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Elementary Analysis Kenneth A. Ross Selected Solutions Angelo Christopher Limnios
Elementary Analysis
Kenneth A. Ross
Selected Solutions
Angelo Christopher Limnios
EXERCISE 1.2
Claim:P (n) = 3 + 11 + + (8n 5) = 4n2 n n N
Proof :
By induction. Let n = 1. Then 3 = 4(1)2 (1) = 3, which will serve as theinduction basis. Now for the induction step, we will assume P (n) holds trueand we need to show that P (n+ 1) holds true. P (n+ 1) = 3 + 11 + + (8n5) + (8(n+ 1) 5) = 4(n+ 1)2 (n+ 1).Now, 3 + 11 + + (8n 5) + (8(n+ 1) 5) = 4n2 n + 8(n + 1) 5. So inorder to show that P (n+ 1) is true, we need to show that 4(n+ 1)2 (n+ 1) =4n2 n+ 8(n+ 1) 5, which, when evaluated is true, as desired.
EXERCISE 1.3
Claim:P (n) = 13 + 23 + + n3 = (1 + 2 + + n)2 n N
Proof :
By induction. Let n = 1. Then, P (1) = 13 = 1 = 12, which will serve as theinduction basis. Now for the induction step, we will assume P (n) holds trueand we need to show that P (n+ 1) holds true.
P (n+ 1) = 13 + 23 + + n3 + (n+ 1)3 = (1 + 2 + + n)2 + (n+ 1)3
= (1 + 2 + + n)2 + [(n+ 1)3 (n+ 1)2] + (n+ 1)2
= (1 + 2 + + n)2 + (n+ 1)2[n+ 1 1] + (n+ 1)2
= (1 + 2 + + n)2 + 2(n+ 1)n(n+ 1)/2 + (n+ 1)2
= (1 + 2 + + n)2 + 2(n+ 1)(1 + 2 + + n) + (n+ 1)2
= 1(1 + 2 + + (n+ 1))2,as desired.
1(a+ b)2 = a2 + 2ab+ b2, where a = 1 + 2 + + n and b = n+ 1.
1
EXERCISE 1.4
(a) For
n = 1, the expression is 1.
n = 2, the expression is 4.
n = 3, the expression is 9.
n = 4, the expression is 16.
We note that 1 + 3 + (2n 1) = n2, which is the proposed formula.(b) Claim:
P (n) = 1 + 3 + (2n 1) = n2 n NProof :
By induction. We already showed the case when n = 1. Now for the inductionstep, we will assume P (n) holds true and we need to show that P (n+ 1) holdstrue.
P (n+ 1) = 1 + 3 + (2(n+ 1) 1) = 1 + 3 + + (2n 1) + (2n+ 1)
= n2 + 2n+ 1 = (n+ 1)2,
as desired.
EXERCISE 1.6
Claim:P (n) = (11)n 4n is divisible by 7 when n N
Proof :
By induction. P (1) = 11 4 = 7 is divisible by 7. Now for the induction step,we will assume P (n) holds true and we need to show that P (n+ 1) holds true.
P (n+ 1) = (11)n+1 4n+1 = 11n+1 4 11n + 4 11n 4n+1
= 11n(11 4) + 4(11n 4n) = 7 11n + 4(11n 4n).Since we assumed that 11n 4n was divisible by 7 because we assumed P (n)was true, we can see that
11n+1 4n+1
is divisible by 7, as desired.
2
EXERCISE 1.9
(a) The inequality does not hold for n = 2, 3, 4. It holds true for all other n N.(b) It is true by by inspection for n = 1 and 24 = 42 also holds for n = 4.Implement the induction step. For n 4, if 2n n2, then 2(n+1) > (n + 1)2.But 2(n+1) = 2 2n 2n2 > (n + 1)2 iff (n + 1) < 2n, for example whenn > 1
21 =
2 + 1, which includes n N : n 4.
EXERCISE 1.10
Claim:
(2n+ 1) + (2n+ 3) + (2n+ 5) + + (4n 1) = 3n2 n N.
Proof :
Using the results from 1.4, we can avoid induction. Observe that
(2n+1)+(2n+3)+ +(4n1) = (1+3+ +(4n1))(1+3+ +(2n1))
= (1 + 3 + + 2(2n) 1) (1 + 3 + + (2n 1))= (2n)2 n2 = 4n2 n2 = 3n2,
as desired.
EXERCISE 1.12 (b) and (c)
(b) Observe the following:
1k
+1
n k + 1 =(n k + 1) + kk(n k + 1) =
n+ 1k(n k + 1) .
Hence,
n!k!(n k)! +
n!(k 1)!(n k + 1)! =
n!(k 1)!(n k)!
[1k
+1
n k + 1]
=n!
(k 1)!(n k)![
n+ 1k(n k + 1)
]=
(n+ 1)!k!(n k + 1)! .
(c) For n = 1 : (a+ b)n = a+ b =(11
)a+
(11
)b. Now, for n 1, we get
(a+ b)n =nk=0
(n
k
)akbnk.
3
Hence,
(a+ b)(n+1) = (a+ b)nl=0
(n
l
)albnl =
n+1k=0
[(n
k 1)
+(n
k
)]akbnk+1
=n+1k=0
(n+ 1k
)akbn1+k.
EXERCISE 2.1
Show that
3 is not rational (The exact same technique can be used to showthe other numbers are not rational).
Claim: 3 / R
Proof :
By Contradiction. If
3 R, then an r Q : pq =
3, where r = pq =(pq
)2= 3 = p2 = 3q2 = p2 is divisible by 3 = p is divisible by 3. Let p =
3k. Then 9k2 = 3q2 = 3k2 = q2 = q2 is divisible by 3 = q is divisible by 3,which is a contradiction.
EXERCISE 2.5
Lets assume [3+
2]23 does represent a rational number. Further, lets call this
rational number q. This implies q3 = [3 +
2]2 = 9 + 6
2 + 2 = 11 + 6
2,which in turn implies
2 = (q
311)6 , which is a rational number. But we know
2 / Q - a contradiction.
4
EXERCISE 3.3
This problem will be done in two parts.
Part I: Show that (a)(b) = ab (a, b) R.Proof:
(a)(b) = ab (1) (a)(b) + (ab) = ab+ (ab) (2) (a)(b) + (a)b = ab+ (ab) (3) (a) [(b) + b] = ab+ (ab) (4) (a) [b+ (b)] = ab+ (ab) (5)
(a)(0) = ab+ (ab) (6)0 = ab+ (ab), (7)
which is true by A4. Hence, (a)(b) = ab2
Part II: ac = bc and c 6= 0 imply a = b (a, b, c) R.Proof:
ac(c1) = bc(c1) (8)
a(cc1) = b(cc1) (9)a(1) = b(1) (10)
Hence, a = b3
EXERCISE 3.6
a. Claim: |a+ b+ c| |a|+ |b|+ |c|
Proof: One iteration of the triangle inequality can be used to construct:
|a+ b+ c| = |(a+ b) + c| |a+ b|+ |c| .A second iteration of the triangle inequality yields:
|a+ b|+ |c| |a|+ |b|+ |c| ,2(1) used (i), (2) used (iii), (3) used DL, (4) used A2, (5) used A4, (7) reintroduced
(1) and the conclusion reused (i).3(8) was built from multiplying both sides of the equality by the same element to preserve
equality, (9) used M1 and (10) used M4.
5
as desired. b. We want to show |a1 + a2 + + an| |a1|+ |a2|+ |an|
We will use the principal of mathematical induction to show the above is true.
Proof:
P1 is certainly true, since |a1| = |a1|, and thus will serve as our basis for induc-tion. Our induction hypothesis is:
Assume an n R : Pn holds, i.e. Pn : |a1 + a2 + + an| |a1|+|a2|+ |an| .
Implementing the induction step and the triangle inequality yield:
|a1 + a2 + + an + an+1| |a1 + a2 + + an|+ |an+1| .
The induction step and O4 yield:
|a1 + a2 + + an + an+1| |a1|+ |a2|+ + |an|+ |an+1| ,
as desired.
EXERCISE 4.2, (a) through (n)
This exercise is asking us to list three lower bounds for the set; if the set is notbounded below, it will be labeled NOT BOUNDED BELOW or NBB. Therule outlined in the text we will follow is Definition 4.2 (b):
If a real number m satisfies m s s S, then m is called a lower bound of Sand the set S is said to be bounded below.
a. S = [0, 1]; 1, 2 and 3 satisfy Definition 4.2 (b).
b. S = [0, 1); 1, 2 and 3 satisfy Definition 4.2 (b).
c. S = {2, 7}; 1, 2 and 3 satisfy Definition 4.2 (b).
d. S = {pi, e}; 1, 2 and 3 satisfy Definition 4.2 (b).
e. S ={1n : n N
}; 1, 2 and 3 satisfy Definition 4.2 (b).
f. S = {0}; 1, 2 and 3 satisfy Definition 4.2 (b).
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g. S = [0, 1] [2, 3]; 1, 2 and 3 satisfy Definition 4.2 (b).
h. S =n=1 [2n, 2n+ 1]; 1, 2 and 3 satisfy Definition 4.2 (b).
i. S =n=1
[ 1n , 1 + 1n]; 1, 2 and 3 satisfy Definition 4.2 (b).j. S =
{1 13n : n N
}; 1, 2 and 3 satisfy Definition 4.2 (b).
k. S ={n+ (1)
n
n : n N}
; 1, 2 and 3 satisfy Definition 4.2 (b).
l. S = {r Q : r < 2}; NBB. The set Q has no maximum or minimum4.S has an upper bound, but since there is no lower bound outlined, this set isunbounded from below.
m. S ={r Q : r2 < 4}; 21 , 31 and 41 satisfy Definition 4.2 (b).
n. S ={r Q : r2 < 2}; 21 , 31 and 41 satisfy Definition 4.2 (b).
EXERCISE 4.4, (a) through (n), ADDITIONALLY, DETERMINE IF THEMINIMUM OF THE SET EXISTS.
This exercise is asking us to give the infima of each set S. The rule outlined inthe text we will follow is Definition 4.3 (b):
If S is bounded below and S has a greatest lower bound, then we will call it theinfimum of S and denote it by infS.
Additionally, we are asked to determine if the minimum of the set (denotedminS) exists. Note that, unlike the minimum of the set S, infS need not belongto the set S5
a. S = [0, 1]; infS = minS = 0.
b. S = (0, 1); infS = 0, minS D.N.E.
c. S = {2, 7}; infS = minS = 2.
d. S = {pi, e}; infS = minS = e.
4Page 20, Example 1 (c).5Page 21, first paragraph below Definition 4.3, first sentence.
7
e. S ={1n : n N
}; infS = 0, however, minS does not exist.
f. S = {0}; infS = minS = 0.
g. S = [0, 1] [2, 3]; infS = minS = 0.
h. S =n=1 [2n, 2n+ 1]; infS = minS = 2.
i. S =n=1
[ 1n , 1 + 1n]; infS = 0, minS D.N.E.j. S =
{1 13n : n N
}; infS = minS = 23 .
k. S ={n+ (1)
n
n : n N}
; infS = minS = 0.
l. S = {r Q : r < 2}; both infS and minS do not exist since S is NBB.
m. S ={r Q : r2 < 4}; infS = 2, however, minS does not exist.
n. S ={r Q : r2 < 2}; infS = 2, however, minS does not exist.
EXERCISE 4.8
Let S and T be non-empty subsets of R, with s t s S and t T .(a) Since it is given that s t s S and t T , we know that any elementof T will bound S from above, so supS. Conversely, since it is given thats t s S and t T , we know that any element of S will bound T frombelow, so infT .
(b) Claim:supS infT
This will be done in two parts.
Proof :
Given s t t T , it follows that s is a lower bound for T . By definition, infTis the greatest lower bound for T . Hence, s infT s S.Since s infT s S, infT is an upper bound for S. It follows that since supSis the least upper bound for S,
supS infT,
8
as desired. (c) Let S = (0, 1] and T = [1, 2). It is readily observable that supS = infT . Itis also readily observable that
S T 6= .
(d) Let S = (0, 1) and T = (1, 2). It is readily observable that supS = infT . Itis also readily observable that
S T = .
EXERCISE 4.12
Claim:
Given a < b, x R \Q : a < x < b.Proof :
Following the hint, we know r +
2 R \ Q when r Q. By contradiction,if r Q and if the number x = r + 2 Q, then 2 = x r would havebeen Q, a contradiction. Now, due to the denseness of Q in R, we find anr Q : a2 < r < b2. Then we have x = r +2 R \Q and we havea < x < b, as desired.
EXERCISE 4.14 (a)
Let A and B be nonempty bounded subsets of R, and let S be the set of allsums a+ b where a A and b B.Claim: supS = supA + supB.
Proof :
The information we are given is that A and B are both subsets of R whichare non-empty and bounded from above (since they are bounded, this impliesthey are bounded from above and below). Since both subsets A and B are bothnon-empty and bounded from above, they both possess a least upper bound,i.e., both supA and supB exist6. If S is to be defined as the set whose elementsare all of the sums a + b, where a A and b B and we are given both Aand B bounded, we know S is bounded; more specifically bounded from belowand above. Hence, by the Completeness Axiom, S is bounded above and as aconsequence supS exists. If supS exists, then a number O R : r 0 r Sand whenever O1 < O, r1 O : r1 < O1..
6Otherwise, the Completeness Axiom wouldnt hold.
9
If both supA and supB exist, then there exists a number M R : s M s Aand there exists a number N R : t N t B; whenever M1 < M s1 A : s1 < M1 and whenever N1 < N t1 B : t1 < N1.As a consequence of the above identities, it is obvious that since S set of allsums a + b where a A and b B, O M + N , and hence supS = supA +supB, as desired.
EXERCISE 4.15
Claim:
Let a, b R. If a b+ 1n n N, then a b.Proof :
By contradiction. Assume a > b. This implies a b > 0. By the ArchimedianProperty of R n N : a b > 1n . Using this specific n, we have a > b+ 1n , acontradiction, as desired.
EXERCISE 4.16
Claim: sup{r Q : r < a} = a for each a R.Proof :
Let S {r Q : r < a}. From this, we can see that S is certainly boundedfrom above, since we can chose any number a + n with n N and Definition4.2 (b) will still be satisfied. Now by the denseness of Q in R, we know that7if (a, b) R with a < b, then an r Q : a < r < b. So although we knowS is bounded and we can certainly povide many upper bounds for this set, ifwe search for a least upper bound for this set, the search will go on indefinitely,since whatever rational upper bound is discovered, it is always possible to findone smaller. Thus, although a R, but a / S, this does not disqualify it frombeing the supS. However, since the search for a least upper bound for S goeson indefinitely due to the denseness of Q, by the definition of the supremum ofa set, sup{r Q : r < a} = a for each a R, as desired.
EXERCISE 7.3 (b), (d), (f), (h), (j), (l), (n), (p), (r), (t)
(b) bn = n2+3n23 converges to 1.
(d) tn = 1 + 2n converges to 1.
(f) sn = (2)1n converges to 1.
(h) dn = (1)nn diverges.7Proof ommitted, as it was done in class
10
(j) 7n3+8n
2n331 converges to72 .
(l) sin(npi2
)diverges.
(n) sin(2npi3
)diverges.
(p) 2n+1+52n7 =
2n(2)+52n7 converges to 2.
(r)(1 + 1n
)2 converges to 1.(t) 6n+49n2+7 converges to 0.
EXERCISE 8.4
If (sn) is a sequence which converges to 0 and (tn) is a bounded sequence, thenthe sequence (sntn) converges to 0.
We will argue the fact that we are given a sequence (sn) which converges to 0so we know that
limn (sn) = 0,
and we are given a sequence (tn) which is a bounded sequence, so we know that
a constant M : |tn| M n N,which means, in a geometric sense, that we can find an interval [M,M ] thatcontains every term in the sequence tn.
With the above given, we can almost assuredly argue with what we know aboutproducts that if one sequence which arbitrarily closes in on the value zero ismultiplied by another sequence which is bounded by a constant we can con-clude that their product will eventually close in on the value zero since zero a constant = zero (eventually).
Claim: Given (sn), a sequence which converges to 0 and (tn), a bounded se-quence,
limn(sntn) 0.
Proof :
Let > 0 be given. Since we know limn sn 0, we can always find ann > N : |sn 0| < < M 8, for some constant M > 1. Further, a constantM : |tn| M n N. Then, for n > N , it holds that
|sntn 0| = |sn| |tn| N : sn = 1 =sn+3 = 1 = |sn s| 1 or |sn+3 s| 1, by the triangle inequality, whichproves this sequence does not converge, as desired.
EXERCISE 8.8 (c)
Claim:
limn[
4n2 + n 2n] = 14 .Proof :
4n2 + n 2n = n4n2 + n+ 2n
=1
2
1 + 14n + 2
If 1 < a, then 1 < a < a2 which then implies
1 limn
1 +
12n limn
[1 +
12n
]= 1 + 0 = 1.
Now, applying the limit theorems, we see that
limn
1
2
1 + 14n + 2=
12 1 + 2 =
14,
as desired.
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EXERCISE 8.10
We are given s > a. Let limn sn = s.
Claim:
a number N : n > N = sn > a.Proof :
s > a = there is an > 0 : s > a9 For such chosen N : n > N =sn (s , s+ ) = n > N = sn > s > a, as desired.
EXERCISE 9.4
Let s1 = 1 and for n 1 let sn+1 =sn + 1.
(a) The first four elements are 1,
2,
2 + 1,
2 + 1 + 1.
(b) Assume sn converges.
Claim: limn sn = 12 (1 +
5).
Proof :
We are given the assumption that sn converges. We will name some asthe element sn converges to. If sn converges, sn+1 inherently converges; morespecifically, sn+1 converges to , as well. From the problem, we have
(sn+1)2 = sn + 1,
which when utilized with the assumption laid out will converge to
()2 10 = + 1 2 1 = 0.
Implementing the quadratic formula11 to find a pair of solutions to the equationyields
1 =1 +
52
and 2 =15
2.
Since the greatest lower bound of the sequence sn is 1, the solution we wantto pick will reflect that the limit will trivially be positive, eliminating 2 as asolution.
Hence, limn sn = 1 = 12 (1 +
5).
9For example := sa2
.10For this, we will use the fact that the limit of a product of sequences is the product of the
limits, i.e. if lim sn s and lim tn t, then lim(sntn) st.11The quadratic formula, learned from principals, is x = b
b24ac2a
.
13
EXERCISE 9.5
Assume lim tn exists and is defined to be t. Then lim tn+1 = t as well. For alln, we get 2tntn+1 = t2n + 2. Implementing the limit theorems, we see that
2t2 = t2 + 2 = t =
2.
Since we are given t1 = 1 and the fact that the sequence stays positive, we caneliminate 2 and conclude that the limit equals 2.
EXERCISE 9.11
(a) Show that if lim sn = + and inf{tn : n N} > , then lim (sn + tn) =+.Let M > 0 and let m = inf {tn : n N}. We want sn+ tn > M for n > N . Thiswill be sufficed by sn + m > M or sn > M m for n > N . So we will choosean N : sn > M m for n > N . Let = M m. From the constraints above,we see that 0 for n > N , wecan conclude that lim (sn + tn) = +.(b) Show that if lim sn = + and lim tn > , then lim (sn + tn) = +.If lim tn > , we can conclude that inf{tn : n N} exists. We can then usethe same argument as above by letting M > 0 and m = inf {tn : n N}. Wewant sn+tn > M for n > N . This will be sufficed by sn+m > M or sn > Mmfor n > N . So we will choose an N : sn > M m for n > N . Let = M m.From the constraints above, we see that 0 < . Since we can always findan N : sn > for n > N , we can conclude that lim (sn + tn) = +.(c) Show that if lim sn = + and if (tn) is a bounded sequence, then lim (sn + tn) =+.Let M > 0. If (tn) is a bounded sequence, then a constant : |tn| n N,which implies tn is bounded from above, but more importantly for this proof,bounded from below by . We want sn + tn > M for n > N . This will besufficed by sn > M or sn > M + for n > N . So we will choose anN : sn > M + for n > N . Let = M + . From the constraints above, wesee that 0 for n > N , we canconclude that lim (sn + tn) = +.
EXERCISE 9.12 (a)
Choose a : L < a < 1. If = aL > 0, there exists an N : n N | sn+1sn L| sn+1sn < L+ = a < 1. More specifically, |sN+1| < a |sN | , |sN+2| < a sN+1 0. Since v + is not an upper bound for S,N : sN < v + . Since sn is nonincreasing, = sN sn n > N . sn v nand hence, n > N = v + > sn v = |sn v| < = lim sn = v, asdesired.
EXERCISE 10.5
Claim: If (sn) is an unbounded non-increasing sequence, then lim sn = .
We want to show that
lim sn = for each M < 0 a number N : n > N = sn < M.
15
Proof :
Let (sn) be an unbounded non-decreasing sequence. Let M < 0. Since the set{sn : n N} is unbounded and it is bounded above by s1, it must be unboundedbelow, since for this sequence to be non-increasing, the condition sn sn+1 mustbe fulfiilled. Hence, for some N N we have sN < M . Clearly, n > N =M > sN sn, so
lim sn = ,as desired.
EXERCISE 10.6
(a) Let (sn) be a sequence such that
|sn+1 sn| < 2n n NClaim: (sn) is a Cauchy sequence and hence a convergent sequence.
Proof :
Given an arbitrary > 0, for n > N , choose N : 2N < 2 . Then m > n > N ,we can see that
|sm sn| m1k=n
sk+1 sk .
Now, m1k=n
sk+1 sk
m1k=n
|sk+1 sk| m1k=n
2k (11)
*Note: The second inequality from line (1) is justified by Let (sn) be a sequencesuch that |sn+1 sn| < 2n n N in the first line of the problem.
m1k=n
2k k=N
2k = 2N+1 < , (12)
as desired. (b) Is the result in (a) true if we only assume that |sn+1 sn| < 1n n N?If we only assume that |sn+1 sn| < 1n n N, our immediate reaction is tosay that the new constraint makes the result in (a) false, since the harmonicseries,
n=1
1n , is taken to be divergent.
As a counter-example, let us choose sn =n=2
1n1 , which will still fulfill
|sn+1 sn| 1n , but will inevitably diverge. Thus, as we travel down the se-quence (N ), arbitrary sms and sns are not encroaching towards eachother,violating the Cauchy criterion.
16
EXERCISE 10.7
Choose n N. Construct sn S : supS sn < 1n and sn > sn1 for n > 1 =sn an increasing sequence converging to supS.Pick s1 S : supS 1 < supS is not an upper bound of S. Implement induc-tion. Assume s1 < < sn1 exists. Since supS / S = sn1 < supS = sn S : supS sn > sn1 and supS sn < 1n . This is possible given thatneither sn1 < supS nor supS 1n < supS is an upper bound for S.
EXERCISE 10.8
Let (sn) be a non-decreasing sequence of positive numbers and define
n =1n
(s1 + s2 + + sn)
Claim: (n) is a non-decreasing sequence.
Proof :
From the assumption of sn being a non-decreasing sequence, we will proclaim
sn sn+1Hence,
nsn nsn+1 (13)= (s1 + s2 + sn) nsn n(sn+1) (14)
= n(s1 + s2 + sn) + (s1 + s2 + sn) n(s1 + s2 + sn) + n(sn+1)(15)
= (n+ 1)(s1 + s2 + sn) n(s1 + s2 + sn + sn+1) (16)= 1
n(s1 + s2 + sn) 1
n+ 1(s1 + s2 + sn + sn+1),
(17)
Thus,n n+1,
as desired.
EXERCISE 11.8
(a) Use Definition 10.6 and Exercise 5.4 to prove that lim inf sn = lim sup(sn).It is readily observable that
lim sup(sn) = limN
sup {(sn) : n > N} = limN
inf {(sn) : n > N} = lim inf(sn).
17
Hence,
lim sup(sn) = lim inf(sn) lim inf(sn) = lim sup(sn),
as desired.
(b) Let (tk) be a monotonic subsequence of (sn) converging to lim sup(sn).Show that (tk) is a monotonic subsequence of (sn) converging to lim inf sn.Observe that this completes the proof of Corollary 11.4.
We are given a monotonic subsequence of (sn), denoted (tk). Then
tk = snk k, (tk monotonic) tk = snk k, (tk monotonic).
When taken with the results from part (a), the desired result is apparent.
EXERCISE 11.9
(a) Let (xn) be a convergent sequence such that a xn b n N =a limn xn b = [a, b] is a closed set.(b) No. We know that the set of any subsequential limits of any set mus beclosed. The interval (0, 1) is not closed.
EXERCISE 11.10
Let (sn) be the sequence of numbers in Figure 11.2 listed in the indicated order.
(a) Find the set S of subsequential limits of (sn).
Laying out the numbers listed in the indicated order gives the sequence
1,12, 1, 1,
12,
13,
14,
13,
12, 1, 1, ...
Rearranging,
1/1,1/2, 1/1,1/1, 1/2, 1/3,1/4, 1/3, 1/2, 1/1,1/1, 1/2, 1/3, 1/4, 1/5,1/6, 1/5, 1/4, 1/3, 1/2, 1/1,1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7,1/8, 1/7, 1/6, 1/5, 1/4, 1/3, 1/2, 1/1,
It seems that 2 sequences, both monotonic, one increasing the other decreas-ing. The even rows of the pyramid represent a monotonically increasing se-quence and the odd rows correspond to a monotonically decreasing sequence.Upon further inspection, we see that with respect to the decreasing sequences,
18
the denominator of the last entry corresponds to the row number of the pyramidfor which the decreasing sequence is located, i.e., the last entry of row 3 has13 as its last entry. Conversely, with respect to the increasing sequences, thedenominator of the entry for which the increasing sequence commences corre-sponds to the row number of the pyramid for which the increasing sequence islocated, i.e., the first entry of row 6 is 16 . Since the odd rows all commence at11 and progressively converge to
1n , we can say that one subsequential limit is
0. Conversely, since the even rows all seem to converge to 1 from a fractionprogressively closer to 1n , we could say the other subsquential limit is 1.
Hence the set of subsequential limits is
S ={
1n
: n N}{
0}
EXERCISE 12.1
Let (sn) and (tn) be sequences and suppose that there exists N0 : sn tn n >N0.
Claim:
lim inf sn lim inf tn and lim sup sn lim sup tn.Proof :
Refer to Exercise 4.8 (b). From there, we see that
sup {sn : n > N} inf {tn : n > N} N N0 =limN
sup {sn : n > N} limN
inf {tn : n > N} lim sup(sn) lim inf(tn) ()
Since
lim inf(sn) lim sup(sn) andlim inf(tn) lim sup(tn),
the desired inequalities are obtained implementing (*), as desired.
19
EXERCISE 12.2
Prove that lim sup |sn| = 0 if and only if lim sn = 0.We will use Theorem 11.7 throughout this proof.
Claim: lim sup |sn| = 0 limsn = 0.Proof :
= :limsn = 0 lim|sn| = 0 lim supsn = 0.12
= :lim sup |sn| = 0 lim inf |sn| = 0, since |sn| 0.
As N , lim sup |sn| lim sn, since Theorem 11.7 tells us that lim sup |sn|is exactly the largest subsequential limit of sn. Hence,
lim sup |sn| = 0 lim sn = 0,
as desired.
EXERCISE 12.3
(a) lim inf sn + lim inf tn = 0 + 0 = 0
(b) lim inf(sn + tn) = 1
(c) lim inf sm + lim sup tn = 0 + 2 = 2
(d) lim sup(sn + tn) = 3
(e) lim sup sn + lim sup tn = 2 + 2 = 4
(f) lim inf sntn = 0
(g) lim sup sntn = 2
EXERCISE 12.4
Claim:
lim sup(sn + tn) lim sup sn + lim sup tn for bounded (sn) and (tn).Proof :
For any n > N ,
sn + tn sup {sn : n > N}+ sup {tn : n > N} =sup {(sn + tn) : n > N} sup {sn : n > N}+ sup {tn : n > N} .
12By Theorem 11.7.
20
Hence,
limN
sup {(sn + tn) : n > N} limN
(sup {sn : n > N}+ sup {tn : n > N})= limN
sup {sn : n > N}+ limN
sup {tn : n > N} ,
where the last equality holds since(sup {sn : n > N}
)N=N0
and(
sup {tn : n > N})N=N0
are both convergent and monotonic, as desired.
EXERCISE 12.8
Let (sn) and (tn) be bounded sequences of nonnegative numbers. Prove that
lim sup sntn (lim sup sn)(lim sup tn).
Give an example where the strict inequality holds.
Claim:lim sup sntn (lim sup sn)(lim sup tn).
Proof :
Since sn and tn are both bounded sequences of nonnegative numbers,
sn sup {sn : n > N} n N
andtn sup {tn : n > N} n N,
by definition. This implies
(sn)(tn) (sup {sn : n > N})(sup {tn : n > N}) n N,
by multiplicativity and since, once again, sn and tn are both bounded andnonnegative. Now since the right hand side of the inequality serves as an upperbound for the left hand side,
sup {sntn : n > N} (sup {sn : n > N})(sup {tn : n > N}) n N
If sn and tn are both sequences which converge to s and t with sn tn n N,we can take limits and conclude
lim sup sntn (lim sup sn)(lim sup tn),
as desired.
21
An example where the strict inequality would hold would be the case when snand tn both converge to s and t, respectively, but sn < tn n N.
EXERCISE 12.10
Claim:
(sn) is bounded lim sup |sn| < +Proof :
lim sup = + = a subsequence (snk) : lim |snk | = + = this subsequenceis unbounded. Conversely, (sn) unbounded = for any k N snk : |snk | > k.Assume that n1 < n2 < < nk < and thus get a subsequence (snk) :lim |snk | = +, i.e., lim sup |sn| = +.
EXERCISE 12.11
Claim:
lim infsn+1sn
lim inf |sn| 1nProof :
Let = lim inf |sn|1n . Let L = lim inf
sn+1sn . We need to show that L forany L1 < L.
L = lim infsn+1sn
= limn(
inf{sn+1sn
: n > N)} > L1= N N : inf
{sn+1sn : n N} > L1
= n N we havesn+1sn
> L1It then follows that
|sn| > LnN1 |sn| , n > N|sn| > Ln1
(LN1 |sn|
) =
, n > N
|sn| > Ln1, n > N = |sn|1n > L1
1n , n > N
Since limn 1n = 1, the result is that lim sup |sn|
1n L1.
22
EXERCISE 13.3 (a)
Let B be the set of all bounded sequences x = (x1, x2, ...) and define d(x,y) =sup {|xj yj | : j = 1, 2, ...}.Show that d is a metric for B.
To show that d is indeed a metric for B, we need to show that it satisfies thethree conditions of Definition 13.1. As we will see the first 2 conditions aretrivial.
D1. d(x, x) = 0, since sup {|xj xj | : j = 1, 2, ...} = sup {0} = 0, andd(x, y) > 0, since for bounded sequences x = (x1, x2, ...) and y = (y1, y2, ...),sup {|xj yj | : j = 1, 2, ...} > 0.D2. d(x,y) = sup {|xj yj | : j = 1, 2, ...} = sup {|yj xj | : j = 1, 2, ...} =d(y,x).
D3. If x, y, z B, then for each j = 1, 2, ..., k,
d(x, z) = sup {|xj zj | : j = 1, 2, ...} (18)= sup {|xj yj + yj zj | : j = 1, 2, ...} (19)= sup {|(xj yj) + (yj zj)| : j = 1, 2, ...} (20) sup {|xj yj | : j = 1, 2, ...}+ sup {|yj zj | : j = 1, 2, ...} (21) d(x, y) + d(y, z), (22)
where the inequality follows from the triangle inequality and the nature of thesupremum. Hence d is a metric.
EXERCISE 13.6
We will prove each part of Proposition 13.9 in turn.
(a) Claim:
The set E is closed if and only if E = E.
Proof :
=Assume E is closed = E is a closed set containing E = the intersection ofall closed sets containing E E = E = E.=Assume E = E. By definition, E is a closed set since any intersection of closedsets is closed (by Definition 13.8). E closed set = E is closed.
23
(b) Claim:
The set E is closed if and only if it contains the limit of every convergentsequence of points in E.
Proof :
Assume E is closed and for purposes of contradiction, assume the limit of everyconvergent sequence of points / E = a convergent sequence of points E = for some arbitrary limit point in E some -ball around the limit pointsuch that all points are inside the ball. But, it then becomes a problem (acontradiction) to have a sequence from E converge to its limit point if for some there does not exist elements of E contained in such a ball. Hence, all limitpoints of E must be contained in E E, as desired. (c) Claim:
An element is in E if and only if it is the limit of some sequence of points inE.
Proof :
=Let x E be arbitrary. If this x is not the limit point = some -ballaround this x: x is the only element of E in this ball = Eo is not the interiorof E because any (s 6= x S) the aforementioned -ball is not contained inEo, which is a contradiction. Now if x E \ Eo = we can create a similar-ball around x: some of the interior of E is missing, another contradiction =any such point in E is a limit point.
=see part (b) above, as desired, lol. (d) Claim:
A point is in the boundary of E if and only if it belongs to the closure of bothE and its complement.
Proof :
Let E Eo = S \ E closed = S \ E contains its boundary = any s boundary of S \E must also be in the boundary of E, otherwise either somex: x is neither E or S \ E, or x both E and S \ E, which both contradictS \ E Ec. Let E E = S \ E is open = S \ E does not contain itsboundary = any s boundary of S \ E must also be the boundary of Eand E = S \ E Ec, as desired.
24
EXERCISE 13.8 (b)
We will verify each assertion step by step.
1. In Rk, open balls {x : d(x,x0) < r} are open sets. We can verify this byshowing that every point in the set is interior to the set. Choose an arbitrary x1in the set. Then |x0|+r represents the boundary. We can then take (|x0|+r)|x1|2 ,which represents the distance between the boundary and x1, but divided by 2.Hence all of the points in the set are interior to the set.
2. In Rk, closed balls {x : d(x,x0) r} are closed sets. We want to show thatthe complement of closed balls are by definition open sets = closed balls areclosed sets. WLOG define our closed ball to be
B(0) := {x : d(x, 0) } ,
WLOG, define the complement of the closed ball by
B(0) := {x : d(x, 0) > } .
To show the complement is open, let be arbitrary. Define := the magnitudeof - . Define a new -ball, with radius:= 2 . All points in this new ball arecontained in the complement of our closed ball.
3. The boundary of each of these sets is
{x : d(x,x0) = r}
The boundary of each of these sets is the boundary of the ball with radius r.By Proposition 13.9 (d),
A point is in the boundary of E if and only if it belongs to the closureof both E and its complement.
Consider the neighborhood of this set
{x : d(x,x0) < r} = (x r,x + r).
whose closure, the intersection of all closed sets containing (x r,x + r), is[x r,x + r]. The complement of {x : d(x,x0) < r} is defined to be {x : d(x,x0) r}.Hence,
[x r,x + r] {x : d(x,x0) r} = {x : d(x,x0) = r} .
Assertion 1{(x1, x2) : x1 > 0} is open.
Discussion: Noting that any element in this set is, by definition, interior to it,we see this is an open set.
25
Assertion 2{(x1, x2) : x1 > 0 and x2 > 0} is open.
Discussion: Once, again, noting the strict inequalties, this defines an open set.Conversely, if non-strict inequalities, i.e., >, we can note that the set willinclude its limit points, hence closing the set.
Assertion 3
{(x1, x2) : x1 > 0 and x2 0} is neither open nor closed.
Discussion: This set fails to be open or closed. As counterexamples, consider(1, 0) which is not interior to this set and (0, 0), which is a limit point, but isnot the set.
EXERCISE 13.10
(a) We want to show that the interior of the set{1n : n N
}= .
Claim:
int
{1n
: n N}
=
Proof :
Define a neighborhood in this set as:
1n+ 1
1n 1n 1 .
Now define an -ball with radius
r =1n 1n+1
2.
This makes -balls around an arbitrary element in the set such that only thearbitrary element is contained in the ball, as desired. (b) Using the fact that Q contains gaps,13 we want to show that
int(Q) =
Proof :
Let q Q and r > 0 be arbitrary. Consider the neighborhood
{s R : |s q| < r} = (s r, s+ r)13This is resolved by the Completeness Axiom.
26
By the denseness of the irrationals14, we know that an irrational numberz (s r, s + r) = the neighborhood (s r, s + r) is not contained in Q.But s was given as arbitrary = q cannot be defined as an interior point. Butq was given as arbitrary = interior of Q = , as desired. (c) Claim:
The interior of the Cantor set is empty.
Proof:
By the definition of the Cantor Set outlined in the text, the Cantor Set theintersection of closed sets = the Cantor Set is closed. Now define the differentarticulations (labeled Tn) of the Cantor Set as follows:
T0 = [0, 1], T1 = [0, 1] \(
13,
23
), ..., T15 = of all Tn
Recalling the above, T is closed = T T, the closure of the Cantor Set.Now,
int(T) =(T c)c
and note that we are still in the metric space [0, 1]. Thus, T c = [0, 1] \ T =T c = ([0, 1] \ T) {x [0, 1] : x is a limit point of [0, 1] \ T}. Since T {x [0, 1] : x is a limit point of [0, 1] \ T} [0, 1], we know
T c = ([0, 1] \ T) T [0, 1].
In other words, the closure of the Cantor Set is T0, the interval from 0 to 1.Hence,
int(T) =(T c)c [0, 1]c ,
as desired.
EXERCISE 13.12
(a) Let (S, d) be any metric space.
Claim:
If E is a closed subset of a compact set F , then E is also compact.
Proof :
Let E be a closed subset of a compact set F . Let a collection of open sets U bean open cover for E. E closed = Ec open. Now, Ec + U an open cover ofF . Given F compact, a finite subcover that covers F . Now given E F andF covered by some finite subcover, E compact.
14For every real numbers x and y, with x < y, a rational and an irrational such thatx < < y and x < < y.
15T THE Cantor Set
27
(b) Let (S, d) be any metric space.
Claim:
The finite union of compact sets S is compact.Proof :
Let C1, C2, ..., Cn : n
Lu and Gl, (supE, infE) E.
EXERCISE 14.2
(a)
n1n2 =
16
nn2
1n2 =
1n
1n2 . Since
1n diverges, the entire
series diverges.
(b) The series
(1)n fails to converge because it doesnt satisfy the Cauchycriterion17. In other words, the terms of the sequence an do not arbitrarily growcloser to eachother as n.(c)
3nn3 =
3n2 = 3
1n2 , which we know is 3 times a convergent series, thus
the series converges.
(d) If an = n3
3n , then an+1/an =(n+1)33n
3n+1n3 , so lim|an+1/an| = 13 . Hence theseries converges by the Ratio Test.
(e) If an = n2
n! , then an+1/an =(n+1)2n!(n+1)!n2 , so lim|an+1/an| = 0. Hence the series
converges by the Ratio Test.
(f) If an = 1nn =(1n
)n, then lim sup|an| 1n = 0. Hence the series converges bythe Root Test.
(g) If an = n2n , then an+1/an =(n+1)2n
2n+1n , so lim|an+1/an| = 12 . Hence the seriesconverges by the Ratio Test.
EXERCISE 14.4
(a) We will use induction and the Comparison Test to show that the series
n=2
1[n+ (1)n]2
converges.
To accomplish this task we need to show
n+ (1)n 12n,
so that n=2
1[n+ (1)n]2
n=2
4n2
= 4n=2
1n2,
since, by the Comparison Test, this would show the series is convergent.16This single series has been split into 2 seperate series. The rule I am follow-
ing is that the sum of two series will converge is both of the sums converge. Hencethe series will diverge, if we can show that one of the sums diverges. Reference:http://www.sosmath.com/calculus/series/examples/examples.html
17Proof ommitted, as it wasnt required.
29
The summand index commences at n = 2, so this will serve as the inductionbasis.
2 + (1)2 12
(2) = 3 1,which is trivial.
Let us now implement the induction step, n + 1, and show the inequality stillholds.
(n+ 1) + (1)n+1 12
(n+ 1) (23)
n+ 1 + (1)n(1) 12n+
12
(24)
n+ 1 (1)n 12n+
12
(25)
12n+
12 (1)n 0 (26)12n+
12 (1)n, (27)
which holds for n 2, as desired. Hence,
n+ (1)n 12n (28)
(n+ (1)n)2 (
12n
)2(29)
1(n+ (1)n)2
1(12n)2 = 4n2 (30)
n=2
1[n+ (1)n]2
n=2
4n2
= 4n=2
1n2, (31)
as desired.
(b) Since[n+ 1n] = 1
n+ 1 +n 1
2n+ 1
1
2
2n=
12
2
1n,
which is a divergent p-series18, the series diverges.
(c) We will show the series n!nn
18A p-series is a series of the form 1
np. Such a series converges if p > 1 and diverges if
p 1.
30
converges via the Ratio Test. We want to show
liman+1an
< 1.liman+1an
= lim (n+ 1)!(n+ 1)n+1 nnn! = lim n!(n+ 1)nn(n+ 1)n(n+ 1)n!
= lim nn(n+ 1)n = lim ( nn+ 1
)n .Since
lim(n+ 1n
)n = lim (1 + 1n)n = e,
this suffices to show that
lim( nn+ 1
)n = 1e < 1.Hence, the series converges by the Ratio Test.
EXERCISE 14.7; assume p Z : p > 1We want to show that if we have a known convergent series
an and raise it
to a power p > 1, it will simply converge quicker. An example would be theconvergent p-series. We know that 1
np
converges for values of p > 1. Now if the series is further raised by a power ofp > 1 the original p will be even greater, and thus wll still be convergent.
Claim:an := convergent and an 0 n N =
(an)p converges.
Proof :
If we know thatan is a convergent series, then
limn an = 0 = N N : an (0, 1) n N = (an)
p < an p > 1 and n N
(an)p < an p > 1 and n N =
(an)p 1 and n N.Hence,
(an)p converges by the Comparison Test, as desired.
31
EXERCISE 14.10
Consider this series: n=0
2(1)n+n
EXERCISE 15.4
Determine which of the following series converge.
(a)n=2
1n logn
. Since log n 1 for n 2.
32
Claim:an := convergent and an 0 n N =
(an)2 converges.
Proof :
If we know thatan is a convergent series, then
limn an = 0 = N N : an (0, 1) n N = (an)
2 < an n N
(an)2 < an n N =
(an)2 0 and x for x < 0.At x = 0, the function f(x) = |x| is continuous because for any > 0 we have:|x 0| < = |f(x) f(0)| = |x| < .
EXERCISE 17.8
(a) Claim:
min(f, g) =12
(f + g) 12|f g|
Proof :
Case 1 : Let f(x) g(x). Then
min(f, g) = f(x) =12
(f + g) 12
(g f) (32)
=12
(f + g) 12|f g| (33)
Case 2 : Let f(x) g(x). Then
min(f, g) = g(x) =12
(f + g) 12
(f g) (34)
=12
(f + g) 12|f g| . (35)
Hence,
min(f, g) =12
(f + g) 12|f g| ,
as desired.
34
(b) Claim:min (f, g) = max (f,g)
Proof :
Case 1 : Let f(x) g(x) = f(x) g(x)= min (f, g) = f(x) = (f(x)) = max(f,g)Case 2 : Let f(x) g(x) = f(x) g(x)= min (f, g) = g(x) = (g(x)) = max(f,g).Hence,
min (f, g) = max (f,g) ,as desired. (c) Claim:
f and g continuous at x0 R = min(f, g) is continuous at x0Proof :
Recall Theorems 17.3 and 17.4 (i):
Theorem 17.3: Let f be a real-valued function with dom(f) R.If f is continuous at x0 in dom(f), then |f | and kf , k R, arecontinuous at x0.
Theorem 17.4 (i): Let f and g be real-valued functions that arecontinuous at x0 R. Then f + g is continuous at x0.
In combination with the results from part (a), i.e.,
min(f, g) =12
(f + g) 12|f g| ,
we see that min(f, g) is simply the sum, difference and composition of functionswhich are continuous at x0, and hence is itself continuous at x0, as desired. .
EXERCISE 17.9
(a) Claim:f(x) = x2 is continuous at x0 = 2
Proof :
Let > 0 be given. We want to show that |x2 4| < provided |x 2| issufficiently small, i.e., less than some . We observe that |x2 4| = |(x+ 2)(x2)| |x + 2| |x 2|. We need to get a bound for |x 2| that doesnt dependon x. We notice that if |x 2| < 1, say, then |x + 2| < 5, so it suffices to get|x2| 5 < . So by setting = min{1, 5}, we see that f(x) = x2 is continuousat x0 = 2, as desired.
35
(b) Claim:f(x) =
x is continuous at x0 = 0
Proof :
Let > 0 be given. We want to show that |x 0| < provided |x 0|is sufficiently small, i.e., less than some . We observe that |x 0| = x.Since we want this to be less than , we set = 2. Then |x 0| < impliesx < = , so
|x 0| < = |f(x) f(0)| < ,as desired. (c) Claim:
f(x) = xsin(
1x
)for x 6= 0 and f(0) = 0 is continuous at x0 = 0
Proof :
Let > 0 be given. We want to show that |xsin ( 1x) 0| < provided |x 0|is sufficiently small, i.e., less than some . We see that |xsin ( 1x) 0| x x.Since we want this to be less than , we set = . Then |x 0| < impliesx < = , so
|x 0| < =xsin( 1x
) 0 < ,
as desired. (d) Claim:
g(x) = x3 is continuous at x0 arbitrary
Proof :
By the hint given,
x3 x30 = (x x0)(x2 + x0x+ x20) = (x x0)[(x2 x0)2 + 3xx0)
]and we know that
|x| = |x x0 + x0| |x x0|+ |x0|,
by the triangle inequality. Hence,
|g(x) g(x0)| = |x3 x30| |x x0|(|x x0|2 + 3|x x0||x0|+ 3x20).
Now let > 0 be given. We show that a = (x0, ) > 0 such that
|x x0| < = |g(x) g(x0)| < .
36
Let = min(
1, 3 ,
g|x0|+1 ,
gx20+1
). Then |x x0| < implies
|g(x) g(x0)| (2 + 3|x0|+ 3x20) 20(1 + 3|x0|+ 3x20)
= + 3|x0|+ 3x20 0, let > 0 be defined as the distance from x to the closest irreducible fraction pq withdenominator q 1 . Then for any x : |x x| < = |f(x) f(x)| < 0 = fcontinuous at x R \Q, as desired.
EXERCISE 18.2
The limit x0 (or y0) of the subsequence xnk (a, b) (or ynk (a, b)) may bean endpoint a or b of the interval and thus lie outside of the domain of thefunction.)
38
EXERCISE 18.4
Let S R and suppose there exists a sequence (xn) in S that converges to anumber x0 / S.Claim:
an unbounded continuous finction on S.Proof :
Let x0 / S. We are given a sequence (xn) S which converges to x0. Then|xn x0| := the distance to x0 is continuous and strictly positive on S. Definef := 1|xnx0| = f is well-defined and continuous on S = limn f = , asdesired.
EXERCISE 18.6
Claim:
x = cosx for some x (0, pi2 ).Proof :
We know f(x) = xcosx is continuous [0, pi2 ], < 0 at x = 0 and > 0 at x = pi2 .Implement the Intermediate Value Theorem. Hence, an x (0, pi2 ) : f(x) = 0,as desired.
EXERCISE 18.8
Suppose that f is a real-valued function continuous on R and that f(a)f(b) < 0for some a, b R.Before commencing, we will state the properties that
0 a = 0 a Zand
a b < 0 a < 0, b > 0 or a > 0, b < 0 a, b ZClaim:
an x between a and b : f(x) = 0.Proof :
Given f(a)f(b) < 0, either
Case 1 : f(a) < 0 = f(b) > 0 = f(a) < 0 < f(b), orCase 2 : f(a) > 0 = f(b) < 0 = f(b) < 0 < f(a).In either case, the Intermediate Value Theorem tells us that x (a, b) : f(x) =0, as desired.
39
EXERCISE 18.10
Suppose that f is continuous on [0, 2] and f(0) = 2.
Claim:
x, y [0, 2] : |x y| = 1 and f(x) = f(y).Proof :
Let g(x) = f(x + 1) f(x) = g is continuous on [0, 1] and g(0) = f(1) f(0) = f(1) f(2) = g(1). Implement the Intermediate Value Theorem.x [0, 1] : g(x) = 0, as desired.
EXERCISE 19.2
(a) Claim:f(x) = 3x+ 1 is uniformly continuous on R.
Proof :
Let > 0 be given. We want |f(x) f(y)| = |(3x + 1) (3y + 1)| < for|x y| < with x, y R. We know |(3x+ 1) (3y+ 1)| = |3x 3y| = 3|x y|.Take := 3 . Then
|xy| < = |xy| < 3
= 3|xy| < = |3x3y| < = |3x3y+11| < =
|3x+ 1 3y 1| < = |(3x+ 1) (3y + 1)| < = |f(x) f(y)| < ,as desired.
(b) Claim:f(x) = x2 is uniformly continuous on [0, 3] .
Proof :
Let > 0 be given. We want |f(x) f(y)| = |x2 y2| < for |x y| < withx, y [0, 3]. We know |x2 y2| = |(x y)(x + y)| = |x y| |x + y|. Withx, y [0, 3], let |x + y| |3 + 3| = 6. Define := 6 . Then for any x, y [0, 3]with |x y| < ,
|f(x) f(y)| = |x2 y2| = |(x y)(x+ y)| = |x y| |x+ y| < |x y| 6 = ,
as desired. (c) Claim:
f(x) =1x
is uniformly continuous on[
12,).
40
Proof :
Let > 0 be given. We want |f(x) f(y)| = | 1x 1y | < for |x y| < , withx, y [ 12 ,). We know | 1x 1y | = |yxxy | = |( 1xy )(y x)| = |x y| | 1xy |. Withx, y [ 12 ,), let 1xy 112 12 = 114 = 4. Define := 4 . Then for any x, y [ 12 ,)with |x y| < ,
|f(x) f(y)| = 1x 1y
= y xxy = |x y| 1xy
< |x y| 4 = ,as desired. .
EXERCISE 19.4(a) Claim:If f is uniformly continuous on a bounded set S, then f is a bounded functionon S.Proof :
By contradiction. Assume that f is uniformly continuous on a bounded set Sand an unbounded function on S. Then a sequence, (xn) in the domain off such that |f(xn)| n. By the Bolzano-Weierstrass Theorem, the sequencecontains a convergent subsequence (xnk), since the domain is bounded. A con-vergent subsequence is Cauchy (by definition) and hence the sequence of valuesf(xnk) is Cauchy by the property of uniformly continuous functions. But thesequence |f(xnk)| nk is unbounded, contradicting our assumption in the out-set of this proof. Hence, if f is uniformly continuous on a bounded set S, thenf is a bounded function on S, as desired.
(b) Claim:1x2
is not uniformly continuous on (0, 1).
Proof :
We want to show that if 1x2 is not a bounded function on (0, 1), then1x2 is not
uniformly continuous on (0, 1)21. So it will suffice to show that
f(x) =1x2
is not a bounded function on (0, 1).
Let M > 1 be arbitrary = 0 < 1M2 < 1. We want to show that 1x2 > M forsome x (0, 1) = x < 1
M. Let x = 1
M+1, since 0 < 1
M+1< 1
M< 1. But
then for f(x) = 1x2 , we see that
1x2
=1(1M+1
)2 = M + 1 > M,21This is the contrapositive of what was proved in part (a).
41
which shows that f(x) = 1x2 is not a bounded function on (0, 1), as desired.
EXERCISE 19.6
(a) Claim:
f(x) =x is uniformly continuous on (0, 1], although f is unbounded.
Proof :
f (x) = 12x + as x 0 = f (x) is unbounded. f continuous on
the closed interval [0, 1] = f uniformly continuous on [0, 1] = f uniformlycontinuous on the subset (0, 1], as desired. Let f(x) =
x for x 1.
(b) Claim:
f is uniformly continuous on [1,).Proof :
Let > 0 be given. We want |f(x) f(y)| = |xy| < for |x y| < withx, y R. We know |xy| = |xy||x+y| . Take := 2. Then
|x y| < = |x y| < 2 = |x y|2
|x y||x+y| = |xy| < ,
as desired.
EXERCISE 19.10
The limit
limx0
g(x)x
= limx0
x sinx1x
= 0,
in other words, g is differentiable at x = 0 (g(0) = 0). At x 6= 0,
g(x) = 2x sin1x
+ x2 cos1x( 1x2
)= 2x sin
1x cos 1
x,
which is bounded. Since |cos y| 6= 1and |sin y| |y| y = for y = 1x :2x sin 1x cos 1x = 2y sin y cos y
2 + 1 = 3.Thus, g is both bounded and defined on R = g is uniformly continuous.
42
EXERCISE 20.11 (b)
Claim:
limxb
xbx b =
12b
Proof :
xbx b =
(xb)
x+b
x+b
x b =xbx+b
x b =1
x+b
Hence,
limxb
1x+b 1
b+b
=1
2b
EXERCISE 20.14
Claim:limx0+
1x
= + and limx0
1x
=
We will split this up into 2 cases/claims.
Claim 1 :limx0+
1x
= +
Proof :
Let M > 0 and := 12M . Then 0 < x < 0 + = 0 < x < =f(x) = 1x >
1 =
11
2M= 2M > M . Hence, limx0+ 1x = +, as desired.
Claim 2 :limx0
1x
=
Proof :
Let M < 0 and := 12M . Then 0 < x < 0 = < x < 0 =f(x) = 1x L2. Corollary 20.8 sayslimxa+
f(x) = L > 0 > 0 : a < x < a+ = |f(x) L| <
Define := 12 (L1 L2) = for > 0 :
1 : x (a, a+ 1), f1(x) > L1 = L1 [
12
(L1 L2)]
=12
(L1 + L2)
2 : x (a, a+ 2), f2(x) < L2 + = L2 +[
12
(L1 L2)]
=12
(L1 + L2).
Let := min {1, 2} =
f1(x) >12
(L1 + L2) > f2(x),
which is a contradiction. Hence, if f1(x) f2(x) x in some interal (a, b), thenL1 L2, as desired. (b) Suppose f1(x) < f2(x) x in some interval (a, b). Can you conclude thatL1 < L2.
Consider f1(x) = x and f2(x) = 2x, x (a, b) = f1(x) < f2(x) x (a, b).Consider a = 0 and b = 1 = (a, b) = (0, 1). We can plainly see that
limx0
f1(x) = limx0
f2(x) = 022,
a contradiction = we cannot conclude that L1 < L2 if f1(x) < f2(x) x insome interval (a, b).
EXERCISE 20.18
Claim:
For f(x) =
1 + 3x2 1x2
, limx0
f(x) =32.
Proof :
Non-formal. Rearranging f(x),
1 + 3x2 1x2
=
1 + 3x2 1x2
1 + 3x2 + 11 + 3x2 + 1
=(1 + 3x2) 1
(
1 + 3x2 + 1)x2=
31 + 3x2 + 1
,
which we see is a composition of continuous functions which happen to behavewell near x = 0. Hence, limx0 f(x) f(0) = 32 , as desired.
22Since both f1 and f2 extend continuously to 0 where they take on the value 0.
44
EXERCISE 21.6
Let (S1, d1), (S2, d2), and (S3, d3) be metric spaces.
Claim:
f : S1 S2 and g : S2 S3 continuous = g f continuous from S1 into S3Proof :
From both the claim and Theorem 21.3, f1(U) is an open subset of S1 for everyopen subset U of S2, i.e., f1(U) = {s S1 : f(s) U} and g1(V ) is an opensubset of S3 for every open subset V of S3, i.e., g1(V ) = {s S3 : f(s) V }.Define a new mapping, h(S1) = g f : S1 S3. Then h1(V ) is an open subsetof S1 for every open subset V of S3, i.e. h1(V ) = {s S1 : h(S) V }, by thecontinuity of composite functions outlined in Theorem 17.5 and the fact thatg and f are both coninuous throughout S. Thus, by Theorems 17.5 and 21.3,h(S1) := (g f)(S1) : S1 S3 is continuous, as desired.
EXERCISE 21.10 (b) and 21.11 (b)
For 21.10 (b), in order to show continuous functions mapping (0, 1) R,consider
f(x) =log(2x)1 x
As x 0, f(x) and as x 1, f(x) +. We see f(x) is the com-position of continuous functions and the denominator 6= 0 through f s domainand is hence continuous, as desired.
For 21.11 (b), we can use informal contradiction:
f : [0, 1] R = R is compact, since [0, 1] is compact, by Theorem 21.4 (i).But R is unbounded and thus cannot be compact, which is a contradiction.Hence, there do not exist continuous functions mapping [0, 1] onto R.
EXERCISE 23.1 (b), (d), (f) and (h)
(b)(
xn
)n = xnnn = ( 1n)n xn. If an = ( 1n)n, then lim sup |an| 1n = 0.Therefore, = 0, R = + and this series has a radius of convergence + andhence an interval of convergence of (,+).(d)
(n33n
)xn. If an = n33n , then
an+1an
= (n+1)3
3n+1 3n
n3 , so liman+1an = 13 . There-
fore, = 13 and R = 3. This series diverges for both x = 3 and x = 3, hencethe radius of convergence is 3 and the interval of convergence is (3, 3).(f)
(1
(n+1)22n
)xn. If an = 1(n+1)22n , then
an+1an
= (n+1)22n
(n+2)22n+1 , so liman+1an =
12 . Therefore, =
12 and R = 2. This series converges at both x = 2 and
45
x = 2, hence the radius of convergence is 2 and the interval of convergence is[2, 2].(h)
( (1)nn24n
)xn. If an =
(1)nn24n , then
an+1an
= (1)n+1
(n+1)24n+1 n24n
(1)n , so liman+1an =
14 . Therefore, =
14 and R = 4. This series converges at both x = 4 and x = 4,
hence the radius of convergence is 4 and the interval of convergence is [4, 4].
EXERCISE 23.2
(a)
nxn. If an =n, then an+1an =
n+1n
, so limn+1n = 1. Therefore,
= 1 and R = 1. This series diverges at x = 1 and x = 1, hence the radiusof convergence is 1 and the interval of convergence is (1, 1).
(b)
1nnx
n. If an = 1nn , then |an|1n =
1n
1n
and lim sup 1n
1n
= 1.At x = 1, the series converges with comparison to
1np with p > 1 and for
x = 1, the series diverges by the alternating series theorem since the limit of|an| approaches 1 and not 0. Hence, the radius on convergence is 1 and theinterval of convergence is (1, 1].(c)
xn!. When |x| 1, the series diverges since limxn! does not tend to
0 as n . Now for |x| < 1, the series converges absolutely by comparisonwith
|x|m23. Therefore, = 1 and R = 1. This series diverges at x = 1 andx = 1, hence the radius of convergence is 1 and the interval of convergence is(1, 1).(d)
3nnx2n+1. For this series, the radius of convergence is:
R =1
lim(3nn
) 12n+1
=13
lim(3n)1
4n+2 =13.
When x = 13, the series explodes in both directions24. Hence, the radius of
convergence is 13
and the interval of convergence is(13, 1
3
).
EXERCISE 23.6 (b)
An example of such a series is n>0
(x)nn
.
23Note thatxn! =
amxm with am = 1 when m = n! and am = 0 when m 6= n!.
24The series turns into 1
3n.
46
This series converges to ln(1 + x) when |x| < 1, diverges at x = 1 and con-verges at x = +1.
EXERCISE 23.8
fn(x) 0 sincen1 sinnx 6= 1n 0 as n. But f n(x) = cosnx (1)n
when x = pi, which has no limit.
EXERCISE 24.2
For x [0,), let fn(x) = xn .(a) For fn(x) = xn , limn fn(x) = limn
xn = 0. Hence, f(x) = 0.
(b) YES.
Claim: fn f uniformly on [0, 1].Proof :
Let > 0 be given. Let N := 1 . Then, for n > N , |fn(x) 0| = xn
1n =1n < , as desired. (c) NO.
Claim: fn does not converge uniformly to f on [0,).Proof :
By contradiction. Using the above, let := 1 = an N : xn < 1 n >N = |x| < n n > N = |x| < n + for some > 0. But since x [0,),x is unbounded, contradicting |x| < n + for some > 0. Hence, fn does notconverge uniformly to f on [0,), as desired.
EXERCISE 24.6
Let fn(x) =(x 1n
)2 for x [0, 1].(a) YES.
Claim:
fn(x) =(x 1n
)2 for x [0, 1] converges pointwise on the set [0, 1].Proof :
fn(x) =(x 1n
)2 = x2 2xn + 1n2 With x [0, 1], let n grow arbitrarily large= fn(x) x2, since with n large, 2xn 0 and 1n2 0 = fn(x) f(x) := x2. Hence, given x arbitrary, fn(x) converges pointwise for x [0, 1],as desired. The limit function will thus be f(x) = x2.
(b) YES.
47
Claim:
fn(x) =(x 1n
)2 for x [0, 1] converges uniformly on the set [0, 1].Proof :
Let > 0 be given and N := 12 . Then for x [0, 1], we have:(x 1
n
)2 x2
=1 2xnn2
1n < 1N = ,as desired. .
EXERCISE 24.14
Let fn(x) = nx1+n2x2(a) Claim:
fn 0 pointwise on R.Proof :
Let x R. Let > 0 be arbitrary. Let N := max{x2,x
}. Then for n > N ,
|fn 0| = nx1 + n2x2
= x1n + nx
2
x1n + n
2
xn2 xN2 < .Hence, fn(x) converges uniformly to zero, as desired. (b) Claim:
fn(x) does not converge to 0 uniformly on [0, 1].
Proof :
Implement Remark 24.4. Hence,
f n(x) =n
1 + n2x2 nxn
22x(1 + n2x2)2
=n+ n3x2 2n3x2
(1 + n2x2)2=
n n3x2(1 + n2x2)2
Assume, for contradiction, nn3x2
(1+n2x2)2 0, which implies x2 = 1n2 which impliesx = 1n . But we see that fn
(1n
)= 12 6= 0, a contradiction, since x [0, 1]
satisfies x = 1n . Hence, fn(x) does not converge to 0 uniformly on [0, 1] asdesired. (c) Claim:
fn(x) converges to 0 uniformly on [1,).
48
Proof :
Implement Remark 24.4. With x [1,), x is unable to always satisfy 1n .Using methods from calculus we can see that fn(x) = nx1+n2x2 assumes its max-imum at x = 1, which is [1,). Since fn(1) = n1+n2 0 as n , fn(x)converges to 0 uniformly on [1,) as desired.
EXERCISE 25.2
Let fn(x) = xn
n .
Claim: (fn) is uniformly convergent on [1, 1].Proof :
Let > 0 be given. Let N := 1 . Then for n > N and x [1, 1],
|fn(x) f(x)| =xnn 0
1n 25 1N = ,as desired. The limit function is fn(x) f(x) = 0 for large n.
EXERCISE 25.6
(a) Show that if |ak| < , then akxk converges uniformly on [0, 1] to a
continuous function.
Implement Theorem 25.5 and the Weierstrass M-test. Since |ak| 0 begiven and let N := M . Then
n m > N =n
k=m
hgk
n
k=m
Mgk
= Mn
k=m
gk
M M = ,as desired.
EXERCISE 27.2
Show that if f is continuous on R, then there exists a sequence (pn) of polyno-mials such that pn f uniformly on each bounded subset of R. Hint : Arrangefor |f(x) pn(x)| < 1n for |x| n.Claim:
If f is continuous on R, then there exists a sequence (pn) of polynomials suchthat pn f uniformly on each bounded subset of R.Proof :
Given |x| n, suppose I = [n, n], a closed and bounded interval; hence x Iand f : I R is a continuous function. Let g : [0, 1] [n, n] be a bijectivemap defined by g(x) = n+ x(2n), and hence continuous, i.e., g(0) = n andg(1) = n. Since f is continuous, the composite function, f g : [0, 1] Ris continuous. Hence, for any > 0,N > 0 such that for any n N , theBernstein Polynomial Bn(f g) satisfies
|(f g)(x)Bn(f g)(x)| < x [0, 1]
Now g is a continuous injective map and so g has a continuous inverse functiondefined by
g1(x) =x+ n
2nx [n, n]
Thus, for all x [n, n], |f(x)BN (f g)(g1(x))| < .Hence, f(x)BN (f g)(x+ n2n
) < x [n, n]
50
SinceBN (fg) is a polynomial function, p(x) = BN (fg)(x+n2n
)is a polynomial
function in x and |f(x) p(x)| < x I. If we let qn(x) = BN (f g)(x+n2n
),
then
qn(x) = BN (f g)(x+ n
2n
)=
nk=0
f g(k
n
)(n
k
)(x+ n
2n
)k (1
(x+ n
2n
))nk(36)
=nk=0
(f g)(k
n
)(n
k
)(x+ n
2n
)k (n x
2n
)nk(37)
=nk=0
f
(a+
k
n(2n)
)(n
k
)(x+ n
2n
)k (n x
2n
)nk(38)
It then follows from |(f g)(x) Bn(f g)(x)| < x [0, 1] that qn funiformly on [n, n], as desired.
EXERCISE 27.6
Claim:
If Bnf f uniformly on [0, 1], then f is continuous on [0, 1].Proof :
If Bn(f) f uniformly on [0, 1], then
> 0 N x [0, 1] n > N : |Bnf(x) f(x)| < 2
Let > 0 be given. Let N := . Then x [0, 1] and |x x0| < =
|f(x)f(x0)| = |Bnf(x)f(x0)+f(x)Bnf(x)| |Bnf(x)f(x0)|+|Bnf(x)f(x)| 2+
2= ,
as desired.
EXERCISE 28.2
Use the definition of the derivative to calculate the derivatives of the followingfunctions at the indicated points.
(a) f(x) = x3 at x = 2.
f (2) = limx2
x3 8x 8 = limx2
(x 2)(x2 + 2x+ 4)x 2 = limx2x
2+2x+4 = (2)2+2(2)+4 = 12
51
(b) g(x) = x+ 2 at x = a.
f (a) = limxa
(x+ 2) (a+ 2)x a = limxa
x+ 2 a 2x a = limxa
x ax a = 1
(c) f(x) = x2 cosx at x = 0.
f (0) = limx0
x2 cosx (0)2 cos(0)x 0 = limx0
x2 cosx 0x 0 = limx0x cosx = 0 1 = 0
(d) r(x) = 3x+42x1 at x = 1
r(1) = limx1
3x+42x1 71x 1 = limx1
11x+112x1x 1 = limx1
11x+ 112x 1
1x 1 = limx1
112x 1 = 11
EXERCISE 28.4
Let f(x) = x2 sin 1x for x 6= 0 and f(0) = 0.(a) Use Theorems 28.3 and 28.4 to show that f is differentiable at each a 6= 0and calculate f (a). Use, without proof, the fact that sinx is differentiable andthat cosx is its derivative.
Using the definition, we see that
f (a) = limxa
x2 sin(1x
) a2 sin ( 1a)x a
= limxax
2 sin(1x
) sin ( 1a)x a + sin
(1a
)x2 a2x a ,
where thesin( 1x )sin( 1a )
xa limit represents sin ( 1x
). We are given that a 6= 0,
which gives the function sin(1a
)some meaning in terms of differentiability and
hence, using Theorems 28.3 and 28.4 and the given fact that sin x = cosx,
f (a) = a2 sin(
1a
)+ sin
(1a
)(a2) = a2 cos
(1a
)(1a2
)+ 2a sin
(1a
)= 2a sin
(1a
) cos
(1a
).
(b) Use the definition to show that f is differentiable at x = 0 and that f (0) =0.
limx0
f(x) f(0)x 0 = limx0
f(x)x
= limx0
x2 sin 1xx
= limx0
x sin1x
= 0 ,
where || 1, and hence the whole expression equals zero, as desired.(c) Show that f is not continuous at x = 0.
52
Implement Theorem 28.4, which implies f is differentiable everywhere. Further,f (x) = x sin
(1x
) cos ( 1x) at x 6= 0. We know x sin ( 1x) 0 as x 0 andcos(1x
)doesnt have a limit at x = 0. Hence, the sum f (x) has no limit at
x 6= 0 and thus is discontinuous at x = 0.
EXERCISE 28.8
Let f(x) = x2 for x rational and f(x) = 0 for x irrational.
(a) Claim:
f is continuous x = 0.
Proof :
Let > 0 be given. Let |x 0| < . Then f(x) is either equal to x2 [0, ) or0. In either of these 2 cases, |f(x) f(0)| < = continuity at 0, as desired.
(b) Claim:
f is discontinuous at all x 6= 0.Proof :
This will be done in 2 cases.
Case I: x 6= 0, x QLet > 0 be given. Let > 0 be given. Further, let := x2. Due to thedenseness of the rationals, an irrational number q in (a , a+ ). But while|x q| < and |f(x) f(q)| = , can be made arbitrarily small (once again,due to the denseness property) = f is not continuous at x, as desired.Case II: x 6= 0, x R \QLet > 0 be given. Let 0 < < |x|2 be given. Further, let :=
x2
10 . Due tothe denseness of the irrationals, a rational number q in (a , a+ ). By thetriangle inequality, |q| > |x|2 = f(q) = x
2
4 = |f(x) f(q)| = x2
4 > . Since can be made arbitrarily small = f is not continuous at x, as desired.
(c) Claim:
f is differentiable at x = 0.
Proof :
Let x = 0 and a 6= 0. Then limxa f(x)f(a)xa = f(a)a , which will will equal a ifa Q and 0 otherwise. Both cases show that the limit 0 as a 0, and hencef is differentiable at x = 0 with derivative equal to 0, as desired.
53
EXERCISE 28.15
Proof of Leibniz Rule
Claim:
(fg)(n)(a) =nk=0
(n
k
)f (k)(a)g(nk)(a)
Proof :
By Induction on n27. Let Leibniz Rule hold for n = m. Then
(fg)(m+1) = (f g + fg)(m) =mk=0
(m
k
)f (k+1)g(mk) +
mk=0
(m
k
)f (k)g(m+1k)
=m+1k=0
[(m
k 1)
+(m
k
)]f (k)g(m+1k)
=m+1k=0
(m+ 1k
)f (k)g(m+1k),
where, using Pascals triangle28 (from the Binomial Theorem handout), the lastequality holds. Hence, Leibniz Rule holds true, as desired.
EXERCISE 29.4
Let f and g be differentiable functions on an open interval I. Suppose that a, bin I satisfy a < b and f(a) = f(b) = 0.
Claim:f (x) + f(x)g(x) = 0 for some x (a, b).
Proof :
Consider the function h(x) = f(x)eg(x). Since f(a) = f(b) = 0 = h(a) =h(b) = 0, implementing Rolles Theorem, we know some x (a, b) : h(x) = 0.Differentiating h(x) yields:
h(x) = f(x)g(x)eg(x) + f (x)eg(x)
= eg(x) (f(x)g(x) + f (x)) .
Given h(x) = 0 for some x (a, b), either eg(x) has to equal 0 for some x (a, b)or f(x)g(x) +f (x) has to equal 0 for some x (a, b). Since we know eg(x) can-not equal 0 for any value g(x), we can safely conclude that f(x)g(x)+f (x) = 0for some x (a, b), as desired.
27For n = 1, Leibniz Rule turns into the product rule, i.e., (fg) = f g + fg.28(m+1k
)=( mk1
)+(mk
).
54
EXERCISE 29.8
Claim:
f is strictly decreasing is f (x) < 0 x (a, b).Proof :
Consider x1, x2 where a < x1 < x2 < b. By the Mean Value Theorem, for somex (x1, x2) we have
f(x2) f(x1)x2 x1 = f
(x) < 0.
Since x2 x1 > 0 and f(x2) f(x1) < 0 = f(x2) < f(x1), as desired.
Claim:
f is increasing if f (x) 0 x (a, b).Proof :
Consider x1, x2 where a < x1 x2 < b. By the Mean Value Theorem, for somex (x1, x2) we have
f(x2) f(x1)x2 x1 = f
(x) 0.
Since x2 x1 > 0 and f(x2) f(x1) 0 = f(x2) f(x1), as desired.
Claim:
f is decreasing if f (x) 0 x (a, b).Proof :
Consider x1, x2 where a < x1 x2 < b. By the Mean Value Theorem, for somex (x1, x2) we have
f(x2) f(x1)x2 x1 = f
(x) 0.
Since x2 x1 > 0 and f(x2) f(x1) 0 = f(x2) f(x1), as desired.
EXERCISE 29.10
Let f(x) = x2 + x2 sin 1x at x 6= 0, and f(0) = 0.
Claim:
f (0) > 0, but f is not increasing on any interval containing 0. Compare thisresult with Theorem 29.7 (i).
55
Proof :
We know
f (0) = limx0
f(x)x
= limx0
[12
+ x sin1x
]=
12
+ 0 > 0
Conversely, at x 6= 0, we get
f (x) = 2x sin1x
+12 cos 1
x.
The first term 0 as x 0. The last term oscillates between +1 and 1infinitely many times in any neighborhood of x = 0. 12 1 < 0 = in anyneighborhod of x = 0, f stay < 0 on some intervals. Implement Theorem 29.7.The function f is therefore decreasing on these intervals = f is not increasingin any neighborhod of x = 0. If f were continuous at x = 0, then it wouldremain positive in some neighborhood of x = 0. Hence, discontinuity of f atx = 0, as desired.
EXERCISE 29.15
We know that (xm) = mxm1 for m 0, ( 1x) = 1x2 and (x 1n) = x 1n1. Wecan thus calculate
(xmn
) using the chain rule as follows:d
dx
(xmn
)=
d
dx(xm)
1n =
y1n1
n
y=xm
mxm1 = mnxm(
1n1)xm1 =
m
nxmn 1.
EXERCISE 29.18
Let f be be differentiable on R with a = sup {|f (x)| : x R} < 1. Select s0 Rand define sn = f(sn1) for n 1. Thus s1 = f(s0), s2 = f(s1),...etcetra.Claim:
(sn) is a convergent sequence.
Proof :
Implement the Mean Value Theorem. Then, for each n > 0,
|sn+1 sn| = |f(sn) f(sn1)| = |f (y)(sn sn1)| a|sn sn1|,which implies
|sn+1 sn| an|s1 s0|, n > 0,by induction. Further,
|sm+1 sn| mk=n
|sk+1 sk| (s1 s0)mk=n
ak, m n > 0.
56
We know the geometric seriesak converges if a < 1 and further, its partial
sums,nk=0 form a Cauchy Sequence. Therefore, by the previous estimate, (sn)
is a Cauchy sequence, and hence converges, as desired.
EXERCISE 30.2
Find the following limits if they exist.
(a) limx0 x3
sin xx =00 , which is an indeterminant form, so we apply LHospitals
rule and get limx0 3x2
cos x1 =00 , which is an indeterminant form, so we apply
LHospitals rule and get limx0 6x sin x = 6 limx0 xsin x = 6 1 = 6.(b) limx0 xx3 =
00 , which is an indeterminant form, so we apply LHospitals
rule and get limx0 sec2 x13x2 =
00 , which is an indeterminant form, so we apply
LHospitals rule and get limx0 2 sec2 x tan x6x =
00 , which is an indeterminant
form, so we apply LHospitals rule and get limx0 sec4 x+2 sec2 x tan2 x
3 =13
(c) limx0[
1sin x 1x
]=, which we can rearrange to suffice for an appli-
cation of LHospitals rule. limx0[
1sin x 1x
]= limx0 xsin xx sin x =
00 , which is an
indeterminant form, so we apply LHospitals rule and get limx0 1cos xx cos x+sin x =00 , which is an indeterminant form, so we apply LHospitals rule and getlimx0 sin x2 cos xx sin x =
02 = 0.
(d) limx0(cosx)1x2 = 1, which when manipulated, can yield an indetermi-
nant form sufficient for the application of LHospitals rule.
Let y = limx0(cosx)1x2 . Then ln y = ln limx0(cosx)
1x2 = limx0 ln(cosx)
1x2 =
limx0ln(cos x)x2 =
00 , which is an indeterminant form, so we apply LHospitals
rule and get limx0 tan x2x =00 , which is an indeterminant form, so we apply
LHospitals rule and get limx0 sec2 x
2 =12 . Thus ln y =
12 = eln y =
e12 = y = 1
e. Therefore, the limit as x 0 = 1
e.
EXERCISE 30.4
Let f be a function defined on some interval (0, a), and define g(y) = f(1y
)for
y (a1,); here we set a1 = 0 if a =.Claim:
limx0+ f(x) exists if and only if limy g(y) exists, in which case they areequal.
Proof :
In two parts.
limx0+ f(x) exists = limy g(y) exists.
57
Assume limx0+ f(x) exists and is equal to L. Then for each > 0 > 0 : 0 0 = U(f) 6= L(f) = f is not integrable on [0, b], since theTheorems of this chapter wont hold for a degenerate interval.
EXERCISE 32.6
Let f be a bounded function on [a, b]. Suppose there exist sequences (Un) and(Ln) of upper and lower Darboux sums for f such that lim(UnLn) = 0. Showf is integrable and
baf = limUn = limLn.
We want to show that limLn = limUn. Assume that limLn < limUn. Thenlim(Un Ln) > 0 which is a contradiction. Observe that Ln L(f) U(f) Un. Taking limits of both sides and implementing the Squeeze Theorem yieldslimLn L(f) U(f) limUn = limLn. Thus L(f) = U(f) by the SqueezeTheorem which implies f is integrable by Theorem 32.9.
29Each interval has length bn
; on the interval [ti1, ti], the maximum value of f is attainedat ti =
in
.30Each interval has length b
n; the supremum of f on [ti1, ti] is ti since you can pick rational
numbers in the interval arbitrarily close to ti.31Since tk > tk1, tk >
tk+tk12
.32Note that this is a telescoping series, lol.
60
EXERCISE 33.3 (a)
A function f on [a, b] is called a step-function if a partition P = {a = u0 < u1 < < um = b}of [a, b] such that f is constant on each interval (uj1, uj), say f(x) = cj for xin (uj1, uj).
Claim:
f is integrable.
Proof :
Consider a sub-partition of P , called P , where
P ={u0 < u1 < u1 < u1 < < un1 < un1 < un1 < un
},
where ui ui < i.We want to show |U(f, P ) S| < and |S L(f, P )| < . From P ,
U(f, P ) L(f, P ) =
(ui ui) |ci ci1| =|ci ci1| max {ci} ,
which will tend to 0 as tends to 0. Hence,
L(f, P ) nj=1
cj(uj uj1) U(f, P ),
and since L(f, P ) = U(f, P ) from above, baf =
nj=1 cj(uj uj1), as de-
sired.
EXERCISE 33.4
Give an example of a function f on [0, 1] that is not integrable for which |f | isintegrable.
Consider the intervale [0, 1] and let f(x) = 1 for rational x [0, 1], and letf(x) = 1 for irrational x [0,1]. For any partition
P = {0 = t0 < t1 < < tn = 1} ,
we have
U(f, P ) =nk=1
M(f, [tk1, tk]) (tk tk1) =nk=1
1 (tk tk1) = 1
and
L(f, P ) =nk=1
(1) (tk tk1) = 1.
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It follows that U(f) 6= L(f) which implies that f is not integrable. However, ifthe absolute value of f is taken, we see that f(x) = 1 x R, and hence willbe integrable since U(f) = L(f).
EXERCISE 33.7
Let f be a bounded function on [a, b], so that B > 0 : |f(x)| B x [a, b].(a)Claim:
U(f2, P ) L(f2, P ) 2B[U(|f |, P ) L(|f |, P )
] partitions P of [a, b]
Proof :nk=1
[M(f2, [tk1, tk])m(f2, [tk1, tk)
] (tk tk1)=
nk=1
([M(|f |, [tk1, tk]) +m(|f |, [tk1, tk)] [M(|f |, [tk1, tk])m(|f |, [tk1, tk)]
) (tk tk1)
2Bnk=1
[M(|f |, [tk1, tk])m(|f |, [tk1, tk])
] (tk tk1),
as desired33. .(b)Claim:
If f is integrable on [a, b], then f2 is integrable on [a, b].
Proof :
From part (a), we know
U(f2, P ) L(f2, P ) 2B[U(|f |, P ) L(|f |, P )
] partitions P of [a, b],
and by Theorem 32.5, if f is integrable on [a, b], then for > 0, U(f, P ) L(f, P ) < .
But,
U(f2, P )L(f2, P ) 2B[U(|f |, P )L(|f |, P )
]< = U(f
2, P ) L(f2, P )2B
< ,
which impliesU(f2, P ) L(f2, P ) < 2B.
So define a new , := 2B > 0 and we thus have
U(f2, P ) L(f2, P ) < ,33The first equality in this proof is from factoring the perfect squares.
62
which implies that if f is integrable on [a, b], then f2 is integrable on [a, b], asdesired.
EXERCISE 33.8
Let f and g be integrable functions on [a, b].
(a) Claim:
fg is integrable on [a, b].
Proof :
Implement Theorem 33.3. Let f and g be integrable on [a, b]. Then, by Theo-rem 33.3, f + g and f g are integrable on [a, b]. Further, if f + g and f gare integrable on [a, b], then (f + g)2 and (f g)2 are both integrable on [a, b],as per the results of exercise 33.7 (b). Since 14
[(f + g)2 (f g)2] = fg, and
Theorem 33.3 tells us a constant times an integrable function is integrable, thisshows that if f and g are integrable on [a, b], then fg is integrable on [a, b], asdesired.
(b) Claim:
max(f, g) and min(f, g) are integrable on [a, b].
Proof :
Implement Theorem(s) 33.3 and 33.5. We know
max(f, g) =12
(f + g) +12|f g|
min(f, g) =12
(f + g) 12|f g|,
which are compositions of functions and constants integrable on [a, b] and henceare integrable by Theorems 33.3 and 33.5, as desired. .
EXERCISE 33.10
Let f(x) = sin(1x
)for x 6= 0 and f(0) = 0. Show that f is integrable on [1, 1].
Let > 0 be given. Since f is piece-wise continuous, by Definition 33.7, on [ 4 , 1], a partition P1 of [ 4 , 1] : U(f1, P )L(f1, P ) < 2 . Similarly, a partition P2 of[1, 4 ] : U(f2, P ) L(f2, P ) < 2 . Define P = P[1 uniondbl P2, a partition of [1, 1].Since {
M(f,[
4,
4
])m(f,
[
4,
4
])}{
4(
4
)}< ,
which, when combined with Theorem 32.5, shows f is integrable on [1, 1].
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EXERCISE 33.14
Suppose f and g are continuous functions on [a, b] and that g(x) 0 x [a, b].(a) Claim:
x [a, b] : ba
f(t)g(t)dt = f(x) ba
g(t)dt.
Proof :
Given that ba
f(t)g(t)dt =i=1
f(ti)g(ti) (ti ti1),
where ti [ui1, ui] and f(ti) = ai and g(ti) = bi. Further,aibi min {ai} bi = ai min {ai}aibi max {ai} bi = ai max {ai} .
Then, ba
f(t)g(t)dt fmin ba
g(t)dt ba
f(t)g(t)dt fmax ba
g(t)dt.
Note: if bi = 0 then the equality holds, trivially. Hence, by the IntermediateValue Theorem, x [a, b] : b
af(t)g(t)dt = f(x)
bag(t)dt., as desired. .
(b) Let g(x) := 1ba . Then
1b a
ba
f(t)dt = ba
f(t)g(t)dt = f(x) ba
g(t)dt = f(x) 1 = f(x).
EXERCISE 34.2
(a)
limx0
1x
x0
et2dt
To calculate this integral, let us use the formulation
F (x) F (x0)x x0 =
1x x0
xx0
f(t)dt for x 6= x0
Hence,
f(x0) =1
x x0
xx0
f(x0)dt.
64
Thus, by the Fundamental Theorem of Calculus II,
limx0
1x
x0
et2dt = e0
2= 1.
(b)
limh0
1h
3+h3
et2dt
Using the same argument as above,
limh0
1h
3+h3
et2dt = lim
h01h
30
e(t+3)2dt = e3
2= e9.
EXERCISE 34.6
Let f be a continuous function on R and define
G(x) = sin x0
f(t)dt for x R.
If f is continuous at x R, then f(sinx) is differentiable on R as a com-position of differentiable functions. Hence, f is differentiable at sinx andG = f(sinx) cosx, per the chain rule.
Let > 0 and B > 0 : |f(sinx)| B x R. Now x R : | sinx sin 0| | sinx| < , we have |F (sinx) F (0)| | sin x
0f | | sinx| < , which shows G
is continuous.
EXERCISE 36.1
Show that if f is integrable on [a, b] as in Definition 32.1, then
limdb
da
f(x)dx = ba
f(x)dx.
It suffices to show that if |f | is bounded by some number B, then da
f(x)dx ba
f(x)dx
B(b d).Hence, we want to show that for |b d| < = |B(b d)| < . Choose := B .Then
|B(b d)| = B|(b d)| < B = .
65
Hence as d approaches b from the left, the difference between the integrals con-verges to zero, showing they are equivalent.
EXERCISE 36.6
Let f and g be continuous functions on (a, b) : 0 f(x) g(x) x (a, b); acan be and b can be +.(a) Claim: b
a
g(x)dx
Proof :
Using the result from part (a),f(x)dx =
g(x)dx
h(x)dx,
and therefore,f(x)dx = =
g(x)dx
h(x)dx = =
g(x)dx =,
as desired. .
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