SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL.

ERNEST YEUNG

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

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Ernest Yeung is supported by Mr. and Mrs. C.W. Yeung, Prof. Robert A. Rosenstone, Michael Drown, Arvid Kingl, Mr .and Mrs. Valerie Cheng, and the Foundation for Polish Sciences, Warsaw University.

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

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

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

Suppose x ∈ Xx ∈ 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 ⊆ YSuppose y ∈ Yy is at least in either (A ∩B) or A ∩ Cthen y ∈ A and either in B or Cy ∈ X,Y ⊆ X

X = Y

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

Suppose x ∈ Xthen x is at least either in A or in (B ∩ C)

if x ∈ A, x ∈ Yif x ∈ (B ∩ C), x ∈ Y x ∈ Y,X ⊆ Y

Suppose y ∈ Ythen y is at least in A or in B and y is at least in A or in Cif y ∈ A, then y ∈ Xif y ∈ A ∩B or y ∈ A ∪ C, y ∈ X (various carvings out of A, simply )

if y ∈ (B ∩ C), y ∈ X y ∈ X,Y ⊆ XX = Y

1

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

If x ∈ A ∩A, then x is in A and in A. Then x ∈ A. So A ∩A ⊆ A. Of course A ⊆ A ∩A.

Exercise 12. Let x ∈ A. y ∈ A ∪B if y is at least in A or in B. x is in A so x ∈ A ∪B. =⇒ A ⊆ A ∪B.

Suppose ∃b ∈ B and b /∈ A. b ∈ A ∪B but b /∈ A. so A ⊆ A ∪B.

Exercise 13. Let x ∈ A ∪ ∅, then x is at least in A or in ∅. If x ∈ ∅, then x is a null element (not an element at all). Then

actual elements must be in A. =⇒ A ∪ ∅ ⊆ A.Let x ∈ A. Then x ∈ A ∪ ∅. A ⊆ A ∪ ∅. =⇒ A = A ∪ ∅.

Exercise 14. From distributivity, A ∪ (A ∩B) = (A ∪A) ∩ (A ∪B) = A ∩ (A ∪B).

If x ∈ A ∩ (A ∪B), x ∈ A and x ∈ A ∪B, i.e. x ∈ A and x is at least in A or in B.=⇒ x is in A and is in B or is not in B. Then x ∈ A. =⇒ A ∩ (A ∪ B) ⊆ A. Of course, A ⊆ A ∩ (A ∪ B).=⇒ A ∩ (A ∪B) = A ∪ (A ∩B) = A.

Exercise 15. ∀a ∈ A, a ∈ C and ∀b ∈ B, b ∈ C. Consider x ∈ A ∪ B. x is at least in A or in B. In either case, x ∈ C.

=⇒ A ∪B ⊆ C.

Exercise 16.

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

Exercise 17.

(1)if A ⊂ B and B ⊂ C then∀a ∈ A, a ∈ B.∀b ∈ B, b ∈ C.then since a ∈ B, a ∈ C,∃c ∈ C such that c /∈ B.∀a ∈ A, a ∈ B so a 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 ∩ Cthen x is not in even at least one B or C

=⇒ x ∈ (A−B) ∪ (A− C)

Suppose x ∈ (A−B) ∪ (A− C)

then x is at least in (A−B) or in (A− C) =⇒ x is at least in A and not in B or in A and not in C

then consider when one of the cases is true and when both cases are true =⇒ x ∈ A− (B ∩ C)

Exercise 19.

Suppose x ∈ B −⋃A∈F

A

then x ∈ B, x /∈⋃A∈F

A

x /∈⋃A∈F

A =⇒ x /∈ A,∀A ∈ F

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

(B −A)

2

Suppose x ∈⋂A∈F

(B −A)

then x ∈ B −A1 and x ∈ B −A2 and . . .

then ∀A ∈ F , x ∈ B, x /∈ Athen x /∈ even at least one A ∈ F

=⇒ x ∈ B −⋃A∈F

A

Suppose x ∈ B −⋂A∈F

A

then x /∈⋂A∈F

A

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

=⇒ x ∈⋃A∈F

(B −A)

Suppose x ∈⋃A∈F

(B −A)

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

=⇒ then x ∈ B −⋂A∈F

A

Exercise 20.

(1) (ii) is correct.

Suppose x ∈ (A−B)− Cthen x ∈ A−B, x /∈ Cthen x ∈ A and x /∈ B and x /∈ Cx /∈ B and x /∈ C =⇒ x /∈ even at least B or C

x ∈ A− (B ∪ C)

Suppose x ∈ A− (B ∪ C)

then x ∈ A, x /∈ (B ∪ C)

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

To show that (i) is sometimes wrong,

Suppose y ∈ A− (B − C)

y ∈ A and y /∈ B − Cy /∈ B − C

then y /∈ B or y ∈ C or y /∈ C(where does this lead to?)

Consider directly,Suppose x ∈ (A−B) ∪ C

then x is at least in A−B or in C

then x is at least in A and /∈ B or in C

Suppose x = c ∈ C and c /∈ A3

(2)If C ⊆ A,

A− (B − C) = (A−B) ∪ C

I 3.3 Exercises - The field axioms. The goal seems to be to abstract these so-called real numbers into just x’s and y’s thatare purely built upon these axioms.

Exercise 1. Thm. I.5. a(b− c) = ab− ac.

Let y = ab− ac;x = a(b− c)Want: x = y

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

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

ac+ x = ac+ ab+ a(−c) = a(c+ (−c)) + ab = a(0 + b) = ab

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

0(a) = a(0) (by commutativity axiom)Given b ∈ R and 0 ∈ R,∃ exactly one − b s.t. b− a = 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 = 1

Want: x = by

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

=⇒ x = by (by Thm. I.7)

Thm. I.10. If a 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.

Thm.I.11. If ab = 0, a = 0 or b = 0.ab = 0 = a(0) =⇒ b = 0 or ab = ba = b(0) =⇒ a = 0. (we used Thm. I.7, cancellation law for multiplication)

Thm. I.12. Want: x = y if x = (−a)b and y = −(ab).

ab+ y = 0

ab+ x = ab+ (−a)b = b(a+ (−a)) = b(a− a) = b(0) = 0

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

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

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

So using b, d 6= 0, which is given, and Thm. I.7, then x+ y = z.4

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

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

b, d 6= 0, so by Thm. I.7, xy = z.Thm.I.15. Want: x = yz, if bx = a, dy = c, (bc)z = ad

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

d 6= 0 so by Thm. I.7, by z = a, byz = abx

b 6= 0 so by Thm. I.7, yz = x

Exercise 2. Consider 0 + z = 0. By Thm. I.2, there exists exactly one z, z = −0. By Axiom 4, z = 0. 0 = −0.

Exercise 3. Consider 1(z)z(1) = 1. Then z = 1−1. But by Axiom 4, there exists distinct 1 such that z(1) = 1, so z = 1.

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

Exercise 5. a+ (−a) = 0, 0 + 0 = 0. Then

a+ (−a) + b+ (−b) = (a+ b) + (−a) + (−b) = 0

−(a+ b) = −a+ (−b) = −a− b

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

a+ (−a) + b+ (−b) = a+ (−b) + (−a) + b = (a− b) + (−a) + b = 0 + 0 = 0

−(a− b) = −a+ b.

Exercise 7.

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

Exercise 8.

(ab)x = 1 (ab)−1 = x

a(bx) = 1 a−1 = bx

b(ax) = 1 b−1 − axa−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 = −(ab

) (ab

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

a+ (−a) = vb+ by = b(v + y) = 0

if b 6= 0, v + y = 0, but v + x = 0

by Thm. I.1 , x = y

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

a+ zb = 0 =⇒ −a = zb = by

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

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

z =ad− bcbd

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

x =a

bbx = a

t =−cd

dt = −c (By Thm. I.3, we know that b− a = b+ (−a) )5

dbx+ bdt = (bd)(x+ y) = ad− bc = (bd)z

b, d 6= 0, so x+ y = z

I 3.5 Exercises - The order axioms.

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

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

Theorem 3 (I.20). If a 6= 0, then a2 > 0

Theorem 4 (I.21). 1 > 0

Theorem 5 (I.22). If a < b and c < 0, then ac > bc.

Theorem 6 (I.23). If a < b and −a > −b. In particular, if a < 0, then −a > 0.

Theorem 7 (I.24). If ab > 0, then both a and b are positive or both are negative.

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

Exercise 1.

(1) By Thm. I.19, −c > 0

a(−c) < b(−c)→ −ac < −bc−bc− (−ac) = ac− bc > 0. Then ac > bc (by definition of > )

(2)a < b→ a+ 0 < b+ 0→ a+ b+ (−b) < b+ a+ (−a)→ (a+ b)− b < (a+ b) + (−a)

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

−b < −a(3)

If a = 0 or b = 0, ab = 0, but 0 ≯ 0

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

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

(4)a < c so a+ b < c+ b = b+ c

b < d so b+ c < d+ c

By Transitive Law , a+ b < d+ c

Exercise 2. If x = 0, x2 = 0. 0 + 1 = 1 6= 0. So x 6= 0.

If x 6= 0, x2 > 0, and by Thm. I.21 , 1 > 0

x2 + 1 > 0 + 0 = 0→ x2 + 1 6= 0

=⇒ @x ∈ R such that x2 + 1 = 0

Exercise 3.

a < 0, b < 0, a+ b < 0 + 0 = 0 ( By Thm. I.25)

Exercise 4. Consider ax = 1.

ax = 1 > 0. By Thm. I.24 , a, x are both positive or a, x are both negative

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

xb− ax = xb− 1 > 0 =⇒ bx > 1 = by

b > 0 so x > y

Exercise 6.6

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

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

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

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

Exercise 7. If a ≤ b and b ≤ c, then a ≤ c. If a = c, then by previous proof, a = b.

Exercise 8. If a ≤ b and b ≤ c, then a ≤ c. If a = c, then by previous proof, a = b.

Exercise 8. If a or b is zero, a2 or b2 = 0. By Thm. I.20, b2 ≥ 0 or a2 ≥ 0, respectively.

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

Exercise 9. Suppose a ≥ x. Then a− x ≥ 0.

If a ∈ R so ∃y ∈ R, such that a− y = 0.Consider y + 1 ∈ R (by closure under addition).

a− (y + 1) = a− y − 1 = 0− 1 < 0 Contradiction that a ≥ y + 1

Exercise 10.If x = 0, done.

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

2.

=⇒ x

2> x Contradiction.

I 3.12 Exercises - Integers and rational numbers, Geometric interpretation of real numbers as points on a line,Upper bound of a set, maximum element, least upper bound (supremum), The least-upper-bound axiom (completenessaxiom), The Archimedean property of the real-number system, Fundamental properties of the suprenum and infimum.We use Thm I.30, the Archimedean property of real numbers, alot.

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

We will use the least upper-bound axiom (completeness axiom) alot for continuity and differentiation theorems later.Apostol states it as an axiom; in real analysis, the existence of a sup for nonempty, bounded sets can be shown with 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 becauseIf ∀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 > 1n .

Exercise 4. x is an arbitrary real number. By Thm. I.29 and well-ordering principle, there exists a smallest n + 1 positive

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

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

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

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

Exercise 6. y − x > 0.

n(y − x) > h, h arbitrary , n ∈ Z+

y > x+ h/n = z > x

Since h was arbitrary, there are infinitely many numbers in between x, y.

Exercise 7. x = ab ∈ Q, y /∈ Q.

x± y =a± byb

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 y

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

Likewise, y 1y = 1.

Exercise 9.

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

y >k

n+ x > x. Let z = x+

k

n, z irrational .

Exercise 10.

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

2m1 + 1 = 2m2 2(m1 −m2) = 1 m1 −m2 =1

2. 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) even

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

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

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

2(2m1m2)− (n1 + n2)− 1 odd, the product of two odd numbers n1, n2 is odd

(4) If n2 even, n is even, since for n = 2m, (2m)2 = 4m2 = 2(2m2) is even.

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

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

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

Thus b is even.8

(5) For pq , If p or q or both are odd, then we’re done.Else, when p, q are both even, p = 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 satisfies the Archimedean property but not the least-upper-bound property.

Since pq ∈ Q ⊆ R, np1q1 >

p2q2

since if q1, q2 > 0,

np1q2

q1q2>q1p2

q1q2np1q2 > q1p2

n exists since (p1q2), (q1p2) ∈ R.The set of rational numbers does not satisfy the least-upper-bound property.

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

Suppose x < b1, x < b2∀x ∈ S.

r

s< b2 < nb1 but likewise

r

s< b1 < mb2, n,m ∈ Z+

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

I 4.4 Exercises - An example of a proof by mathematical induction, The principle of mathematical induction, 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 = 2

N∑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)+ Nbut S = N+ N − 1+ · · ·+ 3+ 2+ 1

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

29

Telescoping series will let you get∑Nj=1 j

2 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)

=⇒ 3

N∑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 =

= 4

N∑j=1

j3 − 6N(N + 1)(2N + 1)

6+ 4

N(N + 1)

2−N = N4

=⇒N∑j=1

j3 =1

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

1

4(N4 + (2N)N(N + 1)−N(N + 1) +N)

=1

4(N4 + 2N3 + 2N2 −N2 −N +N) =

1

4N2(N2 + 2N + 1) =

1

4

(N(N + 1))2

2

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 =1

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

1

8(4k2 + 4k + 1) +

8k + 8

8=

(2k + 3)2

8(2) The n = 1 case isn’t true.(3)

1 + 2 + · · ·+ n =(n+ 1)n

2=n2 + n

2<n2 + n+ 1

4

2

and(

2n+ 1

2

)21

2=

(n+ 1/2)2

2=n2 + n+ 1/4

2

Exercise 7.

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

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

0 > x2 =⇒ Impossible

(1 + x)3 = 1 + 3x+ 3x2 + x3 > 1 + 3x+ 3x2

=⇒ x3 > 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 find, explicitly

(1 + x)n =

n∑j=0

(n

j

)xj = 1 + nx+

n(n− 1)

2x2 +

n∑j=3

(n

j

)xj

10

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 = 1

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

Assume n = qb+ r; 0 ≤ r < b; b ∈ Z+, b fixed

n+ 1 = qb+ r + 1 = qb+ 1 + r = qb+ 1 + b− 1 = (q + 1)b+ 0

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

Assume the k − 1th case. Consider kj , 1 ≤ j ≤ k.If kj ∈ Z

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

+, for some 1 < j < k, kj = c ∈ Z+. c, j < k.Thus k = cj. c, j are products of primes or are primes, by induction hypothesis. Thus k is a product of primes.

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

blue-eyed.

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

(1) n(n+1)2 =

∑4k=1 k = 10

(2)∑5n=2 2n−2 =

∑3n=0 2n = 1 + 14 = 15

(3) 2∑3r=0 22r = 2

∑3r=0 4r = 170

(4)∑4j=1 j

j = 1 + 4 + 27 + 44 = 288

(5)∑5j=0(2j + 1) = 2 5(6)

2 + 6(1) = 36

(6)∑

1k(k+1) =

∑nk=1

(1k −

1k+1

)= 1− 1

n+1 = nn+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+ 1

2(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 10.11

(1) am + am+1 + · · ·+ am+n

(2)

n = 11

2=

1

1− 1

2=

1

2

n+ 1

2(n+1)∑k=n+2

1

k=

2n∑m=1

(−1)m+1

m− 1

n+ 1+

1

2n+ 1+

1

2n+ 2=

2n∑m=1

(−1)m+1

m+− 1

2(n+ 1)+

(−1)2n+1+1

(2n+ 1)

=

2(n+1)∑m=1

(−1)m+1

m

Exercise 13.

n = 12(√

2− 1) < 1 < 2 since1

2>√

2− 1

n case (√n+ 1−

√n)(√n+ 1 +

√n) = n+ 1− n = 1 <

√n+ 1 +

√n

2√n

=1

2(

√1 +

1

n+ 1)

n+ 1 case (√n+ 2−

√n+ 1)(

√n+ 2 +

√n+ 1) = n+ 2− (n+ 1) = 1

√n+ 2 +

√n+ 1

2√n+ 1

=1 +

√1 + 1

n+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

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

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

(2)

| − x| =

{−x if − x ≥ 0

x 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 ≥ 0

xy if − x,−y ≥ 0

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

∣∣∣∣ =

{1y if 1

y ≥ 0−1y if 1

y ≤ 0

1

|y|=

{1y if 1

y ≥ 0−1y if 1

y ≤ 0

12

(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 = − bkak, 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

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

property of products:

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

13

(1) (1 +

1

n

)n=

n∑j=0

(n

k

)(1

n

)j=

n∑k=0

n!

(n− k)!k!

(1

n

)kk−1∏r=0

(1− r

n

)=

k−1∏r=0

(n− rn

)=

(1

nk

)n!

(n− k)!

n∑k=1

1

k!

k−1∏r=0

(1− r

n

)=

(1

nk

)n!

(n− k)!

(2)

(1 +1

n)n = 1 +

n∑k=1

(1

k!

k−1∏r=0

(1− r

n)

)< 1 +

n∑k=1

1

k!< 1 +

n∑k=1

1

2k= 1 +

12 −

(12

)n+1

12

= 1 + (1−(

1

2

)n)

< 3

The first 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 +1

n)n =

n∑k=0

(n

k

)(1

n

)k= 1 +

1

n+

n−1∑k=1

(n

k

)(1

n

)k= 1 +

1

n+

n−1∑k=2

(n

k

)(1

n

)k+n

1

(1

n

)> > 2

Exercise 13.

(1)

S =

p−1∑k=0

(b

a

)k=

1−(ba

)p1− b

a

p−1∑k=0

bkap−1−k = ap−1 1−(ba

)p1− b

a

=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

n = 21p <2p+1

p+ 1< 1p + 2p

p = 1

1 < 22/2 = 2.2 < 1 + 2 = 3

p− 2

1 < 8/3 < 1 + 4 = 5

I 4.10 Miscellaneous exercises involving induction. Exercise 13.

(1)(2)(3) Let n = 2.

2−1∑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

14

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 = 1

23!33 = 2

9 < 1.

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 ≤ n

2 .

(n+ 1)!

(n+ 1)n+1≥

(n+ 1)nn(

12

)k(n+ 1)n+1

=

(n

n+ 1

)n(1

2

)k=

(1− 1

n+ 1

)n(1

2

)k<

1

2

(1

2

)k=

(1

2

)k+1

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

k ≤ n

2=⇒ k +

1

2≤ n+ 1

2=⇒ 1

n+ 1≤ 1

2k + 1<

1

2

Exercise 16.15

a1 = 1 <1 +√

5

2

a2 = 2 <

(1 +√

5

2

)2

=1 + 2

√5 + 5

4=

6 + 2√

5

4

an+1 = an + an−1 <

(1 +√

5

2

)n+

(1 +√

5

2

)n−1

=

(1 +√

5

2

)n(1 +

2

1 +√

5

)=

=

(1 +√

5

2

)n(2(1−

√5)

1− 5+

4

4

)=

(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

(∑xpk

)2

≤(∑

x2pk

)n =⇒

∑(x2pk ) ≥

(∑xpk)

2

nWe define Mp as follows:

Mp =

(∑nk=1 x

pk

n

)1/p

So then

nMpp =

n∑k=1

xpk

M2p =

(∑nk=1 x

2pk

n

)1/2p

nM2p2p =

n∑k=1

x2pk

∑x2pk = nM2p

2p ≥(nMp

p )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 ≥ 64

3

Exercise 19. ak = 1,∑nk=1 1 = n

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

a1 = 1

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

(a1 − 1)(a2 − 1) < 0

a1a2 − 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

16

1.7 Exercises - The concept of area as a set function. We will use the following axioms:Assume a classM 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=1Qj

)= 0

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

a(A ∩B) =√a2 + b2d+ ab− (

1

2ab+

√a2 + b2d) =

1

2ab

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

A trapezoid is simply a rectangle with a right triangle attached to each end of it. Tr = R + T1 + T2. T1, T2 are righttriangles 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).

17

We can compute the area of a trapezoid:

Tr = R+ T1 + T2 =⇒ a(Tr) = a(R) + a(T1) + a(T2)

a(Tr) = hb1 +1

2h(b2 − b1)/2 +

1

2h(b2 − b1)/2 =

1

2h(b1 + h2)

P = R (a parallelogram consists of a right triangle rotated by π and attached to the other side of the same right triangle;the two triangles do not overlap). Since two right triangles are measurable, the parallelogram, P is measurable.Using the Additive Axiom, a(P ) = 2a(T ) = 21

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 polygon

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

and B denotes the number on the boundary.

(1) Consider one side of the rectangle lying on a coordinate axis with one end on the origin. If the rectangle side 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 cotπ/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 letM denote the class of all subsets of A. (There are 32 altogether counting A itself

and the empty set ∅). (My Note: the set of all subsets, in this case, M, is called a power set and is denoted 2A. This isbecause the way to get the total number of elements of this power set, |2A|, or the size, think of assigning to each element a“yes,” if it’s in some subset, or “no”, if it’s not. This is a great way of accounting for all possible subsets and we correctlyget all possible subsets.) For each set S inM, 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}) = 1

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

⋃T , S

⋂T since every element in S

⋃T , S

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

measurable. Since n counts only distinct elements, then n(S⋃T ) = n(S) + a(T ) − a(S

⋂T ), where −a(S

⋂T ) ensures

there is no double counting of distinct elements. Thus, the Additive Property Axiom is satisfied.

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

18

1.11 Exercises - Intervals and ordinate sets, Partitions and step functions, Sum and product of step function. 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+1

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

[x] + [x+1

2] = 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+1

3] + [x+

2

3] = q +

{q if r < 2

3

q + 1 if r > 23

+

{q if r < 1

3

q + 1 if r > 13

=

3q if r < 1

3

3q + 1 if 13 < r < 2

3

3q + 2 if r > 23

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

3q if r < 1

3

3q + 1 if 13 < r < 2

3

3q + 2 if r > 23

Exercise 5. Direct proof.

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

nq if r < 1

n

nq + 1 if 1n < r < 2

n

nq + n− 1 if r > n−1n

Exercise 6.

a(R) = hk = IR +1

2BR − 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.

19

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

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

including points on the hypotenuse

b−1∑n=1

[nab

]+

1

2((a+ 1) + b)− 1 =

ab

2

Now(a− 1)(b− 1)

2=ab

2− a

2− b

2+

1

2

=⇒b−1∑n=1

[nab

]=

(a− 1)(b− 1)

2

(2) a, b ∈ Z+

b−1∑n=1

[nab

]=

b−1∑n=1

[a(b− n)

b

](reverses order of summation)

b−1∑n=1

[a− an

b

]=

{−∑b−1n=1

[anb − a

]if anb − a4 is an integer (but a

(nb − 1

)can’t be!)

−∑b−1n=1

([anb − a

]− 1)

otherwise

= −b−1∑n=1

([anb− a]− 1)

= −b−1∑n=1

([anb

]− a)− (b− 1) =

= −b−1∑n=1

[anb

]+ a(b− 1)− (b− 1)

b−1∑n=1

[nab

]=

(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 ∈ S0 ∀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 definition 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

20

Exercise 3. [x] = y ≤ x so −y ≥ −x.

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

=⇒ [x] + [−x] = y − y − 1 = −1Or 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)

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 =

∫ n1

[t]2dt =∑n−1j=1 j

2 = (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

621

Exercise 8.∫ b+ca+c

f(x)dx =∫ b+c−ca+c−c f(x− (−c))dx =

∫ baf(x+ c)dx

Exercise 9.∫ kbkaf(x)dx = 1

1k

∫ (kb)/k

(ka)/kf(

x1/k

)dx = k

∫ baf(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) = 0

We obtain this formula

f(p) =

{p2 (−1)p+1 p evenp−1

2 (−1)p+1 p evensince

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

∫ p+1

p

s(x)dx ={p−1

2 (−1)p+1 p even + (−1)pp

=

{−p2 p evenp−1

2 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)∫ ba

(s+ t) =∑n3

k=1(s+ t)3k(xk − xk−1) 6=

∫ bas+

∫ bat

(3)∫ bacs =

∑nk=1(cs)3(xk − xk−1) 6= c

∫ bas

(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,∫ bas =

∑nk=1 s

3k(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

22

Then∫ bas <

∫ bat.

Exercise 12.

(1)∫ bas+

∫ cbs =

∑n1

k=1 sk(x2k − x2

k−1) +∑n2

k=n1sk(x2

k − x2k−1) =

∑n3

k=1 sk(x2k − x2

k−1) =∫ cas

(2)∫ ba

(s+t) =∑n3

k=1(s+t)k(x2k−x2

k−1) =∑n3

k=1(sk+tk)(x2k−x2

k−1) =∑n3

k=1 sk(x2k−x2

k−1)+∑n3

k=1 tk(x2k−x2

k−1)

since P3 = {xk} is a finer 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)∫ bacs =

∑nk=1 csk(x2

k − x2k−1) = c

∑nk=1 sk(x2

k − x2k−1) = c

∫ bas

(4)∫ b+ca+c

s(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,∫ basdx =

∑nk=1 sk(x2

k − x2k−1) <

∑nk=1 tk(x2

k − x2k−1) =

∫ batdx

Note that we had shown previously that the integral doesn’t change under finer 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 is

another partition on [a, b] (by partition definition) since xk, yk ∈ R and since real numbers obey transitivity, {xk, yk} can bearranged such that a < z1 < z2 < · · · < zn3−2 < b where zk = xk or yk.

(s+ t)(x) = s(x) + t(x) = sj + tk if xj−1 < x < xj ; yj−1 < x < 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 finer partition doesn’t change the integral R is a finer 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

23

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 define a finer partition from two partitions of the same interval.

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

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

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 apartition 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. Definitions and examples, Integrability of bounded monotonic functions, Calculation of the integralof a bounded monotonic function, Calculation of the integral

∫ b0xpdx when p is a positive integer, The basic properties

of the integral, Integration of polynomials.

Exercise 16.∫ 2

0|(x− 1)(3x− 1)|dx =

24

∫ 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

=4

27∫ 1/3

0

(x− 1)(3x− 1)dx =4

27

So the final 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

= 392

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 =

(1

3x3 − c

4x4

)∣∣∣∣1/c0

=1

12c3∫f − g =

2

3=

1

12c3; c =

1

2√

2

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

∫ 1−a

0

x− x2 − ax =

((1− a)

1

2x2 − 1

3x3

)∣∣∣∣1−a0

= (1− a)3 1

6= 9/2 =⇒ a = −2

Exercise 17. π = 2∫ 1

−1

√1− x2dx

(1) ∫ 3

−3

√9− x2dx = 3

∫ 3

−3

√1−

(x3

)2

= 3(3)

∫ 1

−1

√1− x2 =

9π

2

Now ∫ kb

ka

f(xk

)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−

(x2

)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 isosceles

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

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

c2 = 1 + 1− 2(1)(1) cos 30◦ = 2

(1−√

3

2

)=⇒ c =

√2

√1−√

3

2

25

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)1

2

√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)1

2

2

√√√√1−√

32

1 +√

32

(1) = 12

(2−√

3

2

)> π

Exercise 19.

(1) (x, y) ∈ E if x = ax1, y = by1 such that x21 + y2

1 ≤ 1

=⇒(xa

)2+(yb

)2= 1

(2)

y = b

√1−

(xa

)2

2

∫ a

−ab

√1−

(xa

)2

= 2ba

∫ 1

−1

√1− x2 = ba

π

2(2) = πba

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

Suppose x and y coordinates of S were expanded in different ways x = k1x1, y = k2y1.If f(x1) = y1, g(x) = k2f

(xk1

)= k2y1 = y.

integrating g on [k1a, k1b], ∫ 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) sinπ = 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) cosπ/2 = cos−π/2 = 0

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

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

Exercise 2.

(1) sinπ/2 = 1, sinπ/2(1 + 4j) = 1, j ∈ Z+.(2) cosx = 1, cos 0 = 1, cos 2πj = 1

Exercise 3.

sinx+ π = − sinx+ π/2 + π/2 = cosx+ π/2 = − sinx

cosx+ π = cosx+ π/2 + π/2 = − sinx+ π/2 = − cosx

Exercise 4.26

sin 3x = sin 2x cosx+ sinx cos 2x = 2 sinx cos2 x+ sinx(cos2 x− sin2 x) = 3 cos2 x sinx− sin3 x =

= 3(1− sin2 x) sinx− sin3 x = 3 sinx− 4 sin3 x

cos 3x = cos 2x cosx− sin 2x sinx = (cos2 x− sin2 x) cosx− (2 sinx cosx) sinx = cosx− 4 sin2 x cosx

cos 3x = −3 cosx+ 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 cosx

x = π/6

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

=⇒ cosπ/6 =√

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

(2) sin 2π/6 = 2 cosπ/6 sinπ/6 =√

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

√2 = sinπ/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.

tanx− y =sinx− ycosx− y

=sinx cos y − sin y cosx

cosx cos y + sinx sin y

(1

cos x cos y1

cos x cos y

)=

tanx− tan y

1 + tanx tan y

if tanx tan y 6= −1Similarly,

tanx+ y =sinx+ y

cosx+ y=

sinx cos y + sin y cosx

cosx cos y − sinx sin y=

tanx+ tan y

1− tanx tan y, tanx tan y 6= 1

cotx+ y =cosx+ y

sinx+ y=

cosx cos y − sinx sin y

sinx cos y + sin y cosx=

cotx cot y − 1

cot y + cotx

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

√3

2 cosx) = 32 sinx+ 3

√3

2 cosx

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

A = C cosα,B = C sinα

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

If A 6= 0,

A sinx+B cosx = A(sinx+B

Acosx) == A(sinx+ tanα cosx)

=A

cosα(cosα sinx+ sinα cosx) =

A

cosα(sinx+ α)

where −π/2 < α < π/4, B/A = tanα, C = Acosα

Exercise 10. C sinx+ α = C sinx cosα+ C cosx sinα.

C cosα = −2, C sinα = −2, C = −2√

2, α = π/4

Exercise 11. If A = 0, C = B,α = 0. If B = 0, A = −C,α = π/2. Otherwise,27

A sinx+B cosx = B(cosx+A

Bsinx) =

B

cosβ(cosx cosβ + sinβ sinx) = C cosx+ α

where AB = tanβ, α = −β, C = B

cos β .

Exercise 12.

sinx = cosx =√

1− cos2 x =⇒ cosx = 1/√

2 =⇒ x =π

4Try 5π/4. sin 5π/4 = cos 3π/4 = − sinπ/4 = −1/

√2.

cos 5π/4 = − sin 3π/4 = − cosπ/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.

sinx− cosx = 1 =√

1− cos2 x = 1 + cosx

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

cosx = −1, x = π/2 + 2πn

Exercise 14.

cosx− y + cosx+ y = cosx cos y + sinx sin y + cosx cos y − sinx sin y = 2 cosx cos y

cosx− y − cosx+ y = sinx cos y − sin y cosx+ sinx cos y + sin y cosx = 2 sinx cos y

sinx− y + sinx+ y = sinx cos y − sin y cosx+ sinx cos y + sin y cosx = 2 sinx cos y

Exercise 15.

sinx+ h− sinx

h=

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

h

=sinh/2

h/2cos (x+ h/2)

cosx+ h− cosx

h=

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

h

= − sinh/2

h/2sin (x+ h/2)

Exercise 16.

(1)sin 2x = 2 sinx cosx

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

(2) cosx+ y = cosx cos y − sinx sin y = cosx+ cos y.

cosx cos y − cosx− cos y = sin y√

1− cos2 x

Letting A = cosx,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) sinx+ y = sinx cos y + sin y cosx = sinx+ sin y

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

Checking our result, we find that sin (2πn+ y) = sin 2πn+ sin y(1)28

(4) ∫ y

0

sinxdx = − cosx|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.∫ ba

sinxdx = − 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+ sinx)dx = ( 1

2x2 − cosx)

∣∣π0

= π2

2 − (−1− 1) = π2

2 + 2

Exercise 19.∫ π/2

0(x2 + cosx)dx = ( 1

3x3 + sinx)

∣∣π/20

= 13 (π/2)3 + 1

Exercise 20.∫ π/2

0(sinx− cosx)dx = (− cosx− sinx)|π/20 = −1− (−1) = 0

Exercise 21.∫ π/2

0| sinx− cosx|dx = ( by symmetry )2

∫ π/40

(cosx− sinx)dx = 2(sinx+ cosx)|π/40 = 2(√

2− 1)

Exercise 22.∫ π

0( 1

2 + cos t)dt = ( 12 t+ sin t)

∣∣π0

= π2

Exercise 23. ∫ 2π/3

0

(1

2+ cos t)dt+

∫ π

2π/3

−(1

2+ cos t)dt = (

t

2+ sin t)

∣∣∣∣2π/30

+ (t

2+ sin t)

∣∣∣∣2π/3π

= 2(π

3+

√3

2)− π

2=π

6+√

3

Exercise 24. If −π < x ≤ − 2π3 ,∫ x

−π−(

1

2+ cos t)dt =

∫ −πx

(1

2+ cos t)dt =

(t

2+ sin t

)∣∣∣∣−πx

= −π2− x

2− sinx

If −2π/3 ≤ x ≤ 2π/3,∫ −2π/3

−π−(

1

2+ cos t)dt+

∫ x

−2π/3

(1

2+ cos t)dt =

−π/6+

√3/2 + (t/2 + sin t)|x−2π/3

= x/2 + sinx− π/3−√

3/2 +√

3/2− π/6 =x

2+ sinx− 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− sinx

Exercise 25.∫ x2

x(t2 + sin t)dt = ( 1

3 t3 +− cos t) = x6−x3

3 + cosx− cosx2

Exercise 26.∫ π/2

0sin 2xdx =

(− cos (2x)

2

)∣∣∣π/20

= (−1/2)(−1− 1) = 1

Exercise 27.∫ π/3

0cosx/2dx = 2 sinx/2|π/30 = 2 1

2 = 1

Exercise 28.29

∫ 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)

)∣∣∣x0

=

=cos a

bsin bx+

sin a

b(cos bx− 1) =

1

bsin 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

)∣∣∣∣x0

=

=1

b(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

)∣∣∣∣x0

= −3/4(cosx− 1) +cos 3x− 1

12=

=1

3− 3

4cosx+

1

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

(2) ∫ x

0

cos3 tdt =

∫ x

0

1

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

(1

4

sin 3t

3+

3

4sin t

)∣∣∣∣x0

=

=1

12(sin 2x cosx+ sinx cos 2x) +

3

4sinx =

1

12(2 sinx cosx+ sinx(2 cos2 x− 1)) =

=sinx cos2 x+ 2 sinx

3

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

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

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

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

a

f(x)dx =

∫ r+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

sinnxdx =1

n

∫ 2πn

0

sinxdx =1

n(− cosx)

∣∣∣∣2πn0

= − 1

n(1− 1) = 0∫ 2π

0

cosnxdx =1

n

∫ 2πn

0

cosxdx =1

nsinx

∣∣∣∣2πn0

= 0

(2) ∫ 2π

0

sinnx cosmxdx =

∫ 2π

0

1

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

0

sinnx sinmxdx =

∫ 2π

0

1

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

0

cosnx cosmxdx =

∫ 2π

0

1

2(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

30

Exercise 32. Given that x 6= 2πn; sinx/2 6= 0,

n∑k=1

2 sinx/2 cos kx = 2 sinx/2

n∑k=1

cos kx =

n∑k=1

sin (2k + 1)x

2− sin (2k − 1)

x

2= sin (2n+ 1)

x

2− sinx/2

= sinnx cosx/2 + sinx/2 cosnx− sinx/2 =

= 2 sinnx/2 cosnx/2 cosx/2 + sinx/2(1− 2 sin2 nx/2)− sinx/2 =

= 2(sinnx/2)(cos (n+ 1)x/2)

Exercise 33. Recall that

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

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

= −2 sin kx sinx/2

−2 sinx/2

n∑k=1

sin kx =

n∑k=1

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

= cosnx+ x/2− cosx/2

Nowsinnx/2 sinnx/2 + x/2 = sinnx/2(sinnx/2 cosx/2 + sinx/2 cosnx/2) =

= sin2 nx/2 cosx/2 + sinx/2 cosnx/2 sinnx/2 =

=

(1− cosnx

2

)cosx/2 +

sinnx

2sinx/2 =

=1

2(cosx/2− cosx/2 cosnx+ sinnx sinx/2) =

1

2(cosx/2− cos (nxx/2)

Then

−2 sinx/2

n∑k=1

sin kx = −2 sinnx/2 sin1

2(n+ 1)x

n∑k=1

sin kx =sinnx/2 sin 1

2 (n+ 1)x

sinx/2

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

1

2cosx sinx <

1

2sinx <

x

2=⇒ sinx < x

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

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/20

sinx = 2π

Exercise 6. 1π/2−−π/2

∫cosx = 2/π

Exercise 7. 1π/2−0

∫sin 2x = −1/π(−1− 1) = 2/π

Exercise 8. 1π/4−0

∫sinx cosx = 1

π

31

Exercise 9. 1π/2−0

∫sin2 x = 1

π (x− sin 2x/2)∣∣π0

= 12

Exercise 10. 1π−0

∫cos2 x = 1

2

Exercise 11.

(1) 1a−0

∫x2 = a2/3 = c2 =⇒ c = a/

√3

(2) 1a−0

∫xn = 1

a1

n+1xn+1∣∣∣a0

= an

n+1 = cn =⇒ c = a(n+1)1/n

Exercise 12.

A =

∫wf/

∫w

∫wx2 = k

∫x∫

x3 =1

4x4 = k

1

2x2; k =

1

2, w = x∫

x4 =1

5x5 = k

1

3x3; k =

3

5, w = x2∫

x5 =1

6x6 = k

1

4x4; k =

2

3, w = x3

Exercise 13.

A(f + g) =1

b− a

∫f + g =

1

b− a

∫f +

1

b− a

∫g = A(f) +A(g)

A(cf) =1

b− a

∫cf = c

(1

b− a

)∫f

A(f) =1

b− a

∫f ≤ 1

b− a

∫g = A(g)

Exercise 14.

A(c1f + c2g) =

∫w(c1f + c2g)∫

w=c1∫wf∫w

+c2∫wg∫w

= c1A(f) + c2A(g)

f ≤ g w > 0( nonnegative ),=⇒ wf ≤ wg

Exercise 15.

Aba(f) =1

b− a

∫ b

a

f =1

b− a

(∫ c

a

f +

∫ b

c

f

)=

(c− ab− a

)( ∫ caf

c− a

)+b− a− (c− a)

b− a

∫ baf

b− ca < c < b

0 <c− ab− a

< 1Let t =

c− ab− a

=⇒ Aba(f) = tAca(f) + (1− t)Abc(f)

Aba(f) =

∫ bawf∫ baw

=

∫ caw∫ b

aw

∫ cawf∫ caw

+

(∫ baw −

∫ caw∫ b

aw

) ∫ bcwf∫ bcw

0 <

∫ caw∫ b

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

∫ caw∫ b

aw

=⇒ Aba(f) = tAca(f) + (1− t)Abc(f)

Exercise 16. Recall that xcm =∫xρ∫ρ

or rcm =∫rdmM .

32

xcm =

∫ L0x∫ L

01

=L

2

Icm =

∫r2dm =

∫x2(1) = L3/3

r2 =Icm∫ L0

1= L2/3 =⇒ r =

L√3

Exercise 17.

xcm =

∫ l/20

x+∫ LL/2

2xdx

L2 + 2(L− L/2)

=yL2

12

Icm =

∫ L/2

0

x2 +

∫ L

L/2

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

3L/2=

5L2

12=⇒ r =

√5L

2√

3

Exercise 18. ρ(x) = x for 0 ≤ x ≤ L

xcm =

∫xxdx∫xdx

=13x

3∣∣L0

12x

2∣∣L0

=2

3L

Icm =

∫x2xdx = L4/4

r2 =L4/4

L2/2= L2/2 r =

L√2

Exercise 19.

xcm =

∫xxdx+

∫xL2 dx∫

xdx+∫L/2

=

13x

3∣∣L/20

+ L2 (x2/2)

∣∣LL/2

12x

2∣∣L/20

+ L2 (L− L/2)

= 11L/18

Icm =

∫x2xdx+

∫x2L/2dx = L431/192

r2 = Icm/(L23/8) = L231/72 r =

√31L

6√

2

Exercise 20. ρ(x) = x2 for 0 ≤ x ≤ L

xcm =

∫xx2dx∫x2

= 3L/4

Icm =

∫x2x2dx = L5/5

r2 =Icm13L

3=

3

5L2 r =

√3

5L

Exercise 21.

xcm =

∫ L/20

xx2dx+∫ LL/2

xL2

4 dx∫ L/20

x2dx+∫ LL/2

L2

4 dx= 21L/32

Icm = intL/20 x2x2dx+

∫ L

L/2

x2L2

4dx = 19L5/240

r2 =IcmL3/6

= 19L2/40 =⇒ r =√

19L/2√

10

Exercise 22. Be flexible about how you can choose a convenient origin to evaluate the center-of-mass from33

Let 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+ 1

n+ 2ML =

3ML

4

xcm =

∫xρ

M=

3L

4=⇒

∫xρ =

n+ 1

n+ 2=

3

4=⇒ n = 2

ρ =3M

L3x2

Exercise 23.

(1)1

π/2− 0

∫3 sin 2t =

6

π

(2)1

π/2− 0

∫9 sin2 2t = 9/2 =⇒ vrms = 3

√2/2

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

1

2π

∫ 2π

0

160 sin t2 sin (t− π/6) = 80√

3

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

0(1 + t + t2)dt =

x+ 12x

2 + 13x

3

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

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

Exercise 4.∫ 1−x

1(1− 2t+ 3t2)dt = (t− t2 + t3)

∣∣1−x1

= −2x+ 2x2 − x3

Exercise 5.∫ x−2t4 + t2 = 1

5 t5 + 1

3 t3∣∣x−2

= x5

5 + x3

3 + 403

Exercise 6.∫ x2

xt4 + 2t2 + 1 =

(t5

5 + 23 t

3 + t)∣∣∣x2

x= 1

5 (x10 − x5) + 23 (x6 − x3) + x2 − x

Exercise 7.(

23 t

3/2 + t)∣∣x

1= 2

3 (x3/2 − 1) + (x− 1)

Exercise 8.(

23 t

3/2 + 45 t

5/4)∣∣x2

x= 2

3 (x3 − x3/2) + 45 (x5/2 − x5/4)

Exercise 9. sin t|xiπ = sinx

Exercise 10.(t2 + sin t

)∣∣x2

0= x2

2 + sinx2

Exercise 11.(

12 t+ cos t

)∣∣x2

x= x2−x

2 + cosx2 − cosx

Exercise 12.(

13u

3 +− 13 cos 3u

)∣∣x0

= x3

3 +− 13 (cos 3x− 1)

Exercise 13.(

13v

3 + cos 3v−3

)∣∣∣x2

x= x6−x3

3 + −13 (cos 3x2 − cos 3x)

Exercise 14.∫

1−cos 2x2 + x =

(12x−

sin 2x4 + 1

2x2)∣∣y

0=

y

2− sin 2y

4+y2

234

Exercise 15.(− cos 2w

2 + 2 sin w2

)∣∣x0

= − (cos 2x− 1)

2+ 2 sin

x

2

Exercise 16.∫ x−π( 1

2 +cos t)2dt =∫ x−π

14 +cos t+cos2 t = 1

4 (x+π)+sinx+ 12

(t+ sin 2t

2

)∣∣x−π = 3

4 (x+π)+sinx+ 14 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.

1

4x4 − 1

2x2 =

1

3

(1

2t2 − 1

4t4)∣∣∣∣x√

2

=1

6x2 − 1

12x4

=⇒ 1

3x4 − 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 − 1

2

(1) To show the periodicity, consider

f(x+ 1) = x+ 1− [x+ 1]− 1

2= r − 1

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

x+ 1− [x+ 1] = r − 1

2

(2) P (x) =∫ x

0f(t)dt =

∫ x0

(t− 12 ) = 1

2x2 − 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 =1

2(x2 − x)

∣∣∣∣10

= 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) =1

2(t2 − t)

=⇒∫ 1

0

P (t)dt =1

2

(1

3t3 − 1

2t2)∣∣∣∣1

0

=1

2

−1

6=⇒ c =

1

12

(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

1

2(t2 − t) +

1

12=

1

6x3 − 1

4x2 +

x

12

35

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) =

∫ −x0

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) =

∫ −x0

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!

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)

36

(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 definition of the limit of a function, The definition 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→21x2 = 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 = −1

Exercise 15. limx→02 sin x cos x

x = 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 = 2 Exercise 19.37

limx→0

sin(x+a

2 + x−a2

)− sin

(x+a

2 −(x−a

2

))x− a

=

= limx→0

(sin x+a

2 cos x−a2 + sin x−a2 cos x+a

2 −(sin x+a

2 cos x−a2 − sin x−a2 cos x+a

2

)x− a

)=

= limx→a

2 sin x−a2 cos x+a

2

x− a= cos a

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)=

1

2

Exercise 22. b, c are given.

sin c = ac+ b, a = sin c−bc , 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−bc2 , c 6= 0.

If c = 0, then b = 2, a ∈ R.

Exercise 24.tangent is continuous for x /∈ (2n+ 1)π/2

cotangent is continuous for x /∈ 2nπ

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

Exercise 26.

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

(2)

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

|x|2

2< 2ε/2 = ε

If we had chosen δ0 =√

2ε for a given ε. |x− 0| < δ =√

2ε.(3)

| sinx(cosh− 1) + cosx sinh| ≤ | sinx|| cosh− 1|+ | cosh|| sinh| < ε

2+ε

2= ε

| cosx+ h− cosx| = | cosx cosh− sinx sinh− cosx| = | cosx(cosh− 1)− sinx sinh| ≤

≤ | cosx|| cosh− 1|+ | sinx|| sinh| < ε

2+ε

2= ε

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

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

2; | sinh| < ε

2

Exercise 27. f(x)−A = sin 1x −A.

Let x = 1nπ .

|f(x)−A| = | sinnπ −A| > || sinnπ| − |A|| > |1− |A||Consider |x − 0| = |x| = 1

nπ ≤ 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 ≤ 1n , n ∈ Z

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

f(x) =

[1

x

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

1

mf(x) > M(n)

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

So f(x)→∞ as x→ 0+.38

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

f(x) =

[1

x

]= [n] = n < −M(n)

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

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

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| <δ12< δ1

=⇒ |x− x1| <δ12

+δ12

= 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 infinitely many points that are continuous in (a, b),and at the very least, some or all are “clustered” around some neighborhood about the one point given to make f continuous.

Exercise 32. Given ε = 1n , |f(x)| = |x sin 1

x | = |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 = (1

2x2 − ax)

∣∣∣∣ba

=1

2(b− a)(b+ a)− a(b− a) =

(b− a)2

2

39

(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)− 1

2(c− a)(c+ a) +

1

2(b− c)(b+ c)− c(b− c) =

=1

2((c− a)2 + (b− c)2)

Draw a figure for clear, geometric reasoning.Consider a square of length (b − a) and a 45 − 45 right triangle inside. From the figure, it’s obvious that right

triangles of c− a length and (b− c) length lie within the (b− a) right triangle.

Compare the trapezoid of c− a, b− a bases with the b− a right triangle.

1

2(b− c)(b− a+ c− a) =

1

2(b− c)(b− c+ 2(c− a)) >

1

2(b− c)2

Indeed, the trapezoid and c− a right triangle equals the b− a trapezoid since

1

2(b− c)(b− a+ c− a) +

1

2(c− a)2 =

1

2(b2 − c2 − 2ab+ 2ac+ c2 − 2ca+ a2) =

1

2(b− a)2

=⇒ 1

2(b− a)2 >

1

2(b− c)2 +

1

2(c− a)2

so then

∣∣∣∣∣∫ b

a

f(x)dx− (b− a)f(c)

∣∣∣∣∣ ≤ (b− a)2

2

3.11 Exercises - Bolzano’s theorem for continuous functions, The intermediate-value theorem for continuous func-tions. 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: find the largest c.Let S = { all x ∈ [a, b] s.t. f(x) ≤ 0 }.S is nonempty since f(a) < 0. S is bounded since all S ⊆ [a, b].=⇒ S has a suprenum.Let c = supS.

If f(c) > 0, ∃(c− δ, c+ δ) s.t. f > 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→∞ckx

k

ck−1xk−1 = limx→∞ckck−1

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

k. 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 function

changed direction.

(1) If P (x) = 3x4 − 2x3 − 36x2 + 36x − 8, P (−4) = 168, P (−3) = −143, P (0) = −8, P ( 12 ) = 15

16 , P (1) = −7,P (−3) = −35, P (4) = 200

(2) If P (x) = 2x4 − 14x2 + 14x− 1, P (−4) = 231, P (−3) = −7, P (0) = −1, P ( 12 ) = 1

8 , P ( 32 ) = − 11

8 , P (2) = 2

(3) If P (x) = x4 + 4x3 + x2 − 6x + 2, P (−3) = 2, P (− 52 ) = − 3

16 , P (−2) = 2, P ( 13 ) = 22

81 , P ( 12 ) = − 3

16 , P ( 23 ) =

− 1481 , 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. tanx 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 c

such that g(c) = 0, so that f(c) = c.41

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

Exercise 2. D = R, g(y) = 12 (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

)2if 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.

g preserves monotonicity.=⇒ a1 ≶Mf ≶ an

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

Mh = H

(1

n

n∑i=1

h(ai)

)= H

(1

n

n∑i=1

(af(ai) + b)

)= H

(a

1

n

n∑i=1

f(ai) + b

)The inverse for h is g

(h−ba

)= H(h) = h−1. So then

Mh = g

(1

n

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

and∫ baf(x)g(x)dx∫ bag(x)dx

= f(c)

Exercise 1.

g = x9 > 0 for x ∈ [0, 1]; f =1√

1 + xm =

1√2,M = 1∫ 1

0

x9 =1

10x10

∣∣∣∣10

=1

10

1

10√

2≤∫ 1

0

x9

√1 + x

dx ≤ 1

10

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− 1

3x3)

∣∣∣∣1/20

=11

24

11

24≤∫ 1/2

0

√1− x2dx ≤ 11

24

√4

3

Exercise 3.42

f =1

1 + x6g = 1− x2 + x4

∫ a

0

1− x2 + x4 =

(x− 1

3x3 +

1

5x5

)∣∣∣∣a0

= a− a3

3+a5

5

m =1

1 + a6M = 1

1

1 + a6

(a− a3

3+a5

5

)≤∫ a

0

1

1 + x2dx ≤

(a− a3

3+a5

5

)So if a = 1

10 , (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 =

∫ (1

2t

)(2t sin t2)dt =

1

2c(− cos t2)

∣∣√(n+1)π√nπ

=

=−1

2c((−1)n+1 − (−1)n) =

1

c(−1)n

Exercise 6.∫ ba

(f)(1) = f(c)∫ ba

1 = f(c)(b − a). Then f(c) =∫ baf

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

integrals.

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

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

Let ε =1

2f(c)∃δ > 0 for ε =

1

2f(c)∫ c+δ

c−δf(x)dx >

1

2f(c)(2δ) = f(c)δ > 0

But∫ baf(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 ≥ 0M ≤ 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+ 3

Exercise 4. f ′ = 4x3 + cosx

Exercise 5. f ′ = 4x3 sinx+ x4 cosx

Exercise 6. f ′ = −1(x+1)2

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

Exercise 8. f ′ = x−1−(x)(x−1)2 = −1

(x−1)2

43

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

(2− cosx)2=−2 cosx− 2 sinx+ 1

(2− cosx)2

Exercise 12.

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

(1 + x2)2=

sinx+ x cosx+ x3 cosx− x2 sinx

(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. v016 for Tsec.

(5) f ′′ = −32(6) h = −20t2

Exercise 14. V = s3, dVdS = 3s2

Exercise 15.

(1) dAdr = 2πr = C

(2) dVdr = 4πr2 = A

Exercise 16. f ′ = 12√x

Exercise 17. f ′ = −1(1+√x)2

(1

2√x

)Exercise 18. f ′ = 3

2x1/2

Exercise 19. −32 x−5/2

Exercise 20. f ′ = 12x−1/2 + 1

3x−2/3 + 1

4x−3/4 x > 0

Exercise 21. f ′ = − 12x−3/2 +− 1

3x−4/3 − 1

4x−5/4

Exercise 22. f ′ =12x−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 24.44

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.

(tanx)′ =(cosx

sinx

)′=

cos2 x− (− sinx) sinx

cos2 x= sec2 x

(cotx)′ =(cosx

sinx

)′=− sinx sinx− cosx cosx

sin2 x= − csc2 x

(secx)′ =−1

cos2 x(− sinx) = tanx secx

(cscx)′ =−1

sin2 xcosx = − cotx cscx

Exercise 35.

f ′ =(2ax+ b)(sinx+ cosx)− (cosx− sinx)(ax2 + bx+ c)

(sinx+ cosx)2=

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

(sinx+ cosx)2

Exercise 36.

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

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

Exercise 37.

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

= ax2 cosx− dx2 sinx+ (2a− e)x sinx+ (b+ 2d)x cosx+ (b− f) sinx+ (c+ e) cosx

g = x2 sinx. So d = −1, b = 2, f = 2, a = 0, e = 0, c = 0.

Exercise 38. 1 + x+ x2 + · · ·+ xn = xn+1−1x−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

(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

45

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 − 1

2+

3

2a2

The answer could probably be guessed at, but let’s review some tricks for solving cubics.First, do a translation in the x direction to center the origin on the point of inflection. Find the point of inflection by taking

the second derivative.

f ′′ = 6a+ 3 =⇒ a = −1

2So

a = x− 1

2

=⇒ (x− 1

2)3 +

3

2(x− 1

2)2 − 1

2= x3 =

3

4x− 1

4= 0

46

Then recall this neat trigonometric fact:

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

=⇒ cos3 x =3

4cosx− cos 3x

4= 0

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

what we have for a, a = −1, which we already have in the previous part, and a = 12 . So

f

(1

2

)=

3

8

y(x) =

(1

4x

)+

1

4

Exercise 9.

f(x) =

{x2 if x ≤ cax+ b if x > c

f ′(x) =

{2x if x ≤ ca if x > c

a = 2c; b = −c2

Exercise 10.

f(x) =

{1|x| if |x| > c

a+ bx2 if |x| ≤ cNote 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 = − 1

2c3, a =

3

2c.

Exercise 11.

f ′ =

{cosx if x ≤ ca if a > c

Exercise 12. f(x) =(

1−√

21+√

2

)= 1−A

1+A

A =√xA′ = a =

1

2x−1/2 =

1

2A;A′′ = −1

4x−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+ 1

2x3/2(1 +√x)3

f ′′′ =1

2

( −1A2A

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

(A2(1 +A)3)2

)=−3

4

( 1A + 4 + 5A

A4(1 +A4)

)= −3

4

(1 + 4√x+ 5x)√

x(x+√x)4

Exercise 13.47

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 =

1

2, g = 4

(1)

h′ =f ′g − g′f

g2=f ′

g− g′

g

f

g= 4− 2(

1

8) =

15

4

(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)=

1

2

Exercise 15.

(1) True, by definition of f ′(a).(2)

limh→0

f(a)− f(a− h)

h= − lim

h→0

f(a− h)− f(a)

h= lim−h→0

f(a− h)− f(a)

−h= f ′(a)

True, by definition of f ′(a).(3)

limt→0

f(a+ 2t)− f(a)

t= 2 lim

2t→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+−1

2limt→0

f(a+ t)− f(a)

t=

f ′(a)− 1

2f ′(a) =

1

2f ′(a)

False.

Exercise 16.

(1)

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

limh→0

2FG− 2fg

h= limh→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+ limh→0

2f

G+ glimh→0

G2 − g2

h=

=g

fD∗f +

f

gD∗g

48

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= limh→0

f2(x+h)−f2(x)g2(x+h) + f2(x)

g2(x+h) −f2(x)g2(x)

h=

=D∗f

g2+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 cosx

Exercise 2. x√1+x2

Exercise 3. −2x cosx2 + 2x(x2 − 2) sinx2 + 2 sinx3 + 6x3 cosx3

Exercise 4.

f ′ = cos (cos2 x)(−2 cosx sinx) cos (sin2 x) + sin (cos2 x) sin (sin2 x)(2 sinx cosx) =

= − sin 2x(cos (cos 2x))

Exercise 5.

f ′ = n sinn−1 x cosx cosnx+−n sinnx sinn x

Exercise 6.

f ′ = cos (sin (sinx))(cos (sinx))(cosx)

Exercise 7.

f ′ =2 sinx cosx sinx2 − 2x cosx2 sin2 x

sin2 x2=

sin 2x sinx2 − 2x sin2 x cosx2

sin2 x2

Exercise 8. f ′ = 12 sec2 x

2 + 12 csc2 x

2

Exercise 9. f ′ = 2 sec2 x tanx+−2 csc2 x cotx

Exercise 10. f ′ =√

1 + x2 + x2√

1+x2= 1+2x2√

1+x2

Exercise 11. f ′ = 4(4−x2)3/2

Exercise 12.

f ′ =1

3

(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.49

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.

1

2(x+

√x+√x)−1/2(1 +

1

2(x+

√x)−1/2

(1 +

1

2√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 (−1

x2

)=

1

(x+ 1)2g′ =

−1(1 + 1

f

)2

(−1

f2

)f ′ =

f ′

(f + 1)2

g′ =(x+ 1)−2(xx+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)

dx22x = 2xf ′

(2)g′ = 2 sinx cosxf ′ − 2 cosx sinxf ′ = (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

50

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

s = 5cm 75cm3/sec

s = 10cm 300cm3/sec

s = xcm 3x2cm3/sec

Exercise 21.

l =√x2 + h2

dl

dt=

1

lxdx

dtdx

dt=l

x

dl

dt=

10mi

−√

102 − 82(−4mi/sec) =

20

3

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=

3

512 =

36

5mi/hr

Exercise 24. Given the preliminary information

r

h=

2

5= α, V =

1

3πr2h =

1

3πα2h3

(1)

V =πr2

h2(h2y − hy2 +

1

3y3)

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 =

5

4π

(2)dV

dt= πα2h2 dh

dt,dh

dt=

1

πα2h2

dV

dt=

5

4π

Exercise 25.

α =r

h=

3

2dV

dt= πα2h2 dh

dt

c− 1π9

4(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

51

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− 5

π(2(10)5− 25)3/2

)=

1

15π

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 =1

2

√(y − 1)

36

7y

dA

dt=

1

2

(√36

y

1

2

1√y − 1

y +

√(y − 1)

36

7

)dy

dt=

=1

2

(6

2(7)8 + 6

)(2) =

66

7

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

3 H . So then

V = π/3r2(3r + 3) = πr3 + πr2

dV

dr= πr(3r + 2)

dr

dt

With the given information, we getdr

dt=

1

π(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) dydt = 2xdxdt ; when x = 1

2 , y = 14 ,

dydt = dx

dt

(2) t =π

6

Exercise 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.

1

2

1√x

+1

2√yy′ = 0 y′ =

−√y√x

< 0

Exercise 32.

±√

12− 3x2

452

6x+ 8yy′ = 0 =⇒ y′ =−3x

4y

3 + 4(y′2 + yy′′) = 0

y′′ =

(−3

4− y′2

)1

y=−9

4y3

Exercise 33.

sinxy + x cos2 xy(y + xy′) + 4x = 0

y′x2 cosxy + xy cosxy + sinxy + 4x = 0

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 defined 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)53

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.)�

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 is

If @ 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 ≤ 1

1/x if x ≥ 1f(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−a22

)b− a

=−(a+ b)

2= −c

Note that −1 ≤ f ′ ≤ 0 for 0 ≤ x ≤ 1

For 1 ≤ a < b ≤ 21b −

1a

b− a=−1

ab=−1

c2=⇒ c =

√ab

Note that −1 ≤ f ′ ≤ −1/4

For 0 ≤ a ≤ 1 , 1 ≤ b ≤ 2

1b −

(3−a2

2

)b− a

=2− (3− a2)b

2b(b− a)= −c or

−1

c2

depending upon if 0 ≤ c ≤ 1 or 1 ≤ c ≤ 2, respectively54

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 ′ =−2

3x−1/3 6= 0 for |x| ≤ 1

This is possible since f is not differentiable at x = 0.

Exercise 5. x2 = x sinx+ 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.

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 =

1

2so then lim

h→0θ =

1

2

(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 − xh

√x2 + xh+ h2

3 + x√x2 + xh+ h2

3 + x

=

=x+ h

3

x+√x2 + hx+ h2

3

=⇒ limh→0

θ =1

2

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 definite).55

Exercise 8. Using the mean value theorem

(1)sinx− sin y

x− y= cos c =⇒

∣∣∣∣ sinx− sin y

x− y

∣∣∣∣ = | cos c| ≤ 1

=⇒ | sinx− 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−ynx−y ≤ nx

n−1.

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→a

f(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)56

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 = 32 .

(2) f ′(x) ≷ 0 for x ≷ 32

(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(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− 2x3 =⇒ 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

57

(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+ 13

3

(x− 1)3(x− 3)3

)x2 − 4x+

13

3> 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(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 = π2n

(2)

f ′ > 0 when0 < x <

π

2

πn < x <π

2+ πn

f ′ < 0 whenπ

2+ πn < x < π(n+ 1)

(3)

f ′′ = 2 cos 2xf ′′ > 0 when

−π4

+ πn < x <π

4+ πn

f ′′ < 0 whenπ

4+ πn < x <

3π

4+ πn

(4) See sketch.58

Exercise 12. f(x) = x− sinx

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

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

f ′′ < 0 when 2πn+ π < x < 2π(n+ 1)

(4) See sketch.

Exercise 13. f(x) = x+ cosx

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

(2) f ′ > 0 if x 6= π2 + 2πn

(3)

f ′′ = − cosxf ′′ > 0 when

−π2

+ 2πn < x <π

2+ 2πn

f ′′ < 0 whenπ

2+ 2πn < x <

3π

2+ 2πn

(4) See sketch.

Exercise 14. f(x) = 16x

2 + 112 cos 2x

(1) f ′ = 13x+ − sin 2x

6 f ′(0) = 0(2) f ′ ≷ 0 when x ≷ 0(3) f ′′ = 1

3 −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 +

−Ax2

= 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 > 059

f = x+ y

f ′ = 1 + y′ = 0 = 1 +−xy

= 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′ =−xy

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√

2

2

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 isL√2

+

√L2 − L2

2=

2L√2

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√

2so then the rectangle that has maximum size is a square.

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

A0 = (2x)(2√r2 − x2) = 4x

√r2 − x2 (fix the area to be A0)(

A0

4x

)2= r2 − x2 =⇒ x4 − x2r2 +

A20

16 = 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 “infinity” boundaries, the circumscribing circle would beinfinitely large.

Exercise 10. Given a sphere of radius R, find the radius r and altitude h of the right circular cylinder with the largest lateral60

surface 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 has

radius√

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

(−2

R3

)=⇒ dr

dR= 0 =⇒ R =

√A0

2√π

=⇒ H

2= R

r2 = R2 +R2 = 2R2 =⇒ r =√

2R

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

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

R−r = HR = 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)

sincedA

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 circular

cone of radius R and altitude H .Constraint: h

R−r = HR = α

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

dV

dr= πα(2Rr − 3r2) = rπα(2R− 3r) r =

2R

3

sincedV

dr≷ 0 when r ≶

2R

3, r =

2R

3maximizes volume

h =1

3H

Exercise 14. Given a sphere of radius R. Compute, in terms of R, the radius r and the altitude h of the right circular cone of

maximum volume that can be inscribed in this sphere.

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

361

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 =1

2h(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 turning

up the sides. Find the dimensions of the box of largest volume that can be made in this manner if the material has sides (a)10 and 10; (b) 12 and 18

(1)

(x− 2r)(Y − 2r)r = (xy − 2rx− 2ry + 4r2)r = xyr − 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 − xy6

d2V

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 find 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 ×

203 .

(2) 12 and 18=⇒ 5−

√7× 2 + 2

√7× 8 + 2

√7

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

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 = 300mi x = constant speed. l0x = time spent. K = gas cost = 0.30.62

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)

(−2

x3

))= 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√

6

600

)= 3√

2√

6 + 10D

Remember that there is a speed limit of 60mi/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 = x0

x20+1

=⇒ x0 =1

2y0±

√1

(2y0)2− 1

x2 − x1 =

√1

y20

− 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

(√1

y20

− 4

)= πy0

√1− 4y2

0

dV

dy0= π

(√1− 4y2

0 +y0

2√

1− 4y20

(−8y0)

)= π

(1− 8y2

0√1− 4y2

0

)=⇒ y0 =

1

2√

2

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

Then Vmax = π 182 =

π

4

Exercise 21. Draw a good diagram. Note how the right triangle that you folded is now reflected backwards , so that this

triangle’s right angle is on the left-hand side now.The constraint is that the crease touches the left edge.

w0 = l sinα+ l sinα cos (2α) = l sinα(1 + cos (2α)) =

= 2l sinα cos2 α

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

63

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

l =w0

2 sin (α) cos2 α=w0

2csc (α) sec2 (α)

dl

dα=w0

2

(− cotα cscα sec2 α+ cscα2 secα secα tanα

)=

=w0

2

(−CS

(1

S

)(1

C2

)+

1

S2

(1

C2

)(S

C

))=w0

2

(−1

S2C+

2

C3

)dldα=0−−−−→ sinα =

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

2

11√3

23

=9√

3

2

Exercise 22.

(1) Consider the center of the circle O, the apex of the isosceles triangle that makes an angle 2α, A, and one of its othervertices, 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 figure. The complement of that angle is 2α.Beforehand, we can get the length of the isosceles triangle leg from the law of cosines.

cos (π − 2α) = − cos (2α)

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 cosα+ 2R sin (2α)

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

sinα =− 1

2 ±√

14 − 4(1)(− 1

2 )

2(1)=

1

2> 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 flattened,α = π, triangle “completely disappears.”

(2) I had originally thought to Reuse the constraint equation, Eqn. ( ??). This is wrong!

Think about the problem directly and for what it actually is; less wishful thinking.Consider a fixed perimeter L and imagine L to be a string that can be stretched into an isoceles triangle. A “trivial”isoceles triangle is a collapsed triangle with two sides of lengthL/2 only. Then the radius of the disk needs to beL/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 sinα = 2h(1 + sinα)

=⇒ h =P

2(1 + sinα)

h

2= R cosα

64

We could try to extremize this equation.dR

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

dR

dα= 0 =⇒ cos (2α) = sinα =⇒ sinα =

1√3

cosα =

√2√3

R =3P

4√

2(√

3 + 1)

However, this is the minimized R, minimized radius for the smallest circle fitting a particular isosceles triangle of afixed perimeter. We want to smallest circle with a radius big enough to fit 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(1

2

)=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 its

ends, 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 − yl0 − x

=h/2

l0=⇒

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)2l03h

)(H +

h

2l0

(h−H)2l03h

)2

= πl0(h+ 2H)3

27h

where H = 2, l0 = 12

Exercise 25.

S =n∑k=1

(x− ak)2 =⇒ dS

dx=

n∑k=1

2(x− ak) = 0

=⇒ nx =

n∑k=1

ak =⇒ x =

∑nk=1 akn

Since limx→±∞ S = +∞, x =∑nk=1 akn minimizes S.

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

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

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

61/7 > 0.

f(x =−A1/7

61/7=

A

6x5< 0

Thus, A > 0.

df

dx= 10x− 5Ax−6 = 5

(2x7 −Ax6

)= 0

x =

(A

2

)1/7

d2f

dx2= 10 + 30Ax−7 = 10 + 3−A

(2

A

)= 70 > 0

Thus x =(A2

)1/7minimizes f for A > 0.

f(x =

(A

2

)1/7

) =5(A2

)+A(

A2

)5/7 = 24

=⇒ A = 2

(24

7

)7/2

Exercise 27. Consider f(x) = −x3

3 + t2x over 0 ≤ x ≤ 1.

f(0) = 0, f(1) = −1

3+ t2 =⇒ f(1) ≷ 0 if t2 ≷

1

3

f ′(x) = −x2 + t2 = 0

=⇒ x2 = t2 but x ≥ 0, so x = |t|

f(x2 = t2) = −1

3t2(x) + t2x =

2

3t2x > 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| > 1

3m(t) =

−1

3+ t2 for |t| < 1

3

Exercise 28.

(1)

E(x, t) =|t− x|x

M(t) = max|t− x|x

as x = a→ x = b

|t− x|x

=

{t−xx if t ≥ xx−tx if t < x

d

dt

(|t− x|x

)=

{− tx2 if t ≥ xtx2 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 endpoints

t− aa

= E(a, t),b− tb

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

66

(2)

M(t) =

{t−aa if t−aa > b−t

b i.e. t(b+aab

)> 2

b−tb if b−tb > t−a

a

dM

dt=

{1a if t > 2 ab

a+b−1b if t <

(aba+b

)2

Since dMdt ≷ 0 when t ≷ 2ab

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

a+ b

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 =1

z2x3

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

So

fxx =−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

67

5.5 Exercises - The derivative of an indefinite integral. The first 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

Exercise 6.√

2 23x

3/2 +√

12

23x

3/2 =√

2x3/2

∫ baf =√

2(b3/2 − a3/2)

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

P = 25x

5/2 − 2x3/2 + 7x1/2∫ baf = 2

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

Exercise 8. P = 32x

4/3 − 32x

2/3; x > 0

Exercise 9. P = −3 cosx+ x6

3

Exercise 10. P = 37x

7/3 − 5 sinx

Exercise 11. f ′(x) = 1x

f =

∞∑k=−∞

akxk

f ′ =

∞∑k=−∞

kakxk−1 =

1

x

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

Exercise 12. ∫ x

0

|t|dt =

{∫ x0tdt if x ≥ 0∫ x

0−tdt if x < 0

=

{12x

2 if x ≥ 0−12 x

2 if x < 0=

1

2x|x|

Exercise 13. ∫ x

0

(t+ |t|)2dt =

{∫ x0

(2t)2dt if x ≥ 0

0 if x < 0=

{43x

3 if x ≥ 0

0 if x < 0=

2x2

3(x+ |x|)

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

∫ 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.∫ xcf(t)dt = cosx− 1

2 f(x) = − sinx c = −π6 .

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

0

t sin t− t =

(−t cos t+ sin t− 1

2t2)∣∣∣∣x

0

= sinx− x cosx− 1

2x2

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= −x

18

9+−1

9+x16

8+x18

9+ C

=⇒ C =1

9

Exercise 18. By plugging in x = 0 into the defined 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 + sinx

2 + x2;

f ′(0) =1

2= b

f ′′ =(cosx)(2 + x2)− 2x(1 + sinx)

(2 + x2)2

f ′′(0) =1

2+ 2c; c =

1

4

Exercise 19.

f(x) =1

2

∫ x

0

(x− t)2g(t)dt =1

2

∫ x

0

(x2 − 2xt+ t2)g(t)dt =

=1

2

(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)) +1

2x2g(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

69

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√

2

2

(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) =1

5

Exercise 23.

a3 − 2a cos a+ (2− a2) sin a =

∫ a

0

f2(t)dt

3x2 − 2 cosx+ 2x sinx+−2x sinx+ (2− x2) cosx = 3x2 − x2 cosx = f2(x)

f(x) = x√

3− cosx

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

(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.

70

(1)df

dt=

1 + 2 sinπt 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) +1

2=

(πt2

2− πt+

1

2t

)∣∣∣∣t1

=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

)∣∣∣10

= 3− 1 = 2

f(x) =3x2

2+

2

πsin

πx

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 − cf(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 =1

2∫ a

0

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

If a =1

2, f ′(1/2) ≥ 3

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.

71

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

105

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

3

)′= x2(x+ 1)1/2 +

4x(x+ 1)3/2

3(8x(x+ 1)5/2

15

)′=

4

3x(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 =

∫sinx(1− cos2 x) = − cosx+

1

3cos3 x

Exercise 7. ∫x1/3(1 + x) =

3

4x4/3 +

3

7x7/3 =

3

4(z − 1)4/3 +

3

7(z − 1)7/3

Exercise 8. sin−2 x−2

Exercise 9. (4−sin 2x)3/2

−3

∣∣∣π/40

= 33/2−8−3

Exercise 10. (3 + cosx)−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 =

4

9x9/4 − 4

5x5/4 =

4

9(1 + t)9/4 − 4

5(1 + t)5/4

Exercise 16.∫

(x2 + 1)−3/2dx =?(x√

x2 + 1

)′=

√x2 + 1− x2/

√x2 + 1

x2 + 1=

1

(x2 + 1)3/2

Exercise 17.72

(8x3 + 27)5/3

(3

5

)(1

24

)=

1

40(8x3 + 27)5/3

Exercise 18. 32 (sinx− cosx)2/3

Exercise 19. ∫xdx√

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

∫ 12du√u+ u3/2

=

=1

2

∫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.∫ baf(x)dx =

∫ b+ca+c

f(x− c)dx.

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

a

f(x)dx =1

k

∫ kb

ka

f(xk

)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 =1

k

∫ kb

ka

f(yk

)dy

Exercise 22.

F(xa, 1)

=

∫ x/a

0

up

(u2 + 12)qdu

u =t

a

du =dt

a

F(xa, 1)

=1

a

∫ 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

(1

x

)− F (1)

u =1

t

du =−1

t2dt,−1

u2du = dt∫ x

1

dt

1 + t2=

∫ 1/x

1

−duu2(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− xx = 1− u

Exercise 25.73

cosm x sinm x =

(sin 2x

2

)m= 2−m sinm 2x∫ π

2

0

cosm x sinm xdx = 2−m∫ π/2

0

sinm 2xdx = 2−m∫ π

0

1

2sinm xdx = 2−m−1

∫ π

0

sinm xdx =

= −2−m−1

∫ −π/2π/2

sinm(π

2− x)dx = 2−m−1

∫ π/2

−π/2cosm xdx = 2−m

∫ π/2

0

cosm xdx

Exercise 26.

(1)

u = π − xx = 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 = cosx

du = − sinxdx

∫ π

0

x sinx

1 + cos2 xdx =

∫ π

0

x sinx

2− sin2 x=π

2

∫ π

0

sinx

2− sin2 xdx =

π

2

∫ π

0

sinx

1 + cos2 xdx =

= −π2

∫ −1

1

du

1 + u2=π

2

∫ 1

−1

du

1 + u2= π

∫ 1

−1

dx

1 + x2

Exercise 27.

x = sinu

dx = cosu

∫ 1

0

(1− x2)n−12 dx =

∫ π/2

0

(cos2 u)n−12 cosudu =

∫ π/2

0

cos2n udu

5.10 Exercises - Integration by Parts.Exercise 1.

∫x sinx = −x cosx+ sinx

Exercise 2.∫x2 sinx = −x2 cosx+ 2x sinx+ 2 cosx

Exercise 3.∫x3 cosx = x3 sinx+ 3x2 cosx− 6x sinx+−6 cosx

Exercise 4.∫x3 sinx = −x3 cosx+ 3x2 sinx+ 6x cosx− 6 sinx

Exercise 5.∫

sinx cosx = − 14 cos 2x = − 1

4 (cos2 x− sin2 x)

Exercise 6.∫x sinx cosxdx =

∫x2 sin 2x = −x cos 2x

4 + sin 2x8

Exercise 7.∫

sin2 x =∫

sinx sinx = − 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 = sinxdx∫sinn x = − cosx sinn−1 x+

∫(n− 1) sinn−2 x(1− sin2 x) =

= − cosx sinn−1 x+

∫(n− 1) sinn−2 x− (n− 1)

∫sinn x∫

sinn x =−1

nsinn−1 x cosx+

(n− 1)

n

∫sinn−2 x

Exercise 9.

74

(1) ∫sin2 x =

−1

2sinx cosx+

1

2

∫1 =−1

2sinx cosx+

1

2x∫ π/2

0

sin2 xdx =π

4

(2)∫ π/2

0sin4 x = −1

4 sin3 x cosx∣∣π/20

+ 34

∫ π/20

sin2 x = 3π16

(3)∫ π/2

0sin6 x = 5

6

∫ π/20

sin4 x =5π

32

Exercise 10.

(1) ∫sin3 xdx =

−1

3sin2 x cosx+

2

3

∫sinx = −1

6sin 2x cosx− 2

3cosx =

−3

4cosx+

1

12cos 3x since

−3

4cosx+

1

12cos 3x =

−3

4cosx+

1

12(cosx cos 2x− sin 2x sinx) =

= −3

4cosx+

1

12(cosx(1− 2 sin2 x) +−2 sin2 x cosx) =

−2

3cosx− 1

3sin2 x cosx

(2)∫sin4 xdx =

−1

4sin3 x cosx+

3

4

∫sin2 x =

−1

4sin3 x cosx+

3

4(x

2− sin 2x

4) =−1

4sin3 x cosx+

3x

8− 3 sin 2x

16

Now1

32sin 4x =

1

32(2 sin 2x cos 2x) =

1

8(sinx cosx(1− 2 sin2 x) =

sin 2x

16− 1

4sin3 x cosx

=⇒ −1

4sin3 x cosx+

3x

8− 3 sin 2x

16=

3x

8− 1

4sin 2x+

1

32sin 4x

(3) ∫sin5 xdx =

∫sin4 x sinxdx = − cosx sin4 x+

∫cos2 x4 sin3 x =

= − cosx sin4 x+ 4(

∫sin3 x−

∫sin5 x) = − cosx sin4 x+ 4

∫sin3 x− 4

∫sin5 x

5

∫sin5 dx = − cosx sin4 x+ 4

∫sin3 x

5

∫sin5 dx = − cosx(1− cos2 x)2 + 4(

−3

4cosx+

1

12cos 3x)

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

3cos 3x

= − cosx+ 2 cos3 x− cos5 x− 3 cosx+1

3(cosx cos 2x− sinx sin 2x) =

= −4 cosx+ 2 cos3 x− cos5 x+1

3(4 cos3 x− 3 cosx) = −5 cosx+

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 conflicts 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

75

(2) ∫x sin3 x =

−3x

4cosx+

x

12cos 3x−

∫−3

4cosx+

1

12cos 3x =

=−3x

4cosx+

x

12cos 3x+

3

4sinx+

− sin 3x

36∫sin3 x =

−3

4cosx+

1

12cos 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+

1

2

(−x cos 2x

2+

sin 2x

4

)=

=x3

6− x2 sin 2x

4− x cos 2x

4+

sin 2x

8

Exercise 12. ∫cosn xdx =

∫cosn−1 x cosxdx = cosn−1 x sinx+

∫(n− 1) cosn−2 x sin2 x =

= cosn−1 x sinx+ (n− 1)

∫cosn−2 x−

∫(n− 1) cosn x

=⇒∫

cosn x =cosn−1 x sinx

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 sinx = 34 sinx+ 1

12 sin 3x since

1

12sin 3x =

1

12(sin 2x cosx+ sinx cos 2x) =

1

6sinx cos2 x+

1

12sinx(2 cos2 x− 1) =

=1

3sinx cos2 x− 1

12sinx

(3) ∫cos4 xdx =

cos3 x sinx

4+

3

4

(1

2x+

1

4sin 2x

)=

3

8x+

3

16sin 2x+

cos3 x sinx

4

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

3

8x+

1

4sin 2x+

1

32sin 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 =1

2x√

1− x2 +1

2

∫1√

1− x2

Exercise 15.

76

(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

(2) ∫(a2 − x2)dx =

x(a2 − x2)

3+

2a2

3x =

−x3

3+ a2x∫

(a2 − x2)5/2dx =x(a2 − x2)5/2

6+a25

6

∫(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−

(xa

)2

dx = a2

∫cos2 θdθ =

= a2

∫1 + cos 2θ

2= a2

(θ

2+

sin 2θ

4

)= a2

(arcsin

x

a+

1

2

x

a

√1−

(xa

)2) sin θ =

x

a

cos θdθ =dx

a∫ a

0

√a2 − x2 = a2

(π2− 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

)=

5

16πa6

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∣∣∣20

= 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) =1

5(48− 20(−6 +

10√

5

3)) =

168

5− 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

− 2

3

∫ 3

−1

(4 + t3)1/2 = 2√

31 +2√

3

3− 2

3(11.35)

Exercise 18.77

∫sinn+1 x

cosm+1 xdx =

∫sinn x

(sinx

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 x

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+−mn

∫cosm−1 x

sinn−1 x∫cot2 x =

∫cos1+1 x

sin1+1 x=−1

1

cos1 x

sin1 x− 1

1

∫dx = − cotx− x∫

cot4 xdx =

∫cos3+1 x

sin3+1 x= −1

3

cos3 x

sin3 x−∫

cos3−1 x

sin3−1 x=−1

3cot3 x− (− cotx− x) =

−1

3cot3 x+ cotx+ x

Exercise 20.

(1) ∫ 2

0

tf ′′(t)dt = 2

∫ 1

0

tf ′′(2t)dt n = 2

(2) ∫ 1

0

xf ′′(2x)dx =1

2

∫ 2

0

tf ′′(t)dt =1

2

(tf ′(t)|20 −

∫ 2

0

f ′(t)dt

)=

1

2(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

sinφ(t)

(φ′(t)

φ′(t)

)dt =

1

φ′(a)

∫ c

a

φ′(t) sinφ(t) +1

φ′(b)

∫ b

a

φ′(t) sinφ(t) =

=1

φ′(a)cosφ(t)|ca +

1

φ′(b)(cosφ(b)− cosφ(a)) ≤ 4

m

where1

m≥ 1

φ′(t)∀t ∈ [a, b]

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

a

sin t2dt

∣∣∣∣ ≤ 4

2a= 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) =∑hk=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)

78

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

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 coefficients 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) = sinx, Prove that f (n) = cos (x+ nπ2 ) and g(n)(x) = sin (x+ nπ

2 )

f (n)(x) = cos (x+nπ

2) =

{sinx(−1)j+1 if n = 2j + 1

cosx(−1)j if n = 2j

g(n)(x) = sin (x+nπ

2) =

{cosx(−1)j if n = 2j + 1

sinx(−1)j if n = 2j

f(x) = cosx

f ′(x) = − sinx

f ′′(x) = − cosx

f ′′′(x) = sinx

f ′′′′(x) = cosx

f (2j)(x) = cosx(−1)j

f (2(j+1))(x) = (cosx(−1)j)′ = cosx(−1)j+1

f (2j+1)(x) = sinx(−1)j+1

f (2j+3)(x) = (sinx(−1)j+1) = sinx(−1)j+2

g(x) = sinx

g′(x) = cosx

g′′(x) = − sinx

g′′′(x) = − cosx

g′′′′(x) = sinx

g(2j)(x) = sinx(−1)j

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

g(2j+1)(x) = cosx(−1)j

g(2j+3)(x) = (cosx(−1)j+1)

Exercise 4.79

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)

Nown!

(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 substitution

u = x2 u = 4;x = 2

Exercise 7. dgdu = u3/2; g(u) = 25u

5/2 g(4) = 2525 =

64

5

Exercise 8. ∫ x

0

sin t

t+ 1dt =

1

0 + 1

∫ c

0

sin tdt+1

x+ 1

∫ x

c

sin tdt =

= − cos t|c0 +1

x+ 1− cos t|xc = − cos c+ 1 +

−1

x+ 1(cosx− cos c) =

=−x cos c+ x− cos c+ 1− cosx+ cos c

x+ 1=x(1− cos c) + (1− cos c)

x+ 1> 0

Exercise 9. y = x2 is the curve for C. y = 12x

2 is the curve for C1.80

∫ b

0

(x2 − 1

2x2) =

1

6b3 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)c

3+ b3 − cb2 − b3

3+c3

3=

2b3

3− 2

3cb2 =

2

3b2(b− c)

Now A(A) = A(B)

=⇒ 2

3b2(b− c) =

1

6b3 =⇒ b =

4

3c

so then kx2 =16

9x2

Exercise 10.

(1) |Q(h)− 0| =∣∣∣ f(h)h

∣∣∣ =

{h2

|h| if x is rational

0 if x is irrational.

For now, consider 0 < h < δ; let δ(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.

1

3

∫ 1

0

6x+ 7 + 2

(6x+ 7)3dx =

1

3

(−(6x+ 7)−1

6+−(6x+ 7)−2

6

)∣∣∣∣10

=1

3

(−(

1

13

)(1

6

)+

1

42+−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− xx = 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

265650

Make sure to check your arithmetic.

Exercise 17.∫ 2

1x−2 sin 1

xdx =(cos 1

x

)∣∣21

= cos 12 − cos 1

81

Exercise 18.∫

sin (x− 1)1/4dx

u = (x− 1)1/4

du =1

4(x− 1)−3/4dx =

1

4

1

u3dx

=⇒∫

sin (x− 1)1/4dx =

∫(sinu)4u3du = 4

∫u3 sinudu

∫u3 sinudu = −u3 cosu+ 3u2 sinu+ 6u cosu+ 06 sinu

= −(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 sinx2 cosx2dx = (1/4) sin2 x2 + C

Exercise 20.∫ √

1 + 3 cos2 x sin 2xdx =∫ √

1 + 3 cos2 x2 sinx cosxdx =∫u1/2 du

−3 = 2u3/2

−9 =−2

9(1 + 3 cos2 x)3/2 ,

where we had used this substitution:

u = 1 + 3 cos2 x

du = −6 cosx sinxdx

du

−3= 2 cosx sinxdx

Exercise 21.∫ 2

0375x5(x2 + 1)−4dx

u = x2 + 1

du = 2xdx

(u− 1)2 = x4

=⇒

∫ 2

0

375

2du(u− 1)2u−4 =

375

2

∫ 5

1

du(u2 − 2u+ 1)

u4=

375

2

∫ 5

1

du

(1

u2− 2

u3+

1

u4

)=

=375

2

(−1

u+

1

u2+

1

−3u3

)∣∣∣∣51

= 64 = 26

Exercise 22.∫ 1

0(ax+ b)(x2 + 3x+ 2)−2dx = 3

2

Since(

−1x2+3x+2

)∣∣∣10

= −16 + 1

2 = 23 ,

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+

1

5x5

)∣∣∣∣10

=8

15

I3 =6

7

8

15

16

35

I4 =8

9I3 =

128

315

I5 =10

11I4 =

256

693

Exercise 24. F (m,n) =∫ x

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

82

F (m,n) =tm+1

m+ 1(1 + t)n

∣∣∣∣x0

−∫ 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

)∣∣∣∣x0

=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 tanx =

∫ 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/40

=1

n+ 1

(3)

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

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

1

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

=⇒ 1

n+ 1< 2f(n) <

1

n− 1

Exercise 26. f(0), f(π) = 2∫ π

0(f(x) + f ′′(x)) sinxdx = 5∫ π

0

f ′′ sinx = f ′ sinx−∫f ′ cosx = −

∫f ′ cosx = −f cosx−

∫f sinx∫ π

0

(f + f ′′) sinxdx =

∫f sinx+−(f(π)(−1)− f(0))−

∫f sinx = 2 + f(0) = 5 =⇒ f(0) = 3

Exercise 27. ∫ π/2

0

sinx cosx

x+ 1dx =

∫ π/2

0

sin 2x

2x+ 2dx =

1

2

∫ π

0

sinx

x+ 2dx =

1

2

(− cosx

x+ 2−∫

cosx

(x+ 2)2

)=

=1

2

(1

π + 2+

1

2−A

)=

4 + π

4(π + 2)− A

2

Exercise 28.∫dx

x√a+ bx

=2√a+ bx

bx+

2

b

∫ √a+ bx

x2∫ √a+ bx

x=

23b (a+ bx)3/2

x+

∫ 23b (a+ bx)3/2

x2=

2

3b

√a+ bx

(ax

+ b)

+2

3b

∫ √a+ bx

(a

x2+b

x

)=⇒

∫ √a+ bx

x= a

∫dx

x√a+ bx

+ 2√a+ bx

83

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

=2

bxm(a+ bx)1/2 − 2m

b

∫xm−1(a+ bx)√

a+ bx=

2

bxm(a+ bx)1/2 − 2m

∫xm√a+ bx

− 2ma

b

∫xm−1

√a+ bx∫

xm√a+ bx

dx =1

2m+ 1

2

bxm(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∫

1

xn√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− 1

m− 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

84

Generally, we can say

aj+1 =3

j + 1aj if j ≥ 3

We also havea1(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 find 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 finite terms, then k must be zero. Thus, there’s at most only one polynomial solution P .

Exercise 35. Bernoulli Polynomials.

(1) P1(x) = 1; P ′n(x) = nPn−1(x);∫ 1

0Pn(x)dx = 0, if n ≥ 1

n = 1 (1)(1) = P ′1

∫ 1

0

(x+ c) = (1

2x2 + Cx)

∣∣∣∣10

=1

2+ C = 0 C = −1/2

P1 = x− 1/2

n = 2 2(x− 1/2) = P ′2

∫ 1

0

(x2 − x+ C) =

(1

3x3 − 1

2x2 + Cx

)∣∣∣∣10

=−1

6+ C = 0 C = 1/6

P2 = x2 − x+ 1/6

n = 3 3(x2 − x+1

6) = P ′3

∫ 1

0

(x3 − 3x2

2+x

2+ C) =

1

4− 13

2+

1

4+ C = 0

P3 = x3 − 3x2

2+x

2

n = 4 4(x3 − 3

2x2 +

x

2) = P ′4

∫ 1

0

(x4 − 2x3 + x2 + C) =1

5− 14

2+

1

313 + C = 0 C =

−1

30

P4 = x4 − 2x3 + x2 +−1

30

n = 5 5(x4 − 2x3 + x2 − 1

30) = P ′5

∫ 1

0

(x5 − 5x4

2+

5x3

3− x

6+ C) =

1

6− (1)5

2+

5(1)4

12+−(1)2

12+ C = 0; C = 0

P5 = x5 − 5x4

2+

5x3

3− x

6

(2) The first, second, and up to fifth 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

85

(3) The first, second, and up to fifth 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 j

n =∫ x

0Pn(t)dt = Pn+1(x)−Pn+1(0)

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

Assume the nth case is true.

u = 1− tdu = −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 definitely true.If n+ 1 was odd,

Doing some algebra for the first five cases, we can show that P2j−1(0) = 0 for j = 2, 3. Assume the jth case is86

true. Since P2j−1 is a polynomial and P2j−1(0) = 0, then the form of P2j−1 must be P2j−1 =∑2j−1k=1 akx

j . UsingP ′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

∣∣∣∣x0

=

= 2j

2j−1∑k=1

akk + 1

xk+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 find 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− 1

2) = (−1)2j−1(P2j−1(

1

2))

=⇒ P2j−1(1

2) = 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 definition of the natural logarithm as an integral, The definitionof 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 = 087

(2) log (1 + x) = 1 + log (1− x)

ln (1 + x) = 1 + ln (1− x) = ln (e) + ln (1− x) ln e(1− x)

=⇒ 1 + x = e− ex =⇒ x =e− 1

1 + e

(3) 2 log x = x log 2

lnx2 + ln 2−x = 0 = ln 1 = lnx22−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 − 1

2e

)2

Exercise 3.

f =lnx

x

f ′ =1− lnx

x2

f ′′ =−2

x3−( 1xx

2 − 2x lnx

x4

)=

2 lnx− 3

x3

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 ′ =1

2

−2x

4− x2=−x

4− x2

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

f ′ =1

lnx

(1

x

)=

1

x lnx

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

f ′ = (2 log x+ log log x)′

=2

x+

1

x log x

Exercise 9. f(x) = 14 log x2−1

x2+1

f ′ =1

4

(log x2 − 1− log x2 + 1

)′=

1

4

(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)nn√

1 + x2

Exercise 11. f(x) =√x+ 1− log (1 +

√x+ 1)

88

f ′ =1

2√x+ 1

− 1

1 +√x+ 1

(1

2√x+ 1

)=

1

2(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) = 12√ab

log√a+x√b√

a−x√b

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 (lnx)− cos (lnx) + (cos (lnx) + sin (lnx)) = 2 sin (lnx)

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 lnx

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

Exercise 18.∫x log xdx (

x2 lnx

2

)′= x lnx+

x

2∫x lnx =

x2 lnx

2− x2

4

Exercise 19.∫x log2 xdx (

x2 ln2 x

2

)′= x ln2 x+ x2 lnx

(1

x

)=⇒

∫x log2 x =

x2 ln2 x

2− x2 lnx

2− x2

4

Exercise 20.∫ e2−1

0dt

1+t ∫ e3−1

0

dt

1 + t= ln (1 + t)|e

3−10 = 3

Exercise 21.∫

cotxdx ∫cosx

sinxdx = ln | sinx|

Exercise 22.∫xn log (ax)dx Solve the problem directly.

89

∫xn log ax =

∫xn log a+

∫xn log x =

xn+1

n+ 1log a+

∫xn log x∫

xn lnx =xn+1

n+ 1lnx−

∫xn+1

n+ 1

1

x=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 =

1

3x3 ln2 x−

∫x2

32 lnx =

x3 ln2 x

3− 2

3

(x3 lnx

3− x3

9

)=

x3 ln2 x

3− 2x3 lnx

9+

2x3

27

Exercise 24.∫

dxx log x ∫

dx

x lnx= ln (lnx) + C

Exercise 25.∫ 1−e−2

1log (1−t)

1−t dt

∫ 1−e2

0

ln (1− t)1− t

dt = −1

2(ln (1− t))2

∣∣∣∣1−e−2

0

= −2

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 − 4

3(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 lnx

)=x4 ln3 x

4− 3x4 ln2 x

16+

3x4 lnx

32− 3x4

128

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

(1)

f ′ = 1− 1

x

g′ =1

x− 1

x2=

1

xf ′

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

0 < lnx− 1 +1

xsince 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.

90

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− 1

x< lnx < x− 1 =⇒ 1− 1

x+ 1< lnx+ 1 < x

1− x

1 + x<

lnx+ 1

x< 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)

since

f(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) =1

qf(a)

Exercise 31. lnx =∫ |x|

11t dt

(1) lnx =∫ |x|

11t dt From this definition, 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

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

x and less than rectangular stripscovering over 1

x

(3) ak = 1 + k =⇒ ak−ak−1

ak−1= 1

k

n∑k=1

1

1 + k< ln (n+ 1) <

n∑k=1

1

k=⇒

n+1∑k=2

1

k< ln (n+ 1) <

n∑k=1

1

k

=⇒n∑k=2

1

k< ln (n) <

n−1∑k=1

1

k

Exercise 32.

(1) L(x) = 1ln b lnx = 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 =

1

log10 e=

1

2.302585' 0.43429

Exercise 34. Given∫ xyxf(t)dt = B(y) f(2) = 2, , We Want A(x) =

∫ x1f(t)dt

91

∫ 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 lnx

Exercise 35. Given∫ xy

1f(t)dt = y

∫ x1f(t)dt+ x

∫ y1f(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

∫−→ f(y) = k ln y + C

y=1−−→ f(1) = 0 + C = 3∫ xy1

(k ln t+ 3) = y∫ x

1(k ln t+ 3) + x

∫ y1

(k ln t+ 3)

k ((xy) lnxy − (xy) + 1) + 3(xy − 1) = y (k ((x) lnx− 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 lnx+ 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)

wherex2m+1

2m+ 1< Rm(x) ≤ 2− x

1− xx2m+1

2m+ 1Rm(x) = E2m(x)− E2m(−x)

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

m = 5, x = 13

ln

(4/3

2/3

)= ln 2 ' 2

(1

3+

(13

)33

+

(13

)55

+

(13

)77

+

(13

)99

)' 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

92

x =1

5m = 5

ln

(1 + x

1− x

)= 2

(1

5+

(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

9

1− 19

)= ln

(108

)= ln 5

4 = ln 5− 2 ln 2

For m = 2

2

(1

9+

(19

)53

+

(19

)55

)= 0.2231435

(19

)77' 0.00000003 < R3(x) ≤

17989

(19

)77' 0.000000063

1.609437 < log 5 < 1.609438

Exercise 4. x = 16 ln

(1+ 1

6

1− 16

)= ln

(75

)= ln 7− ln 5.

For m = 3,

ln1 + x

1− x' 2

(1

6+

(16

)33

+

(16

)55

)= 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 = 1.098614

ln 4 = 1.386293

ln 5 = 1.609436 ln 6 = ln 2 + ln 3 = 1.791700

ln 7 = 1.945909

ln 8 = 3 ln 2 = 2.0794404

ln 9 = 2 ln 3 = 2.197228

ln 10 = ln 5 + ln 2 = 1.302577

6.17 Exercises - The exponential function, Exponentials expressed as powers of e, The definition of ex for arbitraryreal x, The definition 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

93

Exercise 8. cosxesin x

Exercise 9. −2 cosx sinxecos2 x

Exercise 10. 1xe

log 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−x

2

.

x2e−x2

= −2x3e−x2

+ 2xe−x2

e−x2

= −2xe−x2

1

−2(x2e−x

2

+ e−x2

)′ =1

−2

(2xe−x

2

+−2x3e−x2

+−2xe−x2)

=x2e−x

2

+ e−x2

−2

Exercise 19. ex = b+∫ baetdt = 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 lnx;f ′

f= lnx+ 1; f ′ = xx(lnx+ 1)

Exercise 22.

lnf ′

f=

1

1 + x+

1

1 + ex2 (2xex2

), f ′ = 1 + ex2

+ 2(1 + x)xex2

Exercise 23. f = ex−e−xex+e−x .

94

ln f = ln (ex − e−x)− ln (ex + e−x)

f ′

f=

1

ex − e−x(ex + e−x)− 1

ex + 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 lnx

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

=⇒ f ′ = xaa−1aa + ax

a+1xa−1 ln a+ (ln a)2ax+ax

Exercise 25.

ef = ln (lnx);

f ′ef =1

lnx

(1

x

)f ′ =

1

ln (lnx)

(1

lnx

)(1

x

)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 lnx

f ′

f= (xx)′ lnx+

xx

x= (xx(lnx+ 1)) lnx+ xx−1

f ′ = xx+xx(lnx+ 1) lnx+ xxxx+x−1

Exercise 28. ln f = x ln (lnx)

f ′

f= ln (lnx) +

1

lnx

f ′ = (lnx)x(ln (lnx) +1

lnx)

Exercise 29. ln f = (lnx) lnx

f ′

f=

2 lnx

x

f ′ = 2xlog x−1 lnx

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

f ′

f= ln (lnx) +

1

lnx− 2

lnx

x

f ′ =(lnx)x

xln x

(ln lnx+

1

lnx− 2 lnx

x

)Exercise 31.

95

ln f1 = cosx ln sinx

f ′1f1

= − sinx ln sinx+cos2 x

sinx

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

sinx

ln f2 = sinx ln cosx

f ′2f2

= cosx ln cosx+− sin2 x

cosx

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

cosx

=⇒ f = sinx cosx(−(ln sinx)2 + (ln cosx)2) +cos3 x ln sinx

sinx+− sin3 x ln cosx

cosx

Exercise 32. ln f = 1x lnx

f ′

f = −1x2 lnx+ 1

x2 =⇒ f ′ = −x1/x−2 lnx+ x1/x−2

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

3 ln (3 + x)

f ′

f=

2

x+

−1

3(3− x)− −1

1− x− 2

3

1

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

3

x2(3− x)1/3

(1− x)(3 + x)5/3

=x(18− 12x+ 4

3x2 + 2

3x3)

(1− x)2(3 + x)5/3(3− x)2/3

Exercise 34. ln f =∑ni=1 bi ln (x− ai)

f ′

f=

n∑i=1

bix− ai

=⇒ f ′ =

n∑j=1

bjx− 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)= r

1

x=⇒ f ′(x) = rxr−1

Exercise 36. Use the definition ax = ex log a

(1) log ax = x log a

Taking the exponential is a well-defined inverse function to log so taking the log of both sides of the definition, 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

96

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

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) =1

2(ax+y + a−(x+y))

f(x+ y) + f(x− y) =1

2

(ax+y + a−(x+y) + ax−y + a−(x−y)

)f(x)f(y) =

1

4(ax + a−x)(ay + a−y) =

1

4

(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 − 1

x= c

(limx→0

ecx − 1

cx

)= c

(limcx→0

ecx − 1

cx

)= c

df(cx)

d(cx)(0) =

d

dx(ecx)(0)

limx→0

ecx − 1

x= f ′(0) = c

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 defined everywhere on the real axis. Suppose also that f satisfies the functional equation

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

(1)

f(0) = f(0)f(0) = f2(0)If f(0) = 0, then we’re done

If f(0) 6= 0 then f(0) = 1 (by dividing both sides by f(0) )(2) Take the derivative with respect to x on both sides of the functional equation.

df(x+ y)

d(x+ y)

d(x+ y)

dx=df(x)

dxf(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.

97

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

f(0) = e0 − 1− 0 = 0

{ex > 1 + x for x > 0

e−x > 1− x for x < 0

(2) ∫ x

0

et = ex − 1 > x+1

2x2 =⇒ ex > 1 + x+

1

2x2

− e−x > −1 + x− x2

2=⇒ e−x < 1− x+

x2

2(3) ∫ x

0

et = ex − 1 > x+1

2x2 +

1

3 ∗ 2x3 =⇒ ex > 1 + x+

1

2x2 +

1

3 ∗ 2x3

− 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!

<∑2mj=0

xj

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)!

<∑2mj=0

xj+1

(j+1)!

= e−x

{<∑2m+2j=0

xj

j!

>∑2m+1j=0

xj

j!

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

x

n

)n=

n∑j=0

(n

j

)1n−j

(xn

)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

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

)ne−x > (1− x/n)n =⇒ ex < (1− x/n)−n

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

∂xf = xyy/x

∂yf = xy lnx

6.19 Exercises - The hyperbolic functions.Exercise 7.

2 sinhx coshx = 2ex − e−x

2

ex + e−x

2=

1

2(e2x − e−x) = sinh 2x

Exercise 8.

cosh2 x+ sinh2 x =

(ex + e−x

2

)2

+

(ex − e−x

2

)2

=1

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

Exercise 9.

coshx+ sinhx =ex + e−x

2+ex − e−x

2= ex

Exercise 10.

coshx− sinhx =ex + e−x

2−(ex − e−x

2

)= e−x

Exercise 11. Use induction.98

(coshx+ sinhx)2 = cosh2 x+ 2 sinhx coshx+ sinh2 x = cosh 2x+ sinh 2x

(coshx+ sinhx)n+1 = (coshx+ sinhx)(coshnx+ sinhnx) =

= coshnx coshx+ coshnx sinhx+ sinhnx coshx+ sinhx sinhnx =

=

enx + e−nx

2

ex + e−x

2+enx + e−nx

2

ex − e−x

2+

+enx − e−nx

2

ex + e−x

2+enx − e−nx

2

ex − 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.

(cosx)′ = − sinx = −√

1− cos2 xD arccosx =1

−√

1− x2− 1 < x < 1

Exercise 2.

(tanx)′ = 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 = − 1

1 + x2

Exercise 4.

(sec y)′ = tan y sec y =√

sec2 y − 1 sec y; | sec y| > 1 ∀y ∈ RIf 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

Exercise 5.

(csc y)′ = − cotx cscx = − csc y(√

csc2 y − 1)

Let y such that−π2

< y <π

2(csc y) < 0

Darccscx =1

−|x|√x2 − 1

Exercise 6.

(xarccotx)′ = arccotx− x

1 + x2(1

2ln (1 + x2)

)′=

1x

(1 + x2)∫arccotx = xarccotx+

1

2ln (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

99

Take a note of this exercise. When dealing with (∓x2 ± 1)2j+1

2 ; j ∈ Z; try x±√x2 ± 1 combinations. It’ll work out.

Exercise 8.

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

1− x2(√1− x2 arcsinx

)′=

−x√1− x2

arcsinx+ 1∫(arcsinx)2 = x(arcsinx)2 + 2

√1− x2 arcsinx− 2x

Exercise 10. (− arcsinx

x

)′=

1

x2arcsinx− 1

x√

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.

(1)

D

(arccotx− arctan

1

x

)=−1

x2 + 1− 1

1 +(

1x

)2 (−1

x2

)= 0

(2) arccotx− arctan 1x = C

Now arccotx = π2 − arctanx.

π

2− arctanx− arctan

1

x= C =⇒ π

2− C = arctanx+ arctan

1

x

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.

100

Exercise 12.

f ′ =1√

1−(x2

)2(

1

2

)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(−1

x2

)=

1√x2 − 1|x|

Exercise 15.

f(x) = arcsin (sinx) =1√

1− sin2 xcosx =

cosx

| cosx|

Exercise 16.1

2 sqrtx− 1

x+ 1

1

2√x

=

√x

2(x+ 1)

Exercise 17.1

1 + 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 x

1

1 + tan4 x

(2 tanx sec2 x

)=

2 tanx 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 sinx cosx)(cosx+ sinx) =

cosx+ sinx√2 sinx 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)=

1

1 + x2

Exercise 24. f = (arccos (x2))−2

f ′ = −2(arccosx2)−3

(−1√

1− x2

)(2x) =

4x(arccosx2)−3

√1− x4

Exercise 25.101

f ′ =

(1

arccos 1√x

) −1√1− 1

x

( −1

2x3/2

)=

1

2 arccos 1√x

√x3 − x2

Exercise 26. dydx = x+yx−y .

(arctan

y

x

)′=

1

1 +(yx

)2 (y′x− yx2

)=

(1

2ln (x2 + y2)

)′=

1

2

1

(x2 + y2)(2x+ 2yy′)

=⇒ y′ =x+ y

x− y

Exercise 27.

ln y = ln (arcsinx)− 1

2ln (1− x2)

y′

y=

1

arcsinx

(1√

1− x2

)− 1

2

1

1− x2(−2x) =

1

arcsinx√

1− x2+

x

1− x2

y =arcsinx√

1− x2

y′ =

(1

1− x2

)+

(arcsinx)x

(1− x2)3/2=

√1− x2 + x(arcsinx)

(1− x2)3/2

ln y′ = ln (√

1− x2 + x arcsinx)− 3

2ln (1− x2)

y′′

y′=

1√1− x2 + x arcsinx

(−x√

1− x2+ arcsinx+

x√1− x2

)− 3

2

(−2x)

1− x2

y′′ = y′arcsinx√

1− x2 + x arcsinx+

y′3x

1− x2=

arcsinx

(1− x2)3/2+

(√

1− x2 + x arcsinx)(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 =⇒ arcsinx

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) =1

aarctan

x

a

Exercise 32.

dx

a(1 +(√

ba√a

)2

)=

1√ba

arctan

√bx√a

Exercise 33. ∫dx(

x− 12

)2+ 7

4

=4

7

∫dx(

2√7

(x− 1

2

))2

+ 1=

2√7

arctan2(x− 1

2

)√

7

102

Exercise 34. (x2 arctanx

2

)′= x arctanx+

x2

2

1

1 + x2= x arctanx+

1

2

(1− 1

1 + x2

)(

1

2(x− arctanx)

)′=

1

2

(1− 1

1 + x2

)∫x arctanx = 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)′ =3

2(−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(arctanx)2 + x2

(1

1 + x2

)arctanx = x(arctanx)2 +

(1−

(1

1 + x2

))arctanx(

(arctanx)2

2

)′=

arctanx

1 + x2

(x arctanx)′ = arctanx+x

1 + x2∫x(arctanx)2dx =

x2(arctanx)2

2−(x arctanx− ln (1 + x2)

2

)+

(arctanx)2

2

Exercise 37.

(arctan√x)′ =

(1

1 + x

)(1

2√x

)(x arctan

√x)′ = arctan

√x+

√x

2(1 + x)= arctan

√x+

1

2

(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 = sinu

∫ √1− x2dx =

∫cos2 udu =

u

2+

sin 2u

4=

arcsinx

2+x√

1− x2

4

Exercise 40.103

∫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

1

2

(xearctan x

√1 + x2

− earctan x

√1 + x2

)′=

xearctan x

(1 + x2)3/2

Exercise 41. From the previous exercise,

1

2

(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 =

−x2(1 + x2)

+1

2arctanx

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− 1

2ln (1 + e2x))′ = e−xarccotex

Exercise 45.

∫ √a+ x

a− xdx =

∫a+ x√a2 − x2

dx =

∫ 1a√1−

(xa

)2 +x√

a2 − x2

dx = a arcsinx

a+−

√a2 − x2

Exercise 46.104

∫ √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+b

2

)(a−b

2

) )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,(arcsinx

2+

1

2x√

1− x2

)′=

1

2

1√1− x2

+

√1− x2

2+

1

4

x(−2x)√1− x2

=√

1− x2

Exercise 47. Wow!∫

dx√(x− a)(b− x)

x− a = (b− a) sin2 u

dx = (b− a)(2) sinu cosudu

b− x = (a− b) sin2 u+ b− a = (b− a)(cos2 u)∫dx√

(x− a)(b− x)=

∫(b− a)(2) sinu cosudu√b− a cosu

√b− a sinu

= 2u = 2 arcsin

√x− ab− a

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

)+(

1x+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

∣∣∣∣∣∣−1/2

2−3/2

A = −1/2, B = 2, C = −3/2

=⇒ −1

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

−3

2ln (x+ 3)

Exercise 3.∫

x(x−2)(x−1) =

∫2

x−2 + −1x−1 = 2 lnx− 2− ln (x− 1)

Exercise 4.∫

x4+2x−6x3+x2−2xdx

105

x4 + 2x− 6

x3 + x2 − 2x= x− 1 +

3(x2 − 2)

x3 + x2 − 2x(do long division)∫

x− 1 +3(x2 − 2)

x(x+ 2)(x− 1)=

1

2x2 − x+ 3

∫x2 − 2

x(x+ 2)(x− 1)∫x2 − 2

x(x+ 2)(x− 1)=

∫1

x+

1/3

x+ 2+−1/3

x− 1= lnx+

1

3lnx+ 2− 1

3lnx− 1

=⇒ 1

2x2 − x+ 3 lnx+ lnx+ 2− lnx− 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 0

1 0 01 00 1

|

12−601

A = 12, B = −6, C = 0, D = 1∫

12

(2x+ 1)3+

−6

(2x+ 1)2+

1

x+ 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

=⇒∫

2

x− 1+

2x+ 1

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

Exercise 7.∫

x4dxx4+5x2+4

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 = 5

4B +D = 4

B =−1

3

D =16

3

=⇒∫

1− 5x2 + 4

(x2 + 1)(x2 + 4)= x−

∫−1/3

x2 + 1+

16/3

x2 + 4= x+

1

3arctanx+ 4/3 arctanx/2 + C

Exercise 8.∫

x+2x(x+1)dx =

∫1

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

∫1x −

1x+1 = − ln |x+ 1|+ 2 lnx

Exercise 9.∫

dxx(x2+1)2 =

∫Ax + Bx+C

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

106

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∫1

x+−x

x2 + 1+

−x(x2 + 1)2

= lnx+− 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∫ (1

x+ 1+

−1

(x+ 1)2

)dx = lnx+ 1 +

1

x+ 1+ C

Exercise 12.∫

dxx(x2−1) =

∫dx

x(x−1)(x+1) =∫Ax + B

x−1Cx+1

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

2= C∫

−1

x+

1/2

x− 1+

1/2

x+ 1= − lnx+

1

2ln |x− 1|+ 1

2ln |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 =−6

5; B =

6

5

−x = (A+B)x− 2A+ 3B

2A = 3B

A =3B

2=⇒

B = −2/5

A− 3/5

=⇒∫

1 +−6/5

x+ 3+

6/5

x− 2+−3/5

x+ 3+−2/5

x− 2=−9

5ln |x+ 3|+ 4

5ln |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.

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)dxx3+3x2+2x =

∫ (x−3)dxx(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)=−1

x+ 2+

1

x+ 1

1

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

x+

B

x+ 2+

C

x+ 1107

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 1

3 1 22 0 0

ABC

=

001

1 1 1

3 1 22

∣∣∣∣∣∣001

=

0 1 00 0 11 0 0

∣∣∣∣∣∣1/2−11/2

=⇒ 1

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

1/2

x+

1/2

x+ 2+−1

x+ 1

=⇒ − ln |x+ 2|+ ln |x+ 1|+−3/2 lnx+−3/2 ln |x+ 2|+ 3 ln |x+ 1| = −5/2 ln |x+ 2|+ 4 ln |x+ 1| − 3/2 lnx

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

−1

x− 1+

1/4

x+ 1+

1/4

(x− 1)2+

1/4

(x+ 1)2∫dx

(x2 − 1)2=

1

4ln

∣∣∣∣x+ 1

x− 1

∣∣∣∣+−1

2

(x

x2 − 1

)Exercise 18. Use the method of partial fractions, where we find that∫

(x+ 1)

x3 − 1dx =

∫x+ 1

(x− 1)(x2 + x+ 1)dx =

∫ − 23x−

13

x2 + x+ 1+

23

x− 1=

= −1

3ln |x2 + x+ 1|+ 2

3ln |x− 1|

where we had used the following partial fraction decomposition for the given integrandAx+B

x2 + x+ 1+

C

x− 1=

x+ 1

x3 − 1

Ax2 +Bx−Ax−B + Cx2 + Cx+ C = x+ 1

2Ax+B −A+ 2Cx+ C = 1 (where we used the trick to take the derivative of the above equation)=⇒ A = −C B −A−A = 1

−B + C = 1; A =2

−3, C =

2

3B = −1

2

Exercise 19.∫

x4+1x(x2+1)2

Again, it helps to be clever here.∫x4 + 1

x(x2 + 1)2=

∫x4 + 2x2 + 1− 2x2

x(x2 + 1)2=

∫(x2 + 1)2

x(x2 + 1)2+

−2x

(x2 + 1)2=

= lnx+ (x2 + 1)−1 + C

Exercise 20.∫

dxx3(x−2)

108

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

1

x3(x− 2)=

(−1/2

x3

)+−1/4

x2+−1/8

x+

1/8

x− 2

So then ∫dx

x3(x− 2)=

1

4x2+

1

4x+−1

8lnx+

1

8ln |x− 2|+ C

Exercise 21. ∫1− x3

x(x2 + 1)= −

∫x3 − 1

x(x2 + 1)= −

∫x2

x2 + 1+

∫1

x(x2 + 1)=

= −∫ (

1− 1

x2 + 1

)+

∫1

x+−x

x2 + 1=

= −x+ arctanx+ lnx− ln |x2 + 1|+ C

Exercise 22. ∫dx

x4 − 1=

∫ (1

x2 + 1

)(1

x2 − 1

)=

∫1/2

x2 − 1− 1/2

x2 + 1=

=

∫1

2

(1/2

x− 1− 1/2

x+ 1

)− 1/2

x2 + 1=

=1

4ln (x− 1)− 1

4ln (x+ 1)− 1

2arctanx+ C

Exercise 23.∫

dx

x4 + 1

I had to rely on complex numbers.Notice that with complex numbers, you can split up polynomial power sums

x4 + 1 = (x2 + i)(x2 − i) = (x+ ie(iπ

4))(x− ie(iπ

4))(x+ e(i

π

4))(x− e(iπ

4)) =

= (x+ e(i3π

4))(x− e(i3π

4))(x+ e(i

π

4))(x− e(iπ

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(i3π4

)) +B(x2 − i)(x+ e(i3π

4)) + C(x2 + i)(x− e(iπ

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(iπ

4) +De(i

π

4) = 0

x1 : −iA− iB + iC + iD = 0

x0 : −e(iπ4

)A+Be(iπ

4) + C(−e(i3π

4)) +D(e(i

3π

4)) = 1

=⇒

1 1 1 1

−e(i 3π4 ) e(i 3π

4 ) −e(iπ4 ) e(iπ4 )−i −i i i−e(iπ4 ) e(iπ4 ) −e(i 3π

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.

109

1 1 1 1↘ ↖ ↙ ↗↓ ↓ ↑ ↑↙ ↗ ↘ ↖

∣∣∣∣∣∣∣∣0001

=

1 1 0 0

2↖ 2↗0 0 1 10 0 2↘ 2↖

∣∣∣∣∣∣∣∣0001

=

1 1 0 0

2↖ 2↗0 0 1 1

0 4↖

∣∣∣∣∣∣∣∣ 01

=

1↖

0 0 1 00 1

∣∣∣∣∣∣∣∣14 ↖14 ↑

14 ↗14 ↙

=

=

1

11

1

∣∣∣∣∣∣∣∣14 ↖14 ↘14 ↗14 ↙

=⇒

A =1

4e(i3π

4)

B = −1

4e(i3π

4)

C =1

4e(i

π

4)

D =−1

4e(i

π

4)

=⇒∫

1/4e( i3π4 )

x+ e(i 3π4 )

+−1/4e(i 3π

4 )

x− e(i 3π4 )

+1/4e(iπ4 )

x+ e(iπ4 )+−1/4e(iπ4 )

x− e(iπ4 )

(1

4

)(e(i

3π

4) ln (x+ e(i

3π

4))− e(i3π

4) ln (x− e(i3π

4)) + e(i

π

4) ln (x+ e(i

π

4))− e(iπ

4) ln (x− e(iπ

4))

)After doing some complex algebra,

=⇒ 1

4√

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?

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) +

1

x2 + 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)2dx = −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+− cosx+ 5=

∫dx

4SC +−C2 + S2 + 5

( 1C2

1C2

)=

∫sec2 x

2dx

4T − 1 + T 2 + 5(1 + T 2)=

=

∫sec2 x

2dx

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

)

=3

5

∫du

9(u+ 13 )2

5 + 1=

1√5

arctan

(3(tan x

2 + 13 )

√5

)110

Exercise 27.∫

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

u = tanx

2

du =sec2 x

2

2dx

,

1

a

∫dx

1a + cosx

=1

a

∫dx

1a + C2 − S2

=1

a

∫sec2 x

2dx1a sec2 x

2 + 1− T 2=

=1

a

∫sec2 x

2dx1a + 1 + T 2( 1

a − 1)=

1

a

∫2du

1a + 1 + u2

(1a − 1

) =2

1 + a

∫du

1 +(u√

1−a1+a

)2 =

2

1 + a

arctan√

1−a1+au√

1−a1+a

==⇒ 2√1− a2

arctan

(√1− a1 + a

tanx

2

)

Exercise 28.∫

dx1+a cos x Half-angle substitution.

∫dx

1 + a cosx=

∫dx

1 + a(C2 − S2)=

∫sec2 x

2dx

sec2 x2 + a(1− T 2)

=

u = tan θ/2 = T

du = sec2 θ/2

(1

2

)dθ

=⇒ =

∫2du

1 + T 2 + a(1− T 2)= 2

∫du

(1− a)T 2 + (1 + a)=

2

1− 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− 1

a+ 1

(1

1− a

)(ln (u−

√a+ 1

a− 1)− ln (u+

√a+ 1

a− 1)

)=

=−1√a2 − 1

ln

tan x2 −

√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=

2

3

∫dx

1− cos 2x3

=2

3

∫dx

1−(c2−s2

3

) =2

3

∫sec2 xdx

sec2 x−(

1−T 2

3

) =

u = tanx

du = sec2 xdx

=⇒ =2

3

∫du

1 + u2 −(

1−u2

3

) =2

3

∫du

23 + 4

3u2

=

∫du

1 + (√

2u)2=

=1√2

arctan√

2 tanx

=⇒∫

sin2 x

1 + sin2 xdx = x− 1√

2arctan (

√2 tanx)

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.111

1

a2s2 + b2c2=

1

a2(1− c2) + b2c2=

1

a2 + (b2 − a2)c2=

1

a2(1 + (kc)2)=

sec2

a2(sec2 +k2)=

sec2

a2(1 + T 2 + k2)=

u = tanx

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=

1

abarctan

(a tanx

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 sinx+ b cosx)2=

1

a2

∫dx

(sinx+ k cosx)2

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

=⇒ 1

a2

∫1

(s+ bac)

2=

−1

(a2 tanx+ 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 θ2 substitution.

∫sinx

1 + cosx+ sinx=

∫2CS

1 + 2CS + C2 − S2=

∫CS

C(S + C)=

∫T

(T + 1)

whereC = cosx/2

S = sinx/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 + 1

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

A = −B C +B = 1 A+ C = 0 =⇒ C =1

2; B =

1

2; A = −1

2

2

∫ (−1/2

u+ 1+

12 (u+ 1)

u2 + 1

)du =

∫−1

u+ 1+

u+ 1

u2 + 1

= − ln |u+ 1|+ 1

2ln |u2 + 1|+ arctanu =

= − ln | tanx/2 + 1|+ 1

2ln | sec2 x/2|+ x

2

=⇒ − (ln |2|) +1

2ln |2|+ π

4= −1

2ln |2|+ π

4

Exercise 33.∫ √

3− x2dx112

∫(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 +

3

2arcsin

x√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

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+

√3

x2− 1

∣∣∣∣∣+√

3

√1− x2

3

Exercise 36.∫ √

1 + 1xdx(

x

√1 +

1

x

)′=

√1 +

1

x+

x

2√

1 + 1x

(−1

x

)=

√1 +

1

x+−1/2√x2 + x(

ln

(x+

1

2+√x2 + x

))′=

1

x+ 12 +√x2 + x

(1 +

x+ 12√

x2 + x

)=

1√x2 + x

=⇒∫ (

1 +1

x

)dx = x

√1 +

1

x+

1

2ln

(x+

1

2+√x2 + x

)Exercise 37.

(x√x2 + 6)′ =

√x2 + 5 +

x2

√x2 + 5

(ln (x+√x2 + b))′ =

(1

x+√x2 + b

)(1 +

x√x2 + b

)=

1√x2 + b∫ √

x2 + 5 =1

2

(x√x2 + 5 + 5 ln (x+

√x2 + 5)

)Exercise 38.

113

(ln (x+

1

2+√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+

1

2+√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+ 12

)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+ 12

)2 − 14

= ln

(2

(x+

1

2

)+√

(2(x+ 1/2))2 − 1

)+ C

Exercise 40.

6.26 Miscellaneous review exercises. Exercise 1.

f(x) =

∫ x

1

log t

t+ 1

f

(1

x

)=

∫ 1x

1

log t

t+ 1dt =

∫ x

1

− ln (u)1u + 1

(−1

u2du

)=

=

∫ x

1

ln (u)

u+ u2du

u =1

t

du = − 1

t2dt

f(x) + f

(1

x

)=

∫ x

1

t ln t+ ln t

t(t+ 1)dt =

∫ x

1

ln t

tdt =

(ln t)2

2

∣∣∣∣x1

=(lnx)2

2

f(2) + f

(1

2

)=

1

2(ln 2)2

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

2ff ′ = f(x)sinx

2 + cosx; =⇒ 2f ′ =

sinx

2 + cosx

At this point, it could be very easy to evaluate the integral by guessing at the solution.

(− ln 2 + cosx)′ =sinx

2 + cosx=⇒ f = − ln |2 + cosx|

2+ C

114

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

u = tanx

2

2du = sec2 x

2dx

C = cosx/2, S = sinx/2

∫sinx

2 + cosxdx =

∫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

(1

2lnT 2 + 1− 1

2lnT 2 + 3

)=

(ln

(T 2 + 1

T 2 + 3

))= ln

(2

4 + 2 cosx

)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|3/20 = 1 .

Exercise 5.

(1)

f =√

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

ln f =1

2ln

(4(x+ 2)

x(x+ 1)(x+ 2)

)=

1

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

f ′

f=

1

2

((4

4x+ 2

)− 1

x− 1

x+ 1− 1

x+ 2

)=⇒ f ′ =

1

2

√4x+ 2

x(x+ 1)(x+ 2)

((4

4x+ 2

)− 1

x− 1

x+ 1− 1

x+ 2

) f ′(1) = − 7

12

(2) ∫ 4

1

π4x+ 2

x(x+ 2)(x+ 1)dx = 2π

∫ 4

1

(2x+ 1)

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

= 2π

(1

2lnx+−3

2ln |x+ 2|+ ln |x+ 1|

)∣∣∣∣41

= πln25

8

since we can find the antiderivative through partial fractions:

A

x+

B

x+ 2+

C

x+ 1=

2x+ 1

x(x+ 2)(x+ 1)

A(x2 + 3x+ 2) +B(x2 + x) + C(x2 + 2x) = 2x+ 11 1 13 1 22 0 0

ABC

=

021

=⇒

0 1 00 0 11 0 0

∣∣∣∣∣∣−3/2

11/2

Exercise 6.

115

(1)

log x =

∫ x

1

1

tdt F (x) =

∫ x

1

et

tdt;

if x > 0

et > 1 for t > 0

If 0 < x < 1

log x =

∫ x

1

1

tdt =

∫ 1

x

−1

tdt > −

∫ 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−∫−e

u

u

)=

= xe1/x − e− F (1/x) where we used the substitutionu =

1

t

du = − 1

t2dt

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= −ex

2 ln 2 + 1

(3)

f(x) = 2f(x)f ′(x); =⇒ f(x) =x

2+ C∫ x

0

(1

2t+ c

)dt =

(t2

4+ ct

)∣∣∣∣x0

=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)

116

(2) Since for f(x) = ex, we defined 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= limh→ 0

g(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

0tet

2

t2+1dt =∫ 1

0

12due

u

u+1 =1

2A

(3)∫ 1

0et

(t+1)2 dt = −et(t+1)

∣∣∣10−∫ 1

0−ett+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.

117

(1) p(x) = c0 + c1x+ c2x2; 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− tdu = −dt

Exercise 16. F (x) =∫ x

0f(t)dt

(1)

F (x) =

∫ x

0

(t+ |t|)2 =

{∫ x0

(2t)2dt = 43x

3 if t, x ≥ 0∫ x0

0dt = 0 if t, x < 0

(2)

F (x) =

∫ x

0

f(t)dt =

{∫ x0

(1− t2)dt if |t| ≤ 1∫ x0

(1− |t|)dt if |t| > 1=

=

(t− 1

3 t3)∣∣x

0= x− x3

3 if |x| ≤ 1{23 +

∫ x1

(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

118

(3) f(t) = e−|t|

F (x) =

∫ x

0

f(t)dt =

∫ x

0

e−|t|dt =

=

{∫ x0e−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| ≤ 1

1 + 13 t

3∣∣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

∫ a0f2 = a2 + a. (

x2 + x

π

)′=

√2x+ 1

π

for∫ a

0

2x+ 1

π=(x2 + x

)∣∣a0

= a2 + a

Exercise 18. f(x) = e−2x.

(1) A(t) =∫ t

0e−2xdx = e−2x

−2

∣∣∣t0

= 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−1

t

= π

where we used the limit limx→0

ecx − 1

x= c

Exercise 19. sinh c = 34

(1)ec = ex +

√e2x + 1

ec − e−c

2=ex +

√e2x + 1− 1

ex+√e2x+1

2=

=e2x + 2ex

√e2x + 1 + e2x + 1− 1

2(ex +√e2x + 1)

= ex =3

4

x = ln 3− 2 ln 2

119

(2)

ec − e−c

2=ex −

√e2x − 1− 1

ex−√e2x−1

2=

=e2x − 2ex

√e2x − 1 + e2x − 1− 1

2(ex −√e2x − 1)

=ex

+

−1

ex −√e2x − 1

= ex − 1

ec=

3

4

=⇒ 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 5

log2 3=

log2 5

(log2 3)2= log2 5

=⇒ 1 = log2 3 False

(3) Use inductionn = 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+1

2>√n2 + n =⇒ n+ 1 >

√n2 + n

=⇒n+1∑k=1

k−1/2 < 2√n+ 1

(4)

f = (coshx− sinhx− 1) =ex + e−x − ex + e−x

2− 1 = e−x − 1 < 0 for x > 0

False.

Exercise 21. For 0 < x < π2 ,

(sinx)′ = cosx > 0 for 0 < x <π

2

(x− sinx)′ = 1− cosx ≥ 0 for 0 < x <π

2(x− sinx)(x = 0) = 0 =⇒ sinx < x

Exercise 22.

1

t<

1

t

(x+ 1

t

)if 0 < x < t < x+ 1∫ x+1

x

1

tdt = ln (x+ 1)− lnx;

∫ x+1

x

x+ 1

t=

1

xSo ln

x+ 1

x<

1

x

Exercise 23.120

(x− sinx)′ = 1− cosx ≥ 0 ∀x > 0

since (x− sinx)(x = 0) = 0; (x− sinx)′(x = 0) = 0, then x− sinx > 0 in general for x > 0

(sinx−(x− x3

6

))′ = cosx− 1 +

x2

2> 0 ∀x > 0

=⇒ x− x3

6< sinx < x

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

(1

nln (1 +An)

))′ = (1 +An)1/n

(−1

n2ln (1 +An) +

1

n

(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 lnAn

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 sinx + b cosx = A(a sinx + b cosx) + B(a cosx − b sinx). Solve for A,B in terms of

121

a1, b1, a, b. Matching up term by term the coefficients for sin and cos separately,

Aa2 − abB = aa1

Ab2 +Bab = b1b

A =aa1 + bb1a2 + b2

−Aab+Bb2 = −a1b

Aab+Ba2 = ab1

B =ab1 − a1b

a2 + b2

So if not both a, b = 0 ,∫a1 sinx+ b1 cosx

a sinx+ b cosx=

∫A(a sinx+ b cosx) +B(a cosx− b sinx)

a sinx+ b cosx=

= Ax+B ln |a sinx+ b cosx|+ C

Exercise 27.

(1)

f ′(x2) =1

xdf

du= u−1/2

f(x2) = 2x− 1

(2)

f ′(sin2 x) = 1− sin2 x f ′(u) = 1− u f = u− 1

2u2 + C =⇒ f(x) = x− x2

2+

1

2

(3)

f ′(sinX) = (1− sin2 x) f(u) = u− 1

3u3 + C f(x) = x− x3

3+

1

3

(4)

f ′(lnx) =

{1 for x ≤ 1

x for x > 1=

{1 for 0 < x ≤ 1

eln x x > 1

f(y) =

{y for y < 0

ey − 1 for y > 0

Exercise 28.

(1)

Li(x) =

∫ x

2

dt

ln tif x ≥ 2

Li(x) =x

lnx− 2

1

ln 2−∫ x

2

−1

(lnx)2dt =

x

lnx+

∫ x

2

dt

(lnx)2− 2

ln 2

(2)

Li(x) =x

lnx− 2

lnx− 2

(ln 2)2+

x

(lnx)2−∫ x

a

−2

(ln t)3dt

Li(x) =x

lnx+

n−1∑k=1

k!x

lnk+1 x+ n!

(x

lnn+1 x−∫ x

2

−(n+ 1)dt

lnn+2 t

)

Li(x) =x

lnx+

n∑k=1

k!x

lnk+1 x+ (n+ 1)!

∫ x

2

dt

ln(n+1)+1 t

C2 =−2

lnx+−

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

122

(3)

Li(x) =

∫ x

2

dt

ln t

u = ln t

du =1

tdt

eudu = dt eu = t

Li(x) =

∫ ln x

ln 2

etdt

t

(4)

c = 1 +1

2ln 2

t = u+ 1 =⇒

∫ x−1

c−1

e2(u+1)du

u= e2

∫ x−1

c−1

e2u

udu =

= e2 1

2

∫ 2(x−1)3

2(c−1)3

et

t2

dt =

= e2Li(e2(x−1))

(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 = RRecall 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) =

∫ xaf(t)dt

Tn+1g(x) =∫ xaTnf(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 finding new Taylor polynomials without having to find the jth derivatives of the desired function.123

Theorem 24. Pn is a polynomial of degree n ≥ 1.Let f, g be 2 functions with derivatives of order n at 0.

(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

Exercise 4. (1

1 + x

)′=

−1

(1 + x)2;

(1

1 + 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.

1

1− 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)jxj+1

j + 1=

n∑j=1

(−1)j+1xj

j

Exercise 7. (log

√1 + x

1− x

)′=

√1− x1 + x

√1− x1 + x

1

(1− x)2=

1

(1 + x)(1− x)=

1

1− 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

)=

1

2Tn

(1

1−(

12

)x

)=

1

2

n∑j=0

(1

2x)j

=

n∑j=0

xj

2j+1

Exercise 9. We can show this in two ways.124

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 coefficients 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(sinx2) = T2n(1

2(1− cos 2x)) =

1

2

1−n∑j=0

(−1)j(2x)2j

(2j)!

=

n∑j=1

(−1)j+122j−1x2j

(2j)!

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) =1

n!

∫ 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 (arctanx)(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).

125

Exercise 4.

(1)

x2 = sinx = x− x3

6=⇒ x3

6+ x2 − x =

x

6

(x−

(−3 +

√15))(

x− (−3−√

15))

x =√

15− 3

(2)

E4(r; 0) =1

4!

∫ r

0

(r − t)4 cos tdt > 0 sin r − r2 = 0 + E4(r) ≤ r5

5!<

r

5!=

3

5(2)(5)(4)(3)(2)1=

34

(5)(4)(2)5<

1

200

Exercise 5.

arctan r − r2 = r − r3

3− r2 + E4(r; 0) = 0 + E4(r; 0)

E4(r, 0) =1

4!

∫ r

0

(x− t)4f (5)(t)dt ≤ M(r5)

5!

r5

5!= 0.065536 <

7

100

E4(r, 0) < Ej(r, 0); j > 4

the 5th degree term isf (5)(0)

5!r5 =

24

5!r5 > 0

so r2 − arctan r = −E4(r, 0) < 0

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 + c1

31x31

∣∣∣∣10

= 1 +c

31

Exercise 7.∫ 1/2

01

1+x4 dx.

1

1 + x4=

∞∑j=0

(−x4)j =

n−1∑j=0

(−x4)j + En = 1 +−x4 + x8 . . .

16

17

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

1

1 + x4' 1

2+−1

5

(1

2

)5

0.493852 < 0.49375 < 0.493858

Exercise 8.

(1)

0 ≤ x ≤ 1

2sinx = x− x3

3!+ E4(x)

|E4(x)| ≤ M |x|5

5!=

sin 13 |x|

5

5!≤

1(

12

)55!

126

(2)

sinx2 = x2 − x6

6+ E4(x2)∫ √

22

0

sinx2 =

(1

3x3 − x7

42

)∣∣∣∣√

2/2

0

=√

2

(1

12− 1

42(16)

)E4(x2) ≤

√2

64(5!)∫ √2

2

0

sinx2 ≤√

2

(55

672+

1

64(5!)

)= 0.1159

Exercise 9.

sinx = x− x3

6+x5

5!E6(x; 0) ≤ (1)x7

7!sinx

x= 1− x2

6+x4

5!+E6(x; 0)

x≤ 1− x2

6+x4

5!+x6

7!∫ 1

0

sinx

xdx = 1− 1

3

(1

6

)+

1

5(5!)+

1

7(7!)= 0.9461 +

1

7(7!)= 0.9461 + 0.0000283

Exercise 10. α = arctan 15 , β = 4α− π

4 .

(1)

tan (A+B) =(tanA+ tanB)

1− tanA tanB;A = B = α; tan 2α =

2 tanα

1− tan2 α=

2/5

24/25= 5/12

tan 4α =2(tan 2α)

1− tan2 (2α)=

2(5/12)

1− (5/12)2=

10/12

119/144=

120

119A = 4α,B = −π

4

tan (4α− π

4) = tanβ =

tan 4α+ tan(−π4)

1− tan 4α tan(−π4) =

120119 +− 119

119

1− 120119 (−1)

=1

239

4 arctan1

5=π

4+ arctan

1

239This is incredible.

(2)

T11(arctanx) =

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 arctan1

5< 3.158328958

(3)

T3(arctanx);x =1

239

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

2!+ o(x2)

Exercise 2.127

x(cosx) = ((x− 1) + 1)(cosx) = (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 sinx− 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)− 1

6

(x3 − 3x4 + 3x5 − x6

)+

1

120

(x5 − 5x6 + 10x7 − 10x8 + 5x9 + x10

)=

= (x− x2)− 1

6x3 − 1

2x4 +

61

120x5 − 25

120x6

Exercise 4.

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

2+

(x− 1)3

3

=⇒ a = 0; b = 1, c =−1

2

Exercise 5.

cosx = 1− 1

2x2 + o(x3) as x→ 0

1− cosx =1

2x2 + o(x3)

1− cosx

x2=

1

2+o(x3)

x2

since1− cosx

x2=

1

2+ o(x),

1− cosx

x2→ 1

2as x→ 0

cosx = 1− 1

2x2 +

x4

4!+ o(x5) =⇒ cos 2x = 1− 2x2 +

2

3x4 + o(x5)

1− cos 2x− 2x2

x4=− 2

3x4 − o(x5)

x4=−2

3− o(x)→ −2

3as x→ 0

Exercise 6.

limx→0

sin ax

sin bx= limx→0

ax− (ax)3

3! + o(x4)

bx+ o(x2)=a

b

Exercise 7.

limx→0

sin 2x

cos 2x sin 3x= limx→0

(2x) + (2x)3

3! + o(x4)(1− (2x)2

2! + (2x)4

4! + o(x5))(

(3x)− (3x)3

3! + o(x4)) =

2

3

Exercise 8.

limx→0

sinx− xx3

= −1

6

Exercise 9.

limx→0

ln 1 + x

e2x − 1= limx→0

x− o(x)

2x+ o(x)=

1

2

Exercise 10. Don’t do the trig. identity.

limx→0

1− cos2 x

x tanx= limx→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 11.128

limx→0

x+−x3

6 + o(x4)

x− x3

3

= 1

Exercise 12.

limx→0

ex ln a − 1

ex 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)=

1

3

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=

1

6

Exercise 16.

limx→0

ln (1 + x)− x1− cosx

= limx→0

x− x2

2 + x3

3 + o(x3)− xx3/2

= −1

Exercise 17.

limx→π/2

cosx

x− π2

=0 +−1(x− π

2 )

x− π2

= −1

Exercise 18. 1/6

Exercise 19.

limx→0

coshx− cosx

x2= limx→0

ex+e−x

2 − cosx

x2=

= limx→0

1 + x+ x2

2 − 2(

1− x2

2

)+ o(x3)

x2= 2

Exercise 20.

limx→0

3 tan 3x− 12 tanx

3 sin 4x− 12 sinx= limx→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

Exercise 21.

limx→0

ax − asinx

x3= limx→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 + sinx 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 sinx− cosx

x4= limx→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=

1

6129

Exercise 23.

limx→ 1x1

1−x = limx→ 1e1

1−x ln x = exp

(limx→ 1

lnx

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

1

xln (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 − ex

= limx→0

e1x ln (1+x) − e

x= limx→0

ex+− x

2

2+o(x2)

x − ex

=

= limx→0

e1− x2 +o(x) − ex

= limx→0

e(1 +−x2 + o(x))− ex

=−e2

Exercise 26.

limx→0

((1 + x)1/x

e

)1/x

= limx→0

(exp

(1

xln (1 + x)

)− 1

)1/x

= limx→0

ex− x

2

2+o(x3)−xx2 = e−1/2

Exercise 27.

(arcsinx)′ =1√

1− x2

(arcsinx)′′ =x

(1− x2)3/2

(arcsinx)′′′ =1

(1− x2)3/2+

3x2

(1− x2)5/2

exp

(limx→0

1

x2

(ln

(arcsinx

x

)))= exp lim

x→0

1

x2ln

(1 +

(arcsinx

x− 1

))=

= exp

(limx→0

1

x2ln

(1 +

x+ x3/6 + o(x4)− xx

))= exp

(limx→0

1

x2ln

(1 +

x2

6+ o(x3)

))=

= exp

(limx→0

1

x2

(x2 + o(x3)

6

))= e1/6

Exercise 28. limx→0

(1x −

1ex−1

)= limx→0

(ex−1−xx(ex−1)

)= limx→0

x2

2 +o(x3)

x2+o(x3) =1

2

Exercise 29.

limx→1

(1

log x− 1

x− 1

)= limx→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)=

1

2

Exercise 30.

limx→0

eax − ex − xx2

= limx→0

1 + ax+ (ax)2

2 + o(x3)− 1− x− x2

2 − xx2

if a = 2, the limit is 2

Exercise 31.

130

(1) Prove∫ x

0f(t)dt = o

(∫ x0g(t)dt

)as x→ 0, given f(x) = o(g(x)).

Consider limx→0

∫ x0f(t)dt∫ x

0g(t)dt

.Since f, g have derivatives in some interval containing 0, f, g continuous and differentiable for |t| ≤ x.

limx→0

∫ x0f(t)dt

limx→0 g(t)dt= limx→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 find that

1

1 + 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 expx ln (e3 + o(1)) = x+ x2 + f(x)

1 expx 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 expx ln (e3 + o(1))

(ln (e3 + o(1)) +

x

e3 + o(1)(o′(1))

)+

+ x expx ln (e3 + o(1))

(ln (e3 + o(1)) +

x

e3 + o(1)o′(1)

)+

x expx 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) = 4

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

131

7.13 Exercises - L’Hopital’s rule for the indeterminate form 0/0. Exercise 1.

limx→2

3x2 + 2x− 16

x2 − x− 2= limx→2

(3x+ 8)(x− 2)

(x− 2)(x+ 1)=

14

3

Exercise 2.

limx→3

x2 − 4x+ 3

2x2 − 13x+ 21= limx→3

(x− 3)(x− 1)

(2x− 1)(x− 3)= −2

Exercise 3.

limx→0

sinhx− sinx

x3=

0

0= limx→0

(coshx− cosx

3x2

)=

0

0= limx→0

(sinhx+ sinx

6x

)= limx→0

(coshx+ cosx

6

)=

1

3

Exercise 4.

limx→0

(2− x)ex − x− 2

x3= limx→0

−ex + (2− x)− 1

3x2= limx→0

(2− x)(1 + x+ x2/2 + x3/6 + o(x3))

x3= −1

6

Exercise 5.

limx→0

log (cos ax)

log (cos bx)= limx→0

− cos bx sin axa

− cos ax sin bxb= limx→0

cos axa2

sin bxb2=a2

b2

Exercise 6. When it doubt, Taylor expand.

limx→0+

x− sinx

(x sinx)3/2= limx→0+

1− cosx32 (x sinx)1/2(sinx+ x cosx)

=2

3limx→0+

1− (1− x2

2 + x4

24 + o(x4))

e12 ln (x sin x)(x− x3

6 + o(x3) + x− x3

2 + o(x3))=

=2

3limx→0+

x2

2 −x4

24 + o(x4)√x(x+ x3

6 + o(x3))(2x− 23x

3 + o(x3))

=

=2

3limx→0+

12 −

x2

24 + o(x2)√1 + x2

6 + o(x2)(2 + −23 x

2 + o(x2))

=2

3limx→0+

(1/2

2

)=

1

6

Notice in the third step how in general we deal with powers, (x sinx)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 first.

limx→a+

√x−√a+√x− a√

x2 − a2= limx→a+

12√x

+ 12√x−a

x√x2−a2

=1

2limx→a+

√x2 − a2

x3/2+

√x+ a

x=

=1

2limx→a+

√x2 − a2 + x1/2

√x+ a

x3/2=

1

2

√2a

a3/2=

√2

2√a

Exercise 8. Do L’Hopital’s at the second step.

limx→1+

exp (x lnx)− x1− x+ lnx

= limx→1+

expx lnx(lnx+ 1)− 1

−1 + 1x

= limx→1+

exp (x lnx)(lnx+ 1)2 + 1x exp (x lnx)

−1/x2=

= limx→1+

x2 exp (x lnx)(lnx+ 1)2 + x exp (x lnx) = 1 + 1 = 2

Exercise 9. Keep doing L’Hopital’s.132

limx→0

arcsin 2x− 2 arcsinx

x3= limx→0

2√1−(2x)3

− 2 1√1−x2

3x2=

=2

3limx→0

− 12 (1− (2x)2)−3/2(−8x)− (− 1

2 )(1− x2)−3/2(−2x)

6x=

=1

9limx→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=

=1

9

4− 1

1=

1

3

Exercise 10.

limx→0

x cotx− 1

x2= limx→0

x cosx− sinx

x2 sinx= limx→0

cosx− x sinx− cosx

2x sinx+ x2 cosx= limx→0

− sinx

2 sinx+ x cosx=

= − limx→0

cosx

2 cosx+ cosx+−x sinx= − lim

x→0

cosx

3 cosx− x sinx=

1

3

Exercise 11.

limx→1

∑nk=1 x

k − nx− 1

= limx→1

∑nk=1 kx

k−1

1=

n∑k=1

k =n(n+ 1)

2

Exercise 12.

limx→0+

1

x√x

(a arctan

√x

a− b arctan

√x

b

)= limx→0+

(a 1

1+ xa2

(1

2a√x

)− b 1

1+ xb2

(1

2b√x

))32x

1/2=

=1

3limx→0+

a2

a2 + x

(1

x

)− b2

b2 + x

(1

x

)=

1

3limx→0+

(b2 + x)a2 − b2(a2 + x)

(a2 + x)(x+ b2)x=

=1

3limx→0+

(a2 − b2)

(a2 − x)(b2 + x)=

1

3

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

2

We 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 =9

2, a = −3 .

Exercise 15.

limx→0

1

bx− sinx

∫ x

0

t2dt√a+ t

=

x2√a+x

b− cosx=

2x1

2√a+x

(1− cosx) +√a+ x(sinx)

=2x√a+ x

12 (1− cosx) + (a+ x) sinx

=

= 2 limx→0

√a+ x+ x

2√a+x

sin x2 + sinx+ (a+ x) cosx

= 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.

133

Exercise 16.

(1)angle ABC is

x

2, length BC is tan

x

2

2 tanx

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 x2sin

x

2= tan

x

2− 1

2sinx

(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 x2 −

12 sinx

x−sin x2

ddx−−→

12 sec2 x

2 −cos x

21−cos x

2

ddx−−→

sec2 x2 tan x

2 + sinx

sinx

ddx−−→

sec2 x2 tan2 x

2 + 12 sec4 x

2 + cosx

cosx→

x→0−−−→ 3

2

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)(−tL

)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; Infinite limits; The behavior of log x andex 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(lnx)2 = lim

x→1−1

2 lnx(

1x

)(−1)

= 0

Exercise 16. Persist in using L’Hopital’s and trying all possibilities systematically.

limx→0+

xxx−1 = lim

x→0+e(xx−1) ln x = lim

x→0+e(ex ln x−1) ln x = exp lim

x→0+

lnx

(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 + 1

ex ln x(x lnx+ x)

)=

= exp− limx→0+

ex ln x(lnx+ 1) + e−x ln x(− lnx− 1)

(lnx+ 1 + 1)= exp− lim

x→0+

ex ln x − e−x ln x

1 + 1ln x+1

= 1

Exercise 17.134

limx→0+

(xxx

− 1) = limx→0+

(exx ln x − 1

)= limx→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 = −1

We usedlimx→0+

xα log x = 0∀α > 0

since

t = 1x , x

α 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/ sinx

)= exp

(limx→0−

11−ex ln 2

(− ln 2ex ln 2

)−1

sin2 xcosx

)=

= exp

(limx→0−

(sin2 x) ln 2ex ln 2

(1− ex ln 2) cosx

)= exp

((ln 2) lim

x→0−

2 sinx cosx

− ln 2ex ln 2

)= 1

Exercise 19.

limx→0+

e1

ln x ln x = e

Exercise 20. At the end, L’Hopital’s could be used to verify that indeed sinx ln sinx→ 0 as x→ 0.

limx→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

(tanx)tan 2x = limx→π

4

etan 2x ln tan x = exp limx→π

4

1

cos 2x(ln sinx− ln cosx) =

= exp limx→π

4

1sin x cosx− 1

cos x (− sinx)

−2 sin 2x cos 2x= exp lim

x→π4

1

− sin2 2x= e−1

Exercise 22.

Exercise 23. Use L’Hopital’s theorem, taking derivatives of top and bottom.

limx→0+

expe

1 + lnxlnx = exp e lim

x→0

lnx

1 + lnx= 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

sinπx/2 ln (2−x)cosπx/2 = exp lim

x→1

(−1)2−x

π2 sinπx/2

= exp−2

π

Exercise 26.

limx→+∞

exp

(x ln

(x+ c

x− c

))= exp lim

x→+∞

ln(

1+c/x1−c/x

)1/x

= exp limx→+∞

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 +

1

x2

)1/2

− x2 = x2

(1 +

1

x2

)1/2

− x2

Let x2 = 1t . So t→ 0 as x→ +∞.

=⇒ (1 + t)1/2 − 1

t=

1 + 12 t− 1 + o(t)

t=

1

2135

Exercise 28.

(x5 + 7x4 + 2)c − x = x5

(1 +

7

x+

2

x5

)c− x

Let 1t = x and guess that c = 1

5 (1

t5+

7

t4+ 2

)c− 1

t=(1 + (7t+ 2t5)

)1/5/t− 1

t=

=1 + 1

5 (7t+ 2t5) + o(t)− 1

t=

7

5

Exercise 29.

g(x) = xex2

g′(x) = ex2

+ 2x2ex2

g′′(x) = 2xex2

+ 4xex2

+ 4x3ex2

= 6xex2

+ 4x3ex2

f(x) =

∫ x

1

g(t)

(t+

1

t

)dt

f ′(x) = g(x)

(x+

1

x

)− 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 =2xex

2

+ 2x3ex2

+ 2xex2

(6xex2 + 4x3ex2)=

=4x+ 2x3

6x+ 4x3=

1

2as 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 − 1

Guessing that c = 1

f ′(x)

g′(x)=e2x(3x2 + 1)1/2 − 1

2xce2x + 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

)mFor each year, there are the just previously shown m compoundings, so for n years,

P(

1 +r

m

)mn(2)

2 = ert

ln 2

r= t = 11.55years

(3)

2P0 = P0

(1 +

r

m

)mtln 2 = mt ln (1 + r/m)

t =ln 2

m ln (1 + r/m)=

ln 2

4 ln (1 + 0.06/4)= 11.64years

136

7.17 Exercises - The symbols +∞ and −∞. Extension of L’Hopital’s rule; Infinite limits; The behavior of log x andex for large x. Exercise 1.

limx→0

e−1/x2

x1000= limu→∞

e−uu500 = limu→∞

u500

eu= 0

u =1

x2

x2 =1

u

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)

Proof. Trick - use the definition of the logarithm as an integral.If c > 0, t ≥ 1, tc ≥ 1 =⇒ tc−1 ≥ t−1.

0 < lnx =

∫ x

1

1

tdt ≤

∫ x

1

tc−1dt =1

c(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= limx→π

2

cosx

cos 3x= limx→π

2

− sinx

−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√ax2

+b

=1√b

Exercise 5. Make the substitution x = 1t .

limx→∞

x4

(cos

1

x− 1 +

1

2x2

)= limt→∞

1

t4

(cos t− 1 +

t2

2

)= limt→∞

t4/4! + o(t4)

t4=

1

4!=

1

120

Exercise 6.

limx→π

ln | sinx|ln | sin 2x|

= limx→π

1sin x cosx2

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)

tanπx= limx→ 1

2−

(1

1−2x

)(−2)

(sec2 πx)π= limx→ 1

2−

−2(cos2 πx)

(1− 2x)π=

= − 2

πlimx→ 1

2−

2 cosπxπ − sinπx

−2= 1

137

Exercise 8.

limx→∞

coshx+ 1

ex= limx→∞

ex+1 + e−x−1

2ex=

1

2limx→∞

e1 +1

e2x+1=

e

2

Exercise 9.

limx→∞

ax

xb= limx→∞

ex ln a

xb→∞; a > 1

since limx→∞

xb

eax(in this case, ln a > 0 )

Exercise 10.

limx→π

2

tanx− 5

secx+ 4= limx→π

2

sec2 x

tanx secx= limx→π

2

secx

tanx= limx→π

2

1

cosx

cosx

sinx= 1

8.5 Exercises - Introduction, Terminology and notation, A first-order differential equation for the exponential func-tion, 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) =∫ xaP (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 for

y(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) =

∫ x1

(−2t

)dt = −2 lnx.

y = e−A(x)

(∫ x

a

Q(t)eA(t)dt+ b

)= e2 ln x(

∫ x1t4e−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(−π2 ,

π2

), with y = 2 when x = 0.

138

A(x) =

∫ x

0

P (t)dt =

∫ x

0

tan tdt = − ln | cos t||x0 = − ln cosx

y = e−A(x)

(∫ x

a

Q(t)eA(t)dt+ b

)= cosx

(∫ b

a

sin 2te− ln cos tdt+ 2

)=

= cosx

(∫ x

a

2 sin tdt+ 2

)= −2 cos2 x+ 4 cosx

Exercise 4. y′ + xy = x3. y = 0, x = 0.

A(x) =1

2x2

y = e−12x

2

(∫ x

0

t3et2

2 dt+ 0

)= e

−x22

(t2e

t2

2 − 2et2

2

)∣∣∣x0

=

= x2 − 2 + 2e−x2

2

Exercise 5. y′ + y = e2t. y = 1, t = 0.

A = x

y(x) = e−x(∫ x

0

e2tetdt+ 1

)= e−x

(e3t

3

∣∣∣∣x0

+ 1

)=

e2x

3+

2

3e−x

Exercise 6. y′ sinx+ y cosx = 1; (0, π). =⇒ y′ + cotxy = cscx

A(x) =

∫ x

a

cot tdt = ln

(sinx

sin a

)y(x) = e− ln ( sin x

sin a )(∫

(csc t)eln ( sin tsin a )dt+ b

)=

(sin a

sinx

)(x− asin a

+ b

)indeed y =

x− asinx

+b sin a

sinx

x→ 0 for y =x− asinx

+b sin a

sinx=x+ b sin a− a

sinx

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′ +1

x(x+ 1)y = (x+ 1)e−x

2

A(x) =

∫ x

a

(1

t− 1

t+ 1

)dt = ln

x

a− ln

x+ 1

a+ 1

eA(x) =(a+ 1)x

a(x+ 1)

y =a(x+ 1)

(a+ 1)x

(∫ x

a

(t+ 1)e−t2

((a+ 1)t

a(t+ 1)

)dt+ b

)=

(x+ 1)

x

(∫ x

a

te−t2

dt+ ba(x+ 1)

(a+ 1)x

)=

=x+ 1

2x

(e−a

2

− 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 +1

x) +

ab

a+ 1(1 +

1

x) =

limx→0

y =1

x

(e−a

2 − e−x2

2+

ab

a+ 1

)a = 0

139

Exercise 8. y′ + y cotx = 2 cosx on (0, π).

A(x) =

∫ x

a

cot tdt = ln

(sinx

sin a

)eA(x) =

sinx

sin a

y =sin a

sinx

(∫2 cos t

sin t

sin a+ b

)=

sin a

sinx

(− cos 2t

2 sin a

∣∣∣∣xa

+ b

)=

=cos (2a) + cos 2x

2 sinx+

sin a

sinx

y = sinx y =cos (2a)− cos (2x)

2 sinx+

sin a

sinx

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− 3

a− 3

∣∣∣∣ ∣∣∣∣a− 2

x− 2

∣∣∣∣If (−∞, 2), (3,+∞),

x− 3 ≶ 0

x− 2 ≶ 0

3− x2− x

=x− 3

x− 2

If (2, 3), x− 3 < 0, but x− 2 > 0

If (−∞, 2), (3,∞) y = b

(x− 2

x− 3

)2(∣∣∣∣a− 3

a− 2

∣∣∣∣)2

+

(x− 2

x− 3

)2(x+

1

x− 2− a− 1

a− 2

)If (2, 3) y = b

(x− 2

x− 3

)2(∣∣∣∣a− 3

a− 2

∣∣∣∣)2

+

(x− 2

3− x

)2 ∣∣∣∣a− 3

a− 2

∣∣∣∣2 ∫ ( t− 1

t− 3

) ∣∣∣∣a− 2

a− 3

∣∣∣∣2(3− tt− 2

)2

=

= b

(x− 2

3− x

)2 ∣∣∣∣a− 3

a− 2

∣∣∣∣2 +x− 2

3− x

(−∫

(t− 1)(3− t)(t− 2)2

)=

= b

(x− 2

x− 3

)2 ∣∣∣∣a− 3

a− 2

∣∣∣∣2 +

(x− 2

x− 3

)2

(x+ (x− 2)−1 − a− (a− 2)−1)

Exercise 10. s(x) = sin xx ;x 6= 0 s(0) = 1, T (x) =

∫ x0s(t)dt f(x) = xT (x)

f ′ = T + x(s(x)) = T + sinx

xf ′ − f = x sinx

y′ − y

x= sinx

A(x) =

∫ x

a

−1

tdt = − ln

x

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) =1

xis not continuous at x = 0

Exercise 11.

f(x) = 1 +1

x

∫ x

1

f(t)dt

xf(x) = x+

∫ x

1

f(t)dt =⇒ f(x) + xf ′ = 1 + f(x) =⇒ f ′ =1

x

=⇒ f(x) = ln |x| − C

ln |x| − C = 1 +1

x

∫ x

1

(ln |t| − c) dt =

= 1 +1

x(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− fx2

)140

So then

f ′ +

(x− 1

x

)f = − 1

x

A(x) =

∫P (t)dt =

∫ (t− 1

t

)dt =

(1

2x2 − lnx

)−(

1

2a2 − ln a

)eA(x) = e

x2−a22

a

x; e−A(x) =

x

aea2−x2

2

f(x) =x

aea2−x2

2

(∫ x

a

−1

t2ae

t2−a22 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

Exercise 14. y′ − 4y = 2exy1/2

n =1

2k = 1− 1

2=

1

2; v = y1/2; v′ +

1

2(−4)v =

1

2(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′ +− 1

xv = 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= 0.141

y′ +y

x= y2x log x

k = 1− n = 1− 2 = −1, v = yk = y−1

v′ + kPv = kQ; v′ +−(

1

x

)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

1

2) + 2

Check:y′ = (C + 2x− 2x log x− x)(−y2)

y2(−C +−2x+ 2x log x+ x+ C + x− x log x)

y =1

2;x = 1 y(x = 1) =

1

2b = 2

y =1

x(x lnx− x+ 3)

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

(xa

)+ (x− a); e

∫ xaPvdt = 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

(1

2ae2t−a

∣∣∣∣xa

+ 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 =1

x

(ex − e−x

)=⇒ sinhx

x

Exercise 19. The Ricatti equation is y′ + P (x)y +Q(x)y2 = R(x).142

If u is a known solution, y = u+ 1v is also a solution if v satisfies a first-order ODE.

(u+ 1/v)′ = u′ +−1

v2v′

y′ + Py +Qy2 = R =⇒ u′ +−v′

v2+ P (u+

1

v) +Q(u2 +

2u

v+

1

v2) = R

=⇒ v′ − Pv = Q(2uv + 1)

Exercise 20. y′ + y + y2 = 2, y = 1,−2.

(1) If −2 ≤ b < 1,

y′ + y + y2 = 2P = 1, Q = 1, R = 2

v′ + (−P − 2Qu)v = Qy = u+1

vu = 1

v′ + (−1− 2(1)(1))v = v′ − 3v = Q = 1

v = e3x

(∫1e−3tdt+ b

)= e3x

(e−3t

−3

∣∣∣∣xa

+ b

)= be3x − 1

3

(1− e3x−3a

)y = 1 +

3

3be3x − (1− e3x−3a)

y(0) = 1 +3

3b− (1− e−3a)=⇒

b = 1 +1

b

b2 − b− 1 = 0

b =1±√

5

2

y = 1 +3

3be3x − (1− e3x)b =

1−√

5

2

(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 +

3

1− e3a + 3b

a=0−−→ y(0) = −2 +3

3b=⇒ 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 first-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

y0.1

kln e = T .

143

(2) f(a) = y0e−ka; f(b) = y0e

−kb.

f(a)

f(b)= e−ka+kb =⇒ ln

f(a)

f(b)= −k(a− b); 1

a− blnf(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= y

1− ab0

f(t) =

(f(a)

(f(b))a/b

) bb−a(f(a)

f(b)

) ta−b

=(f(a))

bb−a

f(b)ab−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− tb− a

; 1− w(t) =b− a− (b− t)

b− a=t− ab− a

Exercise 4. F = mv′ = w0 − 34v

w0 = 192w0

g= 6 = m

v′ =w0

m− 1

8v =⇒ v′ +

1

8v =

w0

m= 32

v(t) = e−t8

(∫ t

0

w0

met8 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′ +12

mv =

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 < 10

16 + 166e20−2t if t > 10

Exercise 7.

(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)

1

Tln

(yf −Mb−M

)= k =

1

T(ln (60)− ln (140)) =

1

Tln

(3

7

)=

1

30(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 − 60

140

)= −kt =⇒ tf =

(ln 140− ln (T − 60))

kfor 60 < T ≤ 200

(3) tf = ln (140)−ln (30)k = 30

ln 7/3 ln 143 = 54 minutes

144

(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

∣∣∣∣t0

− α(ue−ku

−k+−e−ku

k2

)∣∣∣∣t0

)+ bekt =

= −kekt(M0

k(1− e−kt)− α

(te−kt

−k− e−kt

k2+

1

k2

))+ 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 =

1

5ln

(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) saltw(t)

)So then

y′ =−2y

w0 + t=⇒ ln y = −2(ln (w0 + t)− lnw0) = −2 ln

(w0 + t

w0

)is the equation of motion given by the problem.

y(t) = Celn(w0+tw0

)−2

= C

(w0 + t

w0

)−2

y(t) = 50

(100

100 + t

)2

y(t = 60min.) = 50

(100

160

)2

= 5025

64=

625

32' 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)( yw

)+ α

( yw− 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, yw was zero, i.e. water is fresh.

145

y′(t) =−7

3

y

w+ 1; y′ =

7

3

y

w0 + t= 1

P =7/3

w0 + t∫P =

∫7/3

w0 + t=

7

3ln (w0 + t)|t0 =

7

3ln

(w0 + t

w0

)y = e

− ln(w0+tw0

)−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 =

(100

100 + 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 sinωt.

I(t) = I(0)e−Rt/L + e−Rt/L∫ t

0

V (x)

LeRx/Ldx

I(t) = I(0)e−Rt/L + e−Rt/LE

L

∫ t

0

sinωxeRx/Ldx

Using∫eax sin bxdx = aeax sin bx− beax cos bx

I(t) = I(0)e−RtL +

Ee−Rt/L

L

(RL e

RtL sinωt− ωeRtL cosωt(

RL

)2+ ω2

+ω(

RL

)2+ ω2

)I(0) = 0 So

I(t) =E

L

RL sinωt− ω cosωt(

RL

)2+ ω2

+EωL

R2 + (ωL)2e−Rt/L =

E(R sinωt− ωL cosωt)

R2 + (ωL)2+

EωL

R2 + (ωL)2e−Rt/L =

sinα =ωL√

R2 + (ωL2)

=⇒ I(t) =E sin (ωt− α)√R2 + (ωL)2

+EωL

(R2 + ω2L2)e−Rt/L

L = 0 sinα = 0

Exercise 12.

E(t) =

{E if 0 < a ≤ t < b

0 otherwise

I(t) = e−Rt/L∫ t

0

0 = 0 for t < a

I(t) = e−Rt/L∫ t

a

E

LeRx/Ldx =

E

Le−Rt/L

L

R

(eRt/L − eRa/L

)=E

R

(1− e

R(a−t)L

)I(b) =

E

R(1− eR(a−b)/L)

for 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)146

dx

dt= kx(M − x) = kMx− kx2; =⇒ dx

kMx− kx2= dt =

1/kdx

x(M − x)= dt

=⇒ kdt =

(1

x+

1

M − x

)1

Mdx =

lnx+− 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)x2

2 = x1x3(M − x2)2 = x1x3(M2 − 2Mx2 + x22) = x2

2(M2 −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

M∫ ttik(u)du

x =Me

M∫ ttik(u)du

1 + eM∫ ttik(u)du

=M

1 + e−M

∫ ttik(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.

8.14 Exercises - Linear equations of second order with constant coefficients, 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 =√d

2

(3) If d < 0, then u1(x) = cos kx and u2(x) = sin kx; where k = 12

√−d

147

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′′ + 32y′ = 0 y = 1, y′ = 1; x = 0 d = 9

4 > 0

y = e−34 x(c1e

3x4 + c2e

−3x4 ) = c1 + c2e

−3x2

c2 =−2

3=⇒ y =

5

3+−2

3e−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 15

sin 15) = c1

(1

sin 15

) c1 = − sin 15

c2 = − cos 15

y = − sin 15 sin 5x− cos 15 cos 5x

Exercise 13. y′′ − 4y′ − y = 0; y = 2; y′ = −1 when x = 1

d = 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)c1e

2+√

5 + (5− 2√

5)c2e2−√

5 = 0

c1 =2√

5− 5

5 + 2√

5c2e−2√

5(2√

5− 5

5 + 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− 5

2√

5e−2−

√5

y =2√

5− 5

2√

5e−2−

√5e(2+

√5)x +

5 + 2√

5

2√

5e√

5−2e(2−√

5)x

Exercise 14. y′′ + 4y′ + 5y = 0, with y = 2 and y′ = y′′ when x = 0

16− 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 =4

3

y = e−2x(4

3sin 2x+ 2 cos 2x)

148

Exercise 15. y′′ − 4y′ + 29y = 0

d = 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 =

1

2eπ

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 5

6eπsin 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 = 2e

3x2 c1(sinh

(5x

2

))

v = 2b1e−2x(sinh (3x))

u′ = c13

2e

3x2 (e

5x2 − e

−5x2 ) + 2e

3x2 c1

5

2cosh

(5x

2

)u′(0) = 5c1

v′ = −4b1e−2x sinh (3x) + 6b1e

−2x cosh (3x)

v′(0) = 6b1

c1 =6b15

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) = 0

Exercise 18. y′′ + k2y = 0 d = −4k2 < 0√d

2 = 2k2 = k > 0

149

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 sinx+ k2 cosx

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 − b1c2s2c1 − s1c2

s1b2 = k1s1s2 + k2s1c2

−(s2b1 = k1s1s2 + k2c1s2)=⇒ k2 =

s1b2 − s2b1s1c2 − c1s2

y =

(b2 cos a1 − b1 cos a2

sin a2 cos a1 − sin a1 cos a2

)sinx+

(b2 sin a1 − b1 sin a2

sin a1 cos a2 − cos a1 sin a2

)cosx

y =b2 cos a1 − b1 cos a2

sin (a2 − a1)sinx+

b2 sin a1 − b1 sin a2

sin (a1 − a2)cosx if a2 − a1 6= πn

otherwise, if a2 − a1 = πn;b2c1 − b1c2 = 0

b2s1 − b1s2 = 0b2c1 = 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

]=

[b1b2

][AB

]=

[C2 −C1

−S2 S1

](1

S1C2 − C1S2

)[b1b2

]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 − b1a2 − a1

)(x− a1) + b1

Exercise 20.

(1)

u1(x) = ex; u2(x) = e−x u′2 = −e−x u′′2 = e−x =⇒ y′′ − y = 0

150

(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 cosx;

u′1 = −1

2e−x/2 cosx+−e−x/2 sinx

u′′1 =1

4e−x/2 cosx+ e−x/2 sinx+−e−x/2 cosx

=−3

4e−x/2 cosx+ e−x/2 sinx

u2(x) = e−x/2 sinx

u′2 =−1

2e−x/2 sinx+ e−x/2 cosx

u′′2 =1

4e−x/2 sinx+−e−x/2 cosx− e−x/2 sinx =

=−3

4e−x/2 sinx− e−x/2 cosx

u′′2 + u′2 +5

4u2 = 0

y′′ + y′ +5

4y = 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 = coshx

u′1 = sinhx

u′′1 = coshx

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 u2

u1is 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 ∀x151

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 coefficients, 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 = −x

y = c1ex + 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 +−1

3x3 + 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 =1

3B = 0 C = 0

y = c1e−x + c2 +

1

3x3

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 =1

3B =

2

3C =

8

9D =

16

27

y = C1ex sin

√2x+ C2e

x cos√

2x+1

3x3 +

2

3x2 +

8

9x+

16

27

Exercise 5. y′′ − 5y′ + 4y = x2 − 2x+ 1

yh = e5x2

(e

3x2 + e

−3x2

)= c1e

4x + c2ex

152

d =√

25− 4(4) = 3

4(Ax2 +Bx+ C)

− 5(2Ax+B)

2A

4Ax2 + 4Bx+ 4C

−10Ax− 5B

2A

A =1

4B =

1

8

1

2− 5

8+ 4C = 1

C =9

32

y = c1e4x + c2e

x +1

4x2 +

1

8x+

9

32

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 =−1

3B = −1 C =

1

2D =

−7

12

y = C1e−3x + C2e

2x − 1

3x3 +−x2 +

1

2x− 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 = −∫v2

R

W (x)dx; t2(x) =

∫v1R

Wdx

t1 = −∫e−2x e

2x

−4=

1

4x

t2 =

∫e2x e

2x

−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 8.153

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=−1

2

∫e−2x cos 2xdx

t2 =

∫sin 2xe−2x

2=

1

2

∫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

8

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 + 2xexy = c1e

x + c2e−2x +

1

3xe2

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=

1

3ex t2 =

∫exe2x

−3e−x=−1

12e4x

y1 = t1v1 + t2v2 =1

3exex +− 1

12e4xe−2x =

1

4e2x

y = c1ex + c2e

−2x +1

4e2x

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 +1

3xex +

1

4e2x

Exercise 12. y′′ − 2y′ + y = x+ 2xex. d = 4− 4(1) = 0. Recall the definition to be learned for this section of exercises:

Theorem 29. Let d = a2 − 4b be the discriminant of y′′ + 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 =√d

2

(3) If d < 0, u1 = cos kx, u2 = sin kx, k =√−d2

154

yh = ex(c1 + c2x) = c1ex + c2xe

x

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−xx2

)e−2x

= − lnx t2 =

∫ e−x(e−x

x2

)e−2x

=−1

x

W = e−x(e−x +−xe−x)− (−e−x)(xe−x) = e−2x

yp = − lnxe−x + xe−x(−1

x

)= − lnxe−x − e−x

y = c1e−x + c2xe

−x + (− lnx− 1)e−x

Exercise 14. y′′ + y = cot2 x. yh = c1 sinx+ c2 cosx.

t1 =

∫− cosx

−1cot2 xdx =

∫cosx cos2 x

sin2 x=

∫cosx(1− sin2 x)

sin2 x= − 1

sinx+− sinx

t2 =

∫sinx

−1cot2 x = −

∫cos2 x

sinx= −

∫1− sin2 x

sinx= ln | cscx+ cotx|+− cosx

yp = −1− sin2 x+ cosx ln | cscx+ cotx| − cos2 x = −2 + cosx ln | cscx+ cotx|

y = c1 sinx+ c2 cosx− 2 + cosx ln | cscx+ cotx|

Exercise 15. y′′ − y = 21+ex

yh = c1ex + c2e

−x =⇒W (x) = −exe−x − exe−x = −2

t1 = −∫e−x 2

1+ex

−2=

∫e−x

1 + ex=

=

∫1

ex− 1

1 + ex= −e−x −

∫1

1 + ex= −e−x +

∫−e−x

e−x + 1= −e−x + ln (1 + e−x)

t2 =

∫ex 2

1+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

155

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

=1

3

∫1

1 + et=−1

3ln (1 + e−x)

t2 =

∫ et(

et

1+et

)−3e−t

=1

−3

∫e3t

1 + etu=et−−−→ −1

3

∫u2du

1 + u=

=−1

3

∫u+

−uu+ 1

=−1

3

∫u+−1 +

1

u+ 1=−1

3(1

2u2 − u+ lnu+ 1)

=−1

6e2x +

ex

3+−1

3ln (ex + 1)

y1 =−1

3ex ln (1 + e−x) +

−1

6+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

∫ x1−te3tdt =

(−te3t

3 + e3t

9

)∣∣∣x1

= −3xe3x+e3x

9 + 29e

3

a < 1 < 2 < x∫ 2

1−te3tdt = −6e6+e6

9 + 2e3

9 = −5e6

9 + 2e3

9

1 < a < x < 2∫ xa−te3tdt =

(−te3t

3 + e3t9)∣∣∣xa

= −xe3x3 + e3x

9 + ae3a

3 − e3a

9

t2(x) =

∫e−3tf(t)

e−6t=

∫e3tf(t) =

∫ x

a

e3t =1

3

(e3x − e3a

)y1 = e−3x

(−3xe3x + e3x

9+ C

)+x

3

y = c1e−3x + c2xe

−3x +1

9when 1 ≤ x ≤ 2; otherwise y = yh

Exercise 18. Start from y′′ − k2y = R(x). Suppose L(yp) = y′′p − k2yp = R(x).

yh = c1 sinh (kx) + c2 cosh (kx); L(yh) = 0

So consider L(yp + yh) = R(x); yp + yh = y1 is a nother particular solution.

The key to this problem is to apply the integration directly on the ODE itself, not to go the other way around by differenti-ating the supposed particular solution.∫ x

0dt 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) =

1

k

∫ x

0

dtR(t) sinh (k(x− t))dt

Now note that L(yh) = 0, so applying∫ x

0dt sinh (k(x− t)) results in 0 still.

156

With yp(x) = 1k

∫ x0dtR(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) =1

k

∫ x

0

dtR(t) sinh (k(x− t))dt

Now for y′′ − 9y = e3x,

y1(x) =1

3

∫ x

0

dt(e3t) sinh (3(x− t))dt =1

6

∫ x

0

dte3t

(e3x−3t − e−3x+3t

2

)=

1

6

∫ x

0

dt(e3x − e−3x+6t)

=1

6

(e3xx− e−3x 1

6e6t

∣∣∣∣x0

)=

1

6xe3x − e−3x e

6x − 1

36=

1

6(xe3x − e3x

6+e−3x

6)

y′1 =1

6(e3x + 3xe3x − 3

2e3x − 3

2e−3x)

y′′1 =1

6

(3e3x + 3e3x + 9xe3x − 9

2e3x +

9

2e−3x

)=

1

4e3x +

3

4e−3x +

3

2xe3x

y′′1 − 9y1 =1

2e3x +

1

2e−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 satisfies 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.

∫ x0dt 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

=⇒ y(x) =1

k

∫ 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 =1

k

∫ x

0

dtR(t) sin k(x− t)

Now for y′′ + 9y = sin 3x, then k = 3,

y1 =1

3

∫ 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)

2dt =

−1

12(c(6x)− 1)∫ x

0

s2 =

∫ x

0

1− cos (6t)

2=x

2− sin (6t)

12

∣∣∣∣x0

=x

2− sin (6x)

12

y1 =1

3

(sin 3x

−1

12(cos (6x)− 1)− cos 3x

(x

2− sin (6x)

12

))=

sin 3x

18− x cos 3x

6157

It could be shown with some computation that this particular solution satisfies the ODE without having to add or subtractparts of a homogeneous solution.

Exercise 20. y′′ + y = sinx

yh = c1 sinx+ c2 cosx =⇒W (x) = −s2 − c2 = −1

t1 = −∫

cs

−1=− cos 2x

4; t2 =

∫ss

−1= −

∫1− cos 2x

2= −

(x

2− sin 2x

4

)yp = − sinx cos 2x

4+

(sin 2x− 2x

4

)cosx =

sinx cos2 x+ sin3 x− 2x cosx

4

y = c1 sinx+ c2 cosx+sinx cos2 x+ sin3 x− 2x cosx

4

Exercise 21. y′′ + y = cosx

yh = c1 sinx+ c2 cosx = c1S + c2C W (x) = −1

t1 =

∫−CC−1

=

∫1 + cos 2x

2=x+ sin 2x

2

2; t2 =

∫SC

−1=

cos 2x

4

yp =x sinx

2+

sin 2x sinx

4+

cosx cos 2x

4

=⇒ y =x sinx

2+

sin 2x sinx

4+

cosx cos 2x

4+ c1 sinx+ c2 cosx

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 cosx)

−2=

3

2

∫x cosx cos (2x) =

3

2

∫xc(1− 2s2) =

3

2

∫xc− 3

∫xcs2 =

=3

2(xs+ c)− 3

∫x

(c3

3

)′=

3

2(xs+ c)− (xs3 −

∫s3) =

3

2(xs+ c)− xs3 +

∫s(1− c2) =

=3

2(xs+ c)− xs3 +−c+

1

3c3 =

3

2xs+

c

2− xs3 +

1

3c3

t2 =

∫sin (2x)(3x cosx)

−2= −3

∫xsc2 =

∫x(c3)′ = xc3 −

∫c3 = xc3 −

∫c(1− s2) =

= xc3 − s+1

3s3

yp =

(3

2xs+

c

2− xs3 +

1

3c3)

(2sc) + (xc3 − s+s3

3)(1− 2s2) = (lots of algebra) =

= xc2 +2

3s

=⇒ y = c1 sin 2x+ c2 cos 2x+ x sinx− 2

3cosx

Remember, persistence is key to work through the algebra, quickly.

Exercise 23. y′′ + 4y = 3x sinx. 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 =⇒(−2

3c

)′′+ 4

(−2

3c

)= −2c

=⇒ yh = xs− 2

3c

y = x sinx− 2

3cosx+ c1 sin 2x+ c2 cos 2x

158

Exercise 24. y′′ − 3y′ = 2e2x sinx 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 sinx+ cosx)

5=⇒ y = c1 sin

√3x+ c2 cos

√3x+

e2x(3 sinx+ cosx)

−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 sinx+ c2 cosx+e2x cos (3x)− 3e2x sin (3x)

−40

8.19 Exercises - Examples of physical problems leading to linear second-order equations with constant coefficients.In exercises 1-5, a partcile is assumed to be moving in simple harmonic motion, according to the equation y = C sin (kx+ α).The velocity of the particle is defined to be the derivative y′. 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=

1

π=⇒ k = 2

y(x = 0) = C sinα

y′(x = 0) = C cosα=⇒ y(x = 0)

y′(x = 0)=

1

ktanα =

1

2

α =π

4and C = 2

√2

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 =−αk

) = 0 =⇒ y′(x =−αk

) = 140π

Exercise 3.

y = A cos (mx+ β)

y = A cos (mx+ β) = A cosβ cos (mx)−A sinβ sin (mx)

y = C sin kx+ α = C cosα sin kx+ C sinα cos kx=⇒ k = m (since x is arbitrary )

−A sinβ = C cosα

A cosβ = C sinα=⇒ tanα = ± cotβ = ∓ tan

(π2− 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)

159

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 =√v2

0 + y20

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 =

−12

1=⇒ k = 2

√3

y′(0)

y(0)=−kC sin (α)

C cos (α)=

2

1= 2 = −k tan (α) =⇒ α =

−π6

Exercise 7. k = 2π3 y = −C sin (kx)

y = −C sin2πx

3; C > 0

Exercise 8. Let’s first solve the homogenous equation.

y′′ + y = 0

yh = C1 sinx+ C2 cosx

W (x) = −S2 − C2 = 1

t1 =

∫ x

0

− cos t(1)

−1= sinx

t2 =

∫ x

0

(sin t)(1)

−1= (cosx− 1)

for 0 ≤ x ≤ 2π otherwise, for x > 2π, t1 = 0, t2 = 0

y1 = sin2 x+ cos2 x+ (1− cosx)

y′(x) = C1 cosx+ sinx

y(0) = 0 = c2

y′(0) = c1 = 1

y = sinx+ (1− cosx)

=⇒ I(t) = sin t+ (1− cos t) 0 ≤ t ≤ 2π

Exercise 9.

(1) Consider large t. Then I(t) = F (t) +A sin (ωt+ α)→ A sin (ωt+ α)

I = AS(ωt+ α) = A(S(ωt)C(α) + C(ωt)S(α))

I ′ = 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 fixed 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

160

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)

tanα =−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) + 1ddω−−→ 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, and

that it falls along a straight line toward the center of the earth. The rocket fuel is consumed at a constant rate of k pounds 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

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

klnM0

Exercise 11.

Mv′R =−kcg

v′R =−kcg

(1

M0 − ktg

)

vR = c ln (M0 −kt

g)

=⇒ yR = cg

k

((kt

g−M0

)ln

(M0 −

kt

g

)− kt

g

)+M0cg lnM0

k

M0g = w =⇒ yR = cg

k

(kt− wg

ln

(w − ktg

)− kt

g

)+wc

klnw

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+1

2gt2 + c

((t− M0g

k

)ln

(1− kt

M0g

)− t)

161

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)

(−wk

)= − ln (1− kt

w)

vR =v0

1− ktw

=⇒ x(t) = v0

(−wk

)ln (1− kt

w) =−v0w

kln (1− kt

w)

8.22 Exercises - Remarks concerning nonlinear differential equations, Integral curves and direction fields.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′ +

1

y2y′ = 0

y′ =−2xy

1 + y2 − x2

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′ = − tanxy

Exercise 10. arctan y + arcsinx = C

1

1 + y2y′ +

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

162

(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′ +

1

y2y′ = 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 − xy +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)

8.24 Exercises - First-order separate equations.

Exercise 1. y′ = x3/y2

13y

3 = 14x

4 + C =⇒ y3 = 34x

4 + C

Exercise 2. tanx cos y = −y′ tan y

ln | cosx| = 1cos y

Exercise 3. (x+ 1)y′ + y2 = 0

1y = ln (x+ 1) + c

Exercise 4. y′ = (y − 1)(y − 2)(1

y − 2+−1

y − 1

)y′ = 1 =⇒ ln (y − 2)− ln (y − 1) = x

y − 2

y − 1= ex

Exercise 5. y√

1− x2y′ = x

12y

2 = −√

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

163

Exercise 7. (1− x2)1/2y′ + 1 + y2 = 0

arctan y = arccosx+ C

Exercise 8. xy(1 + x2)y′ − (1 + y2) = 0

1

2ln (1 + y2) =

∫ (1

x− x

1 + x2

)+ C = lnx− 1

2 ln |1 + x2|

y2 = k

(x√

1 + x2

)2

Exercise 9. (x2 − 4)y′ = y

ln y =−1

2arctanh

(x2

)since

∫1

x2 − 4dx =

1

4

∫dx(

x2

)2 − 1=

=1

2

∫du

u2 − 1( where u =

x

2)

(tanh (u))′ =cosh2 u− sinh2 u

cosh2 u= 1− tanh2 u

=⇒ y = k exp (−1

2arctanh

(x2

))

Exercise 10. xyy′ = 1 + x2 + y2 + x2y2

12 ln (1 + y2) = lnx+ 1

2x2 + C =⇒ y2 = kx2ex

2

− 1

Exercise 11. yy′ = ex+2y sinx

ye−2y

−2− e−2y

4=ex sinx− ex cosx

2+ C

(2y + 1)e−2y = −2ex(sinx− cosx) + C

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 =2

e

f(x) =2

eex

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) = 0164

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))ddx−−→ f(x) = kf ′(x)

=⇒ f(x) = Ce1kx; C > 0

Exercise 19.∫ xa

(f(t))dt = k(f(x) + f(a)) =⇒ f(x) = Cexk

kCexk − kCe ak = kCe

xk + kCea/k =⇒ 2kCea/k = 0 =⇒ C =

f = 0

Exercise 20.∫ xaf(t)dt = kf(x)f(a); f(x) = kf ′(x)f(a)

1

kf(a)=f ′(x)

f(x)=⇒ ln f(x) =

(1

kf(a)

)x+ C; =⇒ f(x) = C exp

(x

kf(a)

)∫ x

a

f(t)dt =(kf(a)Ce

tkf(a)

)∣∣∣xa

= kf(a)Cex

kf(a) − kf(a)Cea

kf(a) = kCex

kf(a)Cea

kf(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

8.26 Exercises - Homogeneous first-order equations.Exercise 1. f(tx, ty) = f(x, y) homogeneity (or homogeneity of zeroth order).

y′ = f(x, y) =(xv

)′=v − xv′

v2= f(x,

x

v) = f(1,

1

v)

v − v2f(1,1

v) = xv′ =⇒ lnx =

∫dv

v − v2f(1, 1v )

Exercise 2. y′ = −xy =⇒ 1

2y2 = − 1

2x2 + C =⇒ y2 = −x2 + C

Exercise 3. y′ = 1 + yx

yx = v =⇒ y′ = v + xv′ = 1 + v =⇒ v = lnx

y = x(lnx+ C)

Exercise 4. y′ = x2+2y2

xy

165

y′ =x2 + 2y2

xy=x

y+

2y

x

v =y

x=⇒ v + xv′ =

1

v+ 2v =⇒ v′

1v + v

=1

x

1

2ln |1 + v2| = lnx+ 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′ =

1

x

1

2

(1

v+−3v

1 + v2

)v′ =

1

x=⇒ 1

2ln v +

−3

2ln (1 + v2) = lnx+ 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=⇒

(1

v+− 3v

1 + v2

)v′ =

2

x=⇒ ln v +−3

2ln |1 + v2| = 2 lnx+ C

v

(1 + v2)3/2= Cx2 =⇒ y2/x4

(x2 + y2)3= Cx4

=⇒ y2 = C(x2 + y2)3

Exercise 6. xy′ = y −√x2 + y2

=⇒ y′ =y

x=

√1 +

(yx

)2

v =y

xvx = y

v′x+ v = v −√

1 + v2 =⇒

v′x = −√

1 + v2

−v′√1 + v2

=1

x

=⇒ ln (v +√

1 + v2) = lnx+ 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 = 0166

x2y′ = −2y2 − xy =⇒ y′ =−2y2

x2− y

xv =

y

x=⇒

v′x+ v = −2v2 − vv′x = −2(v2 + v)

v′

v(v + 1)=−2

x=

∫v′(

1

v− 1

v + 1

)= −2 lnx+ C =⇒ v

v + 1=C

x2

y =−CxC − x2

Exercise 8. y2 + (x2 − xy + y2)y′ = 0 (yx

)2

+

(1− y

x+(yx

)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 + 1

v(1 + v2)v′ =

−1

x=

(1

v+−1

v2 + 1

)v′ = − 1

x

=⇒ ln v − arctan v = − lnx+ 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)=(yx

)( 1 + yx + 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 +3

v+

1

v2) =−2

x=⇒ v + 3 ln v +

−1

v= −2 lnx+ C

y

x+ 3 ln y − x

y= lnx+ C

Exercise 10. y′ = yx + sin y

x

y

x= x

y′ = v + v′x=⇒

v + v′x = v + sin v

v′

sin v=

1

x

− ln csc v + cot v = lnx+ C =⇒ csc v + cot v =K

x

Exercise 11. x(y + 4x)y′ + y(x+ 4y) = 0

y′ =−y(x+ 4y)

x(y + 4x)=− yx (1 + 4y

x )yx + 4

v= yx−−−→ v + xv′ =

−v(1 + 4v)

v + 4

xv′ =−5v(1 + v)

v + 4=⇒ −5

x=

v + 4

v(1 + v)v′ =

(4

v+−3

1 + v

)v′

∫−→ 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 first-order equations.Exercise 1.

2x+ 3y = C y′ = −2

3g′ =

3

2=⇒ g − 3

2x = C

Exercise 2.

xy = Cd/dx−−−→ y + xy′ = 0 =⇒ y′ = −y/x x 6= 0 =⇒ g′ = x/g =⇒ 1

2g2 =

1

2x2 + C

Exercise 3. x2 + y2 + 2Cy = 1167

x+ yy′ + Cy′ = 0 =⇒ y′(y + C) = −x

y′ =−xy + C

=−x

y + 1−x2−y22y

=−2xy

y2 − x2 + 1

orthogonalcurves−−−−−−−−−−−→ y′ =y2 − x2 + 1

2xy=

(1

2x

)y +

1

2

(1

x− x)y−1

Recognize that this is a Ricatti equation and we know how to solve them.

y′ +−1

2xy = y−1

(−x2

+1

2x

)n = −1

k = 1− n = 1− (−1) = 2

v = yk = y2 v′ + 2

(−1

2x

)v =

2

2

(1

x− x)

A(x) =

∫ x

a

P (t)dt =

∫ x

a

−1

t= ln

a

x

∫ x

a

QeA =

∫ x

a

(1

t− t)a

t=−ax

+ 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

1

2y2 =

x2

4+ C

2y2 − x2 = C

Exercise 6. y = Ce−2x

e2xy = C =⇒ 2e2xy + e2xy′ = 0

y′ = −2yinvert−−−→ y′ =

1

2y

=⇒ y2 = x+ C

Exercise 7. x2 − y2 = C

2x− 2yy′ = 0

y′ =x

y=⇒ y′ =

−yx

=⇒ ln y = − lnx+ C =⇒ y =C

x

Exercise 8. y secx = C

y′ secx+ y tanx secx = 0

y′ = −y tanx(invert)−−−−→ y′ =

1

y tanx

1

2y2 = ln | sinx|+ C

168

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

(1

2x

)v = 2

(x2−1

2x

)= v′ + 1

xv = x2−1x = x− 1/x.

A(x) =

∫ x

a

P (t)dt =

∫ x

a

1

t= ln

x

a∫(t− 1

t) exp

(lnt

a

)dt =

∫ (t2

a− 1

a

)dt =

( 13 t

3

a− t

a

)∣∣∣∣xa

e− ln xa =

a

x

=⇒ y2 = v =13x

3 − x− 13a

3 + 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 − xy +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=

1

x−B=⇒ y = C(x−B)−B

Exercise 11. With (0, Y ) = Q the point that moves up wards along the positive y-axis and

P = (x, y) being the point P that pursues Q,

y′ = Y−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

kY = x.

y′ =

(1k

)x− yx

= f(x, y)

f(x, y) is homogeneous of zero order =⇒ y = vx (try this substitution)

y′ = v′x+ v =1

k− v =⇒ v′

1k − 2v

=1

x

=⇒ −1

2ln

(1

k− 2v

)= lnx+ C =⇒ y =

x

2k− 1

2C2x

(1,0)−−−→ y =x

2k=

1

2kxk =

1

2y = x− 1

x169

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 lnx = 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

2lnx+ C

y = Cxn/2 of 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′x

πv2 − 3v= 1 =⇒ v′

π(v2 − 3vπ )

=1

x=

1

3

(1

v − 3π

− 1

v

)v′

ln (v − 3

π)− ln v = 3 lnx+ C

ln

(v − 3/π

v

)= 3 lnx+ C

vx− 3

πxCvx4

y − 3

πx = Cyx3 =⇒ y =

3x/π

1 + x3

2

Exercise 16. A =∫ a

0f ; B =

∫ 1

af

A−B =∫ a

0f +

∫ a1f = 2f(a) + 3a+ b

d/dx−−−→ 2f(a) = 2f ′(a) + 3 =⇒ 1 = f ′(a)

f(a)− 32

So then

a+ C = ln (y − 3

2); f(a) = Cea +

3

2

f(1) = 0 = Ce1 +3

2; =⇒ C = − 3

2e

f(x) =−3

2ex−1 +

3

2170

To find b,

2

(−3

2ea−1 +

3

2a

)+

3

2e−1 +

3

2− 3

2= 2f(a) + 3a+ b =

= 2

(−3

2ea−1 +

3

2

)+ 3a+ b

=⇒ b =3

2e−1 − 3

Exercise 17.

A(x) =

∫ x

0

(f(t)−

((y(x)− 1

x

)t+ 1

))dt = x3

=

∫ x

0

f +−(y(x)− 1

x

)1

2x2 − x = x3 =

=

∫ x

0

f +−(y(x)− 1

2

)x− x = x3

d/dx−−−→ f(x) +−1

2(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 x

Exercise 18. Assuming no friction at the orifice and energy conservation.

mgh =1

2mv2

f

(imagine how the top layer of water is now at the bottom of the tank (final “potential energy configurations”))Vf =

√2gh (how fast water is rushing out)

A0 = cross-sectional area of the orifice.

dV

dt= A

dh

dt= −c

√2ghA0

2h1/2∣∣∣hfhi

= −c√

2gA0

At

=⇒ T =

√2A

c√gA0

(√hf −

√hi

)= 59.6sec

Note that we included the discharge coefficient C = 0.6.

Exercise 19. dVdt = −c√

2ghA0 + γ0 = Adhdt = −κh1/2 + γ0. κ = c

√2gA0.

171

Adh

γ0 − κh1/2= dt = (A/γ0)

dh

1− κγ0h1/2(

ln(

1− ah1/2))′

=1

1− ah1/2

(−a

2

1

h1/2

)(where a =

κ

γ0)

(h1/2)′ =1

2h1/2

(1− ah1/2

1− ah1/2

)=

(12 −

a2h

1/2)

h1/2(1− ah1/2

)=⇒

∫(A/γ0)

dh

1− κγ0h1/2

= − (A/γ0)

(2γ2

0

κ2ln

(1− κ

γ0h1/2

)+

2γ0h1/2

κ

)∣∣∣∣hfhi

=

= T

=⇒ exp

(−γ0

At− 2γ0

κ

(h

1/2f − h1/2

i

)) κ2

2γ20

=1− κ

γ0h

1/2f

1− κγ0h

1/2i

t→∞−−−→ hf =γ2

0

κ2=

(100in3/s)2

c2(2)(32ft/s2)(5/3in2)2(12in/1ft)= (25/24)

2

Exercise 20.

V0 =1

3πR2

0H0

V (h) = V0 −1

3πh

(hR0

H0

)2

= V0 −1

3πR2

0

H20

h3 = V0 − αh3

mg(H0 − h) =1

2mv2

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 +

2

5y5/2

)∣∣∣∣hfhi

=

= −H0

(2

(1− h

H0

)1/2

− 4

3

(1− h

H0

)3/2

+2

5

(1− h

H0

)5/2)∣∣∣∣∣

hf

hi

=

=cA0

√2gH0/H

20

3(

13π

R20

H20

) T

For hi = 0, hf = H ,

H0(2(1)− 4/3(1) + 2/5) = H0(16/15) = cA0

√2gH0T/(πR

20);T =

1615

√H0πR

20

cA0

√2g

=2

9

πR20

√H0

A0

Exercise 21. m2x−m+ (1− x) = 0 =⇒ (m2 − 1)x+ 1−m = 0, =⇒ m = 1

Exercise 24. Given f s.t. 2f ′(x) = f(

1x

)if x > 0, f(1) = 2 and x2y′′ + axy′ + by = 0

172

(1)

2f ′′(x) =d

dxf

(1

x

)= (f ′)

(−1

x

)=⇒ x2y′′ =

−1

2f ′ =

−1

4f

(1

x

)a = 0; b =

1

4

(2)f(x) = Cxn

f ′ = nCxn−1

f ′′ = n(n− 1)Cxn−2

=⇒ n(n− 1)Cxn =−1

4Cxn

n2 − n+1

4=⇒ n =

1

2

Exercise 28. Choose the units for time to be in days first - we can convert into years later.

If no one died from accidental death, then the population will grow by e. That means, with x = x(t) being the population 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=

1

365x− 1

100x =

100− 365

36500x =

−265

36500

=⇒ x = Ce−26536500 t

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 =365

100

365

2.65(− exp (−2.65t) + 1)

Exercise 29. For constant gravity, ∆K = −∆U

=⇒ −(0−mgh) = 12mv

2f

vf =√

2gh = (6.37× 108 cm)(

1m2.54 cm

) (1 ft12 in

)= 6.93 mi

sec = 24940 mihr

The constant energy formula could also be obtained by considering

F = 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).

=⇒ 23x− 20/9 < f(x)

173

(3) Since f ′(x) ≥ 23 for each x ≥ 10

3 , then f →∞ for x→∞ (otherwise, f would have to decrease somewhere, whichwould contradict the given fact about f ). Rewrite the DE for y′ to be

y′ =2y2 + x

3y2 + 5=⇒ (3y2 + 5)y′ = 2y2 + x =⇒ (3 +

5

y2)y′ = 2 +

x

y2

Consider

y′ =2y2 + x

3y2 + 5=

2 + xy2

3 + 5y2

specifically, 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 infinity faster than linear x. Thus limx→∞ y′ = 23 . So then (3 + 5

y2 )y′x→∞−−−−→ (3 + 0) 2

3 = 2 =

2 + xy2 .

0 =x

y2

(4) y′ =2+ x

y2

3+ 5y2

x→∞−−−−→ 23 . =⇒ y = 2

3x or yx = 23 .

Exercise 31. Given a function f which satisfies 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−xx 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 first and second order derivatives aresimply a combination of infinitely many terms containing powers of x.

Then suppose f =∑∞j=0 ajx

j .

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 =1

2i.e. f =

1

2x2 +

∞∑j=3

ajxj

A =1

2in order for f(x) ≤ Ax2

174

9.6 Exercises - Historical introduction, Definitions and field properties, The complex numbers as an extension ofthe real numbers, The imaginary unit i, Geometric interpretation. Modulus and argument. Exercise 6. Let f be a

polynomial with real coefficients.

(1) Since z1z2

(zn+11 ) = zn1 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 both

Ax. 9 0 /∈ R+

x < y means y − x positive.

Suppose i positive: i(i) = −1 but −1 is not positive.Suppose −i is positive. −i(−i) = −1 but −1 is not positive.i is neither positive nor negative so Ax. 8 is not satisfied.

Exercise 8. Ax. 8 , Ax. 9 are satisfied.

(a+ ib)(c+ id) = (ac− bd, ad+ bc) so Ax. 7 might not be satisfied.

Exercise 9. Ax. 7, Ax. 8 , Ax. 9 are trivially satisfied (all are positive).

Exercise 10. x > y x > y is well defined. Ax. 8 is satisfied.

For Ax. 7, ( 32 , 1), (1, 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.

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|2

If 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 Imz175

9.10 Exercises - Complex exponentials, Complex-valued functions, Examples of differentiation and integration for-mulas.Exercise 7.

(1)

if m 6= n ,

∫ 2π

0

eix(n−m)dx =eix(n−m)

i(n−m)

∣∣∣∣2π0

=1− 1

i(n−m)= 0

if m = n ,

∫ 2π

0

eix(0)dx = 2π

(2)

∫ 2π

0

einxe−imxdx =

∫ 2π

0

(cosnx+ i sinnx)(cosmx− i sinmx) =

=

∫ 2π

0

cosnx cosmx+ sinnx sinmx+ i(sinnx cosmx− sinmx cosnx)∫ 2π

0

e−inxe−imxdx =

∫ 2π

0

(cosnx− i sinnx)(cosmx− i sinmx) =

=

∫ 2π

0

cosnx cosmx− sinnx sinmx+ i(− sinnx cosmx− sinmx cosnx)

Summing the two equations above

0 =

∫ 2π

0

2 cosnx cosmx+ 2

∫ 2π

0

i− sinmx cosnx

=⇒∫ 2π

0

cosnx cosmx = 0,

∫ 2π

0

sinmx cosnx = 0∫ 2π

0

einxe−imxdx =

∫ 2π

0

(cosnx+ i sinnx)(cosmx− i sinmx) =

=

∫ 2π

0

cosnx cosmx+ sinnx sinmx+ i(sinnx cosmx− sinmx cosnx)∫ 2π

0

einxeimxdx =

∫ 2π

0

(cosnx+ i sinnx)(cosmx+ i sinmx) =

=

∫ 2π

0

cosnx cosmx− sinnx sinmx+ i(sinnx cosmx+ sinmx cosnx)

Subtract the two equations above∫ 2π

0

sinnx sinmx− i∫ 2π

0

sinmx cosnx = 0

=⇒∫ 2π

0

sinnx sinmx = 0

176

∫ 2π

0

einxe−inx = 2π =

∫ 2π

0

(cosnx+ i sinnx)(cosnx− i sinnx) =

=

∫ 2π

0

cos2 nx+ sin2 nx∫ 2π

0

einxeinx = 0 =

∫ 2π

0

cos2 nx− sin2 nx+ i(2 cosnx sinnx)

=⇒∫ 2π

0

cos2 nx− sin2 nx = 0

∫ 2π

0

cosnx sinnx = 0

Summing the two results above, we obtain

2π =

∫ 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) = cosu+ v + i sinu+ v =

= (cosu+ i sinu)(cos v + i sin v) = cosu cos v − sinu sin v + i(cos v sinu+ cosu sin v)

=⇒ sinu+ v = cos v sinu+ cosu sin v

=⇒ cosu+ v = cosu cos v − sinu 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 = (cosx+ i sinx)e−y

e−iz = e−i(x+iy) = e−ixey = (cosx− i sinx)ey

Thus it is clear, by mentally adding and subtracting the above results that

=⇒cos z = cosx cosh y − i sinx sinh y

sin z = i cosx sinh y + sinx cosh y

Exercise 10.

(1) Log(−1) = iπ log (i) = ln 1 + iπ2 = iπ2(2) Log(z1z2) = Log(|z1||z2|ei(θ1+θ2)) = ln |z1||z2|+ i(θ1 + θ2 + 2nπ) = Logz1 + Logz2 + i2πn

177

(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.

(1)1i = eiLog1 = ei(i2πn) = e−2πn = 1 if n = 0

ii = eiLogi = ei(iπ2 +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 zw2 ) = 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 tanα =

aω

b− ω2

=⇒ <y =c√

(b− ω2)2 + (aω)2cos (ωx− α)

Exercise 15.

=(y) =c√

(b− ω2)2 + (aω)2sin (ωx+ α)

=⇒ A =c√

(b− ω2)2 + (aω)2; − tanα =

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− 1

n2 + n=−2n −

1n2

1 + 1n

n→∞−−−−→ 0

Exercise 2. Converges to −1.

f(n) =n3 − (n3 + n+ n2 + 1)

(n+ 1)n=n2 − n− 1

n(n+ 1)=−1− 1

n −1n2

1 + 1n

n→∞−−−−→ −1

Exercise 3. Diverges since

|cosnπ2− L| ≥

∣∣∣∣∣∣1 cosnπ

2

∣∣∣− |L|∣∣∣ ≥ |1− |L||Choosing ε1 = |1−|L||

2 , | cos nπ2 − L| > ε1 for n = 4m.178

Exercise 4. f(n) = 15 + 3

5n −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 .

Suppose ε = 3N .

So for n > N, 1N > 1

n , n ≥ N = N(ε) = 3/ε.

|f(n)| =∣∣∣∣1 + (−1)n

n

∣∣∣∣ ≤ 2

n<

3

n<

3

N= 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| − 1

n

∣∣∣∣ =

∣∣∣∣1 + (−1)n

2

∣∣∣∣∣∣∣∣ ≥≥∣∣∣∣∣∣∣∣|L| − 1

n

∣∣∣∣− 1

2

∣∣∣∣Thus, consider ∣∣∣∣∣∣∣∣|L| − 1

n

∣∣∣∣− 1

2

∣∣∣∣ > ∣∣∣∣∣∣∣∣ 1

N− |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 sinn!n+1 .

|f(n)| =∣∣∣∣n2/3 sin (n!)

n+ 1

∣∣∣∣ =

∣∣∣∣ sin (n!)

n1/3 + n−2/3

∣∣∣∣ ≤ ∣∣∣∣ 1

n1/3

∣∣∣∣Thus, for n > N , N(ε) = 1

ε3 , |f(n)| < ε.

Exercise 12. Converges, since

f(n)− 1

3=

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)

)=

=

(5

3

(−2)n

3n+1 + (−2)n+1

)∣∣∣∣f(n)− 1

3

∣∣∣∣ =5

3

∣∣∣∣ (−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

)nFor n > N , consider ε =

(23

)−N, i.e. N = − ln ε

ln 2/3 = N(ε). Thus179

L =1

3;

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√N

and for n > N , 12√n< 1

2√N

= ε.Thus f(n) converges to 0.

Exercise 14.

f(n) = nan = n expn ln a =n

expn 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 > 0

Exercise 16. limn→∞ f(n) = 0

Exercise 17. limn→∞ f(n) = e2.

Exercise 18. ∣∣∣∣1 +n

n+ 1cos

nπ

2− L

∣∣∣∣ ≥ ∣∣∣∣∣∣∣∣1 +n

n+ 1cos

nπ

2

∣∣∣∣− |L|∣∣∣∣ =

∣∣∣∣∣∣∣∣1 +−nn+ 1

∣∣∣∣− |L|∣∣∣∣ =

=

∣∣∣∣ 1

n+ 1− |L|

∣∣∣∣ > |L|Choose ε0 = |L|

2 . For any N , for n > N , |f(n)− L| > ε0.

Exercise 19.

1 +i

2=

√5

2eiα

1 +

(1

2

)2

=5

4

tanα =1

2

(√5

2eiα

)−n,

(2√5

)ne−niα

limn→∞

(2√5

)ne−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 = 1n .

|an| =∣∣∣∣ 1n∣∣∣∣ < 1

N; N(ε) =

1

ε

ε = 1, 0.1, 0.01, 0.001, 0.0001

N = 1, 10, 100, 1000, 10000

Exercise 24. |an − 1| =∣∣∣n+1−nn+1

∣∣∣ = 1n+1 <

1n

180

N(ε) = 1ε = 1, 10, 100, 1000, 10000.

Exercise 25. |an| = 1n

N(ε) = 1, 10, 100, 1000, 10000.

Exercise 26. |an| =∣∣ 1n!

∣∣ ≤ 1expn lnn <

1expn

For n > N , 1expN = 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| =∣∣ 9

10

∣∣n =(

910

)n= en ln 9/10

N(ε) =ln ε

ln(

910

) =− ln 1/ε

ln (9/10)=

= 1, 21, 43, 65, 87

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,

limn→∞

(an −A)(an +A) = limn→∞

(an −A) limn→∞

(an +A) = 0(2A) = 0

2anbn = (an + bn)2 − a2n − b2n

2 limn→∞

anbn = limn→∞

(an + bn)2 − limn→∞

a2n − lim

n→∞b2n =

(limn→∞

(an + bn))2

−A2 −B2 = 2AB

=⇒ limn→∞

anbn = AB

Exercise 33.(αn

)= α(α−1)(α−2)...(α−n+1)

n!

181

(1) (− 1

2

1!

)=−1/2

1(− 1

2

2!

)=

(− 1

2

) (−32

)2!

=5

8(− 1

2

3!

)=

(− 1

2

) (−32

) (−52

)3!

= − 5

16

(− 1

2

4

)=

(− 1

2

) (−32

) (−52

) (−72

)4!

=35

128(− 1

2

5

)=

(− 1

2

) (−32

) (−52

) (−72

) (−92

)5!

=−63

256

(2) an = (−1)n(−1

2n

).

a1 = 12 > 0. a2 = 3

8 > 0

an+1 = (−1)n+1

(− 1

2

n+ 1

)= (−1)n+1

(α

n

)(− 12 − (n+ 1) + 1

n+ 1

)=an(−1)

(−12 − n

)n+ 1

=

=an(n+ 1/2)

n+ 1> 0

an+1 =

(n+ 1/2

n+ 1

)an < an

Exercise 34.

(1)

tn − sn =1

n

n∑k=1

f

(k

n

)− 1

n

n−1∑k=0

f

(k

n

)=

1

n

(f(1) +

n−1∑k=1

f

(k

n

)−

(f(0) +

1

n

n−1∑k=1

f

(k

n

)))=

=1

n(f(1)− f(0))

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 definition of integral)

0 ≤∫ 1

0

f(x)dx− sn ≤ tn − sn =1

n(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 ≤∫ baf(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→∞1n

∑nk=1

(kn

)2=∫ 1

0t2dt = 1

3

(2) limn→∞∑nk=1

1n+k = limn→∞

1n

∑nk=1

11+ k

n

=∫ 1

01xdx = ln 2

182

(3) limn→∞1n

∑nk=1

1

1+( kn )2 =

∫ 1

01

1+x2 dx = arctanx|10 =π

4

(4) limn→∞∑nk=1

1√n2+k2

= limn→∞1n

∑nk=1

1√1+( kn )

2=∫ 1

01√

1+x2dx = ln (x+

√1 + x2)

∣∣10

= ln (1 +√

2)

(5) limn→∞∑nk=1

1n sin kπ

n = limn→∞1n

∑nk=1 sin kπ

n =∫ 1

0sinπxdx =

(− cosπxπ

)∣∣10

= −(−1−1)π =

2

π

(6) limn→∞∑nk=1

1n sin2 kπ

n =∫ n

0sin2 xπ =

1

2

10.9 Exercises - Infinite 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 −

1/22n+1 = 1

2

Exercise 2.∑∞n=1

23n−1 = 2

∑∞n=0

13n = 2 1

1−1/3 = 3

Exercise 3.∑∞n=2

1n2−1 =

∑∞n=2

1/2n−1 −

1/2n+1 =

∑∞n=2

(1/2n−1 −

1/2n

)+−

(1/2n+1 −

1/2n

)= 1

2 + 14 = 3

4

Exercise 4.∑∞n=1

2n+3n

6n =∑∞n=1

(13

)n+∑∞n=1

(12

)n= 1/3

1−1/3 + 1/21−1/2 = 1

2 + 1 = 32

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)+

1

2

(−1

6

)=

=

∞∑n=1

1

1 + 2− 1

n+ 3− 1

12=

1

3− 1

12=

1

4

Exercise 7.∑∞n=1

2n+1n2(n+1)2 =

∑∞n=1

1n2 − 1

(n+1)2 = 1

Exercise 8.∑∞n=1

2n+n2+n2n+1n(n+1) = 1 =

∑∞n=1

12n(n+1) + 1

2n+1 = 12

∑∞n=1

1n −

1n+1 + 1/4

1−1/2 = 12 + 1

2 = 1

Exercise 9.∑∞n=1

(−1)n−1(2n+1)n(n+1) =

∑∞n=1(−1)n−1

(1n + 1

n+1

).

(−1)n−1

(1

n+

1

n+ 1

)+ (−1)n

(1

n+ 1+

1

n+ 2

)= (−1)n

(−1

n+

1

n+ 2

)∞∑j=1

(−1

2j − 1+

1

2j + 1

)=

1

2

∞∑j=1

(−1

(j − 1/2)+

1

(j + 1/2)

)=

1

2

−1

1/2= −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

1

log nn− 1

log (n+ 1)n+1=

=1

2 log 2= log2

√e

since if 12 log 2 = y, then y = log2

√2.

Exercise 11.∑∞n=1 nx

n = x ddx

∑∞n=1 x

n = x(

x1−x

)′= x

(1−x)2 .

Exercise 12.183

∞∑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 n

4x4 = x ddx

∑∞n=1 n

3x3 = x(x3+4x2+x

(1−x)4

)′.

ln

(x3 + 4x2 + x

(1− x)4

)= ln (x3 + 4x2 + x)− 4 ln (1− x)

(ln f)′ =1

ff ′ =

3x2 + 8x+ 1

x3 + 4x2 + x+ 4

1

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=1

xn

n = x∑∞n=1

∫ x0tn−1dt =

∫ x0dt∑∞n=1 t

n−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

(1/2

1− t+

1/2

1 + t

)=

1

2ln

(1 + x

1− x

)Exercise 17.

∑∞n=0(n+ 1)xn =

∑∞n=0

ddxx

n+1 = ddx

(x

1−x

)= 1

(1−x)2

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!=

1

3

d

dx

∞∑n=0

d2

dx2

x(n+1)+2

2=

=1

3

d

dx

∞∑n=1

d2

dx2

xn+2

2=

1

3

d

dx

( ∞∑n=0

d2

dx2

xn+2

2− d2

dx2

(x2

2

))=

=1

3

d

dx

(1

(1− x)3− 1

)=

1

3

−3(−1)

(1− x)4=

1

(1− x)4

Exercise 20.∑∞n=1 n

kxn = Pk(x)(1−x)k+1

184

∞∑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+kk

)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

dxkx(n+1)+k

k!=

=1

k + 1

d

dx

∞∑k=1

dk

dxkxn+k

k!=

(1

k + 1

)d

dx

∞∑n=0

dk

dxkxn+k

k!− dk

dxkxk

k!=

=

(1

k + 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=2

nn! +

∑∞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

−1

n!=

∞∑n=2

n

(n− 1)!+

∞∑n=2

−1

n!=

=

∞∑n=1

n+ 1

n!+

∞∑n=2

− 1

n!=

∞∑n=1

1

(n− 1)!+ 1 =

∞∑n=0

1

n!+ 1 = e+ 1

Exercise 23.

(1) x ddx

(x ddx

∑∞n=1

xn

n!

)= x d

dx

(x∑∞n=1

nxn−1

n!

)= x d

dx

∑∞n=1

nxn

n! =∑∞n=1

n2xn

n! = x ddx

(x ddxe

x)

= 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

)nExercise 25.

(1)

1 + x2 + x4 + · · ·+ x2n + · · · = 1

1− x2if |x| < 1

=⇒ 1 + 0 + x2 + 0 + x4 + · · · = 1

1− x2if |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

185

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, defined 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+−1

4j − 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+−1/8

4j − 1

)∫ n

1

(3/8

4x− 3+−1/8

4x− 1

)dx =

((3/8)

ln (4x− 3)

4+ (−1/8)

ln (4x− 1)

4

)∣∣∣∣n1

=

=1

32ln

(4x− 3)3

4x− 1

∣∣∣∣n1

=1

32ln

(3(4n− 3)3

4n− 1

)limn→∞

1

32ln

(3(4n− 3)3

4n− 1

)= limn→∞

∫ 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)3/2dx =

∫ ((−2)

a(ax+ 1)1/2

)′log (ax+ 1)dx =

=−2

a(ax+ 1)1/2log (ax+ 1)−

∫−2

a(ax+ 1)1/2

(a

ax+ 1

)=

=−2

a(ax+ 1)1/2log (ax+ 1) +

−4

a(ax+ 1)1/2

limn→∞

∫ n

1

log(ax+ 1)

(ax+ 1)3/2dx = lim

n→∞

(−2 log (an+ 1)

a(an+ 1)1/2+

2 log (a+ 1)

a(a+ 1)1/2+−4

a

((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.

Exercise 3.∑∞j=1

j+12j .

∫ n

1

x+ 1

ex ln 2dx =

∫ 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

)∣∣∣∣n1

=ne−n ln 2

− ln 2+e− ln 2

ln 2+−

(1

(ln 2)2+

1

ln 2

)e−n ln 2 +

(1

(ln 2)2+

1

ln 2

)e− ln 2

limn→∞

∫ n

1

x+ 1

ex ln 2dx =

(2

ln 2+

1

(ln 2)2

)(1

2

)186

By integral test,∑∞j=1

j+12j converges.

Exercise 4.∑∞j=1

j2

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

)∣∣∣∣n1

limn→∞

∫ n

1

x2

2xdx = e− ln 2

(1

ln 2+

2

(ln 2)2+

2

(ln 2)3

)=

1

2

(1

ln 2+

2

(ln 2)2+

2

(ln 2)3

)By integral test,

∑∞j=1

j2

2j converges.

Exercise 5.∞∑j=1

| sin jx|j2

=

∞∑j=1

aj ≤∞∑j=1

1

j2∑∞j=1

1j2 converges since

limn→∞

∫ n

1

1

xsdx = lim

n→∞

x−s+1

−s+ 1

∣∣∣∣n1

= limn→∞

(1

1− s

)(1

ns−1− 1

)=

1

s− 1if 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+

3

22j

)=

2(1/4)

1− 1/4+

3(1/4)

1− 1/4=

4

3

Exercise 7.∑∞j=1

j!(j+2)! .

aj =j!

(j + 2)!= aj =

1

(j + 1)(j + 2)≤ 1

j2= 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)∣∣∣n2−∫ n

2

−2x−1/2

xdx =

= (−2)

(log n

n1/2− log 2

21/2

)+−4x−1/2

∣∣∣∣n2

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 = 1

j .

limj→∞

ajbj

= limj→∞

j√j(j + 1)

= limj→∞

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→∞

ajbj

= limj→∞

(1 +√j

(j + 1)3 − 1

)j5/2 = lim

j→∞

j3 + j5/2

(j + 1)3 − 1= limj→∞

1 + 1/j1/2

(1 + 1/j)3 − 1j3

By limit comparison test, since∑bj converges,

∑aj converges.

Exercise 11.∑∞j=2

1(log j)s =

∑aj

187

If s ≤ 0,∑

aj diverges since limj→∞

aj 6= 0

If 0 < s ≤ 1, 1(log j)s >

1js , and since

∑∞j=2

1js diverges for 0 < s < 1, so does

∑1

(log j)s∫(log x)−s =

∫ (1

x(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

10

10j=

∞∑j=0

1

10j=

1

1− 1/10=

10

9

Exercise 13.∑∞j=1

11000j+1 <

∑∞j=1

11000j = 1

1000

∑∞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=3

1j log j(log (log j))s

∫1

x log x(log (log x))s=

∫ ((ln (lnx))−s+1

−s+ 1

)′=

(ln (ln x))−s+1

−s+1 if s > 1, s < 1, s 6= 0

ln (ln (lnx)) if s = 1

ln (lnx) if s = 0

Converges, by integral test, for s > 1

Exercise 16. Converges by integral test since

∫ ∞1

xe−x2

=

(e−x

2

−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 =

2

3x3/2

∣∣∣∣1/n0

=2

3

(1

n

)3/2

∞∑j=2

2

3

(1

j

)3/2

=2

3

∞∑j=2

1

j3/2

So∑

1 n∫ 1/n

0

√x

1+x2 dx converges by comparison test.

Exercise 18.188

(e−√x)′ = e−

√x −1

2√x

(√xe−

√x)′ =

−e−√x

2+

1

2√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

+1

e√x

)∣∣∣∣n+1

n

=

= 2

(√n+ 1

e√n+1

+1

e√n+1−√n

e√n− 1

e√n

)2

∞∑j=1

(√j + 1

e√j+1

−√j

e√j

+1

e√j+1− 1

e√j

)= −1

e

Note the use of telescoping sum in the last step. The series converges.

Exercise 19.∫ n

1f(x)dx =

∫ n1

log xdx = (x lnx− x)|n1 = n lnn− n+ 1.

n−1∑k=1

lnk ≤∫ n

1

lnx ≤n∑k=2

ln (k) =⇒n−1∑k=1

lnk ≤ n lnn− 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 + 1

4j2 + 6j + 2

j→∞−−−→ 1

4

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 + 1

2ej ln 2

j→∞−−−→ 0

Converges by ratio test.

Exercise 3.∑∞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

Converges by ratio test.

Exercise 4.∑∞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.

Exercise 5.∑∞j=1

j!3j .

(j + 1)!

3j+1

3j

j!=j + 1

3

Diverges by ratio test.

Exercise 6.∑∞j=1

j!22j

189

(j + 1)!

22(j+1)

22j

j!=

(j + 1)

4Diverges.

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< limj→∞

ln j

j= 0

=⇒ limj→∞

−1

jln (log j) = 0 so then

limj→∞

exp

(−1

jln (log j)

)= 1∑∞

j=21

(log j)1/jdiverges 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→∞−−−→ 0

Converges by root test.

Exercise 9.∑∞j=1 e

−j2

(e−j2

)1/j = e−jj→∞−−−→ 0

Converges by root test.

Exercise 10. I systematically tried ratio test and then root test. Both were inconclusive.

Consider comparison with∑

1j . (

ej2 − jjej2

)j =

ej2 − jej2

= 1− j

ej2j→∞−−−→ 1

By limit comparison test, since∑

1j diverges, so does

∑(1j −

1ej2

).

Exercise 11.∑∞j=1

(1000)j

j! = e1000

Exercise 12.∑∞j=1

jj+1/j

(j+1/j)j .

a1/jj =

j1/j2

1 + 1j2

limj→∞

a1/jj = lim

j→∞

exp(

1j2 ln j

)1 + 1

j2

= 1

Note that root test is inconclusive.

aj =exp

(1j ln j

)(

1 + 1j

)j ≥exp

(1j ln j

)(

1 + 1j2

)j2limj→∞

aj ≥ limj→∞

exp(

1j ln j

)(

1 + 1j2

)j2 =1

e> 0

Diverges since limj→∞ aj > 0.

Exercise 13.∑∞j=1

j3(√

2+(−1)j)j

3j .190

(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 r

j | 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 ≤aNbNr

(2) cn < 0

bj −bj+1aj+1

aj< 0

ajbj < bj+1aj+1 =⇒ bjbj+1

<aj+1

aj

∑ 1

bjdiverges, so

limj→∞

bjbj+1

≥ 1 by ratio test

j→∞−−−→ 1 ≤ bjbj+1

<aj+1

ajSo 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 finite, so consider s larger than 1 and n going to infinity 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 = M

Ns−1 so that

A = 1 + r +f(n)

ns−1> 1

191

If A > 1, then∑an converges.

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+ 1

2n+ 2

)kn→∞−−−−→ 1

Ratio test fails.

an+1

an=

(2n+ 1

2n+ 2

)k=

(1 +

−1

2n+ 2

)k=

(1 +−1/2

n+ 1

)k=

=

k∑j=0

(k

j

)(−1/2

n+ 1

)j= 1 + k

(−1/2

n+ 1

)+

k∑j=2

(k

j

)(−1/2

n+ 1

)j

Note that for k <∞,k∑j=2

(k

j

)(−1/2

n+ 1

)j<∞

Let

∣∣∣∣∣∣k∑j=2

(k

j

)(−1/2

n+ 1

)j∣∣∣∣∣∣ ≤Mk/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 =

∑nj=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

√j

j+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.192

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.

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 + 1

2j + 2

Absolutely converges.

Exercise 5.∑∞j=1

(−1)j(j−1)/2

2j converges since limj→∞(−1)j(j−1)/2

2j = 0;∑∞j=1

12j = 1/2

1−1/2 = 1 . Absolutely converges.

Exercise 6.∑∞j=1(−1)j

(2j+1003j+1

)j.

exp

(j ln

(2j + 100

3j + 1

))= exp

(j ln

(2

3+

298

9j + 3

))≤ exp

(j

(ln

(2

3

)+

123

(298

9j + 3

)))=

= exp

(j ln

2

3+

146

3 + 1/j

)0 ≤ lim

j→∞exp

(j ln

(2j + 100

3j + 1

))≤ limj→∞

exp

(j ln

2

3+

146

3 + 1/j

)= 0

=⇒ limj→∞

exp

(j ln

(2j + 100

3j + 1

))= 0

So the alternating series converges.(2j + 100

3j + 1

)<

2j + 100

3j<

2.5j

3j=

5

6( for j ≥ 200)

=⇒(

2j + 100

3j + 1

)j<

(5

6

)jfor 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 finding 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 + 12j −

√2j − 2

(√

2j + 1)(√

2j√

1 + 12j − 1)

=

for j large−−−−−−→ '

√2j(

1 + 14j

)−√

2j − 2

(√

2j + 1)(√

2j(

1 + 14j

)− 1)

=−2

j

(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

1

j

(1− 1

4√

2j

2− 12j + 1

2√

2jj3/2

)>

1

j

(1− 1

4√

2j

4− 1√2j

)=

1

4j

193

By comparison test to 1j the series diverges.

Exercise 8. Using the theorem

Theorem 33.Assume

∑|aj | converges

Then∑aj converges and |

∑aj | ≤

∑|aj |.

So using the contrapositive,

If∑aj diverges,∑|aj | diverges.

1

j1/j=

1

e1j ln j

limj→∞

1

j1/j=

1

exp(

limj→∞1j 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 + 2

4j2 − 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)=

1

4

By limit comparison test, with∑

1j3 ,∑∞j=1

j2

1+j2 converges.

Exercise 10.∑∞n=1

(−1)n

log (en+e−n)

limn→∞

1

log (en + e−n)= limn→∞

1

n= 0

The series converges.

limn→∞

n

log (en + e−n)= limn→∞

n

log en + log (1 + e−2n)= limn→∞

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→∞1

j log2 (j+1)= 0 so by Leibniz’s test, the alternating series.

1

n log2 (n+ 1)<

1

n log2 (n)∫1

n log2 n=

∫ (−1

log n

)′=−1

log n

n→∞−−−−→ 1

log 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.194

(−1)

log(

1 + 12j−1

) +1

log(

1 + 12j

) =−1

log(

2j2j−1

) +1

log(

2j+12j

) =− log

(2j+1

2j

)+ log

(2j

2j−1

)log(

2j2j−1

)log(

2j+12j

) =

=log(

(2j)2

4j2−1

)log(

2j2j−1

)(log(

1 + 12j

)) =log(

1 + 14j2−1

)log(

1 + 12j−1

)log(

1 + 12j

) =

=

14j2−1 + o

(1

4j2−1

)(

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=

(1

j + 2

)(1 +

1

j

)37

→ 0

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

1

e2xdx =

e−2x

−2

∣∣∣∣n+1

n

=

(−1

2

)(1

e2(n+1)− 1

e2n

)=

=e2 − 1

2e2n+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 sin

1

j

)= log

(sin 1/j

1/j

)limj→∞

log

(sin 1/j

1/j

)= log

(limj→∞

sin 1/j

1/j

)= log 1 = 0

sin1

j=

∞∑k=0

(1j

)2k+1

(2k + 1)!(−1)k

log

j ∞∑k=0

(1j

)2k+1

(−1)k

(2k + 1)!

= log

1 +−1

6j2+

∞∑k=2

(1j

)2k

(−1)k

(2k + 1)!

≥≥ log

(1 +− 1

6j2

)≥ −1

6j2

The series absolutely converges.

Exercise 17.∑∞j=1(−1)j

(1− j sin 1

j

)195

(1− x sin

1

x

)′= − sin

1

x− x cos

1

x

(−1

x2

)=

= − sin1

x+

1

xcos

1

x=−x sin 1

x + cos 1x

x

sin1

x=

∞∑j=0

(1x

)2j+1(−1)j

(2j + 1)!

−x sin 1x + cos 1

x

x=

=−1 +

( 1x )

3(+1)

3! +∑∞j=2

( 1x )

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 1j

)converges since aj = 1− j sin 1

j is monotonically decreasing sequence with limit 0.

1− j sin1

j= 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.

Exercise 18.∑∞j=1(−1)j

(1− cos 1

j

).

(cos1

x)′ =

(− sin

1

x

(−1

x2

))=−1

x2sin

1

x< 0∑∞

j=1(−1)j(1− cos 1j ) converges since aj = (1− cos 1

j ) is monotonically decreasing to 0

(1− cos1

j) = 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 1

2j+1 .

(arctan

(1

2j + 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

1

1 + x2= (arctanx)′ =

∞∑j=0

(−x2)j =

∞∑j=0

(−1)jx2j

=⇒ arctanx =

∞∑j=0

(−1)jx2j+1

2j + 1

arctan1

2j + 1=

∞∑k=0

(−1)k(

12j+1

)2k+1

(2k + 1)=

1

2j + 1+ (−1)

(1

(2j+1)3

3

)+

∞∑k=2

(−1)k(

12j+1

)2k+1

2k + 1>

>1

2j + 1+ (−1)

1

3(2j + 1)3=

3(4j2 + 4j + 1) + (−1)

3(2j + 1)3=

12j2 + 12j + 2

3(2j + 1)3>

2j

(2j + 1)2>

2

9

1

jfor j > 2

So by comparison test to∑

1j ,∑

arctan 12j+1 diverges absolutely. The series is conditionally convergent.

Exercise 20.∑∞j=1(−1)j

(π2 − arctan log j

)196

(π2− arctan log n

)′=

(−1

1 + (log n)2

)(1

n

)π

2− arctan log n ≥ π

2− arctan (n− 1)→ π

2− arctann just change indices∫ n

0

(π2− arctanx

)dx =

π

2x−

(x arctanx− 1

2ln (1 + x2)

)∣∣∣∣n0

=π

2n−

(n arctann− 1

2

(ln(1 + n2

)))= n

(π2− arctann

)+

1

2ln (1 + n2)

limn→∞

(n(π

2− arctann

)+

1

2ln (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.

Exercise 22.∑∞j=2 sin

(jπ + 1

log j

)

sin

(2jπ +

1

log 2j

)+ sin

((2j + 1)π +

1

log (2j + 1)

)=

= sin (2jπ) cos1

log 2j+ sin

(1

log 2j

)cos (2πj) + sin (2j + 1)π cos

1

log (2j + 1)+ cos (2j + 1)π sin

(1

log 2j + 1

)=

= sin1

log 2j− sin

1

log 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

(1

log 2j

)− sin

(1

log 2j + 1

)=

=

∞∑k=0

(−1)k

(2k + 1)!

((1

log 2j

)2k+1

−(

1

2j + 1

)2k+1)

0 < log 2j < log 2j + 1 so(

1

log 2j

)2k+1

−(

1

log 2j + 1

)2k+1

> 0

and since for j > 1,(

1

log 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 + 12j

)(log (2j))2

≤12j

(log (2j))2∫ n

1

1

2j(log 2j)2= − 1

log 2j

∣∣∣∣n1

n→∞−−−−→ 1

log 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 n

nzn

197

∣∣∣∣∣∑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=0

zj

3j =∑∞j=0

(13

)jzj∑

zj be convergent or∑nj=1 z

j bounded.

|z| < 3 and |z| = 3 if z 6= 3

Exercise 36.∑∞j=1

zj

jj {z} = C since ∣∣∣∣(zj)∣∣∣∣ < 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.

Exercise 38.∑∞j=1

zj√j

log(

2j+1j

).

zj√j

log

(2 +

1

j

)= zj

log(

2 + 1j

)√j

Since limj→∞log (2+ 1

j )√j

= 0 so thatlog (2+ 1

j )√j

is a monotonically convergent sequence.Then by Dirchlet’s test,

∑nj=1 z

j 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 +

1

5j + 1

)j|z|17

)j

limj→∞

(1 +

1

5j + 1

)j≤ limj→∞

(1 +

1/5

j

)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 = 0.198

Exercise 42.∑∞j=1

(2z+3)j

j log (j+1)

limj→∞

1

j log (j + 1)= 0 so

1

j log (j + 1)is a monotonically convergent sequence

∑(2z + 3)j

|2z + 3| < 1

|z +3

2| < 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 +3

2| ≤ 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→∞1

2j−1 = 0 and 12j−1 is monotonically decreasing.

So 12j−1 is a monotonically decreasing convergent sequence of real terms.

For∣∣∣ z−1z+1

∣∣∣ < 1, ∣∣∣∣z − 1

z + 1

∣∣∣∣ < 1 =⇒ |z − 1| < |z + 1|

(u− 1)2 + v2 < (u+ 1)2 + v2

u2 − 2u+ 1 < u2 + 2u+ 1

−u < u

<(z) > 0; z 6= 0

Exercise 44.∞∑j=1

(z

2z + 1

)j=

∞∑j=1

(1

2+−1/2

2z + 1

)j=

∞∑j=1

(1

2

)j (1− 1

2z + 1

)j(

1

2

)jis a monotonically decreasing, convergent sequence

Now we want∣∣∣∣1− 1

2z + 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

−1

4< u =⇒ <(z) > −1

4

if <(z) =−1

4

2(− 1

4 + iv)

2(−1

4 + iv)

+ 1=−12 + 2iv12 + 2iv

=⇒ z 6= −1

4

Exercise 45.∑∞j=1

jj+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

1

2+

−1

2(2z + 1)=

∣∣∣∣ z

2z + 1

∣∣∣∣ < 1∣∣∣∣2z + 1

z

∣∣∣∣ > 1 =⇒ |2 +1

z| > 1

Exercise 46.∑∞j=1

1(1+|z|2)j =

∑∞j=1

(1

1+|z|2

)j199

1

1 + |z|2< 1

1 < 1 + |z|2

0 < |z|2

∀z except z = 0

Exercise 47.∑∞j=1(−1)j 2j sin2j x

j

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 ∈ (−π4

+ nπ,π

4+ nπ), n ∈ Z

Exercise 49.∑aj converges.∑

aj1aj

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 = a2j . |aj | converges, then

limj→∞

|aj | = 0

limj→∞

|aj |2 = 0limj→∞

|aj |2

a2j

= 1

By limit comparison test,∑a2j converges.

Counterexample:∑(

1j

)2

converges, but∑

1j diverges.

Exercise 51. Given∑aj , aj ≥ 0.

∑aj converges.

∑√aj

1

(j)p

limj→∞

aj = 0

limj→∞

√aj =

(limj→∞

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√ajjp =

√ajj .

Exercise 52.

(1)∑aj converges absolutely, then if

∑|aj | converges,

∑a2j converges.

a2j

1 + a2j

= 1 +−1

1 + a2j

a2j

1 + a2j

≤ a2j since

∑a2j converges ,

∑ a2j

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.

200

10.22 Miscellaneous review exercises - Rearrangements of series. Exercise 1.

(1)

aj =√j + 1−

√j =

√j

√1 +

1

j−√j =

√j

(1 +

1

2

(1

j

)+ o

(1

j

)+−1

)=

=√j

(1

2

(1

j

)+ o

(1

j

))limj→∞

aj = 0

(2)

aj = (j + 1)c − jc = jc((

1 +1

j

)c− 1

)= (jc)

(1 + c

(1

j

)+ o

(1

j

)− 1

)=

= jc(c

(1

j

)+ o

(1

j

))=

(cjc−1 + jco

(1

j

))if c > 1, aj divergesif c = 1, lim

j→∞aj = 1

if c < 1, limj→∞

aj = 0

Exercise 2.

(1)

(1 + xn)1n = exp

(1

nln (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−1

2 = an−1+an−2

22 + an−2+an−3

22

10.24 Exercises - Improper integrals.Exercise 1.

∫∞0

x√x4+1

dx

limx→∞

(x√

x4 + 1

)(1

1/x

)= limx→∞

x2

x2√

1 + 1/x4= 1

Since∫∞

11x diverges, so does

∫∞0

x√x4+1

dx

Exercise 2. ∫ ∞−∞

e−x2

dx =

∫ ∞0

e−x2

dx+

∫ −∞0

e−x2

dx =

∫ ∞0

e−x2

dx+−∫ 0

∞e−x

2

dx = 2

∫ ∞0

e−x2

dx∫ ∞0

e−x2

dx ≤∫ ∞

0

e−xdx = −e−x2∣∣∣∞0

= 1

Converges by theorem.

Exercise 3.∫∞

01√x3+1

dx

Exercise 4.∫∞

01√exdx

Exercise 5.∫∞

0+e−√x

√xdx

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

201

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 sufficient condition for uniform convergence, Power series. Circle of convergence. Exercise 1.

∑∞j=0

zj

2j =∑∞j=0

(z2

)jUsing the comparison test,∣∣ z

2

∣∣j ≤ tj ; ∣∣ z2

∣∣ < 1 =⇒ |z| < 2

Suppose |z| = 2, z 6= 2∑∞j=0

(z2

)j=∑∞j=0(e2iθ)j

Now∑nj=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=

sinnθeiθn+iθ

sin θ<

1

sin θ

So∑∞j=0

(z2

)jconverges for |z| = 24, z 6= 2

Exercise 2.∑∞j=0

zj

(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

2

If |z| < 2,∑∞j=0 aj converges, if |z| > 2,

∑aj diverges.

If |z| = 2,∑

zj

(j+1)2j =∑

(e2iθ)j(

1j+1

)Now 1

j+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 + 1

j + 2

)=

(z + 3)

2

(1 + 1/j

1 + 2/j

)j→∞−−−→ z + 3

2

Using Theorem 11.7,

Theorem 34 (Existence of a circle of convergence).

Assume∑

ajzj

converges for at least z1 6= 0

diverges 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

(1j+1

). Since 1

j+1 is a monotonically decreasing sequence of real

numbers 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.

202

(−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 lnM

2j

converges for 0 < 2|z| = M ≤ 1 =⇒ |z| ≤ 1

2

(we use Theorem 11.6 at this point, because real numbers are included in complex numbers).

Theorem 36. Assume∑ajz

j converges for some z = z1 6= 0.

(1)∑ajz

j converges absolutely ∀z with |z| < |z1|.(2)

∑ajz

j converges uniformly on every circular disk with center at 0 and R < |z1|

We had first used Leibniz’s test to find az1 on the real line.

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 first step is to test absolute convergence first; 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| = 12 , z 6= −1

2 , ∑(1− (−2)j)zj =

∑zj −

∑(−e2iθ)j =

= 0−∑

(e2iθ+πi)j <1

sin (θ + π)

If z = −12 ,

∑(−1

2

)j−∑

1j →∞

Exercise 6.∑∞j=1

j!zj

jj

A very big hint is to use Exercise 19 on pp. 399, in the section for Exercises 10.14.203

Since∑n−1j=1 f(j) ≤

∫ n1f(x)dx

n−1∑j=1

ln j ≤∫ n

1

lnx = n lnn− 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| =

=

(1

1 + 1/n

)n|z| n→∞−−−−→ 1

e|z|

Converges for |z| < e .Now try plugging in a complexified 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 + 2

j2 + 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 a

j2zj , 0 < a < 1

Use the root test .(aj

2

zj)1/j = ajzj→∞−−−→ 0

So the series converges ∀ z ∈ C

Exercise 9.∑∞j=1

(j!)2

(2j)!zj

204

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 diverges

Let’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=1

3√jzj

j =∑∞j=1

e√j ln 3zj

j

e√j+1 ln 3zj+1

j + 1

j

e√j ln 3zj

=

(1

j + 1

)e(√j+1−

√j) ln 3z

√j + 1−

√j =

√1 +

1

j− 1 ' 1 +

1

2

(1

j

)− 1 =

1

2j(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)

)31

zj=

(2j + 1

2j + 2

)3

z =

(1 +−1/2

j + 2

)3

z

If |z| < 1, it converges by ratio test, if |z| = 1, then it converges by Gauss test

aj+1

aj=

(1 +−1/2

j + 2

)3

z =

3∑k=0

(3

k

)(−1/2

j + 2

)k|z| =

= |z|

(1 + 3

(−1/2

j + 2

)+ 3

(−1/2

j + 2

)2

+

(−1/2

j + 2

)3)

j→∞−−−→ |z|

diverges for |z| > 1 (by ratio test )

Exercise 12.∑∞j=1

(1 + 1

j

)j2zj

((1 +

1

j

)j2zj

)1/j

=

(1 +

1

j

)jzj→∞−−−→ e1z

|z| < 1

e,∑

(1 +1

j)j

2

zj converges by root test

|z| > 1

e,∑

(1 +1

j)j

2

zj diverges by root test

1

e= r

205

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.

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(eaz)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=1

zj

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

)j1 +

(ab

)j+1

)j→∞−−−→

(zb

)So then |z| ≶ b converges (diverges) by ratio test.

If a = b,∞∑j=1

zj

2aj=

1

2

∞∑j=1

(za

)jBy 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 + 1

j

(a|z|)j=

a|z|1 + 1

j

j→∞−−−→ a|z|

=⇒ |z| < 1

athen the series converges

(b|z|)j+1

(j + 1)2

j2

(b|z|)j= b|z|

(1

1 + 1j

)2

j→∞−−−→ b|z|

=⇒ |z| < 1

b

So if a ≷ b, then r = a; (b)

Exercise 17.∫ 1

0fn(x) =

∫ 1

0nxe−nx

2

= e−nx2

−2

∣∣∣10

= e−n−1−2

n→∞−−−−→ 12

However,limn→∞ nxe−nx

2

= 0This example shows that the operations of integration and limit cannot always be interchanged. We need uniform conver-

gence. Exercise 18. fn(a) = sinnxn limn→∞

sinnxn = 0

f(x) = limn→∞ fn(x)f ′n(x) = n cosnx

n =⇒ limn→∞ f ′n(0) = 1This example shows that differentiation and limit cannot always be interchanged.

Exercise 19.∑∞j=1

sin jxj2 = f(x)

206

sin jx

j2≤ 1

j2∀x ∈ R

by comparison test,∞∑j=1

| sin jx|j2

converges, so∞∑j=1

sin jx

j2converges∣∣∣∣ sin jxj2

∣∣∣∣ ≤ 1

N2∀x ∈ R; ∀j ≥ N

∑N 1N2 converges, so

∑ sin jxj2 uniformly converges.

Then by Thm., since sin jxj2 are continuous,

∑ sin jxj2 is continuous.

Since∑ sin jx

j2 uniformly converges.∫ π

0

∑ sin jx

j2=∑∫ π

0

sin jx

j2=∑ cos jx

−j3

∣∣∣∣π0

=∑ (−1)j − 1

−j3=

= 2

∞∑j=1

1

(2j − 1)3

Exercise 20. It is known that∑∞j=1

cos jxj2 = x2

4 −πx2 + π2

6 if 0 ≤ x ≤ 2π

(1) x = 2π

∞∑j=1

1

j2=

(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/20

=∑ (−1)j+1

(2j − 1)3∫x2

4− πx

2+π2

6=

(1

3

x3

4− πx2

4+π2

6x

)∣∣∣∣π/20

=

= (π)3

(1

12(8)− 1

16+

1

12

)= (π)3 1

32

11.13 Exercises - Properties of functions represented by real power series, The Taylor’s series generated by a function,A sufficient condition for convergence of a Taylor’s series, Power-series expansions for the exponential and trigono-metric functions, Bernstein’s Theorem. Sufficient 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=

1

n!

∫ 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

207

For f(x) =∑nj=0

f(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-infinite value, so En(x)n→∞−−−−→ 0. �

Exercise 1.∑∞j=0(−1)jx2j

Consider absolute convergence. limj→∞(x2)j = 0 If x2 < 1

If |x| = 1, then considerx2(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=0

xj

3j+1 = 13

∑∞j=0

(x3

)jThe 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 > 1

0 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+2

j+1xj =

∑∞j=0(−1)j

(j+2j+1

)(2x)j

limj→∞

(j + 2

j + 1

)(2x)j = lim

j→∞(2x)j if |2x| < 1

So when |x| < 12 , the series converges.(

2j + 2

2j + 1

)−(

2j + 3

2j + 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)/2

2j2 + 3j + 1=

(3j + 1)/2

(2j + 1)(j + 1)

3j + 1

4j2 + 6j + 2<

3j + 1

4j2 + 43

<3(j + 1/3)

4(j2 + 1/3)<

3

4

(1

j

)+

1

12j2

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.

208

(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=1

ej ln 2x

j =⇒ |x| < 1

2it’ll converge, since by comparison test, ej ln 2x

j < 1j2 if

0 < x < 12 .

Exercise 7.∑∞j=0

(−1)j

(2j+1)

(x2

)2j.

∞∑j=0

x2j

2(j + 1/2)22j<

∞∑j=0

x2j

22j+1j=

1

2

∞∑j=0

(x2

)2j(

1

j

)=

1

2

∞∑j=0

e2j ln x2

j

since by comparison test,e2j ln x

2

j<

1

j2if 0 < x < 2

If x = ±2 ,

1

2(2j) + 1− 1

2(2j + 1) + 1=

1

4j + 1− 1

4j + 3=

2

(4j + 1)(4j + 3)≤

≤ 1

8

1

j2(converges by comparison test to

∑ 1

j2)

For |x| < 2 ,∑∞j=0

(−1)j

(2j+1)

(x2

)2jconverges.

Exercise 8.∑∞j=0

(−1)jx3j

j! =∑∞j=0

(−x3)j

j! = e−x3

, which converges ∀x ∈ R

Exercise 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=

1

2

∞∑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− x2

=∑∞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 = −21

2

∞∑j=0

(−2)j

2j=

1

2

∞∑j=0

(−1)j = 0 but1

2− (−2)=

1

4x

209

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 cosx+ sinx cos 2x = 3 sinx− 4 sin3 x.

sin3 x =3 sinx− sin 3x

4=

3

4

∞∑j=0

x2j+1(−1)j

(2j + 1)!−∞∑j=0

(3x)2j+1(−1)j

(2j + 1)!

=

=3

4

∞∑j=0

(−1)jx2j+1(1− 32j+1)

(2j + 1)!

=3

4

∞∑j=0

(−1)j+1(32j+1 − 1)x2j+1

(2j + 1)!

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 + 3

2j + 1

x2j+1

j→∞−−−→ x2

|x2| < 1, converges, with radius of convergence of 1

Exercise 18. 3x1+x−2x2 = 1

1−x −1

1+2x =∑∞j=0

xj

j −∑∞j=0

(−2x)j

j =∑∞j=0

xj

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| < 1

2

For x =1

2x2j

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+

6

6 + x=

∞∑j=0

xj +

∞∑j=0

(−x6

)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− 1

x3 − 1=

1− x1− x3

1

1− x3=

∞∑j=0

(x3)j

1− x1− x3

=

∞∑j=0

((x3)j − x3j+1

)By induction, it could be observed that x3j ,−x3j+1, 0 appear in sequences of 3 terms.

210

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 )

Exercise 21.

1

1− x=

∞∑j=0

xj ;

(1

1− 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=

1

4

∞∑j=0

(xj + (−x)j) + 2

∞∑j=1

jxj−1

x

(1− x)(1− x2)=

1

4

∞∑j=0

xj+1(1 + (−1)j) + 2

∞∑j=1

jxj

=1

4

∞∑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) =5

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

f ′′(x) = 5(2 + x2)3/2 +15

2x(2 + x2)1/2(2x) = 5(2 + x2)3/2 + 15x2(2 + x2)1/2

f ′′′(x) =15

2(2 + x2)1/2(2x) + 30x(2 + x2)1/2 +

15x2

2(2 + x2)−1/2(2x)

f ′′′′(x) = 15(2 + x2)1/2 +15

2(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!+

45

4!

√2x4

Exercise 24. f(x) = e−1/x2

if x 6= 0 and let f(0) = 0

(1)

f(x) =

∞∑j=0

(−1x2

)jj!

= 1 +−1

x2+

∞∑j=2

(−1x2

)jj!

=

∞∑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

211

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 ≥ 1

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 = −3

a3 = 0aj+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=0

x4j

(4j)! ;d4ydx4 = y

212

(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 test

x4j+4

(4j + 4)!

(4j)!

x4j=

x4

(4j + 4)(4j + 3)(4j + 2)(4j + 1)

j→∞−−−→ 0 ∀x ∈ R

So the series converges on R.

Exercise 4. f(x) =∑∞j=0

xj

(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 +

1

1!− 1 =

∞∑j=1

(1

(j!)2− 1

(j!)2

)= 0

Exercise 5. f(x) = 1 +∑∞j=1

1·4·7...(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 + 1

j→∞−−−→ 0 ∀x

By ratio test, f converges ∀x ∈ R.

Exercise 7. f(x) =∑∞j=2

xj

j! y′ = x+ y

213

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 + 1

j→∞−−−→ 0 by ratio test, f converges ∀x ∈ R.

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)24

j→∞−−−→ 0 by ratio test, f converges ∀x ∈ R

x2j+3

(j + 2)!22j+3

j!22j+1

x2j+1=

x2

(j + 2)(j + 1)4

j→∞−−−→ 0 by ratio test, f converges ∀x ∈ R

(2)

J ′0(x) =

∞∑j=1

(−1)jx2j−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)jx2j+1

(j!)222j

j1(x) = xJ1(x) =

∞∑j=0

(−1)jx2j+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 = 0.214

n = 0 =⇒ x2y′′ + xy′ + (x2)y = 0

J0 =

∞∑j=0

(−1)jx2j

(j!)222j=

∞∑j=1

(−1)j−1 x2j−2

((j − 1)!)222j−2;

J ′0 =

∞∑j=1

(−1)jx2j−1

j!(j − 1)!22j−1;

J ′′0 =

∞∑j=1

(−1)jx2j−2

j!(j − 1)!

(2j − 1)

22j−1

∞∑j=1

(−1)j(

(2j − 1)

j!(j − 1)!22j−1+

1

j!(j − 1)!22j−1+

−2j

((j − 1)!)j!22j−1

)= 0

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 =1

2+

∞∑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 + 1

j + 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 = 0.215

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 first 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 =

4

3

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 =1

4

x4 : 5a5 = 0 =⇒ a5 = 0 x5 : 6a6 = 0 =⇒ a6 = 0

x6 : 7a7 = 2a1a4 + 2a2a3 =⇒ a7 =1

14

x7 : 8a8 = 0 + 2a2a4 = 0 =⇒ a8 = 0 x8 : 9a9 = 0 =⇒ a9 = 0

x9 : 10a10 =

(1

4

)2

+ 2(1)1

14=⇒ a10 =

23

1120

Exercise 14. y′ = x+ y2 y = 0 when x = 0 =⇒ a0 = 0216

y′(0) = 0 + 0 = 0 =⇒ a1 = 0

y =

∞∑j=2

ajxj y′ =

∞∑j=2

jajxj−1 =

∞∑j=1

(j + 1)aj+1xj

y2 =

a2x2 + a3x

3 + a4x4 +

∞∑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 =1

2

x2 : 3a3 = 0 =⇒ a3 = 0 x3 : 4a4 = 0 =⇒ a4 = 0

x4 : 5a5 = a22 =⇒ a5 =

1

20x5 : 6a6 = 0 =⇒ a6 = 0x6 : 7a7 = 0 =⇒ a7 = 0

x7 : 8a8 = 2(1

2)(

1

20) =⇒ a8 =

1

160

x8 : 9a9 = 0 =⇒ a9 = 0 x9 : 10a10 = 0 =⇒ a10 = 0

x10 : 11a11 = 2(1

2)(

1

160) +

(1

20

)2

=⇒ a11 =7

8800

Exercise 15. y′ = αy

=⇒∑∞j=0(j + 1)aj+1x

j = 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 = 0217

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)(∫ xaQ(t)eA(t)dt+ y(a)

)where A(x) =

∫ xaP (t)dt,

y = e−2x(x+ 1)

We had obtained the necessary initial conditions to solve this ODE from the information given, that a0 = 1, so that y(0) = 1.By doing some simple computation and comparison of powers with e−2x, then a1 = 2, a2 = −2, a3 = 4/3

Exercise 19. cosx =∑∞j=0 aj(j + 2)xj for f(x) =

∑∞j=0 ajx

j .

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)j1

(2j)!

a5 = 0

a6(6 + 2) =(−1)3

6!=⇒ a6 =

−7

8!

Now notice that for cosx =∑∞j=0 aj(j + 2)xj =

∑∞j=1 jajx

j + 2∑∞j=0 ajx

j is actually a differential equation, cosx =

xy′ + 2y. We can solve this first-order ODE using

y(x) = e−A(x)(∫ xaQ(t)eA(t)dt+ y(a)

)where A(x) =

∫ xaP (t)dt. Then solving y′ + 2y

x = cos xx ,

y =1

x2(x sinx+ cosx− (a sin a+ cos a) + b)

218

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) =sinx

x+

cosx− 1

x2if x 6= 0

So f(0) = 12 and f(π) = −2

π2

Exercise 20.

(1)

(1− x)−1/2 =

∞∑j=0

(−1/2

j

)(−x)j =

= 1 +1

2x+

(−12

) (−32

)2

x2 +

(−12

) (−32

) (−52

)3!

x3 +

(−12

) (−32

) (−52

) (−72

)4!

x4+

+

(−12

) (−32

) (−52

) (−72

) (−92

)5!

x5 + · · · =

= 1 +1

2x+

3

8x2 +

5

8x3 +

35

128x5 +

63

256x5 + . . .

(2) To make the notation clear, (1− x)−1/2 =∑∞j=0

(−1/2j

)(−x)j =

∑∞j=0 bjx

j =∑∞j=0 aj

Now (αj+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 find that

bj+1 < bj1

50

bj+2 < bj+11

50< bj

(1

50

)2

For x = 150 . So by induction, bn+j < bn

(150

)jrn =

∞∑j=1

an+j <

∞∑j=1

an

(1

50

)j= an

1/50

1− 1/50=an49

rn <an49

(3) Note that (1− x)−1/2

=(1− 1

50

)−1/2=(

4950

)−1/2= 5√

27

7

5

(1− 1

50

)−1/2

= 1 +1

100+

3

2

(1

2(50)

)2

+5

2

(1

2(50)

)3

+35

8

(1

2(50)

)4

+63

8

(1

2(50)

)5

√2 ' 1.4142135624

Exercise 21.

(1) 17321000

(1− 176

3000000

)−1/2= 1732

1000

(30000002999824

)1/2Obviously, (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

1732

1000

(1− 176

3000000

)−1/2

=√

3

(2)219

Exercise 22. arcsinx =∫

1√1−x2

(1− x2)−1/2 =∑∞j=0

(αj

)(−x2)j = 1 +

∑∞j=1

(αj

)(−x2)j

=⇒ arcsinx = x+

∞∑j=1

(−1/2

j

)(−1)j

(2j + 1)x2j+1

(−12

) (−32

). . .(−1

2 − j + 1)

j(j − 1) . . . (2)(1)= (−1)j

(1)(3) . . . (1 + 2j − 2)

(2j)!!= (−1)j

(2j − 1)!!

(2j)!!

=⇒ arcsinx = x+

∞∑j=1

(2j − 1)!!

(2j)!!

x2j+1

2j + 1

13.21 Exercises - The conic sections, Eccentricity of conic sections, Polar equations for conic sections.

Exercise 1. F is in the positive half-plane determined by N .

‖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)

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

.220

ed(X,L) = ‖X − F‖ = e|(X − F ) ·N − d| = e|r cos θ − d| = er cos θ − ed = r

−ed1− 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 ,

45

).

L = {x = P + tA}, N ·X = N · P .To find the distance from the focus, at the origin, to the directrix,

dN = P + tA; dN ·N = d = N · P

So for this problem, d = 5.

r = ‖X − F‖ = ed(X,L) = e|(X − F ) ·N − d| = e|X ·N − d| =∣∣∣∣35r cos θ +

4

5r sin θ − 5

∣∣∣∣r =

1

2

(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 ,

35

).

d = 5.

‖X − F‖ = ed(X,L) = e|(X − F ) ·N − d| = e

∣∣∣∣45r cos θ +3

5r sin θ − 5

∣∣∣∣r = 5− r

(4

5cos θ +

3

5sin 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

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/√

2

1 + 2√2

cos θ + 2√2

sin 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.

221

(1)‖X − F‖ = ‖X‖ = ed(X,L) = 1|(X − F ) ·N − d| = d−X ·N = d− r cos

π

3

d = r

(3

2

)=

3

2× 108mi

r =32 × 108mi

1 + cos θθ = 0, r =

3

4× 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) =

1

2× 108mi

r =d

1− cos θ=

12 × 108mi

1− cos θr(θ = π) =

1

4× 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 satisfied

if 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 =4

5.

|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 .

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− 1/41/3 = 1

2 = e. Foci:(±12√

3, 0)

. Center (0, 0).

Vertices (±1/√

3, 0), (0,±1/2).

Exercise 6. Center (−1,−2).222

√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 = 34 . a = 1, e = 3

4 . 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 = 1

Exercise 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)2 x,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.

x2

4 + y2

−12 = 1. y2

12 + 1 = x2

4

x,y→∞−−−−−→ y = ±√

3x.223

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) =⇒ ± 1

a2∓ 4

b2= 1

(−1, 0) =⇒ ± 9

a2∓ 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±√

23

3= 2y

Exercise 25. ±x2

a2 +∓y2

b2 = 1 ±x2

a2 ∓y2

4a2 = 1.

(3,−5)→ ±9∓ 254 = a2; a2 = 11

4 .

=⇒ 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 (confirming the other definition).Let N = ~ey , d = 2c; F = (0, c), e = 1.

x2 + (y − c)2 = ((y − c) + 2c)2 = (y − c)2 + 4c(y − c) + 4c2

x2 = 4cy

Exercise 26. 4c = −8 (0, 0) vertex. y = 0 symmetry axis. x = 5 directrix.

Exercise 27. 4c = 3. Vertex: (0, 0). Symmetry axis: y = 0. Directrix: x = −3/4.

Exercise 28. (y − 1)2 = 12(x− 12 ). 4c = 12, c = 3. Symmetry axis: y = 1. Directrix:

(−52 , 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 = −x2.224

Exercise 33. x2 = 8y.

Exercise 34. (y − 3) = −8(x+ 4)2.

Exercise 35. c = 54 5(x− 7

4 ) = (y + 1)2

Exercise 36. y = ax2 + bx+ c

(0, 1)→ c = 1 (1, 0)→ 0 = a+ b+ 1(2, 0)→ 0 = 4a+ 2b+ 1 a =1

2=⇒ y =

1

2x2 − 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−(xa

)2)

y = 2

∫ a

−ab

√1−

(xa

)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

)2V =

∫ a

−aπb2

(1− x2

a2

)dx = πb2a

∫ 1

−1

(1− x2)dx =4

3π(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 = 43π(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

√1

B

∫ √3/A

−√

3/A

√3−Ax2dx = 2

√3

B

∫ √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

∫ √ 3A+B

−√

3A+B

√√√√√1−

x√3

A+B

2

Equating the two areas after making an appropriate scale change,

2

√3

B

√3

A

∫ 1

−1

√1− x2dx = 2

√3

A−B

√3

A+B

∫ 1

−1

√1− x2dx

225

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

√5

2=B(1 +

√5)

2

Exercise 4. y = − 4hb2 x

2. ∫ 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).

(1)

A = 2

∫ 2

1

√2(x− 1) + 2

∫ 3

2

√2(x− 1)− 2

√x− 2 =

= 2√

22

3(x− 1)3/2

∣∣∣∣21

+ 2√

22

3(x− 1)3/2

∣∣∣32− 4

2

3(x− 2)3/2

∣∣∣32

= 2√

22

3+√

24

3(2)3/2 − 2

√2

2

3− 4

2

3= 8/3

(2) ∫ 2

1

2(x− 1) = 2

(1

2x2 − x

)∣∣∣∣21

= 2

(1

2(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

(−3

80y5 + 3y

)∣∣∣∣20

=

= 2π

(−3(32)

80+ 6

)= 2π

(−96 + 480

80

)= 2π

(384

80

)= π

48

5

Exercise 7. By Apostol’s definition of conic sections, we are basically given the conic section definition 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−

(1

2

)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

226

So by squaring both sides of the vector equation,

x2 + y2 =x2

2+ xy +

y2

2+

1

2+ x+ y

x2

2+y2

2− xy − x− y =

1

2

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)−−−→ 4

a2− 1

a2=

3

a2= 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).

(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) b21 = a21(1− e2) b22 = a2

2(1− e2).

1− b21a2

1

= 1− b22a2

2

;b21a2

1

=b22a2

2

x21

a21

+y2

b21= 1 =

((b2b1

)x)2

a22

+

(b2b1y)2

(b2y)2

(3) ± (tx)2

a2 ∓(ty)2

b2 = 1(bt

)2= −a2(e2−1)

t2 = −(at

)2(e2 − 1).

b21 = a21(e2 − 1) b22 = a2

2(e2 − 1)

b21a2

1

+ 1 = e2 b22a2

2

+ 1 = e2 b21a2

1

=b22a2

2

±(x

a1

)2

∓(y

b1

)2

= ±

(b2b1x)

a2

2

∓

(b2b1y

b2

)2

= 1

227

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 specificcoordinate 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 fit thetangency condition. Thus, in terms of doing the algebra, just eliminate p and q from the relations.

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′ =√p

1√x− a

y′(h) =√p

1√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 locus

in 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 ,

12

)must also be the center. y − 1

2 = −(x+ 1

2

)is the directrix.(

y − 12

)= α

(x+ 1

2

)is the general form of the asymptote.

228

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=

1

cosφ

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 cosφ = x− c√(x− c)2 + y2

x− c=

1

cosφ= e;

(x− c)2 + y2

(x− c)2= e2;

y2

(x− c)2= e2 − 1 =⇒ y =

√e2 − 1x

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+

1

2

)− 5

2

)2

+

(α

(x+

1

2

)− 5

2

)2

= x+ α

(x+

1

2

)+

1

2

=⇒

√(1 + α2)

(x+

1

2

)2

− 5(1 + α)

(x+

1

2

)+

25

2

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 filled: 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√2

d =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 + d2

1

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 =−xd2

229

Point of intersection x20 = 2y0d1 + d2

1 = d22 − 2d2y0

2(d1 + d2)y0 = d22 − d2

1 =⇒ y0 =d2 − d1

2

x20 = d2(d2 − 2y0) = d2d1

=⇒y′1 =

±√d2d1

d1= ±

√d2

d1

y′2 =∓√d2d1

d2= ∓ 1√

d2d1

Exercise 20.

(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)x

ya2

)=

= (a4 +−(a2 − c2)x2 − a2x2)(a2 − c2)

a4

x

y

−xy =−xy2

y= −

x(a2 − c2 −(a2−c2a2

)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

(−1

y′

)2

+ (x2 − y2 − c2)−1

y′− xy = 0 =

−xyy′

+ (x2 − y2 − c2)1

y′+ xy

if y′ 6=0−−−−→ −xy + (x2 − y2 − c2)y′ + xy(y′)2 = 0

Thus S → S since the defining differential equation is invariant under the transformation of the slope.

Exercise 21. For a circle centered at C, then ‖X − C‖ = r20 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 definition d(X,L), thedistance from the circle center to the line. ‖C − F‖ = ‖C −X0‖ is by definition a parabola.

Exercise 22. Consider a circle that’s part of the mentioned family that has its center directly below the given circle with radius

r0, and center Q.

It’s given that the center is equidistant from the point of tangency and the line. This hints at a parabola because the parabola’s230

vertex 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

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 artificial construction.

Let’s show this for any circle C of radius r1 in the family.

Tangent to the circle Q condition: X1 − C = α(Q−X1).

So then ‖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

231

For the line defined 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 =

4

5 + t2

P = αQ =4

5 + t2

((1, 2) + t

(−2√

5,

1√5

))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.

(t

1+t2 ,1

1+t2 ,−2t

(1+t2)2

)Exercise 7. F ′ =

(2

1+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 ,

23 , e

1 − 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+e

2

))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 te

2t + −14 e

2t, t sinh 2t2 − cosh 2t

4 , 2(−12 te

−2t − e−2t

4

))∣∣∣10

=

=(

14e

2 + 14 ,

sinh 22 − cosh 2

4 + 14 , 2

(−34 e−2 + 1

4

)).

A ·B = 12e

2 +−2 sinh 2 + cosh 2 + −32 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) = 0

Exercise 14.232

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 >∑nj=1(Fj(t)−Aj)2 > (Fk(t)−Ak)2 > 0

=⇒ ε > |Fk(t)−Ak| if |t− p| < δ.

Exercise 18. If F is differentiable on I , then

F ′ =

n∑j=1

f ′j~ej f ′j = limh→0

1

h(fj(t+ h)− fj(t))

F ′ =

n∑j=1

limh→0

1

h(fj(t+ h)− fj(t)) = lim

h→0

1

h

n∑j=1

(fj(t+ h)− fj(t))ej = limh→ 01

h(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 =

=∑nj=1 limh→0

1h (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. Thus

F (t) =∑nj=1 cj~ej = C on an open interval I .

Exercise 20. 16 t

3A+ 12 t

2B + 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−∫ xap(t)dt

(bj +

∫ x

a

Qj(t)e∫ tap(u)dudt

)=⇒

n∑j=1

jj(x) = Y (x) = e−∫ xap

(B +

∫ x

a

Qe∫ tapdt

)

Exercise 22.233

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+− 1

x2

∫ x

1

F (t)dt+1

xF (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)e∫ taα

F (t) =

n∑j=1

fj(a)e6

∫ t

a

αej = e∫ taα

n∑j=1

fj(a)ej = u(t)A

14.19 Exercises - Velocity and acceleration in polar coordinates, Plane motion with radial acceleration, Cylindricalcoordinates.

Exercise 1.

v =dr

dt=dr

dt~er + r

dθ

dt~eθ = ~er + r~eθ (θ = t)

a =

(d2r

dt2− r

(dθ

dt

)2)~er +

1

r

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 − 1

2)2 = r2 − r sin θ +

1

4=

1

4234

(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 6.

A =

∫R2(θ)dθ =

∫ 2π

0

1

2e2cθdθ =

e2cθ

4c

∣∣∣∣2π0

=e4πc − 1

4c

Exercise 7.∫ π

0

1

2sin4 θdθ =

1

2

∫ π

0

sin2 θ(1− cos2 θ)dθ =1

2

∫ π

0

((1− cos 2θ

2

)−(

1− cos 4θ

2(4)

))dθ = 3π/16

Exercise 15. Place target at the center (without loss of generality). The strategy is to break up v into the polar coordinate unit

vectors.r = r~er

v =dr

dt~er + r

dθ

dt~eθ

dr

dt= vr = v cos (π − α) = −v cosα

rdθ

dt= v sinα

v sinα

−v cosα= − tanα =

r dθdtdrdt

= rdθ

dr;

1

r

dr

dθ= − tanα

r = e− tanαθ

Exercise 17.

A first 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 find 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, sodt

dt= 0; ωk× r = r

dθ

dt~eθ = ωr~eθ = r

dθ

dt

|ωk× r|2 = ω2r2 = r2

(dθ

dt

)2

ω =

∣∣∣∣dθdt∣∣∣∣ , ω > 0

Exercise 19. (a)235

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 ddt |rp(t)− rq(t)| = 0, which implies

d

dt(r2p − 2rp · rq + r2

q) = 2rp ·drpdt− 2

drpdt· rq − 2rp ·

drqdt

+ 2drqdt· rq = 0

drpdt· (rq − rp) = −drq

dt· (rp − rq)

Supposevp = ωp × rpvq = 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 .

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