Solutions Manual of Mathematical Methods in the Physical Sciences by Boas | 1st edition

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Chapter 1

1.1 1.3 1.6 1.9 1.13 1.16

(2/3)10 = 0.0173 yd; 6(2/3)10 = 0.104 yd (compared to a total of 5 yd) 5/9 1.4 9/11 1.5 7/12 11/18 1.7 5/27 1.8 25/36 6/7 1.10 15/26 1.11 19/28 $1646.99 1.15 Blank area = 1 At x = 1: 1/(1 + r); at x = 0: r/(1 + r); maximum escape at x = 0 is 1/2.

2.1 2.4 2.7

1 ∞ e2

4.1 4.2 4.3 4.4 4.5

= 1/2n → 0; Sn = 1 − 1/2n → 1; Rn = 1/2n → 0 = 1/5n−1 → 0; Sn = (5/4)(1 − 1/5n ) → 5/4; Rn = 1/(4 · 5n−1 ) → 0 = (−1/2)n−1 → 0; Sn = (2/3)[1 − (−1/2)n ] → 2/3; Rn = (2/3)(−1/2)n → 0 = 1/3n → 0; Sn = (1/2)(1 − 1/3n ) → 1/2; Rn = 1/(2 · 3n ) → 0 = (3/4)n−1 → 0; Sn = 4[1 − (3/4)n ] → 4; Rn = 4(3/4)n → 0 1 1 1 an = → 0; Sn = 1 − → 1; Rn = →0 n(n + 1) n+1 n+1

1 1 (−1)n+1 (−1)n an = (−1)n+1 + → 0 ; Sn = 1 + → 1; Rn = →0 n n+1 n+1 n+1

4.6 4.7

2.2 2.5 2.8

1/2 0 0

2.3 2.6 2.9

0 ∞ 1

an an an an an

5.1 5.4 5.7 5.9

D D Test further D

5.2 5.5 5.8 5.10

6.5 Note: 6.7 6.10 6.13 6.19 6.22 6.25 6.28 6.32 6.35

(a) D 6.5 In the following answers, D, I = ∞ 6.8 C, I = π/6 6.11 D, I = ∞ 6.14 C, ρ = 3/4 6.20 C, ρ = 0 6.23 C, ρ = 0 6.26 C, ρ =P 4/27 6.29 D, cf. n−1 6.33 P −2 C, cf. n 6.36

Test further D Test further D

Test further Test further

(b) D R∞ I= an dn; ρ = test ratio.

D, I = ∞ 6.9 C, I = 0 C, I = 0 6.12 C, I = 0 D, I = ∞ 6.18 D, ρ = 2 C, ρ = 0 6.21 D, ρ = 5/4 D, ρ = ∞ 6.24 D, ρ = 9/8 C, ρ = (e/3)3 6.27 D, ρ =P 100 D, ρ =P 2 6.31 D, cf. P n−1 C, cf.

2−n 6.34 C, cf. n−2 P −1/2 D, cf. n

1

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5.3 5.6

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2

7.1 7.5

C C

9.1 9.3 9.5 9.7 9.9 9.11 9.13 9.15 9.17 9.19 9.21 9.22

P −1 D, cf. n C, I =P 0 C, cf. n−2 D, ρ = 4/3 D, ρ = e P D, I = ∞, or cf.P n−1 C, I = 0, or cf. n−2 D, ρ = ∞, an 6→ 0 C, ρ = 1/27 C C, ρ = 1/2 (a) C (b) D

10.1 10.4 10.7 10.10 10.13 10.16 10.19 10.22 10.25

7.2 7.6

D D

7.3 7.7

C C

7.4 7.8

C C

9.2 9.4 9.6 9.8 9.10 9.12 9.14 9.16 9.18 9.20

D, an 6→ 0 P −1 D, I = ∞, or cf. n C, ρ = 1/4 C, ρ = 1/5 D, an 6→ P 0 −2 C, cf. n C, alt.P ser.

C, cf. n−2 C, alt. ser.

C (c) k > e

|x| < √ 1 10.2 |x| < 3/2 10.3 |x| ≤ 1 |x| ≤ 2 10.5 All x 10.6 All x −1 ≤ x < 1 10.8 −1 < x ≤ 1 10.9 |x| < 1 |x| ≤ 1 10.11 −5 ≤ x < 5 10.12 |x| < 1/2 −1 < x ≤ 1 10.14 |x| < 3 10.15 −1 < x < 5 −1 < x < 3 10.17 −2 < x ≤ 0 10.18 −3/4 ≤ x ≤ −1/4 |x| < 3 10.20 All x 10.21 0 ≤ x √ ≤1 No x 10.23 x > 2 or x < −4 10.24 |x| < 5/2 nπ − π/6 < x < nπ + π/6

−1/2 −1/2 (−1)n (2n − 1)!!

13.4 = 1; = 0 n (2n)!!

Answers to part (b), Problems 5 to 19:

∞ n+2 ∞ X X x 1/2 n+1 13.5 − 13.6 x (see Example 2) n n 1 0

∞ ∞ X X (−1)n x2n −1/2 13.7 13.8 (−x2 )n (see Problem 13.4) (2n + 1)! n 0 0 ∞ ∞ X X (−1)n x4n+2 13.9 1 + 2 xn 13.10 (2n + 1)!

1 0 ∞ ∞ n n X X (−1) x (−1)n x4n+1 13.11 13.12 (2n + 1)!

(2n)!(4n + 1) 0 0 ∞ ∞ n 2n+1 X X (−1) x x2n+1 13.13 13.14 n!(2n + 1) 2n + 1 0 0

∞ X −1/2 x2n+1 13.15 (−1)n n 2n + 1 0 ∞ ∞ 2n X X x xn 13.16 13.17 2 (2n)! n 0 oddn

∞ ∞ X (−1)n x2n+1 X −1/2 x2n+1 13.18 13.19 (2n + 1)(2n + 1)! n 2n + 1 0 0 2 3 5 6 13.20 x + x + x /3 − x /30 − x /90 · · · 13.21 x2 + 2x4 /3 + 17x6 /45 · · · 13.22 1 + 2x + 5x2 /2 + 8x3 /3 + 65x4 /24 · · · 13.23 1 − x + x3 − x4 + x6 · · ·

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1 + x2 /2! + 5x4 /4! + 61x6 /6! · · · 1 − x + x2 /3 − x4 /45 · · · 1 + x2 /4 + 7x4 /96 + 139x6 /5760 · · · 1 + x + x2 /2 − x4 /8 − x5 /15 · · · x − x2 /2 + x3 /6 − x5 /12 · · · 1 + x/2 − 3x2 /8 + 17x3 /48 · · · 1 − x + x2 /2 − x3 /2 + 3x4 /8 − 3x5 /8 · · · 1 − x2 /2 − x3 /2 − x4 /4 − x5 /24 · · · x + x2 /2 − x3 /6 − x4 /12 · · · 1 + x3 /6 + x4 /6 + 19x5 /120 + 19x6 /120 · · · x − x2 + x3 − 13x4 /12 + 5x5 /4 · · · 1 + x2 /3! + 7x4 /(3 · 5!) + 31x6 /(3 · 7!) · · · u2 /2 + u4 /12 + u6 /20 · · · −(x2 /2 + x4 /12 + x6 /45 · · · ) e(1 − x2 /2 + x4 /6 · · · ) 4 1 − (x − π/2)2 /2! + (x − π/2) /4! · · · 3 1 − (x − 1) + (x − 1)2 − (x − 1) · · · 3 2 e [1 + (x − 3) + (x − 3) /2! + (x − 3)3 /3! · · · ] 2 −1 + (x − π) /2! − (x − π)4 /4! · · · −[(x − π/2) + (x − π/2)3 /3 + 2(x − π/2)5 /15 · · · ] 5 + (x − 25)/10 − (x − 25)2 /103 + (x − 25)3 /(5 · 104 ) · · ·

14.6 Error < (1/2)(0.1)2 ÷ (1 − 0.1) < 0.0056 14.7 Error < (3/8)(1/4)2 ÷ (1 − 14 ) = 1/32 14.8 For x < 0, error < (1/64)(1/2)4 < 0.001 For x > 0, error < 0.001 ÷ (1 − 12 ) = 0.002 1 14.9 Term n + 1 is an+1 = (n+1)(n+2) , so Rn = (n + 2)an+1 .

14.10 S4 = 0.3052, error < 0.0021 (cf. S = 1 − ln 2 = 0.307) −x4 /24 − x5 /30 · · · ' −3.376 × 10−16 x8 /3 − 14x12 /45 · · · ' 1.433 × 10−16 x5 /15 − 2x7 /45 · · · ' 6.667 × 10−17 x3 /3 + 5x4 /6 · · · ' 1.430 × 10−11 0 15.6 12 15.7 10!

1/2 15.9 −1/6 15.10 −1 4 15.12 1/3 15.13 −1 t − t3 /3, error < 10−6 15.15 32 t3/2 − 52 t5/2 , error < 17 10−7 e2 − 1 15.17 √ cos π2 = 0 ln 2 15.19 2 (a) 1/8 (b) 5e (c) 9/4 (a) 0.397117 (b) 0.937548 (c) 1.291286 (a) π 4 /90 (b) 1.202057 (c) 2.612375 (a) 1/2 (b) 1/6 (c) 1/3 (d) −1/2 (a) −π (b) 0 (c) −1 (d) 0 (e) 0 (f) 0 15.27 (a) 1 − vc = 1.3 × 10−5 , or v = 0.999987c (b) 1 − vc = 5.2 × 10−7 (c) 1 − vc = 2.1 × 10−10 (d) 1 − vc = 1.3 × 10−11 15.28 mc2 + 12 mv 2 15.29 (a) F/W = θ + θ3 /3 · · · (b) F/W = x/l + x3 /(2l3 ) + 3x5 /(8l5 ) · · ·

15.1 15.2 15.3 15.4 15.5 15.8 15.11 15.14 15.16 15.18 15.20 15.21 15.22 15.23 15.24

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3

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Chapter 2 x

y

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20

1 −1 1 √ − 3 0 0 −1 3 −2 √2 3 −2 √0 2 −1 5 1 0 4.69 −2.39

1 1 √ − 3 1 2 −4 0 0 2 −2 1√ −2 3 −1 √ 2 0 0 −1 3 1.71 −6.58

5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18

1/2 −1/2 1 0 2 −1 7/5 1.6 −10.4 −25/17 17 2.65 1.55 1.27 21/29 1.53 −7.35 −0.94

−1/2 −1/2 0 2 √ 2 3 0 −1/5 −2.7 22.7 19/17 −12 1.41 4.76 −2.5 −20/29 −1.29 −10.9 −0.36

r √ √2 2 2 2 2 4 1 3 √ 2√2 2 2 2 4 1 2 1 √5 2 3 5 7 √ 1/√2 1/ 2 1 2 4 √1 2 3.14 p 25 58/17 20.8 3 5 2.8 1 2 13.1 1

5

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θ π/4 3π/4 −π/3 5π/6 π/2 −π/2 π 0 3π/4 −π/4 π/6 −2π/3 3π/2 π/4 −π or π 0 −π/4 π/2 20◦ = 0.35 −110◦ = −1.92 −π/4 −3π/4 or 5π/4 0 π/2 π/3 π −8.13◦ = −0.14 −59.3◦ = −1.04 2 = 114.6◦ 142.8◦ = 2.49 −35.2◦ = −0.615 28◦ = 0.49 2π/5 −1.1 = −63◦ −43.6◦ = −0.76 −40◦ = −0.698 −124◦ = −2.16 201◦ or −159◦, 3.51 or −2.77

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70

4.9 4.12 4.17 4.21

25/102, 25/77, 49/101, 12/25 4.11 0.097, 0.37, 0.67; 13 5 4.14 n!/nn MB: 16, FD: 6, BE: 10 4.18 MB: 125, FD: 10, BE: 35 C(n + 2, n) 4.22 0.135 4.23 0.30 p √ 5.1 µ = 0, σ = √3 5.2 µ = 7, σ = √ 35/6 5.3 µ = 2, σ = p2 5.4 µ = 1, σ = p21/2 5.5 µ = 1, σ = 7/6 p 5.6 µ = 3, σ = 284/13 = 4.67 5.7 µ = 3(2p − 1), σ = 2 3p(1 − p) 5.8 E(x) = $12.25 5.12 E(x) = 7 5.15 x̄ = 3(2p − 1) 2 2 5.17 Problem 5.2: E(x ) = 329/6, σ = 35/6 Problem 5.6: E(x2 ) = 401/13, σ 2 = 284/13 Problem 5.7: E(x2 ) = 24p2 − 24p + 9, σ 2 = 12p(1 − p) √ 6.1 (a) f (x) = π −1 (a2 − x2 )−1/2 (c) x̄ = 0, σ = a/ 2 6.2 e−2 = 0.135√ √ 6.3 f (h) = 1/(2 l l − h ) √ 2 2 √ 6.4 f (x) = αe−α x / π, x̄ = 0, σ = 1/(α 2 ) 6.5 f (t) = λe−λt , F (t) = 1 − e−λt , t̄ = 1/λ, half life = t̄ ln 2 √ 6.6 F (r) = r2 , f (r) = 2r, r̄ = 2/3, σ = 2/6 6.7 (a) F (s) = 2[1 − cos(s/R)], f (s) = (2/R) sin(s/R) (b) F (s) = [1 − cos(s/R)]/[1 − cos(1/R)] ∼ = s2 , f (s) = R−1 [1 − cos(1/R)]−1 sin(s/R) ∼ = 2s p 2 6.8 f (r) = 3r ; r̄ = 3/4, σ = 3/80 = 0.19 6.9 f (r) = 4a−3 r2 e−2r/a

n

Exactly 7h

At most 7h

At least 7h

Most probable number of h

Expected number of h

7.1

7

0.0078

1

0.0078

3 or 4

7/2

7.2

12

0.193

0.806

0.387

6

6

7.3

15

0.196

0.500

0.696

7 or 8

15/2

7.4

18

0.121

0.240

0.881

9

9

7.5

0.263

1 2 µ = 0, σ 2 = kT /m, f (v) = p e−mv /(2kT ) 2πkT /m In (8.11) to (8.20), the first number is the binomial result and the second number is the normal approximation using whole steps at the ends as in Example 2.

8.11 0.0796, 0.0798 8.12 0.03987, 0.03989 8.13 0.9598, 0.9596 8.14 0.9546, 0.9546 8.15 0.03520, 0.03521 8.16 0.4176, 0.4177 8.17 0.0770, 0.0782 8.18 0.372, 0.376 8.19 0.0946, 0.0967 8.20 0.462, 0.455 8.25 C: 38.3%, B and D: 24.2%, A and F: 6.7% In µ + 12 σ and µ + 23 σ, change 12 to 0.5244, and 32 to 1.2816.

8.3

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