SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL PDF

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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 http://igg.me/at/ernestyalumni2014 Facebook : ernestyalumni gmail : ernestyalumni google : ernestyalumni linkedin : ernest...

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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 http://igg.me/at/ernestyalumni2014 Facebook : ernestyalumni gmail : ernestyalumni google : ernestyalumni linkedin : ernestyalumni tumblr : ernestyalumni twitter : ernestyalumni weibo : ernestyalumni youtube : ernestyalumni indiegogo : ernestyalumni 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. S OLUTIONS 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

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). Then

actual 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 ∈ B x ∈ 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 6= c∀a. =⇒ A ⊂ C (2) (3) (4) (5)

If A ⊆ B, B ⊆ C, A ⊆ C since, ∀a ∈ A, a ∈ B, ∀b ∈ B, b ∈ C. Then since a ∈ B, a ∈ C. A ⊆ C A ⊂ B and B ⊆ C. B ⊂ C or B = C. A ⊂ B only. Then A ⊂ C. Yes, since ∀a ∈ A, a ∈ B. No, since x 6= 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

A∈F

then x ∈ B, x ∈ /

[

A

A∈F

x∈ /

[

A =⇒ x ∈ / A, ∀A ∈ F

A∈F

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

\ A∈F

2

(B − A)

\

Suppose x ∈

(B − A)

A∈F

then x ∈ B − A1 and x ∈ B − A2 and . . . then ∀A ∈ F, x ∈ B, x ∈ /A then x ∈ / even at least one A ∈ F [ =⇒ x ∈ B − A A∈F

\

Suppose x ∈ B −

A

A∈F

then x ∈ /

\

A

A∈F

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

Suppose x ∈

[

(B − A)

A∈F

then x is at least in one B − A then for A ∈ F, x ∈ B and x ∈ /A Consider ∀A ∈ F =⇒ then x ∈ B −

\

A

A∈F

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 3

(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 that are 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 = 0 0(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 = 1 since products are uniquely determined, yab = yac =⇒ (ya)b = (ya)c =⇒ 1(b) = 1(c) =⇒ b = c Thm. I.8. Possibility of Division. Given a, b, a 6= 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 so x = z). Thm. I.9. If a 6= 0, then b/a = b(a−1 ). b for ax = b a y = a−1 for ay = 1

Let x =

Want: x = by Now b(1) = b, so ax = b = b(ay) = a(by) =⇒ x = by (by Thm. I.7) −1 −1

Thm. I.10. If a 6= 0, then (a ) = a. Now ab = 1 for b = a−1 . But since b ∈ R and b 6= 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) = 0 0 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 6= 0, which is given, and Thm. I.7, then x + y = z. 4

Thm. I.14. Want: xy = z for bx = a, dy = c, ac = (bd)z. (bd)(xy) = (bx)(dy) = ac = (bd)z b, d 6= 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 6= 0 so by Thm. I.7, by z = a, byz = abx b 6= 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. Then

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

a + (−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 − c Exercise 8.

(ab)x = 1

(ab)−1 = x

a(bx) = 1

a−1 = bx

b(ax) = 1

b−1 − ax

a−1 b−1 = (abx)x = 1(x) = (ab)−1 Exercise 9. Want: x = y = z, if

a −b (−a) y= b  a x=− b z=

a = zt by = u a b

b+t=0 a+u=0

+ x = v + x = 0 vb = a

a + (−a) = vb + by = b(v + y) = 0 if b 6= 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 = by since b 6= 0, z = y so x = y = z Exercise 10. Since b, d 6= 0, Let

ad − bc bd a x= b −c t= d z=

(bd)z = ad − bc by previous exercise or Thm. I.8, the possibility of division bx = a dt = −c (By Thm. I.3, we know that b − a = b + (−a) ) 5

dbx + bdt = (bd)(x + y) = ad − bc = (bd)z b, d 6= 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 6= 0, then a2 > 0 Theorem 4 (I.21). 1 > 0 Theorem 5 (I.22). If a < b and c < 0, then ac > bc. Theorem 6 (I.23). If a < b and −a > −b. In particular, if a < 0, then −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 > 0 a(−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, ab < 0(b) = 0. So if a > 0, then b > 0. If a < 0, then if b > 0, ab < 0(b) = 0. If b < 0, ab > 0(b) = 0. So if a < 0, then b < 0. (4) a < c so a + b < c + b = b + c b < d so b + c < d + c By Transitive Law , a + b < d + c Exercise 2. If x = 0, x2 = 0. 0 + 1 = 1 6= 0. So x 6= 0.

If x 6= 0, x2 > 0, and by Thm. I.21 , 1 > 0 x2 + 1 > 0 + 0 = 0 → x2 + 1 6= 0 =⇒ @x ∈ R such that x2 + 1 = 0 Exercise 3.

a < 0, b < 0, a + b < 0 + 0 = 0 ( By Thm. I.25) Exercise 4. Consider ax = 1.

ax = 1 > 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 = by b > 0 so x > y Exercise 6. 6

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

x > x Contradiction. 2 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 > x so 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 < n Exercise 3. Use Archimedian property.

x > 0 so for 1, ∃n ∈ Z+ such that nx > 1, x >

1 n.

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 smallest element 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 such 7

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 > x Since h was arbitrary, there are infinitely many numbers in between x, y. Exercise 7. x =

a b

x±y =

∈ Q, y ∈ / Q. a ± by b  If a ± by was an integer, say m, then y = ±

xy =

ay ay = b1 b If ay was an integer, ay = n, y = x y

a − mb b

 which is rational. Contradiction.

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

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 y1 = 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 k + x > x. Let z = x + , z irrational . n n

Exercise 10.

(1) Suppose n = 2m1 and n + 1 = 2m2 . 1 . But m1 − m2 can only be an integer. 2 (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 or odd since it doesn’t belong in the above set. So x0 must be odd or even. Contradiction. (3) (2m1 )(2m2 ) = 2(2m1 m2 ) even 2m1 + 1 = 2m2

2(m1 − m2 ) = 1

m1 − m2 =

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) = n1 n2 + n1 + n + 2 + 1 = 2(2m1 m2 ) 2(2m1 m2 ) − (n1 + n2 ) − 1 odd, the product of two odd numbers n1 , n2 is odd 2

(4) If n 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 6= 4n2 Thus b is even. 8

(5) For pq , If p or q or both are odd, then we’re done. Else, when p, q are both even, p = 2l m, q = 2n p, m, p odd. 2l m 2l−n m p = n = and at least m or p odd q 2 p p Exercise 11.

a b

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

a b

cannot be rational.

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

Since

p q

∈ Q ⊆ R, n pq11 >

p2 q2

since if q1 , q2 > 0, np1 q2 q1 p2 > q1 q2 q1 q2

np1 q2 > q1 p2

n exists since (p1 q2 ), (q1 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 r < b2 < nb1 but likewise < b1 < mb2 , n, m ∈ Z+ s s 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 = N2 (N − 1) + (N − 2) + · · · + (N − (N − 1)) + (N − N ) = N 2 −

N X

j=

j=1

N2 + N = 2

N X

j

=⇒

j=1

N X j=1

j=

N (N + 1) 2

An interesting property is that S=

n X

j=

j=m

n X

(n + m − j)

j=m

So that N X

j=

j=1 N X j=m

N X

j+

j=m

j=

m X

j=

j=1

N X

j+

j=m

m(m + 1) N (N + 1) = 2 2

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

Another way to show this is the following. S= 1+ 2+ but S = N + N − 1+

· · · + (N − 2)+ ···+ 3+

2S = (N + 1)N

S= 9

(N − 1)+ N 2+ 1

N (N + 1) 2

N −1 X j=1

j

Telescoping series will let you get

PN

j=1

j 2 and other powers of j.

N X N (N + 1) (2j − 1) = 2 − N = N2 2 j=1

  N N N X X X N (N + 1) − N = N2 (j 2 − (j − 1)2 ) = (j 2 − (j 2 − 2j + 1)) = (2j − 1) = 2 2 j=1 j=1 j=1 N N N X X X 3 3 3 3 3 2 (j − (j − 1) ) = N = (j − (j − 3j + 3j − 1)) = (3j 2 − 3j + 1) j=1

=⇒ 3

N X

j=1

j 2 = −3

j=1 N X

j 4 − (j − 1)4 = N 4 =

=4

j3 − 6

j=1

=⇒

j3 =

j=1

=

N X

j 4 − (j 4 − 4j 3 + 6j 2 − 4j + 1) =

j=1 N X

N

2

2N + 2N − 3N − 3N N (N + 1)(2N + 1) X 2 N (N + 1) + N = N 3 =⇒ = = j 2 2 6 j=1

j=1

N X

j=1 3

N X

4j 3 − 6j 2 + 4j − 1 =

j=1

N (N + 1) N (N + 1)(2N + 1) +4 − N = N4 6 2

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

Exercise 1. Induction proof.

1(1 + 1) 2

N +1 X j=1

j=

n X

j+n+1=

j=1

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

Exercise 6.

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

1 8k + 8 (2k + 3)2 1 (2k + 1)2 + k + 1 = (4k 2 + 4k + 1) + = 8 8 8 8

(2) The n = 1 case isn’t true. (3) n2 + n + 41 (n + 1)n n2 + n 1 + 2 + ··· + n = = < 2 2 2  2 2n + 1 1 (n + 1/2)2 n2 + n + 1/4 and = = 2 2 2 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 > 0 By well-ordering principle, we could argue that n = 3 must be the smallest number such that (1 + x)n > 1 + 2x + 2x2 . Or we could fi...