APPENDIX 2
Answers to Odd-Numbered Problems
Problem Set 1.1, page 8
1.
3. y = ce x
5. y = 2e −x (sin x − cos x + c
7.
9. y = 1.65e −2x + 0.35
11.
13. y = 1/(1 + 3e x )
15. y = 0 and y = 1 because y ′ = 0 for these y
17. exp , t = 1011 (1n 2)/1.4 [sec]
19. Integrate y ″ g twice, y ′(t ) = gt + v 0 , y ′(0) = v 0 = 0 (start from rest), then , where y (0) = y 0 = 0
Problem Set 1.2, page 11
11. Straight lines parallel to the x -axis
13. y = x
15. mv ′ = mg − bv 2 , v ′ = 9.8 − v 2 , v (0) = 10, v ′ = 0 gives the limit/9.8 = 3.1 [meter/sec]
17. Errors of steps 1, 5, 10: 0.0052, 0.0382, 0.1245, approximately
19. x 5 = 0.0286(error 0.0093), x 10 = 0.2196(error 0.0189)
Problem Set 1.3, page 18
1. If you add a constant later, you may not get a solution. Example: y ′ = y , In |y | = x + c , but not e x + c (with c ≠ 0)
3. cos2 y dy = dx ,
5. y 2 + 36x 2 = c , ellipses
7. y = x arctan (x 2 + c )
9. y = x/(c − x)
11. y = 24/x , hyperbola
13. dy /sin2 y = dx/cosh 2 x , −cot y = tanh x + c , c = 0, y = −arccot (tanh x )
15. y 2 + 4x 2 = c = 25
17. y = x arctan (x 3 − 1)
19. y 0e kt = 2y 0, e k = 2 (1 week), e 2k = 22 (2 weeks), e 4k = 24
21. 69.6% of y 0
23. pV = c = const
25. T = 22 − 17e −0.5306t = 21.9 [°C] when t = 9.68 min
27. , e −kt0 − 0.01,
29. No. Use Newton's law of cooling.
31. y = ax , y ′ = g(y/x) = a = const, independent of the point (x,y )
33. ΔS = 0.5S Δφ, ds /dφ = 0.15S , S = S 0e 0.15φ = 1000S 0 , φ = (1/0.15) In 1000 = 7.3 · 2π. Eight times.
Problem Set 1.4, page 26
1. Exact, 2x = 2x , x 2 y = c,y = c/x 2
3. Exact, y = arccos (c /cos x )
5. Not exact,
7.
9. Exact, u = e 2x cos y + k(y) , u y = −e 2x sin y + k ′, k ′ = 0. Ans. e 2x cos y = 1
11. F = sinh x , sinh2 x cos y = c
13. u = e x + k (y ), u y = k ′ = −1 + e y , k = −y + e y . Ans. e x − y + e y = c
15. b = k , ax 2 + 2kxy + ly 2 = c
Problem Set 1.5, page 34
3. y = ce x − 5.2
5. y = (x + c )e −kx
7. y = x 2 (c + e x )
9. y = (x − 2.5/e )e cos x
11. y = 2 + c sin x
13. Separate. y − 2.5 = c cosh4 1.5x
15. (y 1 + y 2 )′ + p (y 1 + y 2 ) = (y 1 ′ + p y 2 ) = 0 + 0 = 0
17. (y 1 + y 2 )′ + p (y 1 + y 2 ) = (y 1 ′ + p y 2 ) = r + 0 = r
19. Solution of cy 1 ′ + pcy 1 + c (y ′1 + py 1 ) = cr
21. . Thus, y = uyh gives (4). We shall see that this method extends to higher-order ODEs (Secs. 2.10 and 3.3 ).
23.
25. y = 1/u , u = ce −3.2x + 10/3.2
27. dx/dy = 6e y − 2x , x = ce −2y + 2e y
31. T = 240e kt + 60, T (10) = 200, k = −0.0539, t = 102 min
33. y ′ = A − ky , y (0) = 0, y = A (1 − e −kt )/k
35. y ′ = 175(0.0001 − y /450), y (0) = 450 · 0.0004 = 0.18, y = 0.135e −0.3889 t + 0.045 = 0.18/2, e −0.3889 t = (0.09 − 0.045)/0.135 = 1/3, t = (In 3)/0.3889 = 2.82. Ans. About 3 years
37. y ′ = y − y 2 − 0.2y , y = 1/(1.25 − 0.75e −0.8t ), limit 0.8, limit 1
39. . y ′ = By (y − A /B ,A > 0,B > 0. Constant solutions y = 0, y = A/B , y ′ > 0 if y > A/B (unlimited growth), y ′ < 0 if 0 < y < A/B (extinction). y = A/(ceAt + B) , y (0) > A/B if c < 0, y (0) < A/B if c > 0.
Problem Set 1.6, page 38
1. x 2 /(c 2 + 9) + Y 2 /c 2 m 1 = 0
3. y − cosh (x − c ) − c = 0
5. , circles
7.
9.
11.
13. y ′ = −4x /9y . Trajectories . Sketch or graph these curves.
15. u = c , ux dx + uy dy = 0, y ′ = −ux /uy . Trajectories. . Now , y ′ = −x /uy . This agrees with the trajectory ODE in u if ux = vy (equal denominators) and uy = −vx (equal numerators). But these are just the Cauchy–Riemann equations.
Problem Set 1.7, page 42
1. y ′ = f(x,y) = r(x) − ∂f /∂y = −p (x ) is continuous and is thus bounded in the closed interval. |x − x 0 | a .
3. In |x − x 0 | < a ; just take b in α = b/k large, namely, b = αk .
5. R has sides 2a and 2b and center (1, 1) since y (1) = 1. = 1. In R , f = 2y 2 2(b + 1)2 = K , α = b/K = b /(2(b + 1)2 2), dα/db = 0 gives b = 1, and . Solution by dy/y 2 2 = 2 dx , etc., y = 1/(3 − 2x ).
7. |1 + y 2 | K = 1 + b 2 , α = b/K , d α/db = 0, b = 1, .
9. No. At a common point (x 1 ,y 1 ) they would both satisfy the “initial condition” y (x 1 ) = y 1 , violating uniqueness.
Chapter 1 Review Questions and Problems, page 43
11. y = ce y −2x
13. y = 1/(ce y −4x + 4)
15. y = ce −x + 0.01 cos 10x + 0.1 sin 10x
17. y = ce −2.5x + 0.640x − 0.256
19. 25y 2 − 4x 2 = c
21. F = x,x3 ey + x2 y = c
23.
25.
27. ek = 1.25, (In 2)/In 1.25 = 3.1, (In 3)/In 1.25 = 4.9 [days]
29. ek = 0.9, 6.6 days. 43.7 days from e kt = 0.5,e kt = 0.01
Problem Set 2.1, page 53
1. F(x,z,z′ ) = 0
3. y = c1 e2x + c 2
5. y = (c1 x + c2 )−1/2
7.
9. y 2 = x 3 In x
11.
13.
15. y = 3 cos 2.5x − sin 2.5x
17. y = −0.75x 3/2 − 2.25x −1/2
19. y = 15e −x − sin x
Problem Set 2.2, page 59
1.
3.
5.
7.
9.
11.
13.
15.
17.
19. y ″ + 4y ′ + 5y = 0
21. y = 4.6 cos 5x − 0.24 sin 5x
23. y = 6e 2x + 4e −3x
25. y = 2e−x
27. y = (4.5 − x )e −πx
29.
31. Independent
33. In x = 0 with x = 1 gives c 1 = 0; then c 2 = 0 for x = 2, say.
35. Dependent since sin 2x = 2 sin x cos x
37. y 1 = e −x , = y 2 = 0.001e x + e −x
Problem Set 2.3, page 61
1. 4e 2x , −e −x + 8e −2x , −cos x − 2 sin x
3. 0, 0, (D − 2I )(−4e −2x = 8e −2x + 8e −2x
5.
7.
9.
11.
15. Combine the two conditions to get L (cy + kw ) = L (cy ) + L (kw ) = cLy + kLw . The converse is simple.
Problem Set 2.4, page 69
1. . y ′ = y 0 cos ω0 t + (v 0 /ω0 ) sin ω0 t . At integer t (if ω0 = π), because of periodicity.
3. (i) Lower by a factor (ii) higher by
5.
7. mL θ″ = −mg sin θ ≈ −mg θ (tangential component of W = mg ), .
9. , where m = 1 kg, ay = π · 0.012 · 2y meter3 is the volume of the water that causes the restoring force. aγy with γ = 9800 nt (= weight/meter3 ). . Frequency .
13.
15.
17. The positive solutions of tan t = 1, that is, π/4 (max), 5π/4 (min). etc
19.
Problem Set 2.5, page 73
3. y = (c 1 + c 2 In x )x −1.8
5.
7.
9. y = (c 1 + c 2 In x )x 0.6
11.
13. y = x −3/2
15. y = (3.6 + 4.0 In x )/x
17. y = cos (In x ) + sin (In x )
19. y = −0.525x 5 + 0.625x −3
Problem Set 2.6, page 79
3. W = −2.2e −3x
5. W = −x 4
7. W = a
9. y ″ + 25y = 0, W = 5, y = 3 cos 5x − sin 5x
11. y ″ + 5y + 6.34 = 0, W = 0.3e −5x , 3e −2.5 cos 0.3x
13. y ″ + 2y ′ = 0, W = −2e −2x , y = 0.5(1 + e −2x
15. y ″ − 3.24y = 0, W = 1.8, y = 14.2 cosh 1.8x + 9.1 sinh 1.8x
Problem Set 2.7, page 84
1.
3.
5.
7.
9.
11.
13.
17. y = e −0.1x (1.5 cos 0.5x − sin 0.5x ) + 2e 0.5x
Problem Set 2.8, page 91
3. y p = 1.0625 cos 2t + 3.1875 sin 2t
5. y p = −1.28 cos 4.5t + 0.36 sin 4.5t
7.
9. y = e −1.5t (A cos t + B sin t ) + 0.8 cos t + 0.4 sin t
11.
13. y = A cos t + B sin t − (cos ωt )/(ω2 − 1)
15.
17.
19.
25. CAS Experiment. The choice of needs experimentation, inspection of the curves obtained, and then changes on a trail-and-error basis. It is interesting to see how in the case of beats the period gets increasingly longer and the maximum amplitude gets increasingly larger as ω/(2π)approaches the resonance frequency.
Problem Set 2.9, page 98
1. RI ′ + I/C = 0, I = ce −t /(RC )
3. LI ′ + RI = E, I = (E/R ) + ce Rt /L = 4.8 + ce −40t
5. I = 2 (cos t − cos 20t )/399
7. I 0 is maximum when S = 0; thus, c = 1/(ω2 L ).
9. I = 0
11. I = 5.5 cos 10t + 16.5 sin 10t A
13. I = e −5t (A cos 10t + B sin 10t ) − 400 cos 25t + 200 sin 25t A
15.
17. E (0) = 600, I ′(0) = 600, I = e −3t (−100 cos 4t + 75 sin 4t ) + 100 cos t
19. R = 2Ω, L = 1 H, C = 1/12 F, E = 4.4 sin 10t V
Problem Set 2.10, page 102
1.
3.
5.
7. y = (c 1 + c 2 x )e 2x + x −2 e 2x
9. y = (c 1 + c 2 x )e x + 4x 7/2 e x
11. y = c 1 x 2 + c 2 x 3 + 1/(2x 4 )
11. y = c 1 x 2 + c 2 x 3 + 1/(2x 4 )
13. y = c 1 x −3 + c 2 x 3 + 3x 5
Chapter 2 Review Questions and Problems, page 102
7. y = c 1 e −4.5x + c 2 e −3.5x
9. y = e −3x (A cos 5x + B sin 5x )
11. y = (c 1 + c 2 x )e 0.8x
13. y = c 1 x −4 + c 2 x 3
15. y = c 1 e 2x + c 2 e −x /2 − 3x + x 2
17. y = (c 1 + c 2 x )e 1.5x + 0.25x 2 e 1.5 x
19.
21. y = −4x + 2x 3 + 1/x
23. I = −0.01093 cos 415t + 0.05273 sin 415t A
25.
27. RLC − circuit with R = 20 Ω, L = 4 H, C = 0.1 F, E = −25 cos 4t V
29.
Problem Set 3.1, page 111
9. Linearly independent
11. Linearly independent
13. Linearly independent
15. Linearly dependent
Problem Set 3.2, page 116
1. y = c 1 + c 2 cos 5x + c 3 sin 5x
3. y = c 1 + c 2 x + c 3 cos 2x + c 4 sin 2x
5. y = A 1 cos x + B 1 sin x + A 2 cos 3x + B 2 sin 3x
7. y = 2.398 + e −1.6x (1.002 cos 1.5x − 1.998 sin 1.5x )
9. y = 4e −x + 5e −x /2 cos 3x
11. y = cosh 5x − cos 4x
13. y = e 0.25x + 4.3e −0.7x + 12.1 cos 0.1x − 0.6 sin 0.1x
Problem Set 3.3, page 122
1.
3. y = c1 cos x + c2 sin x + c3 cos 3x + c4 sin 3x + 0.1 sinh 2x
5.
7.
9.
11. y = e −3x (−1.4 cos x − sin x )
13. y = 2 − 2 sin x + cos x
Chapter 3 Review Questions and Problems, page 122
7. y = c1 + e −2x (A cos 3x + B sin 3x )
9. y = c1 cosh 2x + c2 sinh 2x + c3 cos 2x + c4 sin 2x + cosh x
11. y = (c 1 + c 2 x + c 3 x 2 )e −1.5x
13. y = (c 1 + c 2 x + c 3 x 2 )e −2x + x 2 − 3x + 3
15.
17. y = 2e −2x cos 4x + 0.05 x − 0.06
19. y = 4e −4x + 5e 5x
Problem Set 4.1, page 136
1. Yes
5. y ′1 = 0.02(−y1 + y2 ), y ′2 = 0.02(y 1 − 2y 2 + y 3), y ′3 = 0.02(y 2 − y 3 )
7. c 1 = 1, c 2 = −5
9. c 1 = 10, c 2 = 5
11.
13. y ′1 = y 2 , y ′2 = 24y 1 − 2y 2 , y 1 = c 1 e 4t + c 2 e −6t = y , y 2 = y ′
15. (a) For example, C = 1000 gives −2.39993, −. 0.000167. (b) −2.4, 0. gives the critical case. C about 0.18506.
Problem Set 4.3, page 147
1. y 1 = c 1 e −2t + c 2 e −2t , y 2 = −3c 1 e 2t − c 2
3. y 1 = 2c 1 e 2t + 2c 2 , y 2 = c 1 e 2t − c 2
5.
7.
9.
11.
13. y 1 = 2 sinh t , y 2 = 2 cosh t
15.
17.
19. I 1 = c 1 e −t + 3c 2 e −3t , I 2 = −3c 1 e −t − c 2 e −3t
Problem Set 4.4, page 151
1. Unstable improper node, y 1 = c 1 e t , y 2 = c 2 e 2t
3. Center, always stable, y 1 = A cos 3t + B sin 3t , y 2 = 3B cos 3t − 3A sin 3t
5. Stable spiral, y 1 = e −2t (A cos 2t + B sin 2t ), y 2 = e −2t (B cos 2t − A sin 2t )
7. Saddle point, always unstable, y 1 = c 1 e −t + c 2 e 3t , y 2 = c 1 e −t + c 2 e 3t
9. Unstable node, y 1 = c 1 e 6t + c 2 e 2t , y 2 = 2c 1 e 6t − 2c 2 e 2t
11. y = e −t (A cos t + B sin t ). Stable and attractive spirals
15. p = 0.2 ≠ 0 (was 0) δ < 0, spiral point, unstable.
17. For instance,
Problem Set 4.5, page 159
5. Center at (0, 0). At (2, 0) set . Then . Saddle point at (2, 0).
7. (0, 0), y ′1 = −y 1 + y 2 , y ′2 = −y 1 m y 2 , stable and attractive spiral point; (−2, 2), saddle point
9. (0, 0) saddle point, (−3, 0) and (3, 0) centers.
11. saddle points; centers. Use −cos
13.
15. By multiplication, By integration,
Problem Set 4.6, page 163
3. y 1 = c 1 e −t + c 2 e t , y 2 = −c 1 e −t + c 2 e t − e 3t
5. y 1 = c 1 e 5t + c 2 e 2t − 0.43t − 0.24, y 2 = c 1 e 2t − 2c 2 e 2t + 1.12t + 0.53.
7. y 1 = c 1 e t + 4c 2 e 2t − 3t − 4 − 2e −t − 5c 2 e 2t + 5t + 7.5 + e −t
9. The formula for v shows that these various choices differ by multiples of the eigenvector for λ = −2, which can be absorbed into, or taken out of c 1 , in the general solution. y (h)
11.
13. y 1 = cos 2t + sin 2t + 4 cos t , y 2 = 2 cos 2t − 2 sin 2t + sin t
15. y 1 = 3e −t − 4e t + e 2t , y 2 = −4e −t + t
17.
19. c 1 = 17.948, c 2 = −67.948
Chapter 4 Review Questions and Problems, page 164
11. y 1 = c 1 e 4t + c 2 e −4t , y 2 = 2c 1 e 4t − 2c 2 e −4t . Saddle point.
13. asymptotically stable spiral point
15. y 1 = c 1 e −5t + c 2 e −t , y 2 = c 1 e −5t − c 2 e −t . Stable node
17. y 1 = e −t (A cos 2t + B sin 2t ), y 2 = e −t (B cos 2t − A sin 2t ). Stable and attractive spiral point
19. Unstable spiral point
21. y 1 = c 1 e −4t + c 2 e 4t − 1 − 8t 2 , y 2 = −c 1 e −4t + c 2 e 4t − 4t
23. y 1 = 2c 1 e −t + 2c 2 e 3t + cos t − sin t , y 2 = −c 1 e −t + c 2 e 3t
25. I 1′ + 2.5(I 1 − I 2 ) = 169 sin t , 2.5(I 2 ′ − I 2 ′ − I 1 ′) + 25I 2 = 0, I 1 = (19 + 32.5t )e −5t − 19 cos t + 62.5 sin t , I 2 = (−6 − 32.5t )e −5t + 6 cos t + 2.5 sin t
27. (0, 0) Saddle point; (−1, 0), (1, 0) centers
29. (n π, 0) center when n is even and saddle point when n is odd
Problem Set 5.1, page 174
3.
5.
7.
9.
11.
13.
15.
17.
19. but x = 2 is too large to give good values. Exact: y = (x − 2)2 e x
Problem Set 5.2, page 179
5.
11. Set x = az . y = c 1 P n (x/a ) + c 2 Q n (x/a )
15.
Problem Set 5.3, page 186
3.
5. b 0 = 1, c 0 = 0, r 2 = 0, y 1 = e −x , y 2 = e −x In x
7.
9.
11. y 1 = e x , y 2 = e x /x
13. y 1 = e x , y 2 = e x In x
15.
17.
19.
Problem Set 5.4, page 195
Problem Set 5.5, page 200
1. c 1 J 4 (x ) + c 2 Y 4 (x )
3. c 1 J 2/3 (x 2 ) + c 2 Y 2/3 (x 2 )
5.
7.
9. x 3 (c 1 J 3 (x ) + c 2 Y 3 (x ))
11. Set H (1) = kH (2) and use (10).
13. Use (20) in Sec. 5.4 .
Chapter 5 Review Questions and Problems, page 200
11. cos 2x , sin 2x
13. (x − 1)−5 , (x − 1)7 ; Euler–Cauchy with x − 1 instead of x
15. J /s (x ), J −/5 (x )
17. e x , 1 + x
19.
Problem Set 6.1, page 210
1. 3/s 2 + 12/s
3. s /(s 2 + π2 )
5. 1/((s − 2)2 − 1)
7. (ω cos θ + s sin θ)/(s 2 + ω2 )
9.
11.
13.
15.
19. Use e at = cosh at + sinh at .
23. Set ct = p . Then
25. 0.2 cos 1.8t + sin 1.8t
27.
29. 2t 3 − 1.9t 5
31.
33.
35.
37. πte −πt
39.
41. e −5πt sinh π t
43.
45. (k o + k 1 t )e −at
Problem Set 6.2, page 216
1. y = 1.25e −5.2t − 1.25 cos 2t + 3.25 sin 2t
3. (s − 3)(s + 2) = 11s + 28 − 11 = 11s + 17, y = 10/(s − 3) + 1/(s + 2), y = 10e 3t + e −2t
5.
7.
11. (s + 1.5)2 Y = s + 31.5 + 3 + 54/s 4 + 64/s , Y = 1/(s + 1.5) + 1/(s + 1.5)2 + 24/s 4 + 32/s 2 , y = (1 + t )e −1.5t + 4t 3 − 16t 2 + 32t
13.
15.
17.
19.
21. Answer: (s 2 − 2)/(s 3 − 4a )
23. 12(1 − e −t /4
25. (1 − cos ωt /ω2
27.
29.
Problem Set 6.3, page 223
3.
5.
7.
9.
11. (se −πs /2 + e −πs )/(s 2 + 1)
13. 2[1 + u (t − π)]sin 3t
15. (t − 3)3 u (t − 3)/6
17. e −t cos t (0 < t < 2π)
19.
21.
23.
25. t − sin t (0 < < 1), cos (t − 1) + sin (t − 1) − sin t (t > 1)
27.
29. 0.1i ′ + 25i = 490e −5t [1 − u (t − 1)], i = 20(e −5t − e −250t ) + 20u (t − 1)[−e −5t + e −250t +245 ]
31.
33.
35. i = (10 sin 10t + 100 sin t )(u (t − π) − u (t − 3π))
37.
39.
Problem Set 6.4, page 230
3.
5. sin t (0 < t < π);0(π < t < 2π);−sin t (t > 2)
7.
9. y = 0.1[e t + e −2t (−cos t + 7 sin t )] + 0.1u (t − 10)[−e −t + e 2t +30 (cos (t − 10) − 7 sin (t − 10))]
11.
15. ke −ps /(s − se −ps ) (s > 0)
Problem Set 6.5, page 237
1. t
3. (e t −e −t )/2 = sinh t
5.
7. e t − t − 1
9. y − 1 * y = 1, y = e t
11. y = cos t
13.
17. e 4t − e −1.5t
19. t sin πt
21. (ωt − sin ωt )/ω2
23. 4.5(cosh 3t − 1)
25. 1.5t sin 6t
Problem Set 6.6, page 241
3.
5.
7.
9.
11.
15.
17. In s − In (s − 1);(−1 + e t )/t
19. [In (s 2 + 1) − 2 In (s − 1)]′ = 2s /(s 2 + 1) − 2/(s − 1);2(−cos t + e t )/t
Problem Set 6.7, page 246
3. y 1 = −e −5t + 4e 2t , y 2 = e −5t + 3e 2t
5. y 1 = −cos t + sin t + 1 + u (t − 1)[−1 + cos (t − 1)− sin (t − 1)] y 2 = cos t + sin t − 1 + u (t − 1)[1 − cos (t − 1) − sin (t − 1)]
7.
9. y 1 = (3 + 4t )e 3t , y 2 = (1 − 4t )e 3t
11. y 1 = e t + e 2t , y 2 = e 2t
13. y 1 = −4e t + sin 10t + 4 cos t , y 2 = 4e t − sin 10t + 4 cos t
15. y 1 = e t ,y 2 = e −t ,y 3 = e t − e −t
19.
Chapter 6 Review Questions and Problems, page 251
11.
13.
15.
17. Sec. 6.6 ; 2s 2 /(s 2 + 1)2
19. 12/(s 2 (s + 3))
21. tu(t − 1)
23. sin (ωt + θ)
25. 3t 2 + t 3
27. e −t (3 cos t − 2 sin t )
29. y = e −2t (13 cos t + 11 sin t ) + 10t − 8
31. e −t + u (t − π)[1.2 cos t − 3.6 sin t + 2e −t +π − 0.8e 2t −2π ]
33. 0 ≠ (0 t 2), 1 − 2e −(t −2 ) (t > 2)
35. y 1 = 4e t − e −2t ,y 2 = e t − e −2t
37.
39.
41. 1 − e −t (0 < t < 4), (e 4 − 1)e −t (t > 4)
43.
45.
Problem Set 7.1, page 261
3. 3 × 3, 3 × 4, 3 × 6, 2 × 2, 2 × 3, 3 × 2
5.
7. No, no, yes, no, no
9.
11.
13.
15.
Problem Set 7.2, page 270
5. 10, n (n + 1)/2
7.
11.
13.
15.
17.
19.
25. (d ) AB = (AB )T = B T A T = BA ; etc. (e ) Answer. If AB = −BA .
29. P = [85 62 30]T , ν = [44,920 30,940]T
Problem Set 7.3, page 280
1. x = −2, y = 0.5
3. x = 1, y = 3, z = −5
5. x = 6, y = −7
7. x = −3t , y = t arb., z = 2t
9. x = 3t − 1, y = −t + 4, z = t arb.
11. w = 1, x = t 1 arb., y = 2t 2 − t 1 , z = t 2 arb.
13. w = 3, x = 0, y = 2, z = 6
17. I 1 = 2, I 2 = 6, I 3 = 8
19. I 1 = (R 1 + R 2 )E 0 /(R 1 R 2 )A, I 2 = E 0 /R 1 A, I 3 = E 0 /R 2 A
21 x 2 = 1600 − x 1 , x 3 = 600 + x 1 , x 4 = 1000 − − x 1 . No
23. C:3x 1 − x 3 = 0, H:8x 1 − 2x 4 = 0, O:22 − 2x 3 − x 4 = 0, thus C3 H8 + 5O2 → 3CO2 + 4H2 O
Problem Set 7.4, page 287
1. 1; [2 −1 3]; [2 −1]T
3. 3; {[3 5 0], [0 3 5], [0 0 1]}
5. 3; {[2 −1 4], [0 1 −46], [0 0 1]}; {[2 0 1], [0 3 23], [0 0 1]}
7. 2; [8 0 4 0], [0 2 0 4]; [8 0 4], [0 2 0]
9. 3; [9 0 1 0], [0 9 8 9], [0 0 1 0]
11. (c ) 1
17. No
19. Yes
21. No
23. Yes
25. Yes
27. 2, [−2 0 1], [0 2 1]
29. No
31. No
33.
35.
Problem Set 7.7, page 300
7. cos (α + β)
9. 1
11. 40
13. 289
15. −64
17. 2
19. 2
21. x = 3.5, y = −1.0
23. x = 0, y = 4, z = −1
25. w = 3, x = 0, y = 2, z = −2
Problem Set 7.8, page 308
1.
3.
5.
7. A −1 = A
9.
11.
15. AA −1 = I , (AA −1 )−1 = (A −1 )−1 A −1 = I . Multiply by A from the right.
Problem Set 7.9, page 318
1. [1 0]T , [0 1]T ; [1 0]T , [0 − 1]T ; [0 −1]T ; [1 1]T , [−1 1]T
3. 1, [1 11 −7]T
5. No
7. Dimension 2, basis x e −x , e −x
9.
11. x 1 = 5y 1 − y 2 , x 2 = 3y 1 − y 2
13. x 1 = 2y 1 − 3y 2 , x 2 = −10y 1 + 16y 2 + y 3 , x 3 = −7y 1 + 11y 2 + y 3
15.
17.
19. 1
21. k = −20
23.
25. a = [5 3 2]T , b = [3 2 −1]T , 90 + 14 = 2(38 + 14)
Chapter 7 Review Questions and Problems, page 318
11.
13. [21 −8 −31]T , [21 −8 31]
15. 197, 0
17. −5, det A 2 = (det A )2 = 25, 0
19.
21. x = 4, y = −2, z = 8
23. x = 6, y = 2t + 2, z = t arb.
25. x = 0.4, y = −1.3, z = 1.7
27. x = 10, y = −2
29. Ranks 2, 2, ∞
31. Ranks 2, 2, 1
33. I 1 = 16.5 A, I 2 = 11 A, I 3 = 5.5 A
35. I 1 = 4 A, I 2 = 5 A, I 3 = 1 A
Problem Set 8.1, page 329
1. 3, [1 0]T ; −0.6, [0 1]T
3. −4, [2 9]T ; 3, [1 1]T
5.
7. λ2 = 0, [1 0]T
9. 0.8 + 0.6i , [1 −i ]T ; 0.8 − 0.6i , [1 i ]T
11. −(λ3 − 18λ2 + 99λ − 162)/(λ − 3) = −(λ2 − 15λ + 54); 3, [2 −2 1]T ; 6, [1 2 2]T ; 9, [2 1 −2]T
13. −(λ − 9)3 ; 9, [2 −2 1]T , defect 2
15. (λ + 1)2 (λ2 + 2λ − 15); −1, [1 0 0 0]T , [0 1 0 0]T ; −5, [−3 −3 1 1]T , 3, [3 −3 1 −1]T
17. Eigenvalues i , −i . Corresponding eigenvectors are complex, indicating that no direction is preserved under a rotation.
19. A point onto the x 2 -axis goes onto itself, a point on the x 1 -axis onto the origin.
23. Use that real entries imply real coefficients of the characteristic polynomial.
Problem Set 8.2, page 333
1. 1.5, [1 −1]T , −45°; 4.5, [1 1]T , 45°
3.
5. 0.5, [1 −1]T ; 1.5, [1 1]T ; directions −45° and 45°
7. [5 8]T
9. [11 12 16]T
11. 1.8
13. c [10 18 16]T
15. X = (I − A )−1 y = [0.6747 0.7128 0.7543]T
17. AX j = λj X j (X j ≠ 0), (A − k I )X j = λj X j − k X j = (λj − k )X j .
19. From AX j = λ j X j (X j ≠ 0) and Prob. 18 follows and , integer). Adding on both sides, we see that has the eigenvalue . From this the statement follows.
Problem Set 8.3, page 338
1. 0.8 ± 0.6i , [1 ±i ]T ; orthogonal
3. 2 ± 0.8i , [1 ±i ]. Not skew-symmetric!
5. 1, [0 2 1]T ; 6, [1 0 0]T , [0 1 −2]T ; symmetric
7. 0, ±25i , skew-symmetric
9. 1, [0 1 0]T; i [1 0 i ]T ; −i , [1 0 −2]T , orthogonal
15. No
17. A −1 = (−AT )−1 = −(A −1 )T
19. No since det A = det (A T ) = det (−A ) = (−1)3 det (A ) = −det (A ) = 0.
Problem Set 8.4, page 345
1.
3.
5.
9.
11.
13.
15.
17.
19.
21.
23.
Problem Set 8.5, page 351
1. Hermitian, 5, [−i 1]T , 7, [i 1]T
3. Unitary,
5. Skew-Hermitian, unitary, −i , [0 −1 1]T , i , [1 0 0]T , [0 1 1]T
7. Eigenvalues −1, 1; eigenvectors [1 −1]T , [1 1]T ; [1 −i ]T , [1 i ]T ; [0 1]T , [1 0]T , resp.
9. Hermitian, 16
11. Skew-Hermitian, −6i
13.
15. (H Hermitian, S skew-Hermitian)
19. A T − T A = (H + S )(H − S ) − (H − S )(H + S ) = 2(− HS + SH ) = 0 if and only if HS = SH .
Chapter 8 Review Questions and Problems, page 352
11. 3, [1 1]T ; 2, [1 −1]T
13. 3, [1 5]T ; 7, [1 1]T
15. 0, [2 −2 1]T ; 9i , [−1 + 3i 1 + 3i 4]T ; −9i , [−1 − 3i 1 − 3i 4]T
17.
19.
21.
23.
25.
Problem Set 9.1, page 360
1. 5, 1, 0;
3. 8.5, −4.0, 1.7;
5. 2, 1, −2; , Position vector of Q
7.
9. Q : (0, 0, −8), |ν| = 8
11. [6, 4, 0], , [−3, −2, 0]
13. [1, 5, 8]
15. 7[9, −7, 8] = [63, −49, 56]
17. [12, 8, 0]
21. [4, 9, −3],
23. [0, 0, 5], 5
25.
27. P = [0, 0, −5]
29. V = [ν1 , ν2 , 3], ν1 , ν2 arbitrary
31. k = 10
33. |p + q + u | 18. Nothing
35.
37. u + v + p = [−k , 0] + [l , l ] + [0, −1000] = 0 , −k + l + 0 = 0, 0 + l − 1000 = 0, l = 1000, k = 1000
Problem Set 9.2, page 367
1. 44, 44, 0
3.
5.
7.
9. 300; CF. (5a) and (5b)
13. Use (1) and |cos γ| 1.
15. |a + b |2 + |a − b |2 = a · a + 2a · b + b · b + (a · a − 2a · b + b · b ) = 2|a |2 + 2|b |2
17. [2, 5, 0] · [2, 2, 2] = 14
19. [0, 4, 3] · [−3, −2, 1] = −5 is negative! Why?
21. Yes, because W = (p + q ) · d = p · d + q · d .
23. arccos 0.5976 = 53.3°
27. β − α is the angle between the unit vectors a and b . Use (2).
29.
31.
33.
35. (a + b ) · (a − b ) = |a |2 − |b |2 = 0, |a | = |b |. A square.
37. 0. Why?
39. If |a | = |b | or if a and b are orthogonal.
Problem Set 9.3, page 374
5. −m instead of m , tendency to rotate in the opposite sense.
7. |v | = |[0, 20, 0] × [8, 6, 0]| = |[0, 0, −160]| = 160
9. Zero volume in Fig. 191 , which can happen in several ways.
11. [0, 0, 7], [0, 0, −7], −4
13. [6, 2, 7], [−6, −2, −7]
15. 0
17. [−32, −58, 34], [−42, −63, 19]
19. 1, −1
21.
23. 0, 0, 13
25. m = [−2, −2, 0] × [2, 3, 0] = [0, 0, −10], m = 10 clockwise
27. [6, 2, 0] × [1, 2, 0] = [0, 0, 10]
29.
31. 3x + 2y − z = 5
33. 474/6 = 79
Problem Set 9.4, page 380
1. Hyperbolas
3. Parallel staraight lines (planes in space)
5. Circles, centers on the y -axis
7. Ellipses
9. Parallel planes
11. Elliptic cylinders
13. Paraboloids
Problem Set 9.5, page 390
1. Circle, center (3, 0), radius 2
3. Cubic parabola x = 0, z = y 3
5. Ellipse
7. Helix
9. A “Lissajous curve”
11.
13. r = [2 + t , 1 + 2t , 3]
15. r = [t , 4t − 1, 5t ]
17.
19.
21. Use sin (− α) = −sin α.
25. u = [−sin t , 0, cos t ]. At p , r ′ = [−8, 0, 6]. q (w ) = [6 − 8w , i , 8 + 6w ].
27.
29.
31.
33. Start from r (t ) = [t , f (t )].
35.
37. v (0) = (ω + 1) R i , a (0) = −ω2 R j
39.
41.
43. 1 year = 365 · 86,400 sec, R = 30 · 365 · 86,400/2π = 151 · 1061 [km], |a | = ω2 R = |v |2/R = 5.98 · 10−6 [km/sec2 ]
45.
49. r (t ) = [t , y (t ), 0], r ′ = [1, y ′, 0] r · r ′ = 1 p y ′2 , etc.
51.
53. 3/(1 + 9t 2 + 9t 4 )
Problem Set 9.7, page 402
1. [2y − 1, 2x + 2]
3. [−y > x 2 , 1> x ]
5. [4x 3 , 4y 3 ]
7. Use the chain rule.
9. Apply the quotient rule to each component and collect terms.
11. [y , x ], [5, −4]
13. [2x /(x 2 + y 2 ), 2y /(x 2 + y 2 )], [0.16, 0.12]
15. [8x , 18y , 2z ], [40, −18, −22]
17. For P on the x - and y -axes.
19. [−1.25, 0]
21. [0, −e ]
23. Points with y = 0, ±π, ±2π, ….
25. −ΔT (p ) = [0, 4, −1]
31. Δf = [32x , −2y ], δf (p ) = [160, −2]
33. [12x , 4y , 2z ], [60, 20, 10]
35. [−2x , −2y , 1], [−6, −8, 1]
37.
39.
41.
43. f = xyz
45.
Problem Set 9.8, page 405
1. 2x + 8y + 18z ; 7
3. 0, after simplification; solenoidal
5. 9x 2 y 2 z 2 ; 1296
7. −2e x (cos y )z
9. (b) (fν 1 )x + (fν 2 )y + (fν 3 )z = f [(ν 1 )x + (ν 2)y + (ν 3)z ] + f x ν 1 + f y ν 2 + f z ν 3 , etc.
11. [ν1 , ν2 , ν3 ] = r ′ = [x ′, y ′, z ′] = [y , 0, 0], z ′ = 0, z = c 3 , y ′ = 0, y = c 2 , and x ′ = y = c 2 , x = c 2 t + c 1 . Hence as t increases from 0 to 1, this “shear flow” transforms the cube into a parallelepiped of volume l.
13. div (w × r ) = 0 because ν1 , ν2 ν3 do not depend on x , y , z , respectively.
15. −2 cos 2x + 2 cos 2y
17. 0
19. 2/(x 2 + y 2 + z 2 )2
Problem Set 9.9, page 408
3. Use the definitions and direct calculation.
5. [x (z 2 − y 2 ), y (x 2 − z 2 ), z (y 2 − x 2 )]
7. e −x [cos y , sin y , 0]
9.
11.
13. curl v = 0, irrotational, div v = 1, compressible, r = [c 1 e t , c 2 e t , c 3 e −t ]. Sketch it.
15. [−1, −1, −1], same (why?)
17. −yz −zx −xy , 0 (why?), −y − z − x
19. [−2z −y , −2x −z , −2y −x ], same (why?)
Chapter 9 Review Questions and Problems, page 409
11. −10, 1080, 1080, 65
13. [−10, −30, 0], [10, 30, 0], 0 , 40
15. [−1260, −1830, −300], [−210, 120, −540], undefined
17. −125, 125, −125
19.
21. [−2, −6, −13]
23.
25. [5, 2, 0] · [4 − 1, 3 − 1, 0] = 19
27.
29. [0, 0, −14], tendency of clockwise rotation
31. 4
33. 1, −2y
35. 0, same (why?), 2(y 2 + x 2 − zx )
37. [0, −2, 0]
39.
Problem Set 10.1, page 418
3. 4
5. r = [2 cos t , 2 sin t ],
7. “Exponential helix,” (e 6π − 1)/3
9. 23.5, 0
11.
15.
17. [4 cos t , + sin t , sin t , 4 cos t ], [2, 2, 0]
19. 144t 4 , 1843.2
Problem Set 10.2, page 425
3.
5. e xy sin z , e − 0
7. cosh 1 − 2 = −0.457
9. e x cosh y + e z sinh y , e − (cosh 1 + sinh 1) = 0
13.
15. Dependent, x 2 ≠ −4y 2 , etc.
17. Dependent, 4 ≠ 0, etc.
19. sin (a 2 + 2b 2 + c 2 )
Problem Set 10.3, page 432
3. 8y 3 /3, 54
5.
7.
9. 36 + 27y 2 , 144
11. z = 1 − r 2 , dx dy = r dr d θ, Answer: π/2
13.
15.
17. T x = bh 3 /12, I y = b 3 h /4
19. I x = (a + b )h 3 /24, I y = h (a4 − b 4 )/(48(a − b ))
Problem Set 10.4, page 438
1. (−1 −1) · π/4 = −π/2
3.
5.
7. 0. Why?
9.
13.
15. Δ2 w = 6xy , 3x (10 − x 2 )2 − 3x , 486
17. Δ2 w = 6x − 6y , − 38.4
19.
Problem Set 10.5, page 442
1. Straight lines, k
3. , circles, straight lines, [−cu cos ν, −cu sin ν u ]
5. z = x 2 + y 2 , circles, parabolas, [−2u 2 cos ν, −2u 2 sin ν, u ]
7. x 2 /a 2 + y 2 /b 2 + z 2 /c 2 = 1, [bc cos2 ν cos u , ac cos2 ν sin u , ab sin ν cos ν], ellipses
11.
13. Set x = and y = ν.
15. [2 + 5 cos u , −1 + 5 sin u , ν], [5 cos u , 5 sin u , 0]
17. [a cos ν cos u , −2.8 + a cos ν sin u , 3.2 + a sin ν], a = 1.5; [a 2 cos2 ν cos u , a 2 cos2 ν sin u , a 2 cos ν sin ν]
19. [cosh u , sinh u , ν], [cosh u , −sinh u , 0]
Problem Set 10.6, page 450
1. F (r ) • N = [−u 2 , ν2 , 0] · [−3, 2, 1] = 3u 2 + 2ν2 , 29.5
3. F (r ) • N = cos3 ν cos u sin u from (3), Sec. 10.5 . Answer :
5. F (r ) • N = −u 3 , −128π
7.
9.
13.
15. G (r ) = (1 + 9u 4 )3/2 , |N | = (1 + 9u 4 )1/2 , Answer : 54.4
21.
23. [u cos ν, u sin ν, u ],
25. [cos u cos ν, cos u sin ν, sin u ], dA = (cos u ) du dν , B the z -axis, I B = 8ν/3, I K = I B + 12 · 4π = 20.9.
Problem Set 10.7, page 457
1. 224
3. −e −1−x + e −y −x , −2e −1−x + e −x , 2e −3 − e −2 − 2e −1 + 1
5.
7. [r cos u cos ν, cos u sin ν, r sin u ], dV = r 2 cos u dr du dv , σ = ν, 2π2 a 3 /3
9. div F = 2x + 2z , 48
11. 12(e − 1/e ) = 24 sinh 1
13. div F = −sin z , 0
15.
17. h 4 π/2
19. 8abc (b 2 + c 2 )/3
21. (a 4 /4) · 2π · h = ha 4 π/2
23. h 5 π/10
25. Do Prob. 20 as the last one.
Problem Set 10.8, page 462
1. x = 0, y = 0, z = 0, no contributions. x = a : ∂f /∂n = ∂f /∂x = −2x = −2a , etc. Integrals. x = a : (−2a )bc . y = b : (−2b )ac , z = c : (4c ) ab . Sum 0
3. The volume integral of . The surface integral of f ∂g /∂n = f · 2x = 2f = 8y 2 over x = 1 is . Others 0.
5. The volume integral of 6y 2 · 4 − 2x 2 · 12 is 0; 8(x = 1), −8(y = 1), others 0.
7. F = [x , 0, 0], div F = 1, use (2*), Sec. 10.7 , etc.
9.
10.
Problem Set 10.9, page 468
1. S : z = y (0 x 1, 0 y 4), [0, 2z , −2z ] · [0, −1, 1], ±20
3.
5.
7. [−e x , −e x , −e y ] · [−2x , 0, 1], ±(e 4 − 2e + 1)
9. The sides contribute a , 3a 2 /2, −a , 0.
11. −2π; curl F = 0
13. 5k , 80π
15.
17. r = [cos u , sin u , ν], [−3ν2 , 0, 0] · [cos u , sin u , 0], −1
19. r = [u cos ν, u sin ν, u ], 0 u 1, 0 ν π/2, [−e x , 1, 0] · [−u cos ν −u sin ν, u ]. Answer : 1/2
Chapter 10 Review Questions and Problems, page 469
11. r = [4 − 10t , 2 + 8t ], F (r ) · d r = [2(4 − 10t )2 , −4(2t + 8t 2 )] · [−10, 8] dt ; −4528/3. Or using exactness.
13. Not exact, curl F = (5 cos x )k , ±10
15. 0 since curl F = 0
17. By Stokes, ±18π
19. F = grad (y 2 + xz ), 2π
21.
23.
25.
29. div F = 20 + 6z 2 . Answer: 21
31. 24 sinh 1 = 28.205
33. Direct integration,
35. 72π
Problem Set 11.1, page 482
1.
5. There is no smallest p > 0.
13.
15.
17.
19.
21.
Problem Set 11.2, page 490
1. Neither, even, odd, odd, neither
3. Even
5. Even
9.
11.
13.
15.
17.
19.
23.
25.
27.
29.
Problem Set 11.3, page 494
3. The output becomes a pure cosine series.
5. For A n this is similar to Fig. 54 in Sec. 2.8 , whereas for the phase shift B n the sense is the same for all n .
7. y = C 1 cos ωt + C 2 sin ωt + a (ω) sint , a (ω) = 1/(ω2 − 1) = −1.33, −5.26, 4.76, 0.8, 0.01. Note the change of sign.
11.
13.
15.
17.
19.
3. Set x = ct + k .
5. x = cos θ , dx = −sin θ dθ , etc.
7. λm = (m π/10)2 , m = 1, 2, …; y m = sin (m πx /10)
9. λ = [(2m + 1)π/(2L )]2 , m = 0, 1, …, y m = sin ((2m + 1)πx /(2L ))
11. λm = m 2 , m = 1, 2, …, y m = x sin (m In |x |)
13. p = e 8x , q = 0, r = e 8x , λm = m 2 , y m = e −4x sin mx, m = 1, 2, …
1. 8(p 1 (x ) − p 3 (x ) + p 5 (x ))
3.
9. −0.4775p 1 (x ) − 0.6908p 3 (x ) + 1.844p 5 (x ) − 0.8236p 7 (x ) + 0.1658p 9 (x ) + …, m 0 = 9. Rounding seems to have considerable influence in Probs. 8–13.
11. 0.7854P 0 (x ) − 0.3540P 2 (x ) + 0.0830P 4 (x ) − …, m 0 = 4
13. 0.1212P 0 (x ) − 0.7955P 2 (x ) + 0.9600P 4 (x ) − 0.3360P 6 (x ) + …, m 0 = 8
15.
1. (see Example 3 ), etc.
3.
5.
7.
9.
11.
15. For n = 1, 2, 11, 12, 31, 32, 49, 50 the value of Si (nπ ) − p/2 equals 0.28, −0.15, 0.029, − 0.026, 0.0103, −0.0099, 0.0065, −0.0064 (rounded).
17.
19.
1.
3.
5.
7. Yes. No
9.
11.
13.
Problem Set 11.9, page 533
3.
5.
7.
9.
11.
13. by formula 9
17. No, the assumptions in Theorem 3 are not satisfied.
19. [f 1 + f 2 + f 3 + f 4 , f 1 − if 2 − f 3 + if 4 , f 1 − f 2 + f 3 − f 4 , f 1 + if 2 − f 3 − if 4 ]
21.
Chapter 11 Review Questions and Problems, page 537
11.
13.
15. cosh x , sinh x (−5 < x < 5), respectively
17. Cf. Sec. 11.1 .
19.
21.
23. 0.82, 0.50, 0.36, 0.28, 0.23
25. 0.0076, 0.0076, 0.0012, 0.0012, 0.0004
27.
29.
Problem Set 12.1, page 542
1. L (c 1 u 1 + c 2 u 2 ) = c 1 L (u 1 ) + c 2 L (u 2 ) = c 1 · 0 + c 2 · 0 = 0
3. c = 2
5. c = a/b
7. Any c and ω
9. c = π/25
15. u = 110 − (110/In 100) In (x 2 + y 2 )
17. u = a (y ) cos 4πx + b (y ) sin 4πx
19.
21. u = e −3y (a (x ) cos 2y + b (x ) sin 2y ) + 0.1e 3y
23. u = c 1 (y )x + c 2 (y )/x 2 (Euler–Cauchy)
25. u (x ,y ) = axy + bx + cy + k ; a, b, c, k arbitrary constants
Problem Set 12.3, page 551
5. k cos 3πt sin 3πx
7.
9.
11.
13.
17.
19. (a) u (0, t ) = 0, (b) u (L , t ) = 0, (c) u x (0,t ) = 0, (d) u x (L , t ) = 0. C = −A , D = −B from (a), (c). Insert this. The coefficient determinant resulting from (b), (d) must be zero to have a nontrivial solution. This gives (22).
Problem Set 12.4, page 556
3. c 2 = 300/[0.9/(2 · 9.80)] = 80.832 [m2 /sec2 ]
9. Elliptic, u = f 1 (y + 2ix ) + f 2 (y − 2ix )
11. Parabolic, u = xf 1 (x − y ) + f 2 (x − y )
13. Hyperbolic, u = f 1 (y − 4x ) + f 2 (y − x )
15. Hyperbolic,
17. Elliptic, u = f 1 (y − (2 − i )x ) + f 2 (y − (2 + i )x ). Real or imaginary parts of any function u of this form are solutions. Why?
Problem Set 12.6, page 566
3. u 1 = sin x e −t , u 2 = sin 2x e −4t , u 3 = sin 3x e −9t differ in rapidity of decay.
5.
7.
9. u = u I + u II , where u II = u − u I satisfies the boundary conditions of the text, so that
11. F = A cos px + B sin px , F ′ (0) = Bp = 0, B = 0, F ′ (L ) = −Ap sin pL = 0, p = n π/L , etc.
13. u = 1
15.
17.
19.
21.
23.
25.
Problem Set 12.7, page 574
3.
5.
7.
9. Set w = −ν in (21) to get erf (−x ) = −erf x .
13. In (12) the argument is 0 (the point where f jumps) when ). This gives the lower limit of integration.
15. Set in (21).
Problem Set 12.9, page 584
1. (a) , (b) It is multiplied by . (c ) Half
5. B mn = (−1)n +1 8/(mnπ 2 ) if m odd, 0 if m even
7. B mn = (−1)m+n 4ab /(mn π2 )
11.
13.
17. (corresponding eigenfundtions F 4,16 and F 16,14 ), etc.
19.
Problem Set 12.10, page 591
Problem Set 12.11, page 598
5. A 4 = A 6 = A 8 = A 10 = 0, A 5 = 605/16, A 7 = −4125/128, A 9 = 7315/256
9.
13. u = 320/r + 60 is smaller than the potential in Prob. 12 for 2 < r < 4.
17. u = 1
19.
25. Set 1/r = ρ . Then u (ρ, θ, ø) = r ν(r , θ, ø), u ρ = (ν + r νr )(−1/ρ2 ), u ρρ = (2νr + rν rr (1/ρ4 ) + (ν + rν r )(2/ρ3 ), u ρρ + (2/ρ)u ρ = r 5 (νrr + (2/r )νr ). Substitute this and u øø etc. into (7) [written in terms of ρ ] and divide by r r 5.
Problem Set 12.12, page 602
5.
7.
11. Set x 2 /(4c 2 π) = z 2 . Use z as a new variable of integration. Use erf (∞) = 1.
Chapter 12 Review Questions and Problems, page 603
17. u = c 1 (x )e −3y + c 2 (x )e 2y −3
19. Hyperbolic, f 1 (x ) + f 2 (y + x )
21. Hyperbolic, f 1 (y + 2x ) + f 2 (y − 2x )
23.
25.
29. 100 cos 2x e −4t
39. u = (u 1 − u 0 )(In r )/In (r 1 /r 0 ) + (u 0 In r 1 − u 1 In r 0 )/In (r 1 /r 0 )
Problem Set 13.1, page 612
1. 1/i = i /i 2 = −i , 1/i 3 = i /i 4 = i
3. 4.8 − 1.4i
5. x − iy = −(x + iy ), x = 0
9. − 117, 4
11. −8 − 6i
13. − 120 − 40i
15. 3 − i
17. −4x 2 y 2
19. (x 2 − y 2 )/(x 2 + y 2 ), 2xy /(x 2 + y 2 )
Problem Set 13.2, page 618
1.
3.
5.
7.
9. 3π/4
11.
13. −1024. Answer: π
15. −3i
17. 2 + 2i
21.
23.
25.
27.
29. i , −1 − i
31. ±(1 − i ), ±(2 + 2i )
33.
35.
Problem Set 13.3, page 624
1. Closed disk, center − 1 + 5i , radius
3. Annulus (circular ring), center 4 − 2i , radii π and 3π
5. Domain between the bisecting straight lines of the first quadrant and the fourth quadrant.
7. Half-plane extending from the vertical straight line x = −1 to the right.
11.
15.
17. Yes, because Re z = r cos θ → 0 and 1 − |z | → 1 as r → 0.
19. f ′(z ) = 8(z − 4i )7 . Now z − 4i = 3, hence f ′ (3 + 4i ) = 8 · 37 = 17,496.
21. n (1 − z )−n −1 i , ni
23. 3iz 2 /(z + i )4 , −3i /16
Problem Set 13.4, page 629
1. r x = x /r = cos θ , r y = sin θ , θx = −(sin θ )/r , θ y = (cos θ )/r (a) 0 = u x − νy = u r cos θ + u θ (−sin θ )/r − νr sin θ − νθ (cos θ )/r (b) 0 = u y + νx = u r sin θ + u θ (cos θ )/r + νr cos θ + νθ (−sin θ )/r Multiply (a) by cos θ , (b) by sin θ , and add. Etc.
3. Yes
5.
7. Yes, when z ≠ 0. Use (7).
9. Yes, when z ≠ 0, −2πi , 2πi
11. Yes
13.
15. f (z ) = 1/z + c (c real)
17. f (z ) = z 2 + z + c (c real)
19. No
21. a = π, ν = e πx sin πy
23.
27. f = u + iv implies if = −ν + iu .
29. Use (4), (5), and (1).
Problem Set 13.5, page 632
3. e 2πi e −2π = 0.001867
5. e 2 (−1) = −7.389
7.
9. 5e i arctan (3/4) = 5e 0.644i
11. 6.3e πi
13.
15. exp (x 2 − y 2 ) cos 2xy , exp (x 2 − y 2 ) sin 2xy
17. Re (exp (z 3 )) = exp (x 3 − 3xy 2 ) cos (3x 2 y − y 3 )
19. z = 2n πi , n = 0, 1, …
Problem Set 13.6, page 636
1. Use (11), then (5) for e iy , and simplify.
7. cosh 1 = 1.543, i sinh 1 = 1.175i
9. Both −0.642 − 1.069i . Why?
11. i sinh π = 11.55i , both
15. Insert the definitions on the left, multiply out, and simplify.
17. z = ±(2n + 1)i /2
19. z = ±nπi
Problem Set 13.7, page 640
5. In 11 + πi
7.
9. i arctan (0.8/0.6) = 0.927i
11. In e + πi /2 = 1 + πi /2
13. ±2n πi , n = 0, 1, …
15.
17. In (i 2 ) = In (−1) = (1 ± 2n )πi , 2 In i = (1 ± 4n )±i , n = 0, 1, …
19. e 4−3i = e 4 (cos 3 − i sin 3) = −54.05 − 7.70i
21. e 0.6 e 0.4i = e 0.6 (cos 0.4 + i sin 0.4) = 1.678 + 0.710i
23.
25. e (3−i )(In 3+πi ) = 27e π (cos (3π − In 3) + i sin (3π − In 3)) = −284.2 + 556.4i
27. e (2−i )Ln(−1) = e (2−i )πi = e π = 23.14
Chapter 13 Review Questions and Problems, page 641
1. 2 − 3i
3. 27.46e 0.9929i
11. −5 + 12i
13. 0.16 − 0.12i
15. i
17.
19. 15e −πi /2
21. ±3, ±3i
23.
25. f (z ) = −iz 2 /2
27. f (z ) = e −2z
29.
31. cos 3 cosh 1 + i sin 3 sinh 1 = −1.528 + 0.166i
33. i tanh 1 = 0.7616i
35. cosh π cos π + i sinh π sin π = −11.592
Problem Set 14.1, page 651
1. Straight segment from (2, 1) to (5, 2.5).
3. Parabola y = x 2 from (1, 2) to (2, 8).
5. Circle through (0, 0), center (3, − 1), radius , oriented clockwise.
7. Semicircle, center 2, radius 4.
9. Cubic parabola y = x 3 (−2 x 2)
11. z (t ) = t + (2 + t )i (−1 t 1)
13. z (t ) = 2 − i + 2e it (0 t π)
15. z (t ) = 2 cosh t + i sinh t (− ∞ < t < ∞)
17. Circle z (t ) = −a − ib + re −it (0 t 2π)
19.
21. z (t ) = (1 + i )t (1 t 3), Re z = t , z ′ (t ) = 1 + i . Answer: 4 + 4i
23. e 2πi − e πi = 1 − (−1) = 2
25.
27.
29.
35.
Problem Set 14.2, page 659
1. Use (12), Sec. 14.1 , with m = 2.
3. Yes
5. 5
7. (a) Yes. (b) No, we would have to move the contour across ±2i .
9. 0, yes
11. πi , no
13. 0, yes
15. −π, no
17. 0, no
19. 0, yes
21. 2πi
23. 1/z + 1/(z −1), hence 2πi + 2πi = 4πi .
25. 0 (Why?)
27. 0 (Why?)
29. 0
Problem Set 14.3, page 663
1.
3. 0
5.
7. 2πi (i /2)3 /2 = π/8
11.
13.
15. 2πi cosh (−π2 − πi ) = −2πi cosh π2 = −60,739i since cosh πi = cos π = −1 and sinh πi = i sin π = 0.
17.
19. 2πie 2i /(2i ) = πe 2i
Problem Set 14.4, page 667
1. (2πi /3!)(−cos 0) = −πi /3
3. (2πi /(n − 1)!)e 0
5.
7.
9.
11.
13.
15. 0. Why?
17. 0 by Cauchy's integral theorem for a doubly connected domain; see (6) in Sec. 14.2 .
19.
Chapter 14 Review Questions and Problems, page 668
21.
23. by cauchy's integral formula.
25.
27.
29. −4πi
Problem Set 15.1, page 679
Problem Set 15.2, page 684
1. No! Nonnegative integer powers of z (or z − z 0 ) only!
3. At the center, in a disk, in the whole plane
5.
7. π/2, ∞
9.
11.
13.
15. 2i , 1
17.
Problem Set 15.3, page 689
3.
5. 2
7.
9.
11.
13. 1
15.
Problem Set 15.4, page 697
3.
5.
7.
9.
11. z 3 /(1!3) − z 7 /(3!7) + z 11 /(5!11) − + …, R = ∞
13.
17. Team Project. (a ) (Ln (1 + z ))′ = 1 − z + z 2 − + … = 1/(1 + z ). (c ) Use that the terms of (sin iy )/(iy ) are all positive, so that the sum cannot be zero.
19.
21.
23.
25.
Problem Set 15.5, page 704
3.
5.
7. Nowhere
9. |z − 2i | 2 − ∂, ∂ > 0
11. |z n | 1 ∑ 1/n 2 converges. Use Theorem 5.
13. |sinn |z || 1 for all z , and ∑ 1/n 2 converges. Use Theorem 5.
15. R = 4 by Theorem 2 in Sec. 15.2 ; use Theorem 1.
17.
Chapter 15 Review Questions and Problems, page 706
11. 1
13. 3
15.
17. ∞, e 2x
19. ∞, cosh
21.
23.
25.
27.
29.
Problem Set 16.1, page 714
1.
3.
5.
7.
9.
11.
13.
15.
19.
21.
23. z 8 + z 12 + z 16 + …, |z | < 1, −z 4 − 1 − z −4 − z −8 − …, |z | > 1
25.
1. 0 ± 2π, ±4π, …, fourth order
3. −81i , fourth order
5. ±1, ±2, …, second order
7. ±(2 + 2i ), ±i , simple
9.
11. f (z ) = (z − z 0 )n g (z ), g (z 0 ) ≠ 0, hence f 2 (z ) = (z − z 0 )2n g 2 (z ).
13. Second-order poles at i and −2i
15. Simple pole at ∞, essential singularity at 1 + i
17. Fourth-order poles at ±n πi , n = 0, 1, …, essential singularity at ∞
19. e x (1 − e x ) = 0, e z = 1, z = ±2n πi simple zeros. Answer: at simple poles ±2n πi , essential singularity at ∞
21. 1, ∞ essential singularities, ±2n πi , simple poles
3.
5.
7. 1/π at 0, ±1, …
9. −1 at ±2n πi
11.
15. Simple poles at inside C , residue −1/(2π). Answer: −i
17. Simple poles at π/2, residue e π/2 /(−sin π /2), and at −π/2, residue e −π/2 /sin π/2 = e −π/2 . Answer: −4πi sinh π/2
19.
21. z −2 cos πz = … + π4 /(4!z ) − + …. Answer: 2π5 i /24
23.
25.
Problem Set 16.4, page 733
1.
3.
5. 5π/12
7.
9. 0. Why? (Make a sketch.)
11. π/2
13. 0. Why?
15. π/3
17. 0. Why?
19. Simple poles at
21. Simple poles at 1 and ±2πi , residues i and −. Answer:
23. −π/2
25. 0
27. Let q (z ) = (z − a 1 )(z − a 2 )…(z − a k ). Use (4) in Sec. 16.3 to form the sum of the residues 1/q ′(a 1 ) + … + 1/q ′(a k and show that this sum is 0; here k > 1.
Chapter 16 Review Questions and Problems, page 733
11. 6πi
13. 2πi (− 10 − 10)
15.
17. 0 (n even), (−1)(n −1)/2 2πi /(n − 1)!(n odd)
19. π/6
21. π/60
23. 0. Why?
25.
Problem Set 17.1, page 741
5. Only in size
7. x = c , w = −y + ic ; y = k , w = −k = ix
9. Parallel displacement; each point is moved 2 to the right and 1 up.
11.
13. −5 Re z −2
15. u 1
17.
19. 0 < u < In 4, π/4 < v 3π/4
21.
23.
25. sinh z = 0 at z = 0, ±πi , ±2πi , …
29. M = |z | = 1 on the unit circle, J = |z |2
31. |w ′| = 1/|z |2 on the unit circle, J = 1/|z |4
33. M = e x = 1 for the x = 0, y -axis, J = e 2x
35. M = 1/|z | = on the unit circle, J = 1/|z |2
Problem Set 17.2, page 745
7.
9.
11. z = 0, 1/(a + ib )
13.
15. z = i, 2i
17.
19.
Problem Set 17.3, page 750
3. Apply the inverse g of f on both sides of z 1 = f (z 1 ) to get g (z 1 ) = g (f (z 1 )) = z 1 .
9. w = iz , a rotation. Sketch to see.
11. w = (z + i )/(z − i )
13. w = 1/z , almost by inspection
15. w = 1/z − 1
17. w = (2z − i )/(−i z − 2)
19. w = (z 4 − i )(−i z 4 + 1)
Problem Set 17.4, page 754
1. Circle |w | = e c
3.
5. w -plane without w = 0
7. < |w | < e,v > 0
9. ±(2n + 1)π/2, n = 0, 1, …
11. u 2 /cosh2 2 + v 2 /sinh2 2 < 1, u > 0,v > 0
13. Elliptic annulus bounded by u 2 /cosh2 1 + v 2 /sinh2 1 = 1 and u 2 /cosh2 3 + v 2/sinh 2 3 = 1
15.
17.
19. Hyperbolas u 2 /cos2 c v 2 /sin2 c − sinh2 c − sinh2 c = 1 when c ≠ 0, π, and u 2 /cos2 c − v 2 /sin2 c = cosh2 c − sinh2 c = 1 when c = 0, π.
21. Interior of u 2 /cosh2 2 + v 2 /sinh2 2 = 1 in the fourth quadrant, or map π/2 < x < π, 0 < y < 2 by w = sin z (why?).
23. v < 0
25. The images of the five points in the figure can be obtained directly from the function w .
Problem Set 17.5, page 756
1. w moves once around the circle
3. Four sheets, branch point at z = −1
5. −,i /4, three sheets
7. z 0 , n sheets
9. two sheets
Chapter 17 Review Questions and Problems, page 756
11. 1 < |w | < 4, |arg w | < π/4
13. Horizontal strip −8 < v < 8
15. , same (why?)
17. |w | > 1
19.
21. w = 1 + iv , v < 0
23.
25. Rotation w = i z
27. w = 1/z
29. z = 0
31.
33. z = 0, ±i , ±3i
35. w = e 4x
37. w = i z 2 + 1
39. w = z 2 /(2c )
Problem Set 18.1, page 762
1. 2.5 mm = 0.25 cm; Φ = Re 110 (1 + (Ln z )/ln 4)
3.
5. Φ(x ) = Re(375 + 25z )
7. Φ(r ) = Re (32 − z )
13. Use Fig. 391 in Sec. 17.4 with the z - and w - planes interchanged and cos z = sin .
15. Φ 220 (x 3 − 3xy 2 ) Re (220z 3 )
Problem Set 18.2, page 766
3.
5. (b) x 2 − y 2 = c , x y = c , e x cos y = c
7. See Fig. 392 in Sec. 17.4 . Φ = Re(sin2 z ), sin2 x )y = 0), sin2 x cosh2 1 − cos2 x sinh2 1(y = 1), −sinh2 y (x = 0, π).
9. Φ (x, y ) = cos2 x cosh2 y − sin2 x sinh2 y ; cosh2 y (x = 0), −sinh y cos2 x (y = 0), cos2 x cosh2 1 − sin2 x sinh2 1 (y = 1)
13. Corresponding rays in the w -plane make equal angles, and the mapping is conformal.
15. Apply w = z 2 .
17. z = (2z − i )/(−i Z − 2)(by (3) in Sec. 17.3 .
19.
Problem Set 18.3, page 769
1. (80/d )y + 20. Rotate through π/2.
5.
7.
9.
11.
13.
15.
17. Re F (z ) = 100 + (200/π) Re (arcsin z )
Problem Set 18.4, page 776
1. V (z ) continuously differentiable.
3. |F ′(iy )| = 1 + 1/y 2 , |y | 1, is maximum at y = ±1, namely, 2.
5. Calculate or note that ∇ 2 = div grad and curl grad is the zero vector; see Sec. 9.8 and Problem Set 9.7 .
7. Horizontal parallel flow to the right.
9. F (z ) = z 4
11. Uniform parallel flow upward,
13. F (z ) = z 3
15. F (z ) = z /r 0 + r 0 /z
17. Use that w = arccos z gives z = cos w and interchanging the roles of the z - and w -planes.
19. y /(x 2 + y 2 ) = c or x 2 + (y − k )2 = k 2
Problem Set 18.5, page 781
5.
7.
9. Φ = 3 − 4r 2 cos 2θ + r 4 cos 4θ
11.
13.
15.
17.
Problem Set 18.6, page 784
1.
3. Use (2). F (z 0 + e ix ) = (2 + 3e ix )2 , etc. F (4) = 100
5. No, because |z | is not analytic.
7.
9.
13. |F (z )| = [cos2 x + sinh2 y ]1/2 , z = ±i , Max = [1 + sinh2 1]1/2 = 1.543
15. |F (z )|2 = sinh2 2x cos2 2y + cosh2 2x sin2 2y = sinh2 2x + 1 · sin2 2y , z = 1, Max = sinh 2 = 3.627
17. |F (z )|2 = 4(2 − 2cos 2θ), z = π/2, 3π/2, Max = 4
19. No. Make up a counterexample.
Chapter 18 Review Questions and Problems, page 785
11. Φ = 10(1 − x + y ), F = 10 − 10(1 + i )z
13.
17. 2(1 − (2/π)Arg z )
19. 30(1 − (2/π)Arg (z − 1))
21.
23. T (x ,y ) = x (2y + 1) = const
25.
Problem Set 19.1, page 796
1. 0.84175 · 102 , −0.52868 · 103 , 0.92414 · 10− 3, −0.36201 · 106
3. 6.3698, 6.794, 8.15, impossible
5. Add first, then round.
7. 29.9667, 0.0335; 29.9667, 0.0333704 (6S-exact)
9. 29.97, 0.035; 29.97, 0.03337; 30, 0.0; 30, 0.033
11.
13.
15. (a) 1.38629 − 1.38604 = 0.00025, (b) ln 1.00025 = 0.000249969 is 6S-exact.
19. In the present case, (b) is slightly more accurate than (a) (which may produce nonsensical results; cf. Prob. 20).
21. c 4 · 24 + … + c 0 · 20 = (1 0 1 1 1.)2 , NOT (1 1 1 0 1.)2
23. The algorithm in Prob. 22 repeats 0011 infinitely often.
25. n = 26. The beginning is 0.09375 (n = 1).
27. I 14 = 0.1812 (0.1705 4S-exact), I 13 = 0.1812 (0.1820), I 12 = 0.1951 (0.1951), I 11 = 0.2102 (0.2103), etc.
29. −0.126 · 10−2 , −0.402 · 10−3 ; −0.266 · 10−6 , −0.847 − 10−7
Problem Set 19.2, page 807
3. g = 0.5 cos x , x = 0.450184 (= x 10 , exact to 6S)
5. Convergence to 4.7 for all these starting values;
7. x = x /(e x sin x ); 0.5, 0.63256, …converges to 0.58853 (5S-exact) in 14 steps.
9. x = x 4 − 0.12; x 0 = 0, x 3 = −0.119794(6S-exact)
11. g = 4/x + x 3 /16 − x 5 /576; x 0 = 2, x n = 2.39165 (n 6), 2.405 4S-exact
13. This follows from the intermediate value theorem of calculus.
15. x 3 = 0.450184
17. Convergence to x = 4.7, 4.7, 0.8, −0.5, respectively. Reason seen easily from the graph of f .
19.
21. 1.834243 (= x 4 ), 0.656620(= x 4 ), −2.49086 (= x 4 )
23. x 0 = 4.5, x 4 = 4.73004 (6S-exact)
25. (a) ALGORITHM BISECT (f , a0 ,b 0 , N ) Bisection Method
This algorithm computes the solution c of f (x ) = 0 (f continuous) within the tolerance , given an initial interval [a 0 ,b 0 ] such that f (a 0)f (b 0) < 0.
Note that [a N , b N ] gives (a N + b N )/2 as an approximation of the zero and (b N − a N )/2 as a corresponding error bound.
(b) 0.739085; (c) 1.30980, 0.429494
27. x 2 = 1.5, x 3 = 1.76471, …, x 7 = 1.83424 (6S-exact)
29. 0.904557 (6S-exact)
Problem Set 19.3, page 819
1. L 0 (x ) = −2x + 19, L 1 (x ) = 2x − 18, p 1 (9.3) = L 0 (9.3) · f 0 + L 1 (9.3) · f 1 = 0.1086 · 9.3 + 1.230 = 2.2297
3.
5. 0.8033 (error −0.0245), 0.4872 (error −0.0148); quadratic; 0.7839 (−0.0051), 0.4678 (0.0046)
7. p 2 (x ) = 1.1640x − 0.337x 2 ; −0.5089(error 0.1262), 0.4053(−0.0226), 0.9053(0.0186), 0.9911(−0.0672)
9. p 2 (x ) = −0.44304x 2 + 1.30896x − 0.023220, p 2 (0.75) = 0.70929 (5S-exact 0.71116)
11.
13. 2x 2 − 4x + 2
15. p 3 (x ) = 2.1972 + (x − 9) · 0.1082 + (x − 9)(x − 9.5) · 0.005235
17.
Problem Set 19.4, page 826
9. [−1.39 (x − 5)2 + 0.58(x − 5)3 ]″ = 0.004 at x = 5.8 (due to roundoff; should be 0).
11.
13. 1 − x 2 , −2(x − 1) − (x − 1)2 + 2(x − 1)3 , −1 + 2(x − 2) + 5(x − 2)2
15. 4 + x 2 − x 3 , −8(x − 2) − 5(x − 2)2 + 5(x − 2)3 , 4 + 32(x − 4) + 25(x − 4)2 − 11(x − 4)3
17. Use the fact that the third derivative of a cubic polynomial is constant, so that g ″′ is piecewise constant, hence constant throughout under the present assumption. Now integrate three times.
19. Curvature f ″ /(1 + f ′2 )3/2 ≈ f ″ if |f ′| is small.
Problem Set 19.5, page 839
1. 0.747131, which is larger than 0.746824. Why?
3. 0.5, 0.375, 0.34375, 0.335 (exact)
5. 0.5 ≈ 0.03452 (0.5 = 0.03307), 0.25 ≈ 0.00829 (0.25 = 0.00820)
7. 0.693254 (6S-exact 0.693147)
9. 0.073930 (6S-exact 0.073928)
11. 0.785392 (6S-exact 0.785398)
13. (0.785398126 − 0.785392156)/15 = 0.39792 · 10−6
15. (a) M 2 = 2; |KM 2 | = 2/(12n 2 ) = 10−5 /2,n = 183. (b ) f iv = 24/x 5 ,M 4 = 24, |CM 4 | = 24/(180) · (2m 4 ) = 10−n5 /2, 2m = 12.8, hence 14.
17. 0.94614588, 0.94608693 (8S-exact 0.94608307)
19. 0.9460831 (7S-exact)
21. 0.9774586 (7S-exact 0.9774377)
23.
25.
27. 0.08, 0.32, 0.176, 0.256 (exact)
29.
Chapter 19 Review Questions and Problems, page 841
17. 4.375, 4.50, 6.0, impossible
19. 44.885 s 44.995
21. The same as that of .
23. x 1 = 39.95, x 2 = 0.05, x 2 = 2/39.95 = 0.05006 (erroe less than 1 unit of the last digit)
25. x = x 4 − 0.1, −0.1, −0.9999, −0.99900399
27. 0.824
29. −x + x 3 , 2(x − 1) + 3(x − 1)2 − (x − 1)3
31. 0.26, M 2 = 6, , −0.02 0, 0.01
33. 0.90443, 0.90452 (5S-exact 0.90452)
35. (a) (0.43 − 2 · 0.23 + 0)/0.04 = 1.2, (b) (0.33 − 2 · 0.23 + 0.13 )/0.01 = 1.2 (exact)
Problem Set 20.1, page 851
1. x 1 = 7.3, x 2 = −3.2
3. No solution
5. x 1 = 2, x 2 = 1
7.
9.
11.
13.
15.
Problem Set 20.2, page 857
1.
3.
5.
7.
9.
11.
13. No, since xT (−A)X = −XT Ax < 0; yes; yes; no
15.
17.
19.
Problem Set 20.3, page 863
5. Exact 0.5, 0.5, 0.5
7. x 1 = 2, x 2 = −4, x 3 = 8
9. Exact 2, 1, 4
11. (a) x (3)T = [0.49983 0.50001 0.500017], (b) x (3)T = [0.50333 0.49985 0.49968]
13. 8, −16, 43, 86 steps; spectral radius 0.09, 0.35, 0.72, 0.85, approximately
15. [1.99934 1.00043 3.99684]T (Jacobi, Step 5); [2.00004 0.998059 4.00072]T (Gauss–Seidel)
19.
Problem Set 20.4, page 871
1. 18, 8, [0.125 −0.375 1 0 −0.75 0]
3.
5.
7. ab + bc + ca = 0
9.
11.
13. k = 19 · 13 = 247; ill-conditioned
15. k = 20 · 20 = 400; ill-conditioned
17. 167 21 · 15 = 315
19. [− 4]T , [−144.0 184.0]T , k = 25,921 extremely ill-conditioned
21. Small residual [0.145 0.120], but large deviation of .
23. 27, 748, 28,375, 943,656, 29,070,279
Problem Set 20.5, page 875
1. 1.846 − 1.038x
3. 1.48 + 0.09x
5. s = 90t − 675, v av = 90 km/hr
9. −11.36 + 5.45x − 0.589x 2
11. 1.89 − 0.739x + 0.207x 2
13. 2.552 + 16.23x , −4.114 + 13.73x + 2.500x 2 , 2.730 + 1.466x − 1.778x 2 + 2.852x 3
Problem Set 20.7, page 884
1. 5, 0, 5; radii 6, 5, 6. Spectrum {−1, 4, 9}
3. Centers 0; radii 0.5, 0.7, 0.4. Skew-symmetric, hence λ = i μ, −0.7 μ 0.7.
5.
7. t 11 = 100, t 22 = t 33 = 1
9. They lie in the intervals with endpoints a jj ± (n − 1) · 10−5 . Why?
11.
13.
15.
17. Show that A T = T A .
19. 0 lies in no Gerschgorin disk, by (3) with >; hence det A = λ1 …λn ≠ 0.
Problem Set 20.8, page 887
1. q = 10, 10.9908, 10.9999; | | 3, 3028, 0.0275
3. q ± δ = 4 ± 1.633, 4.786 ± 0.619, 4.917 ± 0.398
5. Same answer as in Prob. 3, possibly except for small roundoff errors.
7. q = 5.5, 5.5738, 5.6018; | | 0.5, 0.3115, 0.1899; eigenvalues (4S) 1.697, 3.382, 5.303, 5.618
9. y = Ax = λx , yT x = λxT x, yT y = λ2 xT x, 2 ≤ yT y/xT x − (yT x/xT x)2 = λ2 − λ2 = 0
11. q = 1, …, −2.8993 approximates −3 (0 of the given matrix), | | 1.633, …, 0.7024 (Step 8)
Problem Set 20.9, page 896
Chapter 20 Review Questions and Problems, page 896
15. [3.9 4.3 1.8]T
17. [−2 0 5]T
19.
21.
23.
25.
27. 30
29. 5
31. 115 · 0.4458 = 51.27
33.
35. 1.514 + 1.129x −0.214x 2
37. Centers 15,35,90; radii 30,35,25, respectively, Eigenvalues (3S) 2.63, 40.8, 96.6
39. Centers 0, −1, −4; radii 9, 6, 7, respectively; eigenvalues 0. 4.446, −9.446
Problem Set 21.1, page 910
1. y = se −0.2x , 0.00458, 0.00830 (errors of y 5 , y 10 )
3. y = x − tanh x (set y − x = u ), 0.00929, 0.01885 (errors of y 5 , y 10 )
5. y = e x , 0.0013, 0.0042 (errors of y 5 , y 10 )
7. y = 1/(1 − x 2 /2), 0.00029, 0.01187 (errors of y 5 , y 10 )
9. Errors 0.03547 and 0.28715 of and y 5 and y 10 much larger
11. y = 1/(1 − x 2 /2); error −10−8 , −4 · 10−8 , …, −6 · 10−7 , +9 · 10−6 ; = 0.0002/15 = 1.3 · 10−5 (use RK with h = 0.2)
13. y = tan x ; error 0.83 · 10−7 , 0.16 · 10−6 , …, −0.56 · 10−6 , +0.13 · 10−5
15. y = 3 cos x − 2 cos2 x ; error · 107 : 0.18, 0.74, 1.73, 3.28, 5.59, 9.04, 14.3, 22.8, 36.8, 61.4
17. y ′ = 1/(2 − x 4 ); error · 109 : 0.2, 3.1, 10.7, 23.2, 28.5, −32.3, −376, −1656, −3489, +80444
19. Errors for Euler–Cauchy 0.02002, 0.06286, 0.05074; for improved Euler–Cauchy −0.000455, 0.012086 0.009601; for Runge–Kutta. 0.0000011, 0.000016, 0.000536
Problem Set 21.2, page 915
1.
3. y = tan x , y 4 , …, y 10 (error · 105 ) 0.422798 (−0.49), 0.546315 (−1.2), 0.684161 (−2.4), 0.842332 (−4.4), 1.029714 (−7.5), 1.260288 (−13), 1.557626 (−22)
5. RK error smaller in absolute value, error · 105 = 0.4, 0.3, 0.2, 5.6 (for x = 0.4, 0.6, 0.8, 1.0)
7. y = exp (x 3 ) − 1, y 4 , …, y 10 (error · 1010 ) 0.008032 (−4), 0.015749 (− 10), 0.027370 (− 17), 0.043810 (− 26), 0.066096 (− 39), 0.095411 (− 54), 0.133156 (− 74)
13. y = exp (x 2 ). Errors · 102 from x = 0.3 to 0.7: −5, −11, −19, −31, −41
15. (a) 0, 0.02, 0.0884, 0.215848, y 4 = 0.417818, y 5 = 0.708887 (poor) (b) By 30−50%
Problem Set 21.3, page 922
1. y 1 = −e−2x + 4e x errors of y 1 (of y 2 ) from 0.002 to 0.5 (from −0.01 to 0.1), monotone
3.
5. , y = 0.8 (error 0.005), 0.61 (0.01), 0.429 (0.012), 0.2561 (0.0142), 0.0905 (0.0160)
7. By about a factor 105 . n (y 1 ) · 106 = −0.082, …, −0.27, n (y 2 ) · 106 = 0.08, …, 0.27
9. Errors of (of y 1 (of y 2 ) from 0.3 · 10−5 to 1.3 · 10−5 (from 0.3 · 10−5 to 0.6 · 10−5 )
11. (y 1 , y 2 ) = (0, 1), (0.20, 0.98), (0.39, 0.92), …, (−0.23, −0.97), (−0.42, −0.91), (−0.59), (−0.81); continuation will give an “ellipse.”
Problem Set 21.4, page 930
3. −3u 11 + u 12 = −200, u 11 − 3u 12 = −100
5. 105, 155, 105, 115; Step 5: 104.94, 154.97, 104.97, 114.98
7. 0, 0, 0, 0. All equipotential lines meet at the corners (why?). Step 5: 0.29298, 0.14649, 0.14649, 0.073245
9. 0.108253, 0.108253, 0.324760, 0.324760; Step 10: 0.108538, 0.108396, 0.324902, 0.324831
11. (a) . u 11 = −u 12 = −66. (b) Reduce to 4 equations by symmetry.
13. u 12 = u 32 = 31.25, u 21 = u 23 = 18.75, u jk = 25 at the others
15. u 21 = u 23 = 0.25, u 12 = u 32 = 0.25, u jk = 0 otherwise
17. u 12 = u 22 = 0.3170. (0.1083, 0.3248 are 4S-values of the solution of the linear system of the problem.)
Problem Set 21.5, page 935
5. u 11 = 0.766, u 21 = 1.109, u 12 = 1.957, u 22 = 3.293
7. A , as in Example 1, right sides −220, −220, −220, −220. Solution u 11 = u 21 = 125.7, u 21 = u 22 = 157.1
13.
15. b = [−200, −100, −100, 0]T ; u 11 = 73.68, u 21 = u 12 = 47.37, u 22 = 15.79 (4S)
Problem Set 21.6, page 941
5. 0, 0.6625, 1.25, 1.7125, 2, 2.1, 2, 1.7125, 1.25, 0.6625, 0
7. Substantially less accurate, 0.15, 0.25 (t = 0.04), 0.100, 0.163 (t = 0.08)
9. Step 5 gives 0, 0.06279, 0.09336, 0.08364, 0.04707, 0.
11. Step 2: 0 (exact 0), 0.0453 (0.0422), 0.0672 (0.0658), 0.0671 (0.0628), 0.0394 (0.0373), 0 (0)
13. 0.3301, 0.5706, 0.4522, 0.2380 (t = 0.04), 0.06538, 0.10603, 0.10565, 0.6543 (t = 0.20)
15. 0.1018, 0.1673, 0.1673, 0.1018 (t = 0.04), 0.0219, 0.0355, …(t = 0.20)
Problem Set 21.7, page 944
1. u (x , 1) = 0, −0.05, −0.10, −0.15, −0.20, 0
3. For x = 0.2, 0.4 we obtain 0.24, 0.40 (t = 0.2), 0.08 0.16 (t = 0.4), −0.08, −0.16 (t = 0.6), etc.
5. 0, 0.354, 0.766, 1.271, 1.679, 1.834, … (t = 0.1); 0, 0.575, 0.935, 1.135, 1.296, 1.357, … (t = 0.2)
7. 0.190, 0.308, 0.308, 0.190, (3S-exact: 0.178, 0.288, 0.288, 0.178)
Chapter 21 Review Questions and Problems, page 945
17. y = e x , 0.038, 0.125 (errors of y 5 and y 10 )
19. y = tan x ; 0 (0), 0.10050 (−0.00017), 0.20304 (−0.00033), 0.30981 (−0.00048), 0.42341 (−0.00062), 0.54702 (−0.00072), 0.68490 (−0.00076), 0.84295 (−0.00066), 1.0299 (−0.0002), 1.2593 (0.0009), 1.5538 (0.0039)
21. 0.1003346(0.8 · 10−7 )0.2027099 (1.6 · 10−7 ), 0.3093360 (2.1 · 10−7 ), 0.4227930 (2.3 · 10−7 ), 0.5463023 (1.8 · 10−7 )
23. y = sin x , y 0.8 = 0.717366, y 1.0 = 0.841496 (errors − 1.0 · 10−5 , −2.5 · 10−5 )
25.
27.
29. 3.93, 15.71, 58.93
31. 0, 0.04, 0.08, 0.12, 0.15, 0.16, 0.15, 0.12, 0.08, 0.04, 0 (t = 0.3 3 time steps)
33. u (p 11 ) = u (p 31 ) = 270, u (p 21 ) = u (p 13 ) = u (p 23 ) = u (p 33 ) = 30., u (p 12 ) = u (p 32 ) = 90, u (p 22 ) = 60
35. 0.043330, 0.077321, 0.089952, 0.058488 (t = 0.04), 0.010956, 0.017720, 0.017747, 0.010964 (t = 0.20)
Problem Set 22.1, page 953
3. f (x ) = 2(x 1 − 1)2 + (x 2 + 2)2 − 6; Step 3; (1.037, −1.926), value −5.992
9 . Step 5: (0.11247, −0.00012), value 0.000016
Problem Set 22.2, page 957
7. No
9. x 3 ,x 4 is the unused time on M 1 , M 2 , respectively.
11. f (2.5, 2.5) = 100
13.
15. f (9,6) = 360
17. 0.5x 1 + 0.75x 2 45 (copper), 0.5x 1 + 0.25x 2 30, f = 120x 1 + 100x 2 , f max = f (45,30) = 8400
19. f = x 1 + x 2 , 2x 1 + 3x 2 1200, 4x 1 + 2x 2 1600, f max = f (300,200) = 500
21. x 1 /3 + x 2 /2 100, x 1 /3 + x 2 /6 80,f = 150x 1 + 100x 2 ,f max = f (210, 60) = 37,500
Problem Set 22.3, page 961
3. f (120/11, 60/11) = 480/11
5. Eliminate in Coloumn 3, so that 20 goes.
7.
9. f max = 6 on the segment from (3, 0, 0) to (0, 0, 2)
11. We minimize! The augmented matrix is
The pivot is 600. The calculation gives
Hence −f has the maximum value −13.5, so that f has the minimum value 13.5, at the point
13. f max = f (5, 4, 6) = 478
Problem Set 22.4, page 968
1. f (6, 3) = 84
3. f (20, 20) = 40
5. f (10, 5) = 5500
7. f (1, 1, 0) = 13
9.
Chapter 22 Review Questions and Problems, page 968
9. Step 5: [0.353 −0.028]T . Slower. Why?
11. Of course! Step 5: [−1.003 1.897]T
17. f (2, 4) = 100
19. f (3, 6) = −54
Problem Set 23.1, page 974
9.
11.
13.
15.
17. If G is complete.
19.
Problem Set 23.2, page 979
1. 5
3. 4
5. The idea is to go backward. There is a v k-1 adjacent to v k and labeled k − 1, etc. Now the only vertex labeled 0 is s . Hence λ(v 0 ) = 0 implies v 0 = s , so thatv o − v 1 − … − v k -1 − v k is a path s → u k that has length k .
15. Delete the edge (2, 4).
17. No
Problem Set 23.3, page 983
1. (1, 2), (2, 4), (4, 3); L 2 = 12, L 3 = 36, L 4 = 28
5. (1, 2), (2, 4), (3, 4), (3, 5); L 2 = 2, L 3 = 4, L 4 = 3, L 5 = 6
7. (1, 2), (2, 4), (3, 4); L 2 = 10, L 3 = 15, L 4 = 13
9. (1, 5), (2, 3), (2, 6), (3, 4), (3, 5); L 2 = 9, L 3 = 7, L 4 = 8, L 5 = 4, L 6 = 14
Problem Set 23.4, page 987
1.
3.
5.
9. Yes
11.
13. New York–Washington–Chicago–Dalles–Denver–Los Angeles
15. G is connected. If G were not a tree, it would have a cycle, but this cycle would provide two paths between any pair of its vertices, contradicting the uniqueness.
19. If we add an edge (u , v ) to T , then since T is connected, there is a path (u → v ) in T which, together with (u , v ), forms a cycle.
Problem Set 23.5, page 990
1. If G is a tree.
3. A shortest spanning tree of the largest connected graph that contains vertex 1.
7. (1, 4), (1, 3), (1, 2), (2, 6), (3, 5); L = 32
9. (1, 4), (4, 3), (4, 2), (3, 5); L = 20
11. (1, 4), (4, 3), (4, 5), (1, 2); L = 12
Problem Set 23.6, page 997
1. {3, 6}, 11 + 3 = 14
3. {4, 5, 6}, 10 + 5 + 13 = 28
5. {3, 6, 7}, 8 + 4 + 4 = 16
7. S {1, 4}, 8 + 6 = 14
9. One is interested in flows from s to t, not in the opposite direction.
13. Δ12 = 5, Δ24 = 8, Δ45 = 2; Δ12 = 5, Δ25 = 3; Δ13 = 4, Δ35 = 9 P 1 : 1 − 2 − 4 − 5, Δf = 2; P 2 : 1 − 2 − 5, Δf = 3; P 3 : 1 − 3 − 5, Δf = 4
15. 1 − 2 − 5, Δf = 2; 1 − 4 − 2 − 5, Δf = 2, etc.
17. f 13 = f 35 = 8, f 14 = f 45 = 5, f 12 = f 24 = f 46 = 4, f 56 = 13, f = 4 + 13 = 17, is f = 17 is unique.
19. For instance, f 12 = 10, f 24 = f 45 = 7, f 13 = f 25 = 5, f 35 = 3, f 32 = 2, f = 3 + 5 + 7 = 15, f = 15 is unique.
3. (2, 3) and (5, 6)
5. By considering only edges with one labeled end and one unlabeled end
7. 1 − 2 − 5, Δt 2; 1 − 4 − 2 − 5, Δt 1; f = 6 + 2 + 1 = 9, where 6 is the given flow
9. 1 − 2 − 4 − 6, Δt 2; 1 − 3 − 5 − 6, Δt 1; f = 4 + 2 + 1 = 7, where 4 is the given flow
15. S = {1, 2, 4, 5}, T = {3, 6}, cap(S , T ) = 14
1. No
3. No
5. Yes, S = {1, 4, 5, 8}
7. Yes, S = {1, 3, 5}
11. 1 − 2 − 3 − 7 − 5 − 4
13. 1 − 2 − 3 − 7 − 5 − 4 is augmenting and gives 1 − 2 − 3 − 7 − 5 − 4 and is of maximum cardinality.
15. 1 − 4 − 3 − 6 − 7 − 8 is augmenting and gives 1 − 4 − 3 − 6 − 7 − 8 and (1, 4), (3, 6), (7, 8) is of maximum cardinality.
19. 3
21. 2
23. 3
25. K 4
Chapter 23 Review Questions and Problems, page 1006
11.
13.
15.
17.
19. (1, 2), (1, 4), (2, 3); L 2 = 2, L 3 = 5, L 4 = 5
23. (1, 6), (4, 5), (2, 3), (7, 8)
1. qL = 19, qM = 20, qU = 20.5
3. qL = 138, qM = 144, qU = 154
5. qL = 199, qM = 201, qU = 201
7. qL = 1.3, qM = 1.4, qU = 1.45
9. qL = 89.9, qM = 91.0, qU = 91.8
11.
13.
15.
17. 3.54, 1.29
1. 23 outcomes: RRR, RRL, RLR, LRR, RLL, LRL, LLR, LLL
3. 62 = 36 outcomes (1, 1), (1, 2), …, (6, 6) first number (second number) referring to the first die (second die)
5. Infinitely many outcomes H TH TTH TTTH … (H = Head, T = Tail)
7. The space of ordered pairs of numbers
9. 10 outcomes: D ND NND … NNNNNNNNND
11. Yes
17. A ∪ B = B implies A ⊆ B by the definition of union. Conversely. A ⊆ B implies that A ∪ B = B because always B ⊆ A ∪ B , and if A ⊆ B , we must have equality in the previous relation.
1. 1 − 4/216 = 98315%, by Theorem 1
3.
5.
7. Small sample from a large population containing many items in each class we are interested in (defectives and nondefectives, etc.)
9.
11.
13. 1 − 0.963 = 11.5%
15. 1 − 0.8754 = 0.4138 < 1 − 0.752 = 0.4375 < 0.5 (c < b < a)
17. A = B ∪(A ∩B c )hence p A = p (B ) + p (A ∩B c ) p (B ) by disjointedness of B and A ∩B c
1. In 10! = 3,628,800 ways
3.
5.
7. 210, 70, 112, 28
9. In 6!/6 = 120 ways
11. 9 · 8 = 72
13. (b ) 1/(12n )
15. P (No two people have a birthday in common ) = 365 · 364 … 346/36520 = 0.59. Answer: 41%, which is surprisingly large.
1.
3.
5. No, because of (6)
7.
9. k = 5; 50%
11. 0.53 = 12.5%
13.
15. X > b , X b , X c , etc.
1.
3. μ = π, σ2 = π2 /3; cf. Example 2
5.
7.
9. 750, 1, 0.002
11. c = 0.073
13. $643.50
15.
17. X = Product of the 2 numbers . E (X ) = 12.25, 12 cents
19. (0 + 1 · 3 + 3 · 8 + 1 · 27)/8 = 54/8 6 · 75
3. 38%
5.
7. 0.265
9. f (x ) = 0.5x e −0.5 /x !,f (0) + f (1) = e −0.5 (1.0 + 0.5) = 0.91. Answer: 9%
11.
13. 42%, 47.2%, 10.5%, 0.3%
15. 1 − e −0.2 = 18%
1. 0.1587, 0.5, 0.6915, 0.6247
3. 45.065, 56.978, 2.022
5. 15.9%
7. 31.1%, 95.4%
9. About 58%
11. t = 1084 hours
13. About 683 (Fig. 521a )
1.
3.
5. f 2 (y ) = 1/(β2 − α2 ) if α2 < y lt; β2
7. 27.45 mm, 0.38 mm
11. 25.26 cm, 0.0078 cm
13. 50%
15. The distributions in Prob. 17 and Example 1
17. No
Chapter 24 Review Questions and Problems, page 1060
11. Q L = 110, Q M = 112, Q U = 115
13.
21. x min x j x max . Sum over j from 1.
17.
19.
21. f (x ) = 2−x , x = 1, 2, …
23.
25. 0.1587, 0.6306, 0.5, 0.4950
1. In Example 1,
3.
5.
7. 7/12
9.
11.
13.
15. Variability larger than perhaps expected
3. Shorter by a factor
5. 4, 16
7.
9. CONF0.99 {63.72 μ 66.28}
11.
13. CONF0.95 {0.023 σ2 0.085}
15. n − 1 = 99 degrees of freedom. F (c 1 ) = 0.025, c 1 = 74.2, F (c 2 ) = 0.975, c 2 = 129.6. Hence k 1 = 12.41, k 2 = 7.10. CONF0.95 {7.10 σ2 12.41}.
17. CONF0.95 {0.74 σ2 5.19}
19. Z = X + Y is normal with mean 105 and variance 1.25. Answer: P (104 Z 106) = 63%
3.
5. c = 6090 > 6019: do not reject the hypothesis.
7. σ2 /n = 1.8, c = 57.8, accept the hypothesis.
9. μ < 58.69 or μ > 61.31
11.
13.
15. 19 · 1.02 /0.82 = 29.69 < c = 30.14 (Table A10 . Appendix 5 ), accept the hypothesis
17.
1. LCL = 1 − 2.58 · 0.02/2 = 0.974, UCL = 1.026
3. 27
5. Choose 4 times the original sample size
9.
11.
13. In about 30% (5%) of the cases
15.
1. 0.9825, 0.9384, 0.4060
3. 0.8187, 0.6703, 0.1353
5. e −25θ (1 + 25θ), P (A; 1.5) = 94.5, α = 5.5%
7. 19.5%, 14.7%
9. (1 − θ)n + n θ(1 − θ)n −1
11.
13.
15.
3.
5.
7.
9. 42 even digits, accept.
13.
15. Combining the last three nonzero values, we have K − r − 1 = 9 (r = 1 since we estimated the mean, Accept the hypothesis.
3. is the probability that 7 cases in 8 trials favor A under the hypothesis that A and B are equally good. Reject.
5.
7. Hypothesis rejected.
9.
11. Consider
13. n = 8;4 transpositions, P (T 4) = 0.007, Assert that fertilizing increases yield.
15. P (T ) 2) = 2.8%. Assert that there is an increase.
1. y = 0.98 + 0.495x
3. y = −11,457.p + 43.2x
5. y = −10 + 0.55x
7. y = 0.5932 + 0.1138x , R = 1/0.1138
9. y = 0.32923 + 0.00032x , y (66) = 0.35035
13. c = 3.18 (Table A9 ), k 1 = 43.2, q 0 = 54,878, K = 1.502, CONF0.95 {41.7 K 1 } 44.7}.
15.
Chapter 25 Review Questions and Problems, page 1111
15.
17. CONF0.99 {27.94 μ 34.81}
19.
21.
23. α = 1 − (1 − θ )6 = 5.85%, when θ = 0.01. For θ = 15% we obtain β = (1 − θ )6 = 37.7%. If n increases, so does α, whereas β decreases.
25. y = 3.4 − 1.85x
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