CALCULUS VOLUME 1 BY TOM APOSTOL. - … · SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL. ERNEST...

SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL.

ERNEST YEUNG, - PRAHA 10, �CESK �A REPUBLIKA

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

I 2.5 Exercises - Introduction to set theory, Notations for designating sets, Subsets, Unions, intersections, complements.Exercise 10. Distributive laws

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

x ∈ A and x ∈ (B ∪ C) =⇒ x ∈ A and x is in at least B or in C

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

Suppose y ∈ Y

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

then y ∈ A and either in B or C

y ∈ X, Y ⊆ X

X = Y

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

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

if x ∈ (B ∩ C), x ∈ Y x ∈ Y, X ⊆ Y

Suppose y ∈ Y

then y is at least in A or in B and y is at least in A or in C

if y ∈ A, then y ∈ X

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

X = Y

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

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

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

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

if C ⊆ A and C ⊆ B, then C ⊆ A ∩B

∀c ∈ C, c ∈ A and c ∈ B

x ∈ A ∩B, x ∈ A and x ∈ B. Then ∀c ∈ C, c ∈ A ∩B. C ⊆ A ∩B1

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) If A ⊆ B, B ⊆ C,A ⊆ C since, ∀a ∈ A, a ∈ B, ∀b ∈ B, b ∈ C. Then since a ∈ B, a ∈ C. A ⊆ C(3) A ⊂ B and B ⊆ C. B ⊂ C or B = C. A ⊂ B only. Then A ⊂ C.(4) Yes, since ∀a ∈ A, a ∈ B.(5) No, since x 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∈FA

then x ∈ B, x /∈⋃

A∈FA

x /∈⋃

A∈FA =⇒ x /∈ A, ∀A ∈ F

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

A∈F(B −A)

Suppose x ∈⋂

A∈F(B −A)

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∈FA

Suppose x ∈ B −⋂

A∈FA

then x /∈⋂

A∈FA

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

=⇒ x ∈⋃

A∈F(B −A)

Suppose x ∈⋃

A∈F(B −A)

then x is at least in one B −A

then for A ∈ F , x ∈ B and x /∈ A

Consider ∀A ∈ F=⇒ then x ∈ B −

⋂

A∈FA

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

2

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 CSuppose x = c ∈ C and c /∈ A

(2)If C ⊆ A,

A− (B − C) = (A−B) ∪ C

I 3.3 Exercises - The �eld axioms. The goal seems to be to abstract these so-called real numbers into just x's and y's thatare purely built upon these axioms.Exercise 1. Thm. I.5. a(b− c) = ab− ac.

Let y = ab− ac; x = a(b− c)Want: x = y

ac + y = ab (by Thm. I.2, possibility of subtraction)Note that by Thm. I.3, a(b− c) = a(b + (−c)) = ab + a(−c) (by distributivity axiom)ac + x = ac + ab + a(−c) = a(c + (−c)) + ab = a(0 + b) = ab

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

0(a) = a(0) (by commutativity axiom)Given b ∈ R and 0 ∈ R, ∃ exactly one − b s.t. b− a = 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 sox = z).

Thm. I.9. If a 6= 0, then b/a = b(a−1).

Let x =b

afor ax = b

y = a−1 for ay = 1Want: x = by

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

Thm. I.10. If a 6= 0, then (a−1)−1 = 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.

3

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

−(a− b) = −a + b.Exercise 7.

(a− b) + (b− c) = a + (−b) + b + (−c) = a + (b + (−b)) + (−c) = a− c

Exercise 8.(ab)x = 1 (ab)−1 = x

a(bx) = 1 a−1 = bx

b(ax) = 1 b−1 − ax

a−1b−1 = (abx)x = 1(x) = (ab)−1

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

z =a

−ba = zt b + t = 0

y =(−a)

bby = u a + u = 0

x = −(a

b

) (a

b

)+ x = v + x = 0 vb = a

a + (−a) = vb + by = b(v + y) = 0if 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) = 0a + zb = 0 =⇒ −a = zb = by

since b 6= 0, z = y so x = y = z4

Exercise 10. Since b, d 6= 0, Let

z =ad− bc

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

x =a

bbx = a

t =−c

ddt = −c (By Thm. I.3, we know that b− a = b + (−a) )

dbx + bdt = (bd)(x + y) = ad− bc = (bd)zb, d 6= 0, so x + y = z

I 3.5 Exercises - The order axioms.Theorem 1 (I.18). If a 0 then ac < bc

Theorem 2 (I.19). If a 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 bc.

Theorem 6 (I.23). If 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 de�nition of > )(2)

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. De�ne 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 > y5

Exercise 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

2> x Contradiction.

I 3.12 Exercises - Integers and rational numbers, Geometric interpretation of real numbers as points on a line, Upperbound of a set, maximum element, least upper bound (supremum), The least-upper-bound axiom (completeness ax-iom), The Archimedean property of the real-number system, Fundamental properties of the suprenum and in�mum.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 analgorithm to zoom into a sup with monotonically increasing and monotonically decreasing sequence of �guesses� and showingits 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; thatis, 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 positiveinteger such that x < n + 1 (consider the set of all m + 1 > x and so by well-ordering principle, there must be a smallestelement of this speci�c 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 suchthat 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 > x6

Since h was arbitrary, there are in�nitely many numbers in between x, y.Exercise 7. x = a

b ∈ Q, y /∈ Q.

x± y =a± by

b

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

a−mb

b

)which is rational. Contradiction.

xy =a

b

y

1=

ay

b

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

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

x

y

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

Likewise, y 1y = 1.

Exercise 9.y − x > 0 =⇒ n(y − x) > k, n ∈ Z+, k arbitrary. Choose k to be irrational. Then k/n irrational.

y >k

n+ x > x. Let z = x +

k

n, z irrational .

Exercise 10.(1) Suppose n = 2m1 and n + 1 = 2m2.

2m1 + 1 = 2m2 2(m1 −m2) = 1 m1 −m2 =12. But m1 −m2 can only be an integer.

(2) By the well-ordering principle, if x ∈ Z+ is neither even and odd, consider the set of all x. There must exist asmallest element x0 of this set. But since x0 ∈ Z+, then there must exist a n < x such that n + 1 = x0. n is even orodd since it doesn't belong in the above set. So x0 must be odd or even. Contradiction.

(3)(2m1)(2m2) = 2(2m1m2) even2m1 + 2m2 = 2(m1 + m2) even(2m1 + 1) + (2m2 + 1) = 2(m1 + m2 + 1) =⇒ sum of two odd numbers is even

(n1 + 1)(n2 + 1) = n1n2 + n1 + n + 2 + 1 = 2(2m1m2)

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

a2 = 2b2. 2(b2) even. a2 even, so a even.If a even a = 2n.a2 = 4n2

If b odd , b2 odd. b has no factors of 2 b2 6= 4n2

Thus b is even.(5) For p

q , If p or q or both are odd, then we're done.Else, when p, q are both even, p = 2lm, q = 2np,m, p odd.

p

q=

2lm

2np=

2l−nm

pand at least m or p odd

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

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

Exercise 12. The set of rational numbers satis�es the Archimedean property but not the least-upper-bound property.Since p

q ∈ Q ⊆ R, np1q1

> p2q2

since if q1, q2 > 0,np1q2

q1q2>

q1p2

q1q2np1q2 > q1p2

n exists since (p1q2), (q1p2) ∈ R.7

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 = r

s ∈ S, x < b.Suppose x < b1, x < b2∀x ∈ S.

r

s< b2 < nb1 but likewise r

s< b1 < mb2, n, m ∈ Z+

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

I 4.4 Exercises - An example of a proof by mathematical induction, The principle of mathematical induction, Thewell-ordering principle. Consider these 2 proofs.

N + N + · · ·+ N = N2

(N − 1) + (N − 2) + · · ·+ (N − (N − 1)) + (N −N) = N2 −N∑

j=1

j =N−1∑

j=1

j

N2 + N = 2N∑

j=1

j =⇒N∑

j=1

j =N(N + 1)

2

An interesting property is that

S =n∑

j=m

j =n∑

j=m

(n + m− j)

So thatN∑

j=1

j =N∑

j=m

j +m∑

j=1

j =N∑

j=m

j +m(m + 1)

2=

N(N + 1)2

N∑

j=m

j =N(N + 1)−m(m + 1)

2=

(N −m)(N + m + 1)2

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

but S = N+ N − 1+ · · ·+ 3+ 2+ 1

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

2

Telescoping series will let you get∑N

j=1 j2 and other powers of j.N∑

j=1

(2j − 1) = 2N(N + 1)

2−N = N2

N∑

j=1

(j2 − (j − 1)2) =N∑

j=1

(j2 − (j2 − 2j + 1)) =N∑

j=1

(2j − 1) = 2(

N(N + 1)2

)−N = N2

N∑

j=1

(j3 − (j − 1)3) = N3 =N∑

j=1

(j3 − (j3 − 3j2 + 3j − 1)) =N∑

j=1

(3j2 − 3j + 1)

=⇒ 3N∑

j=1

j2 = −3N(N + 1)

2+ N = N3 =⇒ 2N3 + 2N − 3N2 − 3N

2=

N(N + 1)(2N + 1)6

=N∑

j=1

j2

N∑

j=1

j4 − (j − 1)4 = N4 =N∑

j=1

j4 − (j4 − 4j3 + 6j2 − 4j + 1) =N∑

j=1

4j3 − 6j2 + 4j − 1 =

= 4N∑

j=1

j3 − 6N(N + 1)(2N + 1)

6+ 4

N(N + 1)2

−N = N4

=⇒N∑

j=1

j3 =14(N4 + N(N + 1)(2N + 1)− 2N(N + 1) + N) =

14(N4 + (2N)N(N + 1)−N(N + 1) + N)

=14(N4 + 2N3 + 2N2 −N2 −N + N) =

14N2(N2 + 2N + 1) =

14

(N(N + 1))2

28

Exercise 1. Induction proof.

1(1 + 1)2

N+1∑

j=1

j =n∑

j=1

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

2+ n + 1 =

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

=(n + 2)(n + 1)

2

Exercise 6.(1)

A(k + 1) = A(k) + k + 1 =18(2k + 1)2 + k + 1 =

18(4k2 + 4k + 1) +

8k + 88

=(2k + 3)2

8(2) The n = 1 case isn't true.(3)

1 + 2 + · · ·+ n =(n + 1)n

2=

n2 + n

2<

n2 + n + 14

2

and(

2n + 12

)2 12

=(n + 1/2)2

2=

n2 + n + 1/42

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. Orwe could �nd, explicitly

(1 + x)n =n∑

j=0

(n

j

)xj = 1 + nx +

n(n− 1)2

x2 +n∑

j=3

(n

j

)xj

andn(n− 1)

2> n

n2 − n > 2n

n2 > 3n

n > 3

Exercise 8.a2 ≤ ca1, a3 ≤ ca2 ≤ c2a1

an+1 ≤ can ≤ ca1cn−1 = a1c

n

Exercise 9.n = 1,

√1 = 1

√12 + 12 =

√2

√(√

2)2 + 12 =√

3√

(√

n)2 + 12 =√

n + 1

Exercise 10.1 = qb + r

q = 0, b = 1, r = 12 = 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 �xedn + 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 k

j , 1 ≤ j ≤ k.If k

j ∈ Z+, only for j = 1, j = k, then k prime.If k

j ∈ 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.9

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 allblue-eyed.

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

(1) n(n+1)2 =

∑4k=1 k = 10

(2)∑5

n=2 2n−2 =∑3

n=0 2n = 1 + 14 = 15(3) 2

∑3r=0 22r = 2

∑3r=0 4r = 170

(4)∑4

j=1 jj = 1 + 4 + 27 + 44 = 288

(5)∑5

j=0(2j + 1) = 2 5(6)2 + 6(1) = 36

(6)∑

1k(k+1) =

∑nk=1

(1k − 1

k+1

)= 1− 1

n+1 = nn+1

Exercise 2.

(1) Want:∑n

k=1(ak + bk) =∑n

k=1 ak +∑n

k=1 bk (additive property)

a1 + b1 = a1 + b1

(a1 + b1) + (a2 + b2) = (a1 + a2) + (b1 + b2)

(a1 + b1) + (a2 + b2) + (a3 + b3) = (a1 + a2 + a3) + (b1 + b2 + b3)n+1∑

k=1

(ak + bk) =n∑

k=1

(ak + bk) + an+1 + bn+1 =n∑

k=1

ak + an+1 +n∑

k=1

bk + bn+1 =n+1∑

k=1

ak +n+1∑

k=1

bk

(2) Want:∑n

k=1(cak) = c∑n

k=1 ak (homogeneous property).

ca1 = (c)a1

ca1 + ca2 = c(a1 + a2)

ca1 + ca2 + ca3 = c(a1 + a2 + a3)n+1∑

k=1

(cak) = c

n∑

k=1

ak + can+1 = c

n+1∑

k=1

ak

(3) Want:∑n

k=1(ak − ak−1) = an − a0 (telescoping property)n∑

k=1

(ak − ak−1) =n∑

k=1

(ak + (−ak−1)) =n∑

k=1

ak +n∑

k=1

(−1)ak−1 =

= an +n−1∑

k=1

ak + (−1)n−1∑

k=1

ak − a0 = an − a0

Exercise 3.∑n

k=1 1 =∑

k=1(k − (k − 1)) = n

Exercise 4.k2 − (k − 1)2 = k2 − (k2 − 2k + 1) = 2k − 1n∑

k=1

(2k − 1) =n∑

k=1

k2 − (k − 1)2 = n2 − 0 = n2

Exercise 5.∑n

k=1 k = n2+n2 = n(n+1)

2

Exercise 6.n∑

k=1

k3 − (k − 1)3 = n3 =n∑

k=1

3k2 − 3k + 1 = 3

(n∑

k=1

(k2)− n(n + 1)2

+n

3

)

=⇒n∑

k=1

k2 =n3

3+

n2

2+

n

610

Exercise 7.k4 − (k − 1)4 = −(−4k3 + 6k2 +−4k + 1) = 4k3 − 6k2 + 4k − 1

n∑

k=1

k4 − (k − 1)4 = n4 = 4n∑

k=1

k3 − 6(

n3

3+

n2

2+

n

6

)+ 4

(n2

2+

n

2

)− n

=⇒ n4 + 2n3 + n2

4=

n∑

k=1

k3

Exercise 8.(1)

−n∑

k=1

(xk+1 − xk) = −(xn+1 − x) =

= (1− x)n∑

k=1

xk = x(1− xn)

=⇒n∑

k=0

xk =x(1− xn)

1− x+

1− x

1− x=

1− xn+1

1− x

(2) If x = 1, the sum equals (n + 1).Exercise 9.

n = 1(−1)(3) + 5 = 2 = 2n

n = 2(−1)(3) + 5 + (−1)7 + 9 = 4 = 2n

n

2n∑

k=1

(−1)k(2k + 1) = 2n

n + 12(n+1)∑

k=1

(−1)k(2k + 1) =2n∑

k=1

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

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

Exercise 10.(1) am + am+1 + · · ·+ am+n

(2)

n = 112

=11− 1

2=

12

n + 12(n+1)∑

k=n+2

1k

=2n∑

m=1

(−1)m+1

m− 1

n + 1+

12n + 1

+1

2n + 2=

2n∑m=1

(−1)m+1

m+− 1

2(n + 1)+

(−1)2n+1+1

(2n + 1)

=2(n+1)∑m=1

(−1)m+1

m

Exercise 13.

n = 12(√

2− 1) < 1 < 2 since 12

>√

2− 1

n case (√

n + 1−√n)(√

n + 1 +√

n) = n + 1− n = 1 <

√n + 1 +

√n

2√

n=

12(

√1 +

1n

+ 1)

n + 1 case (√

n + 2−√n + 1)(√

n + 2 +√

n + 1) = n + 2− (n + 1) = 1

√n + 2 +

√n + 1

2√

n + 1=

1 +√

1 + 1n+1

2> 1

So then, using the telescoping property,n−1∑n=1

2(√

n + 1−√n) = 2(√

m− 1) <

m∑n=1

1√n

<

m∑n=1

2(√

n−√n− 1) = 2(√

m− 1) < 2√

m− 1

11

I 4.9 Exercises - Absolute values and the triangle inequality. Exercise 1.(1) |x| = 0 iff x = 0

If x = 0, x = 0,−x = −0 = 0. If |x| = 0, x = 0,−x = 0.(2)

| − x| ={−x if − x ≥ 0x if − x ≤ 0

=

{x if x ≥ 0−x if x ≤ 0

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

(4) |x|2 =

{(x)2 if x ≥ 0(−x)2 if x ≤ 0

= x2

(5)√

x2 =

{x if x ≥ 0−x if x ≤ 0

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

|xy| ={

xy if xy ≥ 0−xy if xy ≤ 0

=

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

|x||y| ={

x|y| if x ≥ 0−x|y| if x ≤ 0

=

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

(7) By previous exercise, since∣∣∣∣x

y

∣∣∣∣ = |xy−1| = |x||y−1|∣∣∣∣1y

∣∣∣∣ =

{1y if 1

y ≥ 0−1y if 1

y ≤ 01|y| =

{1y if 1

y ≥ 0−1y if 1

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

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

Exercise 4.⇒

If ∀k = 1 . . . n; akx + bk = 0(

n∑

k=1

ak(−xak)

)2

=

(x

n∑

k=1

a2k

)2

=

(n∑

k=1

a2k

) (n∑

k=1

(−xak)2)

=

(n∑

k=1

a2k

)(n∑

k=1

b2k

)

⇐ Proving akx + bk = 0 means x = − bk

ak, ak 6= 0

(a1b1 + a2b2 + · · ·+ anbn)2 =n∑

j=1

a2jb

2j +

n∑

j 6=q

ajakbjbk ==n∑

j=1

a2jb

2j +

n∑

j 6=k

a2jb

2k

=⇒ a2jb

2k − ajakbjbk = ajbk(ajbk − akbj) = 0

if aj , bk 6= 0, ajbk − akbj = 0 =⇒ ak

(bj

−aj

)+ bk = 0

12

Exercise 8. The trick of this exercise is the following algebraic trick (�multiplication by conjugate�) and using telescopingproperty of products:

(1− x2j

)(1 + x2j

) = 1− x2j+2j

= 1− x2j+1

1∏

j=1

1 + x2j−1=

1∏

j=1

1− x2j

1− x2j−1 =1− x2n

1− x

if x = 1, 2n

Exercise 10.x > 1

x2 > x

x3 > x2 > x

xn+1 = xnx > x2 > x

0 < x < 1

x2 < x

X3 < x2 < x

xn+1 = xnx < x2 < x =⇒ xn+1 < x

Exercise 11. Let S = {n ∈ Z+|2n < n!}.By well-ordering principle, ∃ smallest n0 ∈ S. Now24 = 16, 4! = 24. So S starts at n = 4.Exercise 12.

(1) (1 +

1n

)n

=n∑

j=0

(n

k

) (1n

)j

=n∑

k=0

n!(n− k)!k!

(1n

)k

k−1∏r=0

(1− r

n

)=

k−1∏r=0

(n− r

n

)=

(1nk

)n!

(n− k)!

n∑

k=1

1k!

k−1∏r=0

(1− r

n

)=

(1nk

)n!

(n− k)!

(2)

(1 +1n

)n = 1 +n∑

k=1

(1k!

k−1∏r=0

(1− r

n)

)< 1 +

n∑

k=1

1k!

< 1 +n∑

k=1

12k

= 1 +12 −

(12

)n+1

12

= 1 + (1−(

12

)n

)

< 3

The �rst inequality obtained from the fact that if 0 < x < 1, xn < x < 1. The second inequality came from theprevious exercise, that 1

k! < 12k .

(1 +1n

)n =n∑

k=0

(n

k

)(1n

)k

= 1 +1n

+n−1∑

k=1

(n

k

)(1n

)k

= 1 +1n

+n−1∑

k=2

(n

k

) (1n

)k

+n

1

(1n

)> > 2

Exercise 13.(1)

S =p−1∑

k=0

(b

a

)k

=1− (

ba

)p

1− ba

p−1∑

k=0

bkap−1−k = ap−1 1− (ba

)p

1− ba

=bp − ap

b− a

(2)(3) Given

np <(n + 1)p+1 − np+1

p + 1< (n + 1)p

We wantn−1∑

k=1

kp <np+1

p + 1<

n∑

k=1

kp

13

n = 21p <2p+1

p + 1< 1p + 2p

p = 1

1 < 22/2 = 2.2 < 1 + 2 = 3p− 2

1 < 8/3 < 1 + 4 = 5

I 4.10 Miscellaneous exercises involving induction. Exercise 13.(1)(2)(3) Let n = 2.

2−1∑

k=1

kp = 1p = 1,np+1

p + 1=

2p+1

p + 1

2∑

k=1

kp = 1 + 2p

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

1 <21+1

1 + 2= 2 < 1 + 21 = 3

Now assume pth case. Test the p + 1 case.2p+2

p + 2=

2(p + 1)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

p + 1=⇒ (1 + 2p)(p + 1) > 2p+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∑

k=1

kp =n−1∑

k=1

kp + np <np+1

p + 1+ np <

np+1

p + 1+

(n + 1)p+1 − np+1

p + 1=

(n + 1)p+1

p + 1

n+1∑

k=1

kp =n∑

k=1

kp + (n + 1)p >np+1

p + 1+

(n + 1)p+1 − np+1

p + 1=

(n + 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 + a1a2 ≥ 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.

(1 + x)n =n∑

j=0

(n

j

)xj = 1 + nx

Since x and n are arbitrary, we can compare terms of xj's. Then x = 0.Exercise 15. 2!

22 = 12

3!33 = 2

9 < 1.14

So we've shown the n = 2, n = 3 cases. Assume the nth case, that n!nn ≤

(12

)k, where k is the greatest integer ≤ n2 .

(n + 1)!(n + 1)n+1

≥ (n + 1)nn(

12

)k

(n + 1)n+1=

(n

n + 1

)n (12

)k

=(

1− 1n + 1

)n (12

)k

<12

(12

)k

=(

12

)k+1

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

k ≤ n

2=⇒ k +

12≤ n + 1

2=⇒ 1

n + 1≤ 1

2k + 1<

12

Exercise 16.

a1 = 1 <1 +

√5

2

a2 = 2 <

(1 +

√5

2

)2

=1 + 2

√5 + 5

4=

6 + 2√

54

an+1 = an + an−1 <

(1 +

√5

2

)n

+

(1 +

√5

2

)n−1

=

(1 +

√5

2

)n (1 +

21 +

√5

)=

=

(1 +

√5

2

)n (2(1−√5)

1− 5+

44

)=

(1 +

√5

2

)n+1

Exercise 17. Use Cauchy-Schwarz, which says(∑

akbk

)2

≤(∑

a2k

) (∑b2k

)

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

(∑xp

k

)2

≤(∑

x2pk

)n =⇒

∑(x2p

k ) ≥ (∑

xpk)2

n

We de�ne Mp as follows:

Mp =(∑n

k=1 xpk

n

)1/p

So then

nMpp =

n∑

k=1

xpk

M2p =

(∑nk=1 x2p

k

n

)1/2p

nM2p2p =

n∑

k=1

x2pk

∑x2p

k = nM2p2p ≥

(nMpp )2

n= nM2p

p

M2p2p ≥ M2p

p =⇒ M2p ≥ Mp

Exercise 18.(

a4 + b4 + c4

3

)1/4

≥(

a2 + b2 + c2

3

)1/2

=23/2

31/2since

a4 + b4 + c4 ≥ 643

Exercise 19. ak = 1,∑n

k=1 1 = n15

Now consider the case of when not all ak = 1.a1 = 1

a1a2 = 1 and suppose, without loss of generality a1 > 1. Then 1 > a2.(a1 − 1)(a2 − 1) < 0a1a2 − a1 − a2 + 1 < 0 =⇒ a1 + a2 > 2

(consider n + 1 case ) If a1a2 . . . an+1 = 1, then suppose a1 > 1, an+1 < 1 without loss of generalityb1 = a1an+1

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

=⇒ a1 + a2 + · · ·+ an+1 ≥ n + 1

1.7 Exercises - The concept of area as a set function. We will use the following axioms:Assume a class M of measurable sets (i.e. sets that can be assigned an area), set function a, a : M→ R.•

Axiom 2 (Nonnegative property).(1) ∀S ∈M, a(S) ≥ 0

•Axiom 3 (Additive property). If S, T ∈M, then S ∪ T, S ∩ T ∈M and

(2) a(S ∪ T ) = a(S) + a(T )− a(S ∩ T )

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

(3) a(T − S) = a(T )− a(S)

•Axiom 5 (Invariance under congruence). If S ∈M, T = S, then T ∈M, a(T ) = a(S)

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

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

(4) S ⊆ Q ⊆ T

If ∃ only one c such that a(S) ≤ c ≤ a(T ),∀S, T such that they satisfy Eqn. (??)then Q measurable and a(Q) = c

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

to be zero.Let 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⋃

j=1

Qj

=

N∑

j=1

a(Qj)

if Qj's disjoint. Let Qj = {xj}.

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

j=1 Qj

)= 0

16

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

a(A ∩B) =√

a2 + b2d + ab− (12ab +

√a2 + b2d) =

12ab

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 (notethat the triangles and the rectangle don't overlap).

We can compute the area of a trapezoid:Tr = R + T1 + T2 =⇒ a(Tr) = a(R) + a(T1) + a(T2)

a(Tr) = hb1 +12h(b2 − b1)/2 +

12h(b2 − b1)/2 =

12h(b1 + h2)

P = R (a parallelogram consists of a right triangle rotated by π and attached to the other side of the same right triangle;the two triangles do not overlap). Since two right triangles are measurable, the parallelogram, P is measurable.Using the Additive Axiom, a(P ) = 2a(T ) = 2 1

2bh = 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 polygonwhose vertices are lattice points. The area of P is I + 1

2B−1, where I denotes the number of lattice points inside the polygonand B denotes the number on the boundary.

(1) Consider one side of the rectangle lying on a coordinate axis with one end on the origin. If the rectangle side haslength l, then l + 1 lattice points lie on this side (you have to count one more point at the 0 point. Then consider thesame number of lattice points on the opposite side. We have 2(l + 1) lattice points so far, for the boundary.

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

inside the rectangle.

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

(2)(3)

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

My way: I will take, for granted, that we know an equilateral triangle has angles of π/3 for all its angles.Even if we place two of the vertices on lattice points, so that its length is 2L, and put the midpoint and an intersectingperpendicular bisector on a coordinate axis (a picture would help), but the ratio of the perpendicular bisector to the thirdvertex to half the length of the triangle is cot π/3 = 1√

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

�out� to the third vertex by an irrational number√

3L. Thus, the third vertex cannot lie on a lattice point.Exercise 6. Let A = {1, 2, 3, 4, 5} and let M denote the class of all subsets of A. (There are 32 altogether counting A itselfand the empty set ∅). (My Note: the set of all subsets, in this case, M, is called a power set and is denoted 2A. This isbecause the way to get the total number of elements of this power set, |2A|, or the size, think of assigning to each element a�yes,� if it's in some subset, or �no�, if it's not. This is a great way of accounting for all possible subsets and we correctlyget all possible subsets.) For each set S in M, let n(S) denote the number of distinct elements in S. If S = {1, 2, 3, 4} andT = {3, 4, 5},

n(S⋃

T ) = 5

n(S⋂

T ) = 2

n(S − T ) = n({1, 2}) = 2

n(T − S) = n({5}) = 1n satis�es nonnegative property because by de�nition, there's no such thing as a negative number of elements. If S, T aresubsets of A, so are S

⋃T , S

⋂T since every element in S

⋃T , S

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

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

T ) = n(S) + a(T ) − a(S⋂

T ), where −a(S⋂

T ) ensuresthere is no double counting of distinct elements. Thus, the Additive Property Axiom is satis�ed.

For S ⊆ T , then ∀x ∈ T − S, x ∈ T, x /∈ S Now S ⊆ T , so ∀x ∈ S, x ∈ T . Thus T − S is complementary to S �withrespect to� T . n(S) + n(T − S) = n(T ), since n counts up distinct elements.

17

1.11 Exercises - Intervals and ordinate sets, Partitions and step functions, Sum and product of step function. Exercise4.

(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] =

{q1 + q2

q1 + q2 + 1 if r1 + r2 ≥ 1

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

(4)If x is an integer , [2x] = 2x = [x] + [x +

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

[x] + [x +12] = q +

{q if r < 1

2

2q + 1 if r > 12

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

{2q if r < 1

2

2q + 1 if r > 12

(5)

[x] + [x +13] + [x +

23] = q +

{q if r < 2

3

q + 1 if r > 23

+

{q if r < 1

3

q + 1 if r > 13

=

3q if r < 13

3q + 1 if 13 < r < 2

3

3q + 2 if r > 23

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

3q if r < 13

3q + 1 if 13 < r < 2

3

3q + 2 if r > 23

Exercise 5. Direct proof.

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

nq if r < 1n

nq + 1 if 1n < r < 2

n

nq + n− 1 if r > n−1n

Exercise 6.a(R) = hk = IR +

12BR − 1

b∑n=a

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

[f(n)] = g ≤ f(n), g ∈ Z, so that if f(n) is an integer,g = f(n), and if f(n) is not an integer, g is the largest integer suchthat g < f(n), so that all lattice points included and less than g are included.Exercise 7.

(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 a

b so as we move 1 along the base, thehypotenuse (or the y-value, if you will) goes up by a

b . Now

(5)[na

b

]= number of interior points at x = n and below the hypotenuse line of the right triangle of sides a, b,

including points on the hypotenuse18

b−1∑n=1

[na

b

]+

12((a + 1) + b)− 1 =

ab

2

Now (a− 1)(b− 1)2

=ab

2− a

2− b

2+

12

=⇒b−1∑n=1

[na

b

]=

(a− 1)(b− 1)2

(2) a, b ∈ Z+

b−1∑n=1

[na

b

]=

b−1∑n=1

[a(b− n)

b

](reverses order of summation)

b−1∑n=1

[a− an

b

]=

{−∑b−1

n=1

[anb − a

]if an

b − a4 is an integer (but a(

nb − 1

)can't be!)

−∑b−1n=1

([anb − a

]− 1)

otherwise

= −b−1∑n=1

([an

b− a

]− 1

)= −

b−1∑n=1

([an

b

]− a

)− (b− 1) =

= −b−1∑n=1

[an

b

]+ a(b− 1)− (b− 1)

b−1∑n=1

[na

b

]=

(a− 1)(b− 1)2

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

if x ∈ Ik.

Given that χs(x) =

{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.n∑

k=1

ckχIk(x) = cj for x ∈ Ij =⇒

n∑

k=1

ckχIk(x) = f(x)∀x ∈ [a, b]

1.15 Exercises - The de�nition of the integral for step functions, Properties of the integral of a step function, Othernotations for integrals. Exercise 1.

(1)∫ 3

1[x]dx = (−1) + 1 + (2) = 2

(2)∫ 3

−1[x + 1

2 ]dx =∫ 7/2

−1/2[x]dx = (−1) 1

2 + (1)(1) + (2)(1) + 123 = 4

(3)∫ 3

−1([x] + [x + 1

2 ])dx = 6

(4)∫ 3

−12[x]dx = 4

(5)∫ 3

−1[2x]dx = 1

2

∫ 6

−2[x]dx = 1

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

(6)∫ 3

−1[−x]dx = − ∫ −3

1[x]dx =

∫ 1

−3[x]dx = −3 +−2 +−1 = −6

Exercise 2.

s =

{5/2 if 0 < x < 2−1 if 2 < x < 5

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.∫ b

a

([x] + [−x])dx =∫ b

a

[x− x]dx =∫ b

a

(−1)dx = a− b

Exercise 4.(1) n ∈ Z+,

∫ n

0[t]dt =

∑n−1t=0 t = (n−1)(n−1+1)

2 = (n−1)n2

(2)19

Exercise 5.(1)

∫ 2

0[t2]dt =

∫ 2

1[t2]dt = 1(

√2− 1) + 2(

√3−√2) + 3(2−√3) = 5−√2−√3

(2)∫ 3

−3[t2]dt =

∫ 3

0[t2]dt +

∫ 0

−3[t2]dt =

∫ 3

0[t2]dt +− ∫ 0

3[t2]dt = 2

∫ 3

0[t2]dt

∫ 3

2

[t2]dt = 4(√

5− 2) + 5(√

6−√

5) + 6(√

7−√

6) + 7(√

8−√

7) + 8(3−√

8)

16−√

5−√

6−√

7−√

8∫ 2

0

[t2]dt +∫ 3

2

[t2]dt = 21− 3√

2−√

3−√

5−√

6−√

7−√

8

=⇒∫ 3

−3

[t2]dt = 42− 2(3√

2 +√

3 +√

5 +√

6 +√

7)

Exercise 6.(1)

∫ n

0[t]2dt =

∫ n

1[t]2dt =

∑n−1j=1 j2 = (n−1)n(2n−1)

6

(2)∫ x

0[t]2dt =

∑[x−1]j=1 j2 + q2r where x = q + r, q ∈ Z+, 0 ≤ r < 1.

∫ x

0

[t]2dt =q(q − 1)(2q − 1)

6+ q2r = 2(x− 1) = 2(q + r − 1)

=⇒ q(q − 1)(2q − 1) + 6q2r = 12q + 12r − 12

=⇒ x = 1, x = 5/2

Exercise 7.(1)

∫ 9

0

[√

t]dt =∫ 9

1

[√

t]dt = 3(1) + 5(2) = 13∫ 1

0

6[√

t]dt = 3(1) + 5(2) + 7(3) = 34 =(4)(3)(17)

6

Assume∫ n2

0

[√

t]dt =n(n− 1)(4n + 1)

6∫ (n+1)2

0

[√

t]dt =∫ n2

0

[√

t]dt +∫ (n+1)2

n2[√

t]dt =n(n− 1)(4n + 1)

6+ n((n + 1)2 − n2) =

=(n2 − n)(4n + 1) + 6n(2n + 1)

6=

4n3 + n2 − 4n2 − n + 12n2 + 6n

6=

4n3 + 9n2 + 5n

6indeed ,

(n + 1)(n)(4(n + 1) + 1)6

=(n2 + n)(4n + 5)

6=

4n3 + 5n2 + 4n2 + 5n

6

Exercise 8.∫ b+c

a+cf(x)dx =

∫ b+c−c

a+c−cf(x− (−c))dx =

∫ b

af(x + c)dx

Exercise 9.∫ kb

kaf(x)dx = 1

1k

∫ (kb)/k

(ka)/kf

(x

1/k

)dx = k

∫ b

af(kx)dx

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

0s(x)dx.

So for f(3) =∫ 3

0s(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 thenf(3) = (−1)(1) + 2(1) = 1

f(4) = 1 + (−3)(1) = −2

f(f(3)) = f(1) = 020

We obtain this formula

f(p) =

{p2 (−1)p+1 p evenp−12 (−1)p+1 p even

since

f(p + 1) = f(p) +∫ p+1

p

s(x)dx ={

p−12 (−1)p+1 p even + (−1)pp

=

{−p2 p even

p−12 p odd

+

{p

−p=

{p2−p−1

2

=

=

{− (p+1)

2 if p + 1 evenp2 if p + 1 odd

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

(1) ∫ b

a

s(x)dx =n∑

k=1

s3k(xk − xk−1)

∫ b

a

s +∫ c

b

s =n1∑

k=1

s2k(xk − xk−1) +

n2∑

k=n1

s3k(xk − xk−1) =

n2∑

k=1

s3k(xk − xk−1) =

∫ c

a

s(x)dx

(2)∫ b

a(s + t) =

∑n3k=1(s + t)3k(xk − xk−1) 6=

∫ b

as +

∫ b

at

(3)∫ b

acs =

∑nk=1(cs)

3(xk − xk−1) 6= c∫ b

as

(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.

n∑

k=1

s3k(xk − xk−1) =

k∑

k=1

s3k(xk − c− (xk−1 − c)) =

=n∑

k=1

s3k(yk − yk−1) =

∫ b

a

s(y + c)dy

(5) s < t,∫ b

as =

∑nk=1 s3

k(xk − xk−1).

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

if s < 0t, s3 < and t3 > 0

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

t2s < t3

s3 < s2t < t2s < t3

Then∫ b

as <

∫ b

at.

Exercise 12.(1)

∫ b

as +

∫ c

bs =

∑n1k=1 sk(x2

k − x2k−1) +

∑n2k=n1

sk(x2k − x2

k−1) =∑n3

k=1 sk(x2k − x2

k−1) =∫ c

as

(2)∫ b

a(s+t) =

∑n3k=1(s+t)k(x2

k−x2k−1) =

∑n3k=1(sk+tk)(x2

k−x2k−1) =

∑n3k=1 sk(x2

k−x2k−1)+

∑n3k=1 tk(x2

k−x2k−1)

since P3 = {xk} is a �ner partition than the partition for s, P1, t, P2, then consider

sk(y2j − y2

j−1) = sk((x2k+1 − x2

k) + (x2k − x2

k−1)), son3∑

k=1

sk(x2k − x− k − 12) +

n3∑

k=1

tk(x2k − x− k − 12) =

n1∑

j=1

sj(x2j − x− j − 12) +

n2∑

j=1

tj(x2j − x− j − 12) =

=∫ b

a

s +∫ b

a

t

(3)∫ b

acs =

∑nk=1 csk(x2

k − x2k−1) = c

∑nk=1 sk(x2

k − x2k−1) = c

∫ b

as

21

(4)∫ b+c

a+cs(x)dx =

∑nk=1 sk(x2

k − x2k−1) where

s(x) = sk if xk−1 < x < xk

x(y + c) = sk if xk−1 < y + c < xk =⇒ xk−1 − c < y < xk − c =⇒ yk−1 < y < yk

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

∫ b

a

s(y + c)dy =n∑

k=1

sk(y2k − y2

k−1) =

=n∑

k=1

sk((xk − c)2 − (xk−1 − c)2) =n∑

k=1

sk(x2k − 2xkc + c2 − (x2

k−1 − 2xk−1c + c2)) =

=n∑

k=1

sk(x2k − x2

k−1 − 2c(xk − xk−1)) 6=n∑

k=1

sk(x2k − x2

k−1)

(5) Since x2k − x2

k−1 > 0,∫ b

asdx =

∑nk=1 sk(x2

k − x2k−1) <

∑nk=1 tk(x2

k − x2k−1) =

∫ b

atdx

Note that we had shown previously that the integral doesn't change under �ner partition.

Exercise 13.

∫ b

a

s(x)dx

n∑

k=1

sk(xk − xk−1);∫ b

a

t(x)dx =n2∑

k=1

tk(yk − yk−1)

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

Consider P⋃

Q = R. R consists of n3 elements, (since n3 ≤ n + n2 some elements of P and Q may be the same. R isanother partition on [a, b] (by partition de�nition) since xk, yk ∈ R and since real numbers obey transitivity, {xk, yk} can bearranged such that a < z1 < z2 < · · · < zn3−2 < b where zk = xk or yk.

(s + t)(x) = s(x) + t(x) = sj + tk if xj−1 < x < xj ; yj−1 < x < yj

If xj−1 ≶ yj−1, let zl−1 = yj−1, xj−1 andIf 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

∫ b

a

(s(x) + t(x))dx =∫ b

a

((s + t)(x))dx =n3∑

l=1

(sl + t)l)(zl − zl−1) =n3∑

l=1

sl(zl − zl−1) +n3∑

l=1

tl(zl − zl−1)

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

l=1

sl(zl − zl−1) +n∑

l=1

tl(zl − zl−1) =n∑

k=1

sk(xk − xk−1) +n2∑

k=1

tk(yk − yk−1) =∫ b

a

s(x)dx +∫ b

a

t(x)dx

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

c1

∫ b

a

s(x)dx + c2

∫ b

a

t(x)dx = c1

n∑

k=1

sk(xk − xk−1) + c2

n2∑

k=1

tk(xk − xk−1) =

=n3∑

l=1

c1sl(zl − zl−1) +n3∑

l=1

c2tl(zl − zl−1) =n3∑

l=1

(c1sl + c2tl)(zl − zl−1) =

=∫ b

a

(c1s + c2t)(x)dx

We relied on the fact that we could de�ne a �ner partition from two partitions of the same interval.Exercise 15. Prove Theorem 1.5 (the comparison theorem).

22

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

a

s(x)dx =n∑

k=1

sk(xk − xk−1) =n3∑

l=1

sl(zl − zl−1) <

n3∑

l=1

tl(zl − zl−1) =n2∑

k=1

tk(yl − yk−1) =

=∫ b

a

t(x)dx

=⇒∫ b

a

s(x)dx <

∫ b

a

t(x)dx

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].∫ c

a

s(x)dx +∫ b

a

s(x)dx =n1∑

k=1

sl(xk − xk−1) +n2∑

k=1

sk(xk − xk−1) =n3∑

k=1

sk(xk − xk−1) =∫ b

a

s(x)dx

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

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

xk−1 < y − c < 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

a

s(x)dx =n∑

k=1

sk(xk − xk−1) =n∑

k=1

sk(yk = yk−1) =∫ yn

y0

s(y − c)dy =∫ b+c

a+c

s(x− c)dx

1.26 Exercises - The integral of more general functions, Upper and lower integrals, The area of an ordinate set ex-pressed as an integral, Informal remarks on the theory and technique of integration, Monotonic and piecewise mono-tonic functions. De�nitions and examples, Integrability of bounded monotonic functions, Calculation of the integralof a bounded monotonic function, Calculation of the integral

∫ b

0xpdx when p is a positive integer, The basic properties

of the integral, Integration of polynomials.

Exercise 16.∫ 2

0|(x− 1)(3x− 1)|dx =

∫ 2

1

(x− 1)(3x− 1)dx =∫ 2

1

(3x2 − 4x + 1)dx = (x3 − 2x2 + x)∣∣21

= 2∫ 1

1/3

(1− x)(3x− 1)dx = − (x3 − 2x2 + x)∣∣11/3

=427

∫ 1/3

0

(x− 1)(3x− 1)dx =427

So the �nal answer for the integral is 62/27.Exercise 17.

∫ 3

0(2x− 5)3dx = 8

∫ 3

0(x− 5

2 )3dx = 8∫ 3−5/2

−5/2x3dx = 8 1

4x4∣∣1/2

−5/2= 39

2

Exercise 18.∫ 3

−3(x2 − 3)3dx =

∫ 3

0(x2 − 3)3 +

∫ x

−3(x2 − 3)3 =

∫ 3

0(x2 − 3)2 +− ∫ 3

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).

∫f − g =

∫x2 − cx3 =

(13x3 − c

4x4

)∣∣∣∣1/c

0

=1

12c3

∫f − g =

23

=1

12c3; c =

12√

223

Exercise 16. f = x(1− x), g = ax.∫

f − g =∫ 1−a

0

x− x2 − ax =(

(1− a)12x2 − 1

3x3

)∣∣∣∣1−a

0

= (1− a)316

= 9/2 =⇒ a = −2

Exercise 17. π = 2∫ 1

−1

√1− x2dx

(1) ∫ 3

−3

√9− x2dx = 3

∫ 3

−3

√1−

(x

3

)2

= 3(3)∫ 1

−1

√1− x2 =

9π

2

Now ∫ kb

ka

f(x

k

)dx = k

∫ b

a

fdx

(2) ∫ 2

0

√1− 1

4x2dx = 2

∫ 1

0

√1− x2dx =

2π

4=

π

2

(3)∫ 2

−2(x− 3)

√4− x2dx

∫ 2

−2

x√

4− x2dx = (−1)∫ −2

2

−x√

4− x2 =⇒ 2∫ 2

−2

x√

4− x2 = 0

− 3∫ 2

−2

2

√1−

(x

2

)2

dx = (−6)(2)∫ 1

−1

√1− x2 = −6π

Exercise 18. Consider a circle of radius 1 and a twelve-sided dodecagon inscribed in it. Divide the dodecagon by isoscelestriangle pie slices. The interior angle that is the vertex angle of these triangles is 360/12 = 30 degrees.

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

c2 = 1 + 1− 2(1)(1) cos 30◦ = 2

(1−

√3

2

)=⇒ c =

√2

√1−

√3

2

The height is given also by the law of cosines

h = 1 cos 15◦ =

√1 + cos 30◦

2=

√1 +

√3

2

2

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

(12)12

√1 +

√3

2

(1√2

)

√2

√1−

√3

2

= 3

So 3 < π.

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

(12)12

2

√√√√1−√

32

1 +√

32

(1) = 12

(2−

√3

2

)> π

Exercise 19.(1) (x, y) ∈ E if x = ax1, y = by1 such that x2

1 + y21 ≤ 1

=⇒ (xa

)2 +(

yb

)2 = 1(2)

y = b

√1−

(x

a

)2

2∫ a

−a

b

√1−

(x

a

)2

= 2ba

∫ 1

−1

√1− x2 = ba

π

2(2) = πba

24

Exercise 20. Let f be nonnegative and integrable on [a, b] and let S be its ordinate set.Suppose x and y coordinates of S were expanded in different ways x = k1x1, y = k2y1.

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

xk1

)= k2y1 = y.

integrating g on [k1a, k1b],∫ k1b

k1a

g(x)dx =∫ k1b

k1a

k2f

(x

k1

)dx = k2k1

∫ b

a

f(x)dx = k2k1A

2.8 Exercises - The trigonometric functions, Integration formulas for the sine and cosine, A geometric description ofthe sine and cosine functions. Exercise 1.

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

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

Exercise 4.sin 3x = sin 2x cosx + 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

=⇒ cosπ/6 =√

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

(2) sin 2π/6 = 2 cos π/6 sin π/6 =√

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

√2 = sin π/4

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

R =9bc− 27d− 2b3

54

Q =3c− b2

9

S = (R +√

Q3 + R2)1/3

T = (R−√

Q3 + R2)1/3

x1 = S + T − b/3

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.

tan x− y =sin x− y

cosx− y=

sin x cos y − sin y cosx

cos x cos y + sin x sin y

(1

cos x cos y1

cos x cos y

)=

tanx− tan y

1 + tan x tan y

25

if tan x tan y 6= −1Similarly,

tan x + y =sin x + y

cosx + y=

sin x cos y + sin y cosx

cos x cos y − sin x sin y=

tanx + tan y

1− tan x tan y, tanx tan y 6= 1

cot x + y =cos x + y

sinx + y=

cos x cos y − sin x sin y

sin x cos y + sin y cos x=

cot x cot y − 1cot y + cot x

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

√3

2 cos x) = 32 sin x + 3

√3

2 cos x

Exercise 8.C sinx + α = C(sinx cos α + cosx sin α) = C cosα sin x + C sinα cos x

A = C cosα, B = C sin α

Exercise 9. If A = 0, B cosx = B sin π/2 + x = C sinx + α so C = B, α = π/2 if A = 0.If A 6= 0,

A sin x + B cosx = A(sinx +B

Acos x) == A(sinx + tan α cos x)

=A

cosα(cosα sin x + sin α cos x) =

A

cosα(sinx + α)

where −π/2 < α < π/4, B/A = tan α, C = Acos α

Exercise 10. C sin x + α = C sin x cosα + C cos x sin α.C cosα = −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,

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

Bsin x) =

B

cosβ(cosx cosβ + sin β sin x) = C cosx + α

where AB = tan β, α = −β, C = B

cos β .Exercise 12.

sin x = cos x =√

1− cos2 x =⇒ cosx = 1/√

2 =⇒ x =π

4Try 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).

Exercise 13.sin x− cos x = 1 =

√1− cos2 x = 1 + cos x

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

cosx = −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 − cosx + y = sinx 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

h=

sin (x + h/2) cos h/2 + cos (x + h/2) sin h/2− sin (x + h) cos h/2− cosx + h/2 sinh/2h

=sin h/2

h/2cos (x + h/2)

cosx + h− cos x

h=

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

= − sinh/2h/2

sin (x + h/2)

Exercise 16.(1)

sin 2x = 2 sin x cosx

if sin 2x = 2 sin x and x 6= 0, x 6= πn, cos x = 1 but x 6= πn =⇒ x = 2πn

26

(2) cos x + y = cos x cos y − sin x sin y = cos x + cos y.

cosx cos y − cos x− cos y = sin y√

1− cos2 x

Letting A = cos x, B = cos y,

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

A2 + B2 −A2B −AB2 + AB = 1/2

B2(1−A) + B(A−A2) + A2 − 1/2 = 0

B =A(1−A)±

√A2(1−A)2 − 4(1−A)(A2 − 1/2)

1−A= A± 1√

1−A(A2(1−A)− 4(A2 − 1/2))1/2 =

= A± 1√1−A

(−3A2 −A3 + 2)1/2

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

√3,−1 −√3. So suppose cosx = 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 cosx = sin x + sin y

=⇒ sin y(1− cos x) + sin y +− cos x sin y = 0, =⇒ y = 2πn

Checking our result, we �nd that sin (2πn + y) = sin 2πn + sin y(1)(4) ∫ y

0

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

=⇒ 1− cos y =√

1− cos2 y

1− 2 cos y + cos2 y = 1− cos2 y =⇒ cos y(cos y − 1) = 0; y =2(j + 1)π

2, 2πn

Exercise 17.∫ b

asin xdx = − cosx|ba = − cos b + cos a

(1) −√

32 + 1

(2) −√

22 + 1

(3) 12

(4) 1(5) 2(6) 0 We were integrating over one period, over one positive semicircle and over one negative semicircle.(7) 0 We had integrated over two equal parts, though it only shaded in up to x = 1.(8) −

√2

2 +√

32

Exercise 18.∫ π

0(x + sin x)dx = ( 1

2x2 − cosx)∣∣π0

= π2

2 − (−1− 1) = π2

2 + 2

Exercise 19.∫ π/2

0(x2 + cos x)dx = ( 1

3x3 + sin x)∣∣π/2

0= 1

3 (π/2)3 + 1


0(sinx− cos x)dx = (− cosx− sin x)|π/2

0 = −1− (−1) = 0


0| sin x− cosx|dx = ( by symmetry )2

∫ π/4

0(cos x− sin x)dx = 2(sin x + cos x)|π/4

0 = 2(√

2− 1)

Exercise 22.∫ π

0(12 + cos t)dt = ( 1

2 t + sin t)∣∣π0

= π2

Exercise 23.∫ 2π/3

0

(12

+ cos t)dt +∫ π

2π/3

−(12

+ cos t)dt = (t

2+ sin t)

∣∣∣∣2π/3

0

+ (t

2+ sin t)

∣∣∣∣2π/3

π

= 2(π

3+√

32

)− π

2=

π

6+√

3

Exercise 24. If −π < x ≤ − 2π3 ,

∫ x

−π

−(12

+ cos t)dt =∫ −π

x

(12

+ cos t)dt =(

t

2+ sin t

)∣∣∣∣−π

x

= −π

2− x

2− sin x

27

If −2π/3 ≤ x ≤ 2π/3,∫ −2π/3

−π

−(12

+ cos t)dt +∫ x

−2π/3

(12

+ cos t)dt =−π/6

+

√3/2 + (t/2 + sin t)|x−2π/3

= x/2 + sin x− π/3−√

3/2 +√

3/2− π/6 =x

2+ sin x− π/3

If 2π/3 ≤ x ≤ π,√

3/2 +∫ x

2π/3

−(1/2 + cos t)dt =√

3/2 + (t/2 + sin t)|2π/3x = π/3 +

√3− x/2− sin x

Exercise 25.∫ x2

x(t2 + sin t)dt = ( 1

3 t3 +− cos t) = x6−x3

3 + cos x− cos x2


0sin 2xdx =

(− cos (2x)

2

)∣∣∣π/2

0= (−1/2)(−1− 1) = 1


0cosx/2dx = 2 sinx/2|π/3

0 = 2 12 = 1

Exercise 28. ∫ x

0

cos (a + bt)dt =∫ x

0

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

bsin bt− sin a(− cos bt/b)

)∣∣∣x

0=

=cos a

bsin bx +

sin a

b(cos bx− 1) =

1b

sin a + bx− sin a/b∫ x

0

sin (a + bt)dt =∫ x

0

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

sin a

bsin bt− cos a

bcos bt

)∣∣∣∣x

0

=

=1b(cos bx + a + cos a)

Exercise 29.(1)

∫ x

0

sin3 tdt =∫ x

0

3 sin t− sin 3t

4dt =

(−3

4cos t + cos 3t/12

)∣∣∣∣x

0

= −3/4(cos x− 1) +cos 3x− 1

12=

=13− 3

4cosx +

112

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

(2)∫ x

0

cos3 tdt =∫ x

0

14(cos 3t + 3 cos t)dt =

(14

sin 3t

3+

34

sin t

)∣∣∣∣x

0

=

=112

(sin 2x cos x + sin x cos 2x) +34

sin x =112

(2 sin x cosx + sin x(2 cos2 x− 1)) =

=sinx cos2 x + 2 sin x

3Exercise 30. Now using the de�nition 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 ∫ a+p

a

f(x)dx =∫ r+p

r

f(x + np)dx =∫ r+p

r

f(x)dx =∫ p

r

f +∫ r+p

p

f(x)dx =

=∫ p

r

f +∫ r

0

f(x− p)dx =∫ p

r

f +∫ r

0

f =∫ p

0

f

Exercise 31.(1) ∫ 2π

0

sin nxdx =1n

∫ 2πn

0

sin xdx =1n

(− cosx)∣∣∣∣2πn

0

= − 1n

(1− 1) = 0

∫ 2π

0

cosnxdx =1n

∫ 2πn

0

cosxdx =1n

sin x

∣∣∣∣2πn

0

= 0

28

(2)∫ 2π

0

sinnx cos mxdx =∫ 2π

0

12(sin (n + m)x + sin (n−m)x)dx = 0 + 0 = 0

∫ 2π

0

sinnx sin mxdx =∫ 2π

0

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

∫ 2π

0

cosnx cos mxdx =∫ 2π

0

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

While ∫ 2π

0

sin2 nxdx =∫ 2π

0

1− cos 2nx

2dx = π

∫ 2π

0

cos2 nxdx =∫ 2π

0

1 + cos 2nx

2dx = π

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

k=1

2 sinx/2 cos kx = 2 sin x/2n∑

k=1

cos kx =n∑

k=1

sin (2k + 1)x

2− sin (2k − 1)

x

2= sin (2n + 1)

x

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

= cos kx cosx/2− sin kx sin x/2− (cos kx cos x/2 + sin kx sin x/2) =

= −2 sin kx sin x/2

−2 sinx/2n∑

k=1

sin kx =n∑

k=1

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

= cos nx + x/2− cos x/2

Nowsin nx/2 sin nx/2 + x/2 = sin nx/2(sinnx/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 =

=(

1− cosnx

2

)cosx/2 +

sin nx

2sin x/2 =

=12

(cos x/2− cos x/2 cos nx + sin nx sinx/2) =12(cos x/2− cos (nxx/2)

Then

−2 sinx/2n∑

k=1

sin kx = −2 sin nx/2 sin12(n + 1)x

n∑

k=1

sin kx =sinnx/2 sin 1

2 (n + 1)xsin x/2

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

12

cosx sin x <12

sin x <x

2=⇒ sin x < x

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

29

2.11 Exercises - Polar coordinates, The integral for area in polar coordinates.Exercise 1.

(x− 1)2 + y2 = 1

=⇒ x2 − 2x + 1 + y2 = 1=⇒ r2 = 2r cos θ or r = 2 cos θ

Exercise 2.x2 + y2 − x =

√x2 + y2

=⇒ r2 = r(cos θ + 1) or r = 1 + cos θ

Exercise 3.(x2 + y2)2 = x2 − y2; y2 ≤ x2

r4 = r2(cos2 θ − sin2 θ)=⇒

r2 = cos 2θ

r =√

cos 2θ for cos 2θ > 0

Exercise 4.

(x2 + y2)2 = r4 = |x2 − y2| = |r2 cos 2θ| =⇒r2 = | cos 2θ|r =

√| cos 2θ|

Recall that unde�ned the measure of an angle to be 2 times the area of the sector divided by the radius squared (we chosethis because it doesn't change with a change in circle size). Then if r2(θ) is integrable, then A =

∫ θb

θa

12θr2dθ.

See sketches for the following exercises. Otherwise, we compute the area of the sector.Exercise 5. Spiral of Archimedes: f(θ) = θ; 0 ≤ θ ≤ 2π

A =∫ 2π

0

12θ2dθ =

16(2π)3 =

4π3

3

Exercise 6. Circle tangent to y-axis: f(θ) = 2 cos θ; −π/2 ≤ θ ≤ π/2

A =12

∫ π/2

−π/2

4 cos2 θdθ = 2∫ π/2

−π/2

(1 + cos 2θ)2

dθ = π

Exercise 7. Two circles tangent to y-axis: f(θ) = 2| cos θ|; 0 ≤ θ ≤ 2π

12

∫ 2π

0

4 cos2 θdθ = 2∫ 2π

0

1 + cos 2θ

2= 2π

Exercise 8. Circle tangent to x-axis: f(θ) = 4 sin θ ; 0 ≤ θ ≤ π

12

∫ π

0

16 sin2 θdθ = 8∫ π

0

(1− cos 2θ

2

)dθ = 4π

Exercise 9. Two circles tangent to x-axis: f(θ) = 4| sin θ|; 0 ≤ θ ≤ 2π

12

∫ 2π

0

16 sin2 θdθ = 8∫ 2π

0

1− cos 2θ

2dθ = 8π

Exercise 10. Rosepetal: f(θ) = sin 2θ; 0 ≤ θ ≤ π/2

12

∫ π/2

0

sin2 2θdθ =12

∫ π/2

0

dθ

(1− cos 4θ

2

)=

π

8

Exercise 11. Four-leaved rose f(θ) = | sin 2θ|; 0 ≤ θ ≤ 2π

12

∫ 2π

0

sin2 2θdθ =12

∫ 2π

0

dθ

(1− cos 4θ

2

)=

π

230

Exercise 13. Four-leaf clover: f(θ) =√| cos 2θ|; 0 ≤ θ ≤ 2π

12

∫ 2π

0

| cos 2θ|dθ =

=12

(∫ π/4

0

c(2θ)−∫ 3π/4

π/4

c(2θ) +∫ 5π/4

3π/4

c(2θ)−∫ 7π/4

5π/4

c(2θ) +∫ 2π

7π/4

c(2θ)

)=

=12

(s(2θ)

2

∣∣∣∣π/4

0

− s(2θ)2

∣∣∣∣3π/4

π/4

+s(2θ)

2

∣∣∣∣5π/4

3π/4

− s(2θ)2

∣∣∣∣7π/4

5π/4

+s(2θ)

2

∣∣∣∣2π

7π/4

)=

=12

(12−

(−1− 12

)+

(1− (−1)

2

)−

(−1− 12

)+

(−(−1)2

))= 2

Exercise 14. Cardiod: f(θ) = 1 + cos θ; 0 ≤ θ ≤ 2π

12

∫ 2π

0

1 + 2 cos θ + cos2 θ =12(2π +

12(2π)) =

3π

2

Exercise 15. Limacon: f(θ) = 2 + cos θ; 0 ≤ θ ≤ 2π

12

∫ 2π

0

4 + 4 cos θ + cos2 θ =12(4(2π) +

12(2π)) =

9π

2

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

∫x2dx = 1

3 (b2 + ab + a2)

Exercise 2. 11−0

∫x2 + x3 = 7

12

Exercise 3. 14−0

∫x1/2 = 4

3

Exercise 4. 18−1

∫x1/3 = 45

28

Exercise 5. 1π/2−0

∫ π/2

0sin x = 2

π

Exercise 6. 1π/2−−π/2

∫cosx = 2/π


∫sin 2x = −1/π(−1− 1) = 2/π


∫sin x cos x = 1

π


∫sin2 x = 1

π (x− sin 2x/2)∣∣π0

= 12

Exercise 10. 1π−0

∫cos2 x = 1

2

Exercise 11.(1) 1

a−0

∫x2 = a2/3 = c2 =⇒ c = a/

√3

(2) 1a−0

∫xn = 1

a1

n+1xn+1∣∣∣a

0= an

n+1 = cn =⇒ c = a(n+1)1/n

Exercise 12.

A =∫

wf/

∫w

∫wx2 = k

∫x

∫x3 =

14x4 = k

12x2; k =

12, w = x

∫x4 =

15x5 = k

13x3; k =

35, w = x2

∫x5 =

16x6 = k

14x4; k =

23, w = x3

Exercise 13.

A(f + g) =1

b− a

∫f + g =

1b− a

∫f +

1b− a

∫g = A(f) + A(g)

A(cf) =1

b− a

∫cf = c

(1

b− a

) ∫f

A(f) =1

b− a

∫f ≤ 1

b− a

∫g = A(g)

31

Exercise 14.

A(c1f + c2g) =∫

w(c1f + c2g)∫w

=c1

∫wf∫w

+c2

∫wg∫w

= c1A(f) + c2A(g)

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

Exercise 15.

Aba(f) =

1b− a

∫ b

a

f =1

b− a

(∫ c

a

f +∫ b

c

f

)=

(c− a

b− a

) ( ∫ c

af

c− a

)+

b− a− (c− a)b− a

∫ b

af

b− c

a < c < b

0 <c− a

b− a< 1

Let t =c− a

b− a

=⇒ Aba(f) = tAc

a(f) + (1− t)Abc(f)

Aba(f) =

∫ b

awf

∫ b

aw

=

∫ c

aw

∫ b

aw

∫ c

awf∫ c

aw

+

(∫ b

aw − ∫ c

aw

∫ b

aw

) ∫ b

cwf

∫ b

cw

0 <

∫ c

aw

∫ b

aw

< 1 since w is a nonnegative function. Let t =

∫ c

aw

∫ b

aw

=⇒ Aba(f) = tAc

a(f) + (1− t)Abc(f)

Exercise 16. Recall that xcm =R

xρRρ

or rcm =R

rdmM .

xcm =

∫ L

0x

∫ L

01

=L

2

Icm =∫

r2dm =∫

x2(1) = L3/3

r2 =Icm∫ L

01

= L2/3 =⇒ r =L√3

Exercise 17.

xcm =

∫ l/2

0x +

∫ L

L/22xdx

L2 + 2(L− L/2)

=yL2

12

Icm =∫ L/2

0

x2 +∫ L

L/2

2x2 = 5L3/8 r2 =5L3/83L/2

=5L2

12=⇒ r =

√5L

2√

3

Exercise 18. ρ(x) = x for 0 ≤ x ≤ L

xcm =∫

xxdx∫xdx

=13x3

∣∣L0

12x2

∣∣L0

=23L

Icm =∫

x2xdx = L4/4

r2 =L4/4L2/2

= L2/2 r =L√2

Exercise 19.

xcm =

∫xxdx +

∫xL

2 dx∫xdx +

∫L/2

=13x3

∣∣L/2

0+ L

2 (x2/2)∣∣LL/2

12x2

∣∣L/2

0+ L

2 (L− L/2)= 11L/18

Icm =∫

x2xdx +∫

x2L/2dx = L431/192

r2 = Icm/(L23/8) = L231/72 r =√

31L

6√

232

Exercise 20. ρ(x) = x2 for 0 ≤ x ≤ L

xcm =∫

xx2dx∫x2

= 3L/4

Icm =∫

x2x2dx = L5/5

r2 =Icm13L3

=35L2 r =

√35L

Exercise 21.

xcm =

∫ L/2

0xx2dx +

∫ L

L/2xL2

4 dx∫ L/2

0x2dx +

∫ L

L/2L2

4 dx= 21L/32

Icm = intL/20 x2x2dx +

∫ L

L/2

x2 L2

4dx = 19L5/240

r2 =Icm

L3/6= 19L2/40 =⇒ r =

√19L/2

√10

Exercise 22. Be �exible about how you can choose a convenient origin to evaluate the center-of-mass fromLet ρ = cxn

c

∫ L

0

xndx =1

n + 1Ln+1c = M

=⇒ c =(n + 1)M

Ln+1

c

∫ L

0

xxndx = c1

n + 2Ln+2 =

n + 1n + 2

ML =3ML

4

xcm =∫

xρ

M=

3L

4=⇒

∫xρ =

n + 1n + 2

=34

=⇒ n = 2

ρ =3M

L3x2

Exercise 23.(1)

1π/2− 0

∫3 sin 2t =

6π

(2)1

π/2− 0

∫9 sin2 2t = 9/2 =⇒ vrms = 3

√2/2

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

12π

∫ 2π

0

160 sin t2 sin (t− π/6) = 80√

3

2.19 Exercises - The integral as a function of the upper limit. Inde�nite integrals. Exercise 1.∫ x

0(1 + t + t2)dt =

x + 12x2 + 1

3x3

Exercise 2. 2y + 2y2 + 8y3/3Exercise 3. 2x + 2x2 + 8x3/3− (−1 + 1/2 +−1/3) = 2(x + x2 + 4x3/3) + 5/6

Exercise 4.∫ 1−x

1(1− 2t + 3t2)dt = (t− t2 + t3)

∣∣1−x

1= −2x + 2x2 − x3

Exercise 5.∫ x

−2t4 + t2 = 1

5 t5 + 13 t3

∣∣x−2

= x5

5 + x3

3 + 403

Exercise 6.∫ x2

xt4 + 2t2 + 1 =

(t5

5 + 23 t3 + t

)∣∣∣x2

x= 1

5 (x10 − x5) + 23 (x6 − x3) + x2 − x

Exercise 7.(

23 t3/2 + t

)∣∣x1

= 23 (x3/2 − 1) + (x− 1)

Exercise 8.(

23 t3/2 + 4

5 t5/4)∣∣x2

x= 2

3 (x3 − x3/2) + 45 (x5/2 − x5/4)

Exercise 9. sin t|xiπ = sin x

Exercise 10.(

t2 + sin t

)∣∣x2

0= x2

2 + sin x2

33

Exercise 11.(

12 t + cos t

)∣∣x2

x= x2−x

2 + cosx2 − cosx

Exercise 12.(

13u3 +− 1

3 cos 3u)∣∣x

0= x3

3 +− 13 (cos 3x− 1)

Exercise 13.(

13v3 + cos 3v

−3

)∣∣∣x2

x= x6−x3

3 + −13 (cos 3x2 − cos 3x)

Exercise 14.∫

1−cos 2x2 + x =

(12x− sin 2x

4 + 12x2

)∣∣y0

=y

2− sin 2y

4+

y2

2

Exercise 15.(− cos 2w

2 + 2 sin w2

)∣∣x0

= − (cos 2x− 1)2

+ 2 sinx

2

Exercise 16.∫ x

−π( 12 +cos t)2dt =

∫ x

−π14 +cos t+cos2 t = 1

4 (x+π)+sin x+ 12

(t + sin 2t

2

)∣∣x−π

= 34 (x+π)+sinx+ 1

4 sin 2x

Exercise 17.∫ x

0(t3 − t)dt = 1

3

∫ x√2(t− t3)dt

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

2.

14x4 − 1

2x2 =

13

(12t2 − 1

4t4

)∣∣∣∣x

√2

=16x2 − 1

12x4

=⇒ 13x4 − 2

3x2 = 0 =⇒ x = 0, x =

√2

∫ 1

0(t3 − t)dt +

∫√2

1(t3 − t)dt �cancel� each other out.

Exercise 18. f(x) = x− [x]− 12 if x is not an integer; f(x) = 0 if x ∈ Z.

For any real number, x = q + r, 0 ≤ r < 1, q ∈ Z. So then

x− [x] = r

f(x) = r − 12

(1) To show the periodicity, consider

f(x + 1) = x + 1− [x + 1]− 12

= r − 12

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

x + 1− [x + 1] = r − 12

(2) P (x) =∫ x

0f(t)dt =

∫ x

0(t− 1

2 ) = 12x2 − 1

2x because given 0 < x ≤ 1, then q = 0 for x, so we can use r = t.To show periodicity,

P (x + 1) =∫ x+1

0

f(t)dt =∫ 1

0

f(t)dt +∫ x+1

1

f(t)dt = 0 +∫ x

0

f(t + 1)dt =∫ x

0

f(t)dt = P (x)

since∫ 1

0

f(t)dt =12(x2 − x)

∣∣∣∣1

0

= 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)∫ 1

0

(P (t) + c)dt = 0 =⇒∫ 1

0

P (t)dt = −c

0 ≤ t ≤ 1 so P (t) =12(t2 − t)

=⇒∫ 1

0

P (t)dt =12

(13t3 − 1

2t2

)∣∣∣∣1

0

=12−16

=⇒ c =112

34

(5) Q(x) =∫ x

0(P (t) + c)dt

Q(x + 1) =∫ x+1

0

(P (t) + c)dt =∫ 1

0

(P (t) + c)dt +∫ x+1

1

(P (t) + c)dt =

= 0 +∫ x

0

(P (t + 1) + c)dt =∫ x

0

(P (t) + c)dt = Q(x)

so without loss of generality, consider 0 ≤ x < 1

=⇒ Q(x)∫ x

0

12(t2 − t) +

112

=16x3 − 1

4x2 +

x

12

Exercise 19. g(2n) =∫ 2π

0f(t)dt

Consider ∫ 1

−1

f(t)dt =∫ 1

0

f(t)dt +∫ 0

−1

f(t)dt =∫ 1

0

f(t)dt +1−1

∫ 0

1

f(−1t)dt =

=∫ 1

0

f +∫ 0

1

f(t)dt = 0

Consider that∫ 3

1f(t)dt =

∫ 1

−1f(t + 2)dt =

∫ 1

−1f(t)dt = 0. Then, by induction,

∫ 2n+1

1

f =∫ 2n−1

1

f +∫ 2n+1

2n−1

f(t)dt = 0 +∫ 1

−1

f(t + 2n)dt =∫ 1

−1

f(t)dt = 0

(1)

g(2n) =∫ 1

0

f +∫ 2n−1

1

f +∫ 2n

2n−1

f =∫ 1

0

f +∫ 0

−1

f(t)dt =∫ 1

0

f +−∫ 0

1

f(−t)dt

=∫ 1

0

f +∫ 0

1

f = 0

(2)

g(−x) =∫ −x

0

f = −∫ x

0

f(−t)dt =∫ x

0

f(t)dt = g(x)

g(x + 2) =∫ x+2

0

f(t)dt =∫ 2

0

f +∫ x+2

2

f =∫ x

0

f(t + 2)dt =∫ x

0

f(t)dt = g(x)

Exercise 20.(1) g is odd since

g(−x) =∫ −x

0

f(t)dt = −∫ x

0

f(−t)dt = −∫ x

0

f(t)dt = −g(x)

Now

g(x + 2) =∫ x+2

0

f =∫ 2

0

f +∫ x+2

2

f = g(2) +∫ x

0

f(t + 2)dt = g(2) +∫ x

0

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

=⇒ g(x + 2)− g(x) = g(2)

(2)

g(2) =∫ 2

0

f =∫ 2

1

f +∫ 1

0

f =∫ 2

1

f + A =∫ 0

−1

f(t + 2)dt + A =∫ 0

−1

f(t)dt + A =

−∫ 0

1

f(−t)dt + A = 2A

g(5)− g(3) = g(2)

g(3) = g(2) +∫ 3

2

f(t)dt = 2A +∫ 1

0

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

=⇒ g(5) = 3A + 2A = 5A

(3) The key observation is to see that g must repeat itself by a change of 2 in the argument. To make g(1) = g(3) = g(5),they're different, unless A = 0!

35

Exercise 21. From the given, we can derive

g(x) = f(x + 5), f(x) =∫ x

0

g(t)dt

=⇒ f(5) =∫ 5

0

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

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

g(−x) = f(−x + 5) = g(x) = −f(x− 5)

=⇒ −g(x) = f(x− 5)(2)

∫ 5

0

f(t)dt =∫ 0

−5

f(t + 5)dt =∫ 0

−5

g(t)dt = −∫ −5

0

g(t)dt =∫ 5

0

g(−t)dt =∫ 5

0

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

(3)∫ x

0

f(t)dt =∫ x−5

−5

f(t + 5)dt =∫ x−5

−5

g(t)dt =∫ x−5

0

g +∫ 0

−5

g = f(x− 5) +−∫ −5

0

g(t)dt =

f(x− 5) +∫ 5

0

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

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

3.6 Exercises - Informal description of continuity, The de�nition of the limit of a function, The de�nition of continuityof a function, The basic limit theorems. More examples of continuous functions, Proofs of the basic limit theorems.Polynomials are continuous.Exercise 1. limx→2

1x2 = 1

limx→2 x2 = 14

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

Exercise 3. limx→2(x−2)(x+2)

(x−2) = 4

Exercise 4. limx→1(2x−1)(x−1)

x−1 = 1

Exercise 5. limh→0t2+2th+h2−t2

h = 2t

Exercise 6. limx→0(x−a)(x+a)

(x+a)2 = −1

Exercise 7. lima→0(x−a)(x+a)

(x+a)2 = 1

Exercise 8. limx→a(x−a)(x+a)

(x+a)2 = 0

Exercise 9. limt→0 tan t = limt→0 sin tlimx→0 cos t = 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 = −1Exercise 15. limx→0

2 sin x cos xx = 2

Exercise 16. limx→02 sin x cos xcos 2x sin x = 2

Exercise 17. limx→0sin x cos 4x+sin 4x cos x

sin x = 1 + limx→02 sin 2x cos 2x

sin x = 1 + 2(limx→0

2 sin x cos x cos 2xsin x

)= 5

Exercise 18. limx→05 sin 5x

5x − limx→03 sin 3x

3x = 5− 3 = 2Exercise 19.

limx→0

sin(

x+a2 + x−a

2

)− sin(

x+a2 − (

x−a2

))

x− a=

= limx→0

(sin x+a

2 cos x−a2 + sin x−a

2 cos x+a2 − (

sin x+a2 cos x−a

2 − sin x−a2 cos x+a

2

)

x− a

)=

= limx→a

2 sin x−a2 cos x+a

2

x− a= cos a

36

Exercise 20. limx→02 sin2 x/24(x/2)2 = 1

2

(limx→0

sin x/2x/2

)= 1

2

Exercise 21. limx→01−√1−x2

x2

(1+√

1−x2

1+√

1−x2

)= limx→0

1−(1−x2)

x2(1+√

1−x2)=

12

Exercise 22. b, c are given.sin c = ac + b, a = sin c−b

c , c 6= 0.if c = 0, then b = 0, a ∈ R.Exercise 23. b, c are given.2 cos c = ac2 + b, a = 2 cos c−b

c2 , 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 de�ned.Exercise 26.

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

(2)

| cosx− 1| = | − 2 sin2 x/2| = 2| sinx/2|2 < 2|x2|2 =

|x|22

< 2ε/2 = ε

If we had chosen δ0 =√

2ε for a given ε. |x− 0| < δ =√

2ε.(3)

| sinx(cos h− 1) + cos x sin h| ≤ | sin x|| cosh− 1|+ | cosh|| sin h| < ε

2+

ε

2= ε

| cos x + h− cos x| = | cos x cos h− sin x sin h− cosx| = | cosx(cosh− 1)− sin x sin h| ≤≤ | cos x|| cos h− 1|+ | sin x|| sin h| < ε

2+

ε

2= ε

since ∀ε > 0∃δ1, δ2 > 0 such that | cos h− 1| < ε0; | sin h| < ε whenever |h| < min (δ1, δ2)

Choose δ3 such that if |h| < δ3; | cosh− 1| < ε

2; | sinh| < ε

2Exercise 27. f(x)−A = sin 1

x −A.Let x = 1

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

Consider |x − 0| = |x| = 1nπ ≤ δ(n). Consider ε0 = |1−|A||

2 . Then suppose a δ(n) ≥ |x − 0| but |f(x) − A| > ε0. Thus,contradiction.Exercise 28. Consider x ≤ 1

n , n ∈ Z+, n > M(n) (n is a given constant)

f(x) =[

1x

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

1m

f(x) > M(n)

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

So f(x) →∞ as x → 0+.Consider 1

n ≥ x > 0, n ∈ Z−; −n > M(n).

f(x) =[

1x

]= [n] = n < −M(n)

Since integers are unbounded, we can consider n < A, so that|f(x)−A| > ||f | − |A|| = −n− |A| > M(n)− |A|. Choose n such that M(n)− |A| > 0

Exercise 29.|f −A| = |(−1)[1/x] −A| ≥ ||(−1)[1/x]| − |A|| = |1− |A||

Choose ε < |1 − |A||. Then ∀δ > 0 ( such that |x| < δ ), |f − A| > ε. Thus there's no value for f(0) we could choose tomake this function continuous at 0.Exercise 30. Since

|f(x)| = |x||(−1)[1/x]| = |x|So ∀ε, let δ = ε.

37

Exercise 31. f continuous at x0.Choose some ε0, 0 < ε0 < min (b− x0, x0 − a). Then ∃δ0 = δ(x0, ε0).

Consider ε1 = ε02 and δ1 = δ(x0, ε1)

Consider x1 ∈ (x0 − δ1, x0 + δ1), so that |f(x1)− f(x0)| < ε1.Proceed to construct a δ for x1, some δ(x1; ε0)

|x− x1| = |x− x0 + x0 − x1| < |x− x0|+ |x0 − x1|

Without loss of generality, we can specify x1 such that |x0 − x1| < δ12 . Also, �pick� only the x's such that

|x− x0| < δ1

2< δ1

=⇒ |x− x1| < δ1

2+

δ1

2= δ1

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)| < ε1 + ε1 = ε0

So ∀ε0,∃δ1 for x1. f is continuous at x1 ∈ (a, b). Thus, there must be in�nitely 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 ε = 1

n , |f(x)| = |x sin 1x | = |x|| sin 1/x| < |x|(1).

Let δ = δ(n) = 1n , so that |x| < 1

n .=⇒ |f(x)| < 1

n

Exercise 33.

(1) Consider x0 ∈ [a, b].

Choose some ε0, 0 < ε0 < 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(ε0, x0) = ε0 =⇒ |f(x)− f(x0)| < ε0.

Since we didn't specify x0,∀x0 ∈ [a, b], f is continuous at x0.(2) ∣∣∣∣∣

∫ b

a

f(x)dx− (b− a)f(a)

∣∣∣∣∣ =

∣∣∣∣∣∫ b

a

(f(x)− f(a))dx

∣∣∣∣∣ ≤∫ b

a

|f(x)− f(a)|dx ≤

≤∫ b

a

|x− a|dx = (12x2 − ax)

∣∣∣∣b

a

=12(b− a)(b + a)− a(b− a) =

(b− a)2

2

(3)∣∣∣∣∣∫ b

a

f(x)dx− (b− a)f(c)

∣∣∣∣∣ =

∣∣∣∣∣∫ b

a

(f(x)− f(c))dx

∣∣∣∣∣ ≤∫ b

a

|f(x)− f(c)|dx ≤∫ b

a

|x− c|dx =

=∫ c

a

(c− x)dx +∫ b

c

(x− c)dx = c(c− a)− 12(c− a)(c + a) +

12(b− c)(b + c)− c(b− c) =

=12((c− a)2 + (b− c)2)

Draw a �gure for clear, geometric reasoning.Consider a square of length (b − a) and a 45 − 45 right triangle inside. From the �gure, 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.

12(b− c)(b− a + c− a) =

12(b− c)(b− c + 2(c− a)) >

12(b− c)2

38

Indeed, the trapezoid and c− a right triangle equals the b− a trapezoid since12(b− c)(b− a + c− a) +

12(c− a)2 =

12(b2 − c2 − 2ab + 2ac + c2 − 2ca + a2) =

12(b− a)2

=⇒ 12(b− a)2 >

12(b− c)2 +

12(c− a)2

so then∣∣∣∣∣∫ b

a

f(x)dx− (b− a)f(c)

∣∣∣∣∣ ≤(b− a)2

2

3.8 Exercises - Composite functions and continuity.Exercise 1. f(x) = x2 − 2x; g(x) = x + 1

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

h : R→ RExercise 2. f(x) = x + 1; g(x) = x2 − 2x

h = x2 − 2x + 1h : R→ RExercise 3. f(x) =

√x if x ≥ 0, g(x) = x2

h(x) = |x|h : R→ R+

Exercise 4. f(x) =√

x if x ≥ 0 g(x) = −x2

h unde�ned. D(h) = ∅Exercise 5. f(x) = x2; g(x) =

√x if x ≥ 0

h(x) = x if x ≥ 0h : R+ → R+

Exercise 6. f(x) = −x2; g(x) =√

x if x ≥ 0h = −x if x ≥ 0

h : R+ → R−

Exercise 7. f(x) = sin x; g(x) =√

x if x ≥ 0

h(x) = sin√

x if x ≥ 0

h : R+ → [−1, 1]Exercise 8. f(x) =

√x if x ≥ 0, g(x) = sin x

h(x) =√

sin x if 2πj ≤ x ≤ 2πj + π; j ∈ Z+

h : {x|2πj ≤ x ≤ 2πj + π; j ∈ Z+} → [0, 1]Exercise 9. f(x) =

√x if x ≥ 0, g(x) = x +

√x if x > 0

h(x) =√

x +√

x

h : {x > 0} → {x > 0}Exercise 10. f(x) =

√x +

√x if x > 0; g(x) = x +

√x if x > 0

h(x) =

√x +

√x +

√x +

√x

h : {x > 0} → {x > 0}Exercise 11. limx→−2

x3+8x2−4 = limx→−2

(x+2)(x2−2x+4)(x+2)(x−1) = −3

(used limfg = AB and lim fg = A

B )Exercise 12. limx→4

√1 +

√x =

√3. Used continuity of composite function theorem.

Exercise 13. limt→0sin tan t

sin t = limt→0sin (tan t)

tan t

(1

cos t

)= limt→0

sin tan ttan t limt→0

1cos t = 1.

( we used limfg = AB in the last step; limt→0sin (tan t)

tan t is from continuity of composite functionsl this had been a compositefunction f ◦ g of f = sin x and g = tan x).Exercise 14. limx→π/2

sin cos xcos x = 1 (by continuity of composite functions; this was a composite function f ◦ g of f = sinx

and g = cos x).Exercise 15. limt→π

sin (t−π)t−π = 1 (by continuity of composite functions; this was a composite function f ◦ g of f = sin x

and g = t− π).39

Exercise 16. limx→1sin (x2−1)

x−1 = limx→1sin (x2−1)

x2−1 (x + 1) = 2 (we used continuity of composite functions and product oflimits.Exercise 17. Since 0 < x sin 1

x < x; by squeeze principle, limx→0 x sin 1x = 0

Exercise 18. limx→01−cos 2x

x2 = limx→02 sin2 x

x2 = 2

Exercise 19. limx→0

√1+x−√1−x

x

(√1+x+

√1−x√

1+x+√

1−x

)= 1

Exercise 20. limx→01−√1−4x2

x2

(1+√

1−4x2

1+√

1−4x2

)= 2

Exercise 21. f(x) = x+|x|2 g(x) =

{x for x < 0x2 for x ≥ 0

h(x) =

{x2 x ≥ 00 x < 0

h(x) continuous everywhere.

Exercise 22. f(x) =

{1 if |x| ≤ 10 if |x| > 1

g(x) =

{2− x2 if |x| ≤ 22 if |x| > 2

h(x) =

{0 if |x| ≤ 1,

√3 ≤ |x|

1 if 1 ≤ |x| ≤ √3

h cont. for |x| < 1,√

3 < |x|, 1 < |x| < √3

Exercise 23. h(x) = g(f(x))

h(x) =

{0 for x < 0x2 for x ≥ 0

h(x) continuous everywhere.

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) = 0Strategy: �nd 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 > 0c− 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 < 0c + 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.∀ε > 0, ∃δ > 0 s.t. f(c)− ε < f(x) < f(c) + ε if c− δ < x < c + δ (by continuity).Choose δ for ε = f(c)

2 . Thenf(c)2

< f(x) <3f(c)

2∀x ∈ (c− δ, c + δ)

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 − kg(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→∞ ckxk

ck−1xk−1 = limx→∞ ck

ck−1x = ∞ ∃M > 0 such that |cnMn| > |∑n−1

k=0 ckMk. So then

f(M) = cnMn +n−1∑

k=0

ckMk ≶ cn

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 functionchanged 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 (12 ) = 1

8 , P ( 32 ) = − 11

8 , P (2) = 2(3) If P (x) = x4 + 4x3 + x2 − 6x + 2, P (−3) = 2, P (− 5

2 ) = − 316 , P (−2) = 2, P ( 1

3 ) = 2281 , P ( 1

2 ) = − 316 , P ( 2

3 ) =− 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 M2j+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 thatg(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 csuch that g(c) = 0, so that f(c) = c.

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

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

Exercise 4. D = R, g(y) = y1/3

Exercise 5. D = R,

g(y) =

y if y < 1√y if 1 ≤ y ≤ 16(

y8

)2 if y > 16

Exercise 6. f(Mf ) = f(f−1(

1n

∑ni=1 f(ai)

)) = 1

n

∑ni=1 f(ai)

Exercise 7. f(a1) ≶ 1n

∑ni=1 f(ai) ≶ f(an). Since f is strictly monotonic.

41

g preserves monotonicity.=⇒ a1 ≶ Mf ≶ an

Exercise 8. h(x) = af(x) + b, a 6= 0

Mh = H

(1n

n∑

i=1

h(ai)

)= H

(1n

n∑

i=1

(af(ai) + b)

)= H

(a

1n

n∑

i=1

f(ai) + b

)

The inverse for h is g(

h−ba

)= H(h) = h−1. So then

Mh = g

(1n

n∑

i=1

f(ai)

)= Mf

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 func-tions (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

andR b

af(x)g(x)dxR ba

g(x)dx= f(c)

Exercise 1.

g = x9 > 0 for x ∈ [0, 1]; f =1√

1 + xm =

1√2,M = 1

∫ 1

0

x9 =110

x10

∣∣∣∣1

0

=110

110√

2≤

∫ 1

0

x9

√1 + x

dx ≤ 110

Exercise 2.√

1− x2 =1− x2

√1− x2

. f =1√

1− x2g = (1− x2)M =

2√3,m = 1

∫ 1/2

0

(1− x2)dx = (x− 13x3)

∣∣∣∣1/2

0

=1124

1124≤

∫ 1/2

0

√1− x2dx ≤ 11

24

√43

Exercise 3.

f =1

1 + x6g = 1− x2 + x4

∫ a

0

1− x2 + x4 =(

x− 13x3 +

15x5

)∣∣∣∣a

0

= a− a3

3+

a5

5

m =1

1 + a6M = 1

11 + a6

(a− a3

3+

a5

5

)≤

∫ a

0

11 + x2

dx ≤(

a− a3

3+

a5

5

)

So if a = 110 , (a− a3/3 + a5/5) = a− 0.333 . . . a3 + 0.2a5 = 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.

∫sin t2dt =

∫ (12t

)(2t sin t2)dt =

12c

(− cos t2)∣∣√

(n+1)π√nπ

=

=−12c

((−1)n+1 − (−1)n) =1c(−1)n

Exercise 6.∫ b

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

∫ b

a1 = f(c)(b − a). Then f(c) =

R ba

f

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

42

Exercise 7. f nonnegative. Consider f at a point of continuity c, and suppose f(c) > 0. Then 12f(c) > 0.

|f(x)− f(c)| < ε =⇒ f(c)− ε < f(x) < f(c) + ε

Let ε =12f(c) ∃δ > 0 for ε =

12f(c)

∫ c+δ

c−δ

f(x)dx >12f(c)(2δ) = f(c)δ > 0

But∫ b

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

Exercise 8.m

∫g ≤

∫fg ≤ M

∫g =⇒ m

∫g ≤ 0 ≤ M

∫g ∀g

m ≤ 0 ≤ M for∫

g = 1 but also

−m ≤ 0 ≤ −M =⇒ m ≥ 0 M ≤ 0 for∫

g = −1

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 deriva-tives, The algebra of derivatives.Exercise 1. f ′ = 1− 2x, f ′(0) = 1, f ′(1/2) = 0, f ′(1) = −1, f ′(10) = −19

Exercise 2. f ′ = x2 + x− 2(1) f ′ = 0, x = 1,−2(2) f ′(x) = −2, x = 0,−1(3) f ′ = 10, x = −4, 3

Exercise 3. f ′ = 2x + 3Exercise 4. f ′ = 4x3 + cos x

Exercise 5. f ′ = 4x3 sin x + x4 cosx

Exercise 6. f ′ = −1(x+1)2

Exercise 7. f ′ = −1(x2+1)2 (2x) + 5x4 cos x + x5(− sin x)

Exercise 8. f ′ = x−1−(x)(x−1)2 = −1

(x−1)2

Exercise 9. f ′ = −1(2+cos x)2 (− sin x) = sin x

(2+cos x)2

Exercise 10.(2x + 3)(x4 + x2 + 1)− (4x3 + 2x)(x2 + 3x + 2)

(x4 + x2 + 1)2=−2x5 − 9x4 + 12x3 − 3x2 − 2x + 3

(x4 + x2 + 1)2

Exercise 11.f ′ =

(− cos x)(2− cos x)− (sin x)(2− sin x)(2− cosx)2

=−2 cos x− 2 sin x + 1

(2− cos x)2

Exercise 12.f ′ =

(sinx + x cos x)(1 + x2)− 2x(x sin x)(1 + x2)2

=sin x + x cos x + x3 cos x− x2 sin x

(1 + x2)2

Exercise 13.(1)

f(t + h)− f(t)h

=v0h− 32th− 16h2

h= v0 + 32t− 16h

f ′(t) = v0 − 32t

(2) t = v032

(3) −v0

(4) T = v016 , v0 = 16 for 1sec. v0 = 160 for 10sec. v0

16 for Tsec.(5) f ′′ = −32(6) h = −20t2

Exercise 14. V = s3, dVdS = 3s2

Exercise 15.(1) dA

dr = 2πr = C43

(2) dVdr = 4πr2 = A

Exercise 16. f ′ = 12√

x

Exercise 17. f ′ = −1(1+

√x)2

(1

2√

x

)

Exercise 18. f ′ = 32x1/2

Exercise 19. −32 x−5/2

Exercise 20. f ′ = 12x−1/2 + 1

3x−2/3 + 14x−3/4 x > 0

Exercise 21. f ′ = − 12x−3/2 +− 1

3x−4/3 − 14x−5/4

Exercise 22. f ′ =12 x−1/2(1+x)−√x

(1+x)2 = 12√

x(1+x)2

Exercise 23. f ′ =(1+

√x)−x 1

21√x

(1+√

x)2= 1+ 1

2√

x

(1+√

x)2

Exercise 24.g = f1f2

g′ = f ′1f2 + f1f′2

g′

g=

f ′1f1

+f ′2f2

g = f1f2 . . . fnfn+1

g′ = (f1f2 . . . fn)′fn+1 + (f1f2 . . . fn)f ′n+1;

g′

g=

(f1f2 . . . fn)′

f1f2 . . . fn+

f ′n+1

fn+1

=f ′1f1

+f ′2f2

+ · · ·+ f ′nfn

+f ′n+1

fn+1

Exercise 25.(tan x)′ =

(cosx

sin x

)′=

cos2 x− (− sinx) sin x

cos2 x= sec2 x

(cot x)′ =(cos x

sin x

)′=− sin x sin x− cos x cosx

sin2 x= − csc2 x

(sec x)′ =−1

cos2 x(− sin x) = tanx sec x

(csc x)′ =−1

sin2 xcosx = − cot x csc x

Exercise 35.f ′ =

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

=

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

(sinx + cos x)2

Exercise 36.

f ′ = a sin x + (ax + b) cos x + c cosx + (cx + d)(− sinx) = ax cosx + (b + c) cos x + (a− d) sin x− cx sin x

So then a = 1, d = 1, b = d, c = 0.Exercise 37.

g′ = (2ax + b) sin x + (ax2 + bx + c) cos x + (2dx + e) cos x + (dx2 + ex + f)(− sinx) =

= ax2 cos x− dx2 sinx + (2a− e)x sin x + (b + 2d)x cosx + (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)

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

(x− 1)2

=(n + 1)(xn+1 − xn)− xn+1 + 1

(x− 1)2=

nxn+1 − (n1)xn + 1(x− 1)2

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

(x− 1)2

44

(2)(x + 2x2 + · · ·+ nxn)′ = (1 + 22x1 + · · ·+ n2xn−1) =

=(n(n + 2)xn+1 − (n + 1)2xn + 1)(x− 1)2 − 2(x− 1)(nxn+2 − (n + 1)xn+1 + x)

(x− 1)4

x + 22x2 + · · ·+ n2xn =(n(n + 2)xn+2 − (n + 1)2xn+1 + x)(x− 1)− 2(nxn+3 − (n + 1)xn+2 + x2)

(x− 1)3=

=n2xn+3 + (−2n2 − 2n + 1)xn+2 + (n + 1)2xn+1 − x2 − x

(x− 1)3

Exercise 39.f(x + h)− f(x)

h=

(x + h)n − xn

h

(x + h)n =n∑

j=0

(n

j

)xn−jhj

(x + h)n − xn

h=

∑nj=1

(nj

)xn−jhj

h=

n∑

j=1

xn−jhj−1

(n

j

)

limh→0

(x + h)n − xn

h=

(n

1

)xn−1 = nxn−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) = x21 + ax1 + b

f(x2) = x22 + ax2 + b

f(x2)− f(x1)x2 − x1

=x2

2 − x21 + a(x2 − x1)x2 − x1

= x2 + x1 + a

(2)f ′ = 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′ = 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 ′ = 1− 3x2.

f ′(−1) = −2 =⇒ y = −2x− 2

For the other line,f ′(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

=⇒ −2a3 + 1− 3a2 = 0 = a3 − 12

+32a2

The answer could probably be guessed at, but let's review some tricks for solving cubics.45

First, do a translation in the x direction to center the origin on the point of in�ection. Find the point of in�ection by takingthe second derivative.

f ′′ = 6a + 3 =⇒ a = −12

So

a = x− 12

=⇒ (x− 12)3 +

32(x− 1

2)2 − 1

2= x3 =

34x− 1

4= 0

Then recall this neat trigonometric fact:

cos 3x = cos 2x cos x− sin 2x sin x = 4 cos3 x− 3 cos x

=⇒ cos3 x =34

cos x− cos 3x

4= 0

Particularly for this problem, we have cos 3x = 1. So x = 0, 2π/3, 4π/3. cos x = 1,− 12 . 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

f

(12

)=

38

y(x) =(

14x

)+

14

Exercise 9.

f(x) =

{x2 if x ≤ c

ax + b if x > c

f ′(x) =

{2x if x ≤ c

a if x > c

a = 2c; b = −c2

Exercise 10.

f(x) =

{1|x| if |x| > c

a + bx2 if |x| ≤ c

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

f ′(x) =

− 1x2 if x > c

1x2 if x < c

2bx if |x| ≤ c

So b = − 12c3

, a =32c

.

Exercise 11.

f ′ =

{cos x if x ≤ c

a if a > c

46

Exercise 12. f(x) =(

1−√21+√

2

)= 1−A

1+A

A =√

xA′ = a =12x−1/2 =

12A

; A′′ = −14x−3/2 = − 1

4A3

f ′ =−A′(1 + A)−A′(1−A)

(1 + A)2=

−2A′

(1 + A)2=

−1√x(1 +

√x)2

f ′′ =1

(A(1 + A2))2(A′(1 + A)2 + A(2)(1 + A)A′) =

3√

x + 12x3/2(1 +

√x)3

f ′′′ =12

( −1A2 A′(A2(1 + A)3)− (2AA′(1 + A)3 + 3A2(1 + A)2A′)(3 + 1

A )(A2(1 + A)3)2

)

=−34

( 1A + 4 + 5A

A4(1 + A4)

)= −3

4(1 + 4

√x + 5x)√

x(x +√

x)4

Exercise 13.P = ax3 + bx2 + cx + d

P ′ = 3ax2 + 2bx + c

P ′′ = 6ax + 2b

P ′′(0) = 2b = 10 =⇒ b = 5

P ′(0) = c = −1 P (0) = d = −2

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

Exercise 14.

fg = 2,f ′

g′= 2

f ′

g= 4

g′

g= 2 f =

12, g = 4

(1)

h′ =f ′g − g′f

g2=

f ′

g− g′

g

f

g= 4− 2(

18) =

154

(2)k′ = f ′g + fg′ = 4g2 + f2g = 64 + 4 = 68

(3)

limx→0

g′(x)f ′(x)

=limx→0 g′(x)limx→0 f ′(x)

=12

Exercise 15.(1) True, by de�nition of f ′(a).(2)

limh→0

f(a)− f(a− h)h

= − limh→0

f(a− h)− f(a)h

= lim−h→0

f(a− h)− f(a)−h

= f ′(a)

True, by de�nition of f ′(a).(3)

limt→0

f(a + 2t)− f(a)t

= 2 lim2t→0

f(a + 2t)− f(a)2t

= 2f ′(a)

False.(4)

limt→0

f(a + 2t)− f(a) + f(a)− f(a + t)2t

=

lim2t→0

f(a + 2t)− f(a)2t

+−12

limt→0

f(a + t)− f(a)t

=

f ′(a)− 12f ′(a) =

12f ′(a)

False.

Exercise 16.47

(1)

D∗(f + g) = limh→0

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

h= lim

h→0

(F + G)2 − (f + g)2

h=

= D∗f + D∗g + limh→0

2FG− 2fg

h

limh→0

2FG− 2fg

h= lim

h→0

(2(FG)− 2fG)(F + f)(F + f)h

+(2fG− 2fg)(G + g)

(g + G)h=

= limh→0

2g

F + hlimh→0

F 2 − f2

h+ lim

h→0

2f

G + glimh→0

G2 − g2

h=

=g

fD∗f +

f

gD∗g

D∗(f − g) = limh→0

(f(x + h)− g(x + h))2 − (f(x)− g(x))2

h=

= limh→0

(F −G)2 − (f − g)2

h=

= D∗f + D∗g +− limh→0

2FG− 2fg

h

= D∗f + D∗g − g

fD∗f +

f

gD∗g

D∗(fg) = limh→0

((fg)(x + h))2 − ((fg)(x))2

h=

= limh→0

(f2(x + h))(g2(x + h))− f2(x)g2(x + h) + (g2(x + h)− g2(x))f2(x)h

=

= g2D∗f + f2D∗g

D∗(f/g) = limh→0

f2(x+h)g2(x+h) − f2(x)

g2(x)

h= lim

h→0

f2(x+h)−f2(x)g2(x+h) + f2(x)

g2(x+h) − f2(x)g2(x)

h=

=D∗fg2

+f2

g4(−D∗g) when g(x) 6= 0

(2)(3)

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

Exercise 2. x√1+x2

Exercise 3. −2x cosx2 + 2x(x2 − 2) sin x2 + 2 sin x3 + 6x3 cos x3

Exercise 4.f ′ = cos (cos2 x)(−2 cos x sinx) cos (sin2 x) + sin (cos2 x) sin (sin2 x)(2 sin x cos x) =

= − sin 2x(cos (cos 2x))

Exercise 5.f ′ = n sinn−1 x cos x cos nx +−n sinnx sinn x

Exercise 6.f ′ = cos (sin (sin x))(cos (sin x))(cos x)

Exercise 7.

f ′ =2 sinx cosx sin x2 − 2x cosx2 sin2 x

sin2 x2=

sin 2x sin x2 − 2x sin2 x cosx2

sin2 x2

Exercise 8. f ′ = 12 sec2 x

2 + 12 csc2 x

2

Exercise 9. f ′ = 2 sec2 x tan x +−2 csc2 x cot x

Exercise 10. f ′ =√

1 + x2 + x2√1+x2 = 1+2x2√

1+x2

Exercise 11. f ′ = 4(4−x2)3/2

48

Exercise 12.

f ′ =13

(1 + x3

1− x3

)−2/3 (3x2(2)

(1− x3)2

)=

2x2

(1− x3)2

(1 + x3

1− x3

)−2/3

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

f(x) =1√

1 + x2(x +√

1 + x2)=

1√1 + x2(x +

√1 + x2)

(x−√1 + x2

x−√1 + x2

)=

=x−√1 + x2

−√1 + x2= 1− x√

1 + x2

f ′ =

√1 + x2 − x2√

1+x2

1 + x2=

1(1 + x2)3/2

Exercise 14.12(x +

√x +

√x)−1/2(1 +

12(x +

√x)−1/2

(1 +

12√

x

))

Exercise 15.

f ′ = (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)−2/3x2

Exercise 16.

f ′ =−1(

1 + 1x

)2

(−1x2

)=

1(x + 1)2

g′ =−1(

1 + 1f

)2

(−1f2

)f ′ =

f ′

(f + 1)2

g′ =(x + 1)−2

(x

x+1 + 1)2 =

1(2x + 1)2

Exercise 17. h′ = f ′g′

x h h′ k k′

0 f(2) = 0 2(−5) = −10 g(1) = 0 1(5) = 51 f(0) = 1 5(1) = 5 g(3) = 1 −6(−2) = 122 f(3) = 2 4(1) = 4 g(0) = 2 −5(2) = −103 f(1) = 3 −2(−6) = 12 g(2) = 3 1(4) = 4

Exercise 18.g(x) = xf(x2)

g′(x) = f(x2) + x(2x)f ′(x2) = f(x2) + 2x2f ′(x2)

g′′(x) = 2xf ′(x2) + 4xf ′(x2) + 2x2(2x)f ′′(x2) = 6xf ′(x2) + 4x3f ′′(x2)

x g(x) g′(x) g′′(x)0 0 0 01 1 3 102 12 6 + 8(3) = 30 12(3) + 32(0) = 36

Exercise 19.(1)

g′ =df(x2)dx2

2x = 2xf ′

(2)g′ = 2 sin x cos xf ′ − 2 cos x sin xf ′ = (sin 2x)(f ′(sin2 x)− f ′(cos2 x))

(3)

g′ =df(f(x))d(f(x))

f ′

(4)

g′ =df(f(f(x)))d(f(f(x)))

d(f(f(x)))d(f(x))

df

dx

49

Exercise 20. V = s3, s = s(t) dVdt = 3s2 ds

dt .s = 5cm 75cm3/sec

s = 10cm 300cm3/sec

s = xcm 3x2cm3/sec

Exercise 21.

l =√

x2 + h2dl

dt=

1lx

dx

dtdx

dt=

l

x

dl

dt=

10mi

−√102 − 82(−4mi/sec) =

203

mi

sec

(3600sec

1hr

)

Exercise 22.l2 = x2 + s2

2ldl

dt= 2x

dx

dt

dl

dt=

x

l

dx

dtdl

dt

(x =

s

2

)= 20

√5

dl

dt(x = s) = 50

√2

Exercise 23.dl

dt=

x

l

dx

dt=

3512 =

365

mi/hr

Exercise 24. Given the preliminary informationr

h=

25

= α, V =13πr2h =

13πα2h3

(1)

V =πr2

h2(h2y − hy2 +

13y3)

dV

dt=

πr2

h2(h2 − 2hy + y2)

dy

dtdy

dt=

h2

πr2

(1

h2 − 2hy + y2

)dV

dt=

102

π42

(1

102 − 2(10)5 + 25

)5 =

54π

(2)dV

dt= πα2h2 dh

dt,

dh

dt=

1πα2h2

dV

dt=

54π

Exercise 25.

α =r

h=

32

dV

dt= πα2h2 dh

dt

c− 1π94(22)4 = 36π =⇒ c = 36π + 1

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

rdr

dt= (R− h)

dh

dt

V =∫

πr2dh =⇒ dV

dh= πr2 = π(2Rh− h2)

=⇒ dV

dh= π(2(10(5)− 25)) = 50π

50

dV

dt=

dV

dh

dh

dt, =⇒ dh

dt=

dV

dt

(1

π(2Rh− h2)

)

(r

R− h

)dr

dt=

dV

dt

(1

π(2Rh− h2)

)

=⇒dr

dt=

dV

dt

(R− h

rπ(2Rh− h2)

)=

= (5√

3)(

10− 5π(2(10)5− 25)3/2

)=

115π

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

(72

)= 8.

A =12

√(y − 1)

367

y

dA

dt=

12

(√36y

12

1√y − 1

y +

√(y − 1)

367

)dy

dt=

=12

(6

2(7)8 + 6

)(2) =

667

Exercise 28. From the given information, h = 3r + 3. The volume formula is V = πR2

3 H . So thenV = π/3r2(3r + 3) = πr3 + πr2

dV

dr= πr(3r + 2)

dr

dtWith the given information, we get

dr

dt=

1π(6)(20)

Using this, we can plug this back in for the different case:dV

dt= n = π(36)(110)/(120π) = 33

Exercise 29.(1) dy

dt = 2xdxdt ; when x = 1

2 , y = 14 , dy

dt = dxdt

(2) t =π

6Exercise 30.

(1) 3x2 + 3y2y′ = 0 =⇒ x2 + y2y′ = 0(2)

2x + 2yy′2 + y2y′′ = 0 =⇒ y2y′′ = −2(x + yy′2)

=⇒ y′′ = −2(

xy4 + yx4

y6

)= −2xy−5

Exercise 31.12

1√x

+1

2√

yy′ = 0 y′ =

−√y√x

< 0

Exercise 32.

±√

12− 3x2

4

6x + 8yy′ = 0 =⇒ y′ =−3x

4y

3 + 4(y′2 + yy′′) = 0

y′′ =(−3

4− y′2

)1y

=−94y3

Exercise 33.sin xy + x cos2 xy(y + xy′) + 4x = 0

y′x2 cosxy + xy cos xy + sin xy + 4x = 051

Exercise 34. y = x4. yn = xm.

yn = xm, y′nyn−1 = mxm−1; y′ =mxm−1

nyn−1=

m

n

xm−1

xm(1−1/n)=

y′ =m

nxm/n−1

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:

Theorem 13 (Theorem 4.3).Let f be de�ned on I .Assume f has a rel. extrema at an int. pt. c ∈ I .If ∃ f ′(c), f ′(c) = 0; the converse is not true.

Proof. Q(x) = f(x)−f(c)x−c if x 6= c, Q(c) = f ′(c)

∃f ′(c), so Q(x) → Q(c) as x → c so Q is continuous at c.If Q(c) > 0, f(x)−f(c)

x−c > 0. For x− c ≷ 0, f(x) ≷ f(c), thus contradicting the rel. max or rel. min. (no neighborhoodabout c exists for one!)If Q(c) < 0, f(x)−f(c)

x−c < 0. For x− c ≷ 0, f(x) ≶ f(c), thus contradicting the rel. max or rel. min. (no neighborhoodabout 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 ′(x) ∀x ∈ (a, b) and let

f(a) = f(b)

then ∃ at least one c ∈ (a, b), such that f ′(c) = 0.

Proof. Suppose f ′(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 ′(x) 6= 0 ¤

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

(6) f(b)− f(a) = f ′(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 h′(c) = 0 = f ′(c)(b− a)− (f(b)− f(a))

¤

Theorem 16 (Cauchy's Mean-Value Formula). Let f, g cont. on [a, b], ∃f ′, g′ ∀x ∈ (a, b)Then ∃ c ∈ (a, b). x

(7) f ′(c)(g(b)− g(a)) = g′(c)(f(b)− f(a)) (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))

=⇒ h′(c) = f ′(c)(g(b)− g(a))− g′(c)(f(b)− f(a)) = 0 (by Rolle's Thm.)¤

52

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)b− a

=A(b− a)(b + a) + B(b− a)

b− a= A(b + a) + B

y′ = 2Ax + B

y′(

a + b

2

)= A(a + b) + B

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 isIf @ at least one c ∈ (a, b) s.t. f ′(c) = 0,

then f(a) 6= f(b).

g′ = 3x2 − 3 = 3(x2 − 1) =⇒ g′(±1) = 0

Suppose g(B) = 0, B ∈ (−1, 1)

then ∀x ∈ (−1, 1), x 6= B, 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) = 1x if x ≥ 1.

(1) See sketch.(2)

f(x) =

{3−x2

2 if x ≤ 11/x if x ≥ 1

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

f ′(x) =

{−x; f ′(1) = −1 for x ≤ 1−1/x2; f ′(1) = −1 for x > 1

Then f(x) is cont. and diff. on [0, 2].

For 0 ≤ a < b ≤ 13−b2

2 −(

3−a2

2

)

b− a=−(a + b)

2= −c

Note that −1 ≤ f ′ ≤ 0 for 0 ≤ x ≤ 1For 1 ≤ a < b ≤ 2

1b − 1

a

b− a=−1ab

=−1c2

=⇒ c =√

ab

Note that −1 ≤ f ′ ≤ −1/4For 0 ≤ a ≤ 1 , 1 ≤ b ≤ 2

1b −

(3−a2

2

)

b− a=

2− (3− a2)b2b(b− a)

= −c or −1c2

depending upon if 0 ≤ c ≤ 1 or 1 ≤ c ≤ 2, respectively

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

f ′ =−23

x−1/3 6= 0 for |x| ≤ 1

This is possible since f is not differentiable at x = 0.Exercise 5. x2 = x sin x + cosx. g = xS + C − x2. g′ = 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. Since g(0) = 1 and for x → ±∞, g → ∓∞, then we could concludethat g must become zero between 0 and ∞ and −∞ and 0.

53

Exercise 6.f(b)− f(a)

b− a= f ′(c)

b = x + h

b− a = h

a = x

x < x + θh < x + h

=⇒ f(x + h)− f(x) = hf ′(x + θh)

(1) f(x) = x2, f ′ = 2x.(x + h)2 − x2 = 2xh + h2 = h(2(x + θh))

2x + h

2− x = θh =⇒ θ =

12

so then limh→0

θ =12

(2) f(x) = x3, f ′ = 3x2.

(x + h)3 − x3 = 3x2h + 3xh2 + h3 = h3(x + θh)2 =⇒(√

3x2 + 3xh + h2

3− x

)/h = θ

θ =

√3x2 + 3xh + h2

3h2− x

h=

√x2 + xh + h2

3 − x

h

√x2 + xh + h2

3 + x√

x2 + xh + h2

3 + x

=

=x + h

3

x +√

x2 + hx + h2

3

=⇒ limh→0

θ =12

Notice the trick of multiplying by the conjugate on top and bottom to get a way to evaluate the limit.Exercise 7. f(x) = (x− a1)(x− a2) . . . (x− ar)g(x).

(1) a1 < a2 < · · · < ar.Since f(a1) = f(a2) = 0. f ′(c) = 0 for c1 ∈ (a1, a2).Consider that f(a2) = f(a3) = 0 as well as f ′(c2) = 0 for c ∈ (a2, a3).Indeed, since f(aj) = f(aj+1) = 0, f ′(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 de�nite).Exercise 8. Using the mean value theorem

(1)sin x− sin y

x− y= cos c =⇒

∣∣∣∣sin x− sin y

x− y

∣∣∣∣ = | cos c| ≤ 1

=⇒ | sin x− sin y| ≤ |x− y|(2) x ≥ y > 0.

f(z) = zn is monotonically increasing for n ∈ Z.By mean-value theorem,

xn − yn

x− y= ncn−1 for y < c < x

Since 0 < y < c < x; nyn−1 ≤ xn−yn

x−y ≤ nxn−1.54

Exercise 9. Let g(x) =(

f(b)−f(a)b−a

)x +

(bf(a)−af(b)

b−a

).

f − g = hh(a) = h(c)

h(c) = h(b)

so ∃c1 ∈ (a, c), c2 ∈ (c, b) s. t. h′(c1) = h′(c2) = 0 by Rolle's Thm.Let h′ = H

since H(c1) = H(c2) = 0 and H is cont. diff. on (c1, c2). then∃c3 ∈ (c1, c2) s.t. H ′(c3) = h′′(c3) = 0

Now h′′ = (f − g)′′ = f ′′ so f ′′(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) =f(x)− f(a)

x− aif x 6= a; g(a) = f ′(a)

(1) g =(

1x−a

)f − 1

x−af(a). f is cont. on (a, b] since ∃ f ′ ∀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. functionsto get cont. functions because the rules for taking limits allow so).

g is cont. at a since limx→a g = limx→af(x)−f(a)

x−a = f ′(a).

By mean value theorem,(f(x)− f(a)

x− a

)= f ′(c) = g(x) ∀c ∈ (a, x) ∀x ∈ (a, b]

Then ∀c ∈ [a, b], f ′(c) ranges from f ′(a) to g(b) since f ′(c) = g(x) so whatever g(x) ranges from and to, so doesf ′(c).

(2) Let h(x) = f(x)−f(b)x−b if x 6= b; h(b) = f ′(b).

h is cont. on [a, b) since 1x−b is cont., f(x) is cont.

limx→b h = limx→bf(x)−f(b)

x−b = f ′(b) so h is cont. at b.

h is cont. on [a, b] → h takes all values from h(a) to f ′(b) on [a, b] (by intermediate value theorem).

By mean value theorem,

h(x) =f(b)− f(x)

b− x= f ′(c2) for c2 ∈ (x, b) ∀x ∈ [a, b]

So then f ′ ranges from h(a) to f ′(b) just like h.

h(a) = g(b). So then f ′ must range from f ′(a) to f ′(b)

4.19 Exercises - Applications of the mean-value theorem to geometric properties of functions, Second-derivative testfor extrema, Curve sketching.

Exercise 1. f(x) = x2 − 3x + 2(1) f ′(x) = 2x− 3 x0 = 3

2 .(2) f ′(x) ≷ 0 for x ≷ 3

2(3) f ′′ = 2 > 0 for ∀x ∈ R(4) See sketch.

Exercise 2. f(x) = x3 − 4x

(1) f ′ = 3x2 − 4 xc = ± 2√3

(2) f ′ ≷ 0 when |x| ≷ 2√3

(3) f ′′ = 6x f ′′ ≷ 0 when x ≷ 0(4) See sketch.

Exercise 3. f(x) = (x− 1)2(x + 2)(1) f ′ = 3(x− 1)(x + 1) f ′(x) = 0 when x = ±1

55

(2) f ′ ≷ when |x| ≷ 1(3) f ′′ = 3(2x) = 6x f ′′ ≷ 0 when x ≷ 0(4) See sketch.

Exercise 4. f(x) = x3 − 6x2 + 9x + 5

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

f ′(x) > 0 when x < 1, x > 3

f ′(x) < 0 when 1 < x < 3

(3) f ′′ = 6x− 12 = 6(x− 2) f ′′ ≷ 0 when x ≷ 2(4) See sketch.

Exercise 5. f(x) = 2 + (x− 1)4

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

Exercise 6. f(x) = 1/x2

(1) f ′ = −2x3 f ′(x) = 0 for no x

(2) f ′ ≷ 0 when x ≶ 0(3) f ′′ = 6

x4 > 0 ∀x 6= 0(4) See sketch.

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

(1) f ′ = 1 + −2x3 f ′(x) = 0 = 1− 2

x3 =⇒ xc = 21/3

(2)f ′(x) > 0 when x < 0, 0 < x < 21/3

f ′(x) < 0 when x > 21/3

(3) f ′′ = 6x4 > 0 ∀x 6= 0

(4) See sketch.

Exercise 8. f(x) = 1(x−1)(x−3)

(1) f ′ = −1(x−1)2(x−3)2 ((x− 3) + x− 1) = (−2)(x−2)

(x−1)2(x−3)2

f ′(x) = 0 when x = 2(2) f ′ ≷ 0 when x ≶ 2(3)

f ′′ = (−2)(

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

)=

= (6)(

x2 − 4x + 133

(x− 1)3(x− 3)3

)

x2 − 4x +133

> 0 since 144− 4(−3)(−13) = 144 + 12(−13) < 0 so

f ′′ > 0 if x > 3, x < 1

f ′′ < 0 if 1 < x < 3

(4) See sketch.

Exercise 9. f(x) = x/(1 + x2)

(1)

f ′ =(1 + x2)− x(2x)

(1 + x2)2=

1− x2

(1 + x2)2

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

56

(3)

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

(1 + x2)4=

2x(x2 − 3)(1 + x2)3

f ′′ > 0 when x >√

3

f ′′ < 0 when 0 < x <√

3

f ′′ > 0 when −√

3 < x < 0

f ′′ < 0 when x < −√

3(4) See sketch.

Exercise 10. f(x) = (x2 − 4)/(x2 − 9)(1)

f ′ =2x(x2 − 9)− (x2 − 4)(2x)

(x2 − 9)2=

−10x

(x2 − 9)2

f ′(0) = 0(2) f ′ ≷ 0 when x ≶ 0, x 6= ±3(3)

f ′′ = (−10)(

(x2 − 9)2 − 2(x2 − 9)(2x)x(x2 − 9)2

)= (30)

(x2 + 3)(x2 − 9)3

f ′′ ≷ 0 when |x| ≷ 3(4) See the sketch.

Exercise 11. f(x) = sin2 x

(1) f ′ = sin 2x So then f ′ = 0 when x = π2 n

(2)

f ′ > 0 when0 < x <

π

2πn < x <

π

2+ πn

f ′ < 0 when π

2+ πn < x < π(n + 1)

(3)

f ′′ = 2 cos 2xf ′′ > 0 when −π

4+ πn < x <

π

4+ πn

f ′′ < 0 when π

4+ πn < x <

3π

4+ πn

(4) See sketch.Exercise 12. f(x) = x− sin x

(1) f ′ = 1− cosx f ′ = 0 when x = 2πn(2) f ′ > 0 if x 6= 2πn(3)

f ′′ = sin xf ′′ > 0 when 2πn < x < 2πn + π

f ′′ < 0 when 2πn + π < x < 2π(n + 1)(4) See sketch.

Exercise 13. f(x) = x + cos x

(1) f ′ = 1− sin x x = π2 + 2πn f ′(x) = 0

(2) f ′ > 0 if x 6= π2 + 2πn

(3)

f ′′ = − cosxf ′′ > 0 when −π

2+ 2πn < x <

π

2+ 2πn

f ′′ < 0 when π

2+ 2πn < x <

3π

2+ 2πn

(4) See sketch.Exercise 14. f(x) = 1

6x2 + 112 cos 2x

(1) f ′ = 13x + − sin 2x

6 f ′(0) = 0(2) f ′ ≷ 0 when x ≷ 0

57

(3) f ′′ = 13 − cos 2x

3 = 1−cos 2x3

x = πn for f ′′ = 0. Otherwise f ′′ > 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)

P ′ = 2(1− A

x2) = 0

x =√

A

P ′′ =4A

x3> 0 for x > 0 so x =

√A minimizes P

Exercise 2.A = xy L = 2x + y

A = x(L− 2x) = Lx− 2x2 =⇒ dA

dx= L− 4x = 0 when x =

L

4y =

L

2

A′′ = −4 so x =L

4maximizes A

Exercise 3.

A = xy L = 2x + y = 2x +A

x

dL

dx= 2 +

−A

x2= 0 when x =

√A√2

y =√

2√

A

L′′ =2A

x3> 0 for x =

√A

2so x minimizes L

Exercise 4. f = x2 + y2 = x2 + (S − x)2

f ′ = 2x + 2(S − x)(−1) = −2S + 4x =⇒ x = S2

f ′′ = 4 > 0 so x = S2 minimizes f

Exercise 5. x2 + y2 = R > 0f = x + y

f ′ = 1 + y′ = 0 = 1 +−x

y= 0 =⇒ y = x

f ′′ = y′′ =−1− y′2

yfor y > 0, f ′′ < 0 so that f is max. when y = x

Note that

2x + 2yy′ = 0

y′ =−x

y

x + yy′ = 0

1 + y′2 + yy′′ = 0 =⇒ yy′′ = −1− y′2

Exercise 6.l2 = (L− x)2 + x2 = L2 − 2Lx + 2x2 = A

dA

dx= −2L + 4x = 0 =⇒ x =

L

2d2A

dx2= 4 > 0 =⇒ A minimized

l(x =L

2) =

L√

22

Exercise 7.

(x +√

L2 − x2)2 = A

A′ = 2(x +√

L2 − x2)(1 +−x√

L2 − x2) = 0 when L2 − x2 = x2 or x =

L√2

so then the side of the circumscribing and area-maximized square is L√2

+

√L2 − L2

2=

2L√2

58

Exercise 8.

A = (2x)(2√

R2 − x2) = 4x√

R2 − x2

A′ = 4(√

R2 − x2 +−x2

√R2 − x2

) = 4(

R2 − 2x2

√R2 − x2

)=⇒ x =

R√2

since A′ ≷ 0 when x ≶ R√2, so A is maximized at x =

R√2

2x = 2R√2

; 2√

R2 − 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 (�x the area to be A0)(A04x

)2= r2 − x2 =⇒ x4 − x2r2 + A2

016 = 0

=⇒ 0 = 2xr2 + x22rdr

dx− 4x3

dr

dx= 0 (for extrema) =⇒ x =

r√2

and√

r2 − x2 =r√2

We could argue that we had found a minimum because at the �in�nity� boundaries, the circumscribing circle would bein�nitely large.Exercise 10. Given a sphere of radius R, �nd the radius r and altitude h of the right circular cylinder with the largest lateralsurface area 2πrh that can be inscribed in the sphere.

R2 =(

h

2

)2

+ r2

A = 2πrh = 2πr√

4(R2 − r2) = 4πr√

R2 − r2

dA

dr= 4π

(√R2 − r2 +

−r2

√R2 − r2

)= 4π

(R2 − 2r2

√R2 − r2

)=⇒ r =

R√2

=⇒ h =√

2R

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

√2 times that of the cylinder.

A0 = 2πRH (A0 is the total lateral area of the cylinder)

r2 = R2 +(

H

2

)2

= R2 +(

A0

4πR

)2

= R2 +A2

0

16π2R2

2rdr

dR= 2R +

A20

16π2

(−2R3

)=⇒ dr

dR= 0 =⇒ R =

√A0

2√

π=⇒ H

2= R

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 cylinderof largest lateral surface area that can be inscribed in the cone.

hR−r = H

R = α is the constraint (look, directly at the side, at the similar triangles formed)

A = 2πrh = 2πrα(R− r) = 2πα(rR− r2)

dA

dr= 2πα(R− 2r) = 0 =⇒ r =

R

2; h =

H

2

A′′ = 2πα(R− 2r)

since dA

dr≷ 0 when r ≶ R

2, r =

R

2maximizes lateral surface area

Exercise 13. Find the dimensions of the right circular cylinder of maximum volume that can be inscribed in a right circularcone of radius R and altitude H .

59

Constraint: hR−r = H

R = α

V = πr2h = πr2α(R− r) = πr2α(R− r) = πα(Rr2 − r3)

dV

dr= πα(2Rr − 3r2) = rπα(2R− 3r) r =

2R

3

since dV

dr≷ 0 when r ≶ 2R

3, r =

2R

3maximizes volume

h =13H

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 ofmaximum volume that can be inscribed in this sphere.

V =πr2

3

(R +

√R2 − r2

)

dV

dr=

π

3

(2rR + 2r

√R2 − r2 +

r2(−r)√R2 − r2

)=

π

3r(2R

√R2 − r2 + 2R2 − 3r2)√

R2 − r2= 0

=⇒ r =2√

2R

3; h =

4R

3

Considering the geometric or physical constraints, since limV→∞ V = limh→∞ V = 0, so then r = 2√

2R3 must maximize

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

A =√

R2 − x2x

A′ =√

R2 − x2 +−x2

√R2 − x2

= 0 =⇒ x =R√2; h =

R√2

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

A =12h(2

√R2 − h2 + 2R)

dA

dh=

√R2 − h2 + R + h

( −h√R2 − h2

)

dA

dh= 0 =⇒ h =

√3R

2

=⇒ A =5√

3R2

8

√R2 − h2 = 2

√R2 − 3

4R2 = 2

R

2= R

Exercise 17. An open box is made from a rectangular piece of material by removing equal squares at each corner and turningup 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 − 2r2x− 2r2y + 4r3 = V

dV

dr= xy − 4rx− 4ry + 12r2 = 0

=⇒ r =4(x + y)±

√16(x + y)2 − 4(12)xy

2(12)=

(x + y)±√

x2 + y2 − xy

6d2V

dr2= −4x− 4y + 24r = −4(x + y) + 24r

We can plug in our expression for r into the second derivative of V , the volume of the box, to �nd out that we wantto pick the �negative� root from r, in order to maximize the box volume.

Then for x = 10; y = 10, we have r = 53 , so that the box dimensions are 5

3 × 203 × 20

3 .(2) 12 and 18

=⇒ 5−√7× 2 + 2√

7× 8 + 2√

760

Exercise 18. If a and b are the legs of a right triangle whose hypotenuse is 1, �nd the largest value of 2a + b.

L = 2a + b = 2a +√

1− a2

L′ = 2 +−a√1− a2

L′ = 0 =⇒(a

2

)2

= 1− a2 =⇒ a =2√5

L′′ = (−1)

(√1− a2 − −a√

1−a2 a

1− a2

)= (−1)

(1

(1− a2)3/2

)< 0 (so a =

2√5

maximizes L )

Exercise 19. 2 + x2

600 gallons per hour. l0 = 300 mi x = constant speed. l0x = time spent. K = gas cost = 0.30.

C = gas cost + driver labor cost = l0(

2Kx + Kx

600 + Dx

)

dC

dx= l0

(−2K

x2+

K

600− D

x2

)= 0

dCdx =0−−−−→ x =

√2K + D

K10√

6

d2C

dx2= l0

((−2K −D)

(−2x3

))= l0

(2(2K + D)

x3

)> 0

Thus, C is minimized if x =

√2K + D

K10√

6

=⇒ Cmin = (300)

(√2K + D

√K

10√

6+

√K√

2K + D10√

6600

)= 3

√2√

6 + 10D

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) C = 300

(2K60 + K(60)

600 + D60

)= 5(2.4 + D) = 22.00

(4) D = 3, x = 60 C = 27.00(5) D = 4, x = 60 C = 32.00

Exercise 20. y = xx2+1 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 = x0x20+1

=⇒ x0 =1

2y0±

√1

(2y0)2− 1

x2 − x1 =

√1y20

− 4

where x2 − x1 is going to be the base of the rectangle. The volume of the cylinder, V , which is obtained from revolving therectangle about the x axis, is going to be

V = πy20(x2 − x1) = πy2

0

(√1y20

− 4

)= πy0

√1− 4y2

0

dV

dy0= π

(√1− 4y2

0 +y0

2√

1− 4y20

(−8y0)

)= π

(1− 8y2

0√1− 4y2

0

)=⇒ y0 =

12√

2

We could argue that V is maximized, since the �in�nite� boundaries would yield a volume of 0 (imagine stretching andsqueezing the rectangle inside the curve).

Then Vmax = π 182 =

π

4Exercise 21. Draw a good diagram. Note how the right triangle that you folded is now re�ected backwards , so that thistriangle'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 α61

Note that we will obtain a minimum crease because by considering the �physical in�nite� boundary, we could make a bigcrease along the vertical half of the paper or the horizontal half of the paper.

So, isolating l, the length of the crease, and then taking the derivative,

l =w0

2 sin (α) cos2 α=

w0

2csc (α) sec2 (α)

dl

dα=

w0

2(− cot α csc α sec2 α + csc α2 sec α sec α tanα

)=

=w0

2

(−C

S

(1S

)(1

C2

)+

1S

2(

1C2

)(S

C

))=

w0

2

( −1S2C

+2

C3

)

dldα =0−−−−→ sin α =

1√3

or tanα =1√2

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

l =w0

21

1√3

23

=9√

32

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 theisosceles triangle. Find all the angles.

Angle AOB is π − 2α by the geometry or i.e. inspection of the �gure. 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)

s2 = R2 + R2 − 2R2 cos (π − 2α) + 2R2(1 + cos (2α)) = 2R2(2 cos2 α) = 4R2 cos2 α

s = 2R cosα

The constraint equation is

(8) P = 4R cos α + 2R sin (2α)

So thenP ′ = 4R(− sin α) + 4R cos (2α) = 0 =⇒ cos 2α = sin α

sin α =− 1

2 ±√

14 − 4(1)(− 1

2 )

2(1)=

12

> 0

=⇒ 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 �attened,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 �xed 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 thistriangle, P , is then

P = 2h + 2h sin α = 2h(1 + sin α)

=⇒ h =P

2(1 + sin α)h

2= R cos α

62

We could try to extremize this equation.dR

dα(4 cos α + 2 sin (2α)) + R(−4 sin α + 4 cos (2α)) = 0

dR

dα= 0 =⇒ cos (2α) = sin α =⇒ sin α =

1√3

cosα =√

2√3

R =3P

4√

2(√

3 + 1)

However, this is the minimized R, minimized radius for the smallest circle �tting a particular isosceles triangle of a�xed perimeter. We want to smallest circle with a radius big enough to �t all the possible triangles. Thus R = L

4

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

I = Wh +π

2

(W

2

)2 (12

)=

Wp

2− W 2

2− 3π

16W 2

So thendI

dW=

P

2−W − 3π

8W = 0 =⇒ W =

P

2 + 3π4

The height of the rectangle is

h =P − P

2+ 3π4

(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 itsends, where h ≥ 0. Determine, as a function of h, the volume of the largest right circular cylinder that can be cut from thelog, if its axis coincides with that of the log.

Remember to label your diagram carefully.

y

x=

h2 − y

l0 − x=

h/2l0

=⇒h/2−

(xh2l0

)

l0 − x

V = π(H + y)2(l0 − x) = π(H +xh

2l0)2(l0 − x)

Note that V (x = 0) = πH2l0 = π4(12)dV

dx= π(2(H +

hx

2l0)

h

2l0(l0 − x) + (H +

xh

2l0)2(−1)) =

= π(H +hx

2l0)(h−H − 3xh

2l0)

dVdx =0−−−−→ x =

(h−H)2l03h

V (x =(h−H)2l0

3h) = π

(l0 − (h−H)2l0

3h

)(H +

h

2l0

(h−H)2l03h

)2

= πl0(h + 2H)3

27h

where H = 2, l0 = 12Exercise 25.

S =n∑

k=1

(x− ak)2 =⇒ dS

dx=

n∑

k=1

2(x− ak) = 0

=⇒ nx =n∑

k=1

ak =⇒ x =∑n

k=1 ak

n

Since limx→±∞ S = +∞, x =Pn

k=1 ak

n 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 explicitlysoon) and that f must have a minimum somewhere.

If A < 0, then consider f(x) = 5x7+Ax5 . Consider x = −A1/7

61/7 > 0.

f(x =−A1/7

61/7=

A

6x5< 0

63

Thus, A > 0.

df

dx= 10x− 5Ax−6 = 5

(2x7 −A

x6

)= 0

x =(

A

2

)1/7

d2f

dx2= 10 + 30Ax−7 = 10 + 3−A

(2A

)= 70 > 0

Thus x =(

A2

)1/7 minimizes f for A > 0.

f(x =(

A

2

)1/7

) =5

(A2

)+ A

(A2

)5/7= 24

=⇒ A = 2(

247

)7/2

Exercise 27. Consider f(x) = −x3

3 + t2x over 0 ≤ x ≤ 1.

f(0) = 0, f(1) = −13

+ t2 =⇒ f(1) ≷ 0 if t2 ≷ 13

f ′(x) = −x2 + t2 = 0

=⇒ x2 = t2 but x ≥ 0, so x = |t|

f(x2 = t2) = −13t2(x) + t2x =

23t2x > 0 for 1 ≥ x ≥ 0

So the minimum isn't in the interior of [0, 1]. It's on the end points.

m(t) = 0 for |t| > 13

m(t) =−13

+ t2 for |t| < 13

Exercise 28.(1)

E(x, t) =|t− x|

x

M(t) = max|t− x|

xas x = a → x = b

|t− x|x

=

{t−xx if t ≥ x

x−tx if t < x

d

dt

( |t− x|x

)=

{− t

x2 if t ≥ xt

x2 if t < x

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 endpointst− a

a= E(a, t),

b− t

b= E(b, t)

By monotonicity on [a, t) , (t, b], and having shown the relative minimum of |t−x|x at x = t, the maximum occurs

at x = a or x = b, depending upon the relationship E(a, t) ≷ E(b, t).(2)

M(t) =

{t−aa if t−a

a > b−tb i.e. t

(b+aab

)> 2

b−tb if b−t

b > t−aa

dM

dt=

{1a if t > 2 ab

a+b−1b if t <

(ab

a+b

)2

Since dMdt ≷ 0 when t ≷ 2ab

a+b , M is minimized for t =2ab

a + b

64

4.23 Exercises - Partial Derivatives.

Exercise 8. f(x, y) = x√x2+y2

.

fx =1√

x2 + y2+

−x2

(x2 + y2)3/2=

y2

(x2 + y2)3/2

fy =−xy

(x2 + y2)3/2

fxx =−3y2x

(x2 + y2)5/2

fyy = (−x)(

x2 − 2y2

(x2 + y2)5/2

)

fxy = (−y)

((x2 + y2)3/2 − x 3

2 (x2 + y2)1/2(2x)(x2 + y2)3

)=

= (−y)( −2x2 + y2

(x2 + y2)5/2

) fyx =2y

(x2 + y2)3/2+

−3y2y

(x2 + y2)5/2=

(−y)(−2x2 + y2)(x2 + y2)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 + y4)1/2

zx =1z2x3

zy =2y3

z

x(2z)− 4zy = (x− 2y)2√

z = 2z

Exercise 10.f =

xy

(x2 + y2)2, fx =

y

(x2 + y2)2+

−4x2y

(x2 + y2)3=

y3 − 3x2y

(x2 + y2)3

Sofxx =

−6xy(x2 + y2)3 − 3(x2 + y2)2(2x)(y3 − 3x2y)(x2 + y2)6

=

=12xy(x2 − y2)

(x2 + y2)4

By label symmetry,

fxx + fyy =12xy(x2 − y2)

(x2 + y2)4+

12yx(y2 − x2)(x2 + y2)4

= 0

5.5 Exercises - The derivative of an inde�nite integral. The �rst fundamental theorem of calculus, The zero-derivativetheorem, Primitive functions and the second fundamental theorem of calculus, Properties of a function deduced fromproperties 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

(9) A(x) =∫ x

c

f(t)dt if a ≤ x ≤ b

Then ∃A′(x) ∀x ∈ (a, b) where f is continuous at x and(10) A′(x) = f(x)

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

(11) P (x) = P (c) +∫ x

c

f(t)dt

65

Exercise 6.√

223x3/2 +

√12

23x3/2 =

√2x3/2

∫ b

af =

√2(b3/2 − a3/2)

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

P = 25x5/2 − 2x3/2 + 7x1/2

∫ b

af = 2

5 (b5/2 − a5/2)− 2(b3/2 − a3/2) + 7(b1/2 − a1/2)

Exercise 8. P = 32x4/3 − 3

2x2/3; x > 0

Exercise 9. P = −3 cos x + x6

3

Exercise 10. P = 37x7/3 − 5 sin x

Exercise 11. f ′(x) = 1x

f =∞∑

k=−∞akxk

f ′ =∞∑

k=−∞kakxk−1 =

1x

Comparing terms, only k = 0 would work, but the coef�cient is unequivocally 0Exercise 12. ∫ x

0

|t|dt =

{∫ x

0tdt if x ≥ 0∫ x

0−tdt if x < 0

=

{12x2 if x ≥ 0−12 x2 if x < 0

=12x|x|

Exercise 13. ∫ x

0

(t + |t|)2dt =

{∫ x

0(2t)2dt if x ≥ 0

0 if x < 0=

{43x3 if x ≥ 00 if x < 0

=2x2

3(x + |x|)

Exercise 14. Using 1st. fund. thm. of calc.∫ x

0

f(t)dt = A(x)−A(0)

A′(x) = f(x)=⇒ 2x + sin 2x + 2x cos 2x +− sin 2x

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

= 2 + 2 cos 2x− 4x sin 2x

f(π

4

)=

π

2

f ′(π

4

)= 2− π

Exercise 15.∫ x

cf(t)dt = cos x− 1

2 f(x) = − sin x c = −π6 .

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

0

t sin t− t =(−t cos t + sin t− 1

2t2

)∣∣∣∣x

0

= sin x− x cosx− 12x2

So c = 0.Exercise 17. For f(x) = −x2f(x) + 2x15 + 2x17 (found by taking the derivative of

∫ x

0f =

∫ 1

xt2f + x16

8 + x18

9 + C,)Suppose that f = 2x15.

=⇒ x16

8= −x18

9+−1

9+

x16

8+

x18

9+ C

=⇒ C =19

Exercise 18. By plugging in x = 0 into the de�ned f(x), f(x) = 3 +∫ x

01+sin t2+t2 dt, we get for p(x) = a + bx + cx2,

a = 3

Continuing on,

f ′ =1 + sin x

2 + x2;

f ′(0) =12

= b

f ′′ =(cos x)(2 + x2)− 2x(1 + sin x)

(2 + x2)2

f ′′(0) =12

+ 2c; c =14

66

Exercise 19.f(x) =

12

∫ x

0

(x− t)2g(t)dt =12

∫ x

0

(x2 − 2xt + t2)g(t)dt =

=12

(x2

∫ x

0

g − 2x

∫ x

0

tg +∫ x

0

t2g

)

f ′ = x

∫ x

0

g +x2

2g(x)−

∫ x

0

tg − x(xg(x)) +12x2g(x) =

= x

∫ x

0

g −∫ x

0

tg

f ′′ =∫ x

0

g + xg − xg =∫ x

0

g f ′′(1) = 2

f ′′′ = g f ′′′(1) = 5

Exercise 20.(1) (

∫ x

0(1 + t2)−3dt)′ = (1 + x2)−3

(2) (∫ x2

0(1 + t2)−3dt)′ = (1 + x4)−3(2x) = 2x

(1+x4)3

(3) (∫ x2

x3 (1 + t2)−3dt)′ = (1 + x4)−3(2x)− (1 + x6)−3(3x2) = 2x(1+x4)3 − 3x2

(1+x6)3

Exercise 21.

f ′(x) =

(∫ x2

x3

t6

1 + t4dt

)′

=(

x12

1 + x8

)(2x)−

(x18

1 + x12

)3x2

Exercise 22.(1)

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

f(2) = 16

(2) ddx

(∫ b(x)

a(x)f(t)dt

)= f(b)b′ − f(a)a′

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

f(x2) = (1 +3x

2)

f(2) = 1 +3√

22

(3)∫ f(x)

0t2dt = x2(1 + x)

2x + 3x2 = (f(x))2f ′(x)

=⇒ f3(2) = 3(4)(3) = 9(4) = 36

f(2) = 361/3

(4)d

dx

(∫ x2(1+x)

0

f(t)dt

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

x = 1 (f(2))(5) = 1 =⇒ f(2) =15

Exercise 23.a3 − 2a cos a + (2− a2) sin a =

∫ a

0

f2(t)dt

3x2 − 2 cos x + 2x sin x +−2x sin x + (2− x2) cos x = 3x2 − x2 cosx = f2(x)

f(x) = x√

3− cos x

f(a) = a√

3− cos a

Exercise 24. f(t) = t2

2 + 2t sin t

(1)f ′ = 2t + 2 sin t + 2t cos t f ′(π) = 2π − 2π = 0

67

(2)f ′′ = 2 + 2 cos t + 2 cos t +−2t sin t = 2 + 4 cos t− 2t sin t

f ′′(π

2

)= 2− π

(3) f ′′(

3π2

)= 0

(4) f(

5π2

)= 25π2

8 + 5π

(5) f(π) = π2

2

Exercise 25.(1)

df

dt=

1 + 2 sin πt cosπt

1 + t2= v(t)

a(t) =2π(cos (2πt))(1 + t2)− 2t(sin (2πt))

(1 + t2)2

a(t = 2) = a(t = 1) =4π

4= π

(2) v(t = 1) 12

(3) v(t) = π(t− 1) + 12 ; t > 1

(4)

f(t)− f(1) =∫ t

1

v(t)dt =∫ t

1

π(t− 1) +12

=(

πt2

2− πt +

12t

)∣∣∣∣t

1

=πt2

2+−πt +

t

2+

π

2− 1

2

Exercise 26.(1)

f ′′(x) > 0∀x f ′(0) = 1; f ′(1) = 0∫ 1

0

f ′′(t)dt = f ′(1)− f ′(0) = 0− 1 < 0

Thus, it's impossible, since f ′′(x) > 0, so∫ 1

0f ′′(t)dt > 0

(2) ∫ 1

0

(3− π

2sin

πx

2

)dx =

(3x + cos

πx

2

)∣∣∣1

0= 3− 1 = 2

f(x) =3x2

2+

2π

sinπx

2+ C

(3) f ′′(0) > 0∀x f ′(0) = 1; f(x) ≤ 100∀x > 0∫ b

a

f ′′(t)dt = f ′(b)− f ′(a);∫ k

c

f ′(t)dt = f(k)− f(c)

∫ b

0

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

∫ k

c

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

f(k)− f(c) > k − c

f(k)− f(0) > k − 0

f(100)− f(0) > 100

f(x) ≤ 100 is untrue for all x > 0

(4) f ′′(x) = ex > 0 f ′(x) = ex; f ′(0) = 1 f(x) = ex ∀x < 0, ex < 1

Exercise 27. f ′′(t) ≥ 6. b− a = 12 . f ′(0) = 0

∫ b

a

f ′′ = f ′(b)− f ′(a) ≥ 6(b− a) = 3 since b− a =12∫ a

0

f ′′ = f ′(a)− f ′(0) = f ′(a) ≥ 6(a− 0) = 6a

If a =12, f ′(1/2) ≥ 3

68

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

5.8 Exercises - The Leibniz notation for primitives, Integration by substitution.Exercise 1.

∫ √2x + 1dx = 1

3 (2x + 1)3/2.

Exercise 2.∫

x√

1 + 3x = 2x9 (1 + 3x)3/2 +− 4

135 (1 + 3x)5/2

Exercise 3. ∫x2√

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

3− 8x(x + 1)5/2

15+

16(x + 1)7/2

105since

(2x2(x + 1)3/2

3

)′= x2(x + 1)1/2 +

4x(x + 1)3/2

3(

8x(x + 1)5/2

15

)′=

43x(x + 1)3/2 +

8(x + 1)5/2

15(

16(x + 1)7/2

105

)′=

8(x + 1)5/2

15

Exercise 4. ∫xdx√2− 3x

=2x(2− 3x)1/2

−3+

4(2− 3x)3/2

−27∫ 1/3

−2/3

xdx√2− 3x

= −2/9− 4/27− (8/9− 32/27) = −2/27

Exercise 5. ∫(x + 1)dx

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

((x + 1)2 + 1)−2

−4

Exercise 6. ∫sin3 x =

∫sin x(1− cos2 x) = − cos x +

13

cos3 x

Exercise 7. ∫x1/3(1 + x) =

34x4/3 +

37x7/3 =

34(z − 1)4/3 +

37(z − 1)7/3

Exercise 8. sin−2 x−2

Exercise 9. (4−sin 2x)3/2

−3

∣∣∣π/4

0= 33/2−8

−3

Exercise 10. (3 + cos x)−1

Exercise 11. 2 cos−1/2 x

Exercise 12. 2 cos√

x + 1∣∣83

= 2(cos 3− cos 2)

Exercise 13. − cos xn

n

Exercise 14. (1−x6)1/2

−3

Exercise 15. ∫t(1 + t)1/4dt =

∫(x− 1)x1/4dx =

49x9/4 − 4

5x5/4 =

49(1 + t)9/4 − 4

5(1 + t)5/4

Exercise 16.∫

(x2 + 1)−3/2dx =?(

x√x2 + 1

)′=√

x2 + 1− x2/√

x2 + 1x2 + 1

=1

(x2 + 1)3/2

Exercise 17.

(8x3 + 27)5/3

(35

)(124

)=

140

(8x3 + 27)5/3

Exercise 18. 32 (sinx− cosx)2/3

69

Exercise 19. ∫xdx√

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

∫ 12du√

u + u3/2=

=12

∫du√

u√

1 + u1/2= 2(1 + u1/2)1/2 =

= 2(1 +√

1 + x2)1/2 + C

Exercise 20. ∫(x2 − 2x + 1)1/5dx

1− x=

∫ −(x− 1)2/5

x− 1dx = −

∫(x− 1)−3/5dx = −5/2(x− 1)2/5

Exercise 21. Thm. 1.18. invariance under translation.∫ b

af(x)dx =

∫ b+c

a+cf(x− c)dx.

Thm. 1.19. expansion or contraction of the interval of integration.∫ b

a

f(x)dx =1k

∫ kb

ka

f(x

k

)dx

yx + c

dy = dx

∫ b

a

f(x)dx =∫ b+c

a+c

f(y − c)dy

y = kx

dy = kdx

∫ b

a

f(x)dx =1k

∫ kb

ka

f(y

k

)dy

Exercise 22.

F(x

a, 1

)=

∫ x/a

0

up

(u2 + 12)qdu

u =t

a

du =dt

a

F(x

a, 1

)=

1a

∫ x

0

(t/a)pdt((ta

)2 + 12)q =

= a−p−1+2q

∫ x

0

tp

(t2 + 12)qdt = a−p−1+2qF (x, a)

Exercise 23.

∫ 1

x

dt

1 + t2= F (1)− F (x)

∫ x

1

dt

1 + t2= F (x)− F (1)

∫ 1/x

1

dt

1 + t2= F

(1x

)− F (1)

u =1t

du =−1t2

dt,−1u2

du = dt

∫ x

1

dt

1 + t2=

∫ 1/x

1

−du

u2(1 + 1

u2

) =

= −∫ 1/x

1

du

u2 + 1=

∫ 1

1/x

dt

t2 + 1

Exercise 24. ∫ 1

0

xm(1− x)ndx = −∫ 0

1

(1− u)m(un)du =∫ 1

0

(1− x)mxndx usingu = 1− x

x = 1− u

Exercise 25.

cosm x sinm x =(

sin 2x

2

)m

= 2−m sinm 2x

∫ π2

0

cosm x sinm xdx = 2−m

∫ π/2

0

sinm 2xdx = 2−m

∫ π

0

12

sinm xdx = 2−m−1

∫ π

0

sinm xdx =

= −2−m−1

∫ −π/2

π/2

sinm(π

2− x

)dx = 2−m−1

∫ π/2

−π/2

cosm xdx = 2−m

∫ π/2

0

cosm xdx

Exercise 26.70

(1)

u = π − x

x = π − u

∫ π

0

xf(sinx)dx =∫ 0

π

(π − u)f(sin (π − u))(−du) =∫ π

0

(π − u)f(sinu)du =

= π

∫ π

0

f(sinx)dx−∫ π

0

xf(sinx)dx

=⇒∫ π

0

xf(sinx)dx =π

2

∫ π

0

f(sinx)dx

(2)

u = cos x

du = − sin xdx

∫ π

0

x sin x

1 + cos2 xdx =

∫ π

0

x sinx

2− sin2 x=

π

2

∫ π

0

sin x

2− sin2 xdx =

π

2

∫ π

0

sinx

1 + cos2 xdx =

= −π

2

∫ −1

1

du

1 + u2=

π

2

∫ 1

−1

du

1 + u2= π

∫ 1

−1

dx

1 + x2

Exercise 27.x = sin u

dx = cos u

∫ 1

0

(1− x2)n− 12 dx =

∫ π/2

0

(cos2 u)n− 12 cos udu =

∫ π/2

0

cos2n udu

5.10 Exercises - Integration by Parts.Exercise 1.

∫x sin x = −x cos x + sin x

Exercise 2.∫

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

Exercise 3.∫

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

Exercise 4.∫

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

Exercise 5.∫

sin x cosx = − 14 cos 2x = − 1

4 (cos2 x− sin2 x)

Exercise 6.∫

x sin x cos xdx =∫

x2 sin 2x = −x cos 2x

4 + sin 2x8

Exercise 7.∫

sin2 x =∫

sin x sin x = − sinx cosx +∫

cos2 x∫sin2 xdx = −1

4 sin 2x + x2

Exercise 8. ∫sinn xdx = − cosx sinn−1 x +

∫(n− 1) sinn−2 x cos2 x

u = sinn−1 x

dv = sin xdx∫sinn x = − cos x sinn−1 x +

∫(n− 1) sinn−2 x(1− sin2 x) =

= − cos x sinn−1 x +∫

(n− 1) sinn−2 x− (n− 1)∫

sinn x

∫sinn x =

−1n

sinn−1 x cosx +(n− 1)

n

∫sinn−2 x

Exercise 9.(1) ∫

sin2 x =−12

sin x cosx +12

∫1 =

−12

sin x cosx +12x

∫ π/2

0

sin2 xdx =π

4

(2)∫ π/2

0sin4 x = −1

4 sin3 x cosx∣∣π/2

0+ 3

4

∫ π/2

0sin2 x = 3π

16

(3)∫ π/2

0sin6 x = 5

6

∫ π/2

0sin4 x =

5π

32Exercise 10.

(1) ∫sin3 xdx =

−13

sin2 x cos x +23

∫sin x = −1

6sin 2x cos x− 2

3cos x =

−34

cosx +112

cos 3x since

−34

cos x +112

cos 3x =−34

cos x +112

(cosx cos 2x− sin 2x sin x) =

= −34

cos x +112

(cosx(1− 2 sin2 x) +−2 sin2 x cos x) =−23

cosx− 13

sin2 x cosx

71

(2)∫

sin4 xdx =−14

sin3 x cosx +34

∫sin2 x =

−14

sin3 x cos x +34(x

2− sin 2x

4) =

−14

sin3 x cos x +3x

8− 3 sin 2x

16

Now 132

sin 4x =132

(2 sin 2x cos 2x) =18(sinx cosx(1− 2 sin2 x) =

sin 2x

16− 1

4sin3 x cos x

=⇒ −14

sin3 x cos x +3x

8− 3 sin 2x

16=

3x

8− 1

4sin 2x +

132

sin 4x

(3)∫

sin5 xdx =∫

sin4 x sin xdx = − cosx sin4 x +∫

cos2 x4 sin3 x =

= − cos x sin4 x + 4(∫

sin3 x−∫

sin5 x) = − cos x sin4 x + 4∫

sin3 x− 4∫

sin5 x

5∫

sin5 dx = − cos x sin4 x + 4∫

sin3 x

5∫

sin5 dx = − cosx(1− cos2 x)2 + 4(−34

cosx +112

cos 3x)

= − cosx(1− 2 cos2 x + cos4 x) +−3 cos x +13

cos 3x

= − cosx + 2 cos3 x− cos5 x− 3 cos x +13(cos x cos 2x− sinx sin 2x) =

= −4 cos x + 2 cos3 x− cos5 x +13(4 cos3 x− 3 cos x) = −5 cos x +

10 cos3 x

3− cos5 x

∫sin5 xdx = − cosx +

2 cos 3x

3− 1

5cos5 x

My solution to the last part of this exercise con�icts with what's stated in the book.

Exercise 11.

(1)∫

x sin2 xdx = (∫

sin2 x)x−∫

(sin2 t) =x2

2− x sin 2x

4−

(x2

4+

cos 2x

8

)=

=x2

8− x sin 2x

4− cos 2x

8

we had used∫

sin2 x =x

2− sin 2x

4

(2)∫

x sin3 x =−3x

4cosx +

x

12cos 3x−

∫−3

4cosx +

112

cos 3x =

=−3x

4cosx +

x

12cos 3x +

34

sin x +− sin 3x

36∫sin3 x =

−34

cosx +112

cos 3x

(3)∫

x2 sin2 xdx = x2

(x

2− sin 2x

4

)−

∫2x

(x

2− sin 2x

4

)=

x3

2− x2 sin 2x

4− 1

3x3 +

∫x sin 2x

2=

=x3

6− x2 sin 2x

4+

12

(−x cos 2x

2+

sin 2x

4

)=

=x3

6− x2 sin 2x

4− x cos 2x

4+

sin 2x

8

72

Exercise 12.

∫cosn xdx =

∫cosn−1 x cos xdx = cosn−1 x sin x +

∫(n− 1) cosn−2 x sin2 x =

= cosn−1 x sin x + (n− 1)∫

cosn−2 x−∫

(n− 1) cosn x

=⇒∫

cosn x =cosn−1 x sin x

n+

(n− 1

n

) ∫cosn−2 x

Exercise 13.

(1)∫

cos2 x = sin 2x5 + 1

2x

(2)∫

cos3 x = cos2 x sin x3 + 2

3 sin x = 34 sin x + 1

12 sin 3x since

112

sin 3x =112

(sin 2x cosx + sin x cos 2x) =16

sin x cos2 x +112

sin x(2 cos2 x− 1) =

=13

sinx cos2 x− 112

sin x

(3)

∫cos4 xdx =

cos3 x sinx

4+

34

(12x +

14

sin 2x

)=

38x +

316

sin 2x +cos3 x sinx

4

sin 4x = 2 sin 2x cos 2x = 4 sin x cosx(2 cos2 x− 1) = 8 sin x cos3 x− 2 sin 2x then∫cos4 xdx =

38x +

14

sin 2x +132

sin 4x

Exercise 14.

∫ √1− x2dx = x

√1− x2 +

∫x2

√1− x2

dx

∫x2

√1− x2

dx =∫

x2 −+1√1− x2 + 1− 1

= −∫ √

1− x2 +1√

1− x2

x2 = x2 − 1 + 1

=⇒∫ √

1− x2dx =12x√

1− x2 +12

∫1√

1− x2

Exercise 15.

(1)

∫(a2 − x2)ndx = x(a2 − x2)n −

∫n(a2 − x2)n−1(−2x)xdx = x(a2 − x2)n + 2n

∫x2(a2 − x2)n−1dx

∫x2(a2 − x2)n−1dx =

∫((x2 − a2) + a2)(a2 − x2)n−1dx =

∫−(a2 − x2)n + a2(a2 − x2)n−1dx

=⇒∫

(a2 − x2)ndx =x(a2 − x2)n

2n + 1+

2a2n

2n + 1

∫(a2 − x2)n−1dx

73

(2)∫

(a2 − x2)dx =x(a2 − x2)

3+

2a2

3x =

−x3

3+ a2x

∫(a2 − x2)5/2dx =

x(a2 − x2)5/2

6+

a256

∫(a2 − x2)3/2dx

∫(a2 − x2)3/2dx =

x(a2 − x2)3/2

4+

3a2

4

∫(a2 − x2)1/2dx

∫(a2 − x2)1/2 = a

∫ √1−

(x

a

)2

dx = a2

∫cos2 θdθ =

= a2

∫1 + cos 2θ

2= a2

(θ

2+

sin 2θ

4

)= a2

(arcsin

x

a+

12

x

a

√1−

(x

a

)2)

sin θ =x

a

cos θdθ =dx

a

∫ a

0

√a2 − x2 = a2

(π

2− 0

)=

πa2

2∫ a

0

(a2 − x2)3/2dx =3a2

4

(πa2

2

)=

3πa4

8∫ a

0

(a2 − x2)5/2dx =5a2

6

(3πa4

8

)=

516

πa6

Exercise 16. In(x) =∫ x

0tn(t2 + a2)−1/2dt

(1)

In(x) = (t2 + a2)1/2tn−1 −∫

(n− 1)tn−2(t2 + a2)1/2 = tn−1(t2 + a2)1/2 − (n− 1)∫

tn−2(t2 + a2)(t2 + a2)1/2

(n)In = xn−1(x2 + a2)1/2 − a2(n− 1)∫

tn−2

(t2 + a2)1/2= xn−1

√x2 + a2 − (n− 1)a2In−2

(2) n = 5; x = 2; a =√

5.

I1(2) =∫ 2

0

x(x2 + 5)−1/2dx = (x2 + 5)1/2∣∣∣2

0= 3−

√5

5I5(2) =∫ 2

0

t5(t2 + 5)−1/2dt = 25−1(4 + 5)1/2 − 5(5− 1)I3(2) = 48− 20I3(2)

3I3(2) = 22√

4 + 5− 5(3− 1)I1(2) = 12− 10(3−√

5)

I5(2) =15(48− 20(−6 +

10√

53

)) =1685− 40

√5

3

Exercise 17.∫

t3(c + t3)−1/2dt =t2(c + t3)1/2

3−

∫2(c + t3)1/2

3∫ 3

−1

t3(4 + t3)−1/2dt =t2(4 + t3)1/2

3

∣∣∣∣3

−1

− 23

∫ 3

−1

(4 + t3)1/2 = 2√

31 +2√

33

− 23(11.35)

Exercise 18.∫

sinn+1 x

cosm+1 xdx =

∫sinn x

(sin x

cosn+1 x

)dx =

sinn x

m cosm x−

∫n sinn−1 x

m cosm−1 x

=⇒∫

sinn+1 x

cosm+1 xdx =

sinn x

m cosm x− n

m

∫sinn−1 x

cosm−1 x74

Exercise 19.∫

cosm+1 x

sinn+1 xdx =

∫cosm x

(cosxdx

sinn+1 x

)= cosm x

1−n sinn x

−∫

m cosm−1 x

−n sinm x=

=cosm x

−n sinn x+−m

n

∫cosm−1 x

sinn−1 x∫cot2 x =

∫cos1+1 x

sin1+1 x=−11

cos1 x

sin1 x− 1

1

∫dx = − cot x− x

∫cot4 xdx =

∫cos3+1 x

sin3+1 x= −1

3cos3 x

sin3 x−

∫cos3−1 x

sin3−1 x=−13

cot3 x− (− cot x− x) =−13

cot3 x + cot x + x

Exercise 20.(1)

∫ 2

0

tf ′′(t)dt = 2∫ 1

0

tf ′′(2t)dt n = 2

(2)∫ 1

0

xf ′′(2x)dx =12

∫ 2

0

tf ′′(t)dt =12

(tf ′(t)|20 −

∫ 2

0

f ′(t)dt

)=

12

(2f ′(2)− (f(2)− f(0))) = 4

Exercise 21.(1) Recall Theorem 5.5, the second mean-value theorem for integrals:

∫ b

a

f(x)g(x)dx = f(a)∫ c

a

g(x)dx + f(b)∫ b

c

g(x)dx

∫ b

a

sin φ(t)(

φ′(t)φ′(t)

)dt =

1φ′(a)

∫ c

a

φ′(t) sin φ(t) +1

φ′(b)

∫ b

a

φ′(t) sin φ(t) =

=1

φ′(a)cosφ(t)|ca +

1φ′(b)

(cos φ(b)− cosφ(a)) ≤ 4m

where 1m≥ 1

φ′(t)∀t ∈ [a, b]

(2) φ(t) = t2 ; φ′(t) = 2t > 2a if t > a∣∣∣∣∫ x

a

sin t2dt

∣∣∣∣ ≤42a

= 2a

5.11 Miscellaneous review exercises.

Exercise 1. g(x) = xnf(x); f(0) = 1

g′(x) = nxn−1f(x) + xnf ′(x) ; g′(0) = 0 especially if n ∈ Z+ (just note that if negative integer values are included,g′(0) easily blows up)

gj(x) =∑h

k=0

(jk

)n!

(n−k)!xn−kf j−k(x)

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

If j ≥ n,

gj(x) =n∑

k=0

(j

k

)n!

(n− k)!xn−kf (j−k)(x)

gj(0) =j!

(j − n)!f (j−n)(0)

If j = n, gn(0) = n!75

Exercise 2.

P (x) =5∑

j=0

ajxj

P ′(x) =5∑

j=1

jajxj−1

P ′′(x) =5∑

j=2

j(j − 1)ajxj−2

P (0) = 1 = a0

P ′(0) = 0 = a1

P ′′(0) = 0 = 2(1)a2

a1 = a2 = 0

=⇒P (x) = a5x

5 + a4x4 + a3x

3 + 1

P ′(x) = 5a5x4 + 4a4x

3 + 3a3x2

P ′′(x) = 20a5x3 + 12a4x

2 + 6a3x

P (1) = a5 + a4 + a3 + 1 = 2

P ′(1) = 5a5 + 4a4 + 3a3 = 0

P ′′(1) = 20a5 + 12a4 + 6a3 = 0

Solve for the undetermined coef�cients by Gauss-Jordan elimination process

5 4 320 12 61 1 1

a5

a4

a3

=

001

5 4 320 12 61 1 1

∣∣∣∣∣∣

001

=

1 00 1

1 0 0

∣∣∣∣∣∣

−15106

=⇒ a5 = 6 a4 = −15 a3 = 10 P (x) = 6x5 − 15x4 + 10x3 + 1

Exercise 3. If f(x) = cosx and g(x) = sin x, Prove that f (n) = cos (x + nπ2 ) and g(n)(x) = sin (x + nπ

2 )

f (n)(x) = cos (x +nπ

2) =

{sin x(−1)j+1 if n = 2j + 1cosx(−1)j if n = 2j

g(n)(x) = sin (x +nπ

2) =

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

f(x) = cos x

f ′(x) = − sin x

f ′′(x) = − cos x

f ′′′(x) = sin x

f ′′′′(x) = cos x

f (2j)(x) = cosx(−1)j

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

g′(x) = cos x

g′′(x) = − sin x

g′′′(x) = − cos x

g′′′′(x) = sin x

g(2j)(x) = sin x(−1)j

g(2(j+1))(x) = (sin x(−1)j)′′ = sin x(−1)j+1

g(2j+1)(x) = cos x(−1)j

g(2j+3)(x) = (cos x(−1)j+1)

76

Exercise 4.

h′(x) = f ′g + fg′

h′′(x) = f ′′g + 2f ′g′ + fg′′h(n) =

n∑

k=0

(n

k

)f (k)g(n−k)

h(n+1) =n∑

k=0

(n

k

) (f (k+1)g(n−k) + f (k)g(n−k+1)

)=

= f (1)g(n) + fg(n+1) +n−1∑

k=1

(n

k

) (f (k+1)g(n−k) + f (k)g(n−k+1)

)+ f (n+1)g + f (n)g(1)

= f (1)g(n) + fg(n+1) +n∑

k=2

n!(n− k + 1)!(k − 1)!

(f (k)g(n−k+1) +

n−1∑

k=1

n!(n− k)!k!

f (k)g(n−k+1)

)+

+ f (n+1)g + f (n)g(1)

Now n!(n− k + 1)!(k − 1)!

+n!

(n− k)!k!= (k + (n− k + 1))

(n!

(n + 1− k)!(k)!

)=

(n + 1

k

)so then

h(n+1) = fg(n+1) +n∑

k=1

(n + 1

k

)f (k)g(n+1−k) + f (n+1)g =

n+1∑

k=0

(n + 1

k

)f (k)g(n+1−k)

By induction, this formula is true.

Exercise 5.

(1)

f2 + g2 = f(−g′) + gf ′

Y = f2 + g2 Y ′ = 2ff ′ + 2gg′ = 2(gf + 2g(−f)) = 0 =⇒ Y = C

Y = C = f2 + g2 f(0) = 0; g(0) = 1 =⇒ C = 1

(2)

h = (F − f)2 + (G− g)2

= f ′(x) = g(x), g′(x) = −f(x); f(0) = 0; g(0) = 1

h′ = 2(F − f)(F ′ − f ′) + 2(G− g)(G′ − g′); h′(0) = 2(0) + 2(0) = 0

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

h(0) = 0 so C = 0=⇒ F = f ; G = g

Exercise 6. dfdu2x = 3x2 f ′(4) = 3x

2 = 3 where we had used the substitutionu = x2 u = 4; x = 2

Exercise 7. dgdu = u3/2; g(u) = 2

5u5/2 g(4) = 2525 =

645

Exercise 8.∫ x

0

sin t

t + 1dt =

10 + 1

∫ c

0

sin tdt +1

x + 1

∫ x

c

sin tdt =

= − cos t|c0 +1

x + 1− cos t|xc = − cos c + 1 +

−1x + 1

(cos x− cos c) =

=−x cos c + x− cos c + 1− cos x + cos c

x + 1=

x(1− cos c) + (1− cos c)x + 1

> 0

77

Exercise 9. y = x2 is the curve for C. y = 12x2 is the curve for C1.∫ b

0

(x2 − 12x2) =

16b3 P : (b, b2)

Assume C2 is of the form kx2

∫ c

0

(kx2 − x2) +∫ b

c

(b2 − x2) =(k − 1)

3c3 + b2(b− c) +−1

3(b3 − c3)

kc2 = b2

k =b2

c2

=⇒ (b2 − c2)c3

+ b3 − cb2 − b3

3+

c3

3=

2b3

3− 2

3cb2 =

23b2(b− c)

Now A(A) = A(B)

=⇒ 23b2(b− c) =

16b3 =⇒ b =

43c

so then kx2 =169

x2

Exercise 10.

(1) |Q(h)− 0| =∣∣∣ f(h)

h

∣∣∣ =

{h2

|h| if x is rational0 if x is irrational

.

For now, consider 0 < h < δ; let δ(ε;h = 0) = ε

|Q(h)− 0| =∣∣∣∣f(h)

h

∣∣∣∣ =

{h if x is rational0 if x is irrational

< ε

(2) ∣∣∣∣f(h)− f(0)

h− 0

∣∣∣∣ = ε =⇒ f ′(0) = 0

Exercise 11.∫

(2 + 3x) sin bxdx = −25 cos 5x +− 3x

5 cos 5x + 325 sin 5x

Exercise 12. 43 (1 + x2)3/2

Exercise 13. (x2−1)10

20

∣∣∣1

−2= −310

20

Exercise 14.

13

∫ 1

0

6x + 7 + 2(6x + 7)3

dx =13

(−(6x + 7)−1

6+−(6x + 7)−2

6

)∣∣∣∣1

0

=13

(−

(113

) (16

)+

142

+−1

(6)132+

1(6)49

)=

=37

8281

Exercise 15.∫

x4(1 + x5)5dx = (1+x5)6

30

Exercise 16.∫ 1

0

x4(1− x)20dx =u = 1− x

x = 1− u

= −∫ 0

1

(1− u)4u20du =∫ 1

0

(1− u)4u20du =∫ 1

0

(1 + 4(−u) + 6u2 +−4u3 + u4)u20du =

=121

21+−4122

22+

6123

23+−4124

24+

125

25=

=1

265650Make sure to check your arithmetic.

Exercise 17.∫ 2

1x−2 sin 1

xdx =(cos 1

x

)∣∣21

= cos 12 − cos 1

78

Exercise 18.∫

sin (x− 1)1/4dx

u = (x− 1)1/4

du =14(x− 1)−3/4dx =

14

1u3

dx=⇒

∫sin (x− 1)1/4dx =

∫(sinu)4u3du = 4

∫u3 sin udu

∫u3 sin udu = −u3 cosu + 3u2 sinu + 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)1/4dx =

= −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

Exercise 19.∫

x sin x2 cos x2dx = (1/4) sin2 x2 + C

Exercise 20.∫ √

1 + 3 cos2 x sin 2xdx =∫ √

1 + 3 cos2 x2 sin x cosxdx =∫

u1/2 du−3 = 2u3/2

−9 =−29

(1 + 3 cos2 x)3/2 ,

where we had used this substitution:

u = 1 + 3 cos2 x

du = −6 cos x sin xdx

du

−3= 2 cosx sin xdx

Exercise 21.∫ 2

0375x5(x2 + 1)−4dx

u = x2 + 1du = 2xdx

(u− 1)2 = x4

=⇒

∫ 2

0

3752

du(u− 1)2u−4 =3752

∫ 5

1

du(u2 − 2u + 1)

u4=

3752

∫ 5

1

du

(1u2− 2

u3+

1u4

)=

=3752

(−1u

+1u2

+1

−3u3

)∣∣∣∣5

1

= 64 = 26

Exercise 22.∫ 1

0(ax + b)(x2 + 3x + 2)−2dx = 3

2

Since(

−1x2+3x+2

)∣∣∣1

0= −1

6 + 12 = 2

3 ,

then if a = 9/2, b = 272 , we'll obtain 3/2

Exercise 23. In =∫ 1

0(1− x2)ndx

In =∫ 1

0

(1− x2)ndx = x(1− x2)n∣∣10−

∫ 1

0

xn(1− x2)n−1(−2x)dx =

= 2∫ 1

0

x2n(1− x2)n−1dx = 2n

∫ 1

0

((x2 − 1) + 1)(1− x2)n−1dx =

= 2nIn−1 − 2nIn

=⇒ In =(

2n

2n + 1

)In−1

I2 =∫ 1

0

(1− x2)2dx =∫ 1

0

dx(1− 2x2 + x4) =(

x− 2x3

3+

15x5

)∣∣∣∣1

0

=815

I3 =67

815

1635

I4 =89I3 =

128315

I5 =1011

I4 =256693

79

Exercise 24. F (m,n) =∫ x

0tm(1 + t)ndt; m > 0, n > 0

F (m,n) =tm+1

m + 1(1 + t)n

∣∣∣∣x

0

−∫ x

0

tm+1

m + 1n(1 + t)n−1dt =

xm+1(1 + x)n

m + 1− n

m + 1F (m + 1, n− 1)

(m + 1)F (m, n) + nF (m + 1, n− 1) = xm+1(1 + x)n

F (11, 1) =∫ x

0

t11(1 + t)1dt =∫ x

0

t11 + t12 =(

t12

12+

t13

13

)∣∣∣∣x

0

=x12

12+

x13

13

11F (10, 2) + 2(

x12

12+

x13

13

)= x11(1 + x)2

F (10, 2) =x13

13+

x12

6+

x11

11

Exercise 25. f(n) =∫ π/4

0tann xdx

(1)

Use this extremely important fact:∫ b

a

fg = f(b)∫ b

c

g + f(a)∫ c

a

g

f(n + 1) =∫ π/4

0

tann x tan x =∫ π/4

c

tann x <

∫ π/4

0

tann x = f(n)

(2)

f(n + 2) + f(n) =∫ π/4

0

tann x tan2 x +∫ π/4

0

tann x =∫ π/4

0

tann x(sec2 x) =

=tann+1 x

n + 1

∣∣∣∣π/4

0

=1

n + 1

(3)

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

n + 1< f(n + 1) + f(n) < 2f(n)

1n− 1

= f(n− 2) + f(n) > f(n− 1) + f(n) > 2f(n)

=⇒ 1n + 1

< 2f(n) <1

n− 1

Exercise 26. f(0), f(π) = 2∫ π

0(f(x) + f ′′(x)) sin xdx = 5

∫ π

0

f ′′ sin x = f ′ sin x−∫

f ′ cos x = −∫

f ′ cosx = −f cosx−∫

f sin x

∫ π

0

(f + f ′′) sin xdx =∫

f sin x +−(f(π)(−1)− f(0))−∫

f sin x = 2 + f(0) = 5 =⇒ f(0) = 3


0

sin x cos x

x + 1dx =

∫ π/2

0

sin 2x

2x + 2dx =

12

∫ π

0

sin x

x + 2dx =

12

(− cos x

x + 2−

∫cosx

(x + 2)2

)=

=12

(1

π + 2+

12−A

)=

4 + π

4(π + 2)− A

2

Exercise 28.∫

dx

x√

a + bx=

2√

a + bx

bx+

2b

∫ √a + bx

x2

∫ √a + bx

x=

23b (a + bx)3/2

x+

∫ 23b (a + bx)3/2

x2=

23b

√a + bx

(a

x+ b

)+

23b

∫ √a + bx

(a

x2+

b

x

)

=⇒∫ √

a + bx

x= a

∫dx

x√

a + bx+ 2

√a + bx

80

Exercise 29.∫xn

√(ax + b)dx =

2xn(ax + b)3/2

3a−

∫nxn−12(ax + b)3/2

3a=

2xn(ax + b)3/2

3a− 2n

3a

∫xn−1(ax + b)

√ax + b =

=2xn(ax + b)

√ax + b

3a− 2n

3

∫xn√

ax + b− 2nb

3a

∫xn−1

√ax + b

∫xn√

ax + bdx =2

(2n + 3)a

(xn(ax + b)3/2 − nb

∫xn−1

√ax + b

)+ C n 6= −3

2

Exercise 30.∫xm

√a + bx

dx =2xm(a + bx)1/2

b−

∫mxm−12(a + bx)1/2

b

=2bxm(a + bx)1/2 − 2m

b

∫xm−1(a + bx)√

a + bx=

2bxm(a + bx)1/2 − 2m

∫xm

√a + bx

− 2ma

b

∫xm−1

√a + bx

∫xm

√a + bx

dx =1

2m + 12bxm(a + bx)1/2 − 2ma

b(2m + 1)

∫xm−1

√a + bx

Exercise 31.∫

dx

xn√

ax + b= 2

√ax + b

axn+

∫n2√

ax + b

axn+1

√ax + b

ax + b=

2√

ax + b

axn+

2n

a

∫ax + b

xn+1√

ax + b=

=2√

ax + b

axn+ 2n

∫1

xn√

ax + b+

2nb

a

∫1

xn+1√

ax + b

=⇒ (1− 2n)∫

dx

xn√

ax + b− 2

√ax + b

axn=

∫b(

2na

)

xn+1√

ax + b∫

1xn√

ax + b=

−√ax + b

(n− 1)bxn−1− (2n− 3)a

(2n− 2)b

∫1

xn−1√

ax + b

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

(Cm+1S1−n)′ = (m + 1)Cm(−S2−n) + Cm+2(1− n)S−n = −(m + 1)CmS2−n + (1− n)CmS−n(1− S2) =

= −(m + 1 + 1− n)CmS2−n + (1− n)CmS−n = −(m− n + 2)CmS2−n + (1− n)CmS−n

=⇒∫

CmS−n =−(Cm+1S1−n)

n− 1− (m− n + 2)

n− 1

∫CmS2−n

Exercise 33.(Cm−1S1−n)′ = (m− 1)Cm−2(−S2−n) + (1− n)CmS−n

= −(m− 1)Cm−2(S−n)(1− C2) + (1− n)CmS−n =

= −(m− 1)Cm−2S−n + (m− 1)CmS−n + (1− n)CmS−n

Cm−1S1−n = −(m− 1)∫

Cm−2S−n + (m− n)∫

CmS−n

m− 1m− n

∫Cm−2S−n +

Cm−1S1−n

m− n=

∫CmS−n

Exercise 34.(1) P ′(x)− 3P (x) = 4− 5x + 3x2

P =n∑

j=0

ajxj

P ′ =n∑

j=1

ajjxj−1 =

n−1∑

j=0

aj+1(j + 1)xj

=⇒n−1∑

j=0

(aj+1(j + 1)− 3aj)xj = 4− 5x + 3x2

Generally, we can sayaj+1 =

3j + 1

aj if j ≥ 3

81

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 polynomialand we were only asked for one.

Suppose Q s.t. Q′ − 3Q = 4− 5x + 3x2 (another solution). Then

(P −Q)′ − 3(P −Q) = 0 ∀x

=⇒ (P −Q)′

P −Q= 3 =⇒ ln (P −Q) = 3x =⇒ ke3x = P −Q = k

∞∑

j=0

(3x)j

j!

Q = −k

∞∑

j=0

(3x)j

j!+ P

Since we didn't specify what Q has to be, we �nd that, in general, any Q is P plus some �amount� of the homoge-neous solution, ke3x.

(2) If Q(x) is a given polynomial, and suppose P is a polynomial solution to P ′(x) − 3P (x) = Q(x). Suppose R isanother polynomial solution such that R′(x)− 3R(x) = Q(x). Then just like above, P − R = ke3x. If we wantedpolynomial answers of �nite terms, then k must be zero. Thus, there's at most only one polynomial solution P .

Exercise 35. Bernoulli Polynomials.

(1) P1(x) = 1; P ′n(x) = nPn−1(x);∫ 1

0Pn(x)dx = 0, if n ≥ 1

n = 1 (1)(1) = P ′1

∫ 1

0

(x + c) = (12x2 + Cx)

∣∣∣∣1

0

=12

+ C = 0 C = −1/2

P1 = x− 1/2

n = 2 2(x− 1/2) = P ′2

∫ 1

0

(x2 − x + C) =(

13x3 − 1

2x2 + Cx

)∣∣∣∣1

0

=−16

+ C = 0 C = 1/6

P2 = x2 − x + 1/6

n = 3 3(x2 − x +16) = P ′3

∫ 1

0

(x3 − 3x2

2+

x

2+ C) =

14− 13

2+

14

+ C = 0

P3 = x3 − 3x2

2+

x

2

n = 4 4(x3 − 32x2 +

x

2) = P ′4

∫ 1

0

(x4 − 2x3 + x2 + C) =15− 14

2+

1313 + C = 0 C =

−130

P4 = x4 − 2x3 + x2 +−130

n = 5 5(x4 − 2x3 + x2 − 130

) = P ′5

∫ 1

0

(x5 − 5x4

2+

5x3

3− x

6+ C) =

16− (1)5

2+

5(1)4

12+−(1)2

12+ C = 0; C = 0

P5 = x5 − 5x4

2+

5x3

3− x

6

(2) The �rst, second, and up to �fth case has already been proven.Assume the nth case, that Pn(t) = tn +

∑n−1j=0 ajt

j (the general form of a polynomial of degree n).

P ′n+1 = (n + 1)(tn +n−1∑

j=0

ajtj) =⇒ Pn+1 = tn+1 + (n + 1)

n−1∑

j=0

ajtj+1

j + 1+ C

82

(3) The �rst, second, and up to �fth case has already been proven.Assume the nth case, that Pn(0) = Pn(1).

∫P ′n+1 = Pn+1(1)− Pn+1(0) (by the second fundamental theorem of calculus)

P ′n+1 = (n + 1)Pn∫ 1

0

(n + 1)Pn(t) = 0 (by the given properties of Bernoulli polynomials)

=⇒ 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).

P ′n+1 = (n + 1)Pn =⇒∫

P ′n+1 = (n + 1)∫

Pn

Pn+1(x + 1)− Pn+1(x) = Pn+1(a1) + (n + 1)∫ x+1

a1

Pn(t)− (Pn+1(a2) + (n + 1)∫ x

a2

Pn(t))

a1 = 1; a2 = 0; so Pn+1(1)− Pn+1(0) = 0 (from previous problems)

=⇒Pn+1(x + 1)− Pn+1(x) = (n + 1)(

∫ x+1

1

Pn(t)−∫ x

0

Pn(t)) =

= (n + 1)∫ x

0

Pn(t + 1)− Pn(t) = (n + 1)∫ x

0

ntn−1 = (n + 1)xn

(5)∫ k

0

Pn =∫ k

0

P ′n+1

n + 1=

Pn+1(k)− Pn+1(0)n + 1

Pn(x + 1)− Pn(x) = nxn−1

Pn(x + 1)− Pn(x)n

= xn−1

=⇒

k−1∑x=1

Pn+1(x + 1)− Pn+1(x)n + 1

=k−1∑x=1

xn =k−1∑r=1

Pn+1(r + 1)− Pn+1(r)n + 1

=

=k−1∑r=1

rn =Pn+1(k)− Pn+1(0)

n + 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 waythis question is asked), which gave us

∑x−1j=1 jn =

∫ x

0Pn(t)dt = Pn+1(x)−Pn+1(0)

n+1 .Use induction. It can be easily veri�ed, plugging in, that Pn(1 − x) = (−1)nPn(x) is true for n = 0 . . . 5.

Assume the nth case is true.

u = 1− t

du = −dt

∫ x

0

Pn(t)dt =∫ 1−x

1

−Pn(1− u)du = −∫ 1−x

1

Pn(u)(−1)ndu

(since Pn(1− x) = (−1)nPn(x), assumed nth case is true)

= (−1)n+1

∫ 1−x

1

Pn(t)dt =

= (−1)n+1

∫ 1−x

0

Pn(t)dt (since∫ 1

0

Pn = 0 )

=⇒= (−1)n+1

(Pn+1(1− x)− Pn+1(0)

n + 1

)=

Pn+1(x)− Pn+1(0)n + 1

=⇒ Pn+1(1− x) = (−1)n+1Pn+1(x)

In the second to last and last step, we had used (−1)n+1Pn+1(0) = Pn+1(0). For n + 1 even, this is de�nitely true.If n + 1 was odd,

Doing some algebra for the �rst �ve cases, we can show that P2j−1(0) = 0 for j = 2, 3. Assume the jth case istrue. Since P2j−1 is a polynomial and P2j−1(0) = 0, then the form of P2j−1 must be P2j−1 =

∑2j−1k=1 akxj . Using

83

P ′n+1 = (n + 1)Pn,

P2j(x)− P2j(0) = 2j

∫ x

0

P2j−1 = 2j

2j−1∑

k=1

∫ x

0

aktk = 2j

2j−1∑

k=1

ak1

k + 1tk+1

∣∣∣∣x

0

=

= 2j

2j−1∑

k=1

ak

k + 1xk+1 = 2j

2j∑

k=2

ak−1xk

k+ P2j(0)

P2j+1(x)− P2j+1(0) = (2j + 1)∫ x

0

(∑

k=2

(2j)ak−1

ktk + P2j(0)

)= (2j + 1)

2j+1∑

k=3

2jak−2xk

k(k − 1)+ (2j + 1)P2j(0)x

If we take the integral from 0 to 1, then we �nd that P2j+1(0) = 0(7) Using Pn(1− x) = (−1)nPn(x), derived above,

P2j+1(0) = (−1)2j+1P2j+1(1) = (−1)P2j+1(0)

=⇒ P2j+1(0) = 0

P2j−1(1− 12) = (−1)2j−1(P2j−1(

12))

=⇒ P2j−1(12) = 0

Exercise 36. There's a maximum at c for f , so f ′(c) = 0∫ x

a

f ′′(t)dt = f ′(x)− f ′(a)∫ c

0

f ′′(t)dt = f ′(c)− f ′(0) = −f ′(0)∫ a

c

f ′′(t)dt = f ′(a)− f ′(c) = f ′(a)

|f ′(0)| = |∫ 0

c

f ′′(t)dt| ≤∫ c

0

|f ′′|dt ≤ mc

|f ′(a)| ≤ m(a− c)

|f ′(0)|+ |f ′(a)| ≤ ma

6.9 Exercises - Introduction, Motivation for the de�nition of the natural logarithm as an integral, The de�nitionof the logarithm. Basic properties; The graph of the natural logarithm; Consequences of the functional equationL(ab) = L(a) + L(b); Logarithm referred to any positive base b 6= 1; Differentiation and integration formulas involv-ing logarithms;Logarithmic differentiation.

Exercise 1.(1)

log x = c + (ln |t|)|xe = c + ln |x| − 1 =⇒ ln(

x

|x|)

= c− 1

c = 1(2)

f(x) = ln1 + a

1− a+ ln

1 + b

1− b= ln

((1 + a)(1 + b)(1− a)(1− b)

)= ln

(1 + x

1− x

)

=⇒ 1 + x

1− x=

(1 + a)(1 + b)(1− a)(1− b)

=1 + a + b + ab

1− b− a + ab

x =a + b

1 + ab

Exercise 2.(1) log (1 + x) = log (1− x)

=⇒ x = 0(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− 11 + e

84

(3) 2 log x = x log 2

ln x2 + ln 2−x = 0 = ln 1 = ln x22−x =⇒ x = 2

(4) log (√

x +√

x + 1) = 1

√x + 1 = e−√x =⇒ x + 1 = e2 − 2

√xe + x

2√

xe = e2 − 1 =⇒ x =(

e2 − 12e

)2

Exercise 3.

f =ln x

x

f ′ =1− ln x

x2

f ′′ =−2x3

−( 1

xx2 − 2x ln x

x4

)=

2 ln x− 3x3

f ′ < 0 when x > e f ′ > 0 when 0 < x < e

for x3 > 0, 2 lnx− 3 > 0 =⇒ f ′′(x) < 0 (concave), when 0 < x < e3/2 ;

f ′′(x) > 0 (convex), when x > e3/2

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

f ′ =2x

1 + x2

Exercise 5. f(x) = log√

1 + x2

f ′ =x

1 + x2

Exercise 6. f(x) = log√

4− x2

f ′ =12−2x

4− x2=

−x

4− x2

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

f ′ =1

ln x

(1x

)=

1x ln x

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

f ′ = (2 log x + log log x)′ =2x

+1

x log x

Exercise 9. f(x) = 14 log x2−1

x2+1

f ′ =14

(log x2 − 1− log x2 + 1

)′=

14

(2x

x2 − 1− 2x

x2 + 1

)= x

(1

x4 − 1

)

Exercise 10. f(x) = (x +√

1 + x2)n

ln f = n ln (x +√

1 + x2)

f ′

f= n

(1

x +√

1 + x2

)(1 +

x√1 + x2

)=

n√1 + x2

f ′ = (x +√

1 + x2)n n√1 + x2

Exercise 11. f(x) =√

x + 1− log (1 +√

x + 1)

f ′ =1

2√

x + 1− 1

1 +√

x + 1

(1

2√

x + 1

)=

12(1 +

√x + 1)

Exercise 12. f(x) = x log (x +√

1 + x2)−√1 + x2

f ′ = log (x +√

1 + x2) +x

x +√

1 + x2

(1 +

x√1 + x2

)− x√

1 + x2

Note to self: Notice how this had made some of the square root terms disappear.Exercise 13. f(x) = 1

2√

ablog

√a+x

√b√

a−x√

b

85

f =1

2√

ab(ln (

√a + x

√b)− ln (

√a− x

√b))

f ′ = frac12√

ab

(1√

a + x√

b

√b− 1√

a− x√

b(−√

b))

=−x√

b√a(a− bx2)

Exercise 14. f(x) = x(sin (log x)− cos log x)

f ′ = sin (ln x)− cos (lnx) + (cos (ln x) + sin (ln x)) = 2 sin (ln x)

Exercise 15. f(x) = log−1 x

f ′ =−1

x(lnx)2

Exercise 16.∫

dx2+3x = 1

3 ln (2 + 3x)

Exercise 17.∫

log2xdx

(x ln2 x)′ = ln2 x + 2 ln x

(x ln x− x)′ = ln x + 1− 1 = lnx=⇒ xln2x− 2(x ln 2− x) = x ln2 x− 2x ln x + 2x

Exercise 18.∫

x log xdx (x2 ln x

2

)′= x ln x +

x

2∫x ln x =

x2 ln x

2− x2

4

Exercise 19.∫

x log2 xdx (x2 ln2 x

2

)′= x ln2 x + x2 ln x

(1x

)

=⇒∫

x log2 x =x2 ln2 x

2− x2 ln x

2− x2

4

Exercise 20.∫ e2−1

0dt

1+t ∫ e3−1

0

dt

1 + t= ln (1 + t)|e3−1

0 = 3

Exercise 21.∫

cot xdx ∫cos x

sin xdx = ln | sin x|

Exercise 22.∫

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

xn log ax =∫

xn log a +∫

xn log x =xn+1

n + 1log a +

∫xn log x

∫xn ln x =

xn+1

n + 1ln x−

∫xn+1

n + 11x

=xn+1

n + 1log x− xn+1

(n + 1)2

=⇒∫

xn log ax =xn+1

n + 1log a +

xn+1

n + 1log x− xn+1

(n + 1)2

Exercise 23.∫

x2 log2 xdx

∫x2 log2 x =

13x3 ln2 x−

∫x2

32 ln x =

x3 ln2 x

3− 2

3

(x3 ln x

3− x3

9

)=

x3 ln2 x

3− 2x3 ln x

9+

2x3

27

Exercise 24.∫

dxx log x ∫

dx

x ln x= ln (ln x) + C

Exercise 25.∫ 1−e−2

1log (1−t)

1−t dt

∫ 1−e2

0

ln (1− t)1− t

dt = −12(ln (1− t))2

∣∣∣∣1−e−2

0

= −2

86

Exercise 26.∫ log |x|

x√

1+log |x|dx

∫log x

x√

1 + log x=

∫(2(1 + log x)1/2)′ log x = 2(1 + log x)1/2 log x−

∫2(1 + log x)1/2

x=

= 2 log x(1 + log x)1/2 − 43(1 + log x)3/2

Exercise 27. Derive ∫xm logn xdx =

xm+1

m + 1lnn x− n

m + 1

∫xm lnn−1 x

By inspection, we just needed integration by parts.∫

x3 ln3 x =x4

4ln3 x− 3

4

∫x3 ln2 x =

x4

4ln3 x− 3

4

(x4

4ln2 x− 2

4

∫x3 ln x

)=

x4 ln3 x

4− 3x4 ln2 x

16+

3x4 ln x

32− 3x4

128

Exercise 28. Given x > 0, f(x) = x− 1− ln x; g(x) = ln x− 1 + 1x

(1)f ′ = 1− 1

x

g′ =1x− 1

x2=

1x

f ′xg′ = f ′

so then if f ′ > 0, g′ > 0; f ′ < 0, g′ < 0

For f ′ < 0 0 < x < 1 f ′ > 0 x > 1 f ′(1) = g′(1) = 0

f(1) = g(1) = 0

x− 1− ln x > 0 since f(1) = 0 is a rel. min.

0 < ln x− 1 +1x

since g(0) is a rel. min.(2) See sketch.

Exercise 29. limx→0log (1+x)

x = 1(1) L(x) =

∫ x

11t dt ; L′(x) = 1

x ; L′(1) = 1(2) Use this theorem.

Theorem 19 (Theorem I.31).If 3 real numbers a, x, and y satisfy the inequalities

a ≤ x ≤ a +y

n∀n ≥ 1, n ∈ Z, then x = a

1− 1x

< ln x < x− 1 =⇒ 1− 1x + 1

< ln x + 1 < x

1− x

1 + x<

ln x + 1x

< 1 =⇒ ln (1 + x)x

= 1

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

r =p

q=⇒ f(ap/q) = f((a1/q)p) = pf(a1/q) =

p

qf(a)

sincef(a) = f(a); f(a2) = f(aa) = 2f(a)

f(ap+1) = f(ap) + f(a) = (p + 1)f(a)

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

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

Exercise 31. ln x =∫ |x|1

1t dt

(1) ln x =∫ |x|1

1t dt From this de�nition, then for n partitions, a0 = 1, a1 = 1 + b−1

n , . . . , an = b = x

b−an = b−1

n ; so if ak = 1 + k(

b−1n

)n∑

k=1

(ak − ak−1

ak

)< log x <

n∑

k=1

(ak − ak−1

ak−1

)

87

(2) log x is greater than the step function integral consisting of rectangular strips within 1x and less than rectangular strips

covering over 1x

(3) ak = 1 + k =⇒ ak−ak−1ak−1

= 1k

n∑

k=1

11 + k

< ln (n + 1) <

n∑

k=1

1k

=⇒n+1∑

k=2

1k

< ln (n + 1) <

n∑

k=1

1k

=⇒n∑

k=2

1k

< ln (n) <

n−1∑

k=1

1k

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 )

loga x =1

logb alogb x =⇒ logb x = logb a loga x

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

Exercise 33. loge 10 = 2.302585

log10 e =loge e

loge 10=⇒ loge 10 =

1log10 e

=1

2.302585' 0.43429

Exercise 34. Given∫ xy

xf(t)dt = B(y) f(2) = 2, , We Want A(x) =

∫ x

1f(t)dt

∫ xy

x

f(t)dt = B(xy)−B(x)

d

dx

∫ xy

x

f(t)dt =d

dx(B(xy)−B(x)) =

dB(xy)d(xy)

y − dB(x)dx

= f(xy)y − f(x) = 0

=⇒ f(xy) =f(x)

y

A(x) =∫ x

1

f(t)dt =∫ x

1

f((2)(

t

2

))dt =

∫ x

1

f(2)(t/2)

dt = 4 ln x

Exercise 35. Given∫ xy

1f(t)dt = y

∫ x

1f(t)dt + x

∫ y

1f(t)dt, and letting F be the antiderivative of f ,

F (xy)− F (1) = y(F (x)− F (1)) + x(F (y)− F (1))

d/dx−−−→ f(xy)(y) = y(f(x)) +∫ y

0

f(t)dtd/dx−−−→ f ′(xy)y2 = y(f ′(x))

x=1−−→ f ′(y) =(f ′(1))

y

R−→ f(y) = k ln y + C

y=1−−→ f(1) = 0 + C = 3

∫ xy

1(k ln t + 3) = y

∫ x

1(k ln t + 3) + x

∫ y

1(k ln t + 3)

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,

ln1 + x

1− x= 2(x +

x3

3+ · · ·+ x2m−1

2m− 1) + Rm(x)

88

wherex2m+1

2m + 1< Rm(x) ≤ 2− x

1− x

x2m+1

2m + 1Rm(x) = E2m(x)− E2m(−x)

where E2m(x) is the error term for log 1− x

m = 5, x = 13

ln(

4/32/3

)= ln 2 ' 2

(13

+

(13

)3

3+

(13

)5

5+

(13

)7

7+

(13

)9

9

)' 0.693146047

The error for m = 5, x = 13 is

(13

)11

11≤ R5(x) ≤ 5

2

(13

)11

11

Exercise 2. ln(

1+x1−x

)= ln 3

2 = ln 3− ln 2

x =15

m = 5

ln(

1 + x

1− x

)= 2

(15

+(1/5)3

3+

(1/5)7

7+

(1/5)9

9

)= 0.405465104

(1/5)11

11' 0.000000002 < R5(x) ≤ 9

4

(15

)11

11' 0.000000004

=⇒log 3 ' 1.098611

1.098611666 < log 3 < 1.098612438

Exercise 3. x = 19 ln

(1+ 1

91− 1

9

)= ln

(108

)= ln 5

4 = ln 5− 2 ln 2

For m = 2

2

(19

+

(19

)5

3+

(19

)5

5

)= 0.2231435

(19

)7

7' 0.00000003 < R3(x) ≤

17989

(19

)7

7' 0.000000063

1.609437 < log 5 < 1.609438

Exercise 4. x = 16 ln

(1+ 1

61− 1

6

)= ln

(75

)= ln 7− ln 5.

For m = 3,

ln1 + x

1− x' 2

(16

+

(16

)3

3+

(16

)5

5

)= 0.336471193

The error bounds are

(16

)2(3)+1

2(3) + 1= 0.00000051

(16

)2(3)+1

2(3) + 1

(2− 1

6

1− 16

)= 0.000001123

1.945908703 < ln 7 < 1.945910316

Exercise 5.0.6931460 < ln 2 < 0.6931476

ln 3 = 1.098614ln 4 = 1.386293ln 5 = 1.609436 ln 6 = ln 2 + ln 3 = 1.791700

ln 7 = 1.945909ln 8 = 3 ln 2 = 2.0794404ln 9 = 2 ln 3 = 2.197228

ln 10 = ln 5 + ln 2 = 1.302577

89

6.17 Exercises - The exponential function, Exponentials expressed as powers of e, The de�nition of ex for arbitraryreal x, The de�nition of ax for a > 0 and x real, Differentiation and integration formulas involving exponentials.

Exercise 1. f ′ = 3e3x−1 Exercise 2. 8xe4x2 Exercise 3. −2xe−x2 Exercise 4. 12√

xe√

x Exercise 5. −1x2 e−1/x Exercise 6.

ln 22x Exercise 7. (2x ln 2)2x2 Exercise 8. cos xesin x Exercise 9. −2 cos x sinxecos2 x Exercise 10. 1xelog x Exercise 11.

exeex Exercise 12. eeex

(exeex

) Exercise 13.∫

xexdx = xex − ex Exercise 14.∫

xe−xdx = −xe−x + −e−x Exercise 15.∫

x2ex = x2ex − 2xex + 2ex Exercise 16.∫

x2e−2xdx = x2e−2x

−2 + xe−2x

−2 + −e−2x

4

Exercise 17.∫

e√

x = e√

x(2√

x)− ∫1√xe√

x = e√

x(2√

x)− 2e√

x = 2(√

xe√

x − e√

x)

Exercise 18.∫

x3e−x2 .

x2e−x2= −2x3e−x2

+ 2xe−x2

e−x2= −2xe−x2

1−2

(x2e−x2+ e−x2

)′ =1−2

(2xe−x2

+−2x3e−x2+−2xe−x2

)=

x2e−x2+ e−x2

−2

Exercise 19. ex = b +∫ b

aetdt = b + ex − ea, ea = b.

Exercise 20.

A =∫

eax cos bxdx

B =∫

eax sin bxdx

A =eax

acos bx−

∫−b sin bxeax

a=

eax

acos bx +

b

aB =⇒ aA + bB = eax cos bx + C

B =eax

asin bx−

∫eax

ab cos bx =⇒ aB + bA = eax sin bx + C

A =1

a2 + b2(aeax cos bx + beax sin bx)

B =−beax cos bx + aeax sin bx

a2 + b2

Exercise 21.ln f = x ln x;

f ′

f= ln x + 1; f ′ = xx(ln x + 1)

Exercise 22.ln

f ′

f=

11 + x

+1

1 + ex2 (2xex2), f ′ = 1 + ex2

+ 2(1 + x)xex2

Exercise 23. f = ex−e−x

ex+e−x .ln f = ln (ex − e−x)− ln (ex + e−x)

f ′

f=

1ex − e−x

(ex + e−x)− 1ex + e−x

(ex − e−x)

f ′ = 1−(

ex − e−x

ex + e−x

)2

Exercise 24. f ′ = (xaa

)′ + (axa

)′ + (aax

)′.f1 = xaa

ln f1 = aa ln x

f ′1f1

=aa

x

f ′1 = xaa−1aa

f2 = axa

ln f2 = xa ln a

f ′2f2

= axa−1 ln a

f ′2 = axa+1xa−1 ln a

f3 = aax

ln f3 = ax ln a

f ′3f3

= (ln a)2ax

f ′3 = (ln a)2ax+ax

90

=⇒ f ′ = xaa−1aa + axa+1xa−1 ln a + (ln a)2ax+ax

Exercise 25.ef = ln (lnx);

f ′ef =1

ln x

(1x

)

f ′ =1

ln (ln x)

(1

ln x

)(1x

)

Exercise 26. ef = ex +√

1 + e2x

(ef )f ′ = ex +e2x

√1 + e2x

f ′ =√

1 + e2xex + e2x

√1 + e2x(ex +

√1 + e2x)

=ex

√1 + e2x

Exercise 27.ln f = xx ln xf ′

f= (xx)′ ln x +

xx

x= (xx(lnx + 1)) ln x + xx−1

f ′ = xx+xx

(lnx + 1) ln x + xxxx+x−1

Exercise 28. ln f = x ln (ln x)f ′

f= ln (lnx) +

1ln x

f ′ = (ln x)x(ln (ln x) +1

ln x)

Exercise 29. ln f = (ln x) ln xf ′

f=

2 ln x

x

f ′ = 2xlog x−1 ln x

Exercise 30. ln f = x ln (ln x)− ln x ln x = x ln ln x− (ln x)2

f ′

f= ln (ln x) +

1ln x

− 2ln x

x

f ′ =(ln x)x

xln x

(ln lnx +

1ln x

− 2 ln x

x

)

Exercise 31.ln f1 = cos x ln sin x

f ′1f1

= − sin x ln sin x +cos2 x

sinx

f ′1 = − sin x cosx(ln sin x)2 +cos3 x ln sinx

sin x

ln f2 = sinx ln cos x

f ′2f2

= cos x ln cos x +− sin2 x

cos x

f ′2 = sinx cosx(ln cosx)2 − sin3 x ln cos x

cosx

=⇒ f = sin x cos x(−(ln sin x)2 + (ln cos x)2) +cos3 x ln sin x

sin x+− sin3 x ln cos x

cos x

Exercise 32. ln f = 1x ln x

f ′

f = −1x2 ln x + 1

x2 =⇒ f ′ = −x1/x−2 ln x + x1/x−2

Exercise 33. ln f = 2 ln x + 13 ln (3− x)− ln (1− x) + −2

3 ln (3 + x)

f ′

f=

2x

+−1

3(3− x)− −1

1− x− 2

31

3 + x

f ′ = 2x(3− x)1/3

(1− x)(3 + x)2/3+−1

3

(x2(3− x)−2/3

(1− x)(3 + x)2/3

)+

x2(3− x)1/3

(1− x)2(3 + x)2/3− 2

3x2(3− x)1/3

(1− x)(3 + x)5/3

=x(18− 12x + 4

3x2 + 23x3)

(1− x)2(3 + x)5/3(3− x)2/3

91

Exercise 34. ln f =∑n

i=1 bi ln (x− ai)

f ′

f=

n∑

i=1

bi

x− ai=⇒ f ′ =

n∑

j=1

bj

x− aj

n∏

i=1

(x− ai)bi

Exercise 35.(1) Show that f ′ = rxr−1 for f = xr holds for arbitrary real r.

xr = er ln x

(er ln x)′ = er ln x r

x= rxr−1

(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 ′(x)f(x)

= r1x

=⇒ f ′(x) = rxr−1

Exercise 36. Use the de�nition ax = ex log a

(1) log ax = x log a

Taking the exponential is a well-de�ned inverse function to log so taking the log of both sides of the de�nition, weget log ax = x log a

(2) (ab)x = axbx

(ab)x = ex log ab = ex(log a+log b) = axbx

(3) axay = ax+y

ax+y = e(x+y) log a = ex log aey log a = axay

(4) (ax)y = (ay)x = axy

(ax)y = exy log a = (ay)x = axy

(5) If x = loga y,Using the de�nition

logb x =log x

log bif b > 0, b 6= 1, x > 0

so then

loga y =log y

log a= x =⇒

log y = x log a

ex log a = elog y = y = ax

If y = ax,

log y = x log a =⇒ x =log y

log a= loga y

Exercise 37. Let f(x) = 12 (ax + a−x) if a > 0.

f(x + y) =12(ax+y + a−(x+y))

f(x + y) + f(x− y) =12

(ax+y + a−(x+y) + ax−y + a−(x−y)

)

f(x)f(y) =14(ax + a−x)(ay + a−y) =

14

(ax+y + a−x−y + a−(x−y) + a(x−y)

)

Exercise 38.f(x) = ecx; f ′(x) = cecx; f ′(0) = c

limx→0

ecx − 1x

= c

(limx→0

ecx − 1cx

)= c

(lim

cx→0

ecx − 1cx

)= c

df(cx)d(cx)

(0) =d

dx(ecx)(0)

limx→0

ecx − 1x

= f ′(0) = c

92

Exercise 39.g(x) = f(x)e−cx

g′(x) = f ′e−cx +−cg = cg − cg = 0

f = Kekx

Exercise 40. Let f be a function de�ned everywhere on the real axis. Suppose also that f satis�es the functional equation

f(x + y) = f(x)f(y) for all x and y

(1)

f(0) = f(0)f(0) = f2(0)If f(0) = 0, then we're doneIf 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)

d(x + y)dx

=df(x)dx

f(y) =⇒ d(f(x + y))d(x + y)

= f ′(x)f(x + y)

f(x)

Let y = −x + y

df(y)dy

= f ′(x)f(y)f(x)

=⇒ f ′(x)f(y) = f ′(y)f(x)

(3) f ′(x)f(x) = f ′(y)

f(y) ∀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 forthem to both equal a constant=⇒ f ′(y)

f(y) = c

(4) Referring to Exercise 39 of the same section, f = ecx since f ′(0) = 1

Exercise 41.(1)

f = ex − 1− x

f ′ = ex − 1 ≷ 0 if x ≷ 0f(0) = e0 − 1− 0 = 0

{ex > 1 + x for x > 0e−x > 1− x for x < 0

(2) ∫ x

0

et = ex − 1 > x +12x2 =⇒ ex > 1 + x +

12x2

− e−x > −1 + x− x2

2=⇒ e−x < 1− x +

x2

2(3) ∫ x

0

et = ex − 1 > x +12x2 +

13 ∗ 2

x3 =⇒ ex > 1 + x +12x2 +

13 ∗ 2

x3

− e−x > −1 + x− x2

2+

x3

3 ∗ 2=⇒ e−x < 1− x +

x2

2− x3

3 ∗ 2(4) Suppose the nth case is true.

ex >

n∑

j=0

xj

j!e−x =

{>

∑2m+1j=0

xj

j!

<∑2m

j=0xj

j!

ex > 1 +n∑

j=0

xj+1

(j + 1)!= 1 +

n+1∑

j=1

xj

j!=

n+1∑

j=0

xj

j!

− e−x + 1

{>

∑2m+1j=0

xj+1

(j+1)!

<∑2m

j=0xj+1

(j+1)!

= e−x

{<

∑2m+2j=0

xj

j!

>∑2m+1

j=0xj

j!

Exercise 42. Using the result from Exercise 41,(1 +

x

n

)n

=n∑

j=0

(n

j

)1n−j

(x

n

)j

=n∑

j=0

n!(n− j)!j!

xj

nj=

n∑

j=0

n(n− 1) . . . (n− j + 1)j!

xj

nj<

n∑

j=0

xj

j!< ex

93

If you make this clever observation, the second inequality is easy to derive.

x > 0x

n> 0

e−xn > 1− x

n=⇒

(e−x/n

)n

>(1− x

n

)n

e−x > (1− x/n)n =⇒ ex < (1− x/n)−n

Exercise 43. f(x, y) = xy = ey ln x

∂xf = xyy/x

∂yf = xy ln x

6.19 Exercises - The hyperbolic functions.Exercise 7.

2 sinh x cosh x = 2ex − e−x

2ex + e−x

2=

12(e2x − e−x) = sinh 2x

Exercise 8.

cosh2 x + sinh2 x =(

ex + e−x

2

)2

+(

ex − e−x

2

)2

=14(e2x + 2 + e−2x + e2x − 2 + e−2x) = cosh 2x

Exercise 9.

cosh x + sinh x =ex + e−x

2+

ex − e−x

2= ex

Exercise 10.

cosh x− sinhx =ex + e−x

2−

(ex − e−x

2

)= e−x

Exercise 11. Use induction.(coshx + sinh x)2 = cosh2 x + 2 sinh x cosh x + sinh2 x = cosh 2x + sinh 2x

(coshx + sinh x)n+1 = (cosh x + sinh x)(cosh nx + sinh nx) == cosh nx cosh x + cosh nx sinhx + sinh nx cosh x + sinh x sinhnx =

=

enx + e−nx

2ex + e−x

2+

enx + e−nx

2ex − e−x

2+

+enx − e−nx

2ex + e−x

2+

enx − e−nx

2ex − e−x

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.

(cos x)′ = − sinx = −√

1− cos2 xD arccos x =1

−√1− x2− 1 < x < 1

Exercise 2.

(tan x)′ = sec2 x =sin2 x + cos2 x

cos2 x= tan2 x + 1

D arctanx =1

1 + x2

Exercise 3.

(cotx)′ = − csc2 x = − (sin2 x + cos2 x)sin2 x

= −(1 + cot2 x) =⇒ arccotx = − 11 + x2

Exercise 4.(sec y)′ = 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. Then we must make sec y → | sec y|.Darcsecx = 1

|x|√x2−1

94

Exercise 5.(csc y)′ = − cot x csc x = − csc y(

√csc2 y − 1)

Let y such that −π

2< y <

π

2(csc y) < 0

Darccscx =1

−|x|√x2 − 1Exercise 6.

(xarccotx)′ = arccotx− x

1 + x2

(12

ln (1 + x2))′

=1x

(1 + x2)∫arccotx = xarccotx +

12

ln (1 + x2) + C

Exercise 7.(xarcsecx)′ = arcsecx +

x

|x|√x2 − 1

(x

|x| ln |x +√

x2 − 1|)′

=

1+ x√x2−1

|x+√

x2−1| x > 1

−−1+ x√

x2−1

|x+√

x2−1| x < −1

=x

|x|√x2 − 1

=⇒∫

arcsecxdx = xarcsecx− x

|x| log |x +√

x2 − 1|+ C

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.

(xarccscx)′ = arccscx +x

−|x|√x2 − 1(

x

|x| ln |x +√

x2 − 1|)′

=

1x+√

x2−1

(1 + x√

x2−1

)= 1√

x2−1x > 1

−1√x2−1

x < −1

=⇒∫

arccscx = xarccscx +x

|x| ln |x +√

x2 − 1|Exercise 9.

(x(arcsin x)2)′ = (arcsin x)2 +2x arcsin x√

1− x2

(√1− x2 arcsin x

)′=

−x√1− x2

arcsin x + 1

∫(arcsin x)2 = x(arcsin x)2 + 2

√1− x2 arcsinx− 2x

Exercise 10. (− arcsin x

x

)′=

1x2

arcsin x− 1x√

1− x2

I would note how x is in the denominator of the second term. Again, reiterating,

(√

1− x2)′ =−x√1− x2

(y ±√±1∓ x2)(y ∓

√±1∓ x2) = y2 − (±1∓ x2)

Multiply by its �conjugate.� As we see, choose y appropriately to get the desired denominator (that's achieved after differen-tiation). Here, pick y = 1.

(ln (1 +√

1− x2))′ =1

1 +√

1− x2

( −x√1− x2

)=−x(1−√1− x2)√

1− x2(x2)=−(1−√1− x2)

x√

1− x2

=⇒∫

arcsinx

x2= ln

∣∣∣∣∣1−√1− x2

x

∣∣∣∣∣−arcsinx

x+ C

Exercise 11.95

(1)D

(arccotx− arctan

1x

)=

−1x2 + 1

− 1

1 +(

1x

)2

(−1x2

)= 0

(2) arccotx− arctan 1x = C

Now arccotx = π2 − arctanx.

π

2− arctanx− arctan

1x

= C =⇒ π

2− C = arctan x + arctan

1x

x →∞ =⇒ π

2− C =

π

2+ 0 =⇒ C = 0

but x → −∞π

2− C = −π

2+ 0 =⇒ C = π

There are problems with the choice of brances for arccotx, arctan 1x , even though the derivatives work in all cases.

Exercise 12.f ′ =

1√1− (

x2

)2

(12

)

Exercise 13.f ′ =

−1√1−

(1−x√

2

)2

(−1√2

)=

1√1 + 2x− x2

Exercise 14. f = arccos 1x .

f ′ =−1√

1− (1x

)2

(−1x2

)=

1√x2 − 1|x|

Exercise 15.f(x) = arcsin (sin x) =

1√1− sin2 x

cos x =cos x

| cos x|Exercise 16.

12 sqrtx

− 1x + 1

12√

x=

√x

2(x + 1)Exercise 17.

11 + x2

+x2

1 + x6

Exercise 18.1√

1−(√

1−x2

1+x2

)2

( −2x(2)(1 + x2)2

)=

√1 + x2

√1 + x2 − (1− x2)

−4x

(1 + x2)2=

−4(1 + x2)3/2

√2

Exercise 19. f = arctan tan2 x1

1 + tan4 x

(2 tan x sec2 x

)=

2 tan x sec2 x

1 + tan4 x

Exercise 20.

f ′ =1

1 + (x +√

1 + x2)2(1 +

x√1 + x2

) =√

1 + x2 + x

1 + (x +√

1 + x2)2

Exercise 21.f ′ =

1√1− (sin2 x + cos2 x− 2 sin x cosx)

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

Exercise 22.f ′ = (arccos

√1− x2)′ =

1−

√1− (1− x2)

= − 1|x|

Exercise 23.

f ′ =1

1 +(

1+x1−x

)2

(2

(1− x)2

)=

2(1− x)2 + (1 + x)2

=2

(1− 2x + x2 + 1 + 2x + x2)=

11 + x2

96

Exercise 24. f = (arccos (x2))−2

f ′ = −2(arccos x2)−3

( −1√1− x2

)(2x) =

4x(arccosx2)−3

√1− x4

Exercise 25.

f ′ =

(1

arccos 1√x

) −1√

1− 1x

( −12x3/2

)=

12 arccos 1√

x

√x3 − x2

Exercise 26. dydx = x+y

x−y .(arctan

y

x

)′=

1

1 +(

yx

)2

(y′x− y

x2

)=

(12

ln (x2 + y2))′

=12

1(x2 + y2)

(2x + 2yy′)

=⇒ y′ =x + y

x− y

Exercise 27.

ln y = ln (arcsin x)− 12

ln (1− x2)

y′

y=

1arcsin x

(1√

1− x2

)− 1

21

1− x2(−2x) =

1arcsinx

√1− x2

+x

1− x2

y =arcsin x√

1− x2

y′ =(

11− x2

)+

(arcsin x)x(1− x2)3/2

=√

1− x2 + x(arcsin x)(1− x2)3/2

ln y′ = ln (√

1− x2 + x arcsinx)− 32

ln (1− x2)

y′′

y′=

1√1− x2 + x arcsin x

( −x√1− x2

+ arcsin x +x√

1− x2

)− 3

2(−2x)1− x2

y′′ = y′arcsin x√

1− x2 + x arcsin x+

y′3x

1− x2=

arcsin x

(1− x2)3/2+

(√

1− x2 + x arcsin x)(3x)(1− x2)3/2

Exercise 28.

f ′ =1

1 + x2− 1 + x2 =

1− 1− x2 + x4

1 + x2=

x4

1 + x2≥ 0∀x

since f(0) = arctan 0− 0 + 0 = 0, arctanx > x− x3

3, ∀x > 0

Exercise 29. ∫dx√

a2 − x2, a 6= 0 =⇒ arcsin

x

a

Exercise 30. ∫dx√

2− (x + 1)2=

∫dx

√2

√1−

(x+1√

2

)2= arcsin

x + 1√2

Exercise 31. ∫dx

a2(1 +

(xa

)2) =

1a

arctanx

a

Exercise 32.dx

a(1 +(√

ba√a

)2

)=

1√ba

arctan

√bx√a

Exercise 33. ∫dx(

x− 12

)2 + 74

=47

∫dx(

2√7

(x− 1

2

))2

+ 1=

2√7

arctan2

(x− 1

2

)√

7

97

Exercise 34. (x2 arctan x

2

)′= x arctan x +

x2

21

1 + x2= x arctanx +

12

(1− 1

1 + x2

)

(12

(x− arctanx))′

=12

(1− 1

1 + x2

)

∫x arctan x = x arctanx +−1

2(x− arctanx)

Exercise 35. (x3

3arccosx

)′= x2 arccosx +

x3

3−1√1− x2

(x2√

1− x2)′ = 2x√

1− x2 +− x3

√1− x2

((1− x2)3/2)′ =32(−2x)(1− x2)1/2 = −3x(1− x2)1/2

∫x2 arccosx =

x3

3arccosx− 1

3x2

√1− x2 − 9

2(1− x2)3/2

Exercise 36.(

x2(arctanx)2

2

)′= x(arctan x)2 + x2

(1

1 + x2

)arctan x = x(arctan x)2 +

(1−

(1

1 + x2

))arctanx

((arctanx)2

2

)′=

arctan x

1 + x2

(x arctan x)′ = arctan x +x

1 + x2

∫x(arctan x)2dx =

x2(arctanx)2

2−

(x arctanx− ln (1 + x2)

2

)+

(arctan x)2

2

Exercise 37.

(arctan√

x)′ =(

11 + x

)(1

2√

x

)

(x arctan√

x)′ = arctan√

x +√

x

2(1 + x)= arctan

√x +

12

(1√x− 1√

x(1 + x)

)

(x arctan√

x + arctan√

x +−x1/2)′ = arctan√

x + 0∫

arctan√

x = x arctan√

x + arctan√

x− x1/2

Exercise 38. From the previous exercise, ∫arctan

√x√

x(1 + x)dx = (arctan

√x)2

Exercise 39. Let x = sin u∫ √

1− x2dx =∫

cos2 udu =u

2+

sin 2u

4=

arcsinx

2+

x√

1− x2

4

Exercise 40. ∫xearctan x

(1 + x2)3/2

(earctan x

√1 + x2

)′=−xearctan x

(1 + x2)3/2+

earctan x

(1 + x2)3/2

(xearctan x

√1 + x2

)′=

earctan x

√1 + x2

+− x2earctan x

(1 + x2)3/2+

xearctan x

(1 + x2)3/2=

earctan x

(1 + x2)3/2+

xearctan x

(1 + x2)3/2

12

(xearctan x

√1 + x2

− earctan x

√1 + x2

)′=

xearctan x

(1 + x2)3/2

98

Exercise 41. From the previous exercise,12

(xearctan x

√1 + x2

+earctan x

√1 + x2

+ C

)

Exercise 42.

Since −(

x(1 + x2)−1

2

)′=

x2

(1 + x2)2− 1

2(1 + x2)∫x2

(1 + x2)2dx =

−x

2(1 + x2)+

12

arctan x

Exercise 43. arctan ex.Exercise 44. ∫

arccotex

exdx

(arccotex)′ =−ex

1 + e3x

− (e−xarccotex)′ = e−xarccotex +e−x(−1)ex

1 + e2x= e−xarccotex +−

(1− e2x

1 + e2x

)

(ln (1 + e2x))′ =2e2x

1 + e2x

(−e−xarccotex + x− 12

ln (1 + e2x))′ = e−xarccotex

Exercise 45.∫ √

a + x

a− xdx =

∫a + x√a2 − x2

dx =∫

1a√1− (

xa

)2+

x√a2 − x2

dx = a arcsin

x

a+−

√a2 − x2

Exercise 46.∫ √x− a

√b− xdx =

∫ √bx− ab− x2 + axdx =

=∫ √

−(

x−(

a + b

2

))(x−

(a + b

2

))+

a2 + b2

4− 2ab

4=

=∫ √(

a− b

2

)2

−(

x−(

a + b

2

))2

=(

a− b

2

) ∫ √√√√1−(

x− (a+b2

)(

a−b2

))2

dx =

=(

a− b

2

)2 ∫ √1− u2 =

=(

a− b

2

)2 arcsin(

2x−(a+b)a−b

)

2+

2x− (a + b)2(a− b)2

√(a− b)2 − (2x− (a + b))2

Since, recall,(

arcsin x

2+

12x√

1− x2

)′=

12

1√1− x2

+√

1− x2

2+

14

x(−2x)√1− x2

=√

1− x2

Exercise 47. Wow!∫

dx√(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)

∫dx√

(x− a)(b− x)=

∫(b− a)(2) sin u cos udu√b− a cos u

√b− a sin u

= 2u = 2arcsin√

x− a

b− a

99

6.25 Exercises - Integration by partial fractions, Integrals which can be transformed into integrals of rational func-tions.Exercise 1.

∫2x+3

(x−2)(x+5) =∫ (

1x−2

)+

(1

x+5

)= ln (x− 2) + ln (x + 5)

Exercise 2.∫

xdx(x+1)(x+2)(x+3)

A

x + 1+

B

x + 2+

C

x + 3= A(x2 + 5x + 6) + B(x2 + 4x + 3) + C(x2 + 3x + 2)

=⇒

1 1 15 4 36 3 2

ABC

=

010

=⇒

1 1 15 4 36 3 2

∣∣∣∣∣∣

010

=

1 01 00 1

∣∣∣∣∣∣

−1/22

−3/2

A = −1/2, B = 2, C = −3/2

=⇒ −12

ln (x + 1) + 2 ln (x + 2) +−32

ln (x + 3)

Exercise 3.∫

x(x−2)(x−1) =

∫2

x−2 + −1x−1 = 2 ln x− 2− ln (x− 1)

Exercise 4.∫

x4+2x−6x3+x2−2xdx

x4 + 2x− 6x3 + x2 − 2x

= x− 1 +3(x2 − 2)

x3 + x2 − 2x(do long division)

∫x− 1 +

3(x2 − 2)x(x + 2)(x− 1)

=12x2 − x + 3

∫x2 − 2

x(x + 2)(x− 1)∫x2 − 2

x(x + 2)(x− 1)=

∫1x

+1/3

x + 2+−1/3x− 1

= ln x +13

ln x + 2− 13

ln x− 1

=⇒ 12x2 − x + 3 ln x + ln x + 2− ln x− 1

Exercise 5.∫

8x3+7(x+1)(2x+1)3 dx

8x3 + 7(x + 1)(2x + 1)3

=A

(2x + 1)3+

B

(2x + 1)2+

C

(2x + 1)+

D

(x + 1)8x3 + 7 = A(x + 1) + B(2x2 + 3x + 1) + (4x3 + 8x2 + 5x + 1)C + D(8x3 + 12x2 + 6x + 1)

=⇒

0 0 4 80 2 8 121 3 5 61 1 1 1

ABCD

=

8007

=⇒

1 0 0 01 0 0

1 00 1

|12−601

A = 12, B = −6, C = 0, D = 1∫

12(2x + 1)3

+−6

(2x + 1)2+

1x + 1

=−6(2x + 1)−2

2+

6(2x + 1)−1

2+ ln (x + 1)

Exercise 6.∫

4x2+x+1(x−1)(x2+x+1)

4x2 + x + 1(x− 1)(x2 + x + 1)

=A

x− 1+

Bx + C

x2 + x + 1=⇒ A(x2 + x + 1) + (Bx + C)(x− 1) = 4x2 + x + 1

=⇒

1 11 −1 11 0 −1

ABC

=

411

=⇒ A = 2, B = 2, C = 1

=⇒∫

2x− 1

+2x + 1

x2 + x + 1= 2 ln |x− 1|+ ln |x2 + x + 1|

Exercise 7.∫

x4dxx4+5x2+4

100

Doing the long division, x4

x4+5x2+4 = 1 +−(

5x2+4(x2+1)(x2+4)

)

Ax + B

x2 + 1+

Cx + D

x2 + 4=

5x2 + 4(x2 + 1)(x2 + 4)

It could be seen that A + C = 0, 4A + C = 0 so A = C = 0

B + D = 54B + D = 4

B =−13

D =163

=⇒∫

1− 5x2 + 4(x2 + 1)(x2 + 4)

= x−∫ −1/3

x2 + 1+

16/3x2 + 4

= x +13

arctanx + 4/3 arctan x/2 + C

Exercise 8.∫

x+2x(x+1)dx =

∫1

x+1 + 2x(x+1) = ln |x + 1|+ 2

∫1x − 1

x+1 = − ln |x + 1|+ 2 lnx

Exercise 9.∫

dxx(x2+1)2 =

∫Ax + Bx+C

(x2+1) + Dx+E(x2+1)2

A(x4 + 2x2 + 1) + x(Bx + C)(x2 + 1) + Dx2 + Ex

x(x2 + 1)2=

A(x4 + 2x2 + 1) + Bx4 + Cx3 + Bx2 + Cx + Dx2 + Ex

x(x2 + 1)2

=⇒ A = 1; B = −1; D = −1; C = 0; E = 0∫

1x

+−x

x2 + 1+

−x

(x2 + 1)2= ln x +− ln |x2 + 1|

2+

(x2 + 1)−1

2

Exercise 10.∫

dx(x+1)(x+2)2(x+3)3

Exercise 11.∫

x(x+1)2 dx

x

(x + 1)2=

A

x + 1+

B

(x + 1)2=⇒ x = A(x + 1) + B

A = 1; B = −1∫ (

1x + 1

+−1

(x + 1)2

)dx = ln x + 1 +

1x + 1

+ C

Exercise 12.∫

dxx(x2−1) =

∫dx

x(x−1)(x+1) =∫

Ax + B

x−1C

x+1

A(x2 − 1) + Bx(x + 1) + Cx(x− 1) = Ax2 −A + Bx2 + Bx + Cx2 − Cx =⇒ A = −1, B =12

= C∫ −1

x+

1/2x− 1

+1/2

x + 1= − ln x +

12

ln |x− 1|+ 12

ln |x + 1|

Exercise 13.∫

x2dxx2+x−6 =

∫x2dx

(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

x2+x−6 = 1 + 6−xx2+x−6

A

x + 3+

B

x− 2=⇒

6 = A(x− 2) + B(x + 3)

A =−65

; B =65

−x = (A + B)x− 2A + 3B

2A = 3B

A =3B

2=⇒

B = −2/5

A− 3/5

=⇒∫

1 +−6/5x + 3

+6/5

x− 2+−3/5x + 3

+−2/5x− 2

=−95

ln |x + 3|+ 45

ln |x− 2|+ x + C

Exercise 14.∫

x+2(x−2)2 =

∫x−2+4(x−2)2 = ln |x− 2|+ ∫

4(x−2)2 = ln |x− 2|+−4(x− 2)−1

Exercise 15.∫

dx(x−2)2(x2−4x+5)

Consider the denominator with its x2−4x+5. Usually, we would try a partial fraction form such as Ax−2 + B

(x−2)2 + Cx+Dx2−4x+5 ,

but the algebra will get messy. Instead, it helps to be clever here.101

1(x− 2)2(x2 − 4x + 4 + 1)

=1

(x− 2)2((x− 2)2 + 1)=

1(x− 2)2

− 1(x− 2)2 + 1

=⇒∫

dx

(x− 2)2(x2 − 4x + 5)=

∫1

(x− 2)2− 1

(x− 2)2 + 1= −(x− 2)−1 − arctan (x− 2) + C

Exercise 16.∫ (x−3)dx

x3+3x2+2x =∫ (x−3)dx

x(x+2)(x+1)

∫(x− 3)dx

x(x + 2)(x + 1)=

∫1

(x + 2)(x + 1)+−3

∫1

x(x + 2)(x + 1)1

(x + 2)(x + 1)=

−1x + 2

+1

x + 11

x(x + 2)(x + 1)=

A

x+

B

x + 2+

C

x + 1

Now to solve for A,B,C in the last expression, it is useful to use Gaussian elimination for this system of three linearequations:

1 1 13 1 22 0 0

ABC

=

001

1 1 13 1 22

∣∣∣∣∣∣

001

=

0 1 00 0 11 0 0

∣∣∣∣∣∣

1/2−11/2

=⇒ 1x(x + 2)(x + 1)

=1/2x

+1/2

x + 2+

−1x + 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∫

1(x2−1)2 . Then build the sum.

A

(x− 1)2+

B

(x + 1)2+

C

(x− 1)+

D

(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

−1x− 1

+1/4

x + 1+

1/4(x− 1)2

+1/4

(x + 1)2

∫dx

(x2 − 1)2=

14

ln∣∣∣∣x + 1x− 1

∣∣∣∣ +−12

(x

x2 − 1

)

102

Exercise 18. Use the method of partial fractions, where we �nd that∫

(x + 1)x3 − 1

dx =∫

x + 1(x− 1)(x2 + x + 1)

dx =∫ − 2

3x− 13

x2 + x + 1+

23

x− 1=

= −13

ln |x2 + x + 1|+ 23

ln |x− 1|where we had used the following partial fraction decomposition for the given integrand

Ax + B

x2 + x + 1+

C

x− 1=

x + 1x3 − 1

Ax2 + Bx−Ax−B + Cx2 + Cx + C = x + 12Ax + B −A + 2Cx + C = 1 (where we used the trick to take the derivative of the above equation)

=⇒ A = −C B −A−A = 1

−B + C = 1; A =2−3

, C =23

B = −12

Exercise 19.∫

x4+1x(x2+1)2

Again, it helps to be clever here.∫

x4 + 1x(x2 + 1)2

=∫

x4 + 2x2 + 1− 2x2

x(x2 + 1)2=

∫(x2 + 1)2

x(x2 + 1)2+

−2x

(x2 + 1)2=

= ln x + (x2 + 1)−1 + C

Exercise 20.∫

dxx3(x−2)

Working out the algebra for the partial fractions method, we obtain

1x3(x− 2)

=(−1/2

x3

)+−1/4x2

+−1/8

x+

1/8x− 2

So then∫

dx

x3(x− 2)=

14x2

+14x

+−18

ln x +18

ln |x− 2|+ C

Exercise 21.∫

1− x3

x(x2 + 1)= −

∫x3 − 1

x(x2 + 1)= −

∫x2

x2 + 1+

∫1

x(x2 + 1)=

= −∫ (

1− 1x2 + 1

)+

∫1x

+−x

x2 + 1=

= −x + arctan x + ln x− ln |x2 + 1|+ C

Exercise 22.∫

dx

x4 − 1=

∫ (1

x2 + 1

)(1

x2 − 1

)=

∫1/2

x2 − 1− 1/2

x2 + 1=

=∫

12

(1/2

x− 1− 1/2

x + 1

)− 1/2

x2 + 1=

=14

ln (x− 1)− 14

ln (x + 1)− 12

arctanx + C

Exercise 23.∫

dx

x4 + 1

I had to rely on complex numbers.103

Notice that with complex numbers, you can split up polynomial power sums

x4 + 1 = (x2 + i)(x2 − i) = (x + ie(iπ

4))(x− ie(i

π

4))(x + e(i

π

4))(x− e(i

π

4)) =

= (x + e(i3π

4))(x− e(i

3π

4))(x + e(i

π

4))(x− e(i

π

4))

A

(x + e(i 3π4 ))

+B

(x− e(i 3π4 ))

+C

(x + e(iπ4 ))

+D

(x− e(iπ4 ))

=1

x4 + 1

A(x2 − i)(x− e(i3π

4)) + B(x2 − i)(x + e(i

3π

4)) + C(x2 + i)(x− e(i

π

4)) + D(x2 + i)(x + e(i

π

4)) = 1

do the algebra−−−−−−−−→

x3 : A + B + C + D = 0

x2 : −Ae(−3π

4) + Be(i

3π

4)− Ce(i

π

4) + De(i

π

4) = 0

x1 : −iA− iB + iC + iD = 0

x0 : −e(iπ

4)A + Be(i

π

4) + C(−e(i

3π

4)) + D(e(i

3π

4)) = 1

=⇒

1 1 1 1−e(i 3π

4 ) e(i 3π4 ) −e(iπ

4 ) e(iπ4 )

−i −i i i−e(iπ

4 ) e(iπ4 ) −e(i3π

4 ) e(i 3π4 )

ABCD

=

0001

To do the complex algebra for the desired Gaussian elimination procedure, I treated the complex numbers as vectors andadded them and rotated them when multiplied.

1 1 1 1↘ ↖ ↙ ↗↓ ↓ ↑ ↑↙ ↗ ↘ ↖

∣∣∣∣∣∣∣∣

0001

=

1 1 0 02 ↖ 2 ↗

0 0 1 10 0 2 ↘ 2 ↖

∣∣∣∣∣∣∣∣

0001

=

1 1 0 02 ↖ 2 ↗

0 0 1 10 4 ↖

∣∣∣∣∣∣∣∣01

=

1↖

0 0 1 00 1

∣∣∣∣∣∣∣∣

14 ↖14 ↑14 ↗14 ↙

=

=

11

11

∣∣∣∣∣∣∣∣

14 ↖14 ↘14 ↗14 ↙

=⇒

A =14e(

i3π

4)

B = −14e(

i3π

4)

C =14e(i

π

4)

D =−14

e(iπ

4)

=⇒∫

1/4e( i3π4 )

x + e(i3π4 )

+−1/4e(i 3π

4 )x− e(i 3π

4 )+

1/4e(iπ4 )

x + e(iπ4 )

+−1/4e(iπ

4 )x− e(iπ

4 )

(14

)(e(i

3π

4) ln (x + e(i

3π

4))− e(i

3π

4) ln (x− e(i

3π

4)) + e(i

π

4) ln (x + e(i

π

4))− e(i

π

4) ln (x− e(i

π

4))

)

After doing some complex algebra,

=⇒ 14√

2

(ln

∣∣∣∣∣x2 +

√2x + 1

x2 −√2x + 1

∣∣∣∣∣− 2 arctan(

1√2x− 1

)− 2 arctan

(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?

104

Exercise 24.∫

x2dx(x2+2x+2)2

∫x2dx

(x2 + 2x + 2)2=

∫x2 + 2x + 2− 2x− 2

(x2 + 2x + 2)2=

(∫1

x2 + 2x + 2

)+ (x2 + 2x + 2)−1

∫1

x2 + 2x + 2=

∫1

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

∫x2dx

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

1x2 + 2x + 2

+ C

Exercise 25.∫

4x5−1(x5+x+1)2 dx

(−(x5 + x + 1)−1)′ = (x5 + x + 1)−2(5x4 + 1) (doesn't work)(−x(x5 + x + 1)−1)′ = (x5 + x + 1)−2(5x5 + x) +−(x5 + x + 1)−1 = (x5 + x + 1)−2(5x5 + x− x5 − x− 1)

=⇒∫

4x5 − 1(x5 + x + 1)2

dx = −x(x5 + x + 1)−2

Exercise 26.∫

dx2 sin x−cos x+5 (good example of the use of half angle substitution )

∫dx

2 sinx +− cos x + 5=

∫dx

4SC +−C2 + S2 + 5

( 1C2

1C2

)=

∫sec2 x

2dx

4T − 1 + T 2 + 5(1 + T 2)=

=∫

sec2 x2dx

6T 2 + 4T + 4=

∫sec2 x

2dx

6(T + 13 )2 + 10

3

=∫

2du

6(u + 13 )2 + 10

3

(whereu = tan

x

2

du =sec2 x

2

2dx

)

=35

∫du

9(u+ 13 )2

5 + 1=

1√5

arctan(

3(tan x2 + 1

3 )√5

)

Exercise 27.∫

dx1+a cos x (0 < a < 1) Again, using the half-angle substitution,

u = tanx

2

du =sec2 x

2

2dx

,

1a

∫dx

1a + cos x

=1a

∫dx

1a + C2 − S2

=1a

∫sec2 x

2dx1a sec2 x

2 + 1− T 2=

=1a

∫sec2 x

2dx1a + 1 + T 2( 1

a − 1)=

1a

∫2du

1a + 1 + u2

(1a − 1

) =2

1 + a

∫du

1 +(u√

1−a1+a

)2 =

21 + a

arctan√

1−a1+au

√1−a1+a

==⇒ 2√1− a2

arctan

(√1− a

1 + atan

x

2

)

105

Exercise 28.∫

dx1+a cos x Half-angle substitution.

∫dx

1 + a cos x=

∫dx

1 + a(C2 − S2)=

∫sec2 x

2dx

sec2 x2 + a(1− T 2)

=

u = tan θ/2 = T

du = sec2 θ/2(

12

)dθ

=⇒ =∫

2du

1 + T 2 + a(1− T 2)= 2

∫du

(1− a)T 2 + (1 + a)=

21− a

∫du

u2 − a+1a−1

=

=2

1− a

∫ 1

u−√

a+1a−1

− 1

u +√

a+1a−1

1

2√

a+1a−1

=

=

√a− 1a + 1

(1

1− a

) (ln (u−

√a + 1a− 1

)− ln (u +

√a + 1a− 1

)

)=

=−1√a2 − 1

ln

tan x

2 −√

a+1a−1

tan x2 +

√a+1a−1

Exercise 29.∫

sin2 x1+sin2 x

dx

∫s2

1 + s2dx =

∫s2 + 1− 1

1 + s2dx = x +−

∫dx

1 + sin2 x∫dx

1 + sin2 x=

∫dx

1 +(

1−cos 2x2

) =∫

2dx

3− cos 2x=

23

∫dx

1− cos 2x3

=23

∫dx

1− (c2−s2

3

) =23

∫sec2 xdx

sec2 x− (1−T 2

3

) =

u = tan x

du = sec2 xdx

=⇒ =23

∫du

1 + u2 − (1−u2

3

) =23

∫du

23 + 4

3u2=

∫du

1 + (√

2u)2=

=1√2

arctan√

2 tan x

=⇒∫

sin2 x

1 + sin2 xdx = x− 1√

2arctan (

√2 tan x)

It seems like for here, when dealing with squares of trig. functions, �step up� to double angle.

Exercise 30.∫

dxa2 sin2 x+b2 cos2 x

(ab 6= 0) Take note, we need not change the angle to half-angle or double-angle.

1a2s2 + b2c2

=1

a2(1− c2) + b2c2=

1a2 + (b2 − a2)c2

=1

a2(1 + (kc)2)=

sec2

a2(sec2 +k2)=

sec2

a2(1 + T 2 + k2)=

u = tan x

du = sec2 xdx

=⇒ =du

a2(1 + u2 + k2)=

1/a2du

(1 + k2)(1 + u2

1+k2 )=

√1 + k2

a2(1 + k2)arctan

(u√

1 + k2

)

=⇒∫

dx

a2 sin2 x + b2 cos2 x=

1ab

arctan(

a tan x

b

)

Exercise 31.∫

dx(a sin x+b cos x)2 (a 6= 0)

Note it's a good idea to simplify, cleverly, your constants as much as you can.∫

dx

(a sin x + b cosx)2=

1a2

∫dx

(sinx + k cos x)2106

Thus, only one constant, k, is only worried about.

1(s + kc)2

=1

s2 + 2ksc + k2c2=

1/c2

t2 + 2kt + k2=

sec2

(t + k)2=

u = tanx

du = sec2 xdx

=⇒ =du

(u + k)2

=⇒ 1a2

∫1

(s + bac)2

=−1

(a2 tan x + ab)

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 θ

2 substitution.∫

sin x

1 + cosx + sin x=

∫2CS

1 + 2CS + C2 − S2=

∫CS

C(S + C)=

∫T

(T + 1)

whereC = cos x/2

S = sin x/2

u = tan θ/2

du =sec2 θ/2dθ

2∫u

(u + 1)

(2du

u2 + 1

)=

∫2udu

(u2 + 1)(u + 1)A

u + 1+

Bu + C

u2 + 1Au2 + A + Bu2 + Cu + Bu + C = u

A = −B C + B = 1 A + C = 0 =⇒ C =12; B =

12; A = −1

2

2∫ (−1/2

u + 1+

12 (u + 1)u2 + 1

)du =

∫ −1u + 1

+u + 1u2 + 1

= − ln |u + 1|+ 12

ln |u2 + 1|+ arctanu =

= − ln | tanx/2 + 1|+ 12

ln | sec2 x/2|+ x

2

=⇒ − (ln |2|) +12

ln |2|+ π

4= −1

2ln |2|+ π

4

Exercise 33.∫ √

3− x2dx

∫(x)′

√3− x2dx = x

√3− x2 −

∫x(−x)√3− x2

= x√

3− x2 −∫ −x2 + 3− 3√

3− x2= x

√3− x2 −

∫ √3− x2 + 3

∫1√

3− x2

=⇒ 2∫ √

3− x2 = x√

3− x2 +√

3∫

1√1−

(x√3

)2

=⇒∫ √

3− x2 =x

2

√3− x2 +

32

arcsinx√3

Exercise 34.∫

1√3−x2 dx = −(3− x2)1/2 + C.

(arccos

x√3

)′=

1√3

−1√1− x2

3

= − 1√3− x2

(x√

3− x2)′

=√

3− x2 +−x2

√3− x2

x√

3− x2

2+−3

2arccos

x√3

107

Exercise 35.∫ √

3−x2

x dx =∫ √

3x2 − 1dx.

√3

x= sec θ

√3 cos θ = x

dx = − sin θ√

3∫ √sec2 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 ln

∣∣∣∣∣

√3

x+

√3x2− 1

∣∣∣∣∣ +√

3

√1− x2

3

Exercise 36.∫ √

1 + 1xdx

(x

√1 +

1x

)′

=

√1 +

1x

+x

2√

1 + 1x

(−1x

)=

√1 +

1x

+−1/2√x2 + x

(ln

(x +

12

+√

x2 + x

))′=

1x + 1

2 +√

x2 + x

(1 +

x + 12√

x2 + x

)=

1√x2 + x

=⇒∫ (

1 +1x

)dx = x

√1 +

1x

+12

ln(

x +12

+√

x2 + x

)

Exercise 37.

(x√

x2 + 6)′ =√

x2 + 5 +x2

√x2 + 5

(ln (x +√

x2 + b))′ =(

1x +

√x2 + b

)(1 +

x√x2 + b

)=

1√x2 + b∫ √

x2 + 5 =12

(x√

x2 + 5 + 5 ln (x +√

x2 + 5))

Exercise 38.(

ln (x +12

+√

x2 + x + 1))′

=1

x + 12 +

√x2 + x + 1

(1 +

x + 12√

x2 + x + 1

)=

1√x2 + x + 1

∫x√

x2 + x + 1=

∫x + 1

2 − 12√

x2 + x + 1= (x2 + x + 1)1/2 − 1

2ln (x +

12

+√

x2 + x + 1)

The trick is to note how I formed a �conjugate-able� sum from x2 + x + 1's derivative.Exercise 39. ∫

dx√x2 + x

=∫

dx√(x + 1

2

)2 − 14

=∫

2dx√(2(x + 1

2

))2 − 1(ln

(√(2(x + 1/2))2 − 1 + 2(x + 1/2)

))′=

=1√

(2(x + 1/2))2 − 1 + 2(x + 1/2)

(2 +

2(x + 1/2)2√(2(x + 1/2))2 − 1

)=

=2√

(2(x + 1/2))2 − 1∫

dx√(x + 1

2

)2 − 14

= ln(

2(

x +12

)+

√(2(x + 1/2))2 − 1

)+ C

Exercise 40.108

6.26 Miscellaneous review exercises. Exercise 1.

f(x) =∫ x

1

log t

t + 1

f

(1x

)=

∫ 1x

1

log t

t + 1dt =

∫ x

1

− ln (u)1u + 1

(−1u2

du

)=

=∫ x

1

ln (u)u + u2

duu =

1t

du = − 1t2

dt

f(x) + f

(1x

)=

∫ x

1

t ln t + ln t

t(t + 1)dt =

∫ x

1

ln t

tdt =

(ln t)2

2

∣∣∣∣x

1

=(lnx)2

2

f(2) + f

(12

)=

12(ln 2)2

Exercise 2. Take the derivative of both sides, using the (�rst) fundamental theorem of calculus.

2ff ′ = f(x)sin x

2 + cos x; =⇒ 2f ′ =

sin 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)′ =sin x

2 + cos x=⇒ f = − ln |2 + cosx|

2+ C

Otherwise, remember that for rational expressions involving single powers of sin and cos, we can make a u = tan θ/2substitution.

u = tanx

22du = sec2 x

2dx

C = cos x/2, S = sin x/2

∫sin x

2 + cos xdx =

∫2SCdx

2 + C2 − S2=

∫2T

2 sec2 x/2 + 1− T 2

(2du

sec2 x/2

)=

= 4∫

udu

(1 + u2)(3 + u2)= 2

∫T

T 2 + 1− T

T 2 + 3

= 2(

12

ln T 2 + 1− 12

ln T 2 + 3)

=(

ln(

T 2 + 1T 2 + 3

))= ln

(2

4 + 2 cos x

)

where tan2 x

2=

sin2 x2

cos2 x2

=1− cosx

1 + cosx

Exercise 3. ∫ex

xdx = ex −

∫ex(x− 1)

x2xdx = ex −

∫ex(x− 1)

xdx . . .

No way.Exercise 4.

∫ π/2

0ln (ecos x)dx = − cosx|3/2

0 = 1 .Exercise 5.

(1)

f =√

4x + 2x(x + 1)(x + 2)

ln f =12

ln(

4(x + 2)x(x + 1)(x + 2)

)=

12

(ln (4x + 2)− ln x− ln (x + 1)− ln (x + 2))

f ′

f=

12

((4

4x + 2

)− 1

x− 1

x + 1− 1

x + 2

)

=⇒ f ′ =12

√4x + 2

x(x + 1)(x + 2)

((4

4x + 2

)− 1

x− 1

x + 1− 1

x + 2

)f ′(1) = − 7

12

109

(2) ∫ 4

1

π4x + 2

x(x + 2)(x + 1)dx = 2π

∫ 4

1

(2x + 1)x(x + 2)(x + 1)

dx =

= 2π

(12

ln x +−32

ln |x + 2|+ ln |x + 1|)∣∣∣∣

4

1

= πln258

since we can �nd the antiderivative through partial fractions:A

x+

B

x + 2+

C

x + 1=

2x + 1x(x + 2)(x + 1)

A(x2 + 3x + 2) + B(x2 + x) + C(x2 + 2x) = 2x + 1

1 1 13 1 22 0 0

ABC

=

021

=⇒

0 1 00 0 11 0 0

∣∣∣∣∣∣

−3/21

1/2

Exercise 6.(1)

log x =∫ x

1

1tdt F (x) =

∫ x

1

et

tdt;

if x > 0

et > 1 for t > 0If 0 < x < 1

log x =∫ x

1

1tdt =

∫ 1

x

−1t

dt > −∫ 1

x

et

t= F (x)

log x ≤ F (x) for x ≥ 1

(2)

F (x + a)− F (1 + a) =∫ x+a

1

etdt

t−

∫ a+1

1

etdt

t=

∫ x

1−a

et+adt

t + a−

∫ 1

1−a

et+adt

t + a

= ea

∫ x

1

et

t + adt

(3) ∫ x

1

eat

tdt =

∫ ax

a

et

tdt =

∫ ax

1

et

t+

∫ 1

a

et

t=

= F (ax)− F (a)∫ x

1

et

t2= −1

tet −

∫−1

tet = −

(ex

x− e

)+ F (x)

∫ x

1

e1/tdt =∫ 1/x

1

−eu

u2du = −

(−eu

u−

∫−eu

u

)=

= xe1/x − e− F (1/x) where we used the substitutionu =

1t

du = − 1t2

dt

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

∫ x2

0

f(t)dt = f(x2)(2x) = −(2x) ln 2ex2 ln 2 f(x) = − ln 2ex ln 2

∫ x2

0

− ln 2eln 2dt = − et ln 2∣∣x2

0= −ex2 ln 2 + 1

110

(3)f(x) = 2f(x)f ′(x);=⇒ f(x) =

x

2+ C

∫ x

0

(12t + c

)dt =

(t2

4+ ct

)∣∣∣∣x

0

=x2

4+ cx

f2(x)− 1 =x2

4+ Cx + C2 − 1

=⇒ C = ±1, f(x) =x

2+±1

Exercise 8.(1)

f(x + h)− f(x)h

=f(x)f(h)− f(x)

h=

f(x)(hg(h))h

= f(x)g(h)

g(h) → 1 as h → 0 so =⇒ f ′(x) = f(x)(2) Since for f(x) = ex, we de�ned ex such that f ′ = f , if

(ex + g)′ = ex + g′ = ex + g

=⇒ ex + g = Cex =⇒ g = (C − 1)ex but f ′(0) = 1 so g = ex

Exercise 9.(1)

g(2x) = 2exg(x)

g(3x) = exg(2x) + e2xg(x) = ex2exg + e2xg = 3e2xg

(2)Assume g(nx) = ne(n−1)xg

g((n + 1)x) = exg(nx) + e(n+1)xg(x) = nenxg(x) + enxg = (n + 1)enxg

(3) From g(x + y) = eyg(x) + exg(y),g(0) = g(0) + g(0) =⇒ g(0) = 0

g(x + h)− g(x)h

=ehg(x) + exg(h)− g(x)

h= g(x)

(eh − 1

h

)+

exg(h)h

f ′(0) = 2 = limh→0

g(x + h)− g(x)h

= lim h → 0g(h)h

(4) g′(x) = g(x) + 2ex C = 2

Exercise 10.∀x ∈ R, f(x + a) = bf(x); f(x + 2a) = bf(x + a) = b2f(x)

f(x + (n + 1)a) = f(x + na + a) = bf(x + na) = bn+1f(x)

f(x + na) = bnf(x)

f(x) = bx/ag(x) where g is periodic in a

Exercise 11.(ln (fg))′ =

f ′

f+

g′

g=⇒ (fg)′ = f ′g + fg′

(ln

(f

g

))′=

f ′

f− g′

g=⇒ f ′g − g′f

g2=

(f

g

)′

Exercise 12. A =∫ 1

0et

t+1dt

(1) ∫ a

a−1

e−t

t− a− 1dt

u = t− a =⇒∫ 0

−1

e−t−a

t− 1dt = −

∫ 0

1

et−a

−t− 1dt = −e−a

∫ 1

0

et

t + 1= −e−aA

(2)∫ 1

0tet2

t2+1dt =∫ 1

0

12 dueu

u+1 =12A

111

(3)∫ 1

0et

(t+1)2 dt = −et

(t+1)

∣∣∣1

0− ∫ 1

0−et

t+1 =−e1

2+ 1 + A

(4)∫ 1

0et ln (1 + t)dt = et ln (1 + t)− ∫

et

1+t = e ln 2−A

Exercise 13.(1) p(x) = c0 + c1x + c2x

2; f(x) = exp(x) p′ = c1 + 2c2x p′′ = 2C2

f ′ = f + exp′

f (n)(x) =n∑

j=0

(n

j

)ex(p(x))j = f + ex(c1 + 2c2x) +

(n!

(n− 1)!

)+ ex(2c2)

(n!

(n− 2)!2!

)

f (n)(0) = c0 + c1n + n(n− 1)c2

(2) See generalization below.(3)

p =m∑

j=0

ajxj ; p(0) = a0; p(k)(x) =

m∑

j=0

ajj!

(j − k)!xj−k = p(k)(0) = akk!

f (n)(x) =n∑

j=0

(n

j

)exp(j)(x)

f (n)(0) =n∑

j=0

(n

j

)p(j)(0) =

m∑

j=0

(n

j

)ajj! =

m∑

j=0

n!(n− j)!

aj

So for m = 3, then f (n)(0) = a0 + na1 + n(n− 1)a2 + n(n− 1)(n− 2)a3

Exercise 14. f(x) = x sin ax ; f (2) = −a2x sin ax + 2a cos ax

f (2n)(x) = (−1)n(a2nx sin ax− 2na2n−1 cos ax)

f (2n+1)(x) = (−1)n(a2n+1x cos ax + a2n sin ax + 2na2n sin ax)

f (2n+2)(x) = (−1)n(−a2n+2x sin ax + a2n+1 cos ax(2n + 2))

Exercise 15.n∑

k=0

(−1)k

(n

k

)1

k + m + 1=

n∑

k=0

(−1)k

(n

k

) ∫ 1

0

tk+mdt =∫ 1

0

n∑

k=0

(−1)k

(n

k

)tk+mdt =

=∫ 1

0

tmn∑

k=0

(n

k

)(−t)kdt =

∫ 1

0

tm(1− t)ndt =

= −∫ 0

1

(1− u)mundu =∫ 1

0

(1− u)mundu =

=∫ 1

0

∑

j=0

6m

(m

j

)(−u)jundu =

m∑

j=0

(−1)j

(m

j

) ∫ 1

0

tj+ndt

u1− t

du = −dt

Exercise 16. F (x) =∫ x

0f(t)dt

(1)

F (x) =∫ x

0

(t + |t|)2 =

{∫ x

0(2t)2dt = 4

3x3 if t, x ≥ 0∫ x

00dt = 0 if t, x < 0

(2)

F (x) =∫ x

0

f(t)dt =

{∫ x

0(1− t2)dt if |t| ≤ 1∫ x

0(1− |t|)dt if |t| > 1

=

=

(t− 1

3 t3)∣∣x

0= x− x3

3 if |x| ≤ 1{23 +

∫ x

1(1− t)dt = x− x2

2 + 16 if x > 1

−23 +

∫ x

−1(1 + t)dt = x + x2

2 − 16 if x < 1

if |x| ≥ 1

112

(3) f(t) = e−|t|

F (x) =∫ x

0

f(t)dt =∫ x

0

e−|t|dt =

=

{∫ x

0e−tdt = e−t|0x = 1− e−x if x ≥ 0∫ x

0etdt = et|x0 = ex − 1 if x < 0

(4) f(t) = max. of 1 and t2

F (x) =∫ x

0

f(t)dt =

∫ x

01dt = x if |x| ≤ 1

1 +∫ x

1t2dt if x > 1

− ∫ x

0f if x < −1

=

=

x if |x| ≤ 11 + 1

3 t3∣∣x1

= x3

3 + 23 if x > 1

− ∫ −1

xt2 +− ∫ 0

−11 = 1

3 t3∣∣x−1− 1 = x3

3 + 23 if x > 1

Exercise 17.∫

πf2 = π∫ a

0f2 = a2 + a.

(x2 + x

π

)′=

√2x + 1

π

for∫ a

0

2x + 1π

=(x2 + x

)∣∣a0

= a2 + a

Exercise 18. f(x) = e−2x.

(1) A(t) =∫ t

0e−2xdx = e−2x

−2

∣∣∣t

0= e−2t−1

−2

(2) V (t) = π∫ t

0e−4xdx = π

−4 (e−4t − 1)(3)

y = e−2x =⇒ ln y

−2= x

W (t) = π

∫ 1

e−2t

(ln y

−2

)2

dy =π

4(y(ln y)2 − (2(y ln y − y)))

∣∣1e−2t =

=π

4(2− (

e−2t4t2 − (2(e−2t(−2t)− e−2t))))

=

=(π

2− πte−2t − π

2e−2t

)

where the antiderivative used was (y(ln y)2)′ = (ln y)2 + 2 ln y

(4)π4 (1− e−4t)(

1−e−2t

2

) =π

2

(e−4t−1

t

)

e−2t−1t

= π

where we used the limit limx→0

ecx − 1x

= c

Exercise 19. sinh c = 34

(1)ec = ex +

√e2x + 1

ec − e−c

2=

ex +√

e2x + 1− 1ex+

√e2x+1

2=

=e2x + 2ex

√e2x + 1 + e2x + 1− 1

2(ex +√

e2x + 1)= ex =

34

x = ln 3− 2 ln 2

113

(2)ec − e−c

2=

ex −√e2x − 1− 1ex−√e2x−1

2=

=e2x − 2ex

√e2x − 1 + e2x − 1− 1

2(ex −√e2x − 1)

=ex

+−1

ex −√e2x − 1= ex − 1

ec=

34

=⇒ x = ln 5− 2 ln 2

Exercise 20.(1) True. ln (2log 5) = ln 5ln 2 = (ln 2) ln 5.(2) log3 5 = log2 5

log2 3 This is a true fact.log3 5log2 3

=log2 5

(log2 3)2= log2 5

=⇒ 1 = log2 3 False(3) Use induction

n = 1 1−1/2 < 2√

1

n = 2 1 +1√2

< 2√

2

n + 1 case

n+1∑

k=1

k−1/2 =n∑

k=1

k−1/2 +1√

(n + 1)<

< 2√

n

(√(n + 1)√(n + 1)

)+

1√(n + 1)

Now (n + 12 )2 = n2 + n + 1

2 > n2 + n, certainly. So then

n +12

>√

n2 + n =⇒ n + 1 >√

n2 + n

=⇒n+1∑

k=1

k−1/2 < 2√

n + 1

(4)

f = (cosh x− sinhx− 1) =ex + e−x − ex + e−x

2− 1 = e−x − 1 < 0 for x > 0

False.Exercise 21. For 0 < x < π

2 ,(sinx)′ = cos x > 0 for 0 < x <

π

2(x− sin x)′ = 1− cos x ≥ 0 for 0 < x <

π

2(x− sin x)(x = 0) = 0 =⇒ sin x < x

Exercise 22.1t

<1t

(x + 1

t

)if 0 < x < t < x + 1

∫ x+1

x

1tdt = ln (x + 1)− ln x;

∫ x+1

x

x + 1t

=1x

So lnx + 1

x<

1x

Exercise 23.(x− sin x)′ = 1− cosx ≥ 0∀x > 0

since (x− sin x)(x = 0) = 0; (x− sinx)′(x = 0) = 0, then x− sinx > 0 in general for x > 0

(sinx−(

x− x3

6

))′ = cos x− 1 +

x2

2> 0∀x > 0

=⇒ x− x3

6< sin x < x

114

Exercise 24. (xb + yb)1/b < (xa + ya)1/a if x > 0, y > 0 and 0 < a < b

(xn + yn)1/n = x(1 +(

yx

)n)1/n. Without loss of generality, assume x < y.

Consider (1 + An)1/n, A constant.

((1 + An)1/n)′ = (exp(

1n

ln (1 + An))

)′ = (1 + An)1/n

(−1n2

ln (1 + An) +1n

(1

1 + An

)(lnA)An

)=

= (1 + An)1/n

(−(1 + An) ln (1 + An) + n(lnA)An

n2(1 + An)

)=

= (1 + An)1/n

(− ln (1 + An)−An ln (1 + An) + An ln An

n2(1 + An)

)=

=(1 + An)

1−nn

n2

− ln (1 + An) + An ln

(An

1+An

)

n2(1 + An)

< 0 since ln

(An

1 + An

)< 0

=⇒ (xb + yb)1/b < (xa + ya)1/a if b > a

Exercise 25.(1) ∫ x

0

e−tt =(−te−t − e−t

)∣∣x0

= −xe−x − e−x + 1

(2) ∫ t

0

t2e−tdt = −t2e−t∣∣x0−

∫−e−t(2t)dt = −x2e−x + 2

∫te−tdt =

= −x2e−x +−2xe−x − 2e−x + 2

(3) ∫ x

0

t3e−tdt = −t3e−t∣∣x0− 3

∫ x

0

t2(−e−t)dt = −x3e−x + 3∫ x

0

t2e−tdt =

= −x3e−x + 3(2)(e−x)(

ex − 1− x− x2

2!

)

(4) Assume the induction hypothesis, that∫ x

0

tne−tdt = n!e−x

ex −

n∑

j=0

xj

j!

∫ x

0

tn+1e−tdt = −tn+1e−t∣∣x0−

∫(n + 1)t(−e−t) = −xn+1e−x + (n + 1)n!e−x

ex −

n∑

j=0

xj

j!

= (n + 1)!e−x

ex −

n+1∑

j=0

xj

j!

Exercise 26. Consider the hint a1 sin x + b cos x = A(a sin x + b cosx) + B(a cos x − b sin x). Solve for A, B in terms ofa1, b1, a, b. Matching up term by term the coef�cients for sin and cos separately,

Aa2 − abB = aa1

Ab2 + Bab = b1b

A =aa1 + bb1

a2 + b2

−Aab + Bb2 = −a1b

Aab + Ba2 = ab1

B =ab1 − a1b

a2 + b2

So if not both a, b = 0 ,∫

a1 sin x + b1 cosx

a sin x + b cosx=

∫A(a sin x + b cosx) + B(a cosx− b sin x)

a sin x + b cos x=

= Ax + B ln |a sinx + b cos x|+ C

Exercise 27.115

(1)

f ′(x2) =1x

df

du= u−1/2

f(x2) = 2x− 1

(2)

f ′(sin2 x) = 1− sin2 x f ′(u) = 1− u f = u− 12u2 + C =⇒ f(x) = x− x2

2+

12

(3)

f ′(sinX) = (1− sin2 x) f(u) = u− 13u3 + C f(x) = x− x3

3+

13

(4)

f ′(lnx) =

{1 for x ≤ 1x for x > 1

=

{1 for 0 < x ≤ 1eln x x > 1

f(y) =

{y for y < 0ey − 1 for y > 0

Exercise 28.(1)

Li(x) =∫ x

2

dt

ln tif x ≥ 2

Li(x) =x

ln x− 2

1ln 2

−∫ x

2

−1(ln x)2

dt =x

ln x+

∫ x

2

dt

(lnx)2− 2

ln 2(2)

Li(x) =x

ln x− 2

ln x− 2

(ln 2)2+

x

(lnx)2−

∫ x

a

−2(ln t)3

dt

Li(x) =x

ln x+

n−1∑

k=1

k!xlnk+1 x

+ n!(

x

lnn+1 x−

∫ x

2

−(n + 1)dt

lnn+2 t

)

Li(x) =x

ln x+

n∑

k=1

k!xlnk+1 x

+ (n + 1)!∫ x

2

dt

ln(n+1)+1 t

C2 =−2ln x

+−2∑

j=2

2(j − 1)!(ln 2)j

Cn = −21

ln 2−

n∑

j=2

2(j − 1)!(ln 2)j

Cn+1 = −21

ln 2−

n+1∑

j=2

2(j − 1)!(ln 2)j

(3)

Li(x) =∫ x

2

dt

ln t

u = ln t

du =1tdt

eudu = dt eu = t

Li(x) =∫ ln x

ln 2

etdt

t

(4)

c = 1 +12

ln 2

t = u + 1 =⇒

∫ x−1

c−1

e2(u+1)du

u= e2

∫ x−1

c−1

e2u

udu =

= e2 12

∫ 2(x−1)3

2(c−1)3

et

t2

dt =

= e2Li(e2(x−1))

116

(5)

f(x) = e4Li(e2(x−2)) −e2Li(e2(x−1))

=∫ x

c

e2t

t− 2−

∫ x

c

e2t

(t− 1)

=⇒ f ′(x) =e2x

t− 2+− e2x

t− 1= e2x

(1

t2 − 3t + 2

)

Exercise 29.f(x) = log |x| if x < 0. ∀x < 0∃ uniquely ln |x| since f ′ = 1x < 0∀x < 0.

−ey = x(y) = g(y) 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 theinverse 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) =∫ x

0(1 + t3)−1/2dt if x ≥ 0.

(1)f ′(x) =

1√1 + x3

> 0 for x > 0

(2)

g′(x) =1

f ′(x)=

√1 + x3 g′′(x) =

3x2

2√

1 + x3

7.4 Exercises - Introduction, The Taylor polynomials generated by a function, Calculus of Taylor polynomials. Usethe following theorems for the following exercises.

Theorem 22 (Properties of Taylor polynomials, Apostol Vol. 1. Theorem 7.2.). (1) Linearity Tn(c1f+c2g) = c1Tn(f)+c2Tn(g)

(2) Differentiation (Tnf)′ = Tn−1(f ′)(3) Integration. If g(x) =

∫ x

af(t)dt

Tn+1g(x) =∫ x

aTnf(t)dt

Theorem 23 ( Substitution Property, Apostol Vol. 1. Theorem 7.3. ). Let g(x) = f(cx), c is a constant.

(12) Tng(x; a) = Tnf(cx; ca)

This theorem is useful for �nding new Taylor polynomials without having to �nd the jth derivatives of the desired function.

Theorem 24. Pn is a polynomial of degree n ≥ 1.Let f, g be 2 functions with derivatives of order n at 0.

(13) f(x) = Pn(x) + xng(x)

where g(x) 7→ 0 as x 7→ 0.Then Pn = Tn(f, x = 0).

Exercise 3.

Tnf(x) =n∑

j=0

f (j)(a)j!

(x− a)j

ax = ex ln a

(ax)′ = (ax) ln a

(ax)(n+1) = (ax(ln a)n)′ = ax(ln a)n+1

Tn(ax) =n∑

j=0

(ln a)j

j!xj

117

Exercise 4. (1

1 + x

)′=

−1(1 + x)2

;(

11 + x

)′′=

(−1)22(1 + x)3

(1

1 + x

)(n+1)

=(

(−1)nn!(1 + x)n+1

)′=

(−1)n+1(n + 1)!(1 + x)n+2

Tn

(1

1 + x

)=

n∑

j=0

(−1)jxj

Exercise 5. Use Theorem 7.4. . Theorem 7.4 says for f(x) = Pn(x) + xng(x), Pn(x) is the Taylor polynomial.

11− x2

=

n∑

j=0

(x2)j

+

(x2)n+1

1− x2=

n∑

j=0

x2j

+

xnxn+2

1− x2

x

1− x2=

n∑

j=0

x2j+1

+

(x2n+3)1− x2

T2n+1

(x

1− x2

)=

n∑

j=0

x2j+1

Exercise 6.

(ln (1 + x))′ =1

1 + xTn

(1

1 + x

)=

n∑

j=0

(−x)j Tn(ln 1 + x) =n∑

j=0

(−1)j xj+1

j + 1=

n∑

j=1

(−1)j+1 xj

j

Exercise 7. (log

√1 + x

1− x

)′

=

√1− x

1 + x

√1− x

1 + x

1(1− x)2

=1

(1 + x)(1− x)=

11− x2

∫1

1− x2= log

√1 + x

1− xso

∫ n∑

j=0

x2j =n∑

j=0

x2j+1

2j + 1= T2n+1

(ln

√1 + x

1− x

)

Exercise 8.

Tn

(1

2− x

)= Tn

(1/2

1− x/2

)=

12Tn

(1

1− (12

)x

)=

12

n∑

j=0

(12x)j

=

n∑

j=0

xj

2j+1

Exercise 9. We can show this in two ways.We could write out the actual polynomial expansion.

(1 + x)α =n∑

j=0

(α

j

)1α−jxj =

n∑

j=0

(α

j

)xj

or determine each of the coef�cients of the Taylor polynomial.

((1 + x)α)′ = α(1 + x)α−1; ((1 + x)α)′′ = α(α− 1)(1 + x)α−2

((1 + x)α)(n+1) =(

α!(α− n)!

(1 + x)α−n

)′=

α!(α− (n + 1))!

(1 + x)α−(n+1)

Exercise 10. Use the substitution theorem, Apostol Vol.1. Thm. 7.3., to treat cos 2x.

T2n(cosx) =n∑

j=0

(−1)jx2j

(2j)!; T2n(cos 2x) =

n∑

j=0

(−1)j(2x)2j

(2j)!

T2n(sin x2) = T2n(12(1− cos 2x)) =

12

1−

n∑

j=0

(−1)j(2x)2j

(2j)!

=

n∑

j=1

(−1)j+122j−1x2j

(2j)!

118

7.8 Exercises - Taylor's formula with remainder, Estimates for the error in Taylor's formula, Other forms of theremainder 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,

m(x− a)n+1

(n + 1)!≤ En(x) ≤ M

(x− a)n+1

(n + 1)!if x > a(14)

m(a− x)n+1

(n + 1)!≤ (−1)n+1En(x) ≤ M

(a− x)n+1

(n + 1)!if x < a(15)

En(x) =1n!

∫ x

a

(x− t)nf (n+1)(t)dt(16)

Exercise 1. For a = 0, | sin(j)(x)| ≤ 1 for ∀x ∈ R.

En(x) ≤ (x)2n+1

(2n + 1)!if x > 0; (−1)2n+1E2n(x) ≤ (+1)

(−x)2n+1

(2n + 1)!

=⇒ E2n(x)| ≤ |x|2n+1

(2n + 1)!

Exercise 2.

cosx =n∑

k=0

(−1)kx2k

(2k)!+ E2n+1(x) | cos(j) (x)| ≤ 1

E2n+1(x) ≤ x2n+2

(2n + 2)!; (−1)2n+2E2n+1(x) ≤ (1)

(−x)2n+2

(2n + 2)

=⇒ |E2n+1(x)| ≤ |x|2n+2

(2n + 2)!Exercise 3.

arctanx =n−1∑

k=0

(−1)kx2k+1

2k + 1+ E2n(x)

n−1∑

k=0

(−1)kx2k+1

2k + 1

((2k)!(2k)!

)=

n−1∑

k=0

(−1)k(2k)!x2k+1

(2k + 1)!=⇒ f (2k+1)(0) = (−1)k(2k)!

f (2n+1)(0)x2n+1

(2n + 1)!=

(−1)n(2n)!x2n+1

(2n + 1)!=

(−1)nx2n+1

2n + 1≤ x2n+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 asmall 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).Exercise 4.

(1)

x2 = sin x = x− x3

6=⇒ x3

6+ x2 − x =

x

6

(x−

(−3 +

√15

))(x− (−3−

√15)

)

x =√

15− 3(2)

E4(r; 0) =14!

∫ r

0

(r − t)4 cos tdt > 0 sin r − r2 = 0 + E4(r) ≤ r5

5!<

r

5!=

35(2)(5)(4)(3)(2)1

=34

(5)(4)(2)5<

1200

Exercise 5.

arctan r − r2 = r − r3

3− r2 + E4(r; 0) = 0 + E4(r; 0)

E4(r, 0) =14!

∫ r

0

(x− t)4f (5)(t)dt ≤ M(r5)5!

r5

5!= 0.065536 <

7100

E4(r, 0) < Ej(r, 0); j > 4

the 5th degree term is f (5)(0)5!

r5 =245!

r5 > 0

so r2 − arctan r = −E4(r, 0) < 0119

Exercise 6. Apply long division on the fraction in the integrand.∫ 1

0

1 + x30

1 + x60dx =

∫ 1

0

1 +x30 − x60

1 + x60dx = 1 +

∫ 1

0

x30

(1− x30

1 + x60

)dx =

= 1 + c131

x31

∣∣∣∣1

0

= 1 +c

31

Exercise 7.∫ 1/2

01

1+x4 dx.

11 + x4

=∞∑

j=0

(−x4)j =n−1∑

j=0

(−x4)j + En = 1 +−x4 + x8 . . .

1617

x4n+1

(n + 1)!≤ En(x; 0) ≤ 1

(n + 1)!(x4)n+1

=⇒∫ 1/2

0

En =

(12

)4n+5

(n + 1)!

(1

4n + 5

)

∫ 1/2

0

11 + x4

' 12

+−15

(12

)5

0.493852 < 0.49375 < 0.493858

Exercise 8.(1)

0 ≤ x ≤ 12

sin x = x− x3

3!+ E4(x)

|E4(x)| ≤ M |x|55!

=sin 1

3 |x|55!

≤ 1(

12

)5

5!(2)

sinx2 = x2 − x6

6+ E4(x2)

∫ √2

2

0

sin x2 =(

13x3 − x7

42

)∣∣∣∣

√2/2

0

=√

2(

112− 1

42(16)

)

E4(x2) ≤√

264(5!)

∫ √2

2

0

sin x2 ≤√

2(

55672

+1

64(5!)

)= 0.1159

Exercise 9.

sin x = x− x3

6+

x5

5!E6(x; 0) ≤ (1)x7

7!sin x

x= 1− x2

6+

x4

5!+

E6(x; 0)x

≤ 1− x2

6+

x4

5!+

x6

7!∫ 1

0

sin x

xdx = 1− 1

3

(16

)+

15(5!)

+1

7(7!)= 0.9461 +

17(7!)

= 0.9461 + 0.0000283

Exercise 10. α = arctan 15 , β = 4α− π

4 .(1)

tan (A + B) =(tan A + tanB)1− tan A tanB

;A = B = α; tan 2α =2 tan α

1− tan2 α=

2/524/25

= 5/12

tan 4α =2(tan 2α)

1− tan2 (2α)=

2(5/12)1− (5/12)2

=10/12

119/144=

120119

A = 4α, B = −π

4

tan (4α− π

4) = tan β =

tan 4α + tan(−π

4

)

1− tan 4α tan(−π

4

) =120119 +− 119

119

1− 120119 (−1)

=1

239

4 arctan15

=π

4+ arctan

1239

This is incredible.

120

(2)

T11(arctan x) =6−1∑

k=0

(−1)kx2k+1

2k + 1+ E2(6)(x)

= x +−x3

3+

x5

5− x7

7+

x9

9− x11

11. . .

; |E2(6)(x)| ≤ x2(6)+1

2(6) + 1

=⇒ 3.158328957 < 16 arctan15

< 3.158328958

(3)T3(arctanx);x =

1239

−0.016736304 < −4 arctan1

239< −0.016736304

(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 indeterminateforms.Exercise 1.

2x = exp x ln 2 = 1 + (x ln 2) +x2(ln 2)2

2!+ o(x2)

Exercise 2.

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

2+ cos 1+

+ (− sin 1)(x− 1)− cos 1(x− 1)2

2+

sin 1(x− 1)3

3!=

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

2

)(x− 1)2 +

(sin 1− 3 cos 1

3!

)(x− 1)3 + o(x− 1)3

Exercise 3. Just treat the argument of sin x− x2 just like u with u → 0.

sin (x− x2) = (x− x2)− (x− x2)3

3!+

(x− x2)5

5!+ o(x− x2)5 =

= (x− x2)− 16

(x3 − 3x4 + 3x5 − x6

)+

1120

(x5 − 5x6 + 10x7 − 10x8 + 5x9 + x10

)=

= (x− x2)− 16x3 − 1

2x4 +

61120

x5 − 25120

x6

Exercise 4.log x = log (1 + (x− 1)) = (x− 1)− (x− 1)2

2+

(x− 1)3

3

=⇒ a = 0; b = 1, c =−12

Exercise 5.

cosx = 1− 12x2 + o(x3) as x → 0

1− cos x =12x2 + o(x3)

1− cosx

x2=

12

+o(x3)x2

since 1− cos x

x2=

12

+ o(x),1− cosx

x2→ 1

2as x → 0

cos x = 1− 12x2 +

x4

4!+ o(x5) =⇒ cos 2x = 1− 2x2 +

23x4 + o(x5)

1− cos 2x− 2x2

x4=− 2

3x4 − o(x5)x4

=−23− o(x) → −2

3as x → 0

Exercise 6.

limx→0

sin ax

sin bx= lim

x→0

ax− (ax)3

3! + o(x4)bx + o(x2)

=a

b121

Exercise 7.

limx→0

sin 2x

cos 2x sin 3x= lim

x→0

(2x) + (2x)3

3! + o(x4)(1− (2x)2

2! + (2x)4

4! + o(x5))(

(3x)− (3x)3

3! + o(x4)) =

23

Exercise 8.

limx→0

sin x− x

x3= −1

6

Exercise 9.

limx→0

ln 1 + x

e2x − 1= lim

x→0

x− o(x)2x + o(x)

=12

Exercise 10. Don't do the trig. identity.

limx→0

1− cos2 x

x tan x= lim

x→0

1−(1− x2

2! + o(x2))2

x(x + x3

6! + o(x3)) = lim

x→0

1− (1 + x2 + o(x2))x2 + o(x2)

= 1

Exercise 11.

limx→0

x +−x3

6 + o(x4)

x− x3

3

= 1

Exercise 12.

limx→0

ex ln a − 1ex ln b − 1

= limx→0

x ln a + o(x)x ln b + o(x)

= ln a/b

Exercise 13.

limx→1

(x− 1)− (x−1)2

2 + (x−1)3

3 + o(x− 1)4

(x + 2)(x + 1)=

13

Exercise 14. 1 .Exercise 15.

limx→0

x(ex + 1)− 2(ex − 1)x3

= limx→0

x(2 + x + x2

2 )− 2(x + x2/2 + x3/6)x3

= limx→0

x3( 16 )

x3=

16

Exercise 16.

limx→0

ln (1 + x)− x

1− cos x= lim

x→0

x− x2

2 + x3

3 + o(x3)− x

x3/2= −1

Exercise 17.

limx→π/2

cosx

x− π2

=0 +−1(x− π

2 )x− π

2

= −1

Exercise 18. 1/6

Exercise 19.

limx→0

cosh x− cosx

x2= lim

x→0

ex+e−x

2 − cos x

x2=

= limx→0

1 + x + x2

2 − 2(1− x2

2

)+ o(x3)

x2= 2

Exercise 20.

limx→0

3 tan 3x− 12 tan x

3 sin 4x− 12 sin x= lim

x→0

(4x) + (4x)3

3 − 4(x + x3

3

)+ o(x4)

(4x)− (4x)3

3! − 4(x− x3

3!

)+ o(x4)

=

=43 − 4−43+4

2

= −2

122

Exercise 21.

limx→0

ax − asinx

x3= lim

x→0

ex ln a − esin x ln a

x3=

= limx→0

1 + x ln a + (x ln a)2

2! + (x ln a)3

6 −(1 + sin x ln a + sin2 x(ln a)2

2 + sin3 x(ln a)3

3!

)+ o(x4)

x3=

= limx→0

ln a(x−(x− x3

3!

)) + (x2 ln a)2

2! − (ln a)2

2 (x2) + (lna)3

6 (x3 − x3) + o(x4)

x3=

ln a

6

Exercise 22.

limx→0

cos sin x− cos x

x4= lim

x→0

1− sin2 x2! + sin4 x

4! + o(x5)−(1− x2

2! + x4

4!

)

x4=

= limx→0

12

(x2 − (x2 − x4

3 ))

+(

x4−x4

4!

)+ o(x5)

x4=

16

Exercise 23.

limx → 1x1

1−x = lim x → 1e1

1−x ln x = exp(

lim x → 1ln x

1− x

)= exp

(limx → 1

(x− 1) + o(x− 1)2

1− x

)= e−1

Exercise 24.

limx→0

(x + e2x)1/x = exp limx→0

1x

ln (x + e2x)

limx→0

ln (x + e2x)x

= limx→0

ln (1 + x + e2x − 1)x

=

= limx→0

x + e2x − 1 + o(x2)x

= limx→0

3x + o(x2)x

= 3

=⇒ limx→0

(x + e2x)1/x = e3

Exercise 25.

limx→0

(1 + x)1/x − e

x= lim

x→0

e1x ln (1+x) − e

x= lim

x→0

ex+− x2

2 +o(x2)x − e

x=

= limx→0

e1− x2 +o(x) − e

x= lim

x→0

e(1 +−x2 + o(x))− e

x=−e

2Exercise 26.

limx→0

((1 + x)1/x

e

)1/x

= limx→0

(exp

(1x

ln (1 + x))− 1

)1/x

= limx→0

ex− x2

2 +o(x3)−x

x2 = e−1/2

Exercise 27.

(arcsin x)′ =1√

1− x2

(arcsin x)′′ =x

(1− x2)3/2

(arcsin x)′′′ =1

(1− x2)3/2+

3x2

(1− x2)5/2

exp(

limx→0

1x2

(ln

(arcsin x

x

)))= exp lim

x→0

1x2

ln(

1 +(

arcsinx

x− 1

))=

= exp(

limx→0

1x2

ln(

1 +x + x3/6 + o(x4)− x

x

))= exp

(limx→0

1x2

ln(

1 +x2

6+ o(x3)

))=

= exp(

limx→0

1x2

(x2 + o(x3)

6

))= e1/6

Exercise 28. limx→0

(1x − 1

ex−1

)= limx→0

(ex−1−xx(ex−1)

)= limx→0

x22 +o(x3)

x2+o(x3) =12

123

Exercise 29.

limx→1

(1

log x− 1

x− 1

)= lim

x→1

((x− 1)− log x

(x− 1) log x

)=

= limx→1

(x− 1)− ((x− 1)− (x−1)2

2 + o(x− 1)3)(x− 1)((x− 1) + o(x− 1)2)

=12

Exercise 30.

limx→0

eax − ex − x

x2= lim

x→0

1 + ax + (ax)2

2 + o(x3)− 1− x− x2

2 − x

x2

if a = 2, the limit is 2

Exercise 31.(1) Prove

∫ x

0f(t)dt = o

(∫ x

0g(t)dt

)as x → 0, given f(x) = o(g(x)).

Consider limx→0

R x0 f(t)dtR x0 g(t)dt

.Since f, g have derivatives in some interval containing 0, f, g continuous and differentiable for |t| ≤ x.

limx→0

∫ x

0f(t)dt

limx→0 g(t)dt= lim

x→0

(A(x)−A(0)

x

)(

B(x)−B(0)x

) =f(0)g(0)

=limx→0 f(x)limx→0 g(x)

= 0

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

xex = 0. However, limx→0

1ex = 1.

Exercise 32.(1) Use long division to �nd that

11 + g(x)

= 1− g(x) + g2(x) +−g3(x)1 + g(x)

= 1− g(x) + g2(x) + o(g2(x))

(2)

Exercise 33. Given limx→0

(1 + x + f(x)

x

)1/x

= e3, use the hint.

limx→0

g(x) = A, then G(x) = A + o(1) as x → 0

Then

g(x) = e3 + o(1) =(

1 + x +f

x

)1/x

x(e3 + o(1)

)x= x + x2 + f(x) =⇒ f(0) = 0

x exp x ln (e3 + o(1)) = x + x2 + f(x)

1 exp x ln (e3 + o(1)) + x expx ln (e3 + o(1))(

ln e3 + o(1) +x

e3 + o(1)(o′(1))

)= 1 + 2x + f ′(0)

1 + 3(0) + 0 = 1 + 0 + f ′(0) =⇒ f ′(0) = 0

We need to assume that in general o(1) = x + kx2 + o(x2).

2 exp x ln (e3 + o(1))(

ln (e3 + o(1)) +x

e3 + o(1)(o′(1))

)+

+ x exp x ln (e3 + o(1))(

ln (e3 + o(1)) +x

e3 + o(1)o′(1)

)+

x exp x ln (e3 + o(1))(

2o′(1)(e3 + o(1))

+xo′′(1)

e3 + o(1)+− x

(e3 + o(1))(o′(1))2

)= 2 + f ′′(x)

x→0−−−→ 2(3) = 2 + f ′(0) =⇒ f ′(0) = 4124

To evaluate limx→0

(1 + f(x)

x

)1/x

, consider a Taylor series expansion of f .

limx→0

(1 +

f(x)x

)1/x

= limx→0

(1 +

0 + 0 + 4x2

2 + o(x3)x

)1/x

= limx→0

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

= limy→0

limx→0

(1 + x (2 + o(y)))1/x = limy→0

exp 2 + o(y) = ey

7.13 Exercises - L'Hopital's rule for the indeterminate form 0/0. Exercise 1.

limx→2

3x2 + 2x− 16x2 − x− 2

= limx→2

(3x + 8)(x− 2)(x− 2)(x + 1)

=143

Exercise 2.

limx→3

x2 − 4x + 32x2 − 13x + 21

= limx→3

(x− 3)(x− 1)(2x− 1)(x− 3)

= −2

Exercise 3.

limx→0

sinh x− sin x

x3=

00

= limx→0

(cosh x− cosx

3x2

)=

00

= limx→0

(sinhx + sin x

6x

)= lim

x→0

(cosh x + cos x

6

)=

13

Exercise 4.

limx→0

(2− x)ex − x− 2x3

= limx→0

−ex + (2− x)− 13x2

= limx→0

(2− x)(1 + x + x2/2 + x3/6 + o(x3))x3

= −16

Exercise 5.

limx→0

log (cos ax)log (cos bx)

= limx→0

− cos bx sin axa

− cos ax sin bxb= lim

x→0

cos axa2

sin bxb2=

a2

b2

Exercise 6. When it doubt, Taylor expand.

limx→0+

x− sinx

(x sin x)3/2= lim

x→0+

1− cosx32 (x sin x)1/2(sinx + x cos x)

=23

limx→0+

1− (1− x2

2 + x4

24 + o(x4))

e12 ln (x sin x)(x− x3

6 + o(x3) + x− x3

2 + o(x3))=

=23

limx→0+

x2

2 − x4

24 + o(x4)√x(x + x3

6 + o(x3))(2x− 23x3 + o(x3))

=

=23

limx→0+

12 − x2

24 + o(x2)√1 + x2

6 + o(x2)(2 + −23 x2 + o(x2))

=

23

limx→0+

(1/22

)=

16

Notice in the third step how in general we deal with powers, (x sin x)1/2, is to convert it into exponential form, e12 ln (x sin x),

but it wasn't necessary.Exercise 7. Do L'Hopital's �rst.

limx→a+

√x−√a +

√x− a√

x2 − a2= lim

x→a+

12√

x+ 1

2√

x−ax√

x2−a2

=12

limx→a+

√x2 − a2

x3/2+√

x + a

x=

=12

limx→a+

√x2 − a2 + x1/2

√x + a

x3/2=

12

√2a

a3/2=

√2

2√

a

Exercise 8. Do L'Hopital's at the second step.

limx→1+

exp (x ln x)− x

1− x + ln x= lim

x→1+

exp x ln x(ln x + 1)− 1−1 + 1

x

= limx→1+

exp (x ln x)(lnx + 1)2 + 1x exp (x ln x)

−1/x2=

= limx→1+

x2 exp (x ln x)(ln x + 1)2 + x exp (x ln x) = 1 + 1 = 2

125

Exercise 9. Keep doing L'Hopital's.

limx→0

arcsin 2x− 2 arcsin x

x3= lim

x→0

2√1−(2x)3

− 2 1√1−x2

3x2=

=23

limx→0

− 12 (1− (2x)2)−3/2(−8x)− (− 1

2 )(1− x2)−3/2(−2x)6x

=

=19

limx→0

4(1− (2x)2)−3/2 + 4x(− 32 )(1− (2x)2)−5/2(−8x) +−(1− x2)−3/2 − x(− 3

2 )(1− x2)−5/2(−2x)1

=

=19

4− 11

=13

Exercise 10.

limx→0

x cot x− 1x2

= limx→0

x cos x− sin x

x2 sinx= lim

x→0

cosx− x sin x− cosx

2x sin x + x2 cosx= lim

x→0

− sin x

2 sin x + x cosx=

= − limx→0

cos x

2 cos x + cos x +−x sin x= − lim

x→0

cos x

3 cos x− x sin x=

13

Exercise 11.

limx→1

∑nk=1 xk − n

x− 1= lim

x→1

∑nk=1 kxk−1

1=

n∑

k=1

k =n(n + 1)

2

Exercise 12.

limx→0+

1x√

x

(a arctan

√x

a− b arctan

√x

b

)= lim

x→0+

(a 1

1+ xa2

(1

2a√

x

)− b 1

1+ xb2

(1

2b√

x

))

32x1/2

=

=13

limx→0+

a2

a2 + x

(1x

)− b2

b2 + x

(1x

)=

13

limx→0+

(b2 + x)a2 − b2(a2 + x)(a2 + x)(x + b2)x

=

=13

limx→0+

(a2 − b2)(a2 − x)(b2 + x)

=13

a2 − b2

a2b2

Exercise 13.(sin 4x)(sin 3x)

x sin 2x=

(2 sin 2x cos 2x)(sin 3x)x sin 2x

=2(−2 sin 2x sin 3x + cos 2x3 cos 3x)

1= 6 as x → 0 otherwise

2 cos 2x sin 3x

x→ 4

πas x → π

2We used L'Hopital's at the second to last step for x → 0.Exercise 14.

limx→0

(x−3 sin 3x + ax−2 + b) = 0

limx→0

sin 3x + ax + bx3

x3=

3 cos 3x + a + 3bx2

3x2=−9 sin 3x + 6bx

6x=−27 cos 3x + 6b

6=−27 + 6b

6= 0

So b =92, a = −3 .

Exercise 15.

limx→0

1bx− sin x

∫ x

0

t2dt√a + t

=x2√a+x

b− cosx=

2x1

2√

a+x(1− cosx) +

√a + x(sinx)

=2x√

a + x12 (1− cosx) + (a + x) sin x

=

= 2 limx→0

√a + x + x

2√

a+x

sin x2 + sin x + (a + x) cos x

= 2√

a

a= 1

=⇒ 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 alsorearranged the denominator and numerator cleverly at each step.Exercise 16.

126

(1)angle ABC is x

2, length BC is tan

x

22 tan

x

2cos

x

2= 2 sin

x

2is the base length of ABC

tanx

2sin

x

2is the height of triangle ABC

=⇒ T (x) = tanx

2sin2 x

2=

1− cos2 x2

cos x2

sinx

2= tan

x

2− 1

2sin x

(2)S(x) =

( x

2π

)(π(1))− 1

2cos

x

2(2 sin

x

2) =

x

2− sin 2

2(3) Use L'Hopital's theorem.

T (x)S(x)

=tan x

2 − 12 sin x

x−sin x2

ddx−−→

12 sec2 x

2 − cos x2

1−cos x2

ddx−−→ sec2 x

2 tan x2 + sin x

sinx

ddx−−→ sec2 x

2 tan2 x2 + 1

2 sec4 x2 + cos x

cosx→

x→0−−−→ 32

Exercise 17. Use L'Hopital's rule.

I(t) =E

R(1− e−

RtL )

limR→0

I(t) = limR→0

E(−1)(e−Rt/L)(−t

L

)

1=

Et

L

Exercise 18.c− k → 0 since c → k

k − c = u

k − u = c

f(t) =A(sin kt− sin ct)

c2 − k2=

A(sin (kt)− sin (k − u)t)−u(2k − u)

=

=A(cos (k − u)t)(t)−(2k − u) + u

→ −At cos kt

2k

7.17 Exercises - The symbols +∞ and −∞. Extension of L'Hopital's rule; In�nite limits; The behavior of log x and ex

for large x.

Exercise 15. Use L'Hopital's at the second to last step.

limx→1−1

(lnx)(ln (1− x)) = limx→1−1

ln (1− x)1

ln x

= limx→1−1

(1

1−x

)(−1)

−1(ln x)2

(1x

) = limx→1−1

x

1− x(ln x)2 = lim

x→1−1

2 ln x(

1x

)

(−1)= 0

Exercise 16. Persist in using L'Hopital's and trying all possibilities systematically.

limx→0+

xxx−1 = limx→0+

e(xx−1) ln x = limx→0+

e(ex ln x−1) ln x = exp limx→0+

ln x

(ex ln x − 1)−1=

= exp limx→0+

1/x

(−1)(ex ln x − 1)−2(ex ln x(lnx + 1))= exp− lim

x→0+

(e2x ln x − 2ex ln x + 1ex ln x(x ln x + x)

)=

= exp− limx→0+

ex ln x(lnx + 1) + e−x ln x(− ln x− 1)(lnx + 1 + 1)

= exp− limx→0+

ex ln x − e−x ln x

1 + 1ln x+1

= 1

Exercise 17.

limx→0+

(xxx − 1) = limx→0+

(exx ln x − 1

)= lim

x→0+

(eex ln x ln x − 1

)=

(elimx→0+ ex ln x ln x − 1

)=

= eelim

x→0+x ln x

limx→0+ ln x − 1 = 0− 1 = −1127

We usedlim

x→0+xα log x = 0 ∀α > 0

sincet = 1

x , xα log x = − log ttα → 0 as t →∞.

Exercise 18.

limx→0−

esin x ln (1−2x) = exp(

limx→0−

ln (1− ex ln 2)1/ sin x

)= exp

(lim

x→0−

11−ex ln 2

(− ln 2ex ln 2)

−1sin2 x

cos x

)=

= exp(

limx→0−

(sin2 x) ln 2ex ln 2

(1− ex ln 2) cos x

)= exp

((ln 2) lim

x→0−

2 sin x cosx

− ln 2ex ln 2

)= 1

Exercise 19.lim

x→0+e

1ln x ln x = e

Exercise 20. At the end, L'Hopital's could be used to verify that indeed sinx ln sin x → 0 as x → 0.lim

x→0+esin x ln cot x = elimx→0+ sin x(ln cos x−ln sin x) = elimx→0+ − sin x ln sin x = 1

Exercise 21. Rewrite tan into sin and cos and use L'Hopital's.

limx→π

4

(tan x)tan 2x = limx→π

4

etan 2x ln tan x = exp limx→π

4

1cos 2x

(ln sin x− ln cos x) =

= exp limx→π

4

1sin x cosx− 1

cos x (− sin x)−2 sin 2x cos 2x

= exp limx→π

4

1− sin2 2x

= e−1

Exercise 22.Exercise 23. Use L'Hopital's theorem, taking derivatives of top and bottom.

limx→0+

expe

1 + ln xln x = exp e lim

x→0

ln x

1 + ln x= exp e lim

x→0

1/x

1/x= ee

Exercise 24. Rewrite tan into sin and cos and take out sin since we could do the limit before doing L'Hopital's.

limx→1

(2− x)tan (πx/2) = limx→1

etan πx2 ln (2−x) = elimx→1

sin πx/2 ln (2−x)cos πx/2 = exp lim

x→1

(−1)2−x

π2 sinπx/2

= exp−2π

Exercise 26.

limx→+∞

exp(

x ln(

x + c

x− c

))= exp lim

x→+∞

ln(

1+c/x1−c/x

)

1/x= exp lim

x→+∞

1

( 1+c/x1−c/x )

((x−c)−(x+c)

(x−c)2

)

−1x2

=

= 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)

x2

(1 +

1x2

)1/2

− x2 = x2

(1 +

1x2

)1/2

− x2

Let x2 = 1t . So t → 0 as x → +∞.

=⇒ (1 + t)1/2 − 1t

=1 + 1

2 t− 1 + o(t)t

=12

Exercise 28.(x5 + 7x4 + 2)c − x = x5

(1 +

7x

+2x5

)c

− x

Let 1t = x and guess that c = 1

5 (1t5

+7t4

+ 2)c

− 1t

=(1 + (7t + 2t5)

)1/5/t− 1

t=

=1 + 1

5 (7t + 2t5) + o(t)− 1t

=75

128

Exercise 29.

g(x) = xex2

g′(x) = ex2+ 2x2ex2

g′′(x) = 2xex2+ 4xex2

+ 4x3ex2= 6xex2

+ 4x3ex2

f(x) =∫ x

1

g(t)(

t +1t

)dt

f ′(x) = g(x)(

x +1x

)− g(1)2;

f ′′(x) = g′(x)(x + 1/x) + g(x)(1− 1/x2)

f ′′(x)g′′(x)

=(ex2

+ 2x2ex2)(x + 1/x) + xex2

(1− 1x2 )

6xex2 + 4x3ex2 =2xex2

+ 2x3ex2+ 2xex2

(6xex2 + 4x3ex2)=

=4x + 2x3

6x + 4x3=

12

as x →∞

Exercise 30.g(x) = xce2x

g′(x) = cxc−1e2x + 2xce2x

f(x) =∫ x

0

e2t(3t2 + 1)1/2dt

f ′(x) = e2x(3x2 + 1)1/2 − 1Guessing that c = 1

f ′(x)g′(x)

=e2x(3x2 + 1)1/2 − 12xce2x + cxc−1e2x

=

√3x(1 + 1

3x2 )1/2 − e−2x

2x + 1=

√3

2

So c = 1 .Exercise 31.Exercise 32.

(1)P

(1 +

r

m

), P

(1 +

r

m

)2

, . . . P(1 +

r

m

)m

For each year, there are the just previously shown m compoundings, so for n years,

P(1 +

r

m

)mn

(2)2 = ert

ln 2r

= t = 11.55years

(3)2P0 = P0

(1 +

r

m

)mt

ln 2 = mt ln (1 + r/m)

t =ln 2

m ln (1 + r/m)=

ln 24 ln (1 + 0.06/4)

= 11.64years

7.17 Exercises - The symbols +∞ and −∞. Extension of L'Hopital's rule; In�nite limits; The behavior of log x and ex

for large x. Exercise 1.

limx→0

e−1/x2

x1000= lim

u→∞e−uu500 = lim

u→∞u500

eu= 0

u =1x2

x2 =1u

x1000 =1

u500Where we had used Theorem 7.11, which are two very useful limits for log and exp.

Theorem 26. If a, b > 0,

limx→+∞

(log x)b

xa= 0(17)

limx→+∞

xb

eax= 0(18)

129

Proof. Trick - use the de�nition of the logarithm as an integral.If c > 0, t ≥ 1, tc ≥ 1 =⇒ tc−1 ≥ t−1.

0 < ln x =∫ x

1

1tdt ≤

∫ x

1

tc−1dt =1c(xc − 1) <

xc

c

=⇒ 0 <(lnx)b

xa<

xcb−a

cb

Choose c =a

2b,xcb−a

cb=

x−a/2

cb→ 0 as x →∞

then (lnx)b

xa→ 0 as x → 0

For exp, Let t = ex. ln t = x. xb

eax = (ln t)b

ta → 0 as t →∞ as x →∞. ¤

Exercise 2.

limx→0

sin 1x

arctan 1x

= limx→0

1x − o

(1x

)1x − o

(1x

) = 1

Exercise 3. Use L'Hopital's.

limx→π

2

tan 3x

tanx= lim

x→π2

cos x

cos 3x= lim

x→π2

− sin x

−3 sin 3x= −1

3

Exercise 4. Use L'Hopital's.

limx→∞

ln (a + bex)√a + bx2

= limx→∞

1a+bex (bex)

bx√a+bx2

= limx→∞

1ae−x+b

1√a

x2 +b

=1√b

Exercise 5. Make the substitution x = 1t .

limx→∞

x4

(cos

1x− 1 +

12x2

)= lim

t→∞1t4

(cos t− 1 +

t2

2

)= lim

t→∞t4/4! + o(t4)

t4=

14!

=1

120

Exercise 6.

limx→π

ln | sin x|ln | sin 2x| = lim

x→π

1sin x cos x2

sin 2x cos 2x= −1

2limx→π

sin 2x

sinx= −1

2limx→π

2 cos 2x

cosx= 1

Exercise 7.

limx→ 1

2−

ln (1− 2x)tan πx

= limx→ 1

2−

(1

1−2x

)(−2)

(sec2 πx)π= lim

x→ 12−

−2(cos2 πx)(1− 2x)π

=

= − 2π

limx→ 1

2−

2 cos πxπ − sinπx

−2= 1

Exercise 8.

limx→∞

coshx + 1ex

= limx→∞

ex+1 + e−x−1

2ex=

12

limx→∞

e1 +1

e2x+1=

e

2

Exercise 9.

limx→∞

ax

xb= lim

x→∞ex ln a

xb→∞; a > 1

since limx→∞

xb

eax(in this case, ln a > 0 )

Exercise 10.

limx→π

2

tan x− 5sec x + 4

= limx→π

2

sec2 x

tanx sec x= lim

x→π2

sec x

tanx= lim

x→π2

1cosx

cosx

sin x= 1

130

8.5 Exercises - Introduction, Terminology and notation, A �rst-order differential equation for the exponential function,First-order linear differential equations.The ordinary differential equation theorems we will use are

y′ + P (x)y = 0(19)

A(x) =∫ x

a

P (t)dt

y = be−A(x)(20)

Consider y′ + P (x)y = Q(x); A(x) =∫ x

aP (t)dt.

Let h(x) = g(x)eA(x); g a solution.

h′(x) = (g′ + Pg)eA = QeA

2nd. fund. thm. of calc.−−−−−−−−−−−−−→ h(x) = h(a) +∫ x

a

Q(t)eA(t)dt

since h(a) = g(a)

(21) y = g(x) = e−A(x)

(∫ x

a

Q(t)eA(t)dt + b

)

We had done some of these problems previously, using an integration constant C, but following Apostol's notation fory(a) = b for initial conditions is far more advantageous and superior as we seem clearly the dependence upon the initialconditions - so some of the solutions for the exercises will show corrections to the derived formula using Apostol's notationfor y(a) = b initial conditions.Exercise 1. A(x) =

∫ x

0(−3)dt = −3x

y = e3x

(∫ x

0

e2te−3tdt + 0)

= e3x (−e−t)∣∣x0

= −e2x + e3x

Exercise 2. y′ − 2xy = x4. A(x) =

∫ x

1

(−2t

)dt = −2 ln x.

y = e−A(x)

(∫ x

a

Q(t)eA(t)dt + b

)= e2 ln x(

R x1 t4e−2 ln tdt+1) = x2

(∫ x

1

t2dt + 1)

=

= −x2 +x2

3(x3 − 1) =

x5

3+

2x2

3

Exercise 3. y′ + y tanx = sin 2x on(−π

2 , π2

), with y = 2 when x = 0.

A(x) =∫ x

0

P (t)dt =∫ x

0

tan tdt = − ln | cos t||x0 = − ln cos x

y = e−A(x)

(∫ x

a

Q(t)eA(t)dt + b

)= cos x

(∫ b

a

sin 2te− ln cos tdt + 2

)=

= cos x

(∫ x

a

2 sin tdt + 2)

= −2 cos2 x + 4 cos x

Exercise 4. y′ + xy = x3. y = 0, x = 0.

A(x) =12x2

y = e−12 x2

(∫ x

0

t3et22 dt + 0

)= e

−x22

(t2e

t22 − 2e

t22

)∣∣∣x

0=

= x2 − 2 + 2e−x22

Exercise 5. y′ + y = e2t. y = 1, t = 0.

A = x

y(x) = e−x

(∫ x

0

e2tetdt + 1)

= e−x

(e3t

3

∣∣∣∣x

0

+ 1)

=e2x

3+

23e−x

131

Exercise 6. y′ sin x + y cosx = 1; (0, π). =⇒ y′ + cot xy = csc x

A(x) =∫ x

a

cot tdt = ln(

sin x

sin a

)

y(x) = e− ln ( sin xsin a )

(∫(csc t)eln ( sin t

sin a )dt + b

)=

(sin a

sin x

)(x− a

sin a+ b

)

indeed y =x− a

sin x+

b sin a

sin x

x → 0 for y =x− a

sinx+

b sin a

sin x=

x + b sin a− a

sin x

b sin a = a for x → 0

a− b sin a = π for x → π

Exercise 7.

x(x + 1)y′ + y = x(x + 1)2e−x2=⇒ y′ +

1x(x + 1)

y = (x + 1)e−x2

A(x) =∫ x

a

(1t− 1

t + 1

)dt = ln

x

a− ln

x + 1a + 1

eA(x) =(a + 1)xa(x + 1)

y =a(x + 1)(a + 1)x

(∫ x

a

(t + 1)e−t2(

(a + 1)ta(t + 1)

)dt + b

)=

(x + 1)x

(∫ x

a

te−t2dt + ba(x + 1)(a + 1)x

)=

=x + 12x

(e−a2 − e−x2

)+

a(x + 1)b(a + 1)x

It's easy to see that the last equation above goes to 0 as x → −1.

y = (e−a2 − e−x2)(1/2)(1 +

1x

) +ab

a + 1(1 +

1x

) =

limx→0

y =1x

(e−a2 − e−x2

2+

ab

a + 1

)a = 0

Exercise 8. y′ + y cot x = 2 cosx on (0, π).

A(x) =∫ x

a

cot tdt = ln(

sin x

sin a

)eA(x) =

sin x

sin a

y =sin a

sin x

(∫2 cos t

sin t

sin a+ b

)=

sin a

sin x

(− cos 2t

2 sin a

∣∣∣∣x

a

+ b

)=

=cos (2a) + cos 2x

2 sin x+

sin a

sin x

y = sin x y =cos (2a)− cos (2x)

2 sin x+

sin a

sin x

Exercise 9. (x− 2)(x− 3)y′ + 2y = (x− 1)(x− 2). y′ + 2(x−2)(x−3)y = x−1

x−3 .

A(x) =∫ x

a

2dt

(t− 2)(t− 3)= 2

∫ x

a

(1

t− 3− 1

t− 2

)dt = 2 (ln |t− 3| − ln |t− 2|)|xa = 2 ln

∣∣∣∣x− 3a− 3

∣∣∣∣∣∣∣∣a− 2x− 2

∣∣∣∣

If (−∞, 2), (3, +∞),x− 3 ≶ 0x− 2 ≶ 0

3− x

2− x=

x− 3x− 2

If (2, 3), x− 3 < 0, but x− 2 > 0

If (−∞, 2), (3,∞) y = b

(x− 2x− 3

)2 (∣∣∣∣a− 3a− 2

∣∣∣∣)2

+(

x− 2x− 3

)2 (x +

1x− 2

− a− 1a− 2

)

132

If (2, 3) y = b

(x− 2x− 3

)2 (∣∣∣∣a− 3a− 2

∣∣∣∣)2

+(

x− 23− x

)2 ∣∣∣∣a− 3a− 2

∣∣∣∣2 ∫ (

t− 1t− 3

) ∣∣∣∣a− 2a− 3

∣∣∣∣2 (

3− t

t− 2

)2

=

= b

(x− 23− x

)2 ∣∣∣∣a− 3a− 2

∣∣∣∣2

+x− 23− x

(−

∫(t− 1)(3− t)

(t− 2)2

)=

= b

(x− 2x− 3

)2 ∣∣∣∣a− 3a− 2

∣∣∣∣2

+(

x− 2x− 3

)2

(x + (x− 2)−1 − a− (a− 2)−1)

Exercise 10. s(x) = sin xx ; x 6= 0 s(0) = 1, T (x) =

∫ x

0s(t)dt f(x) = xT (x)

f ′ = T + x(s(x)) = T + sin x

xf ′ − f = x sinx

y′ − y

x= sin x

A(x) =∫ x

a

−1t

dt = − lnx

a, e−A(x) =

x

a

y =x

a

(∫ x

a

sin ta

t+ b

)= xT +

bx

a

y(0) = 0 6= 1 since P (x) =1x

is not continuous at x = 0

Exercise 11.

f(x) = 1 +1x

∫ x

1

f(t)dt

xf(x) = x +∫ x

1

f(t)dt =⇒ f(x) + xf ′ = 1 + f(x) =⇒ f ′ =1x

=⇒ f(x) = ln |x| − C

ln |x| − C = 1 +1x

∫ x

1

(ln |t| − c) dt =

= 1 +1x

(t ln |t| − t− ct)|x1 = ln |x| − C +1 + C

x=⇒ C = −1

f(x) = ln |x|+ 1

Exercise 12. Rewrite the second property we want f to have:∫ x

1

f(t)dt =1− f(x)

x=⇒ f(x) = − 1

x2−

(f ′x− f

x2

)

So then

f ′ +(

x− 1x

)f = − 1

x

A(x) =∫

P (t)dt =∫ (

t− 1t

)dt =

(12x2 − ln x

)−

(12a2 − ln a

)

eA(x) = ex2−a2

2a

x; e−A(x) =

x

ae

a2−x22

f(x) =x

ae

a2−x22

(∫ x

a

−1t2

aet2−a2

2 dt + b

)

Exercise 13.

v = yk

v′ = kyk−1y′v′ + kPv = kQ = kyk−1y′ + kPyk = kQ where k = 1− n

=⇒ y′y−n + Py1−n = Q =⇒ y′ + Py = Qyn

133

Exercise 14. y′ − 4y = 2exy1/2

n =12

k = 1− 12

=12; v = y1/2; v′ +

12(−4)v =

12(2ex) = v′ − 2v = ex

A(x) =∫ x

a

P (t)dt = −2(x− a) = −2x

v = e2x

(∫ete−2tdt + b

)= e2x

(−e−x + b)

= be2x − ex

=⇒ y = (1 +√

2)2e4x − 2(1 +√

2)e3x + e2x

y(x) = b2 − 2b + 1 = 2

b = 1 +√

2

Exercise 15. y′ − y = −y2(x2 + x + 1), n = 2 k = 1− n = −1, v = yk v = y−1.

v′ + kPv = kQv′ + (−1)(−1)v = (−1)(−(x2 + x + 1)) = v′ + v = x2 + x + 1

A(x) =∫

P (t)dt =∫ x

0

1dt = x

v = e−x

(∫(t2 + t + 1)etdt + b

)= e−x

(t2et − tet + 2et

)∣∣x0

+ be−x = (x2 − x + 2)− (2e−x) + be−x

y =1

x2 − x + 2− 2e−x

Exercise 16.

v′ +− 1x

v = 2x2

v = eln x

(∫2x2e− ln xdx + C

)= x(x2 + C) = x3 + Cx

Then since v = yk, k = 1− n,y = (x3 + Cx)2 = x2(x2 + C)2; x 6= 0

y = x2(x2 − 1)2

Check:y′ = 2x(x2 − 1)2 + x3(4)(x2 − 1)

2x2 + x4(4)(x2 − 1)− 2x2(x2 − 1)2 = 4x3

Exercise 17.xy′ + y = y2x2 log x on (0, +∞) with y = 12 when x = 1

2 , x 6= 0.

y′ +y

x= y2x log x

k = 1− n = 1− 2 = −1, v = yk = y−1

v′ + kPv = kQ; v′ +−(

1x

)v = −1x log x

v = x

(∫ −x log x

xdt + C

)= x

(∫− log tdt + C

)=

= x(−(x log x− x) + C) = −x2 log x + x2 + Cx

y =1

Cx + x2 − x2 log xC = −1

2(1− log

12) + 2

Check:y′ = (C + 2x− 2x log x− x)(−y2)

y2(−C +−2x + 2x log x + x + C + x− x log x)

y =12;x = 1 y(x = 1) =

12

b = 2

y =1

x(x ln x− x + 3)134

Exercise 18. 2xyy′ + (1 + x)y2 = ex on (0,+∞), y =√

e when x = 1; y = −√e when x = 1.If x 6= 0, y 6= 0,

y′ +(1 + x)

2xy =

e2

2xy−1

k = 1− n = 2; v = yk = y2

v′ + 2Pv = 2Q; v′ + 2(

1 + x

2x

)v =

2ex

2x=⇒ v′ +

1 + x

xv =

ex

x= v′ + Pv = Qv

Av =∫ x

a

Pv =∫ x

a

1 + t

tdt = ln

(x

a

)+ (x− a); e

R xa

Pvdt = eln x/a+x−a =x

aex−a

v =ae−x+a

x

(∫ x

a

et

t

t

aet−adt + bv

)=

a

xe−x+a

(12a

e2t−a

∣∣∣∣x

a

+ bv

)

=a

xe−x+a

(e2x−a

2a− ea

2a+ bv

)

y = ±√

a

xe−x+a

(e2x−a

2a− ea

2a+ bv

)

Now yk = v

y2 = v = e = 11e−1+1

(e2−1

2(1) − e1

2 + bv

)= bv = e

(1) bv = e =⇒ y =√

1xe−x+1

(e2x−1

2 − e2 + e

)

(2) bv = e =⇒ y = −√

1xe−x+1

(e2x−1

2 − e2 + e

)

(3) If we could take a = 0, then limx→0 y = ±√

ex

2x − e−x+2a

2x + bvae−x+a

x = ±√

1+x−e2(0)(1−x)+2bv0e−x+a

2x = ±1

If we consider limx→0 y2, and let a go to 0, then

v =1x

(ex − e−x

)=⇒ sinh x

x

Exercise 19. The Ricatti equation is y′ + P (x)y + Q(x)y2 = R(x).If u is a known solution, y = u + 1

v is also a solution if v satis�es a �rst-order ODE.

(u + 1/v)′ = u′ +−1v2

v′

y′ + Py + Qy2 = R =⇒ u′ +−v′

v2+ P (u +

1v) + Q(u2 +

2uv

+1v2

) = R

=⇒ v′ − Pv = Q(2uv + 1)

Exercise 20. y′ + y + y2 = 2, y = 1,−2.(1) If −2 ≤ b < 1,

y′ + y + y2 = 2 P = 1, Q = 1, R = 2

v′ + (−P − 2Qu)v = Qy = u +1v

u = 1

v′ + (−1− 2(1)(1))v = v′ − 3v = Q = 1

v = e3x

(∫1e−3tdt + b

)= e3x

(e−3t

−3

∣∣∣∣x

a

+ b

)= be3x − 1

3(1− e3x−3a

)

y = 1 +3

3be3x − (1− e3x−3a)

y(0) = 1 +3

3b− (1− e−3a)=⇒

b = 1 +1b

b2 − b− 1 = 0

b =1±√5

2135

y = 1 +3

3be3x − (1− e3x)b =

1−√52

(2)u = −2

v′ + (−1− 2(1)(−2))v = v′ + 3v = 1

v = e−3x

(∫ x

a

e3tdt + b

)= e−3x

(e3x − e3a

3

)+ be−3x =

1− e3a−3x

3+ be−3x

y = −2 +3

1− e3a−3x + 3be−3x; y(0) = −2 +

31− e3a + 3b

a=0−−→ y(0) = −2 +33b

=⇒ b = −1±√

2

b ≥ 1 or b < −2, y = −2 +3

1− e−3x + 3be−3xb = −1±

√2

8.7 Exercises - Some physical problems leading to �rst-order linear differential equations.

Exercise 3.

(1) y′ = −αy(t). y(T ) = y0e−αT = y0

n . n = eαT so the relationship between T and n doesn't depend upon

y0. 1k

ln e = T .

(2) f(a) = y0e−ka; f(b) = y0e

−kb.f(a)f(b)

= e−ka+kb =⇒ lnf(a)f(b)

= −k(a− b);1

a− bln

f(a)f(b)

= −k

f(t) = y0 exp

ln f(a)

f(b) t

a− b

; f(a) = y0

(f(b)y0

)a/b

; =⇒ f(a)(f(b))a/b

= y1− a

b0

f(t) =(

f(a)(f(b))a/b

) bb−a

(f(a)f(b)

) ta−b

=(f(a))

bb−a

f(b)a

b−a

f(a)−t

b−a

f(b)t

a−b

=

=f(a)

b−tb−a

(f(b))a−tb−a

= f(a)b−tb−a f(b)

t−ab−a

w(t) =b− t

b− a; 1− w(t) =

b− a− (b− t)b− a

=t− a

b− a

Exercise 4. F = mv′ = w0 − 34v

w0 = 192w0

g= 6 = m

v′ =w0

m− 1

8v =⇒ v′ +

18v =

w0

m= 32

v(t) = e−t8

(∫ t

0

w0

me

t8 dt + b

)= e

−t8

(8w0

m(et/8 − 1) + 0

)=

(256(1− e−t/8)

)

v(10) = 256(1− e−5/4) = 256(1− 37/128) = 182

F = mv′ = w0 − 12v v′ +12m

v =w0

m= v′ + 2v = 32

v(t) = e−2t

(∫ t

t0

e2x32dx + b

)= e−2t

(16(e2t − e2t0) + b

)=

(16(1− e2(t0−t)) + be−2t

)

v(10) = be−2(10) = 182 =⇒ b = 182e20

v(t) = 16 + 166e20−2t

So then

v(t) =

{256(1− e−t/8) if t < 1016 + 166e20−2t if t > 10

Exercise 7.136

(1)y′(t) = (y −M)k = ky − kM y′ +−ky = −kM

y = ekt(∫−kMe−ktdt + b) = ekt(M(e−kt − 1) + b) M = 60◦

y(0) = b = 200◦

yf = ekT (M(e−kT − 1) + 200) = M + ekT (200−M)

1T

ln(

yf −M

b−M

)= k =

1T

(ln (60)− ln (140)) =1T

ln(

37

)=

130

(ln 3− ln 7)

y(t) = 60 + 140eln 3−ln 7

30 t

(2)

y(t) = 60 + 140e−kt; k =(ln 7− ln 3)

30

ln(

Y − 60140

)= −kt =⇒ tf =

(ln 140− ln (T − 60))k

for 60 < T ≤ 200

(3) tf = ln (140)−ln (30)k = 30

ln 7/3 ln 143 = 54 minutes

(4) M = M(t) = M0 − αt α = 110

y = ekt

(∫−kMe−ktdt + b

)= ekt

(∫ t

0

−k(M0 − αu)e−kudu + b

)=

= −kekt

∫ t

0

(M0e−ku − αueku)du + bekt =

= −ke

(M0e

−ku

−k

∣∣∣∣t

0

− α

(ue−ku

−k+−e−ku

k2

)∣∣∣∣t

0

)+ bekt =

= −kekt

(M0

k(1− e−kt)− α

(te−kt

−k− e−kt

k2+

1k2

))+ bekt =

y(t) = −M0ekt + M0 − αt− α/k + αekt/k + bekt = (−M0 + α/k + b)ekt + (M0 − αt− α/k) =

y(t) = (140 +3

(ln 3− ln 7))e−kt + (60− t

10−

(3

ln 3/7

))

Exercise 8. y′(t) = −k(y −M0); y′ + ky = kM0.

y = e−kt

(∫ tf

ti

kM0ekudu + b

)= e−ktf

(M0(ektf − ekti) + b

)= M0(1− e−k(tf−ti)) + be−ktf

=⇒

y(tf )−M0 = −M0e−k(tf−ti) + be−ktf

65−M0 = −M0e−k(5) + 75e−k(5) = (75−M0)e−5k =⇒ ln

(65−M0

75−M0

)= −5k; k =

15

ln(

75−M0

65−M0

)

60−M0 = −M0e−k(5) + 65e−k(5) = (65−M0)e−5k =⇒ ln

(60−M0

65−M0

)= −5k

=⇒ 65−M0

75−M0=

60−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 =

(2 galmin.

)(y(t) salt

w(t)

)

So then

y′ =−2y

w0 + t=⇒ ln y = −2(ln (w0 + t)− ln w0) = −2 ln

(w0 + t

w0

)

137

is the equation of motion given by the problem.

y(t) = Celn“

w0+tw0

”−2

= C

(w0 + t

w0

)−2

y(t) = 50(

100100 + t

)2

y(t = 60 min.) = 50(

100160

)2

= 502564

=62532

' 19.53

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 tankthat is �coming into� the dissolved salt, adding to the amount of dissolved salt in the mixture. Thus

y′(t) = (−2)( y

w

)+ α

( y

w− 3

); α =

−1 gal3 min

We obtained α easily by considering only the dissolving part and how it dissolves 1 pound of salt per minute if the saltconcentration, y

w was zero, i.e. water is fresh.

y′(t) =−73

y

w+ 1; y′ =

73

y

w0 + t= 1

P =7/3

w0 + t∫P =

∫7/3

w0 + t=

73

ln (w0 + t)|t0 =73

ln(

w0 + t

w0

)

y = e− ln

“w0+t

w0

”−7/3 (∫ t

0

(1)eln“

w0+uw0

”7/3

du + b

)=

=(

w0

w0 + t

)7/3(∫ t

0

(w0 + u

w0

)7/3

du + b

)=

(w0

w0 + t

)7/3(

3w0

10

((w0 + t

w0

)10/3

− 1

)+ b

)=

y =(

100100 + 60

)7/3(

3(100)10

((100 + 60

100

)10/3

− 1

)+ 50

)' 54.78 lbs.

Exercise 11. LI ′(t) + RI(t) = V (t) V (t) = E sin ωt.

I(t) = I(0)e−Rt/L + e−Rt/L

∫ t

0

V (x)L

eRx/Ldx

I(t) = I(0)e−Rt/L + e−Rt/L E

L

∫ t

0

sin ωxeRx/Ldx

Using∫

eax sin bxdx = aeax sin bx− beax cos bx

I(t) = I(0)e−Rt

L +Ee−Rt/L

L

(RL e

RtL sin ωt− ωe

RtL cos ωt

(RL

)2+ ω2

+ω(

RL

)2+ ω2

)

I(0) = 0 So

I(t) =E

L

RL sin ωt− ω cosωt

(RL

)2+ ω2

+EωL

R2 + (ωL)2e−Rt/L =

E(R sin ωt− ωL cosωt)R2 + (ωL)2

+EωL

R2 + (ωL)2e−Rt/L =

sin α =ωL√

R2 + (ωL2)

=⇒ I(t) =E sin (ωt− α)√

R2 + (ωL)2+

EωL

(R2 + ω2L2)e−Rt/L

L = 0 sin α = 0138

Exercise 12.

E(t) =

{E if 0 < a ≤ t b, I(t) = Ke−Rt/L

=⇒ I(t) =Ee−Rt/L

R(e

RbL = e

Ral ) for I(b) = I(b)

Exercise 13. From Eqn. 8.22, dxdt = kx(M − x)

dx

dt= kx(M − x) = kMx− kx2; =⇒ dx

kMx− kx2= dt =

1/kdx

x(M − x)= dt

=⇒ kdt =(

1x

+1

M − x

)1M

dx =ln x +− ln (M − x)

M

Mk(t− ti) = lnx

M − x; eMk(t−ti)(M − x) = x

x(t) =MeMk(t−ti)

1 + eMk(t−ti)=

M

1 + e−Mk(t−ti)

Exercise 14. Note that we are given three equally spaced times.

M = x2 + x2e−α(t2−t0);

M − x2

x2= e−α(t2−t0)

M − x2

x2

(x1

M − x1

)= e−αt2+αt0+αt1−αt0 = e−α(t2−t1)

M − x3

x3

(x2

M − x2

)= e−α(t3−t2) =

M − x2

x2

(x1

M − x1

)

(M − x3)(M − x1)x22 = x1x3(M − x2)2 = x1x3(M2 − 2Mx2 + x2

2) = x22(M

2 −M(x1 + x3) + x1x3)

(x22 − x1x3)M2 = M(x2

2(x1 + x3) +−2x2x1x3) = (−x1(x3 − x2) + x3(x2 − x1))x2

=⇒ M = x2(x3(x2 − x1)− x1(x3 − x2))

x22 − x1x3

Exercise 15.dx

dt= k(t)Mx− k(t)x2 dx

Mx− x2= k(t)dt =⇒ M

∫ t

ti

k(u)du = ln(

x

M − x

)

x

M − x= e

MR t

tik(u)du

x =Me

MR t

tik(u)du

1 + eMR t

tik(u)du

=M

1 + e−M

R tti

k(u)du

Exercise 16.(1) M = 23 92(23−3.9)−3.9(92−23)

232−3.9(92) = 201(2)

M = 122(

150(122− 92)− 92(150− 122)(122)2 − 92(150)

)= 122

(150(30)− 92(28)(122)2 − 92(150)

)= 216

(3) Reject.139

8.14 Exercises - Linear equations of second order with constant coef�cients, Existence of solutions of the equationy′′ + by = 0, Reduction of the general equation to the special case y′′ + by = 0, Uniqueness theorem for the equationy′′ + by = 0, Complete solution of the equation y′′ + by = 0, Complete solution of the equation y′′ + ay′ + by = 0.

Exercise 1. y′′ − 4y = 0 y = c1e2x + c2e

−2x.Exercise 2. y′′ + 4y = 0 y = c1 cos (2x) + c2 sin (2x).

Use Theorem 8.7.

Theorem 27. Let d = a2 − 4b be the discrimnant of y′′ + ay′ + by = 0.Then ∀ solutions on (−∞,∞) has the form

(22) y = e−ax/2(c1u1(x) + c2u2(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 =√

d2

(3) If d < 0, then u1(x) = cos kx and u2(x) = sin kx; where k = 12

√−d

Exercise 3. y′′ − 4y′ = 0; a = −4.

y = e2x(c1e2x + c2e

−2x) = c1e4x + c2

Exercise 4. y′′ + 4y′ = 0

y = e−2x(c1e2x + c2e

−2x) = c1 + c2e−4x

Exercise 5. y′′ − 2y′ + 3y = 0 d = 4− 4(3) = −8 =⇒ y = e−x(c1 sin√

2x + c2 cos√

2x)

Exercise 8. y′′ − 2y′ + 5y = 0 d = −16 y = ex(c1 cos 2x + c2 sin 2x)Exercise 9. y′′ + 2y′ + y = 0 d = 4− 4(1)(1) = 0 y = e−x(1 + x)Exercise 10. y′′ − 2y′ + y = 0 d = 4− 4(1)(1) = 0 y = ex(1 + x)Exercise 11. y′′ + 3

2y′ = 0 y = 1, y′ = 1; x = 0 d = 94 > 0

y = e−34 x(c1e

3x4 + c2e

−3x4 ) = c1 + c2e

−3x2

c2 =−23

=⇒ y =53

+−23

e−3x2

Exercise 12. y′′ + 25y = 0; y = −1, y′ = 0, x = 3.y = c1 sin 5x + c2 cos 5x

y′ = 5c1 cos 5x +−5c2 sin 5x

0 = 5c1 cos 15− 5c2 sin 15c2 sin 15 = c1 cos 15

−1 = c1 sin 15 + c2 cos 15

−1c1(sin 15 +cos2 15sin 15

) = c1

(1

sin 15

) c1 = − sin 15c2 = − cos 15

y = − sin 15 sin 5x− cos 15 cos 5x

Exercise 13. y′′ − 4y′ − y = 0; y = 2; y′ = −1 when x = 1d = 10− 4(1)(−1) = 20

y = c1e(2+

√5)x + c2e

(2−√5)x

y(x = 1) = c1e2+√

5 + c2e2−√5 = 2

y′(x = 1) = (2 +√

5)c1e2+√

5 + (2−√

5)c2e2−√5 = −1

=⇒ (5 + 2√

5)c1e2+√

5 + (5− 2√

5)c2e2−√5 = 0

c1 =2√

5− 55 + 2

√5c2e

−2√

5

(2√

5− 55 + 2

√5

)c2e

2−√5 + c2e2−√5 = 2 =

4√

5c2e2−√5

5 + 2√

5=⇒ 5 + 2

√5

2√

5e√

5−2 = c2 c1 =2√

5− 52√

5e−2−√5

y =2√

5− 52√

5e−2−√5e(2+

√5)x +

5 + 2√

52√

5e√

5−2e(2−√5)x

140

Exercise 14. y′′ + 4y′ + 5y = 0, with y = 2 and y′ = y′′ when x = 016− 4(1)(5) = −4

y = e−2x(c1 sin 2x + c2 cos 2x) y(x = 0) = c2 = 2

y′ = −2e−2x(c1 sin 2x + 2 cos 2x) + 2e−2x(c1 cos 2x− 2 sin 2x)

y′′ = 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) = −2(2) + 2(c1) = −4 + 2c1

y′′(0) = 4(2) + (c1) + 4(−2) = −c1 = −4 + 2c1 c1 =43

y = e−2x(43

sin 2x + 2 cos 2x)

Exercise 15. y′′ − 4y′ + 29y = 0d = 16− 4(1)(29) = −100 =⇒ u = e2x(c1 sin 5x + c2 cos 5x).

v : y′′ + 4y′ + 13y = 0

d = 10− 4(13) = −36 =⇒ v = e−2x(b1 sin 3x + b2 cos 3x)

v(0) = b2

u(0) = 1(0 + c2) = c2 = 0

u = e2xc1 sin 5x

u′ = 2e2xc1 sin 5x + e2xc15 cos 5x

u′(π

2

)= 1 = 2eπc1(1) c1 =

12eπ

u′(0) =1

2eπ5

u′(0) = v′(0) =⇒ b1 =5

6eπ

v = e−2xb1 sin 3x

v′(0) = 3b1

u =1

2eπe2x sin 5x

v = e−2x 56eπ

sin 3x

Exercise 16.

y′′ − 3y′ − 4y = 0 u 9− 4(1)(−4) = 25 u = e3x2 (c1e

5x2 + c2e

− 5x2 )

y′′ + 4y′ − 5y = 0 v 16− 4(1)(−5) = 36 v = e−2x(b1e3x + b2e

−3x)

u(0) = c1 + c2 = 0v(0) = b1 + b2 = 0 =⇒ u = 2e3x2 c1(sinh

(5x

2

))

v = 2b1e−2x(sinh (3x))

u′ = c132e

3x2 (e

5x2 − e

−5x2 ) + 2e

3x2 c1

52

cosh(

5x

2

)

u′(0) = 5c1

v′ = −4b1e−2x sinh (3x) + 6b1e

−2x cosh (3x)

v′(0) = 6b1

c1 =6b1

5

Exercise 17.y′′ + ky = 0

d = −4(k)√

d

2=√−4k

2=√−k

Assume k > 0

y = c1 sin√

kx + c2 cos√

kx

y(0) = c2 = 0

y(1) = c1 sin√

k1 = 0 =⇒√

k = nπ

k < 0; −k = κ > 0

y = c1e√

κx + c2e−√κx; y(0) = c1 + c2 = 0 y = c1 sinh

√κx

y = c1 sinh√

κ1 = 0 c1 = 0

so if k < 0, there are no nontrivial solutions satisfying fk(0) = fk(1) = 0141

Exercise 18. y′′ + k2y = 0 d = −4k2 < 0√

d2 = 2k

2 = k > 0

y = c1 sin kx + c2 cos kx

y′ = kc1 cos kx− c2k sin kx

y(a) = b = c1 sin ka + c2 cos ka

y′(a) = m = kc1 cos ka− c2k sin ka

kb cos ka = kc1 cos ka sin ka + c2k cos2 ka

m sin ka = kc1 cos ka sin ka− c2k sin2 ka

kb cos ka−m sin ka = c2k =⇒ c2 =kb cos ka−m sin ka

k

c1 sin ka = b− c2 cos ka =kb

k−

(kb cos ka−m sin ka

k

)cos ka =

=kb(1− cos2 ka) + m sin ka cos ka

k=

kb sin2 ka + m sin ka cos ka

k

c1 =kb sin ka + m cos ka

k

y =(

kb sin ka + m cos ka

k

)sin kx +

(kb cos ka−m sin ka

k

)cos kx

k = 0 =⇒ y = mx−ma + b

Exercise 19.

(1) y = k1 sin x + k2 cos x

b2 = k1 sin (a2) + k2 cos (a2) = k1s2 + k2c2

b1 = k1 sin (a1) + k2 cos (a1) = k1s1 + k2c1

b2c1 = k1s2c1 + k2c2c1

− (b1c2 = k1s1c2 + k2c1c2)

=⇒b2c1 − b1c2 = k1(s2c1 − s1c2)=⇒ k1 =

b2c1 − b1c2

s2c1 − s1c2

s1b2 = k1s1s2 + k2s1c2

−(s2b1 = k1s1s2 + k2c1s2)=⇒ k2 =

s1b2 − s2b1

s1c2 − c1s2

y =(

b2 cos a1 − b1 cos a2

sin a2 cos a1 − sin a1 cos a2

)sin x +

(b2 sin a1 − b1 sin a2

sin a1 cos a2 − cos a1 sin a2

)cos x

y =b2 cos a1 − b1 cos a2

sin (a2 − a1)sin x +

b2 sin a1 − b1 sin a2

sin (a1 − a2)cos x if a2 − a1 6= πn

otherwise, if a2 − a1 = πn;b2c1 − b1c2 = 0b2s1 − b1s2 = 0

b2c1 = b1(−1)nc1; if c cos (a1) 6= 0, b2 = b1(−1)n

(2) It's true if a1 = a2 = π4 ; b1 = b2.

(3) y′′ + k2y = 0

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

S2 C2

] [AB

]=

[b1

b2

]

[AB

]=

[C2 −C1

−S2 S1

](1

S1C2 − C1S2

)[b1

b2

]

S1C2 − C1S2 = sin ka1 cos ka2 − cos ka1 sin ka2 = sin (k(a1 − a2))

y =b1 cos (ka2)− b2 cos (ka1)

sin (k(a1 − a2))sin ka1 +

−b1 sin (ka2) + b2 sin (ka1)sin (k(a1 − a2))

sin ka2

if k(a1 − a2) 6= πn k = 0; y =(

b2 − b1

a2 − a1

)(x− a1) + b1

Exercise 20.

(1)

u1(x) = ex; u2(x) = e−x u′2 = −e−x u′′2 = e−x =⇒ y′′ − y = 0142

(2)

u1 = e2x

u′1 = 2e2x

u′′1 = 4e2x

u2 = xe2x

u′2 = e2x + 2xe2x

u′′2 = 2e2x + 2e2x + 4xe2x = 4e2x + 4xe2x = 4e2x(1 + x)

u′′2 − 4u′2 + 4u2 = 0 =⇒ y′′ − 4y′ + 4y = 0

(3)

u1(x) = e−x/2 cos x;

u′1 = −12e−x/2 cos x +−e−x/2 sinx

u′′1 =14e−x/2 cos x + e−x/2 sin x +−e−x/2 cos x

=−34

e−x/2 cos x + e−x/2 sin x

u2(x) = e−x/2 sin x

u′2 =−12

e−x/2 sin x + e−x/2 cos x

u′′2 =14e−x/2 sinx +−e−x/2 cos x− e−x/2 sin x =

=−34

e−x/2 sin x− e−x/2 cos x

u′′2 + u′2 +54u2 = 0

y′′ + y′ +54y = 0

(4) u1(x) = sin (2x + 1); u2(x) = sin (2x + 2)

u′1 = 2 cos (2x + 1)

u′′1 = −4 sin (2x + 1)

u′2 = 2 cos (2x + 2)

u′′2 = −4 sin (2x + 2)

y′′ + 4y = 0

(5)u1 = cosh x

u′1 = sinh x

u′′1 = cosh x

y′′ − y = 0

Exercise 21. w = u1u′2 − u2u

′1.

(1) w = 0 ∀x ∈ open interval I ,(

u2

u1

)′=

u′2u1 − u′1u2

u21

= 0 =⇒ u2

u1= c

If u2u1

is not constant, then w(0) 6= 0 for at least one c in I (otherwise, it'd be constant).(2) w′ = u1u

′′2 − u2u

′′1

Exercise 22.

(1) w′ + aw = u1u′′2 − u2u

′′1 + a(u1u

′2 − u2u

′1) = 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 u2

u1constant. If u2

u1constant, w(0) = 0 since from the previous part.

Exercise 23. Recall the properties of the Wronskian.(1) If W (x) = v1(x)v′2 − v2v

′1 = v1v

′2 − v2v

′1 = 0 ∀x ∈ I ,

then v2v1

constant on I

(2) W ′ = v1v′′2 − v2v

′′1

(3) W ′ + aW = 0 if v1, v2 are solutions to y′′ + ay′ + by = 0W (x) = W (0)e−ax

So if W (0) 6= 0, W (x) 6= 0 ∀x143

Consider adding together the solutions and the solution's derivatives into some function f . By the linearity of the differentialequation, we know that f is also a solution since it is a linear superposition of solutions.

y(x) = Av1(x) + Bv2(x)

f(0) = Av1(x) + Bv2(0)

y′(x) = Av′1(x) + Bv′2(x)

f ′(0) = Av′1(0) + Bv′2(0)[v1(0) v2(0)v′1(0) v′2(0)

] [AB

]=

[f(0)f ′(0)

]=⇒ 1

W (0)

[v′2(0) −v2(0)−v′1(0) v1(0)

] [f(0)f ′(0)

]=

[AB

]

since W (0) 6= 0, this division is allowed above

so y(x) =(

v′2(0)f(0)− v2(0)f ′(0)W (0)

)v1(x) +

(v1(0)f ′(0)− v′1(0)f(0)

W (0)

)v2(x)

f(0), f ′(0) are initial conditions for y. f(0), f ′(0) are arbitrary.But since W (0) 6= 0, W (0) = v1(0)v′2(0)− v2(0)v′1(0), we can do things like

f(0) =v′2(0)f(0)− v2(0)f ′(0)v1(0)v′2(0)− v2(0)v′1(0)

v1(0) +v2(0)f ′(0)− v′1(0)f(0)

W (0)v2(0)

8.17 Exercises - Nonhomogeneous linear equations of second order with constant coef�cients, Special methods fordetermining a particular solution of the nonhomogeneous equation y′′ + ay′ + by = R.

Exercise 1. y′′ − y = x homogeneous solution c1ex + c2e

−x. yp = −xy = c1e

x + c2e−x − x

Exercise 2. y′′ − y′ = x2 For the homogeneous solution

y′′ − y′ = 0 d = (−1)2 − 4(1)(0) = 1 yh = ex2

(c1e

x2 + c2e

− x2)

= c2ex + c1

yp = Ax3 + Bx2 + Cx + D

y′p = 3Ax2 + 2Bx + C

y′′p = 6Ax + 2B

y = c1 + c2ex +

−13

x3 + x2

Exercise 3. y′′ + y′ = x2 + 2x

e−x2

(c1e

− x2 + c2e

x2)

= c1e−x + c2

P = Ax3 + Bx2 + Cx + D; P ′ = 3Ax2 + 2Bx + C P ′′ = 6Ax + 2B

3Ax2 + 2Bx + C + 6Ax + 2B = x2 + 2x A =13

B = 0 C = 0

y = c1e−x + c2 +

13x3

Exercise 4. y′′ − 2y′ + 3y = x3 u = ex(c1 sin√

2x + c2 cos√

2x)

3(Ax3 + Bx2 + Cx + D)

2(3Ax2 + 2Bx + C)

(6Ax + 2B)

A =13

B =23

C =89

D =1627

y = C1ex sin

√2x + C2e

x cos√

2x +13x3 +

23x2 +

89x +

1627

Exercise 5. y′′ − 5y′ + 4y = x2 − 2x + 1yh = e

5x2

(e

3x2 + e

−3x2

)= c1e

4x + c2ex

d =√

25− 4(4) = 3

4(Ax2 + Bx + C)

− 5(2Ax + B)2A

4Ax2 + 4Bx + 4C

−10Ax− 5B

2A

A =14

B =18

12− 5

8+ 4C = 1

C =932

y = c1e4x + c2e

x +14x2 +

18x +

932

144

Exercise 6.

y′′ + y′ − 6y = 2x3 + 5x2 − 7x + 2

yh = e−x2 (e

−5x2 + e

5x2 ) = e−3x + e2x

d =√

1− 4(−6) = 5

yp = Ax3 + Bx2 + Cx + D

y′p = 3Ax2 + 2Bx + C

y′′p = 6Ax + 2B

=⇒−6Ax3 − 6Bx2 − 6Cx− 6D

3Ax2 + 2Bx + C

6Ax + 2B

A =−13

B = −1 C =12

D =−712

y = C1e−3x + C2e

2x − 13x3 +−x2 +

12x− 7

12

Exercise 7.

y′′ − 4y = e2x

yh = c1e2x + c2e

−2x

v1 = e2x v′2 = −2e−2x

v′1 = 2e2x v2 = e−2x

Use Theorem 8.9.

Theorem 28. Let v1, v2 be solutions to L(y) = 0 where L(y) = y′′ + ay′ + byLet W = v1v

′2 − v2v

′1. Then L(y) = R where

yp = t1v1 + t2v2

(23) t1 = −∫

v2R

W (x)dx; t2(x) =

∫v1

R

Wdx

t1 = −∫

e−2x e2x

−4=

14x

t2 =∫

e2x e2x

−4=

e4x

−16

w = e2x(−2)e−2x − e−2x2e2x = −4

yp =x

4e2x +

e2x

−16

y = c1e2x + c2e

−2x +x

4e2x +

e2x

−16

Exercise 8.

y′′ + 4y = e−2x

yh = c1 sin 2x + c2 cos 2xw = sin 2x− sin 2x(2)− 2 cos 2x cos 2x = 2

t1 = −∫

cos 2xe−2xdx

2=−12

∫e−2x cos 2xdx

t2 =∫

sin 2xe−2x

2=

12

∫e−2x sin 2xdx

(e−2x cos 2x)′ = −2e−2x cos 2x +−2e−2x sin 2x

(e−2x sin 2x)′ = −2e−2x sin 2x + 2e−2x cos 2x

(e−2x sin 2x− e−2x cos 2x

4

)′= e−2x cos 2x

(e−2x sin 2x + e−2x cos 2x

−4

)′= e−2x sin 2x

t1 =e−2x cos 2x− e−2x sin 2x

8

t2 =e−2x sin 2x + e−2x cos 2x

−8

yp =e−2x sin 2x cos 2x− e−2x sin2 2x

8+

e−2x sin 2x cos 2x + e−2x cos2 2x

−8=

e−2x

8

y = c1 sin 2x + c2 cos 2x +e−2x

8145

Exercise 9. y′′ + y′ − 2y = ex d2 = 1− (4)(1)(−2) = 9

yh = e−1x2

(e

3x2 + e

−3x2

)= ex + e−2x

(xex)′ = ex + xex

+(xex)′′ = +(2ex + xex)

=⇒ 3ex + 2xex

y = c1ex + c2e

−2x +13xe2

Exercise 10. y′′ + y′ − 2y = e2x. yh = e−x2

(c1e

3x2 + c2e

− 3x2

)= c1e

x + c2e−2x.

W (x) = v1v′2 − v2v

′1 = ex(−2)e−2x − e−2xex = −3e−x

t1 = −∫

v2R

Wt2 =

∫v1R

W

t1 = −∫

e−2xe2x

−3e−x=

13ex t2 =

∫exe2x

−3e−x=−112

e4x

y1 = t1v1 + t2v2 =13exex +− 1

12e4xe−2x =

14e2x

y = c1ex + c2e

−2x +14e2x

Exercise 11. y′′ + y′ − 2y = ex + e2x.

Consider solutions to Exercise 9,10.L(ya) = ex; L(yb) = e2x ; L(ya + yb) = ex + e2x

=⇒ y = c1ex + c2e

−2x +13xex +

14e2x

Exercise 12. y′′ − 2y′ + y = x + 2xex. d = 4− 4(1) = 0. Recall the de�nition to be learned for this section of exercises:

Theorem 29. Let d = a2 − 4b be the discriminant of y′′ + ay′ + by = 0. Then every solution of this equation on (−∞,∞)has the form(24) y = e−ax/2(c1u1(x) + c2u2(x))

(1) If d = 0 then u1 = 1, u2 = x

(2) If d > 0, u1 = ekx; u2 = e−kx, k =√

d2

(3) If d < 0, u1 = cos kx, u2 = sin kx, k =√−d

2

yh = ex(c1 + c2x) = c1ex + c2xex

t1 =∫ −xex(2xex)

e2x=

2x3

−3t2 =

∫ex(2xex)

e2x= x2

W (x) = ex(ex + xex)− (xex)(ex) = e2x

yp =2x3

−3ex + x3ex =

x3ex

3

y =x3ex

3+ c1e

x + c2xex

Exercise 13. y′′ + 2y′ + y = e−x

x2

yh = e−x(c1 + c2x) = c!e−x + c2xe−x

t1 =∫ −xe−x

(e−x

x2

)

e−2x= − ln x t2 =

∫ e−x(

e−x

x2

)

e−2x=−1x

W = e−x(e−x +−xe−x)− (−e−x)(xe−x) = e−2x

yp = − ln xe−x + xe−x

(−1x

)= − ln xe−x − e−x

y = c1e−x + c2xe−x + (− ln x− 1)e−x

146

Exercise 14. y′′ + y = cot2 x. yh = c1 sin x + c2 cosx.

t1 =∫ − cosx

−1cot2 xdx =

∫cos x cos2 x

sin2 x=

∫cos x(1− sin2 x)

sin2 x= − 1

sin x+− sin x

t2 =∫

sin x

−1cot2 x = −

∫cos2 x

sinx= −

∫1− sin2 x

sin x= ln | csc x + cot x|+− cosx

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′′ − y = 21+ex

yh = c1ex + c2e

−x =⇒ W (x) = −exe−x − exe−x = −2

t1 = −∫

e−x 21+ex

−2=

∫e−x

1 + ex=

=∫

1ex− 1

1 + ex= −e−x −

∫1

1 + ex= −e−x +

∫ −e−x

e−x + 1= −e−x + ln (1 + e−x)

t2 =∫

ex 21+ex

−2= − ln |1 + ex|

y = −1 + ex ln (1 + e−x) +−e−x ln (1 + ex) + c1ex + c2e

−x

Exercise 16. y′′ + y′ − 2y = ex

1+ex

Discriminant: −1±√

12−4(−2)

2 = −2, 1 =⇒ yh = c1ex + c2e

−2x =⇒ W = ex(−2)e−2x − e−2xex = −3e−x

t1 = −∫ e−2t

(et

1+et

)

−3e−t=

13

∫1

1 + et=−13

ln (1 + e−x)

t2 =∫ et

(et

1+et

)

−3e−t=

1−3

∫e3t

1 + et

u=et

−−−→ −13

∫u2du

1 + u=

=−13

∫u +

−u

u + 1=−13

∫u +−1 +

1u + 1

=−13

(12u2 − u + ln u + 1)

=−16

e2x +ex

3+−13

ln (ex + 1)

y1 =−13

ex ln (1 + e−x) +−16

+e−x

3− e−2x

3ln (ex + 1) + c1e

x + c2e−2x

Exercise 17. y′ + 6y′ + 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 + c2x) = c1e−3x + c2xe−3x

W (x) = e−3x(e−3x − 3xe−3x)− (xe−3x)(−3e−3x) = e−6x

t1(x) =

a < 1 < x∫ x

1−te3tdt =

(−te3t

3 + e3t

9

)∣∣∣x

1= −3xe3x+e3x

9 + 29e3

a < 1 < 2 < x∫ 2

1−te3tdt = −6e6+e6

9 + 2e3

9 = −5e6

9 + 2e3

9

1 < a < x < 2∫ x

a−te3tdt =

(−te3t

3 + e3t9)∣∣∣

x

a= −xe3x

3 + e3x

9 + ae3a

3 − e3a

9

t2(x) =∫

e−3tf(t)e−6t

=∫

e3tf(t) =∫ x

a

e3t =13

(e3x − e3a

)

y1 = e−3x

(−3xe3x + e3x

9+ C

)+

x

3

y = c1e−3x + c2xe−3x +

19

when 1 ≤ x ≤ 2; otherwise y = yh

Exercise 18. Start from y′′ − k2y = R(x). Suppose L(yp) = y′′p − k2yp = R(x).147

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 differenti-ating the supposed particular solution.

R x0 dt sinh (k(x−t))−−−−−−−−−−−−→

∫ x

0

dtd2y

dt2(t) sinh (k(x− t))− k2

∫ x

0

dty(t) sinh (k(x− t)) =∫ x

0

dtR(t) sinh (k(x− t))dt

∫ x

0

y′′ sinh (κ) = −y′(0) sinh (kx)−∫

y′ cosh (κ)(−k) =

= −y′(0) sinh (kx) + k(y(x)− y(0) cosh (kx) + k

∫y(t) sinh (κ)) =

= −y′(0) sinh (kx) + ky(x)− ky(0) cosh (kx) + k2

∫y(t) sinh (κ)

=⇒ y(x)− y′(0) sinh (kx)k

− y(0) cosh (kx) =1k

∫ x

0


Now note that L(yh) = 0, so applying∫ x

0dt sinh (k(x− t)) results in 0 still.

With yp(x) = 1k

∫ x

0dtR(t) sinh (k(x− t))dt + y′(0) sinh (kx)

k + y(0) cosh (kx), we can add a homogeneous solution ofy′(0) sinh (kx)

k + y(0) cosh (kx) to yp(x) to obtain

y1(x) =1k

∫ x

0


Now for y′′ − 9y = e3x,

y1(x) =13

∫ x

0

dt(e3t) sinh (3(x− t))dt =16

∫ x

0

dte3t

(e3x−3t − e−3x+3t

2

)=

16

∫ x

0

dt(e3x − e−3x+6t)

=16

(e3xx− e−3x 1

6e6t

∣∣∣∣x

0

)=

16xe3x − e−3x e6x − 1

36=

16(xe3x − e3x

6+

e−3x

6)

y′1 =16(e3x + 3xe3x − 3

2e3x − 3

2e−3x)

y′′1 =16

(3e3x + 3e3x + 9xe3x − 9

2e3x +

92e−3x

)=

14e3x +

34e−3x +

32xe3x

y′′1 − 9y1 =12e3x +

12e−3x

Thus, we need to add homogeneous parts to our particular solution to make it work. So if

yp =xe3x

6− e3x

y+

e−3x

6

then it could be checked easily with some computation, that this satis�es the ODE.

Exercise 19. Start from y′′ + k2y = R(x)

Again, note that if L(yp) = y′′p + k2yp = R(x), L(yp + yh) = R(x) + 0 = R(x), so y1 = yp + yh is also a particularsolution.

R x0 dt sin k(x−t)−−−−−−−−−−→

∫ x

0

dt sin k(x− t)d2y

dt2+ k2

∫ x

0

dt sin k(x− t)y =∫ x

0

R(t) sin k(x− t)∫ x

0

dt sin k(x− t)d2y

dt2= −y′(0) sin (kx) + k

∫ x

0

y′(t) cos (k(x− t))dt =

= −y′(0) sin (kx) + k(y(x)− y(0) cos (kx)− k

∫ x

0

y(t) sin (k(x− t))dt) =

= ky(x)− ky(0) cos (kx)− y′(0) sin (kx)− k2

∫ x

0

y(t) sin (k(x− t))dt

148

=⇒ y(x) =1k

∫ x

0

dtR(t) sin k(x− t) + y(0) cos (kx) +y′(0) sin (kx)

k

We can add yh with c1 = −y(0), c2 =−y′(0)

k

y1 =1k

∫ x

0

dtR(t) sin k(x− t)

Now for y′′ + 9y = sin 3x, then k = 3,

y1 =13

∫ x

0

sin 3t sin 3(x− t)dt

∫ x

0

sin 3t(sin 3x cos 3t− cos 3x sin 3t)∫ x

0

sc =∫ x

0

s(6t)2

dt =−112

(c(6x)− 1)∫ x

0

s2 =∫ x

0

1− cos (6t)2

=x

2− sin (6t)

12

∣∣∣∣x

0

=x

2− sin (6x)

12

y1 =13

(sin 3x

−112

(cos (6x)− 1)− cos 3x

(x

2− sin (6x)

12

))=

sin 3x

18− x cos 3x

6

It could be shown with some computation that this particular solution satis�es the ODE without having to add or subtractparts of a homogeneous solution.Exercise 20. y′′ + y = sin x

yh = c1 sin x + c2 cos x =⇒ W (x) = −s2 − c2 = −1

t1 = −∫

cs

−1=− cos 2x

4; t2 =

∫ss

−1= −

∫1− cos 2x

2= −

(x

2− sin 2x

4

)

yp = − sin x cos 2x

4+

(sin 2x− 2x

4

)cos x =

sin x cos2 x + sin3 x− 2x cosx

4

y = c1 sin x + c2 cosx +sin x cos2 x + sin3 x− 2x cosx

4

Exercise 21. y′′ + y = cos x

yh = c1 sin x + c2 cos x = c1S + c2C W (x) = −1

t1 =∫ −CC

−1=

∫1 + cos 2x

2=

x + sin 2x2

2; t2 =

∫SC

−1=

cos 2x

4

yp =x sin x

2+

sin 2x sin x

4+

cosx cos 2x

4

=⇒ y =x sin x

2+

sin 2x sin x

4+

cosx cos 2x

4+ c1 sin x + c2 cos x

Exercise 22. y′′ + 4y = 3x cosx

yh = c1 sin 2x + c2 cos 2x W (x) = − sin2 2x(2) +− cos2 2x(2) = −2

t1 =∫ − cos (2x)(3x cos x)

−2=

32

∫x cosx cos (2x) =

32

∫xc(1− 2s2) =

32

∫xc− 3

∫xcs2 =

=32(xs + c)− 3

∫x

(c3

3

)′=

32(xs + c)− (xs3 −

∫s3) =

32(xs + c)− xs3 +

∫s(1− c2) =

=32(xs + c)− xs3 +−c +

13c3 =

32xs +

c

2− xs3 +

13c3

149

t2 =∫

sin (2x)(3x cosx)−2

= −3∫

xsc2 =∫

x(c3)′ = xc3 −∫

c3 = xc3 −∫

c(1− s2) =

= xc3 − s +13s3

yp =(

32xs +

c

2− xs3 +

13c3

)(2sc) + (xc3 − s +

s3

3)(1− 2s2) = (lots of algebra) =

= xc2 +23s

=⇒ y = c1 sin 2x + c2 cos 2x + x sin x− 23

cosx

Remember, persistence is key to work through the algebra, quickly.Exercise 23. y′′ + 4y = 3x sin x. From the work above, we could guess at the solution.(xs)′ = s + xc

(xs)′′ = 2c +−xs(xs)′′ + 4(xs) = 2c− xs + 4xs = 2c + 3xs (c)′′ + 4c = 3c =⇒

(−23

c

)′′+ 4

(−23

c

)= −2c

=⇒ yh = xs− 23c

y = x sin x− 23

cos x + c1 sin 2x + c2 cos 2x

Exercise 24. y′′ − 3y′ = 2e2x sin x Guessing and stitching together the solution seems easier to me.

(e2xc)′ = 2e2xc +−e2xs

(e2xc)′′ = 4e2xc +−4e2xs− e2xc = 3e2xc− 4e2xs

(e2xc)′′ − 3(e2xc)′ = 3e2xc− 4e2xs− 6e2xc + 3e2xs =

= −3e2xc− e2xs

(e2xs)′ = 2e2xs + e2xc

(e2xs)′′ = 4e2xs + 4e2xc +−se2x =

= 3e2xs + 4e2xc

(e2xs)′′ − 3(e2xs)′ = 3e2xs + 4e2xc− 6e2xs− 3e2xc =

= −3e2xs + e2xc

(3e2xs)′′ − 3(3e2xs)′ + (e2xc)′′ − 3(e2xc)′ = −10e2xs

=⇒ yp =e2x(3 sin x + cos x)

5=⇒ y = c1 sin

√3x + c2 cos

√3x +

e2x(3 sin x + cos x)−5

Exercise 25. y′′ + y = e2x cos 3x.

(e2xc(3x))′′ = 4e2xc(3x) +−12e2xs(3x)− 9e2xc(3x) = −5e2xc(3x)− 12e2xs(3x)

(e2xs(3x))′′ = 4e2xs(3x) + 12e2xc(3x) +−9e2xs(3x) = −5e2xs(3x) + 12e2xc(3x)

L(e2xc(3x)− 3e2xs(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 coef�cients.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 de�ned to be the derivative y′. 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′ (when x = 0) are 2 and 4,respectively.

frequency =k

2π=

1π

=⇒ k = 2

y(x = 0) = C sin α

y′(x = 0) = C cos α=⇒ y(x = 0)

y′(x = 0)=

1k

tan α =12

α =π

4and C = 2

√2

150

Exercise 2. Find the velocity when y is zero, given that the amplitude is 7 and the frequency is 10.

y = C sin (kx + α)

y′ = Ck sin (kx + α)C = 7

k

2π= 10 =⇒ k = 20π

y(x =−α

k) = 0 =⇒ y′(x =

−α

k) = 140π

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

=⇒ k = m (since x is arbitrary )

−A sin β = C cosα

A cos β = C sin α=⇒ tan α = ± cot β = ∓ tan

(π

2− β

)= tan (β − π

2)

=⇒ α = β − π

2and |C| = |A|

Exercise 4. 2πT = 4π

y = C cos (kx + α) = C cos (kx) = 3 cos (4πx)

Exercise 5. y = C cos (x + α) y0 = C cos (x0 + α)

y′ = −C sin (x + α) = ±v0

v20 + y2

0 = C2 sin2 (x + α) + C2 cos2 (x0 + α) = C2 =⇒ C =√

v20 + y2

0

Exercise 6.

y = C cos (kx + α)

y(0) = C cos (α) = 1

y′′(0) = −k2C cos (α) = −12

y′ = −kC sin (kx + α)

y′(0) = −kC sin (α) = 2

y′′(0)y(0)

= −k2 =−121

=⇒ k = 2√

3

y′(0)y(0)

=−kC sin (α)C cos (α)

=21

= 2 = −k tan (α) =⇒ α =−π

6

Exercise 7. k = 2π3 y = −C sin (kx)

y = −C sin2πx

3; C > 0

Exercise 8. Let's �rst solve the homogenous equation.

y′′ + y = 0

yh = C1 sin x + C2 cosx

W (x) = −S2 − C2 = 1

t1 =∫ x

0

− cos t(1)−1

= sin x

t2 =∫ x

0

(sin t)(1)−1

= (cos x− 1)

for 0 ≤ x ≤ 2π otherwise, for x > 2π, t1 = 0, t2 = 0

y1 = sin2 x + cos2 x + (1− cosx)

y′(x) = C1 cosx + sin x

y(0) = 0 = c2

y′(0) = c1 = 1

y = sin x + (1− cos x)

=⇒ I(t) = sin t + (1− cos t) 0 ≤ t ≤ 2π

Exercise 9.151

(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 ′ = ωAC(ωt + α) = ωA(C(ωt)C(α)− S(ωt)S(α))

I ′′ = −ω2AS(ωt + α) = −ω2A(S(ωt)C(α) + C(ωt)S(α))

I ′′ + RI ′ + I = I ′′ + I ′ + I =

= A((−ω2C(α) +−ωS(α) + C(α))S(ωt) + (−ω2S(α) + ωC(α) + S(α))C(ωt)) = S(ωt)

=⇒ tan (α) =−ω

1− ω2

With tanα = −ω1−ω2 and the trig identities t2+1 = sec2, 1

C2 = sec2, and S2+C2 = 1, we can getC =

1− ω2

√1− ω2 + ω4

S =−ω√

1− ω2 + ω4

Note that the sign of S is �xed by tan.

A(−ω2C(α) +−ωS(α) + C(α))S(ωt) = S(ωt) =⇒ A =1

(1− ω2)C(α)− ωS(α)=

D

(1− ω2)2 − ω(−ω)

=⇒ A =1√

ω4 − ω2 + 1

We could immediately see that ω = 1√2

, f = 12π√

2will maximize A.

(2) We could have, from the beginning, considered the problem with any R, in general.

A((−ω2C(α) +−RωS(α) + C(α))S(ωt) + (−ω2S(α) + ωRC(α) + S(α))C(ωt)) = S(ωt)

tan α =−ωR

1− ω2

S(α) =−ωR√

(ωR)2 + (1− ω2)2

C(α) =1− ω2

√(ωR)2 + (1− ω2)2

=⇒ A =1

−ω2C(α)− ωRS(α) + C(α)=

1√(ωR)2 + (1− ω2)2

(ωR)2 + (1− ω2)2 = ω4 + ω2(−2 + R2) + 1d

dω−−→ 4ω3 + 2ω(−2 + R2) = 2ω(2ω2 + (−2 + R2)) = 0

=⇒ω = 0

2ω2 = 2−R2

to have resonances−−−−−−−−−−→ R <1√2

Exercise 10. A spaceship is returning to earth. Assume that the only external force acting on it is the action of gravity, andthat it falls along a straight line toward the center of the earth. The rocket fuel is consumed at a constant rate of k pounds persecond 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 + fuel combination at time t. With +y direction being towards earth, then theequation of motion is Fg = +M(t)g, where g = 9.8m/s2.

M(t)v(t) = MvR 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)−MvR = M(vR(t + h)− vR) +−∆m(vR(t + h)− vR) + ∆mc

∆p

∆t= M

(vR(t + h)− vR

∆t

)+−∆m

(vR(t + h)− vR

∆t

)+

(∆m

∆t

)c = M(t)g

152

Mv′R +kc

g= M(t)g

Now M(t) = M0 − kt

g=⇒ v′R = g − kc/g

M= g − kc/g

M0 − kcg

vR = gt− kc

g

g

−kln (M0 − kt

g) = gt + c ln (M0 − kt

g)

yR =gt2

2+ c

g

k

((k

gt−M0

)ln

(M0 − kt

g

)− kt

g

)+

M0cg

kln M0

Exercise 11.

Mv′R =−kc

g

v′R =−kc

g

(1

M0 − ktg

)

vR = c ln (M0 − kt

g)

=⇒ yR = cg

k

((kt

g−M0

)ln

(M0 − kt

g

)− kt

g

)+

M0cg ln M0

k

M0g = w =⇒ yR = cg

k

(kt− w

gln

(w − kt

g

)− kt

g

)+

wc

kln

w

g

We could've also solved this problem with an initial velocity of v0 and gravity. Then

vR(t) = gt + c ln (1− kt

M0g) + v0

y(t) = v0t +12gt2 + c

((t− M0g

k

)ln

(1− kt

M0g

)− t

)

Exercise 12.MvR = (M −∆m)(vR(t + h)) + 0

M(vR(t + h)− vR(t)) = (∆m)vR(t + h)

Mv′R =k

gvR =⇒ v′R

vR=

k

g(M0 − ktg )

=k

M0g(1− ktM0g )

ln vR = (k/w) ln (1− kt

w)(−w

k

)= − ln (1− kt

w)

vR =v0

1− ktw

=⇒ x(t) = v0

(−w

k

)ln (1− kt

w) =

−v0w

kln (1− kt

w)

8.22 Exercises - Remarks concerning nonlinear differential equations, Integral curves and direction �elds.Exercise 1. 2x + 3y = C =⇒ y′ = −2

3

Exercise 2. y = Ce−2x =⇒ y′ = −2y

Exercise 3. x2 − y2 = c =⇒ yy′ = x =⇒ y′ = xy ; y 6= 0

Exercise 4. xy = c =⇒ y′ = −yx ; x 6= 0

Exercise 5. y2 = cx =⇒ y2

x = c =⇒ y′ = y2x x 6= 0

Exercise 6. x2 + y2 + 2Cy = 1

x2

y+ y − 1

y= −2C

2xy − y′x2

y2+ y′ +

1y2

y′ = 0

y′ =−2xy

1 + y2 − x2

153

Exercise 7. y = C(x− 1)ex

y

(x− 1)ex= C

y′(x− 1)ex − (ex + (x− 1)ex)y(x− 1)2e2x

= 0

y′ =xy

x− 1

Exercise 8. y4(x + 2) = C(x− 2)

y4(x + 2)x− 2

= C

(4y3y′(x + 2) + y4)(x− 2)− y4(x + 2)(x− 2)2

= 0 =⇒ 4y3y′(x + 2) + y4 =y4(x + 2)

x− 2

y′ =y

(x− 2)(x + 2)

Exercise 9. y = c cosx =⇒ y′ = − tan xy

Exercise 10. arctan y + arcsinx = C

11 + y2

y′ +1√

1− x2= 0 =⇒ y′ =

−(1 + y2)√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

(1, 0): (1−A)2 + B2 = R2 =⇒ −(1− 2A + A2 + B2 = R2)(−1, 0): (−1−A)2 + B2 = 1 + 2A + A2 + B2 = R2

=⇒ 4A = 0, A = 0 1 + B2 = R2

x2 + (y −±√

R2 − 1)2 = R2

x2 + y2 − 2By + (R2 − 1) = R2 =⇒ x2 + y2 − 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 + y2 − 1 = 2By

x2

y+ y − 1

y= 2B =⇒ 2xy − y′x2

y2+ y′ +

1y2

y′ = 0

y′ =2xy

y2 − x2 + 1

Exercise 12.(x + A)2 + (y + B)2 = r2

(1 + A)2 + (1 + B)2 = 1 + 2A + A2 + 1 + 2B + B2 = r2

− ((−1 + A)2 + (−1 + B)2

)= − (

1− 2A + A2 + 1− 2B + B2 = r2)

=⇒4A + 4B = 0 =⇒ A = −B

(x +−B)2 + (y + B)2 = r2

2(x−B) + 2(y + B)y′ = 0

(y + B)y′ = B − x

=⇒ y′ =B − x

y + B

(1−B)2 + (1 + B)2 = r2 =⇒√

2√

(1 + B2) = r or√

r2

2− 1 = B

=⇒ (so B could be treated as a parameter for the family of curves)154

8.24 Exercises - First-order separate equations.

Exercise 1. y′ = x3/y2

13y3 = 1

4x4 + C =⇒ y3 = 34x4 + C

Exercise 2. tan x cos y = −y′ tan y

ln | cosx| = 1cos y

Exercise 3. (x + 1)y′ + y2 = 01y = ln (x + 1) + c

Exercise 4. y′ = (y − 1)(y − 2)(

1y − 2

+−1

y − 1

)y′ = 1 =⇒ ln (y − 2)− ln (y − 1) = x

y − 2y − 1

= ex

Exercise 5. y√

1− x2y′ = x

12y2 = −√1− x2 =⇒ y2 = −2

√1− x2

Exercise 6. (x− 1)y′ = xy

ln y =∫

1 +1

x− 1= x + ln |x− 1|

y = ex(x− 1) + C

Exercise 7. (1− x2)1/2y′ + 1 + y2 = 0

arctan y = arccos x + C

Exercise 8. xy(1 + x2)y′ − (1 + y2) = 0

12

ln (1 + y2) =∫ (

1x− x

1 + x2

)+ C = ln x− 1

2 ln |1 + x2|

y2 = k

(x√

1 + x2

)2

Exercise 9. (x2 − 4)y′ = y

ln y =−12

arctanh(x

2

)

since∫

1x2 − 4

dx =14

∫dx(

x2

)2 − 1=

=12

∫du

u2 − 1( where u =

x

2)

(tanh (u))′ =cosh2 u− sinh2 u

cosh2 u= 1− tanh2 u

=⇒ y = k exp (−12arctanh

(x

2

))

Exercise 10. xyy′ = 1 + x2 + y2 + x2y2

12 ln (1 + y2) = ln x + 1

2x2 + C =⇒ y2 = kx2ex2 − 1

Exercise 11. yy′ = ex+2y sin x

ye−2y

−2− e−2y

4=

ex sin x− ex cos x

2+ C

(2y + 1)e−2y = −2ex(sinx− cos x) + C155

Exercise 12. xdx + ydy = xy(xdy − ydx)

y(1− x2)dy = x(−y2 − 1)dx

ydy

1 + y2=

xdx

x2 − 1=⇒ 1

2ln |1 + y2| = 1

2ln |x2 − 1|+ C

1 + y2 = (x2 − 1)K =⇒ y2 = K(x2 − 1)− 1

Exercise 13. f(x) = 2 +∫ x

1f(t)dt

f ′(x) = f(x) =⇒ f(x) = Cex

=⇒ Cex = 2 + Cex − Ce1 C =2e

f(x) =2eex

Exercise 14. f(x)f ′(x) = 5x f(0) = 1

f(x)2 = 5x2 + C =⇒ f(x) = ±√

5x2 + C

f(x) =√

5x2 + 1

Exercise 15. f ′(x) + 2xef(x) = 0 f(0) = 0

e−yy′ = −2x =⇒ −e−y = −x2 + C

y = ln (x2 + C)−1 =⇒ y = − ln (x2 + 1)

Exercise 16. f2(x) + (f ′(x))2 = 1

f = −1

y′2 = 1− y2 y′ = ±√

1− y2

± arcsin (y) = x + c =⇒ f(x) = ± sin (x + c)

Exercise 17. ∫ x

a

f(t)dt = K(x− a)

f > 0 ∀x ∈ R=⇒ f(x) = k > 0

Exercise 18. ∫ x

a

f(t)dt = k(f(x)− f(a))d

dx−−→ f(x) = kf ′(x)

=⇒ f(x) = Ce1k x; C > 0

Exercise 19.∫ x

a(f(t))dt = k(f(x) + f(a)) =⇒ f(x) = Ce

xk

kCexk − kCe

ak = kCe

xk + kCea/k =⇒ 2kCea/k = 0 =⇒ C =

f = 0

Exercise 20.∫ x

af(t)dt = kf(x)f(a); f(x) = kf ′(x)f(a)

1kf(a)

=f ′(x)f(x)

=⇒ ln f(x) =(

1kf(a)

)x + C; =⇒ f(x) = C exp

(x

kf(a)

)

∫ x

a

f(t)dt =(kf(a)Ce

tkf(a)

)∣∣∣x

a= kf(a)Ce

xkf(a) − kf(a)Ce

akf(a) = kCe

xkf(a) Ce

akf(a)

f(a)(e

xkf(a) − e

akf(a)

)= Ce

xkf(a) e

akf(a)

x=a−−−→ 0 = Ce2a

kf(a) =⇒ C = 0

=⇒ f = 0

156

8.26 Exercises - Homogeneous �rst-order equations.Exercise 1. f(tx, ty) = f(x, y) homogeneity (or homogeneity of zeroth order).

y′ = f(x, y) =(x

v

)′=

v − xv′

v2= f(x,

x

v) = f(1,

1v)

v − v2f(1,1v) = xv′ =⇒ ln x =

∫dv

v − v2f(1, 1v )

Exercise 2. y′ = −xy =⇒ 1

2y2 = − 12x2 + C =⇒ y2 = −x2 + C

Exercise 3. y′ = 1 + yx

yx = v =⇒ y′ = v + xv′ = 1 + v =⇒ v = ln x

y = x(lnx + C)

Exercise 4. y′ = x2+2y2

xy

y′ =x2 + 2y2

xy=

x

y+

2y

x

v =y

x=⇒ v + xv′ =

1v

+ 2v =⇒ v′1v + v

=1x

12

ln |1 + v2| = ln x + C =⇒ y2 = (Cx2 − 1)x2

Exercise 5. (2y2 − x2)y′ + 3xy = 0

if 2y2 6= x2 , y′ =3xy

x2 − 2y2

y = vx

y′ = v′x + v=⇒ y′ = v′x + v =

3vx2

x2 − 2v2x2=

3v

1− 2v2

=⇒ 1− 2v2

2v(1 + v2)v′ =

1x

12

(1v

+−3v

1 + v2

)v′ =

1x

=⇒ 12

ln v +−32

ln (1 + v2) = ln x + C =⇒ v

(1 + v2)3= Cx2 =

y/x(x2+y2

x2

)3

yx3 = C(x2 + y2)3

However,

y′ =3xy

x2 − 2y2

v =y

xy′ = v′x + v

=⇒ v′x + v =3x2v

x2 − 2v2x2=

3v

1− 2v2

v′x =3v

1− 2v2− (v − 2v3)

1− 2v2=

2(v + v3)1− 2v2

=⇒(

1v

+− 3v

1 + v2

)v′ =

2x

=⇒ ln v +−32

ln |1 + v2| = 2 ln x + C

v

(1 + v2)3/2= Cx2 =⇒ y2/x4

(x2 + y2)3= Cx4

=⇒ y2 = C(x2 + y2)3

157

Exercise 6. xy′ = y −√

x2 + y2

=⇒ y′ =y

x=

√1 +

(y

x

)2

v =y

xvx = y

v′x + v = v −√

1 + v2 =⇒

v′x = −√

1 + v2

−v′√1 + v2

=1x

=⇒ ln (v +√

1 + v2) = ln x + C since

(ln (v +√

1 + v2))′ =1

v +√

1 + v2

(1 +

v√1 + v2

)=

1√1 + v2

v +√

1 + v2 = Cx =⇒ 1 + v2 = C2x2 − 2vCx + v2

=⇒ v =Cx

2− 1

2Cx=⇒ y =

Cx2

2− 1

2C

Exercise 7. x2y′ + xy + 2y2 = 0

x2y′ = −2y2 − xy =⇒ y′ =−2y2

x2− y

xv =

y

x=⇒

v′x + v = −2v2 − v

v′x = −2(v2 + v)

v′

v(v + 1)=−2x

=∫

v′(

1v− 1

v + 1

)= −2 ln x + C =⇒ v

v + 1=

C

x2

y =−Cx

C − x2

Exercise 8. y2 + (x2 − xy + y2)y′ = 0

(y

x

)2

+(

1− y

x+

(y

x

)2)

y′ = 0

Let v =y

x=⇒ −v2

1− v + v2= v′x + v

v′x =−v2

1− v + v2− v =

−v(1 + v2)1− v + v2

=⇒ v2 − v + 1v(1 + v2)

v′ =−1x

=(

1v

+−1

v2 + 1

)v′ = − 1

x

=⇒ ln v − arctan v = − ln x + C =⇒ ln (vx) = arctanx + C ln y = arctany

x+ C

Exercise 9. y′ = y(x2+xy+y2)x(x2+3xy+y2)

y′ =y(x2 + xy + y2)x(x2 + 3xy + y2)

=(y

x

) (1 + y

x + y2

x2

1 + 3yx + y2

x2

)v= y

x−−−→ v′x + v = v

(1 + v + v2

1 + 3v + v2

)= v +

−2v2

v2 + 3v + 1

v′(1 +3v

+1v2

) =−2x

=⇒ v + 3 ln v +−1v

= −2 ln x + C

y

x+ 3 ln y − x

y= ln x + C

Exercise 10. y′ = yx + sin y

x

y

x= x

y′ = v + v′x=⇒

v + v′x = v + sin v

v′

sin v=

1x

− ln csc v + cot v = ln x + C =⇒ csc v + cot v =K

x158

Exercise 11. x(y + 4x)y′ + y(x + 4y) = 0

y′ =−y(x + 4y)x(y + 4x)

=− y

x (1 + 4yx )

yx + 4

v= yx−−−→ v + xv′ =

−v(1 + 4v)v + 4

xv′ =−5v(1 + v)

v + 4=⇒ −5

x=

v + 4v(1 + v)

v′ =(

4v

+−3

1 + v

)v′

R−→ 4 ln v − 3 ln (1 + v) = −5 lnx + C =⇒ (yx)4 = (x + y)3C

8.28 Miscellaneous review exercises - Some geometrical and physical problems leading to �rst-order equations.Exercise 1.

2x + 3y = C y′ = −23

g′ =32

=⇒ g − 32x = C

Exercise 2.xy = C

d/dx−−−→ y + xy′ = 0 =⇒ y′ = −y/x x 6= 0 =⇒ g′ = x/g =⇒ 12g2 =

12x2 + C

Exercise 3. x2 + y2 + 2Cy = 1

x + yy′ + Cy′ = 0 =⇒ y′(y + C) = −x

y′ =−x

y + C=

−x

y + 1−x2−y2

2y

=−2xy

y2 − x2 + 1

orthogonalcurves−−−−−−−−−−−→ y′ =y2 − x2 + 1

2xy=

(12x

)y +

12

(1x− x

)y−1

Recognize that this is a Ricatti equation and we know how to solve them.

y′ +−12x

y = y−1

(−x

2+

12x

)n = −1

k = 1− n = 1− (−1) = 2

v = yk = y2 v′ + 2(−1

2x

)v =

22

(1x− x

)

A(x) =∫ x

a

P (t)dt =∫ x

a

−1t

= lna

x

∫ x

a

QeA =∫ x

a

(1t− t

)a

t=−a

x+ 1− a(x− a)

y2 = v = −1 +x

a− x(x− a) +

bx

a

Exercise 4. y2 = Cx.y2

x= C

d/dx−−−→ 2yy′x− y2

x2= 0

y′ =y

2x=⇒ y′ =

1(−y2x

) =−2x

y

=⇒ y2 + 2x2 = C

Exercise 5. x2y = C.2xy + x2y′ = 0 y′ = −2y

x=⇒ y′ =

x

2y

12y2 =

x2

4+ C

2y2 − x2 = C

Exercise 6. y = Ce−2x

e2xy = C =⇒ 2e2xy + e2xy′ = 0

y′ = −2yinvert−−−→ y′ =

12y

=⇒ y2 = x + C

159

Exercise 7. x2 − y2 = C2x− 2yy′ = 0

y′ =x

y=⇒ y′ =

−y

x

=⇒ ln y = − ln x + C =⇒ y =C

x

Exercise 8. y sec x = Cy′ sec x + y tanx sec x = 0

y′ = −y tanx(invert)−−−−→ y′ =

1y tanx

12y2 = ln | sin x|+ C

Exercise 9. All circles through the points (1, 0) and (−1, 0) From Sec. 8.22, Ex.10, we had obtained y′ = 2xyy2−x2+1

=⇒ y′ = x2−y2−12xy = x2−1

2yx − y2x =⇒ y′ + 1

2xy = x2−12x y−1

Recognize this is a Ricatti equation.

For y′ + Py = Qyn, in this case, n = −1, and so k = 1− n = 1− (−1) = 2.Then v = yk and v′ + kPv = kQ. In this case,v′ + 2

(12x

)v = 2

(x2−12x

)= v′ + 1

xv = x2−1x = x− 1/x.

A(x) =∫ x

a

P (t)dt =∫ x

a

1t

= lnx

a∫

(t− 1t) exp

(ln

t

a

)dt =

∫ (t2

a− 1

a

)dt =

( 13 t3

a− t

a

)∣∣∣∣x

a

e− ln xa =

a

x

=⇒ y2 = v =13x3 − x− 1

3a3 + a

x+

ba

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 + B2 = r2

− ((−1 + A)2 + (−1 + B)2

)= − (

1− 2A + A2 + 1− 2B + B2 = r2)

=⇒4A + 4B = 0 =⇒ A = −B

(x +−B)2 + (y + B)2 = r2

2(x−B) + 2(y + B)y′ = 0

(y + B)y′ = B − x

=⇒ y′ =B − x

y + B

(1−B)2 + (1 + B)2 = r2 =⇒√

2√

(1 + B2) = r or√

r2

2− 1 = B

=⇒ (so B could be treated as a parameter for the family of curves)

y′ =−1(B−xy+B

) =y + B

x−B

y′

y + B=

1x−B

=⇒ y = C(x−B)−B

Exercise 11. With (0, Y ) = Q the point that moves up wards along the positive y-axis andP = (x, y) being the point P that pursues Q,

y′ = Y−yX−x = Y−y

0−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

160

kY = x.

y′ =

(1k

)x− y

x= f(x, y)

f(x, y) is homogeneous of zero order =⇒ y = vx (try this substitution)

y′ = v′x + v =1k− v =⇒ v′

1k − 2v

=1x

=⇒ −12

ln(

1k− 2v

)= ln x + C =⇒ y =

x

2k− 1

2C2x

(1,0)−−−→ y =x

2k=

12kx

k =12

y = x− 1x

Exercise 12.

y =x

2k− 1

2kx

Exercise 13. y = f(x).

n

∫ x

0

f(t)dt = xf(x)−∫ x

0

f(t)dt; (n + 1)∫ x

0

f(t)dt = xy

=⇒ (n1)y = y + xy′ =⇒

ny = xy′

n

x=

y′

y

n ln x = ln y

y = Cxn of y = Cx1/n

Exercise 14.

n

∫ x

0

πf2(t)dt =∫ x

0

(π(y(x))2 − π(f(t))2)dt

(n + 1)∫ x

0

πf2(t)dt = xy2(x) = xy2; (n + 1)f2(x) = y2 + 2xyy′

y′ =ny

2x=⇒ ln y =

n

2ln x + C

y = Cxn/2 or y = Cx1/2n

Exercise 15.

π

∫ x

0

f2(t)dt = x2f(x) πf2(x) = 2xf + x2f ′ =⇒ f ′ =πf2 − 2xf

x2

The left hand side of the last expression shown is homogeneous. Thus do the y = vx substitution.

v′x + v =πv2x2 − 2x2v

x2= πv2 − 2v

v′xπv2 − 3v

= 1 =⇒ v′

π(v2 − 3vπ )

=1x

=13

(1

v − 3π

− 1v

)v′

ln (v − 3π

)− ln v = ln(

v − 3/π

v

)= 3 ln x + C

vx− 3π

xCvx4

y − 3π

x = Cyx3 =⇒ y =3x/π

1 + x3

2

Exercise 16. A =∫ a

0f ; B =

∫ 1

af

A−B =∫ a

0f +

∫ a

1f = 2f(a) + 3a + b

d/dx−−−→ 2f(a) = 2f ′(a) + 3 =⇒ 1 = f ′(a)

f(a)− 32

161

So then

a + C = ln (y − 32); f(a) = Cea +

32

f(1) = 0 = Ce1 +32; =⇒ C = − 3

2e

f(x) =−32

ex−1 +32

To �nd b,

2(−3

2ea−1 +

32a

)+

32e−1 +

32− 3

2= 2f(a) + 3a + b =

= 2(−3

2ea−1 +

32

)+ 3a + b

=⇒ b =32e−1 − 3

Exercise 17.

A(x) =∫ x

0

(f(t)−

((y(x)− 1

x

)t + 1

))dt = x3 =

=∫ x

0

f +−(

y(x)− 1x

)12x2 − x =

∫ x

0

f +−(

y(x)− 12

)x− x

d/dx−−−→ f(x) +−12(y′x + y)− 1

2= 3x2

y′ =−2(3x2 + 1

2 − y/2)x

= −6x− x−1 +y

x

As a leap of faith, try y = vx substitution to solve y′ = −6x− x−1 + yx .

v′x + v = −6x− x−1 + v v′ = −6− x−2; v = −6x + x−1 + C

y = −6x2 + 5x + 1

Exercise 18. Assuming no friction at the ori�ce and energy conservation.

mgh =12mv2

f

(imagine how the top layer of water is now at the bottom of the tank (�nal �potential energy con�gurations�))Vf =

√2gh (how fast water is rushing out)

A0 = cross-sectional area of the ori�ce.

dV

dt= A

dh

dt= −c

√2ghA0

2h1/2∣∣∣hf

hi

= −c√

2gA0

At

=⇒ T =√

2A

c√

gA0

(√hf −

√hi

)= 59.6sec

Note that we included the discharge coef�cient C = 0.6.

Exercise 19. dVdt = −c

√2ghA0 + γ0 = Adh

dt = −κh1/2 + γ0. κ = c√

2gA0.162

Adh

γ0 − κh1/2= dt = (A/γ0)

dh

1− κγ0

h1/2

(ln

(1− ah1/2

))′=

11− ah1/2

(−a

21

h1/2

)(where a =

κ

γ0)

(h1/2)′ =1

2h1/2

(1− ah1/2

1− ah1/2

)=

(12 − a

2h1/2)

h1/2(1− ah1/2

)

=⇒∫

(A/γ0)dh

1− κγ0

h1/2= − (A/γ0)

(2γ2

0

κ2ln

(1− κ

γ0h1/2

)+

2γ0h1/2

κ

)∣∣∣∣hf

hi

=

= T

=⇒ exp(−γ0

At− 2γ0

κ

(h

1/2f − h

1/2i

))κ2

2γ20

=1− κ

γ0h

1/2f

1− κγ0

h1/2i

t→∞−−−→ hf =γ20

κ2=

(100in3/s)2

c2(2)(32ft/s2)(5/3in2)2(12in/1ft)= (25/24)2

Exercise 20.V0 =

13πR2

0H0

V (h) = V0 − 13πh

(h

R0

H0

)2

= V0 − 13π

R20

H20

h3 = V0 − αh3

mg(H0 − h) =12mv2

f ;√

2g(H0 − h) = vf (energy conservation)

cA0vf = cA0

√2gH0

(1− h

H0

)= β

√1− h

H0

dV

dt= −β

√1− h/H = −3αh2 dh

dt=⇒ dh

dt=

β

3αh2

√1− h

H∫h2/H2

0√1− h

H0

=β/H2

0

3αT =

= H0

∫u2du√1− u

= H0

∫(1− y)2(−dy)√

y= −H0

∫1− 2y + y2

√y

=

= −H0

(2y1/2 − 2

3y3/2 +

25y5/2

)∣∣∣∣hf

hi

=

= −H0

(2

(1− h

H0

)1/2

− 43

(1− h

H0

)3/2

+25

(1− h

H0

)5/2)∣∣∣∣∣

hf

hi

=

=cA0

√2gH0/H2

0

3(

13π

R20

H20

) T

For hi = 0, hf = H ,

H0(2(1)− 4/3(1) + 2/5) = H0(16/15) = cA0

√2gH0T/(πR2

0); T =1615

√H0πR2

0

cA0

√2g

=29

πR20

√H0

A0

Exercise 21. m2x−m + (1− x) = 0 =⇒ (m2 − 1)x + 1−m = 0, =⇒ m = 1Exercise 22. Given x + y3 + 6xy2y′ = 0

Notice that (2xy3)′ = 2(y3 + 3xy2y′) = 2y3 + 6xy2y′. So thenx + y3 + 6xy2y′ = (2xy3)′ +−y3 + x = 0

Let u = 2xy3 . So then y3 = u2x . Thus, we have

u′ − u

2x= −x

163

This equation is linear because it is a linear combination of y′ and y. Now we can use the formula for solving �rst-orderlinear differential equations of the form u′ + Pu = Q.

u = e−A

(∫ xf

x0

QeAdx + u(x0))

for A =∫

Pdt

=⇒ A =∫ −1

2tdt =

−12

ln t and u = x1/2

(−23

(x3/2 − x3/20 ) + u(x0)

)=−23

x2 + Cx1/2

=⇒ y3 =−x

3+

C

x1/2for x > 0 or y3 =

−x

3∀x

Exercise 23. Given (1 + y2e2x)y′ + y = 0, let y = uemx , u unknown.

y′ = u′emx + muemx

(1 + u2e2mxe2x)(u′emx + muemx) + uemx = 0 = u′emx + u′(u2e3mx+2x) + muemx + mu3e3mxe2x + uemx

u′(emx + u2e(3m+2)x) + muemx + mu3e3mx+2x + uemx = 0

u′ =−(m + 1)u−mu3e(2m+2)x

1 + u2e2(m+1)x

Let m = −1 .

=⇒ u′ =u3

1 + u2=⇒ 1 + u2

u3du =

(1u3

+1u

)du = dx =⇒ u−2/− 2 + ln u− C = x

y = ue−x or u = yex

=⇒ ln y = x +e−2x

2y2+ C

Exercise 24. Given f s.t. 2f ′(x) = f(

1x

)if x > 0, f(1) = 2 and x2y′′ + axy′ + by = 0

(1)

2f ′′(x) =d

dxf

(1x

)= (f ′)

(−1x

)

=⇒ x2y′′ =−12

f ′ =−14

f

(1x

)

a = 0; b =14

(2)f(x) = Cxn

f ′ = nCxn−1

f ′′ = n(n− 1)Cxn−2

=⇒ n(n− 1)Cxn =−14

Cxn

n2 − n +14

=⇒ n =12

Exercise 28. Choose the units for time to be in days �rst - 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 attime t

dx

dt= x =⇒ x = Cet

which makes sense because if C is the original population number, then after 1 year, x = Ce.

With t in days, we have a decrease of 1100x 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:dx

dt=

1365

x− 1100

x =100− 365

36500x =

−26536500

=⇒ x = Ce−26536500 t

164

Change t units to years by multiplying the �time constant� −26536500 by 365 days.

x = 365 exp (−2.65t)

To get the total fatalities, simply integrate the deaths during each year.∫ t

0

y =365100

3652.65

(− exp (−2.65t) + 1)

Exercise 29. For constant gravity, ∆K = −∆U

=⇒ −(0−mgh) = 12mv2

f

vf =√

2gh = (6.37× 108 cm)(

1 m2.54 cm

) (1 ft12 in

)= 6.93 mi

sec = 24940 mihr

The constant energy formula could also be obtained by consideringF = GMem

−r2 = m dr

dt2 = −∂rU

Exercise 30. Let y = f(x) be the solution to y′ = 2y2+x3y2+5 f(0) = 0

(1) y′(0) = 0 as easily seen. Now y′′ = (4yy′+1)(3y2+5)−(6yy′)(2y2+x)(3y2+5)2 , so then

y′′(0) = 15 > 0. It is a minimum.

(2) f ′(x) ≥ 2/3 ∀x ≥ 10/3. a = 2/3 since f will be above this tangent line.Suppose, in the �worst case,� f ′(x) = 0 for 0 ≤ x ≤ 2/3. Then f(x) = 0 for 0 ≤ x ≤ 2/3. Then the tangent linemust be at y = 0 at x = 10/3 to remain below the graph of f(x).=⇒ 2

3x− 20/9 < f(x)(3) Since f ′(x) ≥ 2

3 for each x ≥ 103 , then f →∞ for x →∞ (otherwise, f would have to decrease somewhere, which

would contradict the given fact about f ). Rewrite the DE for y′ to be

y′ =2y2 + x

3y2 + 5=⇒ (3y2 + 5)y′ = 2y2 + x =⇒ (3 +

5y2

)y′ = 2 +x

y2

Consider

y′ =2y2 + x

3y2 + 5=

2 + xy2

3 + 5y2

speci�cally, xy2 . Now y must, at the very least, have some linear increase because we had already shown that y′ ≥ 2

3 .So y2 would go to in�nity faster than linear x. Thus limx→∞ y′ = 2

3 . So then (3 + 5y2 )y′ x→∞−−−−→ (3 + 0) 2

3 = 2 =2 + x

y2 .

0 =x

y2

(4) y′ =2+ x

y2

3+ 5y2

x→∞−−−−→ 23 . =⇒ y = 2

3x or yx = 2

3 .

Exercise 31. Given a function f which satis�es the differential equation xf ′′(x) + 3x(f ′(x))2 = 1− e−x

(1) c 6= 0 for an extrenum.cf ′′(c) + 3c(f ′(c))2 = cf ′′(c) = 1− e−c =⇒ f ′′(c) = 1−e−c

c > 0(2) Cleverly, consider the limit.

xf ′′(x) + 3x(f ′(x))2 = 1− e−x =⇒ f ′′(x) + 3(f ′(x))2 =(

1− e−x

x

)x→0−−−→ f ′′(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 ′′ + 3(f ′)2 = 1−e−x

x suggests that we consider the Taylor series of e−x.

1− e−x

x=−∑∞

j=1(−x)j

j!

x=

∞∑

j=0

(−1)jxj

(j + 1)!

This further suggests that f itself has a power series representation because its �rst and second order derivatives aresimply a combination of in�nitely many terms containing powers of x.

165

Then suppose f =∑∞

j=0 ajxj .

f ′′ + 3(f ′)2 =1− e−x

x=⇒

f ′ =∞∑

j=0

(j + 1)aj+1xj

f ′′ =∞∑

j=0

(j + 2)(j + 1)aj+2xj

=⇒∞∑

j=0

(j + 2)(j + 1)aj+2xj + 3

∞∑

j=0

∞∑

k=0

(j + 1)(k + 1)aj+1ak+1xj+k =

∞∑

j=0

(−1)jxj

(j + 1)!

If f(0) = 0 a0 = 0

If f ′(0) = 0 a1 = 0

Then f = a2x2 +

∑∞j=3 ajx

j . Consider the x0 terms in the DE. (f ′)2 doesn't contribute, because f ′'s leading orderterm is x1. So then

2(1)a2 + 0 = 1 =⇒ a2 =12

i.e. f =12x2 +

∞∑

j=3

ajxj

A =12

in order for f(x) ≤ Ax2

9.6 Exercises - Historical introduction, De�nitions and �eld properties, The complex numbers as an extension of thereal numbers, The imaginary unit i, Geometric interpretation. Modulus and argument.Exercise 6. Let f be a polynomial with real coef�cients.

(1) Since z1z2

(zn+11 ) = zn

1 z1 = zn1 z1 = zn+1

1

f(z) =∑

ajzj =∑

ajzj = 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 bothAx. 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 satis�ed.Exercise 8. Ax. 8 , Ax. 9 are satis�ed.

(a + ib)(c + id) = (ac− bd, ad + bc) so Ax. 7 might not be satis�ed.Exercise 9. Ax. 7, Ax. 8 , Ax. 9 are trivially satis�ed (all are positive).Exercise 10. x > y x > y is well de�ned. Ax. 8 is satis�ed.

For Ax. 7, (32 , 1), (1, 1

2 ) 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.166

Exercise 12.

w =az + b

cz + d

w =(az + b)(cz + d)(cz + d)(cz + d)

=ac|z|2 + adz + bcz + bd

c2|z|2 + cd(z + z) + d2

w +−w =ac|z|2 + adz + bcz + bd− ac|z|2 − adz − bcz − bd

|cz + d|2 =(ad− bc)(z − z)

|cz + d|2If ad− bc > 0

w − w = 2Imw =ad− bc

|cz + d|2 2Imz;ad− bc

|az + d|2 > 0

So Imw has the same sign as Imz

9.10 Exercises - Complex exponentials, Complex-valued functions, Examples of differentiation and integration formu-las.Exercise 7.

(1)

if m 6= n ,

∫ 2π

0

eix(n−m)dx =eix(n−m)

i(n−m)

∣∣∣∣2π

0

=1− 1

i(n−m)= 0

if m = n ,

∫ 2π

0

eix(0)dx = 2π

(2)∫ 2π

0

einxe−imxdx =∫ 2π

0

(cos nx + i sinnx)(cos mx− i sin mx) =

=∫ 2π

0

cos nx cosmx + sin nx sin mx + i(sinnx cos mx− sin mx cosnx)∫ 2π

0

e−inxe−imxdx =∫ 2π

0

(cos nx− i sinnx)(cos mx− i sin mx) =

=∫ 2π

0

cos nx cosmx− sin nx sin mx + i(− sin nx cos mx− sinmx cos nx)

Summing the two equations above

0 =∫ 2π

0

2 cos nx cos mx + 2∫ 2π

0

i− sin mx cosnx

=⇒∫ 2π

0

cos nx cosmx = 0,

∫ 2π

0

sin mx cosnx = 0

∫ 2π

0

einxe−imxdx =∫ 2π

0

(cos nx + i sinnx)(cos mx− i sin mx) =

=∫ 2π

0

cos nx cosmx + sin nx sin mx + i(sinnx cosmx− sin mx cosnx)∫ 2π

0

einxeimxdx =∫ 2π

0

(cos nx + i sinnx)(cos mx + i sin mx) =

=∫ 2π

0

cos nx cosmx− sin nx sin mx + i(sinnx cosmx + sin mx cosnx)

Subtract the two equations above∫ 2π

0

sin nx sin mx− i

∫ 2π

0

sin mx cos nx = 0

=⇒∫ 2π

0

sin nx sin mx = 0

167

∫ 2π

0

einxe−inx = 2π =∫ 2π

0

(cos nx + i sin nx)(cos nx− i sin nx) =

=∫ 2π

0

cos2 nx + sin2 nx

∫ 2π

0

einxeinx = 0 =∫ 2π

0

cos2 nx− sin2 nx + i(2 cos nx sin nx)

=⇒∫ 2π

0

cos2 nx− sin2 nx = 0∫ 2π

0

cos nx sin nx = 0

Summing the two results above, we obtain2π =

∫ 2π

0

2 cos2 nx

=⇒∫ 2π

0

cos2 nx = π

Then also,∫ 2π

0

sin2 nx = π

Exercise 8.z = reiθ = rei(θ+2πm), m ∈ Z

z1/n = r1/nei(θ/n+2πm/n) m = 0, 1, . . . n− 1

=⇒ z1/n = Reiαεm = z1εm

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.eiueiv = ei(u+v) = cos u + v + i sin u + v =

= (cos u + i sin u)(cos v + i sin v) = cosu cos v − sin u sin v + i(cos v sin u + cos u sin v)=⇒ sin u + v = cos v sinu + cosu sin v

=⇒ cos u + v = cos u cos v − sin u sin v

sin2 z + cos2 z =(

eiz − e−iz

2i

)2

+(

eiz + e−iz

2

)2

=

=−(e2iz + e−2iz − 2) + (e2iz + 2 + e−2iz)

4= 1

cos iy =eiiy + e−iiy

2=

=e−y + ey

2= cosh y

sin iy =eiiy − e−iiy

2i=

=e−y − ey

2i= i sinh y

eiz = ei(x+iy) = eixe−y = (cos x + i sinx)e−y

e−iz = e−i(x+iy) = e−ixey = (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 + iπ

2 = iπ2

(2) Log(z1z2) = Log(|z1||z2|ei(θ1+θ2)) = ln |z1||z2|+ i(θ1 + θ2 + 2nπ) = Logz1 + Logz2 + i2πn

(3) Log(z1/z2) = Log(|z1|/|z2|ei(θ1−theta2)) = ln |z1||z2| + i(θ1 − θ2 + 2nπ) = Logz1 − Logz2 + i2πn

(4) exp (Logz) = exp (ln |z|+ iθ + i2πn) = z

Exercise 11.168

(1)1i = eiLog1 = ei(i2πn) = e−2πn = 1 if n = 0

ii = eiLogi = ei(i π2 +i2πn) = e−

π2−2πn = e−π/2 if n = 0

(−1)i = eiLog−1 = ei(iπ+i2πn) = e−π−2πn = e−π

(2) zazb = eaLogzebLogz = eaLogz+bLogz = e(a+b)Logz = za+bx(3)

(z1z2)w = ewLogz1z2 = ew(Logz1+Logz2+2πmi)

(zw1 zw

2 ) = ewLogz1ewLogz2 = ew(Logz1+Logz2)

m = 0 is the condition required for equality.Exercise 12.

if L(u) = P, L(v) = Q,

L(u + iv) = (u + iv)′′ + a(u + iv)′ + b(u + iv) = u′′ + au′ + bu + i(v′′ + av′ + 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) = −ω2y + aiωy + by = Aeiωx =⇒ (−ω2 + aiω + b)B = A

We cannot let (−ω2 + aiωb) = 0 for a nontrivia solution. Thus b 6= ω2 or aω 6= 0.

B =A

−ω2 + aiω + b

Exercise 14.L(y) = ceiωx; y = Beiωx =

c

−ω2 + aiω + beiωx

=⇒ y =c√

(b− ω2)2 + (aω)2ei(ωx−α) where tan α =

aω

b− ω2

=⇒ <y =c√

(b− ω2)2 + (aω)2cos (ωx− α)

Exercise 15.=(y) =

c√(b− ω2)2 + (aω)2

sin (ωx + α)

=⇒ A =c√

(b− ω2)2 + (aω)2; − tan α =

aω

b− ω2

10.4 Exercises - Zeno's paradox, Sequences, Monotonic sequences of real numbers. Exercise 1. Converges to 0.

f(n) =n

n + 1− n + 1

n=

n2 − (n2 + 2n + 1)n(n + 1)

=−2n− 1n2 + n

=−2n − 1

n2

1 + 1n

n→∞−−−−→ 0

Exercise 2. Converges to −1.

f(n) =n3 − (n3 + n + n2 + 1)

(n + 1)n=

n2 − n− 1n(n + 1)

=−1− 1

n − 1n2

1 + 1n

n→∞−−−−→ −1

Exercise 3. Diverges since|cosnπ

2− L| ≥

∣∣∣∣∣∣1 cos

nπ

2

∣∣∣− |L|∣∣∣ ≥ |1− |L||

Choosing ε1 = |1−|L||2 , | cos nπ

2 − L| > ε1 for n = 4m.Exercise 4. f(n) = 1

5 + 35n − 2

5n2 → limn→∞ f(n) = 15

Exercise 5. f(x) = x2x = x

exp (x ln 2) → 0 since limx→∞ xα

(ex)β .Exercise 6. f(n) = 1 + (−1)n = 0 of 1.

Thus, choosing ε1 = |1−|L||2 ;

|f(n)− L| ≥ ||f(n)| − |L|| = |1− |L|| > ε1 for any n = 2m

Exercise 7. f(n) = 1+(−1)n

n .169

Suppose ε = 3N .

So for n > N, 1N > 1

n , n ≥ N = N(ε) = 3/ε.

|f(n)| =∣∣∣∣1 + (−1)n

n

∣∣∣∣ ≤2n

<3n

<3N

= ε

Exercise 8. f(n) = (−1)n

n + 1+(−1)n

2

|f(n)− L| =∣∣∣∣(−1)n

n+

1 + (−1)n

2− L

∣∣∣∣ ≥∣∣∣∣∣∣∣∣L−

(−1)n

n

∣∣∣∣−∣∣∣∣1 + (−1)n

2

∣∣∣∣∣∣∣∣ ≥

≥∣∣∣∣∣∣∣∣|L| −

∣∣∣∣(−1)n

n

∣∣∣∣∣∣∣∣−

∣∣∣∣1 + (−1)n

2

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣|L| −

1n

∣∣∣∣ =∣∣∣∣1 + (−1)n

2

∣∣∣∣∣∣∣∣ ≥

≥∣∣∣∣∣∣∣∣|L| −

1n

∣∣∣∣−12

∣∣∣∣Thus, consider ∣∣∣∣

∣∣∣∣|L| −1n

∣∣∣∣−12

∣∣∣∣ >

∣∣∣∣∣∣∣∣1N− |L|

∣∣∣∣− 1∣∣∣∣ = ε0 for n > N

Exercise 9. f(x) = exp(

1x ln 2

); limx→∞ f(x) = 0.

Exercise 10.|f(n)− L| = |n(−1)n − L| ≥ ||n(−1)n | − |L|| = ||n| − |L|| = |n| − |L| > N − |L|

Thus, for n > N , N(ε) = ε + |L|, so then |f(n)− L| > ε.Exercise 11. f(n) = n2/3 sin n!

n+1 .

|f(n)| =∣∣∣∣n2/3 sin (n!)

n + 1

∣∣∣∣ =∣∣∣∣

sin (n!)n1/3 + n−2/3

∣∣∣∣ ≤∣∣∣∣

1n1/3

∣∣∣∣Thus, for n > N , N(ε) = 1

ε3 , |f(n)| < ε.Exercise 12. Converges, since

f(n)− 13

=3n+1 + 3(−2)n − 3n+1 − (−2)n+1

3(en+1 + (−2)n+1)=

((−2)n(3 + 2)

3(3n+1 + (−2)n+1)

)=

=(

53

(−2)n

3n+1 + (−2)n+1

)

∣∣∣∣f(n)− 13

∣∣∣∣ =53

∣∣∣∣(−2)n

3n+1 + (−2)n+1

∣∣∣∣ ≤∣∣∣∣

−(−2)n+1

3n+1 + (−2)n+1

∣∣∣∣ =

=

∣∣∣∣∣∣∣−1(

3−2

)n+1

+ 1

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣1

1 +(

3−2

)n+1

∣∣∣∣∣∣∣<

<1(

32

)n+1 <1(32

)n

For n > N , consider ε =(

23

)−N , i.e. N = − ln εln 2/3 = N(ε). Thus

L =13;

Exercise 13.f(n) =

√n + 1−√n

f(n) =(√

n + 1−√n) (√

n + 1 +√

n√n + 1 +

√n

)=

n + 1− n√n + 1 +

√n

=√

1√

n + 1 +√

n

|f(n)| =∣∣∣∣

1√n + 1 +

√n

∣∣∣∣ ≤1

2√

n;

So then ∀ε, we have ε = 12√

Nand for n > N , 1

2√

n< 1

2√

N= ε.

Thus f(n) converges to 0.170

Exercise 14.f(n) = nan = n exp n ln a =

n

exp n ln 1a

→ 0 since limx→∞

xa

(ex)b= 0

Exercise 15. f(n) = loga nn , a > 1. limn→∞ f(n) = 0 since limx→∞

(log x)a

xb = 0 for a > 0, b > 0Exercise 16. limn→∞ f(n) = 0Exercise 17. limn→∞ f(n) = e2.Exercise 18. ∣∣∣∣1 +

n

n + 1cos

nπ

2− L

∣∣∣∣ ≥∣∣∣∣∣∣∣∣1 +

n

n + 1cos

nπ

2

∣∣∣∣− |L|∣∣∣∣ =

∣∣∣∣∣∣∣∣1 +

−n

n + 1

∣∣∣∣− |L|∣∣∣∣ =

=∣∣∣∣

1n + 1

− |L|∣∣∣∣ > |L|

Choose ε0 = |L|2 . For any N , for n > N , |f(n)− L| > ε0.

Exercise 19.

1 +i

2=√

52

eiα

1 +(

12

)2

=54

tanα =12

(√5

2eiα

)−n

,

(2√5

)n

e−niα

limn→∞

(2√5

)n

e−niα = 0

Exercise 20. limn→∞ f(n) = limn→∞ e−πin/2 diverges since

|eiπin/2 − L| ≥ ||e−iπin/2| − |L|| = |1− |L|| > |L|So for ε0 = |L|

2 , for n > N , |f(n)− |L|| > ε0

Exercise 21. f(n) = 1ne−iπn/2 |f(n)| = 1

n .

Suppose N(ε) = 1ε ; then for n > N , |f(n)| < ε0

Exercise 22. |f(n)− L| ≥ ||ne−πin/2| − |L|| = |n− |L||Consider ε0 = |1− |L|| for n > N > 1, |f(n)− L| > ε0.Exercise 23. an = 1

n .

|an| =∣∣∣∣1n

∣∣∣∣ <1N

; N(ε) =1ε

ε = 1, 0.1, 0.01, 0.001, 0.0001N = 1, 10, 100, 1000, 10000

Exercise 24. |an − 1| =∣∣∣n+1−n

n+1

∣∣∣ = 1n+1 < 1

n

N(ε) = 1ε = 1, 10, 100, 1000, 10000.

Exercise 25. |an| = 1n

N(ε) = 1, 10, 100, 1000, 10000.Exercise 26. |an| =

∣∣ 1n!

∣∣ ≤ 1exp n ln n < 1

exp n

For n > N , 1exp N = ε, so that N = ln 1/ε.

N(ε) = 1, 2, 4, 6, 9

Exercise 27. an = 2nn2+1 ; |an| =

∣∣∣ 2n2+1/n

∣∣∣ ≤∣∣ 2n2

∣∣.

N(ε) =√

2√ε

= 1, 4, 14, 44, 141

Exercise 28. |an| =∣∣ 910

∣∣n =(

910

)n = en ln 9/10

N(ε) =ln ε

ln(

910

) =− ln 1/ε

ln (9/10)=

= 1, 21, 43, 65, 87171

Exercise 30. If ∀ε > 0, ∃N ∈ Z+ such that n > N , |an| < ε.

|an|2 < |an|ε < ε2

|a2n| < ε2

So for ∀ε1 > 0, ε1 = ε2 and ∃N = N(ε) = N(ε1), so that |a2n| < ε1.

Exercise 31.|an + bn − (A + B)| = ||an −A||bn −B|| ≤ |an −A|+ |bn −B| < ε + ε = 2ε

∀ε > 0, ∃NA, NB ∈ Z+, |an −A| < ε if n > NA; |bn −B| < ε if n > NB

Consider max (NA, NB) = NA+B

|an + bn − (A + B)| < 2ε

∀ε1 > 0, ε1 = 2ε, then ∃NA+B = NA+B(ε1) ∈ Z+ such that|(an + bn)− (A + B)| < ε1 if n > NA+B

|can − cA| = c|an −A| < cε

∀ε1 > 0, ε1 = cε; then ∃NcA = N(ε) = N(ε1) ∈ Z+ such that|can − cA| < ε1 for n > N(ε1)

Exercise 32. Given limn→∞ an = A,lim

n→∞(an −A)(an + A) = lim

n→∞(an −A) lim

n→∞(an + A) = 0(2A) = 0

2anbn = (an + bn)2 − a2n − b2

n

2 limn→∞

anbn = limn→∞

(an + bn)2 − limn→∞

a2n − lim

n→∞b2n =

(lim

n→∞(an + bn)

)2

−A2 −B2 = 2AB

=⇒ limn→∞

anbn = AB

Exercise 33.(αn

)= α(α−1)(α−2)...(α−n+1)

n!

(1) (− 12

1!

)=−1/2

1(− 12

2!

)=

(− 12

) (−32

)

2!=

58(− 1

2

3!

)=

(− 12

) (−32

) (−52

)

3!= − 5

16

(− 12

4

)=

(− 12

) (−32

) (−52

) (−72

)

4!=

35128(− 1

2

5

)=

(− 12

) (−32

) (−52

) (−72

) (−92

)

5!=−63256

(2) an = (−1)n(−1

2n

).

a1 = 12 > 0. a2 = 3

8 > 0

an+1 = (−1)n+1

( − 12

n + 1

)= (−1)n+1

(α

n

)(− 12 − (n + 1) + 1

n + 1

)=

an(−1)(−1

2 − n)

n + 1=

=an(n + 1/2)

n + 1> 0

an+1 =(

n + 1/2n + 1

)an < an

Exercise 34.(1)

tn − sn =1n

n∑

k=1

f

(k

n

)− 1

n

n−1∑

k=0

f

(k

n

)=

1n

(f(1) +

n−1∑

k=1

f

(k

n

)−

(f(0) +

1n

n−1∑

k=1

f

(k

n

)))=

=1n

(f(1)− f(0))172

Since f(

kn

) ≤ f(t) ≤ f(

k+1n

)for k

n ≤ t ≤ k+1n , by f being monotonically increasing.

=⇒ sn ≤∫ 1

0

f(x)dx ≤ tn (from de�nition of integral)

0 ≤∫ 1

0

f(x)dx− sn ≤ tn − sn =1n

(f(1)− f(0))

(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 ∫ b

a

s(x)dx ≤ I(f) ≤ I(f) ≤∫ b

a

t(x)dx

for all step functions s and t with s ≤ f ≤ t. The function f is integrable on [a, b] iff its upper and lower integralsare equal, ∫ b

a

f(x)dx = I(f) = I(f)

Since f(x) is integrable, then limn→∞ sn = limn→∞ tn =∫ 1

0f(x)dx

(3)(

b−an

)= ∆, sn = 1

∆

∑n−1k=0 f(a + k∆), tn = 1

∆

∑nk=1 f(a + k∆)

So by increasing monotonicity of f , sn ≤∫ b

af(x)dx ≤ tn.

tn − sn =1∆

(f(b) +

n−1∑

k=1

f(a + k∆) =n−1∑

k=1

f(a + k∆)− f(a)

)=

f(b)− f(a)∆

0 ≤∫ b

a

f(x)dx ≤ f(b)− f(a)∆

Exercise 35.(1) limn→∞ 1

n

∑nk=1

(kn

)2=

∫ 1

0t2dt = 1

3

(2) limn→∞∑n

k=11

n+k = limn→∞ 1n

∑nk=1

11+ k

n

=∫ 1

01xdx = ln 2

(3) limn→∞ 1n

∑nk=1

1

1+( kn )2 =

∫ 1

01

1+x2 dx = arctan x|10 =π

4

(4) limn→∞∑n

k=11√

n2+k2 = limn→∞ 1n

∑nk=1

1q1+( k

n )2 =∫ 1

01√

1+x2 dx = ln (x +√

1 + x2)∣∣10

= ln (1 +√

2)

(5) limn→∞∑n

k=11n sin kπ

n = limn→∞ 1n

∑nk=1 sin kπ

n =∫ 1

0sin πxdx =

(− cos πxπ

)∣∣10

= −(−1−1)π =

2π

(6) limn→∞∑n

k=11n sin2 kπ

n =∫ n

0sin2 xπ =

12

10.9 Exercises - In�nite series, The linearity property of convergent series, Telescoping series, The geometric series.Exercise 1.

∑∞n=1

1(2n−1)(2n+1) =

∑∞n=1

1/22n−1 − 1/2

2n+1 = 12

Exercise 2.∑∞

n=12

3n−1 = 2∑∞

n=013n = 2 1

1−1/3 = 3

Exercise 3.∑∞

n=21

n2−1 =∑∞

n=21/2n−1 − 1/2

n+1 =∑∞

n=2

(1/2n−1 − 1/2

n

)+−

(1/2n+1 − 1/2

n

)= 1

2 + 14 = 3

4

Exercise 4.∑∞

n=12n+3n

6n =∑∞

n=1

(13

)n +∑∞

n=1

(12

)n = 1/31−1/3 + 1/2

1−1/2 = 12 + 1 = 3

2

Exercise 5.∑∞

n=1

√n+1−√n√

n2+n=

∑∞n=1

1√n− 1√

n+1= 1

Exercise 6.∞∑

n=1

n

(n + 1)(n + 2)(n + 3)=

∞∑n=1

3/2(n + 2)(n + 3)

+−1/2

(n + 1)(n + 2)=

∞∑n=1

1(n + 2)(n + 3)

+12

(−16

)=

=∞∑

n=1

11 + 2

− 1n + 3

− 112

=13− 1

12=

14

Exercise 7.∑∞

n=12n+1

n2(n+1)2 =∑∞

n=11

n2 − 1(n+1)2 = 1

Exercise 8.∑∞

n=12n+n2+n

2n+1n(n+1) = 1 =∑∞

n=11

2n(n+1) + 12n+1 = 1

2

∑∞n=1

1n − 1

n+1 + 1/41−1/2 = 1

2 + 12 = 1

173

Exercise 9.∑∞

n=1(−1)n−1(2n+1)

n(n+1) =∑∞

n=1(−1)n−1(

1n + 1

n+1

).

(−1)n−1

(1n

+1

n + 1

)+ (−1)n

(1

n + 1+

1n + 2

)= (−1)n

(−1n

+1

n + 2

)

∞∑

j=1

( −12j − 1

+1

2j + 1

)=

12

∞∑

j=1

( −1(j − 1/2)

+1

(j + 1/2)

)=

12−11/2

= −1

Exercise 10.∞∑

n=2

log ((1 + 1n )n(1 + n))

(log nn)(log (n + 1)n+1)=

∞∑n=2

log((

n+1n

)n (1 + n))

log (n + 1)n+1 log nn=

=∞∑

n=2

log(n + 1)n+1 − log nn

log (n + 1)n+1 log nn=

∞∑n=2

1log nn

− 1log (n + 1)n+1

=

=1

2 log 2= log2

√e

since if 12 log 2 = y, then y = log2

√2.

Exercise 11.∑∞

n=1 nxn = x ddx

∑∞n=1 xn = x

(x

1−x

)′= x

(1−x)2 .Exercise 12.

∞∑n=1

n2xn = xd

dx

∞∑n=1

nxn = x

(x

(1− x)2

)′=

= x(1− x)2 + 2(1− x)x

(1− x)4=

x(1− x2)(1− x)4

=x(1 + x)(1− x)3

Exercise 13.∞∑

n=1

n3xn = xd

dx

∞∑n=1

n2xn = x

(x + x2

(1− x)3

)′=

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

(1− x)6= x

(1 + 2x)(1− x) + 3x(x + 1)(1− x)4

=x(x2 + 4x + 1)

(1− x)4

Exercise 14.∑∞

n=1 n4x4 = x ddx

∑∞n=1 n3x3 = x

(x3+4x2+x

(1−x)4

)′.

ln(

x3 + 4x2 + x

(1− x)4

)= ln (x3 + 4x2 + x)− 4 ln (1− x)

(ln f)′ =1f

f ′ =3x2 + 8x + 1x3 + 4x2 + x

+ 41

1− x;

f ′ =(3x2 + 8x + 1)(1− x)

(1− x)5+

4(x3 + 4x2 + x)(1− x)5

=

=3x2 + 8x + 1− 3x3 − 8x2 − x + 4x3 + 16x2 + 4x

(1− x)5=

x3 + 11x2 + 11x + 1(1− x)5

=⇒ x4 + 11x3 − 11x2 + x

(1− x)5

Exercise 15.∑∞

n=1xn

n = x∑∞

n=1

∫ x

0tn−1dt =

∫ x

0dt

∑∞n=1 tn−1 =

∫ x

01

1−t = − ln (1− x).Exercise 16.

∞∑n=1

x2n−1

2n− 1=

∞∑

j=1

∫ x

0

t2j−2dt =∫ x

0

dt

∞∑

j=1

(t2)j−1 =

=∫ x

0

dt

1− t2=

∫ x

0

dt

(1/21− t

+1/21 + t

)=

12

ln(

1 + x

1− x

)

Exercise 17.∑∞

n=0(n + 1)xn =∑∞

n=0ddxxn+1 = d

dx

(x

1−x

)= 1

(1−x)2

174

Exercise 18.∞∑

n=0

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

xn =∞∑

n=0

(xn+2

2

)′′=

d2

dx2

x2

2

(1

1− x

)=

d

dx

(x

1− x+

x2

2(1− x)2

)

=1

1− x+

x

(1− x)2+

x

(1− x)2+

x3

(1− x)3=

1− 2x + x2 + 2x− 2x2 + x2

(1− x)3=

1(1− x)3

Exercise 19.∞∑

n=0

(n + 1)(n + 2)(n + 3)3!

xn =∞∑

n=0

d3

dx3

xn+3

3!=

13

d

dx

∞∑n=0

d2

dx2

x(n+1)+2

2=

=13

d

dx

∞∑n=1

d2

dx2

xn+2

2=

13

d

dx

( ∞∑n=0

d2

dx2

xn+2

2− d2

dx2

(x2

2

))=

=13

d

dx

(1

(1− x)3− 1

)=

13−3(−1)(1− x)4

=1

(1− x)4

Exercise 20.∑∞

n=1 nkxn = Pk(x)(1−x)k+1

∞∑n=1

nk+1xn = xd

dx

∞∑n=1

nkxn = xd

dx

(Pn(x)

(1− x)k+1

)= x

(P ′k(x)(1− x)k+1 + (k + 1)(1− x)kPk(x)

(1− x)2k+2

)=

= x

(P ′k(x)(1− x) + (k + 1)Pk(x)

(1− x)k+2

)=

(k + 1)xPk(x) + x(1− x)P ′k(x)(1− x)k+2

((k + 1)Pk(x) + (1− x)P ′k(x))x has x as its lowest degree term from xP ′x(x) and(k + 1)xk+1 +−kxk+1 = xk+1 highest degree term is obtained from (k + 1)Pk(x) +−xP ′k(x).Exercise 21.

∑∞n=0

(n+k

k

)xn = 1

(1−x)k+1 = dk

dxk

∑∞n=0

xn+k

k! .∞∑

n=0

(n + k + 1

k + 1

)xn =

∞∑n=0

dk+1

dxk+1

xn+k+1

(k + 1)!=

1(k + 1)

d

dx

∞∑n=0

dk

dxk

x(n+1)+k

k!=

=1

k + 1d

dx

∞∑

k=1

dk

dxk

xn+k

k!=

(1

k + 1

)d

dx

∞∑n=0

dk

dxk

xn+k

k!− dk

dxk

xk

k!=

=(

1k + 1

)d

dx

(1

(1− x)k+1− 1

)=

1(1− x)k+2

Exercise 22.(1)

∑∞n=2

n−1n! =

∑∞n=2

1(n−1)! −

∑∞n=2

1n! =

∑∞n=1

1n! −

∑∞n=2

1n! = 1 .

(2)∑∞

n=2nn! +

∑∞n=2

1n! =

∑∞n=2

1(n−1)! +

∑∞n=0

1n! − 1− 1 =

∑∞n=1

1n! +

∑∞n=0

1n! − 2 =

∑∞n=0

2n! − 3 = 2e− 3 .

(3)∞∑

n=2

(n− 1)(n + 1)n!

=∞∑

n=2

n2

n!+

∞∑n=2

−1n!

=∞∑

n=2

n

(n− 1)!+

∞∑n=2

−1n!

=

=∞∑

n=1

n + 1n!

+∞∑

n=2

− 1n!

=∞∑

n=1

1(n− 1)!

+ 1 =∞∑

n=0

1n!

+ 1 = e + 1

Exercise 23.(1) x d

dx

(x d

dx

∑∞n=1

xn

n!

)= x d

dx

(x

∑∞n=1

nxn−1

n!

)= x d

dx

∑∞n=1

nxn

n! =∑∞

n=1n2xn

n! = x ddx

(x d

dxex)

= x2ex + xex

(2) x ddx

(∑∞n=1

n2xn

n!

)=

∑∞n=1

n3xn

n! = x ddx

((x2 + x)ex

)= x

((2x + 1)ex + (x2 + x)ex

)= (x3 + 3x2 + x)ex

x = 1 k = 5

Exercise 24.(1)

∑∞n=2(−1)n =

∑∞n=2(−1)n(n− (n− 1)). Identical.

(2)∑∞

n=2(1− 1) =∑∞

n=2(−1)n. Not identical.(3) Not identical.

∑∞n=2(−1)n vs. (

∑∞n=2(−1 + 1)) + 1.

(4) Identical.∑∞

n=0

(12

)n = 1 +∑∞

n=1

((12

)n−1 − (12

)n)

=∑∞

n=1

(12

)n (2− 1) =∑∞

n=0

(12

)n

Exercise 25.175

(1)1 + x2 + x4 + · · ·+ x2n + · · · = 1

1− x2if |x| < 1

=⇒ 1 + 0 + x2 + 0 + x4 + · · · = 11− x2

if |x| < 1

(2) Thm. 10.2.∑∞

n=1(αan + βbn) = α∑∞

n=1 an + β∑∞

n=1 bn. So then∞∑

j=0

xj −∞∑

j=0

xj + (−x)j

2=

∞∑

j=0

xj − (−x)j

2=

∞∑

j=0

x2j+1 =1

1− x− 1

1− x2=

x

1− x2

(3)∞∑

j=0

(x2)j +∞∑

j=0

xj =∞∑

j=0

(xj − x2j) =x

1− x2

10.14 Exercises - Tests for convergence, Comparison tests for series of nonnegative terms, The integral test. We'll beusing the integral test.

Theorem 31 (Integral Test).Let f be a positive decreasing function, de�ned for all real x ≥ 1.For ∀n ≥ 1, let sn =

∑nk=1 f(k) and tn =

∑n1 f(x)dx.

Then both sequences {sn} and {tn} converge or both diverge.

Exercise 1.3

4j − 3+

−14j − 1

=3(4j − 1) + (−1)(4j − 3)

(4j − 3)(4j − 1)=

8j

(4j − 3)(4j − 1)n∑

j=1

j

(4j − 3)(4j − 1)=

n∑

j=1

(3/8

4j − 3+−1/84j − 1

)

∫ n

1

(3/8

4x− 3+−1/84x− 1

)dx =

((3/8)

ln (4x− 3)4

+ (−1/8)ln (4x− 1)

4

)∣∣∣∣n

1

=

=132

ln(4x− 3)3

4x− 1

∣∣∣∣n

1

=132

ln(

3(4n− 3)3

4n− 1

)

limn→∞

132

ln(

3(4n− 3)3

4n− 1

)= lim

n→∞

∫ n

1

xdx

(4x− 3)(4x− 1)dx = ∞, so

n∑

j=1

j

(4j − 3)(4j − 1)diverges as well

Exercise 2.∞∑

j=1

√2j − 1 log (4j + 1)

j(j + 1)=

∞∑

j=1

aj

aj ≤√

4j + 1 log (4j + 1)j(j + 1/4)

=4 log (4j + 1)j(4j + 1)1/2

((4j + 1)(4j + 1)

)= 4

log (4j + 1)(4 + 1/j)(4j + 1)3/2

≤

≤ 16 log (4j + 1)(4j + 1)3/2

= bj

Now use the integral test on∑

bj to determine the convergence of∑

bj .∫ n

1

log (ax + 1)(ax + 1)3/2

dx =∫ (

(−2)a(ax + 1)1/2

)′log (ax + 1)dx =

=−2

a(ax + 1)1/2log (ax + 1)−

∫ −2a(ax + 1)1/2

(a

ax + 1

)=

=−2

a(ax + 1)1/2log (ax + 1) +

−4a(ax + 1)1/2

limn→∞

∫ n

1

log(ax + 1)(ax + 1)3/2

dx = limn→∞

(−2 log (an + 1)a(an + 1)1/2

+2 log (a + 1)a(a + 1)1/2

+−4a

((1

an + 1

)1/2

− 1(a + 1)1/2

))=

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

+4

a(a + 1)1/2

Then by integral test,∑

bj converges. Since∑

bj converges, then∑

aj converges by comparison test.176

Exercise 3.∑∞

j=1j+12j .

∫ n

1

x + 1ex ln 2

dx =∫ n

1

(xe−x ln 2 + e−x ln 2)dx =(

xe−x ln 2

− ln 2+

e−x ln 2

− ln 2+−e−x ln 2

(ln 2)2

)∣∣∣∣n

1

=ne−n ln 2

− ln 2+

e− ln 2

ln 2+−

(1

(ln 2)2+

1ln 2

)e−n ln 2 +

(1

(ln 2)2+

1ln 2

)e− ln 2

limn→∞

∫ n

1

x + 1ex ln 2

dx =(

2ln 2

+1

(ln 2)2

)(12

)

By integral test,∑∞

j=1j+12j converges.

Exercise 4.∑∞

j=1j2

2j .∫ n

1

x2

2xdx =

∫ n

1

x2

ex ln 2=

(x2e−x ln 2

− ln 2+

2xe−x ln 2

−(− ln 2)2+

2e−x ln 2

(− ln 2)3

)∣∣∣∣n

1

limn→∞

∫ n

1

x2

2xdx = e− ln 2

(1

ln 2+

2(ln 2)2

+2

(ln 2)3

)=

12

(1

ln 2+

2(ln 2)2

+2

(ln 2)3

)

By integral test,∑∞

j=1j2

2j converges.Exercise 5. ∞∑

j=1

| sin jx|j2

=∞∑

j=1

aj ≤∞∑

j=1

1j2

∑∞j=1

1j2 converges since

limn→∞

∫ n

1

1xs

dx = limn→∞

x−s+1

−s + 1

∣∣∣∣n

1

= limn→∞

(1

1− s

)(1

ns−1− 1

)=

1s− 1

if s > 1

∑∞j=1

|sinjx|j2 converges by comparison test and integral test.

Exercise 6. ∞∑

j=1

2 + (−1)j

2j=

∞∑

j=1

(1

22j−1+

322j

)=

2(1/4)1− 1/4

+3(1/4)1− 1/4

=43

Exercise 7.∑∞

j=1j!

(j+2)! .

aj =j!

(j + 2)!= aj =

1(j + 1)(j + 2)

≤ 1j2

= bj

Since∑

bj converges,∑

aj converges, by comparison test.Exercise 8.

∑∞j=2

log jj√

j+1=

∑∞j=2 aj ≤

∑∞j=2

log jj3/2

∫ n

2

log x

x3/2dx =

∫ n

2

(−2x−1/2)′ log x = (−2x−1/2 log x)∣∣∣n

2−

∫ n

2

−2x−1/2

xdx =

= (−2)(

log n

n1/2− log 2

21/2

)+−4x−1/2

∣∣∣∣n

2

limn→∞

∫ n

2

log x

x3/2dx = 21/2 log 2 +

4√2

So∑

aj converges by comparison test.Exercise 9.

∑∞j=1

1√j(j+1)

=∑∞

j=1 aj . Let bj = 1j .

limj→∞

aj

bj= lim

j→∞j√

j(j + 1)= lim

j→∞1√

1 + 1/j= 1

By limit comparison test, since∑

bj diverges,∑

aj diverges.Exercise 10.

∑∞j=1

1+√

j(j+1)3−1 =

∑∞j=1 aj

bj = 1j5/2

limj→∞

aj

bj= lim

j→∞

(1 +

√j

(j + 1)3 − 1

)j5/2 = lim

j→∞j3 + j5/2

(j + 1)3 − 1= lim

j→∞1 + 1/j1/2

(1 + 1/j)3 − 1j3

177

bj converges,∑

aj converges.

Exercise 11.∑∞

j=21

(log j)s =∑

aj

If s ≤ 0,∑

aj diverges since limj→∞

aj 6= 0

If 0 < s ≤ 1, 1(log j)s > 1

js , and since∑∞

j=21js diverges for 0 < s < 1, so does

∑1

(log j)s

∫(log x)−s =

∫ (1x

(log x)−s

)x =

∫ ((log x)−s+1

(1− s)

)′x =

(log x)−s+1

(1− s)x−

∫(log x)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.∑∞

j=1|aj |10j ; |aj | < 10.

∞∑

j=1

|aj |10j

<

∞∑

j=1

1010j

=∞∑

j=0

110j

=1

1− 1/10=

109

Exercise 13.∑∞

j=11

1000j+1 <∑∞

j=11

1000j = 11000

∑∞j=1

1j

The series diverges since∑

1j diverges.

Exercise 14.∞∑

j=1

j cos2 (jπ/3)2j

≤∞∑

j=1

j

2j=

∞∑

j=1

j

ej ln 2

∫ ∞

1

x

ekx=

∫ ∞

1

xe−kx =(

xe−kx

−k− e−kx

(−k)2

)∣∣∣∣∞

1

=e−k

k+

e−k

(−k)2

Exercise 15.∑∞

j=31

j log j(log (log j))s

∫1

x log x(log (log x))s=

∫ ((ln (ln x))−s+1

−s + 1

)′=

(ln (ln x))−s+1

−s+1 if s > 1, s < 1, s 6= 0ln (ln (ln x)) if s = 1ln (ln x) if s = 0

Converges, by integral test, for s > 1

Exercise 16. Converges by integral test since

∫ ∞

1

xe−x2=

(e−x2

−2

)∣∣∣∣∣

∞

1

= 0 +e−1

2

Exercise 17. I drew a picture to help me see what's going on.√

x

1 + x2≤√

x

1∫ 1/n

0

√xdx =

23x3/2

∣∣∣∣1/n

0

=23

(1n

)3/2

∞∑

j=2

23

(1j

)3/2

=23

∞∑

j=2

1j3/2

So∑

1 n∫ 1/n

0

√x

1+x2 dx converges by comparison test.178

Exercise 18.(e−

√x)′ = e−

√x −12√

x

(√

xe−√

x)′ =−e−

√x

2+

12√

xe−√

x(√

xe−√

x + e−√

x)′ =e−√

x

2∫ n+1

n

e−√

xdx = 2 (√

x + 1)e−√

x∣∣∣n+1

n= 2

(√x

e√

x+

1e√

x

)∣∣∣∣n+1

n

=

= 2(√

n + 1e√

n+1+

1e√

n+1−√

n

e√

n− 1

e√

n

)

2∞∑

j=1

(√j + 1

e√

j+1−√

j

e√

j+

1e√

j+1− 1

e√

j

)= −1

e

Note the use of telescoping sum in the last step. The series converges.Exercise 19. f is nonnegative and increasing. f increasing−−−−−−−→ f(k) ≤ f(x) ≤ f(k + 1), k ≤ x ≤ k + 1

∫ k+1

k

f(k)dx ≤∫ k+1

k

f(x)dx ≤∫ k+1

k

f(k + 1)dx =⇒ f(k) ≤∫ k+1

k

f(x)dx ≤ f(k + 1)P−→

P−→

n−1∑

k=1

f(k) ≤n−1∑

k=1

∫ k+1

k

f(x)dx =∫ n

1

f(x)dx ≤n−1∑

k=1

f(k + 1) =n∑

k=2

f(k)

Note that for the summation on the integral, we simply relied on the properties of integrals to get the �nal integral.∫ n

1f(x)dx =

∫ n

1log xdx = (x ln x− x)|n1 = n ln n− n + 1.

n−1∑

k=1

ln k ≤∫ n

1

ln x ≤n∑

k=2

ln (k) =⇒n−1∑

k=1

lnk ≤ n ln n− n + 1 ≤n∑

k=2

ln (k)

exp

(n−1∑

k=1

ln k

)= (n− 1)! ≤ nne−n+1

exp

(n∑

k=2

ln k

)= (n)! ≥ nne−n+1

=⇒ e1/n

e<

(n!)1/n

n<

e1/nn1/n

e

10.16 Exercises - The root test and the ratio test for series of nonnegative terms. Exercise 1.∑∞

j=1(j!)2

(2j)!

((j + 1)!)2

(2j + 2)

((2j)!(j!)2

)=

(j + 1)2

(2j + 2)(2j + 1)=

j2 + 2j + 14j2 + 6j + 2

j→∞−−−→ 14

Converges by ratio test.Exercise 2.

∑∞j=1

(j!)2

2j2 .((j + 1)!)2

2(j+1)2

2j2

(j!)2=

(j + 1)22j2

2j2+2j+1=

j2 + 2j + 12ej ln 2

j→∞−−−→ 0


∑∞j=1

2jj!jj

2j+1(j + 1)!(j + 1)j+1

jj

2jj!=

2(j + 1)(j + 1)

(1

1 + 1/j

)jj→∞−−−→ 2

e< 1


∑∞j=1

3jj!jj

3j+1(j + 1)!(j + 1)j+1

(jj

3jj!

)= 3

(1

(1 + 1/j)j

)j→∞−−−→ 3

e> 1

Diverges by ratio test.179

Exercise 5.∑∞

j=1j!3j .

(j + 1)!3j+1

3j

j!=

j + 13

Diverges by ratio test.Exercise 6.

∑∞j=1

j!22j

(j + 1)!22(j+1)

22j

j!=

(j + 1)4

Diverges.Exercise 7.

∑∞j=2

1(log j)1/j

Draw a picture to see what's going on.∞∑

j=2

1(log j)1/j

=∞∑

j=2

exp(−1

jln log j

)

0 <ln log j

jfor j > 3

ln (log j)j

<ln j

j=⇒ lim

j→∞ln (log j)

j< lim

j→∞ln j

j= 0

=⇒ limj→∞

−1j

ln (log j) = 0 so then

limj→∞

exp(−1

jln (log j)

)= 1

∑∞j=2

1(log j)1/j diverges because the aj term doesn't go to zero.

Exercise 8.∑∞

j=1(j1/j − 1)j

((j1/j − 1)j)1/j = (e1j ln j − 1)

j→∞−−−→ 0Converges by root test.Exercise 9.

∑∞j=1 e−j2

(e−j2)1/j = e−j j→∞−−−→ 0

Converges by root test.Exercise 10. I systematically tried ratio test and then root test. Both were inconclusive.

Consider comparison with∑

1j . (

ej2 − j

jej2

)j =

ej2 − j

ej2 = 1− j

ej2

j→∞−−−→ 1


1j diverges, so does

∑(1j − 1

ej2

).

Exercise 11.∑∞

j=1(1000)j

j! = e1000

Exercise 12.∑∞

j=1jj+1/j

(j+1/j)j .

a1/jj =

j1/j2

1 + 1j2

limj→∞

a1/jj = lim

j→∞

exp(

1j2 ln j

)

1 + 1j2

= 1

Note that root test is inconclusive.

aj =exp

(1j ln j

)

(1 + 1

j

)j≥

exp(

1j ln j

)

(1 + 1

j2

)j2

limj→∞

aj ≥ limj→∞

exp(

1j ln j

)

(1 + 1

j2

)j2 =1e

> 0

Diverges since limj→∞ aj > 0.180

Exercise 13.∑∞

j=1j3(√

2+(−1)j)j

3j .(

j3(√

2 + (−1)j)j

3j

)1/j

=j3/j(

√2 + (−1)j)3

=e

3j ln j(

√2 + (−1)j)3

j→∞−−−→ (√

2 + (−1)j)3

< 1

Converges by root test.Exercise 14.

∑∞j=1 rj | sin jx|.

If 0 < r < 1.∞∑

j=1

rj | sin jx| <∞∑

j=1

rj

so by comparison test,∞∑

j=1

rj | sin jx| converges for 0 < r < 1

If r ≥ 1,

limj→∞

rj | sin jx| 6= 0 so∞∑

j=1

rj | sin jx| diverges, unless jx = πj

Exercise 15.(1) cj = bj − bj+1aj+1

aj> 0 ∀j ≥ N . Then there must be a positive number r that's in between cj and 0.

ajbj − aj+1bj+1 ≥ raj

r

n∑

j=N

aj ≤n∑

j=N

(ajbj − aj+1bj+1) = aNbN − an+1bn+1 ≤ aNbN

=⇒n∑

j=N

aj ≤ aNbN

r

(2) cn < 0

bj − bj+1aj+1

aj< 0

ajbj < bj+1aj+1 =⇒ bj

bj+1<

aj+1

aj

∑ 1bj

diverges, so

limj→∞

bj

bj+1≥ 1 by ratio test

j→∞−−−→ 1 ≤ bj

bj+1<

aj+1

aj

So by ratio test,∑

aj diverges.Exercise 16. bn+1 = n; bn = n− 1.cn = n− 1− nan+1

an≥ r =⇒ an+1

an≤ 1− 1

n − rn .

Using Exercise 15,∑

an converges.∑1bn

diverges since∑

1bn

is a harmonic series of s = 1.

n− 1− nan+1

an≤ 0 =⇒ 1− 1

n≤ an+1

an

Exercise 17. For some N ≥ 1, s > 1, M > 0, and given that

an+1

an= 1− A

n+

f(n)ns

= 1−(

A− f(n)ns−1

n

)

Consider A− f(n)ns−1 .

Since |f(n)| < M , f(n) is �nite, so consider s larger than 1 and n going to in�nity so that f(n)ns−1 → 0.

Using Exercise 16, for∑

aj to converge, A− f(n)ns−1 = 1 + r where r > 0, for all n ≥ N , where N is some positive number.

Let r = MNs−1 so that

A = 1 + r +f(n)ns−1

> 1

If A > 1, then∑

an converges.181

If A = 1, then consider using Exercise 15 and bn = n log n.

cn = bn − bn+1an+1

an= (n− 1) log (n− 1)− n log n

(an+1

an

)

= (n− 1) log (n− 1)− n log n

(1− 1

n+

f(n)ns

)= (n− 1) log

((n− 1)

n

)− n log n

(f(n)ns

)=

= −(n− 1) log(

n

(n− 1)

)− n log n

(f(n)ns

)

since log n

ns−1

n→∞−−−−→ 0,

−(n− 1) log(

n

(n− 1)

)− n log n

(f(n)ns

)< 0 for n large enough

since cn < 0 for n ≥ N for some N > 0, then by Exercise 15,∑

an is divergent.

Given that A < 1, then for A− f(n)ns−1 , choose N > 0 so that M

ns−1 < ε < 1 and that A− f(n)ns−1 ≤ A + M

ns−1 = A + ε ≤ 1.We can always choose ε small enough because there's always a real number in between A and 1 (Axiom of Archimedes).

A− f(n)ns−1

≤ 1 =⇒ −(

A− f(n)ns−1

)≥ −1

=⇒ using Exercise 16, an+1

an= 1− A + f(n)

ns−1

n≥ 1− 1

nfor all n ≥ N

Exercise 18. (1 · 3 · 5 . . . (2n + 1)2 · 4 · 6 . . . (2n + 2)

· 2 · 4 · 6 . . . (2n)1 · 3 · 5 . . . (2n− 1)

)k

=(

2n + 12n + 2

)kn→∞−−−−→ 1

Ratio test fails.an+1

an=

(2n + 12n + 2

)k

=(

1 +−1

2n + 2

)k

=(

1 +−1/2n + 1

)k

=

=k∑

j=0

(k

j

) (−1/2n + 1

)j

= 1 + k

(−1/2n + 1

)+

k∑

j=2

(k

j

) (−1/2n + 1

)j

Note that for k < ∞,

k∑

j=2

(k

j

)(−1/2n + 1

)j

< ∞

Let

∣∣∣∣∣∣

k∑

j=2

(k

j

)(−1/2n + 1

)j∣∣∣∣∣∣≤ M

k/2 = A > 1 or k > 2 means∑

aj converges k/2 = A ≤ 1 or k ≤ 2 means∑

aj diverges

10.20 Exercises - Alternating series, Conditional and absolute convergence, The convergence tests of Drichlet andAbel. We will be using Leibniz's test alot, initially.

Theorem 32 (Leibniz's Rule). If aj is a monotonically decreasing sequence with limit 0,∑∞

j=1(−1)j−1aj converges.

If S =∑∞

j=1 aj , sn =∑n

j=1(−1)j−1aj ,

0 < (−1)j(S − sj) < aj+1

Exercise 1.∑∞

j=1(−1)j+1√

j. limj→∞ 1√

j= 0 Converges conditionally.

Exercise 2.∑∞

j=1(−1)j√

jj+100 limj→∞

√j

j+100 = 0. Converges by Leibniz's test.1√

j+ 100√j

≥ 1101

√j, so by comparison test, the series diverges absolutely. So the alternating series converges conditionally

by comparison test.Exercise 3.

∑∞j=1

(−1)j−1

js If s > 1, then the series absolutely converges. limj→∞ 1js = 0 if s > 0. Converges conditionally

for 0 < s < 1. Otherwise, if s < 0 the series diverges absolutely.182

Exercise 4.∑∞

j=1(−1)j(

1·3·5...(2j−1)2·4·6...(2j)

)3

.

aj+1

aj=

1 · 3 · 5 · (2j + 1)2 · 4 · 6 . . . (2j + 2)

2 · 4 · 6 . . . (2j)1 · 3 · 5 . . . (2j − 1)

=2j + 12j + 2

Absolutely converges.Exercise 5.

∑∞j=1

(−1)j(j−1)/2

2j converges since limj→∞(−1)j(j−1)/2

2j = 0;∑∞

j=112j = 1/2

1−1/2 = 1 . Absolutely converges.

Exercise 6.∑∞

j=1(−1)j(

2j+1003j+1

)j

.

exp(

j ln(

2j + 1003j + 1

))= exp

(j ln

(23

+298

9j + 3

))≤ exp

(j

(ln

(23

)+

123

(298

9j + 3

)))=

= exp(

j ln23

+146

3 + 1/j

)

0 ≤ limj→∞

exp(

j ln(

2j + 1003j + 1

))≤ lim

j→∞

exp

(j ln

23

+146

3 + 1/j

)= 0

=⇒ limj→∞

exp(

j ln(

2j + 1003j + 1

))= 0

So the alternating series converges.(2j + 1003j + 1

)<

2j + 1003j

<2.5j

3j=

56

( for j ≥ 200)

=⇒(

2j + 1003j + 1

)j

<

(56

)j

for j ≥ 200

So the series absolutely converges by comparison with a geometric series.Exercise 7.

∑∞j=2

(−1)j

√j+(−1)j .

limj→∞

1√j + (−1)j

doesn't exist since ???

To show divergence, we usually think of either taking the general term and �nding the limit (and if it goes to a nonzeroconstant, then it diverges), or we use ratio, root, comparison test on the general term. Since this is an alternating series, I'veobserved that the general term is a sum of two adjacent terms, one even and one odd.

(−1)j

√j + (−1)j

(−1)2j

√2j + (−1)2j

+(−1)2j+1

√2j + 1 + (−1)2j+1

=1√

2j + 1+

−1√2j + 1 + 1

=√

2j + 1− 1− (√

2j + 1)(√

2j + 1)(√

2j + 1− 1)=

=√

2j + 1−√2j − 2(√

2j + 1)(√

2j + 1− 1)=

√2j

√1 + 1

2j −√

2j − 2

(√

2j + 1)(√

2j√

1 + 12j − 1)

=

for j large−−−−−−→ '√

2j(1 + 1

4j

)−√2j − 2

(√

2j + 1)(√

2j(1 + 1

4j

)− 1)

=−2j

(1− 1

4√

2j

2− 12j + 1

2√

2j3/2

)

Every term, since we considered any j, will contain −2. So we factor it out. Then

1j

(1− 1

4√

2j

2− 12j + 1

2√

2jj3/2

)>

1j

(1− 1

4√

2j

4− 1√2j

)=

14j

By comparison test to 1j the series diverges.

Exercise 8. Using the theorem

Theorem 33.Assume

∑ |aj | convergesThen

∑aj converges and |∑ aj | ≤

∑ |aj |.183

So using the contrapositive,If

∑aj diverges,∑ |aj | diverges.

1j1/j

=1

e1j ln j

limj→∞

1j1/j

=1

exp(limj→∞ 1

j ln j) = 1

Diverges absolutely.Exercise 9.

∑∞j=1(−1)j j2

1+j2 Diverges absolutely.

(2j)2

1 + (2j)2− (2j − 1)2

1 + (2j − 1)2=

4j2

1 + 4j2

(4j2 − 4j + 24j2 − 4j + 2

)− (4j2 − 4j + 1)

(4j2 − 4j + 2)(1 + 4j2)1 + 4j2

=4j − 1

2(1 + 4j2)(2j2 − 2j + 1)4j − 1

2(1 + 4j2)(2j2 − 2j + 1)(j3) =

4− 1/j

2(4 + 1/j2)(2− 2/j + 1/j2)=

14

By limit comparison test, with∑

1j3 ,

∑∞j=1

j2

1+j2 converges.

Exercise 10.∑∞

n=1(−1)n

log (en+e−n)

limn→∞

1log (en + e−n)

= limn→∞

1n

= 0

The series converges.

limn→∞

n

log (en + e−n)= lim

n→∞n

log en + log (1 + e−2n)= lim

n→∞1(

n+log (1+e−2n)n

) = 1

Since∑

1n diverges,

∑1

log (en+e−n) diverges.

Exercise 11.∑∞

j=1(−1)j

j log2 (j+1)

limj→∞ 1j log2 (j+1)

= 0 so by Leibniz's test, the alternating series.

1n log2 (n + 1)

<1

n log2 (n)∫

1n log2 n

=∫ ( −1

log n

)′=

−1log n

n→∞−−−−→ 1log 2

Converges by comparison test to 1n log2 n

, which converges by integral test. So the series absolutely converges.

Exercise 12.∑∞

j=1(−1)j

log (1+1/j) diverges absolutely.

(−1)

log(1 + 1

2j−1

) +1

log(1 + 1

2j

) =−1

log(

2j2j−1

) +1

log(

2j+12j

) =− log

(2j+12j

)+ log

(2j

2j−1

)

log(

2j2j−1

)log

(2j+12j

) =

=log

((2j)2

4j2−1

)

log(

2j2j−1

)(log

(1 + 1

2j

)) =log

(1 + 1

4j2−1

)

log(1 + 1

2j−1

)log

(1 + 1

2j

) =

=1

4j2−1 + o(

14j2−1

)(

12j−1 + o

(1

2j−1

))(12j + o

(12j

)) ≈

≈ 4j2 − 2j

4j2 − 1=

1− 12j

1− 14j2

j→∞−−−→ 1

So the alternating series diverges.Exercise 13.

∑∞j=1

(−1)jj37

(j+1)! Use the ratio test.

aj+1

aj=

(j + 1)37(j + 2)!

(j + 1)!j37

=(

1j + 2

)(1 +

1j

)37

→ 0

184

Converges for∑ |aj |. Then

∑aj converges. The series absolutely converges.

Exercise 14.∑∞

n=1(−1)n∫ n+1

ne−x

x dx

∫ n+1

n

e−x

xdx ≤

∫ n+1

n

1e2x

dx =e−2x

−2

∣∣∣∣n+1

n

=(−1

2

)(1

e2(n+1)− 1

e2n

)=

=e2 − 12e2n+2

< 1

Converges absolutely.Exercise 15.

∑nj=1 sin (log j)

limj→∞ sin (log j) doesn't exist. So the series is divergent.

Exercise 16.∑∞

j=1 log(j sin 1

j

)Note that

log(

j sin1j

)= log

(sin 1/j

1/j

)

limj→∞

log(

sin 1/j

1/j

)= log

(lim

j→∞sin 1/j

1/j

)= log 1 = 0

sin1j

=∞∑

k=0

(1j

)2k+1

(2k + 1)!(−1)k

log

j

∞∑

k=0

(1j

)2k+1

(−1)k

(2k + 1)!

= log

1 +

−16j2

+∞∑

k=2

(1j

)2k

(−1)k

(2k + 1)!

≥

≥ log(

1 +− 16j2

)≥ −1

6j2

The series absolutely converges.

Exercise 17.∑∞

j=1(−1)j(1− j sin 1

j

)

(1− x sin

1x

)′= − sin

1x− x cos

1x

(−1x2

)=

= − sin1x

+1x

cos1x

=−x sin 1

x + cos 1x

x

sin1x

=∞∑

j=0

(1x

)2j+1 (−1)j

(2j + 1)!

−x sin 1x + cos 1

x

x=

=−1 + ( 1

x )3(+1)

3! +∑∞

j=2( 1

x )2j+1(−1)j

(2j+1)! + 1− (1x

)2/2 +

∑∞j=2

( 1x )2j

(−1)j

(2j)!

x< 0 for x large enough

∑∞j=1(−1)j

(1− j sin 1

j

)converges since aj = 1− j sin 1

j is monotonically decreasing sequence with limit 0.

1− j sin1j

= 1− j

∞∑

k=0

(1j

)2k+1

(−1)k

(2k + 1)!= 1− j

1

j+

∞∑

k=1

(1j

)2k+1

(−1)k

(2k + 1)!

=

= 1−

1 +

∞∑

k=1

(1j

)2k

(−1)k

(2k + 1)!

=

∞∑

k=1

(1j

)2k

(−1)k+1

(2k + 1)!≤ 1

6j2

The series converges absolutely since the term itself is a series that is dominated by 16j2 , so that by comparison test, the series

must converge.185

Exercise 18.∑∞

j=1(−1)j(1− cos 1

j

).

(cos1x

)′ =(− sin

1x

(−1x2

))=−1x2

sin1x

< 0

∑∞j=1(−1)j(1− cos 1

j ) converges since aj = (1− cos 1j ) is monotonically decreasing to 0

(1− cos1j) = 1−

∞∑

k=0

(1/j)2k(−1)k

(2k)!=

∞∑

k=1

(1/j)2k(−1)k+1

(2k)!≤ 1

2j2

So the series converges absolutely, by comparison test with∑

1j2 which converges.

Exercise 19.∑∞

j=1(−1)j arctan 12j+1 .

(arctan(

12j + 1

))′ =

1

1 +(

12j+1

)2

( −1(2j + 1)2

)(2) =

−2(2j + 1)2 + 1

< 0

∑∞j=1(−1)j arctan 1

2j+1 converges, since aj = arctan 12j+1 is monotonically decreasing to 0

11 + x2

= (arctanx)′ =∞∑

j=0

(−x2)j =∞∑

j=0

(−1)jx2j

=⇒ arctanx =∞∑

j=0

(−1)j x2j+1

2j + 1

arctan1

2j + 1=

∞∑

k=0

(−1)k(

12j+1

)2k+1

(2k + 1)=

12j + 1

+ (−1)

(1

(2j+1)3

3

)+

∞∑

k=2

(−1)k(

12j+1

)2k+1

2k + 1>

>1

2j + 1+ (−1)

13(2j + 1)3

=3(4j2 + 4j + 1) + (−1)

3(2j + 1)3=

12j2 + 12j + 23(2j + 1)3

>2j

(2j + 1)2>

29

1j

for j > 2

So by comparison test to∑

1j ,

∑arctan 1

2j+1 diverges absolutely. The series is conditionally convergent.

Exercise 20.∑∞

j=1(−1)j(

π2 − arctan log j

)

(π

2− arctan log n

)′=

( −11 + (log n)2

)(1n

)

π

2− arctan log n ≥ π

2− arctan (n− 1) → π

2− arctan n just change indices

∫ n

0

(π

2− arctanx

)dx =

π

2x−

(x arctanx− 1

2ln (1 + x2)

)∣∣∣∣n

0

=π

2n−

(n arctan n− 1

2(ln

(1 + n2

)))= n

(π

2− arctan n

)+

12

ln (1 + n2)

limn→∞

(n

(π

2− arctann

)+

12

ln (1 + n2))→∞

Then by the integral test,∑

π2 − arctan log n diverges absolutely. So the alternating series is conditionally convergent.

Exercise 21.∑∞

j=1 log(1 + 1

| sin j|)

limj→∞ log(1 + 1

| sin j|)

doesn't exist and log(1 + 1

| sin j|)

> 0 ∀j so the series diverges.186

Exercise 22.∑∞

j=2 sin(jπ + 1

log j

)

sin(

2jπ +1

log 2j

)+ sin

((2j + 1)π +

1log (2j + 1)

)=

= sin (2jπ) cos1

log 2j+ sin

(1

log 2j

)cos (2πj) + sin (2j + 1)π cos

1log (2j + 1)

+ cos (2j + 1)π sin(

1log 2j + 1

)=

= sin1

log 2j− sin

1log 2j + 1

=

=∞∑

k=0

(1

log 2j

)2k+1

(−1)k

(2k + 1)!−

∞∑

k=0

(1

log (2j+1)

)2k+1

(−1)k

(2k + 1)!

sin(

1log 2j

)− sin

(1

log 2j + 1

)=

=∞∑

k=0

(−1)k

(2k + 1)!

((1

log 2j

)2k+1

−(

12j + 1

)2k+1)

0 < log 2j < log 2j + 1 so(

1log 2j

)2k+1

−(

1log 2j + 1

)2k+1

> 0

and since for j > 1,(

1log 2j

)and

(1

log 2j + 1

)are < 1 and so we are adding smaller and smaller amounts

<1

log 2j− log 2j + 1

=

=log (2j + 1)− log 2j

log 2j log 2j + 1≤

log(1 + 1

2j

)

(log (2j))2≤

12j

(log (2j))2∫ n

1

12j(log 2j)2

= − 1log 2j

∣∣∣∣n

1

n→∞−−−−→ 1log 2

So the series converges by using integral test, showing that∑

12j(log 2j)2 converges, so by comparison test, the series con-

verges.Exercise 33.

∑n=1 nnzn

∣∣∣∣∣∑n=1

nnzn

∣∣∣∣∣ =

∣∣∣∣∣∣

∞∑

j=1

(jz)j

∣∣∣∣∣∣=

∑ (eln jz

)j

n∑

j=1

(eln jz)j =

=eln jz − (eln jz)n

1− eln jz−→∞

So z = 0

Exercise 34.∑∞

j=1(−1)jz3j

j! =∑∞

j=1(−z3)j

j! = e−z3 . C .

Exercise 35.∑∞

j=0zj

3j =∑∞

j=0

(13

)jzj

∑zj be convergent or

∑nj=1 zj bounded.

|z| < 3 and |z| = 3 if z 6= 3

Exercise 36.∑∞

j=1zj

jj {z} = C since∣∣∣∣(

z

j

)∣∣∣∣ < 1 for j ≥ N > |z|

Exercise 37.∑∞

j=1(−1)j

z+j

By Leibniz's Rule, since 1z+j

j→∞−−−→ 0, then the series converges. However, z cannot be equal to any negative integer sinceone term in the series will then blow up.

187

Exercise 38.∑∞

j=1zj√

jlog

(2j+1

j

).

zj

√j

log(

2 +1j

)= zj

log(2 + 1

j

)√

j

Since limj→∞log (2+ 1

j )√j

= 0 so that log (2+ 1j )√

jis a monotonically convergent sequence.

Then by Dirchlet's test,∑n

j=1 zj must be bounded. |z| > 1, and |z| = 1 if z 6= 1.

Exercise 39.∑∞

j=1

(1 + 1

5j+1

)j2

|z|17j =∑∞

j=1

(1 + 1

5j+1

)j2

(|z|17)j

((1 +

15j + 1

)j

|z|17)j

limj→∞

(1 +

15j + 1

)j

≤ limj→∞

(1 +

1/5j

)j

= e1/5

e1/5|z|17 < 1 =⇒ |z| < e−1/85

Exercise 40.∑∞

j=0(z−1)j

(j+2)!

∣∣∣∣∣∣

∞∑

j=0

(z − 1)j

(j + 2)!

∣∣∣∣∣∣≤

∞∑

j=0

|(z − 1)|j(j + 2)!

≤∞∑

j=0

|(z − 1)|j+2

(j + 2)!= e|z−1| − 1− |z − 1|

1!

The series converges ∀ z.Exercise 41.

∞∑

j=1

(−1)j(z − 1)j

j=

∞∑

j=1

(1− z)j

j= log (1− (1− z)) = log z

So the series converges ∀ z except for z = 0.

Exercise 42.∑∞

j=1(2z+3)j

j log (j+1)

limj→∞

1j log (j + 1)

= 0 so 1j log (j + 1)

is a monotonically convergent sequence

∑(2z + 3)j

|2z + 3| < 1

|z +32| < 1

2

then∑

(2z + 3)j converges

(1

x log (x + 1)

)′=

−1(x log (x + 1))2

(log (x + 1) +

x

x + 1

)< 0; for x > 0

By Dirichlet's test,∑∞

j=1(2z+3)j

j log (j+1) converges for |z +32| ≤ 1

2; z 6= −1 .

Exercise 43.∑∞

j=1(−1)j

(2j−1)

(1−z1+z

)j

=∑∞

j=1(−1)j/2

j− 12

(1−z1+z

)j

=⇒ ∑∞j=1

( z−1z+1 )

j

2j−1 = limj→∞ 12j−1 = 0 and 1

2j−1 is monotonically decreasing.So 1

2j−1 is a monotonically decreasing convergent sequence of real terms.For

∣∣∣ z−1z+1

∣∣∣ < 1,∣∣∣∣z − 1z + 1

∣∣∣∣ < 1 =⇒ |z − 1| < |z + 1|

(u− 1)2 + v2 < (u + 1)2 + v2

u2 − 2u + 1 < u2 + 2u + 1−u 0; z 6= 0

188

Exercise 44.∞∑

j=1

(z

2z + 1

)j

=∞∑

j=1

(12

+−1/22z + 1

)j

=∞∑

j=1

(12

)j (1− 1

2z + 1

)j

(12

)j

is a monotonically decreasing, convergent sequence

Now we want∣∣∣∣1−

12z + 1

∣∣∣∣ =∣∣∣∣2z + 1− 1

2z+

∣∣∣∣ =∣∣∣∣

2z

2z + 1

∣∣∣∣ < 1

|2z| < |2z + 1||z| < |z + 1/2|

u2 + v2 < (u + 1/2)2 + v2 = u2 + u + 1/4 + v2

−14

 −14

if <(z) =−14

2(− 1

4 + iv)

2(−1

4 + iv)

+ 1=

−12 + 2iv12 + 2iv

=⇒ z 6= −14

Exercise 45.∑∞

j=1j

j+1

(z

2z+1

)j

.(

x

x + 1

)′=

(x + 1)− x

(x + 1)2=

1(x + 1)2

> 0

j

j + 1is a monotonically increasing and convergent sequence

12

+−1

2(2z + 1)=

∣∣∣∣z

2z + 1

∣∣∣∣ < 1

∣∣∣∣2z + 1

z

∣∣∣∣ > 1 =⇒ |2 +1z| > 1

Exercise 46.∑∞

j=11

(1+|z|2)j =∑∞

j=1

(1

1+|z|2)j

11 + |z|2 < 1

1 < 1 + |z|20 < |z|2

∀z except z = 0

Exercise 47.∑∞

j=1(−1)j 2j sin2j xj

Use Dirichlet's Test. 1j is a monotonically decreasing sequence converging to zero. Consider (−2)j sin2j x. The condition

for convergence is|(−2 sin2 x)| < 1 for j ≥ N for some N

=⇒ x ∈ (−π

4+ nπ,

π

4+ nπ), n ∈ Z

Exercise 49.∑

aj converges.∑aj

1aj

diverges.Then since

∑aj is a convergent series (by Abel's test), 1

ajis a divergent sequence.

Then∑

1aj

is divergent (since limj→∞ 1aj

doesn't exist ).Exercise 50.

∑ |aj | converges.∑ |aj | converges, then∑

aj converges.|aj |2 = a2

j . |aj | converges, thenlim

j→∞|aj | = 0

limj→∞

|aj |2 = 0lim

j→∞|aj |2a2

j

= 1

By limit comparison test,∑

a2j converges.

Counterexample:∑ (

1j

)2

converges, but∑

1j diverges.

189

Exercise 51. Given∑

aj , aj ≥ 0.∑

aj converges.

∑√aj

1(j)p

limj→∞

aj = 0

limj→∞

√aj =

(lim

j→∞aj

)1/2

= 0

∫ n−1∑

j=0

xj =n−1∑

j=0

xj+1

j + 1=

n∑

j=1

xj

j=

∫1− xn

1− x

A counterexample would be√

aj

jp =√

aj

j .Exercise 52.

(1)∑

aj converges absolutely, then if∑ |aj | converges,

∑a2

j converges.

a2j

1 + a2j

= 1 +−1

1 + a2j

a2j

1 + a2j

≤ a2j since

∑a2

j converges ,∑ a2

j

1 + a2j

converges

(2)∑

aj converges absolutely, limj→∞ |aj | = 0∑ ∣∣∣ aj

1+aj

∣∣∣ =∑ |aj |

(1

|1+aj |)

.By Abel's test, since

limj→∞

1|1 + aj | =

1|1 + limj→∞ aj | = 1 shows that 1

|1 + aj | ≥ 0 is a monotonically convergent sequence

By Abel's test,∑ aj

1+ajis convergent.

10.22 Miscellaneous review exercises - Rearrangements of series. Exercise 1.(1)

aj =√

j + 1−√

j =√

j

√1 +

1j−

√j =

√j

(1 +

12

(1j

)+ o

(1j

)+−1

)=

=√

j

(12

(1j

)+ o

(1j

))

limj→∞

aj = 0

(2)

aj = (j + 1)c − jc = jc

((1 +

1j

)c

− 1)

= (jc)(

1 + c

(1j

)+ o

(1j

)− 1

)=

= jc

(c

(1j

)+ o

(1j

))=

(cjc−1 + jco

(1j

))

if c > 1, aj divergesif c = 1, lim

j→∞aj = 1

if c < 1, limj→∞

aj = 0

Exercise 2.(1)

(1 + xn)1n = exp

(1n

ln (1 + xn))

= exp

1

n

∞∑

j=1

(xn)j(−1)j−1

j

n→∞−−−−→ 1

(2) limn→∞(an + bn)1/n = limn→∞ a(1 +

(bn

)n)1/n

= a if a > b.

Exercise 3. an+1 = an+an−12 = an−1+an−2

22 + an−2+an−322

190

10.24 Exercises - Improper integrals.Exercise 1.

∫∞0

x√x4+1

dx

limx→∞

(x√

x4 + 1

)(1

1/x

)= lim

x→∞x2

x2√

1 + 1/x4= 1

Since∫∞1

1x diverges, so does

∫∞0

x√x4+1

dx

Exercise 2. ∫ ∞

−∞e−x2

dx =∫ ∞

0

e−x2dx +

∫ −∞

0

e−x2dx =

∫ ∞

0

e−x2dx +−

∫ 0

∞e−x2

dx = 2∫ ∞

0

e−x2dx

∫ ∞

0

e−x2dx ≤

∫ ∞

0

e−xdx = −e−x2∣∣∣∞

0= 1

Converges by theorem.Exercise 3.

∫∞0

1√x3+1

dx

Exercise 4.∫∞0

1√ex dx

Exercise 5.∫∞0+

e−√

x√x

dx

Exercise 6.∫ 1

0+log x√

xdx

Exercise 7.∫ 1−0+

log x1−x dx

Exercise 8.∫∞−∞

xcosh xdx

Exercise 9.∫ 1−0+

dx√x log x

Exercise 10.∫∞ 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 inte-gration, A suf�cient condition for uniform convergence, Power series. Circle of convergence. Exercise 1.

∑∞j=0

zj

2j =∑∞

j=0

(z2

)j

Using the comparison test,∣∣ z2

∣∣j ≤ tj ;∣∣ z2

∣∣ < 1 =⇒ |z| < 2

Suppose |z| = 2, z 6= 2∑∞

j=0

(z2

)j =∑∞

j=0(e2iθ)j

Now∑n

j=0(e2iθ)j ≤ 1

sin θ + 1, sincen∑

j=1

(e2iθ)j =e2iθ − e2iθ(n+1)

1− e2iθ=

einθ − e−iθneiθn+iθ

−e−iθ + eiθ=

sin nθeiθn+iθ

sin θ<

1sin θ

So∑∞

j=0

(z2

)j converges for |z| = 24, z 6= 2

Exercise 2.∑∞

j=0zj

(j+1)2j

Use ratio test .aj+1

aj=

zj+1

(j + 2)2j+1

(j + 1)2j

zj=

z

2(j + 1)(j + 2)

=z

2(1 + 1/j)(1 + 2/j)

j→∞−−−→ z

2If |z| < 2,

∑∞j=0 aj converges, if |z| > 2,

∑aj diverges.

If |z| = 2,∑

zj

(j+1)2j =∑

(e2iθ)j(

1j+1

)

Now 1j+1 is a monotonically decreasing sequence of real terms.∑(e2iθ)j is a bounded series.

By Dirichlet's test,∑

aj converges if |z| = 2, z 6= 2

Exercise 3.∑∞

j=0(z+3)j

(j+1)2j

Use ratio test:aj+1

aj=

((z + 3)j+1

(j + 2)2j+1

)(j + 1)2j

(z + 3)j=

(z + 3)2

(j + 1j + 2

)=

(z + 3)2

(1 + 1/j

1 + 2/j

)j→∞−−−→ z + 3

2191

Using Theorem 11.7,Theorem 34 (Existence of a circle of convergence).

Assume∑

ajzj

converges for at least z1 6= 0diverges for at least one z2 6= 0

∃ r > 0, such that∑

ajzj

absolutely converges for |z| < r

diverges for |z| > r

We can plug in real numbers to satisfy the condition |z+3|2 < 1 for convergence.∑

aj converges for |z + 3| < 2; diverges for |z + 3| > 2.Consider |z + 3| = 2; z 6= 1

∑aj =

∑(e2iθ)j

(1

j+1

). Since 1

j+1 is a monotonically decreasing sequence of realnumbers and

∑(e2iθ)j is a bounded series, by Dirichlet's test,∑

aj converges for |z + 3| = 2; z 6= −1.Exercise 4.

∑∞j=1

(−1)j22jz2j

2j = −∑∞j=1

(−1)j−1(2z)2j

(2j) . Look at what the terms look like.Consider using Leibniz's Rule, Theorem 10.14.

Theorem 35 (Leibniz's rule). If aj is a monotonic decreasing sequence and limj→∞ aj = 0,then

∑∞j=1(−1)j−1aj converges.

(−1)j22jz2j

2j=

(−1)j(2z)2j

2j

Consider∣∣∣∣(2z)2j

2j

∣∣∣∣ =|2z|2j

2j=

(2|z|)2j

2j

Consider 2|z| = M1/2 < ∞

=⇒ (2|z|)2j

2j=

M j

2j=

ej ln M

2j

converges for 0 < 2|z| = M ≤ 1 =⇒ |z| ≤ 12

(we use Theorem 11.6 at this point, because real numbers are included in complex numbers).Theorem 36. Assume

∑ajz

j converges for some z = z1 6= 0.

(1)∑

ajzj converges absolutely ∀z with |z| < |z1|.

(2)∑

ajzj converges uniformly on every circular disk with center at 0 and R < |z1|

We had �rst used Leibniz's test to �nd az1 on the real line.

24jz4j(4j(1− 14z2 )− 2)

4j(4j − 2)=

24jz4j((1− 14z2 )− 2

4j )

(4j − 2)j→∞−−−→ 24jz4j(1− 1

4z2 )(4j − 2)

If |z| > 12 , the series diverges (by aj th general term test).

Exercise 5.∑∞

j=1(1− (−2)j)zj .Try limit comparison test .The �rst step is to test absolute convergence �rst; it's easier.

∣∣(1− (−2)j)zj∣∣

|(−2z)j | =

∣∣∣∣∣(

1−2

)j

− 1

∣∣∣∣∣j→∞−−−→ 1

According to limit comparison test, for∑

(1− (−2)j)zj to converge,∑

(−2z)j must converge.So if |z| < 1

2 , then∑∞

j=1(1− (−2)j)zj absolutely converges.If |z| = 1

2 , z 6= −12 , ∑

(1− (−2)j)zj =∑

zj −∑

(−e2iθ)j =

= 0−∑

(e2iθ+πi)j <1

sin (θ + π)If z = −1

2 ,∑(

−12

)j

−∑

1j →∞192

Exercise 6.∑∞

j=1j!zj

jj

A very big hint is to use Exercise 19 on pp. 399, in the section for Exercises 10.14.Since

∑n−1j=1 f(j) ≤ ∫ n

1f(x)dx

n−1∑

j=1

ln j ≤∫ n

1

ln x = n ln n− n + 1 ≤n∑

j=2

ln j

(n− 1)! ≤ nne−nn ≤ n!n!nn

≥ ne−n

limn→∞

n!nn

≥ limn→∞

ne−n = 0

limn→∞

n!nn

≤ limn→∞

n2

en= 0

=⇒ limn→∞

n!nn

= 0

So then since n!nn is a monotonically decreasing convergent sequence of real terms; if

∑z is a bounded series, then by

Dirichlet's test,∑

n!zn

nn is convergent.|z| < 1; |z| = 1 if z 6= 1Try the ratio test, because it's clear from the results of the ratio test where convergence and divergence begins and ends.

(n + 1)!|z|n+1

(n + 1)n+1

nn

n!|z|n =(

n

n + 1

)n

|z| =

=(

11 + 1/n

)n

|z| n→∞−−−−→ 1e|z|

Converges for |z| < e .Now try plugging in a complexi�ed e:

ej

(jj

j!

) =j!ejei2θj

(j)j≥ jje−jjejei2θj

jjei2θj →∞

So the series diverges for |z| = e.Exercise 7.

∑∞j=0

(−1)j(z+1)j

j2+1

By the ratio test,|z + 1|j+1

j2 + 2j + 2j2 + 1|z + 1|j = |z + 1| 1 + 1/j2

1 + 2/j + 2/j2

j→∞−−−→ |z + 1|

The series absolutely converges for |z + 1| < 1.For z = 0,

∑ (−1)j

j2+1 converges since 1j2+1

j→∞−−−→ 0

For z = −2,∑ (−1)j(−1)j

j2+1 =∑

1j2+1 and 1

j2+1 < 1j2 , but

∑1j2 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 = 1

j2+1 to be the monotonicallydecreasing sequence.

=⇒ |z + 1| ≤ 1 for convergence x

Exercise 8.∑∞

j=0 aj2zj , 0 < a < 1

Use the root test .(aj2

zj)1/j = ajzj→∞−−−→ 0

So the series converges ∀ z ∈ CExercise 9.

∑∞j=1

(j!)2

(2j)!zj

193

Use the ratio test

aj+1

aj=

((j + 1)!)2zj+1

(2(j + 1))!(2j)!

(j!)2zj=

(j + 1)2z(2j + 2)(2j + 1)

j→∞−−−→ 1(z)4

for |z| < 4,∑

aj absolutely converges

for |z| > 4,∑

aj divergesLet's test the boundary for convergence.

(j!)2

(2j)!4j =

(j!)2

(2j)!ej ln 4 =

(j!)2

(2j − 2)!ej ln 4

(2j)(2j − 1)≥ (j!)2

(2j − 2)!ej ln 4

(2j)2≥

≥ (j!)2

(2j − 2)!j2j

(j!)2=

j2j

(2j − 2)!→ ∞

where we had used(n− 1)! ≤ nne−nn ≤ n! =⇒ ne−n ≤ n!

nn

n

en≤ n!

nn=⇒ nn

n!≤ en

n

Exercise 10.∑∞

j=13√

jzj

j =∑∞

j=1e√

j ln 3zj

j

e√

j+1 ln 3zj+1

j + 1j

e√

j ln 3zj=

(1

j + 1

)e(√

j+1−√j) ln 3z

√j + 1−

√j =

√1 +

1j− 1 ' 1 +

12

(1j

)− 1 =

12j

(for large j )

(for large j )(

j

j + 1

)e

12j z → 0

=⇒ Converges ∀ z ∈ C

Exercise 11.∑∞

j=1

(1·3·5...(2j−1)2·4·6...(2j)

)3

zj

aj+1

aj=

(1 · 3 · 5 . . . (2j + 1)2 · 4 · 6 . . . (2j + 2)

)3

zj+1

(2 · 4 · 6 . . . (2j)

1 · 3 · 5 . . . (2j − 1)

)3 1zj

=(

2j + 12j + 2

)3

z =(

1 +−1/2j + 2

)3

z

If |z| < 1, it converges by ratio test, if |z| = 1, then it converges by Gauss test

aj+1

aj=

(1 +

−1/2j + 2

)3

z =3∑

k=0

(3k

)(−1/2j + 2

)k

|z| =

= |z|(

1 + 3(−1/2

j + 2

)+ 3

(−1/2j + 2

)2

+(−1/2

j + 2

)3)

j→∞−−−→ |z|

diverges for |z| > 1 (by ratio test )

Exercise 12.∑∞

j=1

(1 + 1

j

)j2

zj

((1 +

1j

)j2

zj

)1/j

=(

1 +1j

)j

zj→∞−−−→ e1z

|z| < 1e,

∑(1 +

1j)j2

zj converges by root test

|z| > 1e,

∑(1 +

1j)j2

zj diverges by root test

1e

= r

Exercise 13.∑∞

j=0(sin aj)zj a > 0| sin (aj)zj | ≤ |z|jBy comparison test,

∑∞j=0(sin aj)zj converges, since

∑∞j=0 |z|j converges absolutely, for |z| < 1.

194

Note that if a = π, the series is zero.∑∞j=0(sin aj) →∞ for a = 2π so r=1indeed.

Exercise 14.∑∞

j=0(sinh aj)zj =∑∞

j=0

(eaj−e−aj

2

)zj = 1

2

(∑∞j=0(e

az)j −∑∞j=0

(zea

)j)

; a > 0

If |z| < 1ea , then

∑sinh ajzj converges. So then the radius of convergence is r = 1

ea

Exercise 15.∑∞

j=1zj

aj+bj . Assume a < b

zj

bj(1 +(

ab

)j)

(ratio test) zj+1

bj+1(1 +(

ab

)j+1)

bj

(1 +

(ab

)j)

zj

=

z

b

(1 +

(ab

)j

1 +(

ab

)j+1

)j→∞−−−→

(z

b

)

So then |z| ≶ b converges (diverges) by ratio test.If a = b,

∞∑

j=1

zj

2aj=

12

∞∑

j=1

(z

a

)j

By comparison test with xj , if |z| ≶ |a|, the series converges (diverges).

Exercise 16.∑∞

j=1

(aj

j + bj

j2

)zj Use ratio test on each of the sums, separately.

(a|z|)j+1

j + 1j

(a|z|)j=

a|z|1 + 1

j

j→∞−−−→ a|z|

=⇒ |z| < 1a

then the series converges

(b|z|)j+1

(j + 1)2j2

(b|z|)j= b|z|

(1

1 + 1j

)2

j→∞−−−→ b|z|

=⇒ |z| < 1b

So if a ≷ b, then r = a; (b)

Exercise 17.∫ 1

0fn(x) =

∫ 1

0nxe−nx2

= e−nx2

−2

∣∣∣1

0= e−n−1

−2

n→∞−−−−→ 12

However,limn→∞ nxe−nx2

= 0This example shows that the operations of integration and limit cannot always be interchanged. We need uniform conver-

gence. Exercise 18. fn(a) = sin nxn limn→∞ sin nx

n = 0

f(x) = limn→∞ fn(x)f ′n(x) = n cos nx

n =⇒ limn→∞ f ′n(0) = 1This example shows that differentiation and limit cannot always be interchanged.

Exercise 19.∑∞

j=1sin jx

j2 = f(x)

sin jx

j2≤ 1

j2∀x ∈ R

by comparison test,∞∑

j=1

| sin jx|j2

converges, so∞∑

j=1

sin jx

j2converges

∣∣∣∣sin jx

j2

∣∣∣∣ ≤1

N2∀x ∈ R; ∀j ≥ N

∑N 1

N2 converges, so∑ sin jx

j2 uniformly converges.

Then by Thm., since sin jxj2 are continuous,

∑ sin jxj2 is continuous.

195

Since∑ sin jx

j2 uniformly converges.∫ π

0

∑ sin jx

j2=

∑∫ π

0

sin jx

j2=

∑ cos jx

−j3

∣∣∣∣π

0

=∑ (−1)j − 1

−j3=

= 2∞∑

j=1

1(2j − 1)3

Exercise 20. It is known that∑∞

j=1cos jx

j2 = x2

4 − πx2 + π2

6 if 0 ≤ x ≤ 2π

(1) x = 2π

∞∑

j=1

1j2

=(2π)2

4− 2π2

2+

π2

6=

π2

6

(2) As shown in Ex. 19,∑ cos jx

j2 is uniformly convergent on R

∑ ∫cos jx

j2=

∑ (sin jx

j3

)∣∣∣∣π/2

0

=∑ (−1)j+1

(2j − 1)3∫

x2

4− πx

2+

π2

6=

(13

x3

4− πx2

4+

π2

6x

)∣∣∣∣π/2

0

=

= (π)3(

112(8)

− 116

+112

)= (π)3

132

11.13 Exercises - Properties of functions represented by real power series, The Taylor's series generated by a function,A suf�cient condition for convergence of a Taylor's series, Power-series expansions for the exponential and trigono-metric functions, Bernstein's Theorem. Suf�cient Condition for convergence.

Theorem 37 (Bernstein's Theorem). Assume ∀x ∈ [0, r], f(x), f (j)(x) ≥ 0 ∀j ∈ N.Then if 0 ≤ x < r

(25)∞∑ f (k)(0)

k!xk converges

Proof. If x = 0, we're done. Assume 0 < x < r.

f(x) =n∑

k=0

f (k)(0)k!

xk + En(x)

En(x) =xn+1

n!

∫ 1

0

unf (n+1)(x− xu)du

Fn(x) =En(x)xn+1

=1n!

∫ 1

0

unf (n+1)(x− xu)duSince f (n+1) > 0, f (n+1)(x(1− u)) ≤ f (n+1)(r(1− u))

=⇒ Fn(x) ≤ Fn(r) =⇒ En(x)xn+1

≤ En(r)rn+1

For f(x) =∑n

j=0f(j)(0)

j! xj + En(x) =⇒ En(x) ≤ (xr

)n+1En(r)

f(r) =n∑

j=0

f (j)(0)j!

rj + En(r) ≥ En(r) since f (j)(0) ≥ 0 ∀j

So then 0 ≤ En(x) ≤ (xr

)n+1f(r)

n →∞ and f(t) will be some non-in�nite value, so En(x) n→∞−−−−→ 0. ¤

Exercise 1.∑∞

j=0(−1)jx2j

Consider absolute convergence. limj→∞(x2)j = 0 If x2 < 1

196

If |x| = 1, then consider

x2(2j) − x2(2j+1) = x4j(1− x)∞∑

j=0

(1− x)x4j = (1− x)∞∑

j=0

(x4)j

Indeed∞∑

j=0

(−1)jx2j converges for |x| ≤ 1

Exercise 2.∑∞

j=0xj

3j+1 = 13

∑∞j=0

(x3

)j The series converges for |x| < 3

Exercise 3.∑

j = 0∞jxj

∫ x

0

∞∑

j=0

jtj−1 =∞∑

j=0

xj

So the series converges for |x| < 1. Note that we had used the integrability of power series.Exercise 4.

∑∞j=0(−1)jjxj .

jxjjej ln x

limj→∞

jej ln x =

{∞ if x > 10 if 0 < x < 1

If x = 1,

(2j)x2j − (2j + 1)x2j+1 = x2j(2j +−(2j + 1)x) = −1∑

(−1) = ∞

So∑

(−1)jjxj converges only for 0 ≤ x < 1, |x| < 1.

Exercise 5.∑∞

j=0(−2)j j+2j+1xj =

∑∞j=0(−1)j

(j+2j+1

)(2x)j

limj→∞

(j + 2j + 1

)(2x)j = lim

j→∞(2x)j if |2x| < 1

So when |x| < 12 , the series converges.

(2j + 22j + 1

)−

(2j + 32j + 2

)=

(2j + 2)(2j + 2)− (2j + 3)(2j + 1)(2j + 1)(2j + 2)

4j2 + 8j + 4− (4j2 + 5j + 3)4j2 + 6j + 2

=3j + 1

4j2 + 6j + 2=

(3j + 1)/22j2 + 3j + 1

=(3j + 1)/2

(2j + 1)(j + 1)3j + 1

4j2 + 6j + 2<

3j + 14j2 + 4

3

<3(j + 1/3)4(j2 + 1/3)

<34

(1j

)+

112j2

Thus, it diverges, by comparison test with 1j for x = 1

2 .

Theorem 38. Let f be represented by f(x) =∑∞

j=0 aj(x− a)j in the (a− r, a + r) interval of convergence

(1)∑∞

j=1 jaj(x− a)j−1 also has radius of convergence r.(2) f ′(x) exists ∀x ∈ (a− r, a + r) and

(26) f ′(x) =∞∑

j=1

jaj(x− a)j−1

Exercise 6.∑∞

j=1(2x)j

j =∑∞

j=1ej ln 2x

j =⇒ |x| < 12

it'll converge, since by comparison test, ej ln 2x

j < 1j2 if

0 < x < 12 .

197

Exercise 7.∑∞

j=0(−1)j

(2j+1)

(x2

)2j .∞∑

j=0

x2j

2(j + 1/2)22j<

∞∑

j=0

x2j

22j+1j=

12

∞∑

j=0

(x

2

)2j(

1j

)=

12

∞∑

j=0

e2j ln x2

j

since by comparison test, e2j ln x2

j<

1j2

if 0 < x < 2

If x = ±2 ,

12(2j) + 1

− 12(2j + 1) + 1

=1

4j + 1− 1

4j + 3=

2(4j + 1)(4j + 3)

≤

≤ 18

1j2

(converges by comparison test to∑ 1

j2)

For |x| < 2 ,∑∞

j=0(−1)j

(2j+1)

(x2

)2j converges.

Exercise 8.∑∞

j=0(−1)jx3j

j! =∑∞

j=0(−x3)j

j! = e−x3 , which converges ∀x ∈ RExercise 9.

∑∞j=0

xj

(j+3)! = 1x3

∑∞j=0

xj+3

(j+3)! = 1x3

∑∞j=+3

xj

j! = 1x3 (ex − x2/2− x− 1) Thus, it converges for ∀x ∈ R.

Exercise 10.∑∞

j=0(x−1)j

(j+2)! = 1(x−1)2

∑∞j=0

(x−1)j+2

(j+2)! = 1(x−1)2

(∑∞j=2

(x−1)j

j!

)= 1

(x−1)2

(ex−1 − (x− 1)− 1

)=

ex−1 − x

(x− 1)2

Exercise 11. ax = ex log a =∑∞

j=0(log ax)j

j!

(log ax)j+1

(j+1)!j!

(log ax)j = (log ax)j+1

j→∞−−−→ 0. By ratio test,∑∞

j=0(log ax)j

j! converges for all x.Exercise 12.

sinhx =ex − e−x

2=

12

∞∑

j=0

xj

j!−

∞∑

j=0

(−x)j

j!

=

∞∑

j=0

x2j+1

(2j + 1)!

x2j+3

(2j + 3)!(2j + 1)!

x2j+1=

x2

(2j + 3)(2j + 2)j→∞−−−→ 0

Exercise 13.

sin2 x =1− cos 2x

2=

1−∑∞j=0

(2x)2j

(2j)! (−1)j

2=

∞∑

j=1

22j−1x2j(−1)j+1

(2j)!

22j+1x2j+2

(2j + 2)!(2j)!

22j−1x2j=

4x2

(2j + 2)(2j + 1)j→∞−−−→ 0

So the series converges ∀x

Exercise 14. 11− x

2=

∑∞j=0

(x2

)j 12−x =

∑∞j=0

xj

2j+1

xj+1

2j+2

2j+1

xj=

x

2< 1 =⇒ (the series converges for |x| < 2, by ratio test)

If x = 2, the series would diverge

If x = −212

∞∑

j=0

(−2)j

2j=

12

∞∑

j=0

(−1)j = 0 but 12− (−2)

=14

x

Exercise 15. e−x2=

∑∞j=0

(−1)jx2j

j!

x2j+2

(j+1)!j!

x2j = x2

j+1

j→∞−−−→ 0

Exercise 16. sin 3x = sin 2x cos x + sin x cos 2x = 3 sin x− 4 sin3 x.

sin3 x =3 sin x− sin 3x

4=

34

∞∑

j=0

x2j+1(−1)j

(2j + 1)!−

∞∑

j=0

(3x)2j+1(−1)j

(2j + 1)!

=

=34

∞∑

j=0

(−1)jx2j+1(1− 32j+1)(2j + 1)!

=

34

∞∑

j=0

(−1)j+1(32j+1 − 1)x2j+1

(2j + 1)!

198

Exercise 17. log√

1+x1−x = 1

2 (log (1 + x)− log (1− x)) = 12

(∑j=1

(+x)j

j (−1)j−1 −∑j=1

(xj)(−1)j

j

)

ln (1 + x) =∞∑

j=0

(−x)j+1

j + 1(−1) =

∞∑

j=1

(−x)j

j(−1) − ln (1− x) =

∞∑

j=0

xj+1

j + 1=

∞∑

j=1

xj

j=⇒ −

∞∑

j=1

((−1)j + 1)xj

j

=⇒∞∑ x2j+1

2j + 1

x2j+3

2j + 32j + 1x2j+1

j→∞−−−→ x2

|x2| < 1, converges, with radius of convergence of 1

Exercise 18. 3x1+x−2x2 = 1

1−x − 11+2x =

∑∞j=0

xj

j −∑∞j=0

(−2x)j

j =∑∞

j=0xj

j (1− (−2)j)

xj+1|(1− (−2)j+1)|j + 1

j

|xj(1− (−2)j)| = x

∣∣∣∣∣∣

(1

(−2)j + 2)

(1

(−2)j − 1)

∣∣∣∣∣∣j→∞−−−→ 2x < 1

|x| < 12

For x =12

x2j

2j(1− (−2)2j) +

x2j+1

2j + 1(1− (−2)2j+1) =

x2j((2j + 1)(1− 22j) + x(1 + 22j+1)(2j))(2j)(2j + 1)

(12

)2j (2j + 1− (2j + 1)22j + 2jx + 22j+2jx)(2j)(2j + 1)

j→∞−−−→ −(2j + 1) + 4j

(2j)(2j + 1)=

2j − 1(2j)(2j + 1)

→ 0

So 3x1+x−2x2 converges for |x| ≤ 1

2

Exercise 19. 12−5x6−5x−x2 =

∑∞j=0

(1 + (−1)j

6j

)xj (|x| < 1).

12− 5x

6− 5x− x2=

5x− 12(x + 6)(x− 1)

=1

1− x+

66 + x

=∞∑

j=0

xj +∞∑

j=0

(−x

6

)j

=∞∑

j=0

xj(1 +(−1

6

)j

)

|x| < 1 since for x = 1, limj→∞(1 +(−1

6

)j) = 1.

Exercise 20. 1x2+x+1 = 2√

3

∑∞j=0 sin 2π(j+1)

3 xj (|x| < 1)

x3 − 1 = (x− 1)(x2 + x + 1)1

x2 + x + 1=

x− 1x3 − 1

=1− x

1− x3

11− x3

=∞∑

j=0

(x3)j

1− x

1− x3=

∞∑

j=0

((x3)j − x3j+1

)

By induction, it could be observed that x3j ,−x3j+1, 0 appear in sequences of 3 terms.

2√3

sin 2π(j+1)3 accounts for this.

=⇒ 2√3

∑∞j=0 sin 2π(j+1)

3 xj |x| < 1

sin 2π(j+1)3 xj < xj (so by comparison test to

∑xj , the radius of convergence is 1 )

199

Exercise 21.

11− x

=∞∑

j=0

xj ;(

11− x

)′=

1(1− x)2

=∞∑

j=1

jxj−1 =∞∑

j=0

(j + 1)xj

1(1− x)(1− x2)

=1/4

1− x+

1/2(1− x)2

+1/4

1 + x=

14

∞∑

j=0

(xj + (−x)j) + 2∞∑

j=1

jxj−1

x

(1− x)(1− x2)=

14

∞∑

j=0

xj+1(1 + (−1)j) + 2∞∑

j=1

jxj

=

14

∞∑

j=1

xj(1 + (−1)j−1 + 2j)

Exercise 22.

sin(2x +

π

4

)=

∞∑

j=0

(2x)2j+1

(2j + 1)!(−1)j ; cos 2x =

∞∑

j=0

(2x)2j

(2j)!(−1)j

sin 2x =∞∑

j=0

(2x)2j+1

(2j + 1)!(−1)j

cos 2x =∞∑

j=0

(2x)2j

(2j)!(−1)j

∞∑

j=0

ajxj =

√2

2(sin 2x + cos 2x)

For j = 98, aj =298(−1)49

98!

(√2

2

)=

−298

98!

(√2

2

)

Exercise 23.

f(x) = (2 + x2)5/2

f ′(x) =52(2 + x2)3/2(2x) = 5x(2 + x2)3/2

f ′′(x) = 5(2 + x2)3/2 +152

x(2 + x2)1/2(2x) = 5(2 + x2)3/2 + 15x2(2 + x2)1/2

f ′′′(x) =152

(2 + x2)1/2(2x) + 30x(2 + x2)1/2 +15x2

2(2 + x2)−1/2(2x)

f ′′′′(x) = 15(2 + x2)1/2 +152

(2 + x2)−1/2(2x)x + 30(2 + x2)1/2 +30x

2(2 + x2)1/2(2x)+

+ 45x2(2 + x2)−1/2 + 15x3

(−1

2

)(2 + x2)−3/2(2x)

25/2 + 0x +5(23/2)x2

2!+

0x3

3!+

454!

√2x4

Exercise 24. f(x) = e−1/x2 if x 6= 0 and let f(0) = 0(1)

f(x) =∞∑

j=0

(−1x2

)j

j!= 1 +

−1x2

+∞∑

j=2

(−1x2

)j

j!=

∞∑

j=0

(−1)jx−2j

j!

f (k) =∞∑

j=0

(−1)j (−2j)(−2j − 1) . . . (−2j − (k − 1))j!

x−2j =∞∑

j=0

(−1)j (−2j)!(−2j − k)!j!

x−2j

Use ratio test :(−2(j + 1))!

(−2j − 2− k)!(j + 1)!j!(−2j − k)!

(−2j)!x−2(j+1)

x−2j=

(−2j − k)(−2j − k − 1)(j + 1)(−2j)(−2j − 1)

x−2 j→∞−−−→ 0

Thus, by ratio test, every order of derivative exists for every x on the real line since the series representing thederivative converges for every x.

(2) f(x) =∑∞

j=0−x−2j

j! . There are no nonzero terms of positive power, i.e. no xj ; j ≥ 1.

=⇒ f (j)(0) = 0 ∀ j ≥ 1200

11.16 Exercises - Power series and differential equations, binomial series.

Exercise 1. For (1− x2)y′′ − 2xy′ + 6y = 0,

y =∞∑

j=0

ajxj

y′ =∞∑

j=1

jajxj−1

y′′ =∞∑

j=2

j(j − 1)ajxj−2 =

∞∑

j=0

(j + 2)(j + 1)aj+2xj

f(0) = 1 =⇒ a0 = 1 f ′(0) = 0 =⇒ a1 = 0

2(1)a2 + 3(2)a3x +−2(1)a1x + 6a0 + 6a1x +∞∑

j=2

((j + 2)(j + 1)aj+2 − j(j − 1)aj − 2jaj + 6aj)xj =

= 2a2 + 6 + 6a3x +∞∑

j=2

((j + 2)(j + 1)aj+2 + aj(j + 3)(j − 2))xj

=⇒a2 = −3a3 = 0

aj+2 =(j + 3)(j − 2)(j + 2)(j + 1)

aj

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′′ − 2xy′ + 12y = 0

f(0) = 0 =⇒ a0 = 0 f ′(0) = 2 =⇒ a1 = 2

2(1)a2 + 3(2)a3x +−2(1)a1x + 12a0 + 12a1x +∞∑

j=2

((j + 2)(j + 1)aj+2 − j(j − 1)aj − 2jaj + 12aj)xj =

= 2a2 + 6a3x +−4x + 0 + 24x +∞∑

j=2

((j + 2)(j + 1)aj+2 − aj(j + 4)(j − 3))xj

=⇒a2 = 0

a3 = −10/3aj+2 =

(j + 4)(j − 3)(j + 2)(j + 1)

aj

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) =∑∞

j=0x4j

(4j)! ;d4ydx4 = y

(x4)′ = 4x3

(x4)′′ = 12x2

(x4)′′′ = 24x

(x4)′′′′ = 24

(x4j

(4j)!

)′=

x4j−1

(4j − 1)!(

x4j

(4j)!

)′′=

x4j−2

(4j − 2)!(

x4j

(4j)!

)′′′=

x4j−3

(4j − 3)!(

x4j

(4j)!

)′′′′=

x4j−4

(4j − 4)!

∞∑

j=1

x4j−4

(4j − 4)!= y′′′′ =

∞∑

j=0

x4j

(4j)!= f(x)

To test convergence, use the ratio testx4j+4

(4j + 4)!(4j)!x4j

=x4

(4j + 4)(4j + 3)(4j + 2)(4j + 1)j→∞−−−→ 0 ∀x ∈ R

So the series converges on R.201

Exercise 4.f(x) =∑∞

j=0xj

(j!)2 xy′′ + y′ − y = 0

y′ =∞∑

j=1

jxj−1

(j!)2=

∞∑

j=0

(j + 1)xj

((j + 1)!)2

y′′ =∞∑

j=1

(j + 1)jxj−1

((j + 1)!)2

∞∑

j=1

((j + 1)j

((j + 1)!)2+

(j + 1)((j + 1)!)2

− 1(j!)2

)xj +

11!− 1 =

∞∑

j=1

(1

(j!)2− 1

(j!)2

)= 0

Exercise 5. f(x) = 1 +∑∞

j=11·4·7...(3j−2)

(3j)! x3j ; y′′ = xay + b (Find a and b )

f ′ =∞∑

j=1

1 · 4 · 7 . . . (3j − 2)(3j − 1)!

x3j−1

f ′′ =∞∑

j=1

1 · 4 · 7 . . . (3j − 2)(3j − 2)!

x3j−2 =∞∑

j=1

1 · 4 · 7 . . . (3j − 5)(3j − 3)!

x3j−2 =

= x +∞∑

j=2

1 · 4 · 7 . . . (3j − 5)(3j − 3)!

x3j−2 = x +∞∑

j=1

1 · 4 · 7 . . . (3j − 2)(3j)!

x3j+1

xaf = −xa +∞∑

j=1

1 · 4 · 7 . . . (3j − 2)(3j)!

x3j+a

So then a = 1; b = 0 .

1 · 4 · 7 . . . (3j + 1)(3j + 3)!

x3j+3

((3j)!

1 · 4 · 7 . . . (3j − 2)

)1

x3j=

1(3j + 1)(3j − 2)(3j + 3)(3j + 2)(3j + 1)

x3 j→∞−−−→ 0

So the series converges for all x.Exercise 6. f(x) =

∑∞j=0

x2j

j! ; y′ = 2xy.

f ′ =∞∑

j=1

2jx2j−1

j!= 2

∞∑

j=1

x2j−1

(j − 1)!= 2

∞∑

j=0

x2j+1

j!= 2xf

x2j+2

(j + 1)!j!x2j

=x2

j + 1j→∞−−−→ 0 ∀x

By ratio test, f converges ∀x ∈ R.Exercise 7. f(x) =

∑∞j=2

xj

j! y′ = x + y

f ′ =∞∑

j=2

xj−1

(j − 1)!=

∞∑

j=1

xj

j!= x + y

xj+1

(j + 1)!j!xj

=x

j

j→∞−−−→ 0

So the series converges ∀x ∈ R by ratio test.Exercise 8.

f(x) =∞∑

j=0

(−1)j(kx)2j

(2j)!

f ′ =∞∑

j=1

(−1)j(kx)2j−1k

(2j − 1)!=

∞∑

j=0

(−1)j+1(kx)2j+1k

(2j + 1)!

f ′′ =∞∑

j=1

(−1)j+1(kx)2j

(2j)!k2

f ′′ − k2f = 0

(kx)2j+1

(2j + 1)!(2j)!

(kx)2j=

kx

2k + 1j→∞−−−→ 0 by ratio test, f converges ∀x ∈ R.

202

Exercise 9.

f ′′ =∞∑

j=1

(3x)2j−1

(2j − 1)!9 =

∞∑

j=0

9(3x)2j+1

(2j + 1)!

9(f − x) = 9(x +∞∑

j=0

(3x)2j+1

(2j + 1)!− x)

(3x)2j+3

(2j + 3)!(2j + 1)!(3x)2j+1

=9x2

(2j + 3)(2j + 2)j→∞−−−→ 0

(by ratio test, f converges ∀x ∈ R )

Exercise 10. J0(x) =∑∞

j=0(−1)j x2j

(j!)222j J1(x) =∑∞

j=0(−1)j x2j+1

j!(j+1)!22j+1 .

(1)

x2j+2

((j + 1)!)222j+2

(j!)222j

x2j=

x2

(j + 1)24j→∞−−−→ 0 by ratio test, f converges ∀x ∈ R

x2j+3

(j + 2)!22j+3

j!22j+1

x2j+1=

x2

(j + 2)(j + 1)4j→∞−−−→ 0 by ratio test, f converges ∀x ∈ R

(2)

J ′0(x) =∞∑

j=1

(−1)j x2j−1

(j − 1)!(j!)22j−1=

∞∑

j=0

(−1)j+1 x2j+1

j!(j + 1)!22j+1= −J1(x)

(3)

j0(x) = xJ0(x) =∞∑

j=0

(−1)j x2j+1

(j!)222j

j1(x) = xJ1(x) =∞∑

j=0

(−1)j x2j+2

j!(j + 1)!22j+1

j′1 =∞∑

j=0

(−1)jx2j+1

(j!)222j

=⇒ j0 = j′1

Exercise 11. x2y′′ + xy′ + (x2 − n2)y = 0.n = 0 =⇒ x2y′′ + xy′ + (x2)y = 0

J0 =∞∑

j=0

(−1)j x2j

(j!)222j=

∞∑

j=1

(−1)j−1 x2j−2

((j − 1)!)222j−2;

J ′0 =∞∑

j=1

(−1)j x2j−1

j!(j − 1)!22j−1;

J ′′0 =∞∑

j=1

(−1)j x2j−2

j!(j − 1)!(2j − 1)22j−1

∞∑

j=1

(−1)j

((2j − 1)

j!(j − 1)!22j−1+

1j!(j − 1)!22j−1

+−2j

((j − 1)!)j!22j−1

)= 0

203

n = 1 =⇒ x2y′′ + xy′ + (x2 − 1)y = 0

J1(x) =∞∑

j=0

(−1)jx2j+1

j!(j + 1)!22j+1=

x

2+

∞∑

j=1

(−1)jx2j+1

j!(j + 1)!22j+1=

x

2+

∞∑

j=1

(−1)j−1x2j−1

(j − 1)!(j)!22j−1

J ′1 =12

+∞∑

j=1

(−1)j(2j + 1)x2j

j!(j + 1)!22j+1

J ′′1 =∞∑

j=1

(−1)j(2j + 1)(2j)x2j−1

(j!)(j + 1)!22j+1

x2J ′′1 + xJ ′1 + (x2 − 1)J1 =

=∞∑

j=1

x2j+1

(((−1)j(2j + 1)

(j!)(j + 1)!22j+1

)((2j) + (1)) +

(−1)j(−1)(j − 1)!j!

(j + 1j + 1

)(j

j

) (1

22j−1

)(22

22

)− (−1)j

(j + 1)!(j!)22j+1

)+

+x

2− x

2=

=∞∑

j=1

((−1)jx2j−1

(j!)(j + 1)!22j+1

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

Exercise 12. y′ = x2 + y2 ; y = 1 when x = 0.

y′(0) = 0 + 12 = 1

y = a0 + a1x + a2x2 +

∞∑

j=3

ajxj

y2 = a20 + a2

1x2 + a2

2x4 +

∞∑

j=3

ajxj

2

+

+ 2a0a1x + 2a0a2x2 + 2a0

∞∑

j=3

ajxj+

+ 2a1a2x3 + 2a1

∞∑

j=3

ajxj+1 + 2a2

∞∑

j=3

ajxj+2

y′ = a1 + 2a2x +∞∑

j=3

jajxj−1

a1 = 1 since y′(0) = 1

Consider the �rst few terms of x2 + y2

a20 + 2a0a1x + a2

1x2 + 2a0a2x

2 + x2 = a1 + 2a2x + 3a3x2 =⇒

a1 = 1 = a20 =⇒ a0 = 1

2a2 = 2a0a1 =⇒ a2 = 1

3a3 = a21 + 2a0a2 + 1 = 4 =⇒ a3 =

43

204

Exercise 13. y′ = 1 + xy2 with y = 0 when x = 0 =⇒ a0 = 0

y =∞∑

j=1

ajxj

y =∞∑

j=1

jajxj−1 =

∞∑

j=0

(j + 1)aj+1x6j

y = a1x + a2x2 + a3x

3 +∞∑

j=4

ajxj

y2 = a21x

2 + a22x

4 + a3x6 +

∞∑

j=4

ajxj

2

+

+ 2a1a2x3 + 2a1a3x

4 + 2a1

∞∑

j=4

ajxj+1+

+ 2a2a3x5 + 2a2

∞∑

j=4

ajxj+2 + 2a3

∞∑

j=4

ajxj+3

a1 = 1

x : 2a2 = 0 =⇒ a2 = 0 x2 : 3a3 = 0 =⇒ a3 = 0

x3 : 4a4 = 12 =⇒ a4 =14

x4 : 5a5 = 0 =⇒ a5 = 0 x5 : 6a6 = 0 =⇒ a6 = 0

x6 : 7a7 = 2a1a4 + 2a2a3 =⇒ a7 =114

x7 : 8a8 = 0 + 2a2a4 = 0 =⇒ a8 = 0 x8 : 9a9 = 0 =⇒ a9 = 0

x9 : 10a10 =(

14

)2

+ 2(1)114

=⇒ a10 =23

1120

Exercise 14. y′ = x + y2 y = 0 when x = 0 =⇒ a0 = 0y′(0) = 0 + 0 = 0 =⇒ a1 = 0

y =∞∑

j=2

ajxj y′ =

∞∑

j=2

jajxj−1 =

∞∑

j=1

(j + 1)aj+1xj

y2 =

a2x

2 + a3x3 + a4x

4 +∞∑

j=5

ajxj

2

=

= a2x4 + a2

3x6 + a2

4x8 +

∞∑

j=5

ajxj

2

+

+ 2a2a3x5 + 2a2a4x

6 + 2a2x2∞∑

j=5

ajxj + 2a3a4x

y + 2a3x3∞∑

j=5

ajxj + 2a4x

4∞∑

j=5

ajxj

y′ = x + y2

x : 2a + 2 = 1 + 0 =⇒ a2 =12

x2 : 3a3 = 0 =⇒ a3 = 0 x3 : 4a4 = 0 =⇒ a4 = 0

x4 : 5a5 = a22 =⇒ a5 =

120

x5 : 6a6 = 0 =⇒ a6 = 0x6 : 7a7 = 0 =⇒ a7 = 0

x7 : 8a8 = 2(12)(

120

) =⇒ a8 =1

160

x8 : 9a9 = 0 =⇒ a9 = 0 x9 : 10a10 = 0 =⇒ a10 = 0

x10 : 11a11 = 2(12)(

1160

) +(

120

)2

=⇒ a11 =7

8800205

Exercise 15. y′ = αy=⇒ ∑∞

j=0(j + 1)aj+1xj = α

∑∞j=0 ajx

j

aj+1 = αaj

(j+1)

aj =αj

j!a0 x (by induction)

Exercise 16. y′′ = xy

y′′ =∞∑

j=0

(j + 2)(j + 1)aj+2xj =

= 2a2 +∞∑

j=0

(j + 3)(j + 2)aj+3xj+1

=∞∑

j=0

ajxj+1

=⇒ a2 = 0 and aj+3 =aj

(j + 3)(j + 2)

j = 0 a + 3 =a0

3 · 2j = 3 a6 =

a3

6 · 5

j = 1 a4 =a1

4 · 3j = 4 a7 =

a1

7 · 6 · 4 · 3

a3j =a0

(3j)!

j−1∏

k=0

(3k + 1) ; a3j+1 =a1

(3j + 1)!

j−1∏

k=0

(3k + 2)

Exercise 17. y′′ + xy′ + y = 0

y =∞∑

j=0

ajxj

y′ =∞∑

j=1

jajxj−1

y′′ =∞∑

j=2

j(j − 1)ajxj−2

=∞∑

j=0

(j + 2)(j + 1)aj+2xj

y′′ + xy′ + y =∞∑

j=1

xj((j + 2)(j + 1)aj+2 + jaj + aj) = 0 =⇒ aj+2 =−aj

(j + 2)

a2 =−a0

2

a4 =−a2

4

a3 =−a1

3

a5 =−a3

5=

a1

15

a2j =(−1)ja0

(2j)!!a2j+1 =

a1

(2j + 1)!!(−1)j (could be shown by induction)

Exercise 18. Recall that

y =∞∑

j=0

ajxj

y′ =∞∑

j=1

jajxj−1

=∞∑

j=0

(j + 1)aj+1xj

Knowing this, we could cleverly observe that e−2x =∑∞

j=0(2aj + (j + 1)aj+1)xj is actually a differential equation!!!

=⇒ e−2x = y′ + 2y

Solving this ODE using y(x) = e−A(x)(∫ x

aQ(t)eA(t)dt + y(a)

)where A(x) =

∫ x

aP (t)dt,

y = e−2x(x + 1)206

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

Exercise 19. cos x =∑∞

j=0 aj(j + 2)xj for f(x) =∑∞

j=0 ajxj .

Using cosx =∑∞

j=0(x)2j

(2j)! (−1)j representation, we can immediately conclude that for odd terms, a2j+1 = 0 and bymatching powers of x,

a2j(2j + 2) = (−1)j 1(2j)!

a5 = 0

a6(6 + 2) =(−1)3

6!=⇒ a6 =

−78!

Now notice that for cosx =∑∞

j=0 aj(j + 2)xj =∑∞

j=1 jajxj + 2

∑∞j=0 ajx

j is actually a differential equation, cosx =xy′ + 2y. We can solve this �rst-order ODE usingy(x) = e−A(x)

(∫ x

aQ(t)eA(t)dt + y(a)

)where A(x) =

∫ x

aP (t)dt. Then solving y′ + 2y

x = cos xx ,

y =1x2

(x sin x + cosx− (a sin a + cos a) + b)

Plugging 0 as a good guess back into the ODE, cos 0 = 1 = y(0)(2) =⇒ y(0) = 12 With this initial condition, we get

f(x) =sin x

x+

cosx− 1x2

if x 6= 0

So f(0) = 12 and f(π) = −2

π2

Exercise 20.(1)

(1− x)−1/2 =∞∑

j=0

(−1/2j

)(−x)j =

= 1 +12x +

(−12

) (−32

)

2x2 +

(−12

) (−32

) (−52

)

3!x3 +

(−12

) (−32

) (−52

) (−72

)

4!x4+

+

(−12

) (−32

) (−52

) (−72

) (−92

)

5!x5 + · · · =

= 1 +12x +

38x2 +

58x3 +

35128

x5 +63256

x5 + . . .

(2) To make the notation clear, (1− x)−1/2 =∑∞

j=0

(−1/2j

)(−x)j =

∑∞j=0 bjx

j =∑∞

j=0 aj

Now (α

j+1

)(αj

) =α(α− 1) . . . (α− (j + 1) + 1)

(j + 1)!j!

α(α− 1) . . . (α− j + 1)=

(α− j)(j + 1)

So for α = −12 ,

aj+1

aj= −

(1/2 + j

j + 1

)(−x) < x

Using this, we further �nd thatbj+1 < bj

150

bj+2 < bj+1150

< bj

(150

)2

For x = 150 . So by induction, bn+j < bn

(150

)j

rn =∞∑

j=1

an+j <

∞∑

j=1

an

(150

)j

= an1/50

1− 1/50=

an

49

rn <an

49

207

(3) Note that (1− x)−1/2 =(1− 1

50

)−1/2 =(

4950

)−1/2 = 5√

27

75

(1− 1

50

)−1/2

= 1 +1

100+

32

(1

2(50)

)2

+52

(1

2(50)

)3

+358

(1

2(50)

)4

+638

(1

2(50)

)5

√2 ' 1.4142135624

Exercise 21.(1) 1732

1000

(1− 176

3000000

)−1/2 = 17321000

(30000002999824

)1/2

Obviously, (3000000)1/2 = 1000√

3 so that we have 1732 (3/2999824)1/2.With long multiplication, we could show easily that 1732 ∗ 1732 = 2999824 (it's harder to divide). So then

17321000

(1− 176

3000000

)−1/2

=√

3

(2)

Exercise 22. arcsin x =∫

1√1−x2

(1− x2)−1/2 =∑∞

j=0

(αj

)(−x2)j = 1 +

∑∞j=1

(αj

)(−x2)j

=⇒ arcsin x = x +∞∑

j=1

(−1/2j

)(−1)j

(2j + 1)x2j+1

(−12

) (−32

). . .

(−12 − j + 1

)

j(j − 1) . . . (2)(1)= (−1)j (1)(3) . . . (1 + 2j − 2)

(2j)!!= (−1)j (2j − 1)!!

(2j)!!

=⇒ arcsin x = x +∞∑

j=1

(2j − 1)!!(2j)!!

x2j+1

2j + 1

12.4 Exercises - Historical introduction, The vector space of n-tuples of real numbers, Geometric interpretation forn ≤ 3.Exercise 1.A =(1, 3, 6)

B =(4,−3, 3)

C =(2, 1, 5)

(1) (5, 0, 9) = A + B(2) (−3, 6, 3)(3) (3,−1, 4)(4) 7A− 2B − 3C = (7, 21, 42)− (8,−6, 6)− (6, 3, 15) = (−1, 26, 36)− (6, 3, 15) = (−7, 24, 21)

Exercise 2. See sketches.Exercise 3. See sketches.Exercise 4.

(1) See the sketch.(2) I think it'll be a line.(3) It'll �ll a parallelogram with A and B as the sides.(4) It'll �ll the entire real, 2-dim. plane.

Exercise 5.A = (2, 1)

B = (1, 3)

C = (c1, c2)

xA + yB = (2x, x) + (y, 3y) = (2x + y, x + 3y) = (c1, c2)x =

c2 − 3c1

−5

y =c1 − 2c2

−5208

Exercise 6.A =(1, 1, 1)

B =(0, 1, 1)

C =(1, 1, 0)D = xA + yB + zC

(1) D = (x + z, x + y + z, x + y)(2) 0 = x + z = x + y + z =⇒ y = 0

x = 0 z = 0(3) D = (1, 2, 3)

x + y + z = 2 = 3 + z z = −1x− 1 = 1 x = 2 y = 1

Exercise 7.A =(1, 1, 1)

B =(0, 1, 1)

C =(2, 1, 1)D = xA + yB + zC

(1) D = (x + 2z, x + y + z, x + y + z)(2) x = 2, z = −1, y = −1 and D = 0(3) x + y + z = 2 but x + y + z = 3

Exercise 8.A =(1, 1, 1, 0)

B =(0, 1, 1, 1)

C =(1, 1, 0, 0)D = xA + yB + zC

(1) D = (x + z, x + y + z, x + y, y)(2) y = 0 =⇒ x = 0 =⇒ z = 0(3) (1, 5, 3, 4) = D =⇒ y = 4, x = −1, z = 2(4) (1, 2, 3, 4) = D =⇒ y = 4, x = −1, z = −1 but x + z = 1

Exercise 9.A,B parallel =⇒ A = tB t 6= 0C, B parallel =⇒ C = sB s 6= 0

A = tB = tCs and t

s = k 6= 0. So A, C are parallel.Exercise 10. GivenC = A + B

A = sD

If C ‖ D so that C = tDC −A = (t− s)D = B. If t− s 6= 0 then B ‖ C. Even if t = s, then that meant that C = A and B = 0 anyways.If B ‖ D, so that B = uDc = (s + u)D =⇒ C ‖ D

Exercise 11.(1) Theorem 12.1 says that vector addition is commutative and associative.

Let A,B ∈ Vn; A = (a1, . . . , an), B = (b1, . . . , bn)Use this de�nition,

De�nition 1.A = B iff a1 = b1 . . . an = bn

A + B = (a1 + b1, . . . , an + bn)cA = (ca1, . . . , can)

A + B = (a1 + b1, . . . , an + bn) = (b1 + a1, . . . , bn + an) = B + A

A + (B + C) = (a1 + (b1 + c1), . . . , an + (bn + cn)) = ((a1 + b1) + c1, . . . , (an + bn) + cn) =

= (A + B) + C

c(dA) = (c(da1), . . . , c(dan)) = ((cd)a1, . . . , (cd)an) = (cd)A

c(A + B) = (c(a1 + b1), . . . , c(an + bn)) = (ca1 + cb1, . . . , can + cbn) = cA + cB

(c + d)A = ((c + d)a1, . . . , (c + d)an) = (ca1 + da1, . . . , can + dan) = cA + dA209

So commutativity and associativity follows from the associativity, commutativity, and distributivity of the one-dim. reals (or even complex numbers).

(2) See sketch.Exercise 12.

A =B − C

C =B −A

B =A + C

A +12C − 1

2A =

=12A +

12C =

12(A + C)

=12B

It reminds me of how the diagonals of a parallelogram bisect each other.

12.8 Exercises - The dot product, Length or norm of a vector, Orthogonality of vectors.Exercise 1.A = (1, 2, 3, 4)

B = (−1, 2,−3, 0)

C = (0, 1, 0, 1)(1) A ·B = −6(2) B · C = 2(3) A · C = 6(4) A · (B + C) = A · (−1, 3,−3, 1) = 0(5) (A−B) · C = (2, 0, 6, 4) · C = 4

Exercise 2.A = (2, 4,−7)

B = (2, 6, 3)

C = (3, 4,−5)(1) (A ·B)C = 7(3, 4,−5) = (21, 28,−35)(2) A · (B + C) = A · (5, 10,−2) = 64(3) (A + B) · C = (4, 10,−4) · C = 72(4) A(B · C) = (30, 60,−105)(5) A/(B · C) =

(215 , 4

15 , −715

)

Exercise 3. Given A ·B = A · C, A 6= 0,Then ‖A‖‖B‖ cos θAB = ‖A‖‖C‖ cos θAC

Let ‖B‖ = ‖C‖ but θAB = −θAC . Then B 6= C.Exercise 4. Given A ·B = 0 ∀A,‖A‖‖B‖ cos θAB = 0, choose ‖A‖ 6= 0, cos θAB 6= 0‖B‖ = 0Exercise 5.A = (2, 1,−1)

B = (1,−1, 2); �nd a nonzero C in V3 s.t. A · C = B · C = 0

C = (c1, c2, c3) without loss of generality, let c1 = 12 + c2 − c3 = 01 +−c2 + 2c3 = 0

=⇒c2 = −5c3 = −3

Exercise 6.A = (1,−2, 3)

B = (3, 1, 2)C = xA + yB

B ·B = 14C ·B = 0

C ·B = (xA + yB) ·B = x(A ·B) + yB ·B = 0

x(7) + y(14) = 0 =⇒ x + 2y = 0

y = −1, x = 2 C = (−1,−5, 4)Exercise 7.

GivenA = (2,−1, 2)

B = (1, 2,−2)

A = C + D

B ·D = 0C ‖ B B = tC

tC ·D = 0 t 6= 0A ·B = B · C = −4

210

B ·B = tB · C = 9 =⇒ t = 9/− 4C = −4

9 (1, 2,−2)D = A− C = (2,−1, 2)− (

49 , 8

9 , −89

)=

(229 , −1

9 , 109

)

Exercise 8. Given that ifA = (1, 2, 3, 4, 5)

B = (1, 1/2, 1/3, 1/4, 1/5)C, D ∈ V5

B = C + 2D D ·A = 0, C ‖ A; C = tA=⇒ B = tA + 2D

A ·B = 5 = tA ·A + 2A ·D = t(55) t = 1/11C = 1

11 (1, 2, 3, 4, 5)D = 1

2

(B − A

11

)= 1

2

((1, 1/2, 1/3, 1/4, 1/5)− (

111 , 2

11 , 311 , 4

11 , 511

))D =

(511 , 7

44 , 133 , −5

88 , −755

)

Exercise 9.

A = (2,−1, 5)

B = (−1,−2, 3)

C = (1,−1, 1)

‖A + B‖ = ‖(1,−3, 8)‖ =√

74

‖A−B‖ = ‖(3, 1, 2)‖ =√

14

‖A + B − C‖ = ‖(0,−2, 7)‖ =√

53

‖A−B + C‖ = ‖(4, 0, 3)‖ = 5

Exercise 10. B ·A = 0; ‖B‖ = ‖A‖(1) B = (1,−1)(2) A = (1,−1), B = (1, 1)(3) A = (2,−3), B = (3, 2)(4) A = (a, b), B = (b,−a)

Exercise 11.A = (1,−2, 3)

B = (3, 1, 2)

(1) A + B = (4,−1, 5) =⇒ C =(4,−1, 5)√

42

(2) A−B = (−2,−3, 1) C =(−2,−3, 1)√

14

(3) A + 2B = (7, 0, 7) =⇒ C = (1/√

2, 0, 1/√

2)

(4) 2A−B = (−1,−5, 4) C =(−1,−5, 4)√

42

Exercise 12.A = (4, 1,−3)

B = (1, 2, 2)

C = (1, 2,−2)

D = (2, 1, 2)

E = (2,−2, 1)

A ·B = 0C ·D = 0C · E = 0D · E = 0

Exercise 13. The answer is easy to see geometrically. If you do the algebra, the condition of ‖A‖ = ‖B‖ �xes the length ofthe vector and A ·B orthogonality �xes the direction, up to π or 180◦.

(1) A = (1, 2) =⇒ B = (−2, 1), (2,−1)

(2) A = (1,−2) =⇒ B = (2, 1), (−2,−1)

(3) A = (2,−1) =⇒ B = (1, 2), (−1,−2)

(4) A = (−2, 1) =⇒ B = (1, 2), (−1,−2)

Exercise 14.A = (2,−1, 1)

B = (3,−4,−4)

B −A = (1,−3,−5)

A ·B = 6

A + B = (5,−5,−3)

211

We have the condition that we have one right angle for a right triangle: (A−C) · (C −B) = A ·C −C2 −A ·B + B ·C =(A + B) · C − C2 − 6 = 0(5,−5,−3) · C = 6 + C2

5c1 − 5c2 + 0 = 6 + c21 + c2

2

c1=2−−−→ 10− 5c2 = 10 + c22

=⇒ c2 = −5

We just need to �nd one C. Let c3 = 0 for C = (c1, c2, c3). Then we get the above last two statements if we also choosec1 = 2.So we have C = (2,−5, 0)

So in summary, we getB −A = (1,−3,−5)

A− C = (0, 4, 1)

C −B = (−1,−1, 4)and indeed

‖A− C‖ =√

17

‖C −B‖ =√

18

‖B −A‖ =√

35

and (A− C) · (C −B) = 0, as a right triangle should.

Exercise 15.A = (1,−1, 2)

B = (2, 1,−1)=⇒ C = (1,−5,−3)

Exercise 16. Given thatA = (1, 2)

B = (3, 4)

A = P + Q

P ‖ B =⇒ P = sB

Q ⊥ B

=⇒ P =1125

(3, 4) , Q = A− P =(−8

25,

625

)

Exercise 17. Given A = (1, 2, 3, 4), B = (1, 1, 1, 1), then for A = P + Q, P ‖ B, Q ·B = 0,

P =52(1, 1, 1, 1) , Q = A− P =

(−32

,−12

,12,32

)

Exercise 18. GivenA = (2,−1, 1)

B = (1, 2,−1)

C = (1, 1,−2)

D = xB + yC

A ·D = 0

‖D‖ = 1

then D = ± 1√2B +∓ 1√

2C

Exercise 19. ‖A + B‖2 − ‖A−B‖2 = A2 + 2A ·B + B2 − (A2 − 2A ·B + B2) = 4A ·BExercise 20. ‖A + B‖2 + ‖A−B‖2 = A2 + 2A ·B + B2 + (A2 − 2A ·B + B2) = 2A2 + 2B2

The sum of the square of the sides of a parallelogram is equal to the sum of the square of the diagonals of a paralellogram.Exercise 21.�The sum of the squares of the sides of any quadrilateral exceeds the sum of the squares of the diagonals by four times thesquare of the length of the line segment which connects the midpoints of the diagonals.�

Sum of the squares of the sides = A2 + (C −A)2 + (C −B)2 + B2 =

= A2 + C2 − 2AC + A2 + C2 − 2CB + B2 + B2

sum of the squares of the diagonals = C2 + (A−B)2 =

= C2 + A2 − 2AB + B2

four times the square of the length of the line segment which connects the midpoints of the diagonals =

= 4(

C

2−

(A + B

2

))2

=

C2 − 2C(A + B) + A2 + 2AB + B2

So then if we take the sum of the squares of the sides - sum of the squares of the diagonals == four times the square of the length of the line segment which connects the midpoints of the diagonals, then we get the de-sired equality.Exercise 22. Given that ‖A‖ = 6 and for any pair of scalars, x, y,(xA + yB) · (4yA− 9xB) = 4xyA2 − 9x2A ·B + 4y2A ·B − 9xyB2 = 0 =⇒ 144xy + (4y2 − 9x2)A ·B = 9xyB2

212

x = 2, y = 3 without loss of generality.144xy = 9xyB2 =⇒ 16 = B2, ‖B‖ = 4

This implies that A ·B in order for the above statements to be true ∀ x, y‖(2A + 3B)‖2 = 4A2 + 12A ·B + 9B2 = 144 + 144 = (12

√2)2 =⇒ ‖2A + 3B‖ = 12

√2

Exercise 23. Given thatA = (1, 2, 3, 4, 5)

B = (1, 1/2, 1/3, 1/4, 1/5)

B = C + D

D ⊥ A

C ‖ A or C = tA

B ·A = 5 = C ·A + D ·A = tA2 = 55t

C =111

(1, 2, 3, 4, 5)

D = B − C = (1, 1/2, 1/3, 1/4, 1/5)− 111 (1, 2, 3, 4, 5) =

(1011

,722

,233

,−544

,−1455

)

Exercise 24. Given thatA ·B 6= 0

C ‖ A C = tA

D ⊥ A so that A ·D = 0B = C + D

since ‖A‖neq0, if t 6= 0, then C exists.

A ·B = C ·A + D ·A = tA2 =⇒ t = A·BA2 C =

A ·BA2

A

D = B − C = B − A ·BA2

A

Exercise 25.(1) Given that A ·B = 0, ‖A + xB‖2 = A2 + 2xA ·B + x2B2 = A2 + x2B2 ≥ A2 since x2B2 is positive.(2) Given that ‖A + xB‖ ≥ ‖A‖, then A2 + 2xA ·B + x2B2 ≥ A2.

x(2A ·B + xB2) ≥ 0If x > 0

2A ·B + xB2 ≥ 0

xB2 ≥ −2A ·Bthen A ·B ≥ 0

If x < 0

2A ·B + xB2 ≤ 0

xB2 ≤ −2A ·Bthen A ·B ≤ 0

=⇒ A ·B = 0We relied heavily upon the fact that x was any number, so we could make x as small or as big as we want.

12.11 Exercises - Projections. Angle between vectors in n-space, The unit coordinate vectors.

Exercise 1. GivenA = (1, 2, 3)

B = (1, 2, 2)

A ·B = 11B ·B = 9

So the projection of A on B is A ·BB ·B B =

119

(1, 2, 2)

Exercise 2. GivenA = (4, 3, 2, 1)

B = (1, 1, 1, 1)

B ·B = 4A ·B = 10

So the projection of A on B is 52(1, 1, 1, 1)

Exercise 3.

(1) A = (6, 3,−2).

cos a =67

cos b =37

cos c =−27

213

(2) 17 (6, 3,−2)


B = (2, 1,−1)

C = (1, 4, 1)

D = (2, 5, 5)

|A| =√

6

|B| =√

6

|C| =√

18 = 3√

2

|D| =√

54 = 3√

6

A ·B|A||B| =

12

=⇒ θAB =π

6

C ·D|C||B| =

√3

2=⇒ θCD =

π

3Exercise 5. If we letA = (2,−1, 1)

B = (1,−3,−5)

C = (3,−4,−4)

B −A = x1 = (−1,−2,−6) |x1| =√

41

C −B = x2 = (2,−1, 1) |x2| =√

6

A− C = x3 = (−1, 3, 5) |x3| =√

35

x1 · x2 = −6x1 · x3 = −35x2 · x3 = 0

θ12 = −6/(√

246)

θ13 = −35/(√

1435)θ23 = 0

Exercise 6. Given that ‖A‖ = ‖C‖ = 5, ‖B‖ = 1, ‖A−B + C‖ = ‖A + B + C‖, and A·B|A||B| = cos π

8 , then consider

(A−B + C) · (A−B + C) = (A + B + C) · (A + B + C)

((A + C)−B)2 = (A + C + B)2 =⇒ −2(A + C) ·B = 2(A + C) ·B0 = A ·B + B · C =⇒ −A ·B = B · C

=⇒ −5(1) cos π8

(1)5= cos θBC =⇒ θBC =

9π

8

Exercise 7. Given that A·C|A||C| = cos θAC = B·C

|B||C| = cos θBC , then A·C|A| = B·C

|B| or |B|A · C = |A|B · C.So C · (|B|A− |A|B) = |B|(A · C)− |A|B · C = 0Exercise 8. Given that

A = (1, 1, . . . , 1)

B = (1, 2, . . . , n)then

|A|2 = n

|B|2 =n∑

j=1

j2

A ·B =n∑

j=1

j =n(n + 1)

2

To �nd∑n

j=1 j2, recall this trick:

S =n∑

j=1

j3 =n−1∑

j=0

(j + 1)3 =n−1∑

j=0

j3 + 3j2 + 3j + 1 =

= S − n3 + 3n−1∑

j=1

j2 + 3(

(n− 1)n2

)+ n

=⇒n∑

j=1

j2 =(n + 1)(2n + 1)n

6

So now we could get a remarkable limit:

cos θAB =A ·B|A||B| =

n(n+1)2

√n√

(n+1)(2n+1)n6

=√

32

1 + 1/n√(1 + 1/n)(1 + 1/2n)

n→∞−−−−→√

32

or θAB = π/3

Exercise 9. Given that

A = (2, 4, 6, . . . , 2n)

B = (1, 3, 5, . . . , 2n− 1)then

|A|2 =n∑

j=1

(2j)2

|B|2 =n∑

j=1

(2j − 1)2

A ·B =n∑

j=1

2j(2j − 1)

214

Doing the math,

|A|2 =∑

4j2 = 4n∑

j=1

j2 =4(n + 1)(2n + 1)n

6

|B|2 =n∑

j=1

(2j − 1)2 =n∑

j=1

(4j2 − 4j + 1) = 4(n + 1)(2n + 1)n

6− 4

n(n + 1)2

+ n =

= n(4n2 + 6n + 2− 6n− 6 + 3)

3=

n(4n2 − 1)3

A ·B =n∑

j=1

2j(2j − 1) =n∑

j=1

4j2 − 2j = 4(n + 1)(2n + 1)n

6− 2

n(n + 1)2

=

=(2(2n2 + 3n + 1)− 3n− 3)n

3=

(4n2 + 3n− 1)n3

A ·B|A||B| =

n(4n2 + 3n− 1)/3√2(2n2+3n+1)n

3

(n(4n2−1)

3

) =4 + 3/n− 1/n2

√2(2 + 3/n + 1/n)(4− 1/n2)

n→∞−−−−→ 4√2(2)4

= 1

It seems to make sense that θABn→∞−−−−→ 0, as the two vector getting closer and closer.

Exercise 10.

(1)A = (cθ,−sθ)

B = (sθ, cθ)c2 + s2 = 1

A ·B = cs− cs = 0(2) (x, y) ∈ V2,

(x, y) = xA+yB = x(C,−S)+y(S, C) = (xC+yS,−xS+yC) =⇒x = xC + yS

y = −xS + yC=⇒

x(1− C) = yS

y(1− C) = −xSIf θ 6= 2πj and y 6= 0, then

xy(1− C) = y2S

x(−xS) = y2S

−x2S = y2S

If θ 6= πj, then contradiction. If θ = (2j − 1)π, 2x = 0, y = 0. If θ = 2jπ, then ∀x, y ∈ R

Exercise 11. For a rhombus, with adjacent sides A,B, |A| = |B|. The diagonals are given by A + B and A−B.(A + B) · (A−B) = A2 −B2 = 0

Exercise 12. (cos a, sin a) · (cos b, sin b) = cos a cos b + sin a sin b. But (cos a, sin a) · (cos b, sin b) = cos a− b/(1)(1)because the angle between the two vectors is a− b.

Exercise 13.

‖A−B‖2 = (A−B) · (A−B) = A2 + B2 − 2‖A‖‖B‖ cos theta

Exercise 14. Given A ·B =∑n

k=1 |akbk|,

B ·A =n∑

j=1

|bjaj | =n∑

j=1

|ajbj | = A ·B

A · (cB) =n∑

j=1

|ajcbj | =n∑

j=1

|cajbj | = (cA) ·B = |c|n∑

j=1

|ajbj | = |c|(A ·B)

A ·A =n∑

j=1

|aj |2 ≥ 0, A ·A = 0 only if A = 0

215

To prove the Cauchy-Schwarz inequality for this metric, recall the proof for Thm. 1.41.n∑

j=1

(|aj |x + |bj |)2 =n∑

j=1

|aj |2x2 + 2|aj ||bj |x + |bj |2 = x2n∑

j=1

|aj |2 + 2x

n∑

j=1

|ajbj |+n∑

j=1

|bj |2 ≥ 0

x2A′ + 2xB′ + C ′ ≥ 0 where

A′ =n∑

j=1

|aj |2 = A ·A

B′ =n∑

j=1

|ajbj | = A ·B

C ′ =n∑

j=1

|bj |2 = B ·B

=⇒ A′(

x +B′

A′

)2

+A′C ′ −B′2

A′≥ 0

Let x =−B′

A′

=⇒ A′C ′ ≥ B′2 orn∑

j=1

|aj |2n∑

j=1

|b2j | ≥ (

n∑

j=1

|ajbj |)2

Exercise 15. Given A ·B = 2a1b1 + a2b2 + a1b2 + a2b1

B ·A = 2b1a1 + b2a2 + b1a2 + b2a1 = A ·B = 2a1b1 + a2b2 + a1b2 + a2b1

A ·B + A · C = 2a1b1 + a2b2 + a1b2 + a2b1 + 2a1c1 + a2c2 + a1c2 + a2c1 =

= 2a1(b1 + c1) + a2(b2 + c2) + a1(b2 + c2) + a2(b1 + c1) = A · (B + C)

(cA) ·B = 2(ca1)b1 + ca2b2 + ca1b2 + ca2b1 = 2a1(cb1) + a2(cb2) + a1(cb2) + a2(cb1) =

= c(2a1b1 + a2b2 + a1b2 + a2b1) = A · (cB) = c(A ·B)

A ·A = 2a21 + a2

2 + a1a2 + a2a1 = a21 + (a1 + a2)2 ≥ 0, A ·A = 0 only if

a1 = 0a1 + a2 = 0

=⇒ A = 0

For the Cauchy-Schwarz inequality,

A2B2 = (a21 + (a1 + a2)2)(b2

1 + (b1 + b2)2)

(A ·B)2 = (a1b1 + a2(b1 + b2) + a1(b1 + b2))2 = (a1b1 + (a1 + a2)(b1 + b2))2 =

= (a1b1)2 + 2a1b1(a1 + a2)(b1 + b2) + (a1 + a2)2(b1 + b2)2

Now (a1(b1 + b2)− b1(a1 + a2))2 ≥ 0

a21(b1 + b2)2 − 2a1b1(b1 + b2)(a1 + a2) + b2

1(a1 + a2)2 ≥ 0

This fact shows that A2B2 ≥ (A ·B)2

Exercise 16. Given A ·B = 2a1b1 + a2b2 + a3b3 + a1b3 + a3b1

B ·A = 2b1a1 + b2a2 + b3a3 + b1a3 + b3a1 = A ·B = 2a1b1 + a2b2 + a3b3 + a3b1 + a1b3

A ·B + A · C = 2a1b1 + a2b2 + a3b3 + a1b3 + a3b1 + 2a1c1 + a2c2 + a3c3 + a1c3 + a3c1 =

= 2a1(b1 + c1) + a2(b2 + c2) + a3(b3 + c3) + a1(b3 + c3) + a3(b1 + c1) = A · (B + C)

(cA) ·B = 2(ca1)b1 + (ca2)b2 + (ca3)b3 + (ca1)b3 + (ca3)b1 = 2a1(cb1) + a2(cb2) + a3(cb3) + a1(cb3) + a3(cb1) =

= c(2a1b1 + a2b2 + a3b3 + a1b3 + a3b1) = A · (cB) = c(A ·B)

A ·A = 2a21 + a2

2 + a23 + a1a3 + a3a1 = a2

1 + a22 + (a1 + a3)2 ≥ 0, A ·A = 0 only if

a1 = 0a2 = 0a1 + a3 = 0

=⇒ A = 0

For the Cauchy-Schwarz inequality,

A2B2 = (a21 + a2

2 + (a1 + a3)2)(b21 + b2

2 + (b1 + b3)2)

(A ·B)2 = (a1b1 + a2b2 + a3(b1 + b3) + a1(b1 + b3))2 = (a1b1 + a2b2 + (a1 + a3)(b1 + b3))2 =

= (a1b1)2 + a22b

22 + 2a1b1a2b2 + 2(a1b1 + a2b2)(a1 + a3)(b1 + b3) + (a1 + a3)2(b1 + b3)2

216

As you could see, a direct proof has messy algebra. Instead, since this dot product satis�es all the usual dot product properties,then we can just reuse the vector method proof of the Cauchy-Schwarz inequality, without reference to a speci�c coordinatesystem, to conclude that the Cauchy-Schwarz inequality holds. That's the trick - think in terms of generality, of vectors ingeneral, and not speci�c coordinate systems.Exercise 17. Given that ‖A‖ =

∑nj=1 |aj |

(1) ∑|aj | > 0,

∑|aj | = 0 if aj = 0 ∀ j

‖cA‖ =∑

|caj | = |c|∑

|aj | = |c|‖A‖‖A + B‖ =

∑|aj + bj | <

∑|aj |+ |bj | = ‖A‖+ ‖B‖

(2) See sketch. Looks like a diamond.(3) Given ‖A‖ =

∣∣∣∑nj=1 aj

∣∣∣‖A‖ > 0 if A 6= 0 but ‖A‖ = 0 for aj = (−1)j and n = 2m

‖cA‖ =

∣∣∣∣∣∣

n∑

j=1

caj

∣∣∣∣∣∣= |c|

∣∣∣∣∣∣

n∑

j=1

aj

∣∣∣∣∣∣= |c|‖A‖

‖A + B‖ =

∣∣∣∣∣∣

n∑

j=1

(aj + bj)

∣∣∣∣∣∣≤

∣∣∣∣∣∣

n∑

j=1

aj

∣∣∣∣∣∣+

∣∣∣∣∣∣

n∑

j=1

bj

∣∣∣∣∣∣= ‖A‖+ ‖B‖

Exercise 18. Given that ‖A‖ = max1≤j≤n |aj |(1) ‖A‖ ≥ 0 since |aj | ≥ 0 for any aj ∈ R

If A 6= 0 then ∃ at least one aj s.t. aj 6= 0|aj | > 0, so ‖A‖ = 0 only if A = 0.

‖cA‖ = max1≤j≤n |caj | = max1≤j≤n |c||aj |

Consider aJ = max1≤j≤n |aj ||aJ | ≥ |aj | ∀ j 6= JSo |c||aJ | ≥ |c||aj | is valid. |c||aj | = |caj |. Thenmax1≤j≤n |c||aj | = |c|max1≤j≤n |aj |

‖A + B‖ = max1≤j≤n |aj + bj | ≤ max1≤j≤n |aj |+ |bj | ≤ max1≤j≤n |aj |+ max1≤k≤n |bk| = ‖A‖+ ‖B‖

(2) See sketch. Looks like a square.

Exercise 19. Given A = (a1, . . . , an), ‖A‖1 =∑n

j=1 |aj |, ‖A‖2 = max1≤j≤n |aj |, ‖A‖ =√∑n

j=1(aj)2.

Let ‖A‖2 = max1≤j≤n |aj | = |aJ |.‖A‖2 =

∑nj=1 a2

j , ‖A‖2 sum obvious includes aJ term, and all a2j terms are positive.

=⇒ ‖A‖2 ≥ ‖A‖22 or ‖A‖ ≥ ‖A‖2(∑n

j=1 |aj |)2

≥ ∑nj=1(aj)2, because of the cross terms, which are positive from the absolute value function on each

individual term.So then ‖A‖1 ≥ ‖A‖

Note that we had used the fact that if b2 > a2 and a, b > 0, then b > a. This is becauseIf a = b, then a2 = b2. Contradiction.If a < b, then a · a < a · b < b2, from ab < b2. Contradiction.

‖A‖2 ≤ ‖A‖ ≤ ‖A‖1 means that 2-norm will �norm� or �expand� a given coordinate pair the least, assign to it thesmallest norm, so coordinates of bigger values are needed for 2-norm. In contrast, the 1-norm assigns bigger values for thenorm of a given pair of coordinates. The 1-norm requires smaller values for the coordinates to get the same norm value as the2-norm and usual spherical norm.

217

Notice how going from 2 norm, ‖A‖, and 1 norm, we go from a square, to a circle, and �nally to a diamond.Exercise 20. Given that d(A,B) = ‖A−B‖,

d(B,A) = ‖B −A‖ = ‖(−1)(A−B)‖ = | − 1|‖A−B‖ = ‖A−B‖d(A.B) = ‖A−B‖ = 0 only if A−B = 0 =⇒ A = B

d(A,B) = ‖A−B‖ = ‖A− C + C −B‖ ≤ ‖A− C‖+ ‖C −B‖ = d(A, C) + d(C, B)

12.15 Exercises - The linear span of a �nite set of vectors, Linear independence, Bases. Exercise 1. x(i− j) + y(i + j)(1) i; x = 1/2, y = 1/2

(2) j; x = −1/2, y = 1/2

(3) 3i− 5j; x = 4, y = −1(4) 7i + 5j; x = 1, y = 6

Exercise 2.A = (1, 2)

B = (2,−4)

C = (2,−3)C = xA + yB =⇒ x = 1/4, y = 7/8

Exercise 3.A = (2,−1, 1)

B = (1, 2,−1)

C = (2,−11, 7)C = xA + yB

2 1−1 21 −1

∣∣∣∣∣∣

2−117

=⇒

0 1−1 00 1

∣∣∣∣∣∣

−4−3−4

=⇒x = 3y = −4

Exercise 4. If C = (2, 11, 7),

2 1−1 21 −1

∣∣∣∣∣∣

2117

=

0 5−1 20 3

∣∣∣∣∣∣

241129

Exercise 5.(1) A ‖ B =⇒ A = tB, t 6= 0

A− tB = 0 and t 6= 0. So A,B linearly dependent.(2) A ∦ B

A− tB 6= 0 since @ t ∈ R s.t. A = tB. Then A,B linearly independent.Exercise 6. Given (a, b), (c, d), the following statements must be true.

[a cb d

] [xy

]=

[00

]

(1

ad− bc

)[d −c−b a

] [a cb d

] [xy

]=

[xy

]=

[00

]

If ad− bc 6= 0, then we could apply this inverse matrix to obtain x = y = 0.Suppose x = y = 0 and ad− bc = 0. Then ad = bc and we could rewrite (c, d) to be

(adb , bc

a

). Then

−db (a, b) + (c, d) = 0. But we had said that x = y = 0 were the only coef�cients. Then ad− bc 6= 0

Exercise 7. Given (1 + t, 1− t), (1− t, 1 + t), and from above,(1 + t)(1 + t)− (1− t)(1− t) = ad− bc = 4t 6= 0. So for t 6= 0, the two vectors are linearly independent.Exercise 8. i, j, k are linearly independent. i + j + k = L({i, j, k}). By thm, {i, j, k, i + j + k} are linearly dependent(Apostol's Thm. 12.8, any set of k + 1 vectors is dependent, S independent, |S| = k ).

i, j, k are linearly independent and pairwise linearly independent. i + j + k is not in the span of any pair from i, j, kbecause i + j + k is a linear combination of one unit coordinate vector not included in the pair. Then the 3 vectors, includingi + j + k, are linearly independent (if i + j + k is not in the span of two other independent vectors, then there's no linearcombination of the three that would give a nontrivial representation of 0).

218

Exercise 9.(1) c1i + c2(i + j) = 0 c1 = −c2 c2 = 0 Then c1 = 0(2) i + j − i = j ∈ L(S)(3) 3i− 4j = −4(i + j) + 5i(4) i, i + j linearly independent |S| = 2 by Thm. 12.10, since we have 2 linearly independent vectors in V2, S forms a

basis for V2.

Exercise 10. Given A = i, B = i + j, C = i + j + 3k in V3

(1) xA + yB + zC = 0. Then z = 0. Then y = 0. x = 0. A, B,C linearly independent.(2) j = B −A k = C−B

3

(3) 2i− 3j + 5k =−14B + 5C

3+ 5A

(4) {A, B,C} are linearly independent. Since |{A,B, C}| = 3 = n, for Vn = V3, then by Thm. 12.10, {A,B, C} forma basis for V3.

Exercise 11. Let A = (1, 2), B = (2,−4), C = (2,−3), D = (1,−2).Note that B = 2D (B, D are parallel).

A,B, C, D are all each linearly independent, by themselves.{A,D}, {A,B}, {A,C}, {B, C}, {D,C}

Exercise 12.

LetA = (1, 1, 1, 0)

B = (0, 1, 1, 1)

C = (1, 1, 0, 0)(1) A, B,C are linearly independent since for xA + yB + zC = 0, y = 0, x = 0 and so z = 0.(2) −(0, 1, 2, 1) = D would make the set dependent.(3) E = (1, 0, 0, 0)(4) (1, 2, 3, 4) = 4B +−A− C + 3E

Exercise 13.

(1) x(√

3, 1, 0) + y(1,√

3, 1) + z(0, 1,√

3) = 0. Theny = −

√3z

x√

3 = −y

x +√

3y + z = 0

=⇒ −y√3

+√

3y + y

−√3= 0 So y = 0.

Then x = z = 0(2) (

√2, 1, 0)−√2(1,

√2, 1) = (0,−1,−√2). Dependent.

(3) (t, 1, 0) +−t(1, t, 1) = (0, 1, t) =⇒ 1− t2 = 1 so that t = 0 .

Exercise 14.

(1)(1, 0, 1, 0)

(1, 1, 1, 1)

(2, 0,−1, 0)Note that we cannot get any bigger set, since (0, 1, 0, 1) is a linear combination of (1, 0, 1, 0) and (1, 1, 1, 1)

(2) (1, 1, 1, 1), (1,−1, 1, 1) are immediately linearly independent, since they're not parallel to each other.

x(1, 1, 1, 1) + y(1,−1, 1, 1) = (x + y, x− y, x + y, x + y), so the 3rd. and 4th. coordinate of any linear combi-nation of these two vectors will be equal.

Thus(1, 1, 1, 1)

(1,−1, 1, 1)

(1,−1,−1, 1)are linearly independent.

x(1, 1, 1, 1)+

y(1,−1, 1, 1)+

z(1,−1,−1, 1)= (x + y + z, x− y − z, x + y − z, x + y + z)

So any linear combination of the these three vectors, will always have the 1st. and 4th. coordinates equal. Then(1, 1, 1, 1), (1,−1, 1, 1), (1,−1,−1, 1), (1,−1.− 1.− 1) are linearly independent.

219

(3)

(1, 1, 1, 1)

(0, 1, 1, 1)

(0, 0, 1, 1)

(0, 0, 0, 1)

are linearly independent.

Exercise 15.

(1) x(A + B) + y(B + C) + z(A + C) = (x + y + z)A + (x + y)B + (y + z)C = 0=⇒ x = −z = y = −x since A,B,C are linearly independent. Then x = y = z = 0.

(2) x(A−B) + y(B + C) + z(A + C) = (x + z)A + (−x + y)B + (y + z)C = 0 So we havex = y

y = −z

x = −z

If x, y 6= 0, z 6= 0 and condition for linear independence of A,B, C is still satis�ed. Thus, A − B, B + C, A + Clienarly dependent.

Exercise 16.

(1)If S = BV3 ,

S = BV3 means ∀x ∈ V3, x ∈ L(S). Since i, j, k ∈ V3, i, j, k ∈ L(S)

If i, j, k ∈ L(S), then L({i, j, k}) ⊆ L(S) (since every linear combination of i, j, k can be reexpressed as a linearcombination of vectors in S, since each i, j, k is a linear combination of vectors in S).

Then since ∀x ∈ V3, x ∈ L({i, j, k}), x ∈ L(S)So S forms a basis for V3.

(2)If S = BVn ,

S = BVn means ∀x ∈ Vn, x ∈ L(S). Since e1, . . . , en ∈ Vn, e1, . . . , en ∈ L(S)

If e1, . . . , en ∈ L(S), then L({e1, . . . , en}) ⊆ L(S) (since every linear combination of e1, . . . , en can be reex-pressed as a linear combination of vectors in S, since each e1, . . . , en is a linear combination of vectors in S).

Then since ∀x ∈ Vn, x ∈ L({e1, . . . , en}), x ∈ L(S)So S forms a basis for Vn.

Exercise 17. {(0, 1, 1), (1, 1, 1), (0, 0, 1)} and {(0, 1, 1), (1, 1, 1), (0, 0,−1)}Exercise 18. {(0, 1, 1, 1), (1, 1, 1, 1), (0, 0, 0, 1), (0, 0, 1, 0)} and {(0, 1, 1, 1), (1, 1, 1, 1), (0, 0, 0,−1), (0, 0,−1, 0)}Exercise 19. Given thatS = {(1, 1, 1), (0, 1, 2), (1, 0,−1)}T = {(2, 1, 0), (2, 0,−2)}U = {(1, 2, 3), (1, 3, 5)}

(1) Consider x(2, 1, 0) + y(2, 0,−2) = (2x + 2y, x,−2y)

1 0 11 1 01 2 −1

abc

=

2(x + y)x−2y

1 0 11 1 01 2 −1

∣∣∣∣∣∣

2(x + y)x−2y

=

1 0 11 −12 −2

∣∣∣∣∣∣

2(x + y)−x− 2y−2x− 4y

=⇒ a + c = 2(x + y)

b− c = −x− 2y

2(x + y)(1, 1, 1) + (−x− 2y)(0, 1, 2)

220

(2)

1 0 11 1 01 2 −1

∣∣∣∣∣∣

abc

=

000

=⇒

1 0 11 −12 −2

∣∣∣∣∣∣

abc

=⇒

a = −c

b = cso S is linearly dependent

2(2, 1, 0)− 32(2, 0,−2) = (1, 2, 3)

3(2, 1, 0)− 52(2, 0,−2) = (1, 3, 5)

=⇒ L(U) = L(T )

(1, 1, 1) + (0, 1, 2) = (1, 2, 3)

(1, 1, 1) + 2(0, 1, 2) = (1, 3, 5)=⇒ L(S) = L(U) = L(T )

Exercise 20. A,B are �nite subsets.

(1) Consider xA ∈ L(A), xA =∑

cjAJ . Since ∀Aj ∈ B, xA ∈ L(B). Then L(A) ⊆ L(B)(2) Consider cj ∈ A

⋂B so that cj ∈ A and cj ∈ B

∑cjCj ∈ L(A) since ∀Cj ∈ A and

∑cjCj ∈ L(B) since ∀Cj ∈ B

=⇒ ∑cjCj ∈ L(A)

⋂L(B) =⇒ L(A

⋂B) ⊆ L(A)

⋂L(B)

(3) Now L(A⋂

B) ⊆ L(A)⋂

L(B).

IfA = {(1, 1, 0), (1, 0, 1)}B = {(1, 0, 0), (0, 1, 1)}

then we could see that A⋂

B = ∅, but L(A)⋂

L(B) is nonempty; for instance, it'll include (2, 1, 1) = (1, 1, 0) +(1, 0, 1) = 2(1, 0, 0) + (0, 1, 1).

12.16 Exercises - The vector space Vn(C) of n-tuples of complex numbers.

Exercise 1. GivenA =(1, i)

B =(i,−i)

C =(2i, 1)

(1) A ·B = −i− 1(2) B ·A = A ·B = −1 + i(3) (iA) ·B = i(A ·B) = 1− i(4) A · (iB) = −i(A ·B) = −1 + i(5) (iA) · (iB) = i(−i)A ·B = −i− 1(6) (i,−1) · (−1, 1) = B · C = 2 +−i(7) A · C = −2i + i = −i(8) (B + C) ·A = (3i, 1− i) ·A = 3i− i− 1 = 2i− 1(9) (A− C) ·B = (1− 2i, i− 1) ·B = (−i− 2,−1− i) = −3− 2i

(10) (A− iB) · (A + iB) = (A− (−1, 1)) · ((1, i) + (−1, 1)) = (2,−1 + i) · (0, 1 + i) = 2i

Exercise 2. GivenA = (2, 1,−i)

B = (i,−1, 2i)and V3(C)

A · C = (2, 1,−i) · (c1, c2, c3) = 2c1 + c2 − ic3 = 0

B · C = (i,−1, 2i) · (c1, c2, c3) = ic1 +−c2 + 2ic3 = 0=⇒

c2 = c1(−4 + i)

c3 = c1(−2i− 1)

C = (1, i− 4,−1− 2i)

Exercise 3.

‖A + B‖2 = (A + B) · (A + B) = (A + B) ·A + (A + B) ·B = A · (A + B) + B · (A + B) =

= A ·A + A ·B + B ·A + B ·B = ‖A‖2 + ‖B‖2 + A ·B + B ·A =

= ‖A‖2 + ‖B‖2 + A ·B + A ·B221

Exercise 4.‖A + B‖2 − ‖A−B‖2 = (A + B) · (A + B)− (A−B) · (A−B) =

= (A + B) ·A + (A + B) ·B − (A−B) ·A + (A−B) ·B =

= A · (A + B) + B · (A + B)−A · (A−B) + B · (A−B) =

= A ·A + A ·B + B ·A + B ·B −A ·A + A ·B + B ·A−B ·B =

= 2(A ·B + A ·B)

Exercise 5.‖A + B‖2 + ‖A−B‖2 = (A + B) · (A + B) + (A−B) · (A−B) =

= (A + B) ·A + (A + B) ·B + (A−B) ·A− (A−B) ·B =

= A · (A + B) + B · (A + B) + A · (A−B)−B · (A−B) =

= A ·A + A ·B + B ·A + B ·B + A ·A−A ·B −B ·A + B ·B =

= 2‖A‖2 + 2‖B‖2

Exercise 6. A,B ∈ Vn(C)

(1) (A ·B + A ·B) = A ·B + A ·B = A ·B + A ·B = A ·B + A ·B So then A ·B + A ·B must be real.(2)

Now ‖A + B‖2 = ‖A‖2 + ‖B‖2 + A ·B + A ·B, so then

‖ A

‖A‖ +B

‖B‖‖2 = 1 + 1 +

A ·B + A ·B‖A‖‖B‖ ≥ 0

=⇒ A ·B + A ·B‖A‖‖B‖ ≥ −2

Now ‖A−B‖2 = (A−B) · (A−B) = (A−B) ·A− (A−B) ·B = ‖A‖2 −B ·A−A ·B + ‖B‖2 ≥ 0

so then ‖ A

‖A‖ −B

‖B‖‖2 = 2− (A ·B + A ·B)

‖A‖‖B‖ ≥ 0

=⇒ 2 ≥ A ·B + A ·B‖A‖‖B‖

Exercise 7. With this de�nition, that θ = arccos12 (A·B+A·B)

‖A‖‖B‖ ,We had already showed that‖A−B‖2 = (A−B) · (A−B) = (A−B) ·A− (A−B) ·B = ‖A‖2 −B ·A−A ·B + ‖B‖2 ≥ 0Thus ‖A−B‖2 = ‖A‖2 + ‖B‖2 − (A ·B + A ·B) = ‖A‖2 + ‖B‖2 − 2‖A‖‖B‖ cos θ

Exercise 8. Vn(C)A = (1, 0, i, i, i)

B = (i, i, i, 0, i)θ = arccos

12 (A·B+A·B)

‖A‖‖B‖

A ·B = −i + 1 + 1 = 2− i A ·B = 2 + i =⇒ A ·B + A ·B = 2− i + 2 + i = 4‖A‖2 = (1, 0, i, i, i) · (1, 0, i, i, i) = 1 + 1 + 1 + 1 = 4‖B‖2 = (i, i, i, 0, i) · (i, i, i, 0, i) = (1 + 1 + 1 + 1) = 4

θ = arccos12 4

2·2 = arccos 1/2 =⇒ cos θ = 1/2 or θ = π/3

Exercise 9.

Given thatA = (1, 0, 0)

B = (0, i, 0)

C = (1, 1, i)(1) aA + bB + cC = (x, y, z)

c = z/i = −iz Then b = −i(y + iz) and a = x + iz.

If x = y = z = 0, then c = 0 = b = a. So A,B, C does form a basis.(2) (5, 2− i, 2i) = 2C + 3A−B

222

Exercise 10.n∑

j=1

cjej = (x1, . . . , xn) = X

then the cj's simply equal the complex coordinates of X cj = xj

If X = 0, then each cj = 0. So the ej's are linearly independent . =⇒ ej's form a basis for Vn(C)

13.5 Exercises - Introduction, Lines in n-space, Some simple properties of straight lines, Lines and vector-valuedfunctions.Exercise 1. Recall that if X0 ∈ L(P,A), and P, Q ∈ L(P,A), then

X0 = P + tA = P + t(Q− P )

X0 − P = t(Q− P ).

Q− P = (4, 0), P = (−3, 1), Q = (1, 1)(1) (0, 0) no.(2) (0, 1) yes.(3) (1, 2) no(4) (2, 1) yes(5) (−2, 1) yes

Exercise 2. P = (2,−1); Q = (−4, 2), Q− P = (−6, 3)(1) (0, 0) yes(2) (0, 1) no(3) (1, 2) no(4) (2, 1) no(5) (−2, 1) yes.

Exercise 3. P = (−3, 1, 1) A = (1,−2, 3)(1) (0, 0, 0) no(2) (2,−1, 4) no(3) (−2,−1, 4) yes(4) (−4, 3,−2) yes(5) (2,−9, 16) yes

Exercise 4.P = (−3, 1, 1)

Q = (1, 2, 7)and Q− P = (4, 1, 6)

(1) (−7, 0, 5) no(2) (−7, 0,−5) yes(3) (−11, 1, 11) no(4) (−11,−1, 11) no.(5) (−1, 3/2, 4) yes.(6) (−5/3, 4/3, 3) yes.(7) (−1, 3/2,−4) no.

Exercise 5. P = (2, 1, 1), Q = (4, 1,−1), R = (3,−1, 1)(1) Q− P = (2, 0,−2) R− P = (1,−2, 0). No.(2) P = (2, 2, 3), Q = (−2, 3, 1), R = (−6, 4, 1)

Q− P = (−4, 1,−2), R− P = (−8, 2,−2). No.(3) P = (2, 1, 1); Q = (−2, 3, 1), R = (5,−1, 1)

Q− P = (−4, 2, 0) R− P = (3,−2, 0). No.

Exercise 6.A = (2, 1, 1)

B = (6,−1, 1)

C = (−6, 5, 1)Since

B −A = (4,−2, 0)

C −A = (−8, 4, 0)

D = (−2, 3, 1)

D −A = (−4, 2, 0)

F −A = (−6, 3, 0)

E = (1, 1, 1)

F = (−4, 4, 1)

G = (−13, 9, 1)

H = (14,−6, 1){A,B, C, D, F}Exercise 7. P = (1, 1, 1) A = (1, 2, 3)

223

P + tA = Q + uB

(1, 1, 1) + t(1, 2, 3) = (2, 1, 0) + u(3, 8, 13)

1 + t = 2 + 3u

1 + 2t = 1 + 8u=⇒ u = 1, t = 4

(5, 9, 13) is an intersection point.Exercise 8.

(1) X = P + tA = Q + uB =⇒ P −Q = uB − tA

P −Q = aA + bB =⇒ P + (−a)A = Q + bB andP + (−a)A ∈ L(P ;A)

Q + bB ∈ L(Q; B)

(2) L = {(1, 1,−1) + t(−2, 1, 3)}L′ = {(3,−4, 1) + t(−1, 5, 2)}(1, 1,−1) + t(−2, 1, 3) = (3,−4, 1) + u(−1, 5, 2)

1 +−2t = 3− u 1 + t = −4 + 5u 3 = −5 + 9uu = 8/9

t = −5/9

But t = −5/9, u = 8/9, doesn't work for the 3rd. coordinate.

Exercise 9. X(t) = P + tA P = (1, 2, 3) A = (1,−2, 2), Q = (3, 3, 1)(1)‖Q−X(t)‖2 = (Q−X) · (Q−X) = Q2 − 2Q ·X + X2 = 19− 2Q · (P + tA) + P 2 + 2tP ·A + t2A2 =

= 19− 24 +−2t(−1) + 14 + 2t(3) + t29 = 9 + 8t + 9t2

(2) X(t0) ‖Q−X(t)‖ min. ddt‖Q−X(t)‖2 = 18t + 8 = 0 t = −4/9

‖Q−X(t)‖ =√

9 + −329 + 16

9 =√

81−169 =

√659 =

√13·53

(3)(Q−X(t0)) ·A = Q ·A−X ·A = Q ·A− (P + tA) ·A = Q ·A− P ·A− tA ·A =

= −1− 3−(−4

9

)(9) = 0

Exercise 10. Q /∈ L(P ;A)(1) f(t) = ‖Q−X(t)‖2 X(t) = P + tA

(Q−X) · (Q−X) = Q2 − 2Q ·X + X2 = Q2 − 2Q · (P + tA) + P 2 + 2tP ·A + t2A2 =

= Q2 − 2Q · P − 2tQ ·A + P 2 + 2tP ·A + t2A2

ddt (‖Q−X(t)‖2) = 2tA2 + 2(P ·A−Q ·A) = 0 t = (Q−P )·A

A2

(2) (Q−X) ·A = Q ·A− (P + tA) ·A = Q ·A− P ·A− tA2 = (Q− P ) ·A− (Q−P )·AA2 A2 = 0

Exercise 11. L(P ; A), L(Q,A) in Vn

Consider x1 = P + tA.X1 = Q + (P −Q) + t1Aif P −Q = u1A, then X1 ∈ L(Q, A)

Likewise, ifX2 = Q + tA

X2 = P + (Q− P ) + t2A

If P −Q = −u2A, then X2 ∈ L(P ;A)

If P −Q ‖ A, then X1 ∈ L(Q,A) and X2 ∈ L(P, A), so that L(Q,A) = L(P, A)If P −Q ∦ A, then X1 /∈ L(Q,A) and X2 /∈ L(P ; A), ∀X1 ∈ L(P, A), ∀X2 ∈ L(Q,A)

X1 = Q + (P −Q) + t1A 6= Q + (u1 + t1)A =⇒ P −Q 6= u1A so that X1notinL(Q,A)Exercise 12. L(P ; A), L(Q; B), A ∦ B

ConsiderX1 = P + tA

X2 = Q + uB

X1 −X2 = P −Q + tA− uB

X1 −X2 = 0 only if P −Q = u1B − t1A

If P −Q = u2B − t2A, then (u1 − u2)B = (t1 − t2)A

but A ∦ B so t1 = t2, u1 = u2

(only one intersection point). Otherwise, if @u1, t1 s.t. P −Q = u1B − t1A, then there's no intersection.224

13.8 Exercises - Planes in Euclidean n-space, Planes and vector-valued functions.

Exercise 1. Let M = {P + sA + tB}, P = (1, 2,−3), A = (3, 2, 1), B = (1, 0, 4). Remember, ifX = P + sA + tB

X − P = sA + tB

(1) (1, 2, 0) X − P = (0, 0, 3); no(2) (1, 2, 1) X − P = (0, 0, 4); no(3) (6, 4, 6) X − P = (5, 2, 9) = A + 2B, yes(4) (6, 6, 6) X − P = (5, 4, 9) no(5) (6, 6,−5) X − P = (5, 4,−2) = 2A +−B yes.

Exercise 2.P = (1, 1,−1)

Q = (3, 3, 2)

R = (3,−1,−2)so that

Q−R = (0, 4, 4)

P −R = (−2, 2, 1)

(1) (2, 2, 1/2) X −R = (−1, 3, 5/2) yes(2) (4, 0,−1/2) X −R = (1, 1, 3/2) yes(3) (−3, 1,−3) X −R = (−6, 2,−1) yes(4) (3, 1, 3) X −R = (0, 2, 5) no(5) (0, 0, 0) X −R = (−3, 1, 2) no

Exercise 3.

(1) (1, 2, 1) + s(0, 1, 0) + t(1, 1, 4) = (x, y, z) =⇒x = 1 + t

y = 2 + s + t

z = 1 + 4t

(2) (0, 1, 0) + s(1, 1, 1) + t(1, 0, 4)

Exercise 4. (1, 2, 0) + s(1, 1, 2) + t(−2, 4, 1) or N = (−7,−5, 6) (by �nding the cross product of the two spanning vectors),so that −7x− 5y + 6z = −17.

(1) (0, 0, 0) no.(2) (1, 2, 0) yes.(3) (2,−3,−3) yes.

Exercise 5. M = {P + s(Q− P ) + t(R− P )}(1) P + s(Q− P ) + t(R− P ) = (1− s− t)P + sQ + tR

Since 1− s− t = p, for pP , 1 = p + s + t

(2)X = P + s(Q− P ) + t(R− P ) =

= (1− s− t)P + sQ + tRp = 1− s− t, p + s + t = 1

Exercise 6. ax + by + cz = d

(1) (2, 3, 1) spanned by (3, 2, 1) and (−1,−2,−3),

X = P + sA + tB

xyz

=

231

+ s

321

+ t

−1−2−3

=⇒

x = 2 + 3s− t

y = 3 + 2s− 2t

z = 1 + s− 3t

=⇒ x− 2y + z = −3

(2) IfP = (−2,−1,−3)

Q− P = (4, 4, 4)

R− P = (6, 4, 4)(x, y, z) = (−2,−1,−3) + s(1, 1, 1) + t(3, 2, 2)

x = −2 + s + 3t

y = −1 + s + 2t

z = −3 + s + 2t

=⇒ y + 1 = z + 3 or y − z = 2

225

(3) (2, 3, 1) parallel to the plane through the origin, spanned by (2, 0,−2) and (1, 1, 1)P+ sA+tB

xyz

=

231

+ s

20−2

+t

111

=⇒x = 2 + 2s + t

y = 3 + t

z = 1− 2s + t

=⇒ x + 2y + z = −3

Exercise 7. M s.t. 3x− 5y + z = 9

(1) Out of(0,−2,−1)

(−1,−2, 2)

(3, 1,−5),

(0,−2,−1)

(−1,−2, 2)

(2)(0,−2,−1)− (2, 1, 8) = (−2,−3,−9)

(−1,−2, 2)− (2, 1, 8) = (−3,−3,−6)=⇒ (2, 1, 8) + s(2, 3, 9) + t(1, 1, 2)

Exercise 8. Given two planes, M = {P + sA + tB}, M ′ = {Q + sC + tD}, and the speci�c coordinates,P = (1, 1, 1)

A = (2,−1, 3)

B = (−1, 0, 2)

Q = (2, 3, 1)

C = (1, 2, 3)

D = (3, 2, 1)It's easiest to �nd the Cartesian coordinate equations for each of the planes, to �nd the intersection, and the easiest way to�nd the Cartesian coordinate equations is to �nd the normal to each of the planes, using the cross product on A,B and onC, D. A×B = (−2,−7,−1) and C ×D = (−4, 8,−4). Then

M :2x + 7y + z = 10

M ′ :− 4x + 8y +−4z = 12

It's easy to check that these are the correct equations because we could plug in P, Q respectively to check the equation andplug in A,B and C, D, respectively to check that we get 0 (because adding arbitrary amounts of A,B or C,D, respectively,should not affect each respective equation).

Thus, solving for both M and M ′, we get x + 9y = 13. Then two distinct points include (4, 1,−5), (−5, 2, 6)

Exercise 9.(1) Given a plane that

P = (2, 3, 1),A = (1, 2, 3)

B = (3, 2, 1)and x − 2y + z = 0, so that (with a Cartesian coordinate equation, we could easily

plug in numbers to �nd 3 points that this plane contains)

P = (2, 1, 0)

Q = (3, 2, 1)

R = (−1,−1,−1)

Q− P = (1, 1, 1)

R− P = (−3,−2,−1)

X =

210

+ s

111

+ t

−3−2−1

So since these two planes have L(A,B) = L(Q−P,R−

P ) (since (1, 2, 3) + (3, 2, 1) = 4(1, 1, 1) and (3, 2, 1) = −(−3,−2,−1), then the two planes are parallel.)

(2) Given thatM ′ : x− 2y + z = 0M : x + 2y + z = 0

then to �nd the intersection, solve for the two equations to get 2x + 2z = 0 =⇒ x = −z.

Then two points in the intersection, M ′⋂ M ′′ are (1, 0,−1), (−1, 0, 1)

Exercise 10.

L =

111

+ s

2−13

M =

11−2

+ t

213

+ u

011

226

Condition for intersection:

111

+ s

2−13

=

11−2

+ t

213

+ u

011

Then

00−3

= s

2−13

− t

213

− u

011

=

2 −2 0−1 −1 −13 −3 −1

stu

2 −2 0−1 −1 −13 −3 −1

∣∣∣∣∣∣

00−3

=

2 0 00 −2 00 0 1

∣∣∣∣∣∣

−333

=⇒

s =−32

t =3−2

u = 3

=⇒ X =

−25/2−7/2

Exercise 11. L = {(1, 1, 1) + t(2,−1, 3)}(1) M = {(1, 1,−2) + s(2, 1, 3) + t(3/4, 1, 1)}

Suppose (2,−1, 3) = s(2, 1, 3) + t(3/4, 1, 1). Then2 = 2s + 3/4t

− 1 = s + t. Doing the algebra, we get t = −16/5, s =

11/5. But the third coordinate doesn't work, 33/5 − 16/5 = 17/5. So M, L are not parallel since (2,−1, 3) /∈L({(2, 1, 3), (3/4, 1, 1)})

(2) (1, 1− 2), (3, 5, 2), (2, 4,−1).C = (2, 4, 4)

D = (1, 3, 1)sC + tD = (2,−1, 3). Doing the algebra, we get t = −5, s = 7/2

from the �rst two coordinates, but it doesn't agree with the third coordinates, so that 7/2C +−5D 6= (2,−1, 3). No,not parallel.

(3) From, x + 2y + 3z = −3, we get the spanning vectors easily by plugging in numbers that'll make the LHS equal tozero: (−3, 0, 1), (1, 1,−1). Again, we �nd that (2,−1, 3) 6= (−1)(1, 1,−1) + (−1)(−3, 0, 1). No, not parallel.

Exercise 12. P, Q ∈ M .Consider M = {P + SA + tB}. Q ∈ M , so that P + sA + tB = Q. =⇒ Q− P = sA + tB.Now c1(sA + tB) + c2A = (c1s + c2)A + c1tB = 0A,B linearly independent, so c1 = 0, then c2 = 0.So sA + tB = Q− P , A are linearly independent, and Q− P = sA + tB, A span the same space as A,B, so by theorem,M = {P + s(Q− P ) + tA}

For X ∈ L, X = P + s(Q− P ) = P + s(Q− P ) + 0A so L ⊆ M .Exercise 13. For L containing (1, 2, 3) and ‖ to (1, 1, 1), L = {P + sA}.Consider M = {P + sA + tB}, containing Q. Then Q = (2, 3, 5) = P + sA + tB. If we let s = t = 1, then

B = (2, 3, 5)− (1, 2, 3)− (1, 1, 1) = (0, 0, 1)

(0, 0, 1) is linearly independent of (1, 1, 1) and so a possible M is M = (1, 2, 3) + s(1, 1, 1) + t(0, 0, 1) or −x + y = 1.Exercise 14. L = {Q + sA} We have a plane that we want: M = {Q + sA + t(P −Q)}.Consider M ′ s.t. L ⊆ M ′ and P ∈ M ′, M ′ = {R + sB + tC}.Note that P 6= Q + sA ∀ s ∈ R

In general, X = Q + sA = (R + b0B + c0C) + s(b1B + c1C) so that

Q = R + b0B + c0C, b0, c0 6= 0

A = (b1B + c1C), b1, c1 6= 0P = R + b2B + c2C, b2, c2 6= 0, since P ∈ M ′.

So then Q− P = (b0 − b2)B + (c0 − c2)C = b3B + c3C and

b3B + c3C = b3(A− c1C)/b1 + c3C =b3

b1A + (c3 − c1

b1)C = Q− P

C =Q− P − b3

b1A

(c3 − c1b1

)227

X = R + sB + tC

= Q− b0B − c0C + sB + tC =

= Q + (s− b0)B + (t− c0)C =

= Q + (s− b0)((Q− P )− c3C)/b3 + (t− c0)C = Q +(s− b0)

b3(Q− P ) +

−(s− b0)c3 + b3(t− c0)b3

C =

= Q +(s− b0)

b3(Q− P ) +

−(s− b0)c3 + b3(t− c0)b3

Q− P − b3b1

A

(c3 − c1b1

)

=⇒ M ′ ⊆ M

Suppose if c1 = 0 (without loss of generality)A = b1B(B =⇒ A)(without loss of generality)

P −Q = (b2 − b0)A + (c2 − c0)C since (P 6= Q + sA, so (c2 − c0) 6= 0)

R + sB + tC = Q− b0A− c0C + sA + tC =

= Q + (s− b0)A + (t− c0)(

P −Q− (b2 − b0)Ac2 − c0

)=

= Q +(

(s− b0) +(b0 − b2)(t− c0)

c2 − c0

)A +

(t− c0

c2 − c0

)(P −Q) ∈ M

=⇒ M ′ ⊆ M

Q + sA + t(P −Q).

If P = R + b2B + c2C andQ = R + b0B + c0C

A = (b1B + c1C)

Q + sA + t(P −Q) = R + b0B + c0C + s(b1B + c1C) + t(b2B + c2C − b0B − c0C) =

= R + (b0 + sb1 + t(b2 − b0))B + (c0 + sc1 + t(c2 − c0))C

=⇒ M ⊆ M ′

So then M = M ′.

13.11 Exercises - The cross product, The cross product expressed as a determinant.

Exercise 1. GivenA = (−1, 0, 2)

B = (2, 1,−1)

C = (1, 2, 2)

(1) A×B =

∣∣∣∣∣∣

e1 e2 e3

−1 0 22 1 −1

∣∣∣∣∣∣= (−2, 3,−1)

(2) B × C = (4,−5, 3)(3) C ×A = (4,−4, 2)(4) A× (C ×A) = (8, 10, 4)(5) (A×B)× C = (8, 3,−7)(6) A× (B × C) = (10, 11, 5)(7) (A× C)×B = (−2,−8,−12)(8) (A+B)×(A−C) = (A+B)×A−(A+B)×C = B×A−A×C−B×C = (2,−3, 1)+(4,−4, 2)−(4,−5, 3) =

(2,−2, 0)(9) (A×B)× (A× C) = (−2, 0, 4)

Exercise 2.

(1)A = (1, 1, 1)

B = (2, 3,−1)A×B = (−4, 3, 1) =⇒ 1√

26(−4, 3, 1)

(2)A = (2,−3, 4)

B = (−1, 5, 7)A×B = (−41,−18, 7) =⇒ ±1√

2054(−41,−18, 7)

228

(3)A = (1,−2, 3)

B = (−3, 2,−1)A×B = (−4,−8,−4) =⇒ ±(1, 2, 1)√

6

Exercise 3.

(1)A = (0, 2, 2)

B = (2, 0,−1)

C = (3, 4, 0)C −A×B −A = (3, 2,−2)× (2,−2,−3) = (−10, 5,−10)

‖C −A×B −A‖ = 15/2

(2)A = (−2, 3, 1)

B = (1,−3, 4)

C = (1, 2, 1)C −A×B − a = (3,−1, 0)× (3,−6, 3) = (−3,−9,−15)

C −A×B −A =√

315 = 3√

35/2

(3)A = (0, 0, 0)

B = (0, 1, 1)

C = (1, 0, 1)‖C −A×B −A‖ =

√3

2

Exercise 4.A = (2, 5, 3)

B = (2, 7, 4)

C = (3, 3, 6)(A− C)× (B −A) = (−1, 2,−3)× (0, 2, 1) = (8, 1,−2)

Exercise 5. ‖A×B‖2 = ‖A‖2‖B‖2 − (A ·B)2

If ‖A×B‖ = ‖A‖‖B‖, A ·B = 0 so cos θ = 0 or θ = π/2.

If A ·B = 0, then ‖A×B‖ = ‖A‖‖B‖Exercise 6. Given A,B ∈ V3

C = (B ×A)−B

(1) A · (B + C) = A ·B + A · (B ×A)−A ·B = 0(2)

B · C = B · (B ×A)−B ·B = −B2

C · C = ((B ×A)−B) · ((B ×A)−B) = (B ×A)2 + B2

B · C‖B‖‖C‖ = cos θBC =

−B2

B((B ×A)2 + B2)1/2=

−B√(BA sin θAB)2 + B2

=

=−1√

A2 sin2 θAB + 1So cos θBC , since A2 sin2 θAB ≥ 0, ranges from −1 (if sin θAB = 0), to 0 (if ‖A‖ → ∞). So then θBC ranges fromπ/2 to π.

(3) ‖B‖ = 1, ‖B ×A‖ = 2. C2 = 4 + 1 = 5 =⇒ C =√

5

Exercise 7.(1) (A×B)2 = A2B2 −A ·B = 1. A, B,A×B normal.

A · (A×B) = B · (A×B) = 0 orthonormal.c1A + c2B + c3(A×B) = 0

A−→ c1A ·A = 0 c1 = 0B−→ c2B

2 = 0 c2 = 0(A×B)−−−−→ c3(A×B)2 = 0 c3 = 0

A, B,A × B linearly independent. By Thm., A,B, A × B form a basis

for V3.(2) C = (A×B)×A

‖((A×B)×A)‖ =√‖A×B‖2‖A‖2 − ((A×B) ·A)2 = 1

(3) Remember thatA ·B = AjBJ = 0 andA ·A = AjAJ = B ·B = BjBJ = 1

229

(A× (A×B))j = εjklAk(A×B)l = εjklAkεlmnAmBn = −εjklεnmlAkAmBn =

= (notice how the l index is repeated) = −(−AkAjBk) +−(AkAkBj) = 0−Bj

=⇒ (A× (A×B)) = −B

Now A,B are interchangeable labels (the choice of name or label is not special or unique at all), so that we couldreuse what we had just proven

(A×B)×A = B =⇒ (B ×A)×B = A

=⇒ −(A×B)×B = A or (A×B)×B = −A

Exercise 8.

(1) We know thatA ·B = ‖A‖‖B‖ cos θAB

‖A×B‖ = ‖A‖‖B‖ sin θABCertainly, if A×B is 0, then its magnitude is zero. Suppose ‖A‖, ‖B‖ 6=

0 and so sin θAB = 0. But then A ·B = 0 and so cos θAB = 0. But cos and sin must obey 1 = C2 + S2. So then atleast one of A or B must be zero.

(2) Given that A 6= 0

andA×B = A× C

A ·B = A · C , thenA× (B − C) = 0

A · (B − C) = 0so then B − C = 0 or B = C

Exercise 9. Given thatA = (2,−1, 2)

C = (3, 4,−1),

(1) what I did was this: A× C = (−7, 8, 11) = Bp

B must have some of this A×C part to twist A into C, because C is orthogonal to both A, B: (B must not have anypart of C).If B has any part of A, A× cA = 0 anyways.

A× Bp

−9= C

Bp

−9=

(79,−89

,−119

)

Yes, there's more than one solution, since we could attach scalar multiples of A to Bp/− 9.(2) A · Bp

−9 = 14+8−229 = 0

A · cA = 1 =⇒ c = 19 So then the desired vector is Bp

−9 + 19A. Yes, there's only one vector.

Exercise 10. Given that A 6== 0 and C ·A = 0. Suppose that for B, there's A×B = C and A ·B = 1.Suppose ∃B1 s.t. A×B1 = C and A ·B1 = 1. Then

A×B1 = C = A×B and A ·B1 = A ·B=⇒ A× (B1 −B) = 0 and A · (B1 −B) = 0

. Then B1 −B = 0 or B1 = B, from Exercise 8.


B = (−1, 1, 1)

C = (2,−1, 2)(1)

A = (1, 0, 1)

B −A = (−2, 1, 0)

C −B = (3,−2, 1)

D − C = D − (2,−1, 2)

A−D = (1, 0, 1)−D

D − C = −(B −A)

A−D = −(C −B)

D = (4,−2, 2)

A = (1, 0, 1)

C −A = (1,−1, 1)

B − C = −(3,−2, 1)

D −B = D − (−1, 1, 1)

A−D = (1, 0, 1)−D

D −B = −(C −A)

A−D = −(B − C)

D = (−2, 2, 0)

B = (−1, 1, 1)

A−B = −(−2, 1, 0)

C −A = (1,−1, 1)

D − C = D − (2,−1, 2)

B −D = (−1, 1, 1)−D

D − C = −(A−B) = (−2, 1, 0)

B −D = −(C −A) = −(1,−1, 1)

D = (0, 0, 2)

3 is the total number of possible D's, fourth vertex (imagine permutating a 4 vertex ring).(2) B −A× C −A = (−2, 1, 0)× (1,−1, 1) = (1, 2, 1)

‖B −A× C −A‖2

=√

62

230

Exercise 12. Given that A ·B = 2; ‖A‖ = 1, ‖B‖ = 4, and C = 2(A×B)− 3B

A · (B + C) = A ·B + A · C = A ·B − 3A ·B = −4

‖C‖2 = 4(A×B)2 + 9B2 = 4((1)242 84) + 9(16) = 16(3 + 9) = 16 · 12

‖C‖ = 8√

3

B · C‖B‖‖C‖ =

−3B2

BC=

−√32

Exercise 13.(1) c1(A + B) + c2(A−B) = (c1 + c2)A + (c1 − c2)B = 0

c1 + c2 = 0 c1 − c2 = 0, so then c1 = c2 = 0.

=⇒ A + B,A−B linearly independent.

So then A + B, A−B,A×B must be linearly independent (since A×B ⊥ A + B, A−B).

(2)c1(A + B) + c2(A + (A×B)) + c3(B + (A×B)) =

= (c1 + c2)A + (c1 + c3)B + (c2 + c3)(A×B) = 0

c1 + c2 = 0c1 + c3 = 0c2 + c3 = 0

=⇒ c3 = 0 = c2 = c1.

(3) (A + B)× (A− B) = B × A− A× B = 2(B × A). Linearly independent since we already know that A× B isorthogonal to A and B, and A,B are linearly independent.

Exercise 14.(1) If A,B, C lie on a line, then C = A + t(B −A).

(B −A)× (C −A) = (B −A)× (t(B −A)) = 0

If (B − A)× (C − A) = 0, then t(B − A) = C − A (otherwise, at least one is zero, and two points have a linethrough them, and so we'd be done). Then C = A + t(B −A) and so C would be part of a line joining A and B.

(2) A 6= B,L = {A + t(B −A)}

X = A + t(B −A) = B + s(A−B)

X −A = t(B −A)

X −B = s(A−B)=⇒ (X −A)× (X −B) = 0

Exercise 15. Given that A ⊥ B, ‖A‖ = ‖B‖ = 1 and P ×B = A− P ,(1) P ·B = (A− P ×B) ·B = A ·B = 0

P 2 = A2 − 2A · (P ×B) + (P ×B)2 = A2 − 2A · (P ×B) + P 2B2 = 1− 2A · (P ×B) + P 2

=⇒ A · (P ×B) = 1/2

A · (P ×B) = A · (A− P ) = A2 − P ·A = 1/2 =⇒ P ·A = 1/2 = P cos θPA

A× (P ×B) = (A ·B)P − (P ·A)B = −(P ·A)B = −A× P =⇒ (P ·A)B = (A× P )

(P ·A)2B2 = A2P 2 sin2 θPA =14

= P 2 sin2 θPA

P 2S2θPA + P 2C2θPA =14

+14

=12

= P 2

=⇒ P =1√2

Note that we had used the cab-bac rule above, which we'll prove with tensors in the next set of exercises.231

(2) P,B, P ×B.P ·B = 0, so P, B orthogonal. P,B orthogonal to P ×B. By thm., P, B, P ×B form a basis (3 orthogonal vectorsin V3 form a basis).

(3)(B × (B × P ))j = εjklBk(B × P )l = εjklBkεlmnBmPn = εjklεmnlBkBmPn =

= BkBjPk −BkBkPj = (B · P )Bj − (B ·B)Pj = −Pj

=⇒ (B × (B × P )) = ((P ×B)×B) = P

(4)A×B = (P ×B)×B + P ×B = −P + P ×B

(P ×B) + P = A

P =A− (A×B)

213.14 Exercises - The scalar triple product, Cramer's rule for solving a system of three linear equations.Exercise 1.

(1)A = (3, 0, 0)

B = (0, 4, 0)

C = (0, 0, 8)=⇒ 96

(2)A = (2, 3,−1)

B = (3,−7, 5)

C = (1,−5, 2)22− 3 + 8 = 27

(3)A = (2, 1, 3)

B = (−3, 0, 6)

C = (4, 5,−1)−60 + 21 + 45 = 6

Exercise 2. Given(0, 1, t)

(1, t, 1)

(t, 1, 0)=⇒ t + t(1 − t2) = 0 or t = 0,±

√2 . Then, by theorem, since the scalar product is zero, then the vectors are

linearly dependent.Exercise 3. Given(1, 1, 0)

(0, 1, 1)

(1, 0, 1)=⇒ 1 + 1 = 2

Exercise 4. Want A×B = A · (B × e1)e1 + A · (B × e2)e2 + A · (B × e3)e3

Now (A×B)j = εjklAkBl

A · (B × ej) = Al(B × ej)l = AlεlmnBm(ej)n = AlεlmjBm == εjlmAlBm = εjklAkBl

Exercise 5. Prove that i× (A× i) + j × (A× j) + k × (A× k) = 2A.We'll show this in two ways to be instructive.

i× (A× i) =? = −i× (i×A)∣∣∣∣∣∣

e1 e2 e3

1 0 0a1 a2 a3

∣∣∣∣∣∣= (0,−a3, a2)

∣∣∣∣∣∣

e1 e2 e3

1 0 00 −a3 a2

∣∣∣∣∣∣= (0,−a2,−a3)

−i× (i×A) = −i× (0,−a3, a2) = −(−a3e3 − a2e2) = a3e3 + a2e2

We could interchange the labels to get the results for j × (A× j) and k × (A× k).We could also use the tensor notation:

(ej × (A× ej))l = εlmn((ej)m(A× ej)n) = εlmn(ej)mεnojAoej == εljnεnojejejAo = εjklεjkoejejAo = εjklεjklejejAl == Al for l 6= j

So for each j, we run through l 6= j so that we have the Akek and Alel components. Then we get 2A.232

Exercise 6.(1) (a, b, c) · ((0, 0, 1)× (6, 3, 4)) = 3 = (a, b, c) · (3, 6, 0) = −3a + 6b = 3 =⇒ −a + 2b = 1 or a = 2b− 1.

(2b− 1, b, c) = b(2, 1, 0) + (−1, 0, c)(2) We need to minimize a2 + b2 + c2. Since each term in this sum is positive and c is arbitrary, let c = 0.

a2 + b2 + c2 = a2 + b2 = (2b− 1)2 + b2 = 5b2 − 4b + 1

=⇒ b = 2/5 =⇒ a = 2b− 1 =45− 1 = −1/5

=⇒ (−1/5, 2/5, 0)

Exercise 7.(1) (A + B) · (A + B)× C = D · (D × C) = 0(2) Want: A · (B × C) = −B · (A× C)

(A + B) · (A× C + B × C) = 0 =⇒ B · (A× C) = −A · (B × C)(3) Want: A · (B × C) = −A · (C ×B)

A · (B × C) = A · (−C ×B) = −A · (C ×B)(4) A ·B × C = −B ·A× C = B · C ×A = −C ·B ×A

Exercise 9. Tensor notation way:(A× (B × C))j = εjklAk(B × C)l = εjklAkεlmnBmCn = εjklεlmnAkBmCn =

= εjklεmnlAkBmCn = AkBjCk −AkBkCj = (C ·A)Bj − (B ·A)Cj

or(i× (B × C))j = εjkl(e1)k(B × C)l = εj1le1εlmnBmCn =

= εj1lεmnlBmCn = BjC1 −B1Cj

i× (B × C) = c1B − b1C

more generally(et × (B × C))j = εjkl(et)k(B × C)l = εjtlεlmnBmCn =

= εjtlεmnlBmCn = BjCt − cjBt = ctBj − btCj

=⇒ et × (B × C) = ctB − btC

a1i× (B × C) + a2j × (B × C) + a3k × (B × C) = (C ·A)B − (B ·A)C

Exercise 10.(1) (A×B)× (C ×D) = (D ·A×B)C − (A×B · C)D(2)

A× (B × C) + B × (C ×A) + C × (A×B) =

= (C ·A)B − (B ·A)C + (B ·A)C − (B · C)A + (C ·B)A− (C ·A)B = 0

(3) Want: A× (B × C) = (A×B)× C iff B × (C ×A) = 0A× (B × C) + B × (C ×A) + C × (A×B) = 0 (in general).If B × (C ×A) = 0, A× (B × C) = −C × (A×B) = (A×B)× C.If A× (B × C) = (A×B)× C,A× (B × C) + B × (C ×A) +−(A×B)× C = (A×B)× C − (A×B)× C + B × (C ×A) =

= 0 + B × (C ×A) = 0 =⇒ B × (C ×A) = 0(4) Remember, we can cyclically permute the scalar triple:

(A×B) · (C ×D) = C · (D × (A×B)) = C · ((D ·B)A− (D ·A)B) =

= (B ·D)(A · C)− (B · C)(A ·D)

Exercise 11. Given that A, B,C, D ∈ V3

A× C ·B = 5A×D ·B = 3

C + D = (1, 2, 1)

C −D = (1, 0,−1)

(A×B)× (C ×D) = (D · (A×B))C − ((A×B) · C)D = (B ·D ×A)C + (A× C ·B)D =

= (−B ·A×D)C + (A× C ·B)D = (−3)(1, 1, 0) + 5(0, 1, 1) = (−3,−3, 0) + (0, 5, 5) = (−3, 2, 5)233

Exercise 12.Consider (A×B) · (B × C)× (C ×A)

(B × C)× (C ×A) = (A · (B × C))C − (C · (B × C))A = (A · (B × C))C

=⇒ (A×B) · (B × C)× (C ×A) = (A · (B × C))C · (A×B) = (A · (B × C))2

In the last step, we cyclically permutated the scalar triple.Exercise 13. Prove or disprove A× (A× (A×B)) · C = −A2A ·B × C

A× ((A ·B)A−A2B) · C = −A2(A×B) · C = −A2C · (A×B) = −A2A · (B × C)

Exercise 14. A,B, C, D ∈ V3.(1) If we consider A to be the origin, then consider the parallelogram with sides B−A,C −A,D−A → B, C, D. The

total volume of the parallelogram is given by|(B −A) · (C −A)× (D −A) = (D −A) · (B −A)× (C −A)|.

Note that we have a parallelogram from B,C, D vectors. B,C, D themselves are (3 + 1) (1 vertex from A, the�origin�) vertices of this parallelogram, and a parallelogram has 8 vertices.We can translate D vector from the A origin to 3 other unique positions - on B, on C, and on B + C (diagonalacross from A, containing A). This will generate 3 new tetrahedrons that are congruent to the the BCD tetrahedron,because of either the parallelogram has congruent opposite parallel sides or uses or shares the same vertices andvectors as the BCD tetrahedron.

Be careful because the parallelogram is not ��lled up�, its total volume is not accounted for, by these 4 congruenttetrahedrons, even though each tetrahedron shares 3 of its sides each with each of the other 3 tetrahedron.

Indeed, consider a cube of sides a,a,a. Consider a tetrahedron made up of its sides. The diagonal connecting Band C, B − C, is of length a

√2.

Volume of parallelogram = 6× ( Volume of a tetrahedron )

=⇒ Volume of a tetrahedron =16|(B −A) · (C −A)× (D −A)|

(2)

For

A = (1, 1, 1)

B = (0, 0, 2)

C = (0, 3, 0)

D = (4, 0, 0)

14|(B −A) · (C −A)× (D −A)| = 1

4|(−1,−1, 1) · (−1, 2,−1)× (3,−1,−1)| =

=14|(−1,−1, 1) · (−3,−4,−5)| = 1

2

Exercise 15.(1)

‖ C −B

‖C −B‖ × (A−B)‖ = ‖A−B‖ sin θ1

θ1 is the angle between C − B/‖C − B‖ and A − B. From the geometry, it's clear that ‖A − B‖ sin θ1 is theperpendicular distance of A from the line L = {B + t(C −B)}.

(2) Given that A = (1,−2,−5), B = (−1, 1, 1), C = (4, 5, 1)

‖(2,−3,−6)× (5, 4, 0)‖/‖(−5,−4, 0)‖ = ‖(24,−30, 23)‖/√

41 =√

2005/41

Exercise 16.(1) Recall that (A×B)j = εjklAkBl so that

(A×B)2 = εjklAkBlεjmnAmBn = εjklεjmnAkAmBlBn = A2B2 − (A ·B)2

and −2A ·B = ‖A−B‖2 − ‖A‖2 − ‖B‖2 is immediately apparent by doing the algebra. Then

4S2 = 4( area of the triangle )2 = 4(12‖A×B‖)2 = ‖A×B‖2 = a2b2 − (A ·B)2 =

= a2b2 −(−1

2(c2 − a2 − b2)

)2

= a2b2 − 14

(c2 − a2 − b2

)2=

=14(2ab− c2 + a2 + b2)(2ab + c2 − a2 − b2) (just complete the square)

234

(2)16S2 = ((a + b)2 − c2)(−(a− b)2 + c2) = ((a + b) + c)((a + b)− c)(c− (a− b))(c + (a− b))

=⇒ S2 =116

(a + b + c)(a + b− c)(c− a + b)(c + a− b)

Exercise 17.x + 2y + 3z = 52x− y + 4z = 110 +−y + z = 3

=⇒

120

x +

2−1−1

y +

341

z =

5113

Ax + By + Cz = D

Recall thatx =

DBC

ABCy =

DCA

ABCz =

DAB

ABC

A · (B × C) = (1, 2, 0) · (3,−5, 11) = −7

D · (B × C) = (5, 11, 3) · (3,−5, 11) = −7

D · (C ×A) = −D · (2,−1,−2) = −(5, 11, 3) · (2,−1,−2) = 7

D · (A×B) = D · (−2, 1,−5) = (5, 11, 3) · (−2, 1,−5) = −14

=⇒ x = 1, y = −2, z = 2

Exercise 18.x + y + 2z = 43x− y − z = 22x + 5y + 3z3

132

x +

1−15

y +

2−13

z =

423

A · (B × C) = (1, 3, 2) · (2, 7, 1) = 25

D · (B × C) = (4, 2, 3) · (2, 7, 1) = 25

D · (C ×A) = −D · (11, 1,−7) = −(4, 2, 3) · (11, 1,−7) = −25

=⇒ x = 1, y = −1, z = 2

Exercise 19.x + y = 5x + z = 2y + z = 5

110

x +

101

y +

011

z =

525

=⇒ x = 1 = z, y = 4

Exercise 20.P = (1, 1, 1)

A = (2, 1,−1)X = P + tA =⇒

xyz

=

111

+ t

21−1

.

Note that X ·N = P ·N + tA ·N = P ·N .

x− y + z = 1x + y + 3z = 5

3x + y + 7z = 11=⇒

(1,−1, 1) ·X = 1

(1, 1, 3) ·X = 5

(3, 1, 7) ·X = 11We could immediately then see that for all X = P + tA, X satis�es this system of linear equations because X ·N = P ·Nand P ·N is equal to the left hand side (LHS) of each of the equations in this system. Also, A ·N = 0 for each of the normalvectors, as expected.

13.17 Exercises - Normal vectors to planes, Linear Cartesian equations for planes.

Exercise 1. GivenA = (2, 3,−4)

B = (0, 1, 1)

(1) A×B = (7,−2, 2)(2) =⇒ 7x− 2y + 2z = 0(3) 7x− 2y + 2z = 9

Exercise 2. x + 2y − 2z + 7 = 0 or x + 2y − 2z = −7.(1) 1

3 (1, 2,−2)235

(2) x−7 − 2y

7 + 2z7 = 1 so the intercepts are

(−7, 0, 0)

(0,−7/2, 0)

(0, 0, 7/2)

(3) N|N | · P = d =⇒ 1

3 (1, 2,−2) · (0, 0, 7/2) = 7/3

(4) −7/9(1, 2,−2)

Exercise 3. (1, 2,−3) is in 3x− y + 2z = −5 and (2, 0,−1) is in 3x− y + 2z = 4.N|N | = (3,−1,2)√

14N|N | · P1 = (3,−1,2)√

14· (1, 2,−3) = −5√

14N|N | · P2 = (3,−1,2)√

14· (2, 0,−1) = 4√

14

So the distance between the two planes is 9√14

Exercise 4.(1)

x + 2y − 2z = 53x− 6y + 3z = 22x + y + 2z = −1x− 2y + z = 7

(1, 2,−2)

3(1,−2, 1)

(2, 1, 2)

(1,−2, 1)So by de�nition, the second and fourth are parallel to each other and the �rst and the third are perpendicular to eachother, since (1, 2,−2) · (2, 1, 2) = 0

(2) N = (1,−2,1)√6

. (0,−1/3, 0) is on the second plane, 3x − 6y + 3z = 2, and (3, 0, 4) is on the fourth plane,x− 2y + z = 7.

N · (0,−1/3, 0) =2/3√

6

N · (3, 0, 4) =7√6

=⇒ 7/√

6− 2/3√6

=19/3√

6

Exercise 5. Given (1, 1,−1), (3, 3, 2), (3,−1,−2), the spanning vectors for a plane through all three points is(2, 2, 3), (2,−2,−1)

(1) The normal vector to this plane is (4, 8,−8)(2) x + 2y − 2z = 5(3) (1,2,−2)

3 · (1, 1,−1) = 5/3

Exercise 6. Given (1, 2, 3), (2, 3, 4), (−1, 7,−2), the spanning vectors for a plane through all three points are(1, 1, 1), (−2, 5,−5)The normal vector to this plane is (−10, 3, 7).Then the Cartesian equation could easily be obtained: −10x + 3y + 7z = 17Exercise 7.

x + y = 1y + z = 2

(1, 1, 0)

(0, 1, 1)1√2√

2=

12

= cos θ =⇒ θ = π/3 or 2π/3

Exercise 8. (1, 2, 9) =⇒ x + 2y + 9z = −55Exercise 9. 4x− 3y + z = 5 → (4,−3, 1). (4,−3, 1) is perpendicular to this plane.

xyz

=

21−3

+ s

4−31

Exercise 10. X(t) = (1− t, 2− 3t, 2t− 1)

(1) X(t) =

12−1

+ t

−1−32

236

(2) (−1, 1, 1)(3) Plug in (1, 2, 1) and t(−1,−3, 2) into 2x + 3y + 2z to see which t will give −1 on the right hand side. t = 1(4) X(3) = (−2,−7, 5). For a plane to be parallel to another plane, then they each have the same normal vector.

=⇒ (2, 3, 2). So the Cartesian equation is 2x + 3y + 2z = −15(5) X(2) = (−1,−4, 3). So then a plane perpendicular to L and through X(2) would be

−x− 3y + 2z = 19

Exercise 11. Try N = (n1, n2, n3).N · e1

|N | =n1√

n21 + n2

2 + n23

=12

n3√n2

1 + n22 + n2

3

=12

=⇒ n1 = n3

n2√2n2

1 + n22

=1√2

=⇒ n1

n2=

1√2

N =(1,√

2, 1)2

=⇒ x +√

2y + z = 2 +√

2

Exercise 12.x + 2y + 3z = 6x

6+

y

3+

z

2= 1

(6, 0, 0)

(0, 3, 0)

(0, 0, 2)=⇒ 36

6= 6

Exercise 13. x− y + 5z = 1 Then two linearly independent spanning vectors for this could be(1, 1, 0)

(0, 5, 1).

We want this vector to be perpendicular to (1, 2,−3):

(a(1, 1, 0) + b(0, 5, 1)) · (1, 2,−3) = (a + 2(a + 5b)− 3b) = 3a + 7b = 0 =⇒ b = −3a/7

so a((1, 1, 0) +−37

(0, 5, 1)) = a(1,−87

,−37

) =⇒ 1√122

(7,−8,−3)

Exercise 14. Given a plane that needs to span with(1, 1, 0)

(0, 1, 1)since it needs to be parallel to these two vectors, and intercept the

intercept (2, 0, 0)

(1, 1, 0)× (0, 1, 1) = N = (1,−1, 1) =⇒ x− y + z = 2

Exercise 15.3x + y + z = 5

3x + y + 5z = 7x− y + 3z = 3

=⇒

331

x +

11−1

y +

153

z =

573

(1, 1,−1)× (1, 5, 3) = (8,−4, 4) =⇒ A ·B × C = 16

x =D ·B × C

A ·B × C=

2418

=32

(1, 5, 3)× (3, 3, 1) = (−4, 8,−12) =⇒ y = 0 then z = 1/2

x = 3/2, y = 0, z = 1/2237

Exercise 16. We could try to solve for a system of 3 linear equations that represents the 3 planes:a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

If a1, a2, a3 6= 0

=⇒x + b1y + c1z = d1

x + b2y + c2z = d2

x + b3y + c3z = d3

=⇒

1 b1 c1

1 b2 c2

1 b3 c3

∣∣∣∣∣∣

d1

d2

d3

Since the normals of the planes are linearly independent, this means that

1 1 1b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣

000

has a unique solution and we'll be left with the identity matrix. Then for our problem above, we can do the same operations,since the equations are linear, and leave the LHS to be an identity matrix. Then we've solved for a unique point in R3

Exercise 17. Given a line through (1, 2, 3) and parallel to the two planes given by x + 2y + 3z = 4 and 2x + 3y + 4z = 5, to�nd a direction vector for the line that is parallel to both planes, determine a vector that would make each of the two previousequations equal to 0 on the LHS. (1,−2, 1), plugged into the LHS of each of the 2 equations, would make it equal 0. The lineis through (1, 2, 3) so then X = (1, 2, 3) + t(1,−2, 1)

Exercise 18. So given a line L, s.t. X = P + uA and plane M , s.t. X = Q + sB + tC, and thatL ∦ M , so that sB + tC ∦ uA ∀ s, t, u ∈ R,

Consider L⋂

M . Consider X1 ∈ L⋂

M

X1 = P + t1A

X1 − P = t1A =⇒ X = P + tA = P + (X1 − P ) + (t− t1)A = X1 + tA (since t ∈ R )

Since X1 ∈ M as well,X1 = Q + a1B + b1C

X = Q + sB + tC = Q + a1B + b1C + (s− a1)B + (t− b1)C = X1 + sB + tC

so we haveXL = X1 + tA

XM = X1 + sB + tC

Suppose X2 ∈ L⋃

MX2 = X1 + u2A

X2 = X1 + s2B + t2C=⇒ u2A = s2B + t2C

but sB + tC 6= uA ∀ s, t, u ∈ RContradiction, so X2 = X1

Exercise 19.(1) Given ax + by + cz = −d, then the normal vector for this plane, that's normalized, is N

|N | = (a,b,c)√a2+b2+c2 .

Note that the Cartesian equation for the plane is simply X ·N = P ·N .Then the perpendicular distance from the point X0 to a point on the plane, P , is given by (X0−P )·N = X0·N−P ·N .

dist. =|ax0 + by0 + cz0 + d|√

a2 + b2 + c2

(2) 5x− 14y + 2z = −9. The normal to this plane is N|N | = en = (5,−14,2)

15 .

The perpendicular distance from Q to plane is dist. = (Q−P )·N|N | = −234+9

15 = −15So to get this point on the plane, we simply have to go from Q and go +15 distance along the normalized normal

vector en:(−2, 15,−7) + 15en = (3, 1,−5)

238

Exercise 20. Given 2x− y +2z +4 = 0, a plane parallel to this plane will have the same normal vector, N|N | = en = (2,−1,2)

3 .

The distance from Q to this plane is(Q− P ) ·N

|N | =(3, 2,−1) · (2,−1, 2)− (−4)

3= (2 + 4)/3 = 2

To �nd the other plane, simply go a distance 2 along en from Q, to get a point on that other plane.

(3, 2,−1) + 2(

(2,−1, 2)3

)=

(9, 6,−3) + (4,−2, 4)3

=(13, 4, 1)

3= P2

2x− y + 2z =263− 4

3+

23

= P2 ·N = 8 =⇒ 2x− y + 2z = 8

Exercise 21.(1) Given 3 points A,B, C, they determine a plane =⇒ M = {A + s(B −A) + t(C −A)}

The normal vector, normalized, to this plane is N|N | = en = (B−A)×(C−A)

|(B−A)×(C−A)|

(Q− P ) ·N|N |

P=A−−−→ (Q−A) · (B −A)× (C −A)|(B −A)× (C −A)|

(2) GivenQ = (1, 0, 0)

A = (0, 1, 1)

B = (1,−1, 1)

C = (2, 3, 4)

Q−A = (1,−1,−1)

B −A = (1,−2, 0)

C −A = (2, 2, 3)

(Q−A) · (B −A)× (C −A)|(B −A)× (C −A)| =

(1,−1,−1) · (−6,−3, 6)9

= −1

Exercise 22. M, M ′ not parallel. eNm 6= eNm′ .x ∈ M

x ∈ M ′ x + b1y + c1z = d1 x + b2y + c2z = d2 =⇒ (1, b1, c1) 6= (1, b2, c2) so that either

b1 6= b2 or c1 6= c2

Let's solve forx + b1y + c1z = 0x + b2y + c2z = 0

If b2 − b1 6= 0, [1 b1 c1

1 b2 c2

]=⇒

[1 0 c1b2−b1c2

b2−b1b2 − b1 c2 − c1

]

=⇒ (x, y, z) = z

(b2c1 − c2b1

b1 − b2,−(c2 − c1)

b2 − b1, 1

)

Otherwise, if c2 − c1 6= 0[1 b1 c1

0 b2 − b1 c2 − c1

]=⇒

[1 b1(c2−c1)

c2−c1−

(b2−b1c2−c1

)c1 0

0 b2−b1c2−c1

1

]

=⇒ (x, y, z) = y

(b2c1 − b1c2

c2 − c1, 1,

−(b2 − b1)c2 − c1

)

So these two equations with RHS being 0 result in a solution that's completely determined, give a multiplicative factor.We can now �nd at least one point in the intersection by solving the system of equations

x + b1y + c1z = d1

x + b2y + c2z = d2If b2 − b1 6= 0,

[1 b1 c1

1 b2 c2

∣∣∣∣d1

d2

]=⇒

[1 0 c1b2−b1c2

b2−b11 c2−c1

b2−b1

∣∣∣∣∣b2d1−b1d2

b2−b1d2−d1b2−b1

]

=⇒ If z = 1, (x, y, z) =(

b2d1 − b1d2 − (b2c1 − c2b1)b2 − b1

,d2 − d1 − (c2 − c1)

b2 − b1, 1

)

239

Otherwise, if c2 − c1 6= 0[1 b1 c1

0 b2 − b1 c2 − c1

∣∣∣∣d1

d2 − d1

]=⇒

[1 b1c2−c1b2

c2−c10

0 b2−b1c2−c1

1

∣∣∣∣∣d1c2−c1d2

c2−c1d2−d1c2−c1

]

=⇒ If y = 1, (x, y, z) =(

d1c2 − c1d2 − (b1c2 − c1b2)c2 − c1

, 1,d2 − d1 − (b2 − b1)

c2 − c1

)

Thus, it's always possible to �nd a point in M⋂

M ′. This intersection is spanned by only one vector, give a multiplicativefactor, which we have found above. Then the intersection M

⋂M ′ of 2 planes is a line.

Exercise 23. Givenx + 2y + 3z = 42x + y + z = 2 [

1 2 32 1 1

]=⇒

[1 0 −1/30 1 5/3

]

=⇒(

13,−53

, 1)

Let's �nd a point on the intersection, which is a line, for the two planes:[1 2 32 1 1

∣∣∣∣42

]=⇒

[1 0 −1/30 1 5/3

∣∣∣∣02

]

=⇒ k(1,−3, 3)

The normal to this plane containing the intersection line and is parallel to e2 = j is(0, 1, 0)× (1/3,−5/3, 1) = (1, 0, 1/3)

Then x +13z = 2

Exercise 24. Given two equations for the two planes,x + y = 3

2y + 3z = 4, we can immediately get the spanning vector for the line that is the intersection of these planes: (−1, 1,−2/3).

Also, by inspection, we can �nd a point on the intersection of these 2 planes: (1, 2, 0).We want our plane to be parallel to (3,−1, 2). We can now determine the normal to this plane:

(3,−1, 2)× (−1, 1,−2/3) =(−4

3, 0, 2

)

=⇒ −43

x + 2z =−43

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 .‖X − F‖ = ed(X, L)

‖X − F‖ = e|(X − F ) ·N + d|Exercise 2.

(1)‖X − F‖ = ed(X, L)

‖X − F‖ = e|(X − F ) ·N + d|F = 0 =⇒ ‖X‖ = e|(X ·N) + d|; r = e|r cos θ + d|

=⇒ r = e(r cos θ + d) =⇒ r =ed

1− e cos θ

(2) The right branch for the hyperbola is given by r = ed1−e cos θ because X ·N > 0. The left branch for e > 1,

‖X − F‖ = ed(X, L) = e|(X − F ) ·N + d| == e|X ·N + d| = −e(d + r cos θ) = r

r =−ed

(1 + e cos θ)240

Exercise 3. For points below the horizontal directrix,‖X − F‖ = ed(X, L)

F = 0 =⇒ ‖X‖ = 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 + 1if 0 < e ≤ 1

=⇒ r = e(d− r sin θ) =⇒ r =ed

1 + e sin θ

For the �right� or upper-half branch of a hyperbola.‖X − F‖ = e(|(X − F ) ·N − d|) = e(r sin θ − d) = r

r =−ed

1− e sin θ=

ed

e sin θ − 1

Exercise 4. ‖X − F‖ = ed(X, L); ‖X‖ = e|(X − F ) ·N − d| = e|r cos θ − d| = e(d− r cos θ) =⇒ r = ed1+e cos θ

e = 1, d = 2 .

Exercise 5. r = 31+ 1

2 cos θ=

6( 12 )

1+ 12 cos θ

. e = 12 ; d = 6

Exercise 6. r = 63+cos θ = 2

1+ 13 cos θ

. e = 13 ; d = 6.

Exercise 7. r = 1−12 +cos θ

.

ed(X, L) = ‖X − F‖ = e|(X − F ) ·N − d| = e|r cos θ − d| = er cos θ − ed = r

−ed

1− e cos θ= r =

ed

e cos θ − 1

So for r = 22 cos θ−1 , e = 2, d = 1.

Exercise 8. r = 41+2 cos θ e = 2, d = 2.

Exercise 9. r = 41+cos θ e = 1 d = 4.

Exercise 10. 3x + 4y = 25 =⇒ 35x + 4

5y = 5. N =(

35 , 4

5

).

L = {x = P + tA}, N ·X = N · P .To �nd the distance from the focus, at the origin, to the directrix,

dN = P + tA; dN ·N = d = N · PSo for this problem, d = 5.

r = ‖X − F‖ = ed(X, L) = e|(X − F ) ·N − d| = e|X ·N − d| =∣∣∣∣35r cos θ +

45r sin θ − 5

∣∣∣∣

r =12

(5− 3

5r cos θ − 4

5r sin θ

)

r =5/2

1 + 310 cos θ + 4

10 sin θ

Exercise 11. e = 1, 4x + 3y = 25 45x + 3

5y = 5; N =(

45 , 3

5

).

d = 5.‖X − F‖ = ed(X, L) = e|(X − F ) ·N − d| = e

∣∣∣∣45r cos θ +

35r sin θ − 5

∣∣∣∣

r = 5− r

(45

cos θ +35

sin θ

)

r =5

1 + 45 cos θ + 3

5 sin θ

Exercise 12. e = 2, hyperbola, so there's 2 branches.1√2x +

1√2y =

1√2

L = {x = P + tA} X ·N = N · P

dN = P + tA; dN ·N = d = N · P =1√2

241

Note that the sign of d here tells you what side the focus, at the origin, lies on.

‖X − F‖ = ed(X, L) = ‖X‖ = e|(X − F ) ·N − d| = e(d− 1√2r cos θ − 1√

2r sin θ)

r =2/√

21 + 2√

2cos θ + 2√

2sin θ

But for the right side branch,

‖X − F‖ = ed(X,L) = ‖X‖ = e|(X − F ) ·N − d| = −e(d− 1√2r cos θ − 1√

2r sin θ)

r =−2/

√2

1− 2√2

cos θ − 2√2

sin θ

Exercise 13. e = 1 parabola.(1)

‖X − F‖ = ‖X‖ = ed(X, L) = 1|(X − F ) ·N − d| = d−X ·N = d− r cosπ

3

d = r

(32

)=

32× 108mi

r =32 × 108mi

1 + cos θθ = 0, r =

34× 106mi

(2) Focus is in the positive half-plane determined by N .‖X − F‖ = ‖X‖ = ed(X, L) = |(X − F ) ·N + d| = r cos θ + d

d = r(1− cos θ) = 108mi(1− cosθ

3) =

12× 108mi

r =d

1− cos θ=

12 × 108mi

1− cos θr(θ = π) =

14× 108mi

13.24 Exercises - Conic sections symmetric about the origin, Cartesian equations for the conic sections.Quick Review.

Consider symmetry about the origin.‖X − F‖ = ed(X, L) = e|(X − F ) ·N − d| = e|X ·N − F ·N − d| = |eX ·N − e(F ·N + d)|

‖X − F‖2 = ‖X‖2 − 2X · F + ‖F‖2 = e2(X ·N)2 − 2aeX ·N + a2

X → −X; X · F = aeX ·N

X = (F − aeN) = 0 =⇒ F = aeN ; F ·N = ae; a =ed

1− e2F =

e2d

1− e2N

=⇒ ‖X‖2 + (ae)2 = e2(X ·N)2 + a2

if X = ±aN ; ‖X‖2 + (ae)2 = e2(X ·N)2 + a2 is satis�edif X = ±bN ′; b2 + (ae)2 = e2(0) + a2 b2 = a2(1− e2)

Exercise 1. b2 = a2(1− e2) x2

100 + y2

36 = 1√

1− b2

a2 = e =⇒ e =45

.

|F | = |aeN | = 10(

45

)= 8. f = (±8, 0). (0, 0) center. Vertices (±0, 6).

Exercise 2. y2

100 + x2

36 = 1. 45 = 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√

74 =

√7.

√1− b2

a2 =√

1− 916 =

√7

4 = e; (2 +√

7,−3), (2−√7,−3) foci.Vertices (6,−3), (−2,−3), (2, 6), (2,−12).

Exercise 4. x2

( 259 ) + y2 = 1. Center x = (0, 0). |F | = ae =

(53

) (45

)= 4

3 .242

e =√

1− b2

a2 =√

1− 925 = 4

5 . Foci: (± 43 , 0). Vertices (± 5

3 , 0), (0,±1).

Exercise 5. y2

(1/4) + x2

(1/3) = 1

|F | = ae = 1√3

(12

)= 1

2√

3.

√1− 1/4

1/3 = 12 = e. Foci:

(±12√

3, 0

). Center (0, 0).

Vertices (±1/√

3, 0), (0,±1/2).Exercise 6. Center (−1,−2).√

1− b2

a2 =√

1− 1625 = 3

5 = e; |F | = ae = 5 35 = 3.

Foci: (−1,−1), (−1,−5).Vertices: (−1, 3), (−1,−7), (3,−2), (−5,−2).Exercise 7. F = ae = 3

4 . a = 1, e = 34 . b2 = a2(1− e2); b2 = 1

(14

).

x2 + 4y2 = 1 .

Exercise 8. 2a = 4. a2 = 4. 2b = 3. b2 = 9/4. =⇒ (x+3)2

4 + (y−4)2

4 = 1

Exercise 9. (x+3)2

9/4 + (y−4)2

4 = 1.

Exercise 10. 2a = 6, a = 3. (x+4)2

9 + (y−2)2

1 = 1.Exercise 11. 2a = 10, a = 5. |F | = ae = 5e = 4 e = 4/5. b2 = a2(1− e2) = 25

(1− 16

25

)= 9.

Exercise 12. (x−2)2

a2 + (y−1)2

b2 = 1;

a = 4 from (6, 1). b = 2 from (2, 3). =⇒ (x−2)2

42 + (y−1)2

4 = 1Exercise 13. b2 = a2(1− e2).x2

100 − y2

64 = 1; b2 = 100(1− e2) = −64. 1 + 64100 = e2.

Center (0, 0). e = 2√

4110 =

√415 .

Vertices; (±10, 0). F = ae = 2√

41. Foci: (±2√

41, 0).

x2

100 = y2

64 + 1x,y→∞−−−−−→; y = ±4

5 x

Exercise 14. y2

100 − x2

64 = 1; Center (0, 0), a2 = 100; b2 = −64.

b2 = a2(1− e2). e =√

415 . Vertices (0,±10). F = ae = (0,±2

√41).

x2

64 + 1 = y2

100

x,y→∞−−−−−→ ±54 x = y.

Exercise 15. (x+3)2

4 − (y − 3)2 = 1.

Center (−3, 3). e =√

1− b2

a2 =√

1− −14 =

√5

2 .

Foci: ae = 2√

52 =

√5. (−3 +

√5, 3), (−3−√5, 3).

Vertices: (−3, 4), (−3, 2); (1, 3), (−7, 3).(x+3)2

4 = 1 + (y − 3)2x,y→∞−−−−−→ ±(x + 3)

2= y − 3

Exercise 16. x2

144/9 − y2

144/16 = 1 = x2

16 − y2

9 .

e =√

1− −916 = 5

4 . Center (0, 0). |F | = ae = 5. Foci: (5, 0), (−5, 0). Vertices (±4, 0).

Exercise 17. 20 = 5y2 − 4x2. Center (0, 0). |F | = ae = 2(

32

)= 3. Foci: (0,±3). 1 = y2

4 − x2

5 . e =√

1− −54 = 3

2 .Vertices: (0,±2)

Exercise 18. (x−1)2

4 − (y+2)2

9 = 1.

Center (1,−2). e =√

1− −94 =

√132 ; |F | = 2

√132 =

√13. Foci: (1 +

√13,−2), (1−√13,−2).

Vertices: (5,−2), (−3,−2).Exercise 19. F = ae = 2(2) = 4.

243

x2

4 + y2

−12 = 1. y2

12 + 1 = x2

4

x,y→∞−−−−−→ y = ±√3x.

b2 = a2(1− e2) = 4(1− 4) = −12.Exercise 20. F = ae =

√2 = (1)e. b2 = a2(1− e2) = 1(1− 2) = −1. =⇒ y2 − x2 = 1.

Exercise 21. x2

4 − y2

16 = 1

Exercise 22. (y − 4)2 − (x+1)2

−3 = 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

(3,−1) =⇒ ± 1a2∓ 4

b2= 1

(−1, 0) =⇒ ± 9a2∓ 9

b2= 1

=⇒ (y + 3)2

27/8− (x− 2)2

(27/5)= 1

Exercise 24. x2−13 = y2. 2x

3 = 2yy′. yy′ = x3 .

3x− 2y = C. m = 32 =⇒ y0

92 = x0. 81

4 y20 − 1 = 3y2

0 =⇒ y0 = ±2√69

.

The asymptotes of y2 = x2−13 are y = ± x√

3.

3( ±9√

69

)− 2

(± 2√

69

)=±23√

69= C

3x±√

233

= 2y

Exercise 25. ±x2

a2 +∓y2

b2 = 1 ±x2

a2 ∓ y2

4a2 = 1.(3,−5) → ±9∓ 25

4 = a2; a2 = 114 .

=⇒ x2

11/4− y2

11= 1 .

Quick Review of Parabolas.

F on positive half plane to N .‖X − F‖ = e|(X − F ) ·N + d|

Let N = ~ex; d = 2c; F = (c, 0); e = 1.(x− c)2 + y2 = e2((x− c) + 2c)2 = (x− c)2 + 4c(x− c) + 4c2

y2 = 4cx

Thus, for ellipses, the vertex is equidistant to the focus and directrix (con�rming the other de�nition).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− 1

2 ). 4c = 12, c = 3. Symmetry axis: y = 1. Directrix:(−5

2 , 1).

Exercise 29. x2/6 = y. 4c = 16 c = 1

24 . Vertex: (0, 0). Directrix: y = − 124 . Symmetry axis: x = 0.

Exercise 30. x2 + 8y = 0. 4c = −18 ; c = −1

32 . y = 132 directrix; x = 0 axis.

Exercise 31. (x + 2)2 = 4(y + 94 ). 4c = 4; c = 1. Center (−2,−9/4). Directrix: y = −13/4. Axis: x = −2.

Exercise 32. y = −x2.Exercise 33. x2 = 8y.Exercise 34. (y − 3) = −8(x + 4)2.Exercise 35. c = 5

4 5(x− 74 ) = (y + 1)2

244

Exercise 36. y = ax2 + bx + c

(0, 1) → c = 1 (1, 0) → 0 = a + b + 1(2, 0) → 0 = 4a + 2b + 1 a =12

=⇒ y =12x2 − 3

2x + 1

Exercise 37. 4c(x− 1) = (y − 3)2. 4c(−2) = (−4)2 = 16. c = −2. −8(x− 1) = (y − 3)2.Exercise 38. ‖X − F‖ = ed(X, L) = |(X − F ) ·N − d|L = {(x, y)|2x + y = 10; 2√

5x + y√

5= 10√

5}.

d = N = xL dN ·N = d = xL ·N = 10√5

.

F = 0 =⇒ ‖X‖2 = |X ·N − d|2 =(−2√

5x +− y√

5+

10√5

)2

= x2 + y2

5x2 + 5y2 = (−2x− y + 10)2 = 4x2 + y2 + 100 + 4xy − 40x− 20y

=⇒ x2 + 4y2 − 4xy + 40x + 20y − 100 = 0

13.25 Miscellaneous exercises on conic sections.Exercise 1.

y2

b2= 1− x2

a2y2 = b2 −

(bx

a

)2

= b2

(1−

(x

a

)2)

y = 2∫ a

−a

b

√1−

(x

a

)2

dx = 2∫ 1

−1

ab√

1− x2dx = (ab) area of a circle of radius 1

Exercise 2.

(1) Without loss of generality, let the major axis be 2a in the x-axis. y = b

√1− (

xa

)2

V =∫ a

−a

πb2

(1− x2

a2

)dx = πb2a

∫ 1

−1

(1− x2)dx =43π(1)3b2a

(2) If rotated about the minor axis, suppose, without loss of generality, 2a is the minor axis (just note that x2

a2 + b2

a2 = 1have x, y, a, b as dummy labels).=⇒ V = 4

3π(1)3b2a, where 2a is the minor axis, 2b is the major axis.

Exercise 3. x2

(3/A) + y2

(3/B) = 1 By2 = 3−Ax2 =⇒ y2 = 3B − Ax2

B ; y =√

3B − Ax2

B . So the area inside this ellipse is

2

√1B

∫ √3/A

−√

3/A

√3−Ax2dx = 2

√3B

∫ √3/A

−√

3/A

√1− x2

(3A

)

For the other ellipse equation, x2

3/(A+B) + y2

3/(A−B) = 1. y2 =(

3A−B

)(1− x2

(3/(A+B))

); y =

√3

A−B

√1− x2

(3/(A+B)) .Thus, the area inside this ellipse is

2

√3

A−B

∫ √3

A+B

−√

3A+B

√√√√√1− x√

3A+B

2

Equating the two areas after making an appropriate scale change,

2

√3B

√3A

∫ 1

−1

√1− x2dx = 2

√3

A−B

√3

A + B

∫ 1

−1

√1− x2dx

Thus A2 − B2 = AB =⇒ A2 − BA − B2. Simply try treating B as a number and solve the quadratic equation in termsof A.

A =B ±

√B2 − 4(1)(−B2)

2(1)=

B ±B√

52

=B(1 +

√5)

2

Exercise 4. y = − 4hb2 x2.

∫ b/2

−b/2

(4h

−b2x2 + h

)=

4h

−3b2x3

∣∣∣∣b/2

−b/2

+ h

(b

2+

b

2

)=

2hb

3

Exercise 5. y2 = 8x.∫ 2

0π8tdt = 4π(2)2 = 16pi

Exercise 6. y2 = 2(x− 1). y2 = 4(x− 2).245

(1)

A = 2∫ 2

1

√2(x− 1) + 2

∫ 3

2

√2(x− 1)− 2

√x− 2 =

= 2√

223(x− 1)3/2

∣∣∣∣2

1

+ 2√

223

(x− 1)3/2∣∣∣3

2− 4

23

(x− 2)3/2∣∣∣3

2

= 2√

223

+√

243(2)3/2 − 2

√223− 4

23

= 8/3

(2)∫ 2

1

2(x− 1) = 2(

12x2 − x

)∣∣∣∣2

1

= 2(

12(4− 1)− (2− 1)

)= 1

∫ 3

2

(2(x− 1)− 4(x− 2)) dx =∫ 3

2

(−2x + 6)dx = −x2∣∣32

+ 6x|32 = (−9 + 4 + 6(3− 2)) = 1

=⇒ V = π

∫ 2

1

2(x− 1) + π

∫ 3

2

(2(x− 1)− 4(x− 2)) = 2π

(3) y2

2 + 1 = x, y2

4 + 2 = x

2π

∫ 2

0

((y2

4+ 2

)2

−(

y2

2+ 1

)2)

= 2π

∫ 2

0

(−3y4

16+ 3

)dy = 2π

(−380

y5 + 3y

)∣∣∣∣2

0

=

= 2π

(−3(32)80

+ 6)

= 2π

(−96 + 48080

)= 2π

(38480

)= π

485

Exercise 7. By Apostol's de�nition of conic sections, we are basically given the conic section de�nition with e = 12 . So just

plug in the pt. (0, 4).

x2

a2+

y2

b2= 1

(0,4)−−−→ b = 4 b2 = a2(1− e2) = 16 = a2

(1−

(12

)2)

x2

64/3+

y2

16= 1

Exercise 8. F = 0 ‖X − F‖ = ‖X‖ = ed(X, F ) = |X ·N + d| = x√2

+ y√2

+ 1√2

because for the directrix

y + x = −1

N =(

1√2,

1√2

)1√3y +

x√2

= − 1√2

XL = P + tA

XL ·NN · P = −1/√

2dN = XL XL ·N = d = −1/

√2

So by squaring both sides of the vector equation,

x2 + y2 =x2

2+ xy +

y2

2+

12

+ x + y

x2

2+

y2

2− xy − x− y =

12

x2 + y2 − 2xy − 2x− 2y = 1

Exercise 9. Center (1/2, 2) because we equate the asymptotes to see where they intersect: y = 2x + 1 = −2x + 3.

(y − 2)2

a2− (x− 1/2)2

a2/4= 1

(0,0)−−−→ 4a2− 1

a2=

3a2

= 1

(y − 2)2

3− (x− 1/2)2

3/4= 1

Exercise 10. px2 + (p + 2)y2 = p2 + 2p. x2

p+2 + 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 =√

2p+2

F = ae =√

2. (±√2, 0).246

(2) F = ae =√

2 = a(√

3) =⇒ a =√

23 ; b2 = 2

3 (1− 3) = −43 .

x2

2/3− y2

4/3= 1

Exercise 11. e = 1 for an ellipse.

‖X − F‖ = |X ·N − a| = a−X ·N‖−X − F‖ = ‖X + F‖ = | −X ·N − a| = a + X ·N‖X − F‖+ ‖X + F‖ = 2a

Exercise 12.‖X − F‖ = e|(X − F ) ·N − d| = e(d− (X − F ) ·N)

‖X + F‖ = ed(X, L) = e|(X − F ) ·N + d| = e(−d− (X − F ) ·N)

‖X − F‖ − ‖X + F‖ = 2ed

X → −X so for the other branch, ‖X + F‖ − ‖X − F‖ = 2ed

Exercise 13.

(1) (tx)2

a2 + (by)2

b2 = 1(

bt

)2= a2(1−e2)

t2 =(

at

)2 (1− e2)(2) b2

1 = a21(1− e2) b2

2 = a22(1− e2).

1− b21

a21

= 1− b22

a22

;b21

a21

=b22

a22

x21

a21

+y2

b21

= 1 =

((b2b1

)x)2

a22

+

(b2b1

y)2

(b2y)2

(3) ± (tx)2

a2 ∓ (ty)2

b2 = 1(

bt

)2= −a2(e2−1)

t2 = − (at

)2 (e2 − 1).

b21 = a2

1(e2 − 1) b2

2 = a22(e

2 − 1)

b21

a21

+ 1 = e2 b22

a22

+ 1 = e2 b21

a21

=b22

a22

±(

x

a1

)2

∓(

y

b1

)2

= ±

(b2b1

x)

a2

2

∓(

b2b1

y

b2

)2

= 1

Exercise 14. x2

a2 + y2

b2 = 1. =⇒ xa2 + y

b2 y′ = 0

=⇒ y′ = −b2xya2 = −a2(1−e2)x

a2y = (e2−1)xy

Exercise 15.

(1) y = ax2 + bx + c ty = a(tx)2 + btx + c → y = atx2 + bx + c/t = y = Ax2 + b + C(2) y = tx2, t 6= 0

Exercise 16. x− y + 4 = 0 y = 4√

x (y2 = 16x); y′ = 2x−1/2.

y′(x = 4) = 1 (x, y) = (4, 8).Exercise 17.

(1) If we treat the two given parabolas, y2 = 4p(x − a) and x2 = 4qy, as two vector objects free from any speci�ccoordinate system then we observe that we can disregard the sign of q and p and simply state that they are bothpositive. What matters is that we observe that p and q are the distance of the foci to the vertex for each of therespective parabolas.

Second, observe that a is not given. By diagram, if p, q are given, a must be moved along the x-axis to �t thetangency condition. Thus, in terms of doing the algebra, just eliminate p and q from the relations.

247

If (h, k) is the point of contact,

x2 = 4qyx

2q= y′

y′(h) =h

2q

y2 = 4p(x− a) y = 2√

p√

x− a

y′ =√

p1√

x− a

y′(h) =√

p1√

h− a

(Tangent condition)(

h

2q

)2

=p

h− a=⇒ (h2)(h− a) = (2q)2p

(one point of contact condition) with q =h2

4k, p =

k2

4(h− a)

=⇒ h2(h− a) =(

h2

2k

)2k2

4(h− a)=⇒ (h− a)2 =

h2

16

=⇒ h =2a± a/2

15/8= 4a/3

(2)h

2q=

√p√

h− a

2a

3q=

√p√

a/3=√

3p√a

; 2a√

a = 3√

3pq

=⇒ 4a3 = 27pq2

Exercise 18. First hint: Vector methods triumph over algebraic manipulations of Cartesian coordinates. Think of the locusin terms of vector objects that are coordinate-free and the conic section will emerge. I mean, try evaluating ‖P −A‖2 =(x− 2)2 + (x− 3)2 = (x + y)2

A = (2, 3), N = 1√2(1, 1), X = (x, y).

‖X −A‖ = x + y =√

2(X ·N) =√

2(X ·N − (F ·N − d))

where F ·N = d = A ·N = 5√2

d = distance from focus to the directrix .

y = x + 1 (axis of the hyperbola)

d = 5√2

=√

(2− x)2 + (3− y)2 =√

2(2− x) x = − 12 , y = 1

2

(−12 , 1

2

)must also be the center. y − 1

2 = − (x + 1

2

)is the directrix.

(y − 1

2

)= α

(x + 1

2

)is the general form of the asymptote.

Consider asymptotes in general. ‖X − F‖ = ed(X,L).

‖X − F‖d(X,L)

= e =‖X − F‖

|X ·N − (F ·N − d)| =‖X − F‖

(X − F ) ·N + d|For ‖X − F‖ → ∞, ‖X − F‖ > d. To keep ratio of e, X − F must be ultimately directed by N by a ratio of e.

=⇒ e =‖X − F‖

‖X − F‖ cosφ=

1cosφ

e.g. Consider N = ~ex. x2

a2 − y2

b2 = 1 =⇒ y = bax =

√e2 − 1x.

From the vector equation,

(X − F ) ·N = (x− c, y) ·N =√

(x− c)2 + y2 cos φ = x− c√

(x− c)2 + y2

x− c=

1cosφ

= e;(x− c)2 + y2

(x− c)2= e2;

y2

(x− c)2= e2 − 1 =⇒ y =

√e2 − 1x

248

For our problem, consider the conic section approaching the asymptote. Then the conic section will look more like thoselinear asymptotes.

√(x− 2)2 + (y − 3)2 = x + y

y− 12=α(x+ 1

2 )−−−−−−−−−→√((

x +12

)− 5

2

)2

+(

α

(x +

12

)− 5

2

)2

= x + α

(x +

12

)+

12

=⇒√

(1 + α2)(

x +12

)2

− 5(1 + α)(

x +12

)+

252

x→∞−−−−→√

1 + α2

(1 + α)=⇒ α = 0

The asymptotes are y = 12 and x = −1

2 .

In the second part, each quadrant must be checked. So far, I only have that quadrant II is �lled: points in quadrant III andquadrant IV cannot satisfy the given condition. To see this, consider quadrant II.

‖x−A‖ = −x + y =√

2(x, y)(−1√

2,

1√2

)

For quadrant II, N =(−1√

2, 1√

2

). By diagram, (X − F ) ·N > 0 and X ·N > 0.

A ·N =1√2

−A ·N =−1√

2d =

1√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.‖X − F‖ = d(X,L) = |(X − F ) ·N + d|

x2 + y2 = (X ·N + d1)2 = y2 + 2yd1 + d21

F=0−−−→ x2 = 2yd1 + d21 y′1 =

x

d1

‖X − F‖ = |(X − F ) ·N − d2| = d2 − (X − F ) ·NF=0−−−→ ‖X‖ = d2 − y

x2 + y2 = d22 − 2d2y + y2

=⇒ x2 = d2 − 2d2y y′2 =−x

d2

Point of intersection x20 = 2y0d1 + d2

1 = d22 − 2d2y0

2(d1 + d2)y0 = d22 − d2

1 =⇒ y0 =d2 − d1

2x2

0 = d2(d2 − 2y0) = d2d1

=⇒y′1 =

±√d2d1

d1= ±

√d2

d1

y′2 =∓√d2d1

d2= ∓ 1√

d2d1

Exercise 20.249

(1) Use X → −X symmetry.‖X − F‖ = ed(X, L) = e|(X − F ) ·N + d| = e|X ·N − F ·N + d| = |eX ·N − a|

‖X‖2 − 2X · F + ‖F‖2 = e2(X ·N)2 − 2ea(X ·N) + a2

X → −X =⇒ ‖X‖2 + 2X · F + ‖F‖2 = e2(X ·N)2 + 2ea(X ·N) + a2

=⇒ ‖X2‖+ ‖F‖2 = e2(X ·N)2 + a2

x2 + y2 + c2 = e2x2 + a2 |F | = c = ae(

a2 − c2

a2

)x2 + y2 = a2 − c2 =⇒ x2

a2+

y2

a2 − c2= 1

(2) xa2 + yy′

a2−c2 = 0 =⇒ y′ = −(a2−c2)xya2

xy(y′)2 =(

a2 − c2

a2

)2x3

y

(x2 − y2 − c2)y′ =(

x2 +a2 − c2

a2x2 − a2

)(−(a2 − c2)xya2

)=

= (a4 +−(a2 − c2)x2 − a2x2)(a2 − c2)

a4

x

y

−xy =−xy2

y= −

x(a2 − c2 −(

a2−c2

a2

)x2)

y=

= (a2 − c2)(−a4 + a2x2)x/(a4y)

=⇒ xyy′2 + (x2 − y2 − c2)y′ − xy = 0

(3) For y′, consider − 1y′ at every (x, y).

xy

(−1y′

)2

+ (x2 − y2 − c2)−1y′− xy = 0 =

−xy

y′+ (x2 − y2 − c2)

1y′

+ xy

if y′ 6=0−−−−→ −xy + (x2 − y2 − c2)y′ + xy(y′)2 = 0

Thus S → S since the de�ning differential equation is invariant under the transformation of the slope.Exercise 21. For a circle centered at C, then ‖X − C‖ = r2

0 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‖XC − C‖ = r2

0 .Call the point that all the circles pass through F . Then ‖C − F‖ = ‖C −XC‖. ‖C −XC‖ is by de�nition d(X, L), thedistance from the circle center to the line. ‖C − F‖ = ‖C −X0‖ is by de�nition a parabola.Exercise 22. Consider a circle that's part of the mentioned family that has its center directly below the given circle with radiusr0, 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'svertex is equidistant from the focus and the directrix. Thus, we need to show that d(X, L) is equal to the distance from thecircle 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 ‖(Q + r0N)− C‖ = ‖X2 − C‖.

The tangency condition between circle Q and C means that

(X1 − C) = −α(X1 −Q); α > 0 α =r1

r0

Q− r0N − C = Q−X1 − r0N − C + X1

take the magnitude−−−−−−−−−−→ ‖Q−X1‖2 + ‖X1 − C‖2 + r20 + 2(Q−X1)(X1 − C) + 2(X1 −Q)r0N + 2(C −X)r0N

r20 + r2

1 + r20 + 2αr2

0 + 2r0(1 + α)(x1 −Q) ·N2r2

0 + r21 + 2r1r0 + 2(r1 + r0)(X1 −Q) ·N

250

I had thought the key is to use the law of cosines to evaluate (X1 −Q) ·N = 1α (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 keybecause 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 withthe 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‖Q− C2‖ = r2 + r0 = d(C2, L1). It is a clever arti�cial 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 ‖Q− C‖ = r1 + r0

Tangent to the line L2 = B2 + tA2: (X2 − C) ·A2 = 0

‖X2 − C‖ = r1

Consider L1, a line translated by r0 from L2 away from Q.

If L2 = B2 + tA2, L1 = B2 − r0N + tA2.

Since X2 − C = r1(−N) then X2 − r0N − C = (r1 + r0)(−N) will point from C to L1, because (X2 − r0N) =((B2 + tA2)− r0N) ∈ L1.

=⇒ ‖Q− C‖ = r1 + r0 = ‖X2 − r0N − C‖ = d(C,L1).Exercise 23. Without loss of generality, use y2 = 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′ =√

c/√

x y′(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; α > 0

‖P‖ ‖Q‖ = r20 = α ‖Q‖2

For the line de�ned in Cartesian coordinates as x + 2y − 5 = 0, the vector form of this line is given by

XL = B + tA XL ·N = (x, y) ·(

1√5,

2√5

)= N ·B + 0 =

√5

A =(−2√

5,

1√5

)is a vector that's perpendicular to N ;

B = (1, 2) since we can simply plug it in to satisfy the equation

XL = (1, 2) + t

(−2√5,

1√5

)t ∈ R

Q = B + tA =⇒ ‖Q‖2 = B2 + 2tB ·A + t2A2 = 5 + t(0) + t2 = 5 + t2

(5 + t2)(α) = r20 = 4 α =

45 + t2

P = αQ =4

5 + t2

((1, 2) + t

(−2√5,

1√5

))

251

14.4 Exercises - Vector-valued functions of a real variable, Algebraic operations. Components; Limits, derivatives,and integrals.Exercise 1. F ′ = (1, 2t, 3t2, 4t3).Exercise 2. F ′ = (− sin t, 2 sin t cos t, 2 cos 2t, sec2 t)

Exercise 3. F ′ =(

1√1−t2

, −1√1−t2

)

Exercise 4. F ′ = (2et, 3et).Exercise 5. F ′ =

(sinh t, 2 cosh 2t,−3e−3t

)

Exercise 6.(

t1+t2 , 1

1+t2 , −2t(1+t2)2

)

Exercise 7. F ′ =(

21+t2 + −4t2

(1+t2)2 , 4t(1+t2)2 , 0

).

F ′ · F =4t(1 + t2)(1 + t2)3

+−4t2(2t)(1 + t2)3

+4t− 4t3

(1 + t2)3= 0

Exercise 8.(

12 , 2

3 , e1 − 1)

Exercise 9.

(− cos t, sin t,− ln | cos t|)|π/40 =

(−√3

2+ 1,

√2

2,− ln

√2

2

)

Exercise 10. (ln (1 + et), t− ln (1 + et))|10 =(ln

(1+e2

), 1− ln

(1+e2

))

Exercise 11.(tet − et, t2et − 2tet + 2et,−te−t − e−t

)∣∣10

= (1, e− 2,−2e−1 + 1)Exercise 12. (2,−4, 1) = A.∫ 1

0(te2t, t cosh 2t, 2te−2t)dt =

(12 te2t + −1

4 e2t, t sinh 2t2 − cosh 2t

4 , 2(−12 te−2t − e−2t

4

))∣∣∣1

0=

=(

14e2 + 1

4 , sinh 22 − cosh 2

4 + 14 , 2

(−34 e−2 + 1

4

)).

A ·B = 12e2 +−2 sinh 2 + cosh 2 + −3

2 e−2.Exercise 13. F ′(t) = B = 1 = ‖F ′(t)‖ |B| cos θ(t).Given θ(t) = θ0 constant, ‖F ′(t)‖ must be a constant.

‖F ′(t)‖2 = F ′(t) · F ′(t) = g g′ = 2F ′′(t) · F ′(t) = 0 since ‖F ′‖2 constant .=⇒ F ′′(t) · F ′(t) = 0Exercise 14.

F ′ = 2e2tA +−2e−2tB

F ′′ = 4e2tA + 4e−2tB = 4(F )

Exercise 15. G′ = F ′ × F ′ + F × F ′′ = F × F ′′

Exercise 16.G = F · (F ′ × F ′′)

G′ = F ′ · (F ′ × F ′′) + F · (F ′′ × F ′′ + F ′ × F ′′′) = F · (F ′ × F ′′′)

Exercise 17. If limt→p F (t) = A, ∀jth component,

∀√

ε

n> 0, ∃ δj > 0 such that |Fj(t)−Aj | <

√ε

nif |t− p| < δj

Consider minj=1,...n

δj = δ0

n∑

j=1

|Fj(t)−Aj |2 <

n∑

j=1

(√ε

n

)2

= ε whenever |t− p| < δ0

=⇒ limt→p

‖F (t)−A‖ = 0

If limt→p ‖F (t)−A‖ = 0, ∀ε > 0, ∃ δ > 0 such that√∑n

j=1(Fj(t)−Aj)2 < ε if |t− p| < δ.

=⇒ ∑nj=1(Fj(t)−Aj)2 < ε

ε >∑n

j=1(Fj(t)−Aj)2 > (Fk(t)−Ak)2 > 0

=⇒ ε > |Fk(t)−Ak| if |t− p| < δ.252

Exercise 18. If F is differentiable on I , then

F ′ =n∑

j=1

f ′j~ej f ′j = limh→0

1h

(fj(t + h)− fj(t))

F ′ =n∑

j=1

limh→0

1h

(fj(t + h)− fj(t)) = limh→0

1h

n∑

j=1

(fj(t + h)− fj(t))ej = lim h → 01h

(F (t + h)− F (t))

If F ′(t) = limhto01h (F (t + h)− F (t)) = limh→0

1h

∑nj=1(fj(t + h)− fj(t))ej =

=∑n

j=1 limh→01h (fj(t + h)− fj(t))ej =

∑nj=1 f ′j(t)ej

So F ′ is differentiable.Exercise 19. F ′(t) = 0, ∀j = 1 . . . n, f ′j(t) = 0. By one-dimensional zero-derivative theorem, fj(t) = cj constant. ThusF (t) =

∑nj=1 cj~ej = C on an open interval I .

Exercise 20. 16 t3A + 1

2 t2B + Ct + D

Exercise 21. Y ′(x) + p(x)Y (x) = Q(x). Then ∀j = 1, . . . , n

y′j(x) + p(x)yj(x) = Qj(x)

Since p,Q are continuous on I , and given this initial value condition yk(a) = bk,

yj(x) = e−R x

ap(t)dt

(bj +

∫ x

a

Qj(t)eR t

ap(u)dudt

)

=⇒n∑

j=1

jj(x) = Y (x) = e−R x

ap

(B +

∫ x

a

QeR t

apdt

)

Exercise 22.tF ′ = F + tA =⇒ F ′ + tF ′′ = F ′ + A

=⇒ tF ′′ = A

F ′′(t) = A/t

=⇒ F ′(t) = A ln t + B

=⇒ F (t) = A(t ln t− t) + Bt + C

F (1) = A(−1) + B + C = 2A

tF ′ = F + tA =⇒ At ln t + Bt = A(t ln t− t) + Bt + C + tA

C = 0, B = 3A

F (t) = A(t ln t− t) + 3At

F (3) = A(3 ln 3− 3) + 9A = 3A ln 3 + 6A

Exercise 23.

F ′(x) = exA + xexA +− 1x2

∫ x

1

F (t)dt +1x

F (x) =

= exA + xexA + exA− F (x)x

+F (x)

x= 2exA + xexA = (2 + x)exA

F ′(x) = (2 + x)exA; F (x) = 2exA + A(xex − ex) + C = Axex + exA + C∫ x

1

(Atet + etA + C)dt =(A(tet − et) + etA + Ct

)∣∣x1

=

= A(xex − ex) + exA + C(x− 1)− eA = Axex + C(x− 1)− eA

xexA + Aex +C(x− 1)

x− eA

x=⇒ C = eA

Exercise 24. F ′(t) = α(t)F (t)

=⇒ f ′k(t) = α(t)fk(t); ln(

fk(t)fk(a)

)=

∫ t

a

α(x)dx

fk = fk(a)eR t

aα

F (t) =n∑

j=1

fj(a)eR t

aαej = e

R ta

αn∑

j=1

fj(a)ej = u(t)A

253

14.7 Exercises - Applications to curves. Tangency. Applications to curvilinear motion. Velocity, speed, and accelera-tion.Exercise 1.

~r(t) = ((3t− t3), 3t2, (3t + t3))

v = (3− 3t2, 6t, 3 + 3t2) = 3(1− t2, 2t, 1 + t2)

a = 3(−2t, 2, 2t) = 6(−t, 1, t)

s = 3√

(1− 2t2 + t4 + 4t2 + 1 + 2t2 + t4) = 3√

2(1 + t2)

Exercise 2.r = (c, s, et)

v = (−s, c, et)

a = (−c,−s, et)

s =√

1 + e2t

Exercise 3.r = (3tc, 3ts, 4t)

v = (3(c +−ts), 3(s + tc), 4) = 3(c, s, 0) + (−3ts, 3tc, 4)

a = 6(−s, c, 0) + (−3tc,−3ts, 0)

s =√

9(c2 − 2tsc + t2s2) + 9(s2 + 2tsc + t2c2) + 16 =√

9 + 9t2 + 16

Exercise 4.r(t) = ((t− sin t), (1− cos t), 4 sin (t/2))

v = (1− c, s, 2 cos (t/2))

a = (s, c,− sin (t/2))

s =√

(1− 2c + c2 + s2 + 4c2(t/2)) =

√2− 2c + 4

(1 + c

2

)= 2

Exercise 5.r = (3t2, 2t3, 3t)

v = (6t, 6t2, 3)

a = (6, 12t, 0)

s =√

36t2 + 36t4 + 9 = 3√

4t2 + 4t4 + 1

Exercise 6.r = (t, sin t, (1− cos t))

v = (1, c, s)

a = (0,−s, c)

s =√

2

Exercise 7.r = (ac(ωt), as(ωt), bωt)

r′ = (−ωas(ωt), ωac(ωt), bω) = ω(−as, ac, b)

|r′| = ω√

a2 + b2

r′

|r′| · ez =bω

ω√

a2 + b2=

b√a2 + b2

Exercise 8. |r′| = ω√

a2 + b2

a = ω2(−ac,−as, 0), |a| = ω2a

‖v × a‖‖v‖3 =

1(ω√

a2 + b2)3|(ω3(bas,−bac, a2))| = ω3

√b2a2 + a4

ω3(√

a2 + b2)3=

a

a2 + b2

254

Exercise 9. u = (sin (ωt),− cos (ωt), 0)

v × a = ω3(bas,−bac, a2) = ω3bau + ω3a2ez

A = ω3ba

B = ω3a2

Exercise 10.d

dt(v · v) = a · v + v · a = 2v · a

Exercise 11.(1)

v

|v| · c = cos θ;v · ccos θ

= |v|

(r · c)′ = r′ · c = 2e2t

=⇒ 2e2t

cos θ= |v|

(2)4e4t

cos2 θ= |v|2

from the previous exercise, 12

d

dtv2 =

8e4t

cos2 θ= a · v

Exercise 12. r = (a cosh θ, b sinh θ). a = b = 1.Draw a diagram. You'll immediately see that to get sector OAP, we need to take the area of a right triangle, 1

2 cosh θ sin theta =12 ( x-coordinate )( y-coordinate) and subtract away from it the curved piece below the graph of the hyperbola. Thus

12

cosh θ sinh θ −∫ cosh θ

1

√x2 − 1dx = A(θ) with

x2 − y2 = 1√

x2 − 1 = y

ThenA′(θ) =

12(cosh2 θ + sinh2 θ)−

√cosh2 θ − 1 sinh θ =

12

A(θ) =θ

2

Exercise 13.r = (a cosh ωt, b sinh ωt)

v = ω(a sinhωt, b cosh ωt)

a = ω2(a coshωt, b sinhωt) = ω2r

So a is centrifugal, since a = ω2r.Exercise 14. Parabola.

‖X − F‖ = d(X, L) = |(X − (−2a)ex) · ex| = X · ex + 2a

Let F = 0, ‖X − F‖ = ‖X‖Let X = d1u1

X ′ = d′1u1 + d1u′1

X ′ · u1 = d′1 = X ′ · ex

‖X‖ = d1 = X · ex + 2a

d′1 = X ′ · ex

X′‖X′‖=T

−−−−−→T · u1 = T · ex

cos θ1 = cos θ2

Exercise 15.(1) |a| = 4|r|. =⇒ a = −4r.

r(0)=4ex; v(0)=6ey−−−−−−−−−−−−→x = 4 cos (2t)

y = 3 sin (2t)

(2)(x

4

)2

+(y

3

)2

= 1 . Motion is counterclockwise along an ellipse.

255

Exercise 16. Given that x2 + c(y − x) = 0, then y = x− x2

c

r = (x, y) = (x, x− x2

c)

r′ = (x′, x′ − 2xx′

c)

r′′ = (x′′, x′′ − 2c(x′2 + xx′′))

=⇒x′′ = x′′ − 2

c(x′2 + xx′′)

xx′′ + x′2 = 0 or

(xx′)′ = 0 so then xx′ = constant = k

xdx

dt= k =⇒ xdx = kdt

12(x2

f − x2i ) = k(tf − ti) =⇒ 1

2(−c2) = k(T ) so that k =

−c2

2T12

(( c

2

)2

− c2

)=

12

(−3c2

4

)=−c2

2T(tf − 0)

3T

4= tf

Exercise 17. Given that Y (t) = X(u(t)),

Y ′(t) = X ′(u)u′(t)

So Y ′ is a scalar multiple of X ′; Y ′ ‖ X ′

Y ′′ = X ′′(u)(u′(t))2 + X ′(u)u′′(t)

So Y ′′ ‖ X ′′ only if X ′(u) ‖ X ′′(u) as well. This is not necessarily so.

14.9 Exercises - The unit tangent, the principal normal, and the osculating plane of a curve.Exercise 1. t = 2

(1)

T =v

s=

(1− t2, 2t, 1 + t2)√2(1 + t2)

T (t = 2) =(−3, 4, 5)√

2(5)

T ′ =v′

s− s′v

s2=

(−2t, 2, 2t)√2(1 + t2)

−√

2(2t)(1− t2, 2t, 1 + t2)2(1 + t2)2

T ′(t = 2) =√

2(−2, 1, 2)5

−√

2(2)(−3, 4, 5)25

=√

225

((−10, 5, 10)− (−6, 8, 10)) =√

225

(−4,−3, 0)

N(t = 2) =(−4,−3, 0)

5

(2)

a(t = 2) = 6(−2, 1, 2) = 12√

2T (t = 2) + 6(N)

Exercise 2. t = π256

(1)

T =v

s=

(−s, c, et)√1 + e2t

T (t = π) =(0,−1, eπ)√

1 + e2π

T ′ =v′

s− s′v

s2=

(−c,−s, et)√1 + e2t

− e2t

√1 + e2t

(−s, c, et)(1 + e2t)

T ′(t = π) =(1, 0, eπ)√

1 + e2π− e2π

√1 + e2π

(0,−1, eπ

(1 + e2π)=

=(1 + e2π, e2π, eπ)

(1 + e2π)3/2

N(t = π) =(1 + e2π, e2π, eπ)√1 + 3e2π + 2e4π

(2)

a(t = π) =e2π

√1 + e2πT (t = π) +

√1 + 3e2π + 2e4πN(t = π)

1 + e2π

Exercise 3.(1)

T =v

s=

3(1, 0, 0) + (0, 0, 4)5

=(3, 0, 4)

5

T ′ =(v

s

)′=

v′

s− s′v

s2=

=6(−s, c, 0) + (−3tc, 3ts, 0)√

9 + 9t2 + 16−

9t√9+9t2+16

(3(c, s, 0) + (−3ts, 3tc, 4))

9 + 9t2 + 16

T ′(t = 0) =65(0, 1, 0)− 0 =

65(0, 1, 0)

N ′(t = 0) = (0, 1, 0)

(2)a(t = 0) = 6(0, 1, 0) = 0T + 6N ′(t = 0)

Exercise 4.(1)

T =v

s=

12(1− c, s, 2 cos (t/2))

T ′ =12(s, c,− sin (t/2))

|T ′| = 12

√(1 + s2(t/2))

N =T ′

|T ′| =(s, c,− sin (t/2))√

(1 + s2(t/2))

T (t = π) = (1, 0, 0)

N(t = π) = (0,−1/√

2,−1/√

2)

(2)a(t = π) = (0,−1,−1)

a(t = π) = 0T + 2N(t = π)

Exercise 5. t = 1(1)

T =(2t, 2t2, 1)√4t2 + 4t4 + 1

T (t = 1) = 1/3(2, 2, 1)

T ′ =v′

s− s′

s2v =

(2, 4t, 0)√4t2 + 4t4 + 1

− 4t + 8t3

(4t2 + 4t4 + 1)3/2(2t, 2t2, 1)

T ′(t = 1) =(2, 4, 0)

3− 12

27(2, 2, 1) = 1/9(−2, 4,−4)

N(t = 1) = 1/3(−1, 2,−2)

257

(2)a = (6, 12, 0) = 4(3T ) + 2(3N) = 12T + 6N

Exercise 6.(1)

T =v

s=

(1, c, s)√2

T (t = π/4) =(

1√2,12,12

)T ′ =

v′

s− s′v

s2=

(0,−s, c)√2

N(t = π/4) = (0,−1/√

2, 1/√

2)

(2)a(t = π/4) = (0,−1/

√2, 1/

√2) = N

Exercise 7.a = 0

v = v0 = const.r = v0t + r0

Exercise 8.‖v × a‖‖v‖ = ‖ v

‖v‖ × a‖ = ‖T × a‖

a = s′T + sT ′, so the normal component of a is s|T ′| since sT ′ = s|T ′|N .

T × a = sT × T ′

‖T × a‖ = ‖sT × T ′‖ =√|sT |2|T ′|2 − (sT · T ′)2 =

T ·T ′=0−−−−−→ |sT ||T ′| = s|T ′|

Exercise 9.(1)

v = v0

r = v0t + r0 r ∈ line(2) Counterexample: imagine the particle moving at constant speed, but in an S-shape in one plane and then an S shape

in another plane, out of the �rst plane.(3)

a = a0

v = a0t + v0

r =a0

2t2 + v0t + r0

r = a02 t2 + v0t + r0 de�nes a plane.

(4) v · a = 0a = s′T + sT ′

a · T = s′ = 0 =⇒ s constant

=⇒ a = sT ′

v = kT

So s is constant and a = sT ′. We already showed in part (b) of this exercise that there's a counterexample.

Exercise 10.(1)

d

dt(r × v) = v × v + r × a = r × a = 0

=⇒ r × v = const.258

(2) Use what you've learned in these sections: use what you've learned about a being completely in the osculatingplane . The osculating plane idea is very useful.

r × v = c =⇒ r × a = 0

a = s′T + sT ′ =⇒ r × (s′T + sT ′) = 0

Possibilities:r = c(s′T + sT ′) (r is a scalar multiple of s′T + sT ′) =⇒ r ∈ osculating planea = 0, v = v0, r = v0t + r0, r is on a liner = 0, r stationary

(3) If a = |a(t)|r(t)/|r(t)| then r × a = 0 since a is a scalar multiple of r. Then from above, r × a = 0 imples that r ison a plane.

(4) r × v not necessarily constant if r lies in a plane.

Counterexample:

r = (tc, t2s, 0)

v = (c− ts, 2ts + t2c, 0)

r × v = (0, 0, 2t2cs + t3c2 − t2sc− t3s2) = (0, 0, t2sc + t3(c2 − s2))

Exercise 11. Given that r ∈ plane, the v is also in the plane, otherwise, for some time t, r(t+∆t) = r + v∆t out of the plane.(1) We want T × a = v × a = 0.

Given that T · c = cos θ0 = k0, (T · c)′ = 0 = T ′ · c + T · c′ = T ′ · c.c 6= 0, so if T · c 6= 1, T ′ · c = 0 implies T ′ = 0. Then T × a = 0.

If T · c = 1, v = sc a = v′ = s′c

v × a = sc× s′c = ss′(c× c) = 0

(2) Example:r = (c(t), s(t), t3)

v = (−s, c, 3t2)a = (−c,−s, 6t)

Exercise 12.(1) If y = g(t) ≷ 0, vx ≶ 0. counterclockwise(2)

3x2 + y2 = 1

r(t) = (f(t), g(t)) = (x, y) = (x,±√

1− 3x2)

r′ =(

x′,∓ 3x√1− 3x2

x′)

x′ = vx = −g(t) = y

∓3x′x√1− 3x2

=∓3y

yx = ∓3x

so vy = 3x (since we want counterclockwise motion)

(3) x′ =√

1− 3x2 =⇒∫ xf

x0

dx√1− 3x2

=∫ tf

ti

dt = T =arcsin (

√3x)√

3

∣∣∣∣∣

xf

x0

=1√3(π/2−−π/2) = π/

√3

∫ xf

x0

= T =arccos (

√3x)√

3

∣∣∣∣∣

1/√

3

−1/√

3

=0− (−π)√

3= π/

√3 (we need T to be positive)

=⇒ 2π/√

3

259

Exercise 13.r = (x, y) = (x(t), y(t))

r′ = (x′, y′)r · ex = x =

√x2 + y2 cos θ = r cos θ

r′ · ex = x′ = |r′| cosφ

tan φ =43

cot θ =⇒ ±y′

x′=

43

x

y

y′

x′=

dy

dx=−4x

3y(since x

r> 0 for C in �rst quadrant, and dy

dx< 0 in �rst quadrant)

=⇒ y2 =−43

x2 + C

(3/2,1)−−−−→ x2

3+

y2

4= 1

Exercise 14.T =

(x′, y′)√x′2 + y′2

N =(y′,−x′)√x′2 + y′2

(x, y) + (±d)N = (x, y) +(±d)(y′,−x′)√

x′2 + y′2= (x2, 0)

x +((±)d)y′√x′2 + y′2

= x2

y +((±)d)(−x′)√

x′2 + y′2= 0

|x2 − x| = 2 =

∣∣∣∣∣x +±dy′√x′2 + y′2

− x

∣∣∣∣∣ =|d||y′|√x′2 + y′2

Divide the above 2 by the expression for y =⇒ |d||y′|±dx′

=2y

±dy

dx=

2y

=⇒ ±ydy = 2dx

=⇒ ±12(y2

f − y2i ) = 2(xf − xi)

(1,2)−−−→ ±(y2 − 4) = 4(x− 1) =⇒y2 = 4x

y2 = −4x + 8

Exercise 15. r′ = A× r

(1) r′′ = a = A× v =⇒ a ·A = A ·A× v = 0(2)

v = A× r

a = A× v

a = s′T + sT ′, a · T = s′

a · T = (A× v) · T = 0 = s′ =⇒ s constantNow note that

|r|′ = (√

r · r)′ =12

1√r · r (v · r + r · v) =

v · r√r · r

however v · r = (A× r) · r = 0

=⇒ |r|′ = 0So the length of the position vector r is held constant. So if r(0) = B, then |r| = ‖B‖ for all time t > 0.

We have|r′(t)| = s = |A× r| = ‖A‖‖r‖ sin θAr

260

Now s is constant, ‖A‖, ‖B‖ are constant. Then sin θAr must be constant as well. r(0) = B so θAr = θAB .

|r′(t)| = s = |A× r| = ‖A‖‖B‖ sin θAB

(3) See sketch. r starts off as B and r rotates around A like a top.

Exercise 16.

(1)

Y (t) = X[u(t)]

Y ′(t) =dY (t)

dt=

d

dt(X(u(t))) =

dX

du(u)

du

dt

Y ′(t)‖Y ′(t)‖ = TY (t) =

ddt (X(u(t)))‖ d

dt (X(u(t)))‖ =dX(u)

dududt

‖dX(u)du ‖ ∣∣du

dt

∣∣ = TX(u(t))dudt∣∣dudt

∣∣

since for X(u(t)),dX

du(u) = X ′ =⇒ X ′

‖X ′‖ = TX(u) =dX(u)

du

‖dX(u)du ‖

Now∣∣∣∣∣

dudt∣∣dudt

∣∣

∣∣∣∣∣ = 1 always, so for

strictly increasing u ,du

dt> 0 =⇒ du

dt/

∣∣∣∣du

dt

∣∣∣∣ = 1

strictly decreasing u ,du

dt< 0 =⇒ du

dt/

∣∣∣∣du

dt

∣∣∣∣ = −1

strictly increasing u , TY (t) = TX(u)

strictly decreasing u , TY (t) = −TX(u)

(2)

(TY )′ =(

Y ′

|Y ′|)′

=Y ′′

|Y ′| −Y ′

|Y ′|2 |Y′|′

Y ′ = X ′u′

Y ′′ =d

dt

(dX(u)

duu′(t)

)=

d2X(u)du2

u′2 +dX(u)

duu′′ = X ′′u′2 + X ′u′′

Y ′′

|Y ′| =X ′′u′2 + X ′u′′√

X ′u′ ·X ′u′

|Y ′|′ = (√

Y ′ · Y ′)′ =Y ′ · Y ′′√

Y ′ · Y ′ =X ′u′ · (X ′′u′2 + X ′u′′)√

X ′u′ ·X ′u′

−Y ′

|Y ′|2 |Y′|′ =

−(X ′)(X ′ · (X ′′u′2 + X ′u′′))(X ′)2

√X ′u′ ·X ′u′

Y ′′

|Y ′| −Y ′

|Y ′|2 |Y′|′ =

X ′′u′2 + X ′u′′√X ′u′ ·X ′u′

− (X ′)(X ′ ·X ′′u′2 + X ′2u′′)X ′2(

√X ′u′ ·X ′u′)

=X ′′u′2√

X ′u′ ·X ′u′− X ′(X ′ ·X ′′)u′2

X ′2√X ′u′ ·X ′u′=

u′2 = |u′|2√

u′2 = |u′|−−−−−−−−−−→

(X ′′

|X ′| −X ′|X ′|′|X ′|2

)|u′|

261

‖T ′Y ‖2 =(

Y ′′

|Y ′| −Y ′

|Y ′|2 |Y′|′

)·(

Y ′′

|Y ′| −Y ′

|Y ′|2 |Y′|′

)=

Y ′′2

Y ′2 −2Y ′′ · Y ′

|Y ′|3Y ′ · Y ′′

|Y ′| +Y ′2

|Y ′|4(Y ′ · Y ′′)2

Y ′2 =

=Y ′′2

Y ′2 −(Y ′ · Y ′′)2

(Y ′)4

=(X ′′u′2 + X ′u′′)2

(X ′u′)2− (X ′u′ · (X ′′u′2 + X ′u′′))2

(X ′u′)4=

=X ′′2u′4 + 2X ′′ ·X ′u′2u′′ + X ′2u′′2

(X ′u′)2− (X ′ ·X ′′)2u′6 + 2X ′ ·X ′′u′4X ′2u′′ + (X ′2)2u′2u′′2

(X ′u′)4=

=(

X ′′2

X ′2 −(X ′ ·X ′′)2

X ′4

)u′2

T ′Y‖T ′Y ‖

=T ′X |u′|‖T ′X‖|u′|

=T ′X‖T ′X‖

=⇒ NY = NX

Thus, the osculating plane is invariant under a parameter change, since the osculating plane can still be made upof NY = NX and TY = ±TX .

14.13 Exercises - The de�nition of arc length, Additivity of arc length, The arc-length function.Exercise 1.

r(t) = a(1− cos t, t− sin t) 0 ≤ t ≤ 2π, a > 0

r′(t) = a(sin t, 1− cos t)v2 = a2(sin2 t + 1− 2 cos t + cos2 t) = 2a2(1− cos t)

v =√

2a√

1− cos t = 2a sin t/2∫vdt = 2a

∫sin t/2 = −(4a) cos t/2|2π

0 = 8a

Exercise 2.r(t) = (et cos t, et sin t); 0 ≤ t ≤ 2

r′(t) = (et(c− s), et(s + c))

v2 = e2t(c2 + s2 + s2 + c2) = 2e2t =⇒ v =√

2et

∫ 2

0

√2etdt =

√2(e2 − 1)

Exercise 3.r(t) = (a(cos t + t sin t), a(sin t− t cos t)) = a(cos t + t sin t, sin t− t cos t)

r′ = a(−s + s + tc, c− c + ts) = a(tc, ts)v = at

∫ 2π

0

at = a124π2 = 2π2a

Exercise 4.

r(t) =(

c2

acos3 t,

c2

bsin3 t

)= C2

(c3

A,s3

B

) 0 ≤ t ≤ 2π

c2 = a2 − b2, 0 < b < a

r′ = 3C

(−c2s

A,s2c

B

)

v2 = 9C2

(c4s2

A2+

s4c2

B2

)= 9C4c2s2

(c2

A2+

s2

B2

)

v = 3C2|cs|√

c2

A2+

s2

B2= 3C2|cs|

√B2 + C2s2

AB262

∫cs

√B2 + C2s2 =

13C2

(B2 + C2s2)3/2

s =3C2

AB

(∫ π/2

0

cs√

B2 + C2s2 −∫ π

π/2

cs√

B2 + C2s2 +∫ 3π/2

π

cs√

B2 + C2s2 −∫ 2π

3π/2

cs√

B2 + C2s2

)=

=3C2

AB

43C2

((B2 + C2)3/2 −B3) = 4(A3 −B3)/AB

Exercise 5.r(t) = (a(sinh t− t), a(cosh t− 1))

r′(t) = a(cosh t− 1, sinh t)

v2 = a2(cosh2 t− 2 cosh t + 1 + sinh2 t) = a2(2)(cosh t)(cosh t− 1)

Knowing that cosh (2t) = cosh2 t + sinh2 t = 2 cosh2 t− 1 = 1 + 2 sinh2 t

s =∫|v|dt =

∫ T

0

a√

2√

cosh t(cosh t− 1)dt =√

2a

∫ T

0

√2 cosh2 t/2− 1

√2 sinh t/2dt =

u = cosh t/2

du =12

sinh t/2dt−−−−−−−−−−−−−−→ 2a

∫ √2u2 − 12du

√2u = y

du =dy√

2−−−−−−−−−→ 4a√2

∫ √y2 − 1dy

∫ √y2 − 1dy = y

√y2 − 1−

∫y2

√y2 − 1

= y√

y2 − 1−∫

y2 − 1 + 1√y2 − 1

= y√

y2 − 1−∫ √

y2 − 1dy +∫ −1√

y2 − 1

since (ln (y ±√

y2 − 1))′ =1

y ±√

y2 − 1

(1± y√

y2 − 1

)= ± 1√

y2 − 1

=⇒∫ √

y2 − 1dy =12(y

√y2 − 1− ln (y +

√y2 − 1))

s =4a√

212

(√2 cosh t/2

√2 cosh2 t/2− 1− ln (

√2 cosh t/2 +

√2 cosh2 t/2− 1)

)∣∣∣∣T

0

=

= 2a(coshT/2√

coshT − 1) +−√

2a ln

(√2 cosh T/2 +

√coshT√

2 + 1

)

Exercise 6.r(t) = (sin t, t, (1− cos t)); (0 ≤ t ≤ 2π)

r′ = (c, 1, s)

v2 = c2 + 1 + s2 = 2 =⇒∫ 2π

0

√2dt =

√2(2π)

Exercise 7.r(t) = (t, 3t2, 6t3), (0 ≤ t ≤ 2)

r′ = (1, 6t, 18t2)

v2 = 1 + 36t2 + 182t4

s =∫ 2

0

√(18t2)2 + 2(18)t2 + 1dy =

∫ 2

0

(18t2 + 1)dt = (6t3 + t)∣∣20

= 50

Exercise 8.r = (t, log (sec t), log (sec t + tan t)), (0 ≤ t ≤ π

4)

r′ = (1, tan t, sec t)

v2 = 1 + tan2 t + sec2 t = 2 sec2 t

s =∫ π/4

0

√2 sec tdt =

√2 (ln | tan t + sec t|)|π/4

0 =√

2 ln (1 +√

2)

263

Exercise 9.r(t) = (a cos ωt, a sin ωt, bωt), (t0 ≤ t ≤ t1)

r′ = (ω)(−as(ωt), ac(ωt), b)

v2 = ω2(a2 + b2)

s =∫ t1

t0

|ω|√

a2 + b2 = |ω|√

a2 + b2(t1 − t0)

Exercise 10.r = (x, y) = (g(y), y)

r′ =(

dg(y(t))dy

dy

dt,dy

dt

)

r′2 =

((dg

dy

)2

+ 1

) (dy

dt

)2

s =∫ tf

t0

√1 +

(dg(y(t))

dy

)2dy

dtdt =

∫ y(tf )

y(t0)

√1 +

(dg

dy

)2

dy

Exercise 11. Given y2 = x3,

r = (x,±x3/2)dy

dx= ±3

2x1/2

∫ 1

0

√1 +

94x =

827

(1 +94x)3/2

∣∣∣∣1

0

=827

(26√

138

− 2

)

r = (y2/3, y)dg

dy=

23y−1/3

∫ 1

−1

√1 +

(23

)2

y−2/3dy =∫ 1

−1

√y2/3 +

(23

)2

|y|1/3dy = (y2/3 + (2/3)2)3/2

∣∣∣1

0− (y2/3 + (2/3)2)3/2

∣∣∣0

−1=

=(

139

)3/2

−(

23

)3

−((

23

)3

−(

139

)3/2)

= 2

(13√

1327

− 827

)

Exercise 12.r = (cos θ, sin θ)

r′ = (− sin θ, cos θ)

v2 = 1

s =∫ θ1

θ0

1dθ = (θ1 − θ0) = 2(

θ1 − θ0

2

)

So the arc AB has a length equal to twice the area of the sector.Exercise 13.

(1)y = ex, 0 ≤ x ≤ 1

∫ 1

0

√1 + e2xdx

264

(2)(x, y) = (t + log t, t− log t); 1 ≤ t ≤ e

r′ = (1 +1t, 1− 1

t)

r′2 = 2(1 +(

1t

)2

);=⇒√

2

√1 +

(1t

)2

∫ e

1

√2√

1 + (1/t)2dt

t = ex

dt = exdx−−−−−−−−→

∫ 1

0

√2√

1 + e−2xexdx =√

2∫ 1

0

√1 + e2xdx

Exercise 14.y = c cosh (x/c)

dy

dx= sinh (x/c)

∫ a

0

√1 + sinh2 (x/c)dx =

∫ a

0

coshx/adx = c sinhx

c

∣∣∣a

0= c sinh

a

c

So indeed,∫ a

0c cosh (x/a) dx gives the area underneath a curve cosh x/a from x = 0 to x = a.

Exercise 15. Given y = cosh x, (0, 1) and (x, cosh x) is sinhx if x > 0∫ x

0

√1 + sinh2 tdt = sinh t|x0 = sinhx

Exercise 16.

A =∫ b

a

f(x)dx = k

∫ b

a

√1 +

(dy

dx

)2

dx =⇒∫ b

a

k

√1 +

(dy

dx

)2

− f(x)

dx = 0

±√(y

k

)2

− 1 =dy

dx; ±(x− x0) =

∫ y

y0

dy√(yk

)2 − 1∫ y

y0

dyq( y

k )2−1could be evaluated by cosh u = y/k substitution, and then solve for u, (inverse of cosh u), by cosh u =

eu+e−u

2 = y/k and then use the quadratic equation trick to get eu.

±(x− x0) = k ln

(yk +

√y2

k2 − 1)

(yk +

√y2

k2 − 1) Let C0 =

y0

k±

√y20

k2− 1

(C0e

(±(x− x0)k

)− y

k

)2

=y2

k2− 1

C20e

(±2(x− x0)k

)+ 1 = 2C0e

(±(x− x0)k

)y

k

y =k

(C0e

(±(x−x0)

k

)+ e

(∓(x−x0)

k

))

2

If we choose y0 = k, so that C0 = 1, then y = k cosh(

x− x0

k

). Also note that y = k works as well.

Exercise 17.r = (a sin t, b cos t) 0 < b < a

r′ = (ac,−bs)

r′2 = a2c2 + b2s2 = a2(1− s2) + b2s2 = a2

(1−

(a2 − b2

a2

)s2

)

L =∫ 2π

0

|a|√

1− e2 sin2 tdt = 4a

∫ π/2

0

√1− e2 sin2 tdt

265

Exercise 18.

r(t) = (a(t− sin t), a(1− cos t), b sin t/2) = a(t− sin t, (1− cos t),b

asin t/2)

r′ = a(1− cos t, sin t,b

2acos t/2)

r′2 = a2(1− 2 cos t + 1 +(

b

2a

)2

cos2 t/2) = 2a2(1− cos t +(

b

4a

)2

(1 + cos t)) =

= 2a2(2 sin2 t/2 +(

b

4a

)2

(2 +−2 sin2 t/2)) = 4a2

((b

4a

)2

+ k2 sin2 t/2

)= 4a2(1− k2 + k2 sin2 t/2) =

= 4a2(1− k2 cos2 t/2)

|r′| = 2a√

1− k2 cos2 t/2∫ 2π

0

2a√

1− k2 cos2 t/2 dt = 4a

∫ π

0

√1− k2 cos2 t dt

t=π/2−θ−−−−−−→ = (−4a)∫ −π/2

π/2

√1− k2 sin2 t dt = 8a

∫ π/2

0

√1− k2 sin2 t dt

Exercise 19.A ·B = cos π/3 = 1/2

(A×B)2 = A2B2 − (A ·B)2 = 1− 1/4 = 3/4

r = tA + t2B + 2(

23t

)3/2

A×B

r′ = A + 2tB + 2(

23t

)1/2

A×B

r′2 = 1 + 2t

(12

)+

2t

2+ (2t)2 + 4

(2t

3

)34

= 1 + 4t + 4t2 = (2t + 1)2

L =∫ T

0

(2t + 1)dt = T 2 + T = 12 =⇒ (T + 4)(T − 3) = 0 T = 3

Exercise 20.(1)

The circle rotates about its center and the circlesrolls (moves along x-axis) without slipping, translating its origin by how far its circumference moved

through.

θ0 = −π/2a(cos θ,− sin θ) (clockwise rotation of pt. on circumference)a(cos (θ − θ0),− sin (θ − θ0)) = a(− sin θ,− cos θ)(θa, a) is the position of the circle center.

(θa, a) + a(− sin θ,− cos θ) = a(θ − sin θ, 1− cos θ)

(2) dydx = sin θ

1−cos θ = 2 sin θ/2 cos θ/22 sin2 θ/2

= cot θ/2

T =(1, cot θ/2)√1 + cot2 θ/2

= (sin θ/2, cos θ/2)

T · ex = sin θ/2 = cos φ = cos(

π

2− θ

2

)

φ =π − θ

2

We'll show that the tangent line passes through the highest point at each place the marked point rotates through.Marked point on the circle rotated through by θ: P = a(θ − sin θ, 1− cos θ)Tangent line: P + sT = a(θ − sin θ, 1− cos θ) + s(sin θ/2, cos θ/2)

266

Highest pt. on the circle: (aθ, 2a)

equate the y-coordinates: 2a = a(1− cos θ) + s cos θ/2 =⇒ s = 2a| cos θ/2|

equate the x-coordinates:a(θ − sin θ) + s sin θ/2 = aθ − a sin θ + s sin θ/2 =

= aθ − a sin θ + a2 cos θ/2 sin θ/2 =

= aθ

So indeed, the tangent line will always intersect the highest point of the circle.Exercise 21.

dY

dt=

dX

du

du

dt∫ d

c

‖Y ′(t)‖dt =∫ d

c

‖X ′(u(t))‖∣∣∣∣du(t)

dt

∣∣∣∣ dt

g(u) = t

u = u(t)du

dt= 1/

dg

du(u(t))

g(u) = t is strictly increasing as it traces the arc, since c < t < d.Then du

dt > 0

=⇒∫ d

c

‖X ′(u)‖(

du

dt

)dt =

∫ u(d)

u(c)

‖X ′(u)‖du

Exercise 22. For t = 1k , k = 2, . . . , 2n

‖π(P )‖ =2n∑

k=2

‖r(tk)− r(tk−1)‖+ ‖r(t = 0)− r(t2n)‖+ (term to close up this polygon)

‖r(t = 0)− r(t2n)‖ =(

12n

)√2

r = (t, t cos( π

2t

))

r(tk)− r(tk−1) =( −1

k(k − 1),1k

cos(

kπ

2

)− 1

k − 1sin

(kπ

2

))

‖r(tk)− r(tk−1)‖2 =(

1k(k − 1)

)2

+1k2

cos2(

kπ

2

)+−2 cos

(kπ2

)sin

(kπ2

)

k(k − 1)+

1(k − 1)2

sin2

(kπ

2

)=

=(

1k(k − 1)

)2

+1k2

cos2(

kπ

2

)+

1(k − 1)2

sin2

(kπ

2

)since k ∈ Z

k = 2, . . . , 2nk = 2j j = 1, . . . , n

‖r(tk)− r(tk−1)‖2 =(

1k(k − 1)

)2

+1k2

‖r(tk)− r(tk−1)‖ =1k

√1 +

1(k − 1)2

>1k

k = 2j + 1, j = 1, . . . , n− 1

‖r(tk)− r(tk−1)‖2 =(

1k(k − 1)

)2

+(

1k − 1

)2

=(

1k − 1

)2 (1 +

1k2

)

‖r(tk)− r(tk−1)‖ =1

k − 1

√1 +

1k2

>1

k − 1>

1k

These don't seem like very �tight� inequalities, but it'll work for the purposes of this problem.

‖π(P )‖ =2n∑

k=2

‖r(tk)− r(tk−1)‖+ ‖r(t = 0)− r(t2n)‖+ (term to close up this polygon) >

>

2n∑

k=2

1k

+√

22n

+ ‖r(t = 0)− r(t = 1)‖ =2n∑

k=2

1k

+√

22n

+ 1 > >

2n∑

k=1

1k

267

Now, by de�nition, Λ, the arclength, is the least upper bound on all possible polgonal approximations to the length of a curve.∑2nk=2

1k is a series that is unbounded as n →∞. So Λ does not exist.

14.15 Exercises - Curvature of a curve.Exercise 1.

(1)

r = ((3t− t3), 3t2, (3t + t3))

v = (3− 3t2, 6t, 3 + 3t2) = 3(1− t2, 2t, 1 + t2)

a = 3(−2t, 2, 2t) = 6(−t, 1, t)

a(t = 2) = 6(−2, 1, 2); |a| = 18

v(t = 2) = 3(−3, 4, 5); |v| = 15√

2

a · v = 18(6 + 4 + 10) = 360

κ =

√1821522− (18 · 20)2

153√

2 · 2 =175

(2)

r = (c, s, et)

v = (−s, c, et)

a = (−c,−s, et)

a(t = π) = (1, 0, eπ) |a| =√

1 + e2π

v(t = π) = (0,−1, eπ) |v| =√

1 + e2π

a · v = e2π

κ =√

1 + 2e2π + e4π − e4π

(1 + e2π)3/2=

√1 + 2e2π

(1 + e2π)3/2

(3)r = (3tc, 3ts, 4t)

v = (3(c− ts), 3(s + tc), 4)

a = 6(−s, c, 0) + (−3tc, 3ts, 0)

a(t = 0) = 6(0, 1, 0) |a| = 6

v(t = 0) = (3, 0, 4) |v| = 5a · v = 0

κ =625

(4)

r′ = (1− c, s, 2 cos t/2)

r′′ = (s, c,− sin t/2)

r′′ · r′ = s− cs + sc− 2 cos t/2 sin t/2 = 0

v = ‖(2, 0, 0)‖ = 2

a = ‖(0,−1,−1)‖ =√

2

κ =√

24

(5)

r = (3t2, 2t3, 3t)

r′ = (6t, 6t2, 3) = 3(2t, 2t2, 1)

r′′ = 3(2, 4t, 0)

r′′ · r′ = 9(4t + 8t3) t=1−−→ 9(12)

a = 3√

(4 + 16t2)

a(t = 1) = 6√

5

v = 3√

4t2 + 4t4 + 1; v(t = 1) = 9

‖a× v‖v3

=

√(6√

5)292 − (9(12))2

272=

227

(6)

r(t) = (t, sin t, (1− cos t))

r′(t) = (1, cos t, sin t)

r′′ = (0,−s, c)

r′′ · r′ = 0a = 1

κ =‖a× v‖

v3=√

223/2

=12

268

Exercise 2.r = (ac(ωt), as(ωt), bωt)

r′ = ω(−as(ωt), ac(ωt), b)

r′′ = ω2(−ac(ωt),−as(ωt), 0)

r′ · r′′ = 0

|r′| = ω√

a2 + b2

|r′′| = ω2a

‖r′′ × r′‖|r′|3 =

a√

a2 + b2

(a2 + b2)3/2=

a

a2 + b2

Exercise 3. Given A,B, A ·B = cos θ. A,B �xed. |A| = |B| = 1.

r′ = A× r

r′′ = A× r′=⇒ r(0) = B, ‖r(t)‖ = 1

r′′ × r′ = (A× r′)× (A× r) = (A× (A×R))× (A× r) =

= (A× r)× ((A× r)×A) = (A · (A× r))(A× r)− (A× r)2A = −(A× r)2A

‖r′′ × r′‖‖A× r‖3 =

1‖A× r‖ =

1‖B‖ sin θ

Exercise 4.(1) r(t) lies on a plane since a plane M = {P + sA + tB}

r(t) = 4(c, s, c) = 4c(1, 0, 1) + 4s(0, 1, 0) = 4cA + 4sBr(t) ∈ M1 where M1 = {0 + sA + tB} = {sA + tB}

Normalize A,B → A = (1,0,1)√2

; B = (0, 1, 0)

X = 4√

2cA + 4sB

Recall an ellipse: ‖X − F‖ = ed(X, L) = e|(X − (F + dN)) ·N | = e|(X − F ) ·N − d|Suppose ‖F‖ = f and suppose F = fA

X − F = (4√

2c− f)A + 4sB

‖X − F‖2 = 32c2 − 8√

2cf + f2 + 16s2 = 16 + 16c2 − 8√

2cf + f2

e2((X − F ) ·N − d)2 = e2(((X − F ) ·N)2 − 2((X − F ) ·N)d + d2)

((X − F ) ·N)2 = 32c2 − 8√

2cf + f2 e2=1/2−−−−−→ 16c2 − 4√

2cf +12f2

−2e2((X − F ) ·N)d = −d(4√

2c− f) =⇒ d = f

e2d2 =12f2 =⇒ 16 + f2 = 2f2 or f = 4

Indeed for‖X − F1‖ = ‖(4

√2c− 4)A + 4sB‖ =

√32c2 − 32

√2c + 16 + 16s2 =

=√

16c2 − 32√

2c + 32 = 4(√

2− c)

‖X − F2‖ = ‖(4√

2c + 4)A + 4sB‖ = 4(√

2 + c)

=⇒ ‖X − F1‖+ ‖X − F2‖ = 8√

2

A Cartesian equation for the plane containing this ellipse would be −x + z = 0 .(2)

r = 4(c, s, c)

r′ = 4(−s, c,−s)

r′′ = 4(−c,−s,−c) = −4(c, s, c)

r′2 = 16(1 + s2)

r′′2 = 16(1 + c2)

r′ · r′′ = −16(−sc) = 16sc

‖r′′ × r′‖|r′|3 =

√162(1 + c2)(1 + s2)− 162(sc)2

|r′|3 =√

24(1 + s2)3/2

1κ

= 2√

2(1 + s2)3/2

269

Exercise 5.r = (et, e−t,

√2t)

r′ = (et,−e−t,√

2)

r′′ = (et, e−t, 0)

r′′ × r′ = (√

2e−t,√

2et,−1− 1) = (√

2e−t,√

2et,−2)

‖r′′ × r′‖ =√

2e−2t + 2e2t + 4 =√

2(et + e−t)

r′2 = e2t + e−2t + 2 = (et + e−t)2

‖r′′ × r′‖|r′|3 =

√2

et + e−t

Exercise 6.(1)

r(t) = (x(t), y(t))

r′ = (x′, y′)

r′′ = (x′′, y′′)

‖r′′ × r′‖|r′|3 =

√(x′′2 + y′′2)(x′2 + y′2)− (x′′x′ + y′′y′)2

(x′2 + y′2)3/2=

=

√x′′2x′2 + x′′2y′2 + y′′2x′2 + y′′2y′2 − (x′′2x′2 + 2x′′x′y′′y′ + y′′2y′2)

(x′2 + y′2)3/2=

=

√(x′′y′ − y′′x′)2

(x′2 + y′2)3/2=

|x′′y′ − y′′x′|(x′2 + y′2)3/2

(2) For (x, y(x)), then we could treat the x as the parameter. Then x′ = 1 and x′′ = 0. We can reuse our derivedexpression above:

κ =|x′′y′ − y′′x′|(x′2 + y′2)3/2

=|y′′|

(1 + y′2)3/2

Exercise 7. A point moves so its velocity and acceleration vectors always have constant lengths.X

X ′ |X ′| = v0

X ′′ |X ′′| = a0

X ′ = v0T

X ′′ = v0T′ = v2

0κN

‖X ′′‖ = v0κ = a0

since κ =‖T ′‖v0

, so then κ =a0

v20

Exercise 8. If 2 plane curves with y = f(x), y = g(x) have the same tangent at a point (a, b) and the same curvature, prove|f ′′(a)| = |g′′(a)|.

X1 = (x, f(x))

X ′1 = (1, f ′(x))

X ′′1 = (0, f ′′(x))

X2 = (x, g(x))

X ′2 = (1, g′(x))

X ′′2 = (0, g′′(x))

‖X ′′1 ×X ′

1‖(1 + f ′2)3/2

=

√|f ′′|2(1 + f ′2)− (f ′′f ′)2

(1 + f ′2)3/2=

=|f ′′|

(1 + f ′2)3/2

, likewise ‖X′′2 ×X ′

2‖(1 + g′2)3/2

=|g′′|

(1 + g′2)3/2

For (a, b), the 2 plane curves have the same tangent, so then f ′(a) = g′(a). So if the curvature is the same at (a, b), then|f ′′(a)|

(1 + f ′2(a))3/2=

|g′′(a)|(1 + g′2(a))3/2

=⇒ |g′′(a)| = |f ′′(a)|

Exercise 9.(1) This problem could be done without reference to the same curvature or the same tangent line. We only need to use

the fact that they intersect at the same point.Given f = ax(b− x), (x + 2)g = x,

If x = b, f = 0

(b + 2)g = b

If b 6= 2

g =b

b + 2

Same as x = 0, so that f = 0, g = 0270

So one point of intersection for any a, b is (0, 0). Otherwise,

ax(b− x) =x

x + 2x 6== −2( this is okay, plug x = −2 into (x + 2)g = x and you'll get a contradiction )

=⇒ x =−(2− b)

2±

√4 + 4b + b2 − 4/a

2

So we must have, for only one pt. of intersection, (0, 0), b = 2 and√

4 + 4b + b2 − 4/a =√

4 + 8 + 4− 4/a = 0 =⇒ a = 1/4

(2) See sketch. The parabola and (x+2)g = x coincide at (0, 0), have the same tangent line, and have the same directionat this intersection point.

Exercise 10. Remember that the radius of curvature is the reciprocal of the curvature.(1) Without loss of generality, y = 4cx2. Then

y′ = 8cx

y′′ = 8c=⇒ κ = |f ′′|

(1+f ′2)3/2 . Since 1 > 0, 64 + c2x2 > 0, and |8c| constant, for κ maximized, then x = 0 .

(2) Given 2 �xed unit vectors A,B, A ·B = cos θ,

r = tA + t2B

r′ = A + 2tB

r′′ = 2B

r′′ × r′ = 2B × (A + 2tB) = 2B ×A

|r′′ × r′| = 2 sin θ

r′2 = 1 + 4t cos θ + 4t2

(r′2)′ = 8t + 4 cos θ = 0t = − cos θ/2 (r′2)′′ = 8 > 0

So t = − cos θ/2 minimizes r′2 and so maximizes κ. Then r = −cos θ

2A +

cos2 θ

4B

Exercise 11.(1) We have for this plane curve,

r = (x, y)

r′ = (x′, y′)

r′′ = (x′′, y′′)

Given√

x′2 + y′2 = 5, and given(x(0), y(0)) = (0, 0)

(x′(0), y′(0)) = (0, 5)

a× v = ez(x′′y′ − x′y′′)

κ =|a× v|

v3=|x′′y′ − x′y′′|

125= 2t

We want (x′, y′) · ex = 5 cos θ = x′. Then y′ = 5 sin θ. Also, note that x′(0) = 0 =⇒ θ0π/2

=⇒ 250t = | − 5 sin θθ′(5 sin θ)− 5 cos θ5 cos θθ′|10t = |θ′| θ = 5t2 + π/2

(2) v = 5(cos (5t2 + π/2), sin (5t2 + π/2))

Exercise 12.

Given that√

x′2 + y′2 = 2, thenx′ = 2 cos θ

y′ = 2 sin θθ0 = 0 since v0 = 2i

κ(t) =|x′′y′ − x′y′′|

v3=| − 2 sin θθ′2 sin θ − 2 cos θ2 cos θθ′|

23=|θ′|2

= 4tθ′ = 8t

θ = 4t2

v =√

2(1, 1)

14.19 Exercises - Velocity and acceleration in polar coordinates, Plane motion with radial acceleration, Cylindricalcoordinates.

Exercise 1.271

v =drdt

=dr

dt~er + r

dθ

dt~eθ = ~er + r~eθ (θ = t)

a =

(d2r

dt2− r

(dθ

dt

)2)

~er +1r

d

dt

(r2 dθ

dt

)~eθ = −r~er + 2~eθ

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).

r = sin t; θ = t; z = log sec t; θ ≤ t <π

2

(r cos θ)2 + (r sin θ − 12)2 = r2 − r sin θ +

14

=14

(b).

v =dt

dt~er + r

dθ

dt~eθ + log sec t~ez = cos t~er + r~eθ +

tan θ sec θ

sec θ~ez

vz = tan θ; v2 = cos2 t + r2 + tan2 θ = sec2 θ

cosφ =tan θ

sec θ= sin θ = r = sin t

φ = arccos (sin θ)

Exercise 4. Given r = f(θ); a ≤ θ ≤ b ≤ a + 2π

X = rur

X ′ = r′ur + ruθθ′

X ′2 = r′2 + r2θ′2

|X ′| =√

r2 + r′2

s =∫ √

r2 + r′2θ′2dtt=θ−−→ s =

∫ b

a

√r2 +

(dr

dθ

)2

dθ

Exercise 5. Givenr = a(1 + cos θ)

r′ = −a sin θso then r2 + r′2 = a2(1 + 2 cos θ + cos2 θ + sin2 θ) = 4a2 cos2 θ/2

s =∫ √

r2 + r′2dθ = 2a

∫ 2π

0

cos θ/2dθ = 2a

(∫ π

0

cos θ/2dθ +−∫ 2π

π

cos θ/2dθ

)=

= 4a(

sin θ/2|π0 − sin θ/2|2ππ

)= 4a(1 + 1) = 8a

Exercise 6.

A =∫

R2(θ)dθ =∫ 2π

0

12e2cθdθ =

e2cθ

4c

∣∣∣∣2π

0

=e4πc − 1

4c

Exercise 7.∫ π

0

12

sin4 θdθ =12

∫ π

0

sin2 θ(1− cos2 θ)dθ =12

∫ π

0

((1− cos 2θ

2

)−

(1− cos 4θ

2(4)

))dθ = 3π/16

See sketches for exercises 8-11. Exercise 8. Given r = θ; 0 ≤ θ ≤ π

s =∫ π

0

√θ2 + 1dθ =

12

(θ√

θ2 + 1 + ln (θ +√

θ2 + 1))∣∣∣

π

0=

=12(π

√π2 + 1 + ln (π +

√π2 + 1))

272

Exercise 9. Given r = eθ; 0 ≤ θ ≤ π

dr

dθ= eθ

s =∫ π

0

√e2θ + e2θdθ =

√2

∫ π

0

eθdθ =√

2(eπ − 1)

Exercise 10. r = 1 + cos θ ; drdθ = − sin θ; 0 ≤ θ ≤ π

∫ π

0

√r2 +

(dr

dθ

)2

=∫ π

0

√2√

1 + cos θ =√

2∫ π

0

√2 cos θ/2 = 4 (sin θ/2)|π0 = 4

Exercise 11. Given r = 1− cos θ; 0 ≤ θ ≤ 2π, then drdθ = sin θ

∫ 2π

0

√r2 +

(dr

dθ

)2

=∫ 2π

0

√2√

1− cos θ =√

2∫ 2π

0

√2| sin θ/2| = 4 − cos θ/2|2π

0 = 8

Exercise 12. Given r = f(θ), ρ = radius of curvature = 1κ , r′ = f ′(θ), r′′ = f ′′(θ)

X = rur


X ′′ = r′′ur + 2r′uθθ′ +−rθ′ur + rθ′′uθ =

= (r′′ − rθ′)ur +1r(r2θ′)′utheta

ur = (cos θ, sin θ)

uθ = (− sin θ, cos θ)ur × uθ = ez

X ′′ ×X ′ = ((r′′ − rθ′)ur +1r(r2θ′)′uθ)× (r′ur + ruθθ

′) =1r(r2θ′)′r′(−ez) + (r′′ − rθ′)rθ′ez

‖X ′′ ×X ′‖ = r′′rθ′ − r2θ′2 − 2r′2θ′ − rr′θ′′

|X ′|2 = r′2 + r2θ′2

If we let θ = t, since the curve is described completely by r = f(θ), then

κ =|r2 − r′′r + 2r′2|

(r′2 + r2)3/2=⇒ ρ =

(r′2 + r2)3/2

|r2 − r′′r + 2r′2|

Exercise 13. Using ρ = (r2+r′2)3/2

|r2−rr′′+2r′2|(1)

r = θ

r′ = 1

r′′ = 0ρ =

(θ2 + 1)3/2

θ2 + 2

(2)r = eθ

r′ = eθ

r′′ = eθ

ρ =(e2θ + e2θ)3/2

2e2θ=√

2eθ

(3) θ = π/4

r = 1 + cos θ

r′ = − sin θ

r′′ = − cos θ

ρ =(1 + 2c + c2 + s2)3/2

|1 + 2c + c2 + (c + c2) + 2s2| =(2(1 + c))3/2

3(1 + c)=

2√

2 +√

23

(4) θ = π/2r = 1− cos θ

r′ = sin θ

r′′ = cos θ

ρ =(2(1− c))3/2

(3 + 2c)(1− c)=

2√

2(1− c)1/2

3 + 2c

π/2−−→ 2√

23

273

Exercise 14.X = rur


X ′ ·X ′ = r′2 + r2θ′2 = v2

=⇒ X ·X ′ = rr′ = rv cosφ or r′ = v cosφ

v2 = v2 cos2 φ + r2θ′2

Use θ as the parameter for t, since the curve is invariant. =⇒ r = v sin φ

Exercise 15. Place target at the center (without loss of generality). The strategy is to break up v into the polar coordinate unitvectors.

r = r~er

v =dr

dt~er + r

dθ

dt~eθ

dr

dt= vr = v cos (π − α) = −v cosα

rdθ

dt= v sin α

v sin α

−v cos α= − tanα =

r dθdt

drdt

= rdθ

dr;

1r

dr

dθ= − tan α

r = e− tan αθ

Exercise 16. Admittedly, I had looked in the back of the book.

The big trick is that the choice of origin is important!. The clever choice of origin will cut out the algebra indealing with Cartesian coordinates. The clever choice of origin will illuminate the problem entirely. So try �ipping around,re�ecting, and turning around the problem until the right choice of origin is selected.

Let the positive axis be from where the missle, the target missile, m, is sighted to the ground crew. So the origin is at thetarget missile, not the ground crew from which the anti-missile missile, a, is launched! It'll make the algebra much easier.

Let a go 3 miles in the negative x direction, to allow for the possibility that the missile is returning to the ground crew,that is 4 mi. in the positive x direction from the origin, at the time of sighting. Let t = 0 be the time that missile a �nishesgoing these 3 miles and is 1 mile in the positive x direction from the origin.

Let

(xm, ym) = 1(t + 1)(cos θm0, sin θm0) = (t + 1)um0 = Xm

(x′m, y′m) = um0

(xa, ya) = Xa = raua

(x′a, y′a) = X ′a = r′aua + raθ′uθ

|X ′a| =

√r′2a + (raθ′)2 = 3

Xa(t = 0) = Xa0 = (1, 0) = ra(0)(cos θ(0), sin θ(0))

θ(0) = 0, ra(0) = 1

At some point t2, (xm, ym) = (xa, ya), i.e. Xm(t2) = Xa(t2). Note that we only need one t2; θm0 constant, but unknown.

(t2 + 1) cos θm0 = ra cos θ

(t2 + 1) sin θm0 = ra sin θ =

(t2 + 1)2 = r2a; (t2 + 1) = ra(t2); r′a(t2) =? well

(r′a(t2))2 + ((t2 + 1)dθ

dt(t2))2 = 9

Try 1 = r′a(t2).1 + (t2 + 1)2θ′2 = 9

(t2 + 1)(

dθ

dt

)= 2

√2 =⇒ dθ =

2√

2t2 + 1

dt

=⇒ θ = 2√

2 ln (t2 + 1) t2 = e

(θ

2√

2

)− 1

r = e

(θ

2√

2

)

274

Exercise 17.A �rst order differential equation of the form y′ = 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 �nd thatdy

dθ=

dr

dθsin θ + r cos θ

dx

dθ=

dr

dθcos θ − r sin θ

Thusdy

dx=

drdθ sin θ + r cos θdrdθ cos θ − r sin θ

= f(θ)

Exercise 18.v = ωk× r

v =dr

dt~er + r

dθ

dt~eθ

v · ~er = 0, so dt

dt= 0; ωk× r = r

dθ

dt~eθ = ωr~eθ = r

dθ

dt

|ωk× r|2 = ω2r2 = r2

(dθ

dt

)2

ω =∣∣∣∣dθ

dt

∣∣∣∣ , ω > 0

Exercise 19. (a)

v =dr

dt~er + r

dθ

dt~eθ = r

dθ

dt~eθ;

dr

dt= 0; ~eθ = ~ez × ~er

v = rdθ

dt~ez × ~er =

dθ

dt~ez × r~er = ω × r

(b).a = v′ = ω′ × r + ω × r′ = α× r + ω × r′

ω × r′ = ω × (ω × r) = (ω · r)ω − ω2r

(c).Now ω · r = 0 =⇒ a = −ω2r

Exercise 20.The distance |rp(t)− rq(t)| is independent of t, so d

dt |rp(t)− rq(t)| = 0, which implies

d

dt(r2

p − 2rp · rq + r2q) = 2rp · drp

dt− 2

drp

dt· rq − 2rp · drq

dt+ 2

drq

dt· rq = 0

drp

dt· (rq − rp) = −drq

dt· (rp − rq)

Supposevp = ωp × rp

vq = ωq × rq

Thenvp · (rq − rp) = −vq · (rp − rq)ωp × rp · rq = −ωq × rq · rp = = ωq × rp · rq

((ωp − ωq)× rp) · rq = 0

Thus, ωp = ωq .275

14.21 Miscellaneous review Exercises - Applications to planetary motion.Exercise 1. Use polar coordinates if Cartesian coordinates doesn't help!

y2 = x

X = (r cos θ, r sin θ) = rur

X ′ = r′ur + rθ′uθ

|X ′| =√

r′2 + (rθ′)2

T =X ′

|X ′| =r′ur + rθ′uθ√

r′2 + (rθ′)2

x = r cos θ

y = r sin θ=⇒

r2 sin2 θ = r cos θ

r =cos θ

sin2 θ

r′s2 + 2r2scθ′ = −sθ′

r′ = −(

1 + c2

s3

)θ′

r · T =rr′√

r′2 + (rθ′)2= r cosα =⇒ r′√

r′2 + (rθ′)2= cos α

X · er = r cos θ

r′2 + (rθ′)2 =(

1 + 2c2 + c4

s6

)θ′2 +

c2

s4θ′2 =

1 + 3c2

s4

cos α =(1 + c2)/s3

((1 + 3c2)/s6)1/2=

1 + c2

√1 + 3c2

sin2 α = 1− cos2 α =1 + 3c2 − (1 + 2c2 + c4)

(1 + 3c2)=

c2s2

(1 + 3c2)

sin α

cos α= tan α =

cs/√

1 + 3c2

(1 + c2)/√

1 + 3c2=

tan θ

1 + sec2 θ

Exercise 2.

X =(

y2

4c, y

)

X ′ = y′( y

2c, 1

)X ′

|X ′| =

(y2c , 1

)√

1 + (y/2c)2=

1√1 + (y/2c)2

T

N = (2c,−y) =⇒ T ·N = 0

Exercise 3.

(x, y) =(

y2

4c, y

)

X ′ =( y

2c, 1

)y′

|X ′| = y′√

(1 +( y

2c

)2

)

X ′

|X ′| =

(y2c , 1

)√

1 + (y/2c)2; m = 1/(y/2c) =

2c

y

m = m(x0) =2c

y0

y = mx + b =⇒ y0 =2c

y0

y20

4c+ b =

y0

2+ b so that b =

y0

2= c/m

=⇒ y = mx + c/my0 =

2c

m

x0 =4c2

m2

14c

=c

m2

Exercise 4.276

(1)

(y − y0)2 = 4c(x− x0)

(y − y0)2

4c+ x0 = x

=⇒

X =(

(y − y0)2

4c+ x0, y

)

T =X ′

|X ′| =y′

((y−y0)

2c , 1)

|y′|√

1 +(

(y−y0)2c

)2

m =2c

y − y0

m(y1) =2c

y1 − y0

y = m(y1)(x− x0) + b + y0

=⇒y1 =

(2c

y1 − y0

)((y1 − y0)2

4c

)+ b + y0

y1 − y0 =y1 − y0

2+ b

(y − y0) = m(x− x0) +c

m

(2)

(x− x0)2 = 4c(y − y0)

X = (x,(x− x0)2

4c+ y0)

T =

(1, x−x0

2c

)√

1 + (x−x0)2

4c2

m =x− x0

2c= m(x)

y − y0 = m(x− x0) + b

y1 − y0 =(x1 − x0)2

4c=

((x1 − x0)2

2c

)+ b

b =−(x1 − x0)2

4c

=⇒ y − y0 = m(x− x0) +−m2c

x2 = 4cy

X = (x, x2/4c)

T =

(1, x

2c

)x′

|x′|√

1 + (x/2c)2

m = x/2c

y = mx + b

y1 =x2

1

2c+ b =

x21

4c=

x21

2c+ b

b =−x2

1

4c

=⇒ y = mx + cm2

Exercise 5. GivenX =

(y2

4c, y

)

X ′ = y′( y

2c, 1

) =⇒ m = 2cy

y1 =2c

y1x1 + b =

2c

y1

y21

4c=

y1

2+ b =⇒ b =

y1

2

y =2c

y1x +

y1

2; =⇒ y1y = 2cx + 2x1c = 2c(x + x1)

277

Exercise 6. (y − y0)2 = 4c(x− x0) =⇒ m(y) = 2cy−y0

.

y − y0 =(

2c

y1 − y0

)(x− x0) + b

b = (y1 − y0)− 2c(y1 − y0)4c

=c

m

y − y0 =(

2c

y1 − y0

)(x− x0) +

(y1 − y0)2

(y1 − y0)(y − y0) = 2c(x− x0) +(y1 − y0)2

2= 2c((x− x0) + (x1 − x0))

For X = (x, x2

4c ) =⇒ m(x) = x2c

y =x1

2cx + b =⇒ b =

x21

−4c= −cm2

y =x1x

2c− y1 =⇒ 2c(y + y1) = x1x

For X =(x, (x−x0)

2

4c + y0

)=⇒ m = x1−x0

2c

y − y0 =(

x1 − x0

2c

)(x− x0) + b

(x1 − x0)2

4c− (x1 − x0)2

2c=−(x1 − x0)2

4c= b

y − y0 =(

x1 − x0

2c

)(x− x0) +− (x1 − x0)2

4c

y − y0 =(x1 − x0)(x− x0)

2c− (y1 − y0)

(y − y0) + (y1 − y0) =(x1 − x0)(x− x0)

2c

Exercise 7.(1) Given y = x2, then

X = (x, x2)

X ′

‖X ′‖ =(1, 2x)x′

|x′|√1 + 4x2

N =(−2x, 1)x′

|x′|√1 + 4x2

tN + (x, y)

(x, x2) + t(−2x, 1)x′

|x′|√1 + 4x2= (0, yq) = Q

=⇒ x =2xt√

1 + 4x2

t =

√14

+ x2

y =t√

1 + 4x2+ x2 =

√1 + 4x2

2√

1 + 4x2+ x2

=12

+ x2

(2)

Exercise 8. Given y = x2

±√c = x when y = c

Circle condition: ‖C −R‖ = ‖(0, y0)− (±√c, c)‖ =√

c + (y0 − c)2 = y0

=⇒ 1 + c

2= y0

c → 0, then y0 = 1/2 , radius of circle.

Exercise 9. Consider the condition, x2

a2 + y2

b2 = 1, pts. on an ellipse.

x2

a2+

y2

b2T 1 ⇐⇒

b2x2 + a2y2 T a2b2

r2(b2 cos2 θ + a2(1− cos2 θ)) = r2(a2 + (b2 − a2) cos2 θ) =

= r2a2(1− e2 cos2 θ)

⇐⇒ r2(1− e2 cos2 θ) T b2

Suppose for each �xed θ1, we consider all pts. along a line ray from the origin (which is inside the ellipse) to in�nity, radiallyout. Then r2(1− e2 cos2 θ1) T b2 ∀ r T r1

278

Exercise 10. For X = (x, y, ) and given that x2

a2 + y2

b2 = 1, or b2x2 + a2y2 = a2b2, so then

2b2xx′ + 2a2yy′ = 0 =⇒ y′ =−b2xx′

a2y

X = (x, y) =⇒ X ′ = (x′, y′) = x′(1,−b2x

a2y) =

−x′b2

y

(−y

b2,

x

a2

)

T =(−y

b2,

x

a2

)‖ X ′ N =

( x

a2,

y

b2

)⊥ X ′ since N ·X ′ = 0

m =x0/a2

−y0/b2=

b2x0

−a2y0

(y − y0) = m(x− x0) =⇒ y =b2x0

−a2y0x +

b2x20

a2y0+

a2y0

a2y0y0 =

b2x0

−a2y0x +

b2

y0

x0x

a2+

yy0

b2= 1 =⇒ x cos θ0

a+

y sin θ0

b= 1

Exercise 11. Using the work of the previous exercise, Exercise 10, thenx cos θ0

a+

y sin θ0

b= 1 =

xa cos θ0

a2+

yb sin θ0

b2= xx0/a2 + yy0/b2

Exercise 12. Use T =(−y

b2 , xa2

); N =

(xa2 , y

b2

)from the previous problems.

Without loss of generality, center the ellipse on the origin and the major axis on the x axis. Then F = aeex and P = (x, y).∣∣∣∣(F − P ) ·N

|N |

∣∣∣∣ =1− ex

a√x2

a4 + y2

b4

∣∣∣∣(−F − P )·

|N |

∣∣∣∣ =1 + xe

a√x2

a4 + y2

b4

Multiply the above distances together to get(

x2

a4+

y2

b4

)−1/2 (1 +

xe

a

)(1− ex

a

)=

(x2

a4+

y2

b4

)−1/2 (x2

a2+

y2

b2− e2

a2x2

)= b2

Exercise 13. x2 + 4y2 = 8 or x2

8 + y2

2 = 1. We want to �nd the tangent lines parallel to x + 2y = 7 or y = 72 − x

2 . So theslope of this line is − 1

2 . Using T =(−y

b2 , xa2

)=

(−y2 , x

8

), then

x/8−y/2

=x

−4y= −1/2 or x = 2y

plugging back into the ellipse equation , (2, 1), (−2,−1)

Exercise 14. By equidistant property of the circle, the focus length from origin, ae, must equal the minor axis length, b.ae = b = a

√1− e2 =⇒ e = 1/

√2

Exercise 15.

(1) x2

a2− y2

b2= 1 . Recall that b2 = a2(1− e2) = −3a2, so that 1

a2(x2 − y2

3) = 1 .

(2)

A =∫ x

a

b

√(t

a

)2

− 1dt− 12(x− a)b

√(x

a

)2

− 1

r is the length of V P :

Exercise 16. Given x2

a2 − y2

b2 = 1, then y = b

√(xa

)2 − 1

X ′ =

1,

b(

xa2

)√(

xa

)2 − 1

=

(1,

b2(

xa2

)

y

)=

b2

y

( y

b2,

x

a2

)

So X ′ ‖ T ; and X ′ ·N = 0

Exercise 17. m(x0) = b2x0a2y0

from the tangent vector above and so, using the ellipse equation,

(y − y0) =b2

a2

x0

y0(x− x0) =⇒ x0x

a2− y0y

b2= 1

279

Exercise 18. Given X = (x, y),

X ′ = (x′, y′); N ‖ (y′,−x′)

(x, y) + s(y′,−x′) = (x1, 0) =⇒y = sx′

x + sy′ = x1

=⇒ x1 = x + yy′/x′

Isosceles condition:√

x2 + y2 = ‖(x1, 0)− (x, y)‖ =

√(yy′x′

)2

+ y2

x

y=

dy/dt

dx/dt=

dy

dx=⇒ y2

2C− x2

2C= 1

Note there's some ambiguity with the signs.Exercise 19. N ‖ (y′,−x′)

(x, y) + s(y′,−x′) = (x1, 0)

y = sx′

x + sy′ = x1 = x +yy′

x′

(x, y) + t(y′,−x′) = (0, y1)

x = −ty′

y − tx′ = y1

y +xx′

y′= y1

‖(x, y)− (x1, 0)‖ = ‖(x, y)− (0, y1)‖

=⇒ (x− x1)2 + y2 = x2 + (y − y1)2 =⇒(

yy′

x′

)2

+ y2 = x2 +(

xx′

y′

)2

=⇒(

dy

dx

)2

=x2

y2

(4,5)−−−→ y2 − x2 = 9

Exercise 20. Without loss of generality, consider a hyperbola given by x2

a2 − y2

b2 = 1 with asymptotes of yx = ± b

a . Thedistances from a point on the hyperbola to the asymptotes are the following:

(X − P ) ·N1

|N1| =(x, y) · (−b, a)√

a2 + b2=−bx + ay√

a2 + b2

(X − P ) ·N2

|N2| =(x, y) · (b, a)√

a2 + b2=

bx + ay√a2 + b2

( | − bx + ay|√a2 + b2

)(bx + ay√a2 + b2

)=

b2x2 − a2y2

a2 + b2=

a2b2

a2 + b2

Exercise 21. Recall that for X = rur (useful trick to change into polar coordinates),


|X ′| =√

r′2 + (rθ′)2=⇒

∫ √r′2 + (rθ′)2dt

(1)∫ θ

θ0

√r′2 + r2dθ = k(θ − θ0)

√r′2 + r2 − C = k

r′2 = k − r2=⇒

(r′√k

)2

+(

r√k

)2

= 1

r =√

k sin θ or r =√

k

(2)∫ θ

θ0

√(drdθ

)2+ r2dθ = k(r(θ)− r(θ0))

=⇒r′2 + r2 = (kr′)2

r′2 =r2

k2 − 1

lnr

r0=

θ − θ0√k2 − 1

r = Ce

(θ√

k2 − 1

)

280

(3)∫ θ

θ0

√r′2 + r2dθ = k

∫ θ

θ0

12r2dθ

r′2 + r2 =k2

4r4

=⇒ r′

|r|√(

kr2

)2 − 1= 1

sec t =kr

2

tan t sec t =k

2dr

dt−−−−−−−−−−−−−→

∫ 2k tan t sec t

2k | sec t|√sec2 t− 1

= θ − θ0

arcsec

(kr

2

)= θ − C

r =2k

sec (θ − θ0) or r =2k

Exercise 22. Given r′(t) = (r(a) × r(b)), r′(t) = r(b)−r(a)b−a for some t ∈ [a, b] (by mean-value thm. on each of the

components).r′(t) · (r(a)× r(b)) = 0

Geometrically it means that there must be some component of r′(t) that belongs in the plane containing r(b)− r(a) becausethe velocity vector r′(t) must �take� r(a) to r(b).

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CALCULUS VOLUME 1 BY TOM APOSTOL. - … · SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL. ERNEST...

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