Solution Manual For History of Mathematics An Introduction, A, 2025 4th Edition by Victor J. Katz

Solutions

1

CHAPTER 1 1.

The answers are given in the answer section of the text. For the Egyptian hieroglyphics, 375 is three hundreds, seven tens, and five ones, while 4856 is four thousands, eight hundreds, five tens, and six ones. For Babylonian cuneiform, note that 375 = 6 × 60 + 15, while 4856 = 1 × 3600 + 20 × 60 + 56.

2.

1 ′

34

2

68

4

136

8

272

16

544

18

612

′

1

5

10

50′

(multiply by 10)

2

10

(double first line)

4

20

(double third line)

8

40′

(double fourth line)

2

22

10

2

18 2 10

93

′

(halve first line)

′

(invert third line)

3.

1

2 14

2

4 28

4

8 56 1

2

Solutions 4.

1

28

2

56

4

112 16

5.

We multiply 10 by 3 30:

1 ′

3

30

2

1

3

15

4

2

3

10

5

2

10

′

8

30

The total of the two marked lines is then 7, as desired.

6.

1

7

2

4

2

15

2

4

4

31

2

8

63

3

5

8

4

Note that the sum of the three last terms in the second column is 99 2 4. We therefore need to figure out by what to multiply 7 2 4 8 to give 4 so that we get a total of 100.

But since we know from the fourth line that multiplying that value by 8 gives 63, we also know that multiplying it by 63 gives 8. Thus the required number is double 63, which is 42 126. Thus the final result of our division is 12 3 42 126.

Solutions

3

7.

1

7248

2

15 2 4

′

4

31 2

8

63

3

4 3 3 6 12

12 3

98 2 3 3 6 12 99 2 4

′ ′

8.

2 ÷ 11

1

11

3

2 ÷ 23

1

23

73

3

15 3

3

33

3

73

6

136

6

323

66

6

12

1246

276

12

12 276

2

6 66

′

2

′

′

′

9.

x + 17 x = 19. Choose x = 7; then 7 + 17 · 7 = 8. Since 19 ÷ 8 = 2 38 , the correct answer is 2 83 × 7 = 16 58 .

10.

(x + 23 x) − 13 (x + 23 x) = 10. In this case, the “obvious” choice for x is x = 9. Then 9

added to 2/3 of itself is 15, while 1/3 of 15 is 5. When you subtract 5 from 15, you get 10. So in this case our “guess” is correct.

4

Solutions 11.

The equation here is (1 + 13 + 14 )x = 2. Therefore. we can find the solution by dividing 2 by 1 + 13 + 14 . We set up that problem: 1

124

3

1 18

3

2 36

6

4 72

12

8 144

The sum of the numbers in the right-hand column beneath the initial line is 1 141 144 . So 3 we need to find multipliers giving us 144 = 144 72. But 1 3 4 times 144 is 228. It follows that multiplying 1 3 4 by 228 gives 144 and multiplying by 114 gives 72.

Thus, the answer is 1 6 12 114 228.

12.

13.

Since there are 10 hekats to be divided among 10 men, the average for each is 1 hekat.

So to get the largest share, we should add to the average 1/8 times half the number of differences. Since there are 9 differences, we add 1/8 4 21 times, or 9/16. Therefore, 9 the largest share is 1 16 hekats, which the scribe writes as 1 2 16. We then subtract 1/8 9 7 5 from this value 9 times to get the share of each man. The answers are 1 16 , 1 16 , 1 16 , 3 1 15 13 11 9 7 1 16 , 1 16 , 16 , 16 , 16 , 16 , and 16 .

100 Since x must satisfy 100 : 10 = x : 45, we would get that x = 45× 10 ; the scribe breaks 35×100 10×100 this up into a sum of two parts, 10 and 10 .

14.

The ratio of the cross section area of a log of 5 handbreadths in diameter to one of 4 9 handbreadths diameter is 52 : 42 = 25 : 16 = 1 16 . Thus, 100 logs of 5 handbreadths 9 × 100 = 156 41 logs of 4 handbreadths diameter. diameter are equivalent to 1 16

16.

The modern formula for the surface area of a half-cylinder of diameter d and height h is A = 12 πdh. Similarly, the modern formula for the surface area of a hemisphere of diameter d is A = 12 πd2 . These formulas are identical if h = d.

17.

7/5 = 1;24 13/15 = 0;52 11/24 = 0;27,30 33/50 = 0;39,36

18.

0;22,30 = 3/8 0;08,06 = 27/200 0;04,10 = 5/72 0;05,33,20 = 5/54

19.

Since 60 is (2 21 ) × 24, the reciprocal of 24 is 2;30. Since 60 is (1 78 ) × 32, and 78 can be expressed as 0;52,30 the reciprocal of 32 is 1;52,30. Since 60 is (1 13 ) × 45, the 1 1 reciprocal of 45 is 1:20. Since 60 = 1 91 × 54, and 19 can be expressed as 10 + 90 = 6 40 + = 0;06,40, the reciprocal of 54 is 1;06,40.

If the only prime divisors of n are 60 3600 2, 3, 5, then n is a regular sexagesimal.

20.

25 × 1,04 = 1,40 + 25,00 = 26,40. 18 × 1,21 = 6,18 + 18,00 = 24,18. 50 ÷ 18 = 50 × 0;3,20 = 2;30 + 0;16,40 = 2;46,40. 1,21 ÷ 32 = 1,21 × 0;01,52,30 = 1;21 + 1;10,12 + 0;00,40,30 = 2;31,52,30.

21.

1,08,16 × 3,45 = 4,16,0,0. 4,16 × 3,45 = 16,0,0. 16 × 3,45 = 1,0,0.

Solutions

5

22.

Since the length of the circumference C is given by C = 4a, and because C = 6r, it √ follows that r = 23 a. The length T of the long transversal is then T = r 2 = ( 32 a)( 17 12 ) = 17 t 2 17 7 a.

The length t of the short transversal is t = 2 ( r − ) = 2a ( − ) = a.

The area 18 2 3 36 18 A of the barge is twice the difference between the area of a quarter circle and the area of the right triangle formed by the long transversal and two perpendicular radii drawn from the two ends of that line. Thus

2

2 a 2 C r2 2a2 A=2 − =2 − = a2 .

48 2 3 9 9

23.

Since the length of the circumference C is given by C = 3a, and √ because C = 6r, it follows that r = a2 . The length T of the long transversal is then T = r 3 = ( a2 )( 74 ) = 87 a.

The length t of the short transversal is twice the distance from the midpoint of the arc to the center of the long transversal. If we set up our circle so that it is centered on the √ origin, the midpoint of the arc has coordinates ( 2r , 23r ) while the midpoint of the long √ transversal has coordinates ( 4r , 43r ). Thus the length of half of the short transversal is a r 2 and then t = r = 2 . The area A of the bull’s eye is twice the difference between the area of a third of a circle and the area of the triangle formed by the long transversal and radii drawn from the two ends of that line. Thus 2

2

C 1r 9a 1 a 7a 1 7 9 2 A=2 − T =2 − = 2a2 − = a .

36 2 2 36 24 8 4 64 32

24.

If a is the length of one of the quarter-circle arcs defining the concave square, then the diagonal is equal to the diameter of that circle. Since the circumference is equal to 4a, the diameter is one-third of that circumference, or 1 13 a. The transversal is equal to the diagonal of the circumscribing square less the diameter of the circle (which is equal to the side of the square). Since the diagonal of a square is approximated by 17/12 of 5 4 5 the side, the transversal is therefore equal to 5/12 of the diameter, or 12 3 a = 9 a. √

√

25.

3 = 22 − 1 ≈ 2 − 12 · 1 · 12 = 2 − 0;15 = 1;45. Since an approximate rep √ (1;45)2 − 0;03,45 = ciprocal of 1;45 is 0;34,17,09, we get further that 3 = 1;45 − (0;30)(0;03,45)(0;34,17, 09) = 1;45 − 0;01,04,17,09 = 1;43,55,42,51, which we truncate to 1;43,55,42 because we know this value is a slight over-approximation.

26.

v + u = 1;48 = 1 45 and v − u = 0;33,20 = 95 . So 2v = 2;21,20 and v = 1;10,40 = 106 90 .

Similarly, 2u = 1;14,40 and u = 0;37,20 = 56 .

Multiplying by 90 gives x = 56, 90 1 d = 106. In the second part, v + u = 2;05 = 2 12 and v − u = 0;28,48 = 12 .

25 So 769 . Similarly, 2u = 1;36,12 and u = 0;48,06 = 481 2v = 2;33,48 and v = 1;16,54 = 600 600 .

319 Multiplying by 600 gives x = 481, d = 769. Next, if v = 481 and u = , then 360 360 289 161 v + u = 2 29 = 2;13,20. Finally, if v = 240 and u = 240 , then v + u = 1 78 = 1;52,30.

27.

98569 The equations for u and v can be solved to give v = 1:22,08,27 = 295707 216000 = 72000 and 67319 201957 u = 0;56,05,57 = 216000 = 72000 . Thus the associated Pythagorean triple is 67319, 72000, 98569.

28.

The two equations are x2 + y2 = 1525; y = 23 x + 5. If we substitute the second equation into the first and simplify, we get 13x2 + 60x = 13500. The solution is then x = 30, y = 25.

6

Solutions 29.

If we √ guess that the length of the rectangle is 60, then the width is 45 and the diagonal is 602 + 452 = 75. Since this value is 1 87 times the given value of 40, the correct length of the rectangle should be 60 ÷ 1 78 = 32. Then the width is 24.

30.

One way to solve this is to let x and x − 600 be the areas of the two fields. Then the equation is 32 x + 12 (x − 600) = 1100. This reduces to 67 x = 1400, so x = 1200. The second field then has area 600.

31.

1 Let x be the weight of the stone. The equation to solve is then x − 17 x − 13 (x − 17 x) = 60.

1 We do this using false position twice. First, set y = x − 7 x. The equation in y is then 1 1 y − 13 y = 60. We guess y = 13. Since 13 − 13 13 = 12, instead of 60, we multiply our guess by 5 to get y = 65. We then solve x − 17 x = 65. Here we guess x = 7 and calculate the value of the left side as 6. To get 65, we need to multiply our guess by 1 65 5 5 65 6 = 10 6 . So our answer is x = 7 × 6 = 75 6 gin, or 1 mina 15 6 gin.

32.

1 We do this in three steps, each using false position. First, set z = x − 17 x + 11 (x − 17 x).

1 The equation for z is then z − 13 z = 60. We guess 13 for z and calculate the value of the left side to be 12, instead of 60. Thus we must multiply our original guess by 5 1 and put z = 65. Then set y = x − 17 x. The equation for y is y + 11 y = 65. If we now guess y = 11, the result on the left side is 12, instead of 65. So we must multiply our 65 7 1 7 guess by 12 to get y = 715 12 = 59 12 . We now solve x − 7 x = 59 12 . If we guess x = 7, 7 the left side becomes 6 instead of 59 12 . So to get the correct value, we must multiply 715 715 5005 37 37 7 by 715 12 /6 = 72 . Therefore, x = 7 × 72 = 72 = 69 72 gin = 1 mina 9 72 gin.

33.

Start with a square of side x and cut off a strip of width a from the right side. The remaining rectangle then has area x2 − ax, or b. This rectangle can then be thought of as a square of side x − a/2 that is missing a small square of side a/2. If one adds back that small square, then the square of side x − a/2 has area b + (a/2)2 , so we can find x.

34.

2 2 The equation x − 60 x = 7 is equivalent to x − 60 = 7x or to x − 7x = 60. The solution q 7 is then x = ( 72 )2 + 60 + 27 = 17 2 + 2 = 12. Thus the two numbers are 12 and 5.

35.

Given the appropriate coefficients, the equation becomes 49 a2 + a + 34 a = 23 18 , where a is the length of the arc. If we scale up by 94 , we get the equation ( 94 a)2 + 73 ( 49 a) = 46 81 . q 4 46 7 25 7 7 2 The algorithm for this type of equation gives 9 a = ( 6 ) + 81 − 6 = 18 − 6 = 29 .

Thus a = 21 .

36.

The equation is 23 x2 + 13 x = 13 . To solve, we scale by 23 : ( 32 x)2 + 13 ( 23 x) = 92 . The q solution is 23 x = ( 16 )2 + 29 − 16 = 12 − 61 = 13 . Thus x = 12 .

37.

All the triangles in Figure 1.15 are similar to one another, and therefore their sides are all in the ratio 3:4:5. Therefore, AE = 35 AD = 0;36 × 0;36 = 0;21,36. Also, DE = 54 AD = 0;48 × 0;36 = 0;28,48. Also, EF = 45 DE = 0;48 × 0;28,48 = 0;23,02,24.

Finally, DF = 35 DE = 0;36 × 0;28,48 = 0;17,16,48.

38.

If the circumference is 60, then the radius is 10. Thus, if the distance of the chord from the circumference is x, then we have a right triangle of sides 6 and 10 − x, with hypotenuse 10. The Pythagorean theorem leads to the equation 62 + (10 − x)2 = 100, or x2 + 36 = 20x, for which the only valid solution is x = 2.

Solutions 39.

7

The two equations are ℓ + w = 7, ℓw + 12 ℓ + 31 w = 15. To put this into a standard Babylonian form, we can rewrite the second equation in the form (ℓ + 13 )(w + 12 ) = 15 61 .

We can then rewrite the first equation as (ℓ + 31 ) + (w + 12 ) = 7 65 . Then the Babylonian q 47 5 13 47 2 algorithm yields ℓ + 13 = 47 ( 12 ) − 91 12 + 6 = 12 + 12 = 3 . Therefore, ℓ = 4 and so w = 3.

CHAPTER 2 1.

125 = ρκϵ, 62 = ξβ, 4821 = ′ δωκα, 23, 855 = Mβ ′ γωνϵ

2.

8 9 = ∠ γ́ ίή (8/9 = 1/2 + 1/3 + 1/18)

3.

The answer is in the back of the text. The basic idea is that 200/9 = 22 29 = 22 + 1/6 + 1/18.

4.

The average of a and c is 1/4 + 1/16 + 1/64. The average of b and d is 1/2 + 1/4 + 1/8 + 1/16. The product of the two averages is 1/8 + 1/16 + 1/32 + 1/64 + 1/32 + 1/64 + 1/128 + 1/256 + 1/128 + 1/256 + 1/512 + 1/1024, or 1/4 + 59/1024. This is slightly less than the given answer of 1/4 + 1/16.

5.

Since AB = BC; since the two angles at B are equal; and since the angles at A and C are both right angles, it follows by the angle-side-angle theorem that △EBC is congruent to △SBA and therefore that SA = EC.

6.

Because both angles at E are right angles; because AE is common to the two triangles; and because the two angles CAE are equal to one another, it follows by the angle-sideangle theorem that △AET is congruent to △AES. Therefore SE = ET.

7.

The distance from the center of the pyramid to the tip of the shadow is 378 + 342 = 720 feet. Therefore the height of the pyramid is 6/9 = 2/3 of this value, or 480 feet.

8.

Tn = 1 + 2 + · · · + n = n(n2+ 1) . Therefore the oblong number n(n + 1) is double the triangular number Tn .

9.

n2 = (n −2 1)n + n(n2+ 1) , and the summands are the triangular numbers Tn − 1 and Tn .

10.

8n(n + 1) + 1 = 4n2 + 4n + 1 = (2n + 1)2 . (2n + 1)2 − 1 = 4n2 + 4n = 8n(n2+ 1) .

2 2 2 2 2 2

11.

Suppose a + b = c . Suppose a is odd. Then a is odd. If b is odd, then b is odd and c2 is even, so c is even. If b is even, then b2 is even and c2 is odd, so c is odd. A similar result holds if c is odd.

12.

Examples using the first formula are (3,4,5), (5,12,13), (7,24,25), (9,40,41), (11,60,61). Examples using the second formula are (8,15,17), (12,35,37), (16,63,65), (20,99,101), (24,143,145).

13.

Let us assume that the second leg is commensurable to the first and let b, a be numbers representing the two legs (in terms of some unit). We may as well assume that b and a are relatively prime. Since the hypotenuse is double the first leg, we have b2 + a2 = (2a)2 = 4a2 , or b2 = 3a2 . Since b2 is a multiple of 3, it must also be a multiple of 9, so b2 = 9c2 and b = 3c. Then 9c2 = 3a2 , or a2 = 3c2 . This implies that a2 is a multiple of 9, so that a is a multiple of 3. But then both a and b are multiples of 3, contradicting the fact that they are relatively prime.

8

Solutions 14.

Since similar segments are to their corresponding circles in the same ratio, the areas of similar segments are to one another as the squares on the diameters of the circles. Thus, the areas of similar segments are also to one another as the squares on the radii of the circles. But in similar segments, the triangles formed by the two radii and chords are similar triangles. Thus the chord of one segment is to the chord in the similar segment as the radius of the first circle to the radius of the second. That is, the squares on the radii are to one another as the squares on the chords. Therefore, the areas of similar segments are to one another as the squares on their chords.

15.

By Exercise 14, the area of segment BD is the area of segment AB as the square on BD is the square on AB. But this ratio is equal to 3. Thus, the area of segment BD is three times the area of segment AB, or is equal to the sum of the areas of segments AB, AC, and CD. Therefore, the area of lune is equal to the difference between the area of the large segment and the area of segment BD. But this is equal to the difference between the area of the large segment and the areas of the three small segments, which is in turn equal to the area of the trapezoid. To construct the trapezoid, note that one √ can certainly construct a line segment equal to 3 times the length of a given line segment. To place this line segment both parallel to the original one and such that the lines connecting the endpoints of the two segments are each equal to the original line segment, we simply need to find the distance between the two segments. And that can be constructed by using the Pythagorean Theorem applied to the triangle whose hypotenuse is equal to the original segment and one leg of which is equal to half the difference between the new line segment and the original one. To circumscribe a circle around this trapezoid, note that one can construct a circle through three points, say B, A, and C. By the symmetry of the trapezoid, this circle will also go through point D.

21.

If one equates the times of the two runners, where d is the distance traveled by Achilles, the equation is d/10 = (d − 500)/(1/5). This is equivalent to 49d = 25,000, so d = 510.2 yards. Since Achilles is traveling at 10 yards per second, this will take him 51.02 seconds.

CHAPTER 3 1.

One way to do this is to use I-4. Namely, consider the isosceles triangle ABC with equal sides AB and BC also as a triangle CBA. Then the triangles ABC and CBA have two sides equal to two sides and the included angles also equal. Thus, by I-4, they are congruent. Therefore, angle BAC is equal to angle BCA, and the theorem is proved.

2.

Put the point of the compass on the vertex V of the angle and swing equal arcs intersecting the two legs at A and B. Then place the compass at A and B, respectively, and swing equal arcs, intersecting at C. The line segment connecting V to C then bisects the angle. To show that this is correct, note that triangles VAC and VBC are congruent

by SSS. Therefore, the two angles AVC and BVC are equal.

3.

Let the lines AB and CD intersect at E. Then angles AEB and CED are both straight angles, angles equal to two right angles. If one subtracts the common angle CEB from each of these, the remaining angles AEC and BED are equal, and these are the vertical angles of the theorem.

Solutions 4.

9

Suppose the three lines have length a, b, and c, with a ≥ b ≥ c. On the straight line DH of length a + b + c, let the length of DF be a, the length of FG be b, and the length of GH be c. Then draw a circle centered on F with radius a and another circle centered at G with radius c. Let K be an intersection point of the two circles. Then connect FK and GK. Triangle FKG will then be the desired triangle. For FK = FD, and this has length a. Also FG has length b, while GK = GH and this has length c. Note also that we must have b + c > a, for otherwise the two circles would not intersect. That a + b > c and a + c > b are obvious from how we have labeled the three lengths.

F

D E

G

H

J

I

5.

Suppose angle DCE is given, and we want to construct an angle equal to angle DCE at point A of line AB. Draw the line DE so that we now have a triangle DCE. (Here D and E are arbitrary points along the two arms of the given angle.) Then, by the result of Exercise 4, construct a triangle AGF, with AG along line AB, where AG = CE, AF = CD, and FG = DE. By the side-side-side congruence theorem, triangle AGF is congruent to triangle CED. Therefore, angle FAG is equal to angle DCE, as desired.

6.

Let ABC be the given triangle. Extend BC to D and draw CE parallel to AB. By I–29, angles BAC and ACE are equal, as are angles ABC and ECD. Therefore angle ACD equals the sum of the angles ABC and BAC. If we add angle ACB to each of these, we get that the sum of the three interior angles of the triangle is equal to the straight angle BCD. Because this latter angle equals two right angles, the theorem is proved.

A

B

7.

E

C

D

Place the given rectangle BEFG so that BE is in a straight line with AB. Extend FG to H so that AH is parallel to BG. Connect HB and extend it until it meets the extension of

10

Solutions FE at D. Through D draw DL parallel to FH and extend GB and HA so they meet DL in M and L, respectively. Then HD is the diagonal of the rectangle FDLH and so divides it into two equal triangles HFD and HLD. Because triangle BED is equal to triangle BMD and also triangle BGH is equal to triangle BAH, it follows that the remainders, namely rectangles BEFG and ABML, are equal. Thus ABML has been applied to AB and is equal to the given rectangle BEFG. 8.

Because triangles ABN, ABC, and ANC are similar, we have BN : AB = AB : BC, so AB2 = BN · BC, and NC : AC = AC : BC, so AC2 = NC · BC. Therefore AB2 + AC2 = BN · BC + NC · BC = (BN + NC) · BC = BC2 , and the theorem is proved.

9.

In this proof, we shall refer to certain propositions in Euclid’s Book I, all of which are proved before Euclid first uses postulate 5. (That occurs in proposition 29.) First, assume Playfair’s Axiom. Suppose line t crosses lines m and l and that the sum of the two interior angles (angles 1 and 2 in the diagram) is less than two right angles. We know that the sum of angles 1 and 3 is equal to two right angles. Therefore ∠2 < ∠3.

Now on line BB′ and point B′ construct line B′ C′ such that ∠C′ B′ B = ∠3 (Proposition 23). Therefore, line B′ C′ is parallel to line l (Proposition 27). Therefore, by Playfair’s Axiom, line m is not parallel to line l. It therefore meets l. We must show that the two lines meet on the same side as C′ . If the meeting point A is on the opposite side, then ∠2 is an exterior angle to triangle ABB′ , yet it is smaller than ∠3, one of the interior angles, contradicting Proposition 16. We have therefore derived Euclid’s postulate 5.

B’

C‘

2

A?

m 3 1 B

1

P

m

1 n Q

R

1

Second, assume Euclid’s postulate 5. Let l be a given line and P a point outside the line.

Construct the line t perpendicular to l through P (Proposition 12). Next, construct the line m perpendicular to line t at P (Proposition 11). Since the alternate interior angles formed by line t crossing lines m and l are both right and therefore are equal, it follows from Proposition 27 that m is parallel to l. Now suppose n is any other line through P. We will show that n meets l and is therefore not parallel to l. Let ∠1 be the acute angle that n makes with t. Then the sum of angle 1 and angle PQR is less than two right angles. By postulate 5, the lines meet.

Solutions

11

Note that in this proof, we have actually proved the equivalence of Euclid’s postulate 5 to the statement that given a line l and a point P not on l, there is at most one line through P which is parallel to l. The other part of Playfair’s Axiom was proved (in the second part above) without use of postulate 5 and was not used at all in the first part.

10.

One possibility for an algebraic translation: If the line has length a and is cut at a point with coordinate x, then 4ax + (a − x)2 = (a + x)2 . This is a valid identity. Here is a geometric diagram, with AB = ME = a and CB = BD = BK = KR = x: C

A

D

B T

G

M

K

O

Q

R

H

L

S

N P

U

E

11.

F

If ABC is the given acute-angled triangle and AD is perpendicular to BC, then the theorem states that the square on AC is less than the squares on CB and BA by twice the rectangle contained by CB and BD. If we label AC as b, BA as c, and CB as a, then BD = c cos B. Thus the theorem can be translated algebraically into the form b2 = a2 + c2 − 2ac cos B, exactly the law of cosines in this case.

A

B

12.

D

C

Suppose the diameter CD of a circle with center E bisects the chord AB at F. Then join EA and EB, forming triangle EAB. Triangles AEF and BEF are congruent by side-side-side (since AE = BE are both radii of the circle and F bisects AB). Therefore angles EFA and EFB are equal. But the sum of those two angles is equal to two right angles. Hence each is a right angle, as desired. To prove the converse, use the same construction and note that triangle AEB is isosceles, so angle EAF is equal to angle EBF, while both angles EFA and EFB are right by hypothesis. It follows that triangles AEF and BEF are again congruent, this time by angle-angle-side. So AF = BF, and the diameter bisects the chord.

12

Solutions C

E

A

F

B

D

13.

In the circle ABC, let the angle BEC be an angle at the center and the angle BAC be an angle at the circumference which cuts off the same arc BC. Connect ∠EAB. Similarly, ∠FEC is double ∠EAC. Therefore the entire ∠BEC is double the entire ∠BAC. Note that this argument holds as long as line EF is within ∠BEC. If it is not, an analogous argument by subtraction holds.

B

A

E

F C

14.

Let ∠BAC be an angle cutting off the diameter BC of the circle. Connect A to the center E of the circle. Since EB = EA, it follows that ∠EBA = ∠EAB. Similarly, ∠ECA = ∠EAC. Therefore the sum of ∠EBA and ∠ECA is equal to ∠BAC. But the sum of all three angles equals two right angles. Therefore, twice ∠BAC is equal to two right angles, and angle BAC is itself a right angle.

B

A E

C

15.

Let triangle ABC be given. Let D be the midpoint of AB and E the midpoint of AC. Draw a perpendicular at D to AB and a perpendicular at E to AC and let them meet at point F (which may be inside or outside the triangle, or on side BC). Assume first that

Solutions

13

F is inside the triangle, and connect FB, FA, and FC. Since BD = BA, triangles FDB and FDA are congruent by side-angle-side. Therefore FB = FA. Similarly, triangles FEA and FEC are congruent. So FC = FA. Therefore all three lines FA, FB, and FC are equal, and a circle can be drawn with center F and radius equal to FA. This circle will circumscribe the given triangle. Finally, note that the identical construction works if F is on line BC or if F is outside the triangle.

A

A A D D

E F

F B

E

B B

16.

D

E C

C F

C

Let G be the center of the given circle and AGD a diameter. With center at D and radius DG, construct another circle. Let C and E be the two intersections of the two (equal) circles, and connect DC and DE. Then DE and DC are two sides of the desired regular hexagon. To find the other four sides, draw the diameters CGF and EGB. Then CB, BA, AF, and FE are the other sides. To demonstrate that we have in fact constructed a regular hexagon, note that all the triangles whose bases are sides of the hexagon and whose other sides are radii are equilateral; thus all the sides of the hexagon are equal and all the angles of the hexagon are also equal.

H

D E

C G

F

B A

17.

In the circle, inscribe a side AC of an equilateral triangle and a side AB of an equilateral pentagon. Then arc BC is the difference between one-third and one-fifth of the 2 circumference of the circle. That is, arc BC = 15 of the circumference. Thus, if we bisect that arc at E, then lines BE and EC will each be a side of a regular 15-gon.

18.

Let a = s1 b + r1 , b = s2 r1 + r2 , . . ., rk − 1 = sk + 1 rk . Then rk divides rk − 1 and therefore also rk − 2 , . . . , b, a. If there were a greater common divisor of a and b, it would divide r1 , r2 , . . ., rk . Since it is impossible for a greater number to divide a smaller, we have shown that rk is in fact the greatest common divisor of a and b.

14

Solutions 19.

963 = 1 · 657 + 306 657 = 2 · 306 + 45 306 = 6 · 45 + 36 45 = 1 · 36 + 9 36 = 4 · 9 + 0 Therefore, the greatest common divisor of 963 and 657 is 9.

4001 = 1 · 2689 + 1312 2689 = 2 · 1312 + 65 1312 = 20 · 65 + 12 65 = 5 · 12 + 5 12 = 2 · 5 + 2 5 = 2 · 2+1 Therefore, the greatest common divisor of 4001 and 2689 is 1.

20.

46 = 7 · 6 + 4 6 =1 · 4 +2 4 =2 · 2

23 = 7 · 3 + 2 3 =1 · 2 +1 2 =2 · 1

Note that the multiples 7, 1, 2 in the first example equal the multiples 7, 1, 2 in the second.

21.

33 = 2 · 12 + 9 12 = 1 · 9 + 3 9 = 3·3

11 = 2 · 4 + 3 4 =1 · 3 +1 3 =3 · 1

It follows that both ratios can be represented by the sequence (2, 1, 3).

22.

Since 1 − x = x2 , we have 1 = 1 · x + (1 − x) = 1 · x + x2 x = 1 · x2 + (x − x2 ) = 1 · x2 + x(1 − x) = 1 · x2 + x3 x2 = 1 · x3 + (x2 − x3 ) = 1 · x3 + x2 (1 − x) = 1 · x3 + x4 ···

Thus 1 : x can be expressed in the form (1, 1, 1, . . .).

Solutions

15

23.

If d is the diagonal of a square of side s, then the first division gives d = 1s + r. To understand the next steps, √ it is probably √ easiest to set s = 1 and deal with the numerical values. Therefore, d = 2, and r = 2 − 1. Our next division gives s = 2r + t, where √ t = 3 − 2 2. Geometrically, if we lay off s along the diagonal, then r is the remainder d − s. Then draw a square of side r with part of the side of the original square being its diagonal. Note that if we now lay off r along the diagonal of that square, the remainder is t. In other words, r is the difference between the diagonal of a square of side s and s, while t is the difference between the diagonal of a square of side r and r. It follows that if one performs the next division in the process, we will get the same relationship.

That is, r = 2t + u, where now u is the difference between the diagonal of a square of side t and t. Thus, this process will continue indefinitely and the ratio d : s can be expressed as (1, 2, 2, 2, . . .).

24.

Since a > b, there is an integral multiple m of a − b with m(a − b) > c. Let q be the first multiple of c that exceeds mb. Then qc > mb ≥ (q − 1)c, or qc − c ≤ mb < qc.

Since c < ma − mb, it follows that qc ≤ mb + c < ma. But also qc > mb. Thus we have a multiple (q) of c that is greater than a multiple (m) of b, while the same multiple (q) of c is not greater than the same multiple (m) of a. Thus by definition 7 of Book V, c : b > c : a.

25.

Let A : B = C : D = E : F. We want to show that A : B = (A + C + E) : (B + D + F). Take any equimultiples mA and m(A + C + E) of the first and third and any equimultiples nB and n(B + D + F) of the second and fourth. Since m(A + C + E) = mA + mC + mE, and since n(B + D + F) = nB + nD + nF, and since whenever mA > nB, we have mC > nD and mE > nF, it follows that mA > nB implies that m(A + C + E) > n(B + D + F). Since a similar statement holds for equality and for “less than,” the result follows from Eudoxus’s definition. A modern proof would use the fact that a1 bi = b1 ai for every i and then conclude that a1 (b1 + b2 + · · · + bn ) = b1 (a1 + a2 + · · · + an ).

26.

Given that a : b = c : d, we want to show that a : c = b : d. So take any equimultiples ma, mb of a and b and also equimultiples nc, nd of c and d. Now ma : mb = a : b = c : d = nc : nd. Thus if ma > nc, then mb > nd; if ma = nc, then mb = nd; and if ma < nc, then mb < nd. Thus, by the definition of equal ratio, we have a : c = b : d.

27.

Let AB = 9 and BC = 5. Draw a circle with AC as diameter and erect a perpendicular to AC at B, meeting the circle at E. Then BE is the desired length x.

28.

Suppose the first of the equal and equiangular parallelograms has sides of length a and b while the second has sides of length c and d, each pair surrounding an angle equal to α. Since the area of a parallelogram is the product of the two sides with the sine of the included angle, we know that ab sin α = cd sin α. It follows that ab = cd or that a : c = d : b, as desired. Conversely, if a : c = d : b and the parallelograms are equiangular with angle α between each pair of given sides, then ab = cd and ab sin α = cd sin α, so the parallelograms are equal. Euclid’s proof is, of course, different from this modern one. Namely, if the two parallelograms are P1 = ADBF and P2 = BGCE, with equal angles at B, Euclid places them so that FB and BG are in a straight line as are EB and BD. He then completes the third parallelogram P3 = FBEK. Since P1 = P2 , we have P1 : P3 = P2 : P3 . But P1 : P3 = DB : BE, since BF is common, and P2 : P3 = BG : BF, since BE is common. Thus, DB : BE = BG : BF, the desired conclusion. The converse is proved by reversing the steps.

16

Solutions C

E F

A

B

G

D

29

We want to prove that numbers to which the same number has the same ratio are equal.

So suppose a : b = a : c. Therefore, a is the same multiple or the same part or the same parts of both b and c. That is, a = mb and a = mc for some rational number m.

But this means that b = m1 a and c = m1 a, thus b = c.

30.

Suppose a : b = f : g and suppose the numbers c, d, . . ., e are the numbers in continued proportion between a and b. Let r, s, t, . . ., u, v be the smallest numbers in the same ratio as a, c, d, . . ., e, b. Then r, v are relatively prime and r : v = a : b = f : g. It follows that f = mr, g = mv for some integer m and that the numbers ms, mt, . . ., mu are in the same ratio as the original set of numbers. Thus there are at least as many numbers in continued proportion between f and g as there were between a and b. Since the same argument works starting with f and g, it follows that there are exactly as many numbers in continued proportion between f and g as between a and b. Since there is no integer between n and n + 1, it follows that there cannot be a mean proportional between any pair of numbers in the ratio (n + 1) : n.

31.

The number ab is the mean proportional between a2 and b2 .

32.

The numbers a2 b and ab2 are the two mean proportionals between a3 and b3 .

33.

If a2 measures b2 , then b2 = ma2 for some integer m. Since every prime number which divides b2 must divide ma2 and therefore must divide either m or a2 , it follows by counting primes that m must itself be a square. Thus m = n2 and b = na, so a measures b. Conversely, if a measures b, then b = na and b2 = n2 a2 , so a2 measures b2 .

34.

Suppose m factors two different ways as a product of primes: m = pqr · · · s = p′ q′ r′ · · · s′ . Since p divides pqr · · · s, it must also divide p′ q′ r′ · · · s′ . By VII–30, p must divide one of the prime factors, say p′ . But since both p and p′ are prime, we must have p = p′ . After canceling these two factors from their respective products, we can then repeat the argument to show that each prime factor on the left is equal to a prime factor on the right and conversely.

35.

One standard modern proof is as follows. Assume there are only finitely many prime numbers p1 , p2 , p3 , . . ., pn . Let N = p1 p2 p3 · · · pn + 1. There are then two possibilities.

Either N is prime or N is divisible by a prime other than the given ones, since division by any of those leaves remainder 1. Both cases contradict the original hypothesis, which therefore cannot be true.

36.

We keep adding powers of 2 until we get a prime. After 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127, the next sums are 255, 511, 1023, 2047, 4095, and 8191. The first five of these are not prime (note √ that 2047 = 23 × 89). But 8191 is prime (check by dividing by all primes less than 8191). So the next perfect number is 8191 × 4096 = 33,550,336.

Solutions

17

37.

Since BC is the side of a decagon, triangle EBC is a 36-72-72 triangle. Thus ∠ECD = 108◦ . Since CD, the side of a hexagon, is equal to the radius CE, it follows that triangle ECD is an isosceles triangle with base angles equal to 36◦ . Thus triangle EBD is a 3672-72 triangle and is similar to triangle EBC. Therefore BD : EC = EB : BC or BD : CD = CD : BC and the point C divides the line segment BD in extreme and mean ratio.

38.

By Exercise 37, if we set d to be the length of the side√of a decagon, we have (1 + d) : 1 = 1 : d or d2 + d − 1 = 0. It follows that d = 52−1 . The length p of the side p √ of a pentagon is (p. 91) p = 12 10 − 2 5. It is then straightforward to show that p2 = 12 + d2 as asserted. p √ 1 The edge length of the icosahedron in a sphere of diameter 1 is ei = 10 50 − 10 5 = 0.52573111. The circumradius of the equilateral triangle with this edge length is r = 2 sinei60◦ = 0.303531. The edge length of the dodecahedron in the same √ √ sphere is ed = 16 ( 15 − 3) = 0.35682209. The circumradius of the pentagon with this edge length is r = 2 sined36◦ = 0.303531, the same value as in the icosahedron.

39.

40.

We begin with a rectangle of sides x and y. We then lay off y along x, with the remainder being x − y = 7. Divide the rectangle with sides 7 and y in half and move one half to the bottom. We then add the square of side 27 to get a square of side y + 72 = x − 27 with 11 7 11 7 area 18 + ( 72 )2 = 121 4 . Thus y = 2 − 2 = 2 and x = 2 + 2 = 9.

41.

First, to solve the two equations, we set x = a/y and substitute into y2 − αx2 = b. After multiplying by y2 , we get the fourth degree equation y4 − αa2 = by2 . This is quadratic in y2 , so we can solve to get y2 =

b+

√

b2 + 4αa2 .

2

(Note that we only use the plus sign, since y2 must be positive.) Then r y= ±

b+

√

b2 + 4αa2 .

2

We can then solve for x: r√ b2 + 4αa2 − b x= ± .

2α (Of course, since a is assumed positive, we take the positive solution for x when we have the positive one for y, as well as the negative solution for x with the negative one for y.) The asymptotes of the hyperbola xy = a are the lines x = 0 and y = 0, and these are the axes of the hyperbola y2 − αx2 = b. Thus the asymptotes of one hyperbola are the axes of the other.

18

Solutions

CHAPTER 4 1.

If x is the distance to the 14 kg end, then 10(10 − x) = 14x, 100 = 24x, and x = 4 61 m from the 14 kg end.

2.

Since 8 · 10 = 80 and 12 · 8 = 96, the lever inclines toward the 12 kg weight.

3.

Since a weight W of gold displaces a volume V1 of fluid, a weight w1 of gold will displace wW1 · V1 of fluid. Similarly, a weight w2 of silver displaces wW2 · V2 of fluid. Thus the wreath, of total weight W, displaces the sum of these two amounts of fluid. That is, V = wW1 · V1 + wW2 · V2 . Since W = w1 + w2 , it follows that (V − V1 )w1 = (V2 − V)w2 , and the desired result follows.

4.

Lemma 1: DA/DC = OA/OC by Elements VI–3. Therefore DA/OA = DC/OC = (DC + DA)/(OC + OA) = AC/(CO + OA). Also, DO2 = OA2 + DA2 by the Pythagorean Theorem.

Lemma 2: AD/DB = BD/DE = AC/CE = AB/BE = (AB + AC)/(CE + BE) = (AB + AC)/BC. Therefore, AD2 /BD2 = (AB + AC)2 /BC2 . But AD2 = AB2 − BD2 .

So (AB2 − BD2 )/BD2 = (AB + AC)2 /BC2 and AB2 /BD2 = 1 + (AB + AC)2 /BC2 .

5.

Set r = 1, ti and ui as in the text, and Pi the perimeter of the ith circumscribed polygon.

Then the first ten iterations of the algorithm give the following: t1 = 0.577350269 t2 = 0.267949192 t3 = 0.131652497 t4 = 0.065543462 t5 = 0.03273661 t6 = 0.016363922 t7 = 0.0081814134 t8 = 0.004090638249 t9 = 0.002045310568 t10 = 0.001022654214

6.

u1 = 1.154700538 u2 = 1.03527618 u3 = 1.008628961 u4 = 1.002145671 u5 = 1.0005357 u6 = 1.00013388 u7 = 1.000033467 u8 = 1.000008367 u9 = 1.000002092 u10 = 1.000000523

P1 = 3.464101615 P2 = 3.21539031 P3 = 3.159659943 P4 = 3.146086215 P5 = 3.1427146 P6 = 3.141873049 P7 = 3.141662746 P8 = 3.141610175 P9 = 3.141597032 P10 = 3.141593746

Let d be the diameter of the circle, ti the length of one side of the regular inscribed polygon of 3 · 2i sides, and ui the length of the other leg of the right triangle formed from the diameter and the side of the polygon. Then t2i ti + 1 2 = 2 2 d ti + (d + ui )2 or ti + 1 = q

dti t2i + (d + ui )2

ui + 1 =

p d 2 − ti + 1 2 .

Solutions

19

If Pi is the perimeter of the ith inscribed polygon, then Pdi = 3·d2 ti . So let d = 1. Then i

√ t1 = 2d = 0.5 and u1 = 23d = 0.8660254. Then repeated use of the algorithm gives us:

t1 = 0.500000000 t2 = 0.258819045 t3 = 0.130526194 t4 = 0.06540313 t5 = 0.032719083 t6 = 0.016361731 t7 = 0.008181140 t8 = 0.004090604 t9 = 0.002045306 7.

u1 = 0.866025403 u2 = 0.965925826 u3 = 0.991444861 u4 = 0.997858923 u5 = 0.999464587 u6 = 0.999866137 u7 = 0.999966533 u8 = 0.999991633 u9 = 0.999997908

P1 = 3.000000000 P2 = 3.105828542 P3 = 3.132628656 P4 = 3.13935025 P5 = 3.141031999 P6 = 3.141452521 P7 = 3.141557658 P8 = 3.141583943 P9 = 3.141590016

In the diagram the spiral is given by r = a(θ − π2 ) rather than r = aθ as mentioned in the text. The slope of the tangent line to a curve given in polar coordinates is dr sin θ + r cos θ dy . = drdθ dx dθ cos θ − r sin θ π So when θ = 5π 2 , that is when the curve has made one full turn starting at θ = 2 , the 1 a slope of the tangent line is −2πa = − 2π . So the equation of the tangent line through 1 the point L = (0, 2πa) is then y = − 2π x + 2πa. To find the point D, we set y = 0 and 2 find that the x coordinate of D is 4π a. Since L = (0, 2πa), the radius AL of the circle is 2πa and its circumference is 2π · 2πa = 4π2 a, the same as the length of AD.

8.

Let the equation of the parabola be y = −x2 + 1. Then the tangent line at C = (1, 0) has the equation y = −2x + 2. Let the point O have coordinates (−a, 0). Then MO = 2a + 2, OP = −a2 + 1, CA = 2, AO = −a + 1. So MO : OP = (2a + 2) : (1 − a2 ) = 2 : (1 − a) = CA : AO.

9. a.

Draw line AO. Then MS · SQ = CA · AS = AO2 = OS2 + AS2 = OS2 + SQ2 .

b.

Since HA = AC, we have HA : AS = MS : SQ = MS2 : MS · SQ = MS2 : (OS2 + SQ2 ) = MN2 : (OP2 + QR2 ). Since circles are to one another as the squares on their diameters, the latter ratio equals that of the circle with diameter MN to the sum of the circle with diameter OP and that with diameter QR.

c.

Since then HA : AS = (circle in cylinder):(circle in sphere + circle in cone), it follows that the circle placed where it is is in equilibrium about A with the circle in the sphere together with the circle in the cone if the latter circles have their centers at H.

d.

Since the above result is true whatever line MN is taken, and since the circles make up the three solids involved, Archimedes can conclude that the cylinder placed where it is is in equilibrium about A with the sphere and cone together, if both of them are placed with their center of gravity at H. Since K is the center of gravity of the cylinder, it follows that HA : AK = (cylinder):(sphere + cone).

20

Solutions e.

Since HA = 2AK, it follows that the cylinder is twice the sphere plus the cone AEF. But we know that the cylinder is three times the cone AEF. Therefore the cone AEF is twice the sphere. But the cone AEF is eight times the cone ABD, because each of the dimensions of the former are double that of the latter. Therefore, the sphere is four times the cone ABD.

10.

Since BOAPC is a parabola, we have DA : AS = BD2 : OS2 , or HA : AS = MS2 : OS2 . Thus HA : AS = (circle in cylinder):(circle in paraboloid). Thus the circle in the cylinder, placed where it is, balances the circle in the paraboloid placed with its center of gravity at H. Since the same is true whatever cross section line MN is taken, Archimedes can conclude that the cylinder, placed where it is, balances the paraboloid, placed with its center of gravity at H. If we let K be the midpoint of AD, then K is the center of gravity of the cylinder. Thus HA : AK = cylinder : paraboloid. But HA = 2AK. So the cylinder is double the paraboloid. But the cylinder is also triple the volume of the cone ABC. Therefore, the volume of the paraboloid is 3/2 the volume of the cone ABC which has the same base and same height.

11.

Suppose the radius of the base of the cylinder is r and the height is h. The volume of the parallelepiped circumscribing the cylinder is 4r2 h. To find the volume of the segment cut off by the plane, we note that the equation of the cutting plane is z = (h/r)x.

Therefore, the volume of the segment is Z r Z √r2 − x2 2 0

0

h x dy dx = 2 r

Z r 0

r 2h 3 2 2 2h h p2 x r − x2 dx = − (r2 − x2 )3/2 = r = hr . r 3r 3r 3 0

This value is 1/6 of the volume of the parallelpiped, as desired.

12.

Let the parabola be given by y = a − bx2 . Then the area A of the segment cut off by the x axis is given by √a/b R √a/b (a − bx2 ) dx = 2 ax − 31 bx3 0 A =2 0 p p a 4a p a = 2a ab − 2a 3 b = 3 b.

Since the area of the inscribed triangle is a

13.

2

pa

b , the result is established.

Let the equation of the parabola be y = x , and let the straight line defining the segment be the line through the points (−a, a2 ) and (b, b2 ). Thus the equation of this line is (a − b)x + y = ab, and its normal vector is N = (a − b, 1). Also, since the midpoint of a b2 + a2 that line segment is B = ( b− 2 , 2 ), the x coordinate of the vertex of the segment is 2 2 b−a 2 . If S = (x, x ) is an arbitrary point on the parabola, then the vector M from (−a, a ) 2 2 to S is given by (x + a, x − a ). The perpendicular distance from S to the line is then the dot product of M with N, divided by the length of N. Since the length of N is a constant, to maximize the distance it is only necessary to maximize this dot product. The dot product is (x + a, x2 − a2 ) · (a − b, 1) = ax − bx + a2 − ab + x2 − a2 = ax − bx + x2 − ab. a The maximum of this function occurs when a − b + 2x = 0, or when x = b− 2 . And, as we have already noted, the point on the parabola with that x coordinate is the vertex of the segment. So the vertex is the point whose perpendicular distance to the base of the segment is the greatest.

Solutions

21

14.

Let r be the radius of the sphere. Then we know from calculus that the volume of the sphere is VS = 43 πr3 and the surface area of the sphere is AS = 4πr2 . The volume of the cylinder whose base is a great circle in the sphere and whose height equals the diameter is VC = πr2 (2r) = 2πr3 , while the total surface area of the cylinder is AC = (2πr)(2r) + 2πr2 = 6πr2 . Therefore, VC = 23 VS and AC = 32 AS , as desired.

15.

Let A be the area bounded by one complete turn of the spiral and AC the area of the circle. Then A=

16.

17.

18.

1 2

Z 2 0

2π

πa2 θ2 dθ =

4 1 2 θ3 1 1 = π3 a2 = π(2πa)2 = AC . a 2 3 0 3 3 3

Suppose the cylinder P has diameter d and height h, and suppose the cylinder Q is constructed with the same volume but with its height and diameter both equal to f. It follows that d 2 : f 2 = f : h, or that f 3 = d 2 h. It follows that one needs to construct the cube root of the quantity d 2 h, and this can be done by finding two mean proportionals between 1 and d 2 h, or, alternatively, two mean proportionals between d and h (where the first one will be the desired diameter f). p The focus of y2 = px is at ( 4p , 0). The length of the latus rectum is 2 p p4 = p.

The equation of the ellipse can be rewritten as 2 x2 − 2ax + 2a p y = 0, or finally as (x − a)2

a2

+

p 2 2 2a x − px + y

=

0 or as

y2 = 1. pa/2

Therefore the center of the ellipse is at (a, 0) and b2 = pa 2 . The hyperbola can be treated analogously.

19.

Let the parabola be y2 = px and the point C = (x0 , y0 ). Then the tangent line at C has slope 2yp0 , and the equation of the tangent line is y = 2yp0 (x − x0 ) + y0 . If we set y = 0, we can solve this equation for x to get x = −x0 .

20.

Begin with the ellipse given in the form b2 x2 + a2 y2 = a2 b2 . The slope of the tangent 2 line at the point C = (−x0 , y0 ) is then ba2 xy00 , and the equation of the tangent line is 2

y − y0 = ba2 xy00 (x + x0 ). If we set y = 0, then the x coordinate of the point where the 2

2

tangent line intersects the axis is x = −x0 − ab2yx00 . It is then straightforward to show that AH : AG = (a − x) : (−a − x) is equal to BH : BG = (a + x0 ) : (−x0 + a).

21.

We can arrange the axes and origin so that the first conic is given by the equation y2 = ax + bx2 . Then the second conic will be given by the general equation Ax2 + Bxy + y2 + Dx + Ey + F = 0, where we have chosen the coefficient of y2 to be 1 for reasons that will soon √ be clear. To find the intersections, we can solve the first equation for y (y = ± ax2 + bx) and substitute in the second. Since this will involve radicals in the new equation for x, we can put all of them on one side and then square both sides, resulting in a fourth degree equation for x. Such an equation has at most four real solutions. We need to show that these solutions for x lead to no more than four actual intersections. So rewrite the equation for the second conic as y2 = −(Ax2 + Bxy + Dx + Ey + F) and subtract this from the equation for the first conic. This eliminates the y2 term and gives (Bx + E)y + (A + b)x2 + (a + D)x + F = 0,

22

Solutions an equation certainly equivalent to the fourth degree equation. Note that as long as Bx + E is not 0, this equation can be solved for y in terms of x. In other words, each x that is a solution of the fourth degree equation corresponds to exactly one y, thus insuring that there are no more than four intersections. On the other hand, if one of the solutions of the fourth degree equation is x = −E/B, thus giving Bx + E = 0, then the equivalent equation reduces to a quadratic in x, which has at most two other solutions.

In other words, although x = −E/B may correspond to two actual intersections, one with a positive y value and one with a negative y value, there are then only two additional solutions to the original fourth degree equation, each corresponding to exactly one y value. (Note that if B = E = 0, then again there are only two solutions for x, but each may correspond to two different y values.) Thus, there are at most four real intersection points to the two conics.

22.

For two conics to be tangent to one another at a point (x1 , y1 ), the fourth degree equation mentioned in Exercise 21 must have a double root at x1 . Thus, since there can be at most two double roots to a fourth degree equation, and since, by the same argument as in Exercise 21, each double root can correspond to at most one value for y1 , there can be no more than two points at which two conics are tangent. (Note that if x = −E/B is a double root of the fourth degree equation, it corresponds to two different y values and therefore does not give a tangent point.)

23. a.

Let the ellipse be given by the equation b2 x2 + a2 y2 = a2 b2 . Let P have coordinates 2 (x0 , y0 ). Then the slope of the tangent line at P is − ba2 xy0 . Thus the equation of line 0 2

DK is y = − ba2 xy00 x. By solving this equation simultaneously with the equation of

the ellipse, we get the coordinates of the point D as (− ayb0 , bxa0 ). It follows that the 2

ay0 /b slope of the tangent line at D is ab2 bx = yx00 , which is the slope of the diameter 0 /a PG, as desired.

b.

Given that the coordinates of P are (x0 , y0 ), it follows that tan θ = xy00 as before. Sim-

bx0 /a ilarly, since the coordinates of D are (− ayb0 , bxa0 ), it follows that tan α = − ay = 0 /b 2

− ba2 xy0 .

0

c.

Take an arbitrary point S in the plane with rectangular coordinates (x, y) and oblique coordinates (x′ , y′ ). By drawing lines from S parallel to the two original axes and to the two oblique axes, one can show that x = x′ cos θ − y′ cos(180 − α) = x′ cos θ + y′ cos α and that y = x′ sin θ + y′ sin(180 − α) = x′ sin θ + y′ sin α. If we replace x and y in the equation of the ellipse by their values in terms of x′ and y′ , we get the equation specified in the problem, once we notice that b2 cos θ cos α + a2 sin θ sin α = 0, given the values for tan θ and tan α found in part b.

d.

Let y = (tan θ)x be the equation of the diameter PG. If we solve this equation simultaneously with the original equation for the ellipse, we find the coordinates of the point P to be

ab ab tan θ x= √ , y= √ .

2 2 2 2 b + a tan θ b + a2 tan2 θ

Solutions

23

It follows that a′ =

p ab sec θ x2 + y2 = √ . b2 + a2 tan2 θ

Similarly, ab tan α b′ = √ .

2 b + a2 tan2 α Then a′ = 2

a2 b2 sec2 θ a2 b2 = 2 2 2 A b + a tan θ

or

A=

a2 b2 . a′ 2

Similarly, C=

a2 b2 . b′ 2

If we substitute these values for A and C into the equation of the ellipse given in ′2 ′2 part (c), we get the equation ax′ 2 + by′ 2 = 1, or ′2

y =b

′2

a′ − x′ a′ 2 2

2

!

b′ ′ b′ (a − x′ )(a′ + x′ ) = 2 x1 ′ x2 ′ 2 ′ a a′ 2

=

2

as desired. e.

Since PF = a′ sin(α − θ) and CD = b′ , we have PF × CD = 2 2 sin(α−θ) a′ b′ sin(α − θ) = a b √ . But since b2 cos θ cos α + a2 sin θ sin α = 0, it folAC lows that (b2 cos θ cos α + a2 sin θ sin α)2 = 0 and therefore that

AC = a2 b2 (sin α cos θ − cos α sin θ)2 = a2 b2 sin2 (α − θ).

Therefore, PF × CD =

a2 b2 sin(α − θ) = ab ab sin(α − θ)

as claimed.

24.

For simplicity, we will show that the sum of the squares on the halves of two conjugate diameters is equal to the sum of the squares on the halves of the two axes. Using Fig. 4.32, let P = (x0 , y0 ) and let the equation of the ellipse be b2 x2 + a2 y2 = a2 b2 . To find the slope of the tangent line RP, we take differentials: 2b2 x dx + 2a2 y dy = 0, so dy/dx = −b2 x/a2 y. Thus, the slope of RP is −b2 x0 /a2 y0 , and the equation of line DK is y = −(b2 x0 /a2 y0 )x. To find the coordinates of point D, we take the intersection of this line with the ellipse. That is, we substitute for y in the equation for the ellipse: 4 2 b x0 b2 x2 + a2 x2 = a2 b2 . a4 y20

24

Solutions Solving this for x and noting that the x-coordinate of D is negative, we get x = −q

a2 y0 a2 y20 + b2 x20

=−

a2 y0 ay =− 0 ab b

and thus

y=

bx0 . a

Therefore, the square on CD is a2 y20 b2 x20 + 2 . b2 a If we then add this to x20 + y20 , the square on CP, we get b2 x2 a2 y20 + y20 + 2 0 + x20 = (a2 + b2 ) 2 b a

2

y0 x20 + = a2 + b2 . b2 a2

And, of course, the sum of the squares on AC and BC is also a2 + b2 .

25.

By Proposition II–8, if we pass a secant line through the hyperbola xy = 1 which goes through points M and N on that curve and points T and U on the y-axis and x-axis, respectively (the asymptotes), then the segments TM and TN are equal. Thus, if we let M approach N, then the secant line approaches the tangent line at N and therefore the two line segments TN, NU between N and the asymptotes are equal. Therefore, the triangles TSN and NRU are congruent. If the coordinates of N are (x0 , x10 ), then TS = NR = x10 , and NS = x0 . So the slope of the tangent line TNU is 1/x0 1 TS = − 2. =− SN x0 x0

26.

Suppose the coordinates of the point N are (x, 0). Then, if the coordinates of G are (g, 0), the relationship between the two points is given by (g − x) : (a − x) = p : 2a, so g = p(a2a−x) + x. If P = (x, y) and Q = (x′ , y′ ), then the square of the distance from G to Q is given by y′2 +

p(a − x) + x − x′ 2a

2

2

p(a − x) p + x − x′ . = x′ p − x + 2a 2a

If we take the derivative of the right side with respect to x′ and set the result equal to 0, we find that x′ = x. Thus the point P is the closest point on the curve to G. Also, the −px . The slope of line PG is slope of the tangent line to the ellipse at P is given by pa2ay y 2ay − p(a−x)/2a = − pa−px . Thus PG is perpendicular to the tangent line.

27.

Unfortunately, the definition of similarity given is not quite accurate. It should say that two conic sections are similar when a ratio exists between the corresponding abscissas of the two curves such that the ratios of the corresponding ordinates to the abscissas are equal. In modern terms, two conics are similar if there is a similarity transformation between them, a function taking (x, y) to (kx, ky) for some k. Thus, given the parabolas y2 = px and y′2 = qx′ , consider the transformation x → (p/q)x′ , y → (p/q)y′ . The

Solutions

25

transformed parabola is that (p2 /q2 )y′2 = (p2 /q)x′ , or y′2 = qx′ . Thus the similarity transformation has taken the first parabola to the second, and similarity is proved. To use the definition more directly, we can let x1 , x2 be coordinates of two points on the x-axis of the first parabola and x′1 , x′2 coordinates of two points on the x-axis of the second parabola, such that x′1 : x1 = x′2 : x2 = q : p. Let yi and y′i be the corresponding y-coordinates. We need to show that yi : xi = y′i : x′i for i = 1, 2, or that y′i : yi = x′i : xi .

2 We have y′i : yi 2 = qx′i : pxi = q2 : p2 , so y′i : yi = q : p = x′i : xi , as desired.

28.

According to Proposition VI–12, two ellipses are similar if and only if p1 : a1 = p2 : a2 , where pi and ai are the parameter and the length of the semi-major axis, respectively, of the ith ellipse. But pi = 2b2i /ai , where bi is the length of the semiminor axis. So p1 /a1 = 2b21 /a21 and p2 /a2 = 2b2 /a22 , so b21 /a21 = b2 /a22 and therefore b1 /a1 = b2 /a2 . For hyperbolas, we have that two hyperbolas are similar if the ratios of the perpendicular distance from the vertex to the asymptote to the semi-major axis are equal. The proof is virtually identical.

29.

We first need to show that DB · AC = pa/2 = b2 . To do this, suppose E = (x0 , y0 ) and p write the equation of the ellipse in the form y2 = 2a x(2a − x), where B is the origin of the coordinate system. We know from Proposition I–34 that the x coordinate of K is (−t, 0), where t = ax0 /(a − x0 ). The equation of the tangent line is then y=

y0 (x + t), x0 + t

and we can then find the lengths of DB and AC. We find that DB =

ay0 2a − x0

and AC =

ay0 . x0

The product is then a2 y20 a2 y20 pa2 pa = = = = b2 .

2 x0 (2a − x0 ) (2a/p)y0 2a 2 Since AF · FB = b2 , we have AF · FB = DB · AC, or AC : AF = FB : DB. Since the angles at A and B are right angles, triangles FAC and FBD are similar, so angle ACF is equal to angle BFD. But the sum of angles ACF and AFC is a right angle, so the sum of angles BFD and AFC is also a right angle. It follows that angle CFD is a right angle. By a similar argument, angle DGC is also a right angle.

30.

Since angles CFD and DGC are right angles, the circle drawn with diameter CD will pass through both F and G. But then angles DCG and DFG will cut off the same arc of that circle, so they are equal. But also, we know that angle DFG is equal to angle ACF from Exercise 29. Therefore, angle DCG is equal to angle ACF, as desired. By a similar argument, angle CDF is equal to angle BDG.

31.

If EH is not perpendicular to CD, then let HL be drawn from H perpendicular to CD. Since angle CDF equals angle BDG and angle HLD equals angle DBG (both are right), triangle DGB is similar to triangle LHD, so GD : DH = BD : DL. Also, triangles DGH and HFC are similar, having a common angle at H and right angles at G and F by Exercise 29. So GD : DH = FC : CH. Also, since triangles AFC and LCH are

26

Solutions similar by Exercise 30, FC : CH = AC : CL. It follows that BD : DL = AC : CL or BD : AC = DL : CL. But BD : AC = BK : AK, so also DL : CL = BK : AK. Now draw a line from E parallel to AC meeting AK at M. Given that KEC is a tangent line to the ellipse, it follows from Proposition I–34 that BK : KA = BM : AM. But BM : AM = DE : EC. It follows that DL : CL = BK : AK = BM : AM = DE : EC, and this is impossible. Thus EH must be perpendicular to CD. 32.

Since angles DGH and DEH are right angles, the circle drawn with diameter DH will pass through E and G. Therefore, since angles DHG and DEG cut off the same arc on this circle, they are equal. Similarly, using the circle through C, E, H, and F, angle CEF equals angle CHF. But angle CHF equals angle DHG. Therefore, angle CEF is equal to angle DEG as claimed.

33.

If we apply a rectangle equal to one-fourth of the rectangle on the parameter N and the axis AB to the axis AB of a hyperbola that exceeds by a square figure, then the application results in two points F and G on the axis called the foci of the hyperbola.

If the equation of thephyperbola is y2 = x√ (px + (p/2a)x), the coordinates of the two points are x = −a ± a2 + pa/2 = −a ± a2 + b2 . The analogue to Proposition III– 48 is that if E is a point on the hyperbola and a tangent line is drawn at E, then the angles that the lines EF and EG make with the tangent line are equal. The analogue to Proposition III–52 is that the difference of the lengths of the two lines drawn in the previous sentence is equal to the axis of the hyperbola.

34.

Let the parabola have the equation y2 = px, with the focus at ( 4p , 0). Since the slope of p the tangent line at the point P = (x, y) is 2y , it follows that the direction vector T of the tangent line can be written in the form (2y, p). Similarly, the direction vector L of the line parallel to the axis can be written as (1, 0) and the direction vector V of the line from P to the focus can be written as (x − p4 , y). Then the cosine of the angle between T and L is √ 2y2 2 . The cosine of the angle between T and V is given by 4y + p

2y x − p4 + py 2y x + 4p 2xy + py 2y 2 q q =p =p =p . p p

2 2 2 2 4y + p x + 4 4y2 + p2 4y2 + p2 4y2 + p2 x − p4 + y2 x + p4 Since these two cosines are equal, so are the angles.

35. a.

By Elements II–6, with BC the bisected line and CK the added straight line, BK · KC + EC2 = EK2 . If we add EF2 to both sides, we have BK · KC + EC2 + EF2 = EK2 + EF2 , or BK · KC + CF2 = KF2 .

b.

Since MA : AB = ML : LK = BC : CK and AB = 2 AD while BC = 12 GC, if we halve the consequent in the first ratio and double the antecedent in the third ratio, the equality still holds, so MA : AD = GC : CK. But since CH is parallel to GF, we also have GC : CK = FH : HK, so MA : AD = FH : HK. Then (MA + AD) : AD = (FH + HK) : HK, or MD : AD = FK : HK. But AD = HK, so MD = FK and MD2 = FK2 .

c.

Again using Elements II-6, we have MB · MA + AD2 = MD2 = FK2 = BK · KC + CF2 . Since AD2 = CF2 , we have MB · MA = BK · KC.

Solutions d.

27

From the last equality, we get BM : BK = KC : MA. But by similarity, we have BM : BK = CL : CK = MA : AL. So CL : CK = CK : MA = MA : AL. But CL = AB and AL = BC. Therefore, AB : CK = CK : MA = MA : BC.

36.

If the two parallel lines are x = 0 and x = k and the perpendicular line is the x-axis, then the equation of the curve satisfying the problem is y2 = px(k − x) or y2 = kpx − px2 .

This is the equation of a conic section.

37.

Since the square of the distance between a point and a line is a quadratic function of the coordinates x, y of the point, and since the same is true for the product of the distances to two separate lines, the equation defining the locus in the three-line problem will be a quadratic equation in x and y. Thus the locus will be a conic section, possibly a degenerate one.

38.

Since ∠BGA has been bisected, it follows from Elements VI–3 and similarity that AG : BG = AD : BD = AE : EZ. Therefore, triangle ADG is isosceles and BG : EZ = AG : AE = 2 : 1. Therefore, B lies on the hyperbola defined as the locus of points such that the ratio of their distances to the point G and to the line EZ is a constant greater than 1. To do the synthesis, we just have to construct a hyperbola defined in this manner. Then the intersection point B of that hyperbola with the arc AG enables the arc to be trisected.

CHAPTER 5 1.

2.

√

√

◦ ◦ We first 4R2 − R2 = 3R √ need to calculate crd(120 ). We get crd(120 ) = 3 ·60 = 103;55,23. Then the half-angle formulaq gives crd(30◦ ) = = p R(2R − crd(120◦ )) = 31;03,30. Next, crd(150◦ ) = 4R2 − crd2 (30◦ ) = p 115;54,40, so crd(15◦ ) = R(2R − crd(150◦ )) = 15;39,47. q p ◦ crd(165◦ ) = 4R2 − crd2 (15◦ ) = 118;58,25. crd(7 12 ) = R(2R − crd(165◦ )) = q ◦ ◦ 7;50,54. Then, crd(172 21 ) = 4R2 − crd2 (7 12 ) = 119;42,28.

√

3.

Since 672 √ = 4489, 4500 = 67;x,y. Now 4500 − 672 = 11; divide 11 · 60 by 2 · 67 to get x = 4. So 4500 = 67;4, y. Now 4500 − (67;4)2 = 2;03,44 = 7200 √ + 180 + 44 = 7424 seconds. Then 7424 ÷ 2(67;4) = 55 to the nearest integer. So 4500 = 65;04,55.

4.

Use a quadrilateral ABCD with AB = crd α, BC = crd (180 − (α + β)), CD = crd β, AD = 120 (the diameter of the circle). The diagonals are then AC = crd (180 − β) and BD = crd (180 − α). Then apply Ptolemy’s Theorem.

5.

120 crd(72 − 60) = crd(72) crd(120) − crd(60) crd(108). So 120 crd(12) = 70;32,3 × 103;55,23 − 60 × q 97;04,56 = 1505;11,34. It follows that crd(12) = 12;32,36. Then crd(168) = 4 × 602 − crd2 (12) = 119;20,33. Thus crd(6) = p 60(2 × 60 − crd(168)) = 6;16,49. Similarly, crd(3) = 3;08,29; crd(1 21 ) = 1;34,15; and crd( 34 ) = 0;47,07.

6.

Archimedes’ Lemma 2 can be rewritten, using chord notation, in the form (2R)2

crd

2

α 2

=

crd2 (α) + (2R + crd(180 − α))2 . crd2 (α)

28

Solutions By expanding, using the Pythagorean Theorem, and simplifying, this result can be put into the form crd2

α 2

=

R crd2 (α) .

2R + crd(180 − α)

Multiplying numerator and denominator of the right side by 2R − crd(180 − α) and simplifying then gives Hipparchus’s theorem.

7.

Let crd α = AB and crd β = BC. First, bisect the angle at B, and extend the angle bisector until it meets the line AC at E and the circle at D. Then AD = DC, because these chords subtend equal angles. Also, since BE bisects the angle at B, we have by Elements VI–3 that AE : EC = AB : BC. Since AB < BC, we also have AE < EC. Next, drop a perpendicular DF from D to AC. Because D bisects arc ADC, it follows that F is the midpoint of AC. Then a circle of center D and radius DE will cut AD at G between A and D and will cut DF extended at H. Therefore, sector DEH > triangle DEF, sector DEG < triangle DEA, and triangle DEF : triangle DEA < sector DEH : sector DEG. Since the two triangles have the same altitude, the ratio of their areas is the same as the ratio of their bases. Also, the ratio of the sectors is the same as the ratio of the corresponding angles with vertex at D. Thus EF : EA < ∠EDH : ∠EDG. If we add 1 to each side of the inequality, we get the new inequality AF : EA < ∠GDH : ∠EDG. If we double the numerators, we then get AC : EA < ∠ADC : ∠EDG, or, subtracting 1 from both sides, that EC : EA < ∠EDC : ∠EDG. The left side of this inequality is equal to BC : AB, while the right side is equal to the corresponding ratio of arcs, namely β : α. Thus we have the final result that crd β : crd α < β : α, as desired.

B H E

A

C F

G

D ◦

8.

We know that crd(1◦ ) < 43 crd( 34 ). Using the values from Exercise 5, we get crd(1◦ ) < ◦ 1;02,49,53,03. Similarly, since crd(1◦ ) > 23 crd(1 12 ), we calculate that crd(1◦ ) > 1;02,49,48,13.

9.

At the vernal equinox at latitude 40◦ at noon, the sun is 40◦ below the zenith. Using Figure 5.19, CF = crd 80◦ = 77;08,05 and CE = crd(100◦ ) = 91;55,31, so the shadow 60 = 91;55,31 · 77;08,05 = 50;21.

10.

Since the spring and fall paths together total 180◦ , or half a year, and since half a year is 21 (365;14,48) = 182;37,24 days, then the fall path is 182;37,24 − 94;30 = 88;7,24 days, or approximately 88 18 days. Similarly, comparing the winter with summer, the winter is about 90 18 days.

Solutions

29

11.

Note that ϵ is the latitude where the sun is directly overhead at noon on the summer solstice. The angular distance between the noon altitudes of the sun at the summer and winter solstice is, given the assumption that at any given time the sun’s rays to every point on the earth are parallel to each other, equal to the angle between the sun at noon on the summer solstice and the sun at noon on the winter solstice, as viewed from the center of the earth. And this angle, by Figure 5.43, is twice ϵ.

12.

At noon on the summer solstice at latitude 36◦ , the sun is 12 21 below the zenith, so ◦ the length of the shadow is 60 tan 12 12 = 13;18,06. At the winter solstice, the sun is ◦ ◦ 59 21 below the zenith and the shadow length is 60 tan 59 12 = 101;51,35.

◦

13.

When λ = 90◦ , then δ = 23◦ 51′ and α = 90◦ . When λ = 45◦ , we have sin δ = sin(23◦ 51′ ) sin(45◦ ) and δ = 16◦ 37′ . Also tan α = cos(23◦ 51′ ) tan(45◦ ), so α = 42◦ 27′ . By symmetry, the values for the declination at 270◦ and 315◦ are the negatives of the values at 90◦ and 45◦ , respectively.

14.

To calculate ρ(60◦ , 45◦ ), we note that if λ = 60◦ , then δ = 20◦ 30′ and α = 57◦ 44′ .

Since sin σ = tan δ tan 45◦ , we have σ = 21◦ 57′ and ρ = α − σ = 35◦ 47′ . If λ = 90◦ , then δ = 23◦ 51′ and α = 90◦ . So σ = 26◦ 14′ and ρ = α − σ = 63◦ 46′ .

15.

L(λ, ϕ) = 180◦ + 2σ(λ, ϕ). When λ = 60◦ and ϕ = 36◦ , we calculate that sin σ = tan δ tan ϕ = tan(20◦ 30′ ) tan(36◦ ); so σ = 15◦ 46′ and L = 211;32, which corresponds to 14 hours, 6 minutes. Therefore, sunrise is 7 hours, 3 minutes before noon, or 4:57 a.m. and sunrise is at 7:03 p.m.

16.

If the length of day is 15 hours when λ = 90◦ , then 180◦ + 2σ(90◦ , ϕ) = 225◦ . sin(22◦ 30′ ) Therefore σ = 22◦ 30′ and, since sin σ = tan δ tan ϕ, we have tan ϕ = tan , so (23◦ 51′ ) ◦

′

(23 51 ) δ ϕ = 40◦ 53′ . Also, since sin β = sin(sin = sin , we calculate that β = 32◦ 20′ .

90−ϕ) sin(49◦ 7′ ) That is, the sunrise occurs at 32◦ 20′ north of east and sunset at the same angle north of west. By symmetry, the positions of sunrise and sunset at the winter solstice occur 34◦ 20′ south of east and south of west, respectively.

17.

◦

The expression tan δ tan ϕ will be greater than 1 for δ = 23 12 when 0.4348 tan ϕ > 1, ◦ or when tan ϕ > 2.2998, or when ϕ > 66 12 . When that occurs, the formula for L no longer makes sense. Since when tan δ tan ϕ = 1, we know that L = 360◦ or 24 hours, it follows that the sun does not set at all on the summer solstice when the latitude is ◦ greater than 66 21 .

18.

If λ = 45◦ , the δ = 16◦ 37′ ; so SZ = ϕ − δ = 45◦ − 16◦ 37′ = 28◦ 23′ . Similarly, if λ = 90◦ , then δ = 23◦ 51′ and SZ = 21◦ 9′ .

19.

The sun is directly overhead at noon at latitude 20◦ when δ = 20◦ . Since sin λ = ◦ sin δ sin 23.5◦ , we find that λ = 59 . This value for the longitude of the sun occurs at approximately 60 days after the spring equinox and 60 days before the fall equinox, or at approximately May 20 and July 21.

20.

The maximal northerly sunrise point occurs when λ = 90◦ and therefore when δ = 23◦ 51′ . When ϕ = 36◦ , we calculate that sin β = so β = 29◦ 59′ .

sin δ sin 23◦ 51′ sin 23◦ 51′ = = = 0.4998, ◦ sin(90 − ϕ) sin(90 − 36) sin 54◦

30

Solutions 21.

When ϕ = 75◦ , we need to find δ so that sinsin15δ◦ = 1. Clearly, δ = 15◦ , and since sin 15◦ = sin 23◦ 51′ sin λ, it follows that λ = 39◦ 48′ . This value occurs approximately 40 days after the vernal equinox, or about April 30.

22.

The actual distance between Alexandria and Syene (modern Aswan) is approximately 850 km or 520 miles. Aswan is located at 24◦ 5′ N latitude and 32◦ 56′ E longitude, while the coordinates of Alexandria are 31◦ 12′ N and 29◦ 54′ E. Thus Aswan is not on the Tropic of Cancer, but is approximately 15′ north of it (using Ptolemy’s value).

Also, Alexandria and Aswan do not have the same longitude; Aswan is approximately 3◦ east of Alexandria. On the other hand, the difference in degrees of the latitudes of ◦ the two cities is 7◦ 7′ , which is very close to Eratosthenes’s value of 7 51 . If there were 5000 stades between the two cities, then a stade would be approximately 0.17 km = 170 m.

23.

Let the total length of a parallel at latitude α be Cα and the circumference of the earth at the equator be C. Let the radius of the latitude circle at latitude α be rα and the radius of the earth be r. Then Cα : C = rα : r = cos α, a relationship easily seen by constructing the right triangle in a cross section of the earth one of whose sides is rα and whose hypotenuse is r. Thus, Cα = C cos α.

24.

Since the ratio of a degree at latitude α to a degree at the equator is as cos α (by ◦ Exercise 23), we just need to check approximations to cos α. cos 23 65 = 0.9147, ◦ 7 5 while 4 12 = 0.9592, while 4 56 : 5 = 0.9667. : 5 = 0.9167. cos 16 12

25.

The exact area A of an equilateral triangle of side s is A = 43 s2 . Thus the Roman √ surveyor has approximated 3/4 by 1/3 + 1/10. This is the same as approximating √ 3 by 4/3 + 4/10 = 26/15 ≈ 1.733.

26.

Suppose points A and B are on the opposite bank of the river. One possible method to determine the distance from A to B is the following: First determine the distance across the river from a point C on your side (using the method in the text). For simplicity, we will assume that line AB is perpendicular to line AC. Choose a point D on the extension of line AC and construct a line CF perpendicular to AC. Next, determine the point E on CF where the line of sight from D to B intersects CF. Since triangles ABD and CED are similar, and since the distances AD and CE are known, the distance AB can be calculated.

27.

In Figure 5.38, we will take a = 10, b = 7, and c = 4. The method based on Elements 2 2 2 II–12, 13 gives us that CD = a + b2a− c . Thus CD = 6.65, h = AD = 2.19, and the area 1 = 2 ah = 10.93. To use Heron’s formula, we first calculate s = 21 2 = 10.5 and then find √ √ that the area is 10.5 · 6.5 · 3.5 · 0.5 = 119.4375 ≈ 10.93.

28.

Divide the pentagon into five equalp isosceles triangles with base a, side r, and altitude √ 2 p. As on page 91 of the text, a = 2r 10 − 2 5. Since p2 = r2 − a2 , a computation shows that

√

s p=

√

3+ 5 √ a 20 − 4 5

Solutions

31

and therefore, since A5 = 52 pa, that r

√

5+2 5 2 a .

20

5 A5 = 2

This value is roughly approximated by Heron’s value A5 = 53 a2 .

29.

Given that the radius r of the circle is 87 a, where a is the side of the inscribed heptagon, q √ 2 2 3 it follows that the area of the heptagon is A = 7 · 12 207 196 a = 4 23a . Heron has √

therefore approximated 34 23 by 43 12 or

√

√

23 by 43 9 .

8 = 3. Then 12 (3 + 83 ) = 17 6 .

30.

Start with the approximation

31.

Since the regular octahedron is made up of two pyramids, each with square base of side a, the volume is double the volume of one of these pyramids. But the volume of a pyramid is 31 hB, where h is the height and B the area of the base. In this case, B = a2 √

and h is one leg √ of a right triangle whose hypotenuse is 23 a and whose other√leg is 2a .

Therefore, h = 22 a, and the volume of the octahedron is given by V = 2 · 13 22 aa2 = √

2 3 3 a , as asserted by Heron.

32.

We square the hypotenuse 13 to give 169. We double that to get 338. We subtract from 338 the square of the sum of the base and height, 289, to get 49. This is the square of the difference of the base and the height, so that difference is 7. Therefore the base is 12 and the height is 5.

33.

The square on the diagonal plus twice the area is 152 + 2 × 60 = 345, which is the square of the sum of the length and width. The square on the diagonal minus twice the area is 152 − 2 × 60 = 105, which is the square of the difference of the length and √ the√width. Therefore, the sum of the length and width is 345, while the difference is 105. So the length is half of the sum of the two square roots, or 14.41, while the width is half of the difference, or 4.16.

CHAPTER 6 1.

The pentagonal numbers are 1; 1 + 4 = 5; 1 + 4 + 7 = 12; 1 + 4 + 7 + 10 = 22;. . . The nth pentagonal number is therefore nX −1

(3i + 1) = n + 3

i=0

n(n − 1) 2

=

3n2 − n .

2

Similarly, since the hexagonal numbers are 1; 1 + 5 = 6; 1 + 5 + 9 = 15; 1 + 5 + 9 + 13 = 28;. . . it follows that the nth hexagonal number is nX −1

(4i + 1) = n + 4

i=0

n(n − 1) 2

= 2n2 − n.

32

Solutions 2.

The pyramidal numbers with triangular base are 1; 1 + 3 = 4; 1 + 3 + 6 = 10; 1 + 3 + 6 + 10 = 20; . . . Therefore the nth pyramidal number with triangular base is n X k(k + 1) k=1

2

1 n(n + 1)(2n + 1) n(n + 1) + 2 6 2 k=1

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

2 6 2 6

= 12

n P

(k2 + k) =

The pyramidal numbers with square base are 1; 1 + 4 = 5; 1 + 4 + 9 = 14; 1 + 4 + 9 + 16 = 30;. . . Thus the nth pyramidal number with square base is given by the sum of the squares from 1 to n, namely, n(n + 1)(2n + 1) .

6 3.

In a harmonic proportion, c : a = (c − b) : (b − a). It follows that ac − ab = bc − ac or that b(a + c) = 2ac. Thus the sum of the extremes multiplied by the mean equals twice the product of the extremes.

4.

Since 6 : 3 = (5 − 3) : (6 − 5), the numbers 3, 5, 6 are in subcontrary proportion.

Other examples of triples in subcontrary proportion include 4, 10, 12 and 5, 17, 20.

5.

In the example 4, 10, 12 of three numbers in subcontrary proportion, it is not true that the product of the greater and mean terms is twice the product of the mean and smaller.

For in this case, the first product equals 120 while the second equals 80.

6.

In a “fifth proportion,” if a < b < c, then b : a = (b − a) : (c − b). Thus 2, 4, 5 are in fifth proportion, because 4 : 2 = (4 − 2) : (5 − 4). Two other triples in fifth proportion are 3, 6, 7 and 3, 9, 11.

7.

1 Let x = Diophantus’s age at death. Then x = 16 x + 12 x + 71 x + 5 + 12 x + 4. It follows that 9x = 756 and x = 84.

8.

To solve x + y = 20, x2 + y2 = 208, we set x = 10 + z, y = 10 − z. Then 200 + 2z2 = 208; z2 = 4; and z = 2. Thus x = 12 and y = 8.

9.

To solve x + y = 20; x2 − y2 = 80, we set x = 10 + z, y = 10 − z. Then subtracting the squares, we get 40z = 80. So z = 2, x = 12, and y = 8.

10.

To find two squares whose difference is 60, set x2 = smaller square and x2 + 60 = larger square. Then x2 + 60 = (x + 3)2 , where 3 is arbitrarily chosen. This equation reduces 289 1 1 to 6x = 51 and therefore x = 17 2 . The two squares are therefore 4 = 72 4 and 132 4 . In 2 2 2 2 the general case, the two squares are x and x + b = (x + a) , where a < b. It follows a2 that x = b− 2a and the two squares are found to be

11.

b − a2 2a

2

and

b + a2 2a

2 .

Let x2 be the least square and (x + m)2 = x2 + 2mx + m2 be the middle square. The difference is 2mx + m2 . Therefore the largest square is x2 + 2mx + m2 + n(2mx + m2 ) = x2 + (2m + 2mn)x + m2 + nm2 = (x + b)2 = x2 + 2bx + b2 . Provided that m2 (1 + n) < b2 <

Solutions

33

m2 (1 + n)2 , the solution is x=

b2 − m2 − nm2 .

2m + 2mn − 2b

12.

To solve x − 6 = u2 , x − 7 = v2 , we subtract and get u2 − v2 = 1. It follows that (u + v)(u − v) = 2 · 12 . If we set u + v = 2 and u − v = 21 , we get u = 54 , v = 43 , and x = 121 16 .

13.

Let the side of one cube be x, the side of the second cube be 2x, and the side of the square be 6x. Therefore, the sum of the two cubes is 9x3 . This must equal the square on 6x, that is, 9x3 = 36x2 . Then 9x = 36 and x = 4. So the two cubes are 43 = 64 and 83 = 512. Their sum is 576, which is the square of 24.

14.

To solve x − y = 10, x3 − y3 = 2170, set x = z + 5 and y = z − 5. It follows that (z + 5)3 − (z − 5)3 = 2170. This equation reduces to 30z2 = 1920 or z2 = 64 or z = 8.

Thus x = 13 and y = 3. In the general case, if x − y = a and x3 − y3 = b, we set x = z + 2a and y = z − a2 If we substitute for x and y in the second equation, we get an equation in −a3 . It follows that this latter expression must be a square. z which reduces to z2 = 4b12a

15.

To solve x + y = 20, x3 + y3 = 140(x − y)2 , set x = 10 + z, y = 10 − z. Then (10 + z)3 + (10 − z)3 = 140(2z)2 . This equation reduces to 2000 + 60z2 = 560z2 or 500z2 = 2000 or z2 = 4 or z = 2. Thus the solution is x = 12, y = 8. In general, if x + y = a, x3 + y3 = b(x − y)2 , we set x = 2a + z, y = a2 − z. On substituting into the second equation, 3 3 we get 2( 2a )3 + 6 a2 z2 = b(2z)2 , which reduces to a4 + 3az2 = 4bz2 or a4 = (4b − 3a)z2 .

3 Thus z2 = 4(4ba− 3a) . This equation has a rational solution provided that the right side is a square, and that condition is equivalent to Diophantus’s condition that a3 (b − 34 a) is a square.

16.

Simply divide the given square a2 into two squares. This is possible by II–8.

17.

We want to solve x + y = (x3 + y)3 . We set x = 2z and y = 27z3 − 2z (so that x + y = (3z)3 ). Then (x3 + y)3 = (35z3 − 2z)3 = (3z)3 . It follows that 35z2 = 5. This is impossible for rational z. But now note that 35 = 27 + 8 = 33 + 23 and 5 = 3 + 2.

In order that the equation in z be solvable in rationals, we need two numbers a and 3 3 b (to replace the 3 and 2) so that aa ++ bb is a square. So let a + b = 2 (where 2 is arbitrary). Then b = 2 − a and a3 + b3 must equal 2 times a square. This implies that a3 + (2 − a)3 = 8 − 12a + 6a2 = 2(square) or that 4 − 6a + 3a2 is a square. So set 16 4 − 6a + 3a2 = (2 − 4a)2 and solve for a. We get a = 10 13 and therefore b = 13 . Since it is only the ratio of a and b which is important, we can choose a = 5, b = 8 and therefore put x = 5z, y = 512z3 − 5z, and repeat the initial calculation. We then get 267 1 637z3 = 13z and z2 = 49 , so z = 17 . Then x = 75 ; y = 343 is the desired solution.

18.

Since the given number is 6, we must divide 13 into two squares that are close to 6 12 , that is we must find a small number to add to 6 21 to make a square. We can take this small number as a fraction, say y12 . Then 6 21 + y12 is a square, or, multiplying by 4, 26 + z12 is a square. If we multiply by z2 , we can rewrite this as 26z2 + 1 is a square, say (5z + 1)2 . Setting these two expressions equal to each other and solving gives z = 10.

1 1 That is, to make 26 + 100 is a square, or 6 12 + 400 is a square. This latter value is 51 2 2601 400 = ( 20 ) . Thus, we need to divide 13 into two squares whose sides are as close

34

Solutions 2 2 as possible to 51 20 . Now, 13 = 2 + 3 (and this is why we needed the condition that twice the given number + 1 must not be congruent to 3 modulo 4). Now, we need 51 9 11 two numbers r, s such that 3 − r = 20 and 2 + s = 51 20 . Thus r = 20 and s = 20 . But 51 1 the sum of the square of 20 with itself is not 13. So we replace the 20 by x and write 5 13 = (11x + 2)2 + (3 − 9x)2 = 202x2 − 10x + 13. Thus x = 101 and the two sides are 55 257 45 258 + 2 = and 3 − = .

We then need to square each of these and subtract 6 101 101 101 101 257 2 4843 258 2 5358 to get the parts of unity: ( 101 ) − 6 = 10201 and ( 101 ) − 6 = 10201 .

19.

Suppose the right triangle has legs a, b, hypotenuse c, and angle bisector d. Let r be the length of that part of leg a from the right angle to the point where the bisector intersects the leg. To make the right triangle with the angle bisector as hypotenuse a rational triangle, we can set d = 5x and r = 3x. It follows that b = 4x. If we then let a = 3, we have from Elements VI–3 that c : (a − r) = b : 4 or that c : (3 − 3x) = 4x : 3x. Thus c = 4 − 4x and the reason why a was chosen to be 3 is evident. By the Pythagorean theorem, we have (4 − 4x)2 = 32 + (4x)2 or 16 − 32x + 16x2 = 16x2 + 9. Thus 32x = 7 7 and x = 32 . To get integral answers, we can multiply through by 32. Thus the original triangle is (96, 28, 100) and the bisector equals 35.

20.

The diagram for Elements VI–28 is Figure 3.15. Let us assume that the proposed rectangle has been constructed with base AS and area equal to c and that the defect is a square. If we set AB = b, and BS = x, then AS = b − x and x(b − x) = c. Since the maximum of the ”function” f(x) = x(b − x) occurs when x = 2b , and since this maximum is ( b2 )2 , it follows that c cannot exceed the value ( b2 )2 . This means that the area c of the given rectilinear figure must not be greater than the area of the square on half the given line of length b.

21.

Assume that the theorem is true. Then AB2 + BC2 = 3AC2 . But since AB = AC + BC, we have (AC + BC)2 + BC2 = 3AC2 . This reduces to AC2 + 2AC · BC + 2BC2 = 3AC2 or AC · BC + BC2 = AC2 . This in turn implies that BC(AC + BC) = AC2 or that AB · BC = AC2 . But this is precisely the statement that AB is cut in extreme and mean ratio at C.

22.

Suppose that three of the lines have equations x = a, x = b, x = c, that the other two have equations y = d, y = e, and that the fixed line has length k. Then the equation of the locus is (x − a)(x − b)(x − c) = k(y − d)(y − e). Other arrangements of the lines will give somewhat different equations, but in any case the locus is described by a cubic equation in x and y.

23.

Suppose the hexagon has perimeter 6d. Then it is composed of six equilateral triangles √ of√side d. Since such a triangle has area 43 d2 , it follows that the hexagon has area 3 3 2 3 2 d . The square with perimeter 6d has side equal to 2 d and therefore area equal to 9 2 4 d , which is less than the area of the hexagon.

24.

Since the center of gravity of a disk is its center, it follows from Pappus’s theorem that the volume of the given torus is πr2 · 2πR = 2π2 r2 R.

25.

We can write the equation of x, the original number of apples, as 62 x + 18 x + 1 1 4 x + 5 x + 11 = x. The solution is x = 120.

26.

In 12 days the spouts will fill 12 + 6 + 4 + 3 = 25 tanks. Therefore, one tank will be filled in 12 25 of a day.

Solutions 27.

35

This problem can be translated into two equations in two unknowns: x + 10 = 3(y − 10); y + 10 = 5(x − 10). We can write these as x − 3y = −40; −5x + y = −60. The solution is then that A has x = 15 57 coins and B has y = 18 47 coins.

CHAPTER 7 1.

The Chinese forms are given in the Answers in the text.

2.

The largest digit a so that (100a)2 < 142,884 is a = 3. If we subtract 3002 = 90,000 from 142,884, the remainder is 52,884. We then need to find b so that 2(100a)(10b) < 52,884, or 6000b < 52,884. We take b = 7 and check that 6000b+(10b)2 < 52,884. But this inequality reduces to 42,000 + 4900 < 52,884. Since the left side equals 46,900, the inequality is in fact true. Note that if we had taken b = 8, this second inequality would not have been true. We now subtract 46,900 from 52,884 to get 5,984. We now need to find c so that 2(370)c < 5984. We try c = 8 and check that 740c + c2 ≤ 5984.

In fact, 740 · 8 + 82 = 5984, so the desired square root is 378.

3.

To calculate the cube root x of 12,812,904, we note first that it is a three-digit number beginning with 2: x = 200 + 10b + c. We then temporarily ignore the c and calculate (200 + 10b)3 = 8,000,000 + 3 · 40000 · 10b + 3 · 200 · 100b2 + 1000b3 .

This result must be less than or equal to 12,812,904, or b(1,200,000 + 60000b + 1000b2 ) ≤ 4,812,904. We check that the largest b that satisfies this inequality is b = 3.

Thus the answer begins with 23, and we then repeat the steps to find c: (230 + c)3 = 12,167,000 + 3 · 2302 c + 3 · 230c2 + c3 ≤ 12,812,904.

This reduces to c(158,700 + 690c + c2 ) ≤ 645,904, and a calculation shows that this is an equality for c = 4. Thus x = 234.

4.

We add 2 + 4 + 8 + 16 + 32 = 62. So 62 is the divisor. Then on the first day, the weaver weaves (5 × 2) ÷ 62 = 10/62 chi = 100/62 cun = 1 38/62 cun. On each successive day, the weaver doubles her output, so we get 3 14/62 cun on the second day, 6 28/62 cun on the third day, 12 56/62 cun on the fourth day, and 25 50/62 cun on the fifth day. We would probably set x to be the amount woven on the first day and then solve the equation x + 2x + 4x + 8x + 16x = 50, but this essentially amounts to the same procedure.

5.

1 1 It takes 30 of a day to make one arrow and 20 of a day to feather one arrow. Therefore, 1 1 1 it takes 30 + 20 = 12 of a day to make and feather one arrow. Therefore, the person can make and feather 12 arrows in one day.

6.

560 41 350 12 560 + 350 + 180 = 1090. Tx = 1090 × 100 = 51 109 ; Ty = 1090 × 100 = 32 109 ; Tz = 180 56 1090 × 100 = 16 109 .

36

Solutions 7.

In 15 days, the first channel fills the reservoir 45 times, the second channel 15 times, the third 6 times, the fourth 5 times, and the fifth 3 times. It follows that in 15 days the reservoir is filled 74 times. To fill it once then requires 15 74 of a day.

8.

If x is the unknown amount, the conditions show that after the first tax the man had 2 4 2 6 4 2 3 x; after the second, he had 5 · 3 x; and after the third, he had 7 · 5 · 3 x. This amount 175 15 must equal 5. It follows that x = 16 = 10 16 pounds.

9.

We begin with c6 = r = 10 and a6 = 102 − 52 = 8.6603. Therefore, S12 = 12 · p 6 · 10 · 10 = 300. Then c12 = (10/2)2 + (10 − 8.6603)2 = 5.1764, and S24 = 12 · p 12 − (c12 /2)2 = 9.6593. Next, c24 = p · 10 · c12 = 310.5859. We then get a12 = 100 1 (c12 /2)2 + (10 − a12 )2 = 2.6105 and S48 = 2 · 24 · 10 · c24 = 313.2629. Next. p p )2 + (10 − a24 )2 = 1.3081. So a24 = 100 − (c24 /2)2 = 9.9144 and c48 = (c24 /2p 1 S96 = 2 · 48 · 10 · c48 = 313.9350. Finally, we get a48 = 100 − (c48 /2)2 = 9.9786, so p c96 = (c48 /2)2 + (10 − a48 )2 = 0.6544. We then get S192 = 12 · 96 · 10 · c96 = 314.1032.

10.

From Figure 7.6, we see that one-eighth of the volume of the double box-lid is Rr 2 3 3 (r − x2 ) dx = (r2 x − x3 )|r0 = r3 − r3 = 23 r3 . It follows that the entire volume is 0 16 3 3 r .

11.

If n is the number of people and p is the price, and if a1 and a2 are the guesses for the price per person, that is, 8 and 7, respectively, and if f1 , f2 are the surplus and deficiency, respectively, that is, 3 and 4, then, in general, we have the equation a1 n − f1 = a2 n + f2 . + f2 7 This gives n = af11 − a2 = 1 = 7. Since 7 × 8 = 56, is a surplus of 3, the actual price is 53.

12.

If x is the hypotenuse of a right triangle and 10 and d the legs, then x = d + 1 or d = x − 1. Then x2 = (x − 1)2 + 100 and x = 50.5.

13.

The center of the desired circle is the intersection of the angle bisectors of the two acute angles of the triangle. To show this, drop perpendiculars to all three sides of the triangle from that intersection and show that those three lines are all equal by considering the two pairs of congruent right triangles formed by the three perpendiculars and the angle bisectors. Let the length of one of the perpendiculars be r, the radius of the circle. Then, since the two perpendiculars to the legs of the triangle also produce a square of side r, we get that r2 + r(b − r) + r(a − r) = 21 ab. This equation reduces to r(a + b) − r2 = 12 ab and then to 2r(a + b − r) = ab. Since D = 2r, we can rewrite this as

√

D=

ab . a+b − r

But the diagram also shows that c = a − r + b − r = a + b − 2r, so that a + b + c = 2(a + b − r). Therefore, a + b − r = 21 (a + b + c) and, substituting in the formula for D, we get D= as desired.

2ab , a+b+c

Solutions

37

14.

As part of the previous exercise, we have shown that D = 2r = a + b − c = a − (c − b).

Then square both sides. So D2 = a2 − 2a(c − b) + (c − b)2 = a2 p − 2ac + 2ab + c2 − 2 2 2bc + b = 2(c − bc−ac + ab) = 2(c−a)(c−b). It follows that D = 2(c − a)(c − b).

15.

If we set x to be the length of a side of the city, draw a line through the center of the city extending 20 pu north and 14 pu south, extend a line 1775 pu west from the bottom of that line, and connect the end of that new line with the end of the line to the north.

We get a right triangle with legs x + 34 and 1775. Since we also have a similar triangle with legs 20 and 2x , we get the proportion 20 : 2x = (x + 34) : 1775. The resulting equation is x2 + 34x = 71000 and x = 250.

16.

If x is the depth of the well, the similarity relationship gives 5 −x 0.4 = 0.4 5 . Thus x = 5·4.6 0.4 = 57.5.

17.

The simplest way is to set y = DC, x = CE. Then y 10 = x + 13 13 13 31

and

3 93 y = 120 . x+5 5

Simplifying these equations gives 40y = 30x + 400; 200y = 151x + 755. Solving these simultaneously gives the solution x = 1245, y = 943 34 .

18.

We know that 100 cun cost 128 coins. But 1 pi is 400 cun, and 9 chi is 90 cun, so we are asked how much 495 cun cost. Therefore, we use the rule of three to solve 100 : 128 = 495 : x, so x = (128 · 495)/100 = 633 35 coins.

19.

If x is the yield of good grain, y the yield of ordinary grain, and z the yield of worst grain, then the system of equations is 2x + y =1 3y + z = 1 x + 4z = 1 In matrix form we get, in turn 1 0 4 1

0 2 3 1 1 0 1 1

0

−1

8 1

0 2 3 1 1 0 1 1

0 0 25 4

0 2 3 1 1 0 1 1

4 7 18 It follows that 25z = 4 or z = 25 ; 3y + z = 1, 3y = 21 25 , or y = 25 ; 2x + y = 1, 2x = 25 , or 9 x = 25 .

20.

By Theorem, the altitude h of the p lower triangle is given by p the Pythagorean b2 − ( 2c )2 . The area B of that triangle is then B = 2c b2 − ( 2c )2 as stated. Simip larly, the area A of the upper triangle is A = 2c a2 − ( 2c )2 . If x = A + B, then x4 = (A + B)4 = A4 + 4A3 B + 6A2 B2 + 4AB3 + B4 = 2(A4 + 2A3 B + 2A2 B2 + 2AB3 + B4 ) − (A4 − 2A2 B2 + B4 ) = 2(A2 + B2 )(A2 + 2AB + B2 ) − (A2 − B2 )2 = 2(A2 + B2 )x2 − (A2 − B2 )2 . The follows immediately. If a = 39, b = 25, and √ c = 30, we √ equation for x √ have A = 15 1521 − 225 = 15 1296 = 15 · 36 = 540. Similarly, B = 15 400 = 300.

38

Solutions Then 2(A2 + B2 ) = 2(291,600 + 90,000) = 763,200 and (A2 − B2 )2 = (201,600)2 = 40,642,560,000, as desired.

21.

In this case, we see by trial that the solution is between 6 and 7. So we use 6 in the synthetic division procedure: 6|

16

6|

16

6|

16

|16

192 96 288 96 |384

−1863.2

1728

|−135.2

For the next step, we will use decimals. The (positive) solution to 16x2 + 384x − 135.2 is between 0 and 1. Again, we find by trial that the value is between 0.3 and 0.4. So our next chart is as follows: .3 |

16

.3 |

16

.3 |

16

|16

384 4.8 388.8 4.8 |393.6

−135.2

116.64

|−18.56

For a third step, we will try values between 0 and 0.1. Again, the closest value seems to be 0.5, as in the following chart: .05 | 16 393.6 −18.56 0.8 19.72 16 394.4 |0.96 Since the last value is relatively close to 0, we will leave the solution as 6.35. If we wanted to go further, we could have used 0.4 in this last step and continued to find the next decimal place.

22.

Qin’s method gives the following diagrams to produce 234 as the answer. We begin by noting that the answer is a three-digit number beginning with a 2.

200 |

1

0 200

0 40,000

−12,812,904 |−4,812,904

200 |

1

200 200

40,000 80,000

200 |

1

400 200

|120,000

1

|600

|1

8,000,000

Solutions

39

The second digit is a 3.

30 |

1

600 30

120,000 18,900

−4,812,904 |−645,904

4,167,000

30 |

1

630 30

138,900 19,800

30 |

1

660 30

|158,700

1

|690

4| 1

690 4

158,700 −645,904 2,776 645,904

1

694

161,476

|1

The final digit is a 4.

|0

The third order coefficients occur, for example, in 600 = 3 × 200, 120,000 = 3 × 2002 , and in 690 = 3 × 230 and 158,700 = 3 × 2302 . Note that these numbers occur in the solution to Exercise 3 above.

23.

The diagram is as follows, where the solution is x = 23. We begin by noting that the first digit is a 2.

20 |

1

0 20

0 400

0 8,000

−279,841 |−119,841

20 |

1

20 20

400 800

8,000 24,000

20 |

1

40 20

1,200 1,200

|32,000

20 |

1

60 20

|2,400

1

|80

20 |

|1

160,000

The second digit is a 3.

3 | 1 80 3

2,400 249

32,000 7,947

−119,841

1 83

2,649

39,947

|0

119,841

40

Solutions The fourth order coefficients show up in 80 = 4 × 20, 2400 = 6 × 202 , 32,000 = 4 × 203 , and 160,000 = 1 × 204 .

24.

We are given that b2 − [c − (b − a)] = ba, a2 + c + b − a = ac, and a2 + b2 = c2 . Also, x = b and y = a + c. Then y − x = a + c − b = c − (b − a). From a2 + b2 = c2 , we get b2 = c2 − a2 = (c + a)(c − a), so c − a = b2 /(a + c) = x2 /y. Then 2a = (c + a)−(c − a) = 2 2 2 y − xy , so a = 12 (y − xy ). Thus the first equation becomes x2 −[y − x] = x[ 21 (y − xy )]. If we multiply by 2y, we get 2x2 y − 2y2 + 2xy = xy2 − x3 , or x3 + 2yx2 + 2xy − xy2 − 2y2 = 0, 2 2 the first of our desired equations. Next, we have c = y − a = y − 12 (y − xy ) = 12 (y + xy ).

2

2

4

So ac = 21 (y − xy ) 12 (y + xy ) = 41 (y2 − xy2 ). If we now substitute this in the second equation in a, b, and c, we get

2

1 x2 x2 1 2 x4 1 2 x4 x2 y− + +x= y − 2 or y − 2x2 + 2 + 2 y y 4 4 y y y

4 1 2 x +x = y − 2 .

4 y Now multiply by 4y2 . We get y4 − 2x2 y2 + x4 + 4x2 y + 4xy2 = y4 − x4 , or 2x4 − 2x2 y2 + 4x2 y + 4xy2 = 0. If we divide by 2x, we get the second desired equation: x4 + 2yx − xy2 + 2y2 = 0.

25.

Using the notation from the description of Qin Jiushao’s method, we first note that M = 12. Then M1 = 12 ÷ 3 = 4 and M2 = 12 ÷ 4 = 3. Also P3 = 1, and P4 = 3. Therefore, we need to solve two congruences: x1 ≡ 1 (mod 3), and 3x2 ≡ 1 (mod 4). The solutions are x1 = 1 and x2 = 3. Therefore N = 0 · 4 · 1 + 1 · 3 · 3 = 9 ≡ 9 (mod 12).

26.

Using the notation from the description of Qin Jiushao’s method, we first calculate M = 11 · 5 · 9 · 8 · 7 = 27720. Then M1 = M ÷ 11 = 2520; M2 = M ÷ 5 = 5544; M3 = M ÷ 9 = 3080; M4 = M ÷ 8 = 3465; and M5 = M ÷ 7 = 3960. We then calculate that M1 ≡ 1 (mod 11); M2 ≡ 4 (mod 5); M3 ≡ 2 (mod 9); M4 ≡ 1 (mod 8); and M5 ≡ 5 (mod 7). We next need to solve congruences: the solution to 1x1 ≡ 1 (mod 11) is x1 = 1; to 4x2 ≡ 1 (mod 5) is x2 = 4; to 2x3 ≡ 1 (mod 9) is x3 = 5; to 1x4 ≡ 1 (mod 8) is x4 = 1; and to 5x5 ≡ 1 (mod 7) is x5 = 3. Given that 3 of the ri are equal to 0, we calculate N simply as n = 4 · 3080 · 5 + 6 · 3465 · 1 = 82,390. Subtracting off twice M, we get the solution as N = 82,390 − 2 · 27,720 = 26,950, where the answer is taken modulo 27,720.

27.

If s is the price of an inkstone and b is the price of a brush, the conditions give two equations: 7s = 3b + 480, 9b = 3s + 180. Solving these simultaneously gives s = 90 and b = 50.

Solutions

41

CHAPTER 8 1.

The diagram for this procedure is as follows: 1 1 2

1 5

|

8 7

3 8 5 3 1

3

1 2

0

3 6 7 1 5 4 1 1 7 2 2 7 |1 1 4 5 0 1 1 1 0 9 3 4 1 3 3 8 3 3

5

3

)2 )3

)7

5 1 4 3 4 3

√

First digit 2 ≈ 3 12 23 12 = 3 × 22 3 = quotient; (4 is too large) 36 = 3 × 22 × 3 54 = 3 × 2 × 32 33 1587 = 3 × 232 7 = new quotient 11109 = 3 × 232 × 7 3381 = 3 × 23 × 72 73

2.

We have KM = KL = BC = a and KA = b. It follows by the Pythagorean Theorem that AM2 = KM2 − AK2 = a2 − b2 . Thus the square on AM is the difference of the two original squares, the one on AB and the one on PQ.

3.

The rectangle ABCD is transformed into the gnomon AEGFKH of the same area by the indicated construction. This gnomon is equal to the difference of the squares on AE and FK. By the previous exercise, we can construct a square equal to that difference.

This square will therefore be equal to the rectangle, as desired.

4.

In the Indian method, note thatp to cut off a square of side y from a square of side x, the resulting square has a side z = x2 − y2 . If one starts with a rectangle of length ℓ and width w, then one ends up by cutting off a square of side ℓ−2 w from a square of side q √ ℓ+w ( ℓ +2 w )2 − ( ℓ−2 w )2 = ℓw. It clear 2 . So the side of the resulting square is z = from √ the diagram for Elements II–14 that the side of the desired square is also equal to ℓw.

5.

Since AB = s, we have MN = r = 2s + 13 ( s 2 2 − 2s ). Thus sr = 12 + 62 − 16 = 2+6 2 .

Given that the area of the circle of radius MN = r equals the area of the square of side √ 2 √ AB = s, we have πr2 = s2 or sr2 = π1 or sr = √1π . Thus √1π = 2+6 2 , or π = 6√ , or 2+ 2 π = 3.088311755 . . .

6.

If we calculate the sum and difference of the given fractions, we get 0.878681752.

If the square of this side is equal to the area of a circle of diameter 1, then (0.878681752)2 = π4 , or π = 4(0.878681752)2 = 3.088326491.

7.

In this problem, the speeds of the peacock and the snake are assumed equal. Let x be the distance traveled by the snake from its original position to the point where the peacock catches it. So x is also the distance traveled by the peacock. That latter line segment forms the hypotenuse of a right triangle, whose two legs have lengths 9, the height of the pole, and 27 − x, the distance from the meeting point to the hole. By the

√

√

√

42

Solutions Pythagorean Theorem, we have (27 − x)2 + 92 = x2 , so 54x = 810 and x = 15. Thus the distance between the meeting point and the hole is 27 − 15 = 12 hastas.

8. a.

Let ∠ABC = θ. Then ∠ADC = π − θ. Set x = AC. In triangle ABC, we have x2 = a2 + b2 − 2ab cos θ, while in triangle ADC, we have x2 = c2 + d2 − 2cd cos(π − θ) = c2 + d2 + 2cd cos θ. Setting the two expressions for x2 equal, we get a2 + b2 − 2ab cos θ = c2 + d2 + 2cd cos θ and therefore cos θ =

b.

a2 + b2 − c2 − d2 .

2ab + 2cd

We get x2 = a2 + b2 −

2ab(a2 + b2 − c2 − d2 ) cd(a2 + b2 ) + ab(c2 + d2 ) = .

2ab + 2cd ab + cd

c.

cd(a2 + b2 ) + ab(c2 + d2 ) = a2 cd + b2 cd + c2 ab + d2 ab = (ac + bd)(ad + bc).

d.

From parts b and c we get 2

x =

(ac + bd)(ad + bc)

ab + cd

r or

x = AC =

(ac + bd)(ad + bc)

ab + cd

.

Similarly, we have r y = BD = 9. a.

(ac + bd)(ab + cd)

ad + bc

.

From the law of cosines applied to triangle ABC, we get b2 = a2 + x2 − 2ax cos ∠BAE = a2 + x2 − 2ax

AE = a2 + x2 − 2x · AE. a

Therefore, b2 − a2 = x(x − 2AE). b.

Because x = 2AM, we have b2 − a2 = x(2AM − 2AE) = x(2EM) = 2x · EM. There2 − a2 fore, EM = b 2x .

c.

By the same arguments as in parts a and b applied to triangle ADC, we get 2 − c2 FM = d 2x .

d.

Let P be the area of quadrilateral ABCD. Then P is the sum of the areas of triangles ABC and ADC. Therefore, P = 21 x · BE + 12 x · DF = 12 x(BE + DF), and P2 = 41 x2 (BE + DF)2 .

e.

Because BE + DF = BK, we get from part d and the Pythagorean Theorem that P2 = 41 x2 BK2 = 14 x2 (BD2 − DK2 ) = 14 x2 (y2 − EF2 ).

f.

From parts b and c, we have EF = EM + FM =

b2 − a2 d2 − c2 (b2 + d2 ) − (a2 + c2 ) + = .

2x 2x 2x

Solutions

43

Substituting this value into the expression in part e, along with the values for x2 and y2 from Exercise 8, we have

1 (ac + bd)(ad + bc) (ac + bd)(ab + cd) [(b2 + d2 ) − (a2 + c2 )]2 P2 = − 4 ab + cd ad + bc 4x2 i2 1 1 h 2 2 2 2 2 = (ac + bd) − (b + d ) − (a + c ) 4 16 h i 1 = 4(ac + bd)2 − [(b2 + d2 ) − (a2 + c2 )]2 .

16 g.

We have s − a = 12 (a + b + c + d) − a = 12 (b + c + d − a), with a similar result for the three other cases.

h.

First, we calculate (b + c + d − a)(a + c + d − b)(a + b + d − c)(a + b + c − d).

By first multiplying together the first two expressions and then the last two expressions and then multiplying the two resulting expressions together, we find that this expression becomes (2ab + 2cd + c2 + d2 − a2 − b2 ) (2ab + 2cd + a2 + b2 − c2 − d2 ) = 8abcd + 2(a2 b2 + a2 c2 + a2 d2 + b2 c2 + b2 d2 + c2 d2 ) −a4 − b4 − c4 − d4 . On the other hand, if we multiply out the numerator of the expression for P2 from part f, we get 4(ac + bd)2 − [(b2 + d2 ) − (a2 + c2 )]2 = 4(a2 c2 + 2abcd + b2 d2 − 2 2 2 2 2 2 2 [(b + d ) − 2(b + d )(a + c ) + (a2 + c2 )2 ] = 4(a2 c2 + 2abcd + b2 d2 ) − [b4 + 2b2 d2 + d4 − 2a2 b2 − 2b2 c2 − 2a2 d2 − 2c2 d2 + a4 + 2a2 c2 + c4 ] = 8abcd + 2(a2 c2 + b2 d2 + a2 b2 + b2 c2 + a2 d2 + c2 d2 ) − a4 − b4 − c4 − d4 , the same expressionpas in the first calculation. It follows that the area of the quadrilateral is S = (s − a)(s − b)(s − c)(s − d), as asserted.

10.

27 By the rule of three, the amount of musk is (1 12 × 1 15 ) ÷ 8 13 = 125 pala.

11.

If t is the number of days until the second person overtakes the first, the equation is 5(t + 7) = 9t. The solution is t = 8 34 days.

12.

In 1 day, the well is filled 2 + 3 + 4 + 5 = 14 times. Thus the well will be filled once in 1 1 2 14 of a day. In that time period, the first pipe will fill the well 14 · 2 = 14 full. Similarly, 3 4 5 the second pipe will fill 14 of the well, the third 14 , and the fourth 14 . q If x is the number of elephants, then the equation is 31 x + 3 23 x = x − 4. If one trans-

13.

poses and then squares both sides, this becomes 9 · 23 x = 94 x2 − 16 3 x + 16, which simplifies to 2x2 − 51x + 72 = 0. The only integral solution to this quadratic equation is x = 24.

14.

Let p be the number of sets of 3 peacocks, g the number of sets of 4 pigeons, s the number of sets of 5 swans, and b the number of sets of 6 sˉarasa birds. Then the conditions of the problem produce two equations in the four unknowns: 3p + 4g + 5s + 6b = 72 and 2p + 3g + 4s + 5b = 56. If we solve both equations for p and equate the two solu4s − 5b 5s − 6b tions, we get 72 − 4g − = 56 − 3g − , or g = 24 − 2s − 3b. Thus there are two 3 2 arbitrary parameters, but they must be chosen so that all the solutions are non-negative.

If, for example, we assume that b = 1, then g = 21 − 2s and p = 72 − 4(21 −3 2s)−5s − 6 = s − 6. So, for example, if we further take s = 7, we get p = 1 and g = 7. Thus, one solution is that we have 3 peacocks, 28 pigeons, 35 swans, and 6 sˉarasa birds. If we take s = 8, we get p = 2 and g = 5, so we have 6 peacocks, 20 pigeons, 40 swans, and

44

Solutions 6 sˉarasa birds. There are other solutions as well. If we do not worry about having an integral number of sets, but just an integral number of birds, we can find many additional solutions. Unfortunately, the original statement of the problem does not make clear whether such a solution is allowed.

15.

b 4625 − 125 = 300. Then x + y = xy = c = g− 3b = 15 35. Since x − y = 5, we get x = 20, y = 15.

3

√

4c + b2 =

√

1200 + 25 =

√

1225 =

16.

If we substitute x3 for x and y3 for y in the formula (x − y)2 + 4xy = (x + y)2 , we 3 2 get (x3 − yp ) + 4x3 y3 = (x3 + y3 )2 . Therefore (x3 − y3 )2 = (x3 + y3 )2 − 4(xy)3 , or g = 3 3 x −p y = f2 − 4c3 . Letting c = xy = 300 and f = x3 + y3 = 11, 375, we calculate that √ g = 11, 3752 − 4 · 3003 = 21,390,625 = 4625. Therefore x3 = 12 (f + g) = 8000 and x = 20. Similarly, y3 = 12 (f − g) = 3375 and y = 15.

17.

We begin by dividing: 1096 = 365 · 3 + 1. Since we already have a remainder of 1, that is the end of the process. We now have to find v and w such that 1 · v + 808 = 3w, since 3 is the last divisor. A solution by inspection is v = 2, w = 270. Thus we have the table 365 2 270

1000 2

Here, the 1000 at the top of the second column is 2 · 365 + 270. Since we are now finished with the table, the result is x = 2, y = 1000. This gives N = 808 + 1096 · 2 = 0 + 3 · 1000 = 3000. Alternatively, we can write this as follows: 3y = 1096x + 808 x = 3z − 808

y = 1096x3+ 808 = 365x + z 1096x + 808 = 3(365x + z)

By inspection, we get x = 2, z = 270. Then y = 365x + z = 730 + 270 = 1000. Since 2 is already the smallest possible solution for x, the result again is x = 2, y = 1000, and N = 3000.

18.

As in Exercise 17, we divide: 1096 = 365 · 3 + 1. Since we have a remainder of 1, the process ends. We now have to find v and w such that 1 · v + 1 = 3w, since 3 is the last divisor. The easiest solution is v = 2, w = 1. Thus we get the table 365 2 1

731 2

Here, the 731 at the top of the second column is 2 · 365 + 1. Since we are finished with the table, the result is x = 2, y = 731. To solve 1096x + 10 = 3y, we simply multiply everything by 10: x = 20, y = 7310.

19.

We will show, via a generalizable example, that Brahmagupta’s method does give a solution to the equation rx + c = sy, assuming that the greatest common divisor of r and s divides c. In fact, we will assume that the greatest common divisor is equal to 1, although one could generalize the procedure here to the case where it is greater

Solutions

45

than 1. We will apply the Euclidean algorithm to r and s and assume, for simplicity, that it stops after four steps. In other words, we assume that r = q1 s + r1 , s = q2 r1 + r2 , r1 = q3 r2 + r3 , and r2 = q4 r3 + 1. We therefore have the following system: sy = rx + c y = rx s+ c = q1 x + t rx + c = s(q1 x + t) st − c = r1 (q2 t + u) r1 x = st − c x = str−1 c = q2 t + u r2 t = r1 u + c t = r1 ur2+ c = q3 u + v r1 u = c = r2 (q3 u + v) c r3 u = r2 v − c u = r2 vr− = q4 v + w r2 v − c = r3 (q4 v + w) 3 v = r3 w + c

(r − sq1 )x = st − c (s − r1 q2 )t = r1 u + c (r1 − r2 q3 )u = r2 v − c (r2 − r3 q4 )v = r3 w + c

We now set w = 1, v = r3 + c, and solve for the other letters. We get u = q4 v + w = r3 q4 + cq4 + 1, t = q3 u + v = r3 q3 q4 + cq3 q4 + q3 + r3 + c. Then x = q2 t + u = r3 q2 q3 q4 + cq2 q3 q4 + q2 q3 + q2 r3 + q2 c + r3 q4 + cq4 + 1 and y = q1 x + t = q1 x + r3 q3 q4 + cq3 q4 + q3 + r3 + c. To prove that the method does in fact give us a solution to the original equation, we need to substitute these values into that equation. In other words, we must show that rx + c = sy. Substituting the value for r given in the first step of the Euclidean algorithm, we need to show that (q1 s + r1 )x + c = s(q1 x + r3 q3 q4 + cq3 q4 + q3 + r3 + c).

This is equivalent to showing that r1 x + c = s(r3 q3 q4 + cq3 q4 + q3 + r3 + c).

We next substitute for s its value from the second step of the algorithm. At the same time, we substitute for x the value we calculated above. We must therefore show that r1 [q2 (r3 q3 q1 + cq3 q4 + q3 + r3 + c) + r3 q4 + cq4 + 1] + c = (q2 r1 + r2 )(r3 q2 q4 + cq3 q4 + q3 + r3 + c).

This is in turn equivalent to showing that r1 (r3 q4 + cq4 + 1) + c = r2 (r3 q3 q4 + cq3 q4 + q3 + r3 + c).

We next substitute for r1 from the third step of the algorithm. We thus must show that (q3 r2 + r3 )(r3 q4 + cq4 + 1) + c = r2 (r3 q2 q4 + cq3 q4 + q3 + r3 + c).

To demonstrate this equality, it suffices to show that r3 (r3 q4 + cq4 + 1) + c = r2 (r3 + c).

To do this, we finally substitute for r2 its value from the last step of the algorithm. We therefore must demonstrate that r3 (r3 q4 + cq4 + 1) + c = (q4 r3 + 1)(r3 + c).

That this final equation is true comes from multiplying it out. We have thus shown that the values calculated by Brahmagupta’s method in fact satisfy the original equation.

46

Solutions 20.

The Chinese method requires that the moduli be relatively prime. In this problem, we note that if N ≡ 2 (mod 3) and N ≡ 3 (mod 4), then N ≡ 1 (mod 2) and also N ≡ 5 (mod 6). Therefore, we may ignore the first congruence and solve the last three.

Using the notation from Chapter 7, we first calculate that M = 60. Then M1 = 12, M2 = 15, M3 = 20, P1 = 2, P2 = 3, and P3 = 2. We must solve 2x1 ≡ 1 (mod 5); 3x2 ≡ 1 (mod 4) and 2x3 ≡ 1 (mod 3). The solutions are x1 = 3, x2 = 3, x3 = 2.

Therefore, N = 4 · 12 · 3 + 3 · 15 · 3 + 2 · 20 · 2 ≡ 359 (mod 60), and the smallest positive N is 59. For the Indian method, we solve the first two congruences, then use that answer along with the third, and the new answer along with the fourth. The solution of N ≡ 5 (mod 6) ≡ 4 (mod 5) requires solving the equation 6x + 1 = 5y. The solution by the procedure of the previous exercises is x = 4, y = 5 and then N = 4 · 6 + 5 = 29. We next solve N ≡ 29 (mod 30) ≡ 3 (mod 4). We must solve 30x + 26 = 4y. We get x = 1, y = 14, and N = 1 · 30 + 29 = 59. To solve N ≡ 59 (mod 60) ≡ 2 (mod 3), we note that already 59 ≡ 2 (mod 3); so N = 59 is the solution to the entire set of congruences.

21.

This problem is equivalent to finding N to solve N ≡ 0 (mod 17) ≡ 1 (mod 75). In the Chinese method, M1 = 75 and M2 = 17. Then P1 = 7 and P2 = 17. We then solve 7x1 ≡ 1 (mod 17) and 17x2 ≡ 1 (mod 75). The solution to the first congruence is unnecessary, because it will be multiplied by 0. We get the solution to the second congruence by the Euclidean algorithm. We get 75 = 4 · 17 + 7 17 = 2 · 7 + 3 7 = 2 · 3+1 By substitution, we get 1 = 53 · 17 − 12 · 75, so x2 = 53. Then N ≡ 1 · 17 · 53 ≡ 901 (mod 1275). In terms of the original problem, we have 17 · 53 − 1 = 75m, so m = 12 and n = 53.

In the Indian method, we use the Euclidean algorithm in Brahmagupta’s procedure, beginning with the equation 75m + 1 = 17n. The algorithm gives the same results as above, so the three quotients are 4, 2, 2. We then need v and w so that 1 · v + 1 = 3w, so the simplest solution is v = 2, w = 1. Our table is then as follows: 4 2 2 2 1

4 2 5 2

4 53 12 12 5

Note, for example, that the 53 is calculated as 12 · 4 + 5. Thus the solution to the equation is m = 12, n = 53.

22.

D(u0 v1 + u1 v0 )2 + c0 c1 = D(u0 v1 + u1 v0 )2 + (v20 − Du20 )(v21 − Du21 ) = 2Du0 v1 u1 v0 + D2 u20 u21 + v20 v21 = (Du0 u1 + v0 v1 )2 .

23.

To solve 83x2 + 1 = y2 , we begin by noting that (1, 9) is a solution for subtractive 2; that is, 83·12 − 2 = 92 . If we compose this solution with itself, we get y1 = 83·12 + 92 = 164, x1 = 9 + 9 = 18, b1 = 4. Therefore, (18, 164) is a solution for additive 4. But then we can simply divide everything by 4 = 22 to get that (9, 82) is a solution for additive 1: 83 · 92 + 1 = 822 .

Solutions 24.

47

To show that (u1 , v1 ) is a solution, we calculate each side of the equation Du21 + 1 = v21 and show that they are equal. The left side is

L =D

2 1 uv(v2 + 1)(v2 + 3) + 1 2

1 2 2 2 Du v (v + 1)2 (v2 + 3)2 + 1 4 1 = (4 + v2 )v2 (v2 + 1)2 (v2 + 3)2 + 1 4 1 = (v4 + 4v2 )(v2 + 1)2 (v2 + 3)2 + 1.

4 =

The right side is

2 1 2 2 R = (v + 2) (v + 1)(v + 3) − 1 2

1 2 (v + 1)2 (v2 + 3)2 − (v2 + 1)(v2 + 3) + 1 = (v2 + 2)2 4 1 4 2 = (v + 4v + 4)(v2 + 1)2 (v2 + 3)2 − (v2 + 2)2 (v2 + 1)(v2 + 3) + (v2 + 2)2 4 1 = (v4 + 4v2 )(v2 + 1)2 (v2 + 3)2 + (v2 + 1)2 (v2 + 3)2 − (v2 + 2)2 (v2 + 1)(v2 + 3) 4 +(v2 + 2)2 2

=

1 4 (v + 4v2 )(v2 + 1)2 (v2 + 3)2 + (v2 + 1)(v2 + 3)[(v2 + 1)(v2 + 3) − (v2 + 2)2 ] 4 +(v2 + 2)2

1 4 (v + 4v2 )(v2 + 1)2 (v2 + 3)2 + (v2 + 1)(v2 + 3)(−1) + (v2 + 2)2 4 1 = (v4 + 4v2 )(v2 + 1)2 (v2 + 3)2 + 1.

4 =

Thus the two sides are equal, as asserted. Next, note that if u or v is even, then 12 uv is an integer, so u1 is an integer. If both u and v are odd, then v2 + 1 is even, so 12 (v2 + 1) is an integer and u1 is an integer. If v is even, then v2 + 2 is even and (v2 + 2) 21 (v2 + 1) is an integer. If v is odd, then v2 + 1 is even, so 21 (v2 + 1) is an integer. Thus in either case, v1 is an integer.

25.

To solve 13x2 + 1 = y2 , we begin by noting that (1, 3) is a solution for subtractive 4: 13 · 12 − 4 = 32 . By the previous problem, set u1 = 12 · 1 · 3 · 10 · 12 = 180 and v1 = 11[ 12 · 10 · 12 − 1] = 649. Then (180, 649) is the desired solution.

26.

If Du2 + 2 = v2 , then D(uv)2 + 1 = Du2 v2 + 1 = (v2 − 2)v2 + 1 = v4 − 2v2 + 1 = (v2 − 1)2 , as desired. If Du2 − 2 = v2 , then (u1 , v1 ) = (uv, v2 + 1) solves Du21 + 1 = v21 .

The proof is virtually identical to the previous one.

27.

To solve 61x2 + 1 = y2 , we begin by noting that 61 · 12 + 3 = 82 ; that is, that (1, 8) is a solution for additive 3. We then need to solve 1m + 8 = 3n. The general solution is easily seen to be m = 1 + 3t, n = 3 + t. We now choose t so that m2 is “close” 49 to 61: t = 2, m = 7, m2 = 49. Then take u1 = 1·73+ 8 = 5, b1 = − 61 − = −4, and 3

48

Solutions √

28.

v1 = 61 · 25 − 4 = 39. We now have 61 · 52 − 4 = 392 ; that is, (5, 39) is a solution for subtractive 4. Now use the result of Exercise 24. Set u1 = 12 · 5 · 39(392 + 1)(392 + 3) = 226,153,980 and v1 = (392 + 2)[ 21 (392 + 1)(392 + 3) − 1] = 1,766,319,049. Then (u1 , v1 ) is a solution to the original equation.

8 − i + 1) These values are the binomial coefficients 8i = 8·7···( for i = 1,2, . . . ,8. These 1·2 ···i

8 8 8 8 8 numbers are 1 = 8, 2 = 28, 3 = 56, 4 = 70, 5 = 56, 86 = 28, 87 = 8, and

8 8 = 1.

29.

The fourth Sine difference is 225 − 1 − 2 − 3 = 219, because 671 ÷ 225 ≈ 3.

Then s4 = Sin 15◦ = 671 + 219 = 890. Similarly, the fifth Sine difference is 225 − 1 − 2 − 3 − 4 = 215 and s5 = Sin 18◦ 45′ = 890 + 215 = 1105. Also, the sixth Sine difference is 225 − 1 − 2 − 3 − 4 − 5 = 210 and s6 = Sin 22◦ 30′ = 1105 + 210 = 1315.

30.

If we calculate a table of sine values by Bhaskara’s formula (not multiplying by 3438) and compare them to actual sine values, we get the following: Angle

Bhaskara’s sine

Actual sine

0 10 20 30 40 50 60 70 80 90

0.00000 0.17525 0.34317 0.50000 0.64183 0.76471 0.86486 0.93903 0.98461 1.00000

0.00000 0.17365 0.34202 0.50000 0.64279 0.76604 0.86603 0.93969 0.98481 1.00000

An inspection of this table shows that the difference between the two values is greatest at 10 degrees, where the actual difference is 0.00160, which corresponds to a percentage error of less than 1%.

31.

Bhˉaskara’s formula is Sin θ =

4Rθ(180 − θ) .

40, 500 − θ(180 − θ)

To write this in terms of radians, we first set R = 1, replace 180 by π, and then note that the units of 40,500 are “degrees squared.” Thus, we need to convert that value π to “radians squared.” We know that 1 degree is equivalent to 180 radians, so 40,500 2 π degrees squared are equivalent to 40,500 1802 radians squared. We then substitute to get 4x(π − x) 4x(π − x) 16x(π − x) , sin x = 40,500π2 = 5π2 = 2 − 4x(π − x) 5π 4 − x(π − x) 32,400 − x(π − x) as desired.

Solutions 32.

49

Brahmagupta’s procedure gives us 1 12 (219 + 215) − (219 − 215) 3 2(3 4 ) 2(3 34 )2 2 8 = 890 + (434) − (4) = 948 15 225

sin(16) = sin(15 + 1) = sin(15) +

to the nearest integer. Bhˉaskara’s procedure gives sin 16 = 3438 ·

4 · 16 · 164 = 953 40,500 − 16 · 164

to the nearest integer. The exact value is 948 to the nearest integer, so Bhˉaskara’s answer is in excess of the correct answer by approximately 0.5%.

33.

We assume that yi ≈ (is/n) − (is)3 /(6n3 ), given the approximation we have found for y. We put this into the expression for x = cos s and use the formula for the sum of integral cubes:

(n − 1)s ((n − 1)s)3 s s s3 2s (2s)3 − 3 + − + · · · + − n→∞ n n n n 6n 6n3 6n3 2 4 s s = 1 − + lim [13 + 23 + · · · + (n − 1)3 ] 2 n→∞ 6n4 Pn − 1 3 s2 s4 s2 s4 1 i=1 i = 1− + = 1− + · lim 4 2 6 n→∞ 2 6 4 n s2 s4 = 1− + .

2 24

x ≈ 1 − lim

We can similarly calculate a new value of y = sin s, using our knowledge of the double sum formula as well. We have y ≈s−

s3 6

+ lim

s 2 s3

+ 3

(2s)3 s3 + 3 6n 6n3

+ ··· +

(2s)3 ((n − 1)s)3 s3 + + ··· + 3 3 6n 6n 6n3

n→∞ n 6n s3 s5 3 = s − + lim [1 + (13 + 23 ) + · · · + (13 + 23 + · · · + (n − 1)3 )] 6 n→∞ 6n5 s3 s5 = s − + lim [n(13 + 23 + · · · + (n − 1)3 ) − (14 + 24 + · · · + (n − 1)4 )] 6 n→∞ 6n5" Pn − 1 3 Pn − 1 4 # s3 s5 i=1 i i=1 i =s− + lim − 6 6 n→∞ n4 n5

3 5 s s 1 1 =s− + − 6 6 4 5 3 5 s s =s− + .

6 120

Putting the new value for y into the formulas for x and y will lead by a similar argument to the next terms in both the sine and cosine series.

50

Solutions

CHAPTER 9 1.

8023 4638 32 48 0

3 2.

7

2

24 0 8

1

64 0 12 12

0 6 16 18 9

24

0

6

4

7

Al-Khwˉarizmˉi’s rule for solving bx + c = x2 translates to the formula s x=

2 b b +c+ .

2 2

The main point of the geometric proof is that rectangle RBMN is equal to rectangle NKTL. Then, because ( b2 )2 is equal to rectangle KHGT, we get that ( b2 )2 + c is represented by rectangle MAGL, so that the square root in the formula is equal to the side of that square, namely GA. 3. a. b.

x2 +√(10 − x)2 = 58 transforms to x2 + 21 = 10x. The formula gives x = 5 ± 25 − 21 = 7, 3.

The equation is x 10 − x 13 + = .

10 − x x 6 If we multiply both sides by 6x(10 − x) and simplify, we get x2 + 24 = 10x. The solutions are then x = 6 and x = 4.

4. a. b.

Multiplying the equation by 2 gives x2 + 10x = 56. The solution is x =

√

81−5 = 4.

2

Dividing the equation by 2 gives x + 5x = 24. The formula then gives q 5 11 5 x = 121 4 − 2 = 2 − 2 = 3.

5.

Using Figure 9.5, we note that line CH has been divided into equal segments at G and into unequal segments at A. Under these circumstances, Elements II–5 shows that the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half. In terms of the diagram, this translates as AH · AC + AG2 = CG2 . We can rewrite this as AH · AB + GK2 = CG2 , or that GK2 = CG2 − AH · AB = (b/2)2 − c. Therefore, GK is determined and x can be found as al-Khwˉarizmˉi wrote.

6.

Let x be the shorter side. Then x + 2 is the longer side. The equation is then (x + 2)x − (x + 2) = 40, which reduces to x2 + x = 42. The solution is then x = 6. So the shorter side is 6 and the longer side is 8.

Solutions 7. a.

√

The equation is x2 = (10 − x) 10. We can rewrite this as x2 + The formula then gives us that one part is

√

51 √

10x = 10 10.

s r r √ √ 2 √ 10 10 1 √ 1 x= + 10 10 − = 2 + 1000 − 2 .

2 2 2 2 The other part is r y = 10 − x = 10 + b.

1 2 − 2

r

1 √ 2 + 1000.

2 √

If we set y = 10 x− x , then the equation becomes y + 1y = 5. This transforms √ √ to y2 + 1 = 5y. The formula gives us that y = 21 ( 5 ± 1). If we set 10 x− x = √ √ 1 5x. By squaring both sides, 2 ( 5 + 1), then this can be rewritten as 20 − 3x = 2 rearranging, and simplifying, we get x + 100 = 30x, for which √ √ a solution is x = 15 − 125. The other part, namely 10 − x, is then equal to 125 − 5. If we solve this problem directly for x, we get the equation √ 10 − x x + = 5, x 10 − x √

√

which simplifies to (10 − x)2 + x2 = 5x(10 − x) and p to x2 + 100( 5 − 2) = √ 10x. p The formula then gives us two values for x: 5 ± 25 − 100 5 + 200 = √ 5 ± 5 9 − 4 5. It is straightforward to check that these are the same answers as we calculated first.

8. a.

2 If we set x = y2 , the equation becomes [y√ − (2y + 10)]2 = 8y2 . Taking square 2 roots and rearranging gives us y = (2 + 8)y + 10. The formula then gives us p √ √ y = 1 + 2 + 13 + 8, and x = y2 .

b.

If we expand the left side, rearrange, and then square both sides, the √ equation 1 becomes 9x = x2 + 49 8. (We 4 . An application of the formula gives x = 4 2 − note that if we use the plus sign in this result, we do not get a correct answer.) As an alternative, we could set x = 2y2 . The equation then becomes (2y2 + y)2 = 8y2 √ √ √ or 2y2 + y = 8y. The solution to this is y = 82−1 . Then x = 2y2 = 4 12 − 8 as before.

9.

If we set z = 1, then the second and third equations together give y2 = x and x2 + x = 1.

Therefore, sr r √ 5 1 5 1 x= − and y= x= − .

4 2 4 2 Now, returning to the first equation, we note that the sum of the three “false” values is 1 + 2

r

5 + 4

sr

5 1 − , 4 2

52

Solutions instead of 10. To find the correct values, we need to divide 10 by this value and multiply the quotient by the “false” values. Since the false value of z is 1, this means that the correct value for z is z= 1 2 +

q

10 rq

5 4 +

.

1 5 4 − 2

q To simplify, we multiply the denominator by z and set it equal to 10: 12 z + 54 z2 = rq rq q 1 2 1 1 2 5 4 5 2 5 4 4 z − 2 z = 10. We rewrite this as 10 − 2 z − 4z = 4 z − 2 z and then q q q square both sides. We get 100 + 14 z2 + 54 z2 − 10z − 20 54 z2 + 54 z4 = 54 z4 − 21 z2 .

This simplifies to 2z2 + 100 = 10z +

√

500z2 , or z2 + 50 = (5 +

√

125)z, a quadratic q

equation that can be solved by the normal procedure. The result is z = 2 12 + 31 14 − rq 781 14 − 12 21 . Note that the result is expressed essentially as Abˉu Kˉamil did. We normally write square roots of fractions somewhat differently.

10.

2 2 2 100 x8 + 100x4 = If we set y = 10 x and z = x3 , then the equation x + y = z becomes √ 4 4 10, 000. This is a quadratic equation in x . The solution is x = 12, 500 − 50.

Therefore qp 10 100 4 12, 500 − 50, y= x= , z= 3 . x x

11.

If d is the number of ducks, c the number of chickens, and p the number of pigeons, then we have two equations: 2d + 21 c + 31 p = 100 and d + c + p = 100. If we multiply the second equation by 2, then subtract the first equation from it and simplify, we get 9 c. So we can choose c = 10; then p = 51 and d = 39. Or 9c + 10p = 600, or p = 60 − 10 we can choose c = 20; then p = 42 and d = 38. Or we can choose c = 30; then p = 33 and d = 37.

12.

We need to solve the system x + y = 10, 20 + x = z2 , 50 + y = w2 . By substituting y = 10 − x in the third equation and then substituting x = z2 − 20 in the result, we get w2 + z2 = 80. We also know, since all the numbers need to be positive, that 20 < z2 < 80 and 50 < w2 < 80. So we need to divide 80 into two squares such that these last conditions are satisfied. We know that 80 = 42 + 82 , but the first square does not satisfy the inequality for z. So we need to find a small value r to add to 4 so that the square of the sum satisfies the inequality without changing the second inequality. We write 80 −(r + 4)2 = 82 − 8r − r2 and set that equal to (8 − mr)2 and search for an appropriate −8 m. The equation for r becomes 16 − mr = r2 + m2 r2 , or r = 16m . By trial and error, 1 + m2 184 we find that m = 12 works. For that value of m, we get r = 145 . We then calculate 2 2 2 2 583,696 1,098,304 163,196 1048 z = 764 145 , z = 21,025 , w = 80 − z = 21,025 , and w = 145 . Then x = z − 20 = 21,025 47,054 and y = 10 − x = 21,025 .

13.

We need to solve x2 + 10 = u2 , x2 − 8 = v2 . If we subtract the second equation from the first, we get 18 = u2 − v2 = (u + v)(u − v). If we set u + v = 9 and u − v = 2, then 2 2 7 121 81 9 2 9 u = 11 2 and v = 2 . Then x = u − 10 = 4 − 10 = 4 = ( 2 ) , so x = 2 .

Solutions 14.

x2

x

20 30 6 0 30

1 x

1 3 13

5

1 x2

−6 32

1 x3

−10

1 x4

1 x5

13 13

1 x6

20

53

1 x7

−26 32

−40

320 320

480

12

−40 −40

−60 −60

80 80

120 120

−160 −160

−240 −240

If an represents the coefficient of x1n and bn represents the leftmost entry in the row which begins two columns to the left of the column under x1n , then the method of calculation shows that bn + 2 = −2bn and that bn = 6an . But then an + 2 = bn6+ 2 = − 2b6n = −2an .

15.

Begin with the basic equation p n n n X X X X (n + 1) i4 = i5 + i4 i=1

i=1

p=1

! .

i=1

Given the result for the sum of fourth powers, we rewrite this in the form

n n n 5 X X X p4 p3 p p + + − . i5 = (n + 1) i4 − 5 2 3 30 i=1

i=1

p=1

Therefore

n n n 1 X 4 1X 3 1 X n+ i + i i − 2 3 30 i=1 i=1

i=1 5

n 1 n4 n3 n 1 n4 n3 n2 1 n2 n = n+ + + − − + + + + 2 5 2 3 30 3 4 2 4 30 2 2

6X 5 i = 5 n

i=1

=

n6 3n5 n4 n2 + + − .

5 5 2 10

If we multiply throughout by 56 , we get the final result: n X 1 5 1 1 i5 = n 6 + n 5 + n 4 − n 2 .

6 2 12 12 i=1

54

Solutions 16. nX −1

(n4 − 2n2 i2 + i4 ) = (n − 1)n4 − 2n2

i=1

3

n n5 n4 n3 n2 n n − + + − + − 3 2 6 5 2 3 30 4

7 n n = (n − 1)n4 − 15 (n − 1)n4 + 30 − 30

8 1 1 (n − 1)n4 + n4 − n 15 30 30 8 1 1 8 5 n − n4 + n4 − n = 15 15 30 30 8 1 1 = n · n4 − n4 − n 15 2 30 =

17.

Since triangle EGI is right, EH : HG = HG : HI, or HG2 = EH · HI. Also, HG2 = DH · HB. Therefore, EH · HI = DH · HB, or DH : EH = HI : HB. Then (DH − EH) : EH = (HI − HB) : HB or EC : EH = BI : HB. Since EH : HB = AE : GH, we have HB : GH = EH : AE = EH : EC. So BI : HB = EC : EH = GH : HB and BI = GH. But EG = EB. So EI = EB + BI = EG + GH.

18.

To solve x3 + d = cx, rewrite the equation of the given parabola as y = √x c and substitute

2

4

into the equation y2 − x2 + dc x = 0 of the hyperbola. The result is xc − x2 + dc x = 0.

If we multiply by c and divide by x, we get x3 − cx + d = 0, an equation equivalent to our original one. To sketch the two curves, note that the parabola has vertex at the d origin, while the hyperbola has center at ( 2c , 0), vertex of the right-hand branch (the d only relevant one) at ( dc , 0), and asymptote the line y = x − 2c . If one takes c = d = 2, then the parabola does not intersect the asymptote, so cannot intersect the hyperbola. √ 3 If one takes c = 3, d = 2, then the parabola and hyperbola intersect at (1, 3 ) and have the same tangent line there. Therefore the curves are tangent there and that point is the only intersection point. If one takes c = 4, d = 2, one can check by using a graphing calculator that the curves intersect twice, once between x = 12 and x = 34 and once between x = 1 and x = 2.

19.

For three positive solutions to exist for a cubic equation written in modern terms as x3 + qx2 + rx + t = 0, the left side of this equation must factor as (x − m)(x − n)(x − p), where m, n, p are all positive. Expanding this factored form, we get x3 −(m + n + p)x2 + (mn + mp + np)x − mnp. Therefore, the coefficient of x is positive, while the coefficient of x2 and the constant term must be negative. Writing this in al-Khayyˉamˉi’s terms, we get the form x3 + cx = bx2 + d. Now to determine the conditions under which this type of equation will in fact have three positive solutions, we rewrite it in the form x3 − bx2 + cx = d and call the left side of this equation f(x). We note √ that f ′ (x) = 2 3x2 − 2bx + c, and this derivative is 0 at the two critical values x = 3b ± b 3− 3c . For three positive solutions to exist, y = f(x) must cross the line y = d three times in the first quadrant. A consideration of the graph of f(x) shows that it always crosses that line at least once. For it to cross three times, the derivative must in fact equal 0 twice; thus b2 − 3c ≥ 0. Furthermore, the value of f(x) at the leftmost of the two critical values, call it x1 , must be greater than d, that is, f(x1 ) > d, while the value of f(x) at the rightmost critical point, x2 , must be less than d, that is f(x2 ) < d. In the special case

Solutions

55

x3 + 200x = 20x2 + 2000, we have b2 − 3c = 400 − 600 < 0, so this equation only has one positive solution.

20.

To solve x3 + d = bx2 , substitute y = dx into y2 + dx − db = 0. The result is d2 + dx − db = 0. If one multiplies by x2 and divides by d, the result is d + x3 − bx2 = 0, x2 an equation equivalent to the original one. The hyperbola and parabola intersect exactly once when they have identical tangent lines at the intersection point (x0 , y0 ). The tangent line to the hyperbola at that point is y = − xy00 x + 2y0 , while the tangent line to y0 d 0 the parabola there is y = − 2yd0 x + y0 + dx 2y0 . If these lines are identical, then x0 = 2y0 or 2y20 = dx0 . Substituting this value into the equation of the parabola and simpli3d fying shows that x0 = 2b 3 . Then y0 = 2b . By substituting the value for x0 into the 3 original equation, we also get that 4b = 27d. For the curves to have no√intersection, we must have the hyperbola always above the parabola. Thus dx > db − dx; 3 d2 > db − dx; and d > bx2 − x3 for all x. But the maximum of bx2 − x3 is 4b 27 . So x2 4b3 < 27d. The case where the hyperbola and parabola intersect twice is then when 4b3 > 27d.

21.

If y = bx2 − x3 , then y′ = 2bx − 3x2 and y′ = 0 when x = 0 or when x = 2b 3 . The second derivative test shows that x0 = 2b 3 makes y maximal. If we now consider the graph of f(x) = x3 − bx2 + d, we note that it has a maximum at 0 (and f(0) = d) and a 2b 4b2 minimum at 2b 3 (and f( 3 ) = − 27 + d). For this graph to cross the x-axis twice for x positive, this minimum value must be negative. Thus there are two positive solutions 3 3 4b3 to the cubic if − 4b 27 + d < 0 or if 4b > 27d; there is one positive solution if − 27 + 3 d = 0 or if 4b3 = 27d; and there are no positive solutions if − 4b 27 + d > 0 or if 4b3 < 27d. p To solve x3 + d = cx, set f(x) = cx − x3 . Then f ′ (x) = c − 3x2 , which is 0 when x = 3c . p pc (Of course, we only consider the positive value.) Then f( 3c ) = 2c 3 3 is a maximum value for f. The original then has two solutions if this value is greater than pequation d, one solution (at x = 3c ) if this value equals d, and no solutions if this value if less than d. We can rewrite this condition as follows: There are two solutions if 4c3 > 27d2 ; there is one solution if 4c3 = 27d2 ; and there are no solutions if 4c3 < 27d2 .

22.

23.

If we set n = 4 in ibn Qurra’s theorem, we get p3 = 23, p4 = 47, and q4 = 1151. Since these are all prime, the numbers a = 16 × 23 × 47 = 17,296 and b = 16 × 1151 = 18,416 are amicable.

24.

Since the proper divisors of 1184, namely 1, 2, 4, 8, 16, 32, 37, 74, 148, 296, and 592, sum to 1210 and the proper divisors of 1210, namely 1, 2, 5, 10, 11, 22, 55, 110, 121, 242, and 605, sum to 1184, the two numbers are amicable. But 1210 is 2 × 605 and 605 is not equal to q1 ; thus these numbers are not in the form stated in the theorem.

25.

If we set n = 7 in ibn Qurra’s theorem, we get p7 = 383, p6 = 191, and q7 = 73,727.

Each of these numbers is in fact prime. Thus a = 27 × 191 × 383 = 9,363,584 and b = 27 × 73,727 = 9,437,056 are amicable. q q q √ √ √ 26. a. If we square the right side, we get 4 12 ± 2 32 + 12 = 23 2 ± 6 + 12 2 = √

√

2 2± 6 =

√

√

8 ± 6, and this latter value is the square of the left side.

56

Solutions b.

√

√

√

√

√

√

√

√

√

The square of 4 12 ± 3 is 12 ± 2 6 + 3 = 3 3 ± 2 6 = 26 ± 24. Thus the first√two expressions are equal. √ fourth power of the last expression is √ The 51 ± 36 2, which is equal to (3 3 ± 2 6)2 .

27.

The five results listed are standard calculus exercises. For the semicircle of radius 1, the moment about the x-axis is Mx = 23 . The y coordinate of the center of gravity is 4 then Mx divided by the area π2 of the semicircle. Thus this coordinate is 3π . If this 3 28 1 value is to be equal to 7 , then π = 9 = 3 9 .

28.

For the first step, we set y1 = pq = a. Since q = 47, 6; . . . and p = 45, 0, we get a = 1; that is, y1 = 1. For the second step, y2 = a + b. So y2 =

q + y31 p , and, if we substitute for

··· y1 and y2 , we get b = q−app + a = q−pp + 1 = 2,7;8,29,53, = 0; 2, . . . Thus b = 0; 2, and 45,0 3

y2 = 1;2. For the third step, we get y3 = a + b + c = c=

q + y32 q+(a + b)3 . Therefore, p = p

q − (a + b)p + (a + b)3 47,6;8,29,53,37,3,45 − 46,30 + 1;6,12,48 = p 45,0 37;14,42,41,37,3,45 = = 0;0,49, . . .

45,0

Thus c = 0;0,49 and y3 = 1;2,49. For the fourth step, we make a similar calculation q + y3 after setting y4 = a + b + c + d = p 3 . The equation for d is q − p(a + b + c) + (a + b + c)3 p 47,6;8,29,53,37,3,45 − (45,0)(1;2,49) + (1;2,49)3 = .

45,0

d =

It follows that d = 0;0,0,43 and y4 = 1;2,49,43.

29.

From Figure 9.37, we get two equations relating x and y: y = (x + d) tan α2 and y = x tan α1 . The second equation can be rewritten as x = y cot α1 , which is an alternate form of al-Qabˉis. ˉi’s second equation. If we then substitute this value for x into the first equation, we get y = (y cot α1 + d) tan α2 . Solving this for y gives y=

d sin α2 d tan α2 = , 1 − cot α1 tan α2 cos α2 − cot α1 sin α2

a form equivalent to that of al-Qabˉis. ˉi.

30.

Since Mecca has latitude 21◦ 25′ N and longitude 39◦ 49′ E and Rome has latitude 41◦ 53′ N and longitude 12◦ 30′ E, we have (in Figure 9.31) TN = α = 41◦ 53′ , TL = β = 21◦ 25′ , MT = 90 − β = 68◦ 35′ , and ∠MTH = δ − γ = 27◦ 19′ . We then follow the steps outlined on pp. 333–334. First, we get sin MH = 0.42722 and MH = 25◦ 17′ .

So HJ = 64◦ 43′ . Next sin TF = .40383 and TF = 23◦ 49′ . Then FN = 18◦ 4′ and PF = 71◦ 56′ . Then sin PI = 0.85963, so PI = 59◦ 16′ and IQ = 30◦ 44′ . Finally, sin NC = .54871, so NC = 33◦ 17′ and the qibla NK = NC + CK = 123◦ 17′ .

31.

That rα : R = sin(90 − α) follows by considering the right triangle in a cross section of the earth one of whose sides is rα and whose hypotenuse is R. The latitude angle α

Solutions

57

is the angle that the hypotenuse of that triangle makes with the radius of the earth at the equator and is therefore the complement of the angle opposite rα in that triangle.

Therefore, rα = R cos α as desired.

32.

33.

◦ The radius of the latitude circle at latitude 40◦ is r40 = 25,000 2π cos 40 = 3048 miles.

Then, since the difference in latitude between the cities is 108◦ , the actual distance along the latitude circle is 108 360 2πr40 = 5745 miles. The length of the chord between the two cities is 2 × 3048 × sin 54◦ = 4932 miles. On the great circle, with radius /2 25,000 = 3979 miles, this chord subtends an angle β such that sin(β/2) = 4932 2π 3979 = ◦ ◦ 0.6198. Thus β/2 = 38.3 and β = 76.6 . Along a circle of circumference 25,000 miles, an arc of 76.6 360 × 25,000 = 5319 miles.

Let d be the diagonal of an isosceles trapezoid, with upper base a, lower base b (a < b), and where c is the length of each of the other two sides. Also, let h be the altitude a of the trapezoid. If one extends the upper base to the left, say, by length b− 2 and drops a perpendicular from the end of that line segment to the lower base, then two right triangles are formed, one with legs of length a +2 b and h with hypotenuse d, and a the second with legs of length b− 2 and h with hypotenuse c. Then the Pythagorean Theorem gives

d2 =

a+b 2

2

+ h2 =

a+b 2

2

+ c2 −

b−a 2

2 = ab + c2 ,

as desired.

34.

Referring to Figure 9.32, we let New York be at point A and London at point B. Then ◦ the chord of AD is the chord in a circle of radius 25,000 2π cos 41 subtended by an angle ◦ ◦ ◦ 25,000 of 74 . This is equal to 2 2π cos 41 sin 37 = 3614.37. Similarly, the chord of BC is ◦ ◦ equal to 2 25,000 2π cos 52 sin 37 = 2948.46. The chord of AC is the chord in a circle of ◦ 11◦ , so it is equal to 2 25,000 radius 25,000 2π subtended by an angle of 2π sin 5.5 = 762.74. p 2 Then the chord of AB is equal to chord AD · chord BC + chord AC = 3352.4.

The arc of a great circle (radius 25,000 2π ) subtended by a chord of this length is 3352.4/2 ◦ 2 arcsin 25,000/2π = 49.83 . Thus, the distance in miles is 49.83 360 × 25,000 = 3460 miles.

35.

We need to solve the spherical triangle RNM with vertices at Rome, the North Pole, and Mecca. We know from the given information that side m is 48◦ 7′ , side r is 68◦ 15′ , and angle N is 27◦ 19′ . The qibla is then angle R. From the formula of al-Battˉanˉı we get cos n = cos r cos m + sin r sin m cos N = cos(68◦ 15′ ) cos(48◦ 7′ ) + sin(68◦ 15′ ) sin(48◦ 7′ ) cos(27◦ 19′ ) = 0.8618. Therefore side n is 30◦ 29′ . We then calculate R from the law of sines: sin R =

sin N sin r sin(27◦ 19′ ) sin(68◦ 15′ ) = = 0.8403. sin n sin(30◦ 29′ )

A quick glance at a globe should convince you that R is an obtuse angle. Therefore, R = 122◦ 50′ . (Using al-Battˉanˉı’s formula to find cos R will also give you this result.)

36.

We need to solve the spherical triangle NPL with vertices at New York, the North Pole, and London for the side p opposite the pole. The formula gives

58

Solutions cos p = cos n cos ℓ + sin n sin ℓ cos P = cos 38 cos 49 + sin 38 sin 49 cos 74 = 0.6451.

Therefore p = 49.83◦ . To find the distance in miles, we divide this answer by 360 and multiply by 25,000. The result is 3460 miles.

37.

α From Figure 9.38, we have r +r h = cos α. Thus r = r cos α + h cos α and r = 1h−cos cos α . ◦ ′ To calculate r, substitute h = 652;3,18 and α = 0 34 into the formula. The result is 13,331,731 cubits, which equals 19,997,597 feet, or 3787 miles. The procedure is very sensitive to a small change in the measured value of α, and it is difficult to see how α can be measured with much precision.

38.

We have r sin β = sin α = sin(γ − β) = sin γ cos β − cos γ sin β. It follows that (r + cos γ) sin β = sin γ cos β. Therefore,

tan β =

sin γ sin β = . cos β r + cos γ

Since γ and r are known, tan β is known and therefore so is β and then α.

39.

In Figure 9.34, let c = AB = 40◦ , b = AC = 50◦ , and A = 25◦ . We drop a perpendicular BD to AC, and let h = BD. By the law of sines, sin h = sin 25◦ sin 40◦ , so h = 15.76◦ . Now, let e = AD and f = DC. By the cosine relationship for right triangles, we have cos 40◦ = cos h cos e, so cos e = cos 40◦ / cos 15.76◦ = 0.7960, so e = 37.25◦ . Therefore, f = 12.75◦ . Again using the cosine relationship, setting a = BC, we have cos a = cos h cos f = cos 15.76◦ cos 12.75◦ = 0.9387, so a = 20.17◦ . To find angle C, we use the law of sines: sin C = sin 40◦ sin 25◦ / sin 20.17◦ = 0.7878, and C = 51.98◦ . Finally, sin B = sin 50◦ sin 51.98◦ / sin 40◦ = 0.9389, so B = 110.14◦ .

(This final angle must be obtuse, so that the sum of the three angles is greater than 180◦ .)

40.

We are given that AB = 60◦ , AC = 75◦ , and BC = 31◦ as in Figure 9.35. Since AD and AE are quadrants, we know that BD = 30◦ and CE = 15◦ . By the rule of four quantities, sin CF : sin BF = sin CE : sin BD = sin 15 : sin 30 = 0.5176. Since CF = BF − 31◦ , we have that 0.5176 sin BF = sin CF = sin(BF − 31). Therefore, 0.5176 sin BF = sin BF cos 31 − sin 31 cos BF, or 0.5176 sin BF = 0.8571 sin BF − 0.5150 cos BF. It follows that 0.5150 cos BF = 0.3395 sin BF, or that tan BF = 1.5169.

Thus BF = 56◦ 36′ and CF = 25◦ 36′ . To find DF, we use equation 5.6. This implies that cos BF = cos BD cos DF or that cos 56◦ 36′ = cos 30◦ cos DF. Then cos DF = 0.6356 and DF = 50◦ 32′ . Also we have cos CF = cos CE cos EF or cos 25◦ 36′ = cos 15◦ cos EF. Thus cos EF = 0.9336 and EF = 21◦ . Since ∠A = arc DE, we have ∠A = 29◦ 32′ . To find ∠C, we use the sine law: sin BC : sin A = sin AB : sin C. Thus sin C = 0.8280 and C = 55◦ 59′ . Similarly, from sin AC : sin B = sin BC : sin A, we calculate that sin B = 0.9235 and therefore that B = 112◦ 25′ . (Note that because AC > AB, we must have ∠B obtuse.)

41.

Given that A = 75◦ , B = 80◦ , and C = 85◦ , we find that MN = 180◦ − A = 105◦ , LN = 180◦ − B = 100◦ , and ML = 180◦ − C = 95◦ . Although these sides are all greater than quadrants, the method described in the text can be adapted to solve triangle LMN. The results are that L = 106.21◦ , M = 101.77◦ , and N = 97.97◦ . Then BC = 180◦ − L = 73.79◦ , AB = 180◦ − N = 82.03◦ , and AC = 180◦ − M = 78.23◦ .

Solutions

59

CHAPTER 10 1.

Since there are 7200 sextarii in the cask, and since one-third of that amount is 2400, it follows that 2400 + 6(200) = 3600 sextarii flow in through the first pipe. Then 2400 sextarii flow in through the second pipe and 1200 through the third.

2.

One solution is to have the man and the goat go across first, then have the man return.

On his second trip, the man takes the cabbage and returns with the goat. On the third trip, he takes the wolf across and returns alone. Finally, he brings the goat across again.

3.

Since the ratio of 28 to 10 12 equals 8 to 3, a chord of length 6 in a circle of diameter 10 21 corresponds to a chord of length 16 in a circle of diameter 28. According to Abraham’s table, a chord of length 16 corresponds to an arc of 17 parts, 2 minutes, 16 seconds.

To convert this back to a circle of diameter 10 12 , we need to multiply this value by 83 .

This product is 6 parts, 23 minutes, 21 seconds.

4.

The area of the sector is 4 2 = 16.772. To find the area of the corresponding tri√ angle, we can use the Pythagorean Theorem to find the altitude. This is 5.252 − 32 = 4.308. Thus the area of the triangle is 12 (4.308)(6) = 12.924 and the area of the segment is 16.772 − 12.924 = 3.848.

5.

An arc of length 5 12 in a circle of diameter 33 corresponds to an arc of length 4 23 = 4;40 in a circle of diameter 28. If we use linear interpolation in Abraham’s table, we find that the ratio of 4;40−4;0,55 to 5;1,44 − 4;0,55 is equal to 0.64. Therefore, we approximate the corresponding chord by 4.64. Multiplying this by 33 28 gives 5.47 for the desired chord length.

6.

s According to Abraham’s formula, d = 4h + h = 88 + 2 = 10.

7.

Since BH : HA = IE : EA, the lines HE and BG are parallel. Therefore triangle GHE is equal to triangle BHE. Then the region AHG is equal to triangle GHE plus triangle AEH plus section AEG. Since triangle GHE equals triangle BHE, the first two summands are equal to triangle ABE. Therefore, region AHG is equal to triangle ABE plus section AGE. But triangle ABE is half of triangle ABC and section AEG is half of section AGC. Thus the region GHA is half of the total original region.

8.

The nth pentagonal number is 1 + 4 + 7 + 10 + · · · + (3n − 2). That is, this number can be expressed as

5 1 ×6;23,21

2

n X

(3i − 2) =

i=1

2

3n(n + 1) 3n2 − n − 2n = .

2 2 ◦

2

The area of a pentagon with edge length r is 5r tan4 54 . Thus if r = 1, the area is 5 4 · 1.376 = 1.720, while the second pentagonal number is 5. For r = 2, the area is 5 · 1.376 = 6.882 and the third pentagonal number is 12. For r = 3, the area is 45 4 · 1.376 = 15.484 and the fourth pentagonal number is 22.

9.

Since r1 = 25 and r2 = 27 , we have d1 = r1

f

r2 − 1

Therefore, h = r1 d1 = 25 · 125 = 50.

= 7

50

5 −1

=

50 2 5

= 125.

60

Solutions 10.

We first multiply 8 rods, 3 feet, 16 27 unciae by the ratio of 42:10 = 21:5. We get 33 53 rods, 12 35 feet, 68 25 unciae. This is equivalent, first, to 35 35 rods, 3 53 feet, 14 25 unciae.

But 52 of a rod equals 2 25 feet, so we now have 36 rods, 1 15 feet, 14 25 unciae. But 15 of a foot equals 3 35 unciae, which, added to the 14 52 unciae make 1 more foot. Thus our value is 36 rods, 2 feet and this value, according to the table, corresponds to an arc of 42 rods. Multiplying this by the ratio 5:21 gives us a value of 10 rods for the arc in the circle of diameter 10.

11.

If we mark the intersection of ad and bg as e, then since triangle age is a right triangle, the Pythagorean Theorem tells us that ae = 8. But since triangle agd is also a right triangle with altitude 6, we have 8 : 6 = 6 : de, and therefore de = 4 21 . Thus the diameter of the circle is 8 + 4 12 = 12 12 .

12.

Let us take the radius of the circle to be 1. Then the sum formula for chords is given by 2crd (α + β) = crd α crd (180 − β) + crd β crd (180 − α). To calculate a formula for the sum of three chords, say α + β + γ, we use this formula (and the corresponding formula for crd (180 − (α + β))), replacing α by α + β and β by γ. The result, after some simplification, is 4crd (α + β + γ) = crd α crd (180 − β) crd (180 − γ) + crd β crd (180 − α)crd (180 − γ) + crd γ crd (180 − α) crd (180 − β) − crd α crd β crd γ.

This result can be translated into the following formula for the sine of the sum of three angles: sin(α + β + γ) = sin α cos β cos γ + sin β cos α cos γ + sin γ cos α cos β − sin α sin β sin γ.

13.

Suppose we drop a perpendicular from one vertex to the opposite side (and assume that the perpendicular falls inside the triangle). By Elements II–13, we know that the square on the side opposite one of the acute angles is less than the sum of the squares on the other two sides by twice the rectangle contained by one of the sides about the acute angle, namely, that on which the perpendicular falls, and the line segment between the angle and the perpendicular. Since all three sides are known, this result determines the length of that last line segment. By the Pythagorean Theorem, the length of the perpendicular is also known. Therefore, the sine of the acute angle is known and therefore the angle itself. One can repeat this result to determine the other angles.

14.

The icosahedron is made up of 20 equilateral triangles and has 12 vertices, with five triangles meeting at each vertex. So if one cuts off a piece from each triangle at a given vertex, as shown in Figure 10.11, the resultant figure has 12 pentagons. Also, cutting off each of the corners of the equilateral triangles produces three additional sides, thus forming 20 hexagons from the original 20 triangles.

15.

First, formed by five edges of the icosahedron has side d = p the pentagon √ 1 50 − 10 5.

Then the edge s of the decagon formed from this pentagon is given 10 by s = 2 sind72◦ . Therefore the root of 3 times the square on the decagon is the root of

√ 3d2 1 , or 2 sin3d ◦ . Then 3 3 times the area of the equilateral triangle with that value 4 sin2 72 √ 72 √ √ 3 3d 2 5 3 d2 as side is 10 3 4 [ 2 sin 72◦ ] = 8 sin2 72◦ . The square on the edge of the icosahedron is 2 ◦ d2 , so the ratio of that to the previous value is 8 sin√72 = 0.83554916. Also, since the 5 3

Solutions

61

√

edge of the cube is 33 , the area of one face is 13 and the surface area of the entire cube is 2. Also, √ the edge of the icosahedron is d (above), so one face of the icosahedron √ 1 is equal to 43 100 [50 − 10 5]. The surface area of the entire icosahedron is 20 times √

√

16.

that, or 203 [50 − 10 5]. The ratio of 2 to that last value is √ 4 √ = 0.83554916, as 3[5− 5] desired.

Pn − 1 (n − 1)n (1 + 2 + · · · + n) + (1 + 2 + · · · + (n − 1)) = 2 i=1 i + n = 2 2 + n = n2 .

17.

If n is even, break this sum of sums into pairs. The first pair is 1 + (1 + 2), which, by the previous proposition, equals 22 . Similarly, the second pair is (1 + 2 + 3) + (1 + 2 + 3 + 4), which, by the previous proposition, equals 42 . Similarly, the next pair sums to 62 and so on. If n is odd, then break this sum into pairs beginning with the pair (1 + 2) + (1 + 2 + 3). This sum is equal to 32 . Since the first term of the sum is 12 , the entire sum is just the sum of the squares of the odd numbers up to n.

18.

In the sum (1 + 2 + 3 + · · · + n) + (2 + 3 + · · · + n) + (3 + · · · + n) + · · · + n, we have n copies of n, n − 1 copies of n − 1, . . ., 2 copies of 2, and 1 copy of 1. Thus, this sum is equal to 12 + 22 + · · · + n2 as asserted.

19.

Each term in the left pair of square brackets sums with exactly one term in the right pair of square brackets to the value 1 + 2+ · · · + n. Since there are n such pairs of terms (including one for which the corresponding term on the right is empty), the total sum is n(1 + 2 + · · · + n) as claimed.

20.

We suppose that n is even. By Exercises 17, 18, and 19, we have n(1 + 2 + · · · + n) = (12 + 22 + · · · + n2 ) + (12 + 32 + · · · + (n − 1)2 ).

Also, by Exercise 17, (1 + 2 + · · · + n) = (22 + 42 + · · · + n2 ) − [1 + (1 + 2) + · · · + (1 + 2 + · · · + (n − 1))] = (22 + 42 + · · · + n2 ) − 2 2 2 (1 + 3 + · · · + (n − 1) ). By subtracting one copy of 1 + 2 + · · · + n from both sides, it follows that (n − 1)(1 + 2 + · · · + n) = 3(12 + 32 + · · · + (n − 1)2 )

or 1 (n − 1)(1 + 2 + · · · + n) = 12 + 32 + · · · + (n − 1)2 .

3 By subtracting this equation from the first equation, we get that

1 n − (n − 1) (1 + 2 + · · · + n) = 12 + 22 + · · · + n2 , 3

as desired. A similar argument proves the result if n is odd.

62

Solutions 21.

In 15 days, the first hole empties the barrel 3 times, the second hole empties it 5 times, the third hole empties it 18 times, while the fourth hole empties it 30 times. Thus in 15 days the barrel is emptied 56 times. It follows that the barrel can be emptied one time in 15/56 of a day, or approximately 6 hours, 26 minutes.

22.

By taking the least common multiple, we find that 60 dinars will buy 30 litras of the first drug, 20 litras of the second drug, 5 litras of the third drug, and 3 litras of the fourth drug.

23.

From Levi’s answer, we get A + 1 + AB − 1 = (B + 1)A, which is correct. Also, B + 1 + AB − 1 = (A + 1)B, which is also correct.

24.

11 19 In this case, A = 3 19 35 , B = 7 12 , so the three numbers are x = A + 1 = 4 35 , y = B + 1 = 1 11 8 12 , and z = AB − 1 = 27 21 . But y is given as 30. So we need to multiply the x and z by 68,160 30 . Thus the new first number is x = 11,448 749 , while the new third number is z = 749 .

8 11

25.

(1 + 2 + · · · + n)2 − (1 + 2 + · · · + (n − 1))2 = [(1 + 2 + · · · + n) + (1 + 2 + · · · + (n − 1))][(1 + 2 + · · · + n) − (1 + 2 + · · · + (n − 1))] = n2 · n = n3 . (That the first factor

12

in the square brackets equals n2 follows from Exercise 16.)

26.

If we start with 1, we then get 1 · 1 + 1 · 3 = 4, then 1 · 1 + 2 · 3 + 1 · 9 = 16, then 1 · 1 + 3 · 3 + 3 · 9 + 1 · 27 = 64, and so on. In general, suppose we have a geometric series 1, a, a2 , a3 , . . .. If we put this series into the Pascal triangle, we get, first, 1 · 1 = 1, then 1 · 1 + 1 · a = 1 + a, then 1 · 1 + 2 · a + 1 · a2 = (1 + a)2 , then 1 · 1 + 3 · a + 3 · a2 + 1 · a3 = (1 + a)3 , and so on. In other words, the generalization of Jordanus’s formula is simply the binomial theorem.

27.

It is easiest to convert the price to mils and the amount to ounces. Thus 12 ounces of saffron are sold for 37 14 mils. We want to know how much 209 12 ounces sell for. We use the rule of three to get (37 41 × 209 12 ) ÷ 12, or 7803 78 ÷ 12 or 62, 431/96 = 650 31 96 31 mils. We can convert this to bezants: 65 bezants, 96 mils.

28.

5 11 We multiply 7 12 by 21 12 to get 159 24 Pisan denarii. We could rewrite this as 13 solidii, 3 11 denarii.

24

29.

Since 1 roll of pepper costs 4/7 bezant and 1 bezant buys 9/11 pound of saffron, then 9 828 58 23 rolls of pepper cost 92/7 bezants and therefore are worth 92 7 11 = 77 = 10 77 pounds of saffron.

30.

In 60 hours, the lion can eat 15 sheep; the leopard can eat 12 sheep; and the bear can eat 10 sheep. Thus, the three carnivores together can eat 37 sheep in 60 hours. They 23 therefore can eat 1 sheep in 60 37 = 1 37 hours.

31.

If x is the number of denarii of the first man, y that of the second, and p the number in the purse, then we have x + p = 3y and y + p = 4x. Therefore, 3y − x = 4x − y, or 4y = 5x, or y = 54 x. We can choose any value for x, so pick x = 4. Then y = 5 and p = 11.

Solutions 32.

63

Let xi , i = 1, 2, 3, 4, 5, be the amount that each man had originally, and let z be the amount in the purse. The conditions of the problem then produce five equations: 5 (x + x + x + x ) 2 2 3 4 5 10 x2 + z = (x + x + x + x ) 3 1 3 4 5 17 x3 + z = (x + x + x + x ) 4 1 2 4 5 26 x4 + z = (x + x + x + x ) 5 1 2 3 5 37 x5 + z = (x + x + x + x ) 6 1 2 3 4 x1 + z =

Thus, we have five equations in six unknowns and should not expect a unique solution, but instead a one-parameter family of solutions. But since we want a solution in integers, we will simply determine such a solution, indeed the simplest such solution. (Note that if you set this system up in modern matrix form and try to apply Gaussian elimination, the problem becomes computationally very difficult and would certainly require a good calculator, and even then you may find it difficult to get an integer solution. But we want to solve this problem without such sophisticated tools.)

Let us rewrite the first equation in the form 2x1 + 2z = 5x2 + 5x3 + 5x4 + 5x5 and then add 5x1 + 5z to both sides. We get 7(x1 + z) = 5(x1 + x2 + x3 + x4 + x5 + z), or, letting T = x1 + x2 + x3 + x4 + x5 + z, (x1 + z) = 75 T. By applying a similar process to the other four equations, we get the following five equations equivalent to our original five: 5 T 7 10 x2 + z = T 13 17 x3 + z = T 21 26 x4 + z = T 31 37 x5 + z = T 43 x1 + z =

17 If we now add the five equations together, we get T + 4z = αT, where α = 57 + 10 13 + 21 + 26 37 31 + 43 . This equation reduces to 4z = (α − 1)T, or to

T=

4 z. α−1

4 The idea now is to calculate α− 1 as a fraction and then to choose z (which will essentially be our parameter) so that both z and T are integers (and, of course, so that T is divisible by 7, 13, 21, 31, and 43). We first calculate (and you will probably need your calculator for this):

α=

1,452,803 , 363,909

64

Solutions where the denominator is the least common multiple of 7, 13, 21, 31, and 43. Then α−1 =

1,088,894 363,909

4 1,455,636 = . α − 1 1,088,894

and

(One could divide numerator and denominator by 2 here, and get slightly smaller answers, but Leonardo does not do so, so we will leave this fraction as it is.) We therefore set z = 1,088,894 to be the amount of money in the purse. Then T = 1,455,636 and we can calculate xi for i = 1, 2, 3, 4, 5. First. since x1 + z = 75 T, we get x1 + 1,088,894 = 57 · (1,455,636) = 1,039,740 and x1 = −49,154, a debt. Sim10 ilarly, x2 + 1,088,894 = 13 · (1,455,636) = 1,119,720 and x2 = 30,826. Analogous calculations then give x3 = 89, 478, x4 = 131,962, and x5 = 163,630.

33.

We prove this result by induction. It is clearly true for n = 2. Suppose it is true for k = n − 1. We prove the result for k = n: Fn − 1 · Fn + 1 = Fn − 1 (Fn − 1 + Fn ) = F2n − 1 + Fn −1 ·Fn = (Fn · Fn − 2 + (−1)n − 1 ) +Fn − 1 · Fn = (Fn − 2 + Fn − 1 )Fn + (−1)n − 1 = F2n − (−1)n . For the second part, let x = lim

Fn

n→∞ Fn − 1

.

Note that Fn + 1 Fn − 1 Fn + 1 − Fn − 1 − = = 1.

Fn Fn Fn Taking limits in this equation, we have x − 1x = 1, or x2 − 1 = x, or x2 − x − 1 = 0, or, √

finally, x = 1+2 5 , as desired.

34.

If a + b is even, then a and b have the same parity. Suppose both a and b are even.

Then all three factors of ab(a + b)(a − b) are even, so the product is divisible by 8. If either a or b is divisible by 3, then the product is divisible by 24. If neither is divisible by 3, then each is congruent to either 1 or 2 modulo 3. In that case, either the sum or difference of a and b is divisible by 3, so again the product is divisible by 24. Next, suppose a and b are both odd. Then each is congruent to 1 or 3 modulo 4. Thus either their sum or difference is divisible by 4 and the congruous number itself is divisible by 8. But by the same argument as before using divisibility by 3, the product must also be divisible by 3, and the entire number is divisible by 24. If a + b is odd, then a and b have different parities. Therefore ab is even and 4ab(a + b)(a − b) is divisible by 8.

Again, the argument using divisibility by 3 shows that the product is also divisible by 3 and therefore by 24.

35.

We begin by determining a solution to a2 + 24 = b2 , a2 − 24 = c2 . One such solution is a = 5, b = 7, c = 1. Take the two equations 25 + 24 = 49, 25 − 24 = 1, and divide each by 24. The results are 25 49 +1 = ; 24 24

25 1 −1= .

24 24

Solutions

65

Next, square each of these equations to get

25 24

2

+2

25 24

+1 =

49 24

2

;

25 24

2

−2

25 24

+1 =

1 24

2 .

These two equations can be rewritten in the necessary form as

25 24

2 +

25 = 24

49 24

2 −

25 − 1; 24

25 24

2 −

25 = 24

1 24

2 +

25 − 1.

24

35 2 Since the number on the right side of the first equation is ( 24 ) and the number on the 5 2 35 5 right side of the second equation is ( 24 ) , it follows that x = 25 24 , z = 24 , y = 24 is a solution to the original problem.

36.

We have x + y = 9, xy + x − y = 21. A straightforward modern solution would start by setting y = 9 − x and substituting. The second equation then becomes, after some algebraic manipulation, x2 − 11x + 30 = 0. This equation has two solutions, x = 6 and x = 5. Thus there are two answers to the problem: x = 6, y = 3; and x = 5, y = 4. A solution more in keeping with Jordanus’s techniques would be to notice that (x + y)2 − 4(xy + x − y) + 4 = x2 + 2xy + y2 − 4xy − 4(x − y) + 4 = x2 − 2xy + y2 − 4(x − y) + 4 = (x − y)2 − 4(x − y) √ + 4 = [(x − y) − 2]2 . It follows that 2 2 [(x − y) − 2] = a − 4b + 4 or that x − y = 2 ± a2 − 4b + 4. With x − y and x + y known, one can solve for √ x and y as in Jordanus’s Proposition I–1. In this particular case, we have x − y = 2 ± 81 − 84 + 4 = 2 ± 1. When x − y = 3, we get x = 6, y = 3.

When x − y = 1, we get x = 5, y = 4.

37.

Suppose x + y = 10, 3x + 2y = 4. If we multiply the second equation by 6, we have 2x + 3y = 24. Multiplying the first equation by 2 and subtracting from the new one, we then get y = 4 and therefore x = 6. In the general case, multiply the second equation by bc and the first one by c.

38.

If x + y = 9 and x2 y2 = 324, then xy = 18 and 4xy = 72. Since (x + y)2 = x2 + 2xy + y2 = 81, it then follows that (x−y)2 = x2 − 2xy + y2 = (x + y)2 − 4xy = 9. Therefore x−y = 3 and x = 6, y = 3. In general, if we have x + yp= a, x2 y2 = b, we note that (x − y)2 = √ √ use Proposition (x + y)2 − 4xy = a2 − 4 b. Therefore x − y = a2 − 4 b,√and we can √ I–1. In this particular case, the formula gives us x − y = 81 − 72 = 9 = 3, which, together with x + y = 9, gives us x = 6, y = 3.

39.

To divide 3 : 2 by (2 : 1)1/3 , we cube the first ratio to get 27 : 8, then divide that by 2 : 1 to get 27 : 16, and finally take the cube root of that, namely (27 : 16)1/3 , as our answer.

40.

The number of ways of comparing two of the ratios from 2:1 up to 101:1, always

comparing a greater to a smaller, is simply 100 = 4950.

The rational ratios come 2 from the following relationships: (2 : 1)2 = 4 : 1; (2 : 1)3 = 8 : 1; (2 : 1)4 = 16 : 1; (2 : 1)5 = 32 : 1; (2 : 1)6 = 64 : 1; (3 : 1)2 = 9 : 1; (3 : 1)3 = 27 : 1; (3 : 1)4 = 81 : 1; (4 : 1)2 = 16 : 1; (4 : 1)3 = 64 : 1; (5 : 1)2 = 25 : 1; (6 : 1)2 = 36 : 1; (7 : 1)2 = 49 : 1; (8 : 1)2 = 64 : 1; (9 : 1)2 = 81 : 1; (10 : 1)2 = 100 : 1; (4 : 1)3/2 = 8 : 1; (4 : 1)5/2 = 32 : 1; (9 : 1)3/2 = 27 : 1; (8 : 1)4/3 = 16 : 1;

66

Solutions (8 : 1)5/3 = 32 : 1; (16 : 1)5/4 = 32 : 1; (16 : 1)3/2 = 64 : 1; (27 : 1)4/3 = 81 : 1; (32 : 1)6/5 = 64 : 1. There are 25 of these rational ratios.

41.

The distances covered in each of n equal subintervals are in the ratio 1 : 3 : 5 : 7 : · · · : (2n − 1). One can see this by extending the diagram in Figure 10.17. The region over the third subinterval will have area equal to 5 of the triangles AED, the region over the fourth subinterval will have area equal to 7 of these triangles, and so on. One can see this another way by simply noting that the sum of the odd integers 1 + 3 + 5 + · · · + (2n − 1) = n2 and the total distance traveled at the end of the given time is as the square of the time.

42.

Form a rectangle of height 48 and base 1. Over the right half of this rectangle, put two rectangles of height 48 and base 14 . Over the right half of this, put four rectangles of 1 height 48 and base ( 41 )2 = 16 , and so on. The total area of the rectangles is then twice 48, or 96.

43.

The total distance traveled is given by

1 1 1 1 1+2 1 2+4 1 4+8 + · +2 · + · +4 · + · + ··· 2 2 4 8 2 16 32 2 64 1 3 1 3 1 3 = + + + + + + ··· 2 8 4 16 8 32

1 1 1 3 1 1 1 3 7 = + + + ··· + + + + ··· = 1+ = .

2 4 8 4 2 4 8 4 4

D =1·

44.

Collect terms of the harmonic series as follows: 1 + ( 12 ) + ( 13 + 41 ) + ( 15 + 61 + 17 + 81 ) + · · ·. Each collection of terms in parentheses is at least as great as 21 . Thus the series as a whole is greater than the sum of infinitely many terms equal to 12 and thus is itself infinite.

CHAPTER 11 1.

Let t, v, and y represent days in the three cycles of length 13, 20, and 365, respectively.

We know that 1 ≤ t ≤ 13, 1 ≤ v ≤ 20, and 1 ≤ y ≤ 365. In the calendrical dating system m, n, p, q, r, we know that 20 days in the r value make up 1 unit in the q value; then 18 × 20 = 360 days make up 1 unit in the p value. Similarly, 20 × 360 = 7200 days make up 1 unit for the n value and 20 × 7200 = 144,000 days make up 1 unit for the m value. We need to look at the congruence values of these numbers modulo 13, 20, and 365. We find that 20 ≡ 7 (mod 13) ≡ 0 (mod 20) ≡ 20 (mod 365). Also 360 ≡ −4 (mod 13) ≡ 0 (mod 20) ≡ −5 (mod 365). Then 7200 ≡ −2 (mod 13) ≡ 0 (mod 20) ≡ −100 (mod 365) and 144,000 ≡ −1 (mod 13) ≡ 0 (mod 20) ≡ 190 (mod 365). Given t0 , to find the day t which is m, n, p, q, r days later, we add r days, then 20q ≡ 7q days, then 360p ≡ −4p days, then 7200n ≡ −2n days, and finally 144,000m ≡ −m days, where all equivalences are modulo 13. Therefore, t ≡ t0 − m − 2n − 4p + 7q + r (mod 13). Give v0 , to find the day v which is m, n, p, q, r days later, we note that all relevant numbers are congruent to 0 modulo 20. Therefore, we simply add r days to v0 . That is v ≡ v0 + r (mod 20). Finally, to determine y, we add

Solutions

67

r days to y0 , then 20q days, then 360p ≡ −5p days, then 7200n ≡ −100n days, then finally 144,000m ≡ 190m days. So y ≡ y0 + 190m − 100n − 5p + 20q + r (mod 365).

2.

Using the formulas from Exercise 1, we have t = 8 − 2 · 2 − 4 · 3 + 7 · 5 + 10 = 37 ≡ 11 (mod 13). Also, v = 10 + 10 ≡ 20 (mod 20), and y = 193 − 100 · 2 − 5 · 3 + 20 · 5 + 10 ≡ 88 (mod 365). Therefore, the desired Mayan date is (11, 20, 88).

3.

Let ∆t = t1 − t0 (mod 13), ∆v = v1 − v0 (mod 20), and ∆y = y1 − y0 (mod 365). We begin by finding the minimum number of days between (t0 , v0 ) and (t1 , v1 ) in the 260day almanac. Note that every 40 days, the v value remains constant while the t value increases by 1. So first take ∆v days to get the v value correct. This increases t0 to t0 + ∆v. Then take t1 − (t0 + ∆v) = ∆t − ∆v steps of 40 days to get to the correct t value.

That is, we take altogether ∆v + 40(∆t − ∆v) or 40∆t − 39∆v (mod 260) days. Now let ∆a be the difference between these two almanac dates. Note that every 365 days, the almanac date increases by 105, since 365 ≡ 105 (mod 260). Therefore, in two years, the almanac date increases by 210 days, in three years by 55 days, in four years by 160 days, and in five years by 5 days. That is, every 5 years, a increases by 5. So to move from the day (a0 , y0 ) to the day (a1 , y1 ), we first take ∆y days to get the y value correct.

This gives a new a value of a0 + ∆y. We then have to increase the a value by a1 − (a0 + ∆y) = ∆a − ∆y days. We divide this by 5 and take that number of 5-year intervals. That ∆y . In m 5-year intervals (mod 52), we will get to the correct day. But is, let m = ∆a− 5 m 5-year intervals is exactly ∆a − ∆y = ny years. So the total number of days between our two dates is 365ny + ∆y. Putting the two parts of this problem together, we get that the number of days between (t0 , v0 , y0 ) and (t1 , v1 , y1 ) is 365(∆a − ∆y) + ∆y = 365(40∆t − 39∆v − ∆y) + ∆y ≡ 14,600∆t − 14,235∆v − 364∆y (mod 18,980), where 18,980 is the number of days in one calendar round, that is, 52 years.

4.

To calculate the minimum number of days between the two Mayan dates of (8, 20, 13) and (6, 18, 191), we apply the formula worked out in Exercise 3. We have ∆t = −2, ∆v = −2, and ∆y = 178. Therefore, the number of days is 14,600 × (−2) − 14,235 × (−2) − 364 × 178 = −65,522. We want the smallest positive number to which this is equivalent modulo 18,980, so we add 4 × 18,980 to −65,522 to get 10,398 as the minimum number of days between the two given dates. But Pacal lived more than 60 years, or 21,900 days, and less than 100 years, or 36,500 days. So we need to add 18,980 to 10,398 to get 29,378 days as the actual number of days of Pacal’s life. This is equal to 80 years and 178 days.

5.

If the sections are labeled e, m, m2 , f, mf, and m2 f, then the group table is that of the unique non-Abelian group of order 6. That is, it is the group generated by m and f, where m3 = f2 = e and fm = m2 f. Now a woman in section e has a mother in m and a father in f. Therefore, the mother of her father is in section mf, which is the section from which her husband comes. The children will be in that section x such that mx = e, the section of the mother. That means that x = m2 . If a woman is in section m, then her mother is in m2 , her father is in fm = m2 f, and the mother of her father is in mm2 f = f.

Therefore, her husband is in f. The children are in that section x such that mx = m, so x = e. If a woman is in section m2 , then her mother is in m3 = e, her father is in fm2 = mf, and the mother of her father — and therefore her husband — is in m2 f. The children are then in section m. A woman in section f has her mother in section mf, her father in section ff = e, and the mother of her father in section m. Therefore, her

68

Solutions husband is in m. The children are in m2 f, since mm2 f = f, the section of the mother. A woman in section mf has her mother in m2 f, her father in fmf = m2 , and the mother of her father in m3 = e. So her husband is in e. Her children are in f, because mf is the section of the mother. Finally, a woman in section m2 f has her mother in f, her father in m, the mother of her father and therefore her husband in m2 . Her children are then in section mf.

6.

If we put C1 , C2 , C3 , and C4 into matrix form as indicated, we get the matrix 1 2 2 2

1 1 1 2

2 2 1 2

1 1 2 2

Thus, C5 = (1, 1, 2, 1), C6 = (2, 1, 2, 1), C7 = (2, 1, 1, 2), and C8 = (2, 2, 2, 2). Then C9 = C8 ⊕ C7 = (2, 1, 1, 2); C10 = C6 ⊕ C5 = (1, 2, 2, 2); C11 = C4 ⊕ C3 = (2, 1, 1, 2); C12 = C2 ⊕ C1 = (1, 1, 1, 2); C13 = C9 + C10 = (1, 1, 1, 2); C14 = C11 ⊕ C12 = (1, 2, 2, 2); C15 = C13 ⊕ C14 = (2, 1, 1, 2); and C16 = C15 ⊕ C1 = (1, 2, 1, 2).

7.

C13 ⊕ C16 = C14 ⊕ C1 = C11 ⊕ C2 = (2, 1, 2, 2). In general, C13 ⊕ C16 = (C9 ⊕ C10 ) ⊕ (C15 ⊕ C1 ) = C8 ⊕ C7 ⊕ C6 ⊕ C5 ⊕ C3 ⊕ C14 ⊕ C1 = C8 ⊕ C7 ⊕ C6 ⊕ C5 ⊕ C9 ⊕ C10 ⊕ C14 ⊕ C1 = 2C8 ⊕ 2C7 ⊕ 2C6 ⊕ 2C5 ⊕ C14 ⊕ C1 = C14 ⊕ C1 = C11 ⊕ C12 ⊕ C1 = C4 ⊕ C3 ⊕ C2 ⊕ C1 ⊕ C1 = C4 ⊕ C3 ⊕ C2 = C11 ⊕ C2

8.

We need to find the minimum number of days between (25 , 36 , 57 ) and (55 , 26 , 47 ).

Note that the change in the number of days on the five-day calendar is 55 − 25 = 35 , on the six-day calendar is 26 − 36 = 56 , and on the seven-day calendar is 47 − 57 = 67 .

Thus, we need to find N so that N ≡ 3 (mod 5) ≡ 5 (mod 6) ≡ 6 (mod 7). This is a Chinese remainder problem and can be solved most easily by the method of Chapter 7.

In the notation of that chapter, M = 210, M1 = 42, M2 = 35, and M3 = 30. Then P1 = 2, P2 = 5, and P3 = 2. So we need to solve the congruences 2x ≡ 1 (mod 5), 5y ≡ 1 (mod 6), and 2z ≡ 1 (mod 7). The solutions are easily found to be x = 3, y = 5, and z = 4. Then N ≡ 3 · 42 · 3 + 5 · 35 · 5 + 6 · 30 · 4 ≡ 1973 (mod 210). Subtracting 9 · 210 from 1973 gives us the smallest positive value N = 83 days.

CHAPTER 12 1.

We convert 5 lire to 100 soldi, then 112 soldi to 1344 denarii. Thus the gold florin is worth 1350 denarii. We need to know how many florins 13 soldi, 9 denarii, or 165 denarii, are worth. So simply divide 165 by 1350. The result is 11 90 of a florin.

2.

We set up the proportion 8 : 11 = 97 : x. This reduces to 8x = 1067, and x = 133 38 .

3.

The 25 pounds of alloy have 200 ounces of silver, while the 16 pounds have 152 ounces of silver. Thus we have altogether 41 pounds of alloy with 352 ounces of silver. If we

Solutions

69

let x be the number of pounds of copper to be added, there are then 41 + x pounds of alloy with a total of 352 ounces of silver. Therefore, (41 + x) · 7 12 = 352. Thus 41 + x = 46.93 and x = 5.93 pounds.

4.

The courier from Rome to Venice was traveling at the rate of 250 7 miles per day, while the courier from Venice to Rome was traveling at the rate of 250 9 miles per day. If t 250 is the number of days until they meet, then we have the equation 250 7 t + 9 t = 250, 15 or 2250t + 1750t = 15,750, or 4000t = 15,750. It follows that t = 3 16 days. Thus the courier from Rome will have gone 140 85 miles, while the courier from Venice will have gone 109 38 miles.

5

If t is the time when they meet, r1 the speed of the first man, r2 the speed of the second man, and d the distance between Rome and Montpellier, then the conditions give us the d d and r2 = d9 , so the equation becomes 11 t + d9 t = d, equation r1 t + r2 t = d. But r1 = 11 99 or 11t + 9t = 1. The solution is then t = 20 = 4 19 20 days.

6.

Let p be the initial capital. Then the rate of gain is 12 p . So the equation becomes 12 (p + 12) + (p + 12) p = 54. Multiplying by p and simplifying gives p2 + 144 = 30p.

This quadratic equation has two solutions, p = 6 and p = 24, both of which are valid solutions to this problem.

7.

If we let x be the investment of the third partner, then, since each partner should receive a proportionate share of the profits, we get the equation 58 · 368 = 86.

58 + 87 + x This equation reduces to 21,344 = 86(145 + x) or 86x = 8874. The solution is x = 8 103 43 ducats. The second partner will then receive 87 · 368 = 129 ducats.

4 58 + 87 + 103 43 The third partner will receive the difference between 368 and the sum of 129 and 86.

Thus, he receives 368 − 215 = 153 ducats.

8.

The second and third together can complete 6 jobs in 60 days; the first and third together can complete 5 jobs in 60 days; the first and second together can complete 4 jobs in 60 days. By adding these results, we see that all three (doubled) can do 15 jobs in 60 days, or that the first, second, and third working together can do 7 12 jobs in 60 days. Since the second and third can complete 6 jobs in this time, the first worker can do 1 21 jobs in 60 days, or 1 job in 40 days. Similarly, the second worker can do 2 12 jobs in 60 days, or 1 job in 24 days. And finally, the third worker can do 3 12 jobs in 60 days, or 1 job in 17 71 days.

9.

If the two numbers are x + y and x − y, then the two equations become x2 − y = 8 and 2x2 + 2y = 27. If we double the first equation and add it to the second, we get √ 2 2 43 43 11 4x = 43, so x √ = 4 and x = 2 . Then y = 43 4 − 8 = 4 . The two numbers are then √ √ √ 43 43 11 11 2 + 2 and 2 − 2 .

10.

Set the p parts u, v equal p to 5 + x and 5 − x, respectively. The equation is then 97 − (5 + x)2 + 100 − (5 − x)2 = 17. This equation reduces to

√

√

70

Solutions √

√

72 − 10x − x2 + 75 + 10x − x2 = 17. If we move the √ second square root to the right side, then square each side and simplify, we get 34 75 + 10x − x2 = 292 + 20x.

Squaring again gives us 1156(75 + 10x − x2 ) = 85,264 + 11,680x + 400x2 , which in turn reduces to 1556x2 + 120x − 1436 = 0, or 389x2 + 30x − 359 = 0. This equation 359 factors as (389x − 359)(x + 1) = 0, whose solutions are x = 389 and x = −1. Thus the 359 30 parts of 10 are u = 5 389 and v = 4 389 or u = 4 and v = 6.

11.

If we set x as the monthly interest rate in denarii per lire, then the problem leads to the equation

x 4 x 6x2 4x3 x4 100 1 + = 160 = 160 or 100 1 + + + + 20 5 400 8000 16,000 or, finally, to x4 + 80x3 + 2400x2 + 32,000x = 96,000. If we reduce the original equation √ x 4 x to (1 + 20 ) = 1.6, we get 1 + 20 = 4 1.6 = 1.1248 or x = 2.494. On the other hand, if we compare the fourth-degree equation x4 + bx3 + cx2 + dx = e to (x + a)4 = e, or x4 + 4ax3 + 6a2 x2 + 4a3 x + a4 = e, we note that this can be solved by completing the fourth power provided that 4a = b, 6a2 = c, and 4a3 = d. If this is true, then a3 = 4d , q 4 d 2 d and a = 4b , so a2 = db and a = b . If we then add a = ( b ) to both sides of q the original equation, our result is (x + a)4 = ( bd )2 + e. Thus x + a = 4 ( db )2 + e and q q x = 4 ( bd )2 + e − db , as claimed.

12.

2

we get x = 13.

√

10x−x If we let the two numbers be x and 10 − x, the equation becomes 2x 18 or − 10 = √ 2 equation, 10x − x = 18(2x − √10). This √ can be rewritten as a standard quadratic √ √ whose solution is x = 43+5 − 18. Then the second part of 10 is 5+ 18 − 43. On the other hand, if we square both sides of the quadratic equation, we get the equation 100x2 − 20x3 + x4 = 18(4x2 − 40x + 100), which reduces to 720x + 28x2 + x4 = 1800 + 20x3 . If we compare this to the general equation ax + bx2 + cx4 = d + ex3 and use Piero’s formula s r 2 b d e a 4 x= + + − , 4c c 4c 2e

√ 4

72 + 1800 + 5 −

√

18 =

√

43 + 5 −

√

18, the correct solution.

As in the text, the equation becomes 12 n(n + 1) + [ 12 n(n + 1)]2 = 20,400. If we simplify 4 3 2 this, we get n4 + n2 + 3n4 + n2 = 20,400, or, n4 + 2n3 + 3n2 + 2n = 81,600. By trial, we find that if we set n = 16.4, the left side totals 82,001. Again, setting n = 16.38, the left side totals 81,615. So the solution is a bit less than 16.38 days.

14.

Pacioli has combined the terms 79x + 30 into the single term 109x. He then solved the cubic equation 6x3 = 43x2 + 109x, or, dividing by 6, x3 = 7 61 x2 + 18 61 x. Given that we can divide by x, this reduces to the quadratic equation x2 = 7 61 x + 18 16 . Pacioli solves this equation by use of the standard method, but its solution is not a solution of the original cubic.

15.

9 > 6, the next approximation is 2 22 number Since 2 13 6 and 2 20 29 < 49 . Since this last √ √ 31 is also less than 6, the next approximation is 2 69 . Again, this is less than 6, so the

√

√

Solutions

71

√

40 . This is still less than 6. The next approximation is then next approximation is 2 89 √ 49 2 109 . This number is larger than 6, so we combine it with the previous approximation 89 to get 2 198 , the value given by Chuquet.

√

√

16.

We begin with the values 2 14 , which is greater than 5 and 2 15 , which is less than 5.

The next approximation is 2 29 . We continue in this manner, always checking whether √ our value is smaller than or greater than 5, so we know with which previous approx3 4 5 9 17 imation to combine it. The approximations are, in order, 2 13 , 2 17 , 2 21 , 2 38 , 2 13 55 , 2 72 , 21 38 55 72 89 161 2 89 , 2 161 , 2 233 , 2 305 , 2 377 , and finally 2 682 .

17.

20x + 7 3 If x represents the number, the equation is 30x − 9 = 10 . This reduces to 200x + 70 = 90x − 27, or to 110x = −97. Since the solution to this problem is negative, Chuquet said that the problem is impossible.

18.

In 12 hours the largest tap will empty the vessel 4 times, the middle one will empty it 3 times, and the smallest one will empty it 2 times. Thus in 12 hours, all three taps 1 would empty the vessel 9 times. Therefore, all three would empty it 1 time in 12 9 = 13 hours. p √ √ √ √ If we square the equation 27 + 200 = a + b, we get 27 + 10 2 = a2 + b + 2a b.

By inspection, we find that a = 5, b = 2 satisfies this equation. Thus the solution is √ 5 + 2.

19.

20.

The sum of the integers from 1 to n may be expressed as n(n2+ 1) . Therefore, we must find n which satisfies the equation n(n2+ 1) = 3240. This equation can be reduced to n2 + n = 6480, or n2 + n − 6480 = 0. The left side factors as (n − 80)(n + 81). It follows that the (positive) solution to the equation is n = 80, and it takes 80 days to pay off the debt.

21.

The equation √ is x(10 − x) = 13 + 128. This reduces to 10x − x2 = 13 + 8 2 2 or xq+ 13 + 8 2 = 10x. The quadratic formula in this case gives us x = p p √ √ √ 5 ± 25 − (13 + 8 2) = 5 ± 12 − 8 2 = 5 ± 2 3 − 2 2. These two numbers are the two desired parts of 10.

22.

At the first stage, we add just the first odd number. We then skip 1 odd number (where 1 is the first triangular number). At the second stage, we add 4 = 22 odd numbers and then skip 3 odd numbers (where 3 is the second triangular number). At the third stage we add 9 = 32 odd numbers and then skip 6 odd numbers (where 6 is the third triangular number). Therefore, we are always adding up numbers of the form 2n − 1.

We must determine at each stage, exactly which numbers we are adding. At the second stage we add odd numbers 1 + 12 + 1(12+ 1) = 3 to 1 + 12 + 1(12+ 1) + 22 − 1 = 6. At the third stage, we add together odd numbers 1 + 12 + 1(12+ 1) + 22 + 2(22+ 1) = 10 to 1 + 12 + 1(12+ 1) + 22 + 2(22+ 1) + 32 − 1 = 18. In general, at stage k ≥ 2, we begin with the mth odd number, where

√

√

m =1+

kX −1

2

i +

k X

i=1

=1+

i=2

! i 2

(k − 1)k(2k − 1)

6

= 1+ +

(k − 1)k(2k − 1)

6

(k + 1)k(k − 1)

6

=

! k+1 + 3

k3 − k2 + 2 , 2

72

Solutions 3

2

and conclude with the number m + k2 − 1 = k +2 k . Therefore, the sum at stage k is given by k3 + k2 2

Sk =

X

(2n − 1),

3 2 n= k −2k + 2

and we must prove that Sk = k5 . We calculate: k3 + k2 2

Sk = 2

X

n= k −2k + 2 3

2

n − k2 = 2

k2 2

3

k − k2 + 2 k3 + k2 − k2 = k2 (k3 + 1) − k2 = k5 + 2 2

as claimed.

23.

The fourth root of 10,556,001 must be a two-digit number beginning with 5, because 504 = 6,250,000 and 604 = 12,960,000. So, subtract 6,250,000 from the original number to get remainder 4,306,001. We then guess that the next digit is 7. To check, we first subtract 4 × 503 × 7 = 3,500,000 from this remainder to get 806,001. Next, we subtract 6 × 502 × 72 = 735, 000 to get 71,001. Next, we subtract 4 × 50 × 73 = 68,600 to get 2401. Finally, we note that since 74 = 2401, when we subtract this value we get 0. Therefore, 57 is the desired fourth root.

24.

Let d be the number of dukes, e be the number of earls, and s be the number of soldiers.

Since each duke has under him twice as many earls as there are dukes, each duke has under him 2d earls. Since there are d dukes, there are altogether e = 2d2 earls. Since each earl has under him four times as many soldiers as there are dukes, each earl has under him 4d soldiers. Since there are 2d2 earls, there are altogether s = 8d3 soldiers. s = 9d, so that s = 1800d. Combining these last two equations But we also know that 200 3 gives us 1800d = 8d or d2 = 225 or d = 15. It then follows that e = 2d2 = 450 and s = 1800d = 27, 000.

25.

We have x + y = 8 and x + y + x2 + y2 + x3 + y3 = 194. If we set y = 8 − x and substitute in the second equation, we get x2 + (8 − x)2 + x3 + (8 − x)3 = 186, or, expanding, 26x2 − 208x + 576 = 186. This equation reduces to x2 − 8x + 15 = 0, whose solutions are x = 5 and x = 3. Thus the gentleman has 5 crowns in one hand and 3 in the other.

26.

We need to sum the arithmetic progression 1 12 + (1 12 + 61 ) + 2 1 n−1 1 (1 2 + 6 ) + · · · + (1 2 + 6 ) and set this equal to 2955. The sum of the arith2 1 n2 − n metic progression is n · 1 12 + 2n n − = 3n = n +1217n . We therefore get the 6 2 + 12 quadratic equation n2 + 17n = 35, 460. We can solve this by the quadratic formula to get n = 180 days.

27.

We are given that r3 + d = cr and that s3 + d = cs. Also, since the sum of the three roots must be 0 (the coefficient of the x2 term) and their product is −d, it follows that the third root is −(r + s) and that rs(r + s) = d. Thus, t3 = (r + s)3 = r3 + 3r2 s + 3rs2 + s3 = cr − d + 3rs(r + s) + cs − d = c(r + s) + 3d − 2d = c(r + s) + d = ct + d, and t is a root of x3 = cx + d.

Solutions 28.

We are given that t3 = ct + d. Then if r = 2t + 3

t 3

t 2

r

t 2

p

73

c − 3( 2t )2 , we have

t

t 2

+3 c−3 +3 c−3 2 2 2 2 2 r

t 2 t 2 + c−3 c−3 +d 2# " 2 r " # r t 2 t 2 3t t 3 3 = −t + c + c−3 + c−3 + t − ct = c = cr.

2 2 2 2

r +d =

A similar argument holds for s. To solve x3 + 3 = 8x, note that 3 is a rootq of x3 = 8x + 3.

Applying the formula, we have that two roots of x3 + 3 = 8x are 23 ± 8 − 3( 32 )2 = q √ 3± 5 3 ± 8 − 27 . Since the sum of all three roots is 0, and the sum of these two 2 4 = 2 is 3, we know that the third root is −3.

29.

The function y = x3 + cx − d crosses the y-axis at y = −d. Since y′ = 3x2 + c, we know that y′ is never 0 and that the graph of y is always increasing. Thus, it only crosses the x-axis once, and that must be when x is positive.

30.

We apply Cardano’s formula to x3p + 3x = 10. Here c = 3 and d = 10. We get x = p p p √ √ 3 √ 3 √ 3 3 25 + 1 + 5 − 25 + 1 − 5 = 26 + 5 − 26 − 5.

31.

We apply Cardano’s formula to x3 = 6x + 6. Here c = 6 and d = 6. We get x = p p √ √ √ √ 3 3 3 + 9 − 8 + 3 − 9 − 8 = 3 4 + 3 2.

32.

′ 2 Consider the graph of y = x3 − cx − d. We calculate p c that y = 3x − c. It follows that the graph has a local maximum when x = − 3 and a local minimum when p x = 3c . Since the graph crosses the y-axis when y = −d, there must be a real positive solution of x3 − cx − d = 0. There will be two negative solutions p provided that the y-coordinate of the local maximum is positive. But when x = − 3c , we get that y = p p pc pc 2c − 3c 3c + c 3c − d = 2c 3 3 − d. This value is positive provided that 3 3 > d, or, dividing by 2 and then squaring both sides, if ( 3c )3 > ( d2 )2 .

33.

To solve x3 + 21x = 9x2 + 5, we substitute x = y + 3. We get y3 + 9y2 + 27y + 27 + 21y + 63 = 9y2 + 54y + 81 + 5. This reduces to y3 + 4 = 6y. The obvious solution to this equation is y = 2. To find the other solutions, we can √ divide y3 − 6y + 4 by y − 2.

2 We get y + 2y − 2. The roots of that polynomial are −1 ± 3. To find the solutions to the original equation, √ we simply add √ 3 to each of these three solutions. The answers are x = 5, x = 2 + 3, and x = 2 − 3.

34.

To solve x4 + 4x + 8 = 10x2 , we rewrite this as x4 = 10x2 − 4x − 8 and add −2bx2 + b2 to both sides, where b is to be determined. The result is x4 − 2bx2 + b2 = (10 − 2b)x2 − 4x − 8 + b2 . The left side is now the square of x2 − b. For the right side to be a square, we need (−4)2 = 4(10 − 2b)(b2 − 8), or 16 = −8b3 + 40b2 + 64b − 320, or 8b3 − 40b2 − 64b + 336 = 0. We can simplify this equation to b3 − 5b2 − 8b + 42 = 0. One solution to this cubic is b = 3. To determine the others, we can divide√the polynomial by x − 3 to get x2 − 2x − 14. The roots of this polynomial are 1 ± 15.

If b = 3, the fourth-degree polynomial equation becomes x4 − 6x2 + 9 = 4x2 − 4x + 1.

We simplify this to (x2 − 3)2 = (2x − 1)2 , or x2 − 3 = ± (2x − 1). Using the plus

74

Solutions √

sign gives us x2 − 2x − 2 = 0, whose roots are x = √ 1 ± 3. Using the minus sign gives us x2 + 2x − 4 = 0, whose roots are x = −1 ± 5. It is not difficult to check that these √ four quantities are all roots of the original equation. Suppose we choose b = 1 + 15. If we into our√fourth-degree polynomial equation, √ substitute this √ value √ we get x4 − (2 + 2√ 15)x2 + (1 + √ 15)2 = (8 − 2 √ 15)x2 − 4x + 8 + 2 15. This simpli2 fies to (x2 √ − (1 + 15√ ))2 = (√ 8 − 2 15)(x − √ (4 + 15)) . Taking square roots, we get x2 − (1 + 15) = ± ( 5 − 3)( x − ( 4 + 15 )) .

Using the plus √ √ √ sign, we can rewrite √ √ this equation in the form x2 − ( 5 − 3)x − 1 + 3 + 5 − 15 = 0. The solution of this equation is

q √ √ √ √ √ 1 √ ( 5 − 3) ± 8 − 2 15 + 4 − 4 3 − 4 5 + 4 15 2

q √ √ √ √ √ √ √ i 1 h√ 1 √ = 5 − 3 ± 12 − 4 3 − 4 5 + 2 15 = 5 − 3 ± (2 − 5 − 3) 2 √ 2 √ =1− 3 or − 1 + 5.

x =

Note that these roots are two of the roots found earlier.√If we use the minus sign, we get the other two roots. Similarly, if we choose b = 1 − 15, we get the same four roots as before. Thus there are precisely four solutions to our original quartic equation.

35.

Let x be the dowry. Then Francis’s property is x − 100. The equation is then x2 = (x − 100)2 + 400, or x2 = x2 − 200x + 10, 400, or 200x = 10, 400. The solution is x = 52. Thus the dowry is 52 and Francis’s property is −48.

36.

Subtract the first equation, x + y = y3 + 3yx2 , from the second, x + y + 64 = x3 + 3xy2 .

The result is 64 = x3 − 3x2 y + 3xy2 − y3 = (x − y)3 . Thus, x − y = 4 or x = y + 4. If we substitute that value for x into the first equation, we get 2y + 4 = y3 + 3y(y + 4)2 , which, when expanded and rearranged, gives us y3 + 6y2 + 11 21 y − 1 = 0. To solve this cubic, we set y = w − 2. We get (w − 2)3 + 6(w − 2)2 + 11 12 (w − 2) − 1 = 0, which simplifies to w3 = 12 w + 8. Using the appropriate version of Cardano’s formula, we solve for w: s w=

3

r 4+

s r 3 1 1 16 − + 4 − 16 − .

216 216

Since y = w − 2, and x = y + 4 = w + 2, we get s x=

3

s s r s r r 3 215 3 215 215 3 215 15 + 4 − 15 + 2, y = 4 + 15 + 4 − 15 − 2, 216 216 216 216

r 4+

the same solution as Tartaglia.

37.

Since x + y = 8, we have y = 8 − x. Substituting this into the expression xy(x − y) gives us x(8 − x)(2x − 8) = −2x3 + 24x2 − 64x. We can take the derivative of this polynomial to get −6x2 + 48x − 64. To determine the √ maximum, we set this equal to 0: 3x2 − 24x + 32 = 0. The solutions are x = 4 ± 34 3. The maximum occurs when √ √ x = 4 + 43 3 and y = 4 − 34 3.

Solutions 38.

75

The Cardano formula for x3 + 3x = 36 gives us q q p p 3 √ 3 √ 3 √ 3 √ 2 3 2 3 x= 18 + 1 + 18 − 18 + 1 − 18 = 325 + 18 − 325 − 18.

39.

p p √ 3 √ 3 √ 325 + 18 = b + a. Then 325 − 18 = To √ show that this value is in fact 3, set b − a. If we cube both sides of the first equation and set equal the parts not having roots, we get 18 = 3ba + a3 . A solution to this system is a = 23 , b = 13 4 . Thus the q q 13 13 3 3 solution to our original equation is x = ( 4 + 2 ) − ( 4 − 2 ) = 3, as desired. p p √ √ √ √ 3 3 Set 52 + −2209 = a + b −1. Then √52 − −2209 = a − b −1. Multiply3 2 2 2 ing these two equations together gives us 4913 = a + b , or a + b2 = 17. If we cube both sides of the first equation and compare the terms without radicals, we get a3 − 3ab2 = 52. These two equations can be solved by inspection: a = 4, b = 1. Thus p √ √ 3 52 + −2209 = 4 + −1.

40.

Let α be the base angle of the first right triangle and β the base angle of the second.

Then sin α = BZ , cos α = DZ , sin β = XF , and cos β = GX . The sum formula gives us DF BG + DF sin(α + β) = sin α cos β + cos α sin β = BG ZX + ZX = ZX . Thus the base angle of the −BF follows triangle constructed in the text is α + β. That the cosine of this angle is DGZX similarly.

41.

If AE = B and A : E = S : R, then R : S = E : A = (B : A) : A = B : A2 and S : R = A : E = (B : E) : E = B : E2 . It follows that A2 = BS : R and E2 = BR : S. In the case where B = 20, R = 1, S = 5, we get A2 = 20 · 5 : 1 = 100, so A = 10; and E2 = 20 · 1 : 5 = 4, so E = 2.

42.

Let the two numbers be s and t. Then s − t = B, s3 − t3 = D, and s + t = E, where E is unknown. We calculate that 4(s3 − t3 ) − (s − t)3 (s − t)[4(s2 + st + t2 ) − (s2 − 2st + t2 )] 4D − B3 = = 3B 3(s − t) 3(s − t) 3(s2 + 2st + t2 ) 2 2 = = (s + t) = E .

3 Since E2 is known, so is E. And since the sum E and difference B of the two numbers is known, the two numbers can be easily calculated. In the case where B = 6 and − 216 D = 504, we have E2 = 201618 = 100. Therefore, E = 10. Since the sum of the two numbers is 10 and the difference is 6, the two numbers are s = 8, t = 2.

43.

We write 13.395 as 13 0 3 1 9 2 5 3 and 22.8642 as 22 0 8 1 6 2 4 3 2 4 .

If we multiply, we get the result in digits as 3062659590. Since the sum of the right most digits of the original numbers is 7, that is the sign of the right most digit of the answer. So the answer is 306 0 2 1 6 2 5 3 9 4 5 5 9 6 0 7 . Similarly, if we divide the second by the first, we get 1 0 7 1 .

44.

To subtract 59 0 7 1 3 2 9 3 , from 237 0 5 1 7 2 8 3 , we line up the numbers with equal signs under each other and get 177 0 8 1 3 2 9 3 .

76

Solutions

CHAPTER 13 2.

Let us assume, as in Figure 13.4, that d = 3.5b. If one draws a pole of height p at E, then one of height x1 at W1 (the intersection of line EV with the first line above the ground line), then one of height x2 at W2 (the intersection with the second line), and so on, the line connecting the tops of the poles also intersects V. Now draw the perpendicular from V to the ground line, letting Y be its intersection with the ground line, T1 its intersection with the first line above the ground line, T2 its intersection with the second line above the ground line, and so on. Then by similarity, x1 : p = VW1 : VE = VT1 : VY. Because the first line above the ground line has the equation y = hb/(d + b) = h/4.5, we have x1 : p = (3.5h/4.5) : h = 7 : 9. Therefore, x1 = 79 p.

Similarly, x2 : p = VW2 : VE = VT2 : VY. We calculate that the second line above the ground line has the equation y = 2bh/(d + 2b) = 2h/5.5. Therefore x2 : p = 7 7 7 p. Similarly, x3 = 13 p, x4 = 15 p, and (3.5h/5.5) : h = 7 : 11. Therefore x2 = 11 so on.

4.

The construction shows that the difference between the distances of the intersection of the two chords to the two vertices is a constant, namely, the difference between BC and AC. The modern definition of a hyperbola is, in fact, the locus of points such that the difference in their distances to two fixed points is a constant. Thus, Kepler’s construction does give a hyperbola.

5.

To find the distance on the map to the 10◦ parallel, we need to calculate R ◦ 180 10 π sec ϕ dϕ. Since 10◦ corresponds π 0◦ R to 18 radians, we need to integrate the seπ cant between 0 and 18 . Recalling that sec ϕ dϕ = ln(sec ϕ + tan ϕ), we calculate the integral to be 0.17543 and the distance from the equator to the 10◦ parallel to be ◦ ◦ 180 π · 0.17543 = 10.05. We further calculate that the integral of the secant from 0 to 5 ◦ is 0.0873774, so the distance from the equator to the 5 parallel is 5.01; the integral from 10◦ to 15◦ is 0.08941, so the distance from the 10◦ parallel to the 15◦ parallel is 5.12; the integral from 15◦ to 20◦ is 0.091537, so the distance from the 15◦ parallel to the 20◦ parallel is 5.24; the integral from 20◦ to 25◦ is 0.094496, so the distance from the 20◦ parallel to the 25◦ parallel is 5.41; and the integral from 25◦ to 30◦ is 0.098431, so the distance from the 25◦ parallel to the 30◦ parallel is 5.64.

6.

Since 1 cm on the parallel of 40◦ corresponds to 1◦ of longitude on that parallel, the same scale would have sec 40◦ = 1.30 cm correspond to 1◦ on the equator. To find the radius of the corresponding sphere, we multiply this value by 180 π to get 74.794. We need to multiply values calculated by Equation 13.1 by that value. First we need to R 45◦ calculate 40◦ sec ϕ dϕ. This value is 0.11846, so the corresponding value on the map will be 8.86 cm as the distance from the 40◦ parallel to the 45◦ parallel. The integral from 45◦ to 50◦ is 0.12931, so the distance from the 45◦ parallel to the 50◦ parallel is 9.67 cm; the integral from 50◦ to 55◦ is 0.14355, so the distance from the 50◦ parallel to the 55◦ parallel is 10.74 cm; and the integral from 55◦ to 60◦ is 0.16272, so the distance from the 55◦ parallel to the 60◦ parallel is 12.17 cm.

7.

Given the calculations in the text, we need to solve the equation

(10 − x)2 + 25 9 = .

25 (10 + x)2 + 25

Solutions

77

This equation reduces to 9x2 + 180x + 1125 = 25x2 − 500x + 3125, or to 2x2 − 85x√+ 250 = 0. The solution to this quadratic which makes sense in the problem √ is x = 85− 4 5225 = 3.179. Thus AB = 52 + 6.8212 = 8.46 and AG = 5AB 3 = 14.10.

8.

3B Since A : B = 10 : 7, B : C = 7 : 3, and A + B + C = 180, we have A = 10B 7 ,C = 7 , ◦ ◦ ◦ and 20B 7 = 180. The solution is B = 63 , so C = 27 and A = 90 . So if side b = 1, we have side a = sec 27◦ = 1.12 and side c = tan 27◦ = 0.51.

9.

Let us designate side AC by b, side AB by c, side BC by a, and the two segments BD, DC into which AD divides BC by p and q, respectively. Then we have the following four relationships: c2 = 302 + p2 , b2 = 302 + q2 , c − b = 3, and p − q = 12. Substituting c = b + 3 and p = q + 12 into the first equation gives us (b + 3)2 = 900 + (q + 12)2 , or, combining this with the second equation, 6b − 24q = 135, or b = 4q + 45 2 . If we 2 ) = 900 + q2 , or substitute this value for b into the second equation, we get (4q + 45 2 2 2 2 1 1 16q + 180q + 506 4 = 900 + q . This reduces to √q + 12q − 26 4 √ = 0. We solve this −12+ 249 −12+ 144 + 105 equation by the quadratic formula to get q = = = −12 +2 15.78 = 2 2 1.89. Then p = q + 12 = 13.89, b = 4q + 22.5 = 30.06, c = b + 3 = 33.06, and a = p + q = 15.78.

10.

α We are given that α + β = γ and sin sin β = k. We want to find α and β. From the first equation, we get sin α = sin(γ − β) = sin γ cos β − cos γ sin β. Substituting this in the second equation gives

sin γ cos β − cos γ sin β =k sin β

or

sin γ cot β − cos γ = k.

Therefore, cot β =

k + cos γ sin γ

or

tan β =

sin γ . k + cos γ

We can thus find β and therefore α. In the example, we are given that γ = 40◦ and k = 74 . Therefore, tan β =

sin 40◦ = 0.2555, 1.75 + cos 40◦

and β = 14◦ 20′ . It follows that α = 25◦ 40′ .

11.

Since this is a right triangle, we can use the formula cos B = sin A cos b to find b.

We get cos 70◦ = sin 50◦ cos b, so cos b = 0.4465 and b = 63.48 = 63◦ 29′ . By the law of sines, sin a = sin A sin b/ sin B = sin 50◦ sin 63.48◦ / sin 70◦ = 0.7294, and a = 46.84 = 46◦ 50′ . Again by the law of sines, sin c = sin a/ sin A = sin 46.84◦ / sin 50◦ = 0.9522 and c = 72.22 = 72◦ 13′ .

12.

Recall that Versin A = R − R cos A, and note that in the versine formula, the Sines are in circles of radius R. It follows that we can rewrite the versine formula as R2 R − R cos A = (R − R cos a) − (R − R cos(b − c)) R sin bR sin c

78

Solutions or 1 − cos A 1 = . − cos a + cos b cos c + sin b sin c sin b sin c This equation is equivalent to sin b sin c − sin b sin c cos A = − cos a + cos b cos c + sin b sin c or to cos a = cos b cos c + sin b sin c cos A. 13.

Let us first consider △ABC with altitude BF. We have AB2 − AF2 = BC2 − FC2 . If we set AF = p, we get 72 − p2 = 92 − (13 − p)2q , or 49 − p2 = 81 − 169 + 26p − p2 , so

26p = 137 and p = 137 26 . Then the altitude d =

2 49 − ( 137 26 ) = 4.608 and the area of

△ABC = 12 · 4.608 · 13 = 29.95. Similarly, we find that the area of △CDE = 19.96 and the area of △ACE = 71.50. Thus the total area of ABCDE is 121.41.

14.

Since triangles ADE and ABC are similar, we have AB : BC = AD : DE. Since AB = 10, BC = 6, and AD = 5, we have 10 : 6 = 5 : DE, or DE = 3. Thus sin A2 = 3/5 2 = 0.3. Therefore, A2 ≈ 17.5◦ and A ≈ 35◦ . Thus each base angle is 72.5◦ .

15.

We have 12 [cos(α − β) − cos(α + β)] = 12 [cos α cos β + sin α sin β − cos α cos β + sin α sin β] = 12 [2 sin α sin β] = sin α sin β.

16.

Let us suppose that the radius of the circle in which the sines are calculated is 10,000,000. Then we need to find α such that Sin α = 2,189,109 in a circle of radius 10,000,000. This is equivalent, in modern terms, to determining α such that sin α = 0.2189109. By the calculator, α = 12◦ 38′ 42′′ . Then α − β = −14◦ 36′ 40′′ and α + β = 39◦ 54′ 4′′ . So 4,378,218 × Sin(27◦ 15′ 22′′ ) = Cos(α − β) − Cos(α + β) = Cos(14◦ 36′ 40′′ ) − Cos(39◦ 54′ 4′′ ) = 9,676,603 − 7,671,527 = 2,005,076.

If we do the actual multiplication, we get 2,005,087.

17.

By Kepler’s third law, the ratio of the period of Mars to the period of the earth equals the 3/2 power of the ratio of their mean distances. Since the latter ratio is 1.524, we calculate (1.524)3/2 = 1.88. So the period of Mars is 1.88 years = 687 days.

18.

Since equal areas are swept out in equal times, the planet moves faster when it is closest to the sun, that is, at its perigee.

19.

According to the text, Nlog x = r ln xr , where r is Napier’s radius of 10,000,000. Therefore Nlog xy = r ln xyr = r(ln r − ln x − ln y + ln r) − r ln r = r ln xr + r ln yr − r ln r = Nlog x + Nlog y − Nlog 1. Similarly, Nlog yx = r ln x/r y = r(ln r − ln x + ln y − ln r) + r ln r = r ln xr − r ln yr + r ln r = Nlog x − Nlog y + Nlog 1.

20.

Let a = 13, b = 10, γ = 35◦ . Then the law of tangents gives tan 12 (145◦ ) 13 + 10 = .

13 − 10 tan 12 (α − β)

Solutions

79

◦

This reduces to tan 12 (α − β) = 3 tan(72 21 )/23 = 0.4137. Therefore, 12 (α − β) = ◦ 22 21 , so α = 95◦ and β = 50◦ . To find c, we use the law of sines: c = 10 sin 35◦ / sin 50◦ = 7.5.

21.

Let us assume we have a moveable falling along a vertical line and also along an inclined plane which makes an angle α with the horizontal. Let a be the acceleration due to gravity. Therefore, along qthe vertical, the velocity v = at and the distance fallen 2d d = 21 at2 . Thus t2 = 2d a , or t = a . Along the inclined plane, we know that the actual distance traveled s, when the body is at vertical distance d from the initial point, is given by s = d csc α. Also, by hypothesis, the velocity vp at s alongq the plane is the

same as the velocity at d along the vertical. Thus vp (s) = v(d) = a

√

2d a

=

√

2da =

√

2as sin α. But since, in general, vp (t) = kt and s = 12 kt2 , we also have vp (s) = 2ks. It follows that k = a sin α and the velocity vp along the inclined plane is given by vp (t) = (a sin α)t. Then the distance s traveled along the plane is given by s = 21 (a sin α)t2 . To determine the time it takes to move a distance s, when the falling body has reached 2 the point d = s sin α, we just solve d csc α = 12 (a sin α)t2 for t. We get t2 = 2d a csc α, or q t = csc α 2d a . Therefore the ratio of the time to move a distance s along the inclined plane to the time to fall the corresponding distance d is csc α, which is exactly the ratio of the length of the plane to the vertical distance. Since this is true for any two inclined planes, the times of descent along two such planes of the same height are to one another as the lengths of the planes.

22.

By Exercise 21, q the time of descent along the first plane, with height d1 and angle α1 , is t1 = csc α1

2d1 a and the time of descent along the second plane, with height d2 and

q angle α2 , is t2 = csc α2 2da2 . But the lengths of the planes are the same. That means that d1 csc α1 = d2 csc α2√or that csc α√ d1 . The√ratio of 1 : csc α2√= d2 : √ √ the times of descent is t1 : t2 = csc α1 d1 : csc α2 d2 = d2 d1 : d1 d2 = d2 : d1 . Thus this ratio is inversely as the square root of the ratio of the heights of the planes.

23.

Suppose the initial velocity of the projectile is v0 . The horizontal distance x(t) is given by x = (v0 cos α)t, while the vertical distance y(t) is given by y = −at2 + (v0 sin α)t.

If we solve the first equation for the parameter t and substitute in the second, we 2 x get y = −a( v0 cos α ) + (tan α)x. Since y is a quadratic function of x, the graph of the function is a parabola, as claimed.

24.

From Exercise 23, we have the equation expressing y as a function of x. The x coordinate of the point where the projectile reaches the ground, or where y = 0, is found by v2 cos2 α tan α ax solving the equation v2 − + tan α = 0. We get x = 0 a . If α = 45◦ , and the cos2 α 0

v2

v2

distance traveled is 20,000, then 20,000 = 2a0 or a0 = 40,000. Putting in this result into the distance equation for α = 30◦ , we get x = 40,000 cos2 (30◦ ) tan(30◦ ) = 17,321.

We get the same result when α = 60◦ . We note that Galileo’s result is slightly in error.

25.

From Exercise 23, we have the equation expressing y as a function of x. This quadratic 1 2 v0 tan α cos2 α. The y function achieves its maximum when x = − tan α/ v2 −cos2a2 α = 2a 0

1 2 value at this point gives the maximum value, and that is y = 4a v0 tan2 α cos2 α = 2 ◦ 1 2 1 1 2 4a v0 sin α. When α = 45 , this expression is equal to 5000. So 4a · 2 v0 = 5000

80

Solutions 1 2 v0 = 10,000. To find the maximum height when α = 30◦ , we calculate y = and 4a 10,000 sin2 (30◦ ) = 10,000 · 14 = 2,500. To find the maximum height when α = 60◦ , we calculate y = 10,000 sin2 (60◦ ) = 10,000 · 43 = 7,500. We note that Galileo’s result is slightly in error.

26.

Given that d = at2 , we have that the distance fallen in the first time interval is d = a · 12 = a. The distance fallen in the second interval is d = a · 22 − a · 12 = 3a. The distance fallen in the third interval is d = a · 32 − a · 22 = 5a. In general, the distance fallen in the nth interval is d = a · n2 − a · (n − 1)2 = (2n − 1)a. Thus the ratio of the distances traveled in equal intervals is 1 : 3 : 5 : · · · : 2n − 1.

CHAPTER 14 1.

We have (b + a)(c − a)(df − aa) = bcdf − bdfa − dfaa + baaa + cdfa −bcaa − caaa + 2 aaaa. In modern terms, we can rewrite this polynomial as a4 + (b − c)a3 −(df + bc p)a + (cdf − bdf)a + bcdf. The roots of the polynomial are a = −b, a = c, and a = ± df.

2.

Harriot would write this as

ad zz gad + bzz + = . b g bg

3.

If we substitute a = e + r into the equation aaa − 3raa = 2xxx, we get, in modern notation, (e + r)3 − 3r(e + r)2 = 2x3 . If we expand this, we get e3 + 3e2 r + 3er2 + r3 − 3e2 r − 6er2 − 3r3 = 2x3 , or e3 − 3r2 e = 2x3 + 2r3 , an equation with no square term. In Harriot’s notation, this is eee − 3rre = 2xxx + 2rrr. Applying this procedure to aaa − 6aa = 400 gives us eee − 12e = 416. A root of the latter polynomial is e = 8. So a root of the former equation is a = 10.

4.

Substituting a = e + r into the given equation gives (e + r)3 − 3r(e + r) + p2 (e + r) = 2x3 , or e3 + 3e2 r + 3er2 + r3 − 3e2 r − 6er2 − 3r3 + p2 e + p2 r = 2x3 , which reduces to e3 + (p2 − 3r2 )e = 2x3 + 2r3 − p2 r. The equation aaa − 18aa + 87a = 110 thus reduces (with r = 6) to, in modern notation, e3 − 21e = 20. The roots of this latter equation are e = −1, e = 5, and e = −4. Therefore, the roots of the original equation are a = 5, a = 11, and a = 2.

5.

We rewrite the equation x3 = 300x + 432 in the form x3 − 300x = 432. Thus, the first faction is 0; the second faction is −300; and the third faction is 432. Since one solution to the equation is x = 18, we know that the sum of the other two solutions is −18 and their product is 24. These two numbers satisfy the quadratic equation x2 + 18x + 24 = √ 0, whose solutions are x = −9 ± 57.

6.

We see that x = 1 is one solution to x3 = 6x2 − 9x + 4. If we rewrite this in the form Girard uses, we have x3 + 9x = 6x2 + 4. Thus, the first faction is 6, the second is 9, and the third is 4. Since 1 is one solution, we know that the two other solutions have the property that their sum is 5 and their product is 4. Thus, these numbers are x = 1, and x = 4.

Solutions

81

7.

Let the four roots of this polynomial equation be r, s, t, and u. We know that A = r + s + t + u, B = rs + rt + ru + st + su + tu, C = rst + rsu + rtu + stu, and D = rstu. So A is the sum of the roots. Also, r2 + s2 + t2 + u2 = (r + s + t + u)2 − 2(rs + rt + ru + st + su + tu) = A2 − 2B. We also have r3 + s3 + t3 + u3 = (r + s + t + u)3 − 3(r2 s + rs2 + r2 t + rt2 + r2 u + ru2 + s2 t + st2 + s2 u + su2 + t2 u + tu2 + 2rst + 2rsu + 2rtu + 2stu) = A3 − 3(r + s + t + u)(rs + rt + ru + st + su + tu) + 3(rst + rsu + rtu + stu) = A3 − 3AB + 3C. Finally, r4 + s4 + t4 + u4 = (r2 + s2 + t2 + u2 )2 − 2(r2 s2 + r2 t2 + r2 u2 + s2 t2 + s2 u2 + t2 u2 ) = (A2 − 2B)2 − 2(B2 − 2AC + 2D) = A4 − 4A2 B + 4AC + 2B2 − 4D, as desired.

8.

From the diagram, with x = FN, we have the three relations x2 + 42 = AN2 , √ √ 2 2 4 153 (4 − x) + OC = 153, and 4 : OC = AN : 153, or OC = AN . If we substitute this last equation into the second, and then use the first to evaluate AN2 , we get the equation

(4 − x)2 +

16 × 153 = 153. x2 + 16

This reduces to (x2 + 16)(16 − 8x + x2 ) = 153x2 , or x4 − 8x3 − 121x2 − 128x + 256, or, x4 = 8x3 + 121x2 + 128x − 256. One finds that x = 1 and x = 16 are solutions. Since the first faction is 8 and the last is 256, the two other solutions have their sum equal to −9 and their product equal to 16. Thus these two √ solutions satisfy the quadratic equation x2 + 16 = −9x. The solutions are −4 21 ± 12 17. (Note that the three relations that determine FD, FG, and FH, respectively, are somewhat different from the three relations that√determine FN. In particular, the relations for FG and FH require that LG = KH = 153. In each case, however, when the relations are turned into a single equation, we get the same fourth-degree equation as before.)

9.

We rewrite the equation xy + c = rx + sy in the form xy − rx − sy + c = 0, or xy − rx − sy + rs − rs + c = 0, or, finally, (x − s)(y − r) = rs − c. If we let x′ = x − s, y′ = y − r, we can rewrite this equation in the form x′ y′ = rs − c, which is a hyperbola with asymptotes the x′ and y′ axes. In terms of the original equation, the axes of the hyperbola are the lines y = r and x = s.

10.

We add x2 to both sides of the equation b2 − 2x2 = 2xy + y2 to get b2 − x2 = x2 + 2xy + y2 , or b2 − x2 = (x + y)2 , or (x + y)2 + x2 = b2 . If we change coordinates from the original orthogonal x and y coordinates to the new oblique coordinates x′ = x, y′ = x + y, we can rewrite this equation in the form x′2 + y′2 = b2 , which is the equation of an ellipse. (It is not, in general, a circle because the axes are not perpendicular to one another.)

11.

We can rewrite the equation b2 + x2 = ay in the form ay − b2 = x2 , or a(y − ba ) = x2 . If 2 we change coordinates by setting x′ = x, y′ = y − ba , we can rewrite this equation in the form ay′ = x′2 , which is the standard equation of a parabola. Another way of looking 2 at this problem is to rewrite the original equation in the form y = 1a x2 + ba . Thus the graph of this equation comes from the graph of the parabola y = 1a x2 by translation 2 upward by the quantity ba .

2

82

Solutions 12.

If we set A = (−a, 0), B = (a, 0), and P = (x, y), and let m be the given area, then the equation of the locus is given by (x + a)2 + y2 + (x − a)2 + y2 = m. This equation can be rewritten in the form 2x2 + 2y2 + 2a2 = m, or x2 + y2 = m2 − a2 .

13.

If A = (x1 , y1 ), B = (x2 , y2 ), C = (x3 , y3 ), D = (x4 , y4 ), if the general point P = (x, y), and if the given area is m, the equation of the locus can be written as (x − x1 )2 + (y − y1 )2 + (x − x2 )2 + (y − y2 )2 + (x − x3 )2 + (y − y3 )2 + (x − x4 )2 + (y − y4 )2 = m. If we let t = x1 + x2 + x3 + x4 , u = y1 + y2 + y3 + y4 , v2 = x21 + x22 + x23 + x24 , and w2 = y21 + y22 + y23 + y24 , then this equation can be rewritten in the form 4x2 − 2tx + 4y2 − 2uy + v2 + w2 = m.

We can rewrite this further as 1 1 1 1 1 1 1 1 m x2 − tx + t2 + y2 − uy + u2 − t2 − u2 + v2 + w2 = , 2 16 2 16 16 16 4 4 4 or finally as

t 2 u 2 m 1 2 1 2 1 2 1 2 + y− = + t + u − v − w . x− 4 4 4 16 16 4 4 Thus the center of the circle is ( 4t , 4u ), as claimed by Fermat.

14.

By similarity, we have BE : BC = BD : AB = BD, since AB = 1. Therefore, BE = BC · BD. To construct the quotient of BE by BD, we draw those two lengths at a convenient angle, connect DE, lay off AB = 1 along BD, and draw AC parallel to DE. Then by similarity, BC : BE = BA : BD = 1 : BD, so BC = BE : BD, as desired.

15.

First, draw a line through the center N of the circle parallel to LM, and let S be the intersection of this line with the line MR. Connect the radius NR. Then SR2 = NR2 − NS2 = LN2 − LM2 = ( 2a )2 − b2 . The solutions to the equation z2 = az − b2 are p x = a2 ± ( 2a )2 − b2 . These values are equal to LN ± SR = MS ± SR = MR or MQ, as asserted. p The negative solution to z2 = az + b2 is z = a2 − ( a2 )2 + b2 . In terms of the diagram, this is ON − NM or PN − NM. Although we cannot subtract these geometrically, we do notice that NM − PN = MP. So MP, considered as a negative quantity, represents the negative solution of z2 = az + b2 .

16.

17.

Given the notation of p. 507 in the text, we let C = (x, y) be determined so that CB · CF = CH · CD. This translates to y(ey + dek + dex) = (gy + fgl − fgx)(cy + bcx) or to ey2 + deky + dexy = gcy2 + gbcxy + fglcy + fglbcx − fgcxy − fgbcx2 . We can rewrite this as (e − gc)y2 + (de − gbc + fgc)xy + fgbcx2 + (dek + fglc)y − fglbcx = 0, which is the equation of a conic section.

18.

The general equation of a conic section is Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. This is a hyperbola if B2 − 4AC > 0, an ellipse if B2 − 4AC < 0, and a parabola if B2 − 4AC = 0.

In this case, the equation of the curve is αx2 + xy − ay + αax = 0, so B2 − 4AC = 1 > 0.

Thus, the curve is a hyperbola. To find the asymptotes, note first that if we write the +x equation of the curve as y = α · aa− x x, we see that as x approaches a, the value of y grows indefinitely. Thus, the line x = a, namely, the line L1 , is one asymptote of the +x hyperbola. Next, if we write the equation of the curve as yx = α aa− x , we can see that as y x approaches −∞, the slope x approaches −α. Thus the slope of the second asymptote is −α, the same as the slope of the line UPT. To find out where the two asymptotes

Solutions

83

intersect, note that it is on L1 halfway between the minimum y-value of the illustrated branch of the hyperbola and the maximum y-value of the unseen lower branch. To find the maximum and minimum y values, we take the derivative of y = α aa=x −x x to get 2ax−x y′ = α a − . Setting this equal to 0 gives the two x values of the extreme points (a−x)2 2

2

√

√

√

as x = (1 + 2)a. The corresponding y values are (2 2 − 3)aα and −(2 2 + 3)aα.

Thus the average of these two values is −3aα, and that is the y coordinate of the point on the line L1 at which the two asymptotes intersect.

19.

By similarity of triangles OMP and PRQ, OM : MP = PR : RQ or RQ = MP · PR/OM = xy/(2a − y). Then RK = KQ − RQ = a − xy/(2a − y). Also, point P is on the parabola defined so the square on the ordinate is equal to the product of a with the abscissa. Since RK is on the axis of that parabola, we can take that as the abscissa of the point P. The ordinate, is PR = y. We therefore get PR2 = a · RK, or

2

xy 2a − ay − xy 2a3 − a2 y − axy =a = . y2 = a a − 2a − y 2a − y 2a − y It follows that the equation of the curve generated by the intersection of the ruler and the parabola is 2ay2 − y3 = 2a3 − a2 y − axy or y3 − 2ay2 − a2 y + 2a3 = axy.

20.

Setting CB = y, BA = x, and the constants GA = a, KL = b, and NL = c, we note that BK : BC = KL : NL, or BK : y = b : c. Thus BK = (b/c)y. Then BL = BK − KL = (b/c)y − b and AL = BA + BL = x + (b/c)y − b. Since CB : BL = GA : AL, we get y (b/c)y − b

=

a x + (b/c)y − b

or

ab b y − ab = xy + y2 − by, c c

or, after simplification, c y2 = cy − xy + ay − ac. b To show that this is a hyperbola, we can rewrite the equation in the form (a + c − (c/b)x − y)y = ac. Therefore, if we set z = a + c − (c/b)x − y, the equation is of the form zy = ac, namely, the equation of a hyperbola with asymptotes z = 0 and y = 0.

21.

Connect the centers DA, DB, and DC, noting that the line through the centers of tangent circles passes through the points of tangency. Then drop a perpendicular from D to the point E = (x, 0) on the x axis. Also, drop a perpendicular from C to F = (c, 0) and another perpendicular from D to the point G = (c, y) on the line CF. Applying the Pythagorean Theorem to the right triangles ADE, BDE, and CDG, respectively, we get the following equations: x2 + y2 = (r + z)2 (a − x)2 + y2 = (s + z)2 (c − x)2 + (d − y)2 = (t + z)2 Expanding these, we get: x2 + y2 = r2 + 2rz + z2 a − 2ax + x2 + y2 = s2 + 2sz + z2 2 c − 2cx + x2 + d2 − 2dy + y2 = t2 + 2tz + z2 2

84

Solutions Now note that if we subtract the first equation from the second, we get a linear equation in x and z. Similarly, if we subtract the first equation from the third, we get a linear equation in x, y, and z, and the same is true if we subtract the third equation from the second. Thus, our system is equivalent to a system of three linear equations in three variables. The solution to this system can be found by standard methods, but note that each of the three unknowns will be a very complicated function of the known quantities. Nevertheless, since the operations involved in constructing these functions are just the basic operations of arithmetic, each of the values x, y, and z can be constructed according to Descartes’s principles. On the other hand, if you actually work out the three functions giving x, y, and z, you will see how incredibly complicated they are, making an actual construction extraordinarily difficult. (One should note that we have just considered a special case of the Apollonian problem here, where the circle to be found is tangent externally to the three given circles, and those circles are all exterior to each other and do not intersect. There are numerous other solutions to this problem if those conditions are changed.)

22.

q q If we multiply r1 (x)r2 (x) = (x2 − yx + 12 y2 − 12 p − 2y )(x2 + yx + 21 y2 − 21 p + 2y ) and ar2

range the result as a polynomial in x, we get x4 − px2 − qx+ 41 (y4 − 2py2 + p2 − qy2 ). But 2

the condition on y shows that 14 (y4 − 2py2 + p2 − qy2 ) = 4y12 (y6 − 2py4 + p2 y2 − q2 ) = 1 (−4ry2 ) 4y2

= −r. Thus the product r1 (x)r2 (x) = x4 − px2 − qx − r as claimed.

Now, given that y = 4 is a solution to y6 − 34y4 + 313y2 − 400 = 0, the equation x4 − 17x2 − 20x − 6 = 0 can be solved by factoring the right side according to this rule.

We get x4 − 17x2 − 20x − 6 = (x2 − 4x − 3)(x2 + 4x + 2). Thus we need to solve √ the two quadratic √ equations x2 − 4x − 3 = 0 and x2 + 4x + 2 = 0. The roots are x = 2 ± 7 and x = −2 ± 2.

23.

If we substitute x =

√y 3

into the equation x3 −

√

26 3x2 + 27 x−

8√ 27 3 2

= 0, we get

9y − 27y + 26y − 8 = 0. If we then substitute y = 3z , we get z − 9z + 26z − 24 = 0.

The roots of this last equation are z = 2, z = 3, and z = 4. Thus, y = 23 , y = 1, and y = 43 3

2

3

√

are the roots of the equation in y, while x = 2 9 3 , x = the original equation.

24.

In Figure 14.12, we see that the new axes are GC and GF. The latter one is the same as the original y axis. The former has been rotated through an angle α, whose sine is q BC b BG = a . Therefore, cos α =

25.

√ √ 3 4 3 3 , and x = 9 are the roots of

1 − ( ba )2 =

√ a2 − b2 . a

If we set z = y + ba x + c, or y = z − ba x − c, and substitute this into the given equation, 2

fx b b we get (z − ba x − c)2 + 2bx a (z − a x − c) + 2c(z − a x − c) = a + ex + d. This reduces to

2

b f 2cb 2 + z = x + + e x + c2 + d, a a2 a 2

which is the equation of a hyperbola. If we rewrite this as z2 = rx2 + sx + t, then the √ ′ substitution x = rx turns this into an equation of the form z2 = x′2 + ux′ + v. Finally, we can substitute x′′ = x′ + u2 and write this into the simpler form z2 = x′′2 + m, which is also the equation of a hyperbola.

Solutions 26.

The statement is true for n = 2, because 21 = 00 + 10 and 22 = 11 . Now suppose the statement is true for n = m and let us prove it for n = m + 1. Suppose k ≤ m + 1.

Then k − 1 ≤ m, and we have ! m+1 k

27.

28.

85

! =

m k

! +

m k−1

=

m −1 X j=k − 1

! j k−1

! +

m k−1

=

m X j=k − 1

! j k−1

,

and the result is true for n = m + 1. Thus by induction, the result is true.

As Pascal noted, this result is true for n = 1, because 10 : 11 = 1 : 1. So let us assume the result it for n = m + 1. By the induction hypothesis, mis true for n = m and show

m m+1 m : mk = ( mk + k − ) : mk = 1 + m−kk + 1 = k − 1 : k = k : (m−k + 1). Then k 1

m+1 since mk : k m+ 1 = (k + 1) : (m − k), we have mk ++ 11 : mk = m−k + 1 . Also,

k m+1 [ k m+ 1 + mk ] : mk = mk − by dividing the two previous + 1 + 1 = k + 1 . Therefore,

m

m+1 m+1 m+1 m equations we have k : k + 1 = [ k : k ][ k : mk ++ 11 ] = mm−+k +1 1 · mk ++ 11 = k+1 m + 1 − k , and the inductive step is proved. Thus, by mathematical induction, the result is proved for all n.

This assertion is true for n = 1, because 10 : 00 = 1 : 1 = 1 : (1 − 0).

So let us assume that it is true for n = m n = m. We have it for − 1 and mprove

m−1 −1 m−1 m−1 m−1 m−1 m ]. By the result : = 1 + [ ] : + = [ : k k−1 k k−1 k k k k k of the previous exercise, the last ratio is equal to m − 1−( = m−k . It follows that k − 1) m − 1 m k m = 1+ m−k = m−k , and the inductive step is proved. Thus, by mathematical k k : induction, our result is true for all n.

29.

The probability of not throwing a six in a single throw is 65 . Therefore, the probability 625 of not throwing a six in any of the four throws is ( 56 )4 = 1296 . Thus, the probability of 625 671 throwing a least one six in four throws is 1 − 1296 = 1296 . That means that the odds in favor of getting at least one six in four throws are 671 : 625.

30.

The probability of not throwing a one in a single throw is 65 . Therefore, the probability of not throwing a one in any of the three throws is ( 56 )3 = 125 216 . Thus, the probability 91 of getting at least one one in three throws is 1 − 125 = 216 216 . Thus, the odds against throwing a one in three throws are 125 : 91.

31.

Using Pascal’s theorem, with r = 3, s = 4, and n = 6, we see that the first player gets that proportion of the stakes that 60 + 61 + 62 + 63 = 42 is to 26 = 64. Thus, the first 42 22 player gets 64 of the stake and the second player gets 64 of the stake; that is, the stake is split in the proportion 42 : 22, or 21 : 11.

32.

At most three more games are necessary to decide the contest, so we may as well assume that these three are played. Let us represent by A a win by the first player, by B a win by the second player, and by C a win by the third player. There are 27 possible outcomes for three games, where each one could result in a win for A, B, or C. These are as follows: AAA, AAB, AAC, ABA, ABB, ABC, ACA, ACB, ACC, BAA, BAB, BAC, BBA, BBB, BBC, BCA, BCB, BCC, CAA, CAB, CAC, CBA, CBB, CBC, CCA, CCB, CCC. Since the first player wins whenever an A occurs before two Bs or two Cs, and since this happens in 17 out of the 27 cases, we must give 17 27 of the stakes to the first player. Since the portion of the stakes to be given to the other two players must be

86

Solutions 5 of the stakes. That is, the stakes should be divided in the equal, each of them gets 27 ratio 17 : 5 : 5.

33.

In a roll of three dice, a 9 can be achieved in the following six ways, where we will not consider the order: (1, 2, 6), (1, 3, 5), (1, 4, 4), (2, 2, 5), (2, 3, 4), (3, 3, 3). Similarly, to get a 10, the following six ways are possible: (1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4), (3, 3, 4). To calculate probabilities, however, we note that there are altogether 63 = 216 equally likely outcomes in a throw of three dice. Since there are six different ways to arrange three different numbers, any particular outcome which 6 . Any particular outhas different numbers on each of the three dice has probability 216 3 come which has one number repeated twice has probability 216 . Any outcome which 1 has one number repeated three times has probability 216 . Therefore, the probability of 6 6 3 3 6 1 25 throwing a 9 is 216 + 216 + 216 + 216 + 216 + 216 = 216 . The probability of throwing a 6 3 6 3 3 27 6 + 216 + 216 + 216 + 216 + 216 = 216 . Therefore, the probability of throwing a 10 is 216 10 is greater than that of throwing a 9.

34.

We know that the probability of throwing a 7 with two dice is 61 , the probability of 5 25 throwing a 6 is 36 , and the probability of throwing something else is 36 . Thus the 1 25 37 1 expectation of the first player is 1 · 6 + 2 · 36 = 72 . The expectation of the second 5 35 player is 1 · 36 + 12 · 25 36 = 72 .

35.

5 The first player has a probability of 36 of winning on the first throw. If a 6 does not 1 show up then, which happens with probability 31 36 , then I have a probability of 6 of winning on the second throw, for a probability of winning in the first pair of throws of 31 1 31 5 30 36 · 6 = 216 . Thus in the first pair of throws, the first player has probability 36 = 216 of 31 winning, while I have probability 216 . If neither person wins on the first pair of throws, we can simply discount that pair and start over. Thus the ratio of my probability to that of the other player is 31 : 30.

36.

The probability of A winning on the first draw is 13 . The probability of B being able to draw is therefore 23 , so B’s probability of winning is 23 · 31 = 29 . The probability of C 4 being able to draw is 1 − 31 − 29 = 94 . Therefore C’s probability of winning is 94 · 13 = 27 .

2 4 1 Thus, the ratio of the probabilities of A to B to C is 3 to 9 to 27 , or 9 : 6 : 4.

There are 40 ways of picking four cards from a deck of 40. This is equal to 4 40·39·38·37 = 10 · 13 · 19 · 37. Of these ways of picking four cards, there are 104 4! ways of picking a set of cards with one from each suit, for there are 10 possibilities in each suit. Thus the probability of picking a set of cards with one from each suit 4 3 3 is p = 10·1310·19·37 = 13·10 19·37 . Thus a fair wager would have A wagering 10 = 1000 and B wagering 13 · 19 · 37 − 1000 = 8139.

We note first that if p is prime, then pk is a multiple of p for 1 ≤ k ≤ p − 1. After all, this number is an integer expressed as a fraction with p a factor of the numerator and no factor of the denominator can divide p. Thus every term of the expansion of (1 + 1)p is congruent to 0 modulo p except the first term and the last. These terms are both equal to 1, so 2p = (1 + 1)p ≡ 2 (mod p). Now assume that ap ≡ a (mod p). We want to prove that (a + 1)p ≡ a + 1 (mod p). But again because all binomial coefficients except the first and last are divisible by p, we have (a + 1)p ≡ ap + 1 ≡ a + 1 (mod p), and the inductive step is proved.

37.

38.

Solutions

87

39.

We consider the remainders of 1, a, a2 , . . . on division by p. Since there are only p − 1 possible remainders, these remainders must ultimately repeat. So for some n and r we have an + r ≡ ar (mod p), or ar (an − 1) ≡ 0 (mod p). Given that if p divides a product, it must divide one of the factors, and given that p and a are relatively prime, it follows that an − 1 ≡ 0 (mod p), or that an ≡ 1 (mod p). Let n be the smallest positive integer satisfying this congruence. Apply the division algorithm to n and p − 1. We have p − 1 = kn + s, where 0 ≤ s < n. Then 1 ≡ ap − 1 ≡ akn + s ≡ (an )k as ≡ as (mod p). Thus as ≡ 1 (mod p), where 0 ≤ s < n. But n is the smallest positive integer having this property. So s = 0 and n divides p − 1.

40.

Let P be on the conic section. Let Q be a point on the section “infinitesimally” close to P. Pick four additional points on the conic, labeled in order from Q as C, D, E, and F, and connect QC, CD, DE, EF, and FP. We then have a hexagon inscribed in the conic, with one side, namely PQ, an infinitesimal. Lines QC and EF will intersect at a point G, and lines PF and DC will intersect at a point H. (If either of these pairs of lines is parallel, change the points selected so this is not the case.) According to Pascal’s hexagon theorem, the lines ED and PQ will intersect at a point K on line GH. Since the line PQ is the tangent line at P, being drawn through two infinitesimally close points, we can determine that line by finding the point K where ED intersects the line GH and drawing the line from P (or Q) to K.

CHAPTER 15 1.

Place the sphere in a three-dimensional coordinate system with its center at the origin, and let (x, y, z) (x, y, z > 0) be a corner of the rectangular parallelepiped. Let the equation of the sphere be x2 + y2 + z2 = r2 . Then we must maximize 8xyz subject to the condition that x2 + y2 + z2 = r2 . We can use the method of Lagrange multipliers.

4xy We must have 8yz = 2xλ, 8xz = 2yλ, and 8xy = 2zλ. We therefore have 4yz x = z , or 4yz2 = 4x2 y, or x2 = z2 , or x = z. Similarly x = y, and the parallelepiped is a cube. √ In the case where the radius of the sphere is 10, we have 3x2 = 100, or x = 103 3 . √

Thus the side of the cube has length s = 203 3 and the volume of the parallelepiped is √

s3 = 80009 3 = 1539.6.

2.

From the center of the sphere of radius R, draw a line to a point on the circumference of the base of the inscribed cylinder. Assume that line makes an angle α with the line from the center of the sphere to the center of the base of the cylinder. Then the radius r of the cylinder is R sin α and the altitude of the cylinder is 2R cos α. Thus the volume V of the cylinder is given by V = πr2 h = π(R2 sin2 α)(2R cos α) = 2πR3 sin2 α cos α. If 3 3 2 we take the derivative of V with respect to α, we get dV dα = 2πR (2 sin α cos α − sin α).

2 2 To maximize V, we set this derivative equal to 0. Thus sin α(2 cos α − sin α) = 0. If 2 2 sin α = 0, we get V = 0, which is a minimum. Thus we must √ have 2 cos α − sin α = 0, 2 2 2 or 2 cos α = sin α, or tan α = 2, or, finally, tan α = 2. Since the diameter of the base of the cylinder is 2R sin√α, we have the ratio of the diameter of the cylinder to its 2R sin α altitude = 2R 2 : 1, as claimed. cos α = tan α =

3.

Fermat’s initial method for determining maxima and minima for a polynomial p(x) is to set p(x1 ) = p(x2 ), or p(x1 ) − p(x2 ) = 0, divide by x1 − x2 , and then set x1 = x2 . That

88

Solutions is, the maximum or minimum can be found by setting x1 = x2 in the quotient p(x1 ) − p(x2 ) x1 − x2 and setting this result equal to 0. Given that for polynomials, x1 − x2 always divides p(x1 ) − p(x2 ) without remainder, one can determine the limit as x1 approaches x2 in the quotient by simply setting x1 equal to x2 . Thus Fermat’s process is equivalent to determining lim

x1 →x2

p(x1 ) − p(x2 ) x1 − x2

and then setting this limit equal to 0. And this is the same process as determining p′ (x) and setting it equal to 0. The same argument holds for Fermat’s second method, which is equivalent to determining p′ (x) by calculating lim

e→0

p(x + e) − p(x) e

and then setting that equal to 0, again because e will always divide the difference p(x + e) − p(x) without remainder.

4.

We set bx1 − x31 = bx2 − x32 . This is equivalent to bx1 − bx2 = x31 − x32 . If we divide both 2 2 2 sides by xq 1 − x2 , we get b = x1 + x1 x2 + x2 . Setting q x = x1 = x2 now gives us b = 3x , or x = ±

b 2b 3 . Thus the maximum value is 3

b 3 . Since the negative solution to the 3

equation gives a negative value to the expression bx − x , it is the positive solution which gives the maximum.

5.

If p′ (a) = 0, and M = p(a), then lim

x→a

p(x) − p(a) = p′ (a) = 0. x−a

p(a) But the quotient p(x)− is a polynomial q(x), and q(a) = 0. Therefore, q(x) = x−a (x − a)r(x), and p(x) − M = p(x) − p(a) = (x − a)q(x) = (x − a)2 r(x).

3

6.

We adequate (x +x3e) and t +t e . After expanding, we get tx3 + 3tx2 e + 3txe2 + te3 ≈ tx3 + ex3 . After canceling common terms and dividing by e, we have 3tx2 + 3txe + te2 = x3 . Elimination of all terms with e then gives us the relation 3x2 t = x3 , or, finally, t = 3x .

7.

If (x, y) is a point on the curve, and if (x + e, y) is a point on the tangent line, then y−y = yt , or y = t +t e y. Thus, to find the subtangent, and therefore the tangent, we e must adequate f(x, y) and f(x + e, t +t e y). In the case x3 + y3 = pxy, we first rewrite this as x3 + y3 − pxy = 0, and then adequate this expression with (x + e)3 + ( t +t e y)3 − p(x + e) t +t e y. By expanding, we get e2 pe2 e e3 pe y. x3 + y3 − pxy = x3 + 3x2 e + 3xe2 + e3 + y3 + 3 y3 + 3 2 y3 + 3 y3 −pxy− xy−pey− t t t t t

Solutions

89

If we cancel common terms, divide through by e, and then eliminate all remaining 3 pxy − 3y3 terms in e, we get 3x2 + 3yt − pxy t − px = 0. Solving this for t gives us t = 3x2 − py .

8.

Fermat’s method for finding the subtangent t involves setting up the adequality tf(x + e) ≈ (t + e)f(x), then canceling common terms, dividing by e, and removing any remaining terms involving e. We can rewrite this adequality in the form t(f(x + e) − f(x)) ≈ ef(x), or t≈

ef(x) f(x) = . f(x + e) − f(x) f(x + e)−f(x) e

The process of canceling common terms, dividing by e, and removing any remaining terms involving e amounts to, in modern terms, calculating the limit of this expression as e approaches 0. Since the limit of the denominator is f′ (x), we get that t = ff′((xx)) . In the modified method of the previous problem, we have f(x, y) ≈ f(x + e, y+ et y). By modern methods, the expression on the right is approximately equal to f(x, y) + ∂∂xf e + ∂∂yf eyt .

Comparing this result to f(x, y) and dividing through by e gives us the equation ∂∂xf + ∂f y ∂ y t = 0. If we solve this for t, we get t = −y(∂ f/∂ y)/(∂ f/∂ x).

9.

2

2

2

2

2

We adequate ax2 + by2 with (x +a2e) + (1+(eb/2t) )y . By canceling common terms, dividing 2

through by e, and then removing all remaining terms with e, we get 2x + (2/bt2)y = 0. a2 2 2

x . To compare this with Solving for t gives us t = − ab2yx = − a b b−2 xb x = − a − x Apollonius’s result from Chapter 4, we rename the line segments in Figure 4.23. Set AB = t, BG = a + x, BH = a−x, AG = AB−BG = t−a−x, and AH = 2a + AG = t + a−x.

According to Apollonius’s Proposition I–34, AB will be the subtangent if AH : AG = a−x + a−x BH : BG. Therefore tt− a−x = a + x . Thus (t − (a + x))(a − x) = (t + (a − x))(a + x), or 2 2

2 2

2

2

x t((a − x) − (a + x)) = 2(a2 − x2 ), or −2xt = a2 − x2 , or, t = − a − , the same result x calculated via Fermat’s method.

2

2

10.

We need to find a double root to the polynomial (x3/2 )2 + v2 − 2vx + x2 − n2 = x3 + x2 − 2vx + v2 − n2 . Since this is a cubic polynomial, we equate it to (x − x0 )2 (x + b) = x3 + (b − 2x0 )x2 + (x20 − 2x0 b)x + bx20 . Comparing coefficients of like powers of x, we must solve the equations: b − 2x0 = 1; x20 − 2x0 b = −2v; and bx20 = v2 − n2 .

Solving the first equation for b and putting this into the second equation gives us x20 − 2x0 (1 + 2x0 ) = −2v, or −2v = −3x20 − 2x0 , or, finally, v = x0 + 23 x20 .

11.

Rewriting y2 = x as y = x 2 , we need to find a double root to the polynomial (x1/2 )2 + v2 − 2vx + x2 − n2 = x2 + (1 − 2v)x + v2 − n2 . We therefore equate this to (x − x0 )2 = x2 − 2x0 x + x20 by comparing coefficients of like powers. We get the equation 1 − 2v = −2x0 , which reduces to v − x0 = 12 . It follows that the slope of √ √ x0 y0 the normal line is v− −x0 = − 1/2 = −2 x0 . Therefore, the slope of the tangent line is

1

1 √ .

2 x0

12.

To apply Descartes’s rule, we need to find a double root of the polynomial x2n + x2 − 2vx + v2 − m2 . By Hudde’s rule, this root will also be a root of 2nx2n + 2x2 − 2vx or a solution of the equation 2nx2n − 1 + 2x − 2v = 0. Since this root is x0 , we have

90

Solutions −1 v = x0 + nx2n . Since the slope of the tangent line is v−xnx0 , we calculate this to be 0 −1 nx2n 0 = nxn0 − 1 as desired. xn0

0

13.

We multiply the x3 terms by 3, the x term by 1, and set the result equal to 0. We get 2 2 (9a − 3b)x3 − 2b3ca x = 0, or (9a − 3b)x2 − 2b3ca = 0. Thus, the maximum occurs when q 2 /(3c) x = − 2b9aa− 3b and the minimum occurs when x is the positive square root of that expression.

14.

We leave all terms with x on the left and move every term with y to the right. Thus we get x3 − pxy = −y3 + pxy. We then multiply the terms on the left by their x exponent and the terms on the right by their y exponent. We get 3x3 − pxy = −3y3 + pxy. Next, we replace one x in each term on the right by t: 3tx2 − pty = −3y3 + pxy. We finally 3 pxy . solve for t: t = −3x3y2 −+ py

15.

We transfer the y2 term to the right and the bx term to the left and then multiply the terms on the left by their x exponent and the terms on the right by their y exponent.

We get 2x2 − bx = −2y2 . We then replace one x in each term on the right by t to get 2y2 2xt − bt = −2y2 . We solve for t: t = b − 2x . Since t is the subtangent, the slope of the 2xy b − 2x tangent line is given by yt = by − = 2 2y .

2y

16.

Fermat’s procedure begins by adequating tg(x + e) to (t + e)g(x), or t(g(x + e) − g(x)) to eg(x). Let g(x) = an xn + an − 1 xn − 1 + · · · + a1 x + a0 . Then t(g(x + e) − g(x)) = nan xn − 1 te + (n − 1)an − 1 xn − 2 te + a1 te+ terms with a power of e at least 2. When we adequate this with eg(x), divide by e, and remove all remaining terms containing e, we note that on the left side, we have in effect multiplied each term of g(x) by its exponent in x and replaced one x in each term by t, while on the right side we simply have the original polynomial g(x). Sluse’s rule in the case g(x) = y says to do exactly that, since on the right we multiply y by 1 and then, to find t, we replace y by g(x) before dividing. In the general case, recall from Exercise 8 that Fermat’s rule amounts to finding t from the equation t ∂∂xf = −y ∂∂yf . To calculate either partial derivative, we multiply each term by the exponent in x or the exponent in y and then reduce the corresponding exponent by 1. If a term in f(x, y) originally had both x and y factors, we multiply it twice – once by the x exponent and once by the y exponent. Sluse tells us that for the terms in x, we should not reduce the exponent by 1 but instead replace one x in each term by t. That is exactly what Fermat’s rule gives us in the form t ∂∂xf .

Similarly, the expression −y ∂∂yf is what we get by moving all the terms with y to the right side of the equation and simply multiplying each term by its y exponent. Thus Sluse’s rule is equivalent to Fermat’s rule for determining the subtangent to the curve given by f(x, y) = c, where f is a polynomial in x and y.

17.

Let us divide up the sphere of radius r into infinitely many cones of altitude r, each with their vertex at the center of the sphere. The volume of each of these cones is then 1 3 rA, where A is the (infinitesimal) area of the base on the surface of the sphere. If we add all these volumes together, we get 31 rS, where S is the total surface area of the sphere. Since S = 4πr2 , we have V = 43 πr3 .

Solutions 18.

The binomial expression

N+k k+1

91

can be calculated in the usual way to get

(N + k)(N + k − 1)(N + k − 2) · · · N . (k + 1)!

Thus the right side of Fermat’s rule is equal to the right side of the first given expression. Also, since any term in the Pascal triangle is equal to the sum of all the terms in the column to its left up to the row above it, we have ! ! N+k−1 X j N+k . = k−1 k j=k − 1

Thus the left side of Fermat’s rule is equal to the left side of the first given expression.

The right side of the last expression is equal to Nk ++ 1k . The left side is equal to kk +

k+1 + · · · + k + Nk − 1 . Thus, this expression simply notes that a given element in the k Pascal triangle is equal to the sum of the elements in the column to the left up to the row above it, the result proved in Chapter 14, Exercise 26. This result is equivalent to the standard construction rule of the Pascal triangle, which in turn, by the result proved in Chapter 14, Exercise 27, implies Fermat’s rule. On the other hand, Fermat’s rule gives us the standard multiplicative rule for calculating elements in Pascal’s triangle, from which the additive construction rule is easily proved and therefore the last expression.

19.

If we set k = 3, we get N X j(j + 1)(j + 2)

6

j=1

Expanding the left side gives us 16

=

N(N + 1)(N + 2)(N + 3) .

24

P 3 1P 2 1P j +2 j +3 j. Therefore

N(N + 1)(N + 2)(N + 3) 1 N(N + 1)(2N + 1) 1 N(N + 1) 1X 3 j = − − 6 24 2 6 3 2 j=1 2

N + 5N + 6 2N + 1 1 = N(N + 1) − − 24 12 6 2

N + 5N + 6 − 4N − 2 − 4 = N(N + 1) 24 N

=

N2 (N + 1)2 .

24

It follows that

2 N X N(N + 1) j3 = .

2 j=1

20.

If we set k = 4 in the last formula of Exercise 18, we get N X j(j + 1)(j + 2)(j + 3) j=1

24

=

N(N + 1)(N + 2)(N + 3)(N + 4) .

120

92

Solutions 1 This can be rewritten as 24 5 4 3 2 1 120 (N + 10N + 35N + 50N + 24N). Therefore,

PN

4 j=1 j =

P 4 (j + 6j3 + 11j2 + 6j)

=

X 3 X 2 X 1 5 24 N + 2N4 + 7N3 + 10N2 + N − 6 j − 11 j −6 j 5 5 N

N

N

j=1

j=1

j=1

N5 24N 3 4 = + 2N4 + 7N3 + 10N2 + − (N + 2N3 + N2 ) 5 5 2 −

21.

11 N5 N4 N3 N (2N3 + 3N2 + N) − 3(N2 + N) = + + − .

6 5 2 3 30

Let the equation of the parabola be y2 = x and the vertical line BD be x = b2 . Then the volume of the solid formed by rotating the parabola around BD is given by Z b Z b 2 2 2 V = π(b − y ) dy = π (b4 − 2b2 y2 + y4 ) dy 0

0

b

2b2 y3 y5 2b5 b5 8 = π b5 − + + = πb5 . = π b4 y − 3 5 0 3 5 15 The cone with the same base and vertex as the paraboloid has volume 13 bπb4 = 13 πb5 = 5 5 15 πb , so the ratio of the volumes is 8 : 5 as noted by Fermat. Also, the volume of the 8 cylinder circumscribing the paraboloid and cone is πb5 . Thus the paraboloid is 15 of the cylinder, as noted by ibn al-Haytham.

22.

If we divide the interval [0, x0 ] as stated, then the area of the right most circumscribing rectangle is R1 = (1 − mn )x0 pxk0 = (1 − mn )pxk0 + 1 . The area of the next rectangle is ( mn − ( mn )2 )x0 p( mn x0 )k = (1 − mn )p( mn )k + 1 xk0 + 1 = ( mn )k + 1 R1 . Similarly, the area of the third rectangle from the right is ( mn )2(k + 1) R1 , and so on. Thus, the area under all of the circumscribed rectangles is given by

n k + 1 n 2(k + 1) A = R1 1 + + + ··· m m " #

1 1 n k+1 = R1 = 1 − px0

k+1 k+1 m 1− n 1− n =

m 1 2

m

pxk0 + 1 . n k

1+( mn ) + ( mn ) + ··· + ( m )

We get the area under the curve by allowing mn to approach 1. Each of the terms in the denominator approaches 1 and the area is k +1 1 pxk0 + 1 , as desired.

23.

Since 3 is an integer, we use the basic formula for row 3 of the Pascal triangle: a3,n = (n + 1)(n +6 2)(n + 3) . Thus a3,1/2 = (3/2)(5/2)(7/2)/6 = 35/16; a3,3/2 = (5/2)(7/2)(9/2)/6 = 105/16; and a3,5/2 = (7/2)(9/2)(11/2)/6 = 231/16.

24.

We know that a3/2,0 = 1. Therefore, a3/2,1 = 3/21+ 1 · 1 = 25 , and a3/2,2 = 3/22+ 2 · 52 = 5 35 4 7 4 · 2 = 8 . Also, since a3/2,1/2 = a1/2,3/2 , we know that a3/2,1/2 = 3 . Therefore, a3/2,3/2 = 3/23/+23/2 a3/2,1/2 = 83 and a3/2,5/2 = 3/25/+25/2 a3/2,3/2 = 64 15 .

Solutions 25.

93

R b q1 1 The text shows that that the arclength of the parabola is 3 0 4 x + 9 dx. We make a substitution u = 41 x + 19 , so du = 41 dx. Then the integral becomes Z 1 b+ 1 12

26.

4

9

1 9

" 3 32 # 1 1 b 1 2 1 2 32 4 b+ 9 u du = 12 · u 1 = 8 + − 3 4 9 9 9 s s 3 23 3

4 8 4 4 − − = b+ = b+ .

9 9 9 27 1 2

If y4 = x5 , then y = x5/4 and y′ = 54 x1/4 . The arclength L of this curve from 0 to b is given by Z bp Z br 25 L= 1 + y′2 dx = 1 + x1/2 dx.

16 0 0 1/2 −1/2 16 32 1/2 If we set u = 1 + 25 , or x1/2 = 25 (u − 1), then du = 25 dx or dx = 25 x du = 16 x 32 x √ 512 25 625 (u − 1) du. If we set a = 1 + 16 b, then the arclength formula becomes

Z 512 a 3/2 (u − 1)u du = (u − u1/2 ) du 625 1 1

2

a 512 2 5/2 2 3/2 a a 2 1024 √ = u − u − + . = a 625 5 3 625 5 3 15 1

512 L = 625

27.

Z a

1/2

If y = x2 , then y′ = 2x. Thus, the arclength formula in this case gives us L=

Z bp a

1 + (2x)2 dx =

Z b√ 1 + 4x2 dx. a

Thus, to calculate this integral, one needs to find the area under the curve y = or y2 − 4x2 = 1, a hyperbola.

28.

√

1 + 4x2 ,

Recall that, according to Fermat’s method of calculating the subtangent, if f(x) is a x) function, then the subtangent t is found by adequating t to f(x +efe()− , then simplifying. f(x) f(x) It will be convenient to rewrite this as 1t = f(x +efe()− , even though this is not a strict x) equality. Thus, if we have a product of two functions, say zu, we can calculate the subtangent tzu as follows:

z(x + e)u(x + e) − z(x)u(x) 1 = tzu ez(x)u(x) z(x + e)u(x + e) − z(x)u(x + e) + z(x)u(x + e) − z(x)u(x) = ez(x)u(x) [z(x + e) − z(x)]u(x + e) + z(x)[u(x + e) − u(x)] = ez(x)u(x) z(x + e) − z(x) u(x + e) u(x + e) − u(x) z(x) 1 1 · + · = + . = ez(x) u(x) eu(x) z(x) tz tu

94

Solutions Therefore, given the ratio u : v = w : z, we have uz = vw and therefore tuz = tvw .

Therefore t1z + t1u = t1v + t1w , or 1 1 1 1 tu tw + tu tv − tv tw = + − = tz tv tw tu tv tw tu and tz =

tu tv tw tu tw + tu tv − tv tw

as stated. To convert this relationship to formulas involving derivatives, recall that if y = f(x), then y′ = tyy . To determine the product rule, we set u = 1, so z = vw. Since tu is infinite, the rule already derived tells us that tz = tvtv+twtw or that t1z = t1v + t1w . We then have z′ =

z vw vw = + = wv′ + vw′ , tz tv tw

the product rule. Similarly, to determine the quotient rule, we set w = 1, so z = uv .

1 1 1 u tv Since tw is infinite, we get tz = tut− tv or tz = tv − tu . We then have z′ =

z v = tz u

1 1 − tv tu

=

v′ vu′ uv′ − vu′ − 2 = , u u u2

the quotient rule.

29.

We replace y by y + a and x by x + e in the formula x3 + y3 = c3 . We get (x + e)3 + (y + a)3 = c3 , or x3 + 3x2 e + 3xe2 + e3 + y3 + 3y2 a + 3ya2 + a3 = c3 . We then remove all terms containing a power of a or e at least two and also delete the terms forming the original expression. We are left with 3x2 e + 3y2 a = 0. We next substitute y for a and t for e and solve for the ratio y : t. We get 3x2 t + 3y3 = 0, and y : t = −3x2 : 3y2 = −x2 : y2 is the slope of the tangent line.

30.

If we set BG = PM = y and MR p= HG = a, we can rewrite the proportion CL : LF = CB : BH in the form (f + ge) : g2 − 2fge = 1 : (y − a). Squaring both sides gives us 1 f2 + 2fge = , g2 − 2fge y2 − 2ay after neglecting powers of e and a higher than the first. This equation becomes g2 − 2fge = y2 f2 + 2fgy2 e − 2ayf2 . Since y = gf , or y2 f2 = g2 , we have 2fgy2 e + 2fge = 2ayf2 , or, gy2 e + ge = ayf. Now we substitute y for a and t for e. We get (gy2 + g)t = y2 f, 2 so t = g +fygy2 . Since f = gy , we can rewrite this as t=

y BG · CB2 BG · CK2 = = .

1 + y2 CG2 CE2

Since the slope of the tangent line to the curve ANMO is given by yt = 1 + y2 = CG2 = 2 x sec2 x, we have d tan dx = sec x.

Solutions

95

CHAPTER 16 1.

The first term of the square root of 1 + x is 1. We subtract this from 1 + x to get x, then double the 1 and divide into x. This gives us the second term of the square root, 2 namely 2x . We then multiply 2 + 2x by 2x to get x + x4 . We subtract this from x to get 2 2 − x4 . We double the two terms we already have to get 2 + x and divide the 2 into − x4 .

2 2 The result, − x8 , is then the third term of the square root. We then multiply 2 + x − x8 2 2 3 4 2 3 4 x x by − x8 to get − x4 − x8 + 64 . Subtracting this from − x4 gives x8 − 64 . We double the 3 x2 three terms we already have to give us 2 + x − 4 and divide the 2 into x8 . This gives 3 x us 16 for the fourth term of the series. We continue in this way. We therefore have √

1+x = 1 +

x x2 x3 5x4 7x5 − + − + + ··· .

2 8 16 128 256

2.

1

1 + − x ) 1 1 −

x2

+

x4

+

x6

x2 x2 x2

−

x4 x4 x4

−

x6 x6

2

Thus, the power series for 1/(1 − x2 ) is 1 + x2 + x4 + x6 + · · · .

3.

4.

1 6 The power series for (1 − x2 )1/2 is 1 − 12 x2 − 18 x4 − 16 x −· · · . To square this, we need to square each term and also take twice the product of every pair of terms. We therefore 1 8 1 8 1 10 1 12 get 1−x2 − 14 x4 − 18 x6 +· · ·+ 14 x4 + 81 x6 + 16 x + · · ·+ 64 x + 64 x + · · · + 256 x +· · · . We 4 6 8 note that the x and x terms already disappear. To check the x term, we would need 5 8 x , to find the next term in the original power series. That term turns out to be − 128 5 8 which, when doubled, gives − 64 x . This term, combined with the two other terms in x8 already calculated, again give a 0 coefficient. Every term beyond the x2 term does disappear, and the square of the power series is 1 − x2 as desired.

Let the first approximation be x = 1. Then set x = 1 + p and substitute. We get

(1 + p)2 − 2 = 0, or p2 + 2p − 1 = 0. Neglecting the term in p2 gives us 2p = 1 and

p = 0.5. Thus our second approximation is x = 1.5. Next, set p = 0.5 + q and substitute into the equation for p. We get (q + 0.5)2 + 2(q + 0.5) − 1 = 0, or q2 + 3q + 0.25 = 0. If we neglect the term in q2 , we get 3q = −0.25, and q = −0.083333333. Thus our third approximation is x = 1.416666667. Next, set q = −0.083333333 + r and substitute in the equation for q. We get r2 + 2.833333333r + 0.006944444 = 0. Again, neglecting the term in r2 , we get 2.833333333r = −0.006944444 and r = −0.00245098. Therefore, our fourth approximation is x = 1.414215687. Finally, set r = −0.00245098 + s and substitute into the equation for r. This gives us, after neglecting the term in s2 ,

96

Solutions the equation 2.828431373s = −0.000006007, so s = −0.000002124, and our fifth approximation is x = 1.414213563, accurate to 8 decimal places.

5.

As stated in the problem, we substitute p = − 14 x + q into the equation for p. We get 1 + 3(− 14 x + q) +3(− 14 x + q)2 + (− 14 x + q)3 + 1+ (− 14 x + q)− 2 + x+ (− 14 x + q)x − x3 = 0, 1 2 65 3 3 2 or − 16 x − 64 x + 4q + 3q2 + q3 − 12 xq + 16 x q − 43 xq2 = 0. If we neglect terms in q of degree greater than 1 and terms in x of degree greater than 2, as well as products of 1 2 1 2 such terms, we are left with 4q − 16 x = 0, so q = 64 x . Thus the first three terms of 1 2 1 the power series for y are 1 − 4 x + 64 x .

6.

If we substitute y = z + p into the given series, we get 15 (z + p)5 − 41 (z + p)4 + 3 2 1 1 3 (z + p) − 2 (z + p) + z + p − z = 0. We can expand this, but instead we should note that the only terms which will be of degree 1 or 2 in z and 1 in p are among the last four, two of which cancel each other out. We then have − 12 z2 − zp − 12 p2 + p = 0, and, eliminating the second and third terms here, we are left with − 12 z2 + p = 0. Thus p = 21 z2 , and the first two terms of the power series are y = z + 21 z2 . To get the next term, we take p = 12 z2 + q and substitute into the equation for p. Since we again eliminate all terms of degree higher than 1 in q and higher than 3 in z, we are left with 1 3 1 2 1 3 1 2 1 3 1 3 3 z − 2 z − 2 z + 2 z + q = 0. Therefore, − 6 z + q = 0, or q = 6 z . To get the next term, we set q = 16 z3 + r and substitute again. Since as before most terms are eliminated, except those whose degree in z is 4 or less and whose degree in r is 1, we are left with − 14 z4 + 13 z3 + 21 z4 − 12 z2 − 12 z3 − 18 z4 − 16 z4 + 12 z2 + 61 z3 + r = 0. This simplifies 1 4 1 4 to − 24 z + r = 0, or r = 24 z . Therefore, the first four terms of the power series for y 1 3 1 4 1 2 are y = z + 2 z + 6 z + 24 z .

7.

0.001 We have log(1 + x) = x − x2 + x3 − x4 + · · · . So log(1.1) = 1 − 0.01 − 2 + 3 0.0001 0.00001 0.000001 0.0000001 + 5 − + − · · · = 0.09531018. Similarly, log(0.9) = 4 6 7 −0.10536051, log(1.2) = 0.18232156, log(0.8) = −0.22314355, log(1.01) = 0.00995033, log(0.99) = −0.01005034, log(1.02) = 0.01980263, and log(0.98) = 1.2×1.2 −0.02020271. We have log(1) = 0. Since 2 = 0.8 ×0.9 , we have log 2 = 2 log(1.2) − ×2 , we have log 3 = log(1.2) + log(0.8) − log(0.9) = 0.69314718. Since 3 = 1.20.8 log(2) − log(0.8) = 1.09861229. Also, log 4 = log(22 ) = 2 log 2 = 1.38629436, 4 log 5 = log( 0.8 ) = log 4 − log(0.8) = 1.60943791, log 6 = log(2 × 3) = log 2 + log 3 = 1.79175947, log 8 = log(23 ) = 3 log 2 = 2.07944154, log 9 = log(32 ) = 2 log 3 = 2.19722458, and log q (10) = log(2 × 5) = log 2 + log 5 = 2.30258509. To get log 7,

2

we note that 7 = 1.94591015.

8.

3

4

1 1 2 × 100 × 0.98, so log 7 = 2 [− log 2 + 2 log(10) + log(0.98)] =

If we follow Newton’s method but use the progression 4, 3, 2, 1, instead of 3, 2, 1, 0, 3 we get 4x2 ẋ − 3axẋ + 2ayẋ − yx ẋ + x3 ẏy − ax2 ẏy + 2axẏ − 4y2 ẏ = 0. If we factor out the terms with ẋ and ẏ, respectively, and divide, we get x(4y3 − 2axy + ax2 − x3 ) ẋ = . ẏ y(4x3 − 3ax2 + 2axy − y3 )

Solutions

97

Using the original equation, this simplifies to x(4y3 − 2axy + axy − y3 ) ẋ = 3 ẏ y(4x − 3ax2 + 2axy − x3 + ax2 − axy) =

x(3y3 − axy) 3y2 − ax = 2 , 3 2 y(3x − 2ax + axy) 3x − 2ax + ay

the same value as calculated using the original progression. A similar simplification takes place if one uses any other arithmetic progression.

9.

√

We set z = x a2 − x2 . We then have two equations to which to apply Newton’s rules: y2 − a2 − z = 0 and z2 − a2 x2 + x4 = 0. The first equation becomes 2yẏ − ż = 0, so ẏ 2 3 1 ż 2a2 x − 4x3 . Then 2z ż = 2y . The second equation becomes 2zż − 2a xẋ + 4x ẋ = 0, so ẋ = 2 2 ẏ ż ẏ 2a√ x − 4x3 a√ − 2x2 ẋ = ż ẋ = 4xy a2 − x2 = 2y a2 − x2 .

10.

If we replace x by x + 1, we get the fluxional equation ẏ 2 = + 2 − x2 − 2x = 2 − 2x + 2x2 − 2x3 + 2x4 − · · · + 2 − x2 − 2x ẋ x+1 = 4 − 4x + x2 − 2x3 + 2x4 − · · · .

3

4

5

Therefore, y = 4x − 2x2 + x3 − x2 + 2x5 − · · · .

11.

Given x2 + 4y2 = 1, we take fluxions to get 2xẋ + 8yẏ = 0. Since ẋ = 1, this is equivalent to z = ẏ = − 4yx . Then ż = −

4y − 4x(−x/4y) 4y + (x2 /y) 4y − 4xẏ 4y2 + x2 1 =− =− =− =− .

2 2 2 16y 16y 16y 16y3 16y3

The curvature, being a positive number, is then κ = = 12.

1/16y3 ż = (1 + z2 )3/2 (1 + x2 /16y2 )3/2 1/16y3 64y3 4 = = .

(16y2 + x2 /16y2 )3/2 16y3 (16y2 + x2 )3/2 (16y2 + x2 )3/2

n 3/2 2 b 1 n We take the derivative of z = 2a : nc (− 15 c + 5 x )(b + cx )

3/2 2 b 1 n 3 1 n−1 + − nx b + cxn + x (b + cxn )1/2 cnxn − 1 5 15 c 5 2

√ 2a 1 n−1 2 b 1 n 3 = b + cxn nx (b + cxn ) + − + x cnxn − 1 nc 5 15 c 5 2

1 1 3 2a √ 1 n−1 2n − 1 n−1 2n − 1 n bx + cx − bx + cx b + cx = c 5 5 5 10

√ √ 2a 1 2n − 1 = b + cxn cx = ax2n − 1 b + cxn . c 2

z′ =

2a nc

98

Solutions 13.

n−1

n n−1 To integrate y = ax dx. Then e + fxn , we set u = x , so du = nx

Z

axn − 1 1 dx = e + fxn n

Z

a a a du = ln(e + fu) = ln(e + fxn ). e + fu nf nf

R But e +a fu is the area s under the hyperbola v = e +a fu , so Newton’s answer is equivalent to the modern one.

14.

We rewrite 8ags − 4aguv − 2afv 4neg − nf2

z=

as

z=

2a (4gs − 2guv − fv) n(4eg − f 2 )

and take the derivative of the expression in the final parentheses with respect to u. We get p p gu(f + 2gu) f(f + 2gu) dz = 4g e + fu + gu2 − 2g e + fu + gu2 − p − p du e + fu + gu2 2 e + fu + gu2 8g(e + fu + gu2 ) − 4g(e + fu + gu2 ) − 2gu(f + 2gu) − f(f + 2gu) p = 2 e + fu + gu2 2 2 8ge + 8gfu + 8g u − 4ge − 4gfu − 4g2 u2 − 2guf − 4g2 u2 − f 2 − 2fgu p = 2 e + fu + gu2 4ge − f 2 . = p 2 e + fu + gu2 It then follows that dz dz du = dx du dx !

2a 4ge − f 2 p nxn − 1 = n(4eg − f 2 ) 2 e + fu + gu2 axn − 1 =p = y, e + fxn + gx2n as desired.

15.

We begin by using Newton’s substitution u = xn , with du = nxn − 1 dx. Therefore we get Z y=

p

a ax2n − 1 dx = n 2n n e + fx + gx

Z

x2n − 1 a p du = n − 1 2 n x e + fu + gu

Z p

u du. e + fu + gu2

Although this latter integral is not generally in the table of integrals in a standard calculus text, it is available in more extensive integral tables. The result is a n

"p

e + fu + gu2 f − g 2g

Z

# du p e + fu + gu2

=

a af R− ng 2ng

Z

du , R

Solutions

99

p where, to save space, we write R for e + fu + gu2 . Note that this integral is expressed in terms of another one, and we could write out the other one explicitly, but there is no necessity to do so. We want to show that Newton’s answer is equivalent to this. So first note that Newton’s answer also includes an integral, namely s as the area under the curve v = R. By using an integral table, we find that Z Z 2gu + f 4eg − f 2 du s = R du = R+ .

4g 8g R Thus, Newton’s result is 2a [−2fs + fuv + 2ev) n(4eg − f 2 )

Z 2a 2gu + f 4eg − f 2 du = −2f R+ + fuR + 2eR 4g 8g R n(4eg − f 2 )

Z 2a −4fgu − 2f 2 −8feg + 2f 3 du R + + fuR + 2eR = 4g 8g R n(4eg − f 2 )

Z 2 2a −4fgu − 2f + 4fgu + 8eg −4feg + f 3 du = R + 4g 4g R n(4eg − f 2 )

Z 2 2 f(f − 4eg) 2a 4eg − f du = R+ 2g 4g R n(4eg − f 2 ) Z Z a af du a af du = R− = R− . ng 2ng R ng 2ng R

z =

This is the same as the result above from the modern integral tables.

16.

When x becomes x + o, the value 1x becomes x +1 o . Thus the augments are as o to x +1 o − x−(x + o) o 1 1 x = x(x + o) = − x(x + o) . Therefore, the ratio of the augments is as 1 to − x(x + o) , and by letting o vanish, we find that the ratio of the fluxions is as 1 to − x12 .

17.

When x becomes x + o, the value x1n becomes (x +1o)n . Thus the augments are as o to n (n − 1 ) n − 2 2 −nxn − 1 o + x o + ··· 1 1 2 − = . (x + o)n xn xn (x + o)n

Therefore, the ratio of the augments is as 1 to [−nxn − 1 + (n(n − 1)/2)xn − 2 o + · · · ]/[xn (x + o)n ]. By letting o vanish, we find that the ratio of the fluxions is as 1 to n−1 − nxx2n , or as 1 to − xnn+ 1 .

18.

From the diagram, we have CT2 = AT2 + AC2 . Using the basic rules of fluxions and ˙ or AT ˙ = (CT : AT)CT. ˙ = AT · AT, ˙ If z represents noting that AC is fixed, we get CT · CT ˙ = CS : CT. It the arc, we have, by the result demonstrated in section 16.1.6, ż : AT ˙ = CS : CT or that ż : CT ˙ = (CT : AT) · (CS : CT) = follows that ż : (CT : AT)CT CS : AT. In other words, the fluxion of an arc to the fluxion of its secant is as cosine to tangent. If we invert the ratios, the result becomes, in modern terms, that the derivative of the secant is the quotient of the tangent by the cosine, or d(sec z)/dz = tan z/ cos z = tan z sec z.

100

Solutions 19.

˙ = From the geometric result AS2 + CS2 = AC2 , we derive the fluxional result AS · AS ˙ Also, from the geometric result CS : AC = AC : CT, we get CS · CT = AC2 , −CS · CS. ˙ · CT = −CT ˙ = −(CT ˙ · CS, or CS ˙ · CS : CT). We therefore have, using and therefore CS the result from Exercise 18 (and writing everything in fractions rather than ratios): ˙ =− AS

˙ ˙ · CS żAT · CS żCS · AS żCS CS · CS CS · CT = = = = .

AS AS · CT AS · CT AS · AC AC

˙ = AC : CS, or that the fluxion of the arc to the fluxion of the sine It follows that ż : AS is as radius to cosine.

20.

Suppose a planet revolves around the sun at a distance r with a constant linear velocity 2 of v. Then the acceleration to the center is given by vr . But since this acceleration is inversely proportional to the square of the radius, it is also given by rk2 , where k is the q same for all planets. Therefore, v2 = kr , or v = kr . The periodic time T of the planet is the length of the circumference of its orbit divided by its velocity. Therefore √

T=

2πr 2πr r 2π 3/2 = √ =√ r , v k k

and T2 is proportional to r3 , as claimed.

21.

1 1 1 To get the second column, we note that 11 − 12 = 12 , 21 − 13 = 61 , 13 − 14 = 12 , 4 − 15 = 20 , 1 1 1 1 1 1 1 1 1 , and so on. To get the third column, we calculate 2 − 6 = 3 , 6 − 12 = 12 , 12 − 20 = 30 and so on. To calculate a formula for the entries in Leibniz’s triangle, we note that the element in row n and column k is given by dividing n1 by the element in row

−1 n − 1, column k − 1 of the Pascal triangle. Thus that element is equal to 1n ÷ nk − 1 =

(k − 1)!(n−k)! − 1)! = n(n − (1k)···( . n! n−k + 1)

22.

By the result of the previous exercise, kth element in row n of the harmonic triangle − 1)! is n(n − (1k)···( . Thus the denominator of that element is n(n − 1)(n − 2) · · · (n − n−k + 1) k + 2)(n − k + 1)/(k − 1)!. Therefore, the sum of all the denominators of the elements in row n is ! ! ! n n nX −1 X X n−1 n−1 n−1 n =n =n = n2n − 1 , k−1 k−1 k k=1

k=1

k=0

because we know that the sum of the elements in row m of the Pascal triangle is 2m .

23.

By the basic principle of the harmonic triangle, the sum of all the elements of any column will be equal to the first element in the previous column. The sum of the elements in column 4 is the first element in column 3, namely 1/3.

24.

Given that yq = xp , we have qyq − 1 dy = pxp − 1 dx, so xqp −dy1 = ypq −dx1 . If we multiply this py equation by xy1 and then cancel the equal terms yq and xp , we get q ydy = p xdx , or dy dx = qx . q−p dy p To apply the transmutation theorem, we note that z = y−x dx = y−x py qx = y− q y = q y.

Therefore,

Z x0 Z Z x0 1 1 x0 q − p 1 y dx = y dx. x0 y0 + z dx = x0 y0 + 2 2 2 0 q 0 0

Solutions

101

+q −p = p2q , we have Since 1 − q2q

p+q 2q 25.

Z x0 0

1 y dx = x0 y0 2

Z x0 or

y dx = 0

qx0 y0 . p+q

We have d

x x + dx x xy + y dx − xy − x dy y dx − x dy = . − = = y y + dy y y(y + dy) y2

We can ignore the y dy in the denominator because it is infinitesimally small with respect to the y2 term.

26.

d(x3 ) = (x + dx)3 − x3 = x3 + 3x2 dx + 3x(dx)2 + (dx)3 − x3 = 3x2 dx + 3x(dx)2 + (dx)3 = 3x2 dx, since the square and cube of dx are infinitely less than dx.

27.

Using the binomial theorem, we have d(xn ) = (x + dx)n − xn = xn + nxn − 1 dx + n (n − 1 ) n − 2 n(n − 1) x (dx)2 + · · · + (dx)n − xn = nxn − 1 dx + (dx)2 + · · · + (dx)n = 2 2 n−1 nx dx, because all the powers of dx above the first are infinitely less than dx itself.

28.

If y = xx , then log y = x log x. We now take differentials of both sides: d(log y) = log x dx + x d(log x). But we know that d(log x) = dxx . Therefore, we have dyy = log x dx + x dxx = log x dx + dx or dyy = (log x + 1) dx, or finally, dy = xx (log x + 1) dx.

29.

From xx + xc = xy + y, we get d(xx ) + d(xc ) = d(xy ) + dy. Therefore, given the result of Exercise 28 and the calculation on p. 599 of the text, we have xx (log x + 1) dx + cxc − 1 dx = yxy − 1 dx + xy log x dy + dy. Reorganizing the terms, we get (xx log x + xx + cxc − 1 − yxy − 1 ) dx = (xy log x + 1) dy. (Note that in both Exercises 28 and 29, we assume for simplicity that the logarithm is the natural logarithm.)

30.

2 3 Assume that y = a + bx + cx2 + dx3 + ex4 + · · · . Then dy dx = b + 2cx + 3dx + 4ex + · · · .

2 3 1 Since x + 1 = 1 − x + x − x + · · · , we can compare these two power series term by term to get, in turn, b = 1, 2c = −1 or c = − 12 , 3d = 1 or d = 13 , 4e = −1 or e = − 41 , and so on. Since y = 0 when x = 0, we get the power series log(x + 1) = x − 12 x2 + 13 x3 − 14 x4 + · · · .

31.

2 3 We assume that x = a + by + cy2 + dy3 + ey4 + · · · . Then dx dy = b + 2cy + 3dy + 4ey + · · · .

2 3 4 dx But dy also is equal to x + 1 = a + 1 + by + cy + dy + ey + · · · . Since the logarithm of 1 is 0, we know that a = 0. Then we can compare the two series term by term. We get a + 1 = b, so b = 1. Then 2c = b, or 2c = 1, or c = 12 . Next, 3d = c, or 3d = 21 , so d = 16 .

1 Next 4e = d, or 4e = 16 , so e = 24 . Thus the power series for x + 1, the exponential 1 2 1 3 function, is x + 1 = 1 + y + 2! y + 3! y + 4!1 y4 + · · · .

32.

The derivative of the numerator is

3 2a3 − 4x3 √ − √3a 2 2 . If we evaluate this quantity 2 2a3 x−x4 3( a x)

a 4 at x = a, we get −2a2a2 − 3a 2 = − 3 a. Similarly, the derivative of the denominator is 3

3

3 3ax2 √ . Its value at x = a is − 3a = − 34 . If we divide the value of the numerator by 4 4a3 ax3 )3 the value of the denominator, we have the result 34 a · 43 = 16 9 a.

−

4(

102

Solutions 33.

We can rewrite the equation as y =

ax2 . Then dy x2 + a2 2

=

2a3 x dx . Taking another (x2 + a2 )2

differential and noting that d(dx) = 0 and d(x dx) = dx , we get d2 y =

2a3 dx2 (x2 + a2 )2 − 8a3 x2 dx2 (x2 + a2 ) . (x2 + a2 )4

For d2 y to be equal to 0, the numerator of that fraction must be 0. If we divide the 2 2 2 2 numerator q by x + a and simplify, we must solve 3x − a = 0, so the solution is x=a 34.

1 a 3 . Replacing x in the original equation by this value, we find that y = 4 .

If y = xx , then log y = log(xx ) = x log x. Taking fluxions of both sides gives us yẏ = ẋ log x + xxẋ = ẋ log x + ẋ = (1 + log x)ẋ. Thus ẏ = (1 + log x)xx ẋ.

CHAPTER 17 1.

The horizontal force at (x, y) is T(x) cos α, while that at the low point of the cord is T(0) in the opposite direction. Since these forces balance, |T(0)| = |T(x)| cos α. The downward vertical force at (x, y) is ρs, while the upward vertical force is |T(x)| sin α.

Since these balance as well, we have ρs = |T(x)| sin α. Dividing the second equation by the first, we get ρs |T(0)|

=

sin α dy = tan α = . cos α dx

It follows that dy/dx = s/a, where a = |T(0)|/ρ.

2.

If dx = √ a2 dy 2 , then y −a

Z p

x=

a dy . y2 − a2

To integrate this, we substitute y = a cosh u. Then dy = a sinh u du. So Z x=

p

a2 sinh u du a2 cosh2 u − a2

Z =

a2 sinh u du = a sinh u

Z a du = au.

Since u = cosh−1 ay , we have x = a cosh−1 ay , or ax = cosh−1 ay , or cosh ax = ay , or, finally, y = a cosh ax , the standard equation of the catenary.

3.

4.

5.

√ 2 a dy √ , then by squaring both sides we get dx2 + dy2 = a dy . This reduces to x x 2 2 x x dx2 + x dy2 = a dy2 , or x dx2 = (a − x)dy2 , or a− x dx = dy , or, finally, to dy = q x dx a−x , Johann Bernoulli’s differential equation for the brachistochrone.

If ds =

P

( bi − ai ) =

P

bi −

P

ai = B − A. ∫ R dp We can rewrite p = n dx in the form ln p = n dx, or p = e n dx . Thus, if we multiply m dx + ny dx + dy = 0 by p, we get pm dx + pny dx + p dy = pm dx + y dp + p dy. If we

Solutions R

integrate, we have y=

103

pm dx + py = k, or

Z Z ∫ ∫ 1 k − pm dx = e− n dx k − me n dx dx . p

In this particular case, m = −3x and n = 1x . Therefore, p = eln x = x. So y = 1x (k + R 2 3x dx) = kx + x2 .

6.

Given that ur = i cos θ+ j sin θ and uθ = −i sin θ + j cos θ, we get dur /dθ = −isin θ + jcos θ = uθ and duθ /dθ = −icos θ−jsin θ = −ur . Next, since r = rur , the velocity v is given by v = dr/dt = r(dur /dt) + (dr/dt)ur = r(dur /dθ)(dθ/dt) + (dr/dt)ur = r(dθ/dt)uθ + (dr/dt)ur . We next calculate the acceleration: dθ duθ dθ d2 r dr dur dθ dv dr dθ d2 θ ur + = uθ + r 2 uθ + r + dt dt dt dt dθ dt dt2 dt dθ dt dt 2 d2 θ d2 r dθ dr dθ dr dθ u + r 2 uθ − r ur + 2 ur + u = dt dt θ dt dt dt θ dt dt !

2 d2 θ dr dθ d2 r dθ ur + r 2 + 2 uθ . = −r 2 dt dt dt dt dt

a =

It follows that the radial component ar and the transverse component aθ of the acceleration are given respectively by d2 r ar = 2 − r dt

dθ dt

2 and

aθ = r

d2 θ dr dθ +2 . dt dt dt2

Since the force is central, aθ = 0. We multiply the differential equation expressing that fact by r to get r2

d2 θ dr dθ + 2r =0 dt dt dt2

or

dθ d r2 = 0. dt

It follows that r2 dθ dt = k, where k is a constant. But the area swept out by the radius vector between the angles θ0 and θ1 is given by Z θ1 A= θ0

1 2 r dθ.

2

Therefore, dA dA dθ 1 2 dθ 1 = = r = k. dt dθ dt 2 dt 2 Therefore, A = 12 kt + C and the area swept out depends only on time. This is Kepler’s law of areas.

104

Solutions 7.

Triangle SQP has vertices S = (0, 0), P = (x, y), and Q = (x + dx, y + dy). The area of the triangle is therefore 12 ((x + dx)y − (y + dy)x) = 12 (xy + y dx − yx − x dy) = 1 2 (y dx − x dy).

8.

The conversion to differentials of Newton’s description of the central force via QR and the square of the area of triangle SQP was made on p. 620. Thus, the force is proportional to p −b d2 x x2 + y2 , x(y dx − x dy)2 where we have replaced a by b in the expression to avoid confusion with the acceleration. We will show that this expression is equivalent to the expression for ar in Exercise 6, given that aθ = 0. We set x = r cos θ and y = r sin θ.

Then dx = dr cos θ − r sin θ dθ and dy = dr sin θ + r cos θ dθ. So y dx − x dy = r sin θ dr cos θ − r2 sin2 θ dθ − r cos θ dr sin θ − r2 cos2 θ dθ = − r2 dθ. This value is proportional to the infinitesimal area of the triangle SPQ in polar coordinates, and we know that that value is a constant multiple of dt; that is, −r2 dθ = k dt. It follows that (y dx − x dy)2 = k2 dt2 . We next calculate d2 x: d2 x = d2 r cos θ − dr sin θ dθ − dr sin θ − r cos θ(dθ)2 − r sin θ d2 θ = d2 r cos θ − r cos θ(dθ)2 − (r sin θ d2 θ + 2 sin θ dr dθ).

Thus, we can translate Newton’s force F as follows: F =

−b[cos θ(d2 r − r(dθ)2 ) − sin θ(r d2 θ + 2 dr dθ)]r −b

"

2

r cos θk2 dt2 2 2 #

dθ dr dθ d θ b sin θ r 2 +2 + 2 dt dt dt k cos θ dt

d r −r k2 dt2 b sin θ −b aθ = −αar , = 2 ar + 2 k k cos θ =

since aθ = 0.

9.

Given the results of Exercise 6, we see that Hermann’s equation is equivalent to −αar = r12 or to ar = − rk2 , or to d2 r −r dt2

10.

dθ dt

2

k = − 2. r

p If we rewrite the equation a ± cx/b = x2 + y2 in polar coordinates, we get a ± cr cos θ/b = r. This equation can be rewritten as

(c/b)(ab/c) c cos θ a a = 1± r or r = or r = . c cos θ b 1 ± (c/b) cos θ 1± b If we set e = c/b, this latter equation is the standard polar equation of a conic section with eccentricity e. We know in fact that if e = 1, or b = c, we get a parabola; if e < 1, or c < b, we get an ellipse; and if e > 1, or b < c, we get a hyperbola.

Solutions

105

11.

We know that a conic section can be written in polar coordinates as r = de/(1 ± e cos θ), where e is the eccentricity. If we set x = r cos θ, this equation can bep rewritten as r = de/(1 ± exp /r) = rde/(r ± ex). This is equivalent to r ± ex = de or to x2 + y2 = ± ex + de, or to x2 + y2 = αx+β. Note that since |α| = e, we know that if |α| = 1, the conic is a parabola; if |α| > 1, the conic is a hyperbola; and if |α| < 1, the conic is an ellipse. The value β determines the location of the center of the conic section.

12.

If y = ex/a , then dy = 1a ex/a dx, d2 y = a12 ex/a dx2 , and d3 y = a13 ex/a dx3 .

Therefore a3 d3 y = ex/a dx3 = y dx3 . Now we assume that e−x/a (a3 d3 y − y dx3 ) = d[e−x/a (A d2 y + B dy dx + Cy dx2 )]. If we take the differential on the right, we get −a1 e−x/a [A d2 y dx + B dy dx2 + Cy dx3 ] + e−x/a [A d3 y + B d2 y dx + C dy dx2 ] = e−x/a [A d3 y + (B − 1a A)d2 y dx + (C − 1a B)dy dx2 − 1a Cy dx3 ]. If we equate this expression with the one on the left above, we have A = a3 ; B − 1a A = B − a2 = 0, so B = a2 ; and C − a1 B = C − a = 0, so C = a. It follows that the left side above is the differential of e−x/a (a3 d2 y + a2 dy dx + ay dx2 ). But if y is a solution to a3 d3 y − y dx3 = 0, then d[e−x/a (a3 d2 y + a2 dy dx + ay dx2 )] = 0. Thus e−x/a (a3 d2 y + a2 dy dx + ay dx2 ) = k, for some k, or a3 d2 y + a2 dy dx + ay dx2 = kex/a . But if y ̸= cex/a , this is not possible, unless k = 0. Therefore, y satisfies the equation a3 d2 y + a2 dy dx + ay dx2 = 0, or a2 d2 y + a dy dx + y dx2 = 0.

13.

We know that y = ex is a solution of y′′′ − 6y′′ +11y′ − 6y = 0. So assume that e−x (y′′′ − d −x 6y′′ +11y′ − 6y) = dx [e (ay′′ +by′ +cy)] = −e−x (ay′′ +by′ +cy) +e−x (ay′′′ +by′′ +cy′ ).

It follows by comparing coefficients that a = 1; −a + b = −6, so b = −5; and −b + c = d −x ′′ [e (y − 5y′ +6y)] = e−x (y′′′ − 6y′′ +11y′ − 6y). But if y is a 11, so c = 6. Therefore dx d −x ′′ solution to the original differential equation, this implies that dx [e (y −5y′ +6y)] = 0.

So e−x (y′′ − 5y′ + 6y) = k for some k, or y′′ − 5y′ + 6y = kex . But if y ̸= cex , this is not possible, unless k = 0. Therefore, y′′ − 5y′ + 6y = 0, as claimed.

14.

The characteristic polynomial in this case is p3 − 6p2 + 11p − 6 = (p − 1)(p − 2)(p − 3). We could divide everything by −6 and get the factor of the polynomial in the form Euler wrote it as (1 − p)(1 − 21 p)(1 − 31 p). It then follows that independent solutions are Aex , Be2x , and Ce3x , and any solution is simply a combination of these.

15.

Suppose y = ueαx satisfies the differential equation a2 d2 y + a dy dx + y dx2 = 0.

We calculate that dy = eαx du + αueαx dx and that d2 y = αx 2 αx αx 2 αx e d u + αe dx du + αe du dx + α ue dx2 . Therefore, a2 d2 y + ady dx + y dx2 = a2 eαx d2 u + (2αa2 eαx + aeαx ) du dx + (a2 α2 ueαx + aαueαx + ueαx )dx2 . To eliminate the 1 term in du dx, we must have 2αa2 + a = 0, or α = − 2a . Then the coefficient of the dx2 αx 2 1 αx 1 3 term is e (a 4a2 − a 2a + 1)u or 4 ue . Dividing through by eαx , we find that u must be a solution to a2 d2 u + 43 u dx2 = 0.

16.

To solve the equation, we first multiply by du to get a2 du d2 u = − 34 u dudx2 . Since the differential of du2 is 2 du d2 u, we can integrate this equation with respect to u to get 2 2 2 2 2 2 2 2 a2 3 2 2 du = (C − 8 u )dx , or 4a du = (K − 3u )dx . This equation simplifies to 2a dx = √ du.

2 K − 3u2

106

Solutions By making the substitution u = √K sin v, we can integrate this differential equation: 3

Z

Z

K 2a √ cos v dv p 2 − K2 sin2 v K2 − 3u2 3 K √ Z 2a 2a 2a 3u =√ dv = √ v − f = √ arcsin − f.

K 3 3 3

x =

√

2a

du =

2a It follows that x + f = √ arcsin 3

√

√ √ 3u 3 3u K or 2a (x + f) = arcsin K , or, finally, that

u = C sin

√ (x + f) 3

2a

√

17.

.

√

The three cube roots of 1 are 1, −1+2 −3 , and −1−2 −3 . Then,

ln(1) = ± 2nπi,

√ 2 ln −1+2 −3 = ± 2n πi, and 3

2 √ ln −1−2 −3 = − ± 2n πi.

3

18.

For the primitive fifth roots of 1, it is easier to note that they are given by cos ϕ + i sin ϕ, with ϕ = ± 52 π and ϕ = ± 45 π. It follows that the logarithms are ± 25 ± 2n πi and

± 54 ± 2n πi, respectively. p ′2 We begin with the equation F − y′ (∂ f/∂ y′ ) = c, or y 1 + y′2 − √yy ′2 = c. If we 1+y p p p multiply by 1 + y′2 , we get y(1 + y′2 ) − yy′2 = c 1 + y′2 , or y = c 1 + y′2 . Squaring p 2 c2 gives us y2 = c2 (1 + y′2 ) or y′2 = y − , so y′ = 1c y2 − c2 . If we rewrite this in c2 p R differential form, we get dy = 1c y2 − c2 dx, or dx = √ c2dy 2 . Thus x = √ 2c 2 dy. y −c

y −c

We can solve this with the substitution y = c cosh u. Then dy = c sinh u du, so Z x=

c2 sinh u du p = c2 cosh2 u − c2

Z c du = cu.

Thus, x = c cosh−1 cy , or xc = cosh−1 cy , or, finally, y = c cosh xc , the equation of the catenary.

19.

Any given curve f(x, y, α) = 0 in the family of curves satisfies the differential equation ∂f ∂f ∂ x dx + ∂ y dy = 0. The tangent line to a curve orthogonal to the given curve will have a slope that is the negative reciprocal of the slope of the tangent line to the given curve. Therefore, the differential equation it must satisfy is given by ∂∂yf dx − ∂∂xf dy = 0.

In the case of the hyperbolas x2 − y2 = a2 , the original differential equation is 2x dx − 2y dy = 0. Thus the equation of an orthogonal curve is 2y dx + 2x dy = 0, or y dx + x dy = 0. But this differential equation can be rewritten in the form d(xy) = 0, whose solutions are xy = k. Thus all of these hyperbolas are orthogonal to the original ones.

Solutions 20.

107

√

Since the differential equation for the family of brachistochrones is a√− x dy = √ x dx, the differential equation for the orthogonal family is x dy = − a − x dx.

We can rewrite this in the form

√

dy =− dx

r

√

a−x ax − x2 =− . x x

We therefore need to integrate this last expression. We rewrite this as q 2 2 Z Z √ − a4 + ax − x2 + a4 2 ax − x dx = − dx − x x Z p a 2 ( 2 ) − (x − 2a )2 =− dx x Z p a 2 ( 2 ) − u2 =− du, u + a2 where u = x − a2 . We can determine this latter integral by using a trigonometric substitution: u = a2 sin t, du = a2 cos t dt. The integral then becomes −

Z p a 2

( 2 ) − ( a2 sin t)2 a cos t dt = − a 2 2 (sin t + 1)

Z

a =− 2

( a2 )2 cos2 t dt a =− a 2 ( 1 + sin t ) 2

Z

(1 − sin2 t)dt

1 + sin t a (1 − sin t)dt = − (t + cos t).

2

a/2) ) and cos t = Given that t = arcsin (a/u 2) = arcsin( x−( (a/2) solution to the differential equation in the form

a y = − arcsin 2

Z

x − (a/2) a/2

−

√ ax−x2 , we finally have the (a/2)

p ax − x2 .

21.

3x2 y2 + 4x3 y + 3, we have ∂∂Py = R 6xy2 + 12x2 y = ∂∂Qx . Therefore, we define f(x, y) to be (2xy3 + 6x2 y2 + 8x)dx + r(y) = x2 y3 + 2x3 y2 + 4x2 + r(y). Taking the derivative of this function with respect to y and comparing it to Q shows that r′ (y) = 3, so r(y) = 3y and the solution to the differential equation is the equation x2 y3 + 2x3 y2 + 4x2 + 3y = k.

22.

If ax = y2 and by = z2 , then z = (by)1/2 = (b(ax)1/2 )1/2 = b1/2 a1/4 x1/4 . Since y = (ax)1/2 , we have dy = 12 a1/2 x−1/2 dx. Therefore, z dy = 12 b1/2 a3/4 x−1/4 dx, and R R z dy = 21 b1/2 a3/4 x−1/4 dx = 32 b1/2 a3/4 x3/4 . We next integrate this last expression with respect to x between 0 and x0 to get

Since P

=

2xy3 + 6x2 y2 + 8x and Q

Z x0 V= 0

=

8 1/2 3/4 7/4 2 1/2 3/4 3/4 b a x dx = b a x0 .

3 21

108

Solutions The modern method would be to evaluate a triple integral: Z x0 Z √ax Z √by 0

23.

If x = √ t

0

1 + u2

0

Z x0 Z √ax p dz dy dx = by dy dx 0 0 Z x0 √ 2 8 1/2 3/4 7/4 = b(ax)3/4 dx = b a x0 .

3 21 0

, and y = √ tu 2 , then 1+u

∂x ∂y t −tu u ∂x ∂y 1 √ − − =√ 2 )3/2 2 )3/2 2 ∂t ∂u ∂u ∂t ( 1 + u ( 1 + u 1+u 1 + u2

=

t tu2 t + = .

2 2 (1 + u ) (1 + u2 )2 1 + u2

It follows that dx dy =

t dt du .

1 + u2

24.

We suppose that y(t, x) = Ψ(t + x) − Ψ(t − x). Since y(0, x) = f(x), we have f(x) = Ψ(x) − Ψ(−x). Since y(t, 0) = y(t, l) = 0, we have Ψ(t + l) − Ψ(t − l) = 0, or Ψ(t + l) = Ψ(t − l). We note that f(−x) = Ψ(−x) − Ψ(x) = −f(x), so f(x) is an odd function. Also, f(x − l) = Ψ(x − l) − Ψ(−x + l) = Ψ(x + l) − Ψ(−x − l) = f(x + l), so f(x) is periodic of period 2l. Since g(x) = Ψ′ (x) − Ψ′ (−x), we have g(−x) = Ψ′ (−x) − Ψ′ (x) = −g(x), so g(x) is also an odd function. Finally, since Ψ is periodic, so is its derivative. Therefore, g(x + l) = Ψ′ (x + l)− Ψ′ (−x − l) = Ψ′ (x − l)− Ψ′ (−x + l) = g(x − l), and g(x) is periodic of period 2l.

25.

If y = F(t)G(x), then ∂∂yt = F′ (t)G(x), ∂∂ t2y = F′′ (t)G(x), ∂∂ yx = F(t)G′ (x), and ∂∂ xy2 = F(t)G′′ (x). The original partial differential equation then implies that F′′ (t)G(x) = ′′ ′′ F(t)G′′ (x), or that FF = GG . Since the left-hand expression is a function of t and the right hand one is a function of x, each of these quotients must be equal to a constant, say C. So F satisfies the ordinary differential equation F′′ = CF and G also satisfies G′′ = CG. The solution of these equations is the sum of real exponential functions, if C is positive, or the sum of sines and cosines, if C is negative. But since y(t, 0) = y(t, l) = 0, we have F(t)G(0) = 0 and F(t)G(l) = 0. This implies that both G(0) and G(l) = 0, and this could functions.

Thus C < 0, F(t) = √ if G were a sum of real exponential √ √ √ not happen cos −Cx + d sin −Cx. To rewrite the first a cos −Ct + b sin −Ct, and G(x) = c √ of these expressions in the√form A cos( −Ct√− ω), note that this latter expression can be expanded as A cos −Ct cos ω + A sin −Ct sin ω. Therefore, we must have A cos ω = a and A sin ω = b. This means that tan ω = ba and that A2 = a2 + b2 . Thus we can find the appropriate A and ω. A similar argument shows that we can rewrite G(x) √ in the form B sin( −Cx − θ).

26.

Let the sides of the isosceles triangle have length s and the base have length b. If we draw radii from the center of the inscribed circle to the two sides and the base, the radius to the base divides the base into two segments of length b2 , while the radii to the sides divide each side into two segments of length b2 and s − 2b . By drawing

2

2

Solutions

109

lines from the three vertices to the center of the circle, we see that the area of the triangle is given by A = 4 b4 + 2 s−2b/2 = s + b2 . Also, by Heron’s formula, the area q q 2 of the triangle can be expressed as A = (s + b2 )( b2 )( b2 )(s − b2 ) = 2b s2 − b4 . If we solve the first area expression for s = A − b2 and substitute in the second, we get A = q √ 2 b (A − b2 )2 − b4 = b2 A2 − bA. If we square both sides of this equation and simplify, 2 3

we get 4A2 = b2 A2 − b3 A, so (4 − b2 )A2 + b3 A = 0, or A = b2 b− 4 . To minimize A, we

b − 12b take the derivative with respect to b. We get dA db = (b2 − 4)2 . The derivative is 0 when 4

2

√

√

√

b4 − 12b2 = 0, or when b2 = 12. So b = 2 3. Then the area A = 248 3 = 3 3, and the √ side s = A − b2 = 2 3.

27.

The volume V of a cone is given by V = 13 πr2 h, where r is the radius of the base and h is the height. If we designate the slant√height by ℓ, then ℓ2 = h2 + r2 . The surface area S is given by S = 12 ℓ2πr = ℓπr = πr h2 + r2 . If we solve the volume formula for h and substitute in the surface area formula, we get r S = πr

9V2 πr p + r2 = 2 9V2 + π2 r6 = 2 4 π r πr

√

9V2 + π2 r6 . r

To minimize S, we take its derivative and set it equal to 0. We have dS 2√9V2 +π2 r6 (6π r ) − = dr r2 r

2 5

√

9V2 + π2 r6

2 6

.

If the derivative is to be equal to 0, we must have √ 3π2 r 2 6 − 9V +π r

√

9V2 + π2 r6 = 0, or

√ , 3π r − (9V + π r ) = 0, or 2π r = 9V . We can solve this for r to get r3 = 3V π 2 and r √ 3 3 2V r= .

2π 2 6

2

2 6

2 6

2

We then have √

h= 28.

3Vr 3Vrπ 2 √ 3V = 3 = = 2r.

2 3πV πr πr

To find the extreme values of w = (b3 − x3 )(x2 z − z3 )(xy − y2 ), we need to calculate the three partial derivatives of w, set them each equal to zero, and solve the three equations simultaneously. We get ∂w = −3x2 (x2 z − z3 )(xy − y2 ) + (b3 − x3 )(2xz)(xy − y2 ) + (b3 − x3 )(x2 z − z3 )y; ∂x ∂w = (b3 − x3 )(x2 z − z3 )(x − 2y); ∂y ∂w = (b3 − x3 )(x2 − 3z2 )(xy − y2 ). ∂z

110

Solutions If we set ∂∂wy = 0 and ∂∂wz = 0, and assume that at the extreme value w ̸= 0, we get √ x = 2y and x = 3z, or y = 2x and z = √x . Substituting these values into the expression 3

for ∂∂wx and then setting that equal to 0 gives us

−3x2

x3 3

x3 3 3

√ − √

2

2 2

x x2 2x x x2 − + (b3 − x3 ) √ − 2 4 2 4 3

3 3 x x x = 0. + (b3 − x3 ) √ − √ 2 3 3 3

This simplifies to 4

x7 x x4 3 3 − √ + (b − x ) √ + √ =0 2 3 2 3 3 3

− 3x7 + (b3 − x3 )5x4 = 0

or to √

3

√

√ 3

or finally to 8x7 = 5b3 x4 . Thus x3 = 5b8 , and x = 12 b 3 5. Then y = 14 b 3 5 and z = b√5 .

2 3

29.

As noted in the text, if y = cos z, we have żÿ2 = − ay2 . Since y = a when z = 0, we have ˙

Ë = − 1a . Now, taking further fluxions of the above expression, we get żÿ2 = − aẏ2 . But ¨ = − −1/a = 1 . Similarly, we ẏ = 0 when z = 0, so Ë˙ = 0. Then żÿ2 = − aÿ2 . Thus Ë a2 a3 ¨

˙

¨

¨ = 0 and Ë ¨ = − 1 . It follows that the first four non-zero terms of the calculate that Ë a5 1 2 power series for y = cos z are y = a − 2a z + 4!a1 3 z4 − 6!a1 5 z6 .

30.

If the curve is defined by the differential equation a dyy = dx, then we can write the curve in the form x = a ln y. Since x′ = ay and x′′ = − ya2 , the curvature of the curve

is given by K = (1 +−a2a//yy2 )3/2 . To find the maximum curvature, we take the derivative of K with respect to y and set it equal to 0. The numerator of the derivative is (1 + 2 3a a2 1/2 2a2 a2 3/2 2a ) ( y3 ) − 2y ( y3 ). If we factor out (1 + ay2 )1/2 , then the other factor, the 2 (1 + y2 ) y2 2

2

2

3a 2a one which must be equal to 0, is (1 + ay2 )( 2a ) − 2y 2 y3 . After simplification, we need to y3 3

3

√

2

solve the equation 2a = ay5 . After multiplying by ya , we get 2 = ay2 , and y = √a = a 2 2 . y3 2

31.

32.

Let the rectangle be placed so its vertices are at (0, 0), (a, 0), (0, b), and (a, b), and let the line be drawn through (0, 0), intersecting the extensions of the two opposite sides at (−u, b), and (a, −t). Since ub = at , we have u = abt , so the two points at the end of the desired line segment are (− abt , b) and (a, −t). The square of the length of this line segment is given by L = (a+ abt )2 + (−t − b)2 . To find the value of t which minimizes L, we take the derivative: L′ = 2(a + abt )(− ab ) + 2(t + b). The equation L′ = 0 can then t2 4 3 2 2 2 be written as t + bt − a bt − a b = 0. The polynomial on the left factors, so we get (t3 − a2 b)(t + b) = 0.√Since t = −b does not give a solution to the problem, we must √ 3 3 have t3 = a2 b, or t = a2 b. Therefore, u = ab2 , and the desired line segment – which √ √ 3 3 must have a minimum length – goes through the point (a, − a2 b) and (− ab2 , b).

The curve y2 = 4(2 x− x) is defined for 0 < x ≤ 2. For each x value in that interval, there are two symmetrically placed y values, so the curve is symmetric about the x axis.

As x approaches 0, y2 approaches infinity; thus the curve is asymptotic to the y axis.

Solutions

111

To find the area bounded by the curve and the y axis, it is easiest first to solve for x and then integrate. Thus, we rewrite the equation of the curve in the form xy2 = 8 − 4x and solve this as x = y2 8+ 4 . Then the area is given by

∞ 8 1 y π −π dy = 8 =4 −4 arctan = 4π.

2 2 2 −∞ 2 2 −∞ y + 4

Z ∞ A= 33.

Given that 1x is the annual rate of growth, we need to solve the equation 6 1,000,000. Therefore, 1+x = x

1,000,000 6

1 200

or

log

1+x x

=

1 + x 200 = x

1 log(166,667) = 0.0261092.

200

So 1 +x x = 1.061963, or 0.061963x = 1, and x ≈ 16. Thus the annual rate of growth of 1 the population is approximately 16 , or about 6.2%.

2

3

4

34.

If y = ln(1 + x), then y = 1x − x2 + x3 − x4 + · · · , while in general the series for 2 3 4 y = loga (1 + x) is given by y = 1k ( 1x − x2 + x3 − x4 + · · · ). Thus loga (1 + x) = 1k ln(1 + x).

If we take 1 + x = a, then 1 = 1k ln a, or k = ln a.

35.

We have ln(1 + x) = j(1 + x)1/j − j. But 1

1 j

(1 + x) j = 1 + x−

1(j − 1) 2 1(j − 1)(2j − 1) 3 1(j − 1)(2j − 1)(3j − 1) 4 x + x − x +· · · . j · 2j j · 2j · 3j j · 2j · 3j · 4j

1 1 2j−1 2 3j − 1 3 Since j is infinite, j − 2j = 2 , 3j = 3 , 4j = 4 , and so on. Therefore, 1

j(1 + x) j = j +

x x2 x3 x4 − + − + ··· .

1 2 3 4

If we subtract j from this series, we get the standard power series for ln(1 + x).

36.

x

−x

2

4

6

We begin with e +2 e = 1 + x2! + x4! + x6! + · · · . This can be factored as the product

iz −ix 4x2 4x2 4x2 1 + 9π 1 + 25π · · · since the zeros of e +e2 = cos x are given by x = 1 + π2 2 2 2 2 (2n + 1)π , and, in the original infinite series, 1 + (2n 4x is 0 precisely when (2n 4x = 2 + 1) π 2 + 1) π 2 (2n + 1)πi 2 2x −1, or (2n + 1)π = i, or when x = . Therefore the coefficient of the x term 2 4 in the infinite product, which equals π42 + 9π4 2 + 25π 2 + · · · , is equal to the coefficient 2 1 1 1 of x in the infinite series expansion, that is 2 . Therefore, π12 + 9π1 2 + 25π 2 + ··· = 8, 1 1 π2 ex −e−x or 1 + 9 + 25 + · · · = 8 . Next, if we substitute ix for x into the equation = 2 x5 x3 x + 1·2·3 + 1·2·3·4·5 + · · · , we obtain

eix − e−ix ix3 ix5 = ix − + + ··· , 2 1·2·3 1·2·3·4·5 or, since the left-hand side is equal to i sin x, sin x = x −

x3 x5 x2 x4 + −··· = x 1− + − ··· .

1·2·3 1·2·3·4·5 1·2·3 1·2·3·4·5

112

Solutions We can get the infinite product representation by replacing x by ix in the infinite prodx −x uct representation for e −2e or, alternatively, by noting that the roots of sin x are 0, ± π, ± 2π, . . .. In either case, the representation is

x2 x2 x2 1− 2 1− 2 ··· . sin x = x 1 − 2 π 4π 9π We know that the coefficient of x4 in the polynomial is the sum of the products of the coefficients of x2 taken two at a time in the infinite product. Recalling Exercise 7 of Chapter 14, we also know that the sum of the squares of the coefficients of x2 in the infinite product equals the square of the sum of the coefficients of x2 less twice the sum of the products of those coefficients taken two at a time. But these latter sums are coefficients in the polynomial. Therefore, 2 2

∞ X 1 1 1 = − −2 .

6 120 k2 π 2 k=1

Simplifying, we get ∞ X 1 1 1 1 − = = k4 π4 36 60 90 k=1

37.

or

∞ X 1 k=1

k

= 4

π4 .

90

We find ∂∂Vx = 3x2 − 3y + 32 and ∂∂Vy = 2y − 3x. We set each of these expressions equal to 0 and solve the system. Namely, we set y = 32 x and substitute, giving 3x2 − 29 x + 32 = 0. This equation reduces to 2x2 − 3x + 1 = 0, whose solutions are x = 1 and x = 3 1 3 1 2 . Thus, the two possibilities for extreme points are (1, 2 ) and ( 2 , 4 ). To check on whether these points are maxima, minima, or neither, we need to consider the second 2 2 2 derivatives. We have ∂∂ xV2 = 6x, ∂∂x∂Vy = −3, and ∂∂ yV2 = 2. We therefore look at the 2

2

2

2

expression ∂∂ xV2 ∂∂ yV2 − ( ∂∂x∂Vy )2 = 12x − 9. When x = 1, this value is 3; since ∂∂ xV2 = 6, we conclude that (1, 32 ) is a minimum point. On the other hand, when x = 12 , the value is −3, which indicates that the point is neither a maximum nor a minimum.

38.

If with legs 1 and x and hypotenuse √ y = arctan x, then by looking at the right triangle 1 + x2 , we get sin y = sin(arctan x) = √ x 2 and cos y = cos(arctan x) = √ 1 2 .

1+x 1+x p If p = √ x 2 = sin y, then dp = (1 + dx and 1 − p2 = cos y = √ 1 2 . Since x2 )3/2 1+x

y = arcsin p, we also know that dy = √ dp √

1 − p2

1+x

. If we substitute in this expression for

1 + x2 = 1 dx dp, we get dy = (1 + dx . + x2 x2 )3/2 39.

2

2

If y = ax , then dy = ax + dx − ax = ax (adx − 1) = ax (ln a dx + (ln a2) dx + · · · ) = ax ln a dx+ terms in higher powers of dx. Therefore, dy = (ln a)ax dx.

Solutions 40.

113

If y = tan x, then tan x + tan dx − tan x 1 − tan x tan dx 2 tan x + tan dx − tan x + tan x tan dx tan dx + tan2 x tan dx = = 1 − tan x tan dx 1 − tan x tan dx 2 2 (1 + tan x) tan dx (1 + tan x)dx = = = (1 + tan2 x)dx, 1 − tan x tan dx 1 − tan x dx

dy = tan(x + dx) − tan x =

because tan dx = dx and tan x dx is infinitely small with relation to 1.

41.

If we first set 1 − x√+ x2 = a2 − 2abx cos ζ + b2 x2 , we get a = 1, b = 1, and cos ζ = 12 .

Therefore sin ζ = 23 . To integrate the given expression, we also note that A = B = 1, so Z 1 + 1/2 x − 1/2 1+x 1 dx = ln(1 − x + x2 ) + √ arctan √ 2 1 − x + x2 3/2 3 /2 √ 1 2x − 1 2 √ = ln(1 − x + x ) + 3 arctan .

2 3

42.

We first factor: 2 + 5x + 3x2 = (1 + x)(2 + 3x). Using Euler’s notation, we have a = 1, b = 1, f = 2, g = 3. So we set (1 + x)(2 + 3x) = (1 + x)2 z2 , or 2 + 3x = (1 + x)z2 . Thus, x = (z2 − 2)/(3 − z2 ), 1 + x = 1/(3 − z2 ), dx = 2z dz/(3 − z2 )2 , and therefore, √

Z Z Z dx 2 dz 1 dx 3+z √ = = = √ ln √ .

(1 + x)z 3 − z2 3 3−z 2 + 5x + 3x2 Therefore, we get Z

dx 1 √ = √ ln 2 3 2 + 5x + 3x

√ √ √

3 1 + x + 2 + 3x √ √ √ .

3 1 + x − 2 + 3x

43.

According to d’Alembert, a magnitude y is the limit of a magnitude x, when x may approach y within any given magnitude, though it may not exceed y. If we think of x as being a function of t, then we could translate this into the statement that lim x(t) = y if x(t) < y for all t and, given any ϵ > 0, there is a value t such that y − x(t) < ϵ. This is quite different from the modern idea of a limit, partly in that we usually define the limit of x(t) as t approaches some particular quantity and also in that we do not have any such restriction that x(t) is always less than y.

44.

We have P(x, i) =

√ 1 √ i 1 ( x + i − x) = √ √ . i i x+i + x

1 Therefore p = 2√ . Next, x

√

√ 1 1 1 1 x − x+i √ √ √ − = √ √ √ i x + i + x 2 x i 2 x( x + i + x) 1 −i 1 = =− √ √ √ √ √ √ . i 2 x( x + i + x)2 2 x( x + i + x)2

Q(x, i) =

114

Solutions It follows that q = − 8x1√x . Finally, √

√ 1 1 1 1 −4x + ( x + i + x)2 √ √ − √ √ = + √ √ √ i " 2 x( x + i + x)2# 8x x i 8x x( x + i + x)2 p 1 −2x + i + 2 x(x + i) = . √ √ √ i 8x x( x + i + x)2

R(x, i) =

It follows that r = 8x√x(12√x)2 = 32x12 √x .

45.

If f(x) = e−1/x , for x ̸= 0, with f(0) defined to be 0, we find that f ′ (x) = 2x−3 e−1/x for 2 2 x ̸= 0. Similarly, f ′′ (x) = (4x−6 −6x−4 )e−1/x , f ′′′ (x) = (8x−9 −36x−7 +24x−5 )e−1/x , 2 and, in general, f (n) (x) will be a polynomial in x−1 multiplied by e−1/x . We will show for each n that the nth derivative of f(x) at x = 0 is 0. For the first derivative, we look at the absolute value of the difference between 0 and the differential quotient at 0 and show that it approaches 0 with x. This absolute value of the difference is 2

2

2 e−1/x − 0 = x−1 e−1/x . x−0 2

0−

Since ez is larger than 1 + z for z positive, we have for x ̸= 0, 2 1/x2

x e

1 > x 1 + 2 = x2 + 1 > 1, x 2

and so |x| >

1

= x−1 e−1/x .

2

|x|e1/x

2

It follows that the absolute value of the difference between 0 and the differential quotient at 0 is less than |x| and therefore that this difference goes to 0 with x. Thus, the derivative of f(x) is 0 at 0. By a similar argument, we can show that for any positive 2 integer k, x−k e−1/x can be made as close as we want to 0 by taking x sufficiently close to 0. It then follows, since the difference between 0 and the nth differential quotient of 2 f(x) is a polynomial in x−1 multiplied by e−1/x , that the nth derivative of f(x) is 0 for every n. So if we set up the power series for f(x), all of the coefficients of the powers of x would be 0. Nevertheless, the function is not identically 0.

46.

Given that f(x + i) = f(x) + pi + qi2 + ri3 + · · · , we can take the derivative with respect to i. We get f ′ (x + i) = p + 2qi + 3ri2 + · · · . If we set i = 0, we have f ′ (x) = p. Next, take the second derivative. We have f ′′ (x + i) = 2q + 6ri + · · · . Again setting i = 0, we get f ′′ (x) = 2q, or q = f ′′ (x)/2. If we take the third derivative, we have f ′′′ (x + i) = 6r + · · · . If we again set i = 0, we get f ′′′ (x) = 6r, or r = f ′′′ (x)/6. We can continue this process indefinitely.

Solutions

115

CHAPTER 18 1.

We calculate X 8 1 1 2 8·7·6 5 8·7·6·5·4 3 j = n9 + n8 + n7 − n + n + B8 n 9 2 3 2 · 3 · 4 · 30 2 · 3 · 4 · 4 · 6 · 42 1 1 2 7 2 = n9 + n8 + n7 − n5 + n3 + B8 n.

9 2 3 15 9 7 1 − 92 = − 30 . Next, Thus B8 = 1 − 19 − 12 − 23 + 15

X 10 1 11 1 10 5 9 10 · 9 · 8 7 10 · 9 · 8 · 7 · 6 5 j = n + n + n − n + n − 11 2 6 2 · 3 · 4 · 30 2 · 3 · 4 · 5 · 6 · 42 10 · 9 · 8 · 7 · 6 · 5 · 4 3 n + B10 n, 2 · 3 · 4 · 5 · 6 · 7 · 8 · 30 or X 10 1 11 1 10 5 9 1 j = n + n + n − n7 + n5 − n3 + B10 n.

11 2 6 2 1 5 Thus B10 = 1 − 11 − 12 − 65 + 1 − 1 + 21 = 66 . Finally,

X 12 1 13 1 12 11 12 · 11 · 10 9 12 · 11 · 10 · 9 · 8 7 j = n + n +n − n + n − 13 2 2 · 3 · 4 · 30 2 · 3 · 4 · 5 · 6 · 42 12 · 11 · 10 · 9 · 8 · 7 · 6 5 12 · 11 · 10 · 9 · 8 · 7 · 6 · 5 · 4 · 5 3 n + n + B12 n, 2 · 3 · 4 · 5 · 6 · 7 · 8 · 30 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 · 66 or X 12 1 13 1 12 11 11 9 22 7 33 5 5 3 j = n + n + n − n + n − n + n + B12 n.

13 2 6 7 10 3

2.

3.

1 22 33 5 691 Therefore, B12 = 1 − 13 − 12 − 1 + 11 6 − 7 + 10 − 3 = − 2730 .

P 4 1 5 1 4 1 3 1 P 5 1 6 1 5 5·1 4 We have j = 5 n + 2 n + 3 n − 30 n. Also, j = 6 n + 2 n + 2·6 n − 2·53··44··330 n2 = 1 5 5 4 1 2 1 6 off the sum of the tenth powers from the previ6 n + 2 n + 12 n − 12 n . We can read P 5 1 11 5 ous exercise, since B10 = 66 . We get j10 = 11 n + 12 n10 + 56 n9 − n7 + n5 − 12 n3 + 66 n.

33 30 27 21 15 9 3 1 1 5 1 5 If we set n = 1000, we get 11 10 + 2 10 + 6 10 − 10 + 10 − 2 10 + 66 10 . A careful calculation, first dealing with the three terms which themselves are not integers, then adding the two remaining integral terms and finally subtracting the other two terms, gives the solution: 91,409,924,241,424,243,424,241,924,242,500.

The power series with the Bernoulli numbers can be written as X Bi i 1 1 2 1 1 1 5 x = 1− x+ x − x4 + x6 − x8 + x10 i!

2 6 · 2!

30 · 4!

42 · 6!

30 · 8!

66 · 10!

691 x12 + · · · . − 2730 · 12!

116

Solutions The power series for ex − 1 is given by ex − 1 = 1 + x +

1 2 1 3 1 4 x + x + x + ··· .

2!

3!

4!

If we multiply these two power series together, we get x. Note that in the product, the 1 coefficient of the x2 term is 21 − 12 = 0, the coefficient of the x3 term is 61 − 14 + 12 = 0, 4 5 1 1 1 the coefficient of the x term is 24 − 12 + 24 = 0, the coefficient of the x term is 1 1 1 1 120 − 48 + 72 − 720 = 0, and so on.

4.

5.

6.

7.

There is only one way of getting three successes in three tries, namely by being successful each time. Since the probability of one success in one try is a, the probability of three consecutive successes is a3 . There are three ways of getting two successes in three tries, namely, the failure could be on the first, the second, or the third try.

Since the probability of success in any one try is a and the probability of failure is b, the probability of any one of these methods happening is a2 b. Therefore, the probability of two successes in three tries is 3a2 b. The probability of one success in three tries is the same as that of two failures in three tries, so is equal to 3ab2 . Finally, the probability of no successes is the same as that of three failures, namely b3 .

n Since there are n− r ways of getting r successes in n tries, because each set of r successes corresponds to n − r failures, and since the probability of any one of these ways of r successes and n − r failures is ar bn−r , it follows that the total probability for n r n−r . r successes in n tries is n− r a b 1 4 2 6 1 2 If a = 3 , b = 3 , and n = 10, we calculate that P(S = 4) = 10 6 ( 3 ) ( 3 ) = 210 ·

10 1 5 2 5 13,440 64 1 32 8064 1 81 · 729 = 59,049 . Also, P(S = 5) = 5 ( 3 ) ( 3 ) = 252 · 243 · 243 = 59,049 . Finally, 10 1 6 2 4 1 16 3360 P(S = 6) = 6 ( 3 ) ( 3 ) = 210 · 729 · 81 = 59,049 . Therefore, P(4 ≤ S ≤ 6) = 13,440 24,864 8064 3360 59,049 + 59,049 + 59,049 = 59,049 = 0.42.

We calculate log c(s − 1) log(19,000) = = 300.465 log(r + 1) − log r log 31 − log 30 and log c(r − 1) log(29,000) = = 210.597. log(s + 1) − log s log 21 − log 20 Thus, we take m = 301 and n = 211. Then mt +

st(m − 1) 20 · 50 · 300 = 301 · 50 + = 24,727.419 r+1 31

and nt +

rt(n − 1) 30 · 50 · 210 = 211 · 50 + = 25,550. s+1 21

Since the latter number is the larger, we have N(1000) = 25,550.

Solutions 8.

117

We calculate log c(s − 1) log(190,000) = = 370.687 log(r + 1) − log r log 31 − log 30 and log c(r − 1) log(290,000) = = 257.79. log(s + 1) − log s log 21 − log 20 Thus, we take m = 371 and n = 258. Then mt +

st(m − 1) 20 · 50 · 370 = 371 · 50 + = 30,485.5 r+1 31

and nt +

rt(n − 1) 30 · 50 · 257 = 258 · 50 + = 31,257.14. s+1 21

Since the latter number is the larger, we have N(10,000) = 31,258.

9.

Let P(S) be the probability of A winning, given a current score S. We know that there are two possibilities for the next score SA and SB , depending on whether A or B wins the next point. As a general rule, very analogous to Huygens’s third proposition (p. 523), we know that P(S) = P(S → SA )P(SA ) + P(S → SB )P(SB ), where P(S → SA ) is the probability that the next point is won by A, with an analogous definition for P(S → SB ). In this situation, the probability that A wins the next point is 1/2, with the same for B. Thus P(15 : 30) = (1/2)P(30 : 30) + (1/2)P(15 : 40). We know that P(30 : 30) = 1/2, since the players are evenly matched. Then P(15 : 40) = (1/2)P(30 : 40) + (1/2) · 0, since if B wins the next point, B wins the set. We also get P(30 : 40) = (1/2)P(40 : 40) + (1/2) · 0, and P(40 : 40) = 1/2. Putting all this together, we have P(15 : 30) = 1/4 + (1/2)P(15 : 40) = 1/4 + (1/2)(1/2)P(30 : 40) = 1/4 + (1/2)(1/2)(1/2) P(40 : 40) = 1/4 + (1/2)(1/2)(1/2)(1/2) = 1/4 + 1/16 = 5/16.

10.

If A is twice as strong a player as B, this means that at any given score S, P(S → SA ) = 2/3 and P(S → SB ) = 1/3, using the notation of the previous solution.

Let x = P(30 : 30). Then x = P(S → SA )P(40 : 30) + P(S → SB )P(30 : 40) = (2/3) P(40 : 30) + (1/3)P(30 : 40). Now, P(40 : 30) = (2/3) · 1 + (1/3)P(40 : 40), because a win for A at this point wins the game. On the other hand P(40 : 40) = P(30 : 30) = x. So P(40 : 30) = 2/3 + (1/3)x. Similarly, P(30 : 40) = (2/3)x + (1/3) · 0. Putting the pieces together, we have x = (2/3)(2/3 + (1/3)x) + (1/3)(2/3)x = 4/9 + (4/9)x. Therefore, (5/9)x = 4/9 and x = P(30 : 30) = 4/5. For the second case, P(15 : 30) = (2/3)P(30 : 30) + (1/3)P(15 : 40). We already know that P(30 : 30) = 4/5. On the other hand, P(15 : 40) = (2/3)P(30 : 40) + (1/3) · 0, and P(30 : 40) = (2/3)P(40 : 40) + (1/3) · 0 = (2/3)(4/5). Therefore, P(15 : 30) = (2/3)(4/5) + (1/3)(2/3)(2/3)(4/5) = 8/15 + 16/135 = 88/135.

11.

log 2 According to De Moivre’s procedure, the number of trials x is given by log 10 −log 9 = 6.6. For the approximation procedure, we take q = 9, so x = 0.7 × 9 = 6.3. In either case, this means that in 6 trials the odds are slightly less than even, while in 7 trials, they are more than even.

118

Solutions 12.

1 . Therefore, the number of throws The probability of throwing three ones is ( 16 )3 = 216 2 necessary to insure even odds that three ones will occur is given by log 216log −log 215 = 149.37; thus the odds are slightly better than even with 150 throws.

13.

1 The probability of winning the prize on one ticket is 40 . Therefore, to insure even log 2 odds, one must buy x = log 40−log 39 = 27.37, that is, 28 tickets.

14.

Since bx + xabx − 1 represents the number of chances in which the event will succeed x x− 1 . Since we no more than once, the probability that this will happen is given by b (+a xab + b )x

want this probability to be equal to 12 , we get the equation (a + b)x = 2bx + 2xabx − 1 . x If we set a : b = 1 : q, we can rewrite this in the form ( bq + b)x = 2bx + 2xb q or 1 x (1 + q1 )x = 2 + 2x q . Taking logarithms, we get x log(1 + q ) = log 2 + log(1 + q ). If we assume that q is large, then the left side of this equation can be approximated by using just the first term of the power series for the logarithm: x · 1q . Thus, if we write z = qx , we have z = log 2 + log(1 + z). The easiest way to solve this equation is by using a graphing calculator. We find that z ≈ 1.678, so x ≈ 1.678q.

15.

We have col. 1 =

2 s2 + s s2 + s 2X i= = . m m 2 m

Also, 2 X 3 2 col. 2 = i = 3m3 3m3

4

4

s s3 s2 1 s s2 3 + + = +s + .

4 2 4 2 3m3 2

Next,

2 1 6 1 5 5 4 2 X 5 1 2 i = col. 3 = s + s + s − s 2 12 12 5m5 5m5 6 1 6 5 5 4 1 2 1 = s +s + s − s .

6 6 5m5 3 Finally,

2 1 8 1 7 7 6 7 4 1 2 2 X 7 i = s + s + s − s + s col. 4 = 2 12 24 12 7m7 7m7 8 1 1 8 7 7 6 7 4 1 2 s +s + s − s + s . = 6 12 6 7m7 4 16.

The sum of the highest-degree terms from Exercise 15 is

s4 s6 s8 s 1 s3 1 s5 1 s7 s2 + + + + + ··· = s + + + ··· . m 6m3 15m5 28m7 m 2 · 3 m3 3 · 5 m5 4 · 7 m7 If we set x = ms , we can rewrite this series as

s

2x 2x3 2x5 2x7 + + + + ··· 1·2 3·4 5·6 7·8

.

Solutions

119

We note that

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

=

x2 x3 x4 x2 x3 x4 x3 x5 x− + − + · · · − −x − − − − · · · + −x − − − ··· 2 3 4 2 3 4 2 3

1 2x 2x3 2x5 1 + + + ··· . = x + x3 + x5 + · · · = 6 15 1·2 3·4 5·6 It follows that the original series may be written in the form

mx log

1+x 1−x

+ m log(1 − x2 ),

and, if mx = m − 1, in the form (m − 1) log

−1 1 + mm −1 1− mm

! + m log

m−1 m−1 1+ 1− . m m

−1 −1 Since 1 + m m = 2mm− 1 and 1 − m m = m1 , this last expression can be written as

(m − 1)[log(2m − 1) − log m + log m] + m[log(2m − 1) − log m − log m]

= (2m − 1) log(2m − 1) − 2m log m.

17.

The sum of the second-highest-degree terms of each column in Exercise 15 is

1h s s3 s5 s s i 1 s7 + ··· = + 3+ log 1 + − log 1 − = log + 7 5 m 3m 2 m m 2 7m 5m

1 + ms 1 − ms

.

−1 −1 But 1 + ms = 1 + m m = 2mm− 1 and 1 − ms = 1 − m m = m1 . It follows that the 1 last expression above can be rewritten in the form 2 [log(2m − 1) − log m + log m] = 1 2 log(2m − 1).

18.

If in the expression in the text for log M Q , we divide the arguments of the first two logarithm terms by m, this has the effect of subtracting 2m log m from the expression.

1 Since there is already a term −2m log m in the expression, and, since m + mt − 1 = 1+ t− m m−t + 1 t−1 and m = 1 − m , we get that log

M = Q

1 t−1 1 t−1 t m+t − . log 1 + + m−t+ log 1 − +log 1 + 2 m 2 m m

120

Solutions If we assume that m is large, we can approximate this expression by just using the first two terms in the power series for the logarithms. We get

M 1 t − 1 (t − 1)2 1 t − 1 (t − 1)2 log ≈ m+t − − + m−t+ − − Q 2 m 2 m 2m2 2m2

2 t t + − m 2m2 (t − 1)2 2t(t − 1) t−1 t t2 =− + − + − m m m m 2m2 2 2 t 2t 1 2t 2t t 1 t t2 t2 2t2 =− + − + − − + + − 2 ≈ = , m m m m m m m m m m n since we can ignore the term with denominator m2 and replace m by 2n . Thus, log M Q ≈ Q 2t2 2t2 n , and log M ≈ − n .

19.

p p If u = t/ np(1 − p), then du = dt/ np(1 − p) and when t = nϵ, u = p √ p nϵ/ np(1 − p) = nϵ/ p(1 − p). Given that the integrand is symmetric about x = 0, we see that we can write p Z ϵ Z √nϵ/√p(1 − p) 2 2 np(1 − p) 1 −[t2 /2np(1 − p)] Pϵ = p e dt = p e−u /2 du 2πnp(1 − √ p) −nϵ 2πnp(1 − p) 0 Z √nϵ/ p(1 − p) 2 2 =√ e−u /2 du.

2π 0 To calculate this integral when p = 0.6, ϵ = 0.02, and n = 6498, we need to use a graphing utility. The value of the integral in this case is 0.999. With the same values of p and ϵ, to find n so that Pϵ = 0.99 requires making several trials with the graphing utility. By experimenting, we find that n = 3980 solves the problem.

20.

According to Bayes’s theorem, expressed in its integral form, we have R s n n−1 Rs x (1 − x) dx n(xn − 1 − x) dx = R 1r .

P((r < x < s)|X = n − 1) = Rr 1 n −n 1 xn − 1 (1 − x) dx n(xn − 1 − xn ) dx 0 n−1 0 n+1

The integral in the denominator is (xn − nxn + 1 )|10 = n +1 1 , while the integral in the numerator is sn −rn − n +n 1 (sn + 1 −rn + 1 ). Thus the desired probability is P = (n + 1)(sn − rn ) − n(sn + 1 − rn + 1 ). In the particular case indicated, we have n = 11, r = 0.7, and s = 1. Thus, P = 12(1 − 0.711 ) − 11(1 − 0.712 ) = 11.763 − 10.848 = 0.915.

21.

We again use Bayes’s theorem in its integral form. Here p = n, so the quotient of the two integrals, as noted in the text, equals sn + 1 − rn + 1 . If we take s = 1 and r = 12 , then n+1 the probability P that x is greater than 21 is given by P = 1 − 2n1+ 1 = 2 2n +−1 1 . It follows that the odds are 2n + 1 − 1 to 1 for more than an even chance of the event happening again.

22.

If E is the event that the third ball is white and F is the event that the first two balls were white, then P(E|F) = P(E ∩ F)/P(F). We calculate the two probabilities on

Solutions

121

the right. First, note that P(E ∩ F) is the probability of drawing three white balls in three draws. Since we do not know the color composition of the balls in the urn, we can assume there are three equally likely possibilities. Both balls could be white; both could be black; or there could be one white and one black. In the first case, with probability 13 , it is certain that three white balls will be drawn; in the second case, with probability 13 , it is impossible that three white balls will be drawn; and in the third case, with probability 13 , the probability of drawing three white balls is 81 .

5 Therefore. P(E ∩ F) = 13 · 1 + 13 · 18 = 38 . Similarly, P(F) = 13 · 1 + 13 · 41 = 12 . Therefore, 3 5 9 P(E|F) = 8 ÷ 12 = 10 .

23.

According to the formula with P = 1, r = 0.04 and n = 50, the present value is A=

24.

Setting n = 86 − k = 86 − 36 = 50 and A = 21.4822 from Exercise 23, the formula for present value with interest at 4% gives Q=

25.

26.

27.

(1.04)50 − 1 = 21.4822 pounds.

0.04(1.04)50

1 − (1.04)(21.4822)/50 = 13.8293 pounds.

0.04

On a bet of 1 number, you have five chances of winning in a given drawing. The 1 probability of any one number occurring is 90 ; therefore, the probability of a win is 1 5 = .

Thus, the odds against winning are 17 : 1. On a bet on a pair of numbers, 90 18

there are 52 = 10 chances of winning on a single draw. The number of possible pairs

10 1 is 90 2 = 45 · 89. Thus, the probability of a win in this situation is 45·89 = 400.5 , and the odds against winning are 399.5 : 1. Finally, on a bet on a triple of numbers, there are 53 = 10 chances of winning on a single draw. The number of possible triples is

90 10 1 3 = 15 · 89 · 88. Thus, the probability of a win in this case is 15·89·88 = 11,748 , and the odds against winning are 11, 747 : 1.

1) In the text, we showed that the advantage to the player is nt((tn− a + n2t((nn−−1t)) b and this − 1) should equal the cost of a ticket, say 1 ecu. Thus, we need to find a and b. But since this is one equation in two unknowns, we can do this in infinitely many ways. Thus, we choose α and β to be any positive numbers such that α + β = 1 and set the first term in the sum equal to α and the second term equal to β. Solving for a and b, we get (n − 1) a = αnt((tn−−1)1) and b = βn .

2t(t − 1) 1)(t − 2) The probability of matching three numbers is nt((nt − , because the probability − 1)(n − 2) of the first number being chosen is t/n, that of the second is (t − 1)/(n − 1), and that of the third is (t − 2)/(n − 2). To match two numbers, the probability of matching the first is t/n and that of the second is (t − 1)/(n − 1). The probability of not matching the third is (n − t)/(n − 2), because there are n − t numbers that do not match any of the chosen ones. Since we could also not match the first or not match the second, there are three ways a match of two numbers could arise; thus the probability of matching −t) two numbers is n3t(n(t−−11)()(nn− . Similarly, the probability of matching one number is 2) 3t(n−t)(n−t − 1) . n(n − 1)(n − 2)

122

Solutions 28.

If n = 90 and t = 5, then, using the formulas from Exercise 27, we find that the 1 1 4·3 probability of matching one number is 905··89 ·88 = 6·89·22 = 11,748 . The probability of 3·5·4·85 85 85 matching two numbers is 90 ·89·88 = 6·89·22 = 11,748 . The probability of matching 1 3·5·85·84 7·85 595 number is 90·89·88 = 89·44 = 3916 .

29.

Using the results of Exercise 28, we see that the advantage to the player is a 85b 7 · 85c + + .

6 · 89 · 22 6 · 89 · 22 89 · 44 If α, β, γ are three positive numbers such that α + β + γ = 1, then the three prizes are a = 6·89·22α = 11,748α, b = 6·89·22β/85 = 138.21β, and c = 89·44γ/7·85 = 6.58γ.

30.

If we choose α = β = γ = 1/3, then the prizes are a = 11,748α = 3916, b = 138.21β = 46.07, and c = 6.58γ = 2.19. If we choose α = 1/7, β = γ = 3/7, the prizes are a = 11,748α = 1678.3, b = 138.21β = 59.23, and c = 6.58γ = 2.82. Finally, if we choose α = 1/16, β = 6/16, and γ = 9/16, the prizes become a = 11,748α = 734.25, b = 138.21β = 51.83, and c = 6.58γ = 3.70.

31.

The probability of A winning on the first flip is 21 . The probability of A winning on the second flip is ( 12 )2 (since in this case the first flip must have been a head).

Similarly, the probability of A winning on the third flip is ( 12 )3 and the probability of A winning on the nth flip is ( 12 )n . The expectation for A is the sum of the probabilities each multiplied by the payoff for that outcome.

Thus the expectation P 1 i− 1 2 = 21 + 12 + 21 + · · · . is 21 · 1 + ( 12 )2 · 2 + ( 12 )3 · 4 + · · · + ( 12 )n · 2n − 1 + · · · = ∞ i i=0 2 Since this sum is infinite, so is the expectation.

CHAPTER 19 1.

In 24 weeks, A can finish the job 8 times, B can finish it 9 times, and C can finish it 10 times. So in 24 weeks, all three can finish the job 27 times. Therefore, they can do 8 the job 1 time in 24 27 = 9 of a week.

2.

Since 12 cattle eat 3 13 acres in 4 weeks, 36 cattle will eat 10 acres of grass in 4 weeks.

This means that 16 cattle will take 9 weeks to eat the same 10 acres, or that 8 cattle will take 18 weeks, all assuming that the grass does not grow. But because the grass grows, we know that 21 cattle eat 10 acres in 9 weeks. That means that the growth of the grass on the 10 acres in 9 − 4 = 5 weeks is enough to pasture 21 − 16 = 5 cattle during 9 weeks or 2 12 cattle during 18 weeks. Therefore in 18 − 4 = 14 weeks, the growth of the grass will be sufficient to pasture 7 cattle during 18 weeks, because 5 : 14 = 2 21 : 7. Consequently, to the 8 cattle which 10 acres can support in 18 weeks without any growth, we must add 7 cattle which the growth will feed. That is, 15 cattle can live on 10 acres for 18 weeks. Therefore, since 36 : 10 = 54 : 15, we know that 54 cattle will eat up 36 acres in 18 weeks.

3.

If by x and y, then pthe two legs of the triangle are designated p we need to find h = x2 + y2 , given the two equations b2 = 21 xy and a = x + y + x2 + y2 . We have h = a − (x + y), so by squaring each side we get x2 + y2 = a2 − 2a(x + y) + (x + y)2 , or 2 2 2 4b2 a2 − 2a(x + y) + 4b2 = 0. Therefore x + y = a +2a4b and h = a − (x + y) = a − .

2a

Solutions

123

4.

If d is the distance from Edinburgh where the couriers meet, then their times to that d point are equal. Therefore 10 = 3608− d , or 8d = 3600 − 10d, or 18d = 3600. Thus d = 200, and the couriers meet 200 miles from Edinburgh.

5.

We solve each equation for x: x = m−bya −cz = n−eyd −fz = p−hyg −kz . We then equate the first and second expressions to get one equation in y and z and equate the second and third to get a second equation: (bd − ae)y + (cd − af)z = md − an, (dh − ge)y + (dk − gf)z = pd − gn. We then solve the first equation for cd−af)z y = md−anbd−( and substitute in the second equation. After simplifying, we get −ae (dh − ge)(md − an) − (dh − ge)(cd − af)z + (bd − ae)(dk − gf)z = (bd − ae)(pd − gn).

When we multiply out and solve for z, we get z=

bgn + aep + hmd − bdp − ahn − gem . hcd + bgf + aek − haf − gec − bdk

To find y, one could substitute, but it is easier simply to interchange the coefficients of y with those of z in the final expression for z. A similar process then works to find x.

6.

175 If x represents the number of people in the company, then we have x175 −2 = 10 + x .

2 Therefore 175x = 10x(x − 2) + 175(x − 2), or 10x − 20x − 350 = 0, or x2 − 2x − 35 = 0. The positive solution to this quadratic equation is x = 7.

7.

The algebraic rule a b = ab only applies to the positive square root of a positive number. We must note that there are always two square roots to any number, positive √ or negative, but the use of the symbol x by convention designates the positive square root. It is not possible to extend this rule to square roots of negative numbers, although it is true that the product of some square root of −1 times some square root of −4 equals some square root of (−1)(−4) = 4.

8.

Let x be the number of men dining. Then there are 20 − x women dining. So 8x + 7(20 − x) = 145, or x + 140 = 145, and x = 5. So there are 5 men and 15 women dining.

9.

Let x be the purchase price. Then the profit is 119 − x. The equation is then 119 − x x = 100 . This becomes x2 = 11,900 − 100x, or x2 + 100x − 11,900 = 0. The x positive solution to this quadratic equation is x = 70. Thus the purchase price was 70 and the profit, 49, is 70% of the original price.

10.

Let x be the amount of money of the youngest brother, y the amount of the second brother, and z the amount of the eldest. We then have x + 12 y = 100; y + 13 z = 100; and 1 1 4 x + z = 100. We can solve this system by substitution. Namely, set z = 100 − 4 x, so y + 31 (100 − 14 x) = 100. This equation reduces to −x + 12y = 800. Therefore, x = 12y − 800, and 12y − 800 + 12 y = 100. Then 25y = 1800, and y = 72. It follows that x = 64, and z = 84.

11.

We have

√ √

√

√

x4 + a4 = x4 + 2a2 x2 + a4 − 2a2 x2 = (x2 + a2 )2 − (a 2x)2 √

√

= (x2 + a2 + a 2x)(x2 + a2 − a 2x).

12.

We first have x5 − 1 = (x − 1)(x4 + x3 + x2 + x + 1). To factor the second factor, we set x4 + x3 + x2 + x + 1 = (x2 + ax + 1)(x2 + bx + 1). Multiplying out the right-hand side and

124

Solutions comparing coefficients leads to the two equations a + b = 1 and ab = −1. Solving, √ √ 5 5 1 1 we get a = 2 + 2 and b = 2 − 2 . Therefore, the factorization of the original polynomial is √ √

1 5 1 5 2 2 x − 1 = (x − 1) x + + x+1 x + − x+1 .

2 2 2 2 5

13.

In multiplying the two given factors we note that the coefficient of x4 is 1. ptogether, p first √ √ Then, thep coefficient of x3 is −2+ coefficient of p 4 +√2 7√− 2 − 4 + 2 √7 = −4. The √ √ √ √ 2 7+ p 7 + 4 − (4 + 2 7) = 2 + 2 7 − 2 7 = 2. x is 1 − 4 + 2 7 + 7 + 1 + 4 + 2 p √ √ √ Next, the coefficient of x is − 2 + 4 + 2 7 1 − 4 + 2 7 + 7 −

p p √ √ √ 2 − 4 + 2 7 1 + 4 + 2 7 + 7 . When you multiply this out, you get

− 2−2

− 2+2

p

p

√

√

√

p p √ √ √ 4 + 2 7 − 4 − 2 7 + 28 + 14 7

√

p

√

4+2 7 + 2 7 +

4+2 7 + 2 7 −

√

4+2 7 − 4−2 7 −

p √ 28 + 14 7 = −(−4) = 4.

Finally, the constant term is p p √ √ √ √ 1 + 4+2 7 + 7 1 − 4+2 7 + 7 √ 2 √ √ √ = 1 + 7 − 4 + 2 7 = 8 + 2 7 − 4 − 2 7 = 4.

Therefore, the product of the two factors is x4 − 4x3 + 2x2 + 4x + 4 as asserted.

14.

We factor the polynomial as x4 − 2x2 + 8x − 3 = (x2 + ux + α)(x2 − ux + β), and by comparing coefficients, we have α + β − u2 = −2, (β − α)u = 8, and αβ = −3. The equation for u then becomes, by Euler’s procedure, u6 − 4u4 + 16u2 − 64 = 0. We can find a solution for this by inspection, namely u = 2. The equations for α and β then become α + β = 2, 2(β − α) = 8, and αβ = −3. Using the first two, we get β = 3 and α = −1, and these also solve the third equation. Thus, the factorization is x4 − 2x2 + 8x − 3 = (x2 + 2x − 1)(x2 − 2x + 3).

15.

One example is y = x3 − 3x and x = y2 − 2.

16.

Note that equation 4 is the sum of equations 2 and 3, while equation 1 is the sum of twice equation 2 and equation 3. Therefore, the system reduces to two equations in four unknowns. In this case there is a two-parameter family of solutions and not a unique solution.

17.

We know that every solution of xn − 1 = 0, except 1, is a power of the complex root λ r s whose angle is 2π n . So suppose α = λ and β = λ . We need to show that β is a power of α. So set rx ≡ s (mod n). Since n is prime, we know this equation has a solution x = q. But then λrq = λs , or αq = β, as desired.

18.

If v = x1 − x2 , then v2 = (x1 + x2 )2 − 4x1 x2 = t2 − 4c. Thus v satisfies a quadratic equation in t. It is simpler to rewrite this equation in the form v2 = b2 − 4c, so

Solutions √

v = ± b2 − 4c. We also have x1 = (x1 −√2x2 )−b = v−2 b . It follows that x1 = 2 and the second solution x2 is given by − b 2− 4c−b .

19.

125

√ b2 − 4c−b 2

We set x = y + 2y and substitute into the original equation. We get (y + 2y )3 − 6(y + 2y ) − 9 = 0, or y3 + y83 − 9 = 0. If we let r = y3 , then r + 8r − 9 = 0, or r2 − 9r + 8 − 0, and r = 1 or r = 8. In the first case, we have y = 1, y = ω, and y = ω2 .

In the second case, we have y = 2, y = 2ω, and y = 2ω2 . It follows that the three roots of the original equation are x1 = 1 + 2 = 3, x2 = ω + 2ω2 , and x3 = ω2 + 2ω. The six values of y can then be expressed as follows: 1 1 (3 + ω(ω + 2ω2 ) + ω2 (ω2 + 2ω)) = (7 + ω + ω2 ) = 2 3 3 1 1 2 2 2 (ω + 2ω + ω(ω + 2ω) + ω · 3) = (7ω2 + ω + 1) = 2ω2 3 3 1 2 1 (ω + 2ω + ω · 3 + ω2 (ω + 2ω2 )) = (7ω + ω2 + 1) = 2ω 3 3 1 1 (3 + ω(ω2 + 2ω) + ω2 (ω + 2ω2 )) = (5 + 2ω + 2ω2 ) = 1 3 3 1 2 1 2 2 (ω + 2ω + ω(ω + 2ω ) + ω · 3) = (5ω2 + 2ω + 2) = ω2 3 3 1 1 (ω + 2ω2 + ω · 3 + ω2 (ω2 + 2ω)) = (5ω + 2ω2 + 2) = ω 3 3

20.

The three values are x1 x2 + x3 x4 , x1 x3 + x2 x4 , and x1 x4 + x2 x3 . The value of the variable which is multiplied by x1 determines the entire expression, and there are just three possible values for that variable, as indicated.

21.

The relationship between roots and coefficients shows that x1 + x2 + x3 + x4 = − a, x1 x2 + x1 x3 + x1 x4 + x2 x3 + x2 x4 + x3 x4 = b, x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4 = −c, and x1 x2 x3 x4 = d. It follows immediately that α + β + γ = b.

For the second result, note that αβ + αγ + βγ = (x1 x2 + x3 x4 ) (x1 x3 + x2 x4 ) +(x1 x2 + x3 x4 )(x1 x4 + x2 x3 ) + (x1 x3 + x2 x4 )(x1 x4 + x2 x3 ) = x21 x2 x3 +x1 x22 x4 + x1 x23 x4 + x2 x3 x24 + x21 x2 x4 + x1 x22 x3 + x1 x3 x24 +x2 x23 x4 + x21 x3 x4 + x1 x2 x23 + x1 x2 x24 + x22 x3 x4 = (x1 + x2 + x3 + x4 )(x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4 ) − 4x1 x2 x3 x4 = ac − 4d. Finally, αβγ = x31 x2 x3 x4 + x21 x22 x23 + x21 x22 x24 + x1 x32 x3 x4 + x21 x23 x24 + 3 3 2 2 2 2 x1 x2 x3 x4 + x1 x2 x3 x4 + x2 x3 x4 . Also, a d + c2 − 4bd = (x1 + x2 + x3 + x4 )2 x1 x2 x3 x4 + (x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4 )2 − 4(x1 x2 + x1 x3 + x1 x4 + x2 x3 + x2 x4 + x3 x4 ) (x1 x2 x3 x4 ). These two expressions are in fact equal. We note, for example, that the terms in the first expression with cubes come from the first term in the second expression, while the terms in the first expression which have squares of three variables come from the second term in the second expression. The third term in the second expression eliminates all the additional expressions in the first two terms. Since we now know the sum of three values, the sum of the three values taken in pairs, and the product of the three values, we know from the relationship between roots and coefficients of a polynomial that the three values are the roots of the cubic equation y3 −(α+ β + γ)y2 + (αβ + αγ + βγ)y − αβγ = 0 or y3 − by2 + (ac − 4d)y − (a2 d + c2 − 4bd) = 0.

22.

Using the notation of Exercise 21, we have a = 0, b = 0, c = −12, and d = 3.

Therefore, the reduced cubic equation is y3 − 12y − 144 = 0. We can find one

126

Solutions root by trial and error: α = 6. It follows that the other two roots have their sum equal to −6 and their product equal to 24. Thus, they satisfy the quadratic equation √ y2 + 6y + 24 = 0. Therefore, the two additional roots of the cubic are β = −3 + −15 √ and γ = −3 − −15. In particular, if the four roots of the quartic equation are x1 , x2 , x3 , and x4 , we know that x1 x2 + x3 x4 = 6. From the relationship between roots and coefficients, we also know that x1 x2 x3 x4 = 3. Therefore, x1 x2 and √ x3 x4 2 are the roots √ of the quadratic equation z − 6z + 3 = 0. Thus, x1 x2 = 3 + 6 and x3 x4 = 3 − 6. Also, we know that (x1 + x2 ) + (x3 + x4 ) = 0, or x3 + x4 = −(x1 + x2 ).

Given further that x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x√ this equation 3 x4 = 12, we can rewrite √ as x2 x2 (x3 + x4 ) +√ x3 x4 (x1 + x2 ) = 12 or as −(3 + 6)(x1 + x2 ) + (√ 3 − 6)(x1 + x2 ) = 12, = − 6. Therefore, x1 and or, finally, as −2 6(x1 + x2 ) = 12. It follows that √ x1 + x2 √ x2 are the roots of the quadratic equation w2 + 6w + 3 + 6 = 0. We therefore have

x1 =

q √ √ − 6 + 6 − 4(3 + 6) 2

=

p √ √ − 6 + −6 − 4 6 2

=−

1√ 6+ 2

r

√ 1 − 6−1 .

2

Similarly, 1√ x2 = − 6− 2

r

√ 1 − 6−1 .

2

√

√

We also find that x3 and x4 satisfy the quadratic equation w2 − 6w + 3 − 6 = 0. So 1√ x3 = 6+ 2

r

√

1 6−1 2

and

1√ 6− x4 = 2

r

√

1 6−1 .

2

23.

Suppose that p and p2 + 3q2 have a common factor. Then p2 and p2 + 3q2 will have the same factor, and so will 3q2 . But since x and y are relatively prime, so are p and q.

Therefore, the only possible common factor of p2 and 3q2 is 3. But we are assuming that p is not divisible by 3, so this is also impossible.

24.

Assume that p = 3r. Then 14 3r(9r2 + 3q2 ) = 94 r(3r2 + q2 ) is a cube. We first note that the two factors here are relatively prime. Recall that p and q are relatively prime, with p even. Thus r is even. Neither 2 nor 3 can divide 3r2 + q2 because q is odd. If any other number divided 3r2 + q2 and r, it would also divide q2 and therefore p. Since this is impossible, the two factors are relatively √ therefore both are cubes. √ prime and . Since the two√factors As in the text, we can factor q2 + 3r2 as (q + r −3)(q − r −3)√ are relatively prime, both of them are also cubes. Thus q + r −3 = (t + u −3)3 .

Then q = t3 − 9tu2 = t(t2 − 9u2 ) and r = 3u(t2 − u2 ), where t is odd and u is even.

2 2 8 9 2 Also, 49 r is a cube, so 27 4 r = 3 r is a cube. That is, 2u(t − u ) = 2u(t + u)(t − u) is a cube. Again, these factors are relatively prime because u is even. So each is a cube: t + u = f 3 , t − u = g3 , and 2u = h3 . Therefore, f 3 − g3 = h3 , or g3 + h3 = f 3 .

We have therefore found a new sum of cubes equation with each term less than the corresponding numbers in the original sum of cubes equation. The impossibility of a solution is thus proved in this case too by infinite descent.

25.

The residue of 1 is 1; of 5 is 5; of 52 is 12; and of 53 is 8. These are the distinct residues in this sequence, since the residue of 54 is again 1. If we now multiply each

Solutions

127

of the distinct residues by 2, we get the coset {2, 10, 11, 3}. If we multiply each of the residues by 4, we get the coset {4, 7, 9, 6}.

26.

We have, modulo 13, 12 = 1, 22 = 4, 32 = 9, 42 = 3, 52 = 12, and 62 = 10. The squares of the other integers less than 13 are the same as these. Therefore, the quadratic residues modulo 13 are 1, 3, 4, 9, 10, and 12.

27.

Let a be a primitive root modulo q, that is, a number such that no power less than the (q − 1)st is congruent to 1 modulo q. Such a primitive root always exists. Since q−1 aq − 1 ≡ 1 (mod q), we must have a 2 ≡ −1 (mod q), and no other power of a q−1 1 4 less than q − 1 can be congruent to −1 modulo q. If q − has 2 is even, then b = a q−1 2 the property that b ≡ −1 (mod q) and thus −1 is a quadratic residue. But 2 is even if and only if q ≡ 1 (mod 4).

28.

If b is the number of bullocks, c the number of cows, and s the number of sheep, then 1 we have the two equations b + c + s = 100 and 5b + c + 20 s = 100. If we solve each of 1 these equations for c and equate the results, we get 100 − 5b − 20 s = 100 − b − s, or 19 19 s = 4b.

Therefore b = s.

Since a valid solution has only integral values, we can 20 80 choose s = 80. Then b = 19 and c = 1. Thus, there are 80 sheep, 19 bullocks, and 1 cow.

29.

Let the four parts be x, y, z, w. We have the equations x + y + z + w = 60, x + 4 = y − 4 = 4z = w4 . We can pick three equations out of this last list: x − y = −8, 16z − w = 0, and y − 4z = 4. One way to solve this system is by substitution. We have y = 4 + 4z, w = 16z, and x = y − 8 = 4z − 4. Then 4z − 4 + 4z + 4 + z + 16z = 60, or 192 68 28 25z = 60, and z = 12 5 . Then w = 16z = 5 , y = 4 + 4z = 5 , and x = 4z − 4 = 5 .

30.

The ladder is the hypotenuse of two right triangles. One has √ leg 37 and the other has 2 2 leg 23. It follows that the second leg of the √ first triangle is 60 − 37 = 47.233, and the second leg of the second triangle is 602 − 232 = 55.417. Since these two legs together form the width of the street, that width is 102.65 feet.

CHAPTER 20 1.

Connect AD and BC. We show first that triangles ABC and ABD are congruent. We know that they share a common side, AB. Also, AC = DB and ∠CAB = ∠ABD. Thus, the triangles are congruent by side-angle-side. Therefore, AD = BC and ∠DAB = ∠CBA. We now show that triangles ACD and BCD are congruent. We have AD = BC, AC = DB, and ∠DAC = ∠CBD by subtracting equal angles from equal angles.

Therefore, the triangles are congruent by side-angle-side and so ∠C = ∠D.

2.

Let δ(ABC) denote the defect of triangle ABC. Then δ(ABD) + δ(BDC) = (180 − ∠A − ∠ABD − ∠ADB) + (180 − ∠C − ∠CDB − ∠CBD) = 360 − ∠A − ∠C − 180 − ∠ABC = 180 − ∠A − ∠C − ∠ABC = δ(ABC). Note that we have used the fact that ∠ABD + ∠CDB = 180◦ .

3.

Since DE = EF, we have AE = 12 DE = 12 EF = BE. Therefore, △AEB is isosceles.

Similarly, triangles DAC and CBF are isosceles. Also, since AE = AD, BE = DC, and ∠E = ∠D, triangles AEB and DAC are congruent. Similarly, both of these are congruent to triangle CBF. It follows that AB = AC = BC, so △ABC is equilateral.

Next, 3α + γ = δ(CBF) + δ(DAC) + δ(AEB) + δ(ABC) = 180 − ∠F − ∠BCF −

128

Solutions ∠FBC + 180 − ∠D − ∠DCA − ∠CAD + 180 − ∠E − ∠EBA − ∠EAB + 180 − ∠BAC − ∠ABC − ∠BCA = 720 − ∠E − ∠D − ∠F − (∠BAC + ∠CAD + ∠EAB) − (∠ABC + ∠FBC + ∠EBA) − (∠BCA + ∠BCF + ∠DCA) = 180 − ∠E − ∠D − ∠F = δ(DEF) = β.

Finally, if the four smaller triangles were all congruent, then angles BCF, BCA, and ACD would all be equal. Because their sum is 180◦ , each angle would be 60◦ . The same would be true of all the other angles in the four triangles, so the defects would be 0, contrary to our assumption.

4.

Let ϵ(ABC) denote the excess of triangle ABC. Then ϵ(ABD) + ϵ(BDC) = (∠A + ∠ABD + ∠ADB − 180) + (∠C + ∠CDB + ∠CBD − 180) = ∠A + ∠C + ∠ABC − 360 + (∠ABD + ∠CDB) = ∠A + ∠C + ∠ABC − 180 = ϵ(ABC).

5.

We have cosh x = cosh(i(−ix)) = cos(−ix) = cos ix. Also, sinh x = sinh(i(−ix)) = i sin(−ix) = −i sin ix. Then cosh2 x − sinh2 x = cos2 ix − (−i sin ix)2 = cos2 ix + sin2 ix = 1.

6.

If ax = y2 , then y =

√

2

2

dx ax and dy = 2a√dxax , so dy2 = a4axdx = a 4x . Also, since (9/16)az2 = 3/ 2

1/ 2

2

4y 2y dy 2 a y3 , we have 34 a1/2 z = y3/2 , so z = 3a , and dz2 = 4y ady = 4y 1/2 , dz = a 4x dx = a1/2q √ √ √ √ p ax a dx2 y 2 2 a √ . Then dx2 + dy2 + dz2 = 1 + 4x + √ax dx = (1 + 2√ax ) dx. x dx = x dx = x Thus, to determine the length L of the curve, we must integrate. We get Z x0 √ √ √ x √ a L= 1+ √ dx = x + a x 00 = x0 + ax0 = x0 + y0 .

2 x 0

7.

√

dx2 + dy2 . In this case, since x2 − a2 = y2 dz 2 and y2 − a2 = z2 , we have 2x dx = 2y dy, so dy = yx dx and dy2 = yx2 dx2 . Also, 2y dy = 2z dz, so dz = yz dy = yz xy dx = xz dx. Therefore,

From Figure 20.10, the subtangent is Mt =

√

Mt =

z

dx2 + dy2 = dz

p z2

= xy

z

√ 2 z dx2 + x2 dx2

y2 + x2 =

√ 2 = zx

y

x dx z

x2 − 2a2 √ x x2 − a2

2

1 + x2 dx y

dx

√

2x2 − a2 = x −x 2a 2

2

q

2x2 − a2 . x2 − a 2

The tangent is Nt, where √

Nt =

dx2 + dy2 + dz2 = dz

z

r =z 8.

2

√ 2 2 z dx2 + x2 dx2 + x2 dx2 y

z

x dx z

2

= zx

q

2

r 1 1 1 1 1 1 2 2 + + = (x − 2a ) + + . x2 y2 z2 x2 x2 − a2 x2 − 2a2

According to the text, the length of the perpendicular is NO = Nt √ dz 2

dx + dy

NO =

2

1 + xy2 + xz2

√ 2 2 2 dx + dy + dz √ = 2 2

z

dx + dy

. Using the values calculated in the text, we get 2

p by + 4x + a ·

4 √ 4x + a dx 4x

ab dx √ 4 2 3 3 b a x

=

ab

p by + 4x + a √ 4

4 b2 a3 x3

p abx by + 4x + a 4x √ √ = .

4x + a zy 4x + a

Solutions 9.

129

The normal vector to the plane αx + βy + γz = a is v = (α, β, γ). The angle θ between this plane and the xy plane is the same as the angle between the normal vectors, and v·w the normal vector w to the xy plane is (0, 0, 1). Thus cos θ = ||v|||| = √ 2 γ 2 2 . The w|| α +β +γ

cosine of the angle that the plane makes with the xz plane is then √ 2 β 2 cosine of the angle the plane makes with the yz plane is √ 2 α 2 2 . α +β +γ

α +β +γ2

, and the

10.

We have κ ϕ = 12 (κ1 +κ2 )− 21 (κ1 −κ2 ) cos 2ϕ = 12 (κ1 +κ2 )− 12 (κ1 −κ2 )(cos2 ϕ−sin2 ϕ) = 2 2 2 2 2 2 1 1 1 1 2 κ 1 (1 − cos ϕ) + 2 κ 1 sin ϕ + 2 κ 2 (1 − sin ϕ) + 2 κ 2 cos ϕ = κ 1 sin ϕ + κ 2 cos ϕ.

11.

The normal line to the surface z = f(x, y) is a line in the direction of the gradient, namely ( ∂∂xz , ∂∂yz , −1). The normal vector to the plane z = αy − βx + γ is (β, −α, 1).

Thus the plane will contain the normal line if these two vectors are perpendicular, that is, if their dot product is zero. This amounts to the condition β ∂∂xz − α ∂∂xz − 1 = 0, as stated.

12.

Since (A, B, C) is a normal vector to the plane, the normal line is given by the three equations x = At + x0 , y = Bt + y0 , z = Ct + z0 .

13.

Monge’s form of the equations of the normal line are x − x′ + (z − z′ ) ∂∂xz = 0 and ′ y − y′ + (z − z′ ) ∂∂yz = 0. If we set t = ∂xz−/∂x x , then we have x = ∂∂xz t + x′ and, by substituting in the first equation of Monge, t + z − z′ = 0, or z = −t + z′ . Then the second equation of Monge becomes y − y′ − t ∂∂yz = 0, or y = ∂∂xz t + y′ . These three equations form the modern vector equation of the normal line.

14.

We need to show that under either of these circumstances, if we add together the numbers associated to each bridge, that is, if we count the number of letters in the path, the total is 1 more than the number of bridges. We suppose that there are n regions. We designate the number of bridges leading into region i by ki . We first suppose that exactly two of the ki are odd, say k1 and k2 . Then the total number of P letters is k3 kn k1 + 1 k2 + 1 1 P ki + 1. Since the number of bridges is 21 ki , we 2 + 2 + 2 + ··· + 2 = 2 have shown that the number of letters in this case is 1 more than the number of bridges, so the Euler path exists. If all of the ki are even, let us suppose that our pathP starts in k1 + 1. region 1. Then the total number of letters is k21 + 1 + k22 + k23 + · · · + k2n = 12 Thus again the number of letters is 1 more than the number of bridges, and the Euler path exists.

15.

One possible Euler path in the first diagram is EADCBAEC. An Euler path in the second diagram is CBDBADACAC. In each of these, every crossing between the same two regions is on a different bridge.

16.

In a tetrahedron, V = 4, E = 6, and F = 4, so V − E + F = 2. In a cube, V = 8, E = 12, and F = 6, so V − E + F = 2. In an octahedron, V = 6, E = 12, and F = 8, so V − E + F = 2. In a dodecahedron, V = 20, E = 30, and F = 12, so V − E + F = 2. (To find the number of vertices, note that each pentagon has five vertices, but each vertex lies on three different pentagons. Similarly, to find the number of edges, note that each pentagon has five edges, but each edge lies on two different pentagons.) Finally, in an icosahedron, V = 12, E = 30, and F = 20, so again V − E + F = 2.

130

Solutions

CHAPTER 21 1.

We divide p − 1 by m. We get p − 1 = qm + r, where 0 ≤ r < m. Since by Fermat’s Little Theorem, ap − 1 ≡ 1 (mod p), we have 1 ≡ ap − 1 ≡ aqm + r ≡ (am )q ar ≡ 1q ar ≡ ar (mod p). But m is the smallest positive integer such that am ≡ 1 (mod p).

Therefore, r = 0, so m divides p − 1.

2.

We calculate 23 ≡ 1, 36 ≡ 1, 43 ≡ 1, 56 ≡ 1, and 62 ≡ 1 (mod 7). All of the exponents, namely 2, 3, and 6 divide p − 1 = 6, as proved in Exercise 1.

3.

Note that a is a primitive root for p = 13 if and only if 6 is the smallest positive integer such that a6 ≡ −1 (mod 13). Note also that a is a primitive root if and only if −a is.

We then check the integers in turn: Since 1 is not a primitive root, neither is −1 ≡ 12.

Since 26 ≡ −1 and no smaller power of 2 is congruent to −1, we know that 2 is a primitive root for p = 13. It follows that −2 ≡ 11 is also a primitive root. Next, we have 33 ≡ 1, so 3 is not a primitive root, nor is −3 ≡ 10. Also, since 4 = 22 , we know that 46 = 212 = 1, so 4 is not a primitive root, nor is −4 ≡ 9. Also, 52 ≡ −1, so 54 ≡ 1 and 5 is not a primitive root, nor is −5 ≡ 8. Finally, 66 ≡ −1 and no smaller power is congruent to −1. Therefore, 6 is a primitive root and so is −6 ≡ 7. Thus the primitive roots of 13 are 2, 6, 7, and 11.

4.

We assume that there exist integers a, b, c such that a2 ≡ 453 (mod 4), b2 ≡ 453 (mod 3), and c2 ≡ 453 (mod 103). By the Chinese Remainder Theorem, there is an integer d such that d ≡ a (mod 4), d ≡ b (mod 3), and d ≡ c (mod 103). Therefore d2 ≡ 453 modulo each of 4, 3, and 103. So d2 − 453 is divisible by 4, 3, and 103, and therefore is divisible by their product, namely 1236. So x2 ≡ 453 (mod 1236) has a solution. Now we need to show that the original assumptions are true. Since 453 ≡ 1 (mod 4), and 1 is a square modulo 4, we know that 453 is a quadratic residue modulo 4. Similarly, since 453 ≡ 0 (mod 3), and 0 is a square modulo 3, we know 453 )= that 453 is a quadratic residue modulo 3. Since 453 ≡ 41 (mod 103), we have ( 103 41 41 ( 103 ). Finally, as in the text, the quadratic reciprocity theorem shows that ( 103 ) = 1 22 5 5 41 1 (− 41 )( 41 )( 41 ). Another application of the theorem gives ( 41 ) = ( 5 ) = ( 5 ) = 1, so 453 is a quadratic residue modulo 1236.

5.

We show that a + bi (b ̸= 0) is a prime Gaussian integer if and only if N(a + bi) = a2 + b2 is an ordinary prime. First, suppose that a2 + b2 is prime. Then if α = a + bi factored non-trivially as α = βγ, we would have a2 + b2 = N(α) = N(β)N(γ), with neither N(β) nor N(γ) equaling 1. So a2 + b2 would not be prime. Second, let us suppose that a2 + b2 is composite. We may as well assume that (a, b) = 1, for otherwise a + bi would clearly be composite. Thus, at least one of a and b is odd. Checking congruences modulo 4, we see that a2 + b2 must be congruent to 1 or 2. Therefore, a2 + b2 cannot be a power of 2. So let p be an odd prime which divides a2 + b2 . If p ≡ 3 (mod 4), then p must divide both a2 and b2 . For otherwise, a ≡ bu (mod p) is solvable for u, so a2 + b2 ≡ b2 u2 + b2 ≡ b2 (1 + u2 ) ≡ 0 (mod p). Because p does not divide b2 , we have u2 ≡ −1 (mod p). This contradicts the fact that ( −p1 ) = −1. Thus p divides a2 and b2 , so p divides a and b, contradicting the fact that a and b are relatively prime. Thus we must have p ≡ 1 (mod 4). In this case, we know that there are integers α and β such that p = α2 + β2 . Now (aα + bβ)(aα − bβ) = a2 (α2 + β2 ) − β2 (a2 + b2 ). Since p divides the right side of this equation, it must divide the left side. Thus either p divides aα + bβ or p divides aα − bβ. But also, (aα + bβ)2 + (bα − aβ)2 = (aα − bβ)2 + (bα + aβ)2 =

Solutions

131

(a2 + b2 )(α2 + β2 ). Since the right-hand expression is divisible by p2 , so are the other two expressions. Therefore, if p divides aα + bβ, p must also divide bα − aβ. In that

case, a + bi (aα + bβ) + (bα − aβ)i aα + bβ bα − aβ = + i, = α + βi p p α 2 + β2 and this latter expression is a Gaussian integer. On the other hand, if p divides aα − bβ, aα−bβ + bi p must also divide bα + aβ. Then αa− + bα +p aβ i, and the latter expression is βi = p again a Gaussian integer. Therefore, we have shown that if a2 + b2 is composite, then a + bi is divisible by a Gaussian integer whose norm is a prime congruent to 1 modulo 4. Thus, a + bi is composite and the proof is complete.

6.

Suppose p divides abc · · · . Then N(p) divides N(abc · · · ) = N(a)N(b)N(c) · · · . But N(p) is prime. Therefore N(p) must divide one of the factors, say N(a). But N(a) is also prime. Therefore N(p) = N(a) and p = ua, where u is a unit.

7.

N(3 + 5i) = 9 + 25 = 34. Since 34 = 2 × 17, we see that 3 + 5i factors into a product of two primes, one with norm 2 and one with norm 17. We get 3 + 5i = (1 − 4i)(−1 + i), and since the norm of each of the factors is prime, the factors themselves must be prime.

8.

If we cube the residues modulo 7, we get 13 = 1, 23 = 1, 33 = 6, 43 = 1, 53 = 6, and 63 = 6. Thus, the cubes modulo 7 are 1 and 6, and therefore 3 is not a cube. Also, these cubes (or cubic residues) do not differ by 1. It therefore follows from Germain’s theorem that in any solution to the Fermat equation for exponent 3, one of x, y, or z must be divisible by 9.

9.

Suppose p is prime and assume it factors into irreducible integers: p = q1 q2 · · · qn .

Then since p divides the product, it must divide one of the factors, say q1 . But then q1 would not be irreducible, contrary to our hypothesis. Thus p itself must be irreducible.

10.

Each of the four factors involved in the two factorizations is irreducible. To show this, note that the norm of each of the factors is 169. But 169 factors in the domain of integers as 13 × 13. Thus any√non-trivial factor of any of the numbers must have norm equal to 13. But if N(a + b −17) = 13, then a2 + 17b2 = 13. This equation is impossible to solve for integers a and b.

11.

Let mz = s + ti, where s and t are rational numbers. Choose integers q1 and q2 as close as possible to s and t, respectively. Let q = q1 + q2 i and r = z − mq. We need to show that N(r) < N(m). If r = 0, we are done. Otherwise, we note that |s − q1 | ≤ 12 and |t − q2 | ≤ 12 . Therefore, N( mz − q) = N((s + ti) − (q1 + q2 i)) = N((s − q1 ) + (t − q2 )i) ≤ ( 21 )2 + ( 12 )2 = 12 . Therefore, N(r) = N(z − mq) = N(m( mz − q)) = N(m)N( mz − q) ≤ 1 2 N(m) < N(m), as desired.

12.

In the given domain, we have N(a + b −2) = a2 + 2b2 . To show that this domain is Euclidean, we need to show that given any two integers α, β in the domain, there exist two integers σ, ρ such√that α = βσ + ρ, where N(ρ) < N(β). We begin by dividing α by β. Let βα = r + s −2, where r and s are rational numbers. Let q1 and q2 be √ ordinary integers as close as possible to r and s, respectively. Let σ = q1 + q2 −2 and 1 q2 | ≤ 12 . ρ = α − βσ. If ρ = 0, we are done. √ Otherwise, we√see that |r − q1 | ≤ 2 and |s −√ α Therefore, N( β − σ) = N((r + s −2) − (q1 + q2 −2)) = N((r − q1 ) + (s − q2 ) −2) ≤

√

132

Solutions ( 12 )2 + 2( 12 )2 = 34 . It follows that N(ρ) = N(α − βσ) = N(β( βα − σ)) = N(β)N( βα − σ) ≤

N(β) 34 < N(β), as desired. √

13.

We have N(a + b −5) = a2 + 5b2 . Since N(2) = 4, any factor of 2 must have norm equal to 2, and no such number exists. Similarly, since N(3) = 9, any factor of √ 3 must have norm equal to 3, and again there is no such number. Also, N(−2 ± −5) = 9. Again, √ any factor of either of these numbers must have norm equal to 3. And N(1 ± −5) = 6, so a factor of either of these must have norm equal to 2 or 3, both of which are impossible.

14.

2 If of 2 × 2 = 4, 2(1 + √ √ A = (2, 1 +√ −5), then A√ consists√of sums of multiples −5) = 2 + 2 −5, and (1 + −5)(1 + −5) = −4 + 2 −5. Each of these numbers is a multiple in the domain), so A2 ⊂ (2). But also 2 = (−1)(4 + √ of 2 (by some integer √ (−4 + 2 −5) + (−1)(2 + 2 −5)), so 2 ∈ A2 , and therefore (2) ⊂ A2 . Therefore (2) = A2 .

15.

Let us write this group in the form G = {1, α, α2 , . . . , α17 }, where α18 = 1. Then the subgroup H of order 6 is {1, α3 , α6 , α9 , α12 , α15 }. To find one coset, multiply each of these elements by α. We get the coset αH = {α, α4 , α7 , α10 , α13 , α16 }. A second coset comes from multiplying each of the elements of H by α2 . This coset is α2 H = {α2 , α5 , α8 , α11 , α14 , α17 }.

16.

Since 3 is a primitive root modulo 7, we set h = 33 ≡ 6 (mod 7). Then if r is a fixed root of the equation, we set αi = ri + rih , for i = 1, 2, 3. We get α1 = r + r6 , α2 = r2 + r5 , and α3 = r3 + r4 . We now need to find the cubic equation satisfied by the αi . We note that α31 + α21 − 2α1 = r3 + r2 + r + 2 + r6 + r5 + r4 = 1. Therefore, α1 (and also α2 and α3 ) satisfy the equation y3 + y2 − 2y − 1 = 0. We now need to solve this cubic equation.

We simplify it by setting y = z − 13 . We then get (z − 31 )3 + (z − 13 )2 − 2(z − 31 ) − 1 = 0, 7 7 or z3 − 37 z − 27 = 0. If we rewrite this in the form z3 = 73 z + 27 , we can apply Cardano’s formula. We have z = ρ + σ, where v v s s u u 2 3 2 3 u u 3 3 7 7 7 7 7 7 t t ρ= + − − and σ = − .

54 54 9 54 54 9

√

With a bit of algebraic manipulation, we get r r p p 13 7 13 7 (1 + 3 −3) and σ = (1 − 3 −3). ρ= 3 2 3 2 Since y = z − 13 , it follows that the three roots of the equation in y are α1 = 2 2 1 1 1 3 (−1 + ρ + σ), α2 = 3 (−1 + ω ρ + ωσ), α3 = 3 (−1 + ωρ + ω σ), where ω is a complex cube root of 1. To solve the original equation, we note that since α1 = r + r6 , we have r2 +q 1 = α1 r, or r2 − α1 r + 1 = 0. There are two solutions to this equation: α21 − 4). These two solutions are, in fact, the original r and r6 , since 2 6 2 α1 − 4 = (r − r ) . We find that the other powers of r can be written similarly in terms

r = 21 (α1 ±

of α2 and α3 .

17.

P18 i 2 We first note that α1 + α2 + α3 = i=1 r = −1. Thus the coefficient of x in the cubic equation for the αi must be +1. For the remainder of the problem, it

Solutions

133

is convenient to use both negative and positive powers of r in the representation of αi . Thus, for example, α1 = (r + r−1 ) + (r7 + r−7 ) + (r8 + r−8 ). We now multiply the αi in pairs. We find α1 α2 = (r + r−1 ) + 2(r2 + r−2 ) + 2(r3 + r−3 ) + 3(r4 + r−4 ) + 2(r5 + r−5 ) + 3(r6 + r−6 ) + (r7 + r−7 ) + (r8 + r−8 ) + 3(r9 + r−9 ).

Also, α1 α3 = 2(r + r−1 ) + 3(r2 + r−2 ) + 3(r3 + r−3 ) + (r4 + r−4 ) + 3(r5 + r−5 ) + (r6 + r−6 ) + 2(r7 + r−7 ) + 2(r8 + r−8 ) + (r9 + r−9 ). Finally, α2 α3 = 3(r + r−1 ) + (r2 + r−2 ) + (r3 + r−3 ) + 2(r4 + r−4 ) + (r5 + r−5 ) + 2(r6P + r−6 ) + 3(r7 + r−7 ) + 8 −8 9 −9 i 3(r + r ) + 2(r + r ). Therefore, α1 α2 + α1 α3 + α2 α3 = 6 18 i=1 r = −6. It follows that the coefficient of the x term in the cubic equation must be −6. Finally, we calculate α1 α2 α3 by multiplying α1 α2 (already calculated) by α3 . We find that each positive and negative power r occurs 11 times, while r0 = 1 occurs 18 times. Thus Pof 18 i the product is equal to 11 i=1 r + 18 = −11 + 18 = 7. Thus the constant term in the cubic equation must be −7, and the equation itself must be x3 + x2 − 6x − 7 = 0, as asserted.

18.

To show that β1 = r + r18 , β8 = r8 + r11 , and β7 = r7 + r12 are the roots of x3 − α1 x2 + (α1 + α4 )x − (2 + α2 ) = 0, we check the symmetric functions of these three values. We have β1 + β8 + β7 = r + r18 + r8 + r11 + r7 + r12 = α1 , so the sum of the βi is the negative of the coefficient of the x2 term. Also, β1 β8 + β1 β7 + β8 β7 = (r9 + r12 + r7 + r10 ) + (r8 + r13 + r6 + r11 ) + (r15 + r + r18 + r4 ) = α1 + α4 . Therefore, the sum of products of the βi taken two at a time is equal to the coefficient of the x term. Finally, the product β1 β8 β7 = r16 + r14 + r17 + r2 + r5 + r3 + 2 = α2 + 2, and the product of the βi is the negative of the constant term. From the basic properties of the symmetric functions, we have that the βi are the roots of the given polynomial.

19.

2 From the q equation x − β1 x + 1 = 0, where β1

√

= r + r18 , we have x = p 18 1 (r − r18 )2 ) = 2 (r + r ±

1 β21 − 4) = 12 (r + r18 ± r2 − 2 + r17 ) = 2 (β1 ± 18 18 18 1 2 ((r + r ) ± (r − r )). Thus the two roots of this equation are r and r .

3 3

20.

A graphical analysis of y = x + 6x − 6 shows that the equation x + 6x = 6 has one real root and therefore two complex ones. Thus adjoining the real root gives an extension F of Q of degree 3, which does not contain the complex roots. They satisfy an equation of degree 2 over this field, so adjoining one of those roots gives us an additional extension of degree 2. Therefore, the splitting field of the cubic equation had degree 6 over Q. Therefore the Galois group of the equation has order 6. There are only two groups of order 6, S3 and the cyclic group. Every subgroup of the cyclic group is normal. In this case, however, field F is not a normal extension of Q because it does not contain the conjugates of the real root, so the subgroup of G corresponding to F is not normal.

It follows that G = S3 . This group does have a normal subgroup H of degree 3, namely the cyclic group generated by a cyclic permutation of three elements, say (1, 2, 3). The index of H in G is 2, so both the order of H and its index in G are prime, as required.

21.

Let us designate the real root of the equation by x0 and the two complex roots by x1 and x2 . The subgroup of G = S3 leaving the real field F fixed consists of the identity and a permutation interchanging the complex roots and leaving x0 fixed. The nonidentity element can then be expressed in the form xk → xk′ , where k′ ≡ 2k (mod 3), since 2 · 1 ≡ 2 and 2 · 2 ≡ 1. The other four elements in the Galois group can be expressed as xk → xk′ with k′ ≡ ak + b, a = 1, 2 and b = 1, 2. For example, if k′ ≡ k + 1, then this representation is equivalent to the permutation (0, 1, 2). If k′ ≡

134

Solutions k + 2, this representation is equivalent to the permutation (0, 2, 1). If k′ ≡ 2k + 1, this representation is equivalent to the permutation (0, 1), while if k′ ≡ 2k + 2, we get the permutation (0, 2).

22.

Let α be the real fifth root of 2 and ζ a primitive fifth root of unity. Then the zeros of x5 − 2 are α, ζα, ζ2 α, ζ3 α, and ζ4 α. An element of the Galois group is determined by its action on α and on ζ. The group element must take α to one of the roots of the equation and ζ to a power of ζ from 1 to 4. So define the group element σab (1 ≤ a ≤ 4, 0 ≤ b ≤ 4) by σab (ζ) = ζa and σab (α) = ζb α. The mapping which takes the substitution x′ ≡ ax + b (mod 5) to the group element σab is then an isomorphism of groups. To demonstrate this, we show that the mapping preserves the group operation. The composition of the substitutions x′ ≡ ax + b and x′ ≡ cx + d is x′ ≡ a(cx + d) + b ≡ acx + (ad + b). On the other hand, σab σcd (ζ) = σab (ζc ) = ζac and σab σcd (α) = σab (ζd α) = (σab (ζ))d σab (α) = ζad ζb α = ζad + b α. This group element corresponds under the mapping to the substitution x′ ≡ acx + (ad + b) as asserted, where all the operations are taken modulo 5. It follows that the Galois group and the group of substitutions are isomorphic and therefore that the Galois group has 20 elements.

23.

One example is x5 − 3x + 2. To check that this polynomial has three real roots and two complex roots, and therefore does not meet Galois’ criterion for a solvable fifth-degree polynomial, we use calculus. The derivative of f(x) = x5 − 3x + 2 isqf ′ (x) = 5x4 − 3.

24.

25.

Setting this equal to zero and solving gives us two critical values ± 4 35 . We calculate q q q q q q that f(− 4 35 ) = − 35 4 35 + 3 4 35 + 2 > 0 and that f( 4 35 ) = 35 4 35 − 3 4 35 + 2 < 0, where the last result comes from using a calculator. It follows that the graph of f(x) has two turning points, one above the x axis and one below, so that the graph crosses the x axis exactly three times. Thus f(x) has three real roots and two complex roots as asserted.

a b A matrix is in SL(2, p) if ad − bc = 1. We must count the number of solutions c d of this equation modulo p. Let us rewrite the equation as a = (1 + bc)/d. It is evident that if d ̸= 0, then any choices of b and c determine a value for a. There are p − 1 nonzero choices for d and p2 choices for b and c together, thus giving us p2 (p − 1) choices altogether. On the other hand, if d = 0, then our equation becomes bc = −1, and any non-zero choice for b determines a value for c. In this case, in addition, the choice of a is arbitrary. Thus, with d = 0, there are (p − 1)p solutions to the equation. Therefore, the total number of solutions is p2 (p − 1) +p(p − 1) = p(p + 1)(p − 1) = p(p2 − 1), and that is the order of the group SL(2, p). The group PSL(2, p) is the quotient group of SL(2, p) by its subgroup of multiples of the identity matrix. The only multiples of the identity matrix which have determinant 1 are I and (p − 1)I (because 1 and p − 1 are the only solutions to x2 ≡ 1 (mod p)). Therefore, the subgroup has only two elements and the order of the quotient group is half the order of SL(2, p), namely 12 p(p2 − 1).

First, note that the linear fractional transformations do act on P1 (p), because if (x1 , y1 ) ≡ (x2 , y2 ), then the transforms of both z1 = x1 /y1 and z2 = x2 /y2 are the same. Second, note that as mentioned in the solution to Exercise 24, the subgroup of SL(2, p) consisting of multiples of the identity only consists of two elements, I and (p − 1)I. Now let LF(p) denote the group of linear fractional transformations acting on P1 (p), and consider the function ϕ : PSL(2, p) → LF(p) such that if

Solutions

135

a b , then ϕ(M)(z) = (az + b)/(cz + d). We note that ϕ is a group homoc d

a b e f ae + bg af + bh morphism: = , while the composite of the two c d g h ce + dg cf + dh transformations takes z to

ez + f a gz +h + b aez + af + bgz + bh (ae + bg)z + (af + bh)

= = . ez + f c gz + h + d cez + cf + dgz + dh (ce + dg)z + (cf + dh) M =

It is clear that ϕ is an onto homomorphism. We will show that it is one-to-one. Assume ϕ(M) is the identity transformation. That is, assume that (az + b)/(cz + d) = z.

Therefore cz2 + (d − a)z + b = 0 for every z in P1 (p). Thus, c = b = 0 and a = d. But since ad = a2 = 1, we also know that a = 1 or a = p − 1. Therefore, the only elements of SL(2, p) taken to the identity are the elements in the subgroup of multiples of the identity. It follows that the kernel of ϕ is the identity of PSL(2, p), and that ϕ is a group isomorphism, as claimed.

26.

The sum of α + βi and γ + δi as complex numbers is (α + γ) + (β + δ)i, which mirrors Hamilton’s rule for addition of pairs. Also, the normal complex product of α + βi and γ + δi is (αγ − βδ) + (αδ + βγ)i, exactly in accord with Hamilton’s rule for multiplication of pairs.

27.

(3 + 4i + 7j + k)(2 − 3i + j − k) = (6 + 12 − 7 + 1) + (− 9 + 8 − 7 − 1)i + (3 + 14 − 3 + 4)j + (−3 + 2 + 4 + 21)k = 12 − 9i + 18j + 24k. To divide 3 + 4i + 7j + k by 2 − 3i + j − k, we multiply both dividend and divisor by 2 + 3i − j + k. The denominator is then 4 + 9 + 1 + 1 = 15, while the numerator is (6 − 12 + 7 − 1) + (9 + 8 + 7 + 1)i + (− 3 + 14 + 3 − 4)j + (3 + 2 − 4 − 21)k = 25i + 10j − 20k. It follows that the quotient is 53 i + 23 j − 43 k.

28.

In general, (a + bi + cj + dk)(e + fi + gj + hk) = (ae − bf − cg − dh) + (af + be + dg − ch)i + (ag + ce + df − bh)j + (ah + de + bg − cf)k.

To demonstrate that the modulus of the product is the product of the moduli, we must show that (ae − bf − cg − dh)2 + (af + be + dg − ch)2 + (ag + ce + df − bh)2 + (ah + de + bg − cf)2 = (a2 + b2 + c2 + d2 )(e2 + f2 + g2 + h2 ). To do this, note that the 16 squares of the terms on the left side are precisely the 16 products on the right. Also, every other term in the expansion on the left side occurs twice, once positive and once negative, so all of these terms cancel out.

29.

The general form of the expansion of f(x, y, z) is f(1, 1, 1)xyz + f(1, 1, 0)xyz + f(1, 0, 1)xyz + f(1, 0, 0)xyz + f(0, 1, 1)xyz + f(0, 1, 0)xyz + f(0, 0, 1)xyz + f(0, 0, 0)xyz.

Then, if we evaluate V = x−yz on every possible triple consisting of 0’s and 1’s, we find that x − yz = 0xyz + xyz + xyz + xyz − xyz + 0xyz + 0xyz + 0xyz = xyz + xyz + xyz − xyz.

30.

We interpret the equation xyz = 0 to mean there are no beasts which are clean and chew the cud, but do not divide the hoof. We interpret the equation xyz = 0 to mean there are no beasts which are clean but do not chew the cud or divide the hoof. We

136

Solutions interpret the equation xyz = 0 to mean there are no beasts which are not clean, but both chew the cud and divide the hoof.

31.

It is easiest to do this in matrix form. If we let

x X= , Xt = (x, y), and y

B=

a b , b c

then the quadratic form F can be written in the form F = Xt BX, where, in general, the exponent t indicates the transpose of a matrix. If we let

A=

α γ

β , δ

then the substitution can be written in either the form X = AX′ or Xt = X′t At . Thus, we have F′ = X′t At BAX′ . Let C = At BA, so F′ = X′t CX′ . To get back to F, we make the substitution, X′ = A−1 X, or X′t = Xt (A−1 )t . But (A−1 )t = (At )−1 . Since this substitution turns F′ = X′t CX′ into Xt (At )−1 CA−1 X = Xt (At )−1 At BAA−1 X = Xt BX = F, the substitution X′ = A−1 X is the desired inverse substitution. Given that the determinant of A is 1, we know that

δ −β −1 A = . −γ α We can write this substitution in the form x′ = δx − βy, y′ = −γx + αy.

32.

If AB = 0, then 0 = det(AB) = det(A) det(B). Therefore, either det(A) = 0 or det(B) = 0.

33.

The characteristic equation of A may be written as λ2 − (a + d)λ + ad − bc = 0. Thus, we need to show that A2 − (a + d)A + (ad − bc)I = 0. We calculate:

a b c d

2 − (a + d)

a c

b d

+

ad − bc 0 = 0 ad − bc

2 2

a + bc ab + bd a + ad ab + bd ad − bc − + 2 2 0 ac + cd bc + d ac + cd ad + d 34.

0

ad − bc

=

0 0 .

0 0

If we square the matrix L, we get  ab + 2bY + bd  X2 . bc + d2 + 2dY + Y2 X2

 2 a + 2aY + Y2 + bc  X2  ac + 2cY + cd X2

√

√

a(a + 2 ad−bc+d) √ −bc + bc = √ The entry in row 1, column 1 reduces to a + 2aa + dad+−2bc+ad = a. ad−bc a + d + 2 ad−bc Similarly, the entry in row 1, column 2 reduces to b; the entry in row 2, column 1 reduces to c; and the entry in row 2, column 2 reduces to d. It follows that L2 = M. 2

Solutions

137

√

35.

Since (det M)2 = det M, we must have det M p ≥ 0 for a square root to exist. Also, √ from the previous exercise, we must have X = a + d + 2 ad − bc ̸= 0. Therefore, since det M ≥ 0, X can be 0 only if det M = 0 and a + d = 0. Thus, if det M > 0, the square root exists. If det M = 0 and a + d > 0, the square root exists. Finally, if M is the zero matrix, the square root exists. To determine how many square roots exist in general, again we consider the formula from Exercise 34. It is straightforward to show that there are two choices for Y, the positive square root or the negative one, each of which will give a square root of M. Similarly, there are two choices for X as well, the positive square root or the negative one. It follows that as long as ad − bc > 0, then there are four square roots of M.

36.

Suppose PMP−1 = D, where D is a 3 × 3 matrix in Jordan canonical form. If N2 = D, then we set L = P−1 NP. Therefore, L2 = P−1 NPP−1 NP = P−1 N2 P = P−1 DP = M, so L is a square root of M. There are three possibilities for the matrix D. First, suppose that D is a diagonal matrix with λ, µ, ν, along the√diagonal.√So we choose N to be a √ diagonal matrix with the diagonal elements being λ, µ, ν. Thus this square root exists as long as the eigenvalues are non-negative. Second, suppose D is of the form 

λ D = 1 0

 0 0 . µ

0 λ 0

Then we can choose N to be √

λ

N =  2√1 λ 0

 0 0 . √ µ

√0

λ 0

It is straightforward to show that N2 = D, although again we require that λ and µ be positive. Finally, D could be of the form 

λ D = 1 0

 0 0 . λ

0 λ 1

In this case, we choose N to be  √  N= 

λ

1 √ 2 λ − 1√ 8λ λ

0

√

λ

1 √ 2 λ

0



 0 .  √ λ

Again, we can check that N2 = D (and, of course, require that λ > 0).

37.

The characteristic equation belonging to the quadratic form 2x2 + 4xy + 5y2 is (2 − λ)(5 − λ) − 4 = 0, or λ2 − 7λ + 6 = 0. The roots of this equation are λ1 = 1 and λ2 = 6.

To determine the solution (x1 , y1 ) corresponding to λ1 = 1, we solve (2 − 1)x + 2y = 0, or x + 2y = 0. Thus x1 = −2y1 . But since we want x21 + y21 = 1, we have 4y21 + y21 = 1,

138

Solutions or y21 = 15 . We pick y1 = − √1 and therefore x1 = √2 . To determine the solution 5 5 (x2 , y2 ) corresponding to λ2 = 6, we solve (2 − 6)x + 2y = 0, or −4x + 2y = 0. Thus y2 = 2x2 . Since also x22 + y22 = 1, we have x22 + 4x22 = 1, or x22 = √1 . We pick x2 = √1 5

5

and therefore y2 = √2 . Thus the orthogonal substitution which converts the original 5

quadratic form into the form u2 + 6v2 is x = √2 u + √1 v, y = − √1 u + √2 v.

5

38.

5

5

5

By calculating all 3 × 3 determinants, we see that each of them has value 0. On the other hand, 2 5

1 = 1, 3

so the order of the maximal non-vanishing determinant is 2. To solve the system, we solve the third equation for u: u = −v + 8x − y + 12z. We then substitute in the first and second equations. The first equation reduces to −v + 18x − y + 27z = 0, while the second reduces to −2v + 36z − 2y + 54z = 0. Since these equations are equivalent, we can solve either one for v. We get v = 18x − y + 27z. Substituting this value back into the equation for u gives us u = −10x − 15z. We can choose values for x, y, and z arbitrarily.

39.

Since the largest non-vanishing determinant of the matrix of coefficients has order 2, the rank of the matrix is 2. Given the solution to the system worked out in exercise 38, we see that the solution depends on three arbitrary constants, x, y, and z. Thus if we choose x = 1, y = 0, z = 0, we get the solution (−10, 18, 1, 0, 0). If we choose x = 0, y = 1, z = 0, we get the solution (0, −1, 0, 1, 0). If we choose x = 0, y = 0, z = 1, we get the solution (−15, 27, 0, 0, 1). These three vectors form a basis for the set of solutions.

40.

The system associated to the chosen basis for the set of solutions to the original equation is simply the system whose coefficients are those basis elements: −10u + 18v + x =0 −v + y = 0. −15u + 27v +z= 0

Since there is a 3 × 3 non-vanishing determinant, the rank of this system is 3. We solve this system by solving the first equation for x, the second for y, and the third for z. Thus, x = 10u − 18v, y = v, and z = 15u − 27v. A basis for the set of solutions to this system is found by first setting u = 1, v = 0, and then setting u = 0, v = 1.

Thus the basis consists of the two vectors (1, 0, 10, 0, 15) and (0, 1, −18, 1, −27). The system associated to this basis is u

+ 10x + 15z = 0 . v − 18x + y − 27z = 0

The solution to this system, namely u = −10x − 15z, v = 18x − y + 27z, is the same as the solution to the original system, as worked out in Exercise 38.

41.

If we make the substitution x = αx′ + βy′ , y = γx′ + δy′ , we find that the quadratic form f = ax2 + 2bxy + cy2 is transformed into the quadratic

Solutions

139

form f′ = a(αx′ + βy′ )2 + 2b(αx′ + βy′ )(γx′ + δy′ ) + c(γx′ + δy′ )2 = 2 2 ′2 (aα + 2bαγ + cγ )x + 2(aαβ + bαδ + bβγ + cγδ)x′ y′ + (aβ2 + 2bβδ + cδ2 )y′2 .

The discriminant of f′ is then (aαβ + bαδ + bβγ + cγδ)2 − 2 2 2 2 (aα + 2bαγ + cγ )(aβ + 2bβδ + cδ ). If we simplify this, we get 2(ac − b2 )αβγδ + (b2 − ac)α2 δ2 + (b2 − ac)β2 γ2 = (b2 − ac)(α2 δ2 − 2αβγδ + β2 γ2 ) = (b2 − ac)(αδ − βγ)2 = b2 − ac. Thus the discriminant of f ′ is the same as the discriminant of f.

42.

There are three Abelian groups of order 8. The first is the cyclic group of order 8.

The second has two generators, α and β, with α4 = 1 = β2 and αβ = βα. The third has three generators α, β, and γ, with α2 = β2 = γ2 = 1, and with all the generators commuting with each other. There are also two non-Abelian groups of order 8. The first has two generators, α and β, with α4 = 1 = β2 and with αβ = βα3 . The second one has two generators, α and β, with α4 = 1, β2 = α2 , and αβ = βα3 .

43.

Assume that the group G is not cyclic. Then no element of the group has index p2 .

Therefore, every element other than the identity must have index p. Choose one such element α, and let H be the subgroup of G generated by α. According to one of the Sylow theorems, H must be normal in a group of order p2 , therefore in G itself. Now let β be another element of order p not in H, and let K be the subgroup of G generated by β. K is also normal in G, and by counting we see that every element of G can be written as αi β j , 0 ≤ i < p, 0 ≤ j < p. Furthermore, H ∪ K = {1}, because no element of H could generate K and vice-versa. It remains to determine the product βα. We know that αβα−1 β−1 = (αβα−1 )β−1 ∈ K, because K is normal. Similarly αβα−1 β−1 = α(βα−1 β−1 ) ∈ H, because H is normal. Therefore αβα−1 β−1 = 1, or αβ = βα. In other words, the group G is Abelian. Therefore, it is simply the product of two cyclic groups of order p. Thus, there are exactly two groups of order p2 , the cyclic group of order p2 and the product of two cyclic groups of order p. Both are Abelian.

44.

Since S−1 TS = T r , we have S−2 TS2 = S−1 S−1 TSS = S−1 T r S = (S−1 TS)r = (T r )r = 2 j T r . Similarly, forq any j, we have S−qj TSj = T r . In particular, setting j = q, we have −q q r r −1 T = S TS = T . It follows that T = 1 and therefore that r q − 1 is divisible by q p; that is, r ≡ 1 (mod p).

45.

We can take x3 + x + 1 ≡ 0 as the irreducible congruence modulo 5. This is irreducible, because none of the residues modulo 5 satisfy this congruence. If we then assume that α is a solution of this congruence, the set F = {a0 + a1 α + a2 α2 |0 ≤ a0 , a1 , a2 ≤ 4, α3 = −α − 1} is a field under the ordinary operations and has order 53 = 125.

46.

If G satisfies the usual axioms of a group, namely a set with an associative binary operation which has an identity and in which every element has an inverse, then we see that G satisfies Weber’s criteria. First, Weber states the associative law. Second, the equation AX = B will have the unique solution X = A−1 B. And if θr θ = θs θ, then on multiplying both sides on the right by θ−1 , we get θr = θs . Now let us assume that G satisfies Weber’s criteria. We know that the operation is associative. To find the identity, find the unique solution to XA = A for any A. To see that X is the identity, note that since BXA = BA, we can cancel the A to get BX = B, where B is arbitrary.

Similarly, BXB = BB, so by canceling the B on the left, we have XB = B. Thus X is the identity. Therefore, the unique solution Y to the equation AY = X is the inverse of A. Thus, a set satisfying Weber’s criteria also satisfies the modern group conditions.

140

Solutions 47.

The standard modern definition of a field is a set F with two binary operations, + and ·, under the first of which the set is an Abelian group, and under the second of which F − {0} is also an Abelian group. The only further requirement is that the distributive law holds: a(b + c) = ab + ac. Weber has three other axioms in his definition: a(−b) = −ab, (−a)(−b) = ab, and a · 0 = 0. These three axioms can, however, be proved from the other axioms in the field definition.

CHAPTER 22 1.

Given that limx→∞ f(x + 1) − f(x) = ∞, we know that given any positive number M, there exists a positive number N such that if x ≥ N, then f(x + 1)− f(x) > M. Therefore, for i = 1, 2, . . . n, we have f(N + i) − f(N + i − 1) > M. The arithmetic mean of these n expressions also satisfies the same inequality. Thus f(N + n) − f(N) > M, n

or

f(N + n) − f(N) = M + α, n

where α > 0. Let x = N + n. The equation then becomes f(x) − f(N) =M+α x−N

or f(x) = f(N) + (x − N)(M + α).

Therefore, f(x) f(N) = + x x

N 1− (M + α). x

Since N is fixed, we see that as x → ∞, f(xx) approaches M + α. Thus eventually, f(x) x > M and therefore, lim

x→∞

2.

f(x) = ∞. x

We assume that a > 1, Then limx→∞ ax + 1 − ax = limx→∞ ax (a − 1) = ∞. It follows from the theorem in Exercise 1 that lim

x→∞

ax = ∞. x

Also, we have limx→∞ log(x + 1) − log x = limx→∞ log( x +x 1 ) = limx→∞ log(1 + 1x ) = 0. By the theorem in the text, we have lim

x→∞

3.

log x = 0. x

Note that 2 π

Z ∞ 0

2 x2 dt = π t2 + x 2

Z ∞ 0

1 t2 +1 x2

dt.

Solutions

141

Now assume that x > 0 and let u = xt . Then du = 1x dt, or dt = x du. Using this substitution, the integral becomes 2x π

Z ∞ 0

∞

1 2x π 2x arctan u = = x. du = π π 2 u2 + 1 0

If x < 0, then u < 0. Therefore, the integral after the substitution becomes 2x π

Z −∞ 0

∞

1 u2 + 1

du = −

2x π 2x =− arctan u = −x. π π 2 0

Thus, the integral is equal to x if x > 0 and −x if x < 0. It follows that the integral is equal to |x|, as asserted.

4.

Let ϵ > 0. We need to find δ > 0 such that | sin(x + α) − sin x| < ϵ, whenever |α| < δ.

We choose δ = ϵ. Then for |α| < δ, we have 1 2

| sin(x + α) − sin x| = 2 sin α cos x +

1 α 2

≤2·

1 |α| · 1 = |α| < δ = ϵ.

2

Therefore, sin x is continuous at x.

5.

Given any ϵ > 0, we know there exists an h such that if x ≥ h, then k−ϵ<

f(x + 1) < k + ϵ. f(x)

Since this inequality is true for f(hf(+h i+−i)1) for i = 1, 2, . . . , n, it is true for the geometric mean of all these quotients. Therefore, r k−ϵ< n

f(h + 1) f(h + 2) f(h + n) ··· < k + ϵ. f(h) f(h + 1) f(h + n − 1)

It follows that ( f(fh(+h)n) )1/n = k + α, where −ϵ < α < ϵ. We set x = h + n. Then 1/(x−h) f(x) = k + α, or ff((hx)) = (k + α)x−h , or f(x) = f(h)(k + α)x−h . We now raise f(h) each side to the power 1x . We get f(x)1/x = f(h)1/x (k + α)1 − h/x . As x → ∞, the expression f(h)1/x → 1 and the expression (k + α)1 − h/x → (k + α). It follows that f(x)1/x approaches k + α, where −ϵ < α < ϵ. Because ϵ is arbitrary, we see that f(x)1/x approaches k, as desired.

6.

We know that limx→∞ x +x 1 = 1. Therefore, by the theorem of Exercise 5, we have limx→∞ x1/x = 1.

7.

Given ϵ > 0, we can find n such that 2n 1− 2 < ϵ. But n!1 < 2n 1− 1 . So, 1 1 1 1 1 1 1 + + + · · · < n − 1 + n + n + 1 + · · · < n − 2 < ϵ. n! (n + 1)! (n + 2)!

2 2 2 2

142

Solutions Therefore, any finite sum of the reciprocals of the factorials where n! > 2n − 1 is also less than ϵ, and by the Cauchy criterion, the series converges.

8.

Choose ϵ > 0. We need to find n such that |f(ai ) − f(a)| < ϵ, whenever i > n. Since f is continuous at a, we know there exists a δ such that |f(x) − f(a)| < ϵ whenever |x − a| < δ. Since the sequence {ai } converges to a, we know that there is an n so that |ai − a| < δ whenever i > n. But then when i > n, we have |f(ai ) − f(a)| < ϵ, as desired.

9.

Each of the ui (x) is a continuous function in a neighborhood of x = 1, and sk (x) = P∞ Pk k k i=1 ui (x) = x . We let s(x) = i=1 ui (x) = limk→∞ x . We see that for any x < 1, s(x) = 0; for x = 1, s(x) = 1; while for x > 1, s(x) is not defined. Therefore, s(x) is not a continuous function of x in any neighborhood of x = 1. The problem with Cauchy’s proof in this instance is that although given any positive ϵ, for each k there exists δk such that |sk (1 + a) − sk (1)| = |(1 + a)k − 1| < ϵ whenever |a| < δk , there is no δ that will work for all k. In fact, since the curve y = xk gets steeper near 1 with increasing k, for given ϵ, we have lim δk = 0.

10.

We have d sin x dx

= limα→0 sin(x+αα)−sin x = limα→0 = limα→0

11.

sin 12 α cos 1 α 2

2 sin 21 α cos(x+ 12 α) α

x + 21 α = 1 · cos x = cos x.

If y = ax , then ax (ai − 1) ax + i − ax ai − 1 = lim = ax lim . i→0 i→0 i→0 i i i

y′ = lim

To calculate the last limit, we set ai = 1 + β, or i = loga (1 + β), where β approaches 0 with i. We need to determine now limβ→0 log (β1+β) , but it is easier to work with the a

log (1+β)

reciprocal. We have a β = loga (1 + β)1/β . As β approaches 0, the argument of this logarithm function approaches e. Therefore, 1 ai − 1 = , i→0 i loga e lim

12.

and

y′ =

ax . loga e

Given that A < abii < B for each i, we have bi A P < ai < bi BP for eachP i. If we add together these inequalities for i = 1, 2, . . . , n, we get ( b ) A < a < ( bi )B. If we divide i i P by bi , we get Pn ai < B, A < Pi=1 n i=1 bi as desired.

13.

Let us assume that f(xj ) is the smallest value of the f(xi ) and that f(xk ) is the largest of these values. Let us designate by S the sum f(x0 )(x1 − x0 ) + f(x1 )(x2 − x1 ) + · · · + f(xn − 1 )(xn − xn − 1 ). We then have f(xj )(b − a) ≤ S ≤ f(xk )(b − a), or S f(xj ) ≤ b− a ≤ f(xk ). By the intermediate value theorem, there is a value x ∈ [a, b]

Solutions

143

S such that f(x) = b− a . Therefore, S = f(x)(b − a). But since x0 ≤ x ≤ xn , we can write x as x = x0 + θ(b − a), where 0 ≤ θ ≤ 1. Thus we have S = (b − a)f(x0 + θ(b − a)), as desired.

14.

We take the partition 1, 45 , 32 , 47 , 2, 94 , 25 , 11 4 , 3. Since the length of each subinterval is 1 1 5 3 7 9 5 11 , the sum in Exercise 13 is ( f ( 1 ) + f ( 4 4 4 ) + f( 2 ) + f( 4 ) + f(2) + f( 4 ) + f( 2 ) + f( 4 )) = 1 1 1212 4 · 16 (64 + 85 + 108 + 133 + 160 + 189 + 220 + 253) = 64 . We must find x so that 2 1212 2f(x) = 1212 64 , or so that f(x) = 128 = 9.46875. Thus x must satisfy x + 3x = 9.46875, and x ≈ 1.92. Therefore, 1 + 2θ ≈ 1.92, and θ ≈ 0.46.

15.

First, we show that the quantity U has the property that all smaller x have property M. For if there were an x < U such that x did not have property M, then U − x > 2Dq for some q, or x < U − 2Dq . This means that there is an element ui in Bolzano’s sequence such that x < ui . But this contradicts the definition of the ui , that M is valid for all x less than ui . Second, we show that there is no V > U such that all smaller x have property M. If such a V existed, then V − U > 2Dr for some r. Thus U + 2Dr < V and therefore all x smaller than U + 2Dr satisfy property M. But this means that there is a ui in Bolzano’s sequence such that ui > U, contradicting the definition of U as the limit of the (strictly increasing) sequence. Therefore, U is the greatest number of those of which it can be asserted that all smaller x have property M. In other words, U is the least upper bound of all the x which have property M.

16.

Since the number 0.66 . . . 6 is in A no matter how many 6’s are in the expansion, the least upper bound of A must be greater or equal to 0.66 . . . = 32 . But if it were greater than 23 , there would be an element in A greater than 32 , which is impossible. So the least upper bound is equal to 32 .

17.

Since (1 + 1)3 > 3, while 13 < 3, we begin Bolzano’s process by trying 1 + 12 . We check that (1 + 12 )3 > 3, so we next try 1 + 14 . In this case, (1 + 41 )3 < 3, so the first two terms of the desired sequence are 1, 1 + 14 . We next note that (1 + 14 + 81 )3 < 3, and also 1 3 1 (1+ 14 + 18 + 16 ) < 3. So 1+ 41 + 18 and 1+ 14 + 18 + 16 are the next two in the sequence. After 1 1 1 that, we try adding, in turn, 32 , 64 , and 128 , each of which gives a value too large. But 1 1 3 1 1 (1 + 14 + 81 + 16 + 256 ) < 3, so 1 + 14 + 18 + 16 + 256 is the next number in the sequence.

1 1 1 1 1 3 Again, if we add 512 or 1024 , the value is too large. But (1+ 14 + 81 + 16 + 256 + 2048 ) < 3.

1 1 1 + 256 + 2048 is equal to 1.4418945, which to three decimal In decimals, 1 + 41 + 81 + 16 √ places is equal to 3 3.

18.

We calculate ϕ′ (x) = αmemx and ϕ′′ (x) = αm2 emx = Aα emx . Therefore, ϕ ( x) αemx = α mx = A. ′′ ϕ (x) A e Similarly, ψ′ (y) = −βn sin ny and ψ′′ (y) = −βn2 cos ny = − Aβ cos ny. Therefore,

−

ψ(y) β cos ny =− β = A. ψ′′ (y) − A cos ny

144

Solutions We then have v = ae−nx cos ny = ϕ(x)ψ(y). Therefore, ψ(y) ϕ(x) ∂2v ∂2v ψ(y) − ϕ(x) = 0, + = ϕ′′ (x)ψ(y) + ϕ(x)ψ′′ (y) = A A ∂ x2 ∂ y2 so v is a solution to the given partial differential equation.

19.

If m = n, then set u = mx and du = m dx and integrate: Z π

Z π sin2 mx dx =

sin mx sin mx dx = 0

0

=

1 m

Z mπ

sin2 u du

0

mπ 1 u sin 2u 1 mπ π = · − = . m 2 4 m 2 2 0

If m ̸= n, we integrate by parts twice. First, set u = sin mx, dv = sin nx dx. Then du = m cos mx dx and v = − 1n cos nx. The integral is Z π 0

π

m 1 sin mx sin nx dx = − sin mx cos nx + n n 0 Z π m = cos mx cos nx dx. n 0

Z π cos mx cos nx dx 0

Next, we set u = cos mx, dv = cos nx dx, and calculate du = −m sin mx dx and v = 1 n sin nx. Our integral then becomes

π Z m 1 m π sin mx sin nx dx sin nx cos mx + n n n 0 0 Z m π = sin mx sin nx dx. n 0

Z π

sin mx sin nx dx = 0

Therefore, Z

m π sin mx sin nx dx = 0 1− n 0 20.

Z π and

sin mx sin nx dx = 0.

0

We first note that, analogously to Exercise 19, (

Z π cos mx cos nx = 0

0, if m ̸= n; π 2 , if m = n.

Rπ Furthermore, we note that 0 b0 cos nx = 0 for every positive integer n. Therefore, if we multiply the given equation throughout by cos nx for each n in turn and integrate from 0 to π, we get Z π 0

π 1 πϕ(x) cos nx dx = bn , 2 2

Solutions

145

Rπ so bn = 0 ϕ(x) cos nx dx. Also, if we just integrate both sides of the equation without Rπ multiplying by anything, and note that 0 cos nx = 0 for every positive integer n, we Rπ 1 Rπ Rπ find that 0 2 πϕ(x) dx = 0 b0 dx = b0 π, and therefore that b0 = 12 0 ϕ(x) dx.

21.

We write the sine series for 21 x in the form 1 x π = a1 sin x + a2 sin 2x + a3 sin 3x + · · · .

2 π Then we calculate Z π ak = 0

x 1 sin kx dx = π π

Z π x sin kx dx.

0

We integrate by parts by setting u = x, dv = sin kx dx, du = dx, and v = − 1k cos kx.

Then Z

π 1 R π 1 π π x sin kx dx = π1 − kx cos kx 0 + kπ cos kx dx = − 1k cos kπ + k21π sin kx 0 0 π 0 = − 1k cos kπ.

Note that if k is even, this value is − 1k and if k is odd, the value if 1k . Therefore, the sine series is given by 1 1 1 1 x = sin x − sin 2x + sin 3x − sin 4x + · · · .

2 2 3 4 1 1 5π If we replace x by π2 , we get π4 = sin π2 − 12 sin π + 31 sin 3π 2 − 4 sin 2π + 5 sin 2 − · · · , π 1 1 or 4 = 1 − 3 + 5 − · · · . This is the same result we get by setting x = 1 in the power series for arctan x and is also the result which Leibniz found by using his transmutation theorem applied to a circle.

22.

23.

First, we note that the series converges for every x ∈ [0, 1], so f(x) is well-defined.

Next, note that ϕ(y) is continuous for all y except where y is an odd multiple of 21 .

Therefore, f(x) is certainly continuous for any x except possibly where mx is an odd p where p is odd and relatively multiple of 21 for some m, namely, when x is of the form 2n prime to n. When x0 is of that form, however, we note further that the difference in the function values between a number slightly larger than x0 and one slightly smaller is provided by the sum of the terms ϕ((jnjnx)20 ) , where j is odd. In other words, limx→x+0 f(x) = f(x) − [ 21 n12 + 12 321n2 + 12 521n2 + · · · ] and limx→x− f(x) = f(x) + [ 12 n12 + 21 321n2 + 21 521n2 + · · · ].

0 p It follows that f is not continuous at the values of x of the form 2n .

P∞ Pk We let s(x) = n=0 un (x), sk (x) = n=0 un (x), and rk (x) = s(x) − sk (x). We want to show that s(x) is continuous in I. So let x ∈ I and choose ϵ > 0. Then, since the series converges uniformly in I, we can find N such that |rn (x)| < ϵ3 for every n > N and for every x ∈ I. Choose some n > N. Since sn (x) is continuous, there is a δ > 0 such that |sn (x + a) − sn (x)| < ϵ3 whenever |a| < δ and x + a ∈ I. Then, if |a| < δ and x + a ∈ I, we have |s(x + a) − s(x)| = |sn (x + a) + rn (x + a) − sn (x) − rn (x)| ≤ |sn (x + a) − sn (x)| + |rn (x + a)| + |rn (x)| ≤ ϵ3 + ϵ3 + ϵ3 = ϵ. Therefore, s(x) is continuous in I as claimed.

146

Solutions 24.

First, since the denominator in the definition of vk does not vanish as h approaches 0, we have lim vk (h) = uk + lim

2h

h→0 ((k − 1)h + 1)(kh + 1)

h→0

= uk .

Second, we claim that u1 + u2 + · · · + uk = k +k 1 . This is clearly true for k = 1. To prove this by induction, we assume the truth of the statement and calculate the sum up to k + 1: u1 + u2 + · · · + uk + uk + 1 =

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

=

(k + 1)2 k2 + 2k + 1 k+1 = = . (k + 1)(k + 2) (k + 1)(k + 2) k + 2

P Thus the result is true for all k. We then have uk = limk→∞ k +k 1 = 1. Third, we k 2kh claim that v1 (h) + v2 (h) + · · · + vk (h) = k + 1 + kh + 1 . Again, we note that this is true for k = 1. We assume the truth of this statement for k and show that it is also true for k + 1: k+1 X i=1

vi (h) =

k X

vi (h) + vk + 1 (h)

i=1

k 2kh 1 2h + + + k + 1 kh + 1 (k + 1)(k + 2) (kh + 1)((k + 1)h + 1) k + 1 2kh((k + 1)h + 1) + 2h = + k + 2 (kh + 1)((k + 1)h + 1) k + 1 2k(k + 1)h2 + 2kh + 2h + = k + 2 (kh + 1)((k + 1)h + 1) 2h(k + 1)(kh + 1) k+1 = + k + 2 (kh + 1)((k + 1)h + 1) 2(k + 1)h k+1 = + . k + 2 (k + 1)h + 1 =

P k 2kh The result is therefore trueP by induction. Then vk (h) = lim P k→∞ k + 1 + kh + 1 = 1 + 2 =P 3. Finally,P limh→0 vk (h) = limh→0 3 = 3, while uk = 1. Therefore, limh→0 vk (h) ̸= uk , as claimed.

25.

Let A1 be the subset of the rational numbers consisting of all rational cuts in A1 and A2 the subset of the rational numbers consisting of all rational cuts in A2 . Then the set of rational numbers has been split into two subsets A1 and A2 such that every element of A1 is less than every element of A2 . Let α be the cut (or real number) determined by A1 and A2 . Then α belongs to A1 or A2 . Now suppose β < α. Then there is a rational number c between β and α. By the definition of A1 , we know that c ∈ A1 . Therefore, c ∈ A1 . It follows that β ∈ A1 . Similarly, if β > α, then β ∈ A2 . Therefore, if α ∈ A1 , then α is the greatest number in A1 . Similarly, if α ∈ A2 , then it is the smallest element in that set.

Solutions

147

26.

If α = (A1 , A2 ) and β = (B1 , B2 ), we define α < β to mean that A1 is strictly included in B1 ; that is, there are rational numbers in B1 which are not in A1 . Now, this ordering satisfies the transitive law. Suppose α < β and β < γ = (C1 , C2 ). Then A1 is strictly included in B1 and B1 is strictly included in C1 . Therefore A1 is strictly included in C1 , so α < γ. We also show the trichotomy law: for any two real numbers α, β, either α < β or α = β or β < α. So let us suppose that α ̸= β. We compare the sets A1 and B1 .

If one is strictly included in the other, then we are done, according to our definition. So suppose neither is strictly included in the other. Thus, there exists a rational number a with a ∈ A1 and a ∈ / B1 and a rational number b with b ∈ B1 and b ∈ / A1 . By the definition of a cut, we must have a ∈ B2 and b ∈ A2 . Since a ∈ A1 and b ∈ A2 , we have a < b. But since b ∈ B1 and a ∈ B2 , we also have b < a. This is a contradiction, and the trichotomy law holds.

27.

Let α = (A1 , A2 ) and β = (B1 , B2 ). To define α + β, we need to define a particular set C1 of rationals which is the “lower” set of a pair defining a cut. We let C1 be the set of all rational numbers formed by adding any element in A1 to any element in B1 . Let C2 be all the remaining rational numbers. To show that γ = (C1 , C2 ) is a cut, we must show that every number in C1 is less than every number in C2 . So suppose c ∈ C1 and d ∈ C2 . Then c = a + b, where a ∈ A1 and b ∈ B1 . Since d is not the sum of any pair of numbers from A1 and B1 , then d − a ∈ / B1 . Therefore, d − a ∈ B2 . So d − a is larger than any number in B1 . In particular, d − a > b and c = a + b < d. Thus γ = (C1 , C2 ) is a cut. To show that α + β = β + α, we just note that the rational numbers formed by adding any element in A1 to any element in B1 are the same as those formed by adding any element in B1 to any element in A1 .

28.

Suppose {αi } form a bounded increasing sequence of cuts. There is then one cut β, and therefore infinitely many, such that αi < β for all i. So let A2 be the set of all such β and A1 be all cuts not in A2 . Then any element β1 ∈ A1 has the property that αi > β1 for some i. So β1 is less than every cut β2 ∈ A2 . By Exercise 25, there is a cut γ which is either the greatest in A1 or the least in A2 . The first case is impossible, since the sequence is increasing. Therefore γ ∈ A2 and is the desired limit number.

Now take an increasing sequence of real numbers defined by fundamental sequences (a1i ), (a2i ), (a3i ), . . ., which is bounded above by the rational number B. Note that b can be thought of as the fundamental sequence (b, b, b, . . .). Since the sequence is an increasing sequence, we can replace each real number by an equivalent sequence for which all of the components of (a1i ) are less than all of the components of (a2i ) which are in turn less than all of the components of (a3i ), and so on. Now consider the sequence a of rational numbers (a11 , a22 , a33 , . . .). This sequence is an increasing sequence of rational numbers, bounded above by b. This sequence is a fundamental sequence, for if not, then for some ϵ, there would be infinitely many pairs of numbers in the sequence whose distance from each other was greater than ϵ, and thus the sequence would not be bounded. Therefore a is a fundamental sequence of rational numbers which therefore represents a real number, and this number is the limit of the original sequence of real numbers.

29.

Since {bi } is a fundamental sequence not defining the limit 0, there are positive numbers r and L such that r < |bi |, |bj | < L for every i, j. In particular, for every i, j, we have |bi bj | > r2 or |b1b | < r12 . Also, since {ai } is a fundamental sequence, there is a number i j

148

Solutions 2

K such that |ai | < K for every i. So, given ϵ > 0, choose N so that |bi − bj | < r2Kϵ and 2 |ai − aj | < r2Lϵ whenever i, j > N. Then ai aj − bi bj

= <

bj ai − bi aj |bj ai − bi aj | |bj ai − bj aj + bj aj − bi aj | < = bi bj r2 r2 |bj (ai − aj )|

r2

+

|aj (bi − bj )|

r2

<

L r2 ϵ K r2 ϵ ϵ ϵ + = + = ϵ. r2 2L r2 2K 2 2

Therefore, { abii } is a fundamental sequence.

30.

We first show that the product sequence {ai bi } is a fundamental sequence. We know that there exists a number K such that |ai | < K for all i and also a number L such that |bi | < L for all i. So given ϵ > 0, choose N so that for i, j > N, we have ϵ ϵ and also |ai − aj | < 2L . Then |ai bi − aj bj | = |ai bi − ai bj + ai bj − aj bj | ≤ |bi − bj | < 2K ϵ ϵ |ai (bi − bj )| + |bj (ai − aj )| < K 2K + L 2L = ϵ2 + ϵ2 = ϵ. Thus {ai bi } is a fundamental sequence. To show that the product makes sense, we need to show that if {a′i } is any other sequence equivalent to A and if {b′i } is any other sequence equivalent to B, then {a′i b′i } is equivalent to {ai bi }. So given ϵ > 0, choose M so that for i > M, we have ϵ ϵ and |bi − b′i | < 2K , where L and K are chosen as above to work for both |ai − a′i | < 2L the original and primed sequences. Then |ai bi − a′i b′i | = |ai bi − ai b′i + ai b′i − a′i b′i | ≤ ϵ ϵ + L 2L = ϵ, and the two sequences are equivalent. |ai (bi − b′i )| + |b′i (ai − a′i )| < K 2K Finally, note that if C = {ai bi }, then B = {bi } = { aai bi i }, so B = CA , where the division is defined as in Exercise 29.

31.

We can take the set P to be P = { m1 − m(n1+ 1) }, where m and n are positive integers.

Then the set P′ of limit points of P is {1, 21 , 31 , 14 , . . .}. The set P′ only has one limit point, namely 0. Thus P′′ = {0}.

32.

Suppose b were in the original list. Then b = rm for some m. But then consider the digit bm . According to the definition of b, bm ̸= amm . On the other hand, the mth digit of rm is in fact amm . We therefore have a contradiction and b cannot be on the original list. Therefore, such a list cannot be a complete list of real numbers in the interval (0, 1).

33.

x is an odd function, and since the integral of Let f(x) = 1 e+ x2 = cos 1x ++ xi 2sin x . Since 1sin + x2 an odd function over the entire real line must be 0, we have

ix

Z ∞

eix dx = 2 −∞ 1 + x

Z ∞

cos x + i sin x dx = 1 + x2 −∞

Z ∞

cos x dx .

2 −∞ 1 + x

Thus to calculate the desired integral, we will begin by integrating the complex funciz tion f(z) = 1 e+ z2 around the path consisting of that part of the real axis between −r and r (r > 1) and of the semicircle of radius r in the upper half of the complex plane extending from r to −r. Since f(z) has a single pole at z = i in the region bounded by this curve, the integral of f(z) around this path is 2πiR(f, i), where R(f, i) is the residue of f at i. We calculate eiz e−1 eiz eiz = lim = .

R(f, i) = lim(z − i) 2 = lim(z − i) z→i (z + i)(z − i) z→i z + i 2i z + 1 z→i

Solutions

149

−1

Therefore, the integral of f(z) around the path is 2πi e2i = πe . We now calculate the integral of f(z) over the semicircle γ of radius r in the upper half of the complex plane by setting z = reiθ and dz = ireiθ dθ. We get Z

Z f(z) dz = γ

eiz dz = 2 γ z +1

Z π 0

iθ

eire ir dθ . r2 e2iθ + 1

But since |eiz | = |ei(x + iy) | = |eix−y | = |e−y | ≤ 1 for z in the upper half of the complex plane, and since |r2 e2iθ + 1| > r2 − 1, we now have Z π Z dθ πr f(z) dz ≤ r = 2 .

2 −1 r r −1 0 γ As r → ∞, we note that this integral approaches 0. It follows that the integral over the entire path as r → ∞ is given by the limit of the integral over the line segment from z −r to r as r → ∞. And because the integral of g(z) = 1cos over that line segment + z2 is equal to the integral of f(z) over the same segment, and since the latter integral is equal to πe , we have shown that Z ∞ cos x π dx = .

2 e 1 + x −∞ 34.

We consider the integral of the complex function f(z) = 1 +1 z6 over the path γ consisting of that part of the real axis between −r and r (r > 1) and of the semicircle of radius r in the upper half of the complex plane. The function f(z) has three poles in the region bounded by this path, at z1 = eπi/6 , z2 = eπi/2 , and at z3 = e5πi/6 . Therefore, R f(z) dz = 2πi(R(f, z1 ) + R(f, z2 ) + R(f, z3 )). But as r → ∞, the integral of f(z) over γ the semicircle approaches 0 since the Rabsolute value R ∞of the denominator of the function grows without bound. Therefore, I = γ f(z) dz = −∞ f(x) dx = 2πi(R(f, z1 ) +R(f, z2 ) + R(f, z3 )). We must calculate the three residues. √

e−5πi/6 1 z − eπi/6 1 3 1 R(f, z1 ) = lim = lim = = − − i .

6 5 6 6 2 2 z→eπi/6 z + 1 z→eπi/6 6z Similarly, R(f, z2 ) = 61 e−5πi/2 = 16 e−πi/2 = − 61 i and R(f, z3 ) = 61 e−25πi/6 = 16 e−πi/6 =

√ 3 1 1 1 1 i 6 ( 2 − 2 i). Thus the sum of the three residues equals − 6 2i = − 3 i, and I = 2πi(− 3 ) = 2π 3 , as claimed.

35.

We have ∂u ∂u ∂ x dx + ∂ y dy + i

∂v ∂v ∂ x dx + ∂ y dy

dw du + i dv = = dz dx + i dy dx + i dy

∂v ∂v ∂u ∂u + i dx + − i ∂x ∂x ∂y ∂ y i dy = = dx + i dy

∂u + i ∂∂ vx (dx + i dy) ∂ u ∂v = ∂x = +i . dx + i dy ∂x ∂x

∂u ∂v ∂x + i ∂x

dx + ∂∂ux + i ∂∂ vx i dy dx + i dy

150

Solutions 36.

The vector v = ( ∂∂px , ∂∂py , ∂∂pz ) dp is an infinitesimal tangent vector to the surface S in the direction given by the transform of the p axis, while the vector w = ( ∂∂qx , ∂∂qy , ∂∂qz ) dq is an infinitesimal tangent vector to S in the direction given by the transform of the q axis.

Thus, we can consider the element of surface dS to be the infinitesimal parallelogram determined by v and w. But the area of this parallelogram is given by |v × w| and the length of this cross product is "

∂(y, z) ∂(p, q)

2

+

∂(z, x) ∂(p, q)

2

+

∂(x, y) ∂(p, q)

2 #1/2 dp dq,

as claimed.

37.

Given that σ = Ai + Bj + Ck is a vector field with divRσ = 0, we need to Rfind X, Y, Z such B C that Rif τ = Xi + Yj + Zk, then R A σ = curl τ. So set X =R A 3 dz + f1 (x, y) −R B3 dy + f2 (x, z), C Y = 3 dx + g1 (y, z) − 3 dz + g2 (x, y), and Z = 3 dy + h1 (x, z) − 3 dx + h2 (y, z), where the integrations are all partial with respect to the given variable, so that in each case, the “constant of integration” is a function of the remaining variables. Then ∂Z ∂Y A 1 − = − ∂y ∂z 3 3

=

38.

2A 1 + 3 3

Z Z

∂B ∂h 1 dx + 2 − ∂y ∂y 3

Z

∂C ∂g A dx − 1 + ∂z ∂z 3

∂A ∂h ∂g ∂h ∂g dx + 2 − 1 = A + k(y, z) + 2 − 1 . ∂x ∂y ∂z ∂y ∂z

We then need to choose g1 and h2 so that ∂∂gz1 − ∂∂hy2 = k. This is possible by general results on differential equations, at least if the domain is reasonable. A similar calculation shows that B and C are the other two parts of curl τ.

RR R curl σ · da = 0, where C is a closed curve bounding If curl σ = 0, then C σ · dr = A the region A. Therefore, if γ is any curve connecting two points, and if −δ designates any other curve connecting the two points but in the opposite direction, then the curve C consisting first of γ and then of −δ is a closed curve, over which the integral is 0.

But this integral is the difference of the integrals over γ and over δ. It follows that the integrals over γ and δ are equal, so RtheR integral depends R R R only on the endpoints of div σ dV = 0, where S is σ · da = the curve. Similarly, if div σ = 0, then V S a closed surface bounding the solid region V. Therefore, if ρ, τ are any two surfaces with the same boundary curve, the “sum” of the two surfaces, with the second taken in the “opposite” direction, is a closed surface. The integral over this closed surface is zero, which implies that the integrals over the two surfaces ρ and τ are equal. Thus, the integral depends only on the boundary curve and not the particular surface of which the curve is the boundary.

CHAPTER 23 1.

We want to find the equation y = ax + b giving the best linear function representing the data points (2.0, 2.5), (4.0, 4.5), (5.0, 7.0), (6.0, 8.5). Thus, we have four equations to determine the two values a and b. These equations are 2.5 = 2.0a + b,

Solutions

151

or − 2.5 + b + 2.0a = 0; 4.5 = 4.0a + b, or −4.5 + b + 4.0a = 0; 7.0 = 5.0a + b, or −7.0 + b + 5.0a = 0; and 8.5 = 6.0a + b, or −8.5 + b + 6.0a = 0. To use the method of least squares, we want to take the two partial derivatives of the sum of the squares function and set them equal to 0. The sum of the squares is given by S = (− 2.5 + b + 2.0a)2 + (− 4.5 + b + 4.0a)2 + (−7.0 + b + 5.0a)2 + (−8.5 + b + 6.0a)2 . ∂S We have ∂ b = 2(−2.5 + b + 2.0a) + 2(−4.5 + b + 4.0a) + 2(− 7.0 + b + 5.0a) + = 2(− 2.5 + b + 2.0a)2.0 + 2(− 4.5 + b + 4.0a)4.0 + 2(−8.5 + b + 6.0a) and ∂∂ Sa 2(− 7.0 + b + 5.0a)5.0 + 2(− 8.5 + b + 6.0a)6.0. Setting the two equations equal to 0 and simplifying gives the following system of two equations in two unknowns: 4b + 17a = 22.5 and 17b + 81a = 109. Solving this system using Cramer’s rule gives

b=

22.5 17 109 81 4 17

17 81

=

−30.5

35

= −0.87 and

4 22.5 17 109 53.5 a= = = 1.53.

35 35

The desired straight line is then y = 1.53x − 0.87.

2.

√ e−(x /c ) , and y′′ = − If y = c√1 π e−(x /c ) , then y′ = − c32x π 2

2

2

2

c3

2 √

e−(x /c ) + 2

π

2

2 2 2 4x √ e−(x /c ) . An inflection point occurs in this function occurs when y′′ = 0. c5 π 2 2 2 Thus, we must solve c54x√π = c3 √ . We get 2x2 = c2 , or x2 = c2 , or, finally, x = √c . π 2

3.

Using a table of the normal curve, we find that a percentile rank of 75 corresponds to a z-score of 0.675. Therefore, one probable error from the mean, which corresponds to a percentile rank of 75, is at distance approximately 0.675σ from the mean, where σ is the standard deviation. By√the result of Exercise 2, a distance c of one modulus from the mean corresponds to 2σ.

4.

To construct a histogram of children of parent of height 68.5, we can use the following table: < 62.7

62.7 − 63.7

63.7 − 64.7 64.7 − 65.7 65.7 − 66.7 66.7 − 67.7

1

7

11

67.7 − 68.7

68.7 − 69.7

69.7 − 70.7

34

48

21

16

25

70.7 − 71.7 71.7 − 72.7 18

4

31

> 72.7 3

The median value calculated from the original table is 68.2 inches.

Note that it is not possible from the data given to construct a histogram of heights of parents of children of height 69.2. One can construct a histogram of the number of children of height 69.2 inches whose parents are of various heights. This would come

152

Solutions from the following table: < 64

64 − 65 65 − 66 66 − 67 67 − 68

1

2

7

13

38

68 − 69 69 − 70 70 − 71 71 − 72 48

33

18

> 72

5

2

The median value calculated from the original table is 68.5.

5.

The histogram for the heights of all adult children can be constructed from the following table: 61.7 − 62.7 62.7 − 63.7 63.7 − 64.7 64.7 − 65.7

< 61.7 5

7

32

59

48

65.7 − 66.7

66.7 − 67.7

117

138

120

167

70.7 − 71.7

71.7 − 72.7

72.7 − 73.7

> 72.7

64

41

17

14

67.7 − 68.7 68.7 − 69.7 69.7 − 70.7 99

The median value here is 68.2. By calculation, the standard deviation is 2.54.

The histogram for the heights of all parents can be constructed from the following table: < 64

64 − 65

65 − 66

1

5

12

68 − 69

69 − 70

70 − 71

49

41

22

66 − 67 67 − 68 20

33

71 − 72 72 − 73 > 73 11

6

5

The median value here is 68.5. By calculation, the standard deviation is 1.94.

6.

It is actually not entirely clear what Playfair meant. The price of wheat in 1821 was 54 shillings, but the graph for wages does not extend that far. If we assume that the wages are, say, 32, then the ratio of the two prices is 1.69. However, in the period 1785–90, it appears that the same ratio is 42:25=1.68, so wheat would have been relatively cheaper

Solutions

153

at that point. But in any other period, the ratio of the price of wheat to the wages of a mechanic are higher than those two values, so the price of wheat is relatively higher than either the assumed ratio in 1821 or the ratio in 1785–90.

7.

There are four countries in which the line from population to revenue rises. These are Spain, Britain, Portugal, and the United Provinces. But the line for Britain is clearly much steeper than that for the other three countries. It follows that the revenues of Britain in proportion to its population are greater than for any other country. In other words, the people of Britain are excessively taxed.

8.

It appears that the size of the African and European domains are each roughly onequarter of the total area, while the Asiatic domains form about one-half of the total area.

9.

Reading the histogram using the scale of one square equals 10 flowers, we see that approximately 24 flowers had seven petals and 133 had five petals.

CHAPTER 24 1.

The infinitesimal vectors extending from the vertex (x, y) are (dx, dy) and (δx, δy). The area of the parallelogram spanned by those two vectors is dx δx

dy = dx δy − dy δx. δy

Therefore, the area of the triangle is 21 (dx δy − dy δx).

2.

The normal vector to the surface z = z(x, y) is (−zx , −zy , 1). Since we want the end of this vector to lie on the sphere of radius 1, we must normalize it. Thus the coordinates on the unit sphere corresponding to the point (x, y, z(x, y)) are X= p

−zx , 1 + z2x + z2y

−zy Y= p , 1 + z2x + z2y

Z= p

1 .

1 + z2x + z2y

The measure of curvature is given by k = ∂∂Xx ∂∂Yy − ∂∂Xy ∂∂Yx . We must calculate these four values. We get the following by use of the quotient rule and the chain rule and by simplifying: ∂ X zx zy zxy − zxx − zxx zy = ∂x (1 + z2x + z2y )3/2

2

∂ Y zx zy zxy − zyy − zyy z2x = ∂y (1 + z2x + z2y )3/2

∂ X zx zy zyy − zxy − zy zxy = ∂y (1 + z2x + z2y )3/2

∂ Y zx zy zxx − zxy − z2x zxy = ∂x (1 + z2x + z2y )3/2

2

154

Solutions If we multiply appropriately and simplify again, we get k = = 3.

zxx zyy + zxx zyy z2x + zxx zyy z2y − z2xy − z2xy z2x − z2xy z2y (1 + z2x + z2y )3 (zxx zyy − z2xy )(1 + z2x + z2y ) zxx zyy − z2xy = .

(1 + z2x + z2y )3 (1 + z2x + z2y )2

If z = x2 + y2 , then zx = 2x, zy = 2y, zxx = 2, zyy = 2, and zxy = 0. Therefore, k=

2 · 2 − 02 4 = .

(1 + (2x)2 + (2y)2 )2 (1 + 4x2 + 4y2 )2

4.

If x = x(u, v), y = y(u, v), and z = z(u, v), then dx = xu du + xv dv, dy = yu du + yv dv, and dz = zu du + zv dv. Therefore, dx2 + dy2 + dz2 = x2u du2 + 2xu xv du dv +x2v dv2 + y2u du2 + 2yu yv du dv + y2v dv2 + z2u du2 + 2zu zv du dv + z2v dv2 = (x2u + y2u + z2u )du2 + 2(xu xv + yu yv + zu zv ) du dv + (x2v + y2v + z2v ) dv2 = E du2 + 2F du dv + G dv2 .

5.

If x = cos u cos v, y = cos u sin v, z = sin u, then xu = − sin u cos v, xv = − cos u sin v, yu = − sin u sin v, yv = cos u cos v, zu = cos u, and zv = 0. Therefore, we calculate E = x2u + y2u + z2u = sin2 u cos2 v + sin2 u sin2 v + cos2 u = sin2 u + cos2 u = 1, F = xu xv + yu yv + zu zv = sin u cos v cos u sin v − sin u sin v cos u cos v = 0, and G = x2v + y2v + z2v = cos2 u sin2 v + cos2 u cos2 v + 0 = cos2 u. So ds2 = E du2 + 2F du dv + G dv2 = du2 + cos2 u dv2 .

6.

Assuming that angle B is acute, we drop a perpendicular from angle B to the opposite side b, intersecting that side at point D. Designate BD by h, AD by r, and DC by b − r.

In right triangle BDC, we have cos a = cos h cos(b − r). In right triangle BDA, we have tan r cos c cos a cos A = tan c and cos c = cos h cos r. Therefore, cos h = cos r = cos(b−r) . So cos a =

cos c cos(b − r) cos c(cos b cos r + sin b sin r) = = cos c cos b + cos c sin b tan r cos r cos r

= cos c cos b + cos c sin b tan c cos A = cos c cos b + sin b sin c cos A. 7.

Note that in Chapter 7, the triangle formulas assumed that the sides were measured in degrees. Naturally, we can also measure the sides in radians and get the same formulas.

But if we measure the sides in linear measure, then we can only use the formulas as written if the radius of the sphere is 1, for in that case the linear measure of an arc is equal to its radian measure. If the sphere has radius K, then the linear measure x of an arc whose radian measure is r is given by x = Kr. Thus, r = Kx , and we can substitute for the radian measure, in any of the formulas, including the one in Exercise 6, the linear measure divided by K. So the formula of Exercise 6 becomes cos

8.

b c b c a = cos cos + sin sin cos A. K K K K K 2

4

The power series for y = cosh x is given by y = 1 + x2! + x4! + · · · , while the power 3 5 series for y = sinh x is given by y = x + x3! + x5! + · · · . If we substitute the terms from

Solutions

155 2

a the power series through the second power into Taurinus’s formula, we get 1 + 2K 2 = 2

2

2

2

2

2 2

c b c a b c b c 2bc cos A b (1 + 2K . If 2 )(1 + 2K2 ) − K K cos A. This simplifies to K2 = K2 + K2 + 2K4 − K2

9.

we multiply through by K2 and neglect the term which still has denominator K2 – because this term becomes much smaller than the other terms as K → ∞ – we get a2 = b2 + c2 − 2bc cos A, exactly the law of cosines. p x 1 2 e−x Since sin B = cosh 1 − sin2 B = eex − . So x , we have sin B = ex + e−x . Then cos B = + e−x B tan = 2

r

v r u x e−x 1 − eex − 1 − cos B u 2e−x √ −2x + e−x t = e−x = = = e ex −e−x 1 + cos B 2ex 1 + x −x e +e

as desired. This argument works in reverse as well, so the two results are in fact equivalent.

10.

A circle of radius r on a sphere of radius K, where r is measured along the surface of the sphere, has radius measured in the three-dimensional space of the sphere equal to K sin Kr . It follows that the circumference of the circle is C = 2πK sin Kr . If the sphere has imaginary radius iK, we substitute iK for K in the formula for the circumference. We get C = 2πiK sin iKr = 2πiK sin(− Kr i). But sin ix = i sinh x. So C = 2πiK(−i sinh Kr ) = 2πK sinh Kr . If we substitute into this formula the power series 2πr3 r3 for y = sinh x, we get C = 2πK( Kr + 3!K 3 + · · · ) = 2πr + 3!K2 + · · · . As K → ∞, this expression approaches the value C = 2πr.

11.

Let us write y = Π(x). Given that tan 2y = e−x , we have tan2 2y = e−2x , so sec2 2y = e−2x + 1. Therefore, r 1 y 1 e−x y , so sin = 1 − −2x =√ . cos = √ − 2x − 2 2 e +1 e +1 e 2x + 1 Then sin y = 2 sin

2 y 1 2e−x 1 y e−x √ = −2x cos = 2 √ . = x −x = − 2x − 2x 2 2 e + e cosh x e + 1 e +1 e +1

Also. cos y = cos2

1 e−2x 1 − e−2x ex − e−x y y − sin2 = −2x − −2x = −2x = x −x = tanh x.

2 2 e e +e +1 e +1 e +1

To get the power series through the terms of degree 2, note that 1 1 sin Π(x) = (cosh x)−1 = (1 + x2 + · · · )−1 = 1 − x2 2 2 and x + 16 x3 + · · · 1 1 = (x + x3 + · · · )(1 + x2 + · · · )−1 1 2 6 2 1 + 2x + ··· 1 1 = (x + x3 + · · · )(1 − x2 + · · · ) = x + · · · .

6 2

cos Π(x) = tanh x =

156

Solutions 12.

Rewrite the first formula as sin A cot Π(b) = sin B cot Π(a). Then we note that

−1

cos Π(x) x 1 2 1 2 ≈ = x 1− x ≈x 1+ x ≈ x, cot Π(x) = sin Π(x) 2 2 1 − 21 x2 where the smaller the value of x, the better the approximation. If we use this approximation in the first formula, we get sin A · b = sin B · a, or sin A = sinb B , the law of sines. Next, we rewrite the second formula in the form a cos A cos Π(b) cos Π(c) sin Π(a) + sin Π(b) sin Π(c) = sin Π(a). If we then approximate the quantities in the formula by the terms of their power series up to degree 2, we get

1 1 1 1 cos A · b · c 1 − a2 + 1 − b2 1 − c2 = 1 − a2 .

2 2 2 2 If we multiply through by 2 and simplify, we get 2bc cos A − a2 bc cos A − b2 − c2 + 2 1 2 2 2 b c = −a . If we then neglect terms which have at least three linear factors – because we are assuming these are all “small” – we get a2 = b2 + c2 − 2bc cos A, the law of cosines.

13.

From a sin(A + C) = b sin A and the law of sines, we have sinb B = sina A = sin(Ab+ C) .

Therefore, sin(A + C) = sin B, or, interchanging A and B, sin(B + C) = sin A. From cos A + cos(B + C) = 0, we have cos(B + C) = − cos A. The sine result implies that either A = B + C or that A = π − (B + C). The cosine result shows that the second equation is the correct one. Thus, A + B + C = π.

14.

We first note that sinh ix = i sin x, cosh ix = cos x, and tanh ix = i tan x.

We replace a, b, c in Lobachevsky’s formulas with ia, ib, and ic, respectively. Formula 24.3 gives sin A sinh ib = sin B sinh ia, or sin A(i sin b) = A sin B sin B(i sin a), or, finally, sin = sin a sin b , the spherical law of sines. Foria = 1. This transforms to mula 24.4 produces cos A tanh ib tanh ic + coshcosh ib cosh ic cos a − cos A tan b tan c + cos b cos c = 1, or to − cos A tan b tan c cos b cos c + cos a = cos b cos c, or, finally, to cos a = cos b cos c + sin b sin c cos A, the spherical triangle ib formula derived in Exercise 6. Formula 24.5 gives cot A sin C cosh1 ib + cos C = tanh tanh ia . cot A tan b tan a π Setting C = 2 , this formula reduces to cos b = tan a , or to tan A = sin b , one of the standard spherical right triangle formulas. Finally, formula 24.6 reduces to cos A + cos B cos C = sin B sin C cosh ia, or, setting C = π2 , to cos A = sin B cos a. This formula is also a standard formula of spherical trigonometry, which can be derived from formulas 5.4, 5.5, and 5.6.

15.

Consider the plane z = a (a ≥ k) with coordinates u, v, sitting above the hemisphere x2 + y2 + z2 = k2 . If we take an arbitrary point on the plane, say (u, v, a) and connect it by a straight line to the origin, then we can calculate the point (x, y, z), where the line intersects the hemisphere. This line has direction vector (ut, vt, at). Its intersection point (x, y, z) with the hemisphere is found by setting the length of that vector equal to 2 k2 . We get u2 t2 + v2 t2 + a2 t2 = k2 , so t2 = u2 + vk2 + a2 , and t = √ 2 k 2 2 . Thus we have a +u +v

x = ut = √

uk a2 + u2 + v2

,

vk y = vt = √ , 2 a + u2 + v2

and

ak z = at = √ .

2 a + u2 + v2

Solutions 16.

157

To determine the curvature of the sphere and the pseudosphere, we use Gauss’s result that the curvature k(p) of a surface at p is the product zxx (p)zyy (p), where the surface is given by the equation z = z(x, y) and the axes are chosen so that p = (0, 0, 0) and zx (0, 0) = zy (0, 0) = zxy (0, 0) = 0. Since the sphere has constant curvature, we can calculate it just at the origin. Thus, p we write the equation of the sphere in the form x2 + y2 + (z + k)2 = k2 , or z = k2 − x2 − y2 − k. Then zx = − √ 2 x 2 2 and k −x −y

zxx = (k2 −yx2−−k y2 )3/2 . It follows that zxx (0, 0) = − kk3 = − 1k . Similarly, zyy (0, 0) = − 1k .

2

2

2

Therefore, the curvature of the sphere is (− 1k )2 = k12 . If we replace k by ik in this expression (taking a sphere of imaginary radius), then the curvature becomes 1 = − k12 . (ik)2 17.

We have ρ=

1 a+r k ln , 2 a−r

s=

√

1 a+u k ln , 2 a−u

and

t=

1 a2 − u2 + v k ln √ , 2 a2 − u2 − v

where u = r cos θ and v = r sin θ. We calculate

ρ a + r 1/2 tanh = tanh ln = k a−r a+r

r − 2 + aa− +r

= a−ar+ r

=

a−r a−r − a + r

a + r 1/2 − a−r

1 / 2 a+r + a−r 2 2

a−r 1/2 a+r

a−r 1/2 a+r

(a + r)2 − 2(a − r ) + (a − r)2 4r2 r = = .

2 2 4ar a (a + r) − (a − r)

θ and tanh kt = √ 2v A similar calculation shows that tanh ks = ua = r cos a

a − u2

cosh x = √ 1 2 , we have 1−tanh x cosh

. Since

s 1 a =q =√ , 2 2 k a − u2 1 − ua2

t 1 cosh = q k 1−

v2 a2 − u2

√

√

a2 − u2 a2 − u2 =√ = √ , 2 2 2 a −u −v a2 − r2

and cosh

1 a ρ . =q =√ 2 − r2 k r2 a 1 − a2

Therefore, √

cosh 18.

s t a a ρ a2 − u2 √ cosh = √ =√ = cosh . k k k a2 − u2 a2 − r 2 a2 − r 2

Take any two parallel lines AB and CD lying in a vertical plane G. Let H be a horizontal plane perpendicular to G. Let K and L be planes through AB and CD, respectively,

158

Solutions which are perpendicular to both G and H. Let P be a point above H, outside of G, and between the two planes K and L. Then the projection from P onto H has the desired property. Let F be the plane through P parallel to H. We may assume that points A and C are in H and points B and D are in F. Let Q be the point in H directly below P. Then the ray from B through A projects to a ray in H from Q through A. B is sent to infinity, while a point at infinity is sent to Q. Similarly, the ray from D through C projects to a ray in H from Q through C. Both of these projections contain Q, so they intersect, as claimed.

19.

AD AC·DB AB AD We have (AB, CD) = AC CB : DB = CB·AD . Also, 1 − (AC, BD) = 1 − ( BC : DC ) = AB·DC BC·AD−AB·DC 1 − BC·AD = . But BC = BD + DC and AB = AD + DB. Therefore, BC·AD

1 − (AC, BD) = =

(BD + DC) · AD − (AD + DB) · DC BD · AD − DB · DC = BC · AD BC · AD −DB · (AD + DC) −DB · AC AC · DB = = = (AB, CD).

BC · AD BC · AD CB · AD

Also, 1 (AB, DC)

20.

1 DB · AC = = (AB, CD).

AC AD · CB DB : CB

= AD

AB AD ·DC If λ = (AB, CD), we note from Exercise 19 that 1 − λ = (AC, BD) = BC : DC = AB BC·AD .

AD·BC 1 1 But then µ = (AC, DB) = DC·AB = (AC,BD) = 1−λ . Also, if ν = (AD, BC), then

·CD AB·DC AB·DC BC·AD 1 λ−1 ν = AB = 1 − λ1 . Therefore, λ + µ1 = BD·AC = AC·DB = BC·AD · AC·DB = (1 − λ)(− λ ) = λ λ −1+λ 1 1 1 λ + (1 − λ) = 1. Also, µ + 1ν = 1−1 λ + λ − 1 = λ − 1 = 1, and ν + λ = (1 − λ ) + λ = 1. λ−1 1 Finally, −λµν = −λ · 1−λ · λ = 1.

21.

We demonstrate this analytically for the case of a circle. Since an arbitrary conic section is the projective image of a circle, and since projection preserves tangents and therefore poles and polars, the result will be true for any conic. So we assume the circle has equation x2 + y2 = 1 and let the point p have coordinates (0, a). If the two lines from p tangent to the circle are (x, y) and (−x, y), then, because the slope of the tangent to the circle at (x, y) is − xy , the equation of one of the tangent lines is y−a 2 2 2 2 x x = − y , which reduces to y − ay = −x , or to x + y − ay = 0, or to 1 − ay = 0.

Therefore, ay = 1, and y = a1 . Thus the horizontal line y = a1 is the polar π of p. So let us pick a point p′ = (t, 1a ) on π. Then the line from p′ to a point (x′ , y′ ) on the circle is ′

′

′

a)−y a tangent to the circle if (1/t− = − xy′ . This equation reduces to ya − y′2 = −tx′ + x′2 , x′ ′

or to ya + tx′ = 1, or finally to y′ = a − atx′ . Thus, the two points on the circle to which the tangent lines from p′ are drawn are (x′1 , y′1 ) and (x′2 , y′2 ), where the x′ and the y′ are related by that equation. The equation of the polar π′ of p′ is then y − y′1 y′2 − y′1 −at(x′2 − x′1 ) = = −at, = x − x′1 x′2 − x′1 x′2 − x′1

or

y − a + atx′1 = −at. x − x′1

We then note that the point p, with coordinates (0, a) does lie on π′ , as asserted.

Solutions 22.

23.

159

Possible sets of homogeneous coordinates for (3, 4) and (−1, 7) are (3, 4, 1) and (−1, 7, 1), respectively. Any other sets would be of the form (3t, 4t, t) and (−t, 7t, t), respectively.

Since points at infinity have their t coordinates equal to 0, we just need to find a point

(x, y) which satisfies the equation 2x − y = 0 and take the point (x, y, 0). The desired point is (1, 2, 0). Notice that any triple of the form (a, 2a, 0) is also a coordinate set

for the point at infinity on the given line.

24.

The rectangular coordinates of (3, 1, 1) are (3, 1), while the rectangular coordinates of (4, −2, 2) are (2, −1). In each case, we divide the x and y coordinates by the t coordinate.

25.

The homogeneous equation of the desired line is ax + by + ct = 0, with a, b, and c to be determined. Since the line passes through (2, 1, 0), we have 2a + b = 0. Since the line passes through (6, 2, 2), we have 6a + 2b + 2c = 0. Therefore, b = −2a and 6a − 4a + 2c = 0, or 2a + 2c = 0, or a = −c. Then b = 2c. So we can choose a = −1, b = 2, c = 1. The homogeneous equation of the line is then −x + 2y + t = 0. To get the rectangular equation, we set x = Xt and y = Yt and divide by t. Then −Xt + 2Yt + t = 0, and the desired equation is −X + 2Y + 1 = 0.

26.

The rectangular equation of a circle is of the form x2 + bx + y2 + cy + d = 0. If we 2 2 rewrite this in homogeneous coordinates, we get Xt + b Xt + Yt + c Yt + d = 0, or X2 + bXt + Y2 + cYt + dt2 = 0. It is clear that (1, i, 0) and (1, −i, 0) satisfy this equation, whatever the values of b, c, and d. Thus those two points at infinity lie on any circle in the plane.

27.

Denote by K the intersection point of the line QF with the line GP. Then, since the points A, B, P, Q are the images of E, F, K, Q under a central projection from G, we know that the two cross ratios (AB, PQ) and (EF, KQ) are equal. Also, since the points B, A, P, Q are the images of E, F, K, Q under a central projection from H, we know that the cross ratios (EF, KQ) and (BA, PQ) are equal. Therefore, (AB, PQ) = (BA, PQ).

AP·QB −1 BP·QA = λ1 . Therefore, If we set λ = (AB, PQ), we note that (BA, PQ) = PA ·BQ = ( PB·AQ ) λ = λ1 , and λ2 = 1. Since the four points are distinct, we cannot have λ = 1. Therefore (AB, PQ) = λ = −1.

28.

We have

′

QR · SP Q R · SQ + c ln ′ RQ RP · QS ′ · Q S

QR · SP · Q R · SQ SP · Q′ R = c ln = c ln = d(P, Q′ ).

RP · QS · RQ · Q′ S RP · Q′ S

d(P, Q) + d(Q, Q′ ) = c ln

29.

By trial, we see that the path WRSTVJHGBCDFKLMNPOZXW passes through each vertex exactly once and returns to the starting point at W.

31.

Suppose we have a minimal map requiring five colors which contains a digon. Remove one boundary line from the digon, thus merging it with one of its neighbors. Then we have a map with one fewer country than the original one. Thus it can be colored with four colors. Now replace the boundary line. The digon is now bounded by only two countries, each colored differently. Thus there is color available (of the four colors) to color the digon. This contradicts the original hypothesis.

160

Solutions 32.

33.

34.

We first multiply the first two quantities: (2i + 3j − 4k)(3i − j + k) = (−2 − 9)[ij] + (2 + 12)[ik] + (3 − 4)[jk]. We then multiply −11[ij] + 14[ik] − [jk] by i + 2j − k. Since any repeated product of a unit is 0, we get 11[ijk] + 28[ikj] − [jki]. But [ikj] = −[ijk] and [jki] = [ijk]. Thus this sum is equal to (11 − 28 − 1)[ijk] = −18[ijk].

P When we multiply the expressions αji ϵi together, the only non-zero terms are those where we take distinct ϵi from each of the n expressions. In each such case of a product of distinct ϵi , we must permute the factors to get the single unit [ϵ1 ϵ2 · · · ϵn ] of order n. We will then get a positive sign if the permutation is even and a negative one if the permutation is odd. Each coefficient of the product of distinct ϵi will have one entry from each row and each column of the matrix (αij ). Thus the total coefficient of the nth order unit will be the sum of all possible products of n terms, each one having a single factor from each row and column, where the sign of each term is positive if the permutation needed to put that product into the natural order of the column numbers is even and negative if it is odd. But that sum of products is exactly det(αij ), as claimed.

If ω = A dx + B dy + C dz, we have dω = dA dx + dB dy + dC dz

∂A ∂A ∂B ∂B ∂B ∂A dx + dy + dz dx + dx + dy + dz dy = ∂x ∂y ∂z ∂x ∂y ∂z

∂C ∂C ∂C + dx + dy + dz dz ∂x ∂y ∂z ∂A ∂A ∂B ∂B ∂C ∂C = dy dx + dz dx + dx dy + dz dy + dx dz + dy dz ∂y ∂z ∂x ∂z ∂x ∂y

∂C ∂B ∂A ∂C ∂B ∂A = − dy dz + − dz dx + − dx dy. ∂y ∂z ∂z ∂x ∂x ∂y

35.

If ω = A dy dz + B dz dx + C dx dy, then ∂A ∂B ∂C dω = dA dy dz + dB dz dx + dC dx dy = dx dy dz + dy dz dx + dz dx dy ∂x ∂y ∂z

∂A ∂B ∂C = + + dx dy dz. ∂x ∂y ∂z

36.

In Exercise 34, we calculated dω for ω a one-form and in Exercise 35, we calculated dω for ω a two-form. Therefore, if ω is a one-form,

d(dω) =

=

∂ ∂x

∂C ∂B − ∂y ∂z

+

∂ ∂y

∂A ∂C − ∂z ∂x

+

∂ ∂z

∂B ∂A − ∂x ∂y

∂2C ∂2B ∂2A ∂2C ∂2B ∂2A − + − + − ∂ x∂ y ∂ x∂ z ∂ y∂ z ∂ y∂ x ∂ z∂ x ∂ z∂ y

dx dy dz

dx dy dz = 0.

If ω is a two-form, then dω is a three-form and d(dω) is a four-form. But a four-form in three-dimensional space must be 0, because one of the dx, dy, or dz will be repeated.

Solutions 37.

161

We first calculate dω by using the result of Exercise 34: (x2 + y2 + z2 )3/2 − 3x2 (x2 + y2 + z2 )1/2 x ∂ = . ∂ x (x2 + y2 + z2 )3/2 (x2 + y2 + z2 )3

Since there are similar results for the partial derivatives with respect to y and z, we get "

# 3(x2 + y2 + z2 )3/2 − 3(x2 + y2 + z2 )(x2 + y2 + z2 )1/2 dω = dx dy dz = 0. (x2 + y2 + z2 )3 R R R If there were a one-form η with dη = ω, then T ω = T dη = S η = 0, because the boundary S of the unit sphere T is empty. But we can calculate the integral of ω over the unit sphere directly by using the parametrization x = cos u cos v, y = cos u sin v, and z = sin u, where − π2 ≤ u ≤ π2 and 0 ≤ v ≤ 2π. We calculate x dy dz +y dz dx + z dx dy = cos u cos v(− sin u sin v du + cos u cos v dv)(cos u du) + cos u sin v(cos u du)(− sin u cos v du − cos u sin v dv) + sin u(− sin u cos v du − cos u sin v dv)(sin u sin v du + cos u cos v dv) = [cos u cos v(− cos2 u cos v) + cos u sin v(− cos2 u sin v) + sin u(− sin u cos u cos2 v + sin u cos u sin2 v)] du dv = (− cos3 u cos2 v − cos3 u sin2 v − sin2 u cos u) du dv = −(cos3 u + sin2 u cos u) du dv = − cos u du dv.

Also, x2 + y2 + z2 = 1. Therefore, Z

Z 2π Z π 2

ω= T

0

− π2

(− cos u) du dv = −2(2π) = −4π ̸= 0.

This is a contradiction, so such an η cannot exist.

CHAPTER 25 1.

The description of the real number s requires a complete knowledge of the set E in advance. In some sense, the element s represents a subset of E, and sets with subsets as elements are inconsistent.

2.

If a set A is well-ordered and we have two elements α, β with α ̸= β, then consider the subset B = {α, β}. By the well-ordering principle, B has a least element. If this element is α, then α < β. If this element is β, then β < α. In any case, one of the three relations α = β, α < β, or α > β must hold.

3.

The separation axiom requires a definite propositional function to define a set. But the function P defining the set of people whose hair is cut by the barber is not definite,

162

Solutions because the laws of logic do not determine whether P(x) holds for a particular x, namely the barber.

4.

The Heine-Borel theorem in the plane states that if a (countably) infinite set A = {A1 , A2 , . . . , An . . .} of open sets covers a closed rectangle B in the plane, then there is a finite subset of A which has the same property. If the conclusion is not true, then for every n there is a point bn ∈ B such that bn ∈ / Ai for every i ≤ n. If we now bisect B, then the same statement must be true of at least one of the pieces. If we continue this process, one obtains a decreasing nested sequence of closed rectangles Bi , each of which has the same property as B itself. But ∩Bi contains a point p. By hypothesis, p ∈ Ak for some k. Since Ak is open, it must contain one of the rectangles Bi , contradicting the property of Bi .

5.

Suppose a connected set A can be expressed as A = B ∪ C, where B and C are closed and B ∩ C is empty. Let b ∈ B. Then there is a region Ub containing b with Ub ⊂ B, because b cannot be a limit point of C. Similarly, around every point c ∈ C, there is a region Vc containing c with Vc ⊂ C. The union of all the Ub and Vc do not, however, generate a single region since the intersection of any Ub with any Vc is empty. This contradicts the connectedness of A.

6.

Let A = {q ∈ Q|0 ≤ q < 22 } and B = {q ∈ Q| 22 < q ≤ 1}. Then√A ∪ B is equal to the set C of rational numbers in [0, 1]. On the other hand, A = [0, 22 ] ∩ C, so A is closed relative to C. Similarly B is closed relative to C. So C has been divided into two disjoint, non-null sets, both closed relative to C. Therefore, C is not connected.

7.

Suppose that E1 is an infinite subset of a compact set E. If e1 ∈ E1 is a limit point of E1 , we are done. So assume it is not a limit point. Then there is a neighborhood U1 of e1 whose complement contains infinitely many points of E1 . If possible, pick a point e2 in this complement which is not a limit point. We can then find a neighborhood U2 of e2 which also contains e1 and whose complement contains infinitely many points of E1 . Continue in this manner. We either have a limit point or we get an infinite sequence of open sets Ui such that Ui contains Ui − 1 and such that the complement of Ui contains infinitely many points of E1 . It follows that Vi = E − Ui is a closed set containing infinitely many elements of E1 and that Vi ⊂ Vi − 1 . Then the Vi form a nested sequence of closed subsets of E and so have a common point e. This point is a limit point of E1 , as desired.

8.

If we let An = (0, n1 ), then the collection of the An is a nested family of intervals, but its intersection is empty. (Note that the An are not closed sets.) If we let Bn = [n, ∞), then the collection of the Bn is a nested family of (unbounded) intervals, and again the intersection of the family is empty.

9.

Given that f is sequentially continuous on a closed compact set E, we suppose f is not bounded. That means there is a sequence a1 , a2 , . . . in E such that given any M, there exists an integer N such that f(ai ) > M whenever i > N. Thus the sequence {f(ai )} has no limit. By the definition of compactness, the sequence {ai } has a limit point a in E and f(a) = limi→∞ f(ai ). This is a contradiction, so f is bounded. Now let Q be the least upper bound of the set f(E). Then there exists a sequence {f(ai )} whose limit is Q. But the sequence {ai } has a limit point a. Therefore, f(a) = Q and f attains its least upper bound.

√

√

Solutions

163

10.

Suppose {En } is a nested sequence of closed subsets of E. If there is an N such that Em = EN for every m > N, then the intersection is not empty, for it would then equal EN . So suppose there is no such N. Then we can renumber the sets so that each of the Ei is properly contained in the previous one. Then choose an element e1 in E1 − E2 , an element e2 in E2 − E3 , and so on. The sequence e1 , e2 , e3 , . . . is an infinite subset of E, so it has a limit point e, according to Frechet’s second definition. Then I claim that e ∈ ∩n En . For if e ∈ / Ek , then e ∈ / En for n > k. But since e is a limit point of the sequence, any neighborhood of it must contain infinitely many of the ei . Also infinitely many of the ei are in Ek , which, since it is closed, must contain its limit point. This is a contradiction, and the claim in proven.

11.

Since a continuous function on a closed interval is uniformly continuous, the limit function is also continuous. Thus the set E of continuous functions on a closed interval is itself closed. Since every continuous function is the limit of some sequence of functions, say the sequence consisting of infinitely many copies of itself, we note that the set E is perfect. The set of polynomial functions with rational coefficients defined on [a, b] is a countable dense subset of E, so E is separable. Finally, any Cauchy sequence of continuous functions under the maximum norm converges to a continuous function. For if two functions are “close” under the maximum norm, then the function values at every x in [a, b] are also “close.” Thus, we can define the limit function at each x to be the limit of the values of the sequence of functions at that point.

And since all the functions are uniformly continuous, this limiting function is also uniformly continuous. Thus E is complete and therefore normal.

12.

If we denote by x(n) the nth point in a sequence of elements, then x(n) approaches (n ) the point x in the sense of this metric if and only if each coordinate xp approaches the corresponding coordinate xp , in the ordinary sense for numbers. Suppose the latter condition is satisfied. Then (n)

(x, x(n) ) ≤

(n)

| x1 − x1 |

+ ··· + (n)

1 + |x1 − x1 |

1 |xp − xp | + rp , p! 1 + |xp − x(pn) |

where rp =

1 (p + 1)!

+

1 (p + 2)!

+ ···

is independent of n. This latter quantity goes to 0 as p increases. Thus one can fix a number p independent of n, such that rp < ϵ2 . With the number p fixed, the sum of the p terms in the inequality above tends to 0 as n increases, because each of the numerators does. One can thus find a number q such that this sum is less than ϵ2 for n > q. It follows that for n > q, we have (x(n) , x) < ϵ, and therefore the x(n) approach x.

Conversely, if the x(n) approach x, then for each value of p we have (n)

|xp − xp | (n) 1 + |xp − xp |

< p!(x, x(n) ).

Since the right side of this inequality tends to 0 as n increases, the same is true of the (n ) left side, and therefore of the numerator on the left side. Thus each xp approaches

164

Solutions the corresponding coordinate xp as desired. It now follows that if we have a Cauchy sequence {x(n) } of elements, then for each p, the sequence of pth coordinates is a (n) Cauchy sequence {xp } in the ordinary sense and therefore converges to a number xp . Then the original Cauchy sequence converges to the point (x1 , x2 , . . .). Also, every element in this set is a limit, so the set is perfect. Finally, the set is separable, because one can take for a countable dense subset the set of all points with rational coordinates which have only the first n of these coordinates different from zero. Thus the set is normal.

13.

|x −y |

The quotient 1 + |pxp −pyp | is always less than 1. Therefore (x, y) =

∞ X 1

∞

p=1

p=1

X 1 | xp − yp | < < e. p! 1 + |xp − yp | p!

Thus (x, y) < e for all x, y is the metric space.

14.

Suppose A is a closed subset of E and let B = E − A. Suppose x ∈ B. Then x is not an accumulation point of A. So there is a neighborhood U of x which does not contain any point of A. Therefore, U ⊂ B and B is an open set. Conversely, suppose B is an open set in E and let A = E − B. Let x be an accumulation point of A, and suppose x ∈ B. Thus there is a neighborhood U of x which is a subset of B and therefore does not intersect A. This contradicts the definition of an accumulation point, so x ∈ / B and x ∈ A. Therefore, A is closed.

15.

Suppose x and y with x ̸= y are both limit points of the infinite set A. By axiom 4, there are neighborhoods Ux of x and Uy of y which are disjoint. But since x is a limit point of A, Ux contains all but a finite number of points of A. Since Uy is disjoint from Ux , that means that Uy can only contain a finite number of points of A, contradicting the fact that y is also a limit point. Thus an infinite set can have no more than one limit point.

16.

Suppose that f : A → B is continuous at a according to the neighborhood definition and suppose Q is a subset of B which has b = f(a) as an interior point. Then Q is a neighborhood of b, so there exists a set U which is a neighborhood of a whose image is in Q. But then f−1 (Q) contains U, so contains the point a in its interior and f is continuous according to Hausdorff’s second definition. Now suppose that f is continuous at a according to the second definition and let Vb be a neighborhood of b = f(a). Then b is an interior point of Vb , so f −1 (Vb ) contains a as an interior point. We can take this set as the desired neighborhood Ua of a, for then f(Ua ) is a subset of Vb and f is continuous at a according to the neighborhood definition.

17.

Suppose f : U → V is continuous, and suppose A is a compact subset of U. We want to show that f(U) is a compact subset of V. Let F be an infinite subset of f(U).

Then E = f −1 (F) is an infinite subset of U and therefore has a limit point α. Then we claim that β = f(α) is a limit point of F. To show this, we need to show that every neighborhood of β contains all but a finite number of points of F. But the inverse image f −1 (Vβ ) of every neighborhood Vβ of β is a neighborhood of α. This neighborhood contains all but a finite number of points of E, and therefore Vβ contains all but a finite number of points of F as required. Next, suppose that A is a connected subset of U and suppose f(A) is not a connected subset of V. Then f(A) can be divided into two

Solutions

165

disjoint, non-null subsets, both open with respect to V. But the inverse image of each of these open sets is open in U, and then A would be divided into two disjoint, non-null subsets, both open with respect to U, contradicting the connectedness of A. Therefore, f preserves both compactness and connectedness.

18.

There is one independent closed one-dimensional subvariety on a sphere, represented by any simple closed curve on the surface. There are two independent closed onedimensional subvarieties on a torus, one represented by a circle around the total circumference and the other represented by the circle which generates the torus by revolving about an axis beyond the circle.

19.

The boundary of the tetrahedron, according to Alexander’s definition, is V1 V2 V3 − V0 V2 V3 + V0 V1 V3 − V0 V1 V2 . Then the boundary of the boundary is (V2 V3 − V1 V3 + V1 V2 ) − (V2 V3 − V0 V3 + V0 V2 ) + (V1 V3 − V0 V3 + V0 V1 ) − (V1 V2 − V0 V2 + V0 V1 ) = 0.

20.

The boundary of V0 V1 V3 is V1 V3 − V0 V3 + V0 V1 . The boundary of the boundary is (V3 − V1 ) − (V3 − V0 ) + (V1 − V0 ) = 0.

21.

First, take x so that (a + x) + b = b. Then a + (x + b) = b. So we can take y = x + b, and a + y = b. Second, if a + z = a, and b is any element in the set, take y so that a + y = b.

So b = a + y = y + a = y + (a + z) = (y + a) + z = (a + y) + z = b + z, so b + z = b. Third, suppose a + b = a + b′ . Find y so that b′ + y = b. Then a + (b′ + y) = a + b = a + b′ . So (a + b′ ) + y = a + (b′ + y) = a + b′ . Therefore, by the previous result, b′ + y = b′ . Thus, b = b′ . It then follows that the y found in the first part is unique.

22.

First, 1 + 1 = 2, and 2 is not in the set, so the set does not satisfy the first axiom.

But the commutative laws of addition and multiplication are satisfied (axioms 2, 3, 6, and 7) as is the closure law under multiplication (axiom 5). The distributive law is satisfied (axiom 9). Finally, we can check all the cases to see that axioms 4 and 8 are satisfied.

23.

The set S = {r 2|r ∈ Q} certainly satisfies the commutative and associative laws of addition and multiplication as well as the distributive law, because it is a subset of the real numbers. √ The √ set is closed under addition, but not under multiplication, since the product (r √ 2)(s 2) =√2rs ∈ / S. Thus S does not √ satisfy axiom 5. On the other hand, given a = r 2, b = s 2, we note that if x = − r 2, then (a + x) + b = b, so S does √ satisfy axiom 4. Also, if we choose y = 2r1 2, then (ay)b = 1b = b, so S satisfies √ axiom 8, with the sought for value y being equal to 2r1 2.

24.

We multiply as follows:

√

4 .324 3 .403 2 .1332 0 .10133 0 .0021332 2 .2312242

166

Solutions The partial products come from multiplying a digit in the multiplier by the entire multiplicand. For example, to multiply 4.324 by 3, we note that 3 × 4 = 2.2 = 2 + 2 · 5, 3 × 0.3 = 0.41 = 4 · 5 + 1 · 52 , 3 × 0.02 = 0.011 = 1 · 52 + 1 · 53 , and 3 × 0.004 = 0.0022 = 2 · 53 + 2 · 54 . Thus the first partial product is equal to 2.2 + 0.41 + 0.011 + 0.0022 = 2.1332.

25.

We do the long division as follows: 2 .42204220 4.21 )3 .12000000 3 .03 .14444444 .1111 .03334444 .0303 .00304444 .00303 .00001444 Since the 0.144444 from the first remainder reappears at this point, the quotient repeats in the four-digit pattern: 4220.

26.

Since 0.2(4·5−1 + 0 + 1·5) = 2×5(4·5−1 + 0 + 1·5) = 8 + 2·25 = 3 + 1·5 + 2·52 = 3.12, we have 3.12 ÷ 0.2 = 4 · 5−1 + 0 + 1 · 5.

27.

Let α = a0 + a1 p + a2 p2 + · · · be a unit. Thus a0 ̸≡ 0 (mod p). Further, suppose αβ = 1, where β = bm pm + bm + 1 pm + 1 + · · · , with bm ̸≡ 0 (mod p). Then 1 = (a0 + a1 p + a2 p2 + · · · )(bm pm + bm + 1 pm + 1 + · · · ) = a0 bm pm + (a0 bm + 1 + a1 bm )pm + 1 + · · ·. It follows that m = 0 and, since a0 b0 ̸≡ 0 (mod p), we have b0 ̸≡ 0 (mod p). Thus β = b0 + b1 p + b2 p2 + · · · and β is a unit.

28.

First, each point x belongs to Ur (x) and Ur (x) contains x. Second, if Ur (x) and Us (x) are neighborhoods of x, with s > r, then their intersection is Us (x) which is a neighborhood of x. Third, if y ∈ Ur (x), then y ≡ x (mod pr ). Consider the neighborhood Ur (y). If z ∈ Ur (y), then z ≡ y (mod pr ). It follows that z ≡ x (mod pr ), so z ∈ Ur (x).

Therefore, Ur (y) ⊆ Ur (x). Finally, suppose x ̸= y. Then there is an r such that x ≡ y (mod pr ), but x ̸≡ y (mod pr + 1 ). Then Ur + 1 (x) and Ur + 1 (y) are neighborhoods of x and y, respectively, whose intersection is empty. For if z belonged to the intersection, then z ≡ x (mod pr + 1 ) and z ≡ y (mod pr + 1 ), so x ≡ y (mod pr + 1 ), a contradiction.

Thus Qp is a topological space in the sense of Hausdorff.

29.

We need to check the three axioms of a metric. First, note that if x = pα e and y = pβ e′ , where e and e′ are units, and if α ≥ β, then x − y = pβ (pα−β e − e′ ) while y − x = pβ (e′ − pα−β e). In each case, the expression inside the parenthesis is a unit (unless x = y). Therefore, νp (x − y) = νp (y − x) = min(νp (x), νp (y)) and (x, y) = (y, x). Also, we have(x, y) = 0 if and only if νp (x − y) = −∞ if and only if x − y = 0 if and only

Solutions

167

if x = y. Finally, if νp (x) = α, νp (y) = β, and νp (z) = γ, then νp (x − y) = min(α, β), νp (x − z) = min(α, γ), and νp (y − z) = min(β, γ). Suppose α is the smallest of the three integers, then (x, y) = (1/p)α ≤ (1/p)α + (1/p)min(β,γ) = (x, z) + (z, y). A similar result is true if β is the smallest of the three integers. If γ is the smallest of the three, then (x, y) = (1/p)min(α,β) ≤ (1/p)γ + (1/p)γ = (x, z) + (z, y). So in any case, the third axiom is satisfied.

30.

Suppose that α1 , α2 , α3 , . . . is a Cauchy sequence of p-adic numbers. We need to define a limit element by defining its digits up to any given power Q. So choose ϵ = p−Q .

Since the sequence is a Cauchy sequence, , we know there is an integer N such that for n, m > N, we have (αm , αn ) < ϵ. This means that ( p1 )νp (αn −αm ) < ϵ, or νp (αn − αm ) > − logp (ϵ). But νp (αn − αm ) = min(νp (αn ), νp (αm )) < logp ( ϵ1 ) = Q. Therefore, we see that any two elements in the sequence beyond N agree for every power of p up to the power Q and we can define the digits for our limit element to be those digits up to the power Q. Thus, the Cauchy sequence converges.

31.

To show that the tables determine an associative algebra over the real numbers, we must check associativity for the basis elements. For each algebra, that means checking associativity for the triples iij, iji, ijj, jii, jij, and jji. For each of the algebras defined here, associativity does hold for all of those triples. There are other associative algebras of dimension 2 over the real numbers. For example, the complex numbers are such an algebra. If we take the basis for the complex numbers to be i, j, then the four multiplications are i · i = i, i · j = j · i = j, and j · j = −i (where we have represented by i, j the usual basis 1, i).

0 1 The matrix A = satisfies A2 = 0 and is therefore nilpotent.

0 0

1 1 The matrix A = satisfies A2 = A and is therefore idempotent.

0 0 Some further examples are finite sets and functions, topological groups and continuous homomorphisms, Banach spaces and linear transformations of norm at most 1, and finite Abelian groups and group homomorphisms.

= 924, if the woman has no discriminating ability, the probability of Because 12 6 picking six cups correctly is 1/924, and the probability of picking five cups correctly

is 6 65 /924 = 36/924. The total probability is then 37/924 = 0.04.

32.

33.

34.

35.

36.

Under the hypothesis that the lady has no discriminatory ability, that is, that p = 21 , there are 210 possible selections she could make of which cup in each pair had the milk added first. In other words, the probability of her making any given selection is

10 10 1 1 10 which ) .

But of the 2 possible selections she could make, there are = ( 2 210 8 10 have 8 selections correct (and 2 incorrect). Thus P(X = 8|p = 1/2) = 10 8 (1/2) .

37.

Here we are assuming that the lady has probability p of choosing the correct cup of tea in each pair. Thus, the probability of making k correct choices out of 10 and therefore 10 − k incorrect choices is pk (1 − p)10 − k . But of all the selections she could make, for a given k there are 10 k selections which have k correct choices and 10 − k incorrect choices.

Therefore, the probability of getting exactly k correct choices is

10 k 10 − k p ( 1 − p ) .

So if our critical region consists of 8, 9, or 10 correct choices, the k

168

Solutions probability of an observation in that region is the sum of the probabilities for the indi10 9 8 2 10 vidual numbers, namely 10 8 p (1 − p) + 9 p (1 − p) + p . That is, this value is the probability of an observation in the critical region, assuming the lady has probability p of choosing correctly. In other words, this expression is the power P(R|H2 ).

38.

39.

40.

We prove this by induction. We know that the first differences of first-degree polynomial functions are constant. So assume that the nth-order differences of nth-degree polynomial functions are constant. Then suppose we have a polynomial of degree n + 1. If we look at all terms of lower degree, the (n + 1)st order differences are all 0 by the induction hypothesis. So we just need to consider the term xn + 1 . The first differences of that function are given by (x + 1)n + 1 − xn + 1 = (n + 1)xn + terms of lower degree. Since the polynomial on the right is one of degree n, its nth-order differences are constant. Therefore, the (n + 1)st-order differences of the original polynomial xn + 1 are constant, and by induction, the general theorem is proved.

The pyramidal numbers are numbers of the form n3 , (n ≥ 3), and can therefore be expressed in the form of a cubic polynomial: 16 n3 − 21 n2 + 13 n. These numbers are the numbers 1, 4, 10, 20, 35, . . .. Their first differences are the triangular numbers 3, 6, 10, 15, . . .. Their second differences are the integers 3, 4, 5, . . . , and their third differences are all constantly 1. Thus to calculate the pyramidal numbers, one starts with the third differences, notes that the first second difference is 3, the first first difference is 3, and the first pyramidal number is 1. We can then use the Difference Engine to calculate by finding in turn the integers, the triangular numbers, and the pyramidal numbers by repeated addition.

We begin with the equation x ex − 1 2

=1−

x x2 x4 x6 + B2 + B4 + B6 + · · · .

2 2!

4!

6!

3

Note that ex − 1 = x + x2! + x3! + · · · , so that 1 x = . ex − 1 1 + x + x2 + · · · 2!

3!

Therefore,

x x2 x4 x x2 x3 1 = 1 − + B2 + B4 + · · · 1+ + + + ··· .

2 2!

4!

2! 3! 4!

If we perform the indicated multiplication, we get a polynomial in x of the form 1 + D1 x + D2 x2 + · · · , in which every Di = 0. In particular, we can calculate D2n explicitly: 0 = D2n =

1 (2n + 1)!

−

1 1 B2 B4 B + + + · · · + 2n .

2 (2n)! 2!(2n − 1)! 4!(2n − 3)!

(2n)!

Solutions

169

If we multiply every term in this expression by (2n)!, we get

2n(2n − 1)(2n − 2) 1 1 2n 0 = − + B2 + B4 + · · · + B2n 2n + 1 2 2!

4!

2n(2n − 1)(2n − 2) 1 2n − 1 2n + B2 + B4 + · · · + B2n . =− 2 2n + 1 2 4!

41.

1 2 1 1 We first calculate for n = 1: − 12 · 22− + 1 + B2 ( 2! ) = 0. Therefore − 6 + B2 = 0, and B2 = 6 .

1 4·3·2 3 1 1 4−1 Next, we take n = 2: − 2 · 4 + 1 + 6 · 2 + B4 ( 4! ) = 0. Therefore, − 10 + 3 + B4 = 0, 1 1 1 6 1 6·5·4 6·5·4·3·2 and B4 = − 30 . Finally, take n = 3: − 21 · 66− = 0. So + 1 + 6 ( 2 ) − 30 4·3·2 + B6 6!

5 1 1 1 − 14 + 2 − 6 + B6 = 0, and B6 = 42 .

42.

Since the machine begins in state q1 and initially reads a 1, according to instruction (a) the machine then reprints that 1, moves one square to the right, and remains in state q1 . Thus the tape still has the two ones it had initially. Next, the machine reads the second 1, so it reprints that 1, moves one further square to the right, and still remains in state q1 . The tape still has just the two ones. But since the next square to the right is blank, we follow instruction (b). The machine thus prints a 1 in that blank square, moves one further square to the right, and changes to state q2 . There are now three ones on the tape and no further instructions. So the machine stopes with a tape with three ones. Note that if the tape initially had n ones, then it would remain in state q1 through n iterations of instruction (a). After each iteration, the tape will still have n ones. But after these n iterations, the machine will be reading a blank square, so it will follow instruction (b) by printing one further 1 and then stopping. Thus the tape will end with n + 1 ones, and the machine represents the function f(n) = n + 1.

43.

We are to begin with a tape which is all blank except for an initial 0, n 1’s beginning at the second square, and a 0 in the following square. We want to end with a tape in a similar configuration except that there are 2n 1’s. Therefore we must get the machine to copy each of the given 1’s over at the end of the string. One way to do this is to begin by having the machine read the leftmost 1, change it to an x (so that it is marked), then move along the row until it gets to a blank square. It should put a 1 in that blank square.

This is the copy of the first 1. The machine should then move back until it reaches the x, change that back to a 1, move one square to the right, change that 1 to an x, and then continue to the right until it reaches a blank. It should then put a 1 in that square and move back to the left until it again reaches an x. This process should be repeated until all the initial 1’s have been changed into xs and back to 1’s. The configuration will then be a 0 followed by n 1s followed by a 0 followed by another n 1’s. We then must eliminate the middle 0 and put a 0 at the end. The actual states and instructions can now be specified, where q1 is the initial state and where the machine begins by reading the leftmost 1 on the tape. a. If the machine is in state q1 and reads a 1, it prints an x, moves one square to the right, and changes to state q2 . b.

If the machine is in state q2 and reads a 1, it moves one square to the right and remains in state q2 . If the machine is in state q2 and reads a 0, it moves one square to the right, and changes to state q3 .

170

Solutions c.

If the machine is in state q3 and reads a blank, it prints a 1, moves one square to the left, and changes to state q4 .

d,

If the machine is in state q4 and reads either a 1 or a 0, it moves one square to the left and remains in state q4 . If it is in state q4 and reads an x, it changes the x to a 1, moves one square to the right, and changes to state q5 .

e.

If the machine is in state q5 and reads a 1, it prints an x, moves one square to the right, and changes to state q6 . If it is in state q5 and reads a 0, it moves one square to the right and changes to state q7 .

f.

If the machine is in state q6 and reads a 0 or a 1, it moves one square to the right, and remains in state q6 . If it is in q6 and reads a blank, it prints a 1, moves one square to the left and changes to state q4 .

g.

If the machine is in state q7 and reads a 1, it replaces it by a 0, moves one square to the left and changes to state q8 .

h.

If the machine is in state q8 and reads a 0, it replaces it by a 1, moves one square to the right and changes to state q5 .

44.

If we substitute 0 for x, the right side of the desired identity is [f(0, y) + 0][f(1, y) + 1] = f(0, y), because f(1, y) + 1 = 1. Also, if we substitute 1 for x, then the right side becomes [f(0, y) + 1][f(1, y) + 0] = f(1, y). Therefore, whatever the value of x, we have f(x, y) = [f(0, y) + x][f(1, y) + x′ ], as desired.

45.

If we add x to each side of the identity from Exercise 44, we have x + f(x, y) = x + [f(0, y) + x][f(1, y) + x′ ] = [x + f(0, y) + x][x + f(1, y) + x′ ] = [x + f(0, y)][1 + f(1, y)] = [x + f(0, y)][1] = x + f(0, y).

46.

If we substitute 0 for x, the right side of the desired identity is 0f(1, y) + 1f(0, y) = f(0, y). If we substitute 1 for x, the right side is 1f(1, y) + 0f(0, y) = f(1, y). Therefore, whatever the value of x, we have f(x, y) = xf(1, y) + x′ f(0, y). If we then multiply each side of this equation by x, we have xf(x, y) = x[xf(1, y) + x′ f(0, y)] = xxf(1, y) + xx′ f(0, y) = xf(1, y) + 0f(0, y) = xf(1, y), as desired.

47.

The circuit is constructed by using series or parallel switches as outlined. We need a separate piece for each digit desired, but these need to be connected so that the circuit for sj has as one of its inputs the outcome cj .

48.

Let z = x + iy, and assume y > 0. We want to show that the imaginary part of f(z) is also positive. We calculate: f(z) =

a(x + iy) + b ax + b + iay acx2 + (ad + bc)x + bd + acy2 + i(ady − bcy) . = = c(x + iy) + d cx + d + icy (cx + d)2 + c2 y2

Given that ad − bc = 1, we see that the imaginary part of f(z) is (cx + d)y2 + c2 y2 . This value is clearly positive.

49.

To find the sum of P1 = (2, 5) and P2 = (4, 9), we first find the line connecting them. Its equation is y = 2x + 1. To find where this line intersects the curve again, we need to solve the equation (2x + 1)2 = x3 + 17, or x3 − 4x2 − 4x + 16 = 0. Given that x = 2 and x = 4 are solutions, we easily find that x = −2 is the third solution.

Thus the point P′3 = (−2, −3). Therefore the sum is P3 = (−2, 3). To double (−2, 3), we need the tangent line to the curve at that point. We calculate that 2y dy = 3x2 dx,

Solutions

171

so dy/dx = 3x2 /2y = 12/6 = 2. The tangent line is then given by y = 2x + 7. We next find the intersection of this line with the curve by solving (2x + 7)2 = x3 + 17, or x3 − 4x2 − 28x − 32 = 0. The solution to this equation other than the double solution at x = −2 is x = 8. Therefore, the point P′3 = (8, 23) and the point P3 which is double (−2, 3) is the point (8, −23).

50.

We first double (3, 8). The slope of the tangent line to the curve at (3, 8) is found by calculating 2y dy = (3x2 − 43) dx, or dy/dx = 3x2 − 43/2y = −16/16 = −1. The tangent line is then y = −x + 11. To find the second intersection of this line with the curve, we solve (−x + 11)2 = x3 − 43x + 166 or x3 − x2 − 21x + 45 = 0. Since 3 is a double root, the other root is found to be x = −5, and the point of intersection is (−5, 16). Therefore, the point which is equal to 2 × (3, 8) = (−5, −16). To find 3 × (3, 8), we add (3, 8) and (−5, −16). The line connecting these two points is y = 3x − 1. To find the third intersection of this line with the curve, we solve (3x − 1)2 = x3 − 43x + 166, or x3 − 9x2 − 37x + 165 − 0. Given that 3 and −5 are roots, we easily find the third root to be x = 11. The intersection point is then (11, 32) and 3 × (3, 8) = (11, −32). We next add (3, 8) and (11, −32). Again, the line connecting these two points is y = −5x + 23. To find the third intersection of this line with the curve, we solve (−5x + 23)2 = x3 − 43x + 166, or x3 − 25x2 + 187x − 363 = 0. Since 3 and 11 are roots of this equation, we find the third root to be x = −11 and therefore the third intersection point is (11, −32). It follows that 4 × (3, 8) = (11, 32). It is now clear that the sum of 3 × (3, 8) and 4 × (3, 8) is the additive identity, namely, the point at infinity.

Thus the order of (3, 8) is 7. It follows that 5×(3, 8) = (−5, 16) and 6×(3, 8) = (3, −8) (or, of course, one can calculate these values directly).

51.

The subgroup of multiples of the identity in SL(2, 8) consists only of the identity, since in the field with 8 elements, the only non-zero solution to m2 = 1 is m = 1. Thus the order of PSL(2, 8) is equal to the order of SL(2, 8). To calculate that order, note

a b that in any matrix in the group, ad − bc = 1. It follows that, if a ̸= 0, then c d d = (1 + bc)/a. Thus, there are 7 possible choices for a and 8 possible choices for b and c, for a total of 7 × 8 × 8 = 448 choices. On the other hand, if a = 0, we have the equation bc = 1. Thus, there are 8 arbitrary choices for d and 7 arbitrary choices for c (which of course cannot equal 0). Thus in this case there are 8 × 7 = 56 choices. The total number of possibilities is then 448 + 56 = 504, and that is the order of SL(2, 8) and also the order of PSL(2, 8).

52.

To calculate the order of PSL(3, 4), we begin by calculating the order of GL(3, 4). Note that an element of GL(3, 4) can be thought of as a non-singular linear transformation T acting on the three-dimensional vector space V over GF(4). Let {α1 , α2 , α3 } be an ordered basis of V. Any non-singular linear transformation T takes {α1 , α2 , α3 } into an ordered basis {T(α1 ), T(α2 ), T(α3 )}, and given an ordered basis {β1 , β2 , β3 }, there is a non-singular linear transformation T such that T(α1 ) = β1 , T(α2 ) = β2 , and T(α3 ) = β3 . So there are exactly as many elements of GL(3, 4) as there are ordered bases of V. To count ordered bases of V, we note that a vector in V is a three-tuple with elements in GF(4). So there are 43 − 1 choices for the first vector β1 in the basis (because (0, 0, 0) cannot be a basis element). Once β1 is chosen, we can take β2 to be any vector which is not in the subspace spanned by β1 , i.e., any vector which is not a multiple of β1 . So since there are four multiples, there are 43 − 4 choices for β2 . Similarly,

172

Solutions β3 can be any vector which is not in the subspace spanned by β1 and β2 , i.e., not a linear combination of β1 and β2 . Since there are 42 such linear combinations, there are 43 − 42 choices for β3 . So there are (43 − 1)(43 − 4)(43 − 42 ) = 63 · 60 · 48 = 181, 440 ordered bases for V and therefore that many elements in GL(3, 4). Now SL(3, 4) is the subgroup of GL(3, 4) consisting of those elements with determinant 1. Since the set of elements with determinant ρ is a coset of SL(3, 4), and there are three possible nonzero determinants, it follows that the order of SL(3, 4) is 13 the order of GL(3, 4), or 63 · 60 · 16 = 60, 480. Finally, PSL(3, 4) is the factor group of SL(3, 4) by the subgroup of multiples of the identity. Since this subgroup consists of elements ρI, where ρ3 = 1, and since each of the three non-zero elements of GF(4) has the property that its cube is 1, the subgroup of multiples of the identity has order 3. It follows that the order of PSL(3, 4) is 13 the order of SL(3, 4), namely 63 · 20 · 16 = 20,160. The order of A8 is 1 2 8! = 20,160, so these two groups have the same order.

53.

By the normal rules for composing cyclic permutations, we have AB = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11)(5, 6, 4, 10)(11, 8, 3, 7) = (1, 2, 3, 8, 4, 11, 9, 10, 6, 5, 7),

BA = (5, 6, 4, 10)(11, 8, 3, 7)(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) = (1, 2, 7, 3, 10, 8, 9, 5, 4, 6, 11), AC = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11)(1, 12)(2, 11)(3, 6)(4, 8)(5, 9)(7, 10) = (1, 12, 2)(3, 7, 11)(4, 9, 6)(5, 10, 8), and CA = (1, 12)(2, 11)(3, 6)(4, 8)(5, 9)(7, 10)(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) = (1, 11, 12)(2, 6, 10)(3, 8, 5)(4, 9, 7).

54.

The fundamental group of the torus is generated by two loops, one loop going around the “hole” in the donut and the other going around the “arm” of the donut. Since each of these loops can be traversed once, twice, three times, and so on, in either of two directions, each of them generates an infinite cyclic group, that is, a group isomorphic to Z. One can also follow any number of loops in one direction with any number of loops in the other direction. But to show that the fundamental group is actually isomorphic to Z × Z, it is necessary to show that the group is Abelian, that is, if p is a single loop around the hole and q is a single loop around the arm, then p followed by q is equivalent to q followed by p. There is an excellent animation that demonstrates this: http://www.youtube.com/watch?v=nLcr-DWVEto. In essence the idea is the following: Consider a rectangle. Then the path going up the left side of the rectangle and then along the top can easily be deformed into the path going first along the bottom and then up the right side. But we can turn the rectangle into a torus by gluing the top edge to the bottom edge and then the right side to the left. So therefore a loop going first around the arm and then around the hole can be deformed into the path that first goes around the hole and then around the arm.