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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 find, explicitly n   n   X n j n(n − 1) 2 X n j n (1 + x) = x = 1 + nx + x + x j 2 j j=0 j=3 10

and

n(n − 1) >n 2 n2 − n > 2n n2 > 3n n>3

Exercise 8.

a2 ≤ ca1 , a3 ≤ ca2 ≤ c2 a1 an+1 ≤ can ≤ ca1 cn−1 = a1 cn Exercise 9.

√ n = 1, 1 = 1 p √ q√ √ 12 + 12 = 2 ( 2)2 + 12 = 3 q√ √ ( 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 fixed n + 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. P4 (1) n(n+1) = k=1 k = 10 2 P5 P3 n−2 (2) = n=0 2n = 1 + 14 = 15 n=2 2 P P3 3 (3) 2 r=0 22r = 2 r=0 4r = 170 P4 j j = 1 + 4 + 27 + 44 = 288 (4) P5j=1 5(6) (5) j=0 (2j + 1) = 2 2 + 6(1)= 36 P 1 Pn 1 1 1 n (6) k=1 k − k+1 = 1 − n+1 = n+1 k(k+1) = Exercise 9.

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

2n X

(−1)k (2k + 1) = 2n

k=1 2(n+1)

n+1

X

(−1)k (2k + 1) =

k=1

2n X

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

k=1

= 2n + 2 = 2(n + 1) Exercise 10. 11

(1) am + am+1 + · · · + am+n (2) 1 1 1 1 = − = 2 1 2 2 2(n+1) 2n 2n X 1 X (−1)m+1 X 1 1 1 (−1)m+1 1 (−1)2n+1+1 n+1 = − + + = +− + k m=1 m n + 1 2n + 1 2n + 2 m=1 m 2(n + 1) (2n + 1)

n=1

k=n+2

2(n+1)

=

X (−1)m+1 m m=1

Exercise 13.

√ 1 √ n = 12( 2 − 1) < 1 < 2 since > 2 − 1 2 r √ √ √ √ √ √ 1 n+1+ n 1 √ n case ( n + 1 − n)( n + 1 + n) = n + 1 − n = 1 < = ( 1 + + 1) 2 n 2 n √ √ √ √ n + 1 case ( n + 2 − n + 1)( n + 2 + n + 1) = n + 2 − (n + 1) = 1 q √ √ 1 1 + 1 + n+1 n+2+ n+1 √ = >1 2 2 n+1 So then, using the telescoping property, m m X X √ √ √ √ √ √ √ 1 √ < 2( n + 1 − n) = 2( m − 1) < 2( n − n − 1) = 2( m − 1) < 2 m − 1 n n=1 n=1 n=1

n−1 X

I 4.9 Exercises - Absolute values and the triangle inequality. Exercise 1. (1) |x| = 0if f x = 0 If x = 0, x = 0, −x = −0 = 0. If |x| = 0, x = 0, −x = 0. (2) ( | − x| =

−x x

if − x ≥ 0 = if − x ≤ 0

(

x −x

if x ≥ 0 if x ≤ 0

(3) |x − y| ( = |y − x| by previous exercise and (−1)(x − y) = y − x (by distributivity) (x)2 if x ≥ 0 (4) |x|2 = = x2 (−x)2 if x ≤ 0 ( √ x if x ≥ 0 2 (5) x = = |x| −x if x ≤ 0 (6) We want to show that |xy| = |x||y| ( ( xy if xy ≥ 0 xy if x, y ≥ 0 or x, y ≤ 0 |xy| = = −xy if xy ≤ 0 −xy if x, −y ≥ 0 or − x, y ≤ 0  xy if x, y ≥ 0   (  x|y| if x ≥ 0 −xy if x, −y ≥ 0 |x||y| = = −x|y| if x ≤ 0  −xy if − x, y ≥ 0    xy if − x, −y ≥ 0 (7) By previous exercise, since

(1 1 = y y −1 y

x = |xy −1 | = |x||y −1 | y ( 1 if y1 ≥ 0 1 = y−1 1 if y ≤ 0 |y| y 12

if if

1 y 1 y

≥0 ≤0

(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| |y| − |x|

if |x| − |y| ≥ 0 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; ak x + bk = 0 !2 !2 n n X X 2 ak (−xak ) = x ak = k=1

n X

k=1

!

n X

a2k

k=1

! (−xak )

2

=

k=1

n X

! a2k

k=1

n X

! b2k

k=1

bk ⇐ Proving ak x + bk = 0 means x = − , ak 6= 0 ak n n n n X X X X (a1 b1 + a2 b2 + · · · + an bn )2 = a2j b2j + aj ak bj bk == a2j b2j + a2j b2k j=1

j=1

j6=q

j6=k

a2j b2k

− aj ak bj bk = aj bk (aj bk − ak bj ) = 0   bj if aj , bk 6= 0, aj bk − ak bj = 0 =⇒ ak + bk = 0 −aj =⇒

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

property of products: j

j

j

(1 − x2 )(1 + x2 ) = 1 − x2 1 Y

j−1

1 + x2

=

j=1

+2j

= 1 − x2

j+1

n j 1 Y 1 − x2 1 − x2 = 1−x 1 − x2j−1 j=1

if x = 1, 2n Exercise 10.

x>1 x2 > x 3

xn+1 = xn x > x2 > x

2

x >x >x 0<x<1 x2 < x

xn+1 = xn x < x2 < x =⇒ xn+1 < x

X 3 < x2 < x Exercise 11. Let S = {n ∈ Z+ |2n < n!}.

By well-ordering principle, ∃ smallest n0 ∈ S. Now 24 = 16, 4! = 24. So S starts at n = 4. Exercise 12. 13

(1)  1+ k−1 Y

1 n

n    j X n 1

n =

n

k

j=0

k−1 Y

=

n X k=0

n! (n − k)!k!

 k 1 n

 n! 1 n−r = k n n n (n − k)! r=0 r=0   n k−1 X 1 n! 1 Y r = 1− k k! r=0 n n (n − k)! 1−

r

=





k=1

(2) n

X 1 (1 + )n = 1 + n

k=1

! k−1 n n X X r 1 Y 1 1 =1+ (1 − ) < 1 + <1+ k! r=0 n k! 2k k=1

1 2

−

k=1


1 n+1 2 1 2

 n 1 ) = 1 + (1 − 2

<3 The first inequality obtained from the fact that if 0 < x < 1, xn < x < 1. The second inequality came from the 1 previous exercise, that k! < 21k .   n    k n−1    k n−1    k 1 n X n 1 1 X n 1 1 X n 1 n 1 (1 + ) = =1+ + =1+ + + > >2 n k n n k n n k n 1 n k=0

k=1

k=2

Exercise 13.

(1) S=

p−1  k X b k=0

p−1 X

a

bk ap−1−k = ap−1


b p a − ab

1−

=

1−

k=0

1

1
b p

a − ab

=

bp − a p b−a

(2) (3) Given np <

(n + 1)p+1 − np+1 < (n + 1)p p+1

We want n−1 X

n

kp <

k=1

X np+1 < kp p+1 k=1

2p+1 n = 21p < < 1p + 2p p+1 p=1 1 < 22 /2 = 2.2 < 1 + 2 = 3 p−2 1 < 8/3 < 1 + 4 = 5 I 4.10 Miscellaneous exercises involving induction. Exercise 13. (1) (2) (3) Let n = 2. 2−1 X

k p = 1p = 1,

k=1

2 np+1 2p+1 X p = k = 1 + 2p p+1 p+1 k=1

What makes this exercise hard is that we have to use induction on p itself. Let p = 1. 1<

21+1 = 2 < 1 + 21 = 3 1+2 14

Now assume pth case. Test the p + 1 case. 2p+2 2(p + 1) = p+2 p+2



2p+1 p+1

 >1

since p + 2 < 2p + 2 = 2(p + 1) for p ∈ Z+ For the right-hand inequality, we will use the fact just proven, that 2p − (p) > 0 and pth case rewritten in this manner (1 + 2p ) >

2p+1 =⇒ (1 + 2p )(p + 1) > 2p+1 p+1

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

kp =

k=1 n+1 X

n−1 X

k p + np <

k=1

kp =

k=1

n X

np+1 (n + 1)p+1 − np+1 (n + 1)p+1 np+1 + np < + = p+1 p+1 p+1 p+1

k p + (n + 1)p >

k=1

(n + 1)p+1 − np+1 (n + 1)p+1 np+1 + = p+1 p+1 p+1

We had used the inequality proven in part b, np <

(n+1)p+1 −np+1 p+1

< (n + 1)p .

Exercise 14. Use induction to prove a general form of Bernoulli’s inequality.

1 + a1 = 1 + a1 (1 + a1 )(1 + a2 ) = 1 + a2 + a1 + a1 a2 ≥ 1 + a1 + a + 2 Test the n + 1 case (1 + a1 )(1 + a2 ) . . . (1 + an+1 ) ≥ (1 + a1 + a2 + · · · + an )(1 + an+1 ) = = 1 + a1 + a2 + · · · + an + an+1 + an+1 (a1 + a2 + . . . an ) ≥ ≥ 1 + a1 + a2 + · · · + an + an+1 Note that the last step depended upon the given fact that all the numbers were of the same sign. For a1 = a2 = · · · = an = x, then we have (1 + x)n ≥ 1 + nx. n   X n j n (1 + x) = x = 1 + nx j j=0 Since x and n are arbitrary, we can compare terms of xj ’s. Then x = 0. Exercise 15.

2! 22

=

1 3! 2 33

=

2 9

< 1.

So we’ve shown the n = 2, n = 3 cases. Assume the nth case, that (n + 1)nn 12 (n + 1)! ≥ (n + 1)n+1 (n + 1)n+1


k

 =

n n+1

n! nn

≤


1 k , 2

where k is the greatest integer ≤

n  k  n  k  k  k+1 1 1 1 1 1 1 = 1− < = 2 n+1 2 2 2 2

where in the second to the last step, we had made this important observation: k≤

n 1 n+1 1 1 1 =⇒ k + ≤ =⇒ ≤ < 2 2 2 n+1 2k + 1 2

Exercise 16. 15

n 2.

an+1

√ 1+ 5 a1 = 1 < 2 √ !2 √ √ 1+ 5 1+2 5+5 6+2 5 a2 = 2 < = = 2 4 4 √ !n−1 √ !n  √ !n  1+ 5 1+ 5 2 1+ 5 √ + = 1+ = an + an−1 < = 2 2 2 1+ 5 ! √ !n+1 √ √ !n 1+ 5 2(1 − 5) 4 1+ 5 = = + 2 1−5 4 2

Exercise 17. Use Cauchy-Schwarz, which says

X

a k bk

2

≤

X



P 2 X 2p ( xpk ) n =⇒ (xk ) ≥ n

a2k

 X

b2k



Let ak = xpk and bk = 1. Then Cauchy-Schwarz says X

xpk

2

≤

X

x2p k

We define Mp as follows:  Pn

k=1

Mp =

xpk

1/p

n

So then nMpp =

n X

xpk

k=1

Pn

k=1

M2p = 2p nM2p =

x2p k

!1/2p

n n X

X

2p x2p k = nM2p ≥

x2p k

(nMpp )2 = nMp2p n

k=1 2p M2p ≥ Mp2p =⇒ M2p ≥ Mp

Exercise 18.



Exercise 19. ak = 1,

Pn

k=1

a4 + b4 + c4 3

1/4

a2 + b2 + c2 3 64 a4 + b4 + c4 ≥ 3 

≥

1/2 =

23/2 since 31/2

1=n

Now consider the case of when not all ak = 1. a1 = 1 a1 a2 = 1 and suppose, without loss of generality a1 > 1. Then 1 > a2 . (a1 − 1)(a2 − 1) < 0 a1 a2 − a1 − a2 + 1 < 0 =⇒ a1 + a2 > 2 (consider n + 1 case ) If a1 a2 . . . an+1 = 1, then suppose a1 > 1, an+1 < 1 without loss of generality b1 = a1 an+1 b1 a2 . . . an = 1 =⇒ b1 + a2 + · · · + an ≥ n (by the induction hypothesis) (a1 − 1)(an+1 − 1) = a1 an+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 16

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). ∀S ∈ M, a(S) ≥ 0

(1) •

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

(2) •

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

(3) •

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 S⊆Q⊆T

(4)

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 Tl = { line segment containing x0 }, Q = {x0 }. For Q, only ∅ ⊂ Q By Axiom 3, let T = S. a(T − S) = a(∅) = a(T ) − a(T ) = 0 ∅ ⊂ Q ⊆ Tl =⇒ a(∅) ≤ a(Q) ≤ a(Tl ) =⇒ 0 ≤ a(Q) ≤ 0 =⇒ a(Q) = 0 (2)  a

N [

 Qj  =

j=1

N X

a(Qj )

j=1

if Qj ’s disjoint. Let Qj = {xj }. Since a(Qj ) = 0. By previous part, a

S

N 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) =

p

p 1 1 a2 + b2 d + ab − ( ab + a2 + b2 d) = ab 2 2

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. Tr = R + T1 + T2 . T1 , T2 are right triangles and so by the previous problem, T1 , T2 are measurable. Then Tr is measurable by the Additive property axiom (note that the triangles and the rectangle don’t overlap). 17

We can compute the area of a trapezoid: Tr = R + T1 + T2 =⇒ a(Tr ) = a(R) + a(T1 ) + a(T2 ) 1 1 1 a(Tr ) = hb1 + h(b2 − b1 )/2 + h(b2 − b1 )/2 = h(b1 + h2 ) 2 2 2 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 12 bh = 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 polygon and 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 the same 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 + 21 B − 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 intersecting perpendicular bisector on a coordinate axis (a picture would help), but the ratio of the perpendicular bisector to the third vertex to half the length of the triangle is cot π/3 = √13 . 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 is because 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} and T = {3, 4, 5}, [ n(S T) = 5 \ n(S T) = 2 n(S − T ) = n({1, 2}) = 2 n(T − S) = n({5}) = 1 n satisfies nonnegativeSproperty T because by definition, there’s S no such T thing as a negative number of elements. If S, T are subsets of A, so are S T , S T since every element in S ST , S T is in S. Thus n could it, so that it’s T be assigned to T measurable. Since n counts only distinct elements, then n(S T ) = n(S) + a(T ) − a(S T ), where −a(S T ) ensures there 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 “with respect to” T . n(S) + n(T − S) = n(T ), since n counts up distinct elements. 18

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 + n If 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 = q1 + r1 , y = q2 + r2 ; 0 ≤ r1 , r2 < 1. ( = [q1 + q2 + r1 + r2 ] = [x] + [y] = q1 + q2

q1 + q2 q1 + q2 + 1 if r1 + r2 ≥ 1

[x] + [y] + 1 = q1 + q2 + 1

(4) 1 If x is an integer , [2x] = 2x = [x] + [x + ] = [x] + [x] = 2x 2 ( q if r < 12 1 [x] + [x + ] = q + 2 2q + 1 if r > 12 ( 2q if r < 12 [2x] = [2(q + r)] = [2q + 2r] = 2q + 1 if r > 12 (5) ( q 1 2 [x] + [x + ] + [x + ] = q + 3 3 q + 1 if r >   3q [3x] = [3(q + r)] = [3q + 3r] = 3q + 1   3q + 2

if r < 2 3

if r < 13 if 13 < r < if r > 23

2 3

( +

q q+1

if r < if r >

1 3 1 3

  3q = 3q + 1   3q + 2

if r < 31 if 13 < r < if r > 32

2 3

2 3

Exercise 5. Direct proof.

  nq [nx] = [n(q + r)] = nq + 1   nq + n − 1

if r < n1 if n1 < r < if r > n−1 n

2 n

Exercise 6.

1 a(R) = hk = IR + BR − 1 2 b X

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

n=a

[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 such that g < f (n), so that all lattice points included and less than g are included. Exercise 7. 19

(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 h na i = number of interior points at x = n and below the hypotenuse line of the right triangle of sides a, b, (5) b including points on the hypotenuse b−1 h X na i n=1

b

1 ab + ((a + 1) + b) − 1 = 2 2

ab a b 1 (a − 1)(b − 1) = − − + 2 2 2 2 2 b−1 h i X (a − 1)(b − 1) na =⇒ = b 2 n=1

Now

(2) a, b ∈ Z+ b−1 h X na i

 b−1  X a(b − n)

= b b n=1 ( P   b−1 b−1 Xh − n=1 an −a an i 
a− = Pb−1 ban b − n=1 b −a −1 n=1

(reverses order of summation)

n=1

=−

n b


− 1 can’t be!)

b−1 h X an

b−1 h i   X an i −a −1 =− − a − (b − 1) = b b n=1

n=1

=−

if an b − a4 is an integer (but a otherwise

b−1 h X an i n=1

b

+ a(b − 1) − (b − 1) b−1 h X na i n=1

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) = ck

if x ∈ Ik .

(

1 ∀x ∈ S . 0 ∀x ∈ /S If x ∈ [a, b], then x must only lie in one open subinterval Ij , since real numbers obey transitivity. Given that χs (x) =

n X

ck χIk (x) = cj for x ∈ Ij =⇒

k=1

n X

ck χIk (x) = f (x)∀x ∈ [a, b]

k=1

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. R3 (1) 1 [x]dx = (−1) + 1 + (2) = 2 R 7/2 R3 (2) −1 [x + 21 ]dx = −1/2 [x]dx = (−1) 12 + (1)(1) + (2)(1) + 12 3 = 4 R3 (3) −1 ([x] + [x + 12 ])dx = 6 R3 (4) −1 2[x]dx = 4 R3 R6 (5) −1 [2x]dx = 12 −2 [x]dx = 12 ((−2)1 + (−1) + (1) + 2 + 3 + 4 + 5) = 6 R3 R −3 R1 (6) −1 [−x]dx = − 1 [x]dx = −3 [x]dx = −3 + −2 + −1 = −6 Exercise 2.

( 5/2 s= −1

if 0 < x < 2 if 2 < x < 5 20

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. Z b

Z

b

a

Z [x − x]dx =

([x] + [−x])dx = a

b

(−1)dx = a − b a

Exercise 4.

(1) n ∈ Z+ , (2)

Rn 0

[t]dt =

Pn−1 t=0

t=

(n−1)(n−1+1) 2

=

(n−1)n 2

Exercise 5.

√ √ √ √ √ √ R2 R2 (1) 0 [t2 ]dt = 1 [t2 ]dt = 1( 2 − 1) + 2( 3 − 2) + 3(2 − 3) = 5 − 2 − 3 R3 2 R3 2 R0 2 R3 2 R0 2 R3 2 (2) −3 [t ]dt = 0 [t ]dt + −3 [t ]dt = 0 [t ]dt + − 3 [t ]dt = 2 0 [t ]dt Z 3 √ √ √ √ √ √ √ √ [t2 ]dt = 4( 5 − 2) + 5( 6 − 5) + 6( 7 − 6) + 7( 8 − 7) + 8(3 − 8) 2 √ √ √ √ 16 − 5 − 6 − 7 − 8 Z 2 Z 3 √ √ √ √ √ √ 2 [t ]dt + [t2 ]dt = 21 − 3 2 − 3 − 5 − 6 − 7 − 8 0

2

Z

3

=⇒

√ √ √ √ √ [t2 ]dt = 42 − 2(3 2 + 3 + 5 + 6 + 7)

−3

Exercise 6.

(1) (2)

Rn

[t]2 dt R0x 2 [t] dt 0

=

Rn

=

P[x−1]

1

[t]2 dt =

j=1

Pn−1 j=1 2

j2 =

(n−1)n(2n−1) 6

j 2 + q r where x = q + r, q ∈ Z+ , 0 ≤ r < 1. Z x q(q − 1)(2q − 1) + q 2 r = 2(x − 1) = 2(q + r − 1) [t]2 dt = 6 0 =⇒ q(q − 1)(2q − 1) + 6q 2 r = 12q + 12r − 12 =⇒ x = 1, x = 5/2

Exercise 7.

(1) Z

9

Z 9√ √ [ t]dt = [ t]dt = 3(1) + 5(2) = 13

0

Z

1 1

√

6[ t]dt = 3(1) + 5(2) + 7(3) = 34 = 0

Assume

n2

√ n(n − 1)(4n + 1) [ t]dt = 6 0 Z (n+1)2 √ Z n2 √ Z [ t]dt = [ t]dt + Z

0

(4)(3)(17) 6

(n+1)2

√ n(n − 1)(4n + 1) [ t]dt = + n((n + 1)2 − n2 ) = 6 0 (n2 − n)(4n + 1) + 6n(2n + 1) 4n3 + n2 − 4n2 − n + 12n2 + 6n 4n3 + 9n2 + 5n = = = 6 6 6 n2

indeed , (n + 1)(n)(4(n + 1) + 1) (n2 + n)(4n + 5) 4n3 + 5n2 + 4n2 + 5n = = 6 6 6 21

Exercise 8.

R b+c

Exercise 9.

R kb

a+c

ka

f (x)dx =

R b+c−c

1

f (x)dx =

1 k

Rb f (x − (−c))dx = a f (x + c)dx R (kb)/k  x  Rb f 1/k dx = k a f (kx)dx (ka)/k

a+c−c

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

R3

So for f (3) =

0

Rp 0

s(x)dx.

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

So then

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

We obtain this formula

( f (p) = Z

f (p + 1) = f (p) + ( −p 2 p−1 2

= ( =

p p+1 2 (−1) p−1 p+1 2 (−1)

p even since p even

p+1

n p+1 s(x)dx = p−1 2 (−1) p ( ( p p p even 2 = −p−1 + −p p odd 2

− (p+1) 2 p 2

p even + (−1)p p =

if p + 1 even if p + 1 odd

Thus, p = 14, p = 15. Exercise 11.

(1) b

Z

s(x)dx = a b

Z

Z s+

a

n X

s3k (xk − xk−1 )

k=1 n1 X

c

s= b

s2k (xk

− xk−1 ) +

k=1

n2 X

s3k (xk

− xk−1 ) =

k=n1 Rb Rb s+ a a

Rb

n2 X

s3k (xk

Z − xk−1 ) =

Pn3 (s + t) = k=1 (s + t)3k (xk − xk−1 ) 6= t Rb Pn (3) a cs = k=1 (cs)3 (xk − xk−1 ) 6= c a 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 . (2)

Rab

n X

s3k (xk − xk−1 ) =

k=1

=

k X k=1 n X

s3k (xk − c − (xk−1 − c)) = s3k (yk − yk−1 ) =

Rb a

s=

Pn

3 k=1 sk (xk

b

s(y + c)dy a

k=1

(5) s < t,

Z

− xk−1 ). if 0 < s, s3 < s2 t < st2 < t3 if s < 0t, s3 < and t3 > 0 if s < t < 0, s3 < s2 t, s(st) < t(ts) = t2 s

s3 < s2 t < t2 s < t3 22

s(x)dx a

k=1

ts > t2 t2 s < t 3

c

Then

Rb a

s<

Rb a

t.

Exercise 12.

Rb

Rc s+ b s a Rb (s+t) = a

(1) (2)

Rc Pn3 Pn2 Pn1 sk (x2k − x2k−1 ) = a s sk (x2k − x2k−1 ) = k=1 sk (x2k − x2k−1 ) + k=n = k=1 1 Pn3 Pn3 Pn3 Pn3 2 2 2 2 2 2 2 2 k=1 tk (xk −xk−1 ) k=1 sk (xk −xk−1 )+ k=1 (sk +tk )(xk −xk−1 ) = k=1 (s+t)k (xk −xk−1 ) =

since P3 = {xk } is a finer partition than the partition for s, P1 , t, P2 , then consider 2 sk (yj2 − yj−1 ) = sk ((x2k+1 − x2k ) + (x2k − x2k−1 )), so n3 X

sk (x2k

2

−x−k−1 )+

k=1

n3 X

tk (x2k

2

−x−k−1 )=

n1 X

sj (x2j

k=1

Z

b

Z

(4)

tj (x2j − x − j − 12 ) =

b

s+

Pn csk (x2k − x2k−1 ) = c k=1 sk (x2k − x2k−1 ) = Pn s(x)dx = k=1 sk (x2k − x2k−1 ) where

cs =

Rab+c a+c

n2 X j=1

a

Rb

−x−j−1 )+

j=1

= (3)

2

Pn

k=1

t

a Rb c a

s

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 < yk 0

where P = {yk } is a partition on [a, b] b

Z

s(y + c)dy = a

= =

n X k=1 n X k=1 n X

2 sk (yk2 − yk−1 )= n X

sk ((xk − c)2 − (xk−1 − c)2 ) =

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

k=1 n X

sk (x2k − x2k−1 − 2c(xk − xk−1 )) 6=

k=1

sk (x2k − x2k−1 )

k=1

Rb

Pn

b

n X

Rb (5) Since x2k − x2k−1 > 0, a sdx = k=1 sk (x2k − x2k−1 ) < k=1 tk (x2k − x2k−1 ) = a tdx Note that we had shown previously that the integral doesn’t change under finer partition. Pn

Exercise 13.

Z

s(x)dx a

Z sk (xk − xk−1 );

b

t(x)dx = a

k=1

n2 X

tk (yk − yk−1 )

k=1

P = {x0 , x1 , . . . , xn }, Q = {y0 , y1 , . . . , yn } Note that x0 =S 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 is another partition on [a, b] (by partition definition) since xk , yk ∈ R and since real numbers obey transitivity, {xk , yk } can be arranged 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 < yj If 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 Z

b

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

a

b

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

n3 X

(sl + t)l)(zl − zl−1 ) =

l=1

n3 X

sl (zl − zl−1 ) +

l=1

n3 X

tl (zl − zl−1 )

l=1

In general, it was shown (Apostol I, pp. 66) that any finer partition doesn’t change the integral R is a finer partition. So Z b Z b n2 n3 n n X X X X sl (zl − zl−1 ) + tl (zl − zl−1 ) = sk (xk − xk−1 ) + tk (yk − yk−1 ) = s(x)dx + t(x)dx l=1

l=1

k=1

k=1 23

a

a

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

b

Z c1

b

Z s(x)dx + c2

t(x)dx = c1

a

a

n X

sk (xk − xk−1 ) + c2

k=1

=

n3 X

tk (xk − xk−1 ) =

k=1

c1 sl (zl − zl−1 ) +

l=1 Z b

=

n2 X

n3 X

c2 tl (zl − zl−1 ) =

l=1

n3 X

(c1 sl + c2 tl )(zl − zl−1 ) =

l=1

(c1 s + c2 t)(x)dx a

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) b

Z

s(x)dx = a

n X

sk (xk − xk−1 ) =

k=1 Z b

=

n3 X

sl (zl − zl−1 ) <

n3 X

l=1

tl (zl − zl−1 ) =

l=1

n2 X

tk (yl − yk−1 ) =

k=1

t(x)dx a

Z =⇒

b

Z s(x)dx <

a

b

t(x)dx a

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

Use the hint: P1 is a partition of [a, c], P2 is a partition of [c, b], then the points of P1 along with those of P2 form a partition of [a, b]. Z

c

Z s(x)dx +

a

b

s(x)dx = a

n1 X

sl (xk − xk−1 ) +

n2 X

sk (xk − xk−1 ) =

Z sk (xk − xk−1 ) =

b

s(x)dx a

k=1

k=1

k=1

n3 X

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

P 0 = {y0 , y1 , . . . , yn }; yk = xk + c; =⇒xk−1 + c < y < xk + c xk−1 < y − c < xk yk − yk−1 = xk + c − (xk−1 + c) = xk − xk−1 s(y − c) = sk if xk−1 < y − c < xk , k = 1, 2, . . . n b

Z

s(x)dx = a

n X

sk (xk − xk−1 ) =

k=1

n X

Z

yn

y0

k=1

Z

b+c

s(y − c)dy =

sk (yk = yk−1 ) =

s(x − c)dx a+c

1.26 Exercises - The integral of more general functions, Upper and lower integrals, The area of an ordinate set expressed as an integral, Informal remarks on the theory and technique of integration, Monotonic and piecewise monotonic functions. Definitions and examples, Integrability of bounded monotonic functions, Calculation of the integral Rb of a bounded monotonic function, Calculation of the integral 0 xp dx when p is a positive integer, The basic properties of the integral, Integration of polynomials. Exercise 16.

R2 0

|(x − 1)(3x − 1)|dx = 24

2

Z

2

Z (x − 1)(3x − 1)dx = 1

1

Z

2 (3x2 − 4x + 1)dx = (x3 − 2x2 + x) 1 = 2

1

1 4 (1 − x)(3x − 1)dx = − (x3 − 2x2 + x) 1/3 = 27 1/3 Z 1/3 4 (x − 1)(3x − 1)dx = 27 0 So the final answer for the integral is 62/27. 1/2 R3 R3 R 3−5/2 Exercise 17. 0 (2x − 5)3 dx = 8 0 (x − 52 )3 dx = 8 −5/2 x3 dx = 8 14 x4 −5/2 = Exercise 18.

R3 −3

(x2 − 3)3 dx =

R3 0

(x2 − 3)3 +

Rx

(x2 − 3)3 =

−3

R3 0

(x2 − 3)2 + −

39 2

R3 0

(x2 − 3)3 = 0

2.4 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 < 1 (since c > 0). So cx3 < x2 (since x2 > 0).   1/c Z Z 1 3 c 4 1 f − g = x2 − cx3 = x − x = 3 4 12c3 0 Z 2 1 1 f −g = = ; c= √ 3 12c3 2 2 Exercise 16. f = x(1 − x), g = ax.

Z

Z

1−a

x − x2 − ax =

f −g =



0

Exercise 17. π = 2

R1 √ −1

 1−a 1 1 1 (1 − a) x2 − x3 = (1 − a)3 = 9/2 =⇒ a = −2 2 3 6 0

1 − x2 dx

(1) Z

3

Z p 2 9 − x dx = 3

−3

3

r

 x 2

1−

3

−3

Z

1

p

= 3(3)

1 − x2 =

−1

9π 2

Now Z

kb

f ka

Z

x

dx = k

k

b

f dx a

(2) Z 0

2

r

1 1 − x2 dx = 2 4

Z

1

p

0

1 − x2 dx =

2π π = 4 2

√ (3) −2 (x − 3) 4 − x2 dx Z 2 p Z −2 Z 2 p p 2 2 x 4 − x dx = (−1) −x 4 − x =⇒ 2 x 4 − x2 = 0 −2 2 −2 Z 2 r Z 1p  x 2 −3 2 1− dx = (−6)(2) 1 − x2 = −6π 2 −2 −1 R2

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: s √ ! √ √ 3 3 c2 = 1 + 1 − 2(1)(1) cos 30◦ = 2 1 − =⇒ c = 2 1 − 2 2 25

The height is given also by the law of cosines √

s

1 + 23 h = 1 cos 15 = = 2 2 The area of the dodecagon is given by adding up twelve of those isosceles triangles s  s  √  √  √ 1 3 3 1 √  2 1 − (12)  1 + =3 2 2 2 2 r

◦

1+

cos 30◦

So 3 < π. Now consider a dodecagon that’s circumscribing the circle of radius 1.  v √  u √ ! 3 u 1 − 1 t 3 2 √  (1) = 12 2 − 2 (12) >π 3 2 2 1+ 2 Exercise 19.

(1) (x, y) ∈ E if x = ax1 , y = by1 such that x21 + y12 ≤ 1
2
2 =⇒ xa + yb = 1 (2) r

 x 2 1− a Z 1p Z a r  x 2 π = 2ba 2 b 1− 1 − x2 = ba (2) = πba a 2 −1 −a y=b

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 = k1 x1 , y = k2 y1 .  x If f (x1 ) = y1 , g(x) = k2 f k1 = k2 y1 = y. integrating g on [k1 a, k1 b], Z

k1 b

Z

k1 b

g(x)dx = k1 a

 k2 f

k1 a

x k1



Z dx = k2 k1

b

f (x)dx = k2 k1 A a

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

(1) sin π/2 = 1, sin π/2(1 + 4j) = 1, j ∈ Z+ . (2) cos x = 1, cos 0 = 1, cos 2πj = 1 Exercise 3.

sin x + π = − sin x + π/2 + π/2 = cos x + π/2 = − sin x cos x + π = cos x + π/2 + π/2 = − sin x + π/2 = − cos x Exercise 4. 26

sin 3x = sin 2x cos x + sin x cos 2x = 2 sin x cos2 x + sin x(cos2 x − sin2 x) = 3 cos2 x sin x − sin3 x = = 3(1 − sin2 x) sin x − sin3 x = 3 sin x − 4 sin3 x cos 3x = cos 2x cos x − sin 2x sin x = (cos2 x − sin2 x) cos x − (2 sin x cos x) sin x = cos x − 4 sin2 x cos x cos 3x = −3 cos 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), cos 3x = 4 cos3 x − 3 cos x x = π/6 cos 3π/6 = 0 = 4 cos3 π/6 − 3 cos π/6 = cos π/6(4 cos2 π/6 − 3) = 0 √ =⇒ cos π/6 = 3/2, sin π/6 = 1/2( by Pythagorean theorem ) √ (2) sin 2π/6 = 2 cos π/6 sin 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, 9bc − 27d − 2b3 54 3c − b2 Q= 9 x1 = S + T − b/3 R=

p

Q3 + R2 )1/3 p T = (R − Q3 + R2 )1/3 S = (R +

√ x2 = −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.

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

1 cos x cos y 1 cos x cos y

! =

tan x − tan y 1 + tan x tan y

if tan x tan y 6= −1 Similarly, sin x + y sin x cos y + sin y cos x tan x + tan y = = , tan x tan y 6= 1 cos x + y cos x cos y − sin x sin y 1 − tan x tan y cos x + y cos x cos y − sin x sin y cot x cot y − 1 cot x + y = = = sin x + y sin x cos y + sin y cos x cot y + cot x tan x + y =

Exercise 7. 3 sin x + π/3 = A sin x + B cos x = 3(sin x 12 +

√

3 2

cos x) =

3 2

sin x +

√ 3 3 2

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 6= 0, B cos x) == A(sin x + tan α cos x) A A A (cos α sin x + sin α cos x) = (sin x + α) = cos α cos α A where −π/2 < α < π/4, B/A = tan α, C = cos α A sin x + B cos x = A(sin x +

Exercise 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

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

A B

= tan β, α = −β, C =

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

B cos β .

Exercise 12.

p √ π 1 − cos2 x =⇒ cos x = 1/ 2 =⇒ x = 4 √ Try 5π/4. sin 5π/4 = cos 3π/4 = − sin π/4√= −1/ 2. cos 5π/4 = − sin 3π/4 = − cos π/4 = −1/ 2. So sin 5π/4 = cos 5π/4. x = 5π/4 must be the other root. So θ = π/4 + πn (by periodicity of sine and cosine). sin x = cos x =

Exercise 13.

sin x − cos x = 1 =

p

1 − cos2 x = 1 + cos x

=⇒ 1 − cos2 x = 1 + 2 cos x + cos2 x =⇒ 0 = 2 cos x(1 + cos x) cos x = −1, x = π/2 + 2πn Exercise 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 y cos 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 y sin 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 x sin (x + h/2) cos h/2 + cos (x + h/2) sin h/2 − sin (x + h) cos h/2 − cos x + h/2 sin h/2 = h h sin h/2 = cos (x + h/2) h/2 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) cos x + h − cos x = h h sin h/2 =− sin (x + h/2) h/2 Exercise 16.

(1) sin 2x = 2 sin x cos x if sin 2x = 2 sin x and x 6= 0, x 6= πn, cos x = 1 but x 6= πn =⇒ x = 2πn (2) cos x + y = cos x cos y − sin x sin y = cos x + cos y. p cos x cos y − cos x − cos y = sin y 1 − cos2 x Letting A = cos x, B = cos y, 2

2

2

2

A B + A + B − 2A2 B − 2AB 2 + 2AB = 1 − A2 − B 2 + A2 B 2 A2 + B 2 − A2 B − AB 2 + AB = 1/2 B 2 (1 − A) + B(A − A2 ) + A2 − 1/2 = 0 B=

A(1 − A) ±

p

A2 (1 − A)2 − 4(1 − A)(A2 − 1/2) 1 (A2 (1 − A) − 4(A2 − 1/2))1/2 = =A± √ 1−A 1−A

1 (−3A2 − A3 + 2)1/2 1−A 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 =A± √

=⇒ sin y(1 − cos x) + sin y + − cos x sin y = 0, =⇒ y = 2πn Checking our result, we find that sin (2πn + y) = sin 2πn + sin y(1) 28

(4) Z 0

y

y

sin xdx = − cos x|0 = −(cos y − 1) = 1 − cos y = sin y p =⇒ 1 − cos y = 1 − cos2 y

1 − 2 cos y + cos2 y = 1 − cos2 y =⇒ cos y(cos y − 1) = 0; y = Exercise 17.

Rb a

2(j + 1)π , 2πn 2

b

sin xdx = − cos x|a = − cos b + cos a

√

(1) (2) (3) (4) (5) (6) (7) (8)

− 23 + 1 √ − 22 + 1 1 2

1 2 0 We were integrating over one period, over one positive semicircle and over one negative semicircle. 0 We had√integrated over two equal parts, though it only shaded in up to x = 1. √ 2 − 2 + 23 π Rπ 2 2 Exercise 18. 0 (x + sin x)dx = ( 12 x2 − cos x) 0 = π2 − (−1 − 1) = π2 + 2 Exercise 19.

R π/2

Exercise 20.

R π/2

0

0

π/2 (x2 + cos x)dx = ( 13 x3 + sin x) 0 = 31 (π/2)3 + 1 π/2

(sin x − cos x)dx = (− cos x − sin x)|0

R π/2

| sin x − cos x|dx = ( by symmetry )2 π Rπ 1 Exercise 22. 0 ( 2 + cos t)dt = ( 12 t + sin t) 0 = π2

Exercise 21.

0

= −1 − (−1) = 0

R π/4 0

π/4

(cos x − sin x)dx = 2(sin x + cos x)|0

√ = 2( 2 − 1)

Exercise 23. 2π/3

Z 0

1 ( + cos t)dt + 2

2π/3 2π/3 t 1 t + ( + sin t) −( + cos t)dt = ( + sin t) 2 2 2 2π/3 0 π √ π 3 π π √ = 2( + )− = + 3 3 2 2 6 π

Z

Exercise 24. If −π < x ≤ − 2π 3 ,

Z

x

1 −( + cos t)dt = 2 −π

Z

−π

x

1 ( + cos t)dt = 2



 −π π x t + sin t = − − − sin x 2 2 2 x

If −2π/3 ≤ x ≤ 2π/3, Z −2π/3 Z x 1 1 −π/6 √ x −( + cos t)dt + ( + cos t)dt = 3/2 + (t/2 + sin t)|−2π/3 2 + −π −2π/3 2 √ √ x = x/2 + sin x − π/3 − 3/2 + 3/2 − π/6 = + sin x − π/3 2 If 2π/3 ≤ x ≤ π, Z x √ √ √ 2π/3 3/2 + −(1/2 + cos t)dt = 3/2 + (t/2 + sin t)|x = π/3 + 3 − x/2 − sin x 2π/3

Exercise 25.

R x2

Exercise 26.

R π/2

Exercise 27.

R π/3

x

(t2 + sin t)dt = ( 13 t3 + − cos t) =

0

0

sin 2xdx =



− cos (2x) 2

x6 −x3 3

+ cos x − cos x2

 π/2 = (−1/2)(−1 − 1) = 1 0

π/3

cos x/2dx = 2 sin x/2|0

= 2 21 = 1

Exercise 28. 29

x

Z

Z

0

x

(cos a cos bt − sin a sin bt)dt =

cos (a + bt)dt =

 cos a

 x sin bt − sin a(− cos bt/b) =

b 0 0 sin a 1 cos a = sin bx + (cos bx − 1) = sin a + bx − sin a/b b b b  x Z x Z x sin a cos a (sin a cos bt + sin bt cos a)dt = sin (a + bt)dt = sin bt − cos bt = b b 0 0 0 1 = (cos bx + a + cos a) b

Exercise 29.

(1) 0

  x 3 sin t − sin 3t cos 3x − 1 3 sin tdt = dt = − cos t + cos 3t/12 = −3/4(cos x − 1) + = 4 4 12 0 0 1 3 1 = − cos x + (cos 2x cos x − sin 2x sin x) = 2/3 − 1/3 cos x(2 + sin2 x) 3 4 12

Z

x

Z

x

x

Z

3

(2) 0

 x 1 sin 3t 3 + sin t = cos tdt = 4 3 4 0 0 1 3 1 = (sin 2x cos x + sin x cos 2x) + sin x = (2 sin x cos x + sin x(2 cos2 x − 1)) = 12 4 12 sin x cos2 x + 2 sin x = 3 Z

3

x

1 (cos 3t + 3 cos t)dt = 4



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 ∈ Z then Z

a+p

r+p

Z f (x)dx =

a

Z

r+p

f (x + np)dx = r p

=

Z f (x − p)dx =

r

0

Z f+

p

Z f+

r

r

r+p

f (x)dx =

r

r

Z f+

p

f (x)dx =

r

Z

Z

p

Z f=

0

p

f 0

Exercise 31.

(1) 2πn 1 1 (− cos x) = − (1 − 1) = 0 n n 0 0 0 2πn Z 2π Z 1 1 2πn cos xdx = sin x cos nxdx = =0 n 0 n 0 0 Z

2π

sin nxdx =

1 n

2πn

Z

sin xdx =

(2) Z

2π

2π

Z sin nx cos mxdx =

0

Z

2π

Z

0 2π

sin nx sin mxdx = 0

Z

0 2π

Z cos nx cos mxdx =

0

0

While Z

2π

1 (sin (n + m)x + sin (n − m)x)dx = 0 + 0 = 0 2 1 (cos (n − m)x + cos (n + m)x)dx = 0 + 0 = 0 2 1 (cos (n − m)x + cos (n + m)x)dx = 0 + 0 = 0 2

2π

sin2 nxdx =

0

Z 0

Z

2π

1 − cos 2nx dx = π 2

2π

1 + cos 2nx dx = π 2

0 2π

cos2 nxdx =

Z 0

30

Exercise 32. Given that x 6= 2πn; sin x/2 6= 0, n X

2 sin x/2 cos kx = 2 sin x/2

k=1

n X

cos kx =

k=1

n X

sin (2k + 1)

k=1

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

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

n X

sin kx =

k=1

n X

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

k=1

= cos nx + x/2 − cos x/2 Now sin nx/2 sin nx/2 + x/2 = sin nx/2(sin nx/2 cos x/2 + sin x/2 cos nx/2) = = sin2 nx/2 cos x/2 + sin x/2 cos nx/2 sin nx/2 =   sin nx 1 − cos nx sin x/2 = cos x/2 + = 2 2 1 1 = (cos x/2 − cos x/2 cos nx + sin nx sin x/2) = (cos x/2 − cos (nxx /2) 2 2 Then −2 sin x/2

n X k=1

n X

1 sin kx = −2 sin nx/2 sin (n + 1)x 2

sin kx =

k=1

sin nx/2 sin 12 (n + 1)x sin x/2

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

1 x 1 cos x sin x < sin x < 2 2 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. Exercise 2.

1 1−0

R

x2 + x3 =

Exercise 3.

1 4−0

R

x1/2 =

4 3

Exercise 4.

1 8−1

R

x1/3 =

45 28

Exercise 5.

1 π/2−0

Exercise 6.

1 π/2−−π/2

Exercise 7.

1 π/2−0

R

sin 2x = −1/π(−1 − 1) = 2/π

Exercise 8.

1 π/4−0

R

sin x cos x =

R π/2 0

R

1 b−a

7 12

sin x =

2 π

cos x = 2/π

1 π 31

R

x2 dx = 13 (b2 + ab + a2 )

Exercise 9.

1 π/2−0

Exercise 10.

1 π−0

sin2 x =

R

R

cos2 x =

1 π (x

π − sin 2x/2) 0 =

1 2

1 2

Exercise 11.

(1) (2)

1 a−0 1 a−0

√ x2 = a2 /3 = c2 =⇒ c = a/ 3 a R n an 1 x = a1 n+1 xn+1 = n+1 = cn =⇒ c = R

0

a (n+1)1/n

Exercise 12.

Z Z Z A = wf / w wx2 Z 1 1 x3 = x4 = k x2 ; k = 4 2 Z 1 1 x4 = x5 = k x3 ; k = 5 3 Z 1 1 x5 = x6 = k x4 ; k = 6 4

Z =k

x

1 ,w = x 2 3 , w = x2 5 2 , w = x3 3

Exercise 13.

Z Z Z 1 1 1 A(f + g) = f +g = f+ g = A(f ) + A(g) b−a b−a b−a  Z Z 1 1 A(cf ) = cf = c f b−a b−a Z Z 1 1 A(f ) = f≤ g = A(g) b−a b−a Exercise 14.

R

R R w(c1 f + c2 g) c1 wf c2 wg R A(c1 f + c2 g) = = R + R w w w = c1 A(f ) + c2 A(g) f ≤ g w > 0( nonnegative ), =⇒ wf ≤ wg Exercise 15.

Aba (f )

1 = b−a

b

Z a

1 f= b−a

Z

c

Z f+

a

!

b

f

 =

c

c−a b−a



Rc

f c−a a

!

Rb b − a − (c − a) a f + b−a b−c

a
R R xρ ρ

or rcm =

R

rdm M . 32

RL xcm = R0L Z0 Icm =

x

=

1

L 2

r2 dm =

Z

x2 (1) = L3 /3

Icm L r2 = R L = L2 /3 =⇒ r = √ 3 1 0 Exercise 17.

R l/2

x+

RL

0 xcm = L 2 + Z L/2

L/2

2xdx =

yL2 12

2(L − L/2) Z L 2x2 = 5L3 /8 x2 +

Icm =

L/2

0

√ 5L3 /8 5L2 5L r = = =⇒ r = √ 3L/2 12 2 3 2

Exercise 18. ρ(x) = x for 0 ≤ x ≤ L

R 1 3 L x 0 xxdx 2 3 xcm = R = L = L 1 2 3 xdx 2x 0 Z Icm = x2 xdx = L4 /4 r2 =

L4 /4 = L2 /2 L2 /2

L r= √ 2

Exercise 19.

L R 1 3 L/2 + L2 (x2 /2) L/2 xxdx + x L2 dx 3x 0 R xcm = R = = 11L/18 1 2 L/2 L xdx + L/2 x + (L − L/2) 2 2 0 Z Z 2 2 4 Icm = x xdx + x L/2dx = L 31/192 √ 31L r2 = Icm /(L2 3/8) = L2 31/72 r = √ 6 2 R

Exercise 20. ρ(x) = x2 for 0 ≤ x ≤ L

R xcm =

xx2 dx R = 3L/4 x2

Z

x2 x2 dx = L5 /5 r 3 2 3 Icm 2 L r = 1 3 = L r= 5 5 L 3

Icm =

Exercise 21.

R L/2 0

xx2 dx +

xcm = R L/2 0

x2 dx +

r2 =

Icm L3 /6

2 x L4 dx L/2 R L L2 dx L/2 4

= 21L/32

L

L2 dx = 19L5 /240 4 L/2 √ √ = 19L2 /40 =⇒ r = 19L/2 10

L/2 2 2

Icm = int0

RL

Z

x x dx +

x2

Exercise 22. Be flexible about how you can choose a convenient origin to evaluate the center-of-mass from 33

Let ρ = cxn L

Z

1 Ln+1 c = M n + 1 0 (n + 1)M =⇒ c = Ln+1 Z L 1 n+1 3M L xxn dx = c c Ln+2 = ML = n+2 n+2 4 0 R Z xρ 3L n+1 3 = =⇒ xρ = = =⇒ n = 2 xcm = M 4 n+2 4 xn dx =

c

ρ=

3M 2 x L3

Exercise 23.

(1) 1 π/2 − 0 (2) 1 π/2 − 0

Z 3 sin 2t =

6 π

√ 9 sin2 2t = 9/2 =⇒ vrms = 3 2/2

Z

Exercise 24. T = 2π (just look at the functions themselves)

1 2π

2π

Z

√ 160 sin t2 sin (t − π/6) = 80 3

0

2.19 Exercises - The integral as a function of the upper limit. Indefinite integrals. Exercise 1. x + 12 x2 + 13 x3 Exercise 2. 2y + 2y 2 + 8y 3 /3 Exercise 3. 2x + 2x2 + 8x3 /3 − (−1 + 1/2 + −1/3) = 2(x + x2 + 4x3 /3) + 5/6 Exercise 4.

R 1−x

Exercise 5.

Rx

Exercise 6.

R x2

1

−2

x

1−x (1 − 2t + 3t2 )dt = (t − t2 + t3 ) 1 = −2x + 2x2 − x3

t4 + t2 =

1 5 5t

x + 13 t3 −2 =

t4 + 2t2 + 1 =



t5 5

x5 5

+

x3 3

+

40 3

 x2 + 23 t3 + t = 51 (x10 − x5 ) + 23 (x6 − x3 ) + x2 − x x

Exercise 7.


x 2 3/2 + t 1 = 23 (x3/2 − 1) + (x − 1) 3t

Exercise 8.

2 3/2 3t


x2 + 45 t5/4 x = 23 (x3 − x3/2 ) + 45 (x5/2 − x5/4 )

x

Exercise 9. sin t|iπ = sin x


x2 + sin t 0 =

Exercise 10.

t 2

Exercise 11.

1 2t

Exercise 12.

1 3 3u

Exercise 13.

Exercise 14.



R

x2 2


x2 + cos t x =

1 3 3v

+ sin x2

x2 −x 2

+ cos x2 − cos x


x + − 13 cos 3u 0 = +

cos 3v −3

1−cos 2x 2

 x2 = x

+x=

1 2x

x3 3

x6 −x3 3

−

+ − 13 (cos 3x − 1) +

sin 2x 4

−1 2 3 (cos 3x

− cos 3x)


y y sin 2y y 2 + 12 x2 0 = − + 2 4 2 34

Rx 0

(1 + t + t2 )dt =

Exercise 15.

− cos 2w 2

Exercise 16.

Rx

Exercise 17.

Rx

(1 −π 2 0


x (cos 2x − 1) x + 2 sin w2 0 = − + 2 sin 2 2 Rx

1 −π 4

+cos t)2 dt =

(t3 − t)dt =

1 3

Rx

√

2

+cos t+cos2 t = 41 (x+π)+sin x+

1 2

t+

0

(t3 − t)dt +

R √2 1


x = 3 (x+π)+sin x+ 1 sin 2x 4 4 −π

(t − t3 )dt

Note that t3 − t < 0 for 0 < t ≤ 1 and t3 − t > 0 for t > 1. t − t3 < 0 for t >

R1

sin 2t 2

√

2.

  x 1 4 1 2 1 1 1 2 1 4 1 x − x = t − t √ = x2 − x4 4 2 3 2 4 6 12 2 √ 1 4 2 2 =⇒ x − x = 0 =⇒ x = 0, x = 2 3 3 (t3 − t)dt “cancel” each other out.

Exercise 18. f (x) = x − [x] −

1 2

if x is not an integer; f (x) = 0 if x ∈ Z.

For any real number, x = q + r, 0 ≤ r < 1, q ∈ Z. So then x − [x] = r 1 f (x) = r − 2 (1) To show the periodicity, consider f (x + 1) = x + 1 − [x + 1] −

1 1 = r − = f (x) sincex + 1 = q + 1 + r, [x + 1] = q + 1 2 2 1 x + 1 − [x + 1] = r − 2

Rx Rx (2) P (x) = 0 f (t)dt = 0 (t − 12 ) = 12 x2 − 21 x because given 0 < x ≤ 1, then q = 0 for x, so we can use r = t. To show periodicity, Z x+1 Z 1 Z x+1 Z x Z x P (x + 1) = f (t)dt = f (t)dt + f (t)dt = 0 + f (t + 1)dt = f (t)dt = P (x) 0

0

1

Z since 0

1

0

0

1 1 2 f (t)dt = (x − x) = 0 2 0

(3) Since P itself is periodic by 1, then we can consider 0 ≤ x < 1 only. Now x − [x] = r and P (x) = 12 (r2 − r). So P (x) = 12 ((x − [x])2 − (x − [x])). (4) Z 1 Z 1 (P (t) + c)dt = 0 =⇒ P (t)dt = −c 0

0

1 0 ≤ t ≤ 1 so P (t) = (t2 − t) 2   1 Z 1 1 1 3 1 2 1 −1 1 =⇒ P (t)dt = t − t = =⇒ c = 2 3 2 2 6 12 0 0 (5) Q(x) =

Rx 0

(P (t) + c)dt Z x+1 (P (t) + c)dt + (P (t) + c)dt = 0 0 1 Z x Z x =0+ (P (t + 1) + c)dt = (P (t) + c)dt = Q(x) Z

Q(x + 1) =

x+1

Z

1

(P (t) + c)dt =

0

0

so without loss of generality, consider 0 ≤ x < 1 Z x 1 2 1 1 x 1 =⇒ Q(x) (t − t) + = x3 − x2 + 12 6 4 12 0 2 35

Exercise 19. g(2n) =

R 2π

f (t)dt

0

Consider Z

1

1

Z

Z

0

−1

f (t)dt +

−1

0 1

Z

1

Z f (t)dt =

f (t)dt +

f (t)dt =

0

1 −1

Z

0

f (−1t)dt = 1

0

Z

f (t)dt = 0 f+ 1 0 R3 R1 R1 Consider that 1 f (t)dt = −1 f (t + 2)dt = −1 f (t)dt = 0. Then, by induction, Z 2n+1 Z 2n−1 Z 2n+1 Z 1 Z f= f+ f (t)dt = 0 + f (t + 2n)dt = =

1

1

−1

2n−1

1

f (t)dt = 0

−1

(1) 1

Z g(2n) =

2n−1

Z

f=

1 1

=

1

Z

f+

0

Z

2n

Z

f+

0

Z

Z

f+

2n−1

1

−1

0

Z f +−

f (t)dt = 0

0

f (−t)dt 1

0

Z f+

f =0

0

1

(2) −x

Z

g(x + 2) =

f (t)dt = g(x)

0

x+2

2

Z f (t)dt =

0 x+2

Z

x

Z

f+

0

x

Z f (−t)dt =

0

Z

x

Z f =−

g(−x) =

f=

0

x

Z f (t + 2)dt =

2

0

f (t)dt = g(x) 0

Exercise 20.

(1) g is odd since Z

−x

x

Z f (t)dt = −

g(−x) = 0

Z f (−t)dt = −

0

x

f (t)dt = −g(x) 0

Now x+2

Z g(x + 2) =

2

Z f=

x+2

Z f+

0

Z

x

f = g(2) +

0

2

x

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

0

f (t)dt = g(2) + g(x) 0

=⇒ g(x + 2) − g(x) = g(2) (2) Z

2

g(2) =

Z f=

0 Z 0

−

2

1

Z f+

1

Z f=

0

2

Z

0

f +A=

Z

0

f (t + 2)dt + A = −1

1

f (t)dt + A = −1

f (−t)dt + A = 2A 1

Z g(3) = g(2) +

3

g(5) − g(3) = g(2) Z 1 f (t)dt = 2A + f (t + 2)dt = 2A + A = 3A

2

0

=⇒ 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

Z

x

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

g(t)dt 0

Z =⇒ f (5) =

5

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

(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

(2) Z

5

0

Z f (t)dt =

Z

0

−5

0

−5

Z g(t)dt = −

f (t + 5)dt = −5

5

Z g(t)dt =

Z g(−t)dt =

0

0

5

g(t)dt = f (5) = 7 0

(3) Z

x

Z

x−5

f (t)dt =

Z

x−5

f (t + 5)dt = −5

0

Z g(t)dt =

−5

Z

x−5

Z

0

g = f (x − 5) + −

g+ 0

Z

−5

−5

g(t)dt = 0

5

f (x − 5) +

g(−t)dt = f (x − 5) + f (5) = −g(x) + g(0) 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 Exercise 2.

1 x2

=

1 limx→2 x2

limx→0 (25x3 +2) limx→0 (75x7 −2)

=

= −1

Exercise 3. limx→2

(x−2)(x+2) (x−2)

Exercise 4. limx→1

(2x−1)(x−1) x−1

Exercise 5. limh→0

t2 +2th+h2 −t2 h

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 =

1 4

=4 =1 = 2t

limt→0 sin t limx→0 cos t

=

0 1

=0

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

2 sin x cos x x

=2

Exercise 16. limx→0

2 sin x cos x cos 2x sin x

=2

Exercise 17. limx→0

limx→0

5 sin 5x 5x

sin x cos 4x+sin 4x cos x sin x

− limx→0

3 sin 3x 3x

= 1 + limx→0

2 sin 2x cos 2x sin x

= 5 − 3 = 2 Exercise 19. 37

= 1 + 2 limx→0

2 sin x cos x cos 2x sin x




= 5 Exercise 18.

= lim

x→0

x+a 2



x−a − sin x+a 2 − 2 = x→0 x−a
! x−a x−a x+a x+a x−a x−a x+a sin x+a cos + sin cos − sin cos − sin cos 2 2 2 2 2 2 2 2 = x−a lim

sin

+

x−a 2




x+a 2 sin x−a 2 cos 2 = cos a x→a x−a  sin x/2 = 12 x/2

= lim

Exercise 20. limx→0

2 sin2 x/2 4(x/2)2

=

Exercise 21. limx→0

√ 1− 1−x2 x2



1 2



limx→0

 √ 1+√1−x2 2 1+ 1−x

= limx→0

1−(1−x2 ) √ x2 (1+ 1−x2 )

=

1 2

Exercise 22. b, c are given.

sin c = ac + b, a = sin cc−b , c 6= 0. if c = 0, then b = 0, a ∈ R. Exercise 23. b, c are given.

2 cos c = ac2 + b, a = 2 cosc2c−b , c 6= 0. If c = 0, then b = 2, a ∈ R. Exercise 24.

tangent is continuous for x ∈ / (2n + 1)π/2 cotangent 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) x |x|2 | cos x − 1| = | − 2 sin2 x/2| = 2| sin x/2|2 < 2| |2 = < 2/2 =  2 2 √ √ If we had chosen δ0 = 2 for a given . |x − 0| < δ = 2. (3)

  + = 2 2 | cos x + h − cos x| = | cos x cos h − sin x sin h − cos x| = | cos x(cos h − 1) − sin x sin h| ≤   ≤ | cos x|| cos h − 1| + | sin x|| sin h| < + =  2 2 since ∀ > 0∃δ1 , δ2 > 0 such that | cos h − 1| < 0 ; | sin h| <  whenever |h| < min (δ1 , δ2 )   Choose δ3 such that if |h| < δ3 ; | cos h − 1| < ; | sin h| < 2 2 | sin x(cos h − 1) + cos x sin h| ≤ | sin x|| cos h − 1| + | cos h|| sin h| <

Exercise 27. f (x) − A = sin

Let x =

1 x

− A.

1 nπ .

|f (x) − A| = | sin nπ − A| > || sin nπ| − |A|| > |1 − |A|| Consider |x − 0| = |x| = contradiction.

1 nπ

Exercise 28. Consider x ≤

1 n, n

so ∀ > 0, we cannot find δ = So f (x) → ∞ as x → 0+ .

≤ δ(n). Consider 0 =

|1−|A|| . 2

Then suppose a δ(n) ≥ |x − 0| but |f (x) − A| > 0 . Thus,

∈ Z+ , n > M (n) (n is a given constant)   1 1 f (x) = = [n] = n, for m > M (n), x = f (x) > M (n) x m 1 n

such that |f (x) − A| <  for x < δ. 38

Consider

1 n

≥ x > 0, n ∈ Z− ; −n > M (n). f (x) =

  1 = [n] = n < −M (n) x

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| > 0 Exercise 29.

|f − A| = |(−1)[1/x] − A| ≥ ||(−1)[1/x] | − |A|| = |1 − |A|| Choose  < |1 − |A||. Then ∀δ > 0 ( such that |x| < δ ), |f − A| > . Thus there’s no value for f (0) we could choose to make 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 ∃δ0 = δ(x0 , 0 ). Consider 1 = 20 and δ1 = δ(x0 , 1 ) Consider x1 ∈ (x0 − δ1 , x0 + δ1 ), 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 | <

δ1 2 .

Also, “pick” only the x’s such that

δ1 < δ1 2 δ1 δ1 =⇒ |x − x1 | < + = δ1 2 2 |x − x0 | <

Thus, “for these x’s” |f (x) − f (x1 )| = |f (x) − f (x0 ) + f (x0 ) − f (x1 )| < |f (x) − f (x0 )| + |f (x1 ) − f (x0 )| < 1 + 1 = 0 So ∀0 , ∃δ1 for x1 . f is continuous at x1 ∈ (a, b). Thus, there must be infinitely many points that are continuous in (a, b), and at the very least, some or all are “clustered” around some neighborhood about the one point given to make f continuous. Exercise 32. Given  =

1 n,

|f (x)| = |x sin x1 | = |x|| sin 1/x| < |x|(1).

Let δ = δ(n) = n1 , so that |x| < =⇒ |f (x)| < n1

1 n.

Exercise 33.

(1) Consider x0 ∈ [a, b]. Choose some 0 , 0 < 0 < min (b − x0 , x0 − a) 6= 0 , ( x0 could be a or b ) Consider, without loss of generality, only “x’s” such that x ∈ [a, b]. |f (x) − f (x0 )| ≤ |x − x0 | Let δ0 = δ(0 , x0 ) = 0 =⇒ |f (x) − f (x0 )| < 0 . Since we didn’t specify x0 , ∀x0 ∈ [a, b], f is continuous at x0 . (2) Z Z Z b b b f (x)dx − (b − a)f (a) = (f (x) − f (a))dx ≤ |f (x) − f (a)|dx ≤ a a a b Z b 1 2 1 (b − a)2 ≤ |x − a|dx = ( x − ax) = (b − a)(b + a) − a(b − a) = 2 2 2 a a 39

(3) Z Z Z Z b b b b |x − c|dx = |f (x) − f (c)|dx ≤ f (x)dx − (b − a)f (c) = (f (x) − f (c))dx ≤ a a a a Z b Z c 1 1 (x − c)dx = c(c − a) − (c − a)(c + a) + (b − c)(b + c) − c(b − c) = (c − x)dx + = 2 2 c a 1 2 2 = ((c − a) + (b − c) ) 2 Draw a figure for clear, geometric reasoning. Consider a square of length (b − a) and a 45 − 45 right triangle inside. From the figure, it’s obvious that right triangles of c − a length and (b − c) length lie within the (b − a) right triangle. Compare the trapezoid of c − a, b − a bases with the b − a right triangle. 1 1 1 (b − c)(b − a + c − a) = (b − c)(b − c + 2(c − a)) > (b − c)2 2 2 2 Indeed, the trapezoid and c − a right triangle equals the b − a trapezoid since 1 1 1 1 (b − c)(b − a + c − a) + (c − a)2 = (b2 − c2 − 2ab + 2ac + c2 − 2ca + a2 ) = (b − a)2 2 2 2 2 1 1 1 2 2 2 =⇒ (b − a) > (b − c) + (c − a) 2 2 2 Z b (b − a)2 so then f (x)dx − (b − a)f (c) ≤ a 2 3.11 Exercises - Bolzano’s theorem for continuous functions, The intermediate-value theorem for continuous functions. These theorems form the foundation for continuity and will be valuable for differentiation later. Theorem 10 (Bolzano’s Theorem). Let f be cont. at ∀x ∈ [a, b]. Assume f (a), f (b) have opposite signs. Then ∃ at least one c ∈ (a, b) s.t. f (c) = 0. Proof. Let f (a) < 0, f (b) > 0. Want: Fine one value c ∈ (a, b) s.t. f (c) = 0 Strategy: find the largest c. Let S = { all x ∈ [a, b] s.t. f (x) ≤ 0 }. S is nonempty since f (a) < 0. S is bounded since all S ⊆ [a, b]. =⇒ S has a suprenum. Let c = supS. If f (c) > 0, ∃(c − δ, c + δ) s.t. f > 0 c − δ is an upper bound on S but c is a least upper bound on S. Contradiction. If f (c) < 0, ∃(c − δ, c + δ) s.t. f < 0 c + δ is an upper bound on S but c is an upper bound on S. Contradiction.



Theorem 11 (Sign-preserving Property of Continuous functions). Let f be cont. at c and suppose that f (c) 6= 0. then ∃(c − δ, c + δ) s.t. f be on (c − δ, c + δ) has the same sign as f (c). Proof. Suppose f (c) > 0. ∀ > 0, ∃δ > 0 s.t. f (c) −  < f (x) < f (c) +  if c − δ < x < c + δ (by continuity). Choose δ for  = f (c) 2 . Then f (c) 3f (c) < f (x) < ∀x ∈ (c − δ, c + δ) 2 2 Then f has the same sign as f (c). 40



Theorem 12 (Intermediate value theorem). Let f be cont. at each pt. on [a, b]. Choose any x1 , x2 ∈ [a, b] s.t. x1 < x2 . s.t. f (x1 ) 6= f (x2 ). Then f takes on every value between f (x1 ) and f (x2 ) somewhere in (x1 , x2 ). Proof. Suppose f (x1 ) < f (x2 ) Let k be any value between f (x1 ) and f (x2 ) Let g = f − k g(x1 ) = f (x1 ) − k < 0 g(x2 ) = f (x2 ) − k > 0 By Bolzano, ∃c ∈ (x1 , x2 ) s.t. g(c) = 0

=⇒ f (c) = k



Exercise 1. f (0) = c0 . f (0) ≷ 0.

Since limx→∞

ck xk ck−1 xk−1

= limx→∞

ck ck−1 x

= ∞ ∃M > 0 such that |cn M n | > | f (M ) = cn M n +

n−1 X

Pn−1 k=0

ck M k . So then

ck M k ≶ cn

k=0

By Bolzano’s theorem ∃b ∈ (0, M ) such that f (b) = 0. Exercise 2. Try alot of values systematically. I also cheated by taking the derivatives and feeling out where the function

changed direction. (1) If P (x) = 3x4 − 2x3 − 36x2 + 36x − 8, P (−4) = 168, P (−3) = −143, P (0) = −8, P ( 12 ) = 15 16 , P (1) = −7, P (−3) = −35, P (4) = 200 (2) If P (x) = 2x4 − 14x2 + 14x − 1, P (−4) = 231, P (−3) = −7, P (0) = −1, P ( 21 ) = 18 , P ( 23 ) = − 11 8 , P (2) = 2 3 1 3 (3) If P (x) = x4 + 4x3 + x2 − 6x + 2, P (−3) = 2, P (− 52 ) = − 16 , P (−2) = 2, P ( 13 ) = 22 , P ( ) = − 16 , P ( 32 ) = 81 2 14 − 81 , P (1) = 2. Exercise 3. . Consider f (x) = x2j+1 − a. f (0) = −a > 0.

Since a is a constant, choose M < 0 such that M 2j+1 − a < 0. f (M ) < 0. By Bolzano’s theorem, there is at least one b ∈ (M, 0) such that f (b) = b2j+1 − a = 0. Since x2j+1 − a is monotonically increasing, there is exactly one b. Exercise 4. tan x is not continuous at x = π/2. Exercise 5. Consider g(x) = f (x) − x. Then g(x) is continuous on [0, 1] since f is.

Since 0 ≤ f (x) ≤ 1 for each x ∈ [0, 1], consider g(1) = f (1) − 1, so that −1 ≤ g(1) ≤ 0. Likewise 0 ≤ g(0) ≤ 1. If g(1) = 0 or g(0) = 0, we’re done (g(0) = f (0) − 0 = 0. f (0) = 0. Or g(1) = f (1) − 1 = 0, f (1) = 1 ). Otherwise, if −1 ≤ g(1) < 0 and 0 < g(0) ≤ 1, then by Bolzano’s theorem, ∃ at least one c such that g(c) = 0 (g(c) = f (c) − c = 0. f (c) = c). Exercise 6. Given f (a) ≤ a, f (b) ≥ b,

Consider g(x) = f (x) − x ≤ 0. Then g(a) = f (a) − a ≤ 0, g(b) = f (b) − b ≥ 0. Since f is continuous on [a, b] (so is g) and since g(a), g(b) are of opposite signs, by Bolzano’s theorem, ∃ at least one c such that g(c) = 0, so that f (c) = c. 41

3.15 Exercises - The process of inversion, Properties of functions preserved by inversion, Inverses of piecewise monotonic functions. Exercise 1. D = R, g(y) = y − 1 Exercise 2. D = R, g(y) =

1 2 (y

− 5)

Exercise 3. D = R, g(y) = 1 − y Exercise 4. D = R, g(y) = y 1/3 Exercise 5. D = R,

  y √ g(y) = y   y
2

if y < 1 if 1 ≤ y ≤ 16 if y > 16

8

Exercise 6. f (Mf ) = f (f −1 Exercise 7. f (a1 ) ≶

1 n

Pn

i=1

1 n

Pn


i=1 f (ai ) ) =

1 n

Pn

i=1

f (ai )

f (ai ) ≶ f (an ). Since f is strictly monotonic.

g preserves monotonicity. =⇒ a1 ≶ Mf ≶ an Exercise 8. h(x) = af (x) + b, a 6= 0

! n 1X h(ai ) = H n i=1

Mh = H The inverse for h is g

h−b a




! n 1X (af (ai ) + b) = H n i=1

! n 1X a f (ai ) + b n i=1

= H(h) = h−1 . So then Mh = g

! n 1X f (ai ) = Mf n i=1

The average is invariant under translation and expansion in ordinate values. 3.20 Exercises - The extreme-value theorem for continuous functions, The small-span theorem for continuous functions (uniform continuity), The integrability theorem for continuous functions. Since for c ∈ [a, b], m = minx∈[a,b] f ≤ f (c) ≤ maxx∈[a,b] f = M and

Rb f (x)g(x)dx aR b g(x)dx a

= f (c)

Exercise 1.

1 1 m = √ ,M = 1 1+x 2 1 Z 1 1 10 1 x9 = x = 10 10 0 0 Z 1 9 1 x 1 √ ≤ √ dx ≤ 10 1 + x 10 2 0

g = x9 > 0 for x ∈ [0, 1]; f = √

Exercise 2.

p

1 − x2 1 2 1 − x2 = √ .f = √ g = (1 − x2 ) M = √ , m = 1 2 2 3 1−x 1−x 1/2 Z 1/2 1 11 (1 − x2 )dx = (x − x3 ) = 3 24 0 0 r Z 1/2 p 11 11 4 ≤ 1 − x2 dx ≤ 24 24 3 0

Exercise 3. 42

f=

1 g = 1 − x2 + x4 1 + x6

Z

a

1 − x2 + x4 =

0

  a 1 a3 a5 1 x − x3 + x5 = a − + 3 5 3 5 0

1 M =1 m= 1 + a6     Z a 1 a3 1 a3 a5 a5 a− dx ≤ a − + ≤ + 2 1 + a6 3 5 3 5 0 1+x 1 3 5 3 5 So if a = 10 , (a − a /3 + a /5) = a − 0.333 . . . a + 0.2a = 0.099669 Exercise 4. (b) is wrong, since it had chosen g = sin t, but g needed to be nonnegative. Exercise 5. At worst, we could have utilized the fundamental theorem of calculus.

Z

Z 

2

sin t dt = = Exercise 6.

Rb

(f )(1) = f (c) a

Rb a

1 2t



√ 1 2 √ (n+1)π = (2t sin t )dt = (− cos t ) nπ 2c 2

−1 1 ((−1)n+1 − (−1)n ) = (−1)n 2c c

1 = f (c)(b − a). Then f (c) =

Rb

f a b−a

= 0 for some c ∈ [a, b] by Mean-value theorem for

integrals. Exercise 7. f nonnegative. Consider f at a point of continuity c, and suppose f (c) > 0. Then 21 f (c) > 0.

|f (x) − f (c)| <  =⇒ f (c) −  < f (x) < f (c) +  1 1 Let  = f (c) ∃δ > 0 for  = f (c) 2 2 Z c+δ 1 f (x)dx > f (c)(2δ) = f (c)δ > 0 2 c−δ But

Rb a

f (x)dx = 0 and f is nonnegative. f (c) = 0.

Exercise 8.

Z

m

Z

Z Z g =⇒ m g ≤ 0 ≤ M g ∀g Z m ≤ 0 ≤ M for g = 1 but also Z −m ≤ 0 ≤ −M =⇒ m ≥ 0 M ≤ 0 for g = −1 g≤

Z

fg ≤ M

So because of this contradiction, m = M = 0. By intermediate value theorem, f = 0, ∀x ∈ [a, b]. 4.6 Exercises - Historical introduction, A problem involving velocity, The derivative of a function, Examples of derivatives, The algebra of derivatives. Exercise 1. f 0 = 1 − 2x, f 0 (0) = 1, f 0 (1/2) = 0, f 0 (1) = −1, f 0 (10) = −19 Exercise 2. f 0 = x2 + x − 2

(1) f 0 = 0, x = 1, −2 (2) f 0 (x) = −2, x = 0, −1 (3) f 0 = 10, x = −4, 3 Exercise 3. f 0 = 2x + 3 Exercise 4. f 0 = 4x3 + cos x Exercise 5. f 0 = 4x3 sin x + x4 cos x Exercise 6. f 0 =

−1 (x+1)2

Exercise 7. f 0 =

−1 (x2 +1)2 (2x)

Exercise 8. f 0 =

x−1−(x) (x−1)2

=

+ 5x4 cos x + x5 (− sin x)

−1 (x−1)2 43

Exercise 9. f 0 =

−1 (2+cos x)2 (− sin x)

=

sin x (2+cos x)2

Exercise 10.

−2x5 − 9x4 + 12x3 − 3x2 − 2x + 3 (2x + 3)(x4 + x2 + 1) − (4x3 + 2x)(x2 + 3x + 2) = 4 2 2 (x + x + 1) (x4 + x2 + 1)2 Exercise 11.

f0 =

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

Exercise 12.

f0 =

(sin x + x cos x)(1 + x2 ) − 2x(x sin x) sin x + x cos x + x3 cos x − x2 sin x = (1 + x2 )2 (1 + x2 )2

Exercise 13.

(1) f (t + h) − f (t) v0 h − 32th − 16h2 = = v0 + 32t − 16h h h f 0 (t) = v0 − 32t (2) (3) (4) (5) (6)

v0 t = 32 −v0 v0 v0 T = 16 , v0 = 16 for 1sec. v0 = 160 for 10sec. 16 for T sec. 00 f = −32 h = −20t2

Exercise 14. V = s3 ,

dV dS

= 3s2

Exercise 15.

(1) (2)

dA dr dV dr

= 2πr = C = 4πr2 = A

Exercise 16. f 0 =

1 √ 2 x

Exercise 17. f 0 =

−1 √ (1+ x)2

Exercise 18. f 0 =

3 1/2 2x

Exercise 19.



1 √ 2 x



−3 −5/2 2 x

Exercise 20. f 0 =

1 −1/2 2x

+ 13 x−2/3 + 14 x−3/4 x > 0

Exercise 21. f 0 = − 12 x−3/2 + − 13 x−4/3 − 14 x−5/4 Exercise 22. f 0 =

√ 1 −1/2 (1+x)− x 2x (1+x)2

Exercise 23. f 0 =

√ (1+ x)−x 21 √ (1+ x)2

√1 x

=

=

√ 1 2 x(1+x)2

√ 1+ 21 x √ 2 (1+ x)

Exercise 24. 44

g = f1 f2 g 0 = f10 f2 + f1 f20

f0 f0 g0 = 1+ 2 g f1 f2

g = f1 f2 . . . fn fn+1 0 g 0 = (f1 f2 . . . fn )0 fn+1 + (f1 f2 . . . fn )fn+1 ;

f0 g0 (f1 f2 . . . fn )0 = + n+1 g f1 f2 . . . fn fn+1 0 0 f0 f f f0 = 1 + 2 + · · · + n + n+1 f1 f2 fn fn+1 Exercise 25.

cos2 x − (− sin x) sin x = sec2 x sin x cos2 x  cos x 0 − sin x sin x − cos x cos x (cot x)0 = = − csc2 x = sin x sin2 x −1 (sec x)0 = (− sin x) = tan x sec x cos2 x −1 (csc x)0 = cos x = − cot x csc x sin2 x (tan x)0 =

 cos x 0

=

Exercise 35.

(2ax + b)(sin x + cos x) − (cos x − sin x)(ax2 + bx + c) = (sin x + cos x)2 (2ax + b)(sin x + cos x) − (cos x − sin x)(ax2 + bx + c) = (sin x + cos x)2

f0 =

Exercise 36.

f 0 = a sin x + (ax + b) cos x + c cos x + (cx + d)(− sin x) = ax cos x + (b + c) cos x + (a − d) sin x − cx sin x So then a = 1, d = 1, b = d, c = 0. Exercise 37.

g 0 = (2ax + b) sin x + (ax2 + bx + c) cos x + (2dx + e) cos x + (dx2 + ex + f )(− sin x) = = ax2 cos x − dx2 sin x + (2a − e)x sin x + (b + 2d)x cos x + (b − f ) sin x + (c + e) cos x g = x2 sin x. So d = −1, b = 2, f = 2, a = 0, e = 0, c = 0. Exercise 38. 1 + x + x2 + · · · + xn =

xn+1 −1 x−1

(1) (n + 1)xn (x − 1) − (1)(xn+1 − 1) (x − 1)2 (n + 1)(xn+1 − xn ) − xn+1 + 1 nxn+1 − (n1 )xn + 1 = = 2 (x − 1) (x − 1)2 nxn+2 − (n + 1)xn+1 + x x(1 + 2x + · · · + nxn−1 ) = x + 2x2 + · · · + nxn = (x − 1)2

(1 + x + x2 + · · · + xn )0 = 1 + 2x + · · · + nxn−1 =

(2) (x + 2x2 + · · · + nxn )0 = (1 + 22 x1 + · · · + n2 xn−1 ) = (n(n + 2)xn+1 − (n + 1)2 xn + 1)(x − 1)2 − 2(x − 1)(nxn+2 − (n + 1)xn+1 + x) (x − 1)4 (n(n + 2)xn+2 − (n + 1)2 xn+1 + x)(x − 1) − 2(nxn+3 − (n + 1)xn+2 + x2 ) x + 2 2 x2 + · · · + n2 xn = = (x − 1)3 n2 xn+3 + (−2n2 − 2n + 1)xn+2 + (n + 1)2 xn+1 − x2 − x = (x − 1)3 =

45

Exercise 39.

f (x + h) − f (x) (x + h)n − xn = h h n   X n n−j j (x + h)n = x h j j=0
Pn   n n−j j n X h (x + h)n − xn n j=1 j x = = xn−j hj−1 h h j j=1   n n−1 (x + h)n − xn = x = nxn−1 lim h→0 h 1 4.9 Exercises - Geometric interpretation of the derivative as a slope, Other notations for derivatives. Exercise 6. (1) f = x2 + ax + b f (x1 ) = f (x2 ) =

x21 x22

+ ax1 + b + ax2 + b

x2 − x21 + a(x2 − x1 ) f (x2 ) − f (x1 ) = 2 x2 − x1 x2 − x1 = x2 + x1 + a

(2) f 0 = 2x + a m = x2 + x1 + a = 2x + a x =

x2 + x1 2

Exercise 7. The line y = −x as slope −1.

y = x3 − 6x2 + 8x y 0 = 3x2 − 12x + 9 3x2 − 12x + 8 = −1 =⇒ x = 3, 1 The line and the curve meet under the condition −x = x3 − 6x2 + 8x =⇒ x = 3; f (3) = −3 At x = 0, the line and the curve also meet. Exercise 8. f = x(1 − x2 ). f 0 = 1 − 3x2 .

f 0 (−1) = −2 =⇒ y = −2x − 2 For the other line, f 0 (a) = 1 − 3a2 =⇒ y(−1) = 0 = (1 − 3a2 )(−1) + b =⇒ b = 1 − 3a2 Now f (a) = a(1 − a2 ) = a − a3 at this point. The line and the curve must meet at this point. y(a) = (1 − 3a2 )a + (1 − 3a2 ) = = a − 3a3 + 1 − 3a2 = a − a3 1 3 =⇒ −2a3 + 1 − 3a2 = 0 = a3 − + a2 2 2 The answer could probably be guessed at, but let’s review some tricks for solving cubics. First, do a translation in the x direction to center the origin on the point of inflection. Find the point of inflection by taking the second derivative. 1 f 00 = 6a + 3 =⇒ a = − 2 So 1 a=x− 2 1 3 3 1 2 1 3 1 =⇒ (x − ) + (x − ) − = x3 = x − = 0 2 2 2 2 4 4 46

Then recall this neat trigonometric fact: cos 3x = cos 2x cos x − sin 2x sin x = 4 cos3 x − 3 cos x 3 cos 3x =0 =⇒ cos3 x = cos x − 4 4 Particularly for this problem, we have cos 3x = 1. So x = 0, 2π/3, 4π/3. cos x = 1, − 21 . Plugging cos x → x back into what we have for a, a = −1, which we already have in the previous part, and a = 12 . So   1 3 f = 2 8   1 1 y(x) = x + 4 4 Exercise 9.

(

x2 if x ≤ c ax + b if x > c ( 2x if x ≤ c f 0 (x) = a if x > c f (x) =

a = 2c; b = −c2 Exercise 10.

( f (x) =

1 |x|

if |x| > c

a + bx

2

if |x| ≤ c

Note that c ≮ 0 since |x| ≤ c, for the second condition.

0

f (x) =

 1  − x 2 1  x2

 So b = −

2bx

if x > c if x < c if |x| ≤ c

1 3 . ,a= 3 2c 2c

Exercise 11.

( cos x f = a 0

Exercise 12. f (x) =



√  1−√2 1+ 2

=

1−A 1+A

f 00 = f

000

= =

√

1 −1/2 1 1 1 x = ; A00 = − x−3/2 = − 3 2 2A 4 4A −A0 (1 + A) − A0 (1 − A) −2A0 −1 √ = =√ (1 + A)2 (1 + A)2 x(1 + x)2 √ 1 3 x+1 0 2 0 √ (A (1 + A) + A(2)(1 + A)A ) = 3/2 (A(1 + A2 ))2 2x (1 + x)3  −1 0 2 1 A2 A (A (1 + A)3 ) − (2AA0 (1 + A)3 + 3A2 (1 + A)2 A0 )(3 + 2 (A2 (1 + A)3 )2 1  √ −3 A + 4 + 5A 3 (1 + 4 x + 5x) √ √ = − 4 A4 (1 + A4 ) 4 x(x + x)4 A=

f0 =

if x ≤ c if a > c

x A0 = a =

Exercise 13. 47

1  A)

P = ax3 + bx2 + cx + d P 0 = 3ax2 + 2bx + c

P 00 (0) = 2b = 10 =⇒ b = 5 P 0 (0) = c = −1

P 00 = 6ax + 2b

P (0) = d = −2

P (1) = a + 5 + −1 + −2 = a + 2 = −2 =⇒ a = −4 Exercise 14.

f g = 2,

f0 f0 g0 1 = 2 = 4 = 2f = , g = 4 g0 g g 2

(1) h0 =

f0 f 0 g − g0 f g0 f 1 15 = − = 4 − 2( ) = 2 g g g g 8 4

(2) k 0 = f 0 g + f g 0 = 4g 2 + f 2g = 64 + 4 = 68 (3) 1 limx→0 g 0 (x) g 0 (x) = = 0 x→0 f 0 (x) limx→0 f (x) 2 lim

Exercise 15.

(1) True, by definition of f 0 (a). (2) f (a) − f (a − h) f (a − h) − f (a) f (a − h) − f (a) = − lim = lim = f 0 (a) lim h→0 h→0 −h→0 h h −h True, by definition of f 0 (a). (3) lim

t→0

f (a + 2t) − f (a) f (a + 2t) − f (a) = 2 lim = 2f 0 (a) 2t→0 t 2t

False. (4) lim

t→0

f (a + 2t) − f (a) + f (a) − f (a + t) = 2t f (a + 2t) − f (a) 1 f (a + t) − f (a) + − lim = 2t→0 2t 2 t→0 t 1 1 f 0 (a) − f 0 (a) = f 0 (a) 2 2 lim

False. Exercise 16.

(1) (F + G)2 − (f + g)2 (f (x + h) + g(x + h))2 − (f (x) + g(x))2 = lim = h→0 h→0 h h 2F G − 2f g = D∗ f + D∗ g + lim h→0 h 2F G − 2f g (2(F G) − 2f G)(F + f ) (2f G − 2f g)(G + g) lim = lim + = h→0 h→0 h (F + f )h (g + G)h 2g F2 − f2 2f G2 − g 2 = lim lim + lim lim = h→0 F + h h→0 h→0 G + g h→0 h h g f = D∗ f + D∗ g f g

D∗ (f + g) = lim

48

(f (x + h) − g(x + h))2 − (f (x) − g(x))2 = h→0 h (F − G)2 − (f − g)2 = lim = h→0 h 2F G − 2f g = D∗ f + D∗ g + − lim h→0 h g ∗ f ∗ ∗ ∗ =D f +D g− D f + D g f g

D∗ (f − g) = lim

((f g)(x + h))2 − ((f g)(x))2 = h→0 h (f 2 (x + h))(g 2 (x + h)) − f 2 (x)g 2 (x + h) + (g 2 (x + h) − g 2 (x))f 2 (x) = = lim h→0 h = g2 D∗ f + f 2 D∗ g

D∗ (f g) = lim

D∗ (f /g) = lim

f 2 (x+h) g 2 (x+h)

f 2 (x) g 2 (x)

h

h→0 ∗

=

−

= lim

f 2 (x+h)−f 2 (x) g 2 (x+h)

f 2 (x) g 2 (x+h)

+

−

f 2 (x) g 2 (x)

h

h→0

=

D f f2 + 4 (−D∗ g) when g(x) 6= 0 2 g g

(2) (3) 4.12 Exercises - The chain rule for differentiating composite functions, Applications of the chain rule. Related rates and implicit differentiation. Exercise 1. −2 sin 2x − 2 cos x Exercise 2.

√ x 1+x2

Exercise 3. −2x cos x2 + 2x(x2 − 2) sin x2 + 2 sin x3 + 6x3 cos x3 Exercise 4.

f 0 = cos (cos2 x)(−2 cos x sin x) cos (sin2 x) + sin (cos2 x) sin (sin2 x)(2 sin x cos x) = = − sin 2x(cos (cos 2x)) Exercise 5.

f 0 = n sinn−1 x cos x cos nx + −n sin nx sinn x Exercise 6.

f 0 = cos (sin (sin x))(cos (sin x))(cos x) Exercise 7.

f0 = Exercise 8. f 0 =

1 2

sec2

x 2

+

2 sin x cos x sin x2 − 2x cos x2 sin2 x sin 2x sin x2 − 2x sin2 x cos x2 = 2 2 sin x sin2 x2 1 2

csc2

x 2

Exercise 9. f 0 = 2 sec2 x tan x + −2 csc2 x cot x Exercise 10. f 0 = Exercise 11. f 0 =

√

1 + x2 +

2

√x 1+x2

=

1+2x2 √ 1+x2

4 (4−x2 )3/2

Exercise 12.

f0 =

1 3



1 + x3 1 − x3

−2/3 

3x2 (2) (1 − x3 )2

 =

2x2 (1 − x3 )2

Exercise 13. This exercise is important. It shows a neat integration trick. 49



1 + x3 1 − x3

−2/3

! √ x − 1 + x2 √ = x − 1 + x2

1 1 √ √ f (x) = √ =√ 2 2 2 1 + x (x + 1 + x ) 1 + x (x + 1 + x2 ) √ x x − 1 + x2 √ =1− √ = 2 − 1+x 1 + x2 √ x2 1 + x2 − √1+x 2 1 0 f = = 2 1+x (1 + x2 )3/2 Exercise 14.

1 (x + 2

q

√

x+

  √ 1 1 x)−1/2 (1 + (x + x)−1/2 1 + √ ) 2 2 x

Exercise 15.

f 0 = (2 + x2 )1/2 (3 + x3 )1/3 + (1 + x)x(2 + x2 )−1/2 (3 + x2 )1/3 + (1 + x)(2 + x2 )1/2 (3 + x3 )−2/3 x2 Exercise 16. 0

f =



−1 1+


1 2 x

−1 x2



−1 1 g0 =  = 2 (x + 1)2 1 + f1



−1 f2



f0 =

f0 (f + 1)2

(x + 1)−2 1 g0 =  2 = (2x + 1)2 x x+1 + 1 Exercise 17. h0 = f 0 g 0

x 0 1 2 3

h f (2) = 0 f (0) = 1 f (3) = 2 f (1) = 3

h0 2(−5) = −10 5(1) = 5 4(1) = 4 −2(−6) = 12

k k0 g(1) = 0 1(5) = 5 g(3) = 1 −6(−2) = 12 g(0) = 2 −5(2) = −10 g(2) = 3 1(4) = 4

Exercise 18.

g(x) = xf (x2 ) g 0 (x) = f (x2 ) + x(2x)f 0 (x2 ) = f (x2 ) + 2x2 f 0 (x2 ) g 00 (x) = 2xf 0 (x2 ) + 4xf 0 (x2 ) + 2x2 (2x)f 00 (x2 ) = 6xf 0 (x2 ) + 4x3 f 00 (x2 ) x 0 1 2

g(x) 0 1 12

g 0 (x) 0 3 6 + 8(3) = 30

g 00 (x) 0 10 12(3) + 32(0) = 36

Exercise 19.

(1) g0 =

df (x2 ) 2x = 2xf 0 dx2

(2) g 0 = 2 sin x cos xf 0 − 2 cos x sin xf 0 = (sin 2x)(f 0 (sin2 x) − f 0 (cos2 x)) (3) g0 =

df (f (x)) 0 f d(f (x))

(4) g0 =

df (f (f (x))) d(f (f (x))) df d(f (f (x))) d(f (x)) dx 50

Exercise 20. V = s3 , s = s(t)

dV dt

= 3s2 ds dt . s = 5cm 75cm3 /sec s = 10cm 300cm3 /sec s = xcm 3x2 cm3 /sec

Exercise 21.

p 1 dx dl = x x2 + h2 dt l dt   dx 20 mi 3600sec l dl 10mi (−4mi/sec) = = = √ dt x dt 3 sec 1hr − 102 − 82 l=

Exercise 22.

l2 = x2 + s2 2l

dl dx dl x dx = 2x = dt dt dt l dt √ s dl  x= = 20 5 dt 2 √ dl (x = s) = 50 2 dt

Exercise 23.

x dx 3 36 dl = = 12 = mi/hr dt l dt 5 5 Exercise 24. Given the preliminary information

r 2 1 1 = = α, V = πr2 h = πα2 h3 h 5 3 3 (1) 1 πr2 2 (h y − hy 2 + y 3 ) h2 3 dV πr2 dy = 2 (h2 − 2hy + y 2 ) dt h dt     dy h2 1 dV 102 1 5 = 2 = 5= 2 2 2 2 dt πr h − 2hy + y dt π4 10 − 2(10)5 + 25 4π V =

(2) dV dh dh 1 dV 5 = πα2 h2 , = = dt dt dt πα2 h2 dt 4π Exercise 25.

r 3 = h 2 dV 2 2 dh = πα h dt dt α=

9 c − 1π (22 )4 = 36π =⇒ c = 36π + 1 4 Exercise 26. The constraint equation, using Pythagorean theorem on the geometry of a bottom hemisphere, is

r2 = R2 − (R − h)2 = 2Rh − h2 So then

dr dh = (R − h) dt dt Z dV 2 2 V = πr dh =⇒ = πr = π(2Rh − h2 ) dh dV =⇒ = π(2(10(5) − 25)) = 50π dh r

51

  dV dV dh dh dV 1 = , =⇒ = dt dh dt dt dt π(2Rh − h2 )     r dr dV 1 = R − h dt dt π(2Rh − h2 )   dr dV R−h = = dt dt rπ(2Rh − h2 )   =⇒ √ 10 − 5 1 = (5 3) = 3/2 15π π(2(10)5 − 25) Exercise 27. I suppose the area of the triangle is 0 at t = 0.


Now the point on vertex B moves up along the y axis according to y = 1 + 2t. y 27 = 8. r 1 36 A= (y − 1) y 2 7 ! r r dA 1 36 1 1 36 dy √ = y + (y − 1) = dt 2 y 2 y−1 7 dt   6 66 1 8 + 6 (2) = = 2 2(7) 7 Exercise 28. From the given information, h = 3r + 3. The volume formula is V =

πR2 3 H.

V = π/3r2 (3r + 3) = πr3 + πr2 dr dV = πr(3r + 2) dr dt With the given information, we get dr 1 = dt π(6)(20) Using this, we can plug this back in for the different case: dV = n = π(36)(110)/(120π) = 33 dt Exercise 29. dy dt

1 1 dy = 2x dx dt ; when x = 2 , y = 4 , dt = π (2) t = 6

(1)

dx dt

Exercise 30.

(1) 3x2 + 3y 2 y 0 = 0 =⇒ x2 + y 2 y 0 = 0 (2) 2x + 2yy 02 + y 2 y 00 = 0 =⇒ y 2 y 00 = −2(x + yy 02 )  4  xy + yx4 00 =⇒ y = −2 = −2xy −5 y6 Exercise 31.

√ − y 1 1 1 0 0 √ + √ y = 0y = √ < 0 2 x 2 y x Exercise 32.

r ±

12 − 3x2 4 52

So then

6x + 8yy 0 = 0 =⇒ y 0 =

−3x 4y

3 + 4(y 02 + yy 00 ) = 0   −9 3 1 00 02 = 3 y = − −y 4 y 4y Exercise 33.

sin xy + x cos2 xy(y + xy 0 ) + 4x = 0 y 0 x2 cos xy + xy cos xy + sin xy + 4x = 0 Exercise 34. y = x4 . y n = xm .

y n = xm , y 0 ny n−1 = mxm−1 ; y 0 =

m xm−1 mxm−1 = = ny n−1 n xm(1−1/n)

m m/n−1 x n 4.15 Exercises - Applications of differentiation to extreme values of functions, The mean-value theorem for derivatives. Let’s recap what was shown in the past two sections: y0 =

Theorem 13 (Theorem 4.3). Let f be defined on I. Assume f has a rel. extrema at an int. pt. c ∈ I. If ∃ f 0 (c), f 0 (c) = 0; the converse is not true. (c) Proof. Q(x) = f (x)−f if x 6= c, Q(c) = f 0 (c) x−c 0 ∃f (c), so Q(x) → Q(c) as x → c so Q is continuous at c. (c) If Q(c) > 0, f (x)−f > 0. For x − c ≷ 0, f (x) ≷ f (c), thus contradicting the rel. max or rel. min. (no neighborhood x−c about c exists for one!) (c) If Q(c) < 0, f (x)−f < 0. For x − c ≷ 0, f (x) ≶ f (c), thus contradicting the rel. max or rel. min. (no neighborhood x−c about c exists for one!) Converse is not true: e.g. saddle points. 

Theorem 14 (Rolle’s Theorem). Let f be cont. on [a, b], ∃f 0 (x) ∀x ∈ (a, b) and let f (a) = f (b) 0

then ∃ at least one c ∈ (a, b), such that f (c) = 0. Proof. Suppose f 0 (x) 6= 0 ∀x ∈ (a, b). By extreme value theorem, ∃ abs. max (min) M, m somewhere on [a, b]. M, m on endpoints a, b (Thm 4.3). F (a) = f (b), so m = M . f constant on [a, b]. Contradict f 0 (x) 6= 0



Theorem 15 (Mean-value theorem for Derivatives). Assume f is cont. everywhere on [a, b], ∃f 0 (x) ∀x ∈ (a, b). ∃ at least one c ∈ (a, b) such that (6)

f (b) − f (a) = f 0 (c)(b − a)

Proof. h(x) = f (x)(b − a) − x(f (b) − f (a)) h(a) = f (a)b − f (a)a − af (b) + af (a) h(b) = f (b)(b − a) − b(f (b) − f (a)) = bf (a) − af (b) = h(a) =⇒ ∃c ∈ (a, b), such that h0 (c) = 0 = f 0 (c)(b − a) − (f (b) − f (a))  Theorem 16 (Cauchy’s Mean-Value Formula). Let f, g cont. on [a, b], ∃f 0 , g 0 Then ∃ c ∈ (a, b). x (7)

f 0 (c)(g(b) − g(a)) = g 0 (c)(f (b) − f (a)) 53

∀x ∈ (a, b)

(note how it’s symmetrical)

Proof. h(x) = f (x)(g(b) − g(a)) − g(x)(f (b) − f (a)) h(a) = f (a)(g(b) − g(a)) − g(a)(f (b) − f (a)) = f (a)g(b) − g(a)f (b) h(b) = f (b)(g(b) − g(a)) − g(b)(f (b) − f (a)) =⇒ h0 (c) = f 0 (c)(g(b) − g(a)) − g 0 (c)(f (b) − f (a)) = 0

(by Rolle’s Thm.) 

Exercise 1. For any quadratic polynomial y = y(x) = Ax2 + Bx + C,

y(a) = Aa2 + Ba + C y(b) = Ab2 + Bb + C y(b) − y(a) A(b − a)(b + a) + B(b − a) = = A(b + a) + B b−a b−a y 0 = 2Ax + B   a+b = A(a + b) + B y0 2 Thus the chord joining a and b has the same slope as the tangent line at the midpt. Exercise 2. The contrapositive of a theorem is always true. So the contrapositive of Rolle’s Theorem is

If @ at least one c ∈ (a, b) s.t. f 0 (c) = 0, then f (a) 6= f (b). g 0 = 3x2 − 3 = 3(x2 − 1) =⇒ g 0 (±1) = 0 Suppose g(B) = 0, then ∀ x ∈ (−1, 1),

x 6= B,

B ∈ (−1, 1)

g(x) 6= g(B),

so g(x) 6= 0 for x 6= B

so only at most one B ∈ (−1, 1) s.t. g(B) = 0 Exercise 3. f (x) =

3−x2 2

if x ≤ 1,

f (x) =

1 x

if x ≥ 1.

(1) See sketch. (2) ( f (x) =

3−x2 2

1/x (

f 0 (x) =

if x ≤ 1 if x ≥ 1

f (1) = 1 = f (1) = 1/1

−x; f 0 (1) = −1 −1/x2 ; f 0 (1) = −1

for x ≤ 1 for x > 1

Then f (x) is cont. and diff. on [0, 2]. For 0 ≤ a < b ≤ 1 3−b2 2

−



3−a2 2



−(a + b) = = −c b−a 2 0 Note that −1 ≤ f ≤ 0 for 0 ≤ x ≤ 1 For 1 ≤ a < b ≤ 2

√ − a1 −1 −1 = = 2 =⇒ c = ab b−a ab c Note that −1 ≤ f 0 ≤ −1/4

1 b

For 0 ≤ a ≤ 1 ,

1≤b≤2 1 b

−



3−a2 2



2 − (3 − a2 )b −1 = −c or 2 b−a 2b(b − a) c depending upon if 0 ≤ c ≤ 1 or 1 ≤ c ≤ 2, respectively =

54

For instance, for a = 0, b = 2, then

f (b)−f (a) b−a

= −1/2, so c = 1/2 or c =

√

2

Exercise 4.

f (1) = 1 − 12/3 = 0 = f (−1) = 1 − ((−1)2 )3 = 0 −2 −1/3 f0 = x 6= 0 for |x| ≤ 1 3 This is possible since f is not differentiable at x = 0. Exercise 5. x2 = x sin x + cos x. g = xS + C − x2 . g 0 = S + xC − S − 2x = xC − 2x = x(C − 2). Since |C| ≤ 1 then

(C − 2) is negative for all x. Then for x ≷ 0, g 0 ≶ 0. Since g(0) = 1 and for x → ±∞, g → ∓∞, then we could conclude that g must become zero between 0 and ∞ and −∞ and 0. Exercise 6.

f (b) − f (a) = f 0 (c) b−a b=x+h b−a=h

=⇒ f (x + h) − f (x) = hf 0 (x + θh)

a=x x < x + θh < x + h 2

(1) f (x) = x ,

0

f = 2x. (x + h)2 − x2 = 2xh + h2 = h(2(x + θh)) 1 2x + h − x = θh =⇒ θ = 2 2

(2) f (x) = x3 ,

so then lim θ = h→0

1 2

f 0 = 3x2 .

! 3x2 + 3xh + h2 − x /h = θ (x + h) − x = 3x h + 3xh + h = h3(x + θh) =⇒ 3 q q  r 2 2 + xh + h2 − x 2 2 x x2 + xh + h3 + x 3x + 3xh + h x 3 q = θ= − = 2 3h2 h h x2 + xh + h3 + x r

3

3

2

2

3

2

x+

= x+

q

h 3

x2 + hx +

h2 3

1 2 Notice the trick of multiplying by the conjugate on top and bottom to get a way to evaluate the limit. =⇒ lim θ = h→0

Exercise 7. f (x) = (x − a1 )(x − a2 ) . . . (x − ar )g(x).

(1) a1 < a2 < · · · < ar . Since f (a1 ) = f (a2 ) = 0. f 0 (c) = 0 for c1 ∈ (a1 , a2 ). Consider that f (a2 ) = f (a3 ) = 0 as well as f 0 (c2 ) = 0 for c ∈ (a2 , a3 ). Indeed, since f (aj ) = f (aj+1 ) = 0, f 0 (c) = 0 for c ∈ (aj , aj+1 ). Thus, ∃r − 1 zero’s. f (k) has r − k zeros in [a, b]. f (k) = (x − a1 )(x − a2 ) . . . (x − ar−k )gk (x) Since f (a1 ) = f (a2 ) = 0, f (k+1) (c1 ) = 0 for c1 ∈ (a1 , a2 ). f (k) (aj ) = f (k) (aj+1 ) = 0,

f (k+1) (cj ) = 0 for cj ∈ (aj , aj+1 )

=⇒ f (k) (x) has at least r − k zeros in [a, b] We had shown the above by induction. (2) We can conclude that there’s at most r + k zeros for f (since f (k) has exactly r zeros, the intervals containing the r zeros are definite). 55

Exercise 8. Using the mean value theorem

(1) sin x − sin y sin x − sin y = | cos c| ≤ 1 = cos c =⇒ x−y x−y =⇒ | sin x − sin y| ≤ |x − y| (2) x ≥ y > 0. f (z) = z n is monotonically increasing for n ∈ Z. By mean-value theorem, xn − y n = ncn−1 for y < c < x x−y ny n−1 ≤

Since 0 < y < c < x; Exercise 9. Let g(x) =



f (b)−f (a) b−a



x+



xn −y n x−y

≤ nxn−1 .

bf (a)−af (b) b−a



. h(a) = h(c)

f −g =h

h(c) = h(b) 0

so ∃c1 ∈ (a, c), c2 ∈ (c, b) s. t. h (c1 ) = h0 (c2 ) = 0 by Rolle’s Thm. Let h0 = H since H(c1 ) = H(c2 ) = 0 and H is cont. diff. on (c1 , c2 ). then ∃c3 ∈ (c1 , c2 ) s.t. H 0 (c3 ) = h00 (c3 ) = 0 Now h00 = (f − g)00 = f 00 so f 00 (c3 ) = 0 We’ve shown one exists; that’s enough. Exercise 10. Assume f has a derivative everywhere on an open interval I.

g(x) = (1) g =



1 x−a



f−

1 x−a f (a).

f (x) − f (a) if x 6= a; x−a

f is cont. on (a, b] since ∃ f 0

g(a) = f 0 (a)

∀x ∈ (a, b).

1 x−a

is cont. on (a, b]. Then g is cont. on (a, b] (remember, you can add, subtract, multiply, and divide cont. functions to get cont. functions because the rules for taking limits allow so). (a) g is cont. at a since limx→a g = limx→a f (x)−f = f 0 (a). x−a By mean value theorem,   f (x) − f (a) = f 0 (c) = g(x) x−a

∀c ∈ (a, x)

∀x ∈ (a, b]

Then ∀c ∈ [a, b], f 0 (c) ranges from f 0 (a) to g(b) since f 0 (c) = g(x) so whatever g(x) ranges from and to, so does f 0 (c). (b) if x 6= b; h(b) = f 0 (b). (2) Let h(x) = f (x)−f x−b 1 h is cont. on [a, b) since x−b is cont., f (x) is cont. limx→b h = limx→b

f (x)−f (b) x−b

= f 0 (b) so h is cont. at b.

h is cont. on [a, b] → h takes all values from h(a) to f 0 (b) on [a, b] (by intermediate value theorem). By mean value theorem, h(x) =

f (b) − f (x) = f 0 (c2 ) b−x

for c2 ∈ (x, b)

So then f 0 ranges from h(a) to f 0 (b) just like h. h(a) = g(b). So then f 0 must range from f 0 (a) to f 0 (b) 56

∀ x ∈ [a, b]

4.19 Exercises - Applications of the mean-value theorem to geometric properties of functions, Second-derivative test for extrema, Curve sketching. Exercise 1. f (x) = x2 − 3x + 2

(1) (2) (3) (4)

f 0 (x) = 2x − 3 x0 = 32 . f 0 (x) ≷ 0 for x ≷ 23 f 00 = 2 > 0 for ∀ x ∈ R See sketch.

Exercise 2. f (x) = x3 − 4x

(1) (2) (3) (4)

f 0 = 3x2 − 4 xc = ± √23 f 0 ≷ 0 when |x| ≷ √23 f 00 = 6x f 00 ≷ 0 when x ≷ 0 See sketch.

Exercise 3. f (x) = (x − 1)2 (x + 2)

(1) (2) (3) (4)

f 0 = 3(x − 1)(x + 1) f 0 (x) = 0 when x = ±1 0 f ≷ when |x| ≷ 1 f 00 = 3(2x) = 6x f 00 ≷ 0 when x ≷ 0 See sketch.

Exercise 4. f (x) = x3 − 6x2 + 9x + 5

(1) f 0 = 3x2 − 12x + 9 = 3(x − 3)(x − 1) f 0 (x) = 0 when x = 3, 1 (2) f 0 (x) > 0 when x < 1, x > 3 00

(3) f = 6x − 12 = 6(x − 2) (4) See sketch.

f 0 (x) < 0 when 1 < x < 3 f ≷ 0 when x ≷ 2 00

Exercise 5. f (x) = 2 + (x − 1)4

(1) (2) (3) (4)

f 0 (x) = 4(x − 1)3 . f 0 (0) = 0 when x = 1 f 0 (x) ≷ 0 when |x| ≷ 1 f 00 (x) = 12(x − 1)2 > 0 ∀ x 6= 1 See sketch.

Exercise 6. f (x) = 1/x2

(1) (2) (3) (4)

0 f 0 = −2 x3 f (x) = 0 for no x 0 f ≷ 0 when x ≶ 0 f 00 = x64 > 0 ∀x 6= 0 See sketch.

Exercise 7. f (x) = x + 1/x2

(1) f 0 = 1 + (2)

−2 x3

f 0 (x) = 0 = 1 −

2 x3

=⇒ xc = 21/3 f 0 (x) > 0 when x < 0, 0 < x < 21/3 f 0 (x) < 0 when x > 21/3

(3) f 00 = x64 > 0 (4) See sketch. Exercise 8. f (x) =

∀x 6= 0

1 (x−1)(x−3)

(1) f 0 = (x−1)−1 2 (x−3)2 ((x − 3) + x − 1) = 0 f (x) = 0 when x = 2 (2) f 0 ≷ 0 when x ≶ 2

(−2)(x−2) (x−1)2 (x−3)2

57

(3) (x − 1)2 (x − 3)2 − (x − 2)(2(x − 1)(x − 3)2 + 2(x − 3)(x − 1)2 ) f = (−2) (x − 1)4 (x − 3)4  2 13  x − 4x + 3 = (6) (x − 1)3 (x − 3)3 13 x2 − 4x + > 0 since 144 − 4(−3)(−13) = 144 + 12(−13) < 0 so 3 f 00 > 0 if x > 3, x < 1 00



f 00 < 0 if 1 < x < 3 (4) See sketch. Exercise 9. f (x) = x/(1 + x2 )

(1) f0 =

1 − x2 (1 + x2 ) − x(2x) = 2 2 (1 + x ) (1 + x2 )2

f 0 (x) = 0 when x = ±1 (2) f 0 ≷ 0 when |x| ≶ 1 (3) f 00 =

−2x(1 + x2 )2 − 2(1 + x2 )(2x)(1 − x2 ) 2x(x2 − 3) = (1 + x2 )4 (1 + x2 )3 √ f 00 > 0 when x > 3 √ f 00 < 0 when 0 < x < 3 √ f 00 > 0 when − 3 < x < 0 √ f 00 < 0 when x < − 3

(4) See sketch. Exercise 10. f (x) = (x2 − 4)/(x2 − 9)

(1) f0 = f 0 (0) = 0 (2) f 0 ≷ 0 when x ≶ 0, (3)

−10x 2x(x2 − 9) − (x2 − 4)(2x) = 2 (x2 − 9)2 (x − 9)2

x 6= ±3 f 00 = (−10)



(x2 − 9)2 − 2(x2 − 9)(2x)x (x2 − 9)2

 = (30)

(x2 + 3) (x2 − 9)3

f 00 ≷ 0 when |x| ≷ 3 (4) See the sketch. Exercise 11. f (x) = sin2 x

(1) f 0 = sin 2x So then f 0 = 0 when x = (2)

π 2n

π 2 f 0 > 0 when π πn < x < + πn 2 π 0 f < 0 when + πn < x < π(n + 1) 2 0<x<

(3)

−π π + πn < x < + πn 4 4 π 3π 00 f < 0 when + πn < x < + πn 4 4

f 00 > 0 when f 00 = 2 cos 2x (4) See sketch. 58

 =

Exercise 12. f (x) = x − sin x

(1) f 0 = 1 − cos x (2) f 0 > 0 if x 6= 2πn (3)

f 0 = 0 when x = 2πn f 00 > 0 when 2πn < x < 2πn + π

f 00 = sin x

f 00 < 0 when 2πn + π < x < 2π(n + 1)

(4) See sketch. Exercise 13. f (x) = x + cos x

(1) f 0 = 1 − sin x x= (2) f 0 > 0 if x 6= π2 + 2πn (3)

π 2

f 0 (x) = 0

+ 2πn

π −π + 2πn < x < + 2πn 2 2 π 3π 00 f < 0 when + 2πn < x < + 2πn 2 2 f 00 > 0 when

00

f = − cos x (4) See sketch. Exercise 14. f (x) =

1 2 6x

+

1 12

cos 2x

2x f 0 (0) = 0 (1) f 0 = 31 x + − sin 6 0 (2) f ≷ 0 when x ≷ 0 2x (3) f 00 = 13 − cos32x = 1−cos 3 00 x = πn for f = 0. Otherwise f 00 > 0 for x 6= πn (4) See sketch.

4.21 Exercises - Worked examples of extremum problems. Exercise 1.

A = xy P = 2(x + y) = 2(x +

A ) x

A )=0 2 x√ x= A √ 4A P 00 = 3 > 0 for x > 0 so x = A minimizes P x P 0 = 2(1 −

Exercise 2.

A = xy

L = 2x + y

dA L = L − 4x = 0 when x = dx 4 L 00 A = −4 so x = maximizes A 4

A = x(L − 2x) = Lx − 2x2 =⇒

y=

L 2

Exercise 3.

A = xy

√ dL −A A = 2 + 2 = 0 when x = √ dx x 2 r 2A A L00 = 3 > 0 for x = so x minimizes L x 2

L = 2x + y = 2x +

A x

Exercise 4. f = x2 + y 2 = x2 + (S − x)2

f 0 = 2x + 2(S − x)(−1) = −2S + 4x f 00 = 4 > 0 so x = S2 minimizes f

=⇒ x =

S 2

Exercise 5. x2 + y 2 = R > 0 59

y=

√ √ 2 A

f =x+y f 0 = 1 + y0 = 0 = 1 + f 00 = y 00 =

Note that

−x =0 y

=⇒ y = x

−1 − y 02 for y > 0, f 00 < 0 so that f is max. when y = x y 2x + 2yy 0 = 0 −x y0 = 1 + y 02 + yy 00 = 0 =⇒ yy 00 = −1 − y 02 y x + yy 0 = 0

Exercise 6.

l2 = (L − x)2 + x2 = L2 − 2Lx + 2x2 = A dA L = −2L + 4x = 0 =⇒ x = dx 2 d2 A = 4 > 0 =⇒ A minimized dx2 √ L L 2 l(x = ) = 2 2 Exercise 7.

p L2 − x2 )2 = A p −x L A0 = 2(x + L2 − x2 )(1 + √ ) = 0 when L2 − x2 = x2 or x = √ 2 L2 − x2 r L L2 2L so then the side of the circumscribing and area-maximized square is √ + L2 − =√ 2 2 2 (x +

Exercise 8.

p p A = (2x)(2 R2 − x2 ) = 4x R2 − x2  2  p −x2 R − 2x2 R A0 = 4( R2 − x2 + √ )=4 √ =⇒ x = √ 2 R 2 − x2 R 2 − x2 R R since A0 ≷ 0 when x ≶ √ , so A is maximized at x = √ 2 2 2x =

2R √ ; 2

√ 2 R 2 − x2 =

2R √ 2

so then the rectangle that has maximum size is a square.

Exercise 9. Prove that among all rectangles of a given area, the square has the smallest circumscribed circle.

√ √ A0 = (2x)(2 r2 − x2 ) = 4x r2 − x2 (fix the area to be A0 )
A2 A0 2 = r2 − x2 =⇒ x4 − x2 r2 + 160 = 0 4x dr − 4x3 dx p dr r r = 0 (for extrema) =⇒ x = √ and r2 − x2 = √ dx 2 2 =⇒ 0 = 2xr2 + x2 2r

We could argue that we had found a minimum because at the “infinity” boundaries, the circumscribing circle would be infinitely large. Exercise 10. Given a sphere of radius R, find the radius r and altitude h of the right circular cylinder with the largest lateral 60

surface area 2πrh that can be inscribed in the sphere. R2 =

 2 h + r2 2

p 4(R2 − r2 ) = 4πr R2 − r2   2  p R − 2r2 dA −r2 R = 4π √ =⇒ r = √ R2 − r 2 + √ = 4π dr 2 R2 − r 2 R2 − r 2 √ =⇒ h = 2R A = 2πrh = 2πr

p

Exercise 11. Among all right circular cylinders of given lateral surface area, prove that the smallest circumscribed sphere has

radius

√

2 times that of the cylinder. A0 = 2πRH

(A0 is the total lateral area of the cylinder) 2  2 H A0 A20 2 2 2 r =R + =R + = R2 + 2 4πR 16π 2 R2 √   2 A0 −2 dr A0 H dr = 2R + =⇒ = 0 =⇒ R = √ =⇒ =R 2r 2 3 dR 16π R dR 2 2 π √ r2 = R2 + R2 = 2R2 =⇒ r = 2R 

Exercise 12. Given a right circular cone with radius R and altitude H. Find the radius and altitude of the right circular cylinder

of largest lateral surface area that can be inscribed in the cone. h H R−r = R = α is the constraint (look, directly at the side, at the similar triangles formed) A = 2πrh = 2πrα(R − r) = 2πα(rR − r2 ) R dA = 2πα(R − 2r) = 0 =⇒ r = ; dr 2

h=

H 2

A00 = 2πα(R − 2r) since

dA R ≷ 0 when r ≶ , dr 2

r=

R maximizes lateral surface area 2

Exercise 13. Find the dimensions of the right circular cylinder of maximum volume that can be inscribed in a right circular

cone of radius R and altitude H. h Constraint: R−r =H R =α V = πr2 h = πr2 α(R − r) = πr2 α(R − r) = πα(Rr2 − r3 ) 2R dV = πα(2Rr − 3r2 ) = rπα(2R − 3r) r= dr 3 dV 2R 2R since ≷ 0 when r ≶ , r= maximizes volume dr 3 3 1 h= H 3 Exercise 14. Given a sphere of radius R. Compute, in terms of R, the radius r and the altitude h of the right circular cone of

maximum volume that can be inscribed in this sphere.  p πr2  R + R2 − r 2 3 √   2 p dV π r (−r) π (2R R2 − r2 + 2R2 − 3r2 ) 2 2 √ = 2rR + 2r R − r + √ = r =0 dr 3 3 R2 − r 2 R2 − r 2 √ 2 2R 4R =⇒ r = ; h= 3 3 V =

61

Considering the geometric or physical constraints, since limV →∞ V = limh→∞ V = 0, so then r = V.

√ 2 2R 3

must maximize

Exercise 15. Find the rectangle of largest area that can be inscribed in a semicircle, the lower base being on the diameter.

A= A0 =

p

R 2 − x2 + √

p

R 2 − x2 x

−x2 R = 0 =⇒ x = √ ; 2 2 2 R −x

R h= √ 2

Exercise 16. Find the trapezoid of largest area that can be inscribed in a semicircle, the lower base being on the diameter.

p 1 h(2 R2 − h2 + 2R) 2   dA p 2 −h = R − h2 + R + h √ dh R2 − h2 √ 3R dA = 0 =⇒ h = dh 2 r √ p 5 3R2 3 R 2 2 =⇒ A = R − h = 2 R2 − R2 = 2 = R 8 4 2 A=

Exercise 17. An open box is made from a rectangular piece of material by removing equal squares at each corner and turning

up the sides. Find the dimensions of the box of largest volume that can be made in this manner if the material has sides (a) 10 and 10; (b) 12 and 18 (1) (x − 2r)(Y − 2r)r = (xy − 2rx − 2ry + 4r2 )r = xyr − 2r2 x − 2r2 y + 4r3 = V dV = xy − 4rx − 4ry + 12r2 = 0 dr p p 4(x + y) ± 16(x + y)2 − 4(12)xy (x + y) ± x2 + y 2 − xy =⇒ r = = 2(12) 6 2 d V = −4x − 4y + 24r = −4(x + y) + 24r dr2 We can plug in our expression for r into the second derivative of V , the volume of the box, to find out that we want to pick the “negative” root from r, in order to maximize the box volume. 20 Then for x = 10; y = 10, we have r = 53 , so that the box dimensions are 53 × 20 3 × 3 . (2) 12 and 18√ √ √ =⇒ 5 − 7 × 2 + 2 7 × 8 + 2 7 Exercise 18. If a and b are the legs of a right triangle whose hypotenuse is 1, find the largest value of 2a + b.

p  a 2 2 L = 2a + b = 2a + 1 − a2 L0 = 0 =⇒ = 1 − a2 =⇒ a = √ 2 5 −a L0 = 2 + √ 2 1−a √ !   −a 1 − a2 − √1−a a 2 1 2 00 L = (−1) = (−1) < 0 (so a = √ maximizes L ) 2 2 3/2 1−a (1 − a ) 5 Exercise 19. 2 +

x2 600

gallons per hour. l0 = 300 mi x = constant speed. 62

l0 x

= time spent. K = gas cost = 0.30.

C = gas cost + driver labor cost = l0

2K x

+

Kx 600

+

D x




r   dC K −2K D 2K + D √ dC dx =0 + = l0 − 2 = 0 −−−−→ x = 10 6 2 dx x 600 x K      −2 2(2K + D) d2 C = l (−2K − D) = l >0 0 0 dx2 x3 x3 r 2K + D √ Thus, C is minimized if x = 10 6 K √ √ √ √ ! √ √ √ 2K + D K K 2K + D10 6 √ =⇒ Cmin = (300) = 3 2 6 + 10D + 600 10 6 Remember that there is a speed limit of 60 mi/hr. √ √ (1) D = 0, x = 20√3 C = 6 √3 ≈ 10.39 (2) D = 1, x = 40 2 C = 12 2 ≈ 16.97 (3) D = 2, x = 60 (because of the speed limit) (4) D = 3, (5) D = 4, Exercise 20. y =

x = 60 x = 60 x x2 +1

C = 300



2K 60

+

K(60) 600

+

D 60



= 5(2.4 + D) = 22.00

C = 27.00 C = 32.00

Suppose the rectangle starts at x0 on the x axis. Then its y coordinate intersecting the curve, and thus

the height of rectangle, must be y0 =

x0 x20 +1

s 1 1 ± −1 =⇒ x0 = 2y0 (2y0 )2 s 1 x2 − x1 = −4 y02 where x2 − x1 is going to be the base of the rectangle. The volume of the cylinder, V , which is obtained from revolving the rectangle about the x axis, is going to be s

! q 1 V = πy02 (x2 − x1 ) = πy02 − 4 = πy0 1 − 4y02 2 y0 ! ! q y0 1 − 8y02 1 dV 2 (−8y0 ) = π p =⇒ y0 = √ =π 1 − 4y0 + p 2 2 dy0 2 2 2 1 − 4y0 1 − 4y0 We could argue that V is maximized, since the “infinite” boundaries would yield a volume of 0 (imagine stretching and squeezing the rectangle inside the curve). π Then Vmax = π 18 2 = 4 Exercise 21. Draw a good diagram. Note how the right triangle that you folded is now reflected backwards , so that this

triangle’s right angle is on the left-hand side now. The constraint is that the crease touches the left edge. w0 = l sin α + l sin α cos (2α) = l sin α(1 + cos (2α)) = = 2l sin α cos2 α Note that we will obtain a minimum crease because by considering the “physical infinite” boundary, we could make a big crease along the vertical half of the paper or the horizontal half of the paper. 63

So, isolating l, the length of the crease, and then taking the derivative, w0 w0 = csc (α) sec2 (α) 2 sin (α) cos2 α 2
− cot α csc α sec2 α + csc α2 sec α sec α tan α =           1 S −C 1 1 1 w0 −1 2 + 2 = + S S C2 S C2 C 2 S2C C3 l=

dl w0 = dα 2 w0 = 2

dl 1 1 dα =0 −− −−→ sin α = √ or tan α = √ 3 2

where α is the angle of the crease. The corresponding minimum length of the crease will be l=

w0 2

1 √1 2 33

=

√ 9 3 2

Exercise 22.

(1) Consider the center of the circle O, the apex of the isosceles triangle that makes an angle 2α, A, and one of its other vertices, B. Draw a line segment from O to B and simply consider the two triangles making up one half of the isosceles triangle. Find all the angles. Angle AOB is π − 2α by the geometry or i.e. inspection of the figure. The complement of that angle is 2α. Beforehand, we can get the length of the isosceles triangle leg from the law of cosines. cos (π − 2α) = − cos (2α) 2

2

2

2

s = R + R − 2R cos (π − 2α) + 2R2 (1 + cos (2α)) = 2R2 (2 cos2 α) = 4R2 cos2 α s = 2R cos α The constraint equation is P = 4R cos α + 2R sin (2α)

(8) So then

P 0 = 4R(− sin α) + 4R cos (2α) = 0 =⇒ cos 2α = sin α q − 21 ± 14 − 4(1)(− 12 ) 1 sin α = = >0 2(1) 2 √ =⇒ P = 3 3R √ P = 3 3 is a max because Look at the “boundary conditions” imposed on P by the physical-geometry. α = 0, triangle is completely flattened, α = π, triangle “completely disappears.” (2) I had originally thought to Reuse the constraint equation, Eqn. ( ??). This is wrong! Think about the problem directly and for what it actually is; less wishful thinking. Consider a fixed perimeter L and imagine L to be a string that can be stretched into an isoceles triangle. A “trivial” isoceles triangle is a collapsed triangle with two sides of length L/2 only. Then the radius of the disk needs to be L/4.

Consider a general isosceles triangle with 2α as the vertex angle and isosceles sides of h. The perimeter for this triangle, P , is then P = 2h + 2h sin α = 2h(1 + sin α) P =⇒ h = 2(1 + sin α) h = R cos α 2 64

We could try to extremize this equation. dR (4 cos α + 2 sin (2α)) + R(−4 sin α + 4 cos (2α)) = 0 dα √ dR 1 2 = 0 =⇒ cos (2α) = sin α =⇒ sin α = √ cos α = √ dα 3 3 3P R= √ √ 4 2( 3 + 1) However, this is the minimized R, minimized radius for the smallest circle fitting a particular isosceles triangle of a fixed perimeter. We want to smallest circle with a radius big enough to fit all the possible triangles. Thus R = L4 Exercise 23. The constraint equation on perimeter is

 P = 2h + W + π

W 2

 = 2h + W (1 +

π ) 2

Then intensity function, “normalized” is given by π I = Wh + 2



W 2

2   3π 2 1 Wp W2 − − W = 2 2 2 16

So then P 3π P dI = −W − W = 0 =⇒ W = dW 2 8 2 + 3π 4 The height of the rectangle is P− h=

P 2+ 3π 4

(1 + π2 )

2

 =P

4+π 16 + 6π



Exercise 24. A log 12 feet long has the shape of a frustum of a right circular cone with diameters 4 feet and (4 + h) feet at its

ends, where h ≥ 0. Determine, as a function of h, the volume of the largest right circular cylinder that can be cut from the log, if its axis coincides with that of the log. Remember to label your diagram carefully.   xh h h/2 − 2l − y y h/2 0 2 = = =⇒ x l0 − x l0 l0 − x xh 2 2 V = π(H + y) (l0 − x) = π(H + ) (l0 − x) 2l0 Note that V (x = 0) = πH 2 l0 = π4(12) dV hx h xh 2 = π(2(H + ) (l0 − x) + (H + ) (−1)) = dV (h − H)2l0 dx 2l0 2l0 2l0 dx =0 −− −−→ x = hx 3xh 3h = π(H + )(h − H − ) 2l0 2l0   2 (h − H)2l0 (h − H)2l0 h (h − H)2l0 (h + 2H)3 V (x = ) = π l0 − H+ = πl0 3h 3h 2l0 3h 27h where H = 2, l0 = 12 Exercise 25.

S=

n X

n

X dS = 2(x − ak ) = 0 dx k=1 k=1 Pn n X ak =⇒ nx = ak =⇒ x = k=1 n (x − ak )2 =⇒

k=1

Since limx→±∞ S = +∞, x =

Pn

k=1

n

ak

minimizes S.

Exercise 26. Hint: draw a picture . Then observe that for f (x) ≥ 24, A must be greater than 0 (we’ll show that explicitly

soon) and that f must have a minimum somewhere. 65

If A < 0, then consider f (x) =

5x7 +A x5 .

Consider x = f (x =

−A1/7 61/7

> 0.

A −A1/7 = 5 <0 1/7 6x 6

Thus, A > 0. df = 10x − 5Ax−6 = 5 dx  1/7 A x= 2 d2 f = 10 + 30Ax−7 = 10 + 3 − A dx2 Thus x =


A 1/7 2



2x7 − A x6

 =0

  2 = 70 > 0 A

minimizes f for A > 0.  1/7 5 A f (x = )= 2  =⇒ A = 2

A 2 +A
A 5/7 2

24 7




= 24

7/2

3

Exercise 27. Consider f (x) = − x3 + t2 x over 0 ≤ x ≤ 1.

1 1 f (0) = 0, f (1) = − + t2 =⇒ f (1) ≷ 0 if t2 ≷ 3 3 f 0 (x) = −x2 + t2 = 0 =⇒ x2 = t2 but x ≥ 0, so x = |t| 2 1 f (x2 = t2 ) = − t2 (x) + t2 x = t2 x > 0 for 1 ≥ x ≥ 0 3 3 So the minimum isn’t in the interior of [0, 1]. It’s on the end points. m(t) = 0 for |t| >

1 −1 1 m(t) = + t2 for |t| < 3 3 3

Exercise 28.

(1) E(x, t) =

|t − x| x

|t − x| as x = a → x = b x ( t−x if t ≥ x |t − x| x = x−t x if t < x x   ( t − 2 if t ≥ x d |t − x| = tx dt x if t < x x2

M (t) = max

Now t, x ≥ a > 0 (this is an important, given, fact ). So x = t should be a relative minimum. So the maximum occurs at either endpoints t−a b−t = E(a, t), = E(b, t) a b By monotonicity on [a, t) , (t, b], and having shown the relative minimum of at x = a or x = b, depending upon the relationship E(a, t) ≷ E(b, t). 66

|t−x| x

at x = t, the maximum occurs

(2) ( M (t) =

Since

dM dt

t−a a b−t b

≷ 0 when t ≷

if if

2ab a+b ,

t−a a b−t b

> >

b−t b t−a a

i.e. t

b+a ab




M is minimized for t =

> 2 dM dt

(1 =

a −1 b

ab if t > 2a+b 

if t <

ab a+b

2

2ab a+b

4.23 Exercises - Partial Derivatives. Exercise 8. f (x, y) = √

x . x2 +y 2

fx = p

1

+

x2 + y 2 −xy fy = 2 (x + y 2 )3/2

fxy

y2 −x2 = (x2 + y 2 )3/2 (x2 + y 2 )3/2

fyy

(x2 + y 2 )3/2 − x 23 (x2 + y 2 )1/2 (2x) = (−y) (x2 + y 2 )3   −2x2 + y 2 = (−y) (x2 + y 2 )5/2

−3y 2 x (x2 + y 2 )5/2  2  x − 2y 2 = (−x) (x2 + y 2 )5/2

fxx =

! = fyx =

(x2

2y −3y 2 y (−y)(−2x2 + y 2 ) + 2 = 2 3/2 2 5/2 +y ) (x + y ) (x2 + y 2 )5/2

Exercise 9.

(1) z = (x − 2y)2

√ zx = 2(x − 2y) = 2 z

√ zy = 2(x − 2y)(−2) = −4 z

√ x(2z) − 4zy = (x − 2y)2 z = 2z

(2) z = (x4 + y 4 )1/2 1 zx = 2x3 z 2y 3 zy = z

√ x(2z) − 4zy = (x − 2y)2 z = 2z

Exercise 10.

f=

xy y −4x2 y y 3 − 3x2 y , f = + = x (x2 + y 2 )2 (x2 + y 2 )2 (x2 + y 2 )3 (x2 + y 2 )3

So −6xy(x2 + y 2 )3 − 3(x2 + y 2 )2 (2x)(y 3 − 3x2 y) = (x2 + y 2 )6 12xy(x2 − y 2 ) = (x2 + y 2 )4

fxx =

By label symmetry, fxx + fyy =

12xy(x2 − y 2 ) 12yx(y 2 − x2 ) + =0 (x2 + y 2 )4 (x2 + y 2 )4 67

5.5 Exercises - The derivative of an indefinite integral. The first fundamental theorem of calculus, The zero-derivative theorem, Primitive functions and the second fundamental theorem of calculus, Properties of a function deduced from properties of its derivatve. Review the fundamental theorems of calculus, Thm. 5.1 and Thm. 5.3. Note the differences between the two. Theorem 17 (First fundamental theorem of calculus). Let f be integrable on [a, x] ∀x ∈ [a, b] Let c ∈ [a, b] and Z (9) A(x) =

x

if a ≤ x ≤ b

f (t)dt

c

Then ∃ A0 (x) ∀x ∈ (a, b) where f is continuous at x and A0 (x) = f (x)

(10)

Theorem 18 (Second fundamental theorem of calculus). Assume f continuous on open interval I Let P be any primitive of f on I, i.e. P 0 = f ∀x ∈ I Then ∀c, x ∈ I Z

x

P (x) = P (c) +

(11)

f (t)dt c

√ Exercise 6.

Rb a

f=

√

q

2 23 x3/2 +

1 2 3/2 2 3x

=

√

2x3/2

2(b3/2 − a3/2 )

Exercise 7. f = x3/2 − 3x1/2 + 72 x−1/2 ;

P = 25 x5/2 − 2x3/2 + 7x1/2 Rb f = 25 (b5/2 − a5/2 ) − 2(b3/2 − a3/2 ) + 7(b1/2 − a1/2 ) a Exercise 8. P =

3 4/3 2x

− 32 x2/3 ;

Exercise 9. P = −3 cos x + Exercise 10. P =

3 7/3 7x

Exercise 11. f 0 (x) =

x>0

x6 3

− 5 sin x

1 x

f= f0 =

∞ X k=−∞ ∞ X

ak xk kak xk−1 =

k=−∞

1 x

Comparing terms, only k = 0 would work, but the coefficient is unequivocally 0 Exercise 12.

Z

(R x

x

R0x 0

|t|dt = 0

tdt −tdt

if x ≥ 0 = if x < 0

(

1 2 2x −1 2 2 x

if x ≥ 0 1 = x|x| 2 if x < 0

Exercise 13.

Z

x 2

(R x

(t + |t|) dt = 0

(2t)2 dt 0 if x < 0 0

if x ≥ 0

Exercise 14. Using 1st. fund. thm. of calc. 68

( =

4 3 3x

0

if x ≥ 0 2x2 = (x + |x|) 3 if x < 0

x

Z

f (t)dt = A(x) − A(0)

f 0 = 2 + 4 cos 2x + −4x sin 2x − 2 cos 2x

0 0

A (x) = f (x)

= 2 + 2 cos 2x − 4x sin 2x

=⇒ 2x + sin 2x + 2x cos 2x + − sin 2x π π = f 4 2 π 0 f =2−π 4 Exercise 15.

Rx c

f (t)dt = cos x −

1 2

f (x) = − sin x

c = − π6 .

Exercise 16. Suppose f (x) = sin x − 1 and c = 0.

Z

x

 t sin t − t =

0

 x 1 1 −t cos t + sin t − t2 = sin x − x cos x − x2 2 2 0

So c = 0. Exercise 17. For f (x) = −x2 f (x) + 2x1 5 + 2x1 7 (found by taking the derivative of

Rx 0

f =

R1 x

t2 f +

x16 8

+

x18 9

Suppose that f = 2x15 . x18 1 x16 x18 x16 =− +− + + +C 8 9 9 8 9 1 =⇒ C = 9 Rx t 2 Exercise 18. By plugging in x = 0 into the defined f (x), f (x) = 3 + 0 1+sin 2+t2 dt, we get for p(x) = a + bx + cx , =⇒

a=3 Continuing on, 1 + sin x ; 2 + x2 1 f 0 (0) = = b 2 f0 =

(cos x)(2 + x2 ) − 2x(1 + sin x) (2 + x2 )2 1 1 f 00 (0) = + 2c; c = 2 4

f 00 =

Exercise 19.

Z Z 1 x 1 x 2 (x − t)2 g(t)dt = (x − 2xt + t2 )g(t)dt = 2 0 2 0  Z x  Z x Z x 1 = x2 g − 2x tg + t2 g 2 0 0 0 Z x Z x 2 x 1 f0 = x g + g(x) − tg − x(xg(x)) + x2 g(x) = 2 2 0 Z0 x Z x =x g− tg 0 Z x0 Z x 00 f = g + xg − xg = g f 00 (1) = 2

f (x) =

0

f

000

=g

0 000

f (1) = 5

Exercise 20.

Rx

(1 + t2 )−3 dt)0 = (1 + x2 )−3 R0x2 2x (2) ( 0 (1 + t2 )−3 dt)0 = (1 + x4 )−3 (2x) = (1+x 4 )3 R x2 2 −3 0 4 −3 (3) ( x3 (1 + t ) dt) = (1 + x ) (2x) − (1 + x6 )−3 (3x2 ) = (1) (

69

2x (1+x4 )3

−

3x2 (1+x6 )3

+ C,)

Exercise 21.

Z

f 0 (x) =

x2

x3

!0     t6 x12 x1 8 dt = (2x) − 3x2 1 + t4 1 + x8 1 + x12

Exercise 22.

(1) f (x) = 2x(1 + x) + x2 = 2x + 3x2

(2)

d dx

f (2) = 16  b(x) f (t)dt = f (b)b0 − f (a)a0 a(x)

R

2x + 3x2 = f (x2 )(2x) 3x f (x2 ) = (1 + ) 2 √ 3 2 f (2) = 1 + 2 (3)

R f (x) 0

t2 dt = x2 (1 + x) 2x + 3x2 = (f (x))2 f 0 (x) =⇒ f 3 (2) = 3(4)(3) = 9(4) = 36 f (2) = 361/3

(4) d dx

Z

x2 (1+x)

! f (t)dt

= 1 = f (x2 (1 + x))(2x(1 + x) + x2 )

0

x=1

(f (2))(5) = 1 =⇒ f (2) =

1 5

Exercise 23.

a3 − 2a cos a + (2 − a2 ) sin a =

Z

a

f 2 (t)dt

0

3x2 − 2 cos x + 2x sin x + −2x sin x + (2 − x2 ) cos x = 3x2 − x2 cos x = f 2 (x) √ f (x) = x 3 − cos x √ f (a) = a 3 − cos a Exercise 24. f (t) =

t2 2

+ 2t sin t

(1) f 0 = 2t + 2 sin t + 2t cos t

f 0 (π) = 2π − 2π = 0

(2) f 00 = 2 + 2 cos t + 2 cos t + −2t sin t = 2 + 4 cos t − 2t sin t π f 00 =2−π 2
(3) f 00 3π 2
= 0 2 (4) f 5π = 25π 2 8 + 5π 2 (5) f (π) = π2 Exercise 25. 70

(1) df 1 + 2 sin πt cos πt = v(t) = dt 1 + t2 2π(cos (2πt))(1 + t2 ) − 2t(sin (2πt)) a(t) = (1 + t2 )2 4π a(t = 2) = a(t = 1) = =π 4 (2) v(t = 1) 21 (3) v(t) = π(t − 1) + 12 ; t > 1 (4)  2  t Z t Z t 1 πt2 πt 1 t π 1 π(t − 1) + = v(t)dt = f (t) − f (1) = − πt + t = + −πt + + − 2 2 2 2 2 2 2 1 1 1 Exercise 26.

(1) f 00 (x) > 0 ∀x

f 0 (0) = 1; f 0 (1) = 0

1

Z

f 00 (t)dt = f 0 (1) − f 0 (0) = 0 − 1 < 0

0

Thus, it’s impossible, since f 00 (x) > 0, so

R1 0

f 00 (t)dt > 0

(2) 1

Z



3−

0

(3) f 00 (0) > 0 ∀x

 πx  πx  1 π sin dx = 3x + cos =3−1=2 2 2 2 0 3x2 2 πx f (x) = + sin +C 2 π 2

f 0 (0) = 1; f (x) ≤ 100 ∀x > 0 Z b f 00 (t)dt = f 0 (b) − f 0 (a); a

Z

k

f 0 (t)dt = f (k) − f (c)

c

Z

b

f 00 = f 0 (b) − f 0 (0) = f 0 (b) − 1 ≷ 0 if b ≷ 0

0

Z

k

(f 0 (b) − 1)db = f (k) − f (c) − (k − c) > 0 if k > c > 0

c

f (k) − f (c) > k − c

f (x) ≤ 100 is untrue for all x > 0

f (k) − f (0) > k − 0 f (100) − f (0) > 100 (4) f 00 (x) = ex > 0 Exercise 27. f 00 (t) ≥ 6.

f 0 (x) = ex ; b − a = 21 . Z a

b

f 0 (0) = 1

f (x) = ex

∀x < 0, ex < 1

f 0 (0) = 0

1 f 00 = f 0 (b) − f 0 (a) ≥ 6(b − a) = 3 since b − a = 2 Z a f 00 = f 0 (a) − f 0 (0) = f 0 (a) ≥ 6(a − 0) = 6a 0

If a =

1 0 , f (1/2) ≥ 3 2

Then by intermediate value theorem, with f being continuous and f 0 (0) = 0, f 0 (1/2) ≥ 3, f 0 must take on the value of 3 somewhere between 0 and 3. Thus there is an interval [a, b] of length 1/2 where f 0 ≥ 3. 71

5.8 Exercises for primitives, Integration by substitution. R √- The Leibniz1 notation3/2 2x + 1dx = 3 (2x + 1) .

Exercise 1. Exercise 2.

R √ x 1 + 3x =

2x 9 (1

4 + 3x)3/2 + − 135 (1 + 3x)5/2

Exercise 3.

Z

√ 2x2 (x + 1)3/2 8x(x + 1)5/2 16(x + 1)7/2 x2 x + 1 = − + 3 15 105 since  2  3/2 0 2x (x + 1) 4x(x + 1)3/2 = x2 (x + 1)1/2 + 3 3   5/2 0 4 8x(x + 1) 8(x + 1)5/2 = x(x + 1)3/2 + 15 3 15   0 16(x + 1)7/2 8(x + 1)5/2 = 105 15

Exercise 4.

Z Z

1/3

√

−2/3

√

xdx 2x(2 − 3x)1/2 4(2 − 3x)3/2 = + −3 −27 2 − 3x

xdx = −2/9 − 4/27 − (8/9 − 32/27) = −2/27 2 − 3x

Exercise 5.

Z

(x + 1)dx ((x + 1)2 + 1)−2 = ((x + 1)2 + 1)3 −4

Exercise 6.

Z

sin3 x =

Z

sin x(1 − cos2 x) = − cos x +

1 cos3 x 3

Exercise 7.

Z

Exercise 8.

sin−2 x −2

Exercise 9.

π/4 (4−sin 2x)3/2 −3 0

=

x1/3 (1 + x) =

3 3 3 4/3 3 7/3 x + x = (z − 1)4/3 + (z − 1)7/3 4 7 4 7

33/2 −8 −3

Exercise 10. (3 + cos x)−1 Exercise 11. 2 cos−1/2 x Exercise 12. 2 cos

√

8 x + 1 3 = 2(cos 3 − cos 2)

n

Exercise 13. − cosnx Exercise 14.

(1−x6 )1/2 −3

Exercise 15.

Z

Exercise 16.

R

t(1 + t)1/4 dt =

(x2 + 1)−3/2 dx =?  √

Z

(x − 1)x1/4 dx =

x x2 + 1

√

0 =

4 9/4 4 5/4 4 4 x − x = (1 + t)9/4 − (1 + t)5/4 9 5 9 5

√ x2 + 1 − x2 / x2 + 1 1 = 2 x2 + 1 (x + 1)3/2

Exercise 17. 72

(8x3 + 27)5/3 Exercise 18.

3 2 (sin x

   3 1 1 = (8x3 + 27)5/3 5 24 40

− cos x)2/3

Exercise 19.

Z

xdx p = 1 + x2 + (1 + x2 )3/2

Z

√

1 2 du

= u + u3/2 Z 1 du = = 2(1 + u1/2 )1/2 = √ √ 2 u 1 + u1/2 p = 2(1 + 1 + x2 )1/2 + C

Exercise 20.

Z −(x − 1)2/5 dx = − (x − 1)−3/5 dx = −5/2(x − 1)2/5 x−1 Rb R b+c Exercise 21. Thm. 1.18. invariance under translation. a f (x)dx = a+c f (x − c)dx. (x2 − 2x + 1)1/5 dx = 1−x

Z

Z

Thm. 1.19. expansion or contraction of the interval of integration. Z Z b 1 kb  x  f dx f (x)dx = k ka k a Z b+c yx + c Z b f (x)dx = f (y − c)dy dy = dx a a+c y = kx

b

Z

f (x)dx =

dy = kdx

a

1 k

Z

kb

f ka

y k

dy

Exercise 22.

F

F

x a

Z



x/a

,1 = 0

t a dt du = a u=

up du 2 (u + 12 )q

 1Z x (t/a)p dt 
q = ,1 = a a 0 t 2 2 + 1 a Z x tp = a−p−1+2q dt = a−p−1+2q F (x, a) 2 2 q 0 (t + 1 )

x

Exercise 23.

Z

1

Z xx 1 1/x

Z 1

dt = F (1) − F (x) 1 + t2 dt = F (x) − F (1) 1 + t2   dt 1 =F − F (1) 2 1+t x

1 t −1 −1 du = 2 dt, 2 du = dt t u Z 1/x dt −du
= = 2 2 1+t u 1 + u12 1 Z 1/x Z 1 du dt = =− 2+1 2+1 u t 1/x 1 u=

Z 1

x

Exercise 24.

Z 0

1

xm (1 − x)n dx = −

Z

0

(1 − u)m (un )du =

1

Z 0

Exercise 25. 73

1

(1 − x)m xn dx using

u=1−x x=1−u

m sin 2x = 2−m sinm 2x 2 Z π Z π/2 Z π 1 m m m −m m −m −m−1 sin 2xdx = 2 cos x sin xdx = 2 sinm xdx = sin xdx = 2 0 2 0 0 Z −π/2 Z π/2 Z π/2 π  sinm cosm xdx = −2−m−1 cosm xdx = 2−m − x dx = 2−m−1 2 π/2 0 −π/2 cosm x sinm x =

Z

π 2

0



Exercise 26.

(1) π

Z

Z π (π − u)f (sin (π − u))(−du) = (π − u)f (sin u)du = π 0 Z π Z π xf (sin x)dx f (sin x)dx − =π Z π 0 Z 0π π xf (sin x)dx = =⇒ f (sin x)dx 2 0 0 Z

0

xf (sin x)dx =

u=π−x

0

x=π−u

(2) Z u = cos x

π

0

Z Z π π sin x π π x sin x sin x = dx = 2 2 dx = 2 2 1 + cos2 x 2 − sin x 2 − sin x 0 0 0 Z −1 Z 1 Z 1 π du π du dx =− = =π 2 1 1 + u2 2 −1 1 + u2 1 + x2 −1

x sin x dx = 1 + cos2 x

du = − sin xdx

Z

π

Exercise 27.

x = sin u

Z

1

2 n− 21

(1 − x )

dx = cos u

Z

π/2 2

dx =

(cos u)

n− 12

Z cos udu =

0

0

π/2

cos2n udu

0

5.10 Exercises - Integration by Parts. R x sin x = −x cos x + sin x

Exercise 1.

Exercise 2.

R

x2 sin x = −x2 cos x + 2x sin x + 2 cos x

Exercise 3.

R

x3 cos x = x3 sin x + 3x2 cos x − 6x sin x + −6 cos x

Exercise 4.

R

x3 sin x = −x3 cos x + 3x2 sin x + 6x cos x − 6 sin x

Exercise 5.

R

sin x cos x = − 14 cos 2x = − 14 (cos2 x − sin2 x)

Exercise 6.

R

x sin x cos xdx =

Exercise 7.

R

sin2 x =

R

sin2 xdx =

−1 4

R

R

x 2

2x sin 2x = − x cos + 4

sin x sin x = − sin x cos x +

sin 2x +

R

sin 2x 8

cos2 x

x 2

Exercise 8.

Z

sinn xdx = − cos x sinn−1 x + Z

Z

sinn x = − cos x sinn−1 x +

(n − 1) sinn−2 x cos2 x

Z

u = sinn−1 x dv = sin xdx

(n − 1) sinn−2 x(1 − sin2 x) = Z Z = − cos x sinn−1 x + (n − 1) sinn−2 x − (n − 1) sinn x Z Z (n − 1) −1 sinn x = sinn−1 x cos x + sinn−2 x n n

Exercise 9. 74

(1) Z −1 −1 1 1 1= sin x cos x + sin x cos x + x 2 2 2 2 Z π/2 π sin2 xdx = 4 0 R R π/2 4 π/2 π/2 3 (2) 0 sin x = −1 + 34 0 sin2 x = 3π 4 sin x cos x 0 16 R π/2 6 R 5π 4 5 π/2 (3) 0 sin x = 6 0 sin x = 32 Z

sin2 x =

Exercise 10.

(1) −1 2 sin2 x cos x + 3 3 1 −3 −3 cos x + cos 3x = cos x + 4 12 4 3 = − cos x + 4 Z

sin3 xdx =

Z

1 2 −3 1 sin x = − sin 2x cos x − cos x = cos x + cos 3x 6 3 4 12

since

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

(2) Z

Z 3 −1 3 x sin 2x −1 3x 3 sin 2x −1 sin3 x cos x + sin2 x = sin3 x cos x + ( − )= sin3 x cos x + − 4 4 4 4 2 4 4 8 16 1 1 1 sin 2x 1 Now sin 4x = (2 sin 2x cos 2x) = (sin x cos x(1 − 2 sin2 x) = − sin3 x cos x 32 32 8 16 4 3x 3 sin 2x 3x 1 1 −1 3 sin x cos x + − = − sin 2x + sin 4x =⇒ 4 8 16 8 4 32

sin4 xdx =

(3)

5

Z

sin5 xdx =

Z

Z sin4 x sin xdx = − cos x sin4 x + cos2 x4 sin3 x = Z Z Z Z = − cos x sin4 x + 4( sin3 x − sin5 x) = − cos x sin4 x + 4 sin3 x − 4 sin5 x Z Z 5 4 5 sin dx = − cos x sin x + 4 sin3 x

Z

sin5 dx = − cos x(1 − cos2 x)2 + 4(

−3 1 cos x + cos 3x) 4 12 1 = − cos x(1 − 2 cos2 x + cos4 x) + −3 cos x + cos 3x 3 1 = − cos x + 2 cos3 x − cos5 x − 3 cos x + (cos x cos 2x − sin x sin 2x) = 3 1 10 cos3 x = −4 cos x + 2 cos3 x − cos5 x + (4 cos3 x − 3 cos x) = −5 cos x + − cos5 x 3 3 Z 2 cos 3x 1 sin5 xdx = − cos x + − cos5 x 3 5

My solution to the last part of this exercise conflicts with what’s stated in the book. Exercise 11.

(1) Z

 2  Z Z x2 x sin 2x x cos 2x x sin2 xdx = ( sin2 x)x − (sin2 t) = − − + = 2 4 4 8 x2 x sin 2x cos 2x = − − 8 4 Z8 x sin 2x we had used sin2 x = − 2 4 75

(2) Z

−3x cos x + 4 −3x = cos x + 4Z

Z x 3 1 cos 3x − − cos x + cos 3x = 12 4 12 x 3 − sin 3x cos 3x + sin x + 12 4 36 −3 1 sin3 x = cos x + cos 3x 4 12

x sin3 x =

(3) Z

 Z   Z x sin 2x x sin 2x x3 x2 sin 2x 1 3 x sin 2x x sin xdx = x − − 2x − = − − x + = 2 4 2 4 2 4 3 2   x3 x2 sin 2x 1 −x cos 2x sin 2x = − + + = 6 4 2 2 4 2

2

2

=



x3 x2 sin 2x x cos 2x sin 2x − − + 6 4 4 8

Exercise 12.

Z

n

Z

Z

x cos xdx = cos x sin x + (n − 1) cosn−2 x sin2 x = Z Z = cosn−1 x sin x + (n − 1) cosn−2 x − (n − 1) cosn x  Z Z n−1 cosn−1 x sin x =⇒ cosn x = + cosn−2 x n n

cos xdx =

cos

n−1

n−1

Exercise 13.

R (1) cos2 x = R (2) cos3 x =

sin 2x + 12 x 5 2 cos x sin x + 23 3

sin x =

3 4

sin x +

1 12

sin 3x since

1 1 1 1 sin 3x = (sin 2x cos x + sin x cos 2x) = sin x cos2 x + sin x(2 cos2 x − 1) = 12 12 6 12 1 1 = sin x cos2 x − sin x 3 12 (3) Z

cos4 xdx =

cos3 x sin x 3 + 4 4



1 1 x + sin 2x 2 4

 =

3 3 cos3 x sin x x+ sin 2x + 8 16 4

sin 4x = 2 sin 2x cos 2x = 4 sin x cos x(2 cos2 x − 1) = 8 sin x cos3 x − 2 sin 2x Z 3 1 1 cos4 xdx = x + sin 2x + sin 4x 8 4 32 Exercise 14.

Z p Z p x2 2 2 √ 1 − x dx = x 1 − x + dx 1 − x2 Z Z Z p x2 x2 − +1 1 √ √ dx = =− 1 − x2 + √ 2 2 1−x 1−x +1−1 1 − x2 2 2 x =x −1+1 Z Z p 1 p 1 1 2 2 √ =⇒ 1 − x dx = x 1 − x + 2 2 1 − x2 Exercise 15. 76

then

(1) Z

Z Z (a2 − x2 )n dx = x(a2 − x2 )n − n(a2 − x2 )n−1 (−2x)xdx = x(a2 − x2 )n + 2n x2 (a2 − x2 )n−1 dx Z Z Z 2 2 2 n−1 2 2 2 2 2 n−1 x (a − x ) dx = ((x − a ) + a )(a − x ) dx = −(a2 − x2 )n + a2 (a2 − x2 )n−1 dx Z Z x(a2 − x2 )n 2a2 n 2 2 n =⇒ (a − x ) dx = + (a2 − x2 )n−1 dx 2n + 1 2n + 1

(2) −x3 x(a2 − x2 ) 2a2 + x= + a2 x 3 3 3 Z Z a2 5 x(a2 − x2 )5/2 2 2 5/2 (a2 − x2 )3/2 dx + (a − x ) dx = 6 6 Z Z x(a2 − x2 )3/2 3a2 2 2 3/2 (a − x ) dx = + (a2 − x2 )1/2 dx 4 4 Z Z r  x 2 1− dx = a2 cos2 θdθ = =a a ! r   Z  x 2 θ sin 2θ 1 + cos 2θ x 1x 2 2 2 1− =a =a + = a arcsin + 2 2 4 a 2a a Z ap π  πa2 a2 − x2 = a2 −0 = 2 2 0  2 Z a 2 3a πa 3πa4 (a2 − x2 )3/2 dx = = 4 2 8 0   Z a 5 5a2 3πa4 = πa6 (a2 − x2 )5/2 dx = 6 8 16 0 Z

Z

(a2 − x2 )1/2

Exercise 16. In (x) =

Rx 0

(a2 − x2 )dx =

x a dx cos θdθ = a sin θ =

tn (t2 + a2 )−1/2 dt

(1) Z n−2 2 t (t + a2 ) (n − 1)tn−2 (t2 + a2 )1/2 = tn−1 (t2 + a2 )1/2 − (n − 1) 2 (t + a2 )1/2 Z p tn−2 n−1 = x x2 + a2 − (n − 1)a2 In−2 (n)In = xn−1 (x2 + a2 )1/2 − a2 (n − 1) (t2 + a2 )1/2 √ (2) n = 5; x = 2; a = 5. Z 2 2 √ I1 (2) = x(x2 + 5)−1/2 dx = (x2 + 5)1/2 = 3 − 5 In (x) = (t2 + a2 )1/2 tn−1 −

Z

0

0

Z

2

t5 (t2 + 5)−1/2 dt = 25−1 (4 + 5)1/2 − 5(5 − 1)I3 (2) = 48 − 20I3 (2) √ 3I3 (2) = 2 4 + 5 − 5(3 − 1)I1 (2) = 12 − 10(3 − 5) √ √ 1 10 5 168 40 5 I5 (2) = (48 − 20(−6 + )) = − 5 3 5 3

5I5 (2) =

0 √ 2

Exercise 17.

Z t2(c + t3 )1/2 2(c + t3 )1/2 t (c + t ) dt = − 3 3 √ Z 3 Z 3 √ t2(4 + t3 )1/2 2 3 2 3 2 3 3 −1/2 3 1/2 t (4 + t ) dt = (4 + t ) = 2 31 + − (11.35) −3 3 3 3 −1 −1 −1 Z

3

3 −1/2

Exercise 18. 77

Z

  Z Z sinn+1 x sin x sinn x n sinn−1 x n dx = sin x dx = − m+1 n+1 m cos x cos x m cos x m cosm−1 x Z Z sinn+1 x sinn x n sinn−1 x =⇒ dx = − cosm+1 x m cosm x m cosm−1 x

Exercise 19.

cosm+1 x dx = sinn+1 x

 Z cos xdx 1 m cosm−1 x m cos x = cos x − = n n+1 −n sin x −n sinm x sin x Z −m cosm−1 x cosm x + = n −n sin x n sinn−1 x Z Z Z −1 cos1 x 1 cos1+1 x cot2 x = = − dx = − cot x − x 1 sin1 x 1 sin1+1 x Z Z Z −1 cos3+1 x 1 cos3 x cos3−1 x −1 4 cot3 x − (− cot x − x) = cot3 x + cot x + x cot xdx = = − − = 3+1 3 3−1 3 3 3 sin x sin x sin x Z

Z



m

Exercise 20.

(1) Z

2

1

Z

00

tf 00 (2t)dt

tf (t)dt = 2 0

n=2

0

(2) Z

1

xf 00 (2x)dx =

0

1 2

Z

2

tf 00 (t)dt =

0

1 2



2

tf 0 (t)|0 −

Z 0

2

 1 f 0 (t)dt = (2f 0 (2) − (f (2) − f (0))) = 4 2

Exercise 21.

(1) Recall Theorem 5.5, the second mean-value theorem for integrals: Z b Z c Z f (x)g(x)dx = f (a) g(x)dx + f (b) a

Z

b

 sin φ(t)

a

φ0 (t) φ0 (t)

(2) φ(t) = t2 ;



a

b

g(x)dx

c

Z c Z b 1 1 0 φ (t) sin φ(t) + φ0 (t) sin φ(t) = φ0 (a) a φ0 (b) a 1 1 4 c = 0 cos φ(t)|a + 0 (cos φ(b) − cos φ(a)) ≤ φ (a) φ (b) m φ0 (t) = 2t > 2a if t > a Z x 4 2 sin t dt ≤ = 2a 2a

dt =

where

1 1 ≥ 0 m φ (t)

∀t ∈ [a, b]

a

5.11 Miscellaneous review exercises. Exercise 1. g(x) = xn f (x);

f (0) = 1

g 0 (x) = nxn−1 f (x) + xn f 0 (x) ; g 0 (0) easily blows up) g j (x) =

Ph

g 0 (0) = 0 especially if n ∈ Z+ (just note that if negative integer values are included,

j n! n−k j−k f (x) k=0 k (n−k)! x




If j < n , then g j (0), since each term contains some power of x If j ≥ n, j

g (x) =

n   X j k=0

j

g (0) =

n! xn−k f (j−k) (x) k (n − k)!

j! f (j−n) (0) (j − n)! 78

If j = n, g n (0) = n! Exercise 2.

P (x) =

5 X

aj xj

j=0 0

P (x) =

5 X

P (0) = 1 = a0 jaj x

P 0 (0) = 0 = a1

j−1

P 00 (x) =

5 X

a1 = a2 = 0

00

j=1

P (0) = 0 = 2(1)a2 j(j − 1)aj xj−2

j=2

P (x) = a5 x5 + a4 x4 + a3 x3 + 1 0

4

3

=⇒ P (x) = 5a5 x + 4a4 x + 3a3 x

2

P 00 (x) = 20a5 x3 + 12a4 x2 + 6a3 x

P (1) = a5 + a4 + a3 + 1 = 2 P 0 (1) = 5a5 + 4a4 + 3a3 = 0 P 00 (1) = 20a5 + 12a4 + 6a3 = 0

Solve for the undetermined coefficients by Gauss-Jordan elimination process          1 0 −15 a5 5 4 3 0 5 4 3 0 20 12 6 0 =  20 12 6 a4  = 0 0 1 10  1 0 0 6 1 1 1 1 1 1 1 1 a3 =⇒ a5 = 6

a4 = −15

P (x) = 6x5 − 15x4 + 10x3 + 1

a3 = 10

Exercise 3. If f (x) = cos x and g(x) = sin x, Prove that f (n) = cos (x +

nπ 2 )

and g (n) (x) = sin (x +

( sin x(−1)j+1 if n = 2j + 1 cos x(−1)j if n = 2j ( cos x(−1)j if n = 2j + 1 nπ )= g (n) (x) = sin (x + 2 sin x(−1)j if n = 2j

nπ f (n) (x) = cos (x + )= 2

f (x) = cos x f 0 (x) = − sin x f 00 (x) = − cos x f 000 (x) = sin x f 0000 (x) = cos x g(x) = sin x 0

g (x) = cos x g 00 (x) = − sin x g 000 (x) = − cos x g 0000 (x) = sin x

f (2j) (x) = cos x(−1)j f (2(j+1)) (x) = (cos x(−1)j )0 = cos x(−1)j+1 f (2j+1) (x) = sin x(−1)j+1 f (2j+3) (x) = (sin x(−1)j+1 ) = sin x(−1)j+2

g (2j) (x) = sin x(−1)j g (2(j+1)) (x) = (sin x(−1)j )00 = sin x(−1)j+1 g (2j+1) (x) = cos x(−1)j g (2j+3) (x) = (cos x(−1)j+1 )

Exercise 4. 79

nπ 2 )

h0 (x) = f 0 g + f g 0

h(n+1)

n   X n (k) (n−k) = f g k

(n)

h h00 (x) = f 00 g + 2f 0 g 0 + f g 00 k=0 n    X n = f (k+1) g (n−k) + f (k) g (n−k+1) = k k=0 n−1  X n  (1) (n) (n+1) = f g + fg + f (k+1) g (n−k) + f (k) g (n−k+1) + f (n+1) g + f (n) g (1) k k=1

=f

(1) (n)

g

+ fg

(n+1)

n X

+

k=2

n! (n − k + 1)!(k − 1)!

f

(k) (n−k+1)

g

+

n−1 X k=1

n! f (k) g (n−k+1) (n − k)!k!

!

+ f (n+1) g + f (n) g (1)     n! n! n! n+1 Now + = (k + (n − k + 1)) = so then (n − k + 1)!(k − 1)! (n − k)!k! (n + 1 − k)!(k)! k  n  n+1 X X n + 1  n + 1 (k) (n+1−k) h(n+1) = f g (n+1) + f g + f (n+1) g = f (k) g (n+1−k) k k k=1

k=0

By induction, this formula is true. Exercise 5.

(1) f 2 + g 2 = f (−g 0 ) + gf 0 Y 0 = 2f f 0 + 2gg 0 = 2(gf + 2g(−f )) = 0 =⇒ Y = C

Y = f 2 + g2

Y = C = f 2 + g2

f (0) = 0;

g(0) = 1 =⇒ C = 1

(2) h = (F − f )2 + (G − g)2 = f 0 (x) = g(x), 0

g 0 (x) = −f (x);

0

f (0) = 0;

0

0

0

h = 2(F − f )(F − f ) + 2(G − g)(G − g );

g(0) = 1 0

h (0) = 2(0) + 2(0) = 0

= 2(F − f )(G − g) + 2(G − g)(−F + f ) = 0

∀x

2

h(x) = C =⇒ h(x) = (F (x) − f (x)) + (G(x) − g(x))2 h(0) = 0

so C = 0

=⇒ F = f ; G = g Exercise 6.

u = x2 Exercise 7.

df du 2x

f 0 (4) =

= 3x2

3x 2

= 3 where we had used the substitution

u = 4; x = 2 dg du

= u3/2 ;

g(u) = 52 u5/2

64 5

g(4) = 25 25 =

Exercise 8.

Z 0

x

sin t 1 dt = t+1 0+1

c

Z

sin tdt + 0

1 x+1

Z

x

sin tdt = c

1 −1 x − cos t|c = − cos c + 1 + (cos x − cos c) = x+1 x+1 −x cos c + x − cos c + 1 − cos x + cos c x(1 − cos c) + (1 − cos c) = = >0 x+1 x+1 c

= − cos t|0 +

Exercise 9. y = x2 is the curve for C. y =

1 2 2x

is the curve for C1 . 80

+

Z

b

0

Z

c 2

Z

2

(kx − x ) + c

0

b

1 1 (x2 − x2 ) = b3 P : (b, b2 ) 2 6 Assume C2 is of the form kx2

(k − 1) 3 1 (b − x ) = c + b2 (b − c) + − (b3 − c3 ) 3 3 2

(b2 − c2 )c b3 c3 2b3 + b3 − cb2 − + = − 3 3 3 3 Now A(A) = A(B) 1 2 2 =⇒ b (b − c) = b3 =⇒ b = 3 6 16 2 so then kx2 = x 9

=⇒

kc2 = b2

2

k=

b2 c2

2 2 2 cb = b2 (b − c) 3 3 4 c 3

Exercise 10.

(1) |Q(h) − 0| = f (h) h =

(

h2 |h|

if x is rational

. 0 if x is irrational For now, consider 0 < h < δ; let δ(; h = 0) =  ( f (h) = h |Q(h) − 0| = h 0

if x is rational < if x is irrational

(2) f (h) − f (0) − 0 =  =⇒ f 0 (0) = 0 h Exercise 11.

R

Exercise 12.

4 3 (1

Exercise 13.

(x2 −1)10 20

(2 + 3x) sin bxdx =

−2 5

cos 5x + − 3x 5 cos 5x +

3 25

sin 5x

+ x2 )3/2 1 −2

=

−310 20

Exercise 14.

Z

1 3

0

1

6x + 7 + 2 1 dx = (6x + 7)3 3 =

Exercise 15.

R



 1      −(6x + 7)−1 −(6x + 7)−2 1 1 1 1 −1 1 + + + + = = 3 − 13 6 6 6 42 (6)132 (6)49 0

37 8281

x4 (1 + x5 )5 dx =

(1+x5 )6 30

Exercise 16.

Z

u=1−x

1

x4 (1 − x)20 dx =

x=1−u

0

Z =− 1

0

(1 − u)4 u20 du =

Z

1

(1 − u)4 u20 du =

0

Z

1

(1 + 4(−u) + 6u2 + −4u3 + u4 )u20 du =

0

121 −412 2 6123 −4124 125 = + + + + = 21 22 23 24 25 1 = 265650 Make sure to check your arithmetic. Exercise 17.

R2 1


2 x−2 sin x1 dx = cos x1 1 = cos 12 − cos 1 81

Exercise 18.

sin (x − 1)1/4 dx

R

Z Z Z u = (x − 1)1/4 1/4 3 =⇒ sin (x − 1) dx = (sin u)4u du = 4 u3 sin udu 1 1 1 du = (x − 1)−3/4 dx = dx 4 4 u3 Z

u3 sin udu = −u3 cos u + 3u2 sin u + 6u cos u + 06 sin u = −(x − 1)3/4 cos (x − 1)1/4 + 3(x − 1)1/2 sin (x − 1)1/4 + 6(x − 1)1/4 cos (x − 1)1/4 − 6 sin (x − 1)1/4 sin (x − 1)1/4 dx = = −4(x − 1)3/4 cos (x − 1)1/4 + 24(x − 1)1/4 cos (x − 1)1/4 + 12(x − 1)1/4 sin (x − 1)1/4 − 24 sin (x − 1)1/4 x sin x2 cos x2 dx = (1/4) sin2 x2 + C

Exercise 19.

R

Exercise 20.

R√

1 + 3 cos2 x sin 2xdx =

R√

1 + 3 cos2 x2 sin x cos xdx =

R

du u1/2 −3 =

2u3/2 −9

=

−2 (1 + 3 cos2 x)3/2 , 9

where we had used this substitution: du = −6 cos x sin xdx 2

u = 1 + 3 cos x

Exercise 21.

R2 0

375x5 (x2 + 1)−4 dx Z

u = x2 + 1 du = 2xdx

0

=⇒

2

  Z (u2 − 2u + 1) 375 5 1 2 1 du = du − 3+ 4 = u4 2 1 u2 u u 1   5 1 1 375 −1 = 64 = 26 + 2+ = 3 2 u u −3u 1

375 375 du(u − 1)2 u−4 = 2 2

(u − 1)2 = x4 Exercise 22.

Since



R1 0

du = 2 cos x sin xdx −3

(ax + b)(x2 + 3x + 2)−2 dx =

−1 x2 +3x+2

 1 = 0

−1 6

+

1 2

Z

5

3 2

= 23 ,

then if a = 9/2, b = 27 2 , we’ll obtain 3/2 R1 Exercise 23. In = 0 (1 − x2 )n dx 1

Z

2 n

(1 − x ) dx = x(1 −

In = 0

Z =2

1 x2 ) n 0

1

x2 n(1 − x2 )n−1 dx = 2n

0



Z −

1

xn(1 − x2 )n−1 (−2x)dx =

0



1

Z

((x2 − 1) + 1)(1 − x2 )n−1 dx =

=⇒ In =

2n 2n + 1

0

= 2nIn−1 − 2nIn Z 1 Z I2 = (1 − x2 )2 dx = 0

I3 =

6 8 16 7 15 35

I4 =

8 128 I3 = 9 315

I5 =

10 256 I4 = 11 693

Rx

tm (1 + t)n dt;

Exercise 24. F (m, n) =

0

1 2

4

dx(1 − 2x + x ) =

0

m > 0,

n>0 82



 1 1 5 8 2x3 + x = x− 3 5 15 0

 In−1

x Z x m+1 t tm+1 xm+1 (1 + x)n n n F (m, n) = (1 + t) − n(1 + t)n−1 dt = − F (m + 1, n − 1) m+1 m+1 m+1 0 m+1 0 (m + 1)F (m, n) + nF (m + 1, n − 1) = xm+1 (1 + x)n  12  x Z x Z x t x12 t13 x13 11 1 1 12 F (11, 1) = t (1 + t) dt = t 1+t = = + + 12 13 0 12 13 0 0  12  x x13 11F (10, 2) + 2 + = x11 (1 + x)2 12 13 F (10, 2) = Exercise 25. f (n) =

R π/4 0

x13 x12 x11 + + 13 6 11

tann xdx

(1) b

Z a π/4

Z

n

f (n + 1) =

g + f (a) c

π/4 n

tan x tan x = 0

Z

f g = f (b)

Use this extremely important fact: Z

b

Z

Z

g a

π/4

tann x = f (n)

tan x < c

c

0

(2) Z

π/4

f (n + 2) + f (n) =

tann x tan2 x +

0

π/4

Z

tann x =

0

Z

π/4

tann x(sec2 x) =

0

π/4 tann+1 x 1 = = n + 1 0 n+1 (3) 1 < f (n + 1) + f (n) < 2f (n) n+1

f (n + 2) + f (n) =

Exercise 26. f (0),

f (π) = 2 Z

1 = f (n − 2) + f (n) > f (n − 1) + f (n) > 2f (n) n−1 1 1 < 2f (n) < =⇒ n+1 n−1 Rπ (f (x) + f 00 (x)) sin xdx = 5 0

π 00

Z

0

Z

0

0

Z

f sin x = f sin x − f cos x = − f cos x = −f cos x − f sin x Z Z π (f + f 00 ) sin xdx = f sin x + −(f (π)(−1) − f (0)) − f sin x = 2 + f (0) = 5 =⇒ f (0) = 3 0

Z 0

Exercise 27.

Z 0

π/2

  Z Z sin 2x 1 π sin x 1 − cos x cos x dx = dx = − = 2x + 2 2 0 x+2 2 x+2 (x + 2)2 0   1 1 4+π A 1 + −A = − = 2 π+2 2 4(π + 2) 2

sin x cos x dx = x+1

Z

π/2

Exercise 28.

√ Z √ dx 2 a + bx 2 a + bx √ = + bx b x2 x a + bx   Z √ Z 2 Z 2 a  3/2 3/2 √ a + bx 2√ 2 a b 3b (a + bx) 3b (a + bx) = + = a + bx +b + a + bx + x x x2 3b x 3b x2 x √ Z Z √ a + bx dx √ =⇒ =a + 2 a + bx x x a + bx Z

83

Exercise 29.

Z

xn

p

Z Z √ 2xn (ax + b)3/2 2xn (ax + b)3/2 2n nxn−1 2(ax + b)3/2 xn−1 (ax + b) ax + b = − = − 3a 3a 3a 3a √ Z Z n √ √ 2nb 2x (ax + b) ax + b 2n xn ax + b − xn−1 ax + b = − 3a 3 3a   Z Z √ √ 2 −3 n n 3/2 n−1 x ax + bdx = x (ax + b) − nb x ax + b + C n 6= (2n + 3)a 2

(ax + b)dx =

Exercise 30.

Z

√

Z xm 2xm (a + bx)1/2 mxm−1 2(a + bx)1/2 dx = − b b a + bx Z m−1 Z Z 2 2ma 2 2m x (a + bx) xm−1 xm √ √ = xm (a + bx)1/2 − 2m √ − = xm (a + bx)1/2 − b b b b a + bx a + bx a + bx Z Z 2 m 2ma xm 1 xm−1 √ √ x (a + bx)1/2 − dx = 2m + 1 b b(2m + 1) a + bx a + bx

Exercise 31.

Z

r √ √ √ Z Z dx ax + b n2 ax + b ax + b 2 ax + b 2n ax + b √ √ = 2 = + = + n n+1 n n n+1 ax ax ax + b ax a x ax + b x ax + b √ Z Z 2 ax + b 1 1 2nb √ √ = + 2n + n n n+1 ax a x ax + b x ax + b
√ Z Z b 2n dx 2 ax + b a √ √ =⇒ (1 − 2n) − = axn xn ax + b xn+1 ax + b √ Z Z − ax + b (2n − 3)a 1 1 √ √ = − (n − 1)bxn−1 (2n − 2)b xn ax + b xn−1 ax + b

Exercise 32. I derived the formulas for this and Exercise 33 by doing the following trick.

(C m+1 S 1−n )0 = (m + 1)C m (−S 2−n ) + C m+2 (1 − n)S −n = −(m + 1)C m S 2−n + (1 − n)C m S −n (1 − S 2 ) = = −(m + 1 + 1 − n)C m S 2−n + (1 − n)C m S −n = −(m − n + 2)C m S 2−n + (1 − n)C m S −n Z Z −(C m+1 S 1−n ) (m − n + 2) =⇒ C m S −n = − C m S 2−n n−1 n−1 Exercise 33.

(C m−1 S 1−n )0 = (m − 1)C m−2 (−S 2−n ) + (1 − n)C m S −n = −(m − 1)C m−2 (S −n )(1 − C 2 ) + (1 − n)C m S −n = = −(m − 1)C m−2 S −n + (m − 1)C m S −n + (1 − n)C m S −n Z Z m−1 1−n m−2 −n C S = −(m − 1) C S + (m − n) C m S −n Z Z C m−1 S 1−n m−1 m−2 −n C S + = C m S −n m−n m−n Exercise 34.

(1) P 0 (x) − 3P (x) = 4 − 5x + 3x2 n X P = a j xj j=0

P0 =

n X j=1

aj jxj−1 =

n−1 X

=⇒

n−1 X

(aj+1 (j + 1) − 3aj )xj = 4 − 5x + 3x2

j=0

aj+1 (j + 1)xj

j=0 84

Generally, we can say aj+1 =

3 aj j+1

if j ≥ 3

We also have a1 (1) − 3a0 = 4 a2 (2) − 3a1 = −5 a3 (3) − 3a2 = 3

=⇒ a3 − a2 = 1

Then let a2 = −1 and a3 = 0. So we have a1 = 1 and a0 = −1. P (x) = −1 + x + −x2 is one possible polynomial and we were only asked for one. Suppose Q s.t. Q0 − 3Q = 4 − 5x + 3x2 (another solution). Then (P − Q)0 − 3(P − Q) = 0

∀x

0

=⇒

(P − Q) =3 P −Q

=⇒ ln (P − Q) = 3x

=⇒ ke3x = P − Q = k

∞ X (3x)j j=0

Q = −k

∞ X (3x)j j=0

j!

j!

+P

Since we didn’t specify what Q has to be, we find that, in general, any Q is P plus some “amount” of the homogeneous solution, ke3x . (2) If Q(x) is a given polynomial, and suppose P is a polynomial solution to P 0 (x) − 3P (x) = Q(x). Suppose R is another polynomial solution such that R0 (x) − 3R(x) = Q(x). Then just like above, P − R = ke3x . If we wanted polynomial answers of finite terms, then k must be zero. Thus, there’s at most only one polynomial solution P . Exercise 35. Bernoulli Polynomials.

(1) P1 (x) = 1; Pn0 (x) = nPn−1 (x); n=1

R1

Pn (x)dx = 0, if n ≥ 1 1 Z 1 1 1 2 (x + c) = ( x + Cx) = + C = 0 C = −1/2 2 2 0 0

(1)(1) = P10

0

P1 = x − 1/2 n=2

2(x − 1/2) = P20

Z

1

(x2 − x + C) =

0



 1 −1 1 3 1 2 x − x + Cx = + C = 0 C = 1/6 3 2 6 0

2

P2 = x − x + 1/6 Z 1 1 x 1 13 1 3x2 0 n = 3 3(x − x + ) = P3 + + C) = − + +C =0 (x3 − 6 2 2 4 2 4 0 2 3x x P3 = x 3 − + 2 2 Z 1 x 1 −1 3 1 14 n = 4 4(x3 − x2 + ) = P40 (x4 − 2x3 + x2 + C) = − + 13 + C = 0 C = 2 2 5 2 3 30 0 −1 P4 = x4 − 2x3 + x2 + 30 Z 1 1 5x4 5x3 x 1 (1)5 5(1)4 −(1)2 n = 5 5(x4 − 2x3 + x2 − ) = P50 (x5 − + − + C) = − + + + C = 0; 30 2 3 6 6 2 12 12 0 5x4 5x3 x P5 = x 5 − + − 2 3 6 2

(2) The first, second, and up to fifth case has already been proven. Pn−1 Assume the nth case, that Pn (t) = tn + j=0 aj tj (the general form of a polynomial of degree n). 0 Pn+1 = (n + 1)(tn +

n−1 X

aj tj ) =⇒ Pn+1 = tn+1 + (n + 1)

j=0

n−1 X j=0

85

aj tj+1 +C j+1

C=0

(3) The first, second, and up to fifth case has already been proven. Assume the nth case, that Pn (0) = Pn (1). Z 0 Pn+1 = Pn+1 (1) − Pn+1 (0) (by the second fundamental theorem of calculus) 0 Pn+1 = (n + 1)Pn

Z

1

(n + 1)Pn (t) = 0

(by the given properties of Bernoulli polynomials)

0

=⇒ Pn+1 (1) = Pn+1 (0) (4) Pn (x + 1)− = Pn (x) = nxn−1 is true for n = 1, 2, by quick inspection (and doing some algebra mentally). Z Z 0 0 Pn+1 = (n + 1)Pn =⇒ Pn+1 = (n + 1) Pn Z x Z x+1 Pn (t)) Pn (t) − (Pn+1 (a2 ) + (n + 1) Pn+1 (x + 1) − Pn+1 (x) = Pn+1 (a1 ) + (n + 1) a2

a1

a1 = 1; a2 = 0; so Pn+1 (1) − Pn+1 (0) = 0 (from previous problems) Z x+1 Z x Pn+1 (x + 1) − Pn+1 (x) = (n + 1)( Pn (t) − Pn (t)) = 1 0 =⇒ Z x Z x = (n + 1) Pn (t + 1) − Pn (t) = (n + 1) ntn−1 = (n + 1)xn 0

0

(5) k

Z

0 Pn+1 Pn+1 (k) − Pn+1 (0) = n + 1 n+1 0 Pn (x + 1) − Pn (x) = nxn−1 k

Z

Pn = 0

=⇒

Pn (x + 1) − Pn (x) = xn−1 n k−1 k−1 X Pn+1 (x + 1) − Pn+1 (x) k−1 X X Pn+1 (r + 1) − Pn+1 (r) = xn = = n+1 n+1 x=1 x=1 r=1 =

k−1 X

Pn+1 (k) − Pn+1 (0) n+1

rn =

r=1

(telescoping series and Pn+1 (1) = Pn+1 (0) )

(6) This part was fairly tricky. A horrible clue was that this part will rely directly on the last part (because of the way Rx Px−1 n+1 (0) this question is asked), which gave us j=1 j n = 0 Pn (t)dt = Pn+1 (x)−P . n+1 Use induction. It can be easily verified, plugging in, that Pn (1 − x) = (−1)n Pn (x) is true for n = 0 . . . 5. Assume the nth case is true. Z

x

Z

u=1−t du = −dt

Z

1−x

−Pn (1 − u)du = −

Pn (t)dt = 0

1−x

1

Pn (u)(−1)n du

1

(since Pn (1 − x) = (−1)n Pn (x), assumed nth case is true) Z 1−x n+1 = (−1) Pn (t)dt = 1

Z 1 Pn (t)dt (since Pn = 0 ) 0 0   Pn+1 (1 − x) − Pn+1 (0) Pn+1 (x) − Pn+1 (0) =⇒= (−1)n+1 = n+1 n+1 = (−1)n+1

Z

1−x

=⇒ Pn+1 (1 − x) = (−1)n+1 Pn+1 (x) In the second to last and last step, we had used (−1)n+1 Pn+1 (0) = Pn+1 (0). For n + 1 even, this is definitely true. If n + 1 was odd, Doing some algebra for the first five cases, we can show that P2j−1 (0) = 0 for j = 2, 3. Assume the jth case is 86

P2j−1 true. Since P2j−1 is a polynomial and P2j−1 (0) = 0, then the form of P2j−1 must be P2j−1 = k=1 ak xj . Using 0 Pn+1 = (n + 1)Pn , x Z x 2j−1 2j−1 XZ x X 1 k+1 k P2j (x) − P2j (0) = 2j P2j−1 = 2j ak t = 2j t ak = k+1 0 0 0 k=1

= 2j Z

x

P2j+1 (x) − P2j+1 (0) = (2j + 1) 0

2j−1 X

ak k+1 x = 2j k+1

k=1

2j X

k

ak−1 x + P2j (0) k k=1 k=2 ! 2j+1 X X 2jak−2 xk ak−1 k (2j) t + P2j (0) = (2j + 1) + (2j + 1)P2j (0)x k k(k − 1) k=2

k=3

If we take the integral from 0 to 1, then we find that P2j+1 (0) = 0 (7) Using Pn (1 − x) = (−1)n Pn (x), derived above, P2j+1 (0) = (−1)2j+1 P2j+1 (1) = (−1)P2j+1 (0) =⇒ P2j+1 (0) = 0 1 1 P2j−1 (1 − ) = (−1)2j−1 (P2j−1 ( )) 2 2 1 =⇒ P2j−1 ( ) = 0 2 Exercise 36. There’s a maximum at c for f , so f 0 (c) = 0 x

Z

f 00 (t)dt = f 0 (x) − f 0 (a)

a

Z

|f 0 (0)| = |

c

Z 0a

f 00 (t)dt = f 0 (c) − f 0 (0) = −f 0 (0)

Z

0

f 00 (t)dt| ≤

c

Z

c

|f 00 |dt ≤ mc

0

|f 0 (a)| ≤ m(a − c) f 00 (t)dt = f 0 (a) − f 0 (c) = f 0 (a)

c

|f 0 (0)| + |f 0 (a)| ≤ ma 6.9 Exercises - Introduction, Motivation for the definition of the natural logarithm as an integral, The definition of the logarithm. Basic properties; The graph of the natural logarithm; Consequences of the functional equation L(ab) = L(a) + L(b); Logarithm referred to any positive base b 6= 1; Differentiation and integration formulas involving logarithms;Logarithmic differentiation. Exercise 1.

(1) log x = c +

x (ln |t|)|e

 = c + ln |x| − 1

=⇒ ln

x |x|

 =c−1

c=1 (2)     1+a 1+b (1 + a)(1 + b) 1+x + ln = ln = ln 1−a 1−b (1 − a)(1 − b) 1−x 1+x (1 + a)(1 + b) 1 + a + b + ab =⇒ = = 1−x (1 − a)(1 − b) 1 − b − a + ab

f (x) = ln

x=

a+b 1 + ab

Exercise 2.

(1) log (1 + x) = log (1 − x) =⇒ x = 0 87

(2) log (1 + x) = 1 + log (1 − x) ln (1 + x) = 1 + ln (1 − x) = ln (e) + ln (1 − x) ln e(1 − x) =⇒ 1 + x = e − ex

=⇒ x =

e−1 1+e

(3) 2 log x = x log 2

√ √ (4) log ( x + x + 1) = 1

ln x2 + ln 2−x = 0 = ln 1 = ln x2 2−x √

=⇒ x = 2

√ x =⇒ x + 1 = e2 − 2 xe + x  2 2 √ e −1 2 xe = e2 − 1 =⇒ x = 2e

x+1=e−

√

Exercise 3.

ln x x 1 − ln x f0 = x2 1 2  −2 2 ln x − 3 00 x x − 2x ln x f = 3 − = x x4 x3 f=

f 0 < 0 when x > e

f 0 > 0 when 0 < x < e

2 ln x − 3 > 0 =⇒ f 00 (x) < 0 (concave), when 0 < x < e3/2 ;

for x3 > 0,

f 00 (x) > 0 (convex), when x > e3/2

Exercise 4. f (x) = log (1 + x2 )

Exercise 5. f (x) = log

Exercise 6. f (x) = log

√

√

f0 =

2x 1 + x2

f0 =

x 1 + x2

1 + x2

4 − x2 f0 =

1 −2x −x = 2 4 − x2 4 − x2

Exercise 7. f (x) = log (log x)

1 f = ln x 0

  1 1 = x x ln x

Exercise 8. f (x) = log x2 log x 0

f 0 = (2 log x + log log x) = Exercise 9. f (x) =

1 4

log

2 1 + x x log x

x2 −1 x2 +1

f0 = Exercise 10. f (x) = (x +

√


0 1 1 log x2 − 1 − log x2 + 1 = 4 4



2x 2x − x2 − 1 x2 + 1



 =x

1 + x2 )n

p ln f = n ln (x + 1 + x2 )    f0 1 x n √ √ =n 1+ =√ 2 2 f x+ 1+x 1+x 1 + x2 p n f 0 = (x + 1 + x2 )n √ 1 + x2 √ √ Exercise 11. f (x) = x + 1 − log (1 + x + 1) 88

1 x4 − 1



1 1 √ f = √ − 2 x+1 1+ x+1 0

Exercise 12. f (x) = x log (x +

√

1 + x2 ) −

√



1 √ 2 x+1

 =

2(1 +

1 √

x + 1)

1 + x2

  x x x √ 1+ √ −√ x + 1 + x2 1 + x2 1 + x2 Note to self: Notice how this had made some of the square root terms disappear. f 0 = log (x +

Exercise 13. f (x) =

√1 2 ab

log

p 1 + x2 ) +

√ √ a+x√b √ a−x b

√ √ √ √ 1 f = √ (ln ( a + x b) − ln ( a − x b)) 2 ab √   √ √ √ 1 1 −x b √ b− √ √ (− b) = √ f 0 = f rac12 ab √ a(a − bx2 ) a+x b a−x b Exercise 14. f (x) = x(sin (log x) − cos log x)

f 0 = sin (ln x) − cos (ln x) + (cos (ln x) + sin (ln x)) = 2 sin (ln x) Exercise 15. f (x) = log−1 x

f0 = Exercise 16.

R

dx 2+3x

Exercise 17.

R

log 2 xdx

=

1 3

ln (2 + 3x)

(x ln2 x)0 = ln2 x + 2 ln x

=⇒ xln2 x − 2(x ln 2 − x) = x ln2 x − 2x ln x + 2x

(x ln x − x)0 = ln x + 1 − 1 = ln x Exercise 18.

R

−1 x(ln x)2

x log xdx 

x2 ln x 2

0

x 2 Z x2 ln x x2 x ln x = − 2 4

Exercise 19.

R

= x ln x +

x log2 xdx 0   x2 ln2 x 1 2 2 = x ln x + x ln x 2 x Z 2 2 2 x ln x x ln x x2 =⇒ x log2 x = − − 2 2 4 

Exercise 20.

R e2 −1 0

dt 1+t

Z

e3 −1

0

Exercise 21.

R

cot xdx Z

Exercise 22.

R

dt e3 −1 = ln (1 + t)|0 =3 1+t

cos x dx = ln | sin x| sin x

xn log (ax)dx Solve the problem directly. 89

Z Z xn+1 xn log a + xn log x = log a + xn log x n+1 Z Z n+1 n+1 n+1 x x 1 x xn+1 xn ln x = ln x − = log x − n+1 n+1x n+1 (n + 1)2 Z n+1 n+1 x x xn+1 =⇒ xn log ax = log a + log x − n+1 n+1 (n + 1)2

Z

Exercise 23.

Z

R

x2 log2 xdx

x2 log2 x =

Exercise 24.

Z

xn log ax =

1 3 2 x ln x − 3

x3 ln2 x 2 x2 2 ln x = − 3 3 3

Z

R 1−e−2 1

1−e2

0

Exercise 26.

 =

x3 ln2 x 2x3 ln x 2x3 − + 3 9 27

dx = ln (ln x) + C x ln x

log (1−t) 1−t dt

Z

R

x3 ln x x3 − 3 9

dx x log x

R

Z

Exercise 25.



√log |x|

1−e−2 ln (1 − t) 1 2 = −2 dt = − (ln (1 − t)) 1−t 2 0

dx

x

1+log |x|

Z

log x = x 1 + log x √

Z

(2(1 + log x)1/2 )0 log x = 2(1 + log x)1/2 log x −

Z

2(1 + log x)1/2 = x

4 = 2 log x(1 + log x)1/2 − (1 + log x)3/2 3 Exercise 27. Derive

Z n xm+1 n ln x − xm lnn−1 x x log xdx = m+1 m+1 By inspection, we just needed integration by parts.   Z Z Z x4 3 3 x4 3 3 x4 2 2 x4 ln3 x 3x4 ln2 x 3x4 ln x 3x4 x3 ln3 x = ln x− x3 ln2 x = ln x− ln x − x3 ln x = − + − 4 4 4 4 4 4 4 16 32 128 Z

n

m

f (x) = x − 1 − ln x;

Exercise 28. Given x > 0,

g(x) = ln x − 1 +

1 x

(1) 1 x 1 1 1 g0 = − 2 = f 0 x x x so then if f 0 > 0, g 0 > 0; f0 = 1 −

For f 0 < 0

0<x<1

f0 > 0

xg 0 = f 0 f 0 < 0, x>1

g0 < 0 f 0 (1) = g 0 (1) = 0

f (1) = g(1) = 0 x − 1 − ln x > 0 since f (1) = 0 is a rel. min. 1 0 < ln x − 1 + since g(0) is a rel. min. x (2) See sketch. Exercise 29. limx→0

log (1+x) x

Rx (1) L(x) = 1 1t dt ; (2) Use this theorem.

=1 L0 (x) = x1 ;

L0 (1) = 1 90

Theorem 19 (Theorem I.31). If 3 real numbers a, x, and y satisfy the inequalities y n n ∈ Z, then x = a

a≤x≤a+ ∀ n ≥ 1, 1−

1 1 < ln x < x − 1 =⇒ 1 − < ln x + 1 < x x x+1

1−

ln x + 1 x < <1 1+x x

=⇒

ln (1 + x) =1 x

Exercise 30. Using f (xy) = f (x) + f (y),

r=

p q

=⇒ f (ap/q ) = f ((a1/q )p ) = pf (a1/q ) =

p f (a) q

since f (a2 ) = f (aa) = 2f (a)

f (a) = f (a);

f (ap+1 ) = f (ap ) + f (a) = (p + 1)f (a) Exercise 31. ln x =

(1) ln x = b−a n

R |x| 1

R |x|

=

1

1 t dt

b−1 n ;

If f (a) = f ((a1/q )q) = qf (a1/q ) 1 then f (a1/q ) = f (a) q

1 t dt

From this definition, then for n partitions, a0 = 1, a1 = 1 + b−1 n , . . . , an = b = x
so if ak = 1 + k b−1 n   n  n  X X ak − ak−1 ak − ak−1 < log x < ak ak−1 k=1

k=1

(2) log x is greater than the step function integral consisting of rectangular strips within covering over x1 k−1 (3) ak = 1 + k =⇒ aka−a = k1 k−1 n X k=1

1 x

and less than rectangular strips

n n+1 n X X1 X 1 1 1 < ln (n + 1) < =⇒ < ln (n + 1) < 1+k k k k k=1

k=2

k=1

n n−1 X X1 1 < ln (n) < =⇒ k k k=1

k=2

Exercise 32.

(1) L(x) =

1 ln b

ln x = logb x

loga x = c logb x

=⇒ loga a = 1 (There must be a unique real number s.t. L(a) = 1 ) 1 loga x = logb x =⇒ logb x = logb a loga x logb a

(2) Changing labels for a, b: logb x =

loga x loga b

Exercise 33. loge 10 = 2.302585

log10 e = Exercise 34. Given

R xy x

loge e loge 10

f (t)dt = B(y)

=⇒ loge 10 =

f (2) = 2,

1 1 = ' 0.43429 log10 e 2.302585

, We Want A(x) = 91

Rx 1

f (t)dt

Z

xy

f (t)dt = B(xy) − B(x) x

Exercise 35. Given

xy

d dx

Z

R xy

f (t)dt = y

1

d dB(xy) dB(x) (B(xy) − B(x)) = y− = f (xy)y − f (x) = 0 dx d(xy) dx f (x) =⇒ f (xy) = y   Z x Z x Z x t f (2) A(x) = f (t)dt = f ((2) )dt = dt = 4 ln x 2 (t/2) 1 1 1

f (t)dt = x

Rx 1

f (t)dt + x

Ry 1

f (t)dt, and letting F be the antiderivative of f ,

F (xy) − F (1) = y(F (x) − F (1)) + x(F (y) − F (1)) Z y d/dx d/dx −−−→ f (xy)(y) = y(f (x)) + f (t)dt −−−→ f 0 (xy)y 2 = y(f 0 (x)) 0

(f 0 (1)) y=1 −−→ f 0 (y) = − → f (y) = k ln y + C −−→ f (1) = 0 + C = 3 y R xy Rx Ry (k ln t + 3) = y 1 (k ln t + 3) + x 1 (k ln t + 3) 1 R

x=1

k ((xy) ln xy − (xy) + 1) + 3(xy − 1) = y (k ((x) ln x − x + 1) + 3(x − 1)) + x (k (y ln y − y + 1) + 3(y − 1)) =⇒ k − 3 = ky − 3y − kxy + kx + 3xy − 3x y(3 − k) + xy(k − 3) + k − 3 + (3 − k)x = 0 =⇒ k = 3 =⇒ f (x) = 3 ln x + 3 Exercise 36.

6.11 Exercises - Polynomial approximations to the logarithm. Exercise 1.

Theorem 20 (Theorem 6.5). If 0 < x < 1 and if m ≥ 1, ln

1+x x3 x2m−1 = 2(x + + ··· + ) + Rm (x) 1−x 3 2m − 1

where x2m+1 2 − x x2m+1 < Rm (x) ≤ 2m + 1 1 − x 2m + 1 Rm (x) = E2m (x) − E2m (−x) where E2m (x) is the error term for log 1 − x m = 5, x =

1 3

 ln The error for m = 5, x =

4/3 2/3 1 3

 = ln 2 ' 2

1 + 3


1 3 3 3

11 

1+x 1−x



5

+


1 7 3 7

+

is
1 11 3

Exercise 2. ln

+


1 5 3


11 5 13 ≤ R5 (x) ≤ 2 11

= ln 32 = ln 3 − ln 2 92


1 9 3 9

! ' 0.693146047

1 5 m=5 x=

 ln

1+x 1−x



(1/5)11 11

Exercise 3. x =

1 9

ln



1+ 19 1− 19



 1 (1/5)3 (1/5)7 (1/5)9 =2 + + + = 0.405465104 5 3 7 9
11 9 15 ' 0.000000002 < R5 (x) ≤ ' 0.000000004 4 11 log 3 ' 1.098611 =⇒ 1.098611666 < log 3 < 1.098612438

10 8

= ln






= ln 54 = ln 5 − 2 ln 2

For m = 2 2
1 7 9 7

Exercise 4. x =

1 6

ln



1+ 16 1− 16



= ln

7 5

1 + 9


1 5 9 3


1 5 9

+

! = 0.2231435

5

' 0.00000003 < R3 (x) ≤

17 9 8 9


1 7 9

' 0.000000063 7 1.609437 < log 5 < 1.609438




= ln 7 − ln 5.

For m = 3, 1+x '2 ln 1−x

1 + 6


1 3 6


1 5 6

+ 3 5
2(3)+1 1

! = 0.336471193

6 = 0.00000051 2(3) + 1 The error bounds are
 1 2(3)+1  2 − 16 6 = 0.000001123 2(3) + 1 1 − 16 1.945908703 < ln 7 < 1.945910316

Exercise 5.

0.6931460 < ln 2 < 0.6931476 ln 7 = 1.945909

ln 3 = 1.098614

ln 8 = 3 ln 2 = 2.0794404

ln 4 = 1.386293 ln 5 = 1.609436 ln 6

= ln 2 + ln 3 = 1.791700

ln 9 = 2 ln 3 = 2.197228 ln 10 = ln 5 + ln 2 = 1.302577

6.17 Exercises - The exponential function, Exponentials expressed as powers of e, The definition of ex for arbitrary real x, The definition of ax for a > 0 and x real, Differentiation and integration formulas involving exponentials. Exercise 1. f 0 = 3e3x−1 Exercise 2. 8xe4x

2

Exercise 3. −2xe−x

2

Exercise 4.

√ 1 x √ e 2 x

Exercise 5.

−1 −1/x x2 e

Exercise 6. ln 22x Exercise 7. (2x ln 2)2x

2

93

Exercise 8. cos xesin x Exercise 9. −2 cos x sin xecos Exercise 10.

2

x

1 log x xe

Exercise 11. ex ee

x

ex

x

Exercise 12. ee (ex ee ) Exercise 13.

R

xex dx = xex − ex

Exercise 14.

R

xe−x dx = −xe−x + −e−x

Exercise 15.

R

x2 ex = x2 ex − 2xex + 2ex

Exercise 16.

R

x2 e−2x dx =

Exercise 17.

R

e

Exercise 18.

R

x3 e−x .

√

x

√

=e

x

x2 e−2x −2

+

R √ (2 x) −

xe−2x −2

+

√ √1 e x x

−e−2x 4

=e

√ x

√ √ √ √ √ (2 x) − 2e x = 2( xe x − e x )

2

2

2

x2 e−x = −2x3 e−x + 2xe−x 2

2

2

e−x = −2xe−x

2

 2 2 2 2 1 2 −x2 1  x2 e−x + e−x (x e + e−x )0 = 2xe−x + −2x3 e−x + −2xe−x = −2 −2 −2 Exercise 19. ex = b +

Rb a

et dt = b + ex − ea ,

2

ea = b.

Exercise 20.

Z A= Z B=

eax cos bxdx eax sin bxdx

Z b sin bxeax eax b eax cos bx − − = cos bx + B =⇒ aA + bB = eax cos bx + C a a a a Z ax eax e ax B= sin bx − b cos bx =⇒ aB + bA = e sin bx + C a a 1 (aeax cos bx + beax sin bx) A= 2 a + b2 −beax cos bx + aeax sin bx B= a 2 + b2 A=

Exercise 21.

ln f = x ln x;

f0 = ln x + 1; f

f 0 = xx (ln x + 1)

Exercise 22.

ln Exercise 23. f =

f0 1 1 x2 = + ), 2 (2xe x f 1+x 1+e

ex −e−x ex +e−x . 94

2

f 0 = 1 + ex + 2(1 + x)xex

2

ln f = ln (ex − e−x ) − ln (ex + e−x ) 1 1 f0 (ex + e−x ) − x (ex − e−x ) = x −x f e −e e + e−x  x 2 e − e−x f0 = 1 − ex + e−x a

a

x

Exercise 24. f 0 = (xa )0 + (ax )0 + (aa )0 . a

f2 = ax

f1 = xa

ln f1 = aa ln x

a

=⇒ f 0 = xa

−1 a

−1 a

a

ln a + (ln a)2 ax+a

+1 a−1

x

ln f3 = ax ln a

f20 = axa−1 ln a f2

f10 = xa a + ax

x

f3 = a a

ln f2 = xa ln a

f10 aa = f1 x a

a

f20 = ax

a

a

+1 a−1

x

f30 = (ln a)2 ax f3 f30 = (ln a)2 ax+a

ln a

x

Exercise 25.

ef = ln (ln x);   1 1 0 f fe = ln x x    1 1 1 0 f = ln (ln x) ln x x Exercise 26. ef = ex +

√

1 + e2x e2x (ef )f 0 = ex + √ 1 + e2x √ x 2x 2x 1+e e +e ex √ f0 = √ =√ 1 + e2x (ex + 1 + e2x ) 1 + e2x

Exercise 27. ln f = xx ln x

f0 xx = (xx )0 ln x + = (xx (ln x + 1)) ln x + xx−1 f x x

f 0 = xx+x (ln x + 1) ln x + xx

xx

+x−1

Exercise 28. ln f = x ln (ln x)

f0 1 = ln (ln x) + f ln x f 0 = (ln x)x (ln (ln x) +

1 ) ln x

Exercise 29. ln f = (ln x) ln x

f0 2 ln x = f x f 0 = 2xlog x−1 ln x Exercise 30. ln f = x ln (ln x) − ln x ln x = x ln ln x − (ln x)2

f0 1 ln x = ln (ln x) + −2 f ln x x   x (ln x) 1 2 ln x 0 f = ln x ln ln x + − x ln x x Exercise 31. 95

x

ln f2 = sin x ln cos x

ln f1 = cos x ln sin x

f20 − sin2 x = cos x ln cos x + f1 f2 cos x 3 cos x ln sin x sin3 x ln cos x f10 = − sin x cos x(ln sin x)2 + f20 = sin x cos x(ln cos x)2 − sin x cos x 3 3 − sin x ln cos x cos x ln sin x =⇒ f = sin x cos x(−(ln sin x)2 + (ln cos x)2 ) + + sin x cos x f10

2

Exercise 32. ln f = f0 f

=

−1 x2

ln x +

cos x sin x

= − sin x ln sin x +

1 x

ln x =⇒ f 0 = −x1/x−2 ln x + x1/x−2

1 x2

Exercise 33. ln f = 2 ln x +

1 3

ln (3 − x) − ln (1 − x) +

−2 3

ln (3 + x)

2 −1 −1 2 1 f0 = + − − f x 3(3 − x) 1 − x 3 3 + x  2  x(3 − x)1/3 x (3 − x)−2/3 1 x2 (3 − x)1/3 2 x2 (3 − x)1/3 0 f =2 + − + − 3 (1 − x)(3 + x)2/3 3 (1 − x)(3 + x)5/3 (1 − x)(3 + x)2/3 (1 − x)2 (3 + x)2/3 2 3 4 2 x(18 − 12x + 3 x + 3 x ) = (1 − x)2 (3 + x)5/3 (3 − x)2/3 Exercise 34. ln f =

Pn

i=1 bi

ln (x − ai ) n

X bi f0 = f x − ai i=1

0

=⇒ f =

n X j=1

n bj Y (x − ai )bi x − aj i=1

Exercise 35.

(1) Show that f 0 = rxr−1 for f = xr holds for arbitrary real r. xr = er ln x r (er ln x )0 = er ln x = rxr−1 x (2) For x ≤ 0, by inspection of xr = er log x , then if xr > 0, then the equality would remain valid. So then xr = |xr | = |x|r and so ln |f (x)| = r ln |x| f 0 (x) 1 = r =⇒ f 0 (x) = rxr−1 f (x) x Exercise 36. Use the definition ax = ex log a

(1) log ax = x log a Taking the exponential is a well-defined inverse function to log so taking the log of both sides of the definition, we get log ax = x log a (2) (ab)x = ax bx

(ab)x = ex log ab = ex(log a+log b) = ax bx (3) ax ay = ax+y ax+y = e(x+y) log a = ex log a ey log a = ax ay (4) (ax )y = (ay )x = axy (ax )y = exy log a = (ay )x = axy 96

(5) If x = loga y, Using the definition logb x = so then loga y =

log x log b

if b > 0, b 6= 1,

x>0

log y = x log a log y = x =⇒ x log a log a e = elog y = y = ax

If y = ax , log y = x log a =⇒ x = Exercise 37. Let f (x) =

1 x 2 (a

log y = loga y log a

+ a−x ) if a > 0.

1 f (x + y) = (ax+y + a−(x+y) ) 2  1  x+y f (x + y) + f (x − y) = a + a−(x+y) + ax−y + a−(x−y) 2  1 x 1  x+y −x y f (x)f (y) = (a + a )(a + a−y ) = a + a−x−y + a−(x−y) + a(x−y) 4 4 Exercise 38.

f (x) = ecx ; f 0 (x) = cecx ; f 0 (0) = c    ecx − 1 ecx − 1 ecx − 1 df (cx) d cx lim = c lim = c lim =c (0) = (e )(0) x→0 x→0 cx→0 x cx cx d(cx) dx ecx − 1 lim = f 0 (0) = c x→0 x 

Exercise 39.

g(x) = f (x)e−cx g 0 (x) = f 0 e−cx + −cg = cg − cg = 0 f = Kekx Exercise 40. Let f be a function defined everywhere on the real axis. Suppose also that f satisfies the functional equation

f (x + y) = f (x)f (y) for all x and y (1) f (0) = f (0)f (0) = f 2 (0)

If f (0) = 0, then we’re done If f (0) 6= 0 then f (0) = 1 (by dividing both sides by f (0) )

(2) Take the derivative with respect to x on both sides of the functional equation. df (x + y) d(x + y) df (x) = f (y) d(x + y) dx dx

=⇒

d(f (x + y)) f (x + y) = f 0 (x) d(x + y) f (x)

Let y = −x + y df (y) f (y) = f 0 (x) dy f (x) (3)

f 0 (x) f (x)

=

f 0 (y) f (y)

=⇒ f 0 (x)f (y) = f 0 (y)f (x)

∀ x, y

The only way they could do that for any arbitrary x, for any arbitrary y they one could choose on either side, is for them to both equal a constant 0 (y) =⇒ ff (y) =c (4) Referring to Exercise 39 of the same section, f = ecx since f 0 (0) = 1 Exercise 41. 97

(1) f = ex − 1 − x 0

( 0

f (0) = e − 1 − 0 = 0

x

f = e − 1 ≷ 0 if x ≷ 0 (2) Z

ex > 1 + x e−x > 1 − x

for x > 0 for x < 0

x

1 1 et = ex − 1 > x + x2 =⇒ ex > 1 + x + x2 2 2 0 2 2 x x − e−x > −1 + x − =⇒ e−x < 1 − x + 2 2

(3) Z

x

1 1 3 1 1 3 et = ex − 1 > x + x2 + x x =⇒ ex > 1 + x + x2 + 2 3 ∗ 2 2 3 ∗ 2 0 x2 x3 x2 x3 − e−x > −1 + x − + =⇒ e−x < 1 − x + − 2 3∗2 2 3∗2 (4) Suppose the nth case is true. ( P2m+1 j n j X > j=0 xj! x e−x = ex > P2m j j! < j=0 xj! j=0 n n+1 X X xj xj+1 =1+ (j + 1)! j! j=0 j=1 ( P2m+1 j+1 x > j=0 (j+1)! = − e−x + 1 P2m xj+1 < j=0 (j+1)!

ex > 1 +

=

n+1 X j=0

( P2m+2 < j=0 e−x P2m+1 > j=0

xj j!

xj j! xj j!

Exercise 42. Using the result from Exercise 41, n   n n n  X X x n X n n−j  x j X n! xj n(n − 1) . . . (n − j + 1) xj xj 1 1+ = = = < < ex j j j n n (n − j)!j! n j! n j! j=0 j=0 j=0 j=0

If you make this clever observation, the second inequality is easy to derive. x >0 x>0 n    n x x n x =⇒ e−x/n > 1 − e− n > 1 − n n e−x > (1 − x/n)n =⇒ ex < (1 − x/n)−n Exercise 43. f (x, y) = xy = ey ln x

∂x f = xy y/x ∂y f = xy ln x 6.19 Exercises - The hyperbolic functions. Exercise 7.

2 sinh x cosh x = 2

1 ex − e−x ex + e−x = (e2x − e−x ) = sinh 2x 2 2 2

Exercise 8.

cosh2 x + sinh2 x =



ex + e−x 2

2

 +

ex − e−x 2

2 =

1 2x (e + 2 + e−2x + e2x − 2 + e−2x ) = cosh 2x 4

Exercise 9.

cosh x + sinh x =

ex + e−x ex − e−x + = ex 2 2

Exercise 10.

ex + e−x cosh x − sinh x = − 2 Exercise 11. Use induction. 98



ex − e−x 2



= e−x

(cosh x + sinh x)2 = cosh2 x + 2 sinh x cosh x + sinh2 x = cosh 2x + sinh 2x (cosh x + sinh x)n+1 = (cosh x + sinh x)(cosh nx + sinh nx) = = cosh nx cosh x + cosh nx sinh x + sinh nx cosh x + sinh x sinh nx = enx + e−nx ex + e−x enx + e−nx ex − e−x + + 2 2 2 2 = enx − e−nx ex − e−x enx − e−nx ex + e−x + + 2 2 2 2 = cosh (n + 1)x + sinh (n + 1)x Exercise 12.

cosh 2x = cosh2 x + sinh2 x = 1 + 2 sinh2 x 6.22 Exercises - Derivatives of inverse functions, Inverses of the trigonometric functions. Exercise 1.

p (cos x)0 = − sin x = − 1 − cos2 x D arccos x =

1 −1<x<1 − 1 − x2 √

Exercise 2.

sin2 x + cos2 x = tan2 x + 1 cos2 x 1 D arctan x = 1 + x2

(tan x)0 = sec2 x =

Exercise 3.

(cot x)0 = − csc2 x = −

(sin2 x + cos2 x) 1 = −(1 + cot2 x) =⇒ arccotx = − 1 + x2 sin2 x

Exercise 4.

p (sec y)0 = tan y sec y = sec2 y − 1 sec y; | sec y| > 1 ∀y ∈ R If we choose to restrict y such that 0 ≤ y ≤ π, then (sec y)0 > 0. Then we must make sec y → | sec y|. Darcsecx =

√1 |x| x2 −1

Exercise 5.

p (csc y)0 = − cot x csc x = − csc y( csc2 y − 1) −π π Let y such that < y < (csc y) < 0 2 2 1 √ Darccscx = −|x| x2 − 1 Exercise 6.

x (xarccotx)0 = arccotx − 1 + x2  0 1 1x ln (1 + x2 ) = 2 (1 + x2 ) Z 1 arccotx = xarccotx + ln (1 + x2 ) + C 2 Exercise 7.

x √ |x| x2 − 1  1+ √ x  √ x2 −1 x>1   0   |x+ x2 −1| p x x 2 √ ln |x + x − 1| = =  x |x| |x| x2 − 1   −1+ √x2 −1  − |x+√x2 −1| x < −1 Z p x =⇒ arcsecxdx = xarcsecx − log |x + x2 − 1| + C |x|

(xarcsecx)0 = arcsecx +

99

Take a note of this exercise. When dealing with (∓x2 ± 1)

2j+1 2

; j ∈ Z; try x ±

√

x2 ± 1 combinations. It’ll work out.

Exercise 8.

x √ −|x| x2 − 1     0  1 x p √ √ 1 + = √x12 −1 x 2 x2 −1 ln |x + x2 − 1| = x+−1x −1 √ 2 |x| x −1 Z p x =⇒ arccscx = xarccscx + ln |x + x2 − 1| |x|

(xarccscx)0 = arccscx +

x>1 x < −1

Exercise 9.

2x arcsin x (x(arcsin x)2 )0 = (arcsin x)2 + √ 1 − x2 0 p −x 1 − x2 arcsin x = √ arcsin x + 1 1 − x2 Z p (arcsin x)2 = x(arcsin x)2 + 2 1 − x2 arcsin x − 2x Exercise 10.



− arcsin x x

0

1 1 arcsin x − √ 2 x x 1 − x2 I would note how x is in the denominator of the second term. Again, reiterating, p −x ( 1 − x2 ) 0 = √ 1 − x2 p p (y ± ±1 ∓ x2 )(y ∓ ±1 ∓ x2 ) = y 2 − (±1 ∓ x2 ) =

Multiply by its “conjugate.” As we see, choose y appropriately to get the desired denominator (that’s achieved after differentiation). Here, pick y = 1. √ √   p 1 −x −x(1 − 1 − x2 ) −(1 − 1 − x2 ) 0 2 √ √ √ (ln (1 + 1 − x )) = = √ = 1 + 1 − x2 1 − x2 1 − x2 (x2 ) x 1 − x2 Z 1 − √1 − x2 arcsin x arcsin x =⇒ = ln +C − x2 x x Exercise 11.

(1) 

1 D arccotx − arctan x



−1 1 = 2 −
x +1 1+ 1 2 x



−1 x2

 =0

(2) arccotx − arctan x1 = C Now arccotx =

π 2

− arctan x. π 1 π 1 − arctan x − arctan = C =⇒ − C = arctan x + arctan 2 x 2 x π π x → ∞ =⇒ − C = + 0 =⇒ C = 0 2 2 but x → −∞ π π − C = − + 0 =⇒ C = π 2 2

There are problems with the choice of brances for arccotx, arctan x1 , even though the derivatives work in all cases. 100

Exercise 12.

f0 = q

  1
2 x 2

1 1−

2

Exercise 13.

−1  2 √ 1 − 1−x 2



−1 f0 = q
2 1 − x1



f0 = r

−1 √ 2



−1 x2



=√

1 1 + 2x − x2

Exercise 14. f = arccos x1 .

=√

1 x2 − 1|x|

Exercise 15.

f (x) = arcsin (sin x) = p Exercise 16.

1

cos x =

2

1 − sin x

cos x | cos x|

√ 1 1 1 x √ = − 2 sqrtx x + 1 2 x 2(x + 1)

Exercise 17.

x2 1 + 1 + x2 1 + x6 Exercise 18.

r 1−

1 q

 1−x2 1+x2

2

−2x(2) (1 + x2 )2

√

 =p

1 + x2

−4x −4 √ = 2 2 (1 + x2 )3/2 2 1 + x2 − (1 − x2 ) (1 + x )

Exercise 19. f = arctan tan2 x


2 tan x sec2 x 1 2 2 tan x sec x = 1 + tan4 x 1 + tan4 x Exercise 20. 0

f =

1 + (x +

1 √

√ x 1 + x2 + x √ (1 + √ )= 2 2 2 1+x ) 1+x 1 + (x + 1 + x2 )2

Exercise 21.

f0 = q

1 2

1 − (sin x +

cos2

cos x + sin x (cos x + sin x) = √ 2 sin x cos x x − 2 sin x cos x)

Exercise 22.

f 0 = (arccos

p

1 − x2 )0 =

1 −

p

1 − (1 −

x2 )

=−

1 |x|

Exercise 23. 0

f = 1+

1 

1+x 1−x

 2

2 (1 − x)2

 =

2 2 1 = = (1 − x)2 + (1 + x)2 (1 − 2x + x2 + 1 + 2x + x2 ) 1 + x2

Exercise 24. f = (arccos (x2 ))−2 0

2 −3

f = −2(arccos x )



−1 √ 1 − x2

Exercise 25. 101

 (2x) =

4x(arccos x2 )−3 √ 1 − x4

0

f =

Exercise 26.

dy dx

=

!

1 arccos √1x



−1

q

1−

1 x





−1 2x3/2

 =

1 2 arccos

√1 x

√

x3 − x2

x+y x−y .



arctan

y 0 1 =
2 x 1 + xy



y0 x − y x2



 =

0 1 1 1 ln (x2 + y 2 ) = (2x + 2yy 0 ) 2 2 2 (x + y 2 )

=⇒ y 0 =

x+y x−y

Exercise 27.

1 ln y = ln (arcsin x) − ln (1 − x2 ) 2   1 1 1 1 1 x y0 √ √ (−2x) = = − + 2 2 2 y arcsin x 21−x 1 − x2 1−x arcsin x 1 − x arcsin x y=√ 1 − x2 √   1 − x2 + x(arcsin x) 1 (arcsin x)x 0 = y = + 2 2 3/2 1−x (1 − x ) (1 − x2 )3/2 p 3 ln y 0 = ln ( 1 − x2 + x arcsin x) − ln (1 − x2 ) 2   y 00 1 −x x 3 (−2x) √ =√ + arcsin x + √ − 0 2 2 2 y 2 1 − x2 1 − x + x arcsin x 1−x 1−x √ y 0 3x arcsin x arcsin x ( 1 − x2 + x arcsin x)(3x) y 00 = y 0 √ = + + 2 2 3/2 2 (1 − x ) (1 − x2 )3/2 1 − x + x arcsin x 1 − x Exercise 28.

1 1 − 1 − x2 + x4 x4 2 − 1 + x = = ≥ 0 ∀x 1 + x2 1 + x2 1 + x2 x3 since f (0) = arctan 0 − 0 + 0 = 0, arctan x > x − , ∀x > 0 3 f0 =

Exercise 29.

Z

√

dx x , a 6= 0 =⇒ arcsin 2 a −x

a2

Exercise 30.

Z

dx p = 2 − (x + 1)2

Z √

r 2

dx x+1  2 = arcsin √2 √ 1 − x+1 2

Exercise 31.

Z

dx a2

Exercise 32.



1+


x 2 a

=

1 x arctan a a

√ dx bx 1  √ 2 = √ arctan √ a ba a(1 + √ba ) a

Exercise 33.

Z x−

dx
1 2 2

+

7 4

4 = 7

Z

2 x− 2 √ = √ arctan 2
7 7 x − 21 +1 dx



√2 7

102

1 2




Exercise 34.

 

x2 arctan x 2

0 = x arctan x +

x2 1 1 = x arctan x + 2 2 1+x 2 

 1−

1 1 + x2



0  1 1 1 (x − arctan x) = 1− 2 2 1 + x2 Z 1 x arctan x = x arctan x + − (x − arctan x) 2

Exercise 35.



0 x3 −1 x3 √ arccos x = x2 arccos x + 3 3 1 − x2 p p x3 (x2 1 − x2 )0 = 2x 1 − x2 + − √ 1 − x2

3 (−2x)(1 − x2 )1/2 = −3x(1 − x2 )1/2 2 Z x3 1 p 9 x2 arccos x = arccos x − x2 1 − x2 − (1 − x2 )3/2 3 3 2

((1 − x2 )3/2 )0 =

Exercise 36.



0      x2 (arctan x)2 1 1 2 2 2 = x(arctan x) + x arctan x = x(arctan x) + 1 − arctan x 2 1 + x2 1 + x2 0  arctan x (arctan x)2 = 2 1 + x2 x (x arctan x)0 = arctan x + 1 + x2   Z ln (1 + x2 ) x2 (arctan x)2 (arctan x)2 − x arctan x − x(arctan x)2 dx = + 2 2 2

Exercise 37.

√

  1 1 √ (arctan x) = 1+x 2 x   √ √ 0 √ √ x 1 1 1 √ −√ = arctan x + (x arctan x) = arctan x + 2(1 + x) 2 x x(1 + x) √ √ √ 1/2 0 (x arctan x + arctan x + −x ) = arctan x + 0 Z √ √ √ arctan x = x arctan x + arctan x − x1/2 0



Exercise 38. From the previous exercise,

Z

√ √ arctan x √ dx = (arctan x)2 x(1 + x)

Z

√ u sin 2u arcsin x x 1 − x2 cos udu = + = + 2 4 2 4

Exercise 39. Let x = sin u

Z p

1−

x2 dx

=

2

Exercise 40. 103

Z

xearctan x (1 + x2 )3/2

0 earctan x −xearctan x earctan x √ = + (1 + x2 )3/2 (1 + x2 )3/2 1 + x2  arctan x 0 arctan x xe e x2 earctan x xearctan x earctan x xearctan x √ =√ +− + = + (1 + x2 )3/2 (1 + x2 )3/2 (1 + x2 )3/2 (1 + x2 )3/2 1 + x2 1 + x2   arctan x 0 1 xe earctan x xearctan x √ −√ = 2 (1 + x2 )3/2 1 + x2 1 + x2 

Exercise 41. From the previous exercise,

1 2



xearctan x earctan x √ +√ +C 2 1+x 1 + x2



Exercise 42.

0 x(1 + x2 )−1 1 x2 − = 2 2 2 (1 + x ) 2(1 + x2 ) −x 1 x2 dx = + arctan x (1 + x2 )2 2(1 + x2 ) 2 

Since − Z

Exercise 43. arctan ex .

Exercise 44.

Z (arccotex )0 = − (e

−x

arccotex dx ex

−ex 1 + e3x x 0

arccote ) = e

(ln (1 + e2x ))0 =

−x

  e−x (−1)ex e2x −x x arccote + = e arccote + − 1 − 1 + e2x 1 + e2x x

2e2x 1 + e2x

(−e−x arccotex + x −

1 ln (1 + e2x ))0 = e−x arccotex 2

Exercise 45.

Z r

 a+x dx = a−x

Z

√

a+x dx = a2 − x2

Z



p x  q 1a  dx = a arcsin x + − a2 − x2 +√
2 a a2 − x2 1 − xa

Exercise 46. 104

Z

√

√

Z p x − a b − xdx = bx − ab − x2 + axdx =      Z s  a+b a+b a2 + b2 2ab = − x− x− + − = 2 2 4 4 v s
!2  2  2  Z u Z u x − a+b a+b a−b a−b t 2
− x− = = 1− dx = a−b 2 2 2 2  2 Z p a−b = 1 − u2 = 2    2 arcsin 2x−(a+b) a−b a−b 2x − (a + b) p = + (a − b)2 − (2x − (a + b))2 2 2 2(a − b)2 

arcsin x 1 p + x 1 − x2 2 2

Z Exercise 47. Wow!

0

Since, recall, √ p 1 − x2 1 1 1 x(−2x) = √ + √ + = 1 − x2 2 1 − x2 2 4 1 − x2

dx p

(x − a)(b − x)

x − a = (b − a) sin2 u dx = (b − a)(2) sin u cos udu b − x = (a − b) sin2 u + b − a = (b − a)(cos2 u) Z

dx p = (x − a)(b − x)

Z

(b − a)(2) sin u cos udu √ √ = 2u = 2 arcsin b − a cos u b − a sin u

r

x−a b−a

6.25 Exercises - Integration by partial fractions, Integrals which can be transformed into integrals of rational functions. R R 1   1  2x+3 Exercise 1. (x−2)(x+5) = x−2 + x+5 = ln (x − 2) + ln (x + 5)

Exercise 2.

R

xdx (x+1)(x+2)(x+3)

A B + x+1 x+2  1 1 =⇒ 5 4 6 3

C = A(x2 + 5x + 6) + B(x2 + 4x + 3) + C(x2 + 3x + 2) x+3         1 A 0 1 1 1 0 1 0 −1/2 3 B  = 1 =⇒ 5 4 3 1 =  1 0 2  2 C 0 6 3 2 0 0 1 −3/2

+

A = −1/2, =⇒

Exercise 3.

R

x (x−2)(x−1)

Exercise 4.

R

x4 +2x−6 x3 +x2 −2x dx

=

R

2 x−2

+

−1 x−1

B = 2,

C = −3/2

−1 −3 ln (x + 1) + 2 ln (x + 2) + ln (x + 3) 2 2

= 2 ln x − 2 − ln (x − 1)

105

x4 + 2x − 6 3(x2 − 2) =x−1+ 3 (do long division) 3 2 x + x − 2x x + x2 − 2x Z Z 1 x2 − 2 3(x2 − 2) = x2 − x + 3 x−1+ x(x + 2)(x − 1) 2 x(x + 2)(x − 1) Z Z 2 x −2 1 1/3 −1/3 1 1 = + + = ln x + ln x + 2 − ln x − 1 x(x + 2)(x − 1) x x+2 x−1 3 3 1 2 x − x + 3 ln x + ln x + 2 − ln x − 1 2

=⇒

Exercise 5.

R

8x3 +7 (x+1)(2x+1)3 dx

A B C 8x3 + 7 D = + + + 3 3 2 (x + 1)(2x + 1) (2x + 1) (2x + 1) (2x + 1) (x + 1) 8x3 + 7 = A(x + 1) + B(2x2 + 3x + 1) + (4x3 + 8x2 + 5x + 1)C + D(8x3 + 12x2 + 6x + 1)        1 0 0 0 12 8 A 0 0 4 8  0 2 8 12 B  0 1 0 0 −6      |  =⇒  1 3 5 6   C  = 0 =⇒  1 0 0 0 1 1 7 D 1 1 1 1 A = 12, B = −6, C = 0, D = 1 Z 12 6(2x + 1)−1 −6 1 −6(2x + 1)−2 = + + ln (x + 1) + + (2x + 1)3 (2x + 1)2 x+1 2 2 Exercise 6.

R

4x2 +x+1 (x−1)(x2 +x+1)

4x2 + x + 1 A Bx + C = + 2 =⇒ A(x2 + x + 1) + (Bx + C)(x − 1) = 4x2 + x + 1 2 (x − 1)(x + x + 1) x−1 x +x+1      4 A 1 1 =⇒ 1 −1 1  B  = 1 =⇒ A = 2, B = 2, C = 1 1 C 1 0 −1 Z 2 2x + 1 + = 2 ln |x − 1| + ln |x2 + x + 1| =⇒ x − 1 x2 + x + 1 Exercise 7.

R

x4 dx x4 +5x2 +4

Doing the long division,

x4 x4 +5x2 +4

=1+−



5x2 +4



(x2 +1)(x2 +4)

Ax + B Cx + D 5x2 + 4 + 2 = 2 2 x +1 x +4 (x + 1)(x2 + 4) It could be seen that A + C = 0, 4A + C = 0 so A = C = 0 −1 B= B+D =5 3 16 4B + D = 4 D= 3 Z Z 2 5x + 4 −1/3 16/3 1 =⇒ 1 − 2 =x− + = x + arctan x + 4/3 arctan x/2 + C (x + 1)(x2 + 4) x2 + 1 x2 + 4 3 Exercise 8.

R

x+2 x(x+1) dx

Exercise 9.

R

dx x(x2 +1)2

=

=

R

R

1 x+1 A x

+

+

2 x(x+1)

Bx+C (x2 +1)

+

= ln |x + 1| + 2

R

1 x

Dx+E (x2 +1)2 106

−

1 x+1

= − ln |x + 1| + 2 ln x

A(x4 + 2x2 + 1) + Bx4 + Cx3 + Bx2 + Cx + Dx2 + Ex A(x4 + 2x2 + 1) + x(Bx + C)(x2 + 1) + Dx2 + Ex = x(x2 + 1)2 x(x2 + 1)2 =⇒ A = 1; B = −1; D = −1; C = 0; E = 0 Z 1 ln |x2 + 1| (x2 + 1)−1 −x −x = ln x + − + 2 + 2 + x x + 1 (x + 1)2 2 2 R dx

Exercise 10. Exercise 11.

Exercise 12.

(x+1)(x+2)2 (x+3)3

R

R

x (x+1)2 dx

dx x(x2 −1)

x = A(x + 1) + B x A B = =⇒ + A = 1; B = −1 (x + 1)2 x + 1 (x + 1)2  Z  1 1 −1 dx = ln x + 1 + + +C 2 x + 1 (x + 1) x+1 R R dx B C = x(x−1)(x+1) = A x + x−1 x+1

1 A(x2 − 1) + Bx(x + 1) + Cx(x − 1) = Ax2 − A + Bx2 + Bx + Cx2 − Cx =⇒ A = −1, B = = C 2 Z 1/2 1/2 1 1 −1 + + = − ln x + ln |x − 1| + ln |x + 1| x x−1 x+1 2 2 R R 2 dx x2 dx Exercise 13. x2x+x−6 = (x+3)(x−2) The easiest way to approach this problem is to notice that this is an improper fraction and to do long division: x2 6−x x2 +x−6 = 1 + x2 +x−6 −x = (A + B)x − 2A + 3B 6 = A(x − 2) + B(x + 3) B A + =⇒ 6 −6 x+3 x−2 ; B= A= 5 5 Z =⇒

1+

x−2+4 (x−2)2

R

x+2 (x−2)2

Exercise 15.

R

dx (x−2)2 (x2 −4x+5)

R

A=

3B 2

=⇒

B = −2/5 A − 3/5

6/5 −3/5 −2/5 4 −6/5 −9 + + + = ln |x + 3| + ln |x − 2| + x + C x+3 x−2 x+3 x−2 5 5

Exercise 14.

=

2A = 3B

= ln |x − 2| +

R

4 (x−2)2

= ln |x − 2| + −4(x − 2)−1

Consider the denominator with its x2 −4x+5. Usually, we would try a partial fraction form such as but the algebra will get messy. Instead, it helps to be clever here.

B A x−2 + (x−2)2

+ x2Cx+D −4x+5 ,

1 1 1 1 = = − (x − 2)2 (x2 − 4x + 4 + 1) (x − 2)2 ((x − 2)2 + 1) (x − 2)2 (x − 2)2 + 1 Z Z dx 1 1 =⇒ = − = −(x − 2)−1 − arctan (x − 2) + C (x − 2)2 (x2 − 4x + 5) (x − 2)2 (x − 2)2 + 1 R (x−3)dx R (x−3)dx Exercise 16. x3 +3x2 +2x = x(x+2)(x+1) Z

(x − 3)dx = x(x + 2)(x + 1)

Z

Z 1 1 + −3 (x + 2)(x + 1) x(x + 2)(x + 1) 1 −1 1 = + (x + 2)(x + 1) x+2 x+1 1 A B C = + + x(x + 2)(x + 1) x x+2 x+1 107

Now to solve for A, B, C in the last expression, it equations:  1 3 2  1 1 3 1 2 =⇒

is useful to use Gaussian elimination for this system of three linear     1 1 A 0 1 2 B  = 0 0 0 C 1    1 0 0 1 0 1/2 2 0 = 0 0 1 −1  1 1 0 0 1/2

1/2 1/2 −1 1 = + + x(x + 2)(x + 1) x x+2 x+1

=⇒ − ln |x + 2| + ln |x + 1| + −3/2 ln x + −3/2 ln |x + 2| + 3 ln |x + 1| = −5/2 ln |x + 2| + 4 ln |x + 1| − 3/2 ln x Exercise 17. Use partial fraction method to integrate

R

1 (x2 −1)2 .

Then build the sum.

A B C D + + + (x − 1)2 (x + 1)2 (x − 1) (x + 1) Now

(x2 − 1)(x − 1) = x3 − x2 − x + 1 (x2 − 1)(x + 1) = x3 + x2 − x − 1

=⇒ (x2 − 1)(x + 1) − (x2 − 1)(x − 1) = 2x2 − 2 x2 + 2x + 1 x2 − 2x + 1 =⇒ (summing the above two expressions we obtain) 2x2 + 2 −1 1/4 1/4 1/4 + + + x − 1 x + 1 (x − 1)2 (x + 1)2   Z 1 x + 1 1 dx x = ln +− (x2 − 1)2 4 x − 1 2 x2 − 1 Exercise 18. Use the method of partial fractions, where we find that

Z

(x + 1) dx = x3 − 1

Z

x+1 dx = (x − 1)(x2 + x + 1)

Z

2 − 32 x − 31 3 + = x2 + x + 1 x − 1

1 2 = − ln |x2 + x + 1| + ln |x − 1| 3 3 where we had used the following partial fraction decomposition for the given integrand Ax + B C x+1 + = 3 x2 + x + 1 x − 1 x −1 Ax2 + Bx − Ax − B + Cx2 + Cx + C = x + 1 2Ax + B − A + 2Cx + C = 1

(where we used the trick to take the derivative of the above equation) =⇒ A = −C

−B + C = 1; Exercise 19.

R

B−A−A=1 2 2 1 ,C= B=− A= −3 3 2

x4 +1 x(x2 +1)2

Again, it helps to be clever here. Z Z 4 Z x + 2x2 + 1 − 2x2 (x2 + 1)2 −2x x4 + 1 = = + 2 = 2 2 2 2 2 2 x(x + 1) x(x + 1) x(x + 1) (x + 1)2 = ln x + (x2 + 1)−1 + C Exercise 20.

R

dx x3 (x−2) 108

Working out the algebra for the partial fractions method, we obtain   1 −1/4 −1/8 −1/2 1/8 + + = + x3 (x − 2) x3 x2 x x−2 So then Z

1 dx 1 −1 1 = 2+ + ln x + ln |x − 2| + C x3 (x − 2) 4x 4x 8 8

Exercise 21.

Z

1 − x3 =− x(x2 + 1)

Z Z x3 − 1 x2 1 = − + = x(x2 + 1) x2 + 1 x(x2 + 1)  Z Z  −x 1 1 =− + + = 1− 2 x +1 x x2 + 1 Z

= −x + arctan x + ln x − ln |x2 + 1| + C Exercise 22.

Z

=

Z Exercise 23.

Z 

  Z 1 1/2 1 1/2 = − 2 = 2 2 2 x +1 x −1 x −1 x +1   Z 1 1/2 1/2 1/2 = − = − 2 2 x−1 x+1 x +1

dx = 4 x −1

1 1 1 ln (x − 1) − ln (x + 1) − arctan x + C 4 4 2

dx x4 + 1

I had to rely on complex numbers. Notice that with complex numbers, you can split up polynomial power sums π π π π x4 + 1 = (x2 + i)(x2 − i) = (x + ie(i ))(x − ie(i ))(x + e(i ))(x − e(i )) = 4 4 4 4 3π 3π π π = (x + e(i ))(x − e(i ))(x + e(i ))(x − e(i )) 4 4 4 4 B C D 1 A + + + = 4 (x + e(i π4 )) (x − e(i π4 )) x +1 (x + e(i 3π (x − e(i 3π 4 )) 4 )) 3π 3π π π A(x2 − i)(x − e(i )) + B(x2 − i)(x + e(i )) + C(x2 + i)(x − e(i )) + D(x2 + i)(x + e(i )) = 1 4 4 4 4 x3 : A + B + C + D = 0 3π 3π π π x2 : −Ae(− ) + Be(i ) − Ce(i ) + De(i ) = 0 do the algebra 4 4 4 4 −−−−−−−−→ x1 : −iA − iB + iC + iD = 0 π π 3π 3π x0 : −e(i )A + Be(i ) + C(−e(i )) + D(e(i )) = 1 4 4 4      4 1 1 1 1 A 0 −e(i 3π ) e(i 3π ) −e(i π ) e(i π )  B  0 4 4 4 4   =   =⇒   −i −i i i   C  0 3π D 1 −e(i π4 ) e(i π4 ) −e(i 3π 4 ) e(i 4 ) To do the complex algebra for the desired Gaussian elimination procedure, I treated the complex numbers as vectors and added them and rotated them when multiplied. 109



1 &  ↓ .

1 ↓ %

1 . ↑ &

  1 1 0  % 0 = ↑ 0 0 0 - 1

1 20 0

  1 0 0  2 % 0 = 1 0 0 2 - 1

0 1 2&

1 20

  1 0  2 %  = 1 1 0 0 0 4 - 1  1   = 0

1  41 ↑ 4 = 0 1 0 14 % 0 1 14 . 1  41 & 1 41  % 1 4 1 14 .

1 i3π e( ) 4 4 1 i3π Z B = − e( ) 1/4e( i3π −1/4e(i 3π 1/4e(i π4 ) −1/4e(i π4 ) 4 4 4 ) 4 ) =⇒ =⇒ + + + π 3π 3π 1 π x + e(i 4 ) x − e(i π4 ) x + e(i 4 ) x − e(i 4 ) C = e(i ) 4 4 −1 π e(i ) D= 4 4    1 3π 3π 3π π π π π 3π e(i ) ln (x + e(i )) − e(i ) ln (x − e(i )) + e(i ) ln (x + e(i )) − e(i ) ln (x − e(i )) 4 4 4 4 4 4 4 4 4 A=

After doing some complex algebra,    ! x2 + √2x + 1 1 1 1 √ =⇒ √ ln − 2 arctan √ − 2 arctan √ x2 − 2x + 1 4 2 2x − 1 2x + 1 The computation could be done to do the derivative on this, so to check our answer and reobtain the integrand. Is there a way to solve this without complex numbers? R x2 dx Exercise 24.

(x2 +2x+2)2

Z

x2 dx = (x2 + 2x + 2)2 Z

Z  x2 + 2x + 2 − 2x − 2 1 = + (x2 + 2x + 2)−1 (x2 + 2x + 2)2 x2 + 2x + 2 Z 1 1 = = arctan (x + 1) x2 + 2x + 2 (x + 1)2 + 1 Z

Z (x2 Exercise 25.

R

1 x2 dx = arctan (x + 1) + 2 +C + 2x + 2)2 x + 2x + 2

4x5 −1 (x5 +x+1)2 dx

(−(x5 + x + 1)−1 )0 = (x5 + x + 1)−2 (5x4 + 1) 5

(−x(x + x + 1)

Exercise 26.

Z

R

−1 0

−2

(doesn’t work)

) = (x + x + 1) (5x + x) + −(x5 + x + 1)−1 = (x5 + x + 1)−2 (5x5 + x − x5 − x − 1) Z 4x5 − 1 =⇒ dx = −x(x5 + x + 1)−2 (x5 + x + 1)2 5

dx 2 sin x−cos x+5

dx = 2 sin x + − cos x + 5

(good example of the use of half angle substitution ) Z

Z =

=

5

3 5

dx 4SC + −C 2 + S 2 + 5 sec2 x2 dx = 6T 2 + 4T + 4 Z

du 9(u+ 31 )2 5

Z



1 C2 1 C2



Z =

sec2 x2 dx 6(T + 31 )2 +

1 = √ arctan 5 +1



110

sec2 x2 dx = 4T − 1 + T 2 + 5(1 + T 2 ) Z

10 3

=

2du 6(u + 13 )2 +

3(tan x2 + 13 ) √ 5



u = tan 10 3

(where du =

x 2

sec2 2

x 2

) dx

u = tan Exercise 27.

1 a

R

dx 1+a cos x

Z 1 a

(0 < a < 1) Again, using the half-angle substitution,

1 dx = a + cos x

Z

1 a

Z

=

1 a 1 a

2 1+a

Exercise 28.

R

dx 1+a cos x

x 2

sec2 du = 2

x 2

, dx

sec2 x2 dx = 1 2 x 2 a sec 2 + 1 − T Z Z sec2 x2 dx 1 2du 2
= = 1 2 1 −1 a 1+a + 1 + T 2 ( a1 − 1) a +1+u a 1 dx = 2 2 a +C −S

arctan q

q

1−a 1+a u

1−a 1+a

Z

2 arctan ==⇒ √ 1 − a2

r

1−a x tan 1+a 2

du  q 2 = 1−a 1 + u 1+a

!

Half-angle substitution.

Z Z sec2 x2 dx dx dx = = = 1 + a cos x 1 + a(C 2 − S 2 ) sec2 x2 + a(1 − T 2 ) u = tan θ/2 = T Z Z Z 2 2du du du   =⇒ = = 2 = 1 a+1 = 2 2 2 2 2 1 + T + a(1 − T ) (1 − a)T + (1 + a) 1−a u − a−1 du = sec θ/2 dθ 2   Z 2 1 1   q1 q q = − = 1−a a+1 a+1 a+1 u − a−1 u + a−1 2 a−1 ! r r r   1 a−1 a+1 a+1 = ln (u − ) − ln (u + ) = a+1 1−a a−1 a−1 q    x a+1 tan − 2 a−1 −1    q = √ ln a2 − 1 tan x + a+1 Z

2

Exercise 29.

R

a−1

sin2 x dx 1+sin2 x

Z 2 Z s2 s +1−1 dx dx = dx = x + − 2 2 1+s 1+s 1 + sin2 x Z Z Z Z Z Z dx dx 2dx 2 dx 2 dx 2 sec2 xdx

= = = = = 2 2 2
= 2x 3 − cos 2x 3 3 3 1 − cos32x 1 + 1−cos 1 + sin2 x 1 − c −s sec2 x − 1−T 2 3 3 Z

u = tan x du = sec2 xdx

2 =⇒ = 3

Z

du 2 1+u −

2
= 1−u2 3 3

√ 1 = √ arctan 2 tan x 2 Z =⇒

Z 2 3

du = + 43 u2

Z

du √ = 1 + ( 2u)2

√ sin2 x 1 √ dx = x − arctan ( 2 tan x) 1 + sin2 x 2

It seems like for here, when dealing with squares of trig. functions, “step up” to double angle. R dx Exercise 30. a2 sin2 x+b (ab 6= 0) Take note, we need not change the angle to half-angle or double-angle. 2 cos2 x 111

a2 s2

1 1 1 1 sec2 sec2 = 2 = 2 = 2 = 2 = 2 = 2 2 2 2 2 2 2 2 2 2 2 +b c a (1 − c ) + b c a + (b − a )c a (1 + (kc) ) a (sec +k ) a (1 + T 2 + k 2 )

u = tan x

=⇒ =

du = sec2 xdx

√   1/a2 du 1 + k2 u du √ = arctan = u2 a2 (1 + u2 + k 2 ) a2 (1 + k 2 ) 1 + k2 (1 + k 2 )(1 + 1+k 2) Z =⇒

Exercise 31.

R

dx (a sin x+b cos x)2

1 dx = arctan ab a2 sin2 x + b2 cos2 x



a tan x b



(a 6= 0)

Note it’s a good idea to simplify, cleverly, your constants as much as you can. Z Z 1 dx dx = 2 2 (a sin x + b cos x) a (sin x + k cos x)2 Thus, only one constant, k, is only worried about. 1 1 1/c2 sec2 = 2 = 2 = = 2 2 2 2 (s + kc) s + 2ksc + k c t + 2kt + k (t + k)2 u = tan x

=⇒ =

du = sec2 xdx =⇒

du (u + k)2

1 a2

Z

1 −1 = 2 (a tan x + ab) (s + ab c)2

Again, note, we need not always step up or step down a half angle in the substitution. Exercise 32. Note that we have a rational expression consisting of single powers of sin and cos. Then use the tan

Z

Exercise 33.

R√

Z Z 2CS CS T = = 1 + 2CS + C 2 − S 2 C(S + C) (T + 1) u = tan θ/2 C = cos x/2 2 where S = sin x/2 du = sec θ/2dθ 2   Z Z u 2du 2udu = (u + 1) u2 + 1 (u2 + 1)(u + 1) A Bu + C + 2 u+1 u +1 2 Au + A + Bu2 + Cu + Bu + C = u 1 1 1 A = −B C + B = 1 A + C = 0 =⇒ C = ; B = ; A = − 2 2 2  Z  Z 1 (u + 1) −1/2 −1 u + 1 2 + 2 2 du = + u+1 u +1 u + 1 u2 + 1 1 = − ln |u + 1| + ln |u2 + 1| + arctan u = 2 1 x = − ln | tan x/2 + 1| + ln | sec2 x/2| + 2 2 1 π 1 π =⇒ − (ln |2|) + ln |2| + = − ln |2| + 2 4 2 4 sin x = 1 + cos x + sin x

Z

3 − x2 dx 112

θ 2

substitution.

Z

(x)0

p

Z Z Z p Z p p p x(−x) −x2 + 3 − 3 1 √ √ = x 3 − x2 − = x 3 − x2 − 3 − x2 dx = x 3 − x2 − 3 − x2 + 3 √ 2 2 3−x 3−x 3 − x2 Z p Z p √ 1 =⇒ 2 3 − x2 = x 3 − x2 + 3 r  2 1 − √x3 Z p xp 3 x 3 − x2 = 3 − x2 + arcsin √ =⇒ 2 2 3

Exercise 34.

R

√ 1 dx 3−x2

= −(3 − x2 )1/2 + C.  0 x 1 −1 1 arccos √ =√ q = −√ 2 3 3 1− x 3 − x2 3  p 0 p −x2 x 3 − x2 = 3 − x2 + √ 3 − x2 √ 3 x x 3 − x2 + − arccos √ 2 2 3

Exercise 35.

R

√

3−x2 dx x

=

Rq

3 x2

− 1dx. √

3 = sec θ x √ 3 cos θ = x √ dx = − sin θ 3 Z Z p √ √ √ Z 2 sec θ − 1(− sin θ) 3 = tan θ sin θ(− 3) = − 3 (sec θ − cos θ) = √ √ = − 3 ln | sec θ + tan θ| + 3 sin θ = √ 3 r3 √ r √ x2 + = − 3 ln − 1 + 3 1 − 2 x x 3 Exercise 36.

Rq

1 + x1 dx r

!0

  r 1 x −1 1 −1/2 = 1+ + q = 1+ + √ x 2 1+ 1 x x x2 + x x   0   x + 21 1 p 2 1 1 √ ln x + + x + x = 1+ √ =√ 1 2 2 2 2 x+ 2 + x +x x +x x +x r    Z  p 1 1 1 1 =⇒ 1+ dx = x 1 + + ln x + + x2 + x x x 2 2 1 x 1+ x

r

Exercise 37.

p p x2 (x x2 + 6)0 = x2 + 5 + √ x2 + 5    p 1 x 1 0 2 √ √ (ln (x + x + b)) = 1+ =√ 2 2 2 x+ x +b x +b x +b Z p  p  p 1 x x2 + 5 + 5 ln (x + x2 + 5) x2 + 5 = 2 Exercise 38. 113



  0 x + 12 1 1 1 p 2 √ √ 1 + =√ + x + x + 1) = 2 x + 12 + x2 + x + 1 x2 + x + 1 x2 + x + 1 Z Z x + 12 − 21 1 x 1 p √ √ = = (x2 + x + 1)1/2 − ln (x + + x2 + x + 1) 2 2 2 2 x +x+1 x +x+1

ln (x +

The trick is to note how I formed a “conjugate-able” sum from x2 + x + 1’s derivative. Exercise 39.

Z

√

dx = x2 + x

Z

Z

dx q

2dx q



2 2 x + 21 −1  p 0 ln (2(x + 1/2))2 − 1 + 2(x + 1/2) = x+


1 2 2

=

−

1 4

2(x + 1/2)2

1

!

2+ p = =p (2(x + 1/2))2 − 1 + 2(x + 1/2) (2(x + 1/2))2 − 1 2 =p (2(x + 1/2))2 − 1     Z p 1 dx 2−1 +C q = ln 2 x + + (2(x + 1/2))
2 2 x + 12 − 14 Exercise 40.

6.26 Miscellaneous review exercises. Exercise 1. Z f (x) = 1

x

log t t+1

   Z 1  Z x x log t 1 − ln (u) −1 f = dt = du = 1 x u2 1 t+1 1 u +1 1 Z x u= ln (u) t = du 2 1 1 u+u du = − 2 dt t x   Z x Z x 1 t ln t + ln t ln t (ln t)2 (ln x)2 f (x) + f = dt = dt = = x t(t + 1) t 2 1 2 1 1   1 1 f (2) + f = (ln 2)2 2 2

Exercise 2. Take the derivative of both sides, using the (first) fundamental theorem of calculus.

2f f 0 = f (x)

sin x sin x ; =⇒ 2f 0 = 2 + cos x 2 + cos x

At this point, it could be very easy to evaluate the integral by guessing at the solution. (− ln 2 + cos x)0 =

sin x ln |2 + cos x| =⇒ f = − +C 2 + cos x 2 114

Otherwise, remember that for rational expressions involving single powers of sin and cos, we can make a u = tan θ/2 substitution. x u = tan 2 C = cos x/2, S = sin x/2 2 x 2du = sec dx 2 Z   Z Z 2T 2du 2SCdx sin x = dx = = 2 + cos x 2 + C2 − S2 2 sec2 x/2 + 1 − T 2 sec2 x/2 Z Z udu T T =4 =2 − 2 2 2 2 (1 + u )(3 + u ) T +1 T +3    2     1 T +1 2 1 2 2 ln T + 1 − ln T + 3 = ln = ln =2 2 2 T2 + 3 4 + 2 cos x where tan2

sin2 x2 x 1 − cos x = = 2 cos2 x2 1 + cos x

Exercise 3.

Z

ex dx = ex − x

Z

ex (x − 1) xdx = ex − x2

Z

ex (x − 1) dx . . . x

No way. Exercise 4.

R π/2 0

3/2

ln (ecos x )dx = − cos x|0

= 1.

Exercise 5.

(1) √ f = 4x + 2x(x + 1)(x + 2)   4(x + 2) 1 1 ln f = ln = (ln (4x + 2) − ln x − ln (x + 1) − ln (x + 2)) 2 x(x + 1)(x + 2) 2    1 4 1 1 1 f0 7 = − − − f 0 (1) = − f 2 4x + 2 x x+1 x+2 12 s    1 4x + 2 4 1 1 1 =⇒ f 0 = − − − 2 x(x + 1)(x + 2) 4x + 2 x x+1 x+2 (2) Z

4

π 1

4x + 2 dx = 2π x(x + 2)(x + 1)

Z 1

 = 2π

4

(2x + 1) dx = x(x + 2)(x + 1)

 4 1 3 25 ln x + − ln |x + 2| + ln |x + 1| = πln 2 2 8 1

since we can find the antiderivative through partial fractions: A B C 2x + 1 + + = x x+2 x+1 x(x + 2)(x + 1) A(x2 + 3x + 2) + B(x2 + x) + C(x2 + 2x) = 2x + 1        1 1 1 A 0 0 1 0 −3/2 3 1 2 B  = 2 =⇒ 0 0 1 1  2 0 0 C 1 1 0 0 1/2 Exercise 6. 115

(1) x

Z

Z

1 dt t

log x = 1

x

Z log x = 1

x

F (x) = 1

if x > 0 et dt; t t e > 1 for t > 0

If 0 < x < 1 Z 1 Z 1 t 1 −1 e dt = dt > − = F (x) t t x x t log x ≤ F (x) for x ≥ 1

(2) Z

x+a

F (x + a) − F (1 + a) = 1

= ea

Z

x

1

et dt − t

Z

a+1

et dt = t

1

Z

x

1−a

et+a dt − t+a

Z

1

1−a

et+a dt t+a

et dt t+a

(3) x

Z 1

eat dt = t

Z a

ax

et dt = t

Z 1

ax

et + t

1

Z a

et = t

= F (ax) − F (a)  x  Z x t Z e e 1 t 1 t − = − e − − e = − e + F (x) 2 t t x 1 t  u Z  Z x Z 1/x −eu −e eu e1/t dt = du = − − − = u2 u u 1 1 u= = xe1/x − e − F (1/x) where we used the substitution

1 t

du = − Exercise 7.

(1) ex = F (x) − F (0); F (x) = ex + F (0) =⇒ F (0) = 1 + F (0) 0 6= 1. False. (2) d dx

Z

x2

f (t)dt = f (x2 )(2x) = −(2x) ln 2ex

ln 2

f (x) = − ln 2ex ln 2

0

Z 0

(3)

2

x2

x2 2 − ln 2eln 2 dt = − et ln 2 0 = −ex ln 2 + 1

x f (x) = 2f (x)f 0 (x); =⇒ f (x) = + C 2   2  x Z x 1 t x2 t + c dt = + ct = + cx 2 4 4 0 0 x2 + Cx + C 2 − 1 4 x =⇒ C = ±1, f (x) = + ±1 2

f 2 (x) − 1 =

Exercise 8.

(1) f (x + h) − f (x) f (x)f (h) − f (x) f (x)(hg(h)) = = = f (x)g(h) h h h 0 g(h) → 1 as h → 0 so =⇒ f (x) = f (x) 116

1 dt t2

(2) Since for f (x) = ex , we defined ex such that f 0 = f , if (ex + g)0 = ex + g 0 = ex + g =⇒ ex + g = Cex =⇒ g = (C − 1)ex but f 0 (0) = 1 so g = ex Exercise 9.

(1) g(2x) = 2ex g(x) g(3x) = ex g(2x) + e2x g(x) = ex 2ex g + e2x g = 3e2x g (2) Assume g(nx) = ne(n−1)x g g((n + 1)x) = ex g(nx) + e(n+1)x g(x) = nenx g(x) + enx g = (n + 1)enx g (3) From g(x + y) = ey g(x) + ex g(y), g(0) = g(0) + g(0) =⇒ g(0) = 0  h  eh g(x) + ex g(h) − g(x) e −1 ex g(h) g(x + h) − g(x) = = g(x) + h h h h g(x + h) − g(x) g(h) f 0 (0) = 2 = lim = lim h → 0 h→0 h h (4) g 0 (x) = g(x) + 2ex C = 2 Exercise 10.

∀x ∈ R, f (x + a) = bf (x); f (x + 2a) = bf (x + a) = b2 f (x) f (x + (n + 1)a) = f (x + na + a) = bf (x + na) = bn+1 f (x) f (x + na) = bn f (x) f (x) = bx/a g(x) where g is periodic in a Exercise 11.

g0 f0 + =⇒ (f g)0 = f 0 g + f g 0 (ln (f g))0 = f g   0  0 f f0 g0 f 0 g − g0 f f = ln = − =⇒ 2 g f g g g Exercise 12. A =

R1

et dt 0 t+1

(1)

Z

u=t−a

0

=⇒ −1

(2)

R1

t2

te dt 0 t2 +1

=

R1

u 1 2 due

0

u+1

=

a

e−t dt a−1 t − a − 1 Z 0 t−a Z 1 t e−t−a e e −a dt = − dt = −e = −e−a A t−1 −t − 1 t + 1 1 0 Z

1 A 2

−e1 +1+A 2 0 R1 t R et (4) 0 e ln (1 + t)dt = et ln (1 + t) − 1+t = e ln 2 − A (3)

R1

et dt 0 (t+1)2

=

1

−et (t+1)

−

R1

−et 0 t+1

=

Exercise 13. 117

(1) p(x) = c0 + c1 x + c2 x2 ; f (x) = ex p(x) p0 = c1 + 2c2 x p00 = 2C2 0 x 0 f =f +e p     n   X n! n x n! f (n) (x) = + ex (2c2 ) e (p(x))j = f + ex (c1 + 2c2 x) + (n − 1)! (n − 2)!2! j j=0 f (n) (0) = c0 + c1 n + n(n − 1)c2 (2) See generalization below. (3) m X p= aj xj ;

p(k) (x) =

p(0) = a0 ;

j=0

m X

aj

j=0

f (n) (x) =

n   X n j=0

j

j! xj−k = p(k) (0) = ak k! (j − k)!

ex p(j) (x)

m n   m   X X X n! n (j) n (n) aj f (0) = p (0) = aj j! = (n − j)! j j j=0 j=0 j=0

So for m = 3, then f (n) (0) = a0 + na1 + n(n − 1)a2 + n(n − 1)(n − 2)a3 f (2) = −a2 x sin ax + 2a cos ax

Exercise 14. f (x) = x sin ax ;

f (2n) (x) = (−1)n (a2n x sin ax − 2na2n−1 cos ax) f (2n+1) (x) = (−1)n (a2n+1 x cos ax + a2n sin ax + 2na2n sin ax) f (2n+2) (x) = (−1)n (−a2n+2 x sin ax + a2n+1 cos ax(2n + 2)) Exercise 15. n X

(−1)k

k=0

   Z 1   Z 1X n n X n n 1 n k+m = (−1)k tk+m dt = (−1)k t dt = k k+m+1 k k 0 0 k=0 k=0 Z 1 Z 1 n   X n m k = t (−t) dt = tm (1 − t)n dt = k 0 0 k=0 Z 0 Z 1 m n =− (1 − u) u du = (1 − u)m un du = 1 1

Z = 0

0

 Z 1   m X X m j m j n (−1) tj+n dt 6m (−u) u du = j j 0 j=0 j=0

u1 − t du = −dt Exercise 16. F (x) =

Rx 0

f (t)dt

(1) Z F (x) = 0

x

(R x (2t)2 dt = 43 x3 (t + |t|) = R0x 0dt = 0 0 2

if t, x ≥ 0 if t, x < 0

(2) Z F (x) =

(R x

x

f (t)dt = 0

(1 − t2 )dt (1 − |t|)dt 0

R0x

if |t| ≤ 1 = if |t| > 1


1 3 x x3  (t − 3Rt 0 = x − 3 x 2 x2 1 = 3 + 1R (1 − t)dt = x − 2 + 6  x  −2 x2 3 + −1 (1 + t)dt = x + 2 − 118

if |x| ≤ 1 1 6

if x > 1 if x < 1

if |x| ≥ 1

(3) f (t) = e−|t| x

Z F (x) =

x

Z

e−|t| dt =

f (t)dt = 0

0

(R x R0x

=

0

0

e−t dt = e−t |x = 1 − e−x x et dt = et |0 = ex − 1

if x ≥ 0 if x < 0

2

(4) f (t) = max. of 1 and t

Z

x

F (x) = 0

R x  =x  0 1dt Rx 2 f (t)dt = 1 + 1 t dt   Rx − 0 f

  x x 3 = 1 + 13 t3 1 = x3 + 32  R R −1 0  − x t2 + − −1 1 = Exercise 17.

R

πf 2 = π

Ra 0

if |x| ≤ 1 if x > 1 = if x < −1

1 3 x 3 t −1

−1=

x3 3

+

2 3

if |x| ≤ 1 if x > 1 if x > 1

f 2 = a2 + a. 0 r x2 + x 2x + 1 = π π
2x + 1 a = x2 + x 0 = a2 + a π

 Z for 0

a

Exercise 18. f (x) = e−2x .

t e−2x e−2t −1 −2x e dx = −2 = −2 0 0 R t −4x π −4t (2) V (t) = π 0 e − 1) dx = −4 (e (3) (1) A(t) =

Rt

ln y y = e−2x =⇒ =x −2 2 Z 1  1 π ln y W (t) = π dy = (y(ln y)2 − (2(y ln y − y))) e−2t = −2 4 e−2t

π −2t 2 = 2 − e 4t − (2(e−2t (−2t) − e−2t )) =  4 π π − πte−2t − e−2t = 2 2 where the antiderivative used was (y(ln y)2 )0 = (ln y)2 + 2 ln y (4) π (1 − e−4t ) 4  1−e−2t 2

 =

π 2

e−4t −1 t e−2t −1 t

 = π

ecx − 1 =c x→0 x

where we used the limit lim Exercise 19. sinh c =

3 4

(1) ec − e−c = 2

p ec = ex + e2x + 1 √ ex + e2x + 1 − ex +√1e2x +1

= 2 √ e2x + 2ex e2x + 1 + e2x + 1 − 1 3 √ = = ex = 4 2(ex + e2x + 1) x = ln 3 − 2 ln 2 119

(2)

√ ex − e2x − 1 − ex −√1e2x −1 ec − e−c = = 2 2 √ e2x − 2ex e2x − 1 + e2x − 1 − 1 √ = 2(ex − e2x − 1) 3 −1 ex 1 √ = ex − c = = + ex − e2x − 1 e 4 =⇒ x = ln 5 − 2 ln 2

Exercise 20.

(1) True. ln (2log 5 ) = ln 5ln 2 = (ln 2) ln 5. 25 (2) log3 5 = log log 3 This is a true fact. 2

log3 5 log2 5 = = log2 5 log2 3 (log2 3)2 =⇒ 1 = log2 3 False (3) Use induction

√ n = 1 1−1/2 < 2 1 √ 1 n = 21 + √ < 2 2 2 n+1 n X X 1 k −1/2 = k −1/2 + p < (n + 1) k=1 k=1 n + 1 case ! p √ (n + 1) 1 +p <2 n p (n + 1) (n + 1)

Now (n + 21 )2 = n2 + n +

1 2

> n2 + n, certainly. So then n+

p 1 p 2 > n + n =⇒ n + 1 > n2 + n 2 n+1 X √ k −1/2 < 2 n + 1 =⇒ k=1

(4) f = (cosh x − sinh x − 1) =

ex + e−x − ex + e−x − 1 = e−x − 1 < 0 for x > 0 2

False. Exercise 21. For 0 < x <

π 2,

(sin x)0 = cos x > 0 for 0 < x <

π 2

(x − sin x)0 = 1 − cos x ≥ 0 for 0 < x <

π 2

(x − sin x)(x = 0) = 0 =⇒ sin x < x Exercise 22.

1 1 < t t Z

x+1

x



x+1 t



1 dt = ln (x + 1) − ln x; t

if 0 < x < t < x + 1 Z

x+1

x

Exercise 23. 120

x+1 1 x+1 1 So ln = < t x x x

(x − sin x)0 = 1 − cos x ≥ 0 ∀x > 0 since (x − sin x)(x = 0) = 0; (x − sin x)0 (x = 0) = 0, then x − sin x > 0 in general for x > 0   x3 0 x2 (sin x − x − ) = cos x − 1 + > 0 ∀x > 0 6 2 x3 < sin x < x =⇒ x − 6 Exercise 24. (xb + y b )1/b < (xa + y a )1/a if x > 0, y > 0 and 0 < a < b


y n 1/n ) . x

(xn + y n )1/n = x(1 +

Without loss of generality, assume x < y.

Consider (1 + An )1/n , A constant.       1 −1 1 1 n 1/n 0 n 0 n 1/n n n ln (1 + A ) + (ln A)A ((1 + A ) ) = (exp ln (1 + A ) ) = (1 + A ) = n n2 n 1 + An   −(1 + An ) ln (1 + An ) + n(ln A)An = = (1 + An )1/n n2 (1 + An )   − ln (1 + An ) − An ln (1 + An ) + An ln An n 1/n = = (1 + A ) n2 (1 + An )  n   A   n n 1−n (1 + An ) n  − ln (1 + A ) + A ln 1+An  An = < 0 since ln <0 n2 n2 (1 + An ) 1 + An =⇒ (xb + y b )1/b < (xa + y a )1/a

if b > a

Exercise 25.

(1) Z 0

x


x e−t t = −te−t − e−t 0 = −xe−x − e−x + 1

(2) t

Z 0

x t2 e−t dt = −t2 e−t 0 −

Z

−e−t (2t)dt = −x2 e−x + 2

Z

te−t dt =

= −x2 e−x + −2xe−x − 2e−x + 2 (3) Z

x 3 −t



Z −3

x −t

2

3 −x

t (−e )dt = −x e + 3 0   x2 3 −x −x x = −x e + 3(2)(e ) e − 1 − x − 2!

t e dt = 0

x −t3 e−t 0

Z

x

t2 e−t dt =

0

(4) Assume the induction hypothesis, that Z



x

tn e−t dt = n!e−x ex −

0

Z 0

x

n X xj



j!



j=0

 x tn+1 e−t dt = −tn+1 e−t 0 −

Z

(n + 1)t(−e−t ) = −xn+1 e−x + (n + 1)n!e−x ex −

n X xj



j!



j=0

 = (n + 1)!e−x ex −

n+1 X j=0

 xj  j!

Exercise 26. Consider the hint a1 sin x + b cos x = A(a sin x + b cos x) + B(a cos x − b sin x). Solve for A, B in terms of 121

a1 , b1 , a, b. Matching up term by term the coefficients for sin and cos separately,

A=

Z

Aa2 − abB = aa1

−Aab + Bb2 = −a1 b

Ab2 + Bab = b1 b

Aab + Ba2 = ab1

aa1 + bb1 a2 + b2

B=

ab1 − a1 b a2 + b2

So if not both a, b = 0 , Z a1 sin x + b1 cos x A(a sin x + b cos x) + B(a cos x − b sin x) = = a sin x + b cos x a sin x + b cos x = Ax + B ln |a sin x + b cos x| + C

Exercise 27.

(1) f 0 (x2 ) =

1 x

df = u−1/2 du f (x2 ) = 2x − 1 (2) x2 1 1 f 0 (sin2 x) = 1 − sin2 x f 0 (u) = 1 − u f = u − u2 + C =⇒ f (x) = x − + 2 2 2 (3) x3 1 1 f 0 (sin X) = (1 − sin2 x) f (u) = u − u3 + C f (x) = x − + 3 3 3 (4) (

( for x ≤ 1 1 for 0 < x ≤ 1 f 0 (ln x) = = ln x for x > 1 e x>1 ( y for y < 0 f (y) = y e − 1 for y > 0 1 x

Exercise 28.

(1) x dt Li(x) = if x ≥ 2 ln t Z x 2 Z x x 1 −1 x dt 2 Li(x) = −2 − dt = + − 2 2 ln x ln 2 (ln x) ln x (ln x) ln 2 2 2

Z

(2) Z x x 2 2 x −2 Li(x) = − − + − dt ln x ln x (ln 2)2 (ln x)2 (ln t)3 a   Z x n−1 X k!x x x −(n + 1)dt Li(x) = + + n! − ln x lnn+1 x lnn+2 t lnk+1 x 2 k=1 Z x n X x k!x dt Li(x) = + + (n + 1)! k+1 (n+1)+1 ln x ln x t 2 ln k=1 122

C2 =

2 X −2 2(j − 1)! +− ln x (ln 2)j j=2 n

Cn = −2

X 2(j − 1)! 1 − ln 2 j=2 (ln 2)j

Cn+1 = −2

X 2(j − 1)! 1 − ln 2 j=2 (ln 2)j

n+1

(3) Z

x

u = ln t 1 du = dt t eu du = dt eu = t

dt ln t

Li(x) = 2

Z

ln x

Li(x) = ln 2

et dt t

(4) 1 ln 2 2 Z x−1 2u e e2(u+1) du = e2 du = u u c−1 Z 2(x−1) 3 1 et = e2 t dt = 2 2(c−1) 2 3

c=1+ Z

x−1

c−1

t = u + 1 =⇒

= e2 Li(e2(x−1) ) (5)   f (x) = e4 Li(e2(x−2) ) −e2 Li(e2(x−1) ) e2x e2x 1 Z x Z x 2t 2x 0 2t =⇒ + − = e f (x) = e e t−2 t−1 t2 − 3t + 2 − = t − 2 (t − 1) c c Exercise 29. f (x) = log |x| if x < 0. ∀x < 0 ∃ uniquely ln |x| since f 0 =

−ey = x(y) = g(y)

1 x

< 0 ∀x < 0.

D=R

Recall Theorem 3.10. Theorem 21. Assume f is strictly increasing and continuous on an interval [a, b]. Let c = f (a), d = f (b) and let −g be the inverse of f . That is ∀y ∈ [c, d], Let g(y) be that x ∈ [a, b], such that y = f (x). Then (1) −g is strictly increasing on [c, d] (2) −g is continuous on [c, d] Exercise 30. f (x) =

Rx 0

(1 + t3 )−1/2 dt if x ≥ 0.

(1) f 0 (x) = √

1 > 0 for x > 0 1 + x3

(2) g 0 (x) =

p 1 3x2 3 g 00 (x) = √ = 1 + x f 0 (x) 2 1 + x3

7.4 Exercises - Introduction, The Taylor polynomials generated by a function, Calculus of Taylor polynomials. Use the following theorems for the following exercises. Theorem 22 (Properties of Taylor polynomials, Apostol Vol. 1. Theorem 7.2.). c2 Tn (g) (2) Differentiation (Tn f )0 R= Tn−1 (f 0 ) x (3) Integration. If g(x) = a f (t)dt Rx Tn+1 g(x) = a Tn f (t)dt

(1) Linearity Tn (c1 f +c2 g) = c1 Tn (f )+

Theorem 23 ( Substitution Property, Apostol Vol. 1. Theorem 7.3. ). Let g(x) = f (cx), c is a constant. (12)

Tn g(x; a) = Tn f (cx; ca)

This theorem is useful for finding new Taylor polynomials without having to find the jth derivatives of the desired function. 123

Theorem 24. Pn is a polynomial of degree n ≥ 1. Let f, g be 2 functions with derivatives of order n at 0. f (x) = Pn (x) + xn g(x)

(13) where g(x) 7→ 0 as x 7→ 0. Then Pn = Tn (f, x = 0). Exercise 3.

Tn f (x) =

n X f (j) (a)

j!

j=0 x

a =e

(x − a)j

x ln a

x 0

(a ) = (ax ) ln a (ax )(n+1) = (ax (ln a)n )0 = ax (ln a)n+1 Tn (ax ) =

n X (ln a)j

j!

j=0

xj

Exercise 4.

 00 −1 1 (−1)2 2 ; = 2 (1 + x) 1+x (1 + x)3 (n+1)    0 (−1)n n! 1 (−1)n+1 (n + 1)! = = 1+x (1 + x)n+1 (1 + x)n+2  X  n 1 Tn = (−1)j xj 1+x j=0 

1 1+x

0

=

Exercise 5. Use Theorem 7.4. . Theorem 7.4 says for f (x) = Pn (x) + xn g(x), Pn (x) is the Taylor polynomial.

    n n 2 n+1 X X 1 (x ) xn xn+2 2j   (x2 )j  +  = = x + 1 − x2 1 − x2 1 − x2 j=0 j=0   n X x (x2n+3 ) 2j+1   = x + 1 − x2 1 − x2 j=0  T2n+1

x 1 − x2

 =

n X

x2j+1

j=0

Exercise 6.

(ln (1 + x))0 =

1 1+x

 Tn

1 1+x

 =

n X

(−x)j

Tn (ln 1 + x) =

j=0

n X j=0

n

(−1)j

X xj+1 xj = (−1)j+1 j + 1 j=1 j

Exercise 7.

r

!0

r 1−x 1−x 1 1 1 log = = = 2 1 + x 1 + x (1 − x) (1 + x)(1 − x) 1 − x2 ! r r Z Z X n n X 1 1+x x2j+1 1+x 2j = log so x = = T2n+1 ln 1 − x2 1−x 2j + 1 1−x j=0 j=0 1+x 1−x

r

Exercise 8.

 Tn

1 2−x



 = Tn

1/2 1 − x/2



1 = Tn 2

!

1 1−

Exercise 9. We can show this in two ways. 124

1 2




x

  n n 1 X 1 j  X xj = ( x) = 2 j=0 2 2j+1 j=0

We could write out the actual polynomial expansion. n   n   X α α−j j X α j α (1 + x) = 1 x = x j j j=0 j=0 or determine each of the coefficients of the Taylor polynomial. ((1 + x)α )0 = α(1 + x)α−1 ; ((1 + x)α )00 = α(α − 1)(1 + x)α−2 0  α! α! α−n α (n+1) = (1 + x) (1 + x)α−(n+1) ((1 + x) ) = (α − n)! (α − (n + 1))! Exercise 10. Use the substitution theorem, Apostol Vol.1. Thm. 7.3., to treat cos 2x.

T2n (cos x) =

n X (−1)j x2j j=0

; T2n (cos 2x) =

(2j)!

n X (−1)j (2x)2j

(2j)!

j=0



n



n

X (−1)j (2x)2j X (−1)j+1 22j−1 x2j 1 1 = T2n (sin x2 ) = T2n ( (1 − cos 2x)) = 1 − 2 2 (2j)! (2j)! j=0 j=1 7.8 Exercises - Taylor’s formula with remainder, Estimates for the error in Taylor’s formula, Other forms of the remainder in Taylor’s formula. We will use Theorem 7.7, which we learn in the preceding sections, extensively. Theorem 25. If for j = 1, . . . , n + 1, m ≤ f (j) (t) ≤< ∀t ∈ I, I containing a, (14) (15) (16)

(x − a)n+1 (x − a)n+1 ≤ En (x) ≤ M (n + 1)! (n + 1)! n+1 (a − x) (a − x)n+1 m ≤ (−1)n+1 En (x) ≤ M (n + 1)! (n + 1)! Z x 1 En (x) = (x − t)n f (n+1) (t)dt n! a m

if x > a if x < a

Exercise 1. For a = 0, | sin(j) (x)| ≤ 1 for ∀x ∈ R.

En (x) ≤

(−x)2n+1 (x)2n+1 if x > 0; (−1)2n+1 E2n (x) ≤ (+1) (2n + 1)! (2n + 1)! =⇒ E2n (x)| ≤

|x|2n+1 (2n + 1)!

Exercise 2.

cos x =

n X (−1)k x2k

(2k)!

+ E2n+1 (x) | cos(j) (x)| ≤ 1

k=0 2n+2

E2n+1 (x) ≤

x (−x)2n+2 ; (−1)2n+2 E2n+1 (x) ≤ (1) (2n + 2)! (2n + 2) 2n+2 |x| =⇒ |E2n+1 (x)| ≤ (2n + 2)!

Exercise 3.

arctan x =

n−1 X k=0

n−1 X k=0

k 2k+1

(−1) x 2k + 1



(2k)! (2k)!

 =

n−1 X k=0

(−1)k x2k+1 + E2n (x) 2k + 1

(−1)k (2k)!x2k+1 =⇒ f (2k+1) (0) = (−1)k (2k)! (2k + 1)!

f (2n+1) (0)x2n+1 (−1)n (2n)!x2n+1 (−1)n x2n+1 x2n+1 = = ≤ (2n + 1)! (2n + 1)! 2n + 1 2n + 1 Note how jth derivative (arctan x)(j) changes sign with each differentiation for f (2j+1) (0). Then we can always pick a small enough closed interval with a = 0 as a left or right end point to make the f (2j+1) (0) value the biggest for f (2j+1) (t). 125

Exercise 4.

(1) x2 = sin x = x −

 √   √  x3 x3 x x − −3 + 15 x − (−3 − 15) =⇒ + x2 − x = 6 6 6 √ x = 15 − 3

(2) E4 (r; 0) =

1 4!

Z

r

(r − t)4 cos tdt > 0 sin r − r2 = 0 + E4 (r) ≤

0

3 r5 r 3 1 4 < = = < 5! 5! 5(2)(5)(4)(3)(2)1 (5)(4)(2)5 200

Exercise 5.

r3 − r2 + E4 (r; 0) = 0 + E4 (r; 0) arctan r − r2 = r − 3 Z 7 1 r M (r5 ) r5 = 0.065536 < E4 (r, 0) = (x − t)4 f (5) (t)dt ≤ 4! 0 5! 5! 100 E4 (r, 0) < Ej (r, 0); j > 4 f (5) (0) 5 24 5 r = r >0 5! 5! 2 so r − arctan r = −E4 (r, 0) < 0

the 5th degree term is

Exercise 6. Apply long division on the fraction in the integrand.

Z 0

1

1 + x30 dx = 1 + x60

Z

1

  Z 1 x30 − x60 1 − x30 30 dx = 1 + x dx = 1 + x60 1 + x60 0 1 c 1 31 x =1+ =1+c 31 31

1+ 0

0

Exercise 7.

R 1/2 0

1 1+x4 dx. ∞ n−1 X X 1 4 j ) = = (−x (−x4 )j + En = 1 + −x4 + x8 . . . 1 + x4 j=0 j=0

16 x4n+1 1 ≤ En (x; 0) ≤ (x4 )n+1 17 (n + 1)! (n + 1)!
 Z 1/2 1 4n+5  1 2 =⇒ En = (n + 1)! 4n + 5 0  5 Z 1/2 1 1 −1 1 ' + 1 + x4 2 5 2 0 0.493852 < 0.49375 < 0.493858 Exercise 8.

(1) 1 x3 sin x = x − + E4 (x) 2 3!
5 1 12 sin 13 |x|5 M |x|5 |E4 (x)| ≤ = ≤ 5! 5! 5! 0≤x≤

126

(2) x6 + E4 (x2 ) 6    √2/2  √ 1 3 x7 1 1 2 sin x = = 2 x − − 3 42 0 12 42(16) √ 2 E4 (x2 ) ≤ 64(5!)   Z √22 √ 1 55 sin x2 ≤ 2 + = 0.1159 672 64(5!) 0 sin x2 = x2 −

Z

√ 2 2

0

Exercise 9.

x3 x5 (1)x7 + E6 (x; 0) ≤ 6 5! 7! x2 x4 E6 (x; 0) x2 x4 x6 sin x =1− + + ≤1− + + x 6 5! x 6 5! 7!   sin x 1 1 1 1 1 dx = 1 − + + = 0.9461 + = 0.9461 + 0.0000283 x 3 6 5(5!) 7(7!) 7(7!) sin x = x −

Z 0

1

Exercise 10. α = arctan 15 , β = 4α −

π 4.

(1) (tan A + tan B) 2/5 2 tan α = ; A = B = α; tan 2α = = 5/12 1 − tan A tan B 24/25 1 − tan2 α 2(5/12) 10/12 120 π 2(tan 2α) = = = A = 4α, B = − tan 4α = 2 2 1 − (5/12) 119/144 119 4 1 − tan (2α)
π 120 119 tan 4α + tan − 4 +− π 1
= 119 120 119 = tan (4α − ) = tan β = 4 239 1 − 119 (−1) 1 − tan 4α tan − π4 tan (A + B) =

4 arctan

π 1 1 = + arctan This is incredible. 5 4 239

(2) T11 (arctan x) =

6−1 X (−1)k x2k+1 k=0

2k + 1

+ E2(6) (x) ; |E2(6) (x)| ≤

−x3 x5 x7 x9 x11 + − + − ... 3 5 7 9 11 1 =⇒ 3.158328957 < 16 arctan < 3.158328958 5

x2(6)+1 2(6) + 1

=x+

(3) T3 (arctan x); x = −0.016736304 < −4 arctan

1 239

1 < −0.016736304 239

(4) 3.141592625 3.158328972 − 0.016736300 = 3.141592672 7.11 Exercises - Further remarks on the error in Taylor’s formula. The o-notation; Applications to indeterminate forms. Exercise 1.

2x = exp x ln 2 = 1 + (x ln 2) + Exercise 2. 127

x2 (ln 2)2 + o(x2 ) 2!

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

cos 1(x − 1)3 + cos 1+ 2

cos 1(x − 1)2 sin 1(x − 1)3 + = 2 3!     cos 1 sin 1 − 3 cos 1 2 = cos 1 + (cos 1 − sin 1)(x − 1) + − sin 1 − (x − 1) + (x − 1)3 + o(x − 1)3 2 3!

+ (− sin 1)(x − 1) −

Exercise 3. Just treat the argument of sin x − x2 just like u with u → 0.

(x − x2 )3 (x − x2 )5 + + o(x − x2 )5 = 3! 5!

1 3 1 x − 3x4 + 3x5 − x6 + x5 − 5x6 + 10x7 − 10x8 + 5x9 + x1 0 = = (x − x2 ) − 6 120 61 5 25 6 1 3 1 4 2 x − x = (x − x ) − x − x + 6 2 120 120 sin (x − x2 ) = (x − x2 ) −

Exercise 4.

(x − 1)2 (x − 1)3 + 2 3 −1 =⇒ a = 0; b = 1, c = 2

log x = log (1 + (x − 1)) = (x − 1) −

Exercise 5.

1 2 x + o(x3 ) 2 1 − cos x 1 o(x3 ) = + 2 x 2 x2 1 − cos x 1 1 − cos x 1 since = + o(x), → as x → 0 x2 2 x2 2

1 cos x = 1 − x2 + o(x3 ) as x → 0 2

1 − cos x =

1 x4 2 cos x = 1 − x2 + + o(x5 ) =⇒ cos 2x = 1 − 2x2 + x4 + o(x5 ) 2 4! 3 − 23 x4 − o(x5 ) −2 2 1 − cos 2x − 2x2 = = − o(x) → − as x → 0 x4 x4 3 3 Exercise 6. 3

4 ax − (ax) sin ax a 3! + o(x ) lim = lim = 2 x→0 sin bx x→0 bx + o(x ) b

Exercise 7.

sin 2x lim = lim  x→0 cos 2x sin 3x x→0 1−

(2x) + (2x)2 2!

+

(2x)4 4!

(2x)3 3!  + 

+ o(x5 )

o(x4 )

(3x) −

(3x)3 3!

2 = 3 + o(x4 )

Exercise 8.

lim

x→0

sin x − x 1 = − x3 6

Exercise 9.

ln 1 + x x − o(x) 1 = lim = 2x x→0 e x→0 2x + o(x) −1 2 lim

Exercise 10. Don’t do the trig. identity. 2

lim

x→0

1 − cos x = lim x→0 x tan x

 1− 1− x x+

x2 2! x3 6!

2 + o(x2 ) 1 − (1 + x2 + o(x2 ))
= lim =1 x→0 x2 + o(x2 ) + o(x3 )

Exercise 11. 128

3

lim

x + − x6 + o(x4 ) x−

x→0

=1

x3 3

Exercise 12.

ex ln a − 1 x ln a + o(x) = lim = ln a/b x→0 ex ln b − 1 x→0 x ln b + o(x) lim

Exercise 13.

lim

(x − 1) −

(x−1)2 2

3

+ (x−1) + o(x − 1)4 1 3 = (x + 2)(x + 1) 3

x→1

Exercise 14. 1 . Exercise 15. x2 2 )

x(2 + x + x(ex + 1) − 2(ex − 1) = lim lim x→0 x→0 x3

− 2(x + x2 /2 + x3 /6) x3 ( 16 ) 1 = lim = x→0 x3 x3 6

Exercise 16. x2 2

x− ln (1 + x) − x = lim lim x→0 x→0 1 − cos x

+

x3 3 3 + o(x ) x3 /2

−x

= −1

Exercise 17.

0 + −1(x − π2 ) cos x = = −1 π x − π2 x→π/2 x − 2 lim

Exercise 18. 1/6 Exercise 19.

cosh x − cos x lim = lim x→0 x→0 x2

ex +e−x 2

− cos x = x2 

1+x+ = lim

x2 2

−2 1−

x2 2



+ o(x3 ) = 2

x2

x→0

Exercise 20.

(4x) + 3 tan 3x − 12 tan x lim = lim x→0 3 sin 4x − 12 sin x x→0 (4x) − =

43 − 4 −43 +4 2

(4x)3 3 (4x)3 3!

 −4 x+

x3 3



+ o(x4 )

−4 x−


x3

+ o(x4 )

3!

=

= −2

Exercise 21.

ax − asinx ex ln a − esin x ln a = lim = x→0 x→0 x3 x3 lim

1 + x ln a + = lim

(x ln a)2 2!

+

(x ln a)3 6

 − 1 + sin x ln a +

+

sin3 x(ln a)3 3!



ln a(x − x −

 3

x 3!

)+

(x2 ln a)2 2!

−

(ln a)2 2 2 (x )

+

(lna)3 3 6 (x

− x3 ) + o(x4 ) =

x3

x→0

Exercise 22.

lim

x→0



+ o(x4 ) =

x3

x→0

= lim

sin2 x(ln a)2 2

cos sin x − cos x = lim x→0 x4 = lim

1− 1 2



sin2 x 2!

+

sin4 x 4!

 + o(x5 ) − 1 −

x2 2!

+

4  x  4 4 x4 2 2 x − (x − 3 ) + x −x + o(x5 ) 4!

x4

x→0 129

x4 4!



=

= 1 6

ln a 6

Exercise 23.

    1 1 ln x (x − 1) + o(x − 1)2 lim x → 1x 1−x = lim x → 1e 1−x ln x = exp lim x → 1 = exp lim x → 1 = e−1 1−x 1−x Exercise 24.

1 ln (x + e2x ) x ln (x + e2x ) ln (1 + x + e2x − 1) lim = lim = x→0 x→0 x x x + e2x − 1 + o(x2 ) 3x + o(x2 ) = lim = lim =3 x→0 x→0 x x =⇒ lim (x + e2x )1/x = e3 lim (x + e2x )1/x = exp lim

x→0

x→0

x→0

Exercise 25. 2 x+− x +o(x2 ) 2

1

x −e e x ln (1+x) − e e (1 + x)1/x − e = lim = lim = lim x→0 x→0 x→0 x x x x e(1 + − x2 + o(x)) − e e1− 2 +o(x) − e −e = lim = lim = x→0 x→0 x x 2

Exercise 26.

 lim

x→0

(1 + x)1/x e

1/x

 = lim

x→0

 exp

 1/x 2 x− x +o(x3 )−x 2 1 x2 = e−1/2 ln (1 + x) − 1 = lim e x→0 x

Exercise 27.

1 1 − x2 x (arcsin x)00 = (1 − x2 )3/2 3x2 1 + (arcsin x)000 = 2 3/2 (1 − x ) (1 − x2 )5/2        1 arcsin x 1 arcsin x exp lim 2 ln = exp lim 2 ln 1 + −1 = x→0 x x→0 x x x       1 x + x3 /6 + o(x4 ) − x 1 x2 = exp lim 2 ln 1 + = exp lim 2 ln 1 + + o(x3 ) = x→0 x x→0 x x 6    1 x2 + o(x3 ) = exp lim 2 = e1/6 x→0 x 6 (arcsin x)0 = √

Exercise 28. limx→0



1 x

−

1 ex −1



= limx→0



ex −1−x x(ex −1)



2

= limx→0

3 x 2 +o(x ) x2 +o(x3 )

=

1 2

Exercise 29.

 lim

x→1

1 1 − log x x − 1



 = lim

x→1

(x − 1) − log x (x − 1) log x

 = 2

(x − 1) − ((x − 1) − (x−1) + o(x − 1)3 ) 1 2 = lim = x→1 (x − 1)((x − 1) + o(x − 1)2 ) 2 Exercise 30. 2

1 + ax + (ax) + o(x3 ) − 1 − x − eax − ex − x 2 lim = lim x→0 x→0 x2 x2 if a = 2, the limit is 2 Exercise 31. 130

x2 2

−x

(1) Prove

Rx 0

Rx

f (t)dt = o Rx

0


g(t)dt as x → 0, given f (x) = o(g(x)).

f (t)dt

Consider limx→0 R0x g(t)dt . 0 Since f, g have derivatives in some interval containing 0, f, g continuous and differentiable for |t| ≤ x. 

Rx

lim

x→0

A(x)−A(0) x



f (t)dt f (0) limx→0 f (x) 0 = = lim  = =0 limx→0 g(t)dt x→0 B(x)−B(0) g(0) limx→0 g(x) x

We can do the second to last step since f, g have derivatives at 0 and thus are continuous about 0. (2) Consider limx→0 exx = 0. However, limx→0 e1x = 1. Exercise 32.

(1) Use long division to find that −g 3 (x) 1 = 1 − g(x) + g 2 (x) + = 1 − g(x) + g 2 (x) + o(g 2 (x)) 1 + g(x) 1 + g(x) (2) 

Exercise 33. Given limx→0 1 + x +

f (x) x

1/x

= e3 , use the hint.

lim g(x) = A,

x→0

then G(x) = A + o(1) as x → 0

Then g(x) = e3 + o(1) =

 1+x+

f x

1/x


x x e3 + o(1) = x + x2 + f (x) =⇒ f (0) = 0 x exp x ln (e3 + o(1)) = x + x2 + f (x)   x 3 3 3 0 1 exp x ln (e + o(1)) + x exp x ln (e + o(1)) ln e + o(1) + 3 (o (1)) = 1 + 2x + f 0 (0) e + o(1) 1 + 3(0) + 0 = 1 + 0 + f 0 (0) =⇒ f 0 (0) = 0 We need to assume that in general o(1) = x + kx2 + o(x2 ).  2 exp x ln (e + o(1)) ln (e3 + o(1)) + 3

 x 0 (o (1)) + e3 + o(1)   x 0 3 3 + x exp x ln (e + o(1)) ln (e + o(1)) + 3 o (1) + e + o(1)   2o0 (1) xo00 (1) x 3 0 2 x exp x ln (e + o(1)) + +− 3 (o (1)) = 2 + f 00 (x) (e3 + o(1)) e3 + o(1) (e + o(1)) x→0

−−−→ 2(3) = 2 + f 0 (0) =⇒ f 0 (0) = 4  To evaluate limx→0 1 +

f (x) x

1/x

, consider a Taylor series expansion of f .

!1/x 2  1/x 0 + 0 + 4 x2 + o(x3 ) f (x) 1/x lim 1 + = lim 1 + = lim (1 + x (2 + o(x))) = x→0 x→0 x→0 x x 1/x

= lim lim (1 + x (2 + o(y))) y→0 x→0

131

= lim exp 2 + o(y) = ey y→0

7.13 Exercises - L’Hopital’s rule for the indeterminate form 0/0. Exercise 1. 3x2 + 2x − 16 14 (3x + 8)(x − 2) = lim = x→2 x2 − x − 2 x→2 (x − 2)(x + 1) 3 lim

Exercise 2.

x2 − 4x + 3 (x − 3)(x − 1) = lim = −2 2 x→3 2x − 13x + 21 x→3 (2x − 1)(x − 3) lim

Exercise 3.

0 sinh x − sin x = = lim x→0 x3 0 x→0



lim

cosh x − cos x 3x2

 =

0 = lim 0 x→0



sinh x + sin x 6x



 = lim

x→0

cosh x + cos x 6

 =

1 3

Exercise 4.

(2 − x)ex − x − 2 −ex + (2 − x) − 1 (2 − x)(1 + x + x2 /2 + x3 /6 + o(x3 )) 1 = lim = lim = − 3 2 x→0 x→0 x→0 x 3x x3 6 lim

Exercise 5.

− cos bx sin axa cos axa2 a2 log (cos ax) = lim = lim = x→0 − cos ax sin bxb x→0 sin bxb2 x→0 log (cos bx) b2 lim

Exercise 6. When it doubt, Taylor expand. 2

4

1 − (1 − x2 + x24 + o(x4 )) x − sin x 2 1 − cos x lim = lim = lim = 1 3 3 x→0+ e 2 ln (x sin x) (x − x63 + o(x3 ) + x − x23 + o(x3 )) x→0+ (x sin x)3/2 x→0+ (x sin x)1/2 (sin x + x cos x) 2   x2 x4 4 − + o(x ) 2 2 24 = = lim  q 3 x→0+ x3 2 3 3 3 x(x + 6 + o(x ))(2x − 3 x + o(x ))     1 x2 2 − + o(x ) 2 2 24  = 2 lim 1/2 = 1 lim+  q = 2 3 x→0 3 x→0+ 2 6 1 + x + o(x2 )(2 + −2 x2 + o(x2 )) 6

3

1

Notice in the third step how in general we deal with powers, (x sin x)1/2 , is to convert it into exponential form, e 2 ln (x sin x) , but it wasn’t necessary. Exercise 7. Do L’Hopital’s first.

√ lim

x→a+

√ √ 1 x2 − a2 x+a = lim + = 2 x→a+ x3/2 x √ √ √ √ 1 x2 − a2 + x1/2 x + a 1 2a 2 = lim = = √ 2 x→a+ 2 a3/2 2 a x3/2

√ √ x− a+ x−a √ = lim+ x→a x2 − a2

1 √ 2 x

√1 2 x−a √ x x2 −a2

+

Exercise 8. Do L’Hopital’s at the second step.

lim+

x→1

exp (x ln x)(ln x + 1)2 + exp (x ln x) − x exp x ln x(ln x + 1) − 1 = lim+ = lim 1 − x + ln x −1/x2 x→1+ x→1 −1 + x1 = lim+ x2 exp (x ln x)(ln x + 1)2 + x exp (x ln x) = 1 + 1 = 2 x→1

Exercise 9. Keep doing L’Hopital’s. 132

1 x

exp (x ln x)

=

√ 2 3 − 2√ 1 2 1−x arcsin 2x − 2 arcsin x 1−(2x) lim = lim = x→0 x→0 x3 3x2 − 1 (1 − (2x)2 )−3/2 (−8x) − (− 21 )(1 − x2 )−3/2 (−2x) 2 = lim 2 = 3 x→0 6x 4(1 − (2x)2 )−3/2 + 4x(− 32 )(1 − (2x)2 )−5/2 (−8x) + −(1 − x2 )−3/2 − x(− 23 )(1 − x2 )−5/2 (−2x) 1 = lim = 9 x→0 1 14−1 1 = = 9 1 3 Exercise 10.

lim

x→0

x cot x − 1 x cos x − sin x cos x − x sin x − cos x − sin x = lim = lim = lim = 2 2 2 x→0 x→0 x→0 x x sin x 2x sin x + x cos x 2 sin x + x cos x cos x cos x 1 = − lim = = − lim x→0 2 cos x + cos x + −x sin x x→0 3 cos x − x sin x 3

Exercise 11.

Pn lim

x→1

xk − n = lim x→1 x−1

k=1

Pn

n

X kxk−1 n(n + 1) = k= 1 2

k=1

k=1

Exercise 12.

     1 1 1√ 1 √  √ a − b x x 1+ a2 1+ b2 2a x 2b x x x 1 − b arctan = lim a arctan = lim √ 3 1/2 x→0+ x→0+ x x a b x 2     1 a2 (b2 + x)a2 − b2 (a2 + x) 1 b2 1 1 = lim 2 lim = − 2 = 3 x→0+ a + x x b +x x 3 x→0+ (a2 + x)(x + b2 )x 

√

=

(a2 − b2 ) 1 a2 − b2 1 lim = 3 x→0+ (a2 − x)(b2 + x) 3 a2 b2

Exercise 13.

(sin 4x)(sin 3x) (2 sin 2x cos 2x)(sin 3x) 2(−2 sin 2x sin 3x + cos 2x3 cos 3x) = = = 6 as x → 0 otherwise x sin 2x x sin 2x 1 2 cos 2x sin 3x 4 π → as x → x π 2 We used L’Hopital’s at the second to last step for x → 0. Exercise 14.

lim (x−3 sin 3x + ax−2 + b) = 0

x→0

3 cos 3x + a + 3bx2 −9 sin 3x + 6bx −27 cos 3x + 6b −27 + 6b sin 3x + ax + bx3 = = = = =0 3 2 x→0 x 3x 6x 6 6 lim

So b =

9 , a = −3 . 2

Exercise 15.

1 lim x→0 bx − sin x

Z 0

x

√ 2x 2x a + x √ = 1 = √1 (1 − cos x) + a + x(sin x) 2 (1 − cos x) + (a + x) sin x 2 a+x √ √ a + x + 2√xa+x a = 2 lim sin x =2 =1 x→0 a 2 + sin x + (a + x) cos x x2

√ t2 dt a+x √ = = b − cos x a+t

=⇒ a = 4 Note that we had dropped the limit notation in some earlier steps and applied L’Hopital’s a number of times, and we also rearranged the denominator and numerator cleverly at each step. 133

Exercise 16.

(1)

x x , length BC is tan 2 2 x x x 2 tan cos = 2 sin is the base length of ABC 2 2 2 x x tan sin is the height of triangle ABC 2 2 1 − cos2 x2 x x x 1 x sin = tan − sin x =⇒ T (x) = tan sin2 = 2 2 cos x2 2 2 2 angle ABC is

(2) S(x) =

x 1 x x x sin 2 (π(1)) − cos (2 sin ) = − 2π 2 2 2 2 2

(3) Use L’Hopital’s theorem. d tan x2 − 21 sin x dx T (x) −−→ = x−sin x S(x) 2 d dx −− →

sec2

x→0

−−−→

x 2

1 2

x cos x 2 − 2 1−cos x 2 2 x 2 x d sec dx 2 tan 2

sec2

tan x2 + sin x −−→ sin x

+ 12 sec4 cos x

x 2

+ cos x

→

3 2

Exercise 17. Use L’Hopital’s rule. Rt E (1 − e− L ) R E(−1)(e−Rt/L ) lim I(t) = lim R→0 R→0 1

I(t) =

−t L


=

Et L

Exercise 18.

c − k → 0 since c → k k−c=u k−u=c A(sin kt − sin ct) A(sin (kt) − sin (k − u)t) = = 2 2 c −k −u(2k − u) A(cos (k − u)t)(t) −At cos kt = → −(2k − u) + u 2k

f (t) =

7.17 Exercises - The symbols +∞ and −∞. Extension of L’Hopital’s rule; Infinite limits; The behavior of log x and ex for large x. Exercise 15. Use L’Hopital’s at the second to last step.

 lim−1 (ln x)(ln (1 − x)) = lim−1

x→1

ln (1 − x)

x→1

1 ln x

= lim−1 x→1



1 1−x −1 (ln x)2

(−1)
= lim 1 −1 x

x→1


2 ln x x1 x (ln x)2 = lim−1 = 0 1−x (−1) x→1

Exercise 16. Persist in using L’Hopital’s and trying all possibilities systematically.

lim+ xx

x→0

x

−1

ln x = (ex ln x − 1)−1  2x ln x  1/x e − 2ex ln x + 1 = exp lim = exp − lim = ex ln x (x ln x + x) x→0+ (−1)(ex ln x − 1)−2 (ex ln x (ln x + 1)) x→0+

= lim+ e(x

x

−1) ln x

x→0

= exp− lim+ x→0

= lim+ e(e

x ln x

−1) ln x

x→0

= exp lim+ x→0

ex ln x (ln x + 1) + e−x ln x (− ln x − 1) ex ln x − e−x ln x = exp − lim+ = 1 1 (ln x + 1 + 1) x→0 1 + ln x+1

Exercise 17. 134

 x   x ln x    x x ln x ln x ln x lim+ (xx − 1) = lim+ ex ln x − 1 = lim+ ee − 1 = elimx→0+ e −1 =

x→0

x→0

=e

e

x→0

lim x ln x x→0+

limx→0+ ln x

− 1 = 0 − 1 = −1

We used lim xα log x = 0 ∀α > 0

x→0+

since t = x1 , xα log x =

− log t tα

→ 0 as t → ∞.

Exercise 18.

lim e

sin x ln (1−2x )

x→0−


!  1 − ln 2ex ln 2 ln (1 − ex ln 2 ) 1−ex ln 2 = = exp lim− = exp lim− −1 1/ sin x x→0 x→0 cos x sin2 x     (sin2 x) ln 2ex ln 2 2 sin x cos x = 1 = exp lim = exp (ln 2) lim x→0− (1 − ex ln 2 ) cos x x→0− − ln 2ex ln 2 

Exercise 19. 1

lim+ e ln x ln x = e

x→0

Exercise 20. At the end, L’Hopital’s could be used to verify that indeed sin x ln sin x → 0 as x → 0.

lim esin x ln cot x = elimx→0+ sin x(ln cos x−ln sin x) = elimx→0+ − sin x ln sin x = 1

x→0+

Exercise 21. Rewrite tan into sin and cos and use L’Hopital’s.

lim (tan x)tan 2x = limπ etan 2x ln tan x = exp limπ

x→ π 4

x→ 4

= exp limπ

1 sin x

x→ 4

x→ 4

1 (ln sin x − ln cos x) = cos 2x

cos x − cos1 x (− sin x) 1 = exp limπ = e−1 x→ 4 − sin2 2x −2 sin 2x cos 2x

Exercise 22. Exercise 23. Use L’Hopital’s theorem, taking derivatives of top and bottom.

lim exp

x→0+

ln x 1/x e ln x = exp e lim = exp e lim = ee x→0 1 + ln x x→0 1/x 1 + ln x

Exercise 24. Rewrite tan into sin and cos and take out sin since we could do the limit before doing L’Hopital’s.

lim (2 − x)tan (πx/2) = lim etan

x→1

x→1

πx 2

ln (2−x)

= elimx→1

sin πx/2 ln (2−x) cos πx/2

= exp lim

x→1 π 2

(−1) 2−x

sin πx/2

= exp

−2 π

Exercise 26.

    (x−c)−(x+c) 1 1+c/x    2 1+c/x ln 1−c/x (x−c) ( 1−c/x ) x+c = exp lim = exp lim = lim exp x ln −1 x→+∞ x→+∞ x→+∞ x−c 1/x x2 = exp (2c) = 4 =⇒ c = ln 2 Exercise 27.

(1 + x)c = exp (c ln (1 + x)) = exp (c(x − o(x))) = 1 + c(x − o(x)) + o(x − o(x)) = 1 + cx + o(x)  1/2  1/2 1 1 x2 1 + 2 − x2 = x2 1 + 2 − x2 x x Let x2 = 1t . So t → 0 as x → +∞. =⇒

1 + 12 t − 1 + o(t) (1 + t)1/2 − 1 1 = = t t 2 135

Exercise 28. 5

4

7 1 + 4 +2 t5 t

c

c

(x + 7x + 2) − x = x Let

1 t

= x and guess that c =

5



7 2 1+ + 5 x x

c −x

1 5



−


1/5 1 1 /t − = = 1 + (7t + 2t5 ) t t =

1 + 15 (7t + 2t5 ) + o(t) − 1 7 = t 5

Exercise 29. x

  1 g(t) t + dt t 1   1 − g(1)2; f 0 (x) = g(x) x + x Z

f (x) =

x2

g(x) = xe

2

g 0 (x) = ex + 2x2 ex

2

2

2

2

2

g 00 (x) = 2xex + 4xex + 4x3 ex = 6xex + 4x3 ex 2

2

2

2

(ex + 2x2 ex )(x + 1/x) + xex (1 − f 00 (x) = 00 g (x) 6xex2 + 4x3 ex2 =

f 00 (x) = g 0 (x)(x + 1/x) + g(x)(1 − 1/x2 ) 2

1 x2 )

=

2

2

2xex + 2x3 ex + 2xex = (6xex2 + 4x3 ex2 )

4x + 2x3 1 as x → ∞ = 3 6x + 4x 2

Exercise 30.

Z

g(x) = xc e2x

f (x) =

x

e2t (3t2 + 1)1/2 dt

0

g 0 (x) = cxc−1 e2x + 2xc e2x

f 0 (x) = e2x (3x2 + 1)1/2 − 1

Guessing that c = 1 f 0 (x) e2x (3x2 + 1)1/2 − 1 = = g 0 (x) 2xc e2x + cxc−1 e2x

√

3x(1 + 3x1 2 )1/2 − e−2x = 2x + 1

√

3 2

So c = 1 . Exercise 31. Exercise 32.

(1)    r 2 r m r ,P 1 + ,...P 1 + P 1+ m m m For each year, there are the just previously shown m compoundings, so for n years,  r mn P 1+ m (2) 2 = ert ln 2 = t = 11.55years r (3)  r mt 2P0 = P0 1 + m ln 2 = mt ln (1 + r/m) t=

ln 2 ln 2 = = 11.64years m ln (1 + r/m) 4 ln (1 + 0.06/4) 136

7.17 Exercises - The symbols +∞ and −∞. Extension of L’Hopital’s rule; Infinite limits; The behavior of log x and ex for large x. Exercise 1. 1 x2 2 500 e−1/x u 1 lim = lim e−u u5 00 = lim =0 x2 = u→∞ u→∞ eu x→0 x1000 u 1 1000 x = 5 u 00 Where we had used Theorem 7.11, which are two very useful limits for log and exp. u=

Theorem 26. If a, b > 0, (log x)b =0 x→+∞ xa xb lim ax = 0 (18) x→+∞ e Proof. Trick - use the definition of the logarithm as an integral. If c > 0, t ≥ 1, tc ≥ 1 =⇒ tc−1 ≥ t−1 . Z x Z x 1 xc 1 0 < ln x = dt ≤ tc−1 dt = (xc − 1) < c c 1 t 1 cb−a b x (ln x) < =⇒ 0 < xa cb cb−a −a/2 a x x Choose c = , b = → 0 as x → ∞ 2b c cb (ln x)b then → 0 as x → 0 xa lim

(17)

For exp, Let t = ex . ln t = x.

xb eax

=

(ln t)b ta

→ 0 as t → ∞ as x → ∞.



Exercise 2.

sin x1 lim = lim x→0 arctan 1 x→0 x

1 x 1 x

−o −o

1 x
1 x


= 1

Exercise 3. Use L’Hopital’s.

lim

x→ π 2

1 tan 3x cos x − sin x = limπ = limπ = − x→ 2 cos 3x x→ 2 −3 sin 3x tan x 3

Exercise 4. Use L’Hopital’s. 1 x a+bex (be ) bx √ a+bx2

ln (a + bex ) √ = lim x→∞ x→∞ a + bx2 lim

Exercise 5. Make the substitution x =

lim x

4



x→∞

1 ae−x +b 1 x→∞ √ a +b x2

= lim

1 =√ b

1 t.

1 1 cos − 1 + 2 x 2x



1 = lim 4 t→∞ t



t2 cos t − 1 + 2



t4 /4! + o(t4 ) 1 1 = = 4 t→∞ t 4! 120

= lim

Exercise 6.

lim

x→π

ln | sin x| = lim ln | sin 2x| x→π

1 sin x 2 sin 2x

cos x 1 sin 2x 1 2 cos 2x = − lim = − lim = 1 2 x→π sin x 2 x→π cos x cos 2x

Exercise 7.



lim−

x→ 21

1 1−2x



(−2) ln (1 − 2x) −2(cos2 πx) = lim− = lim− = 2 tan πx (sec πx)π (1 − 2x)π x→ 12 x→ 12 =−

2 2 cos πxπ − sin πx lim =1 π x→ 12 − −2 137

Exercise 8.

cosh x + 1 ex+1 + e−x−1 1 1 e = lim = lim e1 + 2x+1 = x x→∞ x→∞ e 2ex 2 x→∞ e 2 lim

Exercise 9.

ex ln a ax = lim → ∞; a > 1 x→∞ xb x→∞ xb b x since lim ax (in this case, ln a > 0 ) x→∞ e lim

Exercise 10.

limπ

x→ 2

tan x − 5 sec2 x sec x 1 cos x = limπ = limπ = limπ = 1 x→ 2 tan x sec x x→ 2 tan x x→ 2 cos x sin x sec x + 4

8.5 Exercises - Introduction, Terminology and notation, A first-order differential equation for the exponential function, First-order linear differential equations. The ordinary differential equation theorems we will use are y 0 + P (x)y = 0 Z A(x) =

(19)

x

P (t)dt

a

y = be−A(x)

(20) Rx

Consider y 0 + P (x)y = Q(x); A(x) =

a

P (t)dt.

Let h(x) = g(x)eA(x) ; g a solution. h0 (x) = (g 0 + P g)eA = QeA Z

2nd. fund. thm. of calc.

x

−−−−−−−−−−−−−→ h(x) = h(a) +

Q(t)eA(t) dt

a

since h(a) = g(a) y = g(x) = e

(21)

−A(x)

Z

x

Q(t)e

A(t)

 dt + b

a

We had done some of these problems previously, using an integration constant C, but following Apostol’s notation for y(a) = b for initial conditions is far more advantageous and superior as we seem clearly the dependence upon the initial conditions - so some of the solutions for the exercises will show corrections to the derived formula using Apostol’s notation for y(a) = b initial conditions. Rx Exercise 1. A(x) = 0 (−3)dt = −3x y = e3x

x

Z

e2t e−3t dt + 0



0

Exercise 2. y 0 −

2 xy

= x4 . A(x) =

y=e

−A(x)

Rx

Z

1

−2 t




dt = −2 ln x.

x

Q(t)e

x = e3x (−e−t ) 0 = −e2x + e3x

A(t)

 dt + b

=e

R 2 ln x( 1x t4 e−2 ln t dt+1)

a

= −x2 +

Z

π 2




, with y = 2 when x = 0. 138

x 2

t dt + 1 1

x2 3 x5 2x2 (x − 1) = + 3 3 3

Exercise 3. y 0 + y tan x = sin 2x on − π2 ,

=x

2

 =

x

Z

y = e−A(x)

x

Z

0 Z x

x

tan tdt = − ln | cos t||0 = − ln cos x

P (t)dt =

A(x) =

0

Q(t)eA(t) dt + b



=

a

x

Z

sin 2te− ln cos t dt + 2

= cos x

a

 2 sin tdt + 2

= cos x

!

b

Z

= −2 cos2 x + 4 cos x

a 0

3

Exercise 4. y + xy = x . y = 0, x = 0.

1 A(x) = x2 2  Z x  t2  x −x2 1 2 t2 t2 y = e− 2 x t3 e 2 dt + 0 = e 2 t2 e 2 − 2e 2 = 0

0

2

2

= x − 2 + 2e

− x2

Exercise 5. y 0 + y = e2t . y = 1, t = 0.

y(x) = e−x

x

Z

e2t et dt + 1



= e−x



0

A=x

x  e3t + 1 3 0

2 e2x + e−x 3 3

=

Exercise 6. y 0 sin x + y cos x = 1; (0, π). =⇒ y 0 + cot xy = csc x

  Z x sin x cot tdt = ln A(x) = sin a a    Z  sin x sin t sin a x−a +b y(x) = e− ln ( sin a ) (csc t)eln ( sin a ) dt + b = sin x sin a x − a b sin a indeed y = + sin x sin x x + b sin a − a x − a b sin a + = x → 0 for y = sin x sin x sin x b sin a = a for x → 0 a − b sin a = π for x → π Exercise 7. 2 2 1 x(x + 1)y 0 + y = x(x + 1)2 e−x =⇒ y 0 + y = (x + 1)e−x x(x + 1)  Z x 1 x x+1 1 A(x) = − dt = ln − ln t t+1 a a+1 a (a + 1)x eA(x) = a(x + 1) Z x    Z x  2 a(x + 1) (a + 1)t (x + 1) a(x + 1) −t −t2 y= (t + 1)e dt + b = te dt + b = (a + 1)x a(t + 1) x (a + 1)x a a  a(x + 1)b 2 x + 1  −a2 e − e−x + = 2x (a + 1)x

It’s easy to see that the last equation above goes to 0 as x → −1. 2

2

y = (e−a − e−x )(1/2)(1 + lim y =

x→0

1 x

2

2

1 ab 1 )+ (1 + ) = x a+1 x !

e−a − e−x ab + 2 a+1 139

a=0

Exercise 8. y 0 + y cot x = 2 cos x on (0, π). x



 sin x A(x) sin x A(x) = cot tdt = ln e = sin a sin a a x  Z   sin t sin a sin a cos 2t 2 cos t +b = y= +b = − cos (2a) − cos (2x) sin a sin x sin a sin x 2 sin a a y = sin x y = + 2 sin x sin x cos (2a) + cos 2x sin a = + 2 sin x sin x Z

Exercise 9. (x − 2)(x − 3)y 0 + 2y = (x − 1)(x − 2). y 0 +

Z

x

A(x) = a

x

2 (x−2)(x−3) y

=

x−1 x−3 .

x − 3 a − 2 x dt = 2 (ln |t − 3| − ln |t − 2|)|a = 2 ln a − 3 x − 2 a x − 3 ≶ 03 − x x−3 If (−∞, 2), (3, +∞), = x − 2 ≶ 02 − x x−2

2dt =2 (t − 2)(t − 3)

Z



1 1 − t−3 t−2



If (2, 3), x − 3 < 0, but x − 2 > 0   2   2   a − 3 2 x−2 1 1 + x−2 If (−∞, 2), (3, ∞) y = b x + − a − a − 2 x−3 x−3 x−2 a−2 2  2 Z  2  2  2  2  a − 3 x − 2 a − 3 t − 1 a − 2 3 − t x−2 = If (2, 3) y = b a − 2 + 3 − x a − 2 x−3 t − 3 a − 3 t − 2 2  2  Z  x − 2 a − 3 x−2 (t − 1)(3 − t) − = =b + 3−x a−2 3−x (t − 2)2 2  2 2  x−2 x − 2 a − 3 + (x + (x − 2)−1 − a − (a − 2)−1 ) =b x − 3 a − 2 x−3 Exercise 10. s(x) =

sin x x ;x

6= 0 s(0) = 1, T (x) =

Rx 0

s(t)dt f (x) = xT (x)

f 0 = T + x(s(x)) = T + sin x xf 0 − f = x sin x y y 0 − = sin x x Z x −1 x x A(x) = dt = − ln , e−A(x) = t a a a Z  x x bx a y= sin t + b = xT + a t a a 1 y(0) = 0 6= 1 since P (x) = is not continuous at x = 0 x Exercise 11.

f (x) = 1 + Z

1 x

Z

x

f (t)dt 1

x

xf (x) = x +

f (t)dt =⇒ f (x) + xf 0 = 1 + f (x) =⇒ f 0 =

1

1 x

=⇒ f (x) = ln |x| − C Z

x

1 (ln |t| − c) dt = x 1 f (x) = ln |x| + 1 1 1+C x = 1 + (t ln |t| − t − ct)|1 = ln |x| − C + =⇒ C = −1 x x

ln |x| − C = 1 +

Exercise 12. Rewrite the second property we want f to have:

Z

x

f (t)dt = 1

1 − f (x) 1 =⇒ f (x) = − 2 − x x 140



f 0x − f x2



So then   1 1 f0 + x − f =− x x      Z Z  1 2 1 1 2 A(x) = P (t)dt = dt = x − ln x − a − ln a t− t 2 2 a2 −x2 x2 −a2 a x ; e−A(x) = e 2 eA(x) = e 2 x a Z x  −1 t2 −a2 x a2 −x2 2 2 ae dt + b f (x) = e 2 a a t Exercise 13.

v = yk v 0 = ky k−1 y 0

v 0 + kP v = kQ = ky k−1 y 0 + kP y k = kQ where k = 1 − n =⇒ y 0 y −n + P y 1−n = Q =⇒ y 0 + P y = Qy n

Exercise 14. y 0 − 4y = 2ex y 1/2

n=

1 1 1 1 1 k = 1 − = ; v = y 1/2 ; v 0 + (−4)v = (2ex ) = v 0 − 2v = ex 2 2 2 Z 2 2 x A(x) = P (t)dt = −2(x − a) = −2x a Z 
v = e2x et e−2t dt + b = e2x −e−x + b = be2x − ex √ √ =⇒ y = (1 + 2)2 e4x − 2(1 + 2)e3x + e2x y(x) = b2 − 2b + 1 = 2 √ b=1+ 2

Exercise 15. y 0 − y = −y 2 (x2 + x + 1), n = 2 k = 1 − n = −1, v = y k v = y −1 .

v 0 + kP v = kQ v 0 + (−1)(−1)v = (−1)(−(x2 + x + 1)) = v 0 + v = x2 + x + 1 Z Z x A(x) = P (t)dt = 1dt = x 0 Z 
x v = e−x (t2 + t + 1)et dt + b = e−x t2 et − tet + 2et 0 + be−x = (x2 − x + 2) − (2e−x ) + be−x y=

x2

1 − x + 2 − 2e−x

Exercise 16.

v = eln x

Z

1 v 0 + − v = 2x2 x 

2x2 e− ln x dx + C

= x(x2 + C) = x3 + Cx

Then since v = y k , k = 1 − n, y = (x3 + Cx)2 = x2 (x2 + C)2 ; x 6= 0 y = x2 (x2 − 1)2 Check:

y 0 = 2x(x2 − 1)2 + x3 (4)(x2 − 1) 2x2 + x4 (4)(x2 − 1) − 2x2 (x2 − 1)2 = 4x3

Exercise 17. xy 0 + y = y 2 x2 log x on (0, +∞) with y =

1 2

when x = 12 , x 6= 0. 141

y = y 2 x log x x k = 1 − n = 1 − 2 = −1, v = y k = y −1   1 0 0 v = −1x log x v + kP v = kQ; v + − x Z  Z  −x log x v=x dt + C = x − log tdt + C = 1 x y= 2 Cx + x − x2 log x = x(−(x log x − x) + C) = −x2 log x + x2 + Cx y0 +

1 1 C = − (1 − log ) + 2 2 2

Check: y 0 = (C + 2x − 2x log x − x)(−y 2 ) y 2 (−C + −2x + 2x log x + x + C + x − x log x) y=

1 1 ; x = 1 y(x = 1) = b = 2 2 2 1 y= x(x ln x − x + 3)

Exercise 18. 2xyy 0 + (1 + x)y 2 = ex on (0, +∞), y =

√

√ e when x = 1; y = − e when x = 1.

If x 6= 0, y 6= 0, (1 + x) e2 −1 y= y 2x 2x k = 1 − n = 2; v = y k = y 2 y0 +

  2ex 1+x ex 1+x v= =⇒ v 0 + v= = v 0 + Pv = Qv v 0 + 2P v = 2Q; v 0 + 2 2x 2x x x Z x Z x x Rx 1+t x Av = Pv = dt = ln + (x − a); e a Pv dt = eln x/a+x−a = ex−a t a a a a x Z x t    −x+a ae e t t−a 1 2t−a a v= e dt + bv = e−x+a e + bv x t a x 2a a a   a −x+a e2x−a ea = e − + bv x 2a 2a s   a −x+a e2x−a ea y=± e − + bv x 2a 2a Now y k = v  1 − e2 + bv = bv = e r   2x−1 (1) bv = e =⇒ y = x1 e−x+1 e 2 − 2e + e r   2x−1 (2) bv = e =⇒ y = − x1 e−x+1 e 2 − 2e + e q ex (3) If we could take a = 0, then limx→0 y = ± 2x −

y 2 = v = e = 11 e−1+1



e2−1 2(1)

e−x+2a 2x

+

bv ae−x+a x

If we consider limx→0 y 2 , and let a go to 0, then v=


sinh x 1 x e − e−x =⇒ x x

Exercise 19. The Ricatti equation is y 0 + P (x)y + Q(x)y 2 = R(x). 142

q 2(0) −x+a v 0e = ± 1+x−e (1−x)+2b = ±1 2x

If u is a known solution, y = u +

1 v

is also a solution if v satisfies a first-order ODE.

−1 0 v v2 0 1 −v 2u 1 y 0 + P y + Qy 2 = R =⇒ u0 + 2 + P (u + ) + Q(u2 + + 2) = R v v v v (u + 1/v)0 = u0 +

=⇒ v 0 − P v = Q(2uv + 1)

Exercise 20. y 0 + y + y 2 = 2, y = 1, −2.

(1) If −2 ≤ b < 1, y 0 + y + y 2 = 2 P = 1, Q = 1, R = 2 1 v 0 + (−P − 2Qu)v = Q y = u + v u=1 v 0 + (−1 − 2(1)(1))v = v 0 − 3v = Q = 1  Z   −3t x
e 1 + b = be3x − v = e3x 1e−3t dt + b = e3x 1 − e3x−3a −3 a 3 3 y =1+ 3be3x − (1 − e3x−3a ) 1 b=1+ b 3 b2 − b − 1 = 0 y(0) = 1 + =⇒ 3b − (1 − e−3a ) √ 1± 5 b= 2 √ 3 1− 5 y =1+ b= 3x 3x 3be − (1 − e ) 2 (2) u = −2 0

0

v + (−1 − 2(1)(−2))v = v + 3v = 1 Z x   3x  e − e3a 1 − e3a−3x v = e−3x e3t dt + b = e−3x + be−3x = + be−3x 3 3 a √ 3 3 3 a=0 y = −2 + ; y(0) = −2 + −−→ y(0) = −2 + =⇒ b = −1 ± 2 1 − e3a−3x + 3be−3x 1 − e3a + 3b 3b √ 3 b ≥ 1 or b < −2, y = −2 + b = −1 ± 2 1 − e−3x + 3be−3x

8.7 Exercises - Some physical problems leading to first-order linear differential equations. Exercise 3.

(1) y 0 = −αy(t). y(T ) = y0 e−αT = 1 y0 . ln e = T . k

y0 n.

n = eαT so the relationship between T and n doesn’t depend upon

143

(2) f (a) = y0 e−ka ;

f (b) = y0 e−kb .

f (a) 1 f (a) f (a) = e−ka+kb =⇒ ln = −k(a − b); ln = −k f (b) f (b) a − b f (b)    a/b ln ff (a) f (a) 1− a (b) t ; f (a) = y0 f (b) ; =⇒ = y0 b f (t) = y0 exp  a/b a−b y0 (f (b))  f (t) =

f (a) (f (b))a/b

b  b−a 

f (a) f (b)

t  a−b

b

t

(f (a)) b−a f (a)− b−a = = a t f (b) b−a f (b) a−b

b−t

=

f (a) b−a a−t b−a

b−t

t−a

= f (a) b−a f (b) b−a

(f (b)) b−t w(t) = ; b−a

1 − w(t) =

b − a − (b − t) t−a = b−a b−a

Exercise 4. F = mv 0 = w0 − 43 v

t

v(t) = e− 8

w0 = 192 1 1 w0 w0 w0 − v =⇒ v 0 + v = = 32 v0 = =6=m m 8 8 m g Z t      −t w0 t 8w0 t/8 e 8 dt + b = e 8 (e − 1) + 0 = 256(1 − e−t/8 ) m 0 m v(10) = 256(1 − e−5/4 ) = 256(1 − 37/128) = 182

12 w0 F = mv 0 = w0 − 12v v 0 + v = = v 0 + 2v = 32 m m Z t  
 −2t 2x v(t) = e e 32dx + b = e−2t 16(e2t − e2t0 ) + b = 16(1 − e2(t0 −t) ) + be−2t v(10) = be

t0 −2(10)

= 182 =⇒ b = 182e20 v(t) = 16 + 166e20−2t

So then ( 256(1 − e−t/8 ) v(t) = 16 + 166e20−2t

if t < 10 if t > 10

Exercise 7.

(1) y 0 (t) = (y − M )k = ky − kM y 0 + −ky = −kM Z y = ekt ( −kM e−kt dt + b) = ekt (M (e−kt − 1) + b) M = 60◦ y(0) = b = 200◦ 1 ln T



yf = ekT (M (e−kT − 1) + 200) = M + ekT (200 − M )    yf − M 1 1 3 1 = k = (ln (60) − ln (140)) = ln = (ln 3 − ln 7) b−M T T 7 30 y(t) = 60 + 140e

ln 3−ln 7 t 30

(2) (ln 7 − ln 3) 30   Y − 60 (ln 140 − ln (T − 60)) ln = −kt =⇒ tf = for 60 < T ≤ 200 140 k y(t) = 60 + 140e−kt ;

(3) tf =

ln (140)−ln (30) k

=

30 ln 7/3

ln 14 3 = 54 minutes 144

k=

1 (4) M = M (t) = M0 − αt α = 10  Z Z t  −kM e−kt dt + b = ekt −k(M0 − αu)e−ku du + b = y = ekt 0

Z

= −kekt

t

(M0 e−ku − αueku )du + bekt =

0

t  −ku  t ! M0 e−ku ue −e−ku kt −α + = −ke + be = −k 0 −k k2 0    −kt −kt M0 1 e te = −kekt + bekt = (1 − e−kt ) − α − 2 + 2 k −k k k y(t) = −M0 ekt + M0 − αt − α/k + αekt /k + bekt = (−M0 + α/k + b)ekt + (M0 − αt − α/k) = 3 t y(t) = (140 + )e−kt + (60 − − (ln 3 − ln 7) 10 Exercise 8. y 0 (t) = −k(y − M0 );

y = e−kt

Z



 3 ) ln 3/7

y 0 + ky = kM0 .

tf

kM0 eku du + b




= e−ktf M0 (ektf − ekti ) + b = M0 (1 − e−k(tf −ti ) ) + be−ktf

ti

y(tf ) − M0 = −M0 e−k(tf −ti ) + be−ktf =⇒

65 − M0 = −M0 e

−k(5)

+ 75e

−k(5)

60 − M0 = −M0 e−k(5) + 65e−k(5)

 65 − M0 = −5k; = (75 − M0 )e =⇒ ln 75 − M0   60 − M0 = (65 − M0 )e−5k =⇒ ln = −5k 65 − M0 65 − M0 60 − M0 =⇒ = 75 − M0 65 − M0 −5k



1 k = ln 5



75 − M0 65 − M0



M0 = 55 Exercise 9.

Let y(t) = absolute amount of salt. Water is leaving according   to w(t)  = w0 + (3 − 2)t = w0 + t.

Salt leaving = So then

2 gal min.

y(t) salt w(t)

  −2y w0 + t y = =⇒ ln y = −2(ln (w0 + t) − ln w0 ) = −2 ln w0 + t w0 is the equation of motion given by the problem.  −2   w0 +t −2 w0 + t ln w0 y(t) = Ce =C w0  2  2 100 100 25 625 y(t) = 50 y(t = 60 min.) = 50 = 50 = ' 19.53 100 + t 160 64 32 0

Exercise 10.

Let y be the dissolved salt (total amount of) at t time. The (total) amount of water at any given time in the tank is w = w0 + t. There is dissolved salt in mixture that is leaving the tank at any minute. There is also salt from undissolved salt in the tank that is “coming into” the dissolved salt, adding to the amount of dissolved salt in the mixture. Thus y y  −1 gal y 0 (t) = (−2) +α −3 ; α= w w 3 min We obtained α easily by considering only the dissolving part and how it dissolves 1 pound of salt per minute if the salt concentration, wy was zero, i.e. water is fresh. 145

y 0 (t) =

−7 y + 1; 3 w

7 y =1 3 w0 + t

y0 =

7/3 w +t   Z Z0 7/3 7 7 w0 + t t P = = ln (w0 + t)|0 = ln w0 + t 3 3 w0 Z    −7/3 7/3 t w0 +u w0 +t ln − ln w0 w0 (1)e y=e du + b = 0 !  ! !  7/3 Z t  7/3 7/3  10/3 w0 + u w0 w0 3w0 w0 + t du + b = −1 +b = = w0 + t w0 w0 + t 10 w0 0 ! !  7/3  10/3 100 3(100) 100 + 60 − 1 + 50 ' 54.78 lbs. y= 100 + 60 10 100 P =

Exercise 11. LI 0 (t) + RI(t) = V (t)

V (t) = E sin ωt. I(t) = I(0)e−Rt/L + e−Rt/L

Z

t

V (x) Rx/L e dx L

0

E I(t) = I(0)e−Rt/L + e−Rt/L L Using

R

Z

t

sin ωxeRx/L dx

0

eax sin bxdx = aeax sin bx − beax cos bx I(t) = I(0)e

I(0) = 0

−Rt L

R Rt L Le

Ee−Rt/L + L

Rt

sin ωt − ωe L cos ωt +
R 2 + ω2 L

ω
R 2 L

!

+ ω2

So

I(t) =

E L

R L

sin ωt − ω cos ωt E(R sin ωt − ωL cos ωt) EωL EωL e−Rt/L = + 2 e−Rt/L = + 2
2 2 2 R 2 2 R + (ωL) R + (ωL) R + (ωL)2 + ω L ωL sin α = p 2 R + (ωL2 ) E sin (ωt − α) EωL =⇒ I(t) = p + 2 e−Rt/L 2 2 (R + ω 2 L2 ) R + (ωL) L=0

sin α = 0

Exercise 12.

( E(t) = I(t) = e−Rt/L

Z

E 0

if 0 < a ≤ t < b otherwise

t

0 = 0 for t < a 0

I(t) = e−Rt/L

Z a

t

 E  R(a−t) E Rx/L E L  Rt/L e dx = e−Rt/L e − eRa/L = 1−e L L L R R

E I(b) = (1 − eR(a−b)/L ) R for t > b, =⇒ I(t) = Exercise 13. From Eqn. 8.22,

dx dt

I(t) = Ke−Rt/L

Ra Ee−Rt/L Rb (e L = e l ) R

= kx(M − x) 146

for I(b) = I(b)

dx 1/kdx dx = dt = = kx(M − x) = kM x − kx2 ; =⇒ = dt 2 dt kM x − kx x(M − x)   1 1 1 ln x + − ln (M − x) =⇒ kdt = + dx = x M −x M M x M k(t−ti ) M k(t − ti ) = ln ; e (M − x) = x M −x M eM k(t−ti ) M = 1 + eM k(t−ti ) 1 + e−M k(t−ti )

x(t) =

Exercise 14. Note that we are given three equally spaced times.

M − x2 M = x2 + x2 e−α(t2 −t0 ) ; = e−α(t2 −t0 ) x2   M − x2 x1 = e−αt2 +αt0 +αt1 −αt0 = e−α(t2 −t1 ) x2 M − x1     M − x3 x2 M − x2 x1 = e−α(t3 −t2 ) = x3 M − x2 x2 M − x1 (M − x3 )(M − x1 )x22 = x1 x3 (M − x2 )2 = x1 x3 (M 2 − 2M x2 + x22 ) = x22 (M 2 − M (x1 + x3 ) + x1 x3 ) (x22 − x1 x3 )M 2 = M (x22 (x1 + x3 ) + −2x2 x1 x3 ) = (−x1 (x3 − x2 ) + x3 (x2 − x1 ))x2 =⇒ M = x2

(x3 (x2 − x1 ) − x1 (x3 − x2 )) x22 − x1 x3

Exercise 15.

dx dx = k(t)M x − k(t)x2 = k(t)dt =⇒ M dt M x − x2 Rt x M k(u)du = e ti M −x Me

x=

M

1+e

Rt

M

ti

k(u)du

Rt ti

k(u)du

t

Z

 k(u)du = ln

ti

x M −x



M

= 1+e

−M

Rt ti

k(u)du

Exercise 16.

(1) M = 23 92(23−3.9)−3.9(92−23) = 201 232 −3.9(92) (2)     150(122 − 92) − 92(150 − 122) 150(30) − 92(28) M = 122 = 122 = 216 (122)2 − 92(150) (122)2 − 92(150) (3) Reject. 8.14 Exercises - Linear equations of second order with constant coefficients, Existence of solutions of the equation y 00 + by = 0, Reduction of the general equation to the special case y 00 + by = 0, Uniqueness theorem for the equation y 00 + by = 0, Complete solution of the equation y 00 + by = 0, Complete solution of the equation y 00 + ay 0 + by = 0. Exercise 1. y 00 − 4y = 0

y = c1 e2x + c2 e−2x .

Exercise 2. y 00 + 4y = 0

y = c1 cos (2x) + c2 sin (2x).

Use Theorem 8.7. Theorem 27. Let d = a2 − 4b be the discrimnant of y 00 + ay 0 + by = 0. Then ∀ solutions on (−∞, ∞) has the form (22)

y = e−ax/2 (c1 u1 (x) + c2 u2 (x))

where (1) If a = 0, then u1 (x) = 1 and u2 (x) = x √ (2) If d > 0, then u1 (x) = ekx and u2 (x) = e−kx , where k = 2d √ (3) If d < 0, then u1 (x) = cos kx and u2 (x) = sin kx; where k = 12 −d 147

Exercise 3. y 00 − 4y 0 = 0;

a = −4. y = e2x (c1 e2x + c2 e−2x ) = c1 e4x + c2

Exercise 4. y 00 + 4y 0 = 0

y = e−2x (c1 e2x + c2 e−2x ) = c1 + c2 e−4x Exercise 5. y 00 − 2y 0 + 3y = 0

d = 4 − 4(3) = −8

Exercise 8. y 00 − 2y 0 + 5y = 0

d = −16

Exercise 9. y 00 + 2y 0 + y = 0

√

2x + c2 cos

2x)

y = e−x (1 + x) y = ex (1 + x)

d = 4 − 4(1)(1) = 0

y = 1, y 0 = 1; x = 0 d = −3 4 x

9 4

>0 −3x

3x

(c1 e 4 + c2 e 4 ) = c1 + c2 e −2 5 −2 −3x c2 = =⇒ y = + e 2 3 3 3

y=e

Exercise 12. y 00 + 25y = 0;

√

y = ex (c1 cos 2x + c2 sin 2x)

d = 4 − 4(1)(1) = 0

Exercise 10. y 00 − 2y 0 + y = 0 Exercise 11. y 00 + 23 y 0 = 0

=⇒ y = e−x (c1 sin

−3x 2

y = −1, y 0 = 0, x = 3.

y = c1 sin 5x + c2 cos 5x

−1 = c1 sin 15 + c2 cos 15

0

y = 5c1 cos 5x + −5c2 sin 5x 0 = 5c1 cos 15 − 5c2 sin 15c2 sin 15

= c1 cos 15

cos2 15 ) = c1 −1c1 (sin 15 + sin 15



1 sin 15



c1 = − sin 15 c2 = − cos 15

y = − sin 15 sin 5x − cos 15 cos 5x Exercise 13. y 00 − 4y 0 − y = 0;

y = 2;

y 0 = −1 when x = 1

d = 10 − 4(1)(−1) = 20 y = c1 e(2+ √

√ 5)x

√

+ c2 e(2−

5)x

√

√ √ √ √ y(x = 1) = c1 e2+ 5 + c2 e2− 5 = 2 =⇒ (5 + 2 5)c1 e2+ 5 + (5 − 2 5)c2 e2− 5 = 0 √ √ √ √ y 0 (x = 1) = (2 + 5)c1 e2+ 5 + (2 − 5)c2 e2− 5 = −1 √ 2 5 − 5 −2√5 √ c2 e c1 = 5+2 5 ! √ √ √ √ √ √ √ 2 5−5 4 5c2 e2− 5 5 + 2 5 √5−2 2 5 − 5 −2−√5 2− 5 2− 5 √ √ √ e √ e c2 e + c2 e =2= =⇒ = c2 c1 = 5+2 5 5+2 5 2 5 2 5 √ √ 2 5 − 5 −2−√5 (2+√5)x 5 + 2 5 √5−2 (2−√5)x √ e √ e e + e y= 2 5 2 5

Exercise 14. y 00 + 4y 0 + 5y = 0, with y = 2 and y 0 = y 00 when x = 0

16 − 4(1)(5) = −4 y = e−2x (c1 sin 2x + c2 cos 2x) y(x = 0) = c2 = 2 y 0 = −2e−2x (c1 sin 2x + 2 cos 2x) + 2e−2x (c1 cos 2x − 2 sin 2x) y 00 = 4e−2x (c1 sin 2x + 2 cos 2x) − 8e−2x (c1 cos 2x − 2 sin 2x) + 4e−2x (−c1 sin 2x − 2 cos 2x) y 0 (0) = −2(2) + 2(c1 ) = −4 + 2c1 y 00 (0) = 4(2) + (c1 ) + 4(−2) = −c1 = −4 + 2c1 4 y = e−2x ( sin 2x + 2 cos 2x) 3 148

c1 =

4 3

Exercise 15. y 00 − 4y 0 + 29y = 0

=⇒ u = e2x (c1 sin 5x + c2 cos 5x).

d = 16 − 4(1)(29) = −100

y 00 + 4y 0 + 13y = 0

v: d = 10 − 4(13) = −36

=⇒ v = e−2x (b1 sin 3x + b2 cos 3x) v(0) = b2

u(0) = 1(0 + c2 ) = c2 = 0 v = e−2x b1 sin 3x

u = e2x c1 sin 5x u0 = 2e2x c1 sin 5x + e2x c1 5 cos 5x π 1 = 1 = 2eπ c1 (1) c1 = π u0 2 2e 1 u0 (0) = π 5 2e 5 u0 (0) = v 0 (0) =⇒ b1 = π 6e

v 0 (0) = 3b1 u=

1 2x e sin 5x 2eπ

v = e−2x

5 sin 3x 6eπ

Exercise 16.

y 00 − 3y 0 − 4y = 0 00

0

y + 4y − 5y = 0

u v

9 − 4(1)(−4) = 25 16 − 4(1)(−5) = 36

u=e

(c1 e

−2x

5x 2

(b1 e

+ c2 e −

3x

5x 2

)

−3x

+ b2 e )   3x 5x u = 2e 2 c1 (sinh ) 2 u(0) = c1 + c2 = 0v(0) = b1 + b2 = 0 =⇒ v = 2b1 e−2x (sinh (3x))   −5x 3x 5 5x 3 3x 5x v 0 = −4b1 e−2x sinh (3x) + 6b1 e−2x cosh (3x) u0 = c1 e 2 (e 2 − e 2 ) + 2e 2 c1 cosh 2 2 2 v 0 (0) = 6b1 u0 (0) = 5c1 c1 =

v=e

3x 2

6b1 5

Exercise 17.

Assume k > 0 √ √ y = c1 sin kx + c2 cos kx

y 00 + ky = 0 d = −4(k) √ d −4k √ = = −k 2 2

√

y(0) = c2 = 0 √ √ y(1) = c1 sin k1 = 0 =⇒ k = nπ

−k = κ > 0

k < 0; √

y = c1 e

κx

+ c2 e

√ y(0) = c1 + c2 = 0 y = c1 sinh κx √ y = c1 sinh κ1 = 0 c1 = 0

√ − κx

;

so if k < 0, there are no nontrivial solutions satisfying fk (0) = fk (1) = 0

Exercise 18. y 00 + k 2 y = 0

√

d = −4k 2 < 0

d 2

=

2k 2

=k>0 149

y = c1 sin kx + c2 cos kx 0

y = kc1 cos kx − c2 k sin kx

y(a) = b = c1 sin ka + c2 cos ka y 0 (a) = m = kc1 cos ka − c2 k sin ka

kb cos ka = kc1 cos ka sin ka + c2 k cos2 ka m sin ka = kc1 cos ka sin ka − c2 k sin2 ka kb cos ka − m sin ka kb cos ka − m sin ka = c2 k =⇒ c2 =  k kb kb cos ka − m sin ka c1 sin ka = b − c2 cos ka = − cos ka = k k kb sin2 ka + m sin ka cos ka kb(1 − cos2 ka) + m sin ka cos ka = k k kb sin ka + m cos ka c1 = k     kb sin ka + m cos ka kb cos ka − m sin ka y= sin kx + cos kx k k k = 0 =⇒ y = mx − ma + b =

Exercise 19.

(1) y = k1 sin x + k2 cos x b2 = k1 sin (a2 ) + k2 cos (a2 ) = k1 s2 + k2 c2 b1 = k1 sin (a1 ) + k2 cos (a1 ) = k1 s1 + k2 c1

b2 c1 = k1 s2 c1 + k2 c2 c1 − (b1 c2 = k1 s1 c2 + k2 c1 c2 ) =⇒b2 c1 − b1 c2 = k1 (s2 c1 − s1 c2 )

s1 b2 = k1 s1 s2 + k2 s1 c2  y= y=

=⇒ k1 =

=⇒ k2 =

−(s2 b1 = k1 s1 s2 + k2 c1 s2 )   sin x +

b2 cos a1 − b1 cos a2 sin a2 cos a1 − sin a1 cos a2

s1 b2 − s2 b1 s1 c2 − c1 s2

b2 sin a1 − b1 sin a2 sin a1 cos a2 − cos a1 sin a2

 cos x

b2 cos a1 − b1 cos a2 b2 sin a1 − b1 sin a2 sin x + cos x if a2 − a1 6= πn sin (a2 − a1 ) sin (a1 − a2 )

otherwise, if a2 − a1 = πn; (2) It’s true if a1 = a2 = (3) y 00 + k 2 y = 0

π 4;

b2 c 1 − b1 c 2 = 0 b2 s1 − b1 s2 = 0

b2 c1 = b1 (−1)n c1 ;

if c cos (a1 ) 6= 0,

b2 = b1 (−1)n

b1 = b2 .

y = A sin kx + B cos kx y(a1 ) = A sin ka1 + B cos ka1 = b1 = AS1 + BC1 y(a2 ) = A sin ka2 + B cos ka2 = b2 = AS2 + BC2

     S1 C1 A b = 1 S2 C2 B b2       1 A C2 −C1 b1 = B −S2 S1 b2 S1 C2 − C1 S2

S1 C2 − C1 S2 = sin ka1 cos ka2 − cos ka1 sin ka2 = sin (k(a1 − a2 )) y=

b2 c1 − b1 c2 s2 c1 − s1 c2

b1 cos (ka2 ) − b2 cos (ka1 ) −b1 sin (ka2 ) + b2 sin (ka1 ) sin ka1 + sin ka2 sin (k(a1 − a2 )) sin (k(a1 − a2 ))   b2 − b1 if k(a1 − a2 ) 6= πn k = 0; y= (x − a1 ) + b1 a2 − a1

Exercise 20.

(1) u1 (x) = ex ;

u2 (x) = e−x

u02 = −e−x 150

u002 = e−x =⇒ y 00 − y = 0

(2) u1 = e2x

u2 = xe2x

u01 = 2e2x

u02 = e2x + 2xe2x

u001 = 4e2x

u002 = 2e2x + 2e2x + 4xe2x = 4e2x + 4xe2x = 4e2x (1 + x) u002 − 4u02 + 4u2 = 0 =⇒ y 00 − 4y 0 + 4y = 0

(3) u2 (x) = e−x/2 sin x u1 (x) = e cos x; −1 −x/2 u02 = e sin x + e−x/2 cos x 1 2 0 −x/2 −x/2 cos x + −e sin x u1 = − e 1 2 u002 = e−x/2 sin x + −e−x/2 cos x − e−x/2 sin x = 1 4 u001 = e−x/2 cos x + e−x/2 sin x + −e−x/2 cos x −3 −x/2 4 e sin x − e−x/2 cos x = −3 −x/2 4 −x/2 = e cos x + e sin x 5 4 u002 + u02 + u2 = 0 4 5 y 00 + y 0 + y = 0 4 −x/2

(4) u1 (x) = sin (2x + 1);

u2 (x) = sin (2x + 2) u01 = 2 cos (2x + 1)

u02 = 2 cos (2x + 2)

u001 = −4 sin (2x + 1)

u002 = −4 sin (2x + 2)

y 00 + 4y = 0 (5) u1 = cosh x u01 = sinh x

y 00 − y = 0

u001 = cosh x Exercise 21. w = u1 u02 − u2 u01 .

(1) w = 0 ∀x ∈ open interval I, 

u2 u1

0 =

u02 u1 − u01 u2 u2 = 0 =⇒ =c 2 u1 u1

If uu12 is not constant, then w(0) 6= 0 for at least one c in I (otherwise, it’d be constant). (2) w0 = u1 u002 − u2 u001 Exercise 22.

(1) w0 + aw = u1 u002 − u2 u001 + a(u1 u02 − u2 u01 ) = u1 (−bu2 ) + −u2 (bu1 ) = 0 w(x) = w(0)e−ax if w(0) 6= 0, then w(x) 6= 0 ∀x. (2) u1 6= 0 If w(0) = 0, w(x) = 0 ∀x, so uu12 constant. If uu12 constant, w(0) = 0 since from the previous part. Exercise 23. Recall the properties of the Wronskian.

(1) If W (x) = v1 (x)v20 − v2 v10 = v1 v20 − v2 v10 = 0 ∀x ∈ I, then vv21 constant on I (2) W 0 = v1 v200 − v2 v100 (3) W 0 + aW = 0 if v1 , v2 are solutions to y 00 + ay 0 + by = 0 W (x) = W (0)e−ax So if W (0) 6= 0, W (x) 6= 0 ∀ x 151

Consider adding together the solutions and the solution’s derivatives into some function f . By the linearity of the differential equation, we know that f is also a solution since it is a linear superposition of solutions. y 0 (x) = Av10 (x) + Bv20 (x)

y(x) = Av1 (x) + Bv2 (x)

f (0) = Av1 (x) + Bv2 (0) f 0 (0) = Av10 (0) + Bv20 (0)  0         1 v2 (0) A v2 (0) −v2 (0) f (0) f (0) A = =⇒ = v20 (0) B f 0 (0) f 0 (0) B W (0) −v10 (0) v1 (0)

 v1 (0) v10 (0)

since W (0) 6= 0,

 so y(x) =

this division is allowed above    v20 (0)f (0) − v2 (0)f 0 (0) v1 (0)f 0 (0) − v10 (0)f (0) v1 (x) + v2 (x) W (0) W (0)

f (0), f 0 (0) are initial conditions for y. f (0), f 0 (0) are arbitrary. But since W (0) 6= 0, W (0) = v1 (0)v20 (0) − v2 (0)v10 (0), we can do things like f (0) =

v20 (0)f (0) − v2 (0)f 0 (0) v2 (0)f 0 (0) − v10 (0)f (0) v1 (0) + v2 (0) 0 0 v1 (0)v2 (0) − v2 (0)v1 (0) W (0)

8.17 Exercises - Nonhomogeneous linear equations of second order with constant coefficients, Special methods for determining a particular solution of the nonhomogeneous equation y 00 + ay 0 + by = R. Exercise 1. y 00 − y = x

homogeneous solution c1 ex + c2 e−x . yp = −x

y = c1 ex + c2 e−x − x Exercise 2. y 00 − y 0 = x2 For the homogeneous solution x x
x yh = e 2 c1 e 2 + c2 e− 2 = c2 ex + c1

y 00 − y 0 = 0 d = (−1)2 − 4(1)(0) = 1 yp = Ax3 + Bx2 + Cx + D yp0 = 3Ax2 + 2Bx + C

y = c1 + c2 ex +

yp00 = 6Ax + 2B

−1 3 x + x2 3

Exercise 3. y 00 + y 0 = x2 + 2x x x x
e− 2 c1 e− 2 + c2 e 2 = c1 e−x + c2

P 0 = 3Ax2 + 2Bx + C P 00 = 6Ax + 2B 1 3Ax2 + 2Bx + C + 6Ax + 2B = x2 + 2x A= B=0 C=0 3 1 y = c1 e−x + c2 + x3 3 P = Ax3 + Bx2 + Cx + D;

Exercise 4. y 00 − 2y 0 + 3y = x3

√ √ u = ex (c1 sin 2x + c2 cos 2x)

3(Ax3 + Bx2 + Cx + D) 2(3Ax2 + 2Bx + C)

A=

1 3

B=

2 3

C=

8 9

D=

16 27

(6Ax + 2B) y = C1 ex sin

√

2x + C2 ex cos

√

Exercise 5. y 00 − 5y 0 + 4y = x2 − 2x + 1

yh = e

5x 2



e

3x 2

+e

−3x 2



= c1 e4x + c2 ex 152

1 2 8 16 2x + x3 + x2 + x + 3 3 9 27

d=

p

25 − 4(4) = 3 4(Ax2 + Bx + C)

4Ax2 + 4Bx + 4C

1 A= 4

−10Ax − 5B

− 5(2Ax + B)

2A

2A

1 5 − + 4C = 1 2 8 9 C= 32

1 B= 8

1 1 9 y = c1 e4x + c2 ex + x2 + x + 4 8 32

Exercise 6.

y 00 + y 0 − 6y = 2x3 + 5x2 − 7x + 2 −5x

−x

5x

yh = e 2 (e 2 + e 2 ) = e−3x + e2x p d = 1 − 4(−6) = 5 yp = Ax3 + Bx2 + Cx + D yp0 = 3Ax2 + 2Bx + C

−6Ax3 − 6Bx2 − 6Cx − 6D 3Ax2 + 2Bx + C

=⇒

yp00 = 6Ax + 2B A=

6Ax + 2B −1 3

B = −1

C=

1 2

D=

−7 12

1 1 7 y = C1 e−3x + C2 e2x − x3 + −x2 + x − 3 2 12

Exercise 7.

y 00 − 4y = e2x

v1 = e2x

yh = c1 e2x + c2 e−2x

v10 = 2e2x

v20 = −2e−2x v2 = e−2x

Use Theorem 8.9. Theorem 28. Let v1 , v2 be solutions to L(y) = 0 where L(y) = y 00 + ay 0 + by Let W = v1 v20 − v2 v10 . Then L(y) = R where y p = t 1 v1 + t 2 v2 Z (23)

t1 = −

v2

R dx; W (x)

e2x 1 t1 = − e−2x = x −4 4 Z e2x e4x t2 = e2x = −4 −16

Z t2 (x) =

v1

R dx W

Z

w = e2x (−2)e−2x − e−2x 2e2x = −4

x 2x e2x e + 4 −16 x e2x y = c1 e2x + c2 e−2x + e2x + 4 −16 yp =

Exercise 8. 153

y 00 + 4y = e−2x

w = sin 2x − sin 2x(2) − 2 cos 2x cos 2x = 2 yh = c1 sin 2x + c2 cos 2x Z Z cos 2xe−2x dx −1 t1 = − e−2x cos 2xdx = (e−2x cos 2x)0 = −2e−2x cos 2x + −2e−2x sin 2x 2 2 Z Z 1 sin 2xe−2x (e−2x sin 2x)0 = −2e−2x sin 2x + 2e−2x cos 2x e−2x sin 2xdx t2 = = 2 2 0  −2x e sin 2x − e−2x cos 2x e−2x cos 2x − e−2x sin 2x = e−2x cos 2x t = 1 4 8  −2x 0 −2x −2x e sin 2x + e−2x cos 2x e sin 2x + e cos 2x −2x t = 2 =e sin 2x −8 −4 yp =

e−2x sin 2x cos 2x − e−2x sin2 2x e−2x sin 2x cos 2x + e−2x cos2 2x e−2x + = 8 −8 8 y = c1 sin 2x + c2 cos 2x +

Exercise 9. y 00 + y 0 − 2y = ex

e−2x 8

d2 = 1 − (4)(1)(−2) = 9  3x  −1x −3x yh = e 2 e 2 + e 2 = ex + e−2x (xex )0 = ex + xex +(xex )00 = +(2ex + xex )

=⇒ 3ex + 2xex x

Exercise 10. y 00 + y 0 − 2y = e2x . yh = e− 2



c1 e

3x 2

+ c2 e −

3x 2



1 y = c1 ex + c2 e−2x + xe2 3 = c1 ex + c2 e−2x .

W (x) = v1 v20 − v2 v10 = ex (−2)e−2x − e−2x ex = −3e−x Z Z v1 R v2 R t2 = t1 = − W W Z −2x 2x Z x 2x e e 1 x e e −1 4x t1 = − = e t2 = = e −x −x −3e 3 −3e 12 1 1 1 y1 = t1 v1 + t2 v2 = ex ex + − e4x e−2x = e2x 3 12 4 1 2x x −2x y = c1 e + c2 e + e 4 Exercise 11. y 00 + y 0 − 2y = ex + e2x .

Consider solutions to Exercise 9,10. L(ya ) = ex ; L(yb ) = e2x ; L(ya + yb ) = ex + e2x 1 1 =⇒ y = c1 ex + c2 e−2x + xex + e2x 3 4 Exercise 12. y 00 − 2y 0 + y = x + 2xex . d = 4 − 4(1) = 0. Recall the definition to be learned for this section of exercises:

Theorem 29. Let d = a2 − 4b be the discriminant of y 00 + ay 0 + by = 0. Then every solution of this equation on (−∞, ∞) has the form (24)

y = e−ax/2 (c1 u1 (x) + c2 u2 (x)) (1) If d = 0 then u1 = 1, u2 = x √ (2) If d > 0, u1 = ekx ; u2 = e−kx , k = 2d (3) If d < 0, u1 = cos kx, u2 = sin kx, k =

√

−d 2 154

yh = ex (c1 + c2 x) = c1 ex + c2 xex Z x 2x3 −xex (2xex ) e (2xex ) t1 = = = x2 t2 = 2x e −3 e2x W (x) = ex (ex + xex ) − (xex )(ex ) = e2x Z

yp =

x3 ex 2x3 x e + x3 ex = −3 3

y=

Exercise 13. y 00 + 2y 0 + y =

x3 ex + c1 ex + c2 xex 3

e−x x2

yh = e−x (c1 + c2 x) = c! e−x + c2 xe−x     Z −xe−x e−x Z e−x e−x x2 x2 −1 t1 = = − ln x t2 = = −2x −2x e e x W = e−x (e−x + −xe−x ) − (−e−x )(xe−x ) = e−2x   −1 yp = − ln xe−x + xe−x = − ln xe−x − e−x x y = c1 e−x + c2 xe−x + (− ln x − 1)e−x

Exercise 14. y 00 + y = cot2 x. yh = c1 sin x + c2 cos x.

Z Z − cos x cos x cos2 x cos x(1 − sin2 x) 1 2 t1 = cot xdx = = =− + − sin x 2 2 −1 sin x sin x sin x Z Z Z sin x cos2 x 1 − sin2 x t2 = cot2 x = − =− = ln | csc x + cot x| + − cos x −1 sin x sin x Z

yp = −1 − sin2 x + cos x ln | csc x + cot x| − cos2 x = −2 + cos x ln | csc x + cot x| y = c1 sin x + c2 cos x − 2 + cos x ln | csc x + cot x|

Exercise 15. y 00 − y =

2 1+ex

yh = c1 ex + c2 e−x

=⇒ W (x) = −ex e−x − ex e−x = −2

Z −x 2 Z e 1+ex e−x t1 = − = = −2 1 + ex Z Z Z 1 1 −e−x 1 −x −x − = −e − = −e + = −e−x + ln (1 + e−x ) = ex 1 + ex 1 + ex e−x + 1 Z x 2 e 1+ex t2 = = − ln |1 + ex | −2 y = −1 + ex ln (1 + e−x ) + −e−x ln (1 + ex ) + c1 ex + c2 e−x Exercise 16. y 00 + y 0 − 2y =

ex 1+ex 155

√

Discriminant:

−1±

12 −4(−2) 2

=⇒ yh = c1 ex + c2 e−2x =⇒ W = ex (−2)e−2x − e−2x ex = −3e−x   Z e−2t et t Z 1+e 1 −1 1 t1 = − = = ln (1 + e−x ) −3e−t 3 1 + et 3   Z Z 2 Z e t et t 1+e 1 e3t u=et −1 u du = −−−→ = t2 = −3e−t −3 1 + et 3 1+u Z Z −u 1 −1 −1 −1 1 2 u+ u + −1 + = = = ( u − u + ln u + 1) 3 u+1 3 u+1 3 2 −1 −1 2x ex e + + ln (ex + 1) = 6 3 3 y1 =

= −2, 1

−1 x −1 e−x e−2x e ln (1 + e−x ) + + − ln (ex + 1) + c1 ex + c2 e−2x 3 6 3 3

Exercise 17. y 0 + 6y 0 + 9y = f (x); where f (x) = 1 for 1 ≤ x ≤ 2. f (x) = 0 for all other x.

d = 36 − 4(1)(9) = 0 yh = e

−3x

(c1 + c2 x) = c1 e−3x + c2 xe−3x

W (x) = e−3x (e−3x − 3xe−3x ) − (xe−3x )(−3e−3x ) = e−6x  x  3t  Rx e3t −3xe3x +e3x −te 3t  a < 1 < x −te dt = + + 92 e3  3 9 = 9 1  1 R2 3 6 6 6 2e3 t1 (x) = a < 1 < 2 < x 1 −te3t dt = −6e9+e + 2e9 = −5e 9 + 9    x R  3x 3x 3a 3t x 1 < a < x < 2 + e3t 9 = −xe + e9 + ae3 − −te3t dt = −te 3 3 a a Z −3t Z Z x
e f (t) 1 3x 3t t2 (x) = = e f (t) = e − e3a e3t = −6t e 3  a 3x −3xe + e3x x −3x +C + y1 = e 9 3 y = c1 e−3x + c2 xe−3x +

e3a 9

1 when 1 ≤ x ≤ 2; otherwise y = yh 9

Exercise 18. Start from y 00 − k 2 y = R(x). Suppose L(yp ) = yp00 − k 2 yp = R(x).

yh = c1 sinh (kx) + c2 cosh (kx);

L(yh ) = 0

So consider L(yp + yh ) = R(x); yp + yh = y1 is a nother particular solution.

The key to this problem is to apply the integration directly on the ODE itself, not to go the other way around by differentiating the supposed particular solution. Z x Z x Z x Rx dt sinh (k(x−t)) d2 y −−0−−−−−−−−−−→ dt 2 (t) sinh (k(x − t)) − k 2 dty(t) sinh (k(x − t)) = dtR(t) sinh (k(x − t))dt dt 0 0 0 Z x Z y 00 sinh (κ) = −y 0 (0) sinh (kx) − y 0 cosh (κ)(−k) = 0 Z 0 = −y (0) sinh (kx) + k(y(x) − y(0) cosh (kx) + k y(t) sinh (κ)) = Z = −y 0 (0) sinh (kx) + ky(x) − ky(0) cosh (kx) + k 2 y(t) sinh (κ) Z 1 x y 0 (0) sinh (kx) − y(0) cosh (kx) = dtR(t) sinh (k(x − t))dt =⇒ y(x) − k k 0 Rx Now note that L(yh ) = 0, so applying 0 dt sinh (k(x − t)) results in 0 still. 156

With yp (x) = y 0 (0) sinh (kx) k

1 k

Rx 0

dtR(t) sinh (k(x − t))dt +

y 0 (0) sinh (kx) k

+ y(0) cosh (kx), we can add a homogeneous solution of

+ y(0) cosh (kx) to yp (x) to obtain y1 (x) =

1 k

x

Z

dtR(t) sinh (k(x − t))dt 0

Now for y 00 − 9y = e3x ,  3x−3t  Z Z Z 1 x 1 x e − e−3x+3t 1 x 3t 3t y1 (x) = dt(e ) sinh (3(x − t))dt = dte dt(e3x − e−3x+6t ) = 3 0 6 0 2 6 0 x   1 e6x − 1 1 e3x e−3x 1 3x −3x 1 6t = xe3x − e−3x e x−e e = (xe3x − + ) = 6 6 6 36 6 6 6 0 1 3 3 y10 = (e3x + 3xe3x − e3x − e−3x ) 6 2 2  1 9 3x 9 −3x 1 3 3 00 3x 3x 3x y1 = 3e + 3e + 9xe − e + e = e3x + e−3x + xe3x 6 2 2 4 4 2 1 1 y100 − 9y1 = e3x + e−3x 2 2 Thus, we need to add homogeneous parts to our particular solution to make it work. So if yp =

e3x e−3x xe3x − + 6 y 6

then it could be checked easily with some computation, that this satisfies the ODE. Exercise 19. Start from y 00 + k 2 y = R(x)

Again, note that if L(yp ) = yp00 + k 2 yp = R(x), L(yp + yh ) = R(x) + 0 = R(x), so y1 = yp + yh is also a particular solution. Z x Z x Z x Rx dt sin k(x−t) d2 y 0 2 −−−−−−−−−−→ dt sin k(x − t) 2 + k dt sin k(x − t)y = R(t) sin k(x − t) dt 0 0 0 Z x Z x d2 y dt sin k(x − t) 2 = −y 0 (0) sin (kx) + k y 0 (t) cos (k(x − t))dt = dt 0 0 Z x 0 = −y (0) sin (kx) + k(y(x) − y(0) cos (kx) − k y(t) sin (k(x − t))dt) = Z 0x = ky(x) − ky(0) cos (kx) − y 0 (0) sin (kx) − k 2 y(t) sin (k(x − t))dt 0 Z 1 x y 0 (0) sin (kx) =⇒ y(x) = dtR(t) sin k(x − t) + y(0) cos (kx) + k 0 k −y 0 (0) We can add yh with c1 = −y(0), c2 = k Z 1 x y1 = dtR(t) sin k(x − t) k 0 Now for y 00 + 9y = sin 3x, then k = 3, Z 1 x y1 = sin 3t sin 3(x − t)dt 3 0 Z x sin 3t(sin 3x cos 3t − cos 3x sin 3t) 0 Z x Z x s(6t) −1 sc = dt = (c(6x) − 1) 2 12 0 0 x Z x Z x 1 − cos (6t) x sin (6t) x sin (6x) s2 = = − = 2 − 12 2 2 12 0 0 0    −1 1 x sin (6x) sin 3x x cos 3x y1 = sin 3x (cos (6x) − 1) − cos 3x − = − 3 12 2 12 18 6 157

It could be shown with some computation that this particular solution satisfies the ODE without having to add or subtract parts of a homogeneous solution. Exercise 20. y 00 + y = sin x

=⇒ W (x) = −s2 − c2 = −1   Z Z cs − cos 2x x sin 2x ss 1 − cos 2x t1 = − = ; t2 = =− =− − −1 4 −1 2 2 4   3 2 sin x cos 2x sin 2x − 2x sin x cos x + sin x − 2x cos x yp = − + cos x = 4 4 4

yh = c1 sin x + c2 cos x

Z

y = c1 sin x + c2 cos x +

sin x cos2 x + sin3 x − 2x cos x 4

Exercise 21. y 00 + y = cos x

yh = c1 sin x + c2 cos x = c1 S + c2 C W (x) = −1 Z Z Z x + sin22x −CC 1 + cos 2x SC cos 2x t1 = = = ; t2 = = −1 2 2 −1 4 x sin x sin 2x sin x cos x cos 2x + + yp = 2 4 4 x sin x sin 2x sin x cos x cos 2x + + + c1 sin x + c2 cos x =⇒ y = 2 4 4 Exercise 22. y 00 + 4y = 3x cos x

yh = c1 sin 2x + c2 cos 2x W (x) = − sin2 2x(2) + − cos2 2x(2) = −2 Z Z Z Z Z − cos (2x)(3x cos x) 3 3 3 2 t1 = = x cos x cos (2x) = xc(1 − 2s ) = xc − 3 xcs2 = −2 2 2 2 Z  3 0 Z Z c 3 3 3 3 3 3 = (xs + c) − 3 x = (xs + c) − (xs − s ) = (xs + c) − xs + s(1 − c2 ) = 2 3 2 2 3 1 3 c 1 = (xs + c) − xs3 + −c + c3 = xs + − xs3 + c3 2 Z 3 2 2 3 Z Z Z Z sin (2x)(3x cos x) 2 3 0 3 3 3 t2 = = −3 xsc = x(c ) = xc − c = xc − c(1 − s2 ) = −2 1 = xc3 − s + s3 3   3 c 1 3 s3 3 yp = xs + − xs + c (2sc) + (xc3 − s + )(1 − 2s2 ) = (lots of algebra) = 2 2 3 3 2 = xc2 + s 3 =⇒ y = c1 sin 2x + c2 cos 2x + x sin x −

2 cos x 3

Remember, persistence is key to work through the algebra, quickly. Exercise 23. y 00 + 4y = 3x sin x. From the work above, we could guess at the solution.

(xs)0 = s + xc (xs)00 = 2c + −xs

(xs)00 + 4(xs) = 2c − xs + 4xs = 2c + 3xs

(c)00 + 4c = 3c =⇒

2 =⇒ yh = xs − c 3 y = x sin x −

2 cos x + c1 sin 2x + c2 cos 2x 3 158



00   −2 −2 c +4 c = −2c 3 3

Exercise 24. y 00 − 3y 0 = 2e2x sin x Guessing and stitching together the solution seems easier to me.

(e2x s)0 = 2e2x s + e2x c

(e2x c)0 = 2e2x c + −e2x s

(e2x s)00 = 4e2x s + 4e2x c + −se2x =

(e2x c)00 = 4e2x c + −4e2x s − e2x c = 3e2x c − 4e2x s 2x

00

2x

0

2x

2x

2x

= 3e2x s + 4e2x c

2x

(e c) − 3(e c) = 3e c − 4e s − 6e c + 3e s =

(e2x s)00 − 3(e2x s)0 = 3e2x s + 4e2x c − 6e2x s − 3e2x c =

= −3e2x c − e2x s

= −3e2x s + e2x c

(3e2x s)00 − 3(3e2x s)0 + (e2x c)00 − 3(e2x c)0 = −10e2x s =⇒ yp =

√ √ e2x (3 sin x + cos x) e2x (3 sin x + cos x) =⇒ y = c1 sin 3x + c2 cos 3x + 5 −5

Exercise 25. y 00 + y = e2x cos 3x.

(e2x c(3x))00 = 4e2x c(3x) + −12e2x s(3x) − 9e2x c(3x) = −5e2x c(3x) − 12e2x s(3x) (e2x s(3x))00 = 4e2x s(3x) + 12e2x c(3x) + −9e2x s(3x) = −5e2x s(3x) + 12e2x c(3x) L(e2x c(3x) − 3e2x s(3x)) = −40e2x cos (3x) y = c1 sin x + c2 cos x +

e2x cos (3x) − 3e2x sin (3x) −40

8.19 Exercises - Examples of physical problems leading to linear second-order equations with constant coefficients. In exercises 1-5, a partcile is assumed to be moving in simple harmonic motion, according to the equation y = C sin (kx + α). The velocity of the particle is defined to be the derivative y 0 . The frequency of the motion is the reciprocal of the period. (Period = 2π/k, frequency = k/2π ) Exercise 1. Find the amplitude C if the frequency is 1/π and if the initial values of y and y 0 (when x = 0) are 2 and 4,

respectively. k 1 = =⇒ k = 2 2π π y(x = 0) = C sin α y(x = 0) 1 1 =⇒ 0 = tan α = 0 y (x = 0) k 2 y (x = 0) = C cos α √ π α = and C = 2 2 4 Exercise 2. Find the velocity when y is zero, given that the amplitude is 7 and the frequency is 10. frequency =

y = C sin (kx + α)

k = 10 =⇒ k = 20π 2π y = Ck sin (kx + α) −α −α y(x = ) = 0 =⇒ y 0 (x = ) = 140π k k 0

C=7

Exercise 3.

y = A cos (mx + β) y = A cos (mx + β) = A cos β cos (mx) − A sin β sin (mx) y = C sin kx + α = C cos α sin kx + C sin α cos kx −A sin β = C cos α A cos β = C sin α

=⇒ tan α = ± cot β = ∓ tan =⇒ α = β −

Exercise 4.

2π T

=⇒ k = m (since x is arbitrary ) π

π and |C| = |A| 2

= 4π

y = C cos (kx + α) = C cos (kx) = 3 cos (4πx) 159

 π − β = tan (β − ) 2 2

Exercise 5. y = C cos (x + α)

y0 = C cos (x0 + α)

y 0 = −C sin (x + α) = ±v0 v02 + y02 = C 2 sin2 (x + α) + C 2 cos2 (x0 + α) = C 2

=⇒ C =

q

v02 + y02

Exercise 6.

y 0 = −kC sin (kx + α)

y = C cos (kx + α) y(0) = C cos (α) = 1 y 00 (0) = −k 2 C cos (α) = −12

y 0 (0) = −kC sin (α) = 2 √ −12 y 00 (0) = −k 2 = =⇒ k = 2 3 y(0) 1

y 0 (0) −kC sin (α) 2 = = = 2 = −k tan (α) y(0) C cos (α) 1 Exercise 7. k =

y = −C sin

2π 3

=⇒ α =

−π 6

y = −C sin (kx)

2πx ; 3

C>0

Exercise 8. Let’s first solve the homogenous equation.

y 00 + y = 0 Z x − cos t(1) t1 = = sin x yh = C1 sin x + C2 cos x −1 0 Z x (sin t)(1) W (x) = −S 2 − C 2 = 1 = (cos x − 1) t2 = −1 0 for 0 ≤ x ≤ 2π otherwise, for x > 2π, t1 = 0, t2 = 0 y1 = sin2 x + cos2 x + (1 − cos x)

y(0) = 0 = c2 y 0 (0) = c1 = 1

0

y (x) = C1 cos x + sin x y = sin x + (1 − cos x) =⇒ I(t) = sin t + (1 − cos t)

0 ≤ t ≤ 2π

Exercise 9.

(1) Consider large t. Then I(t) = F (t) + A sin (ωt + α) → A sin (ωt + α) I = AS(ωt + α) = A(S(ωt)C(α) + C(ωt)S(α)) I 0 = ωAC(ωt + α) = ωA(C(ωt)C(α) − S(ωt)S(α)) I 00 = −ω 2 AS(ωt + α) = −ω 2 A(S(ωt)C(α) + C(ωt)S(α)) I 00 + RI 0 + I = I 00 + I 0 + I = = A((−ω 2 C(α) + −ωS(α) + C(α))S(ωt) + (−ω 2 S(α) + ωC(α) + S(α))C(ωt)) = S(ωt) −ω =⇒ tan (α) = 1 − ω2 1 − ω2 1 − ω2 + ω4 = sec2 , and S 2 +C 2 = 1, we can get −ω S=√ 1 − ω2 + ω4 C=√

With tan α =

−ω 1−ω 2

and the trig identities t2 +1 = sec2 ,

1 C2

Note that the sign of S is fixed by tan. A(−ω 2 C(α) + −ωS(α) + C(α))S(ωt) = S(ωt) =⇒ A = =⇒ A = √

1 (1 −

ω 2 )C(α)

1 ω4 − ω2 + 1

160

− ωS(α)

=

(1 −

D − ω(−ω)

ω 2 )2

We could immediately see that ω = √12 , f = 2π1√2 will maximize A. (2) We could have, from the beginning, considered the problem with any R, in general. A((−ω 2 C(α) + −RωS(α) + C(α))S(ωt) + (−ω 2 S(α) + ωRC(α) + S(α))C(ωt)) = S(ωt) −ωR S(α) = p 2 (ωR) + (1 − ω 2 )2 −ωR tan α = 1 − ω2 1 − ω2 C(α) = p (ωR)2 + (1 − ω 2 )2 1 1 =⇒ A = =p 2 −ω 2 C(α) − ωRS(α) + C(α) (ωR) + (1 − ω 2 )2 d

dω (ωR)2 + (1 − ω 2 )2 = ω 4 + ω 2 (−2 + R2 ) + 1 −− → 4ω 3 + 2ω(−2 + R2 ) = 2ω(2ω 2 + (−2 + R2 )) = 0

ω=0

=⇒

1 to have resonances −−−−−−−−−−→ R < √ 2ω = 2 − R 2 2

2

Exercise 10. A spaceship is returning to earth. Assume that the only external force acting on it is the action of gravity, and

that it falls along a straight line toward the center of the earth. The rocket fuel is consumed at a constant rate of k pounds per second and the exhaust material has a constant speed of c feet per second relative to the rocket. Let M (t) = M be the mass of the rocket + f uel combination at time t. With +y direction being towards earth, then the equation of motion is Fg = +M (t)g, where g = 9.8m/s2 . M (t)v(t) = M vR is the momentum of the rocket. M (t + h) = M (t) − ∆m = M − ∆m is the change in mass of the rocket due to spent fuel. ve = velocity of the exhaust in the lab frame = c + vR (t) ∆p = ∆m(c + vR ) + (M − ∆m)vR (t + h) − M vR = M (vR (t + h) − vR ) + −∆m(vR (t + h) − vR ) + ∆mc       vR (t + h) − vR vR (t + h) − vR ∆m ∆p =M + −∆m + c = M (t)g ∆t ∆t ∆t ∆t kc 0 M vR + = M (t)g g kt kc/g kc/g 0 Now M (t) = M0 − =⇒ vR =g− =g− g M M0 − kc g kt kt kc g ln (M0 − ) = gt + c ln (M0 − ) g −k g g      2 gt g k kt M0 cg kt yR = +c t − M0 ln M0 − − + ln M0 2 k g g g k vR = gt −

Exercise 11. 0 M vR = 0 vR

−kc = g

1 M0 −

vR = c ln (M0 −

−kc g

! kt g

kt ) g

g =⇒ yR = c k

M0 g = w =⇒ yR = c

g k





    kt kt kt M0 cg ln M0 − M0 ln M0 − − + g g g k

kt − w ln g



w − kt g

 −

kt g

 +

wc w ln k g

We could’ve also solved this problem with an initial velocity of v0 and gravity. Then kt vR (t) = gt + c ln (1 − ) + v0 M0 g      1 M0 g kt y(t) = v0 t + gt2 + c t− ln 1 − −t 2 k M0 g 161

Exercise 12.

M vR = (M − ∆m)(vR (t + h)) + 0 M (vR (t + h) − vR (t)) = (∆m)vR (t + h) k v0 k k vR =⇒ R = = kt g vR g(M0 − g ) M0 g(1 − Mkt0 g )   kt −w kt ln vR = (k/w) ln (1 − ) = − ln (1 − ) w k w   −w v0 kt −v0 w kt =⇒ x(t) = v0 vR = ln (1 − ) = ln (1 − ) k w k w 1 − kt w 0 M vR =

8.22 Exercises - Remarks concerning nonlinear differential equations, Integral curves and direction fields. =⇒ y 0 = −2 3

Exercise 1. 2x + 3y = C Exercise 2. y = Ce−2x

=⇒ y 0 = −2y =⇒ yy 0 = x =⇒ y 0 = xy ;

Exercise 3. x2 − y 2 = c Exercise 4. xy = c

=⇒ y 0 =

Exercise 5. y 2 = cx

=⇒

y2 x

−y x ;

y 6= 0

x 6= 0

= c =⇒ y 0 =

y 2x

x 6= 0

Exercise 6. x2 + y 2 + 2Cy = 1

1 x2 + y − = −2C y y 1 2xy − y 0 x2 + y0 + 2 y0 = 0 y2 y −2xy y0 = 1 + y 2 − x2 Exercise 7. y = C(x − 1)ex

y =C (x − 1)ex y 0 (x − 1)ex − (ex + (x − 1)ex )y =0 (x − 1)2 e2x xy y0 = x−1 Exercise 8. y 4 (x + 2) = C(x − 2)

y 4 (x + 2) =C x−2 (4y 3 y 0 (x + 2) + y 4 )(x − 2) − y 4 (x + 2) y 4 (x + 2) = 0 =⇒ 4y 3 y 0 (x + 2) + y 4 = 2 (x − 2) x−2 y y0 = (x − 2)(x + 2) Exercise 9. y = c cos x

=⇒ y 0 = − tan xy

Exercise 10. arctan y + arcsin x = C

1 1 y0 + √ =0 2 1+y 1 − x2 Exercise 11. All circles through the points (1, 0) and (−1, 0).

Start with the circle equation: (x − A)2 + (y − B)2 = R2 162

−(1 + y 2 ) =⇒ y 0 = √ 1 − x2

(1, 0): (1 − A)2 + B 2 = R2 =⇒ −(1 − 2A + A2 + B 2 = R2 ) (−1, 0): (−1 − A)2 + B 2 = 1 + 2A + A2 + B 2 = R2 =⇒ 4A = 0, A = 0 1 + B 2 = R2 x2 + (y − ±

p

R2 − 1)2 = R2

x2 + y 2 − 2By + (R2 − 1) = R2 =⇒ x2 + y 2 − 2By = 1 B depends upon R, the radius of the circles, so we could use B as the parameter for the family of circles. x2 + y 2 − 1 = 2By x2 1 + y − = 2B y y

1 2xy − y 0 x2 + y0 + 2 y0 = 0 y2 y

=⇒ y0 =

y2

2xy − x2 + 1

Exercise 12.

(x + A)2 + (y + B)2 = r2 (1 + A)2 + (1 + B)2 = 1 + 2A + A2 + 1 + 2B + B 2 = r2

− (−1 + A)2 + (−1 + B)2 = − 1 − 2A + A2 + 1 − 2B + B 2 = r2 =⇒4A + 4B = 0 =⇒ A = −B 2

(x + −B) + (y + B)2 = r2 2(x − B) + 2(y + B)y 0 = 0

=⇒ y 0 =

(y + B)y 0 = B − x

B−x y+B

√ p (1 − B) + (1 + B) = r =⇒ 2 (1 + B 2 ) = r or

r

r2 −1=B 2 =⇒ (so B could be treated as a parameter for the family of curves) 2

2

2

8.24 Exercises - First-order separate equations. Exercise 1. y 0 = x3 /y 2 1 3 3y

= 41 x4 + C

=⇒ y 3 = 34 x4 + C

Exercise 2. tan x cos y = −y 0 tan y

ln | cos x| =

1 cos y

Exercise 3. (x + 1)y 0 + y 2 = 0 1 y

= ln (x + 1) + c

Exercise 4. y 0 = (y − 1)(y − 2)



1 −1 + y−2 y−1



y0 = 1

=⇒ ln (y − 2) − ln (y − 1) = x

y−2 = ex y−1 √

Exercise 5. y 1 − x2 y 0 = x 1 2 2y

√ = − 1 − x2

√ =⇒ y 2 = −2 1 − x2

Exercise 6. (x − 1)y 0 = xy

Z ln y =

1+

1 = x + ln |x − 1| x−1

y = ex (x − 1) + C 163

Exercise 7. (1 − x2 )1/2 y 0 + 1 + y 2 = 0

arctan y = arccos x + C Exercise 8. xy(1 + x2 )y 0 − (1 + y 2 ) = 0

1 ln (1 + y 2 ) = 2

Z 

 1 1 x + C = ln x − − x 1 + x2 2 ln |1 + x2 |  2 x y2 = k √ 1 + x2

Exercise 9. (x2 − 4)y 0 = y

x −1 arctanh 2 Z 2 Z 1 1 dx since = dx =
x 2 x2 − 4 4 −1 2 Z 1 x du = ( where u = ) 2 u2 − 1 2 2 2 cosh u − sinh u (tanh (u))0 = = 1 − tanh2 u cosh2 u x 1 =⇒ y = k exp (− arctanh ) 2 2 ln y =

Exercise 10. xyy 0 = 1 + x2 + y 2 + x2 y 2 1 2

2

=⇒ y 2 = kx2 ex − 1

ln (1 + y 2 ) = ln x + 12 x2 + C

Exercise 11. yy 0 = ex+2y sin x

ye−2y e−2y ex sin x − ex cos x − = +C −2 4 2 (2y + 1)e−2y = −2ex (sin x − cos x) + C Exercise 12. xdx + ydy = xy(xdy − ydx)

y(1 − x2 )dy = x(−y 2 − 1)dx ydy 1 1 xdx = 2 =⇒ ln |1 + y 2 | = ln |x2 − 1| + C 1 + y2 x −1 2 2 1 + y 2 = (x2 − 1)K =⇒ y 2 = K(x2 − 1) − 1 Exercise 13. f (x) = 2 +

Rx 1

f (t)dt f 0 (x) = f (x)

=⇒ f (x) = Cex

=⇒ Cex = 2 + Cex − Ce1 f (x) = Exercise 14. f (x)f 0 (x) = 5x

C=

2 e

2 x e e

f (0) = 1 f (x)2 = 5x2 + C

=⇒ f (x) = ± p f (x) = 5x2 + 1

Exercise 15. f 0 (x) + 2xef (x) = 0

f (0) = 0 164

p

5x2 + C

e−y y 0 = −2x =⇒ −e−y = −x2 + C y = ln (x2 + C)−1

=⇒ y = − ln (x2 + 1)

Exercise 16. f 2 (x) + (f 0 (x))2 = 1

f = −1 02

y =1−y

y0 = ±

2

p

1 − y2

± arcsin (y) = x + c =⇒ f (x) = ± sin (x + c) Exercise 17.

Z

x

f (t)dt = K(x − a)

=⇒ f (x) = k > 0

a

f >0

∀x ∈ R

Exercise 18.

Z

x

d

dx f (t)dt = k(f (x) − f (a)) −− → f (x) = kf 0 (x)

a 1

=⇒ f (x) = Ce k x ; Exercise 19. x

Rx a

C>0

x

(f (t))dt = k(f (x) + f (a)) =⇒ f (x) = Ce k

a

x

kCe k − kCe k = kCe k + kCea/k =⇒ 2kCea/k = 0 =⇒ C = f =0 Rx Exercise 20. a f (t)dt = kf (x)f (a); f (x) = kf 0 (x)f (a)     f 0 (x) 1 x 1 = =⇒ ln f (x) = x + C; =⇒ f (x) = C exp kf (a) f (x) kf (a) kf (a) Z x   x x a x a t f (t)dt = kf (a)Ce kf (a) = kf (a)Ce kf (a) − kf (a)Ce kf (a) = kCe kf (a) Ce kf (a) a a  x  a a x kf f (a) e (a) − e kf (a) = Ce kf (a) e kf (a) 2a

x=a

−−−→ 0 = Ce kf (a) =⇒ C = 0 =⇒ f = 0 8.26 Exercises - Homogeneous first-order equations. Exercise 1. f (tx, ty) = f (x, y) homogeneity (or homogeneity of zeroth order).

v − xv 0 1 x = f (x, ) = f (1, ) v v2 v v Z 1 dv v − v 2 f (1, ) = xv 0 =⇒ ln x = v v − v 2 f (1, v1 )

y 0 = f (x, y) =

Exercise 2. y 0 =

−x y

Exercise 3. y 0 = 1 + y x

=v

 x 0

=

=⇒ 12 y 2 = − 12 x2 + C =⇒ y 2 = −x2 + C y x

=⇒ y 0 = v + xv 0 = 1 + v =⇒ v = ln x

y = x(ln x + C) Exercise 4. y 0 =

x2 +2y 2 xy 165

x2 + 2y 2 x 2y = + xy y x 1 =⇒ v + xv 0 = + 2v =⇒ v y0 =

v=

y x

1 ln |1 + v 2 | = ln x + C 2

v0 1 = 1 x v +v

=⇒ y 2 = (Cx2 − 1)x2

Exercise 5. (2y 2 − x2 )y 0 + 3xy = 0

if 2y 2 6= x2 , y 0 = y = vx y0 = v0 x + v

=⇒ y 0 = v 0 x + v =



1 −3v + v 1 + v2



v0 =

3vx2 3v = 2 2 2 x − 2v x 1 − 2v 2

1 − 2v 2 0 1 v = 2v(1 + v 2 ) x

=⇒ 1 2

3xy x2 − 2y 2

1 −3 v y/x 1 =⇒ ln v + ln (1 + v 2 ) = ln x + C =⇒ = Cx2 =  3 x 2 2 (1 + v 2 )3 x2 +y 2 x2

yx3 = C(x2 + y 2 )3 However, y0 = y x y0 = v0 x + v v=

3xy x2 − 2y 2

3x2 v 3v = − 2v 2 x2 1 − 2v 2   3v (v − 2v 3 ) 2(v + v 3 ) 1 3v 2 3 0 vx= − = =⇒ +− v 0 = =⇒ ln v + − ln |1 + v 2 | = 2 ln x + C 1 − 2v 2 1 − 2v 2 1 − 2v 2 v 1 + v2 x 2 y 2 /x4 v 2 = Cx =⇒ = Cx4 (x2 + y 2 )3 (1 + v 2 )3/2 =⇒ v 0 x + v =

x2

=⇒ y 2 = C(x2 + y 2 )3 Exercise 6. xy 0 = y −

p x2 + y 2 r

y =⇒ y = = x 0

1+

 y 2 x p v0 x = − 1 + v2

y v= x vx = y

0

v x+v =v−

p

1+

v+

p

v2

−v 0 1 √ = =⇒ x 1 + v2 p =⇒ ln (v + 1 + v 2 ) = ln x + C since   p 1 v 1 0 2 √ (ln (v + 1 + v )) = 1+ √ =√ 2 2 v+ 1+v 1+v 1 + v2

1 + v 2 = Cx

=⇒ v =

=⇒ 1 + v 2 = C 2 x2 − 2vCx + v 2

Cx 1 − 2 2Cx

=⇒ y =

Exercise 7. x2 y 0 + xy + 2y 2 = 0 166

1 Cx2 − 2 2C

v 0 x + v = −2v 2 − v y −2y 2 y − v = =⇒ x2 x x v 0 x = −2(v 2 + v)   Z v0 −2 1 1 v C = = v0 − = −2 ln x + C =⇒ = 2 v(v + 1) x v v+1 v+1 x

x2 y 0 = −2y 2 − xy

=⇒ y 0 =

−Cx C − x2

y= Exercise 8. y 2 + (x2 − xy + y 2 )y 0 = 0

  y  y 2 0 + 1− + y =0 x x x y −v 2 Let v = = v0 x + v =⇒ x 1 − v + v2   −v 2 −v(1 + v 2 ) v2 − v + 1 0 −1 1 −1 1 v0 x = − v = =⇒ v = = + v0 = − 2 2 2 2 1−v+v 1−v+v v(1 + v ) x v v +1 x y ln y = arctan + C =⇒ ln v − arctan v = − ln x + C =⇒ ln (vx) = arctan x + C x  y 2

Exercise 9. y 0 =

y(x2 +xy+y 2 ) x(x2 +3xy+y 2 )

y y(x2 + xy + y 2 ) y0 = = x(x2 + 3xy + y 2 ) x

1+

y y2 x + x2 3y y2 x + x2

!

y v= x



0

−−−→ v x + v = v

1 + v + v2 1 + 3v + v 2



1+ 3 1 −2 −1 v 0 (1 + + 2 ) = =⇒ v + 3 ln v + = −2 ln x + C v v x v x y + 3 ln y − = ln x + C x y

Exercise 10. y 0 =

y x

=v+

−2v 2 v 2 + 3v + 1

+ sin xy y =x x 0 y = v + v0 x

v + v 0 x = v + sin v =⇒

v0 1 = sin v x

− ln csc v + cot v = ln x + C

=⇒ csc v + cot v =

K x

Exercise 11. x(y + 4x)y 0 + y(x + 4y) = 0 y − y (1 + 4y −y(x + 4y) −v(1 + 4v) x ) v= x = xy −−−→ v + xv 0 = x(y + 4x) + 4 v+4 x   −5v(1 + v) −5 v+4 0 4 −3 0 xv = =⇒ = v = + v0 v+4 x v(1 + v) v 1+v

y0 =

R

− → 4 ln v − 3 ln (1 + v) = −5 ln x + C =⇒ (yx)4 = (x + y)3 C 8.28 Miscellaneous review exercises - Some geometrical and physical problems leading to first-order equations. Exercise 1.

2x + 3y = C

y0 = −

2 3

g0 =

3 3 =⇒ g − x = C 2 2

Exercise 2. d/dx

xy = C −−−→ y + xy 0 = 0 =⇒ y 0 = −y/x x 6= 0 =⇒ g 0 = x/g =⇒ Exercise 3. x2 + y 2 + 2Cy = 1 167

1 2 1 g = x2 + C 2 2

x + yy 0 + Cy 0 = 0 =⇒ y 0 (y + C) = −x −x −x −2xy y0 = = = 2 1−x2 −y 2 y+C y − x2 + 1 y+ 2y     y 2 − x2 + 1 1 1 1 orthogonalcurves 0 −−−−−−−−−−−→ y = = y+ − x y −1 2xy 2x 2 x Recognize that this is a Ricatti equation and we know how to solve them.

y0 +

−1 y = y −1 2x



−x 1 + 2 2x

v = yk = y2 Z x Z x −1 a P (t)dt = A(x) = = ln t x a a



n = −1

k = 1 − n = 1 − (−1) = 2    2 1 −1 0 v= −x v +2 2x 2 x  Z x Z x 1 a −a A Qe = −t = + 1 − a(x − a) t t x a a 

y 2 = v = −1 +

x bx − x(x − a) + a a

Exercise 4. y 2 = Cx. 0 2 y2 d/dx 2yy x − y = C −−−→ =0 x x2 1 y −2x =⇒ y 0 = −y
= y0 = 2x y 2x

=⇒ y 2 + 2x2 = C Exercise 5. x2 y = C.

2xy + x2 y 0 = 0 y 0 = −

2y x =⇒ y 0 = x 2y

1 2 x2 y = +C 2 4 2y 2 − x2 = C Exercise 6. y = Ce−2x

e2x y = C =⇒ 2e2x y + e2x y 0 = 0 1 invert y 0 = −2y −−−→ y 0 = 2y =⇒ y 2 = x + C Exercise 7. x2 − y 2 = C

2x − 2yy 0 = 0 x −y y 0 = =⇒ y 0 = y x =⇒ ln y = − ln x + C =⇒ y =

C x

Exercise 8. y sec x = C

y 0 sec x + y tan x sec x = 0 1 (invert) y 0 = −y tan x −−−−→ y 0 = y tan x 1 2 y = ln | sin x| + C 2 168

Exercise 9. All circles through the points (1, 0) and (−1, 0) From Sec. 8.22, Ex.10, we had obtained y 0 =

=⇒ y 0 =

x2 −y 2 −1 2xy

=

x2 −1 2yx

−

y 2x

=⇒ y 0 +

1 2x y

x2 −1 −1 2x y

=

Recognize this is a Ricatti equation. For y 0 + P y = Qy n , in this case, n = −1, and so k = 1 − n = 1 − (−1) = 2. Then v = y k and v0 + kPv = kQ. In this case,
2 2 1 −1 v 0 + 2 2x = v 0 + x1 v = x x−1 = x − 1/x. v = 2 x 2x Z x Z x 1 x A(x) = P (t)dt = = ln t a a a    1 3  x Z Z  2 1 t t 1 t 3t (t − ) exp ln dt = − dt = − t a a a a a a x a e− ln a = x 1 3 1 3 x − x − ba 3a + a + =⇒ y 2 = v = 3 x x Exercise 10. All circles through the points (1, 1) and (−1, −1).

(x + A)2 + (y + B)2 = r2 (1 + A)2 + (1 + B)2 = 1 + 2A + A2 + 1 + 2B + B 2 = r2

− (−1 + A)2 + (−1 + B)2 = − 1 − 2A + A2 + 1 − 2B + B 2 = r2 =⇒4A + 4B = 0 =⇒ A = −B 2

(x + −B) + (y + B)2 = r2 2(x − B) + 2(y + B)y 0 = 0

=⇒ y 0 =

(y + B)y 0 = B − x

B−x y+B

√ p (1 − B) + (1 + B) = r =⇒ 2 (1 + B 2 ) = r or

r

r2 −1=B 2 =⇒ (so B could be treated as a parameter for the family of curves) −1 y+B = y0 =  B−x x−B 2

2

2

y+B

y0 1 = =⇒ y = C(x − B) − B y+B x−B Exercise 11. With (0, Y ) = Q the point that moves up wards along the positive y-axis and

P = (x, y) being the point P that pursues Q, Y −y −y y 0 = X−x = Y0−x is the slope of the tangent line on a point on the trajectory of P . The condition given, that the distance of P from the y-axis is k the distance of Q from the origin, is

kY = x. 1 k




x−y = f (x, y) x f (x, y) is homogeneous of zero order =⇒ y = vx 0

y =

(try this substitution)

1 v0 1 y 0 = v 0 x + v = − v =⇒ 1 = k x k − 2v   1 1 x 1 =⇒ − ln − 2v = ln x + C =⇒ y = − 2 k 2k 2C 2 x (1,0)

−−−→ y =

x 1 = 2k 2kx 169

k=

1 2

y =x−

1 x

2xy y 2 −x2 +1

Exercise 12.

x 1 − 2k 2kx

y= Exercise 13. y = f (x). x

Z

x

Z f (t)dt = xf (x) −

n 0

Z

x

f (t)dt; (n + 1) 0

f (t)dt = xy 0

ny = xy 0 =⇒ (n1 )y = y + xy 0 =⇒

n y0 = x y n ln x = ln y

y = Cxn of y = Cx1/n Exercise 14. x

Z

Z

2

n

πf (t)dt = 0

x

(π(y(x))2 − π(f (t))2 )dt

0

x

Z

πf 2 (t)dt = xy 2 (x) = xy 2 ;

(n + 1)

(n + 1)f 2 (x) = y 2 + 2xyy 0

0

y0 =

ny 2x

=⇒ ln y =

n ln x + C 2

y = Cxn/2 of y = Cx1/2n Exercise 15. x

πf 2 − 2xf x2 0 The left hand side of the last expression shown is homogeneous. Thus do the y = vx substitution. Z

π

f 2 (t)dt = x2 f (x)

πf 2 (x) = 2xf + x2 f 0 =⇒ f 0 =

πv 2 x2 − 2x2 v = πv 2 − 2v x2   v0 x v0 1 1 1 1 = 1 =⇒ = = − v0 3 πv 2 − 3v x 3 v π(v 2 − 3v ) v − π π 3 ln (v − ) − ln v = 3 ln x + C  π  v − 3/π ln = 3 ln x + C v 3 vx − xCvx4 π 3 3x/π y − x = Cyx3 =⇒ y = 3 π 1 + x2 v0 x + v =

Exercise 16. A =

A−B = d/dx

Ra 0

Ra

f+

0

f;

Ra 1

B=

R1 a

f

f = 2f (a) + 3a + b

−−−→ 2f (a) = 2f 0 (a) + 3 =⇒ 1 = So then

f 0 (a) f (a)− 32

3 3 a + C = ln (y − ); f (a) = Cea + 2 2 3 3 1 f (1) = 0 = Ce + ; =⇒ C = − 2 2e −3 x−1 3 f (x) = e + 2 2 170

To find b,  2

 −3 a−1 3 3 3 3 e + a + e−1 + − = 2f (a) + 3a + b = 2 2 2 2 2   −3 a−1 3 =2 e + + 3a + b 2 2 =⇒ b =

3 −1 e −3 2

Exercise 17.

x

    y(x) − 1 f (t) − t+1 dt = x3 x 0   Z x y(x) − 1 1 2 = x − x = x3 = f +− x 2 0   Z x y(x) − 1 = f +− x − x = x3 2 0 1 1 d/dx −−−→ f (x) + − (y 0 x + y) − = 3x2 2 2 1 2 −2(3x + − y/2) y 2 y0 = = −6x − x−1 + x x Z

A(x) =

As a leap of faith, try y = vx substitution to solve y 0 = −6x − x−1 + xy . v 0 x + v = −6x − x−1 + v

v 0 = −6 − x−2 ;

v = −6x + x−1 + C

y = −6x2 + 5x + 1 x

Exercise 18. Assuming no friction at the orifice and energy conservation.

mgh =

1 mv 2 2 f

(imagine √ how the top layer of water is now at the bottom of the tank (final “potential energy configurations”)) Vf = 2gh (how fast water is rushing out) A0 = cross-sectional area of the orifice. p dV dh =A = −c 2ghA0 dt dt hf p A0 2h1/2 = −c 2g t A hi √ p  2A p =⇒ T = √ hf − hi = 59.6sec c gA0 Note that we included the discharge coefficient C = 0.6. Exercise 19.

dV dt

√ √ 1/2 = −c 2ghA0 + γ0 = A dh + γ0 . κ = c 2gA0 . dt = −κh 171

dh Adh = dt = (A/γ0 ) γ0 − κh1/2 1 − γκ0 h1/2   0   1 a 1 κ = ln 1 − ah1/2 − (where a = ) 1/2 1/2 2h γ0 1 − ah
  1 a 1/2 1 − ah1/2 1 2 − 2h
(h1/2 )0 = 1/2 = 2h 1 − ah1/2 h1/2 1 − ah1/2  2    hf Z dh 2γ0 κ 1/2 2γ0 h1/2 =⇒ (A/γ0 ) = − (A/γ0 ) ln 1 − h + = κ2 γ0 κ 1 − γκ0 h1/2 hi =T  =⇒ exp t→∞

−−−→ hf =

 κ2 1− −γ0 2γ0  1/2 1/2 hf − hi = t− 2 A κ 2γ0 1−

κ 1/2 γ0 hf κ 1/2 γ0 hi

γ02 (100in3 /s)2 2 = 2 = (25/24) 2 κ c (2)(32f t/s2 )(5/3in2 )2 (12in/1f t)

Exercise 20.

1 2 πR H0 3 0  2 R0 1 1 R2 V (h) = V0 − πh h = V0 − π 02 h3 = V0 − αh3 3 H0 3 H0 p 1 mg(H0 − h) = mvf2 ; 2g(H0 − h) = vf (energy conservation) s2 r   h h cA0 vf = cA0 2gH0 1 − =β 1− H0 H0 r p h dh β dV 2 dh 1− = −β 1 − h/H = −3αh =⇒ = 2 dt dt dt 3αh H Z β/H02 h2 /H02 q = T = 3α 1− h V0 =

H0

Z Z (1 − y)2 (−dy) u2 du 1 − 2y + y 2 √ = H0 = H0 = −H0 = √ √ y y 1−u   hf 2 2 = −H0 2y 1/2 − y 3/2 + y 5/2 = 3 5 hi 1/2  3/2  5/2 ! hf  4 h 2 h h − 1− + 1− = −H0 2 1 − = H0 3 H0 5 H0 hi √ 2 cA0 2gH0 /H0   T = R2 3 31 π H02 Z

0

For hi = 0,

hf = H, H0 (2(1) − 4/3(1) + 2/5) = H0 (16/15) = cA0

Exercise 21. m2 x − m + (1 − x) = 0 Exercise 24. Given f s.t. 2f 0 (x) = f

p

2gH0 T /(πR02 ); T =

=⇒ (m2 − 1)x + 1 − m = 0, 1 x




√

√ H0 πR02 2 πR02 H0 √ = 9 A0 cA0 2g

16 15

=⇒ m = 1

if x > 0, f (1) = 2 and x2 y 00 + axy 0 + by = 0 172

(1)     d 1 −1 0 2f (x) = f = (f ) dx x x   −1 −1 1 =⇒ x2 y 00 = f0 = f 2 4 x 00

a = 0;

b=

1 4

(2) f (x) = Cxn f 0 = nCxn−1

=⇒ n(n − 1)Cxn =

f 00 = n(n − 1)Cxn−2 n2 − n +

−1 n Cx 4

1 1 =⇒ n = 4 2

Exercise 28. Choose the units for time to be in days first - we can convert into years later.

If no one died from accidental death, then the population will grow by e. That means, with x = x(t) being the population at time t dx = x =⇒ x = Cet dt which makes sense because if C is the original population number, then after 1 year, x = Ce. 1 With t in days, we have a decrease of 100 x in population each day due to death. Add up the changes from the decrease due to deaths and the increase due to growth for the DE:

1 1 100 − 365 −265 dx = x− x= x= dt 365 100 36500 36500 −265 =⇒ x = Ce 36500 t Change t units to years by multiplying the “time constant”

−265 36500

by 365 days.

x = 365 exp (−2.65t) To get the total fatalities, simply integrate the deaths during each year. Z t 365 365 y= (− exp (−2.65t) + 1) 100 2.65 0 Exercise 29. For constant gravity, ∆K = −∆U

=⇒ −(0 − mgh) = 21 mvf2
 1 ft  √ 1m mi mi vf = 2gh = (6.37 × 108 cm) 2.54 cm 12 in = 6.93 sec = 24940 hr The constant energy formula could also be obtained by considering F =

GMe m −r 2

r

d = m dt 2 = −∂r U

Exercise 30. Let y = f (x) be the solution to y 0 =

2y 2 +x 3y 2 +5 0

f (0) = 0 0

2

2

+5)−(6yy )(2y +x) (1) y 0 (0) = 0 as easily seen. Now y 00 = (4yy +1)(3y(3y , so then 2 +5)2 1 00 y (0) = 5 > 0. It is a minimum. (2) f 0 (x) ≥ 2/3 ∀ x ≥ 10/3. a = 2/3 since f will be above this tangent line. Suppose, in the “worst case,” f 0 (x) = 0 for 0 ≤ x ≤ 2/3. Then f (x) = 0 for 0 ≤ x ≤ 2/3. Then the tangent line must be at y = 0 at x = 10/3 to remain below the graph of f (x).

=⇒ 23 x − 20/9 < f (x) 173

(3) Since f 0 (x) ≥ 23 for each x ≥ 10 3 , then f → ∞ for x → ∞ (otherwise, f would have to decrease somewhere, which would contradict the given fact about f ). Rewrite the DE for y 0 to be y0 =

5 x 2y 2 + x =⇒ (3y 2 + 5)y 0 = 2y 2 + x =⇒ (3 + 2 )y 0 = 2 + 2 2 3y + 5 y y

Consider y0 = specifically,

x y2 .

x y2 5 y2

2+ 2y 2 + x = 2 3y + 5 3+

Now y must, at the very least, have some linear increase because we had already shown that y 0 ≥ 23 .

So y 2 would go to infinity faster than linear x. Thus limx→∞ y 0 = 23 . So then (3 + 2 + yx2 . 0= (4) y 0 =

2+ yx2 3+ y52

x→∞

−−−−→ 32 . =⇒ y = 23 x or

y x

5 0 x→∞ −−−→ y 2 )y −

(3 + 0) 23 = 2 =

x y2

= 32 .

Exercise 31. Given a function f which satisfies the differential equation xf 00 (x) + 3x(f 0 (x))2 = 1 − e−x

(1) c 6= 0 for an extrenum. cf 00 (c) + 3c(f 0 (c))2 = cf 00 (c) = 1 − e−c (2) Cleverly, consider the limit.

=⇒ f 00 (c) =

1−e−c c

xf 00 (x) + 3x(f 0 (x))2 = 1 − e−x =⇒ f 00 (x) + 3(f 0 (x))2 =

>0 

1 − e−x x



x→0

−−−→ f 00 (0) + 0 = 1

So a critical point at x = 0 would be a minimum. (3) We’ll have to “cheat” a little and use the idea of power series early on here. f 00 + 3(f 0 )2 =

1−e−x x

suggests that we consider the Taylor series of e−x . j P∞ ∞ X − j=1 (−x) (−1)j xj 1 − e−x j! = = x x (j + 1)! j=0

This further suggests that f itself has a power series representation because its first and second order derivatives are simply a combination of infinitely many terms containing powers of x. P∞ Then suppose f = j=0 aj xj . f0 = f 00 + 3(f 0 )2 =

1 − e−x =⇒ x

f 00 =

∞ X (j + 1)aj+1 xj j=0 ∞ X

(j + 2)(j + 1)aj+2 xj

j=0 ∞ ∞ X ∞ ∞ X X X (−1)j xj =⇒ (j + 2)(j + 1)aj+2 xj + 3 (j + 1)(k + 1)aj+1 ak+1 xj+k = (j + 1)! j=0 j=0 j=0 k=0

If f (0) = 0 0

If f (0) = 0 P∞

Then f = a2 x2 + term is x1 . So then

j=3

a1 = 0

aj xj . Consider the x0 terms in the DE. (f 0 )2 doesn’t contribute, because f 0 ’s leading order

2(1)a2 + 0 = 1 =⇒ a2 =

A=

a0 = 0

1 2

1 in order for f (x) ≤ Ax2 2 174

i.e. f =

∞ 1 2 X x + aj xj 2 j=3

9.6 Exercises - Historical introduction, Definitions and field properties, The complex numbers as an extension of the real numbers, The imaginary unit i, Geometric interpretation. Modulus and argument. Exercise 6. Let f be a polynomial with real coefficients. (1) Since z1 z2 (z1n+1 ) = z1n z1 = z1n z1 = z n+1 1 f (z) =

X

aj z j =

X

aj z j = f (z)

(2) If f (z) = 0, then f (z) = f (z) = 0 as well. Exercise 7. The three ordering axioms are

Ax. 7 If x, y ∈ R+ , x + y, xy ∈ R+ Ax. 8 ∀ x 6= 0, x ∈ R+ or − x ∈ R+ but not both Ax. 9 0 ∈ / R+ x < y means y − x positive. Suppose i positive: i(i) = −1 but −1 is not positive. Suppose −i is positive. −i(−i) = −1 but −1 is not positive. i is neither positive nor negative so Ax. 8 is not satisfied. Exercise 8. Ax. 8 , Ax. 9 are satisfied.

(a + ib)(c + id) = (ac − bd, ad + bc) so Ax. 7 might not be satisfied. Exercise 9. Ax. 7, Ax. 8 , Ax. 9 are trivially satisfied (all are positive). Exercise 10. x > y x > y is well defined. Ax. 8 is satisfied.

For Ax. 7, ( 32 , 1), (1, 12 ) contradicts Ax. 7 since we required the product to be positive as well if the factors are positive. We found this particular counterexample by considering factors (a, b), (c, d), so the product of the two is (ac − bd, ad + bc) and so we need ac − bd − ad − bc = a(c − d) − b(c + d) < 0 Exercise 11. See sketch. Exercise 12.

az + b cz + d (az + b)(cz + d) ac|z|2 + adz + bcz + bd w= = 2 2 (cz + d)(cz + d) c |z| + cd(z + z) + d2 w=

w + −w =

ac|z|2 + adz + bcz + bd − ac|z|2 − adz − bcz − bd (ad − bc)(z − z) = 2 |cz + d| |cz + d|2 If ad − bc > 0 w − w = 2Imw =

ad − bc 2Imz; |cz + d|2

So Imw has the same sign as Imz 175

ad − bc >0 |az + d|2

9.10 Exercises - Complex exponentials, Complex-valued functions, Examples of differentiation and integration formulas. Exercise 7.

(1)

2π

Z if m 6= n ,

e

ix(n−m)

0

2π eix(n−m) 1−1 dx = = =0 i(n − m) 0 i(n − m)

2π

Z

eix(0) dx = 2π

if m = n , 0

(2)

2π

Z

einx e−imx dx =

0

2π

Z

(cos nx + i sin nx)(cos mx − i sin mx) = 0 2π

Z

cos nx cos mx + sin nx sin mx + i(sin nx cos mx − sin mx cos nx)

= 0 2π

Z

e−inx e−imx dx =

0

2π

Z

(cos nx − i sin nx)(cos mx − i sin mx) = 0 2π

Z

cos nx cos mx − sin nx sin mx + i(− sin nx cos mx − sin mx cos nx)

= 0

Z 0=

Summing the two equations above Z 2π 2π 2 cos nx cos mx + 2 i − sin mx cos nx

0 2π

Z =⇒ 2π

Z

einx e−imx dx =

0

0

Z cos nx cos mx = 0,

0 2π

Z

2π

sin mx cos nx = 0 0

(cos nx + i sin nx)(cos mx − i sin mx) = 0 2π

Z

cos nx cos mx + sin nx sin mx + i(sin nx cos mx − sin mx cos nx)

= 0

Z

2π

einx eimx dx =

0

2π

Z

(cos nx + i sin nx)(cos mx + i sin mx) = 0 2π

Z

cos nx cos mx − sin nx sin mx + i(sin nx cos mx + sin mx cos nx)

= 0

Z

2π

Subtract the two equations above Z 2π sin nx sin mx − i sin mx cos nx = 0

0

0

Z =⇒

2π

sin nx sin mx = 0 0 176

2π

Z

einx e−inx = 2π =

2π

Z

(cos nx + i sin nx)(cos nx − i sin nx) =

0

0 2π

Z

cos2 nx + sin2 nx

= 0

Z

2π inx inx

e

e

Z

2π

=0=

0

cos2 nx − sin2 nx + i(2 cos nx sin nx)

0

Z

2π 2

2

Z

2π

cos nx − sin nx = 0

=⇒ 0

cos nx sin nx = 0 0

Summing the two results above, we obtain Z 2π 2 cos2 nx 2π = Z

0 2π

cos2 nx = π

=⇒ 0

Z

2π

Then also,

sin2 nx = π

0

Exercise 8.

z = reiθ = rei(θ+2πm) , m ∈ Z z 1/n = r1/n ei(θ/n+2πm/n) m = 0, 1, . . . n − 1 =⇒ z 1/n = Reiα m = z1 m The roots are spaced equally by an angle 2π/n i = eiπ/2+i2πn =⇒ i1/3 = eiπ/6 , ei5π/6 , ei3π/2 i1/4 = eiπ/8 , e5iπ/8 , e9iπ/8 , e13iπ/8 −i = e−iπ/2+i2πn =⇒ (−i)1/4 = e−iπ/8 , e3iπ/8 , e7iπ/8 , e11iπ/8 Exercise 9.

eiu eiv = ei(u+v) = cos u + v + i sin u + v = = (cos u + i sin u)(cos v + i sin v) = cos u cos v − sin u sin v + i(cos v sin u + cos u sin v) =⇒ sin u + v = cos v sin u + cos u sin v =⇒ cos u + v = cos u cos v − sin u sin v  iz 2  iz 2 e + e−iz e − e−iz 2 2 sin z + cos z = + = 2i 2 −(e2iz + e−2iz − 2) + (e2iz + 2 + e−2iz ) = =1 4 eiiy + e−iiy eiiy − e−iiy cos iy = = sin iy = = 2 2i e−y + ey e−y − ey = = cosh y = = i sinh y 2 2i eiz = ei(x+iy) = eix e−y = (cos x + i sin x)e−y e−iz = e−i(x+iy) = e−ix ey = (cos x − i sin x)ey Thus it is clear, by mentally adding and subtracting the above results that =⇒

cos z = cos x cosh y − i sin x sinh y sin z = i cos x sinh y + sin x cosh y

Exercise 10.

(1) Log(−1) = iπ log (i) = ln 1 + i π2 = i π2 (2) Log(z1 z2 ) = Log(|z1 ||z2 |ei(θ1 +θ2 ) ) = ln |z1 ||z2 | + i(θ1 + θ2 + 2nπ) = Logz1 + Logz2 + i2πn 177

1| (3) Log(z1 /z2 ) = Log(|z1 |/|z2 |ei(θ1 −theta2 ) ) = ln |z |z2 | + i(θ1 − θ2 + 2nπ) = Logz1 − Logz2 + i2πn (4) exp (Logz) = exp (ln |z| + iθ + i2πn) = z

Exercise 11.

(1) 1i = eiLog1 = ei(i2πn) = e−2πn = 1 if n = 0 π

π

ii = eiLogi = ei(i 2 +i2πn) = e− 2 −2πn = e−π/2 if n = 0 (−1)i = eiLog−1 = ei(iπ+i2πn) = e−π−2πn = e−π (2) z a z b = eaLogz ebLogz = eaLogz+bLogz = e(a+b)Logz = z a+b x (3) (z1 z2 )w = ewLogz1 z2 = ew(Logz1 +Logz2 +2πmi) (z1w z2w ) = ewLogz1 ewLogz2 = ew(Logz1 +Logz2 ) m = 0 is the condition required for equality. Exercise 12.

if L(u) = P, L(v) = Q, 00

0

L(u + iv) = (u + iv) + a(u + iv) + b(u + iv) = u00 + au0 + bu + i(v 00 + av 0 + bv) = L(u) + iL(v) = P + iQ = R if L(f ) = R L(u + iv) = L(u) + iL(v) = P + iQ then, equating real and imaginary parts, L(u) = P, L(v) = Q Exercise 13.

L(y) = −ω 2 y + aiωy + by = Aeiωx =⇒ (−ω 2 + aiω + b)B = A We cannot let (−ω + aiωb) = 0 for a nontrivia solution. Thus b 6= ω 2 or aω 6= 0. A B= −ω 2 + aiω + b 2

Exercise 14.

L(b y ) = ceiωx ; yb = Beiωx = =⇒ yb = p

c (b −

ω 2 )2

+

=⇒
(aω)2

c eiωx −ω 2 + aiω + b

ei(ωx−α) where tan α =

c (b −

ω 2 )2

+ (aω)2

aω b − ω2

cos (ωx − α)

Exercise 15.

=(b y) = p =⇒ A = p

c ω 2 )2

(b − c

+ (aω)2

(b − ω 2 )2 + (aω)2

;

sin (ωx + α) − tan α =

aω b − ω2

10.4 Exercises - Zeno’s paradox, Sequences, Monotonic sequences of real numbers. Exercise 1. Converges to 0. f (n) =

−2 − n12 n→∞ n n+1 n2 − (n2 + 2n + 1) −2n − 1 − = = 2 = n −−−−→ 0 n+1 n n(n + 1) n +n 1 + n1

Exercise 2. Converges to −1.

f (n) =

−1 − n1 − n3 − (n3 + n + n2 + 1) n2 − n − 1 = = (n + 1)n n(n + 1) 1 + n1

Exercise 3. Diverges since

nπ nπ − L| ≥ 1 cos − |L| ≥ |1 − |L|| 2 2 − L| > 1 for n = 4m. |cos

Choosing 1 =

|1−|L|| , 2

| cos nπ 2

178

1 n2

n→∞

−−−−→ −1

Exercise 4. f (n) =

1 5

Exercise 5. f (x) =

x 2x

+

3 5n

−

2 5n2

→ limn→∞ f (n) =

x exp (x ln 2)

=

1 5

→ 0 since limx→∞

xα . (ex )β

Exercise 6. f (n) = 1 + (−1)n = 0 of 1.

Thus, choosing 1 =

|1−|L|| ; 2

|f (n) − L| ≥ ||f (n)| − |L|| = |1 − |L|| > 1 Exercise 7. f (n) =

Suppose  =

for any n = 2m

1+(−1)n . n

3 N.

So for n > N,

1 N

>

1 n,

n ≥ N = N () = 3/. 1 + (−1)n ≤ 2 < 3 < 3 = |f (n)| = n n n N

Exercise 8. f (n) =

(−1)n n

+

1+(−1)n 2

(−1)n (−1)n 1 + (−1)n 1 + (−1)n |f (n) − L| = + − L ≥ L − − ≥ n 2 n 2 (−1)n 1 + (−1)n 1 1 + (−1)n ≥ |L| − − = |L| − n = ≥ n 2 2 1 1 ≥ |L| − − n 2 Thus, consider

|L| − 1 − n

Exercise 9. f (x) = exp

1 x


ln 2 ;

1 1 − 1 = 0 for n > N − |L| > 2 N

limx→∞ f (x) = 0.

Exercise 10. n

n

|f (n) − L| = |n(−1) − L| ≥ ||n(−1) | − |L|| = ||n| − |L|| = |n| − |L| > N − |L| Thus, for n > N , N () =  + |L|, so then |f (n) − L| > . Exercise 11. f (n) =

n2/3 sin n! . n+1

2/3 n sin (n!) sin (n!) 1 = ≤ |f (n)| = n + 1 n1/3 + n−2/3 n1/3 Thus, for n > N ,

N () =

1 3 ,

|f (n)| < .

Exercise 12. Converges, since

  1 3n+1 + 3(−2)n − 3n+1 − (−2)n+1 (−2)n (3 + 2) f (n) − = = = 3 3(en+1 + (−2)n+1 ) 3(3n+1 + (−2)n+1 )   5 (−2)n = 3 3n+1 + (−2)n+1 −(−2)n+1 (−2)n f (n) − 1 = 5 ≤ = 3 3 3n+1 + (−2)n+1 3n+1 + (−2)n+1 −1 1 =  n+1 =  n+1 < 3 3 + 1 1 + −2 −2 < For n > N , consider  =


2 −N , 3

i.e. N =

1
3 n+1 2

− ln  ln 2/3

<

1
3 n 2

= N (). Thus 179

L=

1 ; 3

Exercise 13.

f (n) =

√

n+1−

√

n

√ √  √ √ √ n+1+ n n+1−n √ f (n) = =√ √ √ = 1 n+1+ n n+1+ n n+1+ n 1 1 |f (n)| = √ √ ≤ √ ; n + 1 + n 2 n √

√
n+1− n

So then ∀, we have  = 2√1N and for n > N , Thus f (n) converges to 0.

1 √ 2 n

<

1 √ 2 N

= .

Exercise 14.

f (n) = nan = n exp n ln a = Exercise 15. f (n) =

loga n n ,a

n xa =0 → 0 since lim x→∞ (ex )b exp n ln a1

> 1. limn→∞ f (n) = 0 since limx→∞

(log x)a xb

= 0 for a > 0, b > 0

Exercise 16. limn→∞ f (n) = 0 Exercise 17. limn→∞ f (n) = e2 . Exercise 18.

1 + n cos nπ − L ≥ 1 + n cos nπ − |L| = 1 + −n − |L| = n+1 2 n+1 2 n+1 1 = − |L| > |L| n+1 Choose 0 =

|L| 2 .

For any N , for n > N , |f (n) − L| > 0 .

Exercise 19.

√ i 5 iα 1+ = e 2 2  2 5 1 = 1+ 2 4 1 tan α = 2

√

!−n  n 2 5 iα √ e , e−niα 2 5 n  2 e−niα = 0 lim √ n→∞ 5

Exercise 20. limn→∞ f (n) = limn→∞ e−πin/2 diverges since

|eiπin/2 − L| ≥ ||e−iπin/2 | − |L|| = |1 − |L|| > |L| So for 0 =

|L| 2 ,

for n > N , |f (n) − |L|| > 0

Exercise 21. f (n) =

1 −iπn/2 ne

|f (n)| =

1 n.

Suppose N () = 1 ; then for n > N , |f (n)| < 0 Exercise 22. |f (n) − L| ≥ ||ne−πin/2 | − |L|| = |n − |L||

Consider 0 = |1 − |L|| for n > N > 1, |f (n) − L| > 0 . Exercise 23. an =

1 n.

1 1 1 |an | = < ; N () = n N   = 1, 0.1, 0.01, 0.001, 0.0001 N = 1, 10, 100, 1000, 10000 Exercise 24. |an − 1| = n+1−n n+1 =

1 n+1

<

1 n 180

N () =

1 

= 1, 10, 100, 1000, 10000.

Exercise 25. |an | =

1 n

N () = 1, 10, 100, 1000, 10000. 1 1 ≤ Exercise 26. |an | = n! exp n ln n < For n > N ,

1 exp N

1 exp n

= , so that N = ln 1/. N () = 1, 2, 4, 6, 9

Exercise 27. an =

2n n2 +1

2 2 ; |an | = n2 +1/n ≤ n2 . √ 2 N () = √ = 1, 4, 14, 44, 141 

9 n = Exercise 28. |an | = 10






9 n 10

= en ln 9/10 ln  − ln 1/
= = 9 ln (9/10) ln 10 = 1, 21, 43, 65, 87

N () =

Exercise 30. If ∀ > 0, ∃N ∈ Z+ such that n > N , |an | < .

|an |2 < |an | < 2 |a2n | < 2 So for ∀1 > 0, 1 = 2 and ∃N = N () = N (1 ), so that |a2n | < 1 . Exercise 31.

|an + bn − (A + B)| = ||an − A||bn − B|| ≤ |an − A| + |bn − B| <  +  = 2 ∀ > 0, ∃NA , NB ∈ Z+ , |an − A| < 

if n > NA ; |bn − B| <  if n > NB

Consider max (NA , NB ) = NA+B |an + bn − (A + B)| < 2 ∀1 > 0, 1 = 2, then ∃NA+B = NA+B (1 ) ∈ Z+ such that |(an + bn ) − (A + B)| < 1 if n > NA+B |can − cA| = c|an − A| < c ∀1 > 0, 1 = c;

then ∃NcA = N () = N (1 ) ∈ Z+ such that

|can − cA| < 1 for n > N (1 ) Exercise 32. Given limn→∞ an = A,

lim (an − A)(an + A) = lim (an − A) lim (an + A) = 0(2A) = 0

n→∞

n→∞

n→∞

2an bn = (an + bn )2 − a2n − b2n  2 2 lim an bn = lim (an + bn )2 − lim a2n − lim b2n = lim (an + bn ) − A2 − B 2 = 2AB n→∞

n→∞

n→∞

n→∞

n→∞

=⇒ lim an bn = AB n→∞

Exercise 33.

α n




=

α(α−1)(α−2)...(α−n+1) n! 181

(1) 

− 12 1!



−1/2 1

 1 − 12 −3 −2 5 2 = = 2! 2! 8

−5
 1 − 12 −3 −2 5 2 2 = =− 3! 3! 16 −1
n 2 (2) an = (−1) n . a1 =

1 2

> 0. an+1

=

a2 =

3 8


 1 − 12 −2 = 4
 1 − 12 −2 = 5

−3 2




4!
−3 2

−5 2




−7 2




35 128
−7
−9
−5 2

2

5!

=

2

=

−63 256

>0 


    1 an (−1) −1 − 2 − (n + 1) + 1 − 12 n+1 α 2 −n = (−1) = (−1) = = n+1 n+1 n+1 n an (n + 1/2) >0 = n+1   n + 1/2 an < an an+1 = n+1 n+1

Exercise 34.

(1) n

1X tn − sn = f n k=1

  n−1   k k 1X 1 f − = n n n n

f (1) +

k=0

n−1 X k=1

  k f − n

n−1   k 1X f f (0) + n n

!! =

k=1

1 (f (1) − f (0)) n

Since f nk ≤ f (t) ≤ f k+1 for nk ≤ t ≤ k+1 n n , by f being monotonically increasing. Z 1 =⇒ sn ≤ f (x)dx ≤ tn (from definition of integral) =

0 1

Z 0≤

f (x)dx − sn ≤ tn − sn = 0

1 (f (1) − f (0)) n

(2) Use Theorem 1.9. Theorem 30. Every function f which is bounded on [a, b] has a lower integral I(f ) and an upper integral I(f ) satisfying Z b Z b s(x)dx ≤ I(f ) ≤ I(f ) ≤ t(x)dx a

a

for all step functions s and t with s ≤ f ≤ t. The function f is integrable on [a, b] iff its upper and lower integrals are equal, Z b f (x)dx = I(f ) = I(f ) a

(3)

R1 Since f (x) is integrable, then limn→∞ sn = limn→∞ tn = 0 f (x)dx
Pn−1 Pn b−a 1 1 = ∆, sn = ∆ k=0 f (a + k∆), tn = ∆ k=1 f (a + k∆) n Rb So by increasing monotonicity of f , sn ≤ a f (x)dx ≤ tn . 1 tn − sn = ∆

f (b) +

n−1 X

f (a + k∆) =

k=1

n−1 X

! f (a + k∆) − f (a)

k=1

Z 0≤

b

f (x)dx ≤ a

f (b) − f (a) ∆

Exercise 35.


2 R 1 Pn (1) limn→∞ n1 k=1 nk = 0 t2 dt = 13 Pn Pn 1 (2) limn→∞ k=1 n+k = limn→∞ n1 k=1

1 k 1+ n

=

R1

1 dx 0 x 182

= ln 2

=

f (b) − f (a) ∆

π 1 = arctan x|0 = 4 1 √ √ R1 1 Pn 1 1 √ 1 q = limn→∞ n k=1 = 0 √1+x2 dx = ln (x + 1 + x2 ) 0 = ln (1 + 2) n2 +k2 k 2 1+( n )
R1 P 2 n kπ 1 kπ 1 − cos πx 1 = −(−1−1) = k=1 sin n = 0 sin πxdx = n sin n = limn→∞ n π π 0 π Rn 2 1 2 kπ 1 n sin n = 0 sin xπ = 2

Pn

(3) limn→∞

1 n

1 k=1 1+( k )2 n

(4) limn→∞

Pn

(5) limn→∞

Pn

(6) limn→∞

Pn

k=1

k=1

k=1

R1

1 dx 0 1+x2

=

10.9 Exercises - Infinite series, The linearity property of convergent series, Telescoping series, The geometric series. P∞ P∞ 1/2 1/2 1 1 Exercise 1. n=1 (2n−1)(2n+1) = n=1 2n−1 − 2n+1 = 2 P∞

Exercise 2.

P∞

=2

Exercise 3.

P∞

=

Exercise 4.

P∞

2n +3n 6n

Exercise 5.

P∞

√ √ n+1− n √ 2 n +n

2 n=1 3n−1 1 n=2 n2 −1

n=1

n=1

1 n=0 3n

1 = 2 1−1/3 = 3

P∞

=

1/2 n=2 n−1


1 n 3

P∞

n=1

=

−

P∞

n=1

1/2 n+1

+

√1 n

=

P∞  1/2 n=2


1 n 2

P∞

n=1

−

n−1

√1 n+1

=

−

1/2 n



+

1/2 1−1/2

1/3 1−1/3

+−



1/2 n+1

=

1 2

−

1/2 n

+1=



=

1 2

+

1 4

=

3 4

3 2

=1

Exercise 6.

  ∞ ∞ X X 3/2 1 n −1/2 1 −1 = + = + = (n + 1)(n + 2)(n + 3) n=1 (n + 2)(n + 3) (n + 1)(n + 2) n=1 (n + 2)(n + 3) 2 6 n=1 ∞ X

= Exercise 7.

P∞

Exercise 8.

P∞

Exercise 9.

P∞

2n+1 n=1 n2 (n+1)2

=

2n +n2 +n n=1 2n+1 n(n+1)

∞ X

1 1 1 1 1 1 − − = − = 1 + 2 n + 3 12 3 12 4 n=1

P∞

1 n=1 n2

=1=

(−1)n−1 (2n+1) n=1 n(n+1)

1 (n+1)2

=1

P∞

1 n=1 2n(n+1)

P∞

n−1 n=1 (−1)

1 1 + n n+1



1 2n+1

+



1 n

=

1 n+1

+



1 2

P∞

1 n=1 n

−

1 n+1

+

1/4 1−1/2

=

1 2

+

1 2

=1

.

   1 1 −1 1 + = (−1)n + n+1 n+2 n n+2   ∞  ∞  X X −1 1 1 −1 1 1 −1 + = + = = −1 2j − 1 2j + 1 2 (j − 1/2) (j + 1/2) 2 1/2 j=1 j=1

(−1)n−1



=

−

+ (−1)n



Exercise 10.


n
∞ ∞ X X log n+1 (1 + n) log ((1 + n1 )n (1 + n)) n = = (log nn )(log (n + 1)n+1 ) n=2 log (n + 1)n+1 log nn n=2

since if

1 2 log 2

Exercise 11.

= y, then y = log2

P∞

n=1

d nxn = x dx

√

P∞

=

∞ ∞ X X 1 1 log(n + 1)n+1 − log nn = − = n+1 n n log (n + 1) log n log n log (n + 1)n+1 n=2 n=2

=

√ 1 = log2 e 2 log 2

2.

n=1

xn = x



x 1−x

0

=

x (1−x)2 .

Exercise 12. 183

∞ X

0  ∞ d X n x = n x =x nx = x dx n=1 (1 − x)2 n=1 2 n

=x

x(1 − x2 ) x(1 + x) (1 − x)2 + 2(1 − x)x = = 4 (1 − x) (1 − x)4 (1 − x)3

Exercise 13. ∞ X

n3 xn = x

n=1

=x

Exercise 14.

P∞

n=1

0  ∞ d X 2 n x + x2 = n x =x dx n=1 (1 − x)3 (1 + 2x)(1 − x)3 + 3(1 − x)2 (x)(x + 1) (1 + 2x)(1 − x) + 3x(x + 1) x(x2 + 4x + 1) =x = 6 4 (1 − x) (1 − x) (1 − x)4

d n4 x4 = x dx

P∞

n=1

n3 x3 = x



x3 +4x2 +x (1−x)4

0

.

 x3 + 4x2 + x ln = ln (x3 + 4x2 + x) − 4 ln (1 − x) (1 − x)4 3x2 + 8x + 1 1 1 +4 ; (ln f )0 = f 0 = 3 f x + 4x2 + x 1−x (3x2 + 8x + 1)(1 − x) 4(x3 + 4x2 + x) f0 = + = (1 − x)5 (1 − x)5 3x2 + 8x + 1 − 3x3 − 8x2 − x + 4x3 + 16x2 + 4x x3 + 11x2 + 11x + 1 = = (1 − x)5 (1 − x)5 4 3 2 x + 11x − 11x + x =⇒ (1 − x)5 R x P∞ Rx 1 P∞ R x n−1 xn = − ln (1 − x). dt = 0 dt n=1 tn−1 = 0 1−t n=1 0 t n =x 

Exercise 15.

P∞

n=1

Exercise 16.

Z x X ∞ ∞ Z x ∞ X X x2n−1 = t2j−2 dt = (t2 )j−1 = dt 2n − 1 0 n=1 j=1 0 j=1    Z x Z x  1/2 1+x dt 1 1/2 = + = = ln dt 2 1−t 1+t 2 1−x 0 1−t 0   P∞ P∞ d n+1 d x 1 n Exercise 17. = dx n=0 (n + 1)x = n=0 dx x 1−x = (1−x)2 Exercise 18.

00     ∞ ∞  X (n + 1)(n + 2) n X xn+2 d2 x2 1 d x x2 x = = 2 = + 2! 2 dx 2 1 − x dx 1 − x 2(1 − x)2 n=0 n=0 =

1 x x x3 1 − 2x + x2 + 2x − 2x2 + x2 1 + + + = = 1 − x (1 − x)2 (1 − x)2 (1 − x)3 (1 − x)3 (1 − x)3

Exercise 19. ∞ ∞ ∞ X (n + 1)(n + 2)(n + 3) n X d3 xn+3 1 d X d2 x(n+1)+2 x = = = 3! dx3 3! 3 dx n=0 dx2 2 n=0 n=0

 ! ∞ ∞ 1 d X d2 xn+2 1 d X d2 xn+2 d2 x2 = = − 2 = 3 dx n=1 dx2 2 3 dx n=0 dx2 2 dx 2   1 d 1 1 −3(−1) 1 = −1 = = 3 dx (1 − x)3 3 (1 − x)4 (1 − x)4 Exercise 20.

P∞

n=1

nk xn =

Pk (x) (1−x)k+1 184

∞ X

  0   ∞ Pk (x)(1 − x)k+1 + (k + 1)(1 − x)k Pk (x) d X k n d Pn (x) =x = x =x n x =x dx n=1 dx (1 − x)k+1 (1 − x)2k+2   0 (k + 1)xPk (x) + x(1 − x)Pk0 (x) Pk (x)(1 − x) + (k + 1)Pk (x) = =x k+2 (1 − x) (1 − x)k+2

k+1 n

n

n=1

((k + 1)Pk (x) + (1 − x)Pk0 (x))x has x as its lowest degree term from xPx0 (x) and (k + 1)xk+1 + −kxk+1 = xk+1 highest degree term is obtained from (k + 1)Pk (x) + −xPk0 (x). P∞ n+k
n P∞ xn+k 1 dk Exercise 21. n=0 n=0 k! . k x = (1−x)k+1 = dxk  ∞ ∞  ∞ X 1 d X dk x(n+1)+k n + k + 1 n X dk+1 xn+k+1 = = x = dxk+1 (k + 1)! (k + 1) dx n=0 dxk k! k+1 n=0 n=0   ∞ ∞ 1 d X dk xn+k 1 d X dk xn+k dk xk = = − = k + 1 dx dxk k! k + 1 dx n=0 dxk k! dxk k! k=1     d 1 1 1 −1 = = k + 1 dx (1 − x)k+1 (1 − x)k+2 Exercise 22.

(1) (2) (3)

P∞

P∞ P∞ 1 P∞ 1 P∞ 1 n−1 1 n=2 n! = n=2 (n−1)! − n=2 n! = n=1 n! − n=2 n! = 1 . P∞ n P∞ 1 P∞ P∞ 1 P∞ 1 P∞ 1 1 n=2 n! + n=2 n! = n=2 (n−1)! + n=0 n! − 1 − 1 = n=1 n! + n=0 n!

−2 =

P∞

2 n=0 n!

− 3 = 2e − 3 .

∞ ∞ ∞ ∞ ∞ X X (n − 1)(n + 1) X n2 X −1 X n −1 = + = + = n! n! n! (n − 1)! n! n=2 n=2 n=2 n=2 n=2

=

∞ ∞ ∞ ∞ X X X n+1 X 1 1 1 − = + +1= +1= e+1 n! n! n=1 (n − 1)! n! n=1 n=2 n=0

Exercise 23.

 P 
P∞ xn
P∞ nxn P∞ n2 xn ∞ nxn−1 d d d d d x d (1) x dx x dx = x = x2 ex + xex x = x dx n=1 n! n=1 n! = n=1 n! = x dx x dx e n=1 dx n! P 

P∞ 3 n ∞ n2 x n d d (2) x dx = n=1 n n!x = x dx (x2 + x)ex = x (2x + 1)ex + (x2 + x)ex = (x3 + 3x2 + x)ex n=1 n! x=1

k=5

Exercise 24.

P∞ P∞ (1) Pn=2 (−1)n = Pn=2 (−1)n (n − (n − 1)). Identical. ∞ ∞ (2) = n=2 (−1)n . Not n=2 (1 − 1) P P∞identical. ∞ (3) Not identical. n=2 (−1)n vs. ( n=2 (−1 + 1)) + 1.

n P∞ P∞  1
n−1 P∞ 1 n − 21 = n=1 (4) Identical. n=0 2 = 1 + n=1 2


1 n 2

(2 − 1) =

P∞

n=0


1 n 2

Exercise 25.

(1) 1 if |x| < 1 1 − x2 1 =⇒ 1 + 0 + x2 + 0 + x4 + · · · = if |x| < 1 1 − x2 P∞ P∞ P∞ (2) Thm. 10.2. n=1 (αan + βbn ) = α n=1 an + β n=1 bn . So then 1 + x2 + x4 + · · · + x2n + · · · =

∞ X j=0

xj −

∞ ∞ ∞ X X X xj + (−x)j xj − (−x)j 1 1 x = = x2j+1 = − = 2 2 2 1 − x 1 − x 1 − x2 j=0 j=0 j=0

(3) ∞ X j=0

(x2 )j +

∞ X j=0

xj =

∞ X j=0 185

(xj − x2j ) =

x 1 − x2

10.14 Exercises - Tests for convergence, Comparison tests for series of nonnegative terms, The integral test. We’ll be using the integral test. Theorem 31 (Integral Test). Let f be a positive decreasing function, defined Pn Pnfor all real x ≥ 1. For ∀n ≥ 1, let sn = k=1 f (k) and tn = 1 f (x)dx. Then both sequences {sn } and {tn } converge or both diverge. Exercise 1.

3 −1 3(4j − 1) + (−1)(4j − 3) 8j + = = 4j − 3 4j − 1 (4j − 3)(4j − 1) (4j − 3)(4j − 1)  n n  X X −1/8 j 3/8 = + (4j − 3)(4j − 1) j=1 4j − 3 4j − 1 j=1    n Z n 3/8 −1/8 ln (4x − 3) ln (4x − 1) + dx = (3/8) + (−1/8) = 4x − 3 4x − 1 4 4 1 1   3 n 3 1 (4x − 3) 1 3(4n − 3) = ln = ln 32 4x − 1 1 32 4n − 1   Z n n 3 X 1 3(4n − 3) xdx j lim ln = lim dx = ∞, so diverges as well n→∞ 32 n→∞ 4n − 1 (4j − 3)(4j − 1) 1 (4x − 3)(4x − 1) j=1 Exercise 2. ∞ √ X 2j − 1 log (4j + 1) j=1

√

j(j + 1)

=

∞ X

aj

j=1

  4 log (4j + 1) (4j + 1) log (4j + 1)(4 + 1/j) 4j + 1 log (4j + 1) = =4 ≤ j(j + 1/4) j(4j + 1)1/2 (4j + 1) (4j + 1)3/2 16 log (4j + 1) ≤ = bj (4j + 1)3/2 X X Now use the integral test on bj to determine the convergence of bj . 0 Z  Z n (−2) log (ax + 1) dx = log (ax + 1)dx = 3/2 (ax + 1) a(ax + 1)1/2 1   Z a −2 −2 = log (ax + 1) − = a(ax + 1)1/2 a(ax + 1)1/2 ax + 1 −2 −4 = log (ax + 1) + a(ax + 1)1/2 a(ax + 1)1/2 !!  1/2 Z n log(ax + 1) −2 log (an + 1) 2 log (a + 1) −4 1 1 lim dx = lim + + − = n→∞ 1 (ax + 1)3/2 n→∞ a an + 1 a(an + 1)1/2 a(a + 1)1/2 (a + 1)1/2 aj ≤

2 log (a + 1) 4 + a(a + 1)1/2 a(a + 1)1/2 P P P Then by integral test, bj converges. Since bj converges, then aj converges by comparison test. P∞ j+1 Exercise 3. j=1 2j . =

Z

n

1

Z lim

n→∞

1

n

n

xe−x ln 2 e−x ln 2 −e−x ln 2 + + − ln 2 − ln 2 (ln 2)2 1    ne−n ln 2 e− ln 2 1 1 1 = + +− + e−n ln 2 + + − ln 2 ln 2 (ln 2)2 ln 2 (ln 2)2    x+1 2 1 1 dx = + ex ln 2 ln 2 (ln 2)2 2 x+1 dx = ex ln 2

Z

(xe−x ln 2 + e−x ln 2 )dx =

186



 n 1

1 ln 2



e− ln 2

By integral test, P∞

P∞

j+1 j=1 2j

converges.

2

j j=1 2j .

Exercise 4.

 n x2 e−x ln 2 2e−x ln 2 2xe−x ln 2 + + x ln 2 − ln 2 −(− ln 2)2 (− ln 2)3 1 1 e     x2 1 2 1 2 2 1 2 − ln 2 dx = e + = + + + 2x ln 2 (ln 2)2 (ln 2)3 2 ln 2 (ln 2)2 (ln 2)3

n

Z

x2 dx = 2x

1 n

Z lim

n→∞

By integral test,

1

j2 j=1 2j

P∞

Z

n

x2



=

converges.

Exercise 5. ∞ X | sin jx| j=1

P∞

1 j=1 j 2

converges since Z lim n→∞

P∞

|sinjx| j=1 j2

n

1

j2

=

∞ X

aj ≤

j=1

∞ X 1 2 j j=1

n    1 x−s+1 1 1 1 dx = lim = lim − 1 if s > 1 = s−1 n→∞ −s + 1 n→∞ 1 − s xs n s − 1 1

converges by comparison test and integral test.

Exercise 6. ∞ X 2 + (−1)j j=1

Exercise 7.

2j

=

∞  X j=1

1 22j−1

+

3 22j

 =

3(1/4) 4 2(1/4) + = 1 − 1/4 1 − 1/4 3

P∞

j! j=1 (j+2)! .

1 1 j! = aj = ≤ 2 = bj (j + 2)! (j + 1)(j + 2) j P P Since bj converges, aj converges, by comparison test. P∞ log j P∞ P∞ log j √ Exercise 8. j=2 j j+1 = j=2 aj ≤ j=2 j 3/2 aj =

Z

n

2

Z lim

n→∞

2

n

n

n Z n 2x−1/2 dx = (−2x−1/2 )0 log x = (−2x−1/2 log x) − − x 2 2 2   n log n log 2 −1/2 = (−2) − + −4x n1/2 21/2 2 log x 4 1/2 dx = 2 log 2 + √ x3/2 2

log x dx = x3/2

Z

P

aj converges by comparison test. P∞ P∞ √ 1 Exercise 9. = j=1 aj . Let bj = 1j . j=1 So

j(j+1)

aj j 1 = lim p = lim p =1 j→∞ j→∞ bj j(j + 1) 1 + 1/j P P By limit comparison test, since bj diverges, aj diverges. √ P∞ P∞ 1+ j Exercise 10. j=1 (j+1)3 −1 = j=1 aj lim

j→∞

bj =

1 j 5/2

√   aj 1+ j j 3 + j 5/2 1 + 1/j 1/2 5/2 = lim j = lim = lim lim j→∞ bj j→∞ (j + 1)3 − 1 j→∞ (j + 1)3 − 1 j→∞ (1 + 1/j)3 − 13 j P P By limit comparison test, since bj converges, aj converges. P∞ P 1 Exercise 11. aj j=2 (log j)s = 187

If s ≤ 0,

X

aj diverges since lim aj 6= 0 j→∞ P∞ 1 1 If 0 < s ≤ 1, (log j)s > j s , and since j=2 Z

−s

(log x)

Z  =

1 js

diverges for 0 < s < 1, so does

P

1 (log j)s

0  Z Z  (log x)−s+1 1 (log x)1−s (log x)−s+1 −s x= (log x) x− x= x (1 − s) (1 − s) 1−s

Thus, if s > 1 has any decimal part, or is an integer, its integral will diverge, so that by integral test, the series diverges. Exercise 12.

|aj | j=1 10j ;

P∞

|aj | < 10. ∞ X |aj | j=1

Exercise 13.

P∞

1 j=1 1000j+1

The series diverges since

<

P1 j

∞ ∞ X X 1 10 10 1 < = = = j j 10j 10 10 1 − 1/10 9 j=1 j=0

P∞

1 j=1 1000j

=

1 1000

P∞

1 j=1 j

diverges.

Exercise 14. ∞ X j cos2 (jπ/3)

Z 1

Exercise 15.

∞

∞ ∞ X X j j = j j j 2 2 e ln 2 j=1 j=1 j=1  −kx  ∞ Z ∞ xe e−kx e−k e−k x −kx = xe = − = + kx 2 e −k (−k) k (−k)2 1 1

≤

P∞

1 j=3 j log j(log (log j))s

Z

1 = x log x(log (log x))s

Z 

−s+1

(ln (ln x)) −s + 1

0

 (ln (ln x))−s+1   −s+1 = ln (ln (ln x))   ln (ln x)

Converges, by integral test, for s > 1 Exercise 16. Converges by integral test since

Z

2

∞

xe

−x2

1

=

e−x −2

! ∞ e−1 =0+ 2 1

Exercise 17. I drew a picture to help me see what’s going on.

√

√ x x ≤ 1 + x2 1 1/n  3/2 Z 1/n √ 2 3/2 2 1 xdx = x = 3 3 n 0 0   ∞ ∞ X 2 1 3/2 2X 1 = 3 j 3 j=2 j 3/2 j=2 So

P

1

n

R 1/n 0

√ x 1+x2 dx

converges by comparison test.

Exercise 18. 188

if s > 1, s < 1, s 6= 0 if s = 1 if s = 0

(e−

√

x 0

) = e−

√

x

−1 √ 2 x

√

√

√ √ √ √ √ √ −e− x 1 e− x ( xe− x )0 = + √ e− x ( xe− x + e− x )0 = 2 2 2 x √  n+1 Z n+1 √ √ n+1 √ x 1 √ + √ =2 = e− x dx = 2 ( x + 1)e− x x x n e e n n √  √ n+1 1 n 1 √ =2 + √n+1 − √n − √n n+1 e e e e √  ∞ √ X j+1 j 1 1 1 √ − √j + √j+1 − √j = − 2 j+1 e e e e e j=1

Note the use of telescoping sum in the last step. The series converges. Rn Rn n Exercise 19. 1 f (x)dx = 1 log xdx = (x ln x − x)|1 = n ln n − n + 1. n−1 X k=1

Z

n

lnk ≤

ln x ≤ 1

n X

ln (k) =⇒

k=2

exp

exp

lnk ≤ n ln n − n + 1 ≤

k=1

n−1 X k=1 n X

n X

ln (k)

k=2

! = (n − 1)! ≤ nn e−n+1

ln k !

= (n)! ≥ nn e−n+1

ln k

k=2 1/n

=⇒

n−1 X

e

e

<

(n!)1/n e1/n n1/n < n e

10.16 Exercises - The root test and the ratio test for series of nonnegative terms. Exercise 1. ((j + 1)!)2 (2j + 2)



(2j)! (j!)2

 =

j 2 + 2j + 1 j→∞ 1 (j + 1)2 = 2 −−−→ (2j + 2)(2j + 1) 4j + 6j + 2 4

Converges by ratio test. P∞ (j!)2 Exercise 2. j=1 2j 2 . 2

2

((j + 1)!)2 2j (j + 1)2 2j j 2 + 2j + 1 j→∞ = j 2 +2j+1 = −−−→ 0 2 2 (j+1) (j!) 2ej ln 2 2 2 Converges by ratio test. P∞ 2j j! Exercise 3.

j=1 j j

2j+1 (j + 1)! j j 2(j + 1) = (j + 1)j+1 2j j! (j + 1)



1 1 + 1/j

j

j→∞

−−−→

2 <1 e

Converges by ratio test. P∞ 3j j! Exercise 4.

j=1 j j

3j+1 (j + 1)! (j + 1)j+1



jj 3j j!



 =3

1 (1 + 1/j)j

Diverges by ratio test. P∞ j! Exercise 5. j=1 3j . j+1 (j + 1)! 3j = j+1 3 j! 3 Diverges by ratio test. P∞ j! Exercise 6.

j=1 22j

189



j→∞

−−−→

3 >1 e

(j!)2 j=1 (2j)!

P∞

(j + 1)! 22j (j + 1) = 2(j+1) j! 4 2 Diverges. Exercise 7.

P∞

1 j=2 (log j)1/j

Draw a picture to see what’s going on. ∞ X j=2

∞

X 1 = exp 1/j (log j) j=2 0<

ln (log j) ln j < j j

ln log j j



−1 ln log j j



for j > 3 ln (log j) ln j < lim =0 j→∞ j j

=⇒ lim

j→∞

−1 ln (log j) = 0 so then j   −1 lim exp ln (log j) = 1 j→∞ j

=⇒ lim

j→∞

P∞

1 j=2 (log j)1/j

Exercise 8.

diverges because the aj term doesn’t go to zero.

P∞

j=1 (j

1/j

− 1)j j→∞

1

((j 1/j − 1)j )1/j = (e j ln j − 1) −−−→ 0 Converges by root test. P∞ −j 2 Exercise 9. j=1 e j→∞

2

(e−j )1/j = e−j −−−→ 0 Converges by root test. Exercise 10. I systematically tried ratio test and then root test. Both were inconclusive.

P1

j.

Consider comparison with

2

ej − j jej 2

By limit comparison test, since Exercise 11.

P∞

Exercise 12.

P∞

j=1

j

(1000) j!

P1 j

!

2

ej − j j j→∞ j= = 1 − j 2 −−−→ 1 ej 2 e  P 1 1 diverges, so does j − ej 2 .

= e1000

j j+1/j j=1 (j+1/j)j .

exp

2

1/j

aj

j 1/j = 1 + j12

1/j

lim aj

j→∞

= lim

j→∞



1 j2

1+

ln j 1 j2

Note that root test is inconclusive.     exp 1j ln j exp 1j ln j aj =  j ≥  j 2 1 + 1j 1 + j12   exp 1j ln j 1 lim aj ≥ lim  >0  2 = j j→∞ j→∞ e 1 + j12 Diverges since limj→∞ aj > 0. P∞ j 3 (√2+(−1)j )j Exercise 13.

j=1

3j

. 190

 =1

!1/j √ √ √ √ 3 j 3/j ( 2 + (−1)j ) j 3 ( 2 + (−1)j )j e j ln j ( 2 + (−1)j ) j→∞ ( 2 + (−1)j ) = = − − − → <1 3j 3 3 3 Converges by root test. P∞ j Exercise 14. j=1 r | sin jx|. If 0 < r < 1.

∞ X

rj | sin jx| <

j=1

∞ X

rj

j=1

so by comparison test,

∞ X

rj | sin jx| converges for 0 < r < 1

j=1

If r ≥ 1, lim rj | sin jx| = 6 0 so

j→∞

∞ X

rj | sin jx| diverges, unless jx = πj

j=1

Exercise 15.

(1) cj = bj −

bj+1 aj+1 aj

>0

∀j ≥ N . Then there must be a positive number r that’s in between cj and 0. aj bj − aj+1 bj+1 ≥ raj

r

n X

aj ≤

j=N

n X

(aj bj − aj+1 bj+1 ) = aN bN − an+1 bn+1 ≤ aN bN

j=N

=⇒

n X

aj ≤

j=N

aN bN r

(2) cn < 0 X 1 bj+1 aj+1 diverges, so <0 bj aj bj aj+1 bj lim ≥ 1 by ratio test < aj bj < bj+1 aj+1 =⇒ j→∞ bj+1 bj+1 aj bj aj+1 j→∞ −−−→ 1 ≤ < bj+1 aj P So by ratio test, aj diverges. bj −

Exercise 16. bn+1 = n; bn = n − 1. nan+1 an

≥ r =⇒ aan+1 ≤ 1 − n1 − nr . n P Using 15, anPconverges. P Exercise 1 1 diverges since bn bn is a harmonic series of s = 1. cn = n − 1 −

n−1−

nan+1 1 an+1 ≤ 0 =⇒ 1 − ≤ an n an

Exercise 17. For some N ≥ 1, s > 1, M > 0, and given that

an+1 A f (n) =1− + s =1− an n n Consider A −

(n) A − nfs−1 n

!

f (n) ns−1 .

(n) Since |f (n)| < M , f (n) is finite, so consider s larger than 1 and n going to infinity so that nfs−1 → 0. P (n) Using Exercise 16, for aj to converge, A − nfs−1 = 1 + r where r > 0, for all n ≥ N , where N is some positive number. M Let r = N s−1 so that f (n) A = 1 + r + s−1 > 1 n 191

P If A > 1, then an converges. If A = 1, then consider using Exercise 15 and bn = n log n.   an+1 an+1 = (n − 1) log (n − 1) − n log n an an       (n − 1) 1 f (n) f (n) = (n − 1) log = = (n − 1) log (n − 1) − n log n 1 − + s − n log n n n n ns     n f (n) = −(n − 1) log − n log n (n − 1) ns log n n→∞ since s−1 −−−−→ 0, n     n f (n) −(n − 1) log < 0 for n large enough − n log n (n − 1) ns X since cn < 0 for n ≥ N for some N > 0, then by Exercise 15, an is divergent.

cn = bn − bn+1

(n) f (n) M Given that A < 1, then for A − nfs−1 , choose N > 0 so that nM s−1 <  < 1 and that A − ns−1 ≤ A + ns−1 = A +  ≤ 1. We can always choose  small enough because there’s always a real number in between A and 1 (Axiom of Archimedes).   f (n) f (n) A − s−1 ≤ 1 =⇒ − A − s−1 ≥ −1 n n

=⇒ using Exercise 16,

(n) A + nfs−1 1 an+1 =1− ≥ 1 − for all n ≥ N an n n

Exercise 18.



1 · 3 · 5 . . . (2n + 1) 2 · 4 · 6 . . . (2n) · 2 · 4 · 6 . . . (2n + 2) 1 · 3 · 5 . . . (2n − 1)

k

 =

2n + 1 2n + 2

k

n→∞

−−−−→ 1

Ratio test fails. k  k  k −1 −1/2 2n + 1 = 1+ = 1+ = 2n + 2 2n + 2 n+1 j   X j k   k   X k −1/2 k −1/2 −1/2 = =1+k + j j n+1 n+1 n+1 j=0 j=2

an+1 = an



Note that for k < ∞,

j k   X k −1/2

<∞ n+1 X j   k k −1/2 ≤M Let n + 1 j=2 j X X k/2 = A > 1 or k > 2 means aj converges k/2 = A ≤ 1 or k ≤ 2 means aj diverges j=2

j

10.20 Exercises - Alternating series, Conditional and absolute convergence, The convergence tests of Drichlet and Abel. We will be using Leibniz’s test alot, initially. Theorem 32 (Leibniz’s Rule). If aj is a monotonically decreasing sequence with limit 0, P∞ j−1 aj converges. j=1 (−1) If S =

P∞

j=1

aj ,

sn =

Pn

j=1 (−1)

j−1

aj , 0 < (−1)j (S − sj ) < aj+1

Exercise 1.

P∞

Exercise 2.

P∞

√ 1 100 j+ √j

j=1

(−1)j+1 √ . j

limj→∞

1√ , 101 j

= 0 Converges conditionally.

√

j j j=1 (−1) j+100

≥

√1 j

√

limj→∞

j j+100

= 0. Converges by Leibniz’s test.

so by comparison test, the series diverges absolutely. So the alternating series converges conditionally

by comparison test. 192

P∞

Exercise 3.

j=1

(−1)j−1 js

If s > 1, then the series absolutely converges. limj→∞

1 js

= 0 if s > 0. Converges conditionally

for 0 < s < 1. Otherwise, if s < 0 the series diverges absolutely. 3  P∞ j 1·3·5...(2j−1) . (−1) Exercise 4. j=1 2·4·6...(2j) aj+1 1 · 3 · 5 · (2j + 1) 2 · 4 · 6 . . . (2j) 2j + 1 = = aj 2 · 4 · 6 . . . (2j + 2) 1 · 3 · 5 . . . (2j − 1) 2j + 2 Absolutely converges. P∞ (−1)j(j−1)/2 Exercise 5. Exercise 6.

j=1

P∞

2j

j j=1 (−1)

 exp j ln





converges since limj→∞

2j+100 3j+1

j

2j + 100 3j + 1

(−1)j(j−1)/2 2j

= 0;

P∞

1 j=1 2j

=

1/2 1−1/2

= 1 . Absolutely converges.

.



So the alternating series converges. 





       2 298 2 1 298 = exp j ln + ≤ exp j ln + 2 = 3 9j + 3 3 9j + 3 3   2 146 = exp j ln + 3 3 + 1/j    2j + 100 ≤ lim 0 ≤ lim exp j ln j→∞ j→∞ 3j + 1   2 146 exp j ln + =0 3 3 + 1/j    2j + 100 =0 =⇒ lim exp j ln j→∞ 3j + 1 

2.5j 5 2j + 100 < = ( for j ≥ 200) 3j 3j 6  j  j 2j + 100 5 =⇒ < for j ≥ 200 3j + 1 6

2j + 100 3j + 1

<

So the series absolutely converges by comparison with a geometric series. P∞ (−1)j √ Exercise 7. j=2 j+(−1)j . 1 doesn’t exist since ??? j + (−1)j To show divergence, we usually think of either taking the general term and finding the limit (and if it goes to a nonzero constant, then it diverges), or we use ratio, root, comparison test on the general term. Since this is an alternating series, I’ve observed that the general term is a sum of two adjacent terms, one even and one odd. lim √

j→∞

√ √

(−1)j j + (−1)j

√ √ (−1)2j (−1)2j+1 1 −1 2j + 1 − 1 − ( 2j + 1) √ √ √ √ √ + = + = = 2j + (−1)2j 2j + 1 + (−1)2j+1 2j + 1 2j + 1 + 1 ( 2j + 1)( 2j + 1 − 1) q √ √ 1 √ √ 2j 1 + 2j − 2j − 2 2j + 1 − 2j − 2 √ = √ = √ = √ q 1 ( 2j + 1)( 2j + 1 − 1) ( 2j + 1)( 2j 1 + 2j − 1)  √ √  ! 1 2j 1 + − 2j − 2 1 − 4√12j 4j −2 for j large  −−−−−−→ ' √ = √  1 1 1 j 2 − 2j + 2√2j 3/2 ( 2j + 1)( 2j 1 + 4j − 1)

Every term, since we considered any j, will contain −2. So we factor it out. Then ! 1 ! 1 − 4√12j 1 1 1 − 4√2j 1 > = 1 1 1 √ j 2 − 2j j 4j + 2√2jj 4 − 3/2 2j 193

By comparison test to

1 j

the series diverges.

Exercise 8. Using the theorem

Theorem P33. Assume |aj | converges P P P Then aj converges and | aj | ≤ |aj |. So using the contrapositive, P If Paj diverges, |aj | diverges. 1 j 1/j

1

= e

1 j

ln j

1

lim

=

j→∞ j 1/j

1 

exp limj→∞

1 j

ln j

 =1

Diverges absolutely. Exercise 9.

2 j j j=1 (−1) 1+j 2

P∞

Diverges absolutely.

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

 (4j 2 − 4j + 1) (1 + 4j 2 ) 4j 2 − 4j + 2 − 4j 2 − 4j + 2 (4j 2 − 4j + 2) 1 + 4j 2 4j − 1 = 2(1 + 4j 2 )(2j 2 − 2j + 1) 4j − 1 4 − 1/j 1 (j 3 ) = = 2(1 + 4j 2 )(2j 2 − 2j + 1) 2(4 + 1/j 2 )(2 − 2/j + 1/j 2 ) 4

By limit comparison test, with Exercise 10.

P

1 j3 ,

j2 j=1 1+j 2

P∞



converges.

(−1)n n=1 log (en +e−n )

P∞

lim

1 1 = lim =0 −n + e ) n→∞ n

n→∞ log (en

The series converges. n 1 n  =1 = lim = lim  −2n ) −n n −2n n+log (1+e + e ) n→∞ log e + log (1 + e ) n→∞

lim

n→∞ log (en

n

Since

P

1 n

Exercise 11.

limj→∞

diverges,

P

1 log (en +e−n )

diverges.

(−1)j j=1 j log2 (j+1)

P∞

1 j log2 (j+1)

= 0 so by Leibniz’s test, the alternating series. 1 1 < n log2 (n + 1) n log2 (n) 0 Z Z  1 −1 −1 n→∞ 1 = = −−−−→ 2 log n log n log 2 n log n

Converges by comparison test to Exercise 12.

(−1)j j=1 log (1+1/j)

P∞

1 , n log2 n

which converges by integral test. So the series absolutely converges.

diverges absolutely. 194

  2j + log 2j−1  +  = +  =     = 2j 2j 1 1 log 1 + 2j−1 log 1 + 2j log 2j−1 log 2j+1 log 2j−1 log 2j+1 2j 2j     (2j)2 log 4j log 1 + 4j 21−1 2 −1    =   =   = 2j 1 1 1 log 1 + 2j log 1 + 2j log 2j−1 log 1 + 2j−1   1 1 4j 2 −1 + o 4j 2 −1      ≈ = 1 1 1 1 + o + o 2j−1 2j−1 2j 2j (−1)

−1 

1

≈

− log

1



2j+1 2j



1 1 − 2j 4j 2 − 2j j→∞ = 1 −−−→ 1 4j 2 − 1 1 − 4j 2

So the alternating series diverges. Exercise 13.

P∞

j=1

(−1)j j 37 (j+1)!

Use the ratio test. (j + 1)3 7 (j + 1)! aj+1 = = aj (j + 2)! j 3 7



1 j+2

 37 1 1+ →0 j

P |aj |. Then aj converges. The series absolutely converges. R P∞ n n+1 e−x Exercise 14. n=1 (−1) x dx n Converges for

P

Z

n+1

n

e−x dx ≤ x

Z

n+1

n

n+1    e−2x −1 1 1 1 dx = = − = e2x −2 n 2 e2n e2(n+1) =

e2 − 1 <1 2e2n+2

Converges absolutely. Pn Exercise 15. j=1 sin (log j) limj→∞ sin (log j) doesn’t exist. So the series is divergent.   P∞ 1 Exercise 16. j=1 log j sin j Note that     1 sin 1/j log j sin = log j 1/j     sin 1/j sin 1/j lim log = log lim = log 1 = 0 j→∞ 1/j j→∞ 1/j ∞ X

 2k+1 1

j 1 sin = (−1)k j (2k + 1)! k=0      2k+1  2k 1 1 k k ∞ ∞ (−1) (−1) −1 X j  X j    log j  = log 1 + 2 + ≥ (2k + 1)! 6j (2k + 1)! k=0

k=2

 ≥ log 1 + − The series absolutely converges.   P∞ 1 j Exercise 17. (−1) 1 − j sin j=1 j 195

1 6j 2

 ≥

−1 6j 2

  0  1 1 1 −1 = 1 − x sin = − sin − x cos x x x x2 = − sin ∞

1 X sin = x j=0

−x sin x1 + cos x1 1 1 1 + cos = x x x x


1 2j+1 x

(−1)j (2j + 1)!

−x sin x1 + cos x1 = x 3

−1 + = P∞

j j=1 (−1)



( x1 )

(+1) 3!

+

P∞ ( x1 )2j+1 (−1)j j=2

(2j+1)!

+1−


1 2 x

/2 +

P∞ ( x1 )2j (−1)j (2j)!

x

< 0 for x large enough



1 − j sin 1j converges since aj = 1 − j sin 1j is monotonically decreasing sequence with limit 0.    2k+1  2k+1 1 1 k k ∞ ∞ (−1) (−1) X j 1 1 X j  =1−j + 1 − j sin = 1 − j = j (2k + 1)! j (2k + 1)! k=0



k=1

∞

 2k

X  = 1 − 1 + k=1

1 j



∞ (−1)k  X = (2k + 1)!

 2k 1 j

(−1)k+1

1 (cos )0 = x j j=1 (−1) (1

≤

(2k + 1)!

k=1

The series converges absolutely since the term itself is a series that is dominated by must converge.   P∞ 1 j Exercise 18. (−1) 1 − cos j=1 j .

P∞

j=2

1 6j 2 ,

1 6j 2

so that by comparison test, the series

   1 −1 −1 1 − sin = 2 sin < 0 x x2 x x

− cos 1j ) converges since aj = (1 − cos 1j ) is monotonically decreasing to 0

∞ ∞ X X (1/j)2k (−1)k (1/j)2k (−1)k+1 1 1 = ≤ 2 (1 − cos ) = 1 − j (2k)! (2k)! 2j k=0 k=1 P 1 So the series converges absolutely, by comparison test with j 2 which converges. P∞ 1 j Exercise 19. j=1 (−1) arctan 2j+1 .     1 1 −1 −2 (arctan )0 = (2) = <0  2 2 2j + 1 (2j + 1) (2j + 1)2 + 1 1 1 + 2j+1 P∞ 1 1 j j=1 (−1) arctan 2j+1 converges, since aj = arctan 2j+1 is monotonically decreasing to 0 ∞

∞

X X 1 = (arctan x)0 = (−x2 )j = (−1)j x2j 2 1+x j=0 j=0 ∞ X x2j+1 =⇒ arctan x = (−1)j 2j + 1 j=0

 2k+1 1 ∞ (−1)k X 2j+1 1 1 arctan = = + (−1) 2j + 1 (2k + 1) 2j + 1 k=0

1 (2j+1)3

3

! +

∞ (−1)k X k=2



1 2j+1

2k + 1

2k+1 >

1 1 3(4j 2 + 4j + 1) + (−1) 12j 2 + 12j + 2 2j 21 + (−1) = = > > for j > 2 3 3 3 2 2j + 1 3(2j + 1) 3(2j + 1) 3(2j + 1) (2j + 1) 9j P1 P 1 arctan 2j+1 diverges absolutely. The series is conditionally convergent. So by comparison test to j,
P∞ π j Exercise 20. j=1 (−1) 2 − arctan log j >

196

π 2

− arctan log n

0

 =

−1 1 + (log n)2

  1 n

π π π − arctan log n ≥ − arctan (n − 1) → − arctan n just change indices 2 2 2  n Z n  π π 1 − arctan x dx = x − x arctan x − ln (1 + x2 ) 2 2 2 0 0   π  1

π 1 2 = n − n arctan n − ln 1 + n =n − arctan n + ln (1 + n2 ) 2 2 2 2     π 1 lim n − arctan n + ln (1 + n2 ) → ∞ n→∞ 2 2 Pπ Then by the integral test, 2 − arctan log n diverges absolutely. So the alternating series is conditionally convergent. Exercise 21.

 log 1+ j=1

P∞

 limj→∞ log 1 + Exercise 22.

1 | sin j|



1 | sin j|

 doesn’t exist and log 1 +

 sin jπ + j=2

P∞



1 log j

1 | sin j|



>0

∀j so the series diverges.



    1 1 sin 2jπ + + sin (2j + 1)π + = log 2j log (2j + 1)     1 1 1 1 = sin (2jπ) cos + sin + cos (2j + 1)π sin cos (2πj) + sin (2j + 1)π cos = log 2j log 2j log (2j + 1) log 2j + 1 1 1 − sin = = sin log 2j log 2j + 1   2k+1 2k+1 1 1 k ∞ ∞ (−1) (−1)k X X log 2j log (2j+1) = − (2k + 1)! (2k + 1)! k=0 k=0     1 1 − sin = sin log 2j log 2j + 1  2k+1  2k+1 ! ∞ X (−1)k 1 1 = − (2k + 1)! log 2j 2j + 1 k=0  2k+1  2k+1 1 1 0 < log 2j < log 2j + 1 so − >0 log 2j log 2j + 1     1 1 and since for j > 1, and are < 1 and so we are adding smaller and smaller amounts log 2j log 2j + 1 1 log 2j + 1 < − log 2j =   1 1 log 1 + 2j log (2j + 1) − log 2j 2j = ≤ ≤ log 2j log 2j + 1 (log (2j))2 (log (2j))2 n Z n 1 1 n→∞ 1 =− −−−−→ 2 log 2j 1 log 2 1 2j(log 2j) So the series converges by using integral test, showing that verges. Exercise 33.

P

n=1

P

nn z n 197

1 2j(log 2j)2

converges, so by comparison test, the series con-

∞ X X X ln j
j j n n e z n z = (jz) = n=1

j=1

n X

(eln j z)j =

j=1

=

e

ln j

z − (eln j z)n −→ ∞ 1 − eln j z

So z = 0 Exercise 34.

P∞

j=1

(−1)j z 3j j!

=

P∞

j=1

(−z 3 )j j!

3

= e−z . C .


j P∞ = j=0 13 z j P j Pn z be convergent or j=1 z j bounded. Exercise 35.

zj j=0 3j

P∞

|z| < 3 and |z| = 3 if z 6= 3 Exercise 36.

zj j=1 j j

P∞

{z} = C since   z j < 1 for j ≥ N > |z|

Exercise 37.

(−1)j j=1 z+j

P∞

j→∞

1 By Leibniz’s Rule, since z+j −−−→ 0, then the series converges. However, z cannot be equal to any negative integer since one term in the series will then blow up.   P∞ zj √ log 2j+1 . Exercise 38. j=1 j j   1   log 2 + j j z 1 √ log 2 + √ = zj j j j log (2+ 1j ) log (2+ 1j ) √ √ Since limj→∞ = 0 so that is a monotonically convergent sequence. j j Pn j Then by Dirchlet’s test, j=1 z must be bounded. |z| > 1, and |z| = 1 if z 6= 1. j 2 j 2 P∞  P∞  1 1 17j |z| = 1 + (|z|17 )j Exercise 39. 1 + j=1 j=1 5j+1 5j+1

!j j 1 17 1+ |z| 5j + 1  j  j 1 1/5 lim 1 + ≤ lim 1 + = e1/5 j→∞ j→∞ 5j + 1 j 

e1/5 |z|17 < 1 =⇒ |z| < e−1/85 Exercise 40.

(z−1)j j=0 (j+2)!

P∞

∞ ∞ ∞ X X (z − 1)j X |(z − 1)|j+2 |z − 1| |(z − 1)|j ≤ ≤ = e|z−1| − 1 − (j + 2)! (j + 2)! (j + 2)! 1! j=0 j=0 j=0 The series converges ∀ z. Exercise 41. ∞ X (−1)j (z − 1)j j=1

j

=

∞ X (1 − z)j j=1

j

So the series converges ∀ z except for z = 0. 198

= log (1 − (1 − z)) = log z

Exercise 42.

(2z+3)j j=1 j log (j+1)

P∞

1 1 = 0 so is a monotonically convergent sequence j log (j + 1) j log (j + 1) |2z + 3| < 1 X X then (2z + 3)j converges (2z + 3)j 1 3 |z + | < 2 2   0  1 −1 x = log (x + 1) + < 0; for x > 0 x log (x + 1) (x log (x + 1))2 x+1 lim

j→∞

By Dirichlet’s test, Exercise 43.

P∞

j=1

P∞ (

z−1 z+1

3 1 converges for |z + | ≤ ; 2 2     j j j P∞ (−1)j /2 1−z 1−z = j=1 (−1) (2j−1) 1+z 1+z j− 1 (2z+3)j j=1 j log (j+1)

P∞

z 6= −1 .

2

j

)

1 1 = 0 and 2j−1 is monotonically decreasing. =⇒ j=1 2j−1 = limj→∞ 2j−1 1 So 2j−1 is a monotonically decreasing convergent sequence of real terms. For z−1 z+1 < 1, z − 1 z + 1 < 1 =⇒ |z − 1| < |z + 1|

(u − 1)2 + v 2 < (u + 1)2 + v 2 u2 − 2u + 1 < u2 + 2u + 1

<(z) > 0;

z 6= 0

−u < u Exercise 44. ∞  X

j X j X j ∞  ∞  j  z −1/2 1 1 1 + = = 1− 2z + 1 2 2z + 1 2 2z + 1 j=1 j=1  j 1 is a monotonically decreasing, convergent sequence 2 1 2z + 1 − 1 2z Now we want 1 − = = <1 2z + 1 2z+ 2z + 1

j=1

|2z| < |2z + 1| |z| < |z + 1/2| 2

u + v 2 < (u + 1/2)2 + v 2 = u2 + u + 1/4 + v 2 −1 1 < u =⇒ <(z) > − 4 4 Exercise 45.

P∞

j j=1 j+1



z 2z+1

j



Exercise 46.

P∞

1 j=1 (1+|z|2 )j

=

−1 if <(z) = 4
−1 2 − 14 + iv 2 + 2iv
= 1 −1 2 4 + iv + 1 2 + 2iv =⇒ z 6=

−1 4

.

0 x (x + 1) − x 1 = = >0 x+1 (x + 1)2 (x + 1)2 j is a monotonically increasing and convergent sequence j+1 z 1 −1 <1 + = 2 2(2z + 1) 2z + 1 2z + 1 1 z > 1 =⇒ |2 + z | > 1

P∞  j=1

1 1+|z|2

j 199

1 <1 1 + |z|2 ∀z except z = 0

1 < 1 + |z|2 0 < |z|2 Exercise 47.

P∞

j=1 (−1)

Use Dirichlet’s Test. for convergence is

j 2j sin2j x j 1 j

is a monotonically decreasing sequence converging to zero. Consider (−2)j sin2j x. The condition |(−2 sin2 x)| < 1 for j ≥ N for some N −π π =⇒ x ∈ ( + nπ, + nπ), n ∈ Z 4 4

Exercise 49.

P

P

aj converges.

aj a1j

diverges. P Then since aj is a convergent series (by Abel’s test), a1j is a divergent sequence. P 1 1 Then aj is divergent (since limj→∞ aj doesn’t exist ). P Exercise 50. |aj | converges. P P |aj | converges, then aj converges. |aj |2 = a2j . |aj | converges, then lim |aj | = 0 j→∞ |aj |2 lim =1 j→∞ a2 lim |aj |2 = 0 j j→∞ P By limit comparison test, a2j converges. P1 P  1 2 converges, but Counterexample: j j diverges. P P Exercise 51. Given aj , aj ≥ 0. aj converges. lim aj = 0

X√

j→∞

1 aj p (j)

lim

√

j→∞

Z n−1 X

j

x =

j=0

j=0 √

A counterexample would be

aj jp

=

q

n−1 X

1/2

 aj =

lim aj

j→∞

n

X xj xj+1 = = j + 1 j=1 j

Z

=0 1 − xn 1−x

aj j .

Exercise 52.

(1)

P

aj converges absolutely, then if

a2j 1 + a2j (2)

P

|aj | converges,

P

a2j converges.

a2j −1 =1+ 1 + a2j 1 + a2j X X ≤ a2j since a2j converges ,

a2j converges 1 + a2j

P

aj converges absolutely, lim   j→∞ |aj | = 0 P aj P 1 1+aj = |aj | |1+aj | . By Abel’s test, since 1 1 = =1 j→∞ |1 + aj | |1 + limj→∞ aj | P aj By Abel’s test, 1+aj is convergent. lim

shows that

1 ≥ 0 is a monotonically convergent sequence |1 + aj |

200

10.22 Miscellaneous review exercises - Rearrangements of series. Exercise 1. (1) r       p p p p 1 p 1 1 1 aj = j + 1 − j = j 1 + − j = j 1 + +o + −1 = j 2 j j      p 1 1 1 = j +o 2 j j lim aj = 0

j→∞

(2)

    c     1 1 1 − 1 = (j c ) 1 + c +o −1 = aj = (j + 1)c − j c = j c 1+ j j j         1 1 1 = jc c +o = cj c−1 + j c o j j j if c > 1,

aj diverges

if c = 1,

lim aj = 1

j→∞

if c < 1,

lim aj = 0

j→∞

Exercise 2.

(1)    ∞ n j j−1 X 1 1 1 (x ) (−1) n→∞ − (1 + xn ) n = exp ln (1 + xn ) = exp  −−−→ 1 n n j=1 j 
n 1/n (2) limn→∞ (an + bn )1/n = limn→∞ a 1 + nb = a if a > b. 

an +an−1 2

Exercise 3. an+1 =

an−1 +an−2 22

=

+

an−2 +an−3 22

10.24 Exercises R ∞ -xImproper integrals. √ dx 0 x4 +1

Exercise 1.

 √

lim

x4 + 1 x √ dx x4 +1

x→∞

Since

R∞ 1

1 x

diverges, so does

R∞ 0



x

1 1/x

x2

 = lim

x→∞

x2

p

1 + 1/x4

=1

Exercise 2.

Z

∞

e

−x2

Z dx =

−∞

∞

e

−∞

Z

−x2

dx +

0

e

−x2

Z dx =

0

Z 0

∞

2

e−x dx ≤

R∞

Exercise 5.

R∞

Exercise 6.

R1

Exercise 7.

R 1−

Exercise 8.

R∞

Exercise 9.

R 1−

0

0+

0+

0+

√

x

x

log √ x dx x log x 1−x dx

√ dx x log x 201

e ∞

2

∞ = 1 0

dx

0

dx + −

e−x dx = −e−x

0

x dx −∞ cosh x

0+

Z

∞

√1 dx ex e− √

e

−x2

0

Z

Converges by theorem. R∞ Exercise 3. 0 √ 13 dx x +1 Exercise 4.

∞

−x2

Z dx = 2 0

∞

2

e−x dx

Exercise 10.

R∞

dx x(log x)s

11.7 Exercises - Pointwise convergence of sequences of functions, Uniform convergence of sequences of functions, Uniform convergence of sequences of functions, Uniform convergence and continuity, Uniform convergence and inteP∞ zj gration, A sufficient condition for uniform convergence, Power series. Circle of convergence. Exercise 1. j=0 2j = P∞ z
j j=0

2

Using the comparison test, z j z ≤ tj ; < 1 =⇒ |z| < 2 2 2
j P∞ P∞ Suppose |z| = 2, z 6= 2 j=0 z2 = j=0 (e2iθ )j Pn Now j=0 (e2iθ )j ≤ sin1 θ + 1, since n X j=1

So

P∞

j=0

Exercise 2.

(e2iθ )j =

einθ − e−iθn eiθn+iθ sin nθeiθn+iθ e2iθ − e2iθ(n+1) 1 = = < 1 − e2iθ −e−iθ + eiθ sin θ sin θ


z j

converges for |z| = 24, z 6= 2 2 P∞ zj j=0 (j+1)2j

Use ratio test .

(j + 1)2j z (j + 1) z (1 + 1/j) j→∞ z z j+1 aj+1 = = −−−→ = j+1 aj (j + 2)2 zj 2 (j + 2) 2 (1 + 2/j) 2 P∞ P If |z| < 2, j=0 aj converges, if |z| > 2, aj diverges. P zj P 2iθ j  1  If |z| = 2, = (e ) j+1 j (j+1)2 1 Now j+1 is a monotonically decreasing sequence of real terms. P 2iθ j (e ) is a bounded series. P By Dirichlet’s test, aj converges if |z| = 2, z 6= 2 P∞ (z+3)j

Exercise 3.

j=0 (j+1)2j

Use ratio test: aj+1 = aj



(z + 3)j+1 (j + 2)2j+1



(z + 3) (j + 1)2j = (z + 3)j 2



j+1 j+2

 =

(z + 3) 2



1 + 1/j 1 + 2/j



j→∞

−−−→

z+3 2

Using Theorem 11.7, Theorem 34 (Existence of a circle of convergence). X converges for at least z1 6= 0 Assume aj z j diverges for at least one z2 6= 0 X absolutely converges for |z| < r ∃ r > 0, such that aj z j diverges for |z| > r |z+3|

P We can plug in real numbers to satisfy the condition 2 < 1 for convergence. aj converges for |z + 3| < 2; diverges for |z + 3| > 2.  P P 1 1 Consider |z + 3| = 2; z 6= 1 aj = (e2iθ )j j+1 . Since j+1 is a monotonically decreasing sequence of real P 2iθ j numbers and (e ) is a bounded series, by Dirichlet’s test, P aj converges for |z + 3| = 2; z 6= −1. j−1 2j P∞ (−1)j 22j z2j P∞ Exercise 4. = − j=1 (−1) (2j)(2z) . Look at what the terms look like. j=1 2j Consider using Leibniz’s Rule, Theorem 10.14. Theorem (Leibniz’s rule). If aj is a monotonic decreasing sequence and limj→∞ aj = 0, P35 ∞ then j=1 (−1)j−1 aj converges. 202

(−1)j 22j z 2j (−1)j (2z)2j = 2j 2j (2z)2j |2z|2j (2|z|)2j = Consider = 2j 2j 2j Consider 2|z| = M 1/2 < ∞ =⇒

(2|z|)2j Mj ej ln M = = 2j 2j 2j

converges for 0 < 2|z| = M ≤ 1 =⇒ |z| ≤

1 2

(we use Theorem 11.6 at this point, because real numbers are included in complex numbers). Theorem 36. Assume

P

aj z j converges for some z = z1 6= 0.

P (1) P aj z j converges absolutely ∀z with |z| < |z1 |. (2) aj z j converges uniformly on every circular disk with center at 0 and R < |z1 | We had first used Leibniz’s test to find az1 on the real line. 24j z 4j ((1 − 4z12 ) − 24j z 4j (4j(1 − 4z12 ) − 2) = 4j(4j − 2) (4j − 2)

2 4j ) j→∞

−−−→

24j z 4j (1 − 4z12 ) (4j − 2)

If |z| > 21 , the series diverges (by aj th general term test). Exercise 5.

P∞

j=1 (1

− (−2)j )z j .

Try limit comparison test . The first step is to test absolute convergence first; it’s easier. (1 − (−2)j )z j |(−2z)j |

 1 j j→∞ = − 1 −−−→ 1 −2

P P According to limit comparison test, for (1 − (−2)j )z j to converge, (−2z)j must converge. P ∞ So if |z| < 21 , then j=1 (1 − (−2)j )z j absolutely converges. 1 −1 If |z| = 2 , z 6= 2 , X

If z =

X zj − (−e2iθ )j = X 1 =0− (e2iθ+πi )j < sin (θ + π)

(1 − (−2)j )z j =

X

−1 2 ,

X  1 j X − − 1j → ∞ 2

Exercise 6.

j!z j j=1 j j

P∞

A very big hint is to use Exercise 19 on pp. 399, in the section for Exercises 10.14. 203

Since

Pn−1 j=1

f (j) ≤

Rn 1

f (x)dx n−1 X j=1

Z

n

ln x = n ln n − n + 1

ln j ≤ 1

≤

n X

ln j

j=2

(n − 1)! ≤ nn e−n n

≤ n!

n! ≥ ne−n nn n! lim ≥ lim ne−n = 0 n→∞ nn n→∞ n! n2 lim n ≤ lim n = 0 n→∞ n n→∞ e n! =⇒ lim n = 0 n→∞ n P So then since nn!n is a monotonically decreasing convergent sequence of real terms; if z is a bounded series, then by P n!zn Dirichlet’s test, nn is convergent. |z| < 1; |z| = 1 if z 6= 1 Try the ratio test, because it’s clear from the results of the ratio test where convergence and divergence begins and ends.  n (n + 1)!|z|n+1 nn n = |z| = (n + 1)n+1 n!|z|n n+1 n  1 n→∞ 1 |z| −−−−→ |z| = 1 + 1/n e Converges for |z| < e . Now try plugging in a complexified e: ej j!ej ei2θj j j e−j jej ei2θj i2θj  j = ≥ e →∞ j (j)j jj j!

So the series diverges for |z| = e. P∞ (−1)j (z+1)j Exercise 7.

j 2 +1

j=0

By the ratio test, |z + 1|j+1 j 2 + 1 1 + 1/j 2 j→∞ = |z + 1| −−−→ |z + 1| 2 j j + 2j + 2 |z + 1| 1 + 2/j + 2/j 2 The series absolutely converges for |z + 1| < 1. P (−1)j j→∞ 1 For z = 0, j 2 +1 converges since j 2 +1 −−−→ 0 P (−1)j (−1)j P 1 P 1 1 1 For z = −2, = j 2 +1 j 2 +1 and j 2 +1 < j 2 , but j 2 is convergent (by integral test). So the series converges for z = −2. By Dirichlet’s test, the series converges as well, if we treat aj = (−1)j (z + 1)j and bj = j 21+1 to be the monotonically decreasing sequence. =⇒ |z + 1| ≤ 1 for convergence x Exercise 8.

P∞

j=0

2

aj z j , 0 < a < 1

Use the root test . j→∞

2

(aj z j )1/j = aj z −−−→ 0 So the series converges ∀ z ∈ C P∞ (j!)2 j Exercise 9. j=1 (2j)! z 204

Use the ratio test aj+1 (j + 1)2 z ((j + 1)!)2 z j+1 (2j)! j→∞ 1(z) = = −−−→ 2 j aj (2(j + 1))! (j!) z (2j + 2)(2j + 1) 4 X for |z| < 4, aj absolutely converges X for |z| > 4, aj diverges Let’s test the boundary for convergence. (j!)2 j ej ln 4 (j!)2 ej ln 4 (j!)2 j ln 4 (j!)2 ≥ 4 = e ≥ = (2j)! (2j)! (2j − 2)! (2j)(2j − 1) (2j − 2)! (2j)2 (j!)2 j 2j j 2j ≥ = → ∞ (2j − 2)! (j!)2 (2j − 2)! where we had used (n − 1)! ≤ nn e−n n ≤ n! n n! ≤ n en n Exercise 10.

P∞

√

3

j=1

j j

z

j

=

P∞

√

=⇒ ne−n ≤

=⇒

n! nn

nn en ≤ n! n

j ln 3 j

z

e

j=1

j √

j+1 ln 3 j+1

 √  √ j 1 z √ e( j+1− j) ln 3 z = j ln 3 j j+1 j+1 e z r   p p 1 1 1 1 j+1− j = 1+ −1'1+ −1= (for large j ) j 2 j 2j   1 j (for large j ) e 2j z → 0 j+1 =⇒ Converges ∀ z ∈ C e

Exercise 11.

P∞  1·3·5...(2j−1) 3 j=1

aj+1 = aj

2·4·6...(2j)



zj

3  3 3  3  2 · 4 · 6 . . . (2j) 1 −1/2 1 · 3 · 5 . . . (2j + 1) 2j + 1 j+1 z z = 1+ z = 2 · 4 · 6 . . . (2j + 2) 1 · 3 · 5 . . . (2j − 1) zj 2j + 2 j+2 If |z| < 1, it converges by ratio test, if |z| = 1, then it converges by Gauss test  3 k 3   X 3 −1/2 aj+1 −1/2 z= |z| = = 1+ k aj j+2 j+2 k=0    2  3 ! −1/2 −1/2 −1/2 j→∞ = |z| 1 + 3 +3 + −−−→ |z| j+2 j+2 j+2 diverges for |z| > 1 (by ratio test )

Exercise 12.

j 2 P∞  1 1 + zj j=1 j 

1 1+ j

1 , e 1 |z| > , e

|z| <

!1/j

j 2 z

j

 =

1 1+ j

j

j→∞

z −−−→ e1 z

X 1 2 (1 + )j z j converges by root test j X 1 j2 j (1 + ) z diverges by root test j 1 =r e 205

Exercise 13.

P∞

j=0 (sin aj)z

j

a>0

| sin (aj)z j | ≤ |z|j P P∞ ∞ By comparison test, j=0 (sin aj)z j converges, since j=0 |z|j converges absolutely, for |z| < 1. Note that if a = π, the series is zero. P∞ j=0 (sin aj) → ∞ for a = 2π so r=1indeed. P∞ P∞  eaj −e−aj  j j z = Exercise 14. (sinh aj)z = j=0 j=0 2 1 ea ,

then

Exercise 15.

P∞

If |z| <

P

1 2

P

∞ a j j=0 (e z)

−




a j b
a j+1 b

!

j=0

sinh ajz j converges. So then the radius of convergence is r =

zj j=1 aj +bj .


j z ea

P∞

a>0

;

1 ea

Assume a < b zj


j bj (1 + ab )  
j   bj 1 + ab z j+1 = z (ratio test)
j+1  j a z b ) bj+1 (1 + b

1+ 1+

j→∞

−−−→

z  b

So then |z| ≶ b converges (diverges) by ratio test. If a = b, ∞ ∞ X 1 X  z j zj = 2aj 2 j=1 a j=1

By comparison test with xj , if |z| ≶ |a|, the series converges (diverges).  P∞  aj bj Exercise 16. + z j Use ratio test on each of the sums, separately. 2 j=1 j j (a|z|)j+1 j a|z| j→∞ −−−→ a|z| = j j + 1 (a|z|) 1 + 1j 1 then the series converges a !2 (b|z|)j+1 j 2 1 j→∞ = b|z| −−−→ b|z| 1 (j + 1)2 (b|z|)j 1+ j =⇒ |z| <

=⇒ |z| <

1 b

So if a ≷ b, then r = a; (b) Exercise 17.

R1 0

fn (x) =

R1 0

2

nxe−nx =

2

e−nx −2

1 = 0

e−n −1 n→∞ 1 −−−−→ 2 −2

However, 2 limn→∞ nxe−nx = 0 This example shows that the operations of integration and limit cannot always be interchanged. We need uniform converlimn→∞ sinnnx = 0 gence. Exercise 18. fn (a) = sinnnx f (x) = limn→∞ fn (x) nx fn0 (x) = n cos =⇒ limn→∞ fn0 (0) = 1 n This example shows that differentiation and limit cannot always be interchanged. P∞ sin jx Exercise 19. j=1 j 2 = f (x) 206

1 sin jx ≤ 2 ∀x ∈ R j2 j ∞ ∞ X X | sin jx| sin jx by comparison test, converges, so converges 2 j j2 j=1 j=1 sin jx 1 ∀ x ∈ R; ∀j ≥ N j2 ≤ N 2 P

N N12 converges, so

Then by Thm., since

Since

P sin jx j2

P sin jx

uniformly converges. P sin jx are continuous, j 2 is continuous. j2

sin jx j2

uniformly converges. π Z Z πX X (−1)j − 1 sin jx X π sin jx X cos jx = = = = j2 j2 −j 3 0 −j 3 0 0 = 2

∞ X j=1

Exercise 20. It is known that

P∞

j=1

cos jx j2

x2 4

=

−

1 (2j − 1)3

πx 2

+

π2 6

if 0 ≤ x ≤ 2π

(1) x = 2π ∞ X 1 (2π)2 2π 2 π2 π2 = − + = 2 j 4 2 6 6 j=1

(2) As shown in Ex. 19,

P cos jx j2

is uniformly convergent on R

X Z cos jx X  sin jx  π/2 X (−1)j+1 = = j2 j3 (2j − 1)3 0   Z 2 π/2 πx π 2 1 x3 πx2 π 2 x = − + = − + x 4 2 6 3 4 4 6 0   1 1 1 1 3 = (π) − + = (π)3 12(8) 16 12 32 11.13 Exercises - Properties of functions represented by real power series, The Taylor’s series generated by a function, A sufficient condition for convergence of a Taylor’s series, Power-series expansions for the exponential and trigonometric functions, Bernstein’s Theorem. Sufficient Condition for convergence. Theorem 37 (Bernstein’s Theorem). Assume ∀x ∈ [0, r], f (x), f (j) (x) ≥ 0 Then if 0 ≤ x < r ∞ X f (k) (0)

(25)

k!

xk

∀j ∈ N.

converges

Proof. If x = 0, we’re done. Assume 0 < x < r. f (x) = En (x) =

n X f (k) (0) k=0 n+1

k! Z 1

n!

0

x

xk + En (x)

un f (n+1) (x − xu)du

En (x) 1 Fn (x) = n+1 = x n!

Z

Since f (n+1) > 0,

1 n (n+1)

u f 0

f (n+1) (x(1 − u)) ≤ f (n+1) (r(1 − u))

(x − xu)du =⇒ Fn (x) ≤ Fn (r) =⇒ 207

En (x) En (r) ≤ n+1 n+1 x r

For f (x) =

Pn

j=0

f (j) (0) j j! x

+ En (x) =⇒ En (x) ≤ f (r) =

n X f (j) (0)

j!

j=0


x n+1 r

En (r)

rj + En (r) ≥ En (r) since f (j) (0) ≥ 0

∀j


n+1 f (r) So then 0 ≤ En (x) ≤ xr n→∞ n → ∞ and f (t) will be some non-infinite value, so En (x) −−−−→ 0. Exercise 1.

P∞

j=0 (−1)



j 2j

x

Consider absolute convergence. limj→∞ (x2 )j = 0 If x2 < 1 If |x| = 1, then consider x2(2j) − x2(2j+1) = x4j (1 − x) ∞ ∞ X X (1 − x)x4j = (1 − x) (x4 )j j=0

Indeed

j=0 ∞ X (−1)j x2j converges for |x| ≤ 1 j=0

xj j=0 3j+1

Exercise 2.

P∞

Exercise 3.

P

=

1 3

P∞

j=0


x j 3

The series converges for |x| < 3

j = 0∞ jxj Z 0

∞ xX

jtj−1 =

j=0

∞ X

xj

j=0

So the series converges for |x| < 1. Note that we had used the integrability of power series. P∞ j j Exercise 4. j=0 (−1) jx . jxj jej ln x ( ∞ j ln x lim je = j→∞ 0

if x > 1 if 0 < x < 1

If x = 1, (2j)x2j − (2j + 1)x2j+1 = x2j (2j + −(2j + 1)x) = −1

X (−1) = ∞

(−1)j jxj converges only for 0 ≤ x < 1, |x| < 1.   P∞ P∞ j j+2 j j j+2 j (−2) Exercise 5. x = (−1) j=0 j=0 j+1 j+1 (2x) So

P

 lim

j→∞

j+2 j+1



(2x)j = lim (2x)j j→∞

if |2x| < 1

So when |x| < 21 , the series converges.     2j + 2 2j + 3 (2j + 2)(2j + 2) − (2j + 3)(2j + 1) − = 2j + 1 2j + 2 (2j + 1)(2j + 2) 2 2 4j + 8j + 4 − (4j + 5j + 3) 3j + 1 (3j + 1)/2 (3j + 1)/2 = 2 = 2 = 4j 2 + 6j + 2 4j + 6j + 2 2j + 3j + 1 (2j + 1)(j + 1)   3j + 1 3j + 1 3(j + 1/3) 3 1 1 < < < + 4j 2 + 6j + 2 4(j 2 + 1/3) 4 j 12j 2 4j 2 + 34 for x = 21 . P∞ Theorem 38. Let f be represented by f (x) = j=0 aj (x − a)j in the (a − r, a + r) interval of convergence P∞ j−1 (1) also has radius of convergence r. j=1 jaj (x − a) Thus, it diverges, by comparison test with

1 j

208

(2) f 0 (x) exists ∀x ∈ (a − r, a + r) and f 0 (x) =

(26)

∞ X

jaj (x − a)j−1

j=1

P∞

Exercise 6.

j=1

(2x)j j

0 < x < 21 . P∞

(−1)j j=0 (2j+1)

Exercise 7.

P∞

=

j=1

ej ln 2x j

=⇒ |x| <

1 it’ll converge, since by comparison test, 2

ej ln 2x j

<

1 j2

if


x 2j . 2 ∞

∞ X

∞

X x2j 1 X  x 2j x2j < = 2j 2j+1 2(j + 1/2)2 2 j 2 j=0 2 j=0

j=0

  x ∞ 1 1 X e2j ln 2 = j 2 j=0 j

x

e2j ln 2 1 if 0 < x < 2 < 2 j j 1 1 1 1 2 − = − = ≤ 2(2j) + 1 2(2j + 1) + 1 4j + 1 4j + 3 (4j + 1)(4j + 3) If x = ±2 , X 1 1 1 (converges by comparison test to ) ≤ 2 8j j2 P∞ (−1)j x
2j For |x| < 2 , j=0 (2j+1) converges. 2 since by comparison test,

Exercise 8.

P∞

Exercise 9.

P∞

j=0

Exercise 10.

(−1)j x3j j!

xj j=0 (j+3)!

=

(x−1)j j=0 (j+2)!

P∞

Exercise 11. ax = ex log a = (log ax)j+1 j! (j+1)! (log ax)j

=

j=0

1 x3

=

(−x3 )j j!

P∞

=

3

= e−x , which converges ∀ x ∈ R

xj+3 j=0 (j+3)!

P∞

1 (x−1)2

P∞

j=0

P∞

j=0

=

1 x3

xj j=+3 j!

P∞

(x−1)j+2 (j+2)!

=

=

1 (x−1)2

1 x x3 (e

− x2 /2 − x − 1) Thus, it converges for ∀ x ∈ R.

P

∞ (x−1)j j=2 j!



=

1 (x−1)2


ex−1 − x ex−1 − (x − 1) − 1 = (x − 1)2

(log ax)j j!

(log ax) j→∞ j+1 −−−→

0. By ratio test,

P∞

j=0

(log ax)j j!

converges for all x.

Exercise 12.

  ∞ ∞ ∞ ex − e−x 1 X xj X (−x)j  X x2j+1 = sinh x = = − 2 2 j=0 j! j=0 j! (2j + 1)! j=0 x2 x2j+3 (2j + 1)! j→∞ = −−−→ 0 (2j + 3)! x2j+1 (2j + 3)(2j + 2) Exercise 13.

1− 1 − cos 2x sin x = = 2

(2x)2j j j=0 (2j)! (−1)

P∞

2

2

=

∞ X 22j−1 x2j (−1)j+1 j=1

(2j)!

22j+1 x2j+2 (2j)! 4x2 j→∞ −−−→ 0 = 2j−1 2j (2j + 2)! 2 x (2j + 2)(2j + 1) So the series converges ∀ x P∞ Exercise 14. 1−1 x = j=0 2


x j 2

1 2−x

=

xj j=0 2j+1

P∞

xj+1 2j+1 x = <1 2j+2 xj 2

=⇒ (the series converges for |x| < 2, by ratio test) If x = 2, the series would diverge

If x = −2

1 2

∞ X j=0

∞

(−2)j 1X 1 1 = (−1)j = 0 but = x j 2 2 j=0 2 − (−2) 4 209

2

Exercise 15. e−x = x2j+2 j! (j+1)! x2j

j=0

x2 j→∞ j+1 −−−→

=

(−1)j x2j j!

P∞

0

Exercise 16. sin 3x = sin 2x cos x + sin x cos 2x = 3 sin x − 4 sin3 x.

  ∞ ∞ 2j+1 j 2j+1 j X X 3 sin x − sin 3x 3 x (−1) (3x) (−1)  sin3 x = = =  − 4 4 j=0 (2j + 1)! (2j + 1)! j=0     ∞ ∞ j 2j+1 2j+1 j+1 2j+1 2j+1 X X 3 (−1) x (1 − 3 ) 3  (−1) (3 − 1)x  =  = 4 j=0 (2j + 1)! 4 j=0 (2j + 1)! Exercise 17. log

ln (1 + x) =

q

1+x 1−x

= 12 (log (1 + x) − log (1 − x)) =

∞ X (−x)j+1 j=0

j+1

(−1) =

∞ X (−x)j

j

j=1

=

1 1−x

x

−

j+1

1 1+2x

=

j=1

(+x)j j−1 j (−1)

P

−

j=1

(xj )(−1)j j

∞ ∞ X X xj+1 xj = j + 1 j=1 j j=0



=⇒ −

∞ X ((−1)j + 1)xj j j=1

∞ X x2j+1 2j + 1

x2j+3 2j + 1 j→∞ 2 −−−→ x 2j + 3 x2j+1 converges, with radius of convergence of 1

|x2 | < 1, 3x 1+x−2x2

P

− ln (1 − x) =

(−1)

=⇒

Exercise 18.

1 2

xj j=0 j

P∞

j+1

|(1 − (−2) j+1

−

P∞

j=0

(−2x)j j

=

xj j=0 j (1

P∞

− (−2)j )

  1 + 2 j j→∞ (−2) j )| −−−→ 2x < 1   = x j j 1 |x (1 − (−2) )| (−2) j − 1 |x| <

1 2

1 2 x2j x2j+1 x2j ((2j + 1)(1 − 22j ) + x(1 + 22j+1 )(2j)) 2j 2j+1 (1 − (−2) ) + (1 − (−2) )= 2j 2j + 1 (2j)(2j + 1)
1 2j (2j + 1 − (2j + 1)22j + 2jx + 22j+2 jx) j→∞ −(2j + 1) + 4j 2j − 1 2 −−−→ = →0 (2j)(2j + 1) (2j)(2j + 1) (2j)(2j + 1) For x =

converges for |x| ≤ 12 P∞  12−5x Exercise 19. 6−5x−x 2 = j=0 1 +

So

3x 1+x−2x2

(−1)j 6j



xj

(|x| < 1). ∞

∞

X X 12 − 5x 5x − 12 1 6 j = = + = x + 6 − 5x − x2 (x + 6)(x − 1) 1 − x 6 + x j=0 j=0



−x 6

j =

∞ X

j

x (1 +

j=0


j ) = 1. |x| < 1 since for x = 1, limj→∞ (1 + −1 6 P ∞ 1 Exercise 20. x2 +x+1 = √23 j=0 sin 2π(j+1) xj (|x| < 1) 3 ∞

X 1 = (x3 )j 3 1−x j=0

x3 − 1 = (x − 1)(x2 + x + 1) 1 x−1 1−x = 3 = x2 + x + 1 x −1 1 − x3

∞

X
1−x = (x3 )j − x3j+1 3 1−x j=0

By induction, it could be observed that x3j , −x3j+1 , 0 appear in sequences of 3 terms. 210



−1 6

j )

√2 3

sin 2π(j+1) accounts for this. 3

=⇒

√2 3

P∞

j=0

sin 2π(j+1) xj 3

|x| < 1

sin 2π(j+1) xj < xj (so by comparison test to 3

P

xj , the radius of convergence is 1 )

Exercise 21. ∞



X 1 = xj ; 1 − x j=0

∞

0

∞

X X 1 = jxj−1 = (j + 1)xj 2 (1 − x) j=1 j=0   ∞ ∞ X 1/4 1/4 1/2 1 X j 1 + = + = (x + (−x)j ) + 2 jxj−1  (1 − x)(1 − x2 ) 1 − x (1 − x)2 1+x 4 j=0 j=1   ∞ ∞ ∞ X 1 X j+1 1X j x j j = x = x (1 + (−1)j−1 + 2j) (1 + (−1) ) + 2 jx (1 − x)(1 − x2 ) 4 j=0 4 j=1 j=1 1 1−x

=

Exercise 22. ∞  π  X (2x)2j+1 = (−1)j ; sin 2x + 4 (2j + 1)! j=0 ∞ X (2x)2j+1 sin 2x = (−1)j (2j + 1)! j=0

∞ X

∞ X (2x)2j

j=0

cos 2x =

j=0

(2j)!

(−1)j

∞ X (2x)2j j=0

(2j)!

(−1)j

√ aj xj =

2 (sin 2x + cos 2x) 2

√ ! 2 −298 = 2 98!

298 (−1)49 aj = 98!

For j = 98,

cos 2x =

√ ! 2 2

Exercise 23.

f (x) = (2 + x2 )5/2 5 f 0 (x) = (2 + x2 )3/2 (2x) = 5x(2 + x2 )3/2 2 15 00 f (x) = 5(2 + x2 )3/2 + x(2 + x2 )1/2 (2x) = 5(2 + x2 )3/2 + 15x2 (2 + x2 )1/2 2 15 15x2 f 000 (x) = (2 + x2 )1/2 (2x) + 30x(2 + x2 )1/2 + (2 + x2 )−1/2 (2x) 2 2 30x 15 (2 + x2 )1/2 (2x)+ f 0000 (x) = 15(2 + x2 )1/2 + (2 + x2 )−1/2 (2x)x + 30(2 + x2 )1/2 + 2 2   1 + 45x2 (2 + x2 )−1/2 + 15x3 − (2 + x2 )−3/2 (2x) 2 25/2 + 0x +

5(23/2 )x2 0x3 45 √ 4 + + 2x 2! 3! 4!

2

Exercise 24. f (x) = e−1/x if x 6= 0 and let f (0) = 0

(1) f (x) =

∞ X j=0

f (k) =

∞ X j=0

(−1)j


−1 j x2 j!

∞

−1 X =1+ 2 + x j=2


−1 j x2 j!

=

∞ X (−1)j x−2j j=0 ∞ X

(−2j)(−2j − 1) . . . (−2j − (k − 1)) −2j x = j! 211

j=0

j! (−1)j

(−2j)! x−2j (−2j − k)!j!

Use ratio test : (−2(j + 1))! (−2j − k)(−2j − k − 1) −2 j→∞ j!(−2j − k)! x−2(j+1) = x −−−→ 0 −2j (−2j − 2 − k)!(j + 1)! (−2j)! x (j + 1)(−2j)(−2j − 1) Thus, by ratio test, every order of derivative exists for every x on the real line since the series representing the derivative converges for every x. P∞ −2j (2) f (x) = j=0 −xj! . There are no nonzero terms of positive power, i.e. no xj ; j ≥ 1. =⇒ f (j) (0) = 0

∀j ≥ 1

11.16 Exercises - Power series and differential equations, binomial series.

Exercise 1. For (1 − x2 )y 00 − 2xy 0 + 6y = 0,

y=

∞ X

y0 =

aj xj

j=0 ∞ X

y 00 =

jaj xj−1

j=1 ∞ X

f (0) = 1 =⇒ a0 = 1

j(j − 1)aj xj−2 =

j=2

f 0 (0) = 0 =⇒ a1 = 0

∞ X (j + 2)(j + 1)aj+2 xj j=0

2(1)a2 + 3(2)a3 x + −2(1)a1 x + 6a0 + 6a1 x +

∞ X

((j + 2)(j + 1)aj+2 − j(j − 1)aj − 2jaj + 6aj )xj =

j=2 ∞ X ((j + 2)(j + 1)aj+2 + aj (j + 3)(j − 2))xj = 2a2 + 6 + 6a3 x + j=2

=⇒

a2 = −3 a3 = 0

aj+2 =

(j + 3)(j − 2) aj (j + 2)(j + 1)

For j = 2, a4 = 0, so then aj+2 = 0 for j = 2, 4, . . . . Likewise, since a3 = 0, then aj+2 = 0 for j = 3, 5, . . . . =⇒ f (x) = 1 − 3x2 Exercise 2. Using the work from above, then for (1 − x2 )y 00 − 2xy 0 + 12y = 0

f 0 (0) = 2 =⇒ a1 = 2

f (0) = 0 =⇒ a0 = 0 2(1)a2 + 3(2)a3 x + −2(1)a1 x + 12a0 + 12a1 x +

∞ X

((j + 2)(j + 1)aj+2 − j(j − 1)aj − 2jaj + 12aj )xj =

j=2 ∞ X = 2a2 + 6a3 x + −4x + 0 + 24x + ((j + 2)(j + 1)aj+2 − aj (j + 4)(j − 3))xj j=2

=⇒

a2 = 0 a3 = −10/3

aj+2 =

(j + 4)(j − 3) aj (j + 2)(j + 1)

For j = 3, a5 = 0, so then aj+2 = 0 for j = 3, 5, . . . . Likewise, since a2 = 0, then aj+2 = 0 for j = 2, 4, . . . . =⇒ f (x) = −10/3x3 + 2 Exercise 3. f (x) =

x4j j=0 (4j)! ;

P∞

d4 y dx4

=y 212

 (x4 )0 = 4x3 4 00

(x ) = 12x

 2

(x4 )000 = 24x



(x4 )0000 = 24 

x4j (4j)!

0

x4j (4j)!

00

x4j (4j)!

000

x4j (4j)!

0000

=

x4j−1 (4j − 1)!

=

x4j−2 (4j − 2)!

=

x4j−3 (4j − 3)!

=

x4j−4 (4j − 4)!

∞ ∞ X X x4j−4 x4j = f (x) = y 0000 = (4j − 4)! (4j)! j=1 j=0

To test convergence, use the ratio test 4j+4

x x4 (4j)! j→∞ = −−−→ 0 4j (4j + 4)! x (4j + 4)(4j + 3)(4j + 2)(4j + 1) So the series converges on R. P∞ xj Exercise 4. f (x) = j=0 (j!) 2

∀x ∈ R

xy 00 + y 0 − y = 0 y0 =

∞ X jxj−1 j=1 ∞ X

(j!)2

=

∞ X (j + 1)xj ((j + 1)!)2 j=0

(j + 1)jxj−1 ((j + 1)!)2 j=1   ∞  ∞  X X (j + 1)j (j + 1) 1 1 1 1 j + − − x + −1= =0 ((j + 1)!)2 ((j + 1)!)2 (j!)2 1! (j!)2 (j!)2 j=1 j=1 y 00 =

Exercise 5. f (x) = 1 +

P∞

j=1

f0 =

1·4·7...(3j−2) 3j x ; (3j)!

∞ X 1 · 4 · 7 . . . (3j − 2)

(3j − 1)!

j=1

f 00 =

∞ X 1 · 4 · 7 . . . (3j − 2)

(3j − 2)!

j=1

=x+

xa f = −xa +

(Find a and b )

x3j−1 x3j−2 =

∞ X 1 · 4 · 7 . . . (3j − 5) j=1

(3j − 3)!

x3j−2 =

∞ X 1 · 4 · 7 . . . (3j − 5) 3j−2 1 · 4 · 7 . . . (3j − 2) 3j+1 x =x+ x (3j − 3)! (3j)! j=1

∞ X j=2

∞ X 1 · 4 · 7 . . . (3j − 2)

(3j)!

j=1

So then a = 1;

y 00 = xa y + b

x3j+a

b=0.

1 · 4 · 7 . . . (3j + 1) 3j+3 x (3j + 3)!



(3j)! 1 · 4 · 7 . . . (3j − 2)



1 1(3j + 1) j→∞ = x3 −−−→ 0 x3j (3j − 2)(3j + 3)(3j + 2)(3j + 1)

So the series converges for all x. P∞ 2j Exercise 6. f (x) = j=0 xj! ; y 0 = 2xy. f0 =

∞ X 2jx2j−1 j=1

j!

=2

∞ ∞ X X x2j−1 x2j+1 =2 = 2xf (j − 1)! j! j=1 j=0

x2j+2 j! x2 j→∞ = −−−→ 0 (j + 1)! x2j j+1 By ratio test, f converges ∀ x ∈ R. P∞ j Exercise 7. f (x) = j=2 xj! y 0 = x + y 213

∀x

∞ ∞ X X xj−1 xj f = = =x+y (j − 1)! j=1 j! j=2 0

x j→∞ xj+1 j! = −−−→ 0 (j + 1)! xj j So the series converges ∀ x ∈ R by ratio test. Exercise 8.

f (x) = 0

f =

∞ X (−1)j (kx)2j j=0 ∞ X j=1

f 00 =

(2j)! ∞

(−1)j (kx)2j−1 k X (−1)j+1 (kx)2j+1 k = (2j − 1)! (2j + 1)! j=0

∞ X (−1)j+1 (kx)2j

(2j)!

j=1

f 00 − k 2 f = 0

k2

2j+1

(kx) (2j)! kx j→∞ = −−−→ 0 (2j + 1)! (kx)2j 2k + 1

by ratio test, f converges ∀x ∈ R.

Exercise 9.

f 00 =

∞ ∞ X X (3x)2j−1 9(3x)2j+1 9= (2j − 1)! (2j + 1)! j=1 j=0

9(f − x) = 9(x +

∞ X (3x)2j+1 − x) (2j + 1)! j=0

9x2 (3x)2j+3 (2j + 1)! j→∞ = −−−→ 0 2j+1 (2j + 3)! (3x) (2j + 3)(2j + 2) (by ratio test, f converges ∀x ∈ R ) Exercise 10. J0 (x) =

j x2j j=0 (−1) (j!)2 22j

P∞

J1 (x) =

x2j+1 j j=0 (−1) j!(j+1)!22j+1 .

P∞

(1) x2j+2 (j!)2 22j x2 j→∞ = −−−→ 0 2 2j+2 2j ((j + 1)!) 2 x (j + 1)2 4 x2j+3 j!22j+1 x2 j→∞ = −−−→ 0 (j + 2)!22j+3 x2j+1 (j + 2)(j + 1)4

by ratio test, f converges ∀x ∈ R by ratio test, f converges ∀x ∈ R

(2) J00 (x) =

∞ X

∞

(−1)j

j=1

X x2j−1 x2j+1 = (−1)j+1 = −J1 (x) 2j−1 (j − 1)!(j!)2 j!(j + 1)!22j+1 j=0

(3)

j0 (x) = xJ0 (x) =

∞ X

j1 (x) = xJ1 (x) = (−1)j

j=0

x2j+1 (j!)2 22j

∞ X (−1)j j=0

j10 =

∞ X j=0

=⇒ j0 = j10 Exercise 11. x2 y 00 + xy 0 + (x2 − n2 )y = 0. 214

j 2j+1

(−1) x (j!)2 22j

x2j+2 j!(j + 1)!22j+1

n = 0 =⇒ x2 y 00 + xy 0 + (x2 )y = 0

J0 =

∞ X

∞

(−1)j

j=0 ∞ X J00 = (−1)j j=1

J000 =

∞ X

(−1)j

j=1 ∞ X

(−1)

j



j=1

X x2j x2j−2 = ; (−1)j−1 2 2j (j!) 2 ((j − 1)!)2 22j−2 j=1 x2j−1 ; j!(j − 1)!22j−1 x2j−2 (2j − 1) j!(j − 1)! 22j−1

(2j − 1) 1 −2j + + 2j−1 2j−1 j!(j − 1)!2 j!(j − 1)!2 ((j − 1)!)j!22j−1

 =0

n = 1 =⇒ x2 y 00 + xy 0 + (x2 − 1)y = 0

J1 (x) =

∞ X j=0

∞

∞

(−1)j x2j+1 x X (−1)j x2j+1 x X (−1)j−1 x2j−1 = + = + 2j+1 2j+1 j!(j + 1)!2 2 j=1 j!(j + 1)!2 2 j=1 (j − 1)!(j)!22j−1 ∞

J10 = J100 =

1 X (−1)j (2j + 1)x2j + 2 j=1 j!(j + 1)!22j+1 ∞ X (−1)j (2j + 1)(2j)x2j−1 j=1

(j!)(j + 1)!22j+1

x2 J100 + xJ10 + (x2 − 1)J1 =      2   ∞ X (−1)j (2j + 1) (−1)j (−1) j + 1 j 1 2 (−1)j = x2j+1 ((2j) + (1)) + − + (j!)(j + 1)!22j+1 (j − 1)!j! j + 1 j 22j−1 22 (j + 1)!(j!)22j+1 j=1 x x + − = 2 2  ∞  X (−1)j x2j−1 = ((2j + 1)(2j + 1) + (−1)(2j)(2j + 2) − 1) = 0 (j!)(j + 1)!22j+1 j=1

Exercise 12. y 0 = x2 + y 2 ;

y = 1 when x = 0. 215

2

y = a0 + a1 x + a2 x +

∞ X

aj xj

j=3

 2 ∞ X y 2 = a20 + a21 x2 + a22 x4 +  a j xj  + y 0 (0) = 0 + 12 = 1

j=3

+ 2a0 a1 x + 2a0 a2 x2 + 2a0

∞ X

aj xj +

j=3

+ 2a1 a2 x3 + 2a1

∞ X

aj xj+1 + 2a2

j=3

y 0 = a1 + 2a2 x +

∞ X

∞ X

aj xj+2

j=3

jaj xj−1

j=3

a1 = 1 since y 0 (0) = 1 Consider the first few terms of x2 + y 2 a1 = 1 = a20 =⇒ a0 = 1 a20 + 2a0 a1 x + a21 x2 + 2a0 a2 x2 + x2 = a1 + 2a2 x + 3a3 x2

2a2 = 2a0 a1 =⇒ a2 = 1

=⇒

3a3 = a21 + 2a0 a2 + 1 = 4

Exercise 13. y 0 = 1 + xy 2 with y = 0 when x = 0

=⇒ a3 =

=⇒ a0 = 0

y = a1 x + a2 x2 + a3 x3 +

∞ X

aj xj

j=4

y=

∞ X

 aj xj

y 2 = a21 x2 + a22 x4 + a3 x6 + 

j=1

y=

∞ X

∞ X

2 a j xj  +

j=4

jaj xj−1 =

j=1

∞ X

(j + 1)aj+1 x6j

+ 2a1 a2 x3 + 2a1 a3 x4 + 2a1

j=0

∞ X

aj xj+1 +

j=4

+ 2a2 a3 x5 + 2a2

∞ X

aj xj+2 + 2a3

j=4

∞ X

aj xj+3

j=4

a1 = 1 x:

2a2 = 0

=⇒ a2 = 0

4a4 = 12

=⇒ a4 =

x4 :

5a5 = 0

=⇒ a5 = 0

x6 :

7a7 = 2a1 a4 + 2a2 a3

x7 :

8a8 = 0 + 2a2 a4 = 0 =⇒ a8 = 0  2 1 1 23 10a10 = + 2(1) =⇒ a10 = 4 14 1120

Exercise 14. y 0 = x + y 2

y = 0 when x = 0

3a3 = 0

=⇒ a3 = 0

x5 :

6a6 = 0

=⇒ a6 = 0

x8 :

9a9 = 0

=⇒ a9 = 0

1 4

x3 :

x9 :

x2 :

=⇒ a7 =

1 14

=⇒ a0 = 0 216

4 3

y 0 (0) = 0 + 0 = 0

=⇒ a1 = 0 y=

∞ X

y0 =

aj xj

j=2

∞ X

jaj xj−1 =

j=2

 y 2 = a2 x2 + a3 x3 + a4 x4 +

∞ X

∞ X (j + 1)aj+1 xj j=1

2 aj xj  =

j=5

 2 ∞ X = a2 x4 + a23 x6 + a24 x8 +  aj xj  + j=5

+ 2a2 a3 x5 + 2a2 a4 x6 + 2a2 x2

∞ X

aj xj + 2a3 a4 xy + 2a3 x3

j=5

y =x+y

j=5

x3 : 4a4 = 0 =⇒ a4 = 0 1 20

x5 : 6a6 = 0 =⇒ a6 = 0x6 : 7a7 = 0 =⇒ a7 = 0

x8 : 9a9 = 0 =⇒ a9 = 0 x

: 11a11

x9 : 10a10 = 0 =⇒ a10 = 0

1 1 = 2( )( )+ 2 160



1 20

2 =⇒ a11 =

7 8800

Exercise 15. y 0 = αy

=⇒

P∞

j=0 (j

aj+1 = aj =

+ 1)aj+1 xj = α

aj xj

2

1 1 1 x7 : 8a8 = 2( )( ) =⇒ a8 = 2 20 160

10

∞ X

1 2

x2 : 3a3 = 0 =⇒ a3 = 0 x4 : 5a5 = a22 =⇒ a5 =

aj xj + 2a4 x4

j=5 0

x : 2a + 2 = 1 + 0 =⇒ a2 =

∞ X

P∞

j=0

aj xj

αaj (j+1)

αj a0 x (by induction) j!

Exercise 16. y 00 = xy

y 00 =

∞ X (j + 2)(j + 1)aj+2 xj = j=0

∞ X = 2a2 + (j + 3)(j + 2)aj+3 xj+1

=

∞ X

aj xj+1

j=0

j=0

=⇒ a2 = 0 and aj+3 = j=0

a0 3·2 a3 a6 = 6·5

a+3=

j=3 a3j =

j−1 a0 Y (3k + 1) ; (3j)!

aj (j + 3)(j + 2)

j=1 j=4 a3j+1 =

k=0

a4 = a7 =

a1 7·6·4·3

j−1 Y a1 (3k + 2) (3j + 1)! k=0

Exercise 17. y 00 + xy 0 + y = 0 217

a1 4·3

y= y0 = y 00 =

∞ X j=0 ∞ X j=1 ∞ X

aj xj jaj xj−1

∞ X

y 00 + xy 0 + y =

xj ((j + 2)(j + 1)aj+2 + jaj + aj ) = 0 =⇒ aj+2 =

j=1

j(j − 1)aj xj−2

−aj (j + 2)

j=2 ∞ X = (j + 2)(j + 1)aj+2 xj j=0

−a1 3 −a3 a1 a5 = = 5 15

−a0 2 −a2 a4 = 4

a3 =

a2 =

a2j =

(−1)j a0 (2j)!!

a1 (−1)j (could be shown by induction) (2j + 1)!!

a2j+1 =

Exercise 18. Recall that

y= y0 =

∞ X j=0 ∞ X

aj xj jaj xj−1

j=1 ∞ X = (j + 1)aj+1 xj j=0

Knowing this, we could cleverly observe that e

Solving this ODE using y(x) = e−A(x)

−2x

=

P∞

j=0 (2aj

+ (j + 1)aj+1 )xj is actually a differential equation!!!

=⇒ e−2x = y 0 + 2y
Rx Rx Q(t)eA(t) dt + y(a) where A(x) = a P (t)dt, a y = e−2x (x + 1)

We had obtained the necessary initial conditions to solve this ODE from the information given, that a0 = 1, so that y(0) = 1. By doing some simple computation and comparison of powers with e−2x , then a1 = 2, a2 = −2, a3 = 4/3 P∞ P∞ Exercise 19. cos x = j=0 aj (j + 2)xj for f (x) = j=0 aj xj .

P∞ Using cos x = j=0 matching powers of x,

(x)2j j (2j)! (−1)

representation, we can immediately conclude that for odd terms, a2j+1 = 0 and by a2j (2j + 2) = (−1)j

1 (2j)!

a5 = 0 (−1)3 −7 =⇒ a6 = 6! 8! P∞ P P ∞ ∞ Now notice that for cos x = j=0 aj (j + 2)xj = j=1 jaj xj + 2 j=0 aj xj is actually a differential equation, cos x = 0 xy + 2y. We can solve this first-order ODE using
Rx Rx cos x y(x) = e−A(x) a Q(t)eA(t) dt + y(a) where A(x) = a P (t)dt. Then solving y 0 + 2y x = x , a6 (6 + 2) =

y=

1 (x sin x + cos x − (a sin a + cos a) + b) x2 218

Plugging 0 as a good guess back into the ODE, cos 0 = 1 = y(0)(2)

1 2

and f (π) =

1 2

With this initial condition, we get

sin x cos x − 1 + if x 6= 0 x x2

f (x) = So f (0) =

=⇒ y(0) =

−2 π2

Exercise 20.

(1)  ∞  X −1/2 (−x)j = j j=0
−3
−1 1 2 2 x2 + =1+ x+ 2
2




(1 − x)−1/2 =

+

−1 2

−3 2

−5 2

−7 2

−1 2

−3 2







−5 2




3!

−9 2

3

x +

−1 2




−3 2




−5 2




−7 2

4!


x4 +


x5 + · · · =

5!

1 3 5 35 5 63 5 = 1 + x + x2 + x3 + x + x + ... 2 8 8 128 256
P P∞ P∞ ∞ (2) To make the notation clear, (1 − x)−1/2 = j=0 −1/2 (−x)j = j=0 bj xj = j=0 aj j Now α j+1
α j




So for α =

=

j! (α − j) α(α − 1) . . . (α − (j + 1) + 1) = (j + 1)! α(α − 1) . . . (α − j + 1) (j + 1)

−1 2 ,

aj+1 =− aj



1/2 + j j+1

 (−x) < x

Using this, we further find that bj+1 < bj bj+2 < bj+1 For x =

1 50 .

So by induction, bn+j < bn rn =

∞ X

1 50

1 < bj 50



1 50

2


1 j 50

an+j <

j=1

∞ X j=1

 an

1 50

rn <

j = an

an 1/50 = 1 − 1/50 49

an 49

√

−1/2 1 −1/2 49 −1/2 (3) Note that (1 − x) = 1 − 50 = 50 = 572  −1/2  2  3  4  5 7 1 1 3 1 5 1 35 1 63 1 1− =1+ + + + + 5 50 100 2 2(50) 2 2(50) 8 2(50) 8 2(50) √ 2 ' 1.4142135624

Exercise 21.

(1)


−1/2
176 3000000 1/2 = 1732 1 − 3000000 1000 √ 2999824 1/2 Obviously, (3000000)1/2 = 1000 3 so that we have 1732 (3/2999824) . With long multiplication, we could show easily that 1732 ∗ 1732 = 2999824 (it’s harder to divide). So then 1732 1000

1732 1000



176 1− 3000000

(2) 219

−1/2 =

√

3

Exercise 22. arcsin x =

(1 − x2 )−1/2 =

√ 1 1−x2

R

P∞

α j

j=0




(−x2 )j = 1 +

P∞

α j




j=1

(−x2 )j

 ∞  X −1/2 (−1)j 2j+1 x =⇒ arcsin x = x + (2j + 1) j j=1
−3

−1 . . . −1 (1)(3) . . . (1 + 2j − 2) (2j − 1)!! 2 2 2 −j+1 = (−1)j = (−1)j j(j − 1) . . . (2)(1) (2j)!! (2j)!! ∞ X (2j − 1)!! x2j+1 =⇒ arcsin x = x + (2j)!! 2j + 1 j=1

13.21 Exercises - The conic sections, Eccentricity of conic sections, Polar equations for conic sections. Exercise 1. F is in the positive half-plane determined by N .

kX − F k = ed(X, L) kX − F k = e|(X − F ) · N + d| Exercise 2.

(1) kX − F k = ed(X, L) kX − F k = e|(X − F ) · N + d| F = 0 =⇒ kXk = e|(X · N ) + d|;

r = e|r cos θ + d|

ed 1 − e cos θ because X · N > 0. The left branch for e > 1,

=⇒ r = e(r cos θ + d) =⇒ r = (2) The right branch for the hyperbola is given by r =

ed 1−e cos θ

kX − F k = ed(X, L) = e|(X − F ) · N + d| = = e|X · N + d| = −e(d + r cos θ) = r r=

−ed (1 + e cos θ)

Exercise 3. For points below the horizontal directrix,

kX − F k = ed(X, L) F = 0 =⇒ kXk = ed(X, L) = e(|(X − F ) · N − d|) = e|X · N − d| = e|r sin θ − d| Now Thm. 13.18 says r =

ed e cos θ + 1

=⇒ r = e(d − r sin θ) =⇒ r =

if 0 < e ≤ 1

ed 1 + e sin θ

For the “right” or upper-half branch of a hyperbola. kX − F k = e(|(X − F ) · N − d|) = e(r sin θ − d) = r r=

−ed ed = 1 − e sin θ e sin θ − 1

Exercise 4. kX − F k = ed(X, L); kXk = e|(X − F ) · N − d| = e|r cos θ − d| = e(d − r cos θ)

e = 1, d = 2 . Exercise 5. r =

3 1+ 12 cos θ

Exercise 6. r =

6 3+cos θ

Exercise 7. r =

=

=

1 −1 2 +cos θ

6( 12 ) . 1+ 12 cos θ

2 . 1+ 13 cos θ

e = 21 ; d = 6

e = 31 ; d = 6.

. 220

=⇒ r =

ed 1+e cos θ

ed(X, L) = kX − F k = e|(X − F ) · N − d| = e|r cos θ − d| = er cos θ − ed = r −ed ed =r= 1 − e cos θ e cos θ − 1 So for r =

2 2 cos θ−1 ,

e = 2, d = 1.

Exercise 8. r =

4 1+2 cos θ

Exercise 9. r =

4 1+cos θ

e = 2, d = 2. e = 1 d = 4. 3 5x

Exercise 10. 3x + 4y = 25 =⇒

+ 45 y = 5. N =

3 4 5, 5




.

L = {x = P + tA}, N · X = N · P . To find the distance from the focus, at the origin, to the directrix, dN = P + tA;

dN · N = d = N · P

So for this problem, d = 5. 3 4 r = kX − F k = ed(X, L) = e|(X − F ) · N − d| = e|X · N − d| = r cos θ + r sin θ − 5 5 5   1 4 3 r= 5 − r cos θ − r sin θ 2 5 5 5/2 r= 3 4 1 + 10 cos θ + 10 sin θ Exercise 11. e = 1,

4x + 3y = 25

4 5x

+ 35 y = 5;

N=

4 3 5, 5




.

d = 5. 4 3 kX − F k = ed(X, L) = e|(X − F ) · N − d| = e r cos θ + r sin θ − 5 5 5   4 3 r =5−r cos θ + sin θ 5 5 5 r= 1 + 54 cos θ + 35 sin θ Exercise 12. e = 2, hyperbola, so there’s 2 branches.

1 1 1 √ x+ √ y = √ 2 2 2 dN = P + tA;

L = {x = P + tA}

X ·N =N ·P

1 dN · N = d = N · P = √ 2

Note that the sign of d here tells you what side the focus, at the origin, lies on. 1 1 kX − F k = ed(X, L) = kXk = e|(X − F ) · N − d| = e(d − √ r cos θ − √ r sin θ) 2 2 √ 2/ 2 r= 1 + √22 cos θ + √22 sin θ But for the right side branch, 1 1 kX − F k = ed(X, L) = kXk = e|(X − F ) · N − d| = −e(d − √ r cos θ − √ r sin θ) 2 2 √ −2/ 2 r= 1 − √22 cos θ − √22 sin θ Exercise 13. e = 1 parabola. 221

(1)

π kX − F k = kXk = ed(X, L) = 1|(X − F ) · N − d| = d − X · N = d − r cos 3   3 3 d=r = × 108 mi 2 2 3 2

× 108 mi θ = 0, 1 + cos θ (2) Focus is in the positive half-plane determined by N . r=

r=

3 × 106 mi 4

kX − F k = kXk = ed(X, L) = |(X − F ) · N + d| = r cos θ + d θ 1 d = r(1 − cos θ) = 108 mi(1 − cos ) = × 108 mi 3 2 r=

1 × 108 mi d = 2 1 − cos θ 1 − cos θ

r(θ = π) =

1 × 108 mi 4

13.24 Exercises - Conic sections symmetric about the origin, Cartesian equations for the conic sections. Quick Review. Consider symmetry about the origin. kX − F k = ed(X, L) = e|(X − F ) · N − d| = e|X · N − F · N − d| = |eX · N − e(F · N + d)| 2

2

2

kX − F k = kXk − 2X · F + kF k = e2 (X · N )2 − 2aeX · N + a2 X → −X;

X · F = aeX · N

ed 1 − e2 2 2 2 2 =⇒ kXk + (ae) = e (X · N ) + a2

X = (F − aeN ) = 0 =⇒ F = aeN ;

q

2

kXk + (ae)2 = e2 (X · N )2 + a2 is satisfied

if X = ±bN 0 ;

b2 + (ae)2 = e2 (0) + a2

+

y2 36

b2 = a2 (1 − e2 )

=1

4 . 5
|F | = |aeN | = 10 45 = 8. f = (±8, 0). (0, 0) center. Vertices (±0, 6).

1−

b2 a2

= e =⇒ e =

Exercise 2.

y2 100

+

x2 36

= 1.

4 5

= e;

f = (0, ±8).

(0, 0) center; vertices (±6, 0), (0, ±10). Exercise 3.

(x−2)2 16

+

(y−3)2 9

√

= 1. Center (2, −3). |F | = ae = 4

7 4

=

√

7.

q √ √ √ 2 9 1 − ab 2 = 1 − 16 = 47 = e; (2 + 7, −3), (2 − 7, −3) foci. Vertices (6, −3), (−2, −3), (2, 6), (2, −12).

2 Exercise 4. x25 + y 2 = 1. Center x = (0, 0). |F | = ae = 35 45 = 43 . (9) q q 2 9 e = 1 − ab 2 = 1 − 25 = 45 . Foci: (± 43 , 0). Vertices (± 35 , 0), (0, ±1). q

Exercise 5.

y2 (1/4)

+

x2 (1/3)

F =

if X = ±aN ;

x2 100

Exercise 1. b2 = a2 (1 − e2 )

F · N = ae; a =

=1

1 |F | = ae = √13 2 = 2√ . 3 q   1 ±1 √ , 0 . Center (0, 0). 1 − 1/4 = = e. Foci: 1/3 2 2 3 √ Vertices (±1/ 3, 0), (0, ±1/2).


1

Exercise 6. Center (−1, −2). 222

e2 d N 1 − e2

q

b2 a2

1−

q 1−

=

16 25

=

3 5

= e;

|F | = ae = 5 35 = 3.

Foci: (−1, −1), (−1, −5). Vertices: (−1, 3), (−1, −7), (3, −2), (−5, −2). 3 4.

Exercise 7. F = ae =

a = 1, e = 34 . b2 = a2 (1 − e2 );

b2 = 1

1 4




.

x2 + 4y 2 = 1 . Exercise 8. 2a = 4. a2 = 4. 2b = 3. b2 = 9/4. (x+3)2 9/4

Exercise 9.

(y−4)2 4

+

=⇒

(x+3)2 4

+

(y−4)2 4

=1

= 1. (x+4)2 9

Exercise 10. 2a = 6, a = 3.

+

(y−2)2 1

= 1.

Exercise 11. 2a = 10, a = 5. |F | = ae = 5e = 4 e = 4/5. b2 = a2 (1 − e2 ) = 25 1 − Exercise 12.

(x−2)2 a2

(y−1)2 b2

+

2




= 9.

= 1;

a = 4 from (6, 1). b = 2 from (2, 3). =⇒ 2

16 25

(x−2)2 42

+

(y−1)2 4

=1

2

Exercise 13. b = a (1 − e ). 2

x2 100

64 − y64 = 1; b2 = 100(1 − e2 ) = −64. 1 + 100 = e2 . √ √ Center (0, 0). e = 2 1041 = 541 . √ √ Vertices; (±10, 0). F = ae = 2 41. Foci: (±2 41, 0). x2 100

y2 64

=

x,y→∞

+ 1 −−−−−→;

Exercise 14.

y2 100

x2 64

−

+1=

±4 5 x

= 1; Center (0, 0), a2 = 100;

b2 = −64.

√ (0, ±10). F = ae = (0, ±2 41).

√

41 5 . Vertices 2 x,y→∞ y −−−−→ ±5 100 − 4 x = y.

b2 = a2 (1 − e2 ). x2 64

y=

e=

(x+3)2 4

− (y − 3)2 = 1. q q √ 2 5 Center (−3, 3). e = 1 − ab 2 = 1 − −1 4 = 2 . √ √ √ √ Foci: ae = 2 25 = 5. (−3 + 5, 3), (−3 − 5, 3). Exercise 15.

Vertices: (−3, 4), (−3, 2); (x+3) 4

2

(1, 3), (−7, 3). ±(x + 3) x,y→∞ = 1 + (y − 3)2 −−−−−→ =y−3 2

Exercise 16.

e=

q

1−

x2 144/9 −9 16

−

y2 144/16

=1=

x2 16

−

y2 9 .

= 54 . Center (0, 0). |F | = ae = 5. Foci: (5, 0), (−5, 0). Vertices (±4, 0).

Exercise 17. 20 = 5y 2 − 4x2 . Center (0, 0). |F | = ae = 2

3 2




= 3. Foci: (0, ±3). 1 =

y2 4

−

x2 5 .

Vertices: (0, ±2) (x−1)2 4

2

− (y+2) = 1. 9 q √ 13 Center (1, −2). e = 1 − −9 4 = 2 ; Exercise 18.

√

|F | = 2

13 2

=

√

13. Foci: (1 +

Vertices: (5, −2), (−3, −2). Exercise 19. F = ae = 2(2) = 4. x2 4

+

y2 −12

= 1.

y2 12

+1=

x2 x,y→∞ −−−−→ 4 −

√ y = ± 3x. 223

√

13, −2), (1 −

√

13, −2).

e=

q 1−

−5 4

=

3 2.

b2 = a2 (1 − e2 ) = 4(1 − 4) = −12. √ Exercise 20. F = ae = 2 = (1)e. b2 = a2 (1 − e2 ) = 1(1 − 2) = −1. =⇒ y 2 − x2 = 1. Exercise 21.

x2 4

−

y2 16

=1 (x+1)2 −3

Exercise 22. (y − 4)2 −

= 1 where

F = ae = | − 2| = ae. b2 = a2 (1 − e2 ) = 1(1 − 4) = −3 Exercise 23. ±

(x−2)2 a2

(y+3)2 b2

∓

=1 1 4 ∓ 2 =1 (y + 3)2 (x − 2)2 a2 b =⇒ − =1 9 9 27/8 (27/5) (−1, 0) =⇒ ± 2 ∓ 2 = 1 a b (3, −1) =⇒ ±

Exercise 24.

x2 −1 3

2x 3

= y2 . 3 2

3x − 2y = C. m =

The asymptotes of y 2 =

Exercise 25.

±x2 a2

= 2yy 0 .

=⇒ y0 92 = x0 . x2 −1 3

2

25 4

81 2 4 y0

− 1 = 3y02 =⇒ y0 =

±2 √ . 69

are y = ± √x3 .     ±9 2 ±23 3 √ − 2 ±√ = √ =C 69 69 69 r 23 = 2y 3x ± 3 ±x2 a2 2

+ ∓ yb2 = 1

(3, −5) → ±9 ∓

yy 0 = x3 .

= a2 ;

∓

a =

y2 4a2 11 4 .

= 1.

y2 x2 − =1. 11/4 11 Quick Review of Parabolas.

=⇒

F on positive half plane to N . kX − F k = e|(X − F ) · N + d| Let N = ~ex ; d = 2c; F = (c, 0); e = 1. (x − c)2 + y 2 = e2 ((x − c) + 2c)2 = (x − c)2 + 4c(x − c) + 4c2 y 2 = 4cx Thus, for ellipses, the vertex is equidistant to the focus and directrix (confirming the other definition). Let N = ~ey , d = 2c; F = (0, c), e = 1. x2 + (y − c)2 = ((y − c) + 2c)2 = (y − c)2 + 4c(y − c) + 4c2 x2 = 4cy Exercise 26. 4c = −8

(0, 0) vertex. y = 0 symmetry axis. x = 5 directrix.

Exercise 27. 4c = 3. Vertex: (0, 0). Symmetry axis: y = 0. Directrix: x = −3/4. Exercise 28. (y − 1)2 = 12(x − 21 ). 4c = 12, c = 3. Symmetry axis: y = 1. Directrix: Exercise 29. x2 /6 = y. 4c =

1 6

Exercise 30. x2 + 8y = 0.

4c =

c=

1 24 .

−1 8 ;

−5 2 ,1




.

1 Vertex: (0, 0). Directrix: y = − 24 . Symmetry axis: x = 0.

c=

Exercise 31. (x + 2)2 = 4(y + 94 ). 4c = 4;

−1 32 .

y=

1 32

directrix; x = 0 axis.

c = 1. Center (−2, −9/4). Directrix: y = −13/4. Axis: x = −2.

Exercise 32. y = −x2 . 224

Exercise 33. x2 = 8y. Exercise 34. (y − 3) = −8(x + 4)2 . Exercise 35. c =

5 4

5(x − 74 ) = (y + 1)2

Exercise 36. y = ax2 + bx + c

(0, 1) → c = 1

(1, 0) → 0 = a + b + 1(2, 0) → 0 = 4a + 2b + 1

Exercise 37. 4c(x − 1) = (y − 3)2 . 4c(−2) = (−4)2 = 16. c = −2.

a=

1 2

=⇒ y =

1 2 3 x − x+1 2 2

−8(x − 1) = (y − 3)2 .

Exercise 38. kX − F k = ed(X, L) = |(X − F ) · N − d|

L = {(x, y)|2x + y = 10; √25 x + d = N = xL

√y 5

=

10 √ }. 5

dN · N = d = xL · N =

10 √ . 5

2 −2 y 10 √ x + −√ + √ = x2 + y 2 5 5 5 5x2 + 5y 2 = (−2x − y + 10)2 = 4x2 + y 2 + 100 + 4xy − 40x − 20y

F = 0 =⇒ kXk2 = |X · N − d|2 =



=⇒ x2 + 4y 2 − 4xy + 40x + 20y − 100 = 0 13.25 Miscellaneous exercises on conic sections. Exercise 1.

 2   x 2  y2 x2 bx 2 2 2 = 1 − y = b − = b 1 − b2 a2 a a Z a r Z 1 p  x 2 dx = 2 b 1− ab 1 − x2 dx = (ab) area of a circle of radius 1 y=2 a −a −1 Exercise 2.

q
2 (1) Without loss of generality, let the major axis be 2a in the x-axis. y = b 1 − xa   Z 1 Z a 4 x2 V = (1 − x2 )dx = π(1)3 b2 a πb2 1 − 2 dx = πb2 a a 3 −a −1 (2) If rotated about the minor axis, suppose, without loss of generality, 2a is the minor axis (just note that have x, y, a, b as dummy labels).

x2 a2

+

b2 a2

=1

=⇒ V = 34 π(1)3 b2 a, where 2a is the minor axis, 2b is the major axis. Exercise 3.

x2 (3/A)

2

y + (3/B) =1

By 2 = 3 − Ax2 r 2 2

1 B

=⇒ y 2 =

3 B

2

− Ax B ;

y=

q

3 B

−

Ax2 B .

So the area inside this ellipse is

r Z √3/A s Z √3/A p x2 3 3 − Ax2 dx = 2 1− 3
√ √ B −

−

3/A 2

3/A

 3 y 2 = A−B 1−

y x For the other ellipse equation, 3/(A+B) + 3/(A−B) = 1. Thus, the area inside this ellipse is r 3 Z √ A+B 3 2 3 A − B −√ A+B



v  u u x u t1 −  q

A

2

x (3/(A+B))

3 A+B



;

y=

2 

Equating the two areas after making an appropriate scale change, r r Z 1 r r Z 1p p 3 3 3 3 2 2 1 − x dx = 2 1 − x2 dx B A −1 A − B A + B −1 225

q

3 A−B

q

1−

x2 (3/(A+B)) .

Thus A2 − B 2 = AB of A.

=⇒ A2 − BA − B 2 . Simply try treating B as a number and solve the quadratic equation in terms B±

A=

p √ √ B 2 − 4(1)(−B 2 ) B±B 5 B(1 + 5) = = 2(1) 2 2

2 Exercise 4. y = − 4h b2 x .

Z

b/2



−b/2

R2

Exercise 5. y 2 = 8x.

0

b/2    b 4h 3 4h 2 b 2hb + h x + h = x + = 2 2 −b −3b 2 2 3 −b/2

π8tdt = 4π(2)2 = 16pi

Exercise 6. y 2 = 2(x − 1). y 2 = 4(x − 2).

(1) Z A=2

2

p

3

Z 2(x − 1) + 2

1

p √ 2(x − 1) − 2 x − 2 =

2

2 3 3 √ 2 √ 2 2 3/2 = 2 2 (x − 1) + 2 2 (x − 1)3/2 − 4 (x − 2)3/2 3 3 3 2 2 1 √ 2 √ 4 3/2 √ 2 2 = 2 2 + 2 (2) − 2 2 − 4 = 8/3 3 3 3 3 (2)   2  1 1 2 x − x = 2 (4 − 1) − (2 − 1) = 1 2 2 1 1 Z 3 Z 3 3 3 (2(x − 1) − 4(x − 2)) dx = (−2x + 6)dx = −x2 2 + 6x|2 = (−9 + 4 + 6(3 − 2)) = 1 Z

2



2(x − 1) = 2

2

2

Z

2

2(x − 1) + π

=⇒ V = π 1

(3)

3

Z

(2(x − 1) − 4(x − 2)) = 2π 2

2

y2 2

+ 1 = x, y4 + 2 = x 2  2 2 !    2 Z 2  2 Z 2 y y −3y 4 −3 5 2π +2 − +1 = 2π + 3 dy = 2π y + 3y = 4 2 16 80 0 0 0       −3(32) −96 + 480 384 48 = 2π + 6 = 2π = 2π =π 80 80 80 5

Exercise 7. By Apostol’s definition of conic sections, we are basically given the conic section definition with e =

plug in the pt. (0, 4). x2 y2 (0,4) + = 1 −−−→ b = 4 2 2 a b

2

2

2

2

 2 ! 1 1− 2

√1 2

because for the directrix

b = a (1 − e ) = 16 = a x2 y2 + =1 64/3 16

Exercise 8. F = 0

kX − F k = kXk = ed(X, F ) = |X · N + d| =

x √ 2

+

√y 2

+

y + x = −1  1 1 1 x 1 √ y + √ = −√ N = √ ,√ 2 2 3 2 2 XL = P + tA √ √ dN = XL XL · N = d = −1/ 2 XL · N N · P = −1/ 2 

226

1 2.

So just

So by squaring both sides of the vector equation, x2 y2 1 + xy + + +x+y 2 2 2 x2 y2 1 + − xy − x − y = 2 2 2 x2 + y 2 − 2xy − 2x − 2y = 1

x2 + y 2 =

Exercise 9. Center (1/2, 2) because we equate the asymptotes to see where they intersect: y = 2x + 1 = −2x + 3.

(y − 2)2 (x − 1/2)2 1 3 (0,0) 4 − = 1 −−−→ 2 − 2 = 2 = 1 2 a a2 /4 a a a (x − 1/2)2 (y − 2)2 − =1 3 3/4 x2 p+2

Exercise 10. px2 + (p + 2)y 2 = p2 + 2p.

y2 p

+

= 1.

(1) Since p + 2 > p, the foci must lie on the x axis. a2 = p + 2; b2 = a2 (1 − e2 ) = p = (p + 2)(1 − e2 ). e = √ √ F = ae = 2. (± 2, 0). q √ √ (2) F = ae = 2 = a( 3) =⇒ a = 23 ; b2 = 23 (1 − 3) = −4 3 . x2 y2 − =1 2/3 4/3 Exercise 11. e = 1 for an ellipse.

kX − F k = |X · N − a| = a − X · N k−X − F k = kX + F k = | − X · N − a| = a + X · N kX − F k + kX + F k = 2a Exercise 12.

kX − F k = e|(X − F ) · N − d| = e(d − (X − F ) · N ) kX + F k = ed(X, L) = e|(X − F ) · N + d| = e(−d − (X − F ) · N ) kX − F k − kX + F k = 2ed X → −X

so for the other branch, kX + F k − kX − F k = 2ed

Exercise 13.


2 2 2 (by)2 ) b 2 (1) (tx) = a (1−e = a2 + b2 = 1 t t2 2 2 2 2 2 2 (2) b1 = a1 (1 − e ) b2 = a2 (1 − e ).


a 2 t

(1 − e2 )

b21 b22 b21 b22 = 1 − ; = a21 a22 a21 a22   2  2 b2 b2 x y 2 2 b1 b1 x1 y + 2 =1= + a21 b1 a22 (b2 y)2

2 2 2 −1) b 2 = −a (e = − at (e2 − 1). t t2 1−

2

(3) ± (tx) a2 ∓

(ty)2 b2

=1

b21 = a21 (e2 − 1) b21 + 1 = e2 a21  ±

x a1

2

 ∓

y b1

b22

b22 +1 a22 

2

= ± 227

= a22 (e2 − 1) = e2 b2 b1 x

a2

 2  ∓

b21 b2 = 22 2 a1 a2 ! 2 b2 b1 y b2

=1

q

2 p+2

=⇒ y 0 =

y2 b2

x2 a2

+

−b2 x ya2

=

Exercise 14.

= 1. =⇒

−a2 (1−e2 )x a2 y

x a2

=

+

y 0 b2 y

=0

(e2 −1)x y

Exercise 15.

(1) y = ax2 + bx + c (2) y = tx2 , t 6= 0

ty = a(tx)2 + btx + c → y = atx2 + bx + c/t = y = Ax2 + b + C √

Exercise 16. x − y + 4 = 0 y = 4 x

y 0 (x = 4) = 1

(y 2 = 16x); y 0 = 2x−1/2 .

(x, y) = (4, 8).

Exercise 17.

(1) If we treat the two given parabolas, y 2 = 4p(x − a) and x2 = 4qy, as two vector objects free from any specific coordinate system then we observe that we can disregard the sign of q and p and simply state that they are both positive. What matters is that we observe that p and q are the distance of the foci to the vertex for each of the respective parabolas. Second, observe that a is not given. By diagram, if p, q are given, a must be moved along the x-axis to fit the tangency condition. Thus, in terms of doing the algebra, just eliminate p and q from the relations. If (h, k) is the point of contact, √ √ y 2 = 4p(x − a) y = 2 p x − a x2 = 4qy x 1 √ = y0 y0 = p √ 2q x−a h 1 √ y 0 (h) = y 0 (h) = p √ 2q h−a  2 p h = =⇒ (h2 )(h − a) = (2q)2 p (Tangent condition) 2q h−a h2 k2 (one point of contact condition) with q = ,p= 4k 4(h − a)  2 2 2 h k h2 =⇒ h2 (h − a) = =⇒ (h − a)2 = 2k 4(h − a) 16 2a ± a/2 =⇒ h = = 4a/3 15/8 (2) √ p h √ = 2q h−a √ √ p √ p 2a 3p p = = √ ; 2a a = 3 3pq 3q a a/3 =⇒ 4a3 = 27pq 2 Exercise 18. First hint: Vector methods triumph over algebraic manipulations of Cartesian coordinates. Think of the locus 2

in terms of vector objects that are coordinate-free and the conic section will emerge. I mean, try evaluating kP − Ak = (x − 2)2 + (x − 3)2 = (x + y)2 A = (2, 3), N = √12 (1, 1), X = (x, y). √ √ kX − Ak = x + y = 2(X · N ) = 2(X · N − (F · N − d)) where F · N = d = A · N = √52 d = distance from focus to the directrix . y = x + 1 (axis of the hyperbola) p √ d = √52 = (2 − x)2 + (3 − y)2 = 2(2 − x) x = − 12 , y = 12

−1 1 1 1 2 , 2 must also be the center. y − 2 = − x + 2 is the directrix.

y − 12 = α x + 12 is the general form of the asymptote. 228

Consider asymptotes in general. kX − F k = ed(X, L). kX − F k kX − F k kX − F k =e= = d(X, L) |X · N − (F · N − d)| (X − F ) · N + d| For kX − F k → ∞,

kX − F k > d. To keep ratio of e, X − F must be ultimately directed by N by a ratio of e. kX − F k 1 = kX − F k cos φ cos φ √ = 1 =⇒ y = ab x = e2 − 1x. =⇒ e =

x2 a2

e.g. Consider N = ~ex .

−

y2 b2

From the vector equation, p (X − F ) · N = (x − c, y) · N = (x − c)2 + y 2 cos φ = x − c p (x − c)2 + y 2 1 (x − c)2 + y 2 = e2 ; = = e; x−c cos φ (x − c)2 p y2 = e2 − 1 =⇒ y = e2 − 1x 2 (x − c) For our problem, consider the conic section approaching the asymptote. Then the conic section will look more like those linear asymptotes. p (x − 2)2 + (y − 3)2 = x + y s  2    2   y− 12 =α(x+ 21 ) 1 5 1 5 1 1 −−−−−−−−−→ x+ − + α x+ − =x+α x+ + 2 2 2 2 2 2 s √ 2    1 25 x→∞ 1 1 + α2 − 5(1 + α) x + + −−−−→ =⇒ (1 + α2 ) x + 2 2 2 (1 + α) =⇒ α = 0 The asymptotes are y =

1 2

and x =

−1 2 .

In the second part, each quadrant must be checked. So far, I only have that quadrant II is filled: points in quadrant III and quadrant IV cannot satisfy the given condition. To see this, consider quadrant II.   √ −1 1 kx − Ak = −x + y = 2(x, y) √ , √ 2 2   −1 1 For quadrant II, N = √2 , √2 . By diagram, (X − F ) · N > 0 and X · N > 0. 1 −1 1 A·N = √ −A·N = √ d= √ 2 2 2 |(X − F ) · N + d| = (X − F ) · N + d The equation for the axis of the conic section is y = −(x − 5). By taking the asymptotic limit like above, we can show that α = 0 again. We only sketch the part of the hyperbola in quadrant II. By similar procedure, I found that quadrant III, IV cannot satisfy the condition. Exercise 19.

kX − F k = d(X, L) = |(X − F ) · N + d| x2 + y 2 = (X · N + d1 )2 = y 2 + 2yd1 + d21 x F =0 −−−→ x2 = 2yd1 + d21 y10 = d1 kX − F k = |(X − F ) · N − d2 | = d2 − (X − F ) · N F =0

−−−→ 2

kXk = d2 − y

2

x + y = d22 − 2d2 y + y 2 =⇒ x2 = d2 − 2d2 y 229

y20 =

−x d2

Point of intersection x20 = 2y0 d1 + d21 = d22 − 2d2 y0 2(d1 + d2 )y0 = d22 − d21 =⇒ y0 =

d2 − d1 2

x20 = d2 (d2 − 2y0 ) = d2 d1 r √ d2 ± d2 d1 0 y1 = =± d d1 √1 =⇒ ∓ d d 1 2 1 y20 = = ∓q d2 d2 d1

Exercise 20.

(1) Use X → −X symmetry. kX − F k = ed(X, L) = e|(X − F ) · N + d| = e|X · N − F · N + d| = |eX · N − a| 2

2

kXk − 2X · F + kF k = e2 (X · N )2 − 2ea(X · N ) + a2 2

2

X → −X =⇒ kXk + 2X · F + kF k = e2 (X · N )2 + 2ea(X · N ) + a2 =⇒ kX 2 k + kF k2 = e2 (X · N )2 + a2 

(2)

x a2

+

yy 0 a2 −c2

= 0 =⇒ y 0 =

x2 + y 2 + c2 = e2 x2 + a2 |F | = c = ae  2 2 a −c y2 x2 2 2 2 2 + =1 x + y = a − c =⇒ a2 a2 a2 − c2

−(a2 −c2 )x ya2

 2 2 3 a − c2 x xy(y 0 )2 = a2 y    2 2 −(a2 − c2 )x a −c 2 2 (x2 − y 2 − c2 )y 0 = x2 + x − a = a2 ya2 (a2 − c2 ) x = (a4 + −(a2 − c2 )x2 − a2 x2 ) a4 y  2 2 a −c x2 ) x(a2 − c2 − a2 −xy 2 −xy = =− = y y = (a2 − c2 )(−a4 + a2 x2 )x/(a4 y) =⇒ xyy 02 + (x2 − y 2 − c2 )y 0 − xy = 0 (3) For y 0 , consider − y10 at every (x, y).  2 −1 −1 −xy 1 xy + (x2 − y 2 − c2 ) 0 − xy = 0 = 0 + (x2 − y 2 − c2 ) 0 + xy 0 y y y y if y 0 6=0

−−−−→ −xy + (x2 − y 2 − c2 )y 0 + xy(y 0 )2 = 0 Thus S → S since the defining differential equation is invariant under the transformation of the slope. Exercise 21. For a circle centered at C, then kX − Ck = r02 for all points X on that circle.

For the condition of being tangent to a given line, L = P + tA, then (XC − C) · A = 0 and the point lies on the circle so kXC − Ck = r02 . Call the point that all the circles pass through F . Then kC − F k = kC − XC k. kC − XC k is by definition d(X, L), the distance from the circle center to the line. kC − F k = kC − X0 k is by definition a parabola. Exercise 22. Consider a circle that’s part of the mentioned family that has its center directly below the given circle with radius

r0 , and center Q. It’s given that the center is equidistant from the point of tangency and the line. This hints at a parabola because the parabola’s 230

vertex is equidistant from the focus and the directrix. Thus, we need to show that d(X, L) is equal to the distance from the circle center C to the bottom point of Q. Let N be a unit normal vector pointing from the line towards the focus, placing the focus in the positive half-plane. Let C be the center of an arbitrary circle in the family and r1 its radius. Let X1 be the point of tangency between circle Q and circle C. We want k(Q + r0 N ) − Ck = kX2 − Ck. The tangency condition between circle Q and C means that r1 (X1 − C) = −α(X1 − Q); α > 0 α= r0 Q − r0 N − C = Q − X1 − r0 N − C + X1 take the magnitude

2

2

−−−−−−−−−−→ kQ − X1 k + kX1 − Ck + r02 + 2(Q − X1 )(X1 − C) + 2(X1 − Q)r0 N + 2(C − X)r0 N r02 + r12 + r02 + 2αr02 + 2r0 (1 + α)(x1 − Q) · N 2r02 + r12 + 2r1 r0 + 2(r1 + r0 )(X1 − Q) · N I had thought the key is to use the law of cosines to evaluate (X1 − Q) · N =

1 α (C

− X1 ) · N .

Length l = d(X, L) = d(C, L). But that just gets us back to the same place. I had found the solution by a clever construction. But to come to that conclusion it required me to be “unstuck” - if something doesn’t work, move onto the next - don’t try to make something work and go in circles. And persistence is key because there can be many false eurekas. Again, consider a particular circle with its center C2 right below the given Q circle that just makes C2 tangent with the given line L2 . The directrix is not going to be L2 but L1 , a line translated below L2 , line of tangency, by r0 , so that kQ − C2 k = r2 + r0 = d(C2 , L1 ). It is a clever artificial construction. Let’s show this for any circle C of radius r1 in the family. Tangent to the circle Q condition: X1 − C = α(Q − X1 ). So then kQ − Ck = r1 + r0 Tangent to the line L2 = B2 + tA2 : (X2 − C) · A2 = 0 kX2 − Ck = r1 Consider L1 , a line translated by r0 from L2 away from Q. If L2 = B2 + tA2 ,

L1 = B2 − r0 N + tA2 .

Since X2 − C = r1 (−N ) then X2 − r0 N − C = (r1 + r0 )(−N ) will point from C to L1 , because (X2 − r0 N ) = ((B2 + tA2 ) − r0 N ) ∈ L1 . =⇒ kQ − Ck = r1 + r0 = kX2 − r0 N − Ck = d(C, L1 ). Exercise 23. Without loss of generality, use y 2 = 4cx.

The latus rectum intersect the parabola at (c, +2c), (c, −2c). Thus 4c = length of latus rectum = 2d = 2( distance from focus to directrix ). √ √ √ y = 2 cx y 0 = c/ x y 0 (c) = ±1 Tangent lines: y = ±(x + c). intersection

−−−−−−→ +(x + c) = −(x + c) x = −c (at the directrix) Exercise 24. Center of circle is given to be 0.

Collinear with center and center not between them: P = αQ; kP k kQk

= r02 231

α>0 2

= α kQk

For the line defined in Cartesian coordinates as x + 2y − 5 = 0, the vector form of this line is given by   √ 1 2 XL = B + tA XL · N = (x, y) · √ , √ =N ·B+0= 5 5 5   −2 1 A= √ ,√ is a vector that’s perpendicular to N ; 5 5 B = (1, 2) since we can simply plug it in to satisfy the equation   −2 1 t∈R XL = (1, 2) + t √ , √ 5 5 2

Q = B + tA =⇒ kQk = B 2 + 2tB · A + t2 A2 = 5 + t(0) + t2 = 5 + t2 4 (5 + t2 )(α) = r02 = 4 α = 5 + t2    4 −2 1 P = αQ = (1, 2) + t √ , √ 5 + t2 5 5 14.4 Exercises - Vector-valued functions of a real variable, Algebraic operations. Components; Limits, derivatives, and integrals. Exercise 1. F 0 = (1, 2t, 3t2 , 4t3 ). Exercise 2. F 0 = (− sin t, 2 sin t cos t, 2 cos 2t, sec2 t) Exercise 3. F 0 =



−1 √ 1 , √1−t 2 1−t2



Exercise 4. F 0 = (2et , 3et ). Exercise 5. F 0 = sinh t, 2 cosh 2t, −3e−3t Exercise 6.



1 −2t t 1+t2 , 1+t2 , (1+t2 )2

Exercise 7. F 0 =



2 1+t2

+




 

−4t2 4t (1+t2 )2 , (1+t2 )2 , 0

F0 · F = Exercise 8.

1 2 1 2, 3, e

. 4t − 4t3 4t(1 + t2 ) −4t2 (2t) + + =0 (1 + t2 )3 (1 + t2 )3 (1 + t2 )3


−1

Exercise 9. π/4 (− cos t, sin t, − ln | cos t|)|0 1

1+e 2

Exercise 10. (ln (1 + et ), t − ln (1 + et ))|0 = ln




=

, 1 − ln

√ √ ! √ − 3 2 2 + 1, , − ln 2 2 2 1+e 2





1 Exercise 11. tet − et , t2 et − 2tet + 2et , −te−t − e−t 0 = (1, e − 2, −2e−1 + 1)




Exercise 12. (2, −4, 1) = A.

R1 0

=

(te2t , t cosh 2t, 2te−2t )dt =

1 2t 2 te

+

−1 2t t sinh 2t 4 e , 2

2 2 −2 + 14 , sinh + 14 − cosh + 14 , 2 −3 2 4 4 e 1 2 −3 −2 B = 2 e + −2 sinh 2 + cosh 2 + 2 e . 0 0

1 2 4e

A·







−

cosh 2t ,2 4



−1 −2t 2 te

−

e−2t 4

.

Exercise 13. F (t) = B = 1 = kF (t)k |B| cos θ(t).

Given θ(t) = θ0 constant, kF 0 (t)k must be a constant. 2

kF 0 (t)k = F 0 (t) · F 0 (t) = g

2

g 0 = 2F 00 (t) · F 0 (t) = 0 since kF 0 k constant .

=⇒ F 00 (t) · F 0 (t) = 0 Exercise 14. 232

 1 = 0

F 0 = 2e2t A + −2e−2t B F 00 = 4e2t A + 4e−2t B = 4(F ) Exercise 15. G0 = F 0 × F 0 + F × F 00 = F × F 00 Exercise 16.

G = F · (F 0 × F 00 ) G0 = F 0 · (F 0 × F 00 ) + F · (F 00 × F 00 + F 0 × F 000 ) = F · (F 0 × F 000 ) Exercise 17. If limt→p F (t) = A,

∀jth component,

r  ∀ > 0, ∃ δj > 0 n

r such that |Fj (t) − Aj | <

 n

if |t − p| < δj

Consider min δj = δ0 j=1,...n

n X

|Fj (t) − Aj |2 <

j=1

n r X j=1

 n

2 =

whenever |t − p| < δ0

=⇒ lim kF (t) − Ak = 0 t→p

qP

− Aj )2 < 

fj0 = lim

1 (fj (t + h) − fj (t)) h

If limt→p kF (t) − Ak = 0, ∀ > 0, ∃ δ > 0 such that Pn =⇒ j=1 (Fj (t) − Aj )2 <  Pn  > j=1 (Fj (t) − Aj )2 > (Fk (t) − Ak )2 > 0

n j=1 (Fj (t)

if |t − p| < δ.

=⇒  > |Fk (t) − Ak | if |t − p| < δ. Exercise 18. If F is differentiable on I, then

F0 =

n X

fj0 ~ej

h→0

j=1

F0 =

n X

n

1 1X 1 (fj (t + h) − fj (t)) = lim (fj (t + h) − fj (t))ej = lim h → 0 (F (t + h) − F (t)) h→0 h h→0 h h j=1 j=1 lim

Pn If F 0 (t) = limhto0 h1 (F (t + h) − F (t)) = limh→0 h1 j=1 (fj (t + h) − fj (t))ej = Pn P n = j=1 limh→0 h1 (fj (t + h) − fj (t))ej = j=1 fj0 (t)ej So F 0 is differentiable. Exercise 19. F 0 (t) = 0, ∀j = 1 . . . n, fj0 (t) = 0. By one-dimensional zero-derivative theorem, fj (t) = cj constant. Thus

F (t) =

Pn

ej j=1 cj ~

Exercise 20.

1 3 6t A

= C on an open interval I.

+ 12 t2 B + Ct + D

Exercise 21. Y 0 (x) + p(x)Y (x) = Q(x). Then ∀j = 1, . . . , n

yj0 (x) + p(x)yj (x) = Qj (x) Since p, Q are continuous on I, and given this initial value condition yk (a) = bk ,   Z x R Rt − ax p(t)dt p(u)du a yj (x) = e bj + Qj (t)e dt a

=⇒

n X

jj (x) = Y (x) = e−

Rx a

p



Z B+

Exercise 22. 233

Rt

Qe a

j=1

x

a

p

 dt

tF 0 = F + tA =⇒ F 0 + tF 00 = F 0 + A tF 0 = F + tA =⇒ At ln t + Bt = A(t ln t − t) + Bt + C + tA

=⇒ tF 00 = A F 00 (t) = A/t

C = 0,

B = 3A

=⇒ F (t) = A ln t + B

F (t) = A(t ln t − t) + 3At

=⇒ F (t) = A(t ln t − t) + Bt + C

F (3) = A(3 ln 3 − 3) + 9A = 3A ln 3 + 6A

0

F (1) = A(−1) + B + C = 2A Exercise 23.

Z

x

1 F (x) = x 1 F (x) F (x) = ex A + xex A + ex A − + = 2ex A + xex A = (2 + x)ex A x x F 0 (x) = (2 + x)ex A; F (x) = 2ex A + A(xex − ex ) + C = Axex + ex A + C Z x
x (Atet + et A + C)dt = A(tet − et ) + et A + Ct 1 = 1 F (x) = e A + xe A + − 2 x 0

x

x

F (t)dt +

1

= A(xex − ex ) + ex A + C(x − 1) − eA = Axex + C(x − 1) − eA xex A + Aex +

C(x − 1) eA − =⇒ C = eA x x

Exercise 24. F 0 (t) = α(t)F (t)

=⇒ fk0 (t) = α(t)fk (t);

 ln

fk (t) fk (a)



Z

t

=

α(x)dx a

Rt

fk = fk (a)e a α Z t n n Rt X X fj (a)e6 F (t) = αej = e a α fj (a)ej = u(t)A j=1

a

j=1

14.19 Exercises - Velocity and acceleration in polar coordinates, Plane motion with radial acceleration, Cylindrical coordinates. Exercise 1.

dr dr dθ = ~er + r ~eθ = ~er + r~eθ (θ = t) dt dt dt  2 !   d2 r dθ 1 d 2 dθ a= −r ~er + r ~eθ = −r~er + 2~eθ dt2 dt r dt dt

v=

v = ~er + t~eθ = cos t~ex + sin t~ey + −t sin t~ex + t cos t~ey = = (cos t − t sin t)~ex + (sin t + t cos t)~ey a = (−t cos t − 2 sin t)~ex + (−t sin t + 2 cos t)~ey Exercise 2.

v = ~er + r~eθ + ~ez = (cos t − t sin t)~ex + (sin t + t cos t)~ey + ~ez a = (−t cos t − 2 sin t)~ex + (t sin t + 2 cos t)~ey Exercise 3. (a).

π 2 1 2 1 1 2 2 (r cos θ) + (r sin θ − ) = r − r sin θ + = 2 4 4 r = sin t; θ = t; z = log sec t; θ ≤ t <

234

(b). v=

dt dθ tan θ sec θ ~er + r ~eθ + log sec t~ez = cos t~er + r~eθ + ~ez dt dt sec θ vz = tan θ; v 2 = cos2 t + r2 + tan2 θ = sec2 θ cos φ =

tan θ = sin θ = r = sin t sec θ φ = arccos (sin θ)

Exercise 6.

Z A=

R2 (θ)dθ =

Z

2π

0

2π 1 2cθ e4πc − 1 e2cθ = e dθ = 2 4c 0 4c

Exercise 7.

Z

π

0

1 1 sin4 θdθ = 2 2

Z 0

π

1 sin θ(1 − cos θ)dθ = 2 2

2

Z

π



0

1 − cos 2θ 2



 −

1 − cos 4θ 2(4)

 dθ

= 3π/16

Exercise 15. Place target at the center (without loss of generality). The strategy is to break up v into the polar coordinate unit

vectors.

r = r~er dθ dr ~er + r ~eθ dt dt dr = vr = v cos (π − α) = −v cos α dt dθ r = v sin α dt r dθ v sin α dθ 1 dr dt = − tan α = dr = − tan α =r ; −v cos α dr r dθ dt v=

r = e− tan αθ Exercise 17.

A first order differential equation of the form y 0 = f (x, y) is homogeneous if f (tx, ty) = f (x, y). Then f (r cos θ, r sin θ) = f (cos θ, sin θ) = f (θ) We find that

dy dr = sin θ + r cos θ dθ dθ dx dr = cos θ − r sin θ dθ dθ

Thus dy = dx

dr dθ dr dθ

sin θ + r cos θ cos θ − r sin θ

= f (θ)

Exercise 18.

v = ωk × r dr dθ ~er + r ~eθ dt dt dt dθ dθ v · ~er = 0, so = 0; ωk × r = r ~eθ = ωr~eθ = r dt dt dt  2 dθ 2 |ωk × r| = ω 2 r2 = r2 dt dθ ω = , ω > 0 dt v=

Exercise 19. (a) 235

v=

(b).

dr dθ dθ dr ~er + r ~eθ = r ~eθ ; = 0; ~eθ = ~ez × ~er dt dt dt dt dθ dθ v = r ~ez × ~er = ~ez × r~er = ω × r dt dt a = v0 = ω0 × r + ω × r0 = α × r + ω × r0 ω × r0 = ω × (ω × r) = (ω · r)ω − ω 2 r

(c). Now ω · r = 0 =⇒ a = −ω 2 r Exercise 20.

The distance |rp (t) − rq (t)| is independent of t, so

d dt |rp (t)

− rq (t)| = 0, which implies

d 2 drp drp drq drq (r − 2rp · rq + rq2 ) = 2rp · −2 · rq − 2rp · +2 · rq = 0 dt p dt dt dt dt drp drq · (rq − rp ) = − · (rp − rq ) dt dt Suppose vp = ωp × rp vq = ωq × rq Then

vp · (rq − rp ) = −vq · (rp − rq ) ωp × rp · rq = −ωq × rq · rp =

= ωq × rp · rq

((ωp − ωq ) × rp ) · rq = 0 Thus, ωp = ωq .

236