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SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL.

ERNEST YEUNG

Fund Science! & Help Ernest finish his Physics Research! : quantum super-A-polynomials - a thesis by Ernest Yeung

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Ernest Yeung is supported by Mr. and Mrs. C.W. Yeung, Prof. Robert A. Rosenstone, Michael Drown, Arvid Kingl, Mr .

and Mrs. Valerie Cheng, and the Foundation for Polish Sciences, Warsaw University.

SOLUTIONS TO VOLUME 1 One-Variable Calculus, with an Introduction to Linear Algebra

I 2.5 Exercises - Introduction to set theory, Notations for designating sets, Subsets, Unions, intersections, complements.

Exercise 10. Distributive laws

Let X = A

∩(B

∪C ), Y = (A

∩B)

∪(A

∩C )

Suppose x ∈ X x ∈ A and x ∈ (B ∪ C ) =⇒ x ∈ A and x is in at least B or in C then x is in at least either (A ∩ B) or (A ∩ C )

x ∈ Y, X ⊆ Y Suppose y ∈ Y

y is at least in either (A ∩ B) or A ∩ C then y ∈ A and either in B or C

y ∈ X, Y ⊆ X X = Y

Let X = A ∪ (B ∩ C ), Y = (A ∪ B) ∩ (A ∪ C )

Suppose x ∈ X then x is at least either in A or in (B ∩ C )if x ∈ A, x ∈ Y if x ∈ (B ∩ C ), x ∈ Y x ∈ Y, X ⊆ Y

Suppose y ∈ Y then y is at least in A or in B and y is at least in A or in C

if y ∈ A, then y ∈ X if y ∈ A ∩ B or y ∈ A ∪ C, y ∈ X (various carvings out of A, simply )if y ∈ (B ∩ C ), y ∈ X y ∈ X, Y ⊆ X

X = Y 1

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Exercise 11. If x ∈ A ∪ A, then x is at least in A or in A. Then x ∈ A. So A ∪ A ⊆ A. Of course A ⊆ A ∪ A.If x ∈ A ∩ A, then x is in A and in A. Then x ∈ A. So A ∩ A ⊆ A. Of course A ⊆ A ∩ A.

Exercise 12. Let x ∈ A. y ∈ A ∪ B if y is at least in A or in B. x is in A so x ∈ A ∪ B. =⇒ A ⊆ A ∪ B.Suppose ∃b ∈ B and b /∈ A. b ∈ A ∪ B but b /∈ A. so A ⊆ A ∪ B.

Exercise 13. Let x ∈ A ∪ ∅, then x is at least in A or in ∅. If x ∈ ∅, then x is a null element (not an element at all). Thenactual elements must be in A. =⇒ A ∪ ∅ ⊆ A.

Let x ∈ A. Then x ∈ A ∪ ∅. A ⊆ A ∪ ∅. =⇒ A = A ∪ ∅.Exercise 14. From distributivity, A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) = A ∩ (A ∪ B).If x ∈ A ∩ (A ∪ B), x ∈ A and x ∈ A ∪ B, i.e. x ∈ A and x is at least in A or in B.=⇒ x is in A and is in B or is not in B. Then x ∈ A. =⇒ A ∩ (A ∪ B) ⊆ A. Of course, A ⊆ A ∩ (A ∪ B).=⇒ A ∩ (A ∪ B) = A ∪ (A ∩ B) = A.Exercise 15. ∀a ∈ A, a ∈ C and ∀b ∈ B, b ∈ C . Consider x ∈ A ∪ B. x is at least in A or in B . In either case, x ∈ C .=⇒ A ∪ B ⊆ C .Exercise 16.

if C ⊆ A and C ⊆ B, then C ⊆ A ∩ B∀c ∈ C, c ∈ A and c ∈ Bx ∈ A ∩ B, x ∈ A and x ∈ B. Then ∀c ∈ C, c ∈ A ∩ B. C ⊆ A ∩ B

Exercise 17.

(1)if A ⊂ B and B ⊂ C then

∀a ∈ A, a ∈ B.∀b ∈ B, b ∈ C.then since a ∈ B, a ∈ C, ∃c ∈ C such that c /∈ B.∀a ∈ A, a ∈ B so a = c∀a. =⇒ A ⊂ C

(2) If A ⊆ B, B ⊆ C, A ⊆ C

since,∀a ∈ A, a ∈ B, ∀b ∈ B, b ∈ C.

Then since a ∈ B, a ∈ C

. A ⊆ C (3) A ⊂ B and B ⊆ C . B ⊂ C or B = C . A ⊂ B only. Then A ⊂ C .

(4) Yes, since ∀a ∈ A, a ∈ B.(5) No, since x = A (sets as elements are different from elements)

Exercise 18. A − (B ∩ C ) = (A − B) ∪ (A − C )

Suppose x ∈ A − (B ∩ C )then x ∈ A and x /∈ B ∩ C =⇒ x /∈ B ∩ C then x is not in even at least one B or C

=⇒ x ∈ (A − B) ∪ (A − C )Suppose x ∈ (A − B) ∪ (A − C )

then x is at least in (A − B) or in (A − C ) =⇒ x is at least in A and not in B or in A and not in C then consider when one of the cases is true and when both cases are true =⇒ x ∈ A − (B ∩ C )

Exercise 19.

Suppose x ∈ B −A∈F

A

then x ∈ B,x /∈A∈F

A

x /∈A∈F

A =⇒ x /∈ A, ∀A ∈ F

since ∀A ∈ F , x ∈ B , x /∈ A, then x ∈A∈F

(B − A)2


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Suppose x ∈A∈F

(B − A)

then x ∈ B − A1 and x ∈ B − A2 and . . .then ∀A ∈ F , x ∈ B , x /∈ Athen x /

∈ even at least one A

∈ F =⇒ x ∈ B − A∈F

A

Suppose x ∈ B −A∈F

A

then x /∈A∈F

A

then at most x ∈ A for ∀A ∈ F but onethen x is at least in one B − A=⇒ x ∈

A∈F

(B − A)

Suppose x ∈ A∈F

(B − A)then x is at least in one B − Athen for A ∈ F , x ∈ B and x /∈ AConsider ∀A ∈ F

=⇒ then x ∈ B −A∈F

A

Exercise 20.

(1) (ii) is correct.

Suppose x∈

(A−

B)−

C

then x ∈ A − B , x /∈ C then x ∈ A and x /∈ B and x /∈ C

x /∈ B and x /∈ C =⇒ x /∈ even at least B or C x ∈ A − (B ∪ C )

Suppose x ∈ A − (B ∪ C )then x ∈ A,x /∈ (B ∪ C )then x ∈ A and x /∈ B and x /∈ C =⇒ x ∈ (A − B) − C

To show that (i) is sometimes wrong,

Suppose y ∈ A − (B − C )y ∈ A and y /∈ B − C

y /∈ B − C then y /∈ B or y ∈ C or y /∈ C

(where does this lead to?)

Consider directly,

Suppose x ∈ (A − B) ∪ C then x is at least in A − B or in C then x is at least in A and /∈ B or in C

Suppose x = c

∈C and c /

∈A

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(2)If C ⊆ A,

A − (B − C ) = (A − B) ∪ C I 3.3 Exercises - The field axioms. The goal seems to be to abstract these so-called real numbers into just x’s and y’s thatare purely built upon these axioms.

Exercise 1. Thm. I.5. a(b − c) = ab − ac.Let y = ab − ac; x = a(b − c)Want: x = y

ac + y = ab (by Thm. I.2, possibility of subtraction)

Note that by Thm. I.3, a(b − c) = a(b + (−c)) = ab + a(−c) (by distributivity axiom)ac + x = ac + ab + a(−c) = a(c + (−c)) + ab = a(0 + b) = ab

But there exists exactly one y or x by Thm. I.2. x = y.Thm. I.6. 0 · a = a · 0 = 0.

0(a) = a(0) (by commutativity axiom)

Given b ∈ R and 0 ∈ R, ∃ exactly one − b s.t. b − a = 00(a) = (b + (−b))a = ab − ab = 0 (by Thm. I.5. and Thm. I.2)

Thm. I.7.

ab = ac

By Axiom 4, ∃y ∈ R s.t. ay = 1since products are uniquely determined, yab = yac =⇒ (ya)b = (ya)c =⇒ 1(b) = 1(c)

=⇒ b = cThm. I.8. Possibility of Division.

Given a, b, a = 0, choose y such that ay = 1.Let x = yb.

ax = ayb = 1(b) = b

Therefore, there exists at least one x such that ax = b. But by Thm. I.7, there exists only one x (since if az − b, and sox = z).

Thm. I.9. If a = 0, then b/a = b(a−1).

Let x = b

a for ax = b

y = a−1 for ay = 1

Want: x = by

Now b(1) = b, so ax = b = b(ay) = a(by)

=⇒ x = by (by Thm. I.7)Thm. I.10. If a = 0, then (a−1)−1 = a.

Now ab = 1 for b = a−1

. But since b ∈ R and b = 0 (otherwise 1 = 0, contradiction), then using Thm. I.8 on b,ab = b(a) = 1; a = b−1.Thm.I.11. If ab = 0, a = 0 or b = 0.

ab = 0 = a(0) =⇒ b = 0 or ab = ba = b(0) =⇒ a = 0. (we used Thm. I.7, cancellation law for multiplication)Thm. I.12. Want: x = y if x = (−a)b and y = −(ab).

ab + y = 0

ab + x = ab + (−a)b = b(a + (−a)) = b(a − a) = b(0) = 00 is unique, so ab + y = ab + x implies x = y( by Thm. I.1 )

Thm. I.13. Want: x + y = z , if a = bx, c = dy, (ad + bc) = (bd)z.

(bd)(x + y) = bdx + bdy = ad + bc = (bd)z

So using b, d= 0, which is given, and Thm. I.7, then x + y = z .

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Thm. I.14. Want: xy = z for bx = a,dy = c,ac = (bd)z.

(bd)(xy) = (bx)(dy) = ac = (bd)z

b, d = 0, so by Thm. I.7, xy = z .Thm.I.15. Want: x = yz , if bx = a,dy = c, (bc)z = ad

(bc)z = b(dy)z = d(byz) = da

d = 0 so by Thm. I.7, by z = a, byz = abxb = 0 so by Thm. I.7, yz = x

Exercise 2. Consider 0 + z = 0. By Thm. I.2, there exists exactly one z, z = −0. By Axiom 4, z = 0. 0 = −0.Exercise 3. Consider 1(z)z(1) = 1. Then z = 1−1. But by Axiom 4, there exists distinct 1 such that z(1) = 1, so z = 1.

Exercise 4. Suppose there exists x such that 0x = 1, but 0x = 0 and 0 and 1 are distinct, so zero has no reciprocal .

Exercise 5. a + (−a) = 0, 0 + 0 = 0. Thena + (−a) + b + (−b) = (a + b) + (−a) + (−b) = 0

−(a + b) =

−a + (

−b) =

−a

−b

Exercise 6. a + (−a) = 0, b + (−b) = 0, soa + (−a) + b + (−b) = a + (−b) + (−a) + b = (a − b) + (−a) + b = 0 + 0 = 0

−(a − b) = −a + b.Exercise 7.

(a − b) + (b − c) = a + (−b) + b + (−c) = a + (b + (−b)) + (−c) = a − cExercise 8.

(ab)x = 1 (ab)−1 = x

a(bx) = 1 a−1 = bx

b(ax) = 1 b−1

− axa−1b−1 = (abx)x = 1(x) = (ab)−1

Exercise 9. Want: x = y = z , if

z = a

−b a = zt b + t = 0

y = (−a)

b by = u a + u = 0

x = −a

b

ab

+ x = v + x = 0 vb = a

a + (−a) = vb + by = b(v + y) = 0if b

= 0, v + y = 0, but v + x = 0

by Thm. I.1 , x = y

b + t = 0, then z(b + t) = zb + zt = zb + a = z(0) = 0

a + zb = 0 =⇒ −a = zb = bysince b = 0, z = y so x = y = z

Exercise 10. Since b, d = 0, Let

z = ad − bc

bd (bd)z = ad − bc by previous exercise or Thm. I.8, the possibility of division

x = a

b bx = a

t = −c

d

dt =

−c (By Thm. I.3, we know that b

−a = b + (

−a) )

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dbx + bdt = (bd)(x + y) = ad − bc = (bd)zb, d = 0, so x + y = z

I 3.5 Exercises - The order axioms.

Theorem 1 (I.18). If a < b and c > 0 then ac < bc

Theorem 2 (I.19). If a < b and c > 0 , then ac < bc

Theorem 3 (I.20). If a = 0 , then a2 > 0Theorem 4 (I.21). 1 > 0

Theorem 5 (I.22). If a < b and c bc.

Theorem 6 (I.23). If a < b and −a > −b. In particular, if a 0.Theorem 7 (I.24). If ab > 0 , then both a and b are positive or both are negative.

Theorem 8 (I.25). If a < c and b < d , then a + b < c + d.

Exercise 1.

(1) By Thm. I.19, −c > 0a(−c) < b(−c) → −ac < −bc

−bc − (−ac) = ac − bc > 0. Then ac > bc (by definition of > )(2)

a < b → a + 0 < b + 0 → a + b + (−b) < b + a + (−a) → (a + b) − b < (a + b) + (−a)By Thm.I.18 (a + b) + −(a + b) + (−b) < (a + b) − (a + b) + (−a)

−b < −a(3)

If a = 0 or b = 0, ab = 0, but 0 ≯ 0

If a > 0, then if b > 0, ab > 0(b) = 0. If b 0.

If a 0, ab 0 + 0 = 0 → x2 + 1 = 0=⇒ x ∈ R such that x2 + 1 = 0

Exercise 3.

a 0. By Thm. I.24 , a , x are both positive or a, x are both negative

Exercise 5. Define x, y such that ax = 1, by = 1. We want x > y when b > a.

xb − ax = xb − 1 > 0 =⇒ bx > 1 = byb > 0 so x > y

Exercise 6.

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If a = b and b = c, then a = c

If a = b and b < c, then a < c

If a < b and b = c, then a < c

If a < b and b < c, then a < c (by transitivity of the inequality)

=

⇒a

≤c

Exercise 7. If a ≤ b and b ≤ c, then a ≤ c. If a = c, then by previous proof, a = b.Exercise 8. If a ≤ b and b ≤ c, then a ≤ c. If a = c, then by previous proof, a = b.Exercise 8. If a or b is zero, a2 or b2 = 0. By Thm. I.20, b2 ≥ 0 or a2 ≥ 0, respectively.

Otherwise, if neither are zero, by transitivity, a2 + b2 > 0.

Exercise 9. Suppose a ≥ x. Then a − x ≥ 0.If a ∈ R so ∃y ∈ R, such that a − y = 0.

Consider y + 1 ∈ R (by closure under addition).a − (y + 1) = a − y − 1 = 0 − 1 < 0 Contradiction that a ≥ y + 1

Exercise 10.If x = 0, done.

If x > 0, x is a positive real number. Let h = x

2 .

=⇒ x2

> x Contradiction.

I 3.12 Exercises - Integers and rational numbers, Geometric interpretation of real numbers as points on a line,

Upper bound of a set, maximum element, least upper bound (supremum), The least-upper-bound axiom (completeness

axiom), The Archimedean property of the real-number system, Fundamental properties of the suprenum and infimum.

We use Thm I.30, the Archimedean property of real numbers, alot.

Theorem 9 (I.30). If x > 0 and if y is an arbitrary real number, there exists a positive integer n such that nx > y.

We will use the least upper-bound axiom (completeness axiom) alot for continuity and differentiation theorems later.

Apostol states it as an axiom; in real analysis, the existence of a sup for nonempty, bounded sets can be shown with an

algorithm to zoom into a sup with monotonically increasing and monotonically decreasing sequence of “guesses” and showing

its difference is a Cauchy sequence.

Axiom 1 (Least upper-bound axiom). Every nonempty set S of real numbers which is bounded above has a suprenum; that is, there’s a real number B s.t. B = supS .

Exercise 1. 0 < y − x.=⇒ n(y − x) > h > 0, n ∈ Z+, h arbitrary

y − x > h/n =⇒ y > x + h/n > xso let z = x + h/n Done.

Exercise 2. x ∈ R so ∃n ∈ Z+

such that n > x (Thm. I.29).Set of negative integers is unbounded below because

If ∀m ∈ Z−, −x > −m, then −x is an upper bound on Z+. Contradiction of Thm. I.29. =⇒ ∃m ∈ Z such that m < x < nExercise 3. Use Archimedian property.

x > 0 so for 1, ∃n ∈ Z+ such that nx > 1, x > 1n .Exercise 4. x is an arbitrary real number. By Thm. I.29 and well-ordering principle, there exists a smallest n + 1 positive

integer such that x < n + 1 (consider the set of all m + 1 > x and so by well-ordering principle, there must be a smallestelement of this specific set of positive integers).

If x = n for some positive integer n, done.Otherwise, note that if x < n, then n + 1 couldn’t have been the smallest element such that m > x. x > n.

Exercise 5. If x = n, done. Otherwise, consider all m > x. By well-ordering principle, there exists a smallest element n such7


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that n > x.If x + 1 < n, then x < n − 1, contradicting the fact that n is the smallest element such that x < n. Thus x + 1 > n.

Exercise 6. y − x > 0.n(y − x) > h, h arbitrary , n ∈ Z+

y > x + h/n = z > xSince h was arbitrary, there are infinitely many numbers in between x, y.

Exercise 7. x = ab ∈ Q, y /∈ Q.

x ± y = a ± byb

If a ± by was an integer, say m, then y = ±

a − mbb

which is rational. Contradiction.

xy = a

b

y

1 =

ay

b

If ay was an integer, ay = n, y = n

a, but y is irrational. =⇒ xy is irrational.

xy

y is not an integer

Exercise 8. Proof by counterexamples. We want that the sum or product of 2 irrational numbers is not always irrational. If y

is irrational, y + 1 is irrational, otherwise, if y + 1 ∈ Q, y ∈ Q by closure under addition.=⇒ y + 1 − y = 1

Likewise, y 1y = 1.

Exercise 9.

y − x > 0 =⇒ n(y − x) > k, n ∈ Z+, k arbitrary. Choose k to be irrational. Then k/n irrational.

y > k

n + x > x. Let z = x +

k

n , z irrational .

Exercise 10.

(1) Suppose n = 2m1 and n + 1 = 2m2.

2m1 + 1 = 2m2 2(m1 − m2) = 1 m1 − m2 = 12

. But m1 − m2 can only be an integer.(2) By the well-ordering principle, if x ∈ Z+ is neither even and odd, consider the set of all x. There must exist a

smallest element x0 of this set. But since x0 ∈ Z+, then there must exist a n < x such that n + 1 = x0. n is even orodd since it doesn’t belong in the above set. So x0 must be odd or even. Contradiction.

(3)

(2m1)(2m2) = 2(2m1m2) even

2m1 + 2m2 = 2(m1 + m2) even

(2m1 + 1) + (2m2 + 1) = 2(m1 + m2 + 1) =⇒ sum of two odd numbers is even(n1 + 1)(n2 + 1) = n1n2 + n1 + n + 2 + 1 = 2(2m1m2)

2(2m1m2) − (n1 + n2) − 1 odd, the product of two odd numbers n1, n2 is odd(4) If n2 even, n is even, since for n = 2m, (2m)2 = 4m2 = 2(2m2) is even.

a2 = 2b2. 2(b2) even. a2 even, so a even.

If a even a = 2n.a2 = 4n2

If b odd , b2 odd. b has no factors of 2 b2 = 4n2

Thus b is even.8


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(5) For pq , If p or q or both are odd, then we’re done.

Else, when p, q are both even, p = 2lm, q = 2n p, m, p odd.

p

q =

2lm

2n p =

2l−nm p

and at least m or p odd

Exercise 11. ab can be put into a form such that a or b at least is odd by the previous exercise.

However, a2 = 2b2, so a even, b even, by the previous exercise, part (d) or 4th part. Thus ab cannot be rational.

Exercise 12. The set of rational numbers satisfies the Archimedean property but not the least-upper-bound property.

Since pq ∈ Q ⊆ R, n p1q1 > p2q2

since if q 1, q 2 > 0,

np1q 2q 1q 2

> q 1 p2q 1q 2

np1q 2 > q 1 p2

n exists since ( p1q 2), (q 1 p2) ∈ R.The set of rational numbers does not satisfy the least-upper-bound property.

Consider a nonempty set of rational numbers S bounded above so that ∀x = rs ∈ S,x < b.

Suppose x < b1, x < b2∀x ∈ S .r

s < b2 < nb1 but likewise

r

s < b1 < mb2, n, m ∈ Z+

So it’s possible that b1 > b2, but also b2 > b1.

I 4.4 Exercises - An example of a proof by mathematical induction, The principle of mathematical induction, The

well-ordering principle. Consider these 2 proofs.

N + N + · · · + N = N 2

(N − 1) + (N − 2) + · · · + (N − (N − 1)) + (N − N ) = N 2 −N j=1

j =N −1j=1

j

N 2 + N = 2N j=1

j =⇒N j=1

j = N (N + 1)

2

An interesting property is that

S =nj=m

j =nj=m

(n + m − j)

So thatN j=1

j =

N j=m

j +

mj=1

j =

N j=m

j + m(m + 1)

2 =

N (N + 1)

2

N j=m

j = N (N + 1) − m(m + 1)

2 =

(N − m)(N + m + 1)2

Another way to show this is the following.

S = 1+ 2+ · · · + (N − 2)+ (N − 1)+ N but S = N + N − 1+ · · · + 3+ 2+ 1

2S = (N + 1)N S = N (N + 1)

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Telescoping series will let you getN j=1 j

2 and other powers of j .

N j=1

(2 j − 1) = 2 N (N + 1)2

− N = N 2

N j=1

( j2 − ( j − 1)2) =N j=1

( j2 − ( j2 − 2 j + 1)) =N j=1

(2 j − 1) = 2N (N + 1)2 − N = N 2N j=1

( j3 − ( j − 1)3) = N 3 =N j=1

( j3 − ( j3 − 3 j2 + 3 j − 1)) =N j=1

(3 j2 − 3 j + 1)

=⇒ 3N j=1

j2 = −3 N (N + 1)2

+ N = N 3 =⇒ 2N 3 + 2N − 3N 2 − 3N

2 =

N (N + 1)(2N + 1)

6 =

N j=1

j2

N j=1

j4 − ( j − 1)4 = N 4 =N j=1

j4 − ( j4 − 4 j3 + 6 j2 − 4 j + 1) =N j=1

4 j3 − 6 j2 + 4 j − 1 =

= 4N

j=1 j3 − 6N (N + 1)(2N + 1)

6 + 4

N (N + 1)

2 −N = N 4

=⇒N j=1

j3 = 1

4(N 4 + N (N + 1)(2N + 1) − 2N (N + 1) + N ) = 1

4(N 4 + (2N )N (N + 1) − N (N + 1) + N )

= 1

4(N 4 + 2N 3 + 2N 2 − N 2 − N + N ) = 1

4N 2(N 2 + 2N + 1) =

1

4

(N (N + 1))2

2

Exercise 1. Induction proof.

1(1 + 1)

2

N +1j=1

j =nj=1

j + n + 1 = n(n + 1)

2 + n + 1 =

n(n + 1) + 2(n + 1)

2 =

(n + 2)(n + 1)

2

Exercise 6.

(1)

A(k + 1) = A(k) + k + 1 = 1

8(2k + 1)2 + k + 1 =

1

8(4k2 + 4k + 1) +

8k + 8

8 =

(2k + 3)2

8(2) The n = 1 case isn’t true.(3)

1 + 2 + · · · + n = (n + 1)n2

= n2 + n

2 <

n2 + n + 142

and

2n + 1

2

21

2 =

(n + 1/2)2

2 =

n2 + n + 1/4

2

Exercise 7.

(1 + x)2 > 1 + 2x + 2x2

1 + 2x + x2 > 1 + 2x + 2x2

0 > x2 =⇒ Impossible(1 + x)3 = 1 + 3x + 3x2 + x3 > 1 + 3x + 3x2

=⇒ x3 > 0By well-ordering principle, we could argue that n = 3 must be the smallest number such that (1 + x)n > 1 + 2x + 2x2. Orwe could find, explicitly

(1 + x)n =n

j=0n

jxj = 1 + nx +

n(n − 1)2

x2 +n

j=3n

jxj

10


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andn(n − 1)

2 > n

n2 − n > 2nn2 > 3n

n > 3

Exercise 8.

a2 ≤ ca1, a3 ≤ ca2 ≤ c2a1an+1 ≤ can ≤ ca1cn−1 = a1cn

Exercise 9.

n = 1,√

1 = 1 12 + 12 =

√ 2

(√

2)2 + 12 =√

3 (√

n)2 + 12 =√

n + 1

Exercise 10.

1 = qb + r

q = 0, b = 1, r = 1

2 = qb + r, q = 0, r = 2, b = 1, 2 or r = 0, q = 2; q = 1, r = 0

Assume n = qb + r; 0 ≤ r < b; b ∈ Z+, b fixedn + 1 = qb + r + 1 = qb + 1 + r = qb + 1 + b − 1 = (q + 1)b + 0

Exercise 11. For n > 1, n = 2, 3 are prime. n = 4 = 2(2), a product of primes.

Assume the k − 1th case. Consider kj , 1 ≤ j ≤ k.If kj ∈ Z+, only for j = 1, j = k , then k prime.If kj ∈ Z+, for some 1 < j < k , kj = c ∈ Z+. c,j < k.

Thus k = cj . c, j are products of primes or are primes, by induction hypothesis. Thus k is a product of primes.

Exercise 12. n = 2. G1, G2 are blonde. G1 has blue eyes. Consider G2. G2 may not have blue eyes. Then G1, G2 are not all

blue-eyed.

I 4.7 Exercises - Proof of the well-ordering principle, The summation notation. Exercise 1.

(1) n(n+1)

2 =4k=1 k = 10

(2)5n=2 2

n−2 =3n=0 2

n = 1 + 14 = 15

(3) 23r=0 2

2r = 23r=0 4

r = 170

(4)4j=1 j

j = 1 + 4 + 27 + 44 = 288

(5)5j=0(2 j + 1) = 2

5(6)2 + 6(1) = 36

(6) 1k(k+1) =

nk=1

1k − 1

k+1 = 1 − 1n+1 =

nn+1

Exercise 9.

n = 1(−1)(3) + 5 = 2 = 2nn = 2(−1)(3) + 5 + (−1)7 + 9 = 4 = 2n

n2nk=1

(−1)k(2k + 1) = 2n

n + 1

2(n+1)k=1

(−1)k(2k + 1) =2nk=1

(−1)k(2k + 1) + (−1)2n+1(4n + 3) + (−1)2n+2(4n + 5) =

= 2n + 2 = 2(n + 1)

Exercise 10.

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(1) am + am+1 + · · · + am+n(2)

n = 11

2 =

1

1 − 1

2 =

1

2

n + 1

2(n+1)

k=n+2

1

k =

2n

m=1

(

−1)m+1

m − 1

n + 1 +

1

2n + 1 +

1

2n + 2 =

2n

m=1

(

−1)m+1

m + − 1

2(n + 1) +

(

−1)2n+1+1

(2n + 1)

=

2(n+1)m=1

(−1)m+1m

Exercise 13.

n = 12(√

2 − 1) < 1 < 2 since 12

>√

2 − 1

n case (√

n + 1 − √ n)(√ n + 1 + √ n) = n + 1 − n = 1 <√

n + 1 +√

n

2√

n =

1

2(

1 +

1

n + 1)

n + 1 case (√

n + 2

−

√ n + 1)(

√ n + 2 +

√ n + 1) = n + 2

−(n + 1) = 1

√ n + 2 +

√ n + 1

2√

n + 1=

1 +

1 + 1n+1

2 > 1

So then, using the telescoping property,

n−1n=1

2(√

n + 1 − √ n) = 2(√ m − 1) <mn=1

1√ n

<mn=1

2(√

n − √ n − 1) = 2(√ m − 1) < 2√ m − 1

I 4.9 Exercises - Absolute values and the triangle inequality. Exercise 1.

(1) |x| = 0iffx = 0If x = 0, x = 0, −x = −0 = 0. If |x| = 0, x = 0, −x = 0.

(2)

| − x| =

−x if − x ≥ 0x if − x ≤ 0 =

x if x ≥ 0−x if x ≤ 0

(3) |x − y| = |y − x| by previous exercise and (−1)(x − y) = y − x (by distributivity)(4) |x|2 =

(x)2 if x ≥ 0(−x)2 if x ≤ 0 = x

2

(5)√

x2 =

x if x ≥ 0−x if x ≤ 0 = |x|

(6) We want to show that |xy| = |x||y|

|xy

|=

xy if xy ≥ 0−xy if xy ≤ 0

= xy if x, y ≥ 0 or x, y ≤ 0−xy if x, −y ≥ 0 or − x, y ≤ 0

|x||y| =

x|y| if x ≥ 0−x|y| if x ≤ 0 =

xy if x, y ≥ 0−xy if x, −y ≥ 0−xy if − x, y ≥ 0xy if − x, −y ≥ 0

(7) By previous exercise, since xy = |xy−1| = |x||y−1|

1

y

=

1y if

1y ≥ 0

−1y if

1y

≤0

1

|y

|

=

1y if

1y ≥ 0

−1y if

1y

≤0

12


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(8) We know that |a − b| ≤ |a − c| + |b − c|.Let c = 0 =⇒ |x − y| ≤ |x| + |y|

(9) x = a − b, b − c = −y.

|x

| ≤ |x

−y

|+

| −y

| |x

| − |y

| ≤ |x

−y

|(10)

||x| − |y|| =

|x| − |y| if |x| − |y| ≥ 0|y| − |x| if |x| − |y| ≤ 0

|x| ≤ |x − y| + | − y| =⇒ |x| − |y| ≤ |x − y||y| ≤ |y − x| + | − x| =⇒ |y| − |x| ≤ |y − x| = |x − y|

Exercise 4.

⇒If ∀k = 1 . . . n; akx + bk = 0

nk=1

ak(−xak)2 = x nk=1

a2k2 = nk=1

a2k nk=1

(−xak)2 = nk=1

a2k nk=1

b2k⇐ Proving akx + bk = 0 means x = − bk

ak, ak = 0

(a1b1 + a2b2 + · · · + anbn)2 =nj=1

a2jb2j +

nj=q

ajakbjbk ==nj=1

a2jb2j +

nj=k

a2jb2k

=⇒ a2jb2k − ajakbjbk = ajbk(ajbk − akbj) = 0

if aj , bk = 0, ajbk − akbj = 0 =⇒ ak

bj−aj

+ bk = 0

Exercise 8. The trick of this exercise is the following algebraic trick (“multiplication by conjugate”) and using telescoping

property of products:

(1 − x2j )(1 + x2j ) = 1 − x2j+2j = 1 − x2j+11j=1

1 + x2j−1

=

1j=1

1 − x2j1 − x2j−1 =

1 − x2n1 − x

if x = 1, 2n

Exercise 10.

x > 1

x2 > x

x3 > x2 > x

xn+1 = xnx > x2 > x

0 < x < 1

x2 < x

X 3 < x2 < x

xn+1 = xnx < x2 < x =⇒ xn+1 < x

Exercise 11. Let S = {n ∈ Z+|2n < n!}.By well-ordering principle, ∃ smallest n0 ∈ S . Now24 = 16, 4! = 24. So S starts at n = 4.

Exercise 12.

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(1) 1 +

1

n

n=

nj=0

n

k

1

n

j=

nk=0

n!

(n − k)!k!

1

n

kk−1

r=0 1 − r

n =k−1

r=0n − r

n = 1

nk

n!

(n − k)!nk=1

1

k!

k−1r=0

1 − r

n

=

1

nk

n!

(n − k)!(2)

(1 + 1

n)n = 1 +

nk=1

1

k!

k−1r=0

(1 − rn

)


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Now assume pth case. Test the p + 1 case.

2 p+2

p + 2 =

2( p + 1)

p + 2

2 p+1

p + 1

> 1

since p + 2 < 2 p + 2 = 2( p + 1) for p ∈ Z+

For the right-hand inequality, we will use the fact just proven, that 2 p− ( p) > 0 and pth case rewritten in this manner

(1 + 2 p) > 2 p+1

p + 1 =⇒ (1 + 2 p)( p + 1) > 2 p+1

So

( p + 2)(1 + 2 p+1) = ( p + 2) + (( p + 1) + 1)2 p(2) = ( p + 2) + 2( p + 1)2 p + 2 p(2) >

> ( p + 2) + 2(2 p+1 − ( p + 1)) + 2 p(2) = − p + 2 p+2 + 2 p+1 > 2 p+2

So the n = 2 case is true for all p ∈ Z+.Assume nth case is true. We now prove the n + 1 case.

n

k=1 k p =n−1

k=1 k p + n p < n p+1

p + 1

+ n p < n p+1

p + 1

+ (n + 1) p+1 − n p+1

p + 1

= (n + 1) p+1

p + 1n+1k=1

k p =

nk=1

k p + (n + 1) p > n p+1

p + 1 +

(n + 1) p+1 − n p+1 p + 1

= (n + 1) p+1

p + 1

We had used the inequality proven in part b, n p < (n+1)p+1−np+1 p+1


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a1 = 1 < 1 +

√ 5

2

a2 = 2 <

1 +

√ 5

2

2=

1 + 2√

5 + 5

4 =

6 + 2√

5

4

an+1 = an + an−1 < 1 + √ 52

n +1 + √ 52

n−1 = 1 + √ 52

n1 + 21 +

√ 5 =

=

1 +

√ 5

2

n2(1 − √ 5)

1 − 5 + 4

4

=

1 +

√ 5

2

n+1

Exercise 17. Use Cauchy-Schwarz, which saysakbk

2≤

a2k

b2k

Let ak = x

pk and bk = 1. Then Cauchy-Schwarz says

x pk2 ≤ x2 pk n =⇒(x2 pk ) ≥ (x pk)2nWe define M p as follows:

M p =

nk=1 x

pk

n

1/pSo then

nM p p =nk=1

x pk

M 2 p =

nk=1 x

2 pk

n

1/2 p

nM 2 p2 p =

nk=1

x2 pk x2 pk = nM

2 p2 p ≥

(nM p p )2

n = nM 2 p p

M 2 p2 p ≥ M 2 p p =⇒ M 2 p ≥ M pExercise 18.

a4 + b4 + c4

3

1/4≥

a2 + b2 + c2

3

1/2=

23/2

31/2 since

a4 + b4 + c4 ≥ 643

Exercise 19. ak = 1,nk=1 1 = n

Now consider the case of when not all ak = 1.

a1 = 1

a1a2 = 1 and suppose, without loss of generality a1 > 1. Then 1 > a2.

(a1 − 1)(a2 − 1) < 0a1a2 − a1 − a2 + 1 < 0 =⇒ a1 + a2 > 2

(consider n + 1 case ) If a1a2 . . . an+1 = 1, then suppose a1 > 1, an+1 < 1 without loss of generality

b1 = a1an+1

b1a2 . . . an = 1 =⇒ b1 + a2 + · · · + an ≥ n (by the induction hypothesis)(a1 − 1)(an+1 − 1) = a1an+1 − a1 − an+1 + 1 < 0, b1 < a1 + an+1 − 1=⇒ a1 + an+1 − 1 + a2 + · · · + an > b1 + a2 + · · · + an ≥ n=

⇒a1 + a2 +

· · ·+ an+1

≥n + 1

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1.7 Exercises - The concept of area as a set function. We will use the following axioms:

Assume a class M of measurable sets (i.e. sets that can be assigned an area), set function a, a : M → R.•

Axiom 2 (Nonnegative property).

(1) ∀S ∈ M, a(S ) ≥ 0•

Axiom 3 (Additive property). If S, T ∈ M , then S ∪ T, S ∩ T ∈ M and (2) a(S ∪ T ) = a(S ) + a(T ) − a(S ∩ T )

•Axiom 4 (Difference property). If S, T ∈ M, S ⊆ T then T − S ∈ M and

(3) a(T − S ) = a(T ) − a(S )•

Axiom 5 (Invariance under congruence). If S

∈ M, T = S , then T

∈ M , a(T ) = a(S )

•Axiom 6 (Choice of scale). ∀ rectangle R ∈ M , if R has edge lengths h, k then a(R) = hk

•Axiom 7 (Exhaustion property). Let Q such that

(4) S ⊆ Q ⊆ T If ∃ only one c such that a(S ) ≤ c ≤ a(T ), ∀S, T such that they satisfy Eqn. (??)

then Q measurable and a(Q) = c

Exercise 1.

(1) We need to say that we consider a line segment or a point to be a special case of a rectangle allowing h or k (or both)

to be zero.Let T l = { line segment containing x0 }, Q = {x0}.

For Q, only ∅ ⊂ QBy Axiom 3, let T = S .

a(T − S ) = a(∅) = a(T ) − a(T ) = 0∅ ⊂ Q ⊆ T l =⇒ a(∅) ≤ a(Q) ≤ a(T l) =⇒ 0 ≤ a(Q) ≤ 0

=⇒ a(Q) = 0(2)

a

N j=1

Qj

=

N j=1

a(Qj)

if Qj’s disjoint. Let Qj = {xj}.

Since a(Qj) = 0. By previous part, aN

j=1 Qj

= 0

Exercise 2. Let A, B be rectangles. By Axiom 5, A, B are measurable. By Axiom 2, A ∩ B measurable.

a(A ∩ B) =

a2 + b2d + ab − ( 12

ab +

a2 + b2d) = 1

2ab

Exercise 3. Prove that every trapezoid and every parallelogram is measurable and derive the usual formulas for their areas.

A trapezoid is simply a rectangle with a right triangle attached to each end of it. T r = R + T 1 + T 2. T 1, T 2 are righttriangles and so by the previous problem, T 1, T 2 are measurable. Then T r is measurable by the Additive property axiom (notethat the triangles and the rectangle don’t overlap).

17


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We can compute the area of a trapezoid:

T r = R + T 1 + T 2 =⇒ a(T r) = a(R) + a(T 1) + a(T 2)a(T r) = hb1 +

1

2h(b2 − b1)/2 + 1

2h(b2 − b1)/2 = 1

2h(b1 + h2)

P = R (a parallelogram consists of a right triangle rotated by π and attached to the other side of the same right triangle;the two triangles do not overlap). Since two right triangles are measurable, the parallelogram, P is measurable.Using the Additive Axiom, a(P ) = 2a(T ) = 2 12bh = bh

Exercise 4. A point (x, y) in the plane is called a lattice point if both coordinates x and y are integers. Let P be a polygon

whose vertices are lattice points. The area of P is I + 12 B −1, where I denotes the number of lattice points inside the polygonand B denotes the number on the boundary.

(1) Consider one side of the rectangle lying on a coordinate axis with one end on the origin. If the rectangle side has

length l, then l + 1 lattice points lie on this side (you have to count one more point at the 0 point. Then consider thesame number of lattice points on the opposite side. We have 2(l + 1) lattice points so far, for the boundary.

The other pair of sides will contribute 2(h−1) lattice points, the −1 to avoid double counting. Thus 2(l +h) = B .I = (h − 1)(l − 1) by simply considering multiplication of (h − 1) rows and (l − 1) columns of lattice points

inside the rectangle.

I + 12B − 1 = hl − h − l + 1 + (l + h) − 1 = hl = a(R)(2)

(3)

Exercise 5. Prove that a triangle whose vertices are lattice points cannot be equilateral.

My way: I will take, for granted, that we know an equilateral triangle has angles of π/3 for all its angles.

Even if we place two of the vertices on lattice points, so that its length is 2L, and put the midpoint and an intersectingperpendicular bisector on a coordinate axis (a picture would help), but the ratio of the perpendicular bisector to the thirdvertex to half the length of the triangle is cot π/3 = 1√

3. Even if we go down by an integer number L, L steps down, we go

“out” to the third vertex by an irrational number√

3L. Thus, the third vertex cannot lie on a lattice point.

Exercise 6. Let A = {1, 2, 3, 4, 5} and let M denote the class of all subsets of A. (There are 32 altogether counting A itself and the empty set ∅). (My Note: the set of all subsets, in this case, M, is called a power set and is denoted 2A. This isbecause the way to get the total number of elements of this power set, |2A|, or the size, think of assigning to each element a“yes,” if it’s in some subset, or “no”, if it’s not. This is a great way of accounting for all possible subsets and we correctly

get all possible subsets.) For each set S in M, let n(S ) denote the number of distinct elements in S . If S = {1, 2, 3, 4} andT = {3, 4, 5},

n(S T ) = 5n(S T ) = 2n(S − T ) = n({1, 2}) = 2n(T − S ) = n({5}) = 1

n satisfies nonnegative property because by definition, there’s no such thing as a negative number of elements. If S, T aresubsets of A, so are S

T , S

T since every element in S

T , S

T is in S . Thus n could be assigned to it, so that it’s

measurable. Since n counts only distinct elements, then n(S

T ) = n(S ) + a(T ) − a(S T ), where −a(S T ) ensuresthere is no double counting of distinct elements. Thus, the Additive Property Axiom is satisfied.

For S ⊆ T , then ∀ x ∈ T − S , x ∈ T, x /∈ S Now S ⊆ T , so ∀ x ∈ S , x ∈ T . Thus T − S is complementary to S “withrespect to” T . n(S ) + n(T

−S ) = n(T ), since n counts up distinct elements.

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1.11 Exercises - Intervals and ordinate sets, Partitions and step functions, Sum and product of step function. Exercise

4.

(1)

[x + n] = y ≤ x + n, y ∈ Z; y − n ≤ x

[x] + n = z + n ≤ x + nIf y − n < z, then y < z + n ≤ x + n. then y wouldn’t be the greatest integer less than x + n

=⇒ y = z + n(2)

= y2 ≤ x − [x] = −y2 ≥ −x − y2 − 1 ≤ x−x ≥ y1 = [−x] = −y2 − 1 = −[x] − 1; ( and y1 = −y2 − 1 since −y2 > −x )

If x is an integer −[x] = [−x](3) Let x = q 1 + r1, y = q 2 + r2; 0 ≤ r1, r2 < 1.

= [q 1 + q 2 + r1 + r2] =

q 1 + q 2

q 1 + q 2 + 1 if r1 + r2 ≥ 1[x] + [y] = q 1 + q 2 [x] + [y] + 1 = q 1 + q 2 + 1

(4)

If x is an integer , [2x] = 2x = [x] + [x + 1

2] = [x] + [x] = 2x

[x] + [x + 1

2] = q +

q if r < 122q + 1 if r > 12

[2x] = [2(q + r)] = [2q + 2r] =

2q if r < 122q + 1 if r > 12

(5)

[x] + [x + 13

] + [x + 23

] = q + q if r < 23q + 1 if r > 23

+q if r < 13q + 1 if r > 13

= 3q if

r <

1

33q + 1 if 13 < r <

23

3q + 2 if r > 23

[3x] = [3(q + r)] = [3q + 3r] =

3q if r < 133q + 1 if 13 < r <

23

3q + 2 if r > 23

Exercise 5. Direct proof.

[nx] = [n(q + r)] =

nq if r < 1nnq + 1 if 1n < r <

2n

nq + n − 1 if r > n−1nExercise 6.

a(R) = hk = I R + 1

2BR − 1

bn=a

[f (n)] = [f (a)] + [f (a + 1)] + · · · + [f (b)]

[f (n)] = g ≤ f (n), g ∈ Z, so that if f (n) is an integer,g = f (n), and if f (n) is not an integer, g is the largest integer suchthat g < f (n), so that all lattice points included and less than g are included.

Exercise 7.

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(1) Consider a right triangle with lattice points as vertices. Consider b + 1 lattice points as the base with b length.Start from the vertex and move across the base by increments of 1.The main insight is that the slope of the hypotenuse of the right triangle is ab so as we move 1 along the base, the

hypotenuse (or the y-value, if you will) goes up by ab . Now

(5) nab = number of interior points at x = n and below the hypotenuse line of the right triangle of sides a, b,including points on the hypotenuse

b−1n=1

nab

+

1

2((a + 1) + b) − 1 = ab

2

Now (a − 1)(b − 1)

2 =

ab

2 − a

2 − b

2 +

1

2

=⇒b−1n=1

nab

=

(a − 1)(b − 1)2

(2) a, b ∈ Z+b

−1

n=1

nab = b−1

n=1

a(b − n)b

(reverses order of summation)b−1n=1

a − an

b

=

−b−1n=1 anb − a if anb − a4 is an integer (but a nb − 1 can’t be!)−b−1n=1 anb − a− 1 otherwise

= −b−1n=1

anb − a

− 1

= −

b−1n=1

anb

− a

− (b − 1) =

= −b−1n=1

anb

+ a(b − 1) − (b − 1)

b−1

n=1 na

b = (a − 1)(b − 1)

2

Exercise 8. Recall that for the step function f = f (x), there’s a partition P = {x0, x1, . . . , xn} of [a, b] such that f (x) = ckif x ∈ I k.

Given that χs(x) =

1 ∀x ∈ S 0 ∀x /∈ S .

If x ∈ [a, b], then x must only lie in one open subinterval I j , since real numbers obey transitivity.nk=1

ckχI k(x) = cj for x ∈ I j =⇒nk=1

ckχI k(x) = f (x)∀x ∈ [a, b]

1.15 Exercises - The definition of the integral for step functions, Properties of the integral of a step function, Other

notations for integrals. Exercise 1.

(1) 31 [x]dx = (−1) + 1 + (2) = 2

(2) 3−1[x +

12 ]dx =

7/2−1/2[x]dx = (−1)12 + (1)(1) + (2)(1) + 12 3 = 4

(3) 3−1([x] + [x +

12 ])dx = 6

(4) 3−1 2[x]dx = 4

(5) 3−1[2x]dx =

12

6−2[x]dx =

12 ((−2)1 + (−1) + (1) + 2 + 3 + 4 + 5) = 6

(6) 3−1[−x]dx = −

−31

[x]dx = 1−3[x]dx = −3 + −2 + −1 = −6

Exercise 2.

s =

5/2 if 0 < x


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Exercise 3. [x] = y ≤ x so −y ≥ −x.

−y − 1 ≤ −x, otherwise if −y − 1 ≥ −x, y + 1 ≤ x and so y wouldn’t be the largest integer ≤ x.=⇒ [x] + [−x] = y − y − 1 = −1

Or use Exercise 4(c), pp. 64. ba

([x] + [−x])dx = ba

[x − x]dx = ba

(−1)dx = a − b

Exercise 4.

(1) n ∈ Z+, n0

[t]dt =n−1t=0 t =

(n−1)(n−1+1)2 =

(n−1)n2

(2)

Exercise 5.

(1) 20

[t2]dt = 21

[t2]dt = 1(√

2 − 1) + 2(√ 3 − √ 2) + 3(2 − √ 3) = 5 − √ 2 − √ 3(2)

3

−3[t2]dt =

3

0 [t2]dt +

0

−3[t2]dt =

3

0 [t2]dt + −

0

3 [t2]dt = 2

3

0 [t2]dt

32

[t2]dt = 4(√

5 − 2) + 5(√ 6 − √ 5) + 6(√ 7 − √ 6) + 7(√ 8 − √ 7) + 8(3 − √ 8)

16 −√

5 −√

6 −√

7 −√

8 20

[t2]dt +

32

[t2]dt = 21 − 3√

2 −√

3 −√

5 −√

6 −√

7 −√

8

=⇒ 3−3

[t2]dt = 42 − 2(3√

2 +√

3 +√

5 +√

6 +√

7)

Exercise 6.

(1)

n

0 [t]2dt =

n

1 [t]2dt =

n−1j=1 j

2 = (n−1)n(2n−1)

6

(2) x0 [t]2dt = [x−1]j=1 j2 + q 2r where x = q + r, q ∈ Z+, 0 ≤ r


22/236

Exercise 8. b+ca+c f (x)dx =

b+c−ca+c−c f (x − (−c))dx =

ba f (x + c)dx

Exercise 9. kbka

f (x)dx = 11k

(kb)/k(ka)/k

f x1/k

dx = k

ba

f (kx)dx

Exercise 10. Given s(x) = (−1)nn if n ≤ x < n + 1; n = 0, 1, 2, . . . p − 1; s( p) = 0, p ∈ Z+. f ( p) = p

0 s(x)dx.

So for f (3) = 30 s(x)dx, we need to consider n = 0, 1, 2.s(0 ≤ x < 1) = 0s(1 ≤ x < 2) = (−1)(1)s(2 ≤ x < 3) = 2;s(3 ≤ x < 4) = −3

So thenf (3) = (−1)(1) + 2(1) = 1f (4) = 1 + (−3)(1) = −2f (f (3)) = f (1) = 0

We obtain this formula

f ( p) = p2 (−1) p+1 p even p−12 (−1) p+1 p even

since

f ( p + 1) = f ( p) +

p+1 p

s(x)dx = p−1

2 (−1) p+1 p even + (−1) p p

=

− p2 p even p−1

2 p odd

+

p

− p = p2− p−1

2

=

=

− ( p+1)2 if p + 1 even p2 if p + 1 odd

Thus, p = 14, p = 15.

Exercise 11.

(1) ba

s(x)dx =nk=1

s3k(xk − xk−1) ba

s +

cb

s =

n1k=1

s2k(xk − xk−1) +n2k=n1

s3k(xk − xk−1) =n2k=1

s3k(xk − xk−1) = ca

s(x)dx

(2) ba (s + t) =

n3k=1(s + t)

3k(xk − xk−1) =

ba

s + ba

t

(3) ba

cs =nk=1(cs)

3(xk − xk−1) = c ba

s(4) Consider these facts that are true, that xk−1 < x < xk, s(x) = sk; x0 = a + c, xn = b + c,

xk−1 − c < x − c < xl − c =⇒ yk−1 < y < yk so then s(y + c) = sk.nk=1

s3k(xk − xk−1) =kk=1

s3k(xk − c − (xk−1 − c)) =

=nk=1

s3k(yk − yk−1) = ba

s(y + c)dy

(5) s < t, ba

s =nk=1 s

3k(xk − xk−1).

if 0 < s, s3 < s2t < st2 < t3

if s 0

if s < t < 0, s3 < s2t, s(st) < t(ts) = t2sts > t2

t2s < t3

s3 < s2t < t2s < t3

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Then ba s <

ba t.

Exercise 12.

(1) ba

s + cb

s =n1k=1 sk(x

2k − x2k−1) +

n2k=n1

sk(x2k − x2k−1) =

n3k=1 sk(x

2k − x2k−1) =

ca

s

(2) b

a (s+t) = n3k=1(s+t)k(x

2k−x2k−1) =

n3k=1(sk+tk)(x

2k−x2k−1) =

n3k=1 sk(x

2k−x2k−1)+

n3k=1 tk(x

2k−x2k−1)

since P 3 = {xk} is a finer partition than the partition for s, P 1, t , P 2, then considersk(y

2j − y2j−1) = sk((x2k+1 − x2k) + (x2k − x2k−1)), so

n3k=1

sk(x2k − x − k − 12) +

n3k=1

tk(x2k − x − k − 12) =

n1j=1

sj(x2j − x − j − 12) +

n2j=1

tj(x2j − x − j − 12) =

=

ba

s +

ba

t

(3) ba

cs =nk=1 csk(x

2k − x2k−1) = c

nk=1 sk(x

2k − x2k−1) = c

ba

s

(4)

b+c

a+c s(x)dx =

nk=1 sk(x

2k − x2k−1) where

s(x) = sk

if xk−1

< x < xk

x(y + c) = sk if xk−1 < y + c < xk =⇒ xk−1 − c < y < xk − c =⇒ yk−1 < y < ykwhere P = {yk} is a partition on [a, b] ba

s(y + c)dy =nk=1

sk(y2k − y2k−1) =

=nk=1

sk((xk − c)2 − (xk−1 − c)2) =nk=1

sk(x2k − 2xkc + c2 − (x2k−1 − 2xk−1c + c2)) =

=

nk=1

sk(x2k − x2k−1 − 2c(xk − xk−1)) =

nk=1

sk(x2k − x2k−1)

(5) Since x2k − x2k−1 > 0, ba sdx = nk=1 sk(x2k − x2k−1) < nk=1 tk(x2k − x2k−1) = ba tdxNote that we had shown previously that the integral doesn’t change under finer partition.

Exercise 13.

ba

s(x)dxnk=1

sk(xk − xk−1); ba

t(x)dx =

n2k=1

tk(yk − yk−1)

P = {x0, x1, . . . , xn}, Q = {y0, y1, . . . , yn}Note that x0 = y0 = a; xn = yn2 = b.

Consider P

Q = R. R consists of n3 elements, (since n3 ≤ n + n2 some elements of P and Q may be the same. R isanother partition on [a, b] (by partition definition) since xk, yk ∈ R and since real numbers obey transitivity, {xk, yk} can bearranged such that a < z1 < z2 <

· · ·< zn3−2 < b where zk = xk or yk.

(s + t)(x) = s(x) + t(x) = sj + tk if xj−1 < x < xj ; yj−1 < x < yjIf xj−1 ≶ yj−1, let zl−1 = yj−1, xj−1 and

If xj ≶ yj , let zl = xj , yj

Let sj = sl; tk = tl

(s + t)(x) = s(x) + t(x) = sl + tl, if zl−1 < x < zl ba

(s(x) + t(x))dx =

ba

((s + t)(x))dx =

n3l=1

(sl + t)l)(zl − zl−1) =n3l=1

sl(zl − zl−1) +n3l=1

tl(zl − zl−1)

In general, it was shown (Apostol I, pp. 66) that any finer partition doesn’t change the integral R is a finer partition. Son3

l=1sl(zl − zl−1) +

n

l=1tl(zl − zl−1) =

n

k=1sk(xk − xk−1) +

n2

k=1tk(yk − yk−1) =

b

a

s(x)dx +

b

a

t(x)dx

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Exercise 14. Prove Theorem 1.4 (the linearity property).

c1

ba

s(x)dx + c2

ba

t(x)dx = c1

n

k=1sk(xk − xk−1) + c2

n2

k=1tk(xk − xk−1) =

=n3l=1

c1sl(zl − zl−1) +n3l=1

c2tl(zl − zl−1) =n3l=1

(c1sl + c2tl)(zl − zl−1) =

=

ba

(c1s + c2t)(x)dx

We relied on the fact that we could define a finer partition from two partitions of the same interval.

Exercise 15. Prove Theorem 1.5 (the comparison theorem).

s(x) < t(x) ∀x ∈ [a, b]; s(x)(zl − zl−1) < t(x)(zl − zl−1) (zl − zl−1 > 0)

ba

s(x)dx =

nk=1

sk(xk − xk−1) =n3l=1

sl(zl − zl−1) <n3l=1

tl(zl − zl−1) =n2k=1

tk(yl − yk−1) =

=

ba

t(x)dx

=⇒ ba

s(x)dx <

ba

t(x)dx

Exercise 16. Prove Theorem 1.6 (additivity with respect to the interval).

Use the hint: P 1 is a partition of [a, c], P 2 is a partition of [c, b], then the points of P 1 along with those of P 2 form apartition of [a, b].

ca

s(x)dx + ba

s(x)dx =

n1k=1

sl(xk − xk−1) +n2k=1

sk(xk − xk−1) =n3k=1

sk(xk − xk−1) = ba

s(x)dx

Exercise 17. Prove Theorem 1.7 (invariance under translation).

P = {y0, y1, . . . , yn}; yk = xk + c;=⇒xk−1 + c < y < xk + c

xk−1 < y − c < xkyk − yk−1 = xk + c − (xk−1 + c) = xk − xk−1

s(y − c) = sk if xk−1 < y − c < xk , k = 1, 2, . . . n ba

s(x)dx =nk=1

sk(xk − xk−1) =nk=1

sk(yk = yk−1) = yny0

s(y − c)dy = b+ca+c

s(x − c)dx

1.26 Exercises - The integral of more general functions, Upper and lower integrals, The area of an ordinate set ex-

pressed as an integral, Informal remarks on the theory and technique of integration, Monotonic and piecewise mono-

tonic functions. Definitions and examples, Integrability of bounded monotonic functions, Calculation of the integral

of a bounded monotonic function, Calculation of the integral b0

x pdx when p is a positive integer, The basic propertiesof the integral, Integration of polynomials.

Exercise 16. 20 |(x − 1)(3x − 1)|dx = 24


25/236

21

(x − 1)(3x − 1)dx = 21

(3x2 − 4x + 1)dx = (x3 − 2x2 + x)21

= 2 11/3

(1 − x)(3x − 1)dx = − (x3 − 2x2 + x)

1

1/3 =

4

27 1/30

(x − 1)(3x − 1)dx = 427

So the final answer for the integral is 62/27.

Exercise 17. 30 (2x − 5)3dx = 8

30 (x − 52)3dx = 8

3−5/2−5/2 x

3dx = 8 14x41/2−5/2 =

392

Exercise 18. 3−3(x

2 − 3)3dx = 30 (x2 − 3)3 + x−3(x2 − 3)3 = 30 (x2 − 3)2 + − 30 (x2 − 3)3 = 02.4 Exercises - Introduction, The area of a region between two graphs expressed as an integral, Worked examples.

Exercise 15. f = x2, g = cx3, c > 0

For 0 < x < 1c

, cx 0). So cx3 < x2 (since x2 > 0). f − g =

x2 − cx3 =

1

3x3 − c

4x41/c

0

= 1

12c3 f − g = 2

3 =

1

12c3; c =

1

2√

2

Exercise 16. f = x(1 − x), g = ax. f − g =

1−a0

x − x2 − ax =

(1 − a) 12

x2 − 13

x31−a

0

= (1 − a)3 16

= 9/2 =⇒ a = −2

Exercise 17. π = 2 1−1√

1−

x2dx

(1) 3−3

9 − x2dx = 3

3−3

1 −

x3

2= 3(3)

1−1

1 − x2 = 9π

2

Now kbka

f x

k

dx = k

ba

f dx

(2) 20

1 − 1

4x2dx = 2

10

1 − x2dx = 2π

4 =

π

2

(3) 2

−2(x

−3)

√ 4

−x2dx 2−2

x

4 − x2dx = (−1) −22

−x

4 − x2 =⇒ 2 2−2

x

4 − x2 = 0

− 3 2−2

2

1 −

x2

2dx = (−6)(2)

1−1

1 − x2 = −6π

Exercise 18. Consider a circle of radius 1 and a twelve-sided dodecagon inscribed in it. Divide the dodecagon by isosceles

triangle pie slices. The interior angle that is the vertex angle of these triangles is 360/12 = 30 degrees.

Then the length of the bottom side of each triangle is given by the law of cosines:

c2 = 1 + 1 − 2(1)(1) cos 30◦ = 2

1 −√

3

2 =⇒ c =

√ 2

1 −

√ 3

2

25


26/236

The height is given also by the law of cosines

h = 1cos15◦ =

1 + cos 30◦

2 =

1 +

√ 3

2

2

The area of the dodecagon is given by adding up twelve of those isosceles triangles

(12)1

2

1 + √ 32

1√ 2

√ 2 1 − √ 32

= 3So 3 < π.

Now consider a dodecagon that’s circumscribing the circle of radius 1.

(12)1

2

2 1 − √ 32

1 +√

32

(1) = 122 − √ 32

> π

Exercise 19.

(1) (x, y)

∈E if x = ax1, y = by1 such that x21 + y

21

≤1

=⇒ xa2 + yb 2 = 1(2)

y = b

1 −

xa

22

a−a

b

1 −

xa

2= 2ba

1−1

1 − x2 = ba π

2(2) = πba

Exercise 20. Let f be nonnegative and integrable on [a, b] and let S be its ordinate set.

Suppose x and y coordinates of S were expanded in different ways x = k1x1, y = k2y1.

If f (x1) = y1, g(x) = k2f

xk1

= k2y1 = y.

integrating g on [k1a, k1b], k1bk1a

g(x)dx =

k1bk1a

k2f

x

k1

dx = k2k1

ba

f (x)dx = k2k1A

2.8 Exercises - The trigonometric functions, Integration formulas for the sine and cosine, A geometric description of

the sine and cosine functions. Exercise 1.

(1) sin π = sin 0 = 0. sine is periodic by 2π, so by induction, sin nπ = 0.

sin2(n + 1)π = sin 2πn + 2π = sin 2πn = 0

sin (2(n + 1) + 1)π = sin(2n + 3)π = sin ((2n + 1)π + 2π) = sin(2n + 1)π = 0

(2) cos π/2 = cos

−π/2 = 0

by induction, cos π/2 + 2πj = cos π/2(1 + 4 j)

cos −π/2 + 2πj = cos(4 j − 1)π/2, j ∈ Z+

Exercise 2.

(1) sin π/2 = 1, sin π/2(1 + 4 j) = 1, j ∈ Z+.(2) cos x = 1, cos0 = 1, cos2πj = 1

Exercise 3.

sin x + π = − sin x + π/2 + π/2 = cos x + π/2 = − sin xcos x + π = cos x + π/2 + π/2 = − sin x + π/2 = − cos x

Exercise 4.

26


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sin3x = sin 2x cos x + sin x cos2x = 2 sin x cos2 x + sin x(cos2 x − sin2 x) = 3cos2 x sin x − sin3 x == 3(1 − sin2 x)sin x − sin3 x = 3 sin x − 4sin3 x

cos3x = cos 2x cos x − sin2x sin x = (cos2 x − sin2 x)cos x − (2sin x cos x)sin x = cos x − 4sin2 x cos x

cos3x = −3cos x + 4 cos3

x

Exercise 5.

(1) This is the most direct solution. Using results from Exercise 4 (and it really helps to choose the cosine relationship,

not the sine relationship),

cos3x = 4 cos3 x − 3cos xx = π/6

cos3π/6 = 0 = 4cos3 π/6 − 3cos π/6 = cos π/6(4cos2 π/6 − 3) = 0=⇒ cos π/6 =

√ 3/2, sin π/6 = 1/2( by Pythagorean theorem )

(2) sin2π/6 = 2cos π/6sin π/6 =√

32, cos π/3 = 1/2 (by Pythagorean theorem)(3) cos2π/4 = 0 = 2 cos π/4

−1, cos π/4 = 1/

√ 2 = sin π/4

Note that the most general way to solve a cubic is to use this formula. For x3 + bx2 + cx + d = 0,

R = 9bc − 27d − 2b3

54

Q = 3c − b2

9

S = (R +

Q3 + R2)1/3

T = (R −

Q3 + R2)1/3

x1 = S + T − b/3x2 = −1/2(S + T ) − b/3 + 1/2

√ −3(S − T )x3 = −1/2(S + T ) − b/3 − 1/2

√ −3(S − T )Exercise 6.

tan x − y = sin x

−y

cos x − y = sin x cos y

−sin y cos x

cos x cos y + sin x sin y 1

cosx cos y

1cosx cos y = tan x − tan y1 + tan x tan y

if tan x tan y = −1Similarly,

tan x + y = sin x + y

cos x + y =

sin x cos y + sin y cos x

cos x cos y − sin x sin y = tan x + tan y

1 − tan x tan y , tan x tan y = 1

cot x + y = cos x + y

sin x + y =

cos x cos y − sin x sin ysin x cos y + sin y cos x

= cot x cot y − 1

cot y + cot x

Exercise 7. 3sin x + π/3 = A sin x + B cos x = 3(sin x12 +√

32 cos x) =

32 sin x +

3√

32 cos x

Exercise 8.

C sin x + α = C (sin x cos α + cos x sin α) = C cos α sin x + C sin α cos x

A = C cos α, B = C sin α

Exercise 9. If A = 0, B cos x = B sin π/2 + x = C sin x + α so C = B, α = π/2 if A = 0.

If A = 0,A sin x + B cos x = A(sin x +

B

A cos x) == A(sin x + tan α cos x)

= A

cos α(cos α sin x + sin α cos x) =

A

cos α(sin x + α)

where −π/2 < α < π/4, B/A = tan α, C = AcosαExercise 10. C sin x + α = C sin x cos α + C cos x sin α.

C cos α = −2, C sin α = −2, C = −2√

2, α = π/4

Exercise 11. If A = 0, C = B, α = 0. If B = 0, A =−

C, α = π/2. Otherwise,27


28/236

A sin x + B cos x = B(cos x + A

B sin x) =

B

cos β (cos x cos β + sin β sin x) = C cos x + α

where AB = tan β , α = −β , C = Bcos β .Exercise 12.

sin x = cos x = 1 − cos2 x =⇒ cos x = 1/√

2 =⇒

x = π

4Try 5π/4. sin5π/4 = cos3π/4 = − sin π/4 = −1/√ 2.cos5π/4 = − sin3π/4 = − cos π/4 = −1/√ 2. So sin 5π/4 = cos5π/4. x = 5π/4 must be the other root.

So θ = π/4 + πn (by periodicity of sine and cosine).

Exercise 13.

sin x − cos x = 1 =

1 − cos2 x = 1 + cos x=⇒ 1 − cos2 x = 1 + 2 cos x + cos2 x =⇒ 0 = 2cos x(1 + cos x)

cos x = −1, x = π/2 + 2πnExercise 14.

cos x − y + cos x + y = cos x cos y + sin x sin y + cos x cos y − sin x sin y = 2 cos x cos ycos x − y − cos x + y = sin x cos y − sin y cos x + sin x cos y + sin y cos x = 2 sin x cos ysin x − y + sin x + y = sin x cos y − sin y cos x + sin x cos y + sin y cos x = 2 sin x cos y

Exercise 15.

sin x + h − sin xh

= sin(x + h/2) cos h/2 + cos (x + h/2)sin h/2 − sin(x + h)cos h/2 − cos x + h/2sin h/2

h

= sin h/2

h/2 cos(x + h/2)

cos x + h − cos xh

= cos(x + h/2)cos h/2 − sin(x + h/2)sin h/2 − (cos(x + h/2) cos h/2 + sin (x + h/2)sin h/2)

h

=

−

sin h/2

h/2

sin(x + h/2)

Exercise 16.

(1)sin2x = 2 sin x cos x

if sin2x = 2 sin x and x = 0, x = πn, cos x = 1 but x = πn =⇒ x = 2πn(2) cos x + y = cos x cos y − sin x sin y = cos x + cos y.

cos x cos y − cos x − cos y = sin y

1 − cos2 xLetting A = cos x, B = cos y,

A2B2 + A2 + B2 − 2A2B − 2AB2 + 2AB = 1 − A2 − B2 + A2B2A2 + B2 − A2B − AB2 + AB = 1/2

B2(1 − A) + B(A − A2) + A2 − 1/2 = 0

B = A(1 − A) ±

A2(1 − A)2 − 4(1 − A)(A2 − 1/2)

1 − A = A ± 1√ 1 − A (A

2(1 − A) − 4(A2 − 1/2))1/2 =

= A ± 1√ 1 − A (−3A

2 − A3 + 2)1/2

Note that −1 ≤ B ≤ 1, but for |A| ≤ 1.Solve for the roots of −3A2 − A3 + 2, A0 = −1, −1 +

√ 3, −1 − √ 3. So suppose cos x = 9/10. Then there is

no real number for y such that cos y would be real and satisfy the above equation.(3) sin x + y = sin x cos y + sin y cos x = sin x + sin y

=⇒ sin y(1 − cos x) + sin y + − cos x sin y = 0, =⇒ y = 2πnChecking our result, we find that sin (2πn + y) = sin 2πn + sin y(1)

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(4) y0

sin xdx = − cos x|y0 = −(cos y − 1) = 1 − cos y = sin y

=⇒ 1 − cos y =

1 − cos2 y

1 − 2cos y + cos2

y = 1 − cos2

y =⇒ cos y(cos y − 1) = 0; y = 2( j + 1)π

2 , 2πn

Exercise 17. ba

sin xdx = − cos x|ba = − cos b + cos a

(1) −√

32 + 1

(2) −√

22 + 1

(3) 12(4) 1(5) 2(6) 0 We were integrating over one period, over one positive semicircle and over one negative semicircle.(7) 0 We had integrated over two equal parts, though it only shaded in up to x = 1.

(8) −√

22 +

√ 3

2

Exercise 18. π0 (x + sin x)dx = (12x2 − cos x)π0 = π22 − (−1 − 1) = π22 + 2Exercise 19.

π/20

(x2 + cos x)dx = ( 13x3 + sin x)

π/20

= 13(π/2)3 + 1

Exercise 20. π/20 (sin x − cos x)dx = (− cos x − sin x)|π/20 = −1 − (−1) = 0

Exercise 21. π/20 | sin x − cos x|dx = ( by symmetry )2

π/40 (cos x − sin x)dx = 2(sin x + cos x)|π/40 = 2(

√ 2 − 1)

Exercise 22. π0

( 12 + cos t)dt = (12

t + sin t)π0

= π2

Exercise 23.

2π/3

0

(1

2

+ cos t)dt + π

2π/3 −(

1

2

+ cos t)dt = (t

2

+ sin t)2π/3

0

+ (t

2

+ sin t)2π/3

π

= 2(π

3 +

√ 3

2 ) − π

2 =

π

6 +

√ 3

Exercise 24. If −π < x ≤ −2π3 , x−π

−( 12

+ cos t)dt =

−πx

(1

2 + cos t)dt =

t

2 + sin t

−πx

= −π2 − x

2 − sin x

If −2π/3 ≤ x ≤ 2π/3, −2π/3−π

−( 12

+ cos t)dt +

x−2π/3

(1

2 + cos t)dt =

−π/6+

√ 3/2 + (t/2 + sin t)|x−2π/3

= x/2 + sin x−

π/3−

√ 3/2 +

√ 3/2

−π/6 =

x

2 + sin x

−π/3

If 2π/3 ≤ x ≤ π,√

3/2 +

x2π/3

−(1/2 + cos t)dt =√

3/2 + (t/2 + sin t)|2π/3x = π/3 +√

3 − x/2 − sin x

Exercise 25. x2x (t

2 + sin t)dt = ( 13 t3 + − cos t) = x6−x33 + cos x − cos x2

Exercise 26. π/20 sin2xdx =

− cos (2x)

2

π/20

= (−1/2)(−1 − 1) = 1

Exercise 27. π/30

cos x/2dx = 2 sin x/2|π/30 = 2 12 = 1Exercise 28.

29


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x0

cos(a + bt)dt =

x0

(cos a cos bt − sin a sin bt)dt =cos a

b sin bt − sin a(− cos bt/b)

x0

=

= cos a

b sin bx +

sin a

b (cos bx − 1) = 1

b sin a + bx − sin a/b

x

0

sin(a + bt)dt = x

0

(sin a cos bt + sin bt cos a)dt = sin a

b

sin bt

− cos a

b

cos btx

0

=

= 1

b(cos bx + a + cos a)

Exercise 29.

(1) x0

sin3 tdt =

x0

3sin t − sin3t4

dt =

−3

4 cos t + cos 3t/12

x0

= −3/4(cos x − 1) + cos3x − 112

=

= 1

3 − 3

4 cos x +

1

12(cos2x cos x − sin2x sin x) = 2/3 − 1/3cos x(2 + sin2 x)

(2)

x0 cos3 tdt = x

0

1

4 (cos 3t + 3 cos t)dt = 14 sin3t3 + 34 sin tx

0 =

= 1

12(sin2x cos x + sin x cos2x) +

3

4 sin x =

1

12(2sin x cos x + sin x(2 cos2 x − 1)) =

= sin x cos2 x + 2 sin x

3

Exercise 30. Now using the definition of a periodic function,

f (x) = f (x + p); f (x + (n + 1) p) = f (x + np + p) = f (x + np) = f (x)

and knowing that we could write any real number in the following form,

a = np + r; 0 ≤< p, r ∈ R; n ∈ Zthen a+ p

a

f (x)dx = r+ pr

f (x + np)dx = r+ pr

f (x)dx = pr

f + r+ p p

f (x)dx =

=

pr

f +

r0

f (x − p)dx = pr

f +

r0

f =

p0

f

Exercise 31.

(1) 2π0

sin nxdx = 1

n

2πn0

sin xdx = 1

n(− cos x)

2πn0

= − 1n

(1 − 1) = 0 2π0

cos nxdx = 1

n

2πn0

cos xdx = 1

n sin x

2πn

0

= 0

(2) 2π0

sin nx cos mxdx = 2π0

1

2(sin(n + m)x + sin (n − m)x)dx = 0 + 0 = 0 2π

0

sin nx sin mxdx =

2π0

1

2(cos(n − m)x + cos (n + m)x)dx = 0 + 0 = 0 2π

0

cos nx cos mxdx =

2π0

1

2(cos (n − m)x + cos (n + m)x)dx = 0 + 0 = 0

While 2π0

sin2 nxdx =

2π0

1 − cos2nx2

dx = π

2π

0

cos2 nxdx =

2π

0

1 + cos 2nx

2 dx = π

30


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Exercise 32. Given that x = 2πn;sin x/2 = 0,nk=1

2sin x/2cos kx = 2 sin x/2nk=1

cos kx =nk=1

sin(2k + 1)x

2 − sin (2k − 1) x

2 = sin(2n + 1)

x

2 − sin x/2

= sin nx cos x/2 + sin x/2cos nx

−sin x/2 =

= 2 sin nx/2cos nx/2cos x/2 + sin x/2(1 − 2sin2 nx/2) − sin x/2 == 2(sin nx/2)(cos(n + 1)x/2)

Exercise 33. Recall that

cos(2k + 1)x/2 − cos(2k − 1)x/2 = cos kx + x/2 − cos kx − x/2 == cos kx cos x/2 − sin kx sin x/2 − (cos kx cos x/2 + sin kx sin x/2) == −2sin kx sin x/2

−2sin x/2nk=1

sin kx =

nk=1

(cos (2k + 1)x/2 − cos(2k − 1)x/2) = cos (2n + 1)x/2 − cos x/2 =

= cos nx + x/2

−cos x/2

Now

sin nx/2sin nx/2 + x/2 = sin nx/2(sin nx/2cos x/2 + sin x/2cos nx/2) =

= sin2 nx/2cos x/2 + sin x/2cos nx/2sin nx/2 =

=

1 − cos nx

2

cos x/2 +

sin nx

2 sin x/2 =

= 1

2 (cos x/2 − cos x/2cos nx + sin nx sin x/2) = 1

2(cos x/2 − cos(nxx/2)

Then

−2sin x/2nk=1

sin kx = −2sin nx/2sin 12

(n + 1)x

nk=1

sin kx = sin nx/2sin 1

2(n + 1)x

sin x/2

Exercise 34. Using triangle OAP, not the right triangle, if 0 < x < π/2

1

2 cos x sin x <

1

2 sin x <

x

2=⇒ sin x < x

Now if 0 > x > −π/2, sin x < 0,| sin x| = − sin x = sin −x = sin |x| < |x|

2.17 Exercises - Average value of a function. Exercise 1. 1b−a

x2dx = 13(b

2 + ab + a2)

Exercise 2. 11−0 x2 + x3 = 712

Exercise 3. 14−0

x1/2 = 43

Exercise 4. 18−1

x1/3 = 4528

Exercise 5. 1π/2−0 π/20

sin x = 2π

Exercise 6. 1π/2−−π/2

cos x = 2/π

Exercise 7. 1π/2−0

sin2x = −1/π(−1 − 1) = 2/π

Exercise 8. 1

π/4−

0 sin x cos x = 1π 31


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Exercise 9. 1π/2−0

sin2 x = 1π (x − sin2x/2)π0

= 12

Exercise 10. 1π−0

cos2 x = 12

Exercise 11.

(1) 1a−0

x2 = a2/3 = c2 =⇒ c = a/√ 3(2) 1a−0

xn = 1a

1n+1x

n+1a0

= an

n+1 = cn =⇒ c = a

(n+1)1/n

Exercise 12.

A =

wf /

w

wx2 = k

x

x3 = 1

4x4 = k

1

2x2; k =

1

2, w = x

x4 =

1

5x5 = k

1

3x3; k =

3

5, w = x2

x5 = 16

x6 = k 14

x4; k = 23

, w = x3

Exercise 13.

A(f + g) = 1

b − a

f + g = 1

b − a

f + 1

b − a

g = A(f ) + A(g)

A(cf ) = 1

b − a

cf = c

1

b − a

f

A(f ) = 1

b − a

f ≤ 1b − a

g = A(g)

Exercise 14.

A(c1f + c2g) =

w(c1f + c2g)

w =

c1

wf w

+ c2

wg w

= c1A(f ) + c2A(g)

f ≤ g w > 0( nonnegative ), =⇒ wf ≤ wg

Exercise 15.

Aba(f ) = 1

b − a ba

f = 1

b − a

ca

f +

bc

f

=

c − ab − a

ca

f

c − a

+

b − a − (c − a)b − a

ba

f

b − ca < c < b

0 < c − ab − a


33/236

xcm =

L0

x L0 1

= L

2

I cm =

r2dm =

x2(1) = L3/3

r2

=

I cm L0 1

= L2

/3 =⇒ r = L

√ 3

Exercise 17.

xcm =

l/20

x + LL/2 2xdx

L2 + 2(L − L/2)

= yL2

12

I cm =

L/20

x2 +

LL/2

2x2 = 5L3/8 r2 = 5L3/8

3L/2 =

5L2

12 =⇒ r =

√ 5L

2√

3

Exercise 18. ρ(x) = x for 0 ≤ x ≤ L

xcm = xxdx xdx

=1

3x3L0

12x

2L0

= 23

L

I cm =

x2xdx = L4/4

r2 = L4/4

L2/2 = L2/2 r =

L√ 2

Exercise 19.

xcm =

xxdx +

x L2 dx

xdx +

L/2

=

13x

3L/20

+ L2 (x2/2)

LL/2

12

x2

L/2

0 + L2 (L − L/2)

= 11L/18

I cm = x2xdx + x2L/2dx = L431/192r2 = I cm/(L

23/8) = L231/72 r =

√ 31L

6√

2

Exercise 20. ρ(x) = x2 for 0 ≤ x ≤ L

xcm =

xx2dx

x2 = 3L/4

I cm =

x2x2dx = L5/5

r2 = I cm

13L3 =

3

5L2 r = 35 L

Exercise 21.

xcm =

L/20

xx2dx + LL/2

xL2

4 dx L/2

0 x2dx +

LL/2

L2

4 dx

= 21L/32

I cm = intL/20 x

2x2dx +

LL/2

x2L2

4 dx = 19L5/240

r2 = I cmL3/6

= 19L2/40 =⇒ r =√

19L/2√

10

Exercise 22. Be flexible about how you can choose a convenient origin to evaluate the center-of-mass from

33


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Let ρ = cxn

c

L0

xndx = 1

n + 1Ln+1c = M

=⇒ c = (n + 1)M Ln+1

c L0

xxndx = c 1n + 2

Ln+2 = n + 1n + 2

ML = 3ML4

xcm =

xρ

M =

3L

4 =⇒

xρ =

n + 1

n + 2 =

3

4 =⇒ n = 2

ρ = 3M

L3 x2

Exercise 23.

(1)1

π/2 − 0

3sin2t = 6

π

(2)

1π/2 − 0

9sin2 2t = 9/2 =⇒ vrms = 3√ 2/2Exercise 24. T = 2π (just look at the functions themselves)

1

2π

2π0

160sin t2sin(t − π/6) = 80√

3

2.19 Exercises - The integral as a function of the upper limit. Indefinite integrals. Exercise 1. x0

(1 + t + t2)dt =

x + 12x2 + 13x

3

Exercise 2. 2y + 2y2 + 8y3/3

Exercise 3. 2x + 2x2 + 8x3/3 − (−1 + 1/2 + −1/3) = 2(x + x2 + 4x3/3) + 5/6Exercise 4.

1−x1 (1 − 2t + 3t2)dt = (t − t2 + t3)

1−x1

= −2x + 2x2 − x3

Exercise 5. x−2 t

4 + t2 = 15 t5 + 13 t

3x−2 =

x5

5 + x3

3 + 403

Exercise 6. x2x t

4 + 2t2 + 1 =t5

5 + 23 t

3 + tx2x

= 15 (x10 − x5) + 23(x6 − x3) + x2 − x

Exercise 7.23 t

3/2 + tx

1 = 23(x

3/2 − 1) + (x − 1)

Exercise 8.

23

t3/2 + 45 t5/4

x2

x = 23(x

3 − x3/2) + 45(x5/2 − x5/4)

Exercise 9. sin t|xiπ = sin x

Exercise 10.t2 + sin t

x20

= x2

2 + sin x2

Exercise 11.12 t + cos t

x2x

= x2−x2 + cos x

2 − cos x

Exercise 12.13

u3 + − 13 cos 3ux

0 = x

3

3 + − 13(cos3x − 1)

Exercise 13.

13

v3 + cos3v−3x2x

= x6−x33 +

−13 (cos3x

2 − cos3x)

Exercise 14. 1−cos 2x

2 + x = 12x − sin 2x4 + 12x2

y

0 =

y

2 − sin2y

4 +

y2

234


35/236

Exercise 15.− cos2w

2 + 2 sin w2

x0

= − (cos2x − 1)2

+ 2 sin x

2

Exercise 16.

x

−π(12 +cos t)

2dt =

x

−π14 +cos t+cos

2 t = 14(x+π)+sin x+ 12

t + sin2t2

x

−π = 34(x+π)+sin x+

14 sin2x

Exercise 17. x0 (t3 − t)dt = 13 x√ 2(t − t3)dtNote that t3 − t < 0 for 0 < t ≤ 1 and t3 − t > 0 for t > 1. t − t3 √ 2.

1

4x4 − 1

2x2 =

1

3

1

2t2 − 1

4t4x√

2

= 1

6x2 − 1

12x4

=⇒ 13

x4 − 23

x2 = 0 =⇒ x = 0, x =√

2 10 (t

3 − t)dt + √ 21 (t3 − t)dt “cancel” each other out.Exercise 18. f (x) = x − [x] − 12 if x is not an integer; f (x) = 0 if x ∈ Z.

For any real number, x = q + r, 0≤

r


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Exercise 19. g(2n) = 2π0

f (t)dt

Consider 1−1

f (t)dt =

10

f (t)dt +

0−1

f (t)dt =

10

f (t)dt + 1

−1 01

f (−1t)dt =

= 10 f + 0

1 f (t)dt = 0

Consider that 31

f (t)dt = 1−1 f (t + 2)dt =

1−1 f (t)dt = 0. Then, by induction, 2n+1

1

f =

2n−11

f +

2n+12n−1

f (t)dt = 0 +

1−1

f (t + 2n)dt =

1−1

f (t)dt = 0

(1)

g(2n) =

10

f +

2n−11

f +

2n2n−1

f =

10

f +

0−1

f (t)dt =

10

f + − 01

f (−t)dt

=

10

f +

01

f = 0

(2)

g(−x) = −x0

f = − x0

f (−t)dt = x0

f (t)dt = g(x)

g(x + 2) =

x+20

f (t)dt =

20

f +

x+22

f =

x0

f (t + 2)dt =

x0

f (t)dt = g(x)

Exercise 20.

(1) g is odd since

g(−x) = −x0

f (t)dt = − x0

f (−t)dt = − x0

f (t)dt = −g(x)Now

g(x + 2) =

x+2

0

f =

2

0

f +

x+2

2

f = g(2) +

x

0

f (t + 2)dt = g(2) +

x

0

f (t)dt = g(2) + g(x)

=⇒ g(x + 2) − g(x) = g(2)(2)

g(2) =

20

f =

21

f +

10

f =

21

f + A =

0−1

f (t + 2)dt + A =

0−1

f (t)dt + A =

− 01

f (−t)dt + A = 2Ag(5) − g(3) = g(2)

g(3) = g(2) +

32

f (t)dt = 2A +

10

f (t + 2)dt = 2A + A = 3A

=⇒ g(5) = 3A + 2A = 5A(3) The key observation is to see that

g must repeat itself by a change of

2 in the argument. To make

g(1) = g(3) = g(5),

they’re different, unless A = 0!

Exercise 21. From the given, we can derive

g(x) = f (x + 5), f (x) =

x0

g(t)dt

=⇒ f (5) = 50

g(t)dt = g(0) = 7

(1) The key insight I uncovered was, when stuck, one of the things you can do, is to think geometrically and draw

a picture.g(−x) = f (−x + 5) = g(x) = −f (x − 5)

=

⇒ −g(x) = f (x

−5)

36


37/236

(2) 50

f (t)dt =

0−5

f (t + 5)dt =

0−5

g(t)dt = − −50

g(t)dt =

50

g(−t)dt = 50

g(t)dt = f (5) = 7

(3)

x0

f (t)dt = x−5−5

f (t + 5)dt = x−5−5

g(t)dt = x−50

g + 0−5

g = f (x − 5) + − −50

g(t)dt =

f (x − 5) + 50

g(−t)dt = f (x − 5) + f (5) = −g(x) + g(0)

where we’ve used f (x − 5) = −g(x) in the second and third to the last step.

3.6 Exercises - Informal description of continuity, The definition of the limit of a function, The definition of continuity

of a function, The basic limit theorems. More examples of continuous functions, Proofs of the basic limit theorems.

Polynomials are continuous.

Exercise 1. limx→2 1x2 = 1

limx→2 x2 = 14

Exercise 2. limx→0(25x3+2)limx→0(75x7−2) = −1

Exercise 3. limx→2(x−2)(x+2)

(x−2) = 4

Exercise 4. limx→1(2x−1)(x−1)

x−1 = 1

Exercise 5. limh→0 t2+2th+h2−t2

h = 2t

Exercise 6. limx→0(x−a)(x+a)

(x+a)2 = −1

Exercise 7. lima→0(x

−a)(x+a)

(x+a)2 = 1

Exercise 8. limx→a(x−a)(x+a)

(x+a)2 = 0

Exercise 9. limt→0 tan t = limt→0 sin tlimx→0 cos t = 01 = 0

Exercise 10. limt→0(sin 2t + t2 cos5t) = limt→0 sin 2t + limt→0 t2 limt→0 cos 5t = 0 + 0 = 0

Exercise 11. limx→0+|x|x = 1

Exercise 12. limx→0−|x|x = −1

Exercise 13. limx→0+√ x2

x = +1

Exercise 14. limx→0−√ x2

x = −1

Exercise 15. limx→0 2sin x cosxx = 2

Exercise 16. limx→0 2sin x cosxcos2x sinx = 2

Exercise 17. limx→0 sin x cos 4x+sin 4x cos xsinx = 1 + limx→02sin2x cos2x

sinx = 1 + 2

limx→0 2sin x cosx cos2xsinx

= 5 Exercise 18.

limx→

05sin5x

5x −limx

→0

3sin3x

3x

= 5−

3 = 2 Exercise 19.37


38/236

limx→0

sinx+a2 +

x−a2

− sin x+a2 − x−a2 x − a =

= limx→0

sin x+a2 cos

x−a2 + sin

x−a2 cos

x+a2 −

sin x+a2 cos

x−a2 − sin x−a2 cos x+a2

x − a

=

= limx→a2sin x−a

2 cos x+a

2x − a = cos a

Exercise 20. limx→02sin2 x/24(x/2)2 =

12

limx→0

sinx/2x/2

= 12

Exercise 21. limx→0 1−√

1−x2x2

1+

√ 1−x2

1+√

1−x2

= limx→01−(1−x2)

x2(1+√

1−x2) = 1

2

Exercise 22. b, c are given.

sin c = ac + b, a = sin c−bc , c = 0.if c = 0, then b = 0, a ∈ R.Exercise 23. b, c are given.

2cos c = ac2 + b, a = 2cos c−bc2 , c = 0.If c = 0, then b = 2, a ∈ R.Exercise 24.

tangent is continuous for x /∈ (2n + 1)π/2cotangent is continuous for x /∈ 2nπ

Exercise 25. limx→0 f (x) = ∞. No f (0) cannot be defined.

Exercise 26.

(1) | sin x − 0| = | sin x| < |x|. Choose δ = for a given .Then ∀ > 0, ∃δ > 0 such that | sin x − 0| < when |x| < δ .

(2)

| cos x − 1| = | − 2sin2 x/2| = 2| sin x/2|2 0 such that | cos h − 1| < 0; | sin h| < whenever |h| || sin nπ| − |A|| > |1 − |A||Consider |x − 0| = |x| = 1nπ ≤ δ (n). Consider 0 = |1−|A||2 . Then suppose a δ (n) ≥ |x − 0| but |f (x) − A| > 0. Thus,contradiction.

Exercise 28. Consider x ≤ 1n , n ∈ Z+, n > M (n) (n is a given constant)

f (x) =

1

x

= [n] = n, for m > M (n), x =

1

mf (x) > M (n)

so ∀ > 0, we cannot find δ = 1n such that |f (x) − A| < for x < δ .So f (x)

→ ∞as x

→0+.

38


39/236

Consider 1n ≥ x > 0, n ∈ Z−; −n > M (n).

f (x) =

1

x

= [n] = n < −M (n)

Since integers are unbounded, we can consider n < A, so that

|f (x) − A| > ||f | − |A|| = −n − |A| > M (n) − |A|. Choose n such that M (n) − |A| > 0Exercise 29.

|f − A| = |(−1)[1/x] − A| ≥ | |(−1)[1/x]| − |A|| = |1 − |A||Choose 0 ( such that |x| < δ ), |f − A| > . Thus there’s no value for f (0) we could choose tomake this function continuous at 0.

Exercise 30. Since

|f (x)| = |x||(−1)[1/x]| = |x|So ∀, let δ = .Exercise 31. f continuous at x0.

Choose some 0, 0 < 0 < min (b − x0, x0 − a). Then ∃δ 0 = δ (x0, 0).Consider 1 =

02 and δ 1 = δ (x0, 1)

Consider x1 ∈ (x0 − δ 1, x0 + δ 1), so that |f (x1) − f (x0)| < 1.Proceed to construct a δ for x1, some δ (x1; 0)

|x − x1| = |x − x0 + x0 − x1| < |x − x0| + |x0 − x1|Without loss of generality, we can specify x1 such that |x0 − x1| < δ12 . Also, “pick” only the x’s such that

|x − x0|


40/236

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