the right-hand sides of these expressions using Taylor’s expansions. We
choose here a slightly different presentations, which actually shows better how we construct such approxi-
mations by looking for cancellations when we combine Taylor expansions.
We write the expansions for f(x+ h) and f(x− h):
f(x+ h) = f(x) + f ′(x)h+
f ′′(x)
2
h2 +
f ′′′(x)
6
h3 +O(h4), (1)
f(x− h) = f(x)− f ′(x)h+ f
′′(x)
2
h2 − f
′′′(x)
6
h3 +O(h4). (2)
The first expansion readily gives
f(x+ h)− f(x)
h
= f ′(x) +
f ′′(x)
2
h+
f ′′′(x)
6
h2 +O(h3) = f ′(x) +O(h).
Moving the last term to the left and using −O(h) = O(h) (why?) concludes the proof of the first formula
for f ′(x).
Subtracting (2) from (1), the terms involving h2 cancel out and we are left with
f(x+ h)− f(x− h) = 2hf ′(x) + f
′′′(x)
3
h2 +O(h4)
(beware, O(h4)−O(h4) = O(h4), not 0... why?). Dividing by 2h leads to the second formula for f ′(x).
Adding up (1) and (2) cancels out the terms involving h and h3 and gives
f(x+ h) + f(x− h) = f ′′(x)h2 +O(h4).
Divide by h2 and use O(h4)/h2 = O(h2) to get the formula for f ′′(x).
3. Let fn(x) =
nx
1+nx2 on the domain [−1, 1].
(a) Find the pointwise limit function f on [−1, 1].
Model solution:
Consider first x = 0. Then fn(0) = 0 so (fn(0))n∈N converges to 0.
If x 6= 0 then limn→∞ fn(x) is indeterminate of the form ±∞/+∞. We can however write
fn(x) =
x
1
n + x
2
.
Since x 6= 0, 1n + x2 → x2 6= 0 so we can apply the ALT and we find limn→∞ fn(x) = xx2 = 1x .
Defining f : [−1, 1]→ R by f(0) = 0 and f(x) = 1/x if x 6= 0, we have proved that, for any x ∈ [−1, 1],
fn(x)→ f(x) as n→∞. This precisely means that (fn)n∈N converges pointwise to f .
(b) Show that limn→∞
∫ 1
−1 fn(x) dx exists. Is it equal to
∫ 1
−1 f(x) dx?
Model solution:
fn is continuous on [−1, 1], so it is integrable. Moreover, it is odd and [−1, 1] is symmetric with respect
to 0, so
∫ 1
−1 fn(x) dx = 0. Hence, the limit of these integrals exists and is equal to 0.
However, f is not continuous on [−1, 1], and not even integrable (as an improper integral, we already
have
∫ 1
0
f(x) dx = +∞). Hence, the limit above cannot be equal to ∫ 1−1 f(x) dx, which is not defined.
4. State and prove an OLT for sequences of functions that converge pointwise.
Model solution:
Theorem: let (fn)n∈N and (gn)n∈N be sequences of functions on A that converge pointwise towards f and g,
respectively. If, for all n ∈ N and all x ∈ A, fn(x) ≤ gn(x), then f(x) ≤ g(x) for all x ∈ A.
Proof: let x ∈ A. For all n ∈ N, fn(x) ≤ gn(x). Since limn→∞ fn(x) = f(x) and limn→∞ gn(x) = g(x), the
OLT for sequences of real numbers show that f(x) ≤ g(x), as required.
5. Let f : R→ R be continuous and define the sequence of functions (fn)n∈N by fn(x) = f(x+ 1n ) for all n ∈ N and
x ∈ R. Show that fn → f pointwise.
Model solution:
For all x ∈ R, x+ 1n → x by ALT and thus, f being continuous, by sequential characterisation of continuity
we have fn(x) = f(x+
1
n )→ f(x).
Supplementary exercises
A. Let f : [a, b]→ R be continuous. Show that there exists c ∈ (a, b) such that
f(c) =
1
b− a
∫ b
a
f.
Model solution:
The function F (x) =
∫ x
a
f is continuous on [a, b] and differentiable on (a, b) with F ′ = f by fundamental
theorem of calculus. The mean value theorem then gives c ∈ (a, b) such that
F ′(c) =
F (b)− F (a)
b− a ,
which translates into
f(c) =
1
b− a
(∫ b
a
f −
∫ a
a
f
)
=
1
b− a
∫ b
a
f.
B. Let f : R→ R be indefinitely differentiable, and such that f (n)(x) ≥ 0 for x ∈ R and all n ∈ N.
(a) Show that, for any n ∈ N and any x ≥ 0,
f(0) + f ′(0)x+
f ′′(0)
2!
x2 + · · ·+ f
(n)(0)
n!
xn ≤ f(x).
Model solution:
Let An(x) be the term on the left-hand side. The Taylor formula with integral remainder, between 0
and x (instead of x and x+ h as in Theorem 5.21 in the lecture notes) shows that
f(x) = An(x) +
1
n!
∫ x
0
f (n+1)(ξ)(x− ξ)n dξ.
The integrand is positive, since f (n+1)(ξ) ≥ 0 and x− ξ ≥ 0. Hence, An(x) ≤ f(x).
(b) Deduce that
∑ f(n)(0)
n! x
n converges for any x ≥ 0.
Model solution:
This is a series with positive terms, whose partial sums An(x) are bounded by f(x). So by the behaviour
of series with positive terms, we know that this series converges.
♦ More exercises from Abbott’s textbook: 7.5.1, 7.5.2, 7.5.4, 7.5.9
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 11, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. Give an example of each of the following or argue that such a request is impossible.
(a) A sequence of functions fn : [0, 1]→ R which converges uniformly to a discontinuous function f .
Model solution:
For example, fn = f discontinuous; say f(x) = 1/2 if x ≤ 1/2 and f(x) = 0 if x > 1/2.
(b) A sequence of continuous functions fn : [0, 1]→ R which converges pointwise, but not uniformly, to a continuous
function f .
Solution:
For example, fn the “hat” function, linear on [0, 1/n], [1/n, 2/n] and [2/n, 1] with fn(0) = 0, fn(1/n) =
1, fn(2/n) = 0, fn(1) = 0 (a graph is perfectly fine to describe fn). fn → f = 0 pointwise but not
uniformy. Indeed, for any n, sup[0,1] |fn − f | ≥ |fn(1/n)− f(1/n)| = 1.
(c) A sequence of differentiable functions fn : (−1, 1)→ R which converges uniformly to a discontinuous function
f .
Model solution:
Impossible, as each fn is continuous and the uniform limit of a sequence of continuous function is
continuous.
(d) A sequence of differentiable functions fn : (−1, 1) → R which converges uniformly to a non-differentiable
function f .
Model solution:
Already seen in the lectures: fn(x) =
√
x2 + 1n , that converges uniformly to f(x) = |x|.
2. Let (xn)n∈N ⊂ [0, 1] which converges to x, and let (fn)n∈N be a sequence of functions [0, 1] → R which converges
pointwise to f .
(a) Do we have fn(xn)→ f(x) as n→∞? Justify your answer.
Model solution:
No. Take (fn) the “hat” function from Exercise 6.3 (2) in the lecture notes, and xn = 1/n. We have
fn(xn) = n which clearly does not converge.
(b) And if (fn)n∈N converges uniformly to f? Justify your answer.
Model solution:
The result is also false in general. Let fn = f be discontinuous at 1/2, say f(x) = 0 if x ≤ 1/2 and
f(x) = 1 if x > 1/2. Let xn = 1/2 + 1/n, which converges to 1/2. Then fn(xn) = f(xn) = 1 does not
converge to 0 = f(1/2).
(c) State assumptions that ensure that fn(xn)→ f(x) as n→∞.
Model solution:
If we assume the uniform convergence of (fn)n∈N and the continuity of f , then the result holds. We
write
|fn(xn)− f(x)| ≤ |fn(xn)− f(xn)|+ |f(xn)− f(x)|. (1)
For any ε > 0, the uniform convergence of (fn)n∈N gives N such that if n ≥ N we have
|fn(y)− f(y)| ≤ ε for any y ∈ [0, 1]. (2)
The continuity of f gives δ > 0 such that |x− y| ≤ δ implies |f(x)− f(y)| ≤ ε. Since xn → x, we can
find N ′ such that, for any n ≥ N ′, |xn − x| ≤ δ, and therefore
|f(xn)− f(x)| ≤ ε. (3)
Taking n ≥ max(N,N ′) and plugging (2) and (3) into (1) gives |fn(xn)− f(x)| ≤ 2ε and concludes the
proof.
3. We have encountered in Problem Set 10 the functions fn defined on [−1, 1] by fn(x) = nx1+nx2 .
(a) Does (fn)n∈N converge uniformly on [−1, 1]?
Model solution:
Each function fn is continuous on [−1, 1], while the limit function f is discontinuous. By theorem of
continuity of uniform limits, the convergence is not uniform on [−1, 1].
(b) Let a ∈ (0, 1). Prove that the convergence is uniform on [a, 1].
Model solution:
For x ∈ [a, 1] we have x ≥ a and 1 + nx2 ≥ 1 + na so
|fn(x)− f(x)| =
∣∣∣∣ nx1 + nx2 − 1x
∣∣∣∣ = 1x (1 + nx2) ≤ 1a(1 + na2) .
Thus, supx∈[a,2] |fn(x)− f(x)| ≤ 1a(1+na2) . By the ALT, the right-hand side goes to 0 as n→∞. The
squeeze theorem thus shows that supx∈[a,2] |fn(x)− f(x)| → 0 as n→∞. The convergence is therefore
uniform on [a, 2].
Note: it’s also ok to invoke the ”technique” seen in the lecture, whereby a bound |fn(x)− f(x)| ≤ ωn
(with ωn independent of x and converging to 0) yields the uniform convergence.
4. Study the definition, continuity and differentiability of the limits of the following series.
(a)
∑ sin(nx)
2n
Model solution:
Let fn(x) =
sin(nx)
2n . We have, for all x ∈ R,
|fn(x)| ≤ 1
2n
and the series of real numbers
∑
1/2n is convergent as a geometric series with ratio 1/2. The M-test
therefore shows that
∑
fn converges uniformly on R and, since each fn is continuous (ACT), that the
limit
∑
n≥0 fn is continuous on R.
Let us now consider the series of derivatives (each fn is differentiable by ADT). We have, for all x ∈ R,
|f ′n(x)| =
∣∣∣∣n cos(nx)2n
∣∣∣∣ ≤ n2n =: Mn.
We study the convergence of
∑
Mn using the ratio test. We have
Mn+1
Mn
=
n+ 1
n
2n
2n+1
=
(
1 +
1
n
)
1
2
→ 1
2
(ALT)
So by ratio test,
∑
Mn converges and, by M-test,
∑
f ′n converges uniformly on R. This shows, by
differentiation of series, that
∑
fn is differentiable on R.
(b)
∑ arctan(nx)
n2+1 .
Solution:
The function arctan is differentiable on R, and | arctan | ≤ pi/2 on R. Hence, for all x ∈ R, | arctan(nx)n2+1 | ≤
pi/2
n2+1 ≤ pi/2n2 =: Mn. The series of real numbers
∑
Mn converges by ALT and because
∑
1/n2 is a
converging series (p-series with p = 2). By M-test,
∑ arctan(nx)
n2+1 converges uniformly on R and, since
each term is continuous, the limit is continuous on R.
To study the differentiability of the limit we consider the series of derivatives:∑ d
dx
arctan(nx)
n2 + 1
=
∑ n
1 + (nx)2
1
n2 + 1
=:
∑
hn(x).
For all x ∈ R, since n/(n2 + 1) ≤ n/n2 = 1/n,
|hn(x)| ≤ 1
n(1 + n2x2)
.
We would like to bound this by a number Rn not depending on x and such that
∑
Rn converges. Let
a > 0. If |x| ≥ a then 1 + n2x2 ≥ 1 + n2a2 ≥ n2a2 and thus
|hn(x)| ≤ 1
n3a2
=: Rn.
The series
∑
Rn converges by the ALT and using the p-series with p = 3, so
∑
hn converges uniformly
on R\(−a, a) by the M-test and thus, by differentiation of series, ∑ arctan(nx)n2+1 is differentiable on this
set. Since this is true for all a > 0 this shows that this series is differentiable on R\{0}.
Supplementary exercises
A. (a) Let k ∈ N. Prove that ∑nke−n converges.
Model solution:
We use the ratio test:
(n+ 1)ke−n−1
nke−n
=
(n+ 1)k
nk
e−1 =
(
1 +
1
n
)k
e−1.
By the ALT, this last quantity tends to e−1 < 1 and thus the ratio test ensures that
∑
nke−n converges.
(b) Prove that the series
∑
cos(nx)n2e−n−sin(nx) converges uniformly on R towards a continuous function f .
Model solution:
For any x ∈ R, since | cos(nx)| ≤ 1, − sin(nx) ≤ 1 and exp is increasing,∣∣∣cos(nx)n2e−n2−sin(nx)∣∣∣ ≤ n2e−ne− sin(nx) ≤ n2e−ne =: Mn.
Mn is independent of x ∈ R and by Part (a) with k = 2,
∑
Mn = e
∑
n2e−n converges. The M-test
then ensures that
∑
cos(nx)n2e−n−sin(nx) converges uniformly on R to some f . Since each term of this
series is continuous on R, the continuity of the uniform limit shows that f is continuous.
(c) Prove that the function f in Part (b) is differentiable on R and that f ′ is continuous.
Model solution:
By the theorem of differentiation for series, the differentiability of f follows if we prove that the series
of derivatives converge uniformly on R. We have
(cos(nx)n2e−n−sin(nx))′ = −n sin(nx)n2e−n−sin(nx) − cos(nx)n2e−n−sin(nx)(−n cos(nx)). (4)
Thus,
|(cos(nx)n2e−n−sin(nx))′| ≤ n3e−ne+ n3e−ne = 2en3e−n := Kn.
The series
∑
Kn converges by Part (a) with k = 3, and thus, by the M-test,
∑
(cos(nx)n2e−n−sin(nx))′
converges uniformly on R. Hence, f is differentiable on R.
We also have f ′(x) =
∑∞
n=1(cos(nx)n
2e−n−sin(nx))′. Since each term of this series is continuous on R
(see (4)), the continuity of uniform limit ensures that f ′ is continuous on R.
B. [MTH3140] We consider the series
∑ ln(n+x2)
n2+x2 .
(a) Show that this series converges uniformly on any compact interval of the form [−a, a].
Model solution:
Let hn(x) =
ln(n+x2)
x2+n2 . For x ∈ [−a, a] we have, since ln is increasing and n2 + x2 ≥ n2,
|hn(x)| ≤ ln(n+ a
2)
n2
=
ln(n+ a2)
n1/2
1
n3/2
.
By a classical indeterminate limit, ln(n + a2)/n1/2 → 0 as n → ∞. Hence, (ln(n + a2)/n1/2)n∈N is
bounded, say by some Ra. Then,
|hn(x)| ≤ Ra
n3/2
.
The series
∑
1/n3/2 converges (p-series with p = 3/2 > 1) and the bound Ra/n
3/2 is indepenent of
x ∈ [−a, a]. By the M-test, ∑hn converges uniformly on [−a, a].
(b) Call h(x) the limit of the series at x. On what set is h continuous?
Model solution:
Each function hn is continuous on R and in particular on [−a, a]. Moreover the series converges
uniformly on [−a, a]. It follows by the continuity of uniformly convering series, that h is a continuous
function on [−a, a]. But a is an arbitrary positive number so h is continuous at every point x ∈ R.
(c) On what set is h differentiable?
Solution:
We have
h′n(x) =
2x
(n+ x2)(n2 + x2)
− 2x ln(n+ x
2)
(n2 + x2)2
.
Fix a > 0 and take x ∈ [−a, a]. Then, with the same Ra as in Part (a), ln(n + x2) ≤ Ran1/2. Hence,
using n+ x2 ≥ n and n2 + x2 ≥ n2,
|h′n(x)| ≤
2a
n3
+
2aRan
1/2
n4
≤ 2a
n3
+
2aMa
n7/2
=: Mn.
Mn does not depend on x ∈ [−a, a], and
∑
Mn converges (by the ALT for series, since
∑
1/n3 and∑
1/n7/2 converge as p-series with p > 1). Hence,
∑
h′n converge uniformly on [−a, a] by the M-test.
The differentiation theorem for series then ensures that h =
∑∞
n=1 hn is differentiable on (−a, a). Since
this is true for any a, this shows that h is differentiable on R.
♦ More exercises from Abbott’s textbook: 6.2.1, 6.2.2, 6.2.3, 6.2.4, 6.2.5, 6.2.8, 6.2.9, 6.3.2, 6.3.4, 6.4.3, 6.4.5
MTH2140-3140 Real Analysis
School of Mathematical Sciences, Monash University
Problem Set 12, 2021
Solutions
Model solutions: these solutions present all the arguments, and no more, that are required to attract full marks in an assignment. Because they are
in a sense minimal, they do not always explain the ideas that lead to the solution... this is the purpose of the tutorial and of working with your tutors
and fellow students.
Solutions: they usually present all the arguments, but also contain additional comments, intuitions or ideas that would not be required to get full
marks in an assignment, but that should help you understand how we get to that solution. From time to time, some small arguments are omitted (and
indicated by texts such as “prove it” or “write the details”).
Core exercises
1. (Abbott, exercise 6.4.1) Prove that if
∑
gn converges uniformly then (gn)n∈N converges uniformly to 0. (Hint:
uniform Cauchy criterion for series. Alternatively, think about the proof of the n-th term test for series of real
numbers.)
Solution:
The uniform Cauchy criterion on
∑
gn states that, for ε > 0, there is N ∈ N such that, if n ≥ p ≥ N , for
all x in the domain,
|gp(x) + · · ·+ gn(x)| ≤ ε.
Writing this with p = n yields |gn(x)| ≤ ε for all x and all n ≥ N , which is what we wanted.
Alternative approach (longer if done properly): consider Sn =
∑n
k=1 gk the partial sum of the series, and
write that gn = Sn − Sn−1. We know that (Sn)n∈N converges uniformly to some S. A “subsequence limit
theorem for uniformly convergent sequences of functions” (to be written and proved!) shows that (Sn−1)n∈N
also converges uniformly to S. Then, an ALT for uniformly convergent sequences of functions (also to be
written and proved!) yields gn = Sn − Sn−1 → 0 uniformly.
2. Prove that g(x) =
∑
n≥1
sin(2nx)
n2 x
n is defined and continuous on [−1, 1].
Solution:
g is not a power series, but we can apply the M -test to prove that it converges uniformly on [−1, 1]. Indeed,
| sin(2nx)xn/n2| ≤ 1/n2 for any x ∈ [−1, 1]. Since each term of the series is continuous on [−1, 1], the
continuity of g follows from the uniform convergence of the series.
3. Consider the function g defined by g(x) =
∑∞
n=1(
−1
2 )
n+1 1
nx
n.
(a) Find the radius of convergence R of this series.
Model solution:
We use Lemma 6.28 (a direct consequence of the ratio test). Here, an = (
−1
2 )
n+1 1
n so∣∣∣∣an+1ann
∣∣∣∣ =
∣∣∣∣∣ (−12 )n+2 1n+1(−12 )n+1 1n
∣∣∣∣∣ =
∣∣∣∣ −n2(n+ 1)
∣∣∣∣ = ∣∣∣∣ 12(1 + 1n )
∣∣∣∣→ ` := 12 as n→∞, by ALT.
By Lemma 6.28, the radius of convergence is R = 1/` = 2.
(b) Find the interval of convergence I of this series. (Check whether the endpoints are included).
Model solution:
If x = 2, the series is
∑
(−12 )
n+1 1
n2
n = 12
∑
(−1)n+1 1n . This is half times the alternating harmonic
series which we know converges (by the alternating series test).
If x = −2, the series is ∑(−12 )n+1 1n (−2)n = −12 1n . This is negative half times the harmonic series which
we know diverges (a p-series with p = 1).
Thus the interval of convergence is I = (−2, 2].
(c) Where is g differentiable?
Solution:
Note that g is the sum of the series so it is only defined on the interval of convergence I. The sum
function of a power series is differentiable on its open interval of convergence, so g is differentiable on
the interval (−2, 2).
Note: it turns out (from part (e) below) that g is also differentiable at x = 2, provided we use the
definition of “one-sided derivative”.
(d) Find g ′(x) in its simplest form.
Model solution:
Since we have a power series, termwise differentiation is valid on the open interval of convergence (−2, 2)
and, on this interval,
g ′(x) =
d
dx
∞∑
n=1
(
−1
2
)n+1
1
n
xn =
∞∑
n=1
(
−1
2
)n+1xn−1 = (
−1
2
)2
∞∑
n=1
(
−x
2
)n−1 =
1
4
∞∑
k=0
(
−x
2
)k.
We recognise a geometric series, whose sum is known (notice that |x/2| < 1):
g′(x) =
1
4
1
1− (−x/2) =
1
4
2
2 + x
=
1
4 + 2x
.
(e) Find g(x) in its simplest form.
Solution:
Taking the anti-derivatives in Part (d) we get, for x ∈ (−2, 2), g(x) = 12 ln(4 + 2x) + c. But g(0) = 0 so
c = − 12 ln(4) = − ln 2 and g(x) = 12 ln(4 + 2x)− ln 2 = 12 ln (2 + x)− 12 ln 2.
For x = 2 things are a bit more complicated. The exercise was a bit open here and we were not
necessarily expecting you to deal with this specific case, but let’s see how we can do.
A theorem (the Abel theorem for series) claims that if a power series converges at one of the endpoints
of the interval of convergence, then it is continuous at this endpoint; here, this would enable us to see
that g is continuous at x = 2 and thus that the expression g(x) = 12 ln (2 + x)− 12 ln 2 can be extended
and is valid at x = 2 (because the limit as x→ 2 of each side exists).
To get around using this theorem, that we haven’t covered, an option is to substitute x = 2 in the
series defining g, as we already did, to find g(2) = 12
∑∞
n=1
(−1)n+1
n . For y ∈ R, the partial sum∑N
n=1
(−1)n+1yn
n = y − y
2
2 +
y3
3 + · · · + (−1)
N+1yN
N is the Taylor expansion of ln(1 + y). By Corollary
5.23, the remainder of this expansion is bounded, for y = 1, by 1(N+1)! supξ∈[0,1] | d
N+1
dyN+1
(ln(1 + y))|.
Calculating the (N + 1)-th derivative of ln(1 + y) is not difficult, and plugging the obtained expression
inside the supremum shows that the remainder for y = 1 tends to 0 as N →∞. Hence, ∑Nn=1 (−1)n+1n →
ln(1+1) = ln(2), which shows that g(2) = 12 ln(2) (which is indeed
1
2 ln(2+2)− 12 ln(2) since ln(2+2) =
ln(22) = 2 ln(2)).
4. We consider a power series
∑
anx
n such that all odd coefficients vanish (a1 = a3 = . . . = 0), that none of the even
coefficients vanish, and that, for some m ≥ 0,
lim
n→∞
|a2n+2|
|a2n| = m (1)
(note that the ratio is that of two consecutive non-zero coefficients in the power series). Find the radius of conver-
gence of the series.
Solution:
The property above looks very much like the one in Lemma 6.28 in the lecture notes. The lemma itself
cannot be used because every other coefficient vanish, but we could think about following the arguments in
the proof of the lemma and adapt them to the present situation. This would require to establish a ratio test
for series of real numbers with every other coefficient vanishing.
There is however a faster way, which consists in noting that the series is written
a0 + a2x
2 + a4x
2 + · · · = a0 + a2(x2) + a4(x2)2 + · · · .
In other words, it is a series in x2 whose coefficients a0, a2, a4, etc. do not vanish. Lemma 6.28 is applicable
to this series. Let’s make that rigorous.
Define the power series, in the variable y, such that
∑
bny
n with bn = a2n. Then none of the bk vanish,
and (1) shows that |bn+1|/|bn| → m as n → ∞. Hence, by Lemma 6.28,
∑
bny
n has radius of convergence
1/m (or +∞ if m = 0). This means that for any |y| < 1/m, ∑ bnyn converges, and that for any |y| > 1/m,∑
bny
n diverges.
Now consider y = x2. If |x| < 1/√m then |y| < 1/m and ∑ bnyn = ∑ a2nx2n converges; but this series is
precisely
∑
anx
n since the odd coefficients vanish. Hence, for |x| < 1/√m the series ∑ anxn converges. The
same reasoning (based on the divergence of
∑
bny
n when |y| > 1/m) shows that for |x| > 1/√m the series
diverges, and thus that its radius of convergence is exactly 1/
√
m.
5. Let
∑
anx
n be a power series such that all coefficients are non-zero and, for some m ≥ 0,
lim
n→∞
|an+2|
|an| = m. (2)
Show that the radius of convergence of the series is 1/
√
m (or +∞ if m = 0) (Hint: Exercise 4 might be helpful).
Solution:
Let us split the series in two, with even indices on one side and odd indices on the other side:
∑
anx
n =∑
a2nx
2n+
∑
a2n+1x
2n+1. Introducing zero coefficients as odd (resp. even) terms in the first (resp. second)
series, we can come back to classical series without ‘gaps’.
Precisely, let us define
a˜n =
{
an if n is even,
0 if n is odd
ân =
{
0 if n is even,
an if n is odd
Then
∑
anx
n =
∑
a˜nx
n +
∑
ânx
n.
The subsequence limit theorem (with the subsequence of even indices) applied to the limit (2) shows that
|a˜2n+2|/|a˜2n| = |a2n+2|/|a2n| → m as n→∞. Hence, since a˜k = 0 for odd k, Exercise 4 shows that
∑
a˜nx
n
has radius of convergence 1/
√
m.
Applying now the SLT to (2) with the subsequence of odd indices shows that |â2n+3|/|â2n+1| =
|a2n+3|/|a2n+1| → m as n → ∞. Since all âk corresponding to even k vanish, an easy modification of
the first item above (or a shift of the series by one index, transforming odd indices into even indices) shows
that
∑
ânx
n has radius of convergence 1/
√
m.
Hence,
∑
anx
n =
∑
a˜nx
n +
∑
ânx
n has radius of convergence 1/
√
m, as the sum of two power series with
the same radius of convergence is a power series with the same radius of convergence (we use here the ALT...).
6. Legendre’s equation is the ODE (1− x2)y′′(x)− 2xy′(x) + ν(ν + 1)y(x) = 0, where ν is a positive real number.
(a) Find a recurrence relation for the coefficients of all power series solutions to this equation.
Model solution:
Plug y =
∑
n≥0 anx
n into the equation:
(1− x2)
∑
n≥2
n(n− 1)anxn−2 − 2x
∑
n≥1
nanx
n−1 + ν(ν + 1)
∑
n≥0
anx
n = 0.
Develop the products:∑
n≥2
n(n− 1)anxn−2 −
∑
n≥2
n(n− 1)anxn −
∑
n≥1
2nanx
n + ν(ν + 1)
∑
n≥0
anx
n = 0.
Shift the index in the first sum (set k = n− 2):∑
k≥0
(k + 2)(k + 1)ak+2x
k −
∑
n≥2
n(n− 1)anxn −
∑
n≥1
2nanx
n + ν(ν + 1)
∑
n≥0
anx
n = 0.
Notice that all series can actually start at the index 0 (since the terms corresponding to n = 0, 1 in the
second series, and the term corresponding to n = 0 in the third series, vanish); after changing k into n
(just a change of ‘name’ of the index) and gathering the terms:∑
n≥0
[(n+ 2)(n+ 1)an+2 − n(n− 1)an − 2nan + ν(ν + 1)an]xn = 0.
The recurrence relation is: for all n ≥ 0,
(n+ 2)(n+ 1)an+2 − n(n− 1)an − 2nan + ν(ν + 1)an = 0
which gives, after a bit of algebra,
an+2 =
n(n+ 1)− ν(ν + 1)
(n+ 2)(n+ 1)
an =
(n− ν)(n+ 1 + ν)
(n+ 2)(n+ 1)
an
(the second form is not necessary, but pleasant to work with). Note that a0 and a1 are free, and
completely determine all the coefficients.
(b) If ν is a natural number, show that some of these solution are (non-zero) polynomials.
Model solution:
If ν ∈ N, then the recurrence relation shows that aν+2 = 0, and therefore aν+4 = aν+6 = · · · = 0: all
coefficients with an index greater than ν and having the same parity vanish. Assume that ν is even.
Selecting a1 = 0, the recurrence relation show that all odd coefficients vanish. Hence, if a0 6= 0, the
solution is y(x) = a0 + a2x
2 + · · ·+ aνxν/2, and each a2l here is actually non-zero. (Note: if a0 = 1, y
is called the ν-th Legendre polynomial.)
Similarly, if ν is odd, a0 = 0 and a1 is not zero, y(x) = a1x + a3x
3 + · · · + aνx(ν−1)/2 is a non-zero
polynomial. (Note: this is the ν-th Legendre polynomial if a1 = 1.).
(c) Find the radius of convergence of all the power series solutions to the Legendre equation (Hint: Exercises 4
and 5 can be useful).
Solution:
If a0 and a1 are both non-zero and ν is not a natural number, all coefficients are non-zero and Exercise
5 can be applied. We have, after division by n2 of the numerator and denominator,
|an+2|
|an| =
n(n+ 1)− ν(ν + 1)
(n+ 2)(n+ 1)
=
1(1 + 1n )− ν(ν+1)n2
(1 + 2n )(1 +
1
n )
→ 1 by the ALT.
Hence, the radius of convergence is 1/
√
1 = 1.
If a0 or a1 is zero, and ν is not a natural number, every other coefficient vanish and every other coefficient
is non-zero, so Exercise 4 can be applied, giving again 1 as radius of convergence.
If ν is a natural number, the solution is either a pure polynomial (with radius of convergence +∞), or
the sum of a polynomial and one of the non-polynomial solutions above, giving a radius of convergence
of 1 (same as the one for the non-polynomial solution above).
Supplementary exercises
A. This is a follow up to the supplementary exercise A in Problem Set 10. As in this exercise, we take f : R → R
indefinitely differentiable, and such that f (n)(x) ≥ 0 for x ∈ R and all n ∈ N. Prove that ∑ f(n)(0)n! xn converges
uniformly on [−a, a], for any a > 0.
Solution:∑ f(n)(0)
n! x
n is a power series and we proved in Exercise A of Problem Set 10 that it converges for any x > 0.
Hence,
∑ f(n)(0)
n! (a+ 1)
n converges. By the convergence of power series,
∑ f(n)(0)
n! x
n converges uniformly on
[−a, a].
With a little bit more reasoning (working on the Taylor remainder), we can show that the limit of this power
series is actually f .
B. [MTH3140] Let (an)n∈N be a sequence of reals such that |an|1/n → ` ∈ [0,∞) as n → ∞. Prove that the
power series
∑
anx
n has radius of convergence R = 1/` (hint: start by assuming that ` > 0 and notice that, if
0 ≤ `′ < ` < `′′, for n large enough we have (`′)n ≤ |an| ≤ (`′′)n).
Model solution:
If ` > 0 and 0 ≤ `′ < ` < `′′ then, for n large enough, `′ < |an|1/n < `′′ and therefore (`′)n ≤ |an| ≤ (`′′)n.
We deduce, for any x ∈ R,
|`′x|n ≤ |anxn| ≤ |`′′x|n. (3)
If |x| < R then ∑ anxn converges absolutely and therefore so does ∑(`′x)n, which means that |`′x| < 1, or
|x| < 1/`′. Since this is true for any |x| < R, we deduce that R ≤ 1/`′ and, letting `′ → `, that R ≤ 1/`.
Equation (3) also shows that if |x| < 1/`′′, then ∑ anxn converges absolutely (because ∑ |`′′x|n converges).
This shows that R ≥ 1/`′′, and letting `′′ → ` that R ≥ 1/`. The proof is therefore complete if ` > 0.
If ` = 0, then the same reasoning can be done without `′ (i.e. using only `′′) and shows that, for any `′′ > 0,
R ≥ 1/`′′. Letting `′′ → 0 gives R =∞, as required.
♦ More exercises from Abbott’s textbook: 6.5.2, 6.5.4, 6.5.6, 6.5.8, 6.5.10

欢迎咨询51作业君