Answers to Odd-Numbered Problems - Purdue Universitycaiz/MA527-cai/Appendix2.pdf · 2012-09-19 · A6 App. 2 Answers to Odd-Numbered Problems Problem Set 1.7, page 42 1. ; hence is

A4

A P P E N D I X 2

Answers to Odd-Numbered Problems

Problem Set 1.1, page 8

1. 3.

5. 7.

9. 11.

13. 15.

17. exp

19. Integrate (start from rest), then


11. Straight lines parallel to the x-axis 13.

15. gives the limit

[ ]

17. Errors of steps 1, 5, 10: 0.0052, 0.0382, 0.1245, approximately

19. (error 0.0093), (error 0.0189)


1. If you add a constant later, you may not get a solution.

Example:

3.

5. , ellipses 7.

9. 11. , hyperbola

13.

15. 17.

19.

21. 23.

25.

27.

29. No. Use Newton’s law of cooling.

31.

33.

. Eight times.� ϭ (1>0.15) ln 1000 ϭ 7.3 # 2p¢S ϭ 0.15S¢�, dS>d� ϭ 0.15S, S ϭ S0e0.15� ϭ 1000S0,

y ϭ ax, yr ϭ g(y>x) ϭ a ϭ const, independent of the point (x, y)

e�k #10 ϭ 12, k ϭ 1

10, ln 12, e�kt0 ϭ 0.01, t ϭ (ln 100)>k ϭ 66 [min]

T ϭ 22 Ϫ 17e�0.5306t ϭ 21.9 3°C4 when t ϭ 9.68 min

PV ϭ c ϭ const69.6% of y0

y0ekt ϭ 2y0, ek ϭ 2 (1 week), e2k ϭ 22 (2 weeks), e4k ϭ 24

y ϭ x arctan (x3 Ϫ 1)y2 ϩ 4x2 ϭ c ϭ 25

dy>sin2 y ϭ dx>cosh2 x, Ϫcot y ϭ tanh x ϩ c, c ϭ 0, y ϭ Ϫarccot (tanh x)

y ϭ 24>xy ϭ x>(c Ϫ x)

y ϭ x arctan (x2 ϩ c)y2 ϩ 36x2 ϭ c

cos2 y dy ϭ dx, 12 y ϩ 1

4 sin 2y ϩ c ϭ x

yr ϭ y, ln ƒ y ƒ ϭ x ϩ c, y ϭ exϩc ϭ �cex but not ex ϩ c (with c � 0)

x10 ϭ 0.2196x5 ϭ 0.0286

meter>sec

>9.8 ϭ 3.1vr ϭ 0mvr ϭ mg Ϫ bv2, vr ϭ 9.8 Ϫ v2, v(0) ϭ 10,

y ϭ x

y(t) ϭ 12 gt 2 ϩ y0, where y(0) ϭ y0 ϭ 0

ys ϭ g twice, yr(t) ϭ gt ϩ v0, yr(0) ϭ v0 ϭ 0

(Ϫ1.4 # 10�11t) ϭ 12, t ϭ 1011(ln 2)>1.4 [sec]

y ϭ 0 and y ϭ 1 because yr ϭ 0 for these yy ϭ 1>(1 ϩ 3e�x)

y ϭ (x ϩ 12)exy ϭ 1.65e�4x ϩ 0.35

y ϭ1

5.13 sinh 5.13x ϩ cy ϭ 2e�x(sin x Ϫ cos x) ϩ c

y ϭ cexy ϭ1

p cos 2px ϩ c


1. Exact, 3. Exact,

5. Not exact, 7.

9. Exact, . Ans.

11.

13. . Ans.

15.


3. 5.

7. 9.

11. 13.

15.

17.

19. Solution of

21.

. Thus, gives (4). We shall

see that this method extends to higher-order ODEs (Secs. 2.10 and 3.3).

23.

25.

27.

31.

33.

35.

. Ans. About 3 years

37.

39. . Constant solutions


1. 3.

5.

7. 9.

11.

13. .

Sketch or graph these curves.

15. . Trajectories . Now

. This agrees with the trajectory ODE

in u if . But these

are just the Cauchy–Riemann equations.

ux ϭ vy (equal denominators) and uy ϭ Ϫvx (equal numerators)

v ϭ c�, vx dx ϩ vy dy ϭ 0, yr ϭ Ϫvx>vy

y�r ϭ uy�>uxu ϭ c, ux dx ϩ uy dy ϭ 0, yr ϭ Ϫux>uy

yr ϭ Ϫ4x>9y. Trajectories y�r ϭ 9y�>4x, y� ϭ c�x9>4 (c� Ͼ 0)

y� ϭ c�x

yr ϭ Ϫ2xy, y�r ϭ 1>(2xy�), x ϭ c�ey�2

2y�2 Ϫ x2 ϭ c�y>x ϭ c, yr>x ϭ y>x2, yr ϭ y>x, y�r ϭ Ϫx>y�, y�2 ϩ x2 ϭ c�, circles

y Ϫ cosh (x Ϫ c) Ϫ c ϭ 0x2>(c2 ϩ 9) ϩ y2>c2 Ϫ 1 ϭ 0

y ϭ A>(ceAt ϩ B), y(0) Ͼ A>B if c Ͻ 0, y(0) Ͻ A>B if c Ͼ 0.(extinction).

yr Ͼ 0 if y Ͼ A>B (unlimited growth), yr Ͻ 0 if 0 Ͻ y Ͻ A>By ϭ A>B,

y ϭ 0,yr ϭ By2 Ϫ Ay ϭ By(y Ϫ A>B), A Ͼ 0, B Ͼ 0

yr ϭ y Ϫ y2 Ϫ 0.2y, y ϭ 1>(1.25 Ϫ 0.75e�0.8t), limit 0.8, limit 1

t ϭ (ln 3)>0.3889 ϭ 2.82

e�0.3889t ϭ (0.09 Ϫ 0.045)>0.135 ϭ 1>3,

y ϭ 0.135e�0.3889t ϩ 0.045 ϭ 0.18>2,

yr ϭ 175(0.0001 Ϫ y>450), y(0) ϭ 450 � 0.0004 ϭ 0.18,

yr ϭ A Ϫ ky, y(0) ϭ 0, y ϭ A(1 Ϫ e�kt)>kT ϭ 240ekt ϩ 60, T(10) ϭ 200, k ϭ Ϫ0.0539, t ϭ 102 min

dx>dy ϭ 6ey Ϫ 2x, x ϭ ce�2y ϩ 2ey

y ϭ 1>u, u ϭ ce�3.2x ϩ 10>3.2

y2 ϭ 1 ϩ 8e�x2

y ϭ uyhϭ r, ur ϭ r>y* ϭ re�p dx, u ϭ �e�p dx r dx ϩ c

ury* ϩ uy*r ϩ puy* ϭ ury* ϩ u(y*r ϩ py*) ϭ ury* ϩ u # 0y ϭ uy*, yr ϩ py ϭ

cy1r ϩ pcy1 ϭ c(yr1 ϩ py1) ϭ cr

(y1 ϩ y2)r ϩ p(y1 ϩ y2) ϭ (y1r ϩ py1) ϩ (y2r ϩ py2) ϭ r ϩ 0 ϭ r

(y1 ϩ y2)r ϩ p(y1 ϩ y2) ϭ (y1r ϩ py1) ϩ (y2r ϩ py2) ϭ 0 ϩ 0 ϭ 0

Separate. y Ϫ 2.5 ϭ c cosh4 1.5xy ϭ 2 ϩ c sin x

y ϭ (x Ϫ 2.5>e)ecos xy ϭ x2(c ϩ ex)

y ϭ (x ϩ c)eϪkxy ϭ cex Ϫ 5.2

b ϭ k, ax2 ϩ 2kxy ϩ ly2 ϭ c

ex Ϫ y ϩ ey ϭ cu ϭ ex ϩ k(y), uy ϭ kr ϭ Ϫ1 ϩ ey, k ϭ Ϫy ϩ ey

F ϭ sinh x, sinh2 x cos y ϭ c

e2x cos y ϭ 1u ϭ e2x cos y ϩ k(y), uy ϭ Ϫe2x sin y ϩ kr, kr ϭ 0

F ϭ ex2

, ex2

tan y ϭ cy ϭ 2x2 ϩ cx

y ϭ arccos (c>cos x)2x ϭ 2x, x2y ϭ c, y ϭ c>x2

App. 2 Answers to Odd-Numbered Problems A5

A6 App. 2 Answers to Odd-Numbered Problems


1. ; hence is continuous and is thus

bounded in the closed interval .

3. In ; just take b in large, namely,

5. R has sides 2a and 2b and center . In

and . Solution by , etc., .

7. .

9. No. At a common point they would both satisfy the “initial condition”

, violating uniqueness.

Chapter 1 Review Questions and Problems, page 43

11. 13.

15.

17.

19. 21.

23. 25.

27. [days]

29. . 43.7 days from


1. 3.

5.

7.

9. 11.

13. 15.

17. 19.


1. 3.

5. 7.

9. 11.

13. 15.

17. 19.

21. 23.

25. 27.

29. 31. Independent

33. , say.

Hence independent

35. Dependent since

37. y1 ϭ e�x, y2 ϭ 0.001ex ϩ e�x

sin 2x ϭ 2 sin x cos x

c1x2 ϩ c2x2 ln x ϭ 0 with x ϭ 1 gives c1 ϭ 0; then c2 ϭ 0 for x ϭ 2

y ϭ1

1p e�0.27x sin (1px)

y ϭ (4.5 Ϫ x)e�pxy ϭ 2e�x

y ϭ 6e2x ϩ 4e�3xy ϭ 4.6 cos 5x Ϫ 0.24 sin 5x

ys ϩ 4yr ϩ 5y ϭ 0ys ϩ 215yr ϩ 5y ϭ 0

y ϭ e�0.27x (A cos (1px) ϩ B sin (1px))y ϭ (c1 ϩ c2x)e5x>3y ϭ c1e�x>2 ϩ c2e3x>2y ϭ c1e�2.6x ϩ c2e0.8x

y ϭ c1 ϩ c2e�4.5xy ϭ (c1 ϩ c2x)e�px

y ϭ c1e�2.8x ϩ c2e�3.2xy ϭ c1e�2.5x ϩ c2e2.5x

y ϭ 15eϪx Ϫ sin xy ϭ Ϫ0.75x3>2 Ϫ 2.25x �1>2y ϭ 3 cos 2.5x Ϫ sin 2.5xy(t) ϭ c1e�t ϩ kt ϩ c2

y ϭ c1e2x ϩ c2y2 ϭ x3 ln x

(dz>dy)z ϭ Ϫz3 sin y, Ϫ1>z ϭ Ϫdx>dy ϭ cos y ϩ c�1, x ϭ Ϫsin y ϩ c1y ϩ c2

y ϭ (c1x ϩ c2)Ϫ1>2y ϭ c1eϪx ϩ c2F(x, z, zr) ϭ 0

ekt ϭ 0.5, ekt ϭ 0.01ek ϭ 0.9, 6.6 days

ek ϭ 1.25, (ln 2)>ln 1.25 ϭ 3.1, (ln 3)>ln 1.25 ϭ 4.9

3 sin x ϩ 13 sin y ϭ 0y ϭ sin (x ϩ 1

4p)

F ϭ x, x3ey ϩ x2y ϭ c25y2 Ϫ 4x2 ϭ c

y ϭ ceϪ2.5x ϩ 0.640x Ϫ 0.256

y ϭ ceϪx ϩ 0.01 cos 10x ϩ 0.1 sin 10x

y ϭ 1>(ceϪ4x ϩ 4)y ϭ ceϪ2x

y(x1) ϭ y1

(x1, y1)

ƒ1 ϩ y2 ƒ Ϲ K ϭ 1 ϩ b2, a ϭ b>K, da>db ϭ 0, b ϭ 1, a ϭ 12

y ϭ 1>(3 Ϫ 2x)dy>y2 ϭ 2 dxaopt ϭ b>K ϭ 18

da>db ϭ 0 gives b ϭ 1,f ϭ 2y2 Ϲ 2(b ϩ 1)2 ϭ K, a ϭ b>K ϭ b>(2(b ϩ 1)2),

R, (1, 1) since y(1) ϭ 1

b ϭ aK.a ϭ b>Kƒ x Ϫ x0 ƒ Ͻ a

ƒ x Ϫ x0 ƒ Ϲ a

0f>0y ϭ Ϫp(x)yr ϭ f (x, y) ϭ r(x) Ϫ p(x)y


1.

3.

5.

7.

9.

11.

15. Combine the two conditions to get .

The converse is simple.


1. . At integer t (if ), because of periodicity.

3. (i) Lower by a factor , (ii) higher by

5. 0.3183, 0.4775,

7. (tangential component of ),

9. where is the volume of the

water that causes the restoring force .

. Frequency .

13. ;

(ii)

15.

17. The positive solutions , that is, (max), (min). etc

19. from .


3. 5.

7. 9.

11.

13. 15.

17. 19.


3. 5. 7.

9.

11.

13.

15.


1. 3.

5. 7.

9.

11. y ϭ cos(13x) ϩ 6x2 Ϫ 4

y ϭ c1e4x ϩ c2eϪ4x ϩ 1.2xe4x Ϫ 2ex

y ϭ c1e�x>2 ϩ c2e�3x>2 ϩ 45 ex ϩ 6x Ϫ 16y ϭ (c1 ϩ c2x)e�2x ϩ 1

2 e�x sin x

y ϭ c1e�2x ϩ c2e�x ϩ 6x2 Ϫ 18x ϩ 21y ϭ c1e�x ϩ c2e�4x Ϫ 5e�3x

ys Ϫ 3.24y ϭ 0, W ϭ 1.8, y ϭ 14.2 cosh 1.8x ϩ 9.1 sinh 1.8x

ys ϩ 2yr ϭ 0, W ϭ Ϫ2e�2x, y ϭ 0.5(1 ϩ e�2x)

ys ϩ 5y ϩ 6.34 ϭ 0, W ϭ 0.3eϪ5x, 3eϪ2.5 cos 0.3x

ys ϩ 25y ϭ 0, W ϭ 5, y ϭ 3 cos 5x Ϫ sin 5x

W ϭ aW ϭ Ϫx4W ϭ Ϫ2.2e�3x

y ϭ Ϫ0.525x5 ϩ 0.625xϪ3y ϭ cos (ln x) ϩ sin (ln x)

y ϭ (3.6 ϩ 4.0 ln x)>xy ϭ xϪ3>2y ϭ x2(c1 cos (16 ln x) ϩ c2 sin (16 ln x))

y ϭ (c1 ϩ c2 ln x)x0.6y ϭ c1x2 ϩ c2x3

1x (c1 cos (ln x) ϩ c2 sin (ln x))y ϭ (c1 ϩ c2 ln x)x�1.8

exp (Ϫ10 � 3c>2m) ϭ 120.0231 ϭ (ln 2)>30 3kg>sec4

5p>4p>4of tan t ϭ 1

v* ϭ 3v02 Ϫ c2>(4m2)41>2 ϭ v031 Ϫ c2>(4mk)41>2 � v0(1 Ϫ c2>8mk) ϭ 2.9583

v0 ϭ Ϫ2, Ϫ32, Ϫ4

3, Ϫ54, Ϫ6

5

y ϭ [y0 ϩ (v0 ϩ ay0) t]eϪa˛t, y ϭ [1 ϩ (v0 ϩ 1)t]eϪt

v0>2p ϭ 0.4 3sec�14ys ϩ v02 y ϭ 0, v02 ϭ ag>m ϭ ag ϭ 0.000628gagy with g ϭ 9800 nt (ϭ weight>meter3)

m ϭ 1 kg, ay ϭ p # 0.012 # 2y meter3mys ϭ Ϫa�gy,

us ϩ v02u ϭ 0, v0>(2p) ϭ 1g>L >(2p)

W ϭ mgmLus ϭ Ϫmg sin u � Ϫmgu

1(k1 ϩ k2)>m>(2p) ϭ 0.5738

1212

v0 ϭ pyr ϭ y0 cos v0t ϩ (v0>v0) sin v0t

L(cy ϩ kw) ϭ L(cy) ϩ L(kw) ϭ cLy ϩ kLw

(D Ϫ 1.6I )(D Ϫ 2.4I ), y ϭ c1e1.6x ϩ c2e2.4x

(D Ϫ 2.1I )2, y ϭ (c1 ϩ c2x)e2.1x

(2D Ϫ I )(2D ϩ I ), y ϭ c1e0.5x ϩ c2e�0.5x

0, 5e2x, 0

0, 0, (D Ϫ 2I )(Ϫ4e�2x) ϭ 8e�2x ϩ 8e�2x

4e2x, Ϫe�x ϩ 8e2x, Ϫcos x Ϫ 2 sin x



13. 15.

17.


3.

5.

7.

9.

11.

13.

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 approaches the resonance frequency.


1.

3.

5.

7.

9. 11.

13.

15.

17.

19.


1.

3. 5.

7. 9.

11. 13.


7. 9.

11. 13.

15. 17.

19. 21.

23. I ϭ Ϫ0.01093 cos 415t ϩ 0.05273 sin 415t A

y ϭ Ϫ4x ϩ 2x3 ϩ 1>xy ϭ 5 cos 4x Ϫ 34 sin 4x ϩ ex

y ϭ (c1 ϩ c2x)e1.5x ϩ 0.25x2e1.5xy ϭ c1e2x ϩ c2e�x>2 Ϫ 3x ϩ x2

y ϭ c1x�4 ϩ c2x3y ϭ (c1 ϩ c2x)e0.8x

y ϭ eϪ3x(A cos 5x ϩ B sin 5x)y ϭ c1e�4.5x ϩ c2e�3.5x

y ϭ c1xϪ3 ϩ c2x3 ϩ 3x5y ϭ c1x2 ϩ c2x3 ϩ 1>(2x4)

y ϭ (c1 ϩ c2x)ex ϩ 4x7>2exy ϭ (c1 ϩ c2x)e2x ϩ x �2e2x

y ϭ A cos x ϩ B sin x ϩ 12x (cos x ϩ sin x)y ϭ c1x ϩ c2x2 Ϫ x sin x

y ϭ A cos 3x ϩ B sin 3x ϩ 19 (cos 3x) ln ƒ cos 3x ƒ ϩ 1

3 x sin 3x

R ϭ 2 �, L ϭ 1 H, C ϭ 112 F, E ϭ 4.4 sin 10t V

E(0) ϭ 600, Ir(0) ϭ 600, I ϭ e�3t(Ϫ100 cos 4t ϩ 75 sin 4t) ϩ 100 cos t

R Ͼ Rcrit ϭ 22L>C is Case I, etc.

I ϭ e�5t(A cos 10t ϩ B sin 10t) Ϫ 400 cos 25t ϩ 200 sin 25t A

I ϭ 5.5 cos 10t ϩ 16.5 sin 10t AI ϭ 0

I0 is maximum when S ϭ 0; thus, C ϭ 1>(v2L).

I ϭ 2(cos t Ϫ cos 20t)>399

LIr ϩ RI ϭ E, I ϭ (E>R) ϩ ce�Rt>L ϭ 4.8 ϩ ce�40t

I ϭ ce�t>(RC)RIr ϩ I>C ϭ 0,

v>(2p)

v

y ϭ e�t(0.4 cos t ϩ 0.8 sin t) ϩ eϪt>2(Ϫ0.4 cos 12 t ϩ 0.8 sin 12 t)

y ϭ 13 sin t Ϫ 1

15 sin 3t Ϫ 1105 sin 5t

y ϭ e�2t(A cos 2t ϩ B sin 2t) ϩ 14 sin 2t

y ϭ A cos t ϩ B sin t Ϫ (cos vt)>(v2 Ϫ 1)

y ϭ A cos 12t ϩ B sin 12t ϩ t (sin 12t Ϫ cos 12t)>(212)

y ϭ e�1.5t(A cos t ϩ B sin t) ϩ 0.8 cos t ϩ 0.4 sin t

yp ϭ 25 ϩ 43 cos 3t ϩ sin 3t

yp ϭ Ϫ1.28 cos 4.5t ϩ 0.36 sin 4.5t

yp ϭ 1.0625 cos 2t ϩ 3.1875 sin 2t

y ϭ e�0.1x (1.5 cos 0.5x Ϫ sin 0.5x) ϩ 2e0.5x

y ϭ ln xy ϭ ex>4 Ϫ 2ex>2 ϩ 15 e�x ϩ ex

25.

27.

29.


9. Linearly independent 11. Linearly independent

13. Linearly independent 15. Linearly dependent


1. 3.

5.

7.

9. 11.

13.


1.

3.

5.

7.

9. 11.

13.


7.

9.

11. 13.

15. 17.

19.


1. Yes

5.

7. 9.

11.

13.

15. (a) For example, gives , . (b) , 0.

(d) gives the critical case. C about 0.18506.a22 ϭ Ϫ4 ϩ 216.4 ϭ 1.05964

Ϫ2.4Ϫ0.000167Ϫ2.39993C ϭ 1000

y1r ϭ y2, y2r ϭ 24y1 Ϫ 2y2, y1 ϭ c1e4t ϩ c2e�6t ϭ y, y2 ϭ yr

y1r ϭ y2, y2r ϭ y1 ϩ ϩ 154 y2, y ϭ c1[1 4]T e4t ϩ c2 [1 Ϫ1

4 ]T e �t>4c1 ϭ 10, c2 ϭ 5c1 ϭ 1, c2 ϭ Ϫ5

y1r ϭ 0.02(Ϫy1 ϩ y2), y2r ϭ 0.02(y1 Ϫ 2y2 ϩ y3), y3r ϭ 0.02(y2 Ϫ y3)

y ϭ 4e�4x ϩ 5e�5x

y ϭ 2eϪ2x cos 4x ϩ 0.05 x Ϫ 0.06y ϭ c1x ϩ c2x1>2 ϩ c3x3>2 Ϫ 103

y ϭ (c1 ϩ c2x ϩ c3x2)e�2x ϩ x2 Ϫ 3x ϩ 3y ϭ (c1 ϩ c2x ϩ c3x2)e�1.5x

y ϭ c1 cosh 2x ϩ c2 sinh 2x ϩ c3 cos 2x ϩ c4 sin 2x ϩ cosh x

y ϭ c1 ϩ e�2x(A cos 3x ϩ B sin 3x)

y ϭ 2 Ϫ 2 sin x ϩ cos x

y ϭ eϪ3x(Ϫ1.4 cos x Ϫ sin x)y ϭ cos x ϩ 12 sin 4x

y ϭ (c1 ϩ c2x ϩ c3x2)e3x Ϫ 14 (cos 3x Ϫ sin 3x)

y ϭ c1x2 ϩ c2x ϩ c3xϪ1 Ϫ 112 xϪ2

y ϭ c1 cos x ϩ c2 sin x ϩ c3 cos 3x ϩ c4 sin 3x ϩ 0.1 sinh 2x

y ϭ (c1 ϩ c2x ϩ c3x2)e�x ϩ 18 ex Ϫ x ϩ 2

y ϭ e0.25x ϩ 4.3eϪ0.7x ϩ 12.1 cos 0.1x Ϫ 0.6 sin 0.1x

y ϭ cosh 5x Ϫ cos 4xy ϭ 4e�x ϩ 5e�x>2 cos 3x

y ϭ 2.398 ϩ e�1.6x(1.002 cos 1.5x Ϫ 1.998 sin 1.5x)

y ϭ A1 cos x ϩ B1 sin x ϩ A2 cos 3x ϩ B2 sin 3x

y ϭ c1 ϩ c2x ϩ c3 cos 2x ϩ c4 sin 2xy ϭ c1 ϩ c2 cos 5x ϩ c3 sin 5x

v ϭ 3.1 is close to v0 ϭ 2k>m ϭ 3, y ϭ 25(cos 3t Ϫ cos 3.1t).

RLC - circuit with R ϭ 20 �, L ϭ 4 H, C ϭ 0.1 F, E ϭ Ϫ25 cos 4t V

I ϭ 173 (50 sin 4t Ϫ 110 cos 4t) A



1.

3.

5.

7.

9.

11.

13.

15.

17.

, ,

. Note that .

19. ,


1. Unstable improper node, ,

3. Center, always stable, ,

5. Stable spiral, ,

7. Saddle point, always unstable, ,

9. Unstable node, ,

11. . Stable and attractive spirals

15. (was 0), , spiral point, unstable.

17. For instance, (a) , (b) , (c) , (d) , (e) 4.


5. Center at . At set . Then . Saddle point at .

7. , , , stable and attractive spiral point;

, , , , saddle point

9. saddle point, and centers

11. saddle points; centers.

Use .

13. centers; saddle points

15. By multiplication, . By integration,

, where .


3.

5. y2 ϭ c1e5t Ϫ 2c2e2t ϩ 1.12t ϩ 0.53y1 ϭ c1e5t ϩ c2e2t Ϫ 0.43t Ϫ 0.24,

y2 ϭ Ϫc1e�t ϩ c2et Ϫ e3ty1 ϭ c1e�t ϩ c2et,

c* ϭ 12 c2 Ϫ 8y2

2 ϭ 4y12 Ϫ 1

2 y14 ϩ c* ϭ 1

2 (c ϩ 4 Ϫ y12)(c Ϫ 4 ϩ y1

2)

y2y2r ϭ (4y1 Ϫ y13)y1r

y1 ϭ (2n ϩ 1)p ϩ y~1r, (p Ϯ 2np, 0)(Ϯ2np, 0)

Ϫcos (Ϯ12p ϩ y~1) ϭ sin (Ϯy~1) � Ϯy~1

(Ϫ12p Ϯ 2np, 0)(1

2p Ϯ 2np, 0)

(3, 0)(Ϫ3, 0)(0, 0)

y~2r ϭ Ϫy~1 Ϫ y~2y~1r ϭ Ϫy~1 Ϫ 3y~2y2 ϭ 2 ϩ y~2y1 ϭ Ϫ2 ϩ y~1

(Ϫ2, 2),y2r ϭ Ϫy1 Ϫ y2y1r ϭ Ϫy1 ϩ y2(0, 0)

(2, 0)y~2r ϭ y~1y1 ϭ 2 ϩ y~1(2, 0)(0, 0)

ϭ1ϭ Ϫ 12Ϫ1Ϫ2

¢Ͻ0p ϭ 0.2 � 0

y ϭ e�t (A cos t ϩ B sin t)

y2 ϭ 2c1e6t Ϫ 2c2e2ty1 ϭ c1e6t ϩ c2e2t

y2 ϭ Ϫc1e�t ϩ c2e3ty1 ϭ c1e�t ϩ c2e3t

y2 ϭ e�2t(B cos 2t Ϫ A sin 2t)y1 ϭ e�2t(A cos 2t ϩ B sin 2t)

y2 ϭ 3B cos 3t Ϫ 3A sin 3ty1 ϭ A cos 3t ϩ B sin 3t

y2 ϭ c2e2ty1 ϭ c1et

I2 ϭ Ϫ3c1e�t Ϫ c2e�3tI1 ϭ c1e�t ϩ 3c2e�3t

r 2 ϭ y12 ϩ y2

2 ϭ e�2t(A2 ϩ B2)y2 ϭ y1r ϩ y1 ϭ e�t(B cos t Ϫ A sin t)

y1 ϭ e�t(A cos t ϩ B sin t)y1s ϩ 2y1r ϩ 2y1 ϭ 0

y2 ϭ y1r ϩ y1, y2r ϭ y1s ϩ y1r ϭ Ϫy1 Ϫ y2 ϭ Ϫy1 Ϫ (y1r ϩ y1),

y2 ϭ 12 et

y1 ϭ 12 et

y1 ϭ 2 sinh t, y2 ϭ 2 cosh t

y2 ϭ 4et Ϫ 4e�t>2y1 ϭ Ϫ20et ϩ 8e�t>2y3 ϭ c1e�18t Ϫ 2c2e9t Ϫ 1

2 c3e18ty2 ϭ c1e�18t ϩ c2e9t ϩ c3e18t

y1 ϭ 12 c1e�18t ϩ 2c2e9t Ϫ c3e18t

y3 ϭ c2 cos 12t Ϫ c3 sin 12t ϩ c1

y2 ϭ c212 sin 12t ϩ c312 cos 12t

y1 ϭ Ϫc2 cos 12t ϩ c3 sin 12t ϩ c1

y2 ϭ Ϫ2c1 ϩ 5c2e14.5t

y1 ϭ 5c1 ϩ 2c2e14.5t

y1 ϭ 2c1e2t ϩ 2c2, y2 ϭ c1e2t Ϫ c2

y1 ϭ c1e�2t ϩ c2e2t, y2 ϭ Ϫ3c1e�2t ϩ c2e2t


7. ,

9. The formula for v shows that these various choices differ by multiples of the eigen-

vector for , which can be absorbed into, or taken out of, in the general

solution .

11. ,

13. ,

15. ,

17. ,

19. ,


11. , . Saddle point

13. ;

asymptotically stable spiral point

15. , . Stable node

17. , . Stable and attractive

spiral point

19. Unstable spiral point

21.

23.

25.

27. saddle point; centers

29. center when n is even and saddle point when n is odd


3.

5.

7.

9.

11.

13.

15.

17.

19. ; but is too large to give good

values. Exact:


5. ,

P7(x) ϭ 116 (429x7 Ϫ 693x5 ϩ 315x3 Ϫ 35x)

P6(x) ϭ 116 (231x6 Ϫ 315x4 ϩ 105x2 Ϫ 5)

y ϭ (x Ϫ 2)2exx ϭ 2s ϭ 4 Ϫ x2 Ϫ 1

3 x3 ϩ 130 x5, s˛(2) ϭ Ϫ8

5

s ϭ 1 ϩ x Ϫ x2 Ϫ 56 x3 ϩ 2

3 x4 ϩ 1124 x5, s˛(1

2) ϭ 923768

a�

mϭ1

(m ϩ 1)(m ϩ 2)

(m ϩ 1)2 ϩ 1xm, a

�

mϭ5

(m Ϫ 4)2

(m Ϫ 3)!xm

a0(1 Ϫ 12 x2 Ϫ 1

24 x4 ϩ 13720 x6 ϩ Á ) ϩ a1(x Ϫ 1

6 x3 Ϫ 124 x5 ϩ 5

1008 x7 ϩ Á )

a0(1 Ϫ 112 x4 Ϫ 1

60 x5 Ϫ Á ) ϩ a1(x ϩ 12 x2 ϩ 1

6 x3 ϩ 124 x4 Ϫ 1

24 x5 Ϫ Á )

y ϭ a0 ϩ a1x Ϫ 12 a0x2 Ϫ 1

6 a1x3 ϩ Á ϭ a0 cos x ϩ a1 sin x

y ϭ a0(1 Ϫ x2 ϩ x4>2! Ϫ x6>3! ϩ Ϫ Á ) ϭ a0e�x2

23>22 ƒ k ƒ

(np, 0)

(Ϫ1, 0), (1, 0)(0, 0)

I2 ϭ (Ϫ6 Ϫ 32.5t)e�5t ϩ 6 cos t ϩ 2.5 sin t

I1 ϭ (19 ϩ 32.5t)e�5t Ϫ 19 cos t ϩ 62.5 sin t,

2.5(I2r Ϫ I1r) ϩ 25I2 ϭ 0,I1r ϩ 2.5(I1 Ϫ I2) ϭ 169 sin t,

y2 ϭ Ϫc1e�t ϩ c2e3ty1 ϭ 2c1e�t ϩ 2c2e3t ϩ cos t Ϫ sin t,

y2 ϭ Ϫc1e�4t ϩ c2e4t Ϫ 4ty1 ϭ c1e�4t ϩ c2e4t Ϫ 1 Ϫ 8t 2,

y2 ϭ e�t(B cos 2t Ϫ A sin 2t)y1 ϭ e�t(A cos 2t ϩ B sin 2t)

y2 ϭ c1e�5t Ϫ c2e�ty1 ϭ c1e�5t ϩ c2e�t

y1 ϭ e�4t(A cos t ϩ B sin t), y2 ϭ 15 e�4t[(B Ϫ 2A) cos t Ϫ (A ϩ 2B) sin t]

y2 ϭ 2c1e4t Ϫ 2c2e�4ty1 ϭ c1e4t ϩ c2e�4t

c2 ϭ Ϫ67.948c1 ϭ 17.948

l1 ϭ Ϫ0.9 ϩ 10.41, l2 ϭ Ϫ0.9 Ϫ 10.41

I2 ϭ (1.1 ϩ 10.41)c1el1t ϩ (1.1 Ϫ 10.41)c2el2t,

I1 ϭ 2c1el1t ϩ 2c2el2t ϩ 100

y2 ϭ Ϫ4e�t ϩ ty1 ϭ 4e�t Ϫ 4et ϩ e2t

y2 ϭ 2 cos 2t Ϫ 2 sin 2t ϩ sin ty1 ϭ cos 2t ϩ sin 2t ϩ 4 cos t

y2 ϭ Ϫ83 sinh t Ϫ 4

3 cosh t ϩ 43 e2ty1 ϭ Ϫ8

3 cosh t Ϫ 43 sinh t ϩ 11

3 e2t

y(h)c1l ϭ Ϫ2

y2 ϭ Ϫc1et Ϫ 5c2e2t ϩ 5t ϩ 7.5 ϩ e�ty1 ϭ c1et ϩ 4c2e2t Ϫ 3t Ϫ 4 Ϫ 2e�t


11. Set .

15. , , ,


3. ,

5.

7.

9. ,

11. ,

13. ,

15.

17.

19.


3.

5.

7.

9.

13. implies and

somewhere between and by Rolle’s theorem.

Now use (21b) to get there. Conversely, ,

thus implies in between by Rolle’s

theorem and (21a) with .

15. By Rolle, at least once between two zeros of . Use by (21b)

with . Together at least once between two zeros of . Also use

by (21a) with and Rolle.

19. Use (21b) with (21a) with (21d) with respectively.

21. Integrate (21a).

23. Use (21a) with partial integration, (21b) with partial integration.

25. Use (21d) to get


1.

3.

5. c1J0(1x) ϩ c2Y0(1x)

c1J2>3(x2) ϩ c2Y2>3(x2)

c1J4(x) ϩ c2Y4(x)

ϭ Ϫ2J4(x) Ϫ 2J2(x) Ϫ J0(x) ϩ c.

�J5(x)dx ϭ Ϫ2J4(x) ϩ�J3(x)dx ϭ Ϫ2J4(x) Ϫ 2J2(x) ϩ �J1(x)dx

� ϭ 0,� ϭ 1,

� ϭ 2,� ϭ 1,� ϭ 0,

� ϭ 1(xJ1)r ϭ xJ0

J0J1 ϭ 0� ϭ 0

Jr0 ϭ ϪJ1J0Jr0 ϭ 0

� ϭ n ϩ 1

Jn(x) ϭ 0x3nϩ1Jnϩ1(x3) ϭ x4

nϩ1Jnϩ1(x4) ϭ 0

Jnϩ1(x3) ϭ Jnϩ1(x4) ϭ 0Jnϩ1(x) ϭ 0

x2x1[x�nJn(x)]r ϭ 0

x1�nJn(x1) ϭ x2

�nJn(x2) ϭ 0Jn(x1) ϭ Jn(x2) ϭ 0

x��(c1J�(x) ϩ c2J��(x)), � � 0, Ϯ1, Ϯ2, Ác1J1>2(1

2 x) ϩ c2J�1>2(12 x) ϭ x�1>2(c�1 sin 12 x ϩ c�2 cos 12 x)

c1J�(lx) ϩ c2J��(lx), � � 0, Ϯ1, Ϯ2, Ác1J0(1x)

y ϭ c1F(2, Ϫ2, Ϫ12; t Ϫ 2) ϩ c2(t Ϫ 2)3>2F(7

2, Ϫ12, 52; t Ϫ 2)

y ϭ A(1 Ϫ 8x ϩ 325 x2) ϩ Bx3>4F(7

4, Ϫ54, 74; x)

y ϭ AF(1, 1, Ϫ12; x) ϩ Bx3>2F(5

2, 52, 52; x)

y2 ϭ ex ln xy1 ϭ ex

y2 ϭ ex>xy1 ϭ ex

y2 ϭ 1 ϩ xy11x

1120 x5 Ϫ 1

120 x6 ϩ ÁϪ 112 x4 ϩy2 ϭ x ϩ 1

6 x3

124 x4 Ϫ 1

30 x5 ϩ 1144 x6 Ϫ Á ,y1 ϭ 1 ϩ 1

2 x2 Ϫ 16 x3 ϩ

b0 ϭ 1, c0 ϭ 0, r 2 ϭ 0, y1 ϭ e�x, y2 ϭ e�x ln x

y2 ϭ1x

Ϫx

2!ϩ

x3

4!Ϫ ϩ Á ϭ

cos xx

y1 ϭ 1 Ϫx2

3!ϩ

x4

5!Ϫ ϩ Á ϭ

sin xx

P42 ϭ (1 Ϫ x2)(105x2 Ϫ15)>2

P22 ϭ 3(1 Ϫ x2)P2

1 ϭ 3x21 Ϫ x2P11 ϭ 21 Ϫ x2

y ϭ c1Pn(x>a) ϩ c2Qn(x>a)x ϭ az


7.

9.

11. Set and use (10).

13. Use (20) in Sec. 5.4.


11.

13. ; Euler–Cauchy with instead of x

15.

17.

19.


1. 3.

5. 7.

9. 11.

13. 15.

19. Use .

23. Set .

25. 27.

29. 31.

33. 35.

37. 39.

41. 43.

45.


1.

3.

5.

7. 9.

11.

13. t ϭ t� Ϫ 1, Y�

ϭ 4>(s Ϫ 6), y� ϭ 4e6t, y ϭ 4e6(tϩ1)

y ϭ (1 ϩ t)e�1.5t ϩ 4t 3 Ϫ 16t 2 ϩ 32t

24>s4 Ϫ 32>s3 ϩ 32>s2,Y ϭ 1>(s ϩ 1.5) ϩ 1>(s ϩ 1.5) 2 ϩ

(s ϩ 1.5)2Y ϭ s ϩ 31.5 ϩ 3 ϩ 54>s4 ϩ 64>s,

y ϭ et Ϫ e3t ϩ 2ty ϭ 12 e3t ϩ 5

2 e�4t ϩ 12 e�3t

(s2 Ϫ 14)Y ϭ 12s, y ϭ 12 cosh 12 t

y ϭ 10e3t ϩ e�2t(s Ϫ 3)(s ϩ 2) ϭ 11s ϩ 28 Ϫ 11 ϭ 11s ϩ 17, Y ϭ 10>(s Ϫ 3) ϩ 1>(s ϩ 2),

y ϭ 1.25e�5.2t Ϫ 1.25 cos 2t ϩ 3.25 sin 2t

(k0 ϩ k1t)e�at

e3t(2 cos 3t ϩ 53 sin 3t)e�5pt sinh pt

72 t 3e�t22pte�pt

0.5 # 2p

(s ϩ 4.5)2 ϩ 4p2

2

(s ϩ 3)3

l�1a 4

s Ϫ 2Ϫ

3

s ϩ 1b ϭ 4e2t Ϫ 3e�t2t 3 Ϫ 1.9t 5

1

L2 cos

npt

L0.2 cos 1.8t ϩ sin 1.8t

�ϱ

0

e�(s>c)pf (p) dp>c ϭ F(s>c)>cct ϭ p. Then l( f (ct)) ϭ �ϱ

0

e�stf (ct) dt ϭ

eat ϭ cosh at ϩ sinh at

e�s Ϫ 1

2s2Ϫ

e�s

2sϩ

1

s

(1 Ϫ eϪs)2

s

1 Ϫ e�bs

s2Ϫ

be�bs

s

1

sϩ

e�s Ϫ 1

s2

(v cos u ϩ s sin u)>(s2 ϩ v2)1>((s Ϫ 2)2 Ϫ 1)

s>(s2 ϩ p2)3>s2 ϩ 12>s

1x J1(1x), 1xY1(1x)

ex, 1 ϩ x

J25 (x), J�25(x)

x Ϫ 1(x Ϫ 1)�5, (x Ϫ 1)7cos 2x, sin 2x

H (1) ϭ kH (2)

x3(c1J3(x) ϩ c2Y3(x))

1x (c1J1>4(12 kx2) ϩ c2Y1>4(1

2 kx2))


15.

17. 19.

21.

23. 25.

27. 29.


3.

5.

7.

9.

11. 13.

15. 17.

19. 21.

23.

25.

27.

29.

31.

33.

35.

37.

39.


3.

5.

7.

9.

e�2tϩ30(cos (t Ϫ 10) Ϫ 7 sin (t Ϫ 10))40.1u(t Ϫ 10)3Ϫe�t ϩy ϭ 0.1[et ϩ e�2t(Ϫcos t ϩ 7 sin t)] ϩ

y ϭ e�t ϩ 4e�3t sin 12 t ϩ 12 u(t Ϫ 1

2)e�3(t�1>2) sin (12 t Ϫ 1

4)

sin t (0 Ͻ t Ͻ p); 0 (p Ͻ t Ͻ 2p); Ϫsin t (t Ͼ 2p)

y ϭ 8 cos 2t ϩ 12 u(t Ϫ p) sin 2t

i ϭ 1000e�t sin t Ϫ 1000u(t Ϫ 2)e�tϩ2 sin (t Ϫ 2)

ir ϩ 2i ϩ 2�t

0

i(t) dt ϭ 1000(1 Ϫ u(t Ϫ 2)), I ϭ 1000(1 Ϫ e�2s)>(s2 ϩ 2s ϩ 2),

i ϭ 4 cos t Ϫ 4 cos 240t Ϫ 4u(t Ϫ p)[cos t ϩ cos (140 (t Ϫ p))]

(0.5s2 ϩ 20)I ϭ 78s(1 ϩ e�ps)>(s2 ϩ 1),

i ϭ (10 sin 10t ϩ 100 sin t)(u(t Ϫ p) Ϫ u(t Ϫ 3p))

if t Ͼ 21 Ϫ e�10(t�2)

10I ϩ100

sI ϭ

100

s2e�2s, I ϭ e�2sa1

sϪ

1

s ϩ 10b, i ϭ 0 if t Ͻ 2 and

R(sQ Ϫ CV0) ϩ Q>C ϭ 0, q ϭ CV0e�t>(RC)Rqr ϩ q>C ϭ 0, Q ϭ l(q), q(0) ϭ CV0, i ϭ qr(t),

ϩ 20u(t Ϫ 1)[Ϫe�5t ϩ e�250tϩ245]i ϭ 20(e�5t Ϫ e�250t)

0.1ir ϩ 25i ϭ 490e�5t[1 Ϫ u(t Ϫ 1)],

ϩ 49 cos (2t Ϫ 10) ϩ 10 sin (2t Ϫ 10) if t Ͼ 5cos 2t

t ϭ 1 ϩ t�, y�s ϩ 4y� ϭ 8(1 ϩ t�)2(1 Ϫ u( t� Ϫ 4)), cos 2t ϩ 2t 2 Ϫ 1 if t Ͻ 5,

t Ϫ sin t (0 Ͻ t Ͻ 1), cos (t Ϫ 1) ϩ sin (t Ϫ 1) Ϫ sin t (t Ͼ 1)

et Ϫ sin t (0 Ͻ t Ͻ 2p), et Ϫ 12 sin 2t (t Ͼ 2p)

sin 3t ϩ sin t (0 Ͻ t Ͻ p); 43 sin 3t (t Ͼ p)13(et Ϫ 1)3e�5t

e�t cos t (0 Ͻ t Ͻ 2p)(t Ϫ 3)3u(t Ϫ 3)>62[1 ϩ u(t Ϫ p)] sin 3t(se�ps>2 ϩ e�ps)>(s2 ϩ 1)

e�3s>2 a 2

s3ϩ

3

s2ϩ

94

sb

1

s ϩ p(e�2(sϩp) Ϫ e�4(sϩp))

aeta1 Ϫ uat Ϫ1

2pbbb ϭ

1

s Ϫ 1 (1 Ϫ e�ps>2ϩp>2)

l((t Ϫ 2)u(t Ϫ 2)) ϭ e�2s>s2

1

a2(e�at Ϫ 1) ϩ

t

a19 (1 ϩ t Ϫ cos 3t Ϫ 1

3 sin 3t)

(1 Ϫ cos vt)>v212(1 Ϫ e�t>4)

l( f r) ϭ l(sinh 2t) ϭ sl( f ) Ϫ 1. Answer: (s2 Ϫ 2)>(s3 Ϫ 4s)

2v2

s(s2 ϩ 4v2)

1

(s ϩ a)2

t ϭ t� ϩ 1.5, (s Ϫ 1)(s ϩ 4)Y�

ϭ 4s ϩ 17 ϩ 6>(s Ϫ 2), y ϭ 3et�1.5 ϩ e2(t�1.5)


11.

15.


1. t 3.

5. 7.

9. 11.

13.

17. 19.

21. 23.

25.


3. 5.

7. 9.

11. 15.

17.

19.


3.

5.

7.

9.

11.

13.

15.

19.


11. 13.

15. 17. Sec. 6.6; 2s2>(s2 ϩ 1)2e�3sϩ3>2>(s Ϫ 12)

12(1 Ϫ cos pt), p2>(2s3 ϩ 2p2s)

5s

s2 Ϫ 4Ϫ

3

s2 Ϫ 1

i2 ϭ Ϫ26e�2t ϩ 8e�8t ϩ 18 cos t ϩ 12 sin t

i1 ϭ Ϫ26e�2t Ϫ 16e�8t ϩ 42 cos t ϩ 15 sin t,

4i1 ϩ 8(i1 Ϫ i2) ϩ 2i1r ϭ 390 cos t, 8i2 ϩ 8(i2 Ϫ i1) ϩ 4i2r ϭ 0,

y1 ϭ et, y2 ϭ e�t, y3 ϭ et Ϫ e�t

y1 ϭ Ϫ4et ϩ sin 10t ϩ 4 cos t, y2 ϭ 4et Ϫ sin 10t ϩ 4 cos t

y1 ϭ et ϩ e2t, y2 ϭ e2t

y1 ϭ (3 ϩ 4t)e3t, y2 ϭ (1 Ϫ 4t)e3t

y2 ϭ Ϫe�2t ϩ et ϩ 13 u(t Ϫ 1)(Ϫe3�2t ϩ et)

y1 ϭ Ϫe�2t ϩ 4et ϩ 13 u(t Ϫ 1)(Ϫe3�2t ϩ et),

y2 ϭ cos t ϩ sin t Ϫ 1 ϩ u(t Ϫ 1)[1 Ϫ cos (t Ϫ 1) Ϫ sin (t Ϫ 1)]

y1 ϭ Ϫcos t ϩ sin t ϩ 1 ϩ u(t Ϫ 1)[Ϫ1 ϩ cos (t Ϫ 1) Ϫ sin (t Ϫ 1)]

y1 ϭ Ϫe�5t ϩ 4e2t, y2 ϭ e�5t ϩ 3e2t

3ln (s2 ϩ 1) Ϫ 2 ln (s Ϫ 1)4r ϭ 2s>(s2 ϩ 1) Ϫ 2>(s Ϫ 1); 2(Ϫcos t ϩ et)>tln s Ϫ ln (s Ϫ 1); (Ϫ1 ϩ et)>t

F(s) ϭ Ϫ1

2a 1

s2 Ϫ 9br, f (t) ϭ 1

6 t sinh 3t4s2 Ϫ p2

(s2 ϩ 14p

2)2

p(3s2 Ϫ p2)

(s2 ϩ p2)3

2s3 ϩ 24s

(s2 Ϫ 4)3

s2 Ϫ v2

(s2 ϩ v2)2

12

(s ϩ 3)2

1.5t sin 6t

4.5(cosh 3t Ϫ 1)(vt Ϫ sin vt)>v2

t sin pte4t Ϫ e�1.5t

y(t) ϩ 2�t

0

et�t y(t) dt ϭ tet, y ϭ sinh t

y ϭ cos ty Ϫ 1 * y ϭ 1, y ϭ et

et Ϫ t Ϫ 112 t sin vt

(et Ϫ e�t)>2 ϭ sinh t

ke�ps>(s Ϫ se�ps) (s Ͼ 0)

u(t Ϫ 2)(e�2(t�2) Ϫ e�3(t�2))

y ϭ Ϫe�3t ϩ e�2t ϩ 16 u(t Ϫ 1)(1 Ϫ 3e�2(t�1) ϩ 2e�3(t�1)) ϩ


19. 21.

23. 25.

27. 29.

31.

33.

35.

37.

39.

41.

43.

45.


3.

5.

7. No, no, yes, no, no

9. , undefined

11. , same, , same

13. , same, , undefined

15. , same, undefined, undefined 17.


5. 10,

7. 0, I, c10

0

0d , c1

0

1

0d

n(n ϩ 1)>2

D Ϫ4.5

Ϫ27.0

9.0

TD 5.5

33.0

Ϫ11.0

TϪDD 70

Ϫ28

14

28

56

0

TD 5.4

Ϫ4.2

Ϫ0.6

0.6

2.4

0.6

TD 0

34

28

26

32

Ϫ10

TD 0

18

3

6

15

0

12

15

Ϫ9

T , D 0

2.5

Ϫ1

2.5

1.5

2

1

2

Ϫ1

T , D 0

20.5

2

8.5

16.5

2

13

17

Ϫ10

TB ϭ 1

5 A, 110 A

3 � 3, 3 � 4, 3 � 6, 2 � 2, 2 � 3, 3 � 2

i1 ϭ Ϫ8e�2t ϩ 5e�0.8t ϩ 3, i2 ϭ Ϫ4e�2t ϩ 4e�0.8t5i1r ϩ 20(i1 Ϫ i2) ϭ 60, 30i2r ϩ 20(i2r Ϫ i1r) ϩ 20i2 ϭ 0,

i(t) ϭ e�4t( 326 cos 3t Ϫ 10

39 sin 3t) Ϫ 326 cos 10t ϩ 8

65 sin 10t

1 Ϫ e�t (0 Ͻ t Ͻ 4), (e4 Ϫ 1)e�t (t Ͼ 4)

y1 ϭ (1>110) sin 110t, y2 ϭ Ϫ(1>110) sin 110t

y2 ϭ Ϫsin t Ϫ 2u(t Ϫ p) cos2 12 t ϩ u(t Ϫ 2p) sin t

y1 ϭ cos t Ϫ u(t Ϫ p) sin t ϩ 2u(t Ϫ 2p) sin2 12 t,

y1 ϭ 4et Ϫ e�2t, y2 ϭ et Ϫ e�2t

0 (0 Ϲ t Ϲ 2), 1 Ϫ 2e�(t�2) ϩ e�2(t�2) (t Ͼ 2)

e�t ϩ u(t Ϫ p)[1.2 cos t Ϫ 3.6 sin t ϩ 2e�tϩp Ϫ 0.8e2t�2p]

y ϭ e�2t(13 cos t ϩ 11 sin t) ϩ 10t Ϫ 8e�2t(3 cos t Ϫ 2 sin t)

3t 2 ϩ t 3sin (vt ϩ u)

tu(t Ϫ 1)12>(s2(s ϩ 3))


11. , same, , same

13. , undefined,

15. Undefined, , same

17. , undefined, , undefined

19. Undefined, , same

25. (d) etc.

(e) Answer. If

29.


1. 3.

5. 7. arb.,

9. arb.

11. arb., arb.

13. 17.

19.

21. No

23. , thus


1. 3.

5.

[0 0 1]}

3; {[2 Ϫ1 4], [0 1 Ϫ46], [0 0 1]}; {[2 0 1], [0 3 23],

3; {[3 5 0], [0 3 5], [0 0 1]}1; [2 Ϫ1 3]; [2 Ϫ1]T

C3H8 ϩ 5O2 : 3CO2 ϩ 4H2O

C: 3x1 Ϫ x3 ϭ 0, H: 8x1 Ϫ 2x4 ϭ 0, O: 2x2 Ϫ 2x3 Ϫ x4 ϭ 0

x2 ϭ 1600 Ϫ x1, x3 ϭ 600 ϩ x1, x4 ϭ 1000 Ϫ x1.

I1 ϭ (R1 ϩ R2)E0>(R1R2) A, I2 ϭ E0>R1 A, I3 ϭ E0>R2 A

I1 ϭ 2, I2 ϭ 6, I3 ϭ 8w ϭ 4, x ϭ 0, y ϭ 2, z ϭ 6

y ϭ 2t2 Ϫ t1, z ϭ t2w ϭ 1, x ϭ t1

x ϭ 3t Ϫ 1, y ϭ Ϫt ϩ 4, z ϭ t

z ϭ 2tx ϭ Ϫ3t, y ϭ tx ϭ 6, y ϭ Ϫ7

x ϭ 1, y ϭ 3, z ϭ Ϫ5x ϭ Ϫ2, y ϭ 0.5

p ϭ [85 62 30]T, v ϭ [44,920 30,940]T

AB ϭ ϪBA.

AB ϭ (AB)T ϭ BTAT ϭ BA;

D 10.5

0

Ϫ3

T , D 7

Ϫ3

1

TD 22

4

Ϫ12

TDϪ30

45

5

Ϫ18

9

Ϫ7

TD 8

Ϫ4

Ϫ3

T , [7 Ϫ1 3]

cϪ9

Ϫ5

3

Ϫ1

4

0dD12

0

2

13

Ϫ6

0

Ϫ6

4

T , DϪ9

3

4

Ϫ5

Ϫ1

0

TD 10

Ϫ14

Ϫ2

Ϫ5

7

Ϫ4

Ϫ15

Ϫ33

Ϫ4

TD 10

Ϫ5

Ϫ5

Ϫ14

7

Ϫ1

Ϫ6

Ϫ12

Ϫ4

TApp. 2 Answers to Odd-Numbered Problems A17

7.

9.

11. (c) 1 17. No

19. Yes 21. No

23. Yes 25. Yes

27.

29. No 31. No

33. 1, solution of the given system , basis

35.


7. 9. 1

11. 40 13. 289

15. 17. 2

19. 2 21.

23. 25.


1. 3.

5. 7.

9. 11.

15. . Multiply by A from the right.


1.

3. 5. No

7. Dimension 2, basis 9. 3; basis

11.

13. x1 ϭ 2y1 Ϫ 3y2, x2 ϭ Ϫ10y1 ϩ 16y2 ϩ y3, x3 ϭ Ϫ7y1 ϩ 11y2 ϩ y3

x1 ϭ 5y1 Ϫ y2, x2 ϭ 3y1 Ϫ y2

c10

0

Ϫ1d , c0

0

1

0d , c0

1

0

0dxe�x, e�x

1, [1 11 Ϫ7]T

[1 0]T, [0 1]T; [1 0]T, [0 Ϫ1]T; [1 1]T, [Ϫ1 1]T

AA�1 ϭ I, (AA�1)�1 ϭ (A�1)�1A�1 ϭ I

(A2)�1 ϭ (A�1)2 ϭ c3.760

2.400

22.272

15.280dD018

0

0

0

14

12

0

0

TA�1 ϭ AD 1

Ϫ2

3

0

1

Ϫ4

0

0

1

TD 54

2

Ϫ30

0.9

0.2

Ϫ0.5

Ϫ3.4

Ϫ0.2

2

Tc1.20

0.50

4.64

3.60d

w ϭ 3, x ϭ 0, y ϭ 2, z ϭ Ϫ2x ϭ 0, y ϭ 4, z ϭ Ϫ1

x ϭ 3.5, y ϭ Ϫ1.0

Ϫ64

cos (a ϩ b)

1, [4 2 43 1]

[1 103 3]c[1 10

3 3]

2, [Ϫ2 0 1], [0 2 1]

3; [9 0 1 0], [0 9 8 9], [0 0 1 0]

2; [8 0 4 0], [0 2 0 4]; [8 0 4], [0 2 0]


15. 17.

19. 1 21.

23.

25.


11.

13.

15. 197, 0

17.

19. 21.

23. arb. 25.

27. 29. Ranks 2, 2,

31. Ranks 2, 2, 1 33.

35.


1. 3, 3.

5.

7.

9.

11.

13. defect 2

15.

17. Eigenvalues i, . Corresponding eigenvectors are complex,

indicating that no direction is preserved under a rotation.

19. A point onto the -axis goes onto itself,

a point on the -axis onto the origin.

23. Use that real entries imply real coefficients of the characteristic polynomial.

x1

x2c0 0

0 1d ; 1, c0

1d ; 0, c1

0d .

Ϫic0 Ϫ1

1 0d .

Ϫ5, [Ϫ3 Ϫ3 1 1]T, 3, [3 Ϫ3 1 Ϫ1]T

(l ϩ 1)2(l2 ϩ 2l Ϫ 15); Ϫ1, [1 0 0 0]T, [0 1 0 0]T;

Ϫ(l Ϫ 9)3; 9, [2 Ϫ2 1]T,

6, [1 2 2]T; 9, [2 1 Ϫ2]T

Ϫ(l3 Ϫ 18l2 ϩ 99l Ϫ 162)>(l Ϫ 3) ϭ Ϫ(l2 Ϫ 15l ϩ 54); 3, [2 Ϫ2 1]T;

0.8 ϩ 0.6i, [1 Ϫi]T; 0.8 Ϫ 0.6i, [1 i]T

l2 ϭ 0, [1 0]T

Ϫ3i, [1 Ϫi]; 3i, [1 i], i ϭ 1Ϫ1

Ϫ4, [2 9]T; 3, [1 1]T[1 0]T; Ϫ0.6, [0 1]T

I1 ϭ 4 A, I2 ϭ 5 A, I3 ϭ 1 A

I1 ϭ 16.5 A, I2 ϭ 11 A, I3 ϭ 5.5 A

ϱx ϭ 10, y ϭ Ϫ2

x ϭ 0.4, y ϭ Ϫ1.3, z ϭ 1.7x ϭ 6, y ϭ 2t ϩ 2, z ϭ t

x ϭ 4, y ϭ Ϫ2, z ϭ 8D Ϫ2

Ϫ12

Ϫ12

Ϫ12

16

Ϫ9

Ϫ12

Ϫ9

Ϫ14

TϪ5, det A2 ϭ (det A)2 ϭ 25, 0

[21 Ϫ8 Ϫ31]T, [21 Ϫ8 31]

D Ϫ1

Ϫ18

Ϫ13

6

8

Ϫ2

1

Ϫ7

Ϫ7

T , D 1

Ϫ6

Ϫ1

18

Ϫ8

7

13

2

7

Ta ϭ [5 3 2]T, b ϭ [3 2 Ϫ1]T, 90 ϩ 14 ϭ 2(38 ϩ 14)

a ϭ [3 1 Ϫ4]T, b ϭ [Ϫ4 8 Ϫ1]T, �a ϩ b � ϭ 2107 Ϲ 5.099 ϩ 9

k ϭ Ϫ20

25226



1. 1.5,

3. 1,

5. 0.5, directions and

7.

9.

11. 1.8

13.

15.

17. .

19. From and Prob. 18 follows and

. Adding on both sides, we see that

has the eigenvalue . From this the statement follows.


1.

3. !

5.

7.

9.

15. No 17.

19. No since .


1.

3.

5.

9.

11. cϪ2

3

1

Ϫ1d A c1

3

1

2d ϭ c2

0

0

Ϫ5d

c 15

Ϫ25

25

15

d A c12

Ϫ2

1d ϭ c5

0

0

0d

D400

3

Ϫ5

Ϫ5

Ϫ9

15

15

T , 0, D031

T ; 4, D100

T ; 10, DϪ1

1

1

T ; x ϭ D301

T , D010

T , D 1

Ϫ1

1

Tc3.008

5.456

Ϫ0.544

6.992d , 4, cϪ17

31d ; 6, cϪ2

11d ; x ϭ c25

25d , c10

5d

cϪ25

Ϫ50

12

25d , Ϫ5, c3

5d ; 5, c2

5d ; x ϭ cϪ2

4d , c2

1d

det A ϭ det (AT) ϭ det (ϪA) ϭ (Ϫ1)3det (A) ϭ Ϫdet (A) ϭ 0

A�1 ϭ (ϪAT)�1 ϭ Ϫ(A�1)T

1, [0 1 0]T; i, [1 0 i]T; Ϫi, [1 0 Ϫi]T, orthogonal

0, Ϯ25i, skew–symmetric

1, [0 2 1]T; 6, [1 0 0]T, [0 1 Ϫ2]T; symmetric

2 Ϯ 0.8i, [1 Ϯi]. Not skew–symmetric

0.8 Ϯ 0.6i, [1 Ϯi]T; orthogonal

kpljp ϩ kqlj

qkpAp ϩ kqAqkqAqxj ϭ kqlj

qxj (p м 0, q м 0, integer)

kpApxj ϭ kpljpxjAxj ϭ ljxj (xj � 0)

Axj ϭ ljxj (xj � 0), (A Ϫ kI)xj ϭ ljxj Ϫ kxj ϭ (lj Ϫ k)xj

x ϭ (I Ϫ A)Ϫ1y ϭ [0.6747 0.7128 0.7543]T

c [10 18 25]T

[11 12 16]T

[5 8]T

45°Ϫ45°[1 Ϫ1]T; 1.5, [1 1]T;

[Ϫ1>16 1]T, 112.2°; 8, [1 1>16]T, 22.2°

[1 Ϫ1]T, Ϫ45°; 4.5, [1 1]T, 45°


13.

15.

17.

19.

21. , hyperbola

23. , hyperbolaC ϭ cϪ11

42

42

24d , 52y1

2 Ϫ 39y22 ϭ 156, x ϭ

1

213c23

3

Ϫ2dy

C ϭ c 1

Ϫ6

Ϫ6

1d , 7y1

2 Ϫ 5y22 ϭ 70, x ϭ

1

22cϪ1

1

1

1dy

C ϭ c 3

11

11

3d , 14y1

2 Ϫ 8y22 ϭ 0, x ϭ

1

22c11

1

Ϫ1dy; pair of straight lines

C ϭ c73

3

7d , 4y1

2 ϩ 10y22 ϭ 200, x ϭ

1

22c 1

Ϫ1

1

1dy, ellipse

D 13

Ϫ13

0

13

16

Ϫ12

13

16

12

T A D111

Ϫ2

1

1

0

Ϫ1

1

T ϭ D10

0

0

0

1

0

0

0

5

TD 1

Ϫ2

1

0

1

Ϫ2

0

0

1

T A D123

0

1

2

0

0

1

T ϭ D400

0

Ϫ2

0

0

0

1



1. Hermitian, 5,

3. Unitary,

5. Skew-Hermitian, unitary,

7. Eigenvalues eigenvectors

9. Hermitian, 16 11. Skew-Hermitian,

13.

15. (H Hermitian, S skew-Hermitian)

19.

if and only if HS ϭ SH.

AAT

Ϫ ATA ϭ (H ϩ S)(H Ϫ S) Ϫ (H Ϫ S)(H ϩ S) ϭ 2(ϪHS ϩ SH) ϭ 0

A ϭ H ϩ S, H ϭ 12 (A ϩ A

T), S ϭ 1

2 (A Ϫ AT)

(ABC)˛

Tϭ C

TB

TA

Tϭ C�1(ϪB)A

Ϫ6i

[1 0]T, resp.[0 1]T,

[1 Ϫ1]T, [1 1]T; [1 Ϫi]T, [1 i]T;Ϫ1, 1;

Ϫi, [0 Ϫ1 1]T, i, [1 0 0]T, [0 1 1]T

(1 Ϫ i13)>2, [Ϫ1 1]T; (1 ϩ i13)>2, [1 1]T

[Ϫi 1]T, 7, [i 1]T


11.

13.

15.

17. Ϫ1, 1; A ϭ1

16c 5

Ϫ3

Ϫ3

5d c23

39

2

1d ϭ

1

8cϪ1

63

1

1d

0, [2 Ϫ2 1]T; 9i, [Ϫ1 ϩ 3i 1 ϩ 3i 4]T; Ϫ9i, [Ϫ1 Ϫ 3i 1 Ϫ 3i 4]T

3, [1 5]T; 7, [1 1]T

3, [1 1]T; 2, [1 Ϫ1]T

19.

21.

23. , hyperbola

25. , ellipse


1.

3.

5. , position vector of Q

7. 9.

11. 13.

15. 17. [12, 8, 0]

21. 23. [0, 0, 5], 5

25. 27.

29. arbitrary 31.

33. Nothing

35.

37.


1. 44, 44, 0 3.

5.

7. ; cf. (6)

9. 300; cf. (5a) and (5b) 13. Use (1) and

15.

17.

19. is negative! Why?

21. Yes, because 23. arccos

27. is the angle between the unit vectors a and b. Use (2).

29. and

31. 33.

35. A square.

37. 0. Why?

39. If or if a and b are orthogonal.ƒa ƒ ϭ ƒb ƒ

(a ϩ b) • (a Ϫ b) ϭ ƒa ƒ 2 Ϫ ƒb ƒ 2 ϭ 0, ƒa ƒ ϭ ƒb ƒ .

Ϯ[35, Ϫ 4

5]a1 ϭ Ϫ 283

123.7°g ϭ arccos (12>(6113)) ϭ 0.9828 ϭ 56.3°

b Ϫ a

0.5976 ϭ 53.3°W ϭ (p ϩ q) • d ϭ p • d ϩ q • d.

[0, 4, 3] • [Ϫ3, Ϫ2, 1] ϭ Ϫ5

[2, 5, 0] • [2, 2, 2] ϭ 14

ϭ 2 ƒa ƒ 2 ϩ 2 ƒb ƒ 2ƒa ϩ b ƒ 2 ϩ ƒa Ϫ b ƒ 2 ϭ a • a ϩ 2a • b ϩ b • b ϩ (a • a Ϫ 2a • b ϩ b • b)

ƒ cos g ƒ Ϲ 1.

ƒϪ24 ƒ ϭ 24, ƒa ƒ ƒ c ƒ ϭ 135186 ϭ 13010 ϭ 54.86

ƒ [2, 9, 9] ƒ ϭ 1166 ϭ 12.88 Ͻ 180 ϩ 186 ϭ 18.22

135, 1320, 186

l ϭ 1000, k ϭ 10000 ϩ l Ϫ 1000 ϭ 0,

Ϫk ϩ l ϩ 0 ϭ 0,u ϩ v ϩ p ϭ [Ϫk, 0] ϩ [l, l] ϩ [0, Ϫ1000] ϭ 0,

vB Ϫ vA ϭ [Ϫ19, 0] Ϫ [22>12, 22>12] ϭ [Ϫ19 Ϫ 22>12, Ϫ22>12]

ƒp ϩ q ϩ u ƒ Ϲ 18.

k ϭ 10v ϭ [v1, v2, 3], v1, v2

p ϭ [0, 0, Ϫ5][6, 2, Ϫ14] ϭ 2u, 1236

[4, 9, Ϫ3], 1106

7[9, Ϫ7, 8] ϭ [63, Ϫ49, 56]

[1, 5, 8][6, 4, 0], [32, 1, 0], [Ϫ3, Ϫ2, 0]

Q : (0, 0, Ϫ8), ƒv ƒ ϭ 8Q: (4, 0, 12), ƒv ƒ ϭ 116.25

2, 1, Ϫ2; u ϭ [ 23, 13, Ϫ2

3 ]

8.5, Ϫ4.0, 1.7; 191.14, [0.890, Ϫ0.419, 0.178]

5, 1, 0; 126; [5>126, 1>126, 0]

C ϭ c3.7

1.6

1.6

1.3d , 4.5y1

2 ϩ 0.5y22 ϭ 4.5, x ϭ

1

15c21

1

Ϫ2dy

C ϭ c 4

12

12

Ϫ14d , 10y1

2 Ϫ 20y22 ϭ 20, x ϭ

1

15c21

1

�2dy

1

3D110

1

Ϫ1

1

Ϫ1

0

1

T A D 1

1

Ϫ1

2

Ϫ1

1

1

1

2

T ϭ D400

0

Ϫ20

0

0

0

22

T1

3c 2

Ϫ1

Ϫ1

2d A c2

1

1

2d ϭ cϪ0.9

0

0

0.6d



5. instead of m, tendency to rotate in the opposite sense.

7.

9. Zero volume in Fig. 191, which can happen in several ways.

11. 13.

15. 0 17.

19.

21.

23. 0, 0, 13

25. clockwise

27. 29.

31. 33.


1. Hyperbolas

3. Parallel straight lines (planes in space)

5. Circles, centers on the y-axis

7. Ellipses 9. Parallel planes

11. Elliptic cylinders 13. Paraboloids


1. Circle, center , radius 2 3. Cubic parabola

5. Ellipse 7. Helix

9. A “Lissajous curve” 11.

13. 15.

17. 19.

21. Use

25. At P,

27. 29.

31. 33. Start from .

35.

37.

39.

, and

41.

and

43.

45. ,

[mph]

49. , etc.r(t) ϭ [t, y(t), 0], rr ϭ [1, yr, 0] r • rr ϭ 1 ϩ yr226.61 # 108 ϭ 25,700 [ft>sec] ϭ 17,500

ƒv ƒ ϭ 1gR ϭg ϭ ƒa ƒ ϭ v2R ϭ ƒv ƒ 2>RR ϭ 3960 ϩ 80 mi ϭ 2.133 # 107 ft,

ƒa ƒ ϭ v2R ϭ ƒv ƒ 2>R ϭ 5.98 # 10�6 [km>sec2]

R ϭ 30 # 365 # 86,400>2p ϭ 151 # 106 [km],1 year ϭ 365 # 86,400 sec,

atan ϭ

12 sin 2t

4 ϩ sin2 tv.Ϫ4 cos 2t],a ϭ [Ϫcos t, Ϫ4 sin 2t,

ƒv ƒ 2 ϭ 4 ϩ sin2 t,v ϭ [Ϫsin t, 2 cos 2t, Ϫ2 sin 2t],

atan ϭ6 sin 3t

5 Ϫ 4 cos 3tv.Ϫsin t ϩ 4 sin 2t]a ϭ [Ϫcos t Ϫ 4 cos 2t,

ƒv ƒ 2 ϭ 5 Ϫ 4 cos 3t,cos t Ϫ 2 cos 2t],v ϭ [Ϫsin t Ϫ 2 sin 2t,

v(0) ϭ (v ϩ 1) Ri, a(0) ϭ Ϫv2Rj

v ϭ rr ϭ [1, 2t, 0], ƒv ƒ ϭ 21 ϩ 4t 2, a ϭ [0, 2, 0]

r(t) ϭ [t, f (t)]2rr • rr ϭ a, l ϭ ap>22rr • rr ϭ cosh t, l ϭ sinh l ϭ 1.175q(w) ϭ [2 ϩ w, 12 Ϫ 1

4w, 0]

rr ϭ [Ϫ8, 0, 6]. q(w) ϭ [6 Ϫ 8w, i, 8 ϩ 6w].u ϭ [Ϫsin t, 0, cos t].

sin (Ϫa) ϭ Ϫsin a.

r ϭ [cosh t, (13>2) sinh t, Ϫ2]r ϭ [12 cos t, sin t, sin t]

r ϭ [t, 4t Ϫ 1, 5t]r ϭ [2 ϩ t, 1 ϩ 2t, 3]

r ϭ [3 ϩ 113 cos t, 2 ϩ 113 sin t, 1]

x ϭ 0, z ϭ y3(3, 0)

y ϭ 34 x ϩ c

474>6 ϭ 793x ϩ 2y Ϫ z ϭ 5

12 ƒ [Ϫ12, 2, 6] ƒ ϭ 146[6, 2, 0] � [1, 2, 0] ϭ [0, 0, 10]

m ϭ [Ϫ2, Ϫ2, 0] � [2, 3, 0] ϭ [0, 0, Ϫ10], m ϭ 10

[Ϫ48, Ϫ72, Ϫ168], 121248 ϭ 189.0, 189.0

1, Ϫ1

[Ϫ32, Ϫ58, 34], [Ϫ42, Ϫ63, 19]

[6, 2, 7], [Ϫ6, Ϫ2, Ϫ7][0, 0, 7], [0, 0, Ϫ7], Ϫ4

ƒv ƒ ϭ ƒ [0, 20, 0] � [8, 6, 0] ƒ ϭ ƒ [0, 0, Ϫ160] ƒ ϭ 160

Ϫm


51.

53.


1. 3.

5. 7. Use the chain rule.

9. Apply the quotient rule to each component and collect terms.

11.

13.

15. 17. For P on the x- and y-axes.

19. 21.

23. Points with 25.

31. 33.

35. 37.

39.

41. 43.

45.


1. 3. 0, after simplification; solenoidal

5. 7.

9. (b) , etc.

11. , and

. Hence as t increases from 0 to 1, this “shear flow”

transforms the cube into a parallelepiped of volume l.

13. because do not depend on x, y, z, respectively.

15. 17. 0

19.


3. Use the definitions and direct calculation.

5. 7.

9. incompressible,

11. incompressible,

13. irrotational, compressible, Sketch it.

15. same (why?)

17. (why?),

19. same (why?)[Ϫ2z Ϫ y, Ϫ2x Ϫ z, Ϫ2y Ϫ x],

Ϫy Ϫ z Ϫ xϪyz Ϫ zx Ϫ xy, 0

[Ϫ1, Ϫ1, Ϫ1],

r ϭ [c1et, c2et, c3e�t].div v ϭ 1,curl v ϭ 0,

x2 ϩ 12y2 ϭ c, z ϭ c3

xr ϭ y, yr ϭ Ϫ2x, 2xxr ϩ yyr ϭ 0,curl v ϭ [0, 0, Ϫ3],

y ϭ 3c32t ϩ c2yr ϭ 3z2 ϭ 3c3

2,z ϭ c3,

x ϭ c1,v ϭ rr ϭ [xr, yr, zr] ϭ [0, 3z2, 0],curl v ϭ [Ϫ6z, 0, 0]

e�x[cos y, sin y, 0][x (z2 Ϫ y2), y (x2 Ϫ z2), z (y2 Ϫ x2)]

2>(x2 ϩ y2 ϩ z2)2

Ϫ2 cos 2x ϩ 2 cos 2y

v1, v2, v3div (w � r) ϭ 0

xr ϭ y ϭ c2, x ϭ c2t ϩ c1

[v1, v2, v3] ϭ rr ϭ [xr, yr, zr] ϭ [y, 0, 0], zr ϭ 0, z ϭ c3, yr ϭ 0, y ϭ c2

( fv1)x ϩ ( fv2)y ϩ ( fv3)z ϭ f [(v1)x ϩ (v2)y ϩ (v3)z] ϩ fxv1 ϩ fyv2 ϩ fzv3

Ϫ2ex (cos y)z9x2y2z2; 1296

2x ϩ 8y ϩ 18z; 7

f ϭ �v1 dx ϩ �v2 dy ϩ �v3 dz

f ϭ xyz28>3[1, 1, 1] • [Ϫ3>125, 0, Ϫ4>125]>13 ϭ Ϫ7>(12513)

[2, 1] • [1, Ϫ1]>15 ϭ 1>15[Ϫ2x, Ϫ2y, 1], [Ϫ6, Ϫ8, 1]

[12x, 4y, 2z], [60, 20, 10]f ϭ [32x, Ϫ2y], f (P) ϭ [160, Ϫ2]

Ϫ�T(P) ϭ [0, 4, Ϫ1]y ϭ 0, Ϯp, Ϯ2p, Á .

[0, Ϫe][Ϫ1.25, 0]

[8x, 18y, 2z], [40, Ϫ18, Ϫ22]

[2x>(x2 ϩ y2), 2y>(x2 ϩ y2)], [0.16, 0.12]

[y, x], [5, Ϫ4]

[4x3, 4y3]

[Ϫy>x2, 1>x][2y Ϫ 1, 2x ϩ 2]

3>(1 ϩ 9t 2 ϩ 9t 4)

dr

dsϭ

dr

dt >>ds

dt,

d2r

ds2 ϭd2r

dt 2 >>ads

dtb2 ϩ Á ,

d3r

ds3 ϭd3r

dt 3 >>ads

dtb3 ϩ Á



11. 1080, 1080, 65

13.

15. undefined

17. 125,

19.

21.

23.

25. 27.

29. tendency of clockwise rotation 31. 4

33. 1,

35. 0, same (why?),

37. 39.


3. 4

5.

7. “Exponential helix,” 9. 23.5, 0

11. 15.

17. 19.


3. 5.

7.

9.

13. 15. Dependent, etc.

17. Dependent, etc. 19.


3. 5.

7. 9.

11.

13. 15.

17.

19.


1. 3.

5.

7. 0. Why? 9.

13. 2w ϭ cosh x, y ϭ x>2 Á 2, 12 cosh 4 Ϫ 1

2

165

2x Ϫ 2y, 2x(1 Ϫ x2) Ϫ (2 Ϫ x2)2 ϩ 1, x ϭ Ϫ1 Á 1, Ϫ 5615

9(e2 Ϫ 1) Ϫ 83 (e3 Ϫ 1)(Ϫ1 Ϫ 1) � p>4 ϭ Ϫp>2

Ix ϭ (a ϩ b)h3>24, Iy ϭ h(a4 Ϫ b4)>(48(a Ϫ b))

Ix ϭ bh3>12, Iy ϭ b3h>4x ϭ 0, y ϭ 4r>3px ϭ 2b>3, y ϭ h>3

z ϭ 1 Ϫ r 2, dx dy ϭ r dr du, Answer: p>236 ϩ 27y2, 144cosh 2x Ϫ cosh x, 1

2 sinh 4 Ϫ sinh 2

�1

0

[x Ϫ x3 Ϫ (x2 Ϫ x5)] dx ϭ1

128y3>3, 54

sin (a2 ϩ 2b2 ϩ c2)4 � 0,

x2 � Ϫ4y2,ea2

cos 2b

ex cosh y ϩ ez sinh y, e Ϫ (cosh 1 ϩ sinh 1) ϭ 0

cosh 1 Ϫ 2 ϭ Ϫ0.457

exy sin z, e Ϫ 0sin 12 x cos 2y, 1 Ϫ 1>12 ϭ 0.293

144t 4, 1843.2[4 cos t, ϩ sin t, sin t, 4 cos t], [2, 2, 0]

18p, 43 (4p)3, 18p2e�t ϩ 2te�t2

, Ϫ2e�2 Ϫ e�4 ϩ 3

(e6p Ϫ 1)>3r ϭ [2 cos t, 2 sin t], 0 Ϲ t Ϲ p>2; 8

5

9>1225 ϭ 35[0, Ϫ2, 0]

2(y2 ϩ x2 Ϫ xz)

Ϫ2y

[0, 0, Ϫ14],

v • w> ƒw ƒ ϭ 22>18 ϭ 7.78[5, 2, 0] • [4 Ϫ 1, 3 Ϫ 1, 0] ϭ 19

g1 ϭ arccos (Ϫ10>165 # 40) ϭ 1.7682 ϭ Ϫ101.3°, g2 ϭ 23.7°

[Ϫ2, Ϫ6, Ϫ13]

[70, Ϫ40, Ϫ50], 0, 2352 ϩ 202 ϩ 252 ϭ 12250

Ϫ125Ϫ125,

[Ϫ1260, Ϫ1830, Ϫ300], [Ϫ210, 120, Ϫ540],

[Ϫ10, Ϫ30, 0], [10, 30, 0], 0, 40

Ϫ10,


15. 17.

19.


1. Straight lines, k

3. , circles, straight lines,

5. , circles, parabolas,

7. ,

ellipses

11.

13. Set and .

15.

17.

19.


1.

3. from (3), Sec. 10.5. Answer:

5.

7.

9. Integrate to

get

13.

15. . Answer: 54.4

21.

23.

25. B the z-axis,

.


1. 224

3.

5.

7.

9. 11.

13. 15.

17. 19.

21. 23.

25. Do Prob. 20 as the last one.

h5p>10(a4>4) � 2p � h ϭ ha4p>28abc(b2 ϩ c2)>3h4p>21>p ϩ 5

24 ϭ 0.5266div F ϭ Ϫsin z, 0

12(e Ϫ 1>e) ϭ 24 sinh 1div F ϭ 2x ϩ 2z, 48

[r cos u cos v, cos u sin v, r sin u], dV ϭ r 2 cos u dr du dv, s ϭ v, 2p2a3>312 (sin 2x) (1 Ϫ cos 2x), 1

8, 34

Ϫe�1�z ϩ e�y�z, Ϫ2e�1�z ϩ e�z, 2e�3 Ϫ e�2 Ϫ 2e�1 ϩ 1

IK ϭ IB ϩ 12 # 4p ϭ 20.9

IB ϭ 8p>3,[cos u cos v, cos u sin v, sin u], dA ϭ (cos u) du dv,

[u cos v, u sin v, u], �2p

0

�h

0

u2⅐ u12 du dv ϭ

p

12h4

Ixϭy ϭ ��S

[12 (x Ϫ y)2 ϩ z2] s dA

G(r) ϭ (1 ϩ 9u4)3>2, ƒN ƒ ϭ (1 ϩ 9u4)1>27p3>16 ϭ 88.6

2(1 Ϫ 1>12)(cosh 5 Ϫ 1) ϭ 42.885.

2 sinh v sin ur ϭ [2 cos u, 2 sin u, v], 0 Ϲ u Ϲ p>4, 0 Ϲ v Ϲ 5.

F • N ϭ [0, sin u, cos v] • [1, Ϫ2u, 0], 4 ϩ (Ϫ2 ϩ p2>16 Ϫ p>2)12 ϭ Ϫ0.1775

F(r) • N ϭ Ϫu3, Ϫ128p

13F(r) • N ϭ cos3 v cos u sin u

F(r) • N ϭ [Ϫu2, v2, 0] • [Ϫ3, 2, 1] ϭ 3u2 ϩ 2v2, 29.5

[cosh u, sinh u, v], [cosh u, Ϫsinh u, 0]

a2 cos2 v sin u, a2 cos v sin v][a2 cos2 v cos u,

[a cos v cos u, Ϫ2.8 ϩ a cos v sin u, 3.2 ϩ a sin v], a ϭ 1.5;

[2 ϩ 5 cos u, Ϫ1 ϩ 5 sin u, v], [5 cos u, 5 sin u, 0]

y ϭ vx ϭ u

[u~, v~, u~2, ϩ v~2], N~

ϭ [Ϫ2u~, Ϫ2v~, 1]

x2>a2 ϩ y2>b2 ϩ z2>c2 ϭ 1, [bc cos2 v cos u, ac cos2 v sin u, ab sin v cos v]

[Ϫ2u2 cos v, Ϫ2u2 sin v, u]z ϭ x2 ϩ y2

[Ϫcu cos v, Ϫcu sin v, u]z ϭ c2x2 ϩ y2

ƒgrad w ƒ 2 ϭ e2x, 52 (e4 Ϫ 1)

2w ϭ 6x Ϫ 6y, Ϫ 38.42w ϭ 6xy, 3x(10 Ϫ x2)2 Ϫ 3x, 486



1. , no contributions. etc.

Integrals . Sum 0

3. The volume integral of is . The surface

integral of over is Others 0.

5. The volume integral of is 0; others 0.

7. use Sec. 10.7, etc.

9. and

11.


1.

3.

5.

7.

9. The sides contribute a, 0.

11. 13. 5k,

15.

17.

19.

Answer:


11.

Or using exactness.

13. Not exact, 15. 0 since

17. By Stokes, 19.

21.

23.

25. 27.

29. Answer: 21 31.

33. Direct integration, 35.


1. 5. There is no smallest

13.

15.

17. p

2ϩ

4

pacos x ϩ

1

9 cos 3x ϩ

1

25 cos 5x ϩ Á b

13 sin 3x ϩ Á )

43p

2 ϩ 4 (cos x ϩ 14 cos 2x ϩ 1

9 cos 3x ϩ Á ) Ϫ 4p (sin x ϩ 12 sin 2x ϩ

4

p (cos x ϩ

1

9 cos 3x ϩ

1

25 cos 5x ϩ Á ) ϩ 2 (sin x ϩ

1

3 sin 3x ϩ

1

5 sin 5x ϩ Á )

p Ͼ 0.2p, 2p, p, p, 1, 1, 12, 12

72p2243

24 sinh 1 ϭ 28.205div F ϭ 20 ϩ 6z2.

288(a ϩ b ϩ c)pM ϭ 4k>15, x ϭ 516, y ϭ 4

7

M ϭ 6320, x ϭ 8

7 ϭ 1.14, y ϭ 11849 ϭ 2.41

M ϭ 8, x ϭ 85, y ϭ 16

5

F ϭ grad (y2 ϩ xz), 2pϮ18p

curl F ϭ 0curl F ϭ (5 cos x)k, Ϯ10

Ϫ4528>3.

r ϭ [4 Ϫ 10t, 2 ϩ 8t], F(r) • dr ϭ [2(4 Ϫ 10t)2, Ϫ4(2t ϩ 8t)2] • [Ϫ10, 8] dt;

1>2[Ϫez, 1, 0] • [Ϫu cos v, Ϫu sin v, u].

r ϭ [u cos v, u sin v, u], 0 Ϲ u Ϲ 1, 0 Ϲ v Ϲ p>2,

r ϭ [cos u, sin u, v], [Ϫ3v2, 0, 0] • [cos u, sin u, 0], Ϫ1

[0, Ϫ1, 2x Ϫ 2y] • [0, 0, 1], 13

80pϪ2p; curl F ϭ 0

3a2>2, Ϫa,

[Ϫez, Ϫex, Ϫey] • [Ϫ2x, 0, 1], Ϯ(e4 Ϫ 2e ϩ 1)

[0, 2z, 32 ] • [0, 0, 1] ϭ 32 , Ϯ3

2 a2

[2e�z cos y, Ϫe�z, 0] • [0, Ϫy, 1] ϭ ye�z, Ϯ(2 – 2>1e)

S: z ϭ y (0 Ϲ x Ϲ 1, 0 Ϲ y Ϲ 4), [0, 2z, Ϫ2z] • [0, Ϫ1, 1], Ϯ20

r ϭ a, � ϭ 0, cos � ϭ 1, v ϭ 13 a � (4pa2)

Ϫ2p �12 (a2 Ϫ r 2)3>2

�23 ƒ

0

aϭ 2

3pa3

dx dy ϭ r dr du,z ϭ 2a2 Ϫ x2 Ϫ y2 ϭ 2a2 Ϫ r 2,z ϭ 0

(2*),F ϭ [x, 0, 0], div F ϭ 1,

8(x ϭ 1), Ϫ8(y ϭ 1),6y2� 4 Ϫ 2x2

� 12

8y3>3 ϭ 83.x ϭ 1f 0g>0n ϭ f � 2x ϭ 2f ϭ 8y2

8y3>3 ϭ 838y2 ϩ [0, 8y] � [2x, 0] ϭ 8y2

x ϭ a: (Ϫ2a)bc, y ϭ b: (Ϫ2b)ac, z ϭ c: (4c) ab

x ϭ a: 0f>0n ϭ 0f>0x ϭ Ϫ2x ϭ Ϫ2a,x ϭ 0, y ϭ 0, z ϭ 0


19.

21.


1. Neither, even, odd, odd, neither 3. Even 5. Even

9. Odd,

11. Even,

13. Rectifier,

15. Odd,

17. Even,

19.

23. (a) 1, (b)

25. (a)

(b)

27. (a)

(b)

29. Rectifier,

(a) (b)


3. The output becomes a pure cosine series.

5. For this is similar to Fig. 54 in Sec. 2.8, whereas for the phase shift

the sense is the same for all n.

BnAn

sin x2

pϪ

4

pa 1

1 # 3 cos x ϩ

1

3 # 5 cos 3x ϩ

1

5 # 7 cos 5x ϩ Á b ,

L ϭ p,

a1

5ϩ

2

25pb sin 5x ϩ

1

6 sin 6x ϩ Á

a1 ϩ2

pb sin x ϩ

1

2sin 2x ϩ a1

3Ϫ

2

9pb sin 3x ϩ

1

4sin 4x ϩ

1

18 cos 6x ϩ

1

49 cos 7x ϩ

1

81 cos 9x Ϫ

1

50 cos 10x ϩ

1

121 cos 11x ϩ Á b

3p

8ϩ

2

pacos x Ϫ

1

2cos 2x ϩ

1

9 cos 3x ϩ

1

25 cos 5x ϪL ϭ p,

2 (sin x ϩ 12 sin 2x ϩ 1

3 sin 3x ϩ 14 sin 4x ϩ Á )

p

2ϩ

4

pacos x ϩ

1

9 cos 3x ϩ

1

25 cos 5x ϩ Á b ,L ϭ p,

4

pasin px

4ϩ

1

3 sin

3px

4ϩ

1

5 sin

5px

4ϩ Á bL ϭ 4,

38 ϩ 1

2 cos 2x ϩ 18 cos 4x

L ϭ 1,1

2ϩ

4

p2 acos px ϩ1

9 cos 3px ϩ

1

25 cos 5px ϩ Á b

L ϭ p,4

pasin x Ϫ

1

9 sin 3x ϩ

1

25 sin 5x Ϫ ϩ Á b

1

pa1

2sin 2px Ϫ

1

4 sin 4px ϩ

1

6 sin 6px Ϫ

1

8 sin 8px ϩ Ϫ Á b

L ϭ1

2,

1

8Ϫ

1

p2 acos 2px ϩ1

9 cos 6px ϩ

1

25 cos 10px ϩ Á b ϩ

L ϭ 1,1

3Ϫ

4

p2 acos px Ϫ1

4cos 2px ϩ

1

9 cos 3px Ϫ ϩ Á b

L ϭ 2, 4

pasin px

2ϩ

1

3 sin

3px

2ϩ

1

5 sin

5px

2ϩ Á b

2(sin x ϩ 12 sin 2x ϩ 1

3 sin 3x ϩ 14 sin 4x ϩ 1

5 sin 5x ϩ Á )

1

3 sin 3x Ϫ ϩ Á

p

4Ϫ

2

pacos x ϩ

1

9 cos 3x ϩ

1

25 cos 5x ϩ Á b ϩ sin x Ϫ

1

2sin 2x ϩ


7.

Note the change of sign.

11.

13.

15. (n odd),

with as in

Prob. 13.

17.

19.

Section 11.4, page 498

3.

5. 0.6243,

7.

Why is so large?


3. Set 5. etc.

7.

9.

11.

13.


1.

3.

9.

Rounding seems to have considerable influence in Probs. 8–13.m0 ϭ 9.

Ϫ0.4775P1(x) Ϫ 0.6908P3(x) ϩ 1.844P5(x) Ϫ 0.8236P7(x) ϩ 0.1658P9(x) ϩ Á ,

45 P0(x) Ϫ 4

7 P2(x) Ϫ 835 P4(x)

8 (P1(x) Ϫ P3(x) ϩ P5(x))

p ϭ e8x, q ϭ 0, r ϭ e8x, lm ϭ m2, ym ϭ e�4x sin mx, m ϭ 1, 2, Álm ϭ m2, m ϭ 1, 2, Á , ym ϭ x sin (m ln ƒ x ƒ )

l ϭ [(2m ϩ 1)p>(2L)]2, m ϭ 0, 1, Á , ym ϭ sin ((2m ϩ 1)px>(2L))

lm ϭ (mp>10)2, m ϭ 1, 2, Á ; ym ϭ sin (mpx>10)

x ϭ cos u, dx ϭ Ϫsin u du,x ϭ ct ϩ k.

E*E* ϭ 674.8, 454.7, 336.4, 265.6, 219.0.

F ϭ 2[(p2 Ϫ 6) sin x Ϫ 18 (4p2 Ϫ 6) sin 2x ϩ 1

27 (9p2 Ϫ 6) sin 3x Ϫ ϩ Á ];

0.4206 (0.1272 when N ϭ 20)

F ϭ4

pasin x ϩ

1

3 sin 3x ϩ

1

5 sin 5x ϩ Á b , E* ϭ 1.1902, 1.1902, 0.6243,

0.0748, 0.0119, 0.0119, 0.0037

F ϭp

2Ϫ

4

pacos x ϩ

1

9 cos 3x ϩ

1

25 cos 5x ϩ Á b , E* ϭ 0.0748,

Bn ϭ (Ϫ1)nϩ124,000

nDn

, Dn ϭ (10 Ϫ n2)2 ϩ 100n2

I (t) ϭ a

�

nϭ1

(An cos nt ϩ Bn sin nt), An ϭ (Ϫ1)nϩ12400 (10 Ϫ n2)

n2Dn

,

Bn ϭ 10nan>Dn, an ϭ Ϫ400>(n2p), Dn ϭ (n2 Ϫ 10)2 ϩ 100n2An ϭ (10 Ϫ n2)an>Dn,I ϭ 50 ϩ A1 cos t ϩ B1 sin t ϩ A3 cos 3t ϩ B3 sin 3t ϩ Á ,

DnBn ϭ (Ϫ1)nϩ1 # 12(1 Ϫ n2)>(n3Dn)An ϭ (Ϫ1)n # 12nc>n3Dn,

y ϭ a

�

nϭ1

(An cos nt ϩ Bn sin nt),bn ϭ (Ϫ1)nϩ1 # 12 >n3

Bn ϭ [(1 Ϫ n2)bn ϩ ncan]>Dn, Dn ϭ (1 Ϫ n2)2 ϩ n2c2

y ϭ a

N

nϭ1

(An cos nt ϩ Bn sin nt), An ϭ [(1 Ϫ n2)an Ϫ nbnc]>Dn,

1

v2 Ϫ 121

sin 5t ϩ Á by ϭ C1 cos vt ϩ C2 sin vt ϩ

4

pa 1

v2 Ϫ 9

sin t ϩ1

v2 Ϫ 49

sin 3t ϩ

0.8, 0.01.Ϫ5.26, 4.76,

y ϭ C1 cos vt ϩ C2 sin vt ϩ a (v) sin t, a (v) ϭ 1>(v2 Ϫ 1) ϭ Ϫ1.33,


11.

13.

15.


1. gives

(see Example 3), etc.

3. Use (11);

5.

7.

9.

11.

15. For the value of equals 0.28,

0.029, 0.0103, 0.0065, (rounded).

17.

19.


1.

3.

5.

7. Yes. No 9.

11.

13.


3. if otherwise

5.

7. 9.

11. 13. by formula 9e�w2>2i12>p (cos w Ϫ 1)>w12>p(cos w ϩ w sin w Ϫ 1)>w2(e�iaw(1 ϩ iaw) Ϫ 1)>(12pw2)

[e(1�iw)a Ϫ e�(1�iw)a]>(12p(1 Ϫ iw))

a Ͻ b; 0i (e�ibw Ϫ e�iaw)>(w12p)

fs(e�x) ϭ

1

waϪfc(e�x) ϩB

2

p# 1b ϭ

1

waB

2

p# 1

w2 ϩ 1ϩ B

2

pb ϭ B

2

p

w

w2 ϩ 1

12>p ((2 Ϫ w2) cos w ϩ 2w sin w Ϫ 2)>w3

12>p w>(a2 ϩ w2)

fcˆ (w) ϭ B

2

p

(w2 Ϫ 2) sin w ϩ 2w cos w

w3

fcˆ (w) ϭ 1(2>p) (cos 2w ϩ 2w sin 2w Ϫ 1)>w2

(2 sin w Ϫ sin 2w)>wfcˆ (w) ϭ 1(2>p)

2

p ��

0

w Ϫ e (w cos w Ϫ sin w)

1 ϩ w2 sin xw dw

2p �

�

0

1 Ϫ cos ww

sin xwdw

Ϫ0.0064Ϫ0.0099,Ϫ0.026,

Ϫ0.15,Si (np) Ϫ p>2n ϭ 1, 2, 11, 12, 31, 32, 49, 50

2

p ��

0

cos pw ϩ 1

1 Ϫ w2 cos xw dw

A (w) ϭ2

p ��

0

cos wv

1 ϩ v2 dv ϭ e�w (w Ͼ 0)

2p �

�

0

sin w cos xw

wdw

B (w) ϭ2

p �1

0

1

2pv sin wvdv ϭ

sin w Ϫ w cos w

w2

B ϭ2

p ��

0

p

2 sin wvdv ϭ

1 Ϫ cos pw

w

A ϭ ��

0

e�v cos wv dv ϭ1

1 ϩ w2, B ϭ

w

1 ϩ w2f (x) ϭ pe�x(x Ͼ 0)

(c) am ϭ (2>J 12(a0,m)) (J1(a0,m)>a0,m) ϭ 2>(a0,mJ1(a0,m))

0.1212P0(x) Ϫ 0.7955P2(x) ϩ 0.9600P4(x) Ϫ 0.3360P6(x) ϩ Á , m0 ϭ 8

0.7854P0(x) Ϫ 0.3540P2(x) ϩ 0.0830P4(x) Ϫ Á , m0 ϭ 4


17. No, the assumptions in Theorem 3 are not satisfied.

19.

21.


11.

13.

15. 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.


1.

3. 5.

7. Any c and 9.

15. 17.

19.

21.

23. (Euler–Cauchy)

25. a, b, c, k arbitrary constants


5.

7.

9. 0.8

p2 acos pt sin px Ϫ1

9 cos 3pt sin 3px ϩ

1

25 cos 5pt sin 5px Ϫ ϩ Á b

8k

p3 acos pt sin px ϩ1

27 cos 3pt sin 3px ϩ

1

125 cos 5pt sin 5px ϩ Á b

k cos 3pt sin 3px

u(x, y) ϭ axy ϩ bx ϩ cy ϩ k;

u ϭ c1(y)x ϩ c2(y)>x2

u ϭ e�3y(a(x) cos 2y ϩ b(x) sin 2y) ϩ 0.1e3y

u ϭ c(x) e�y3>3u ϭ a(y) cos 4px ϩ b(y) sin 4pxu ϭ 110 Ϫ (110>ln 100) ln (x2 ϩ y2)

c ϭ p>25v

c ϭ a>bc ϭ 2

L(c1u1 ϩ c2u2) ϭ c1L(u1) ϩ c2L(u2) ϭ c1# 0 ϩ c2

# 0 ϭ 0

12>p (cos aw Ϫ cos w ϩ aw sin aw Ϫ w sin w)>w2

1

p ��

0

(cos w ϩ w sin w Ϫ 1)cos wx ϩ (sin w Ϫ w cos w) sin wx

w2 dw

Ϫ1

16# cos 4t

v2 Ϫ 16ϩ Ϫ Á b

y ϭ C1 cos vt ϩ C2 sin vt ϩp2

v2Ϫ 12a cos t

v2 Ϫ 1Ϫ

1

4# cos 2t

v2 Ϫ 4ϩ

1

9# cos 3t

v2 Ϫ 9

1

2Ϫ

4

p2 acos px ϩ1

9 cos 3px ϩ Á b , 2

pasin px Ϫ

1

2sin 2px ϩ Ϫ Á b

cosh x, sinh x (Ϫ5 Ͻ x Ͻ 5),

1

pasin px Ϫ

1

2sin 2px ϩ

1

3 sin 3px Ϫ ϩ Á b

1

4Ϫ

2

p2 acos px ϩ1

9 cos 3px ϩ

1

25 cos 5px ϩ Á b ϩ

1 ϩ4

pasin px

2ϩ

1

3 sin

3px

2ϩ

1

5 sin

5px

2ϩ Á b

c1 1

1 Ϫ1d c f1

f2d ϭ c f1 ϩ f2

f1 Ϫ f2d

[ f1 ϩ f2 ϩ f3 ϩ f4, f1 Ϫ if2 Ϫ f3 ϩ if4, f1 Ϫ f2 ϩ f3 Ϫ f4, f1 ϩ if2 Ϫ f3 Ϫ if4]


11.

13.

No terms with

17.

19. (a) (b) (c) (d)

from (a), (c). Insert this. The coefficient determinant resulting from (b), (d) must be

zero to have a nontrivial solution. This gives (22).


3.

9. Elliptic,

11. Parabolic,

13. Hyperbolic,

15. Hyperbolic,

17. Elliptic, Real or imaginary parts of any

function u of this form are solutions. Why?


3. differ in rapidity of decay.

5.

7.

9. where satisfies the boundary conditions of the text,

so that

11.

etc.

13.

15.

17.

19.

21. u ϭ100

p a�

nϭ1

1

(2n Ϫ 1) sinh (2n Ϫ 1)p sin

(2n Ϫ 1)px

24 sinh

(2n Ϫ 1)py

24

u ϭ 1000 (sin 12px sinh 12py)>sinh p

ϪKp

L a�

nϭ1

nBn e�ln2t

1

2ϩ

4

p2 acos x e�t ϩ1

9 cos 3x e�9t ϩ

1

25 cos 5x e�25t ϩ Á b

u ϭ 1

p ϭ np>L,

F ϭ A cos px ϩ B sin px, Fr(0) ϭ Bp ϭ 0, B ϭ 0, Fr(L) ϭ ϪAp sin pL ϭ 0,

uII

ϭ a�

nϭ1

Bn sin npx

Le�(cnp>L)2t, Bn ϭ

2

L �L

0

[ f (x) Ϫ uI(x)] sin

npx

Ldx.

uII ϭ u Ϫ uIu ϭ uI ϩ uII,

u ϭ800

p3asin 0.1px e�0.01752p2t ϩ

1

33 sin 0.3px e�0.01752(3p)2t ϩ Á b

u ϭ sin 0.1px e�1.752p2t>100

u1 ϭ sin x e�t, u2 ϭ sin 2x e�4t, u3 ϭ sin 3x e�9t

u ϭ f1(y Ϫ (2 Ϫ i)x) ϩ f2(y Ϫ (2 ϩ i)x).

xyr2 ϩ yyr ϭ 0, y ϭ v, xy ϭ w, uw ϭ z, u ϭ1y

f1(xy) ϩ f2(y)

u ϭ f1(y Ϫ 4x) ϩ f2(y Ϫ x)

u ϭ x f1(x Ϫ y) ϩ f2(x Ϫ y)

u ϭ f1(y ϩ 2ix) ϩ f2(y Ϫ 2ix)

c2 ϭ 300>[0.9>(2 # 9.80)] ϭ 80.832 [m2>sec2]

ux(L, t) ϭ 0. C ϭ ϪA, D ϭ ϪBux(0, t) ϭ 0,u(L, t) ϭ 0,u(0, t) ϭ 0,

u ϭ8L2

p3 acos c c apLb2t d sin

px

Lϩ

1

33 cos c c a3p

Lb2t d sin

3px

Lϩ Á b

n ϭ 4, 8, 12, Á .ϩ4 Ϫ 5p

125 cos 5pt sin 5px ϩ Á b .

4

p3 a(4 Ϫ p) cos pt sin px ϩ cos 2pt sin 2px ϩ4 ϩ 3p

27 cos 3pt sin 3px

ϩ1

25 (2 ϩ 12) cos 5pt sin 5px Ϫ ϩ Á b

2

p2 a(2 Ϫ 12) cos pt sin px Ϫ1

9 (2 ϩ 12) cos 3pt sin 3px


23.

25.


3.

5. etc.

7. if and 0 if

9. Set in (21) to get erf

13. In (12) the argument is 0 (the point where f jumps) when

This gives the lower limit of integration.

15. Set in (21).


1. (a), (b) It is multiplied by (c) Half

5. if m odd, 0 if m even

7.

11.

13.

17. (corresponding eigenfunctions and ), etc.

19.


5.

7. 55p Ϫ440

p (r cos u ϩ

1

9r 3 cos 3u ϩ

1

25r 5 cos 5u ϩ Á )

110 ϩ440

p (r cos u Ϫ

1

3r 3 cos 3u ϩ

1

5r 5 cos 5u Ϫ ϩ Á )

cos aptB36

a2ϩ

4

b2b sin

6px

a sin

4py

b

F16,14F4,16cp1260

6.4

p2 a

�

mϭ1

a

�

nϭ1m,n odd

1

m3n3 cos (t2m2 ϩ n2) sin mx sin ny

u ϭ 0.1 cos 120t sin 2x sin 4y

Bmn ϭ (Ϫ1)mϩn4ab>(mnp2)

Bmn ϭ (Ϫ1)nϩ18>(mnp2)

12.

w ϭ s>12

z ϭ Ϫx>(2c1t).x ϩ 2cz1t

(Ϫx) ϭ Ϫerf x.w ϭ Ϫv

u ϭ �1

0

cos px e�c2p2t dp

p Ͼ 1,0 Ͻ p Ͻ 1A ϭ2

p ��

0

sin v

v cos pv dv ϭ

2

p# p

2ϭ 1

A ϭ2

p �1

0

v cos pv dv ϭ2

p# cos p ϩ p sin p Ϫ 1

p2 ,

A ϭ2

p�

�

0

cos pv

1 ϩ v2dv ϭ

2

p# p

2e�p, u ϭ �

�

0

e�p�c2p2t cos px dp

a�

nϭ1

An sin npx

a sinh

np(b Ϫ y)

a, An ϭ

2

a sinh (npb>a) �a

0

f (x) sin npx

adx

A0 ϭ1

242 �24

0

f (y) dy, An ϭ1

12�

24

0

f (y) cos npy

24dy

u ϭ A0x ϩ a�

nϭ1

An

sinh (npx>24)

sinh np cos

npy

24,


11. Solve the problem in the disk subject to (given) on the upper semicircle

and on the lower semicircle.

13. Increase by a factor 15.

17. No 25. See Table A1 in App. 5.


5.

9.

13. is smaller than the potential in Prob. 12 for

17.

19.

25. Set Then

Substitute this and etc. into (7) [written in terms of ] and divide by


5.

7. take to get and from

11. Set Use z as a new variable of integration. Use


17. 19. Hyperbolic,

21. Hyperbolic, 23.

25.

27.

29.

39.


1. 3.

5. 9.

11. 13.

15. 17.

19.


1.

3. 2(cos 12p ϩ i sin 12p), 2(cos 12p Ϫ i sin 12p)

12 (cos 14p ϩ i sin 14p)

(x2 Ϫ y2)>(x2 ϩ y2), 2xy>(x2 ϩ y2)

Ϫ4x2y23 Ϫ i

Ϫ120 Ϫ 40iϪ8 Ϫ 6i

Ϫ117, 4x Ϫ iy ϭ Ϫ(x ϩ iy), x ϭ 0

4.8 Ϫ 1.4i1>i ϭ i>i2 ϭ Ϫi, 1>i3 ϭ i>i4 ϭ i

u ϭ (u1 Ϫ u0)(ln r)>ln (r1>r0) ϩ (u0 ln r1 Ϫ u1 ln r0)>ln (r1>r0)

100 cos 2x e�4t

34 sin 0.01px e�0.001143t Ϫ 1

4 sin 0.03px e�0.01029t

sin 0.01px e�0.001143t

34 cos 2t sin x Ϫ 1

4 cos 6t sin 3xf1(y ϩ 2x) ϩ f2(y Ϫ 2x)

f1(x) ϩ f2(y ϩ x)u ϭ c1(x)e�3y ϩ c2(x)e2y Ϫ 3

erf (ϱ) ϭ 1.x2>(4c2t) ϭ z2.

w(x, 0) ϭ x(c Ϫ 1) ϭ 0.

c ϭ 1g ϭ ce�t ϩ t Ϫ 1f (x) ϭ xw ϭ f (x)g(t), xf rg ϩ fg#ϭ xt,

W ϭc(s)

x sϩ

x

s2(s ϩ 1), W(0, s) ϭ 0, c(s) ϭ 0, w(x, t) ϭ x(t Ϫ 1 ϩ e�t)

r 5.ru�� ϭ rv��

urr ϭ (2vr ϩ rvrr)(1>r4) ϩ (v ϩ rvr)(2>r3), urr ϩ (2>r)ur ϭ r 5(vrr ϩ (2>r)vr).

u(r, u, �) ϭ rv(r, u, �), ur ϭ (v ϩ rvr)(Ϫ1>r2), 1>r ϭ r.

cos 2� ϭ 2 cos2 � Ϫ 1, 2w2 Ϫ 1 ϭ 43 P2(w) Ϫ 1

3, u ϭ 43r 2P2(cos �) Ϫ 1

3

u ϭ 1

2 Ͻ r Ͻ 4.u ϭ 320>r ϩ 60

ur ϭ c~>r 2, u ϭ c>r ϩ k

2u ϭ us ϩ 2ur>r ϭ 0, us>ur ϭ Ϫ2>r, ln ƒur ƒ ϭ Ϫ2 ln ƒ r ƒ ϩ c1,

A4 ϭ A6 ϭ A8 ϭ A10 ϭ 0, A5 ϭ 605>16, A7 ϭ Ϫ4125>128, A9 ϭ 7315>256

a11>(2p) ϭ 0.6098;

T ϭ 6.826rR2f 1212

u ϭ4u0

pa r

a sin u ϩ

1

3a3 r 3 sin 3u ϩ1

5a5r 5 sin 5u ϩ Á b

Ϫu0

u0r Ͻ a


5. 7.

9. 11.

13. 15.

17. 21.

23.

25.

27.

29. 31.

33. Multiply out and use

(Prob. 34).

Hence Taking

square roots gives (6).

35.


1. Closed disk, center radius

3. Annulus (circular ring), center radii and

5. Domain between the bisecting straight lines of the first quadrant and the fourth

quadrant.

7. Half-plane extending from the vertical straight line to the right.

11.

15. Yes, since

17. Yes, because and as

19. Now hence

21. 23.


1.

(a)

(b)

Multiply (a) by (b) by , and add. Etc.

3. Yes 5. No,

7. Yes, when Use (7). 9. Yes, when

11. Yes 13. c real

15. (c real) 17. (c real)

19. No 21.

23. 27. implies

29. Use (4), (5), and (1).


3. 5.

7. 9.

11. 13. 12epi>46.3epi

5ei arctan (3>4) ϭ 5e0.644ie12i ϭ 4.113i

e2(Ϫ1) ϭ Ϫ7.389e2pie�2p ϭ e�2p ϭ 0.001867

if ϭ Ϫv ϩ iu.f ϭ u ϩ iva ϭ 0, v ϭ 12 b(y2 Ϫ x2) ϩ c

a ϭ p, v ϭ epx sin py

f (z) ϭ z2 ϩ z ϩ cf (z) ϭ 1>z ϩ c

f (z) ϭ Ϫ12 i(z2 ϩ c),

z � 0, Ϫ2pi, 2piz � 0.

f (z) ϭ (z2)

sin ucos u,

0 ϭ uy ϩ vx ϭ ur sin u ϩ uu(cos u)>r ϩ vr cos u ϩ vu(Ϫsin u)>r0 ϭ ux Ϫ vy ϭ ur cos u ϩ uu(Ϫsin u)>r Ϫ vr sin u Ϫ vu (cos u)>r

rx ϭ x>r ϭ cos u, ry ϭ sin u, ux ϭ Ϫ(sin u)>r, uy ϭ (cos u)>r

3iz2>(z ϩ i)4, Ϫ3i>16n(1 Ϫ z) �n�1i, ni

f r(3 ϩ 4i) ϭ 8 # 37 ϭ 17,496.z Ϫ 4i ϭ 3,f r(z) ϭ 8(z Ϫ 4i)7.

r: 0.1 Ϫ ƒ z ƒ : 1Re z ϭ r cos u: 0

Im( ƒ z ƒ 2>z) ϭ Im( ƒ z ƒ 2 z>(zz)) ϭ Im z ϭ Ϫr sin u: 0.

v(x, y) ϭ y((1 Ϫ x)2 ϩ y2), v(1, Ϫ1) ϭ Ϫ1

u(x, y) ϭ (1 Ϫ x)>((1 Ϫ x)2 ϩ y2), u(1, Ϫ1) ϭ 0,

x ϭ Ϫ1

3pp4 Ϫ 2i,

32Ϫ1 ϩ 5i,

[(x1 ϩ x2)2 ϩ (y1 ϩ y2)2] ϩ [(x1 Ϫ x2)2 ϩ (y1 Ϫ y2)2] ϭ 2(x12 ϩ y1

2 ϩ x22 ϩ y2

2)

ƒ z1 ϩ z1 ƒ 2 Ϲ ( ƒ z1 ƒ ϩ ƒ z2 ƒ )2.ϭ ( ƒ z1 ƒ ϩ ƒ z2 ƒ )2.ϩ 2 ƒ z1 ƒ ƒ z2 ƒ ϩ ƒ z2 ƒ 2z1z1 ϩ z1z2 ϩ z2z1 ϩ z2z2 ϭ ƒ z1 ƒ 2 ϩ 2 Re z1z2 ϩ ƒ z2 ƒ 2 Ϲ ƒ z1 ƒ 2Re z1z2 Ϲ ƒ z1z2 ƒƒ z1 ϩ z2 ƒ 2 ϭ (z1 ϩ z2)( z1 ϩ z2 ) ϭ (z1 ϩ z2)( z 1 ϩ z 2).

Ϯ(1 Ϫ i), Ϯ(2 ϩ 2i)i, Ϫ1 Ϫ i

cos 15p Ϯ i sin 15p, cos 35p Ϯ i sin 35p, Ϫ1

cos (18p ϩ 1

2 kp) ϩ i sin (18p ϩ 1

2 kp), k ϭ 0, 1, 2, 3

6, Ϫ3 Ϯ 313i

26 2 (cos 112 kp ϩ i sin 1

12p), k ϭ 1, 9, 172 ϩ 2i

Ϫ3iϪ1024. Answer: p

Ϯarctan ( 43 ) ϭ Ϯ0.92733p>4

21 ϩ 14p

2 (cos arctan 12p ϩ i sin arctan 12p)12 (cos p ϩ i sin p)


15.

17.

19.


1. Use (11), then (5) for and simplify. 7.

9. Both Why? 11. both

15. Insert the definitions on the left, multiply out, and simplify.

17. 19.


5. 7.

9. 11.

13.

15.

17.

19.

21.

23.

25.

27.


1. 3.

11. 13.

15. i 17.

19. 21.

23. 25.

27. 29.

31.

33.

35.


1. Straight segment from (2, 1) to (5, 2.5).

3. Parabola from (1, 2) to (2, 8).

5. Circle through (0, 0), center radius oriented clockwise.

7. Semicircle, center 2, radius 4.

9. Cubic parabola

11.

13. z(t) ϭ 2 Ϫ i ϩ 2eit (0 Ϲ t Ϲ p)

z(t) ϭ t ϩ (2 ϩ t)i (Ϫ1 Ϲ t Ϲ 1)

y ϭ x3 (Ϫ2 Ϲ x Ϲ 2)

110,(3, Ϫ1),

y ϭ x2

cosh p cos p ϩ i sinh p sin p ϭ Ϫ11.592

i tanh 1 ϭ 0.7616i

cos 3 cosh 1 ϩ i sin 3 sinh 1 ϭ Ϫ1.528 ϩ 0.166i

f (z) ϭ e�z2>2f (z) ϭ e�2z

f (z) ϭ Ϫiz2>2(Ϯ1 Ϯ i)>12

Ϯ3, Ϯ3i15e�pi>2412e�3pi>40.16 Ϫ 0.12iϪ5 ϩ 12i

27.46e0.9929i, 7.616e1.976i2 Ϫ 3i

e(2�i) Ln(�1) ϭ e(2�i)pi ϭ ep ϭ 23.14

e(3�i)(ln 3ϩpi) ϭ 27ep(cos (3p Ϫ ln 3) ϩ i sin (3p Ϫ ln 3)) ϭ Ϫ284.2 ϩ 556.4i

e(1�i) Ln (1ϩi) ϭ eln12ϩpi>4�i ln12ϩp>4 ϭ 2.8079 ϩ 1.3179i

e0.6e0.4i ϭ e0.6 (cos 0.4 ϩ i sin 0.4) ϭ 1.678 ϩ 0.710i

e4�3i ϭ e4 (cos 3 Ϫ i sin 3) ϭ Ϫ54.05 Ϫ 7.70i

ln (i2) ϭ ln (Ϫ1) ϭ (1 Ϯ 2n)pi, 2 ln i ϭ (1 Ϯ 4n)pi, n ϭ 0, 1, Á

ln ƒ ei ƒ ϩ i arctan sin 1

cos 1Ϯ 2npi ϭ 0 ϩ i ϩ 2npi, n ϭ 0, 1, Á

Ϯ2npi, n ϭ 0, 1, Áln e ϩ pi>2 ϭ 1 ϩ pi>2i arctan (0.8>0.6) ϭ 0.927i

12 ln 32 Ϫ pi>4 ϭ 1.733 Ϫ 0.785iln 11 ϩ pi

z ϭ Ϯnpiz ϭ Ϯ(2n ϩ 1)i>2

i sinh p ϭ 11.55i,Ϫ0.642 Ϫ 1.069i.

cosh 1 ϭ 1.543, i sinh 1 ϭ 1.175ieiy,

z ϭ 2npi, n ϭ 0, 1, ÁRe (exp (z3)) ϭ exp (x3 Ϫ 3xy2) cos (3x2y Ϫ y3)

exp (x2 Ϫ y2) cos 2xy, exp (x2 Ϫ y2) sin 2xy


15.

17. Circle

19.

21.

23.

25.

27.

29. on the axes.

integrated:

35.


1. Use (12), Sec. 14.1, with 3. Yes 5. 5

7. (a) Yes. (b) No, we would have to move the contour across

9. 0, yes 11. no

13. 0, yes 15. no

17. 0, no 19. 0, yes

21. 23. hence

25. 0 (Why?) 27. 0 (Why?)

29. 0


1. 3. 0

5. 7.

11. 13.

15. since

and

17.

19.


1. 3.

5.

7.

9.

11.

ϭ 2.821i

ϭ 12pi(sin 12 ϩ 3

2 cos 12 )

2pi

4((1 ϩ z)sin z)r `

zϭ1>2ϭ 1

2pi(sin z ϩ (1 ϩ z) cos z) ƒ zϭ1>2

Ϫ2pi(tan pz)r `zϭ0

ϭϪ2pi � p

cos2 pz`zϭ0

ϭ Ϫ2p2i

(2pi>(2n)!) (cos z)(2n) ƒ zϭ0 ϭ (2pi>(2n)!)(Ϫ1)n cos 0 ϭ (Ϫ1)n2pi>(2n)!

2pi

3!(cosh 2z)t ϭ

pi

3� 8 sinh 1 ϭ 9.845i

(2pi>(n Ϫ 1)!)e0(2pi>3!)(Ϫcos 0) ϭ Ϫpi>3

2pie2i>(2i) ϭ pe2i

2piLn (z ϩ 1)

z ϩ i`zϭi

ϭ 2piLn (1 ϩ i)

2iϭ p(ln12 ϩ ip>4) ϭ 1.089 ϩ 2.467i

sinh pi ϭ i sin p ϭ 0.

cosh pi ϭ cos p ϭ Ϫ12pi cosh (Ϫp2 Ϫ pi) ϭ Ϫ2pi cosh p2 ϭ Ϫ60,739i

2pi(z ϩ 2) ƒ zϭ2 ϭ 8pi2pi �1

z ϩ 2i`zϭ2i

ϭp

2

2pi(i>2)3>2 ϭ p>82pi(cos 3z)>6 ƒ zϭ0 ϭ pi>32piz2>(z Ϫ 1) ƒ zϭ�1 ϭ Ϫpi

2pi ϩ 2pi ϭ 4pi.1>z ϩ 1>(z Ϫ 1),2pi

Ϫp,

pi,

Ϯ2i.

m ϭ 2.

ƒRe z ƒ ϭ ƒ x ƒ Ϲ 3 ϭ M on C, L ϭ 18

(Ϫ1 ϩ i)>3.y(Ϫ1 ϩ i)(Im z2)#z ϭ 2(1 Ϫ t)

z ϭ 1 ϩ (Ϫ1 ϩ i)t (0 Ϲ t Ϲ 1),Im z2 ϭ 2xy ϭ 0

tan 14pi Ϫ tan 14 ϭ i tanh 14 Ϫ 1

12 exp z2 ƒ

i

1 ϭ 12 (e�1 Ϫ e1) ϭ Ϫsinh 1

e2pi Ϫ epi ϭ 1 Ϫ (Ϫ1) ϭ 2

z(t) ϭ (1 ϩ i)t (1 Ϲ t Ϲ 3), Re z ϭ t, zr (t) ϭ 1 ϩ i. Answer: 4 ϩ 4i

z(t) ϭ t ϩ (1 Ϫ 14 t 2)i (Ϫ2 Ϲ t Ϲ 2)

z(t) ϭ Ϫa Ϫ ib ϩ re�it (0 Ϲ t Ϲ 2p)

z(t) ϭ 2 cosh t ϩ i sinh t (Ϫϱ Ͻ t Ͻ ϱ)


13. 15. 0. Why?

17. 0 by Cauchy’s integral theorem for a doubly connected domain; see (6) in Sec. 14.2.

19.


21.

23. by Cauchy’s integral formula.

25.

27. 0 since and

29.


1. bounded, divergent,

3. by algebra; convergent to

5. Bounded, divergent,

7. Unbounded, hence divergent

9. Convergent to 0, hence bounded

17. Divergent; use 19. Convergent; use

21. Convergent 23. Convergent

25. Divergent

29. By absolute convergence and Cauchy’s convergence principle, for given we

have for every and

hence by Sec. 13.2, hence convergence by Cauchy’s

principle.


1. No! Nonnegative integer powers of z (or only!

3. At the center, in a disk, in the whole plane

5. hence

7. 9. 11.

13. 15. 2i, 1 17.


3. Apply l’Hôpital’s rule to

5. 2 7. 9.

11. 13. 1 15.


3. 2z2 Ϫ(2z2)3

3!ϩ Á ϭ 2z2 Ϫ

4

3z6 ϩ

4

15z10 Ϫ ϩ Á , R ϭ ϱ

3427

3

1>1213

ln f ϭ (ln n)>n.f ϭ 2n n.

1>12Ϫi, 12

0,2265i,13p>2, ϱ

ƒ z ƒ Ͻ 1R.Sanz2n ϭ San(z2)n, ƒ z2 ƒ Ͻ R ϭ lim ƒan>anϩ1 ƒ ;

z Ϫ z0)

(6*),ƒ znϩ1 ϩ Á ϩ znϩp ƒ Ͻ P

ƒ znϩ1 ƒ ϩ Á ϩ ƒ znϩp ƒ Ͻ P,

p ϭ 1, 2, Án Ͼ N(P)

P Ͼ 0

S1>n2.1>ln n Ͼ 1>n.

Ϯ1 ϩ 10i

Ϫpi>2zn ϭ Ϫ12pi>(1 ϩ 2>(ni))

Ϯ1, Ϯizn ϭ (2i>2)n;

Ϫ4pi

y ϭ xz2 ϩ z Ϫ 2 ϭ 2(x2 Ϫ y2)

Ϫ2pi(tan pz)r|zϭ1 ϭ Ϫ2p2i>cos2 pz|zϭ1 ϭ Ϫ2p2i

2pi(ez)(4) ƒ zϭ0 ϭ iez>12 ƒ zϭ0 ϭ pi>12

12 cosh (Ϫ1

4p2) Ϫ 1

2 ϭ 2.469

(2pi>2!)4�3(e3z)s ƒ zϭpi>4 ϭ Ϫ9p(1 ϩ i)>(6412)

2pi # 1z`zϭ2

ϭ pi


5.

7.

9.

11.

13.

17. Team Project. (a)

(c) Use that the terms of are all positive, so that the sum cannot be zero.

19.

21.

23.

25.


3.

5.

7. Nowhere

9.

11. and converges. Use Theorem 5.

13. for all z, and converges. Use Theorem 5.

15. by Theorem 2 in Sec. 15.2; use Theorem 1.

17. use Theorem 1.


11. 1 13. 3

15. 17.

19.

21.

23.

25.

27.

29. ln 3 ϩ1

3(z Ϫ 3) Ϫ

1

2 � 9(z Ϫ 3)2 ϩ

1

3 � 27(z Ϫ 3)3 Ϫ ϩ Á , R ϭ 3

cos [(z Ϫ 12p) ϩ 1

2p] ϭ Ϫ(z Ϫ 12p) ϩ 1

6(z Ϫ 12p)3 Ϫ ϩ Á ϭ Ϫsin (z Ϫ 1

2p)

a�

nϭ1

(Ϫ1)nϩ1

n!z2n�2, R ϭ ϱ

1

2ϩ

1

2cos 2z ϭ 1 ϩ

1

2 a�

nϭ1

(Ϫ1)n

(2n)! (2z)2n, R ϭ ϱ

a�

nϭ0

z4n

(2n ϩ 1)!, R ϭ ϱ

ϱ, cosh 1z

ϱ, e2z12

R ϭ 1>1p Ͼ 0.56;

R ϭ 4

S1>n2ƒ sinn ƒ z ƒ ƒ Ϲ 1

S1>n2ƒ zn ƒ Ϲ 1

ƒ z Ϫ 2i ƒ Ϲ 2 Ϫ d, d Ͼ 0

ƒ z ϩ 12 i ƒ Ϲ 1

4 Ϫ d, d Ͼ 0

ƒ z ϩ i ƒ Ϲ 13 Ϫ d, d Ͼ 0

2az Ϫ1

2ib ϩ

23

3!az Ϫ

1

2ib

3

ϩ25

5!az Ϫ

1

2ib

5

ϩ Á , R ϭ ϱ

Ϫ14 Ϫ 2

8 i(z Ϫ i) ϩ 316 (z Ϫ i)2 ϩ 4

32 i(z Ϫ i)3 Ϫ 564 (z Ϫ i)4 ϩ Á , R ϭ 2

1 Ϫ1

2! az Ϫ

1

2pb

2

ϩ1

4! az Ϫ

1

2pb

4

Ϫ1

6! az Ϫ

1

2pb

6

ϩ Ϫ Á , R ϭ ϱ

12 ϩ 1

2 i ϩ 12 i(z Ϫ i) ϩ (Ϫ1

4 ϩ 14 i)(z Ϫ i)2 Ϫ 1

4 (z Ϫ i)3 ϩ Á , R ϭ 12

(sin iy)>(iy)

(Ln (1 ϩ z))r ϭ 1 Ϫ z ϩ z2 Ϫ ϩ Á ϭ 1>(1 ϩ z).

(2>1p)(z Ϫ z3>3 ϩ z5>(2!5) Ϫ z7>(3!7) ϩ Á ), R ϭ ϱ

z3>(1!3) Ϫ z7>(3!7) ϩ z11>(5!11) Ϫ ϩ Á , R ϭ ϱ

�z

0

a1 Ϫ1

2t 2 ϩ

1

8t 4 Ϫ ϩ Áb dt ϭ z Ϫ

1

6z3 ϩ

1

40z5 Ϫ ϩ Á , R ϭ ϱ

1

2ϩ

1

2cos z ϭ 1 Ϫ

1

2 � 2!z2 ϩ

1

2 � 4!z4 Ϫ

1

2 � 6!z6 ϩ Ϫ Á , R ϭ ϱ

12 Ϫ 1

4 z4 ϩ 18 z8 Ϫ 1

16 z12 ϩ 132 z16 Ϫ ϩ Á , R ϭ 24 2




1.

3.

5.

7.

9.

11.

13.

15.

19.

21.

23.

25.


1. fourth order 3. fourth order

5. second order 7. simple

9. simple

11. hence

13. Second-order poles at i and

15. Simple pole at essential singularity at

17. Fourth-order poles at essential singularity at

19. simple zeros. Answer: simple poles at

essential singularity at

21. essential singularities, simple poles


3. at 0 5.

7. at 9. at

11. at

15. Simple pole at inside C, residue Answer:

17. Simple poles at residue and at residue

Answer:

19.

21. Answer: 2p5i>24z�5 cos pz ϭ Á ϩ p4>(4!z) Ϫ ϩ Á .

2pi (sinh 12 i)>2 ϭ Ϫp sin 12

Ϫ4pi sinh p>2e�p>2>sin p>2 ϭ e�p>2.

Ϫp>2,ep>2>(Ϫsin p>2),p>2,

ϪiϪ1>(2p).14

z ϭ pi(ez)s>2! ƒ zϭpi ϭ Ϫ12

Ϯ2npiϪ10, Ϯ1, Á1>pϮ4i at ϯ i

415

Ϯ2npi, n ϭ 0, 1, Á ,1, ϱ

ϱ

Ϯ2npi,ez(1 Ϫ ez) ϭ 0, ez ϭ 1, z ϭ Ϯ2npi

ϱϮnpi, n ϭ 0, 1, Á ,

1 ϩ iϱ,

Ϫ2i

f 2(z) ϭ (z Ϫ z0)2ng2(z).f (z) ϭ (z Ϫ z0)ng(z), g(z0) � 0,

12 sin 4z, z ϭ 0, Ϯp>4, Ϯp>2, Á ,

Ϯ(2 ϩ 2i), Ϯi,Ϯ1, Ϯ2, Á ,

Ϫ81i,0 Ϯ 2p, Ϯ4p, Á ,

i

(z Ϫ i)2 ϩ1

z Ϫ iϩ i ϩ (z Ϫ i)

ƒ z ƒ Ͼ 1z8 ϩ z12 ϩ z16 ϩ Á , ƒ z ƒ Ͻ 1, Ϫz4 Ϫ 1 Ϫ z�4 Ϫ z�8 Ϫ Á ,

ƒ z ϩ 12p ƒ Ͼ 0

Ϫ(z ϩ 12p)�1 cos (z ϩ 1

2p) ϭ Ϫ(z ϩ 12p)�1 ϩ 1

2 (z ϩ 12p) Ϫ 1

24 (z ϩ 12p)3 ϩ Ϫ Á ,

ƒ z ƒ Ͼ 1a�

nϭ0

z2n, ƒ z ƒ Ͻ 1, Ϫ a�

nϭ0

1

z2nϩ2,

0 Ͻ ƒ z Ϫ p ƒ Ͻ ϱ

(Ϫcos (z Ϫ p))(z Ϫ p)�2 ϭ Ϫ(z Ϫ p)�2 ϩ 12 Ϫ 1

24 (z Ϫ p)2 ϩ Ϫ Á ,

0 Ͻ ƒ z Ϫ i ƒ Ͻ 1Ϫ3(z Ϫ i)�1 Ϫ 6i ϩ 10(z Ϫ i) ϩ Á ,

ϭ i(z Ϫ i)�2i�3 a1 ϩz Ϫ i

ib

�3

(z Ϫ i)�2 ϭ a�

nϭ0

aϪ3

nb i�3�n(z Ϫ i)n�2

[pi ϩ (z Ϫ pi)]2

(z Ϫ pi)4ϭ

(pi)2

(z Ϫ pi)4ϩ

2pi

(z Ϫ pi)3ϩ

1

(z Ϫ pi)2

0 Ͻ ƒ z Ϫ 1 ƒ Ͻ ϱ

exp [1 ϩ (z Ϫ 1)] (z Ϫ 1)�2 ϭ e � [(z Ϫ 1)�2 ϩ (z Ϫ 1)�1 ϩ 12 ϩ 1

6 (z Ϫ 1) ϩ Á ],

z3 ϩ 12 z ϩ 1

24 z�1 ϩ 1720 z3 ϩ Á , 0 Ͻ ƒ z ƒ Ͻ ϱ

z�2 ϩ z�1 ϩ 1 ϩ z ϩ z2 ϩ Á , 0 Ͻ ƒ z ƒ Ͻ 1

z�3 ϩ z�1 ϩ 12 z ϩ 1

6 z3 ϩ 124 z5 ϩ Á , 0 Ͻ ƒ z ƒ Ͻ ϱ

z�4 Ϫ 12 z�2 ϩ 1

24 Ϫ 1720 z2 ϩ Ϫ Á , 0 Ͻ ƒ z ƒ Ͻ ϱ

23. Residues at at Answer:

25. Simple poles inside C at residues

respectively. Answer:


1. 3.

5. 7.

9. 0. Why? (Make a sketch.) 11.

13. 0. Why? 15.

17. 0. Why?

19. Simple poles at (and

21. Simple poles at 1 and residues i and Answer:

23. 25. 0

27. Let Use (4) in Sec. 16.3 to form the sum of

the residues and show that this sum is 0; here


11. 13.

15. 17. 0 (n even), (n odd)

19. 21.

23. 0. Why? 25. Answer:


5. Only in size

7.

9. Parallel displacement; each point is moved 2 to the right and 1 up.

11. 13.

15. 17. Annulus

19.

21.

23.

25. at

29. on the unit circle,


33. for the y-axis,



7. 9.

11. 13. z ϭ 0, Ϯ12 , Ϯ ϭ Ϯi>2z ϭ 0, 1>(a ϩ ib)

z ϭ4w ϩ i

Ϫ3iw ϩ 1z ϭ

w ϩ i

2w

J ϭ 1> ƒ z ƒ 2M ϭ 1> ƒ z ƒ ϭ 1

J ϭ e2xx ϭ 0,M ϭ ex ϭ 1

J ϭ 1> ƒ z ƒ 4ƒwr ƒ ϭ 1> ƒ z ƒ 2 ϭ 1

J ϭ ƒ z ƒ 2M ϭ ƒ z ƒ ϭ 1

z ϭ 0, Ϯpi, Ϯ2pi, Ásinh z ϭ 0

z ϭ (Ϫ1 Ϯ 13)>2z3 ϩ az2 ϩ bz ϩ c, z ϭ Ϫ 1

3 (a Ϯ 2a2 Ϫ 3b)

0 Ͻ u Ͻ ln 4, p>4 Ͻ v Ϲ 3p>412 Ϲ ƒw ƒ Ϲ 4u м 1

Ϫ5 Ϲ Re z Ϲ Ϫ2ƒw ƒ Ϲ 14 , Ϫp>4 Ͻ Arg w Ͻ p>4

x ϭ c, w ϭ Ϫy ϩ ic; y ϭ k, w ϭ Ϫk ϩ ix

p>e.Reszϭi

eiz>(z2 ϩ 1) ϭ 1>(2ie).

p>60p>6(Ϫ1)(n�1)>22pi>(n Ϫ 1)!2pi(25z2)r ƒ zϭ5 ϭ 500pi

2pi(Ϫ10 Ϫ 10)6pi

k Ͼ 1.1>qr(a1) ϩ Á ϩ 1>qr(ak)

q(z) ϭ (z Ϫ a1)(z Ϫ a2) Á (z Ϫ ak).

Ϫp>2

p

5 (cos 1 Ϫ e�2)Ϫi.Ϯ2pi,

Ϫi); 2pi �14 i ϩ pi(Ϫ1

4 ϩ 14 ) ϭ Ϫ1

2pϮ1, i

p>3p>22ap>2a2 Ϫ 15p>12

p>122p>2k2 Ϫ 1

2pi �410

110 , 1

10 , 110 , 1

10 ,

(2i cosh 2i)>(4z3 ϩ 26z) ƒ zϭ2i ϭ2i, Ϫ2i, 3i, Ϫ3i,

5piz ϭ 13 .z ϭ 1

2 , 212


15. 17. 19.


3. Apply the inverse g of f on both sides of to get

9. a rotation. Sketch to see. 11.

13. almost by inspection 15.

17. 19.


1. Circle 3. Annulus

5. w-plane without 7.

9.

11.

13. Elliptic annulus bounded by and

15.

17. is the image of R under Answer:

19. Hyperbolas when and

(thus when

21. Interior of in the fourth quadrant, or map

by (why?).

23.

25. The images of the five points in the figure can be obtained directly from the

function w.


1. w moves once around the circle

3. Four sheets, branch point at

5. three sheets

7. n sheets

9. two sheets


11. 13. Horizontal strip

15. same (why?) 17.

19. 21.

23. 25. Rotation

27. 29.

31. 33.

35. 37.

39. w ϭ z2>(2c)

w ϭ iz2 ϩ 1w ϭ e4z

z ϭ 0, Ϯi, Ϯ3iz ϭ 2 Ϯ16

z ϭ 0w ϭ 1>zw ϭ izw ϭ

10z ϩ 5i

z ϩ 2i

w ϭ 1 ϩ iv, v Ͻ 013 Ͻ ƒw ƒ Ͻ 1

2 , v Ͻ 0

ƒw ƒ Ͼ 1u ϭ 1 Ϫ 14 v2,

Ϫ8 Ͻ v Ͻ 81 Ͻ ƒw ƒ Ͻ 4, ƒ arg w ƒ Ͻ p>4

1z (z Ϫ i)(z ϩ i), 0, Ϯi,

z0,

Ϫi>4,

z ϭ Ϫ1

ƒw ƒ ϭ 12 .

v Ͻ 0

w ϭ sin zp>2 Ͻ x Ͻ p, 0 Ͻ y Ͻ 2

u2>cosh2 2 ϩ v2>sinh2 2 ϭ 1

c ϭ 0, p.ƒu ƒ м 1), v ϭ 0u ϭ Ϯcosh y

c � 0, p,u2>cos2 c Ϫ v2>sin2 c ϭ cosh2 c Ϫ sinh2 c ϭ 1

et ϭ ez2>2.t ϭ z2>2.0 Ͻ Im t Ͻ p

cosh z ϭ cos iz ϭ sin (iz ϩ 12p)



u2>cosh2 2 ϩ v2>sinh2 2 Ͻ 1, u Ͼ 0, v Ͼ 0

Ϯ(2n ϩ 1)p>2, n ϭ 0, 1, Á1 Ͻ ƒw ƒ Ͻ e, v Ͼ 0w ϭ 0

1>1e Ϲ ƒw ƒ Ϲ 1eƒw ƒ ϭ ec

w ϭ (z4 Ϫ i)(Ϫiz4 ϩ 1)w ϭ (2z Ϫ i)>(Ϫiz Ϫ 2)

w ϭ 1>z Ϫ 1w ϭ 1>z,

w ϭ (z ϩ i)>(z Ϫ i)w ϭ iz,

g(z1) ϭ g( f (z1)) ϭ z1.z1 ϭ f (z1)

w ϭaz ϩ b

Ϫbz ϩ aw ϭ

az

cz ϩ az ϭ i, 2i



1.

3.

5.

7.

13. Use Fig. 391 in Sec. 17.4 with the z- and w-planes interchanged and

15.


3. maps R onto the strip and

5. (a) (b)

7. See Fig. 392 in Sec. 17.4.

9.

13. Corresponding rays in the w-plane make equal angles, and the mapping is conformal.

15. Apply

17. by (3) in Sec. 17.3.

19.


1. Rotate through

5.

7.

9.

11.

13. from and Prob. 11.

15.

17. Re (arcsin z)


1. V(z) continuously differentiable.

3. is maximum at namely, 2.y ϭ Ϯ1,ƒFr(iy) ƒ ϭ 1 ϩ 1>y2, ƒ y ƒ м 1,

Re F(z) ϭ 100 ϩ (200>p)

Ϫ20 ϩ (320>p) Arg z ϭ Re aϪ20 Ϫ320i

p Ln zbw ϭ z2100

p [Arg (z2 Ϫ 1) Ϫ Arg (z2 ϩ 1)]

100p (Arg (z Ϫ 1) Ϫ Arg (z ϩ 1)) ϭ Re a100i

p Ln z ϩ 1

z Ϫ 1b

T1

p aarctan y

x Ϫ bϪ arctan

y

x Ϫ ab ϭ Re a iT1

p Ln z Ϫ a

z Ϫ bb

T1 ϩ2p (T2 Ϫ T1) arctan

y

xϭ Re aT1 Ϫ

2i

p (T2 Ϫ T1) Ln zb80p arctan

y

xϭ Re aϪ 80i

p Ln zbp>2.(80>d )y ϩ 20.

£ ϭ5p Arg (z Ϫ 2), F ϭ Ϫ

5i

p Ln (z Ϫ 2)

z ϭ (2Z Ϫ i)>(ϪiZ Ϫ 2)

w ϭ z2.

cos2 x (y ϭ 0), cos2 x cosh2 1 Ϫ sin2 x sinh2 1 (y ϭ 1)

Ϫsinh y (x ϭ p2 ),£ (x, y) ϭ cos2 x cosh2 y Ϫ sin2 x sinh2 y; cosh2 y (x ϭ 0),

sinh2 1 (y ϭ 1), Ϫsinh2 y (x ϭ 0, p).

sin2 x cosh2 1 Ϫ cos2 xsin2 x (y ϭ 0),£ ϭ Re (sin2 z),

x2 Ϫ y2 ϭ c, xy ϭ c, ex cos y ϭ c(x Ϫ 2) (2x Ϫ 1) ϩ 2y2

(x Ϫ 2)2 ϩ y2ϭ c,

U2 ϩ (U1 Ϫ U2) (1 Ϫ xy).

£* ϭ U2 ϩ (U1 Ϫ U2) (1 ϩ 12 u) ϭϪ2 Ϲ u Ϲ 0;w ϭ iz2

£ ϭ 220 (x3 Ϫ 3xy2) ϭ Re (220z3)

cos z ϭ sin (z ϩ 12p).

£ (r) ϭ Re (32 Ϫ z)

£ (x) ϭ Re (375 ϩ 25z)

£ ϭ Re a30 Ϫ20

ln 10 Ln zb

2.5 mm ϭ 0.25 cm; £ ϭ Re 110 (1 ϩ (Ln z)>ln 4)


5. Calculate or note that 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.

11. Uniform parallel flow upward,

13.

15.

17. Use that gives and interchanging the roles of the z- and

w-planes.

19. or


5.

7.

9.

11.

13.

15.

17.


1. Use (2). etc.

3. Use (2). etc.

5. No, because is not analytic.

7.

9.

13.

15.

17.

19. No. Make up a counterexample.

ƒF(z) ƒ 2 ϭ 4(2 Ϫ 2 cos 2u), z ϭ p>2, 3p>2, Max ϭ 4

Max ϭ sinh 2 ϭ 3.627

1 � sin2 2y, z ϭ 1,ƒF(z) ƒ 2 ϭ sinh2 2x cos2 2y ϩ cosh2 2x sin2 2y ϭ sinh2 2x ϩ

ƒF(z) ƒ ϭ [cos2 x ϩ sinh2 y]1>2, z ϭ Ϯi, Max ϭ [1 ϩ sinh2 1]1>2 ϭ 1.543

ϭ1p �

32

� 2p

£ (1, 1) ϭ 3 ϭ1p �

1

0

�2p

0

(3 ϩ r cos a ϩ r sin a ϩ r 2 cos a sin a)r dr da

ϭ1

p �1

0

�2p

0

(Ϫ3r ϩ Á ) dr da ϭ

1

paϪ 3

2b � 2p

£ (2, Ϫ2) ϭ Ϫ3 ϭ1p �

1

0

�2p

0

(1 ϩ r cos a) (Ϫ3 ϩ r sin a)r dr da

ƒ z ƒ

F(4) ϭ 100F(z0 ϩ eia) ϭ (2 ϩ 3eia)2,

F(52) ϭ 343

8F(z0 ϩ eia) ϭ (72 ϩ eia)3,

£ ϭ1

3Ϫ

4

p2 ar cos u Ϫ1

4r 2 cos 2u ϩ

1

9r 3 cos 3u Ϫ ϩ Á b

£ ϭ1

2ϩ

2

par cos u Ϫ

1

3r 3 cos 3u ϩ

1

5r 5 cos 5u Ϫ ϩ Á b

£ ϭ2

pr sin u ϩ

1

2r 2 sin 2u Ϫ

2

9pr 3 sin 3u Ϫ

1

4r 4 sin 4u ϩ ϩ Ϫ Á

£ ϭ2

par sin u Ϫ

1

2r 2 sin 2u ϩ

1

3r 3 sin 3u Ϫ ϩ Á b

£ ϭ 3 Ϫ 4r 2 cos 2u ϩ r 4 cos 4u

£ ϭ 12 a ϩ 1

2 ar 8 cos 8u

£ ϭ 32 r 3 sin 3u

x2 ϩ (y Ϫ k)2 ϭ k2y>(x2 ϩ y2) ϭ c

z ϭ cos ww ϭ arccos z

F(z) ϭ z>r0 ϩ r0>zF(z) ϭ z3

V ϭ F r ϭ iK, V1 ϭ 0, V2 ϭ K

F(z) ϭ z4

2 ϭ



11.

13.

17.

19.

21. parallel flow

23.

25.


1.

3. 6.3698, 6.794, 8.15, impossible

5. Add first, then round.

7. (6S-exact)

9.

11.

13.

hence

15. is 6S-exact.

19. In the present case, (b) is slightly more accurate than (a) (which may produce

nonsensical results; cf. Prob. 20).

21.

23. The algorithm in Prob. 22 repeats 0011 infinitely often.

25. The beginning is 0.09375

27.

etc.

29.


3.

5. Convergence to 4.7 for all these starting values.

7. converges to 0.58853 (5S-exact) in 14 steps.

9. (6S-exact)

11. 4S-exact

13. This follows from the intermediate value theorem of calculus.

15.

17. Convergence to respectively. Reason seen easily from the

graph of f.

x ϭ 4.7, 4.7, 0.8, Ϫ0.5,

x3 ϭ 0.450184

g ϭ 4>x ϩ x3>16 Ϫ x5>576; x0 ϭ 2, xn ϭ 2.39165 (n м 6), 2.405

x ϭ x4 Ϫ 0.12; x0 ϭ 0, x3 ϭ Ϫ0.119794

x ϭ x>(ex sin x); 0.5, 0.63256, Á

g ϭ 0.5 cos x, x ϭ 0.450184 (ϭ x10, exact to 6S)

Ϫ0.126 # 10�2, Ϫ0.402 # 10�3; Ϫ0.266 # 10�6, Ϫ0.847 # 10�7

I11 ϭ 0.2102 (0.2103),

I12 ϭ 0.1951 (0.1951),I14 ϭ 0.1812 (0.1705 4S-exact), I13 ϭ 0.1812 (0.1820),

(n ϭ 1).n ϭ 26.

c4# 24 ϩ Á ϩ c0

# 20 ϭ (1 0 1 1 1.)2, NOT (1 1 1 0 1.)2

(a) 1.38629 Ϫ 1.38604 ϭ 0.00025, (b) ln 1.00025 ϭ 0.000249969

` aa1

a2Ϫ

a�1

a�2b^ ` a1

a2` � ` P1

a1Ϫ

P2

a2` Ϲ ƒ Pr1 ƒ ϩ ƒ Pr2 ƒ Ϲ br1 ϩ br2

a1

a2

ϭa�1 ϩ P1

a�2 ϩ P2

ϭa�1 ϩ P1

a�2

a1 ϪP2

a�2

ϩP2

2

a�22 Ϫ ϩ Á b �

a�1

a�2

ϩP1

a�2

ϪP2

a�2

#a�1

a�2

,

Ϲ ƒ Px ƒ ϩ ƒ Py ƒ ϭ bx ϩ by

ƒ P ƒ ϭ ƒ x ϩ y Ϫ (x� ϩ y�) ƒ ϭ ƒ (x Ϫ x�) ϩ (y Ϫ y�) ƒ ϭ ƒ Px ϩ Py ƒ

29.97, 0.035; 29.97, 0.03337; 30, 0.0; 30, 0.033

29.9667, 0.0335; 29.9667, 0.0333704

0.84175 # 102, Ϫ0.52868 # 103, 0.92414 # 10�3, Ϫ0.36201 # 106

Fr(z) ϭ z ϩ 1 ϭ x ϩ 1 Ϫ iy

T(x, y) ϭ x (2y ϩ 1) ϭ const

£ ϭ x ϩ y ϭ const, V ϭ Fr(z) ϭ 1 Ϫ i,

30(1 Ϫ (2>p) Arg (z Ϫ 1))

2(1 Ϫ (2>p) Arg z)

£ ϭ Re (220 Ϫ 95.54 Ln z) ϭ 220 Ϫ220

ln 10 ln r ϭ 220 Ϫ 95.54 ln r.

£ ϭ 10(1 Ϫ x ϩ y), F ϭ 10 Ϫ 10(1 ϩ i)z


19.

21.

23. (6S-exact)

25. (a) ALGORITHM BISECT Bisection Method

This algorithm computes the solution c of ( f continuous) within the

tolerance given an initial interval such that

INPUT: Continuous function f, initial interval tolerance maximum

number of iterations N.

OUTPUT: A solution c (within the tolerance or a message of failure.

For do:

If then OUTPUT c Stop. [Procedure completed]

Else if then set and

Else set and

If then OUTPUT c. Stop. [Procedure completed]

End

OUTPUT and a message “Failure”. Stop.

[Unsuccessful completion; N iterations did not give an interval of length not

exceeding the tolerance.]

End BISECT

Note that gives as an approximation of the zero and

as a corresponding error bound.

(b) 0.739085; (c) 1.30980, 0.429494

27. (6S-exact)

29. 0.904557 (6S-exact)


1.

3.

5.

7.

9.

11.

13.

15.

17.

ϭ 0.3293

r ϭ Ϫ1.5, p2(0.3) ϭ 0.6039 ϩ (Ϫ1.5) # 0.1755 ϩ 12 (Ϫ1.5)(Ϫ0.5) # (Ϫ0.0302)

p3(x) ϭ 2.1972 ϩ (x Ϫ 9) # 0.1082 ϩ (x Ϫ 9)(x Ϫ 9.5) # 0.005235

2x2 Ϫ 4x ϩ 2

p2(0.5) ϭ 0.943654, p3(1.5) ϭ 0.510116, p3(2.5) ϭ Ϫ0.047991

p3(x) ϭ 1 ϩ 0.039740x Ϫ 0.335187x2 ϩ 0.060645x3;L3 ϭ 16 x(x Ϫ 1)(x Ϫ 2);

L0 ϭ Ϫ16 (x Ϫ 1)(x Ϫ 2)(x Ϫ 3), L1 ϭ 1

2 x(x Ϫ 2)(x Ϫ 3), L2 ϭ Ϫ12 x(x Ϫ 1)(x Ϫ 3),

(5S-exact 0.71116)

p2(x) ϭ Ϫ0.44304x2 ϩ 1.30896x Ϫ 0.023220, p2(0.75) ϭ 0.70929

0.9053 (0.0186), 0.9911 (Ϫ0.0672)

p2(x) ϭ 1.1640x Ϫ 0.3357x2; Ϫ0.5089 (error 0.1262), 0.4053 (Ϫ0.0226),

0.4678 (0.0046)

0.8033 (error Ϫ0.0245), 0.4872 (error Ϫ0.0148); quadratic: 0.7839 (Ϫ0.0051),

ϩ(x Ϫ 1)(x Ϫ 1.02)

0.04 # 0.02# 0.9784 ϭ x2 Ϫ 2.580x ϩ 2.580; 0.9943, 0.9835

p2(x) ϭ(x Ϫ 1.02)(x Ϫ 1.04)

(Ϫ0.02)(Ϫ0.04)# 1.0000 ϩ

(x Ϫ 1)(x Ϫ 1.04)

0.02 (Ϫ0.02)# 0.9888

ϭ 0.1086 # 9.3 ϩ 1.230 ϭ 2.2297

L0(x) ϭ Ϫ2x ϩ 19, L1(x) ϭ 2x Ϫ 18, p1(9.3) ϭ L0(9.3) # f0 ϩ L1(9.3) # f1

x2 ϭ 1.5, x3 ϭ 1.76471, Á , x7 ϭ 1.83424

(bN Ϫ aN)>2(aN ϩ bN)>2[aN, bN]

[aN, bN]

ƒanϩ1 Ϫ bnϩ1 ƒ Ͻ P ƒ c ƒ

bnϩ1 ϭ bn.anϩ1 ϭ c,

bnϩ1 ϭ c.anϩ1 ϭ anf (an) f (bn) Ͻ 0

f (c) ϭ 0

c ϭ 12 (an ϩ bn)

n ϭ 0, 1, Á , N Ϫ 1

P),

P,[a0, b0],

f (a0) f (b0) Ͻ 0.[a0, b0]P,

f (x) ϭ 0

( f, a0, b0, P, N )

x0 ϭ 4.5, x4 ϭ 4.73004

1.834243 (ϭ x4), 0.656620 (ϭ x4), Ϫ2.49086 (ϭ x4)

0.5, 0.375, 0.377968, 0.377964; (b) 1>17



9. at (due to roundoff;

should be 0).

11.

13.

15.

17. Use the fact that the third derivative of a cubic polynomial is constant, so that

is piecewise constant, hence constant throughout under the present assumption.

Now integrate three times.

19. Curvature if is small.


1. 0.747131, which is larger than 0.746824. Why?

3. (exact)

5.

7. 0.693254 (6S-exact 0.693147)

9. 0.073930 (6S-exact 0.073928)

11. 0.785392 (6S-exact 0.785398)

13.

15. (a) (b)

hence 14.

17. 0.94614588, (8S-exact 0.94608307)

19. 0.9460831 (7S-exact)

21. 0.9774586 (7S-exact 0.9774377)

23. Set

25.

27.

29.


17.

19.

21. The same as that of

23.

(error less than 1 unit of the last digit)

25.

27. 0.824

29.

31.

33.

35. (a) (b) (exact)(0.33 Ϫ 2 � 0.23 ϩ 0.13)>0.01 ϭ 1.2(0.43 Ϫ 2 � 0.23 ϩ 0)>0.04 ϭ 1.2,

0.90443, 0.90452 (5S-exact 0.90452)

0.26, M2 ϭ 6, M*2 ϭ 0, Ϫ0.02 Ϲ P Ϲ 0, 0.01

Ϫx ϩ x3, 2(x Ϫ 1) ϩ 3(x Ϫ 1)2 Ϫ (x Ϫ 1)3

x ϭ x4 Ϫ 0.1, Ϫ0.1, Ϫ0.999, Ϫ0.99900399

ϭ 0.05006

x ϭ 20 Ϯ 1398 ϭ 20.00 Ϯ 19.95, x1 ϭ 39.95, x2 ϭ 0.05, x2 ϭ 2>39.95

�a.

44.885 Ϲ s Ϲ 44.995

4.375, 4.50, 6.0, impossible

5(0.1040 Ϫ 12 � 0.1760 ϩ 1

3 � 0.1344 Ϫ 14 � 0.0384) ϭ 0.256

0.08, 0.32, 0.176, 0.256 (exact)

x ϭ 12 (t ϩ 1), dx ϭ 1

2 dt, 0.746824127 (9S-exact 0.746824133)

x ϭ 12 (t ϩ 1), 0.2642411177 (10S-exact), 1 Ϫ 2>e

0.94608693

ƒCM4 ƒ ϭ 24>(180 � (2m)4) ϭ 10�5>2, 2m ϭ 12.8,

f iv ϭ 24>x5, M4 ϭ 24,M2 ϭ 2, ƒKM2 ƒ ϭ 2>(12n2) ϭ 10�5>2, n ϭ 183.

(0.785398126 Ϫ 0.785392156)>15 ϭ 0.39792 # 10�6

P0.5 � 0.03452 (P0.5 ϭ 0.03307), P0.25 � 0.00829 (P0.25 ϭ 0.00820)

0.5, 0.375, 0.34375, 0.335

ƒ f r ƒf s>(1 ϩ f r2)3>2 � f s

gt

4 ϩ 32(x Ϫ 4) ϩ 25(x Ϫ 4)2 Ϫ 11(x Ϫ 4)34 ϩ x2 Ϫ x3, Ϫ8(x Ϫ 2) Ϫ 5(x Ϫ 2)2 ϩ 5(x Ϫ 2)3,

Ϫ 6(x Ϫ 2)3Ϫ1 ϩ 2(x Ϫ 2) ϩ 5(x Ϫ 2)21 Ϫ x2, Ϫ2(x Ϫ 1) Ϫ (x Ϫ 1)2 ϩ 2(x Ϫ 1)3,

1 Ϫ 54 x2 ϩ 1

4 x4

x ϭ 5.8[Ϫ1.39 (x Ϫ 5)2 ϩ 0.58 (x Ϫ 5)3]s ϭ 0.004



1. 3. No solution 5.

7.

9.

11.

13.

15.


1.

3.

5. D1 0 0

6 1 0

3 9 1

T D3 9 6

0 Ϫ6 3

0 0 Ϫ3

T , x1 ϭ Ϫ 115

x2 ϭ 415

x3 ϭ 25

D1 0 0

2 1 0

2 5 1

T D5 4 1

0 1 2

0 0 3

T , x1 ϭ 0.4

x2 ϭ 0.8

x3 ϭ 1.6

c1 0

3 1d c4 5

0 Ϫ1d , x1 ϭ Ϫ4

x2 ϭ 6

x1 ϭ 4.2, x2 ϭ 0, x3 ϭ Ϫ1.8, x4 ϭ 2.0

EϪ1 Ϫ3.1 2.5 0 Ϫ8.7

0 2.2 1.5 Ϫ3.3 Ϫ9.3

0 0 Ϫ1.493182 Ϫ0.825 1.03773

0 0 0 6.13826 12.2765

Ux1 ϭ 0.142856, x2 ϭ 0.692307, x3 ϭ Ϫ0.173912

D5 0 6 Ϫ0.329193

0 Ϫ4 Ϫ3.6 Ϫ2.143144

0 0 2.3 Ϫ0.4

Tx1 ϭ t1 arbitrary, x2 ϭ (3.4>6.12)t1, x3 ϭ 0

D3.4 Ϫ6.12 Ϫ2.72 0

0 0 4.32 0

0 0 0 0

Tx1 ϭ 6.78, x2 ϭ Ϫ11.3, x3 ϭ 15.82

D13 Ϫ8 0 178.54

0 6 13 137.86

0 0 Ϫ16 Ϫ253.12

Tx1 ϭ 3.908, x2 ϭ Ϫ1.998, x3 ϭ 2.557

DϪ3 6 Ϫ9 Ϫ46.725

0 9 Ϫ13 Ϫ51.223

0 0 Ϫ2.88889 Ϫ7.38689

Tx1 ϭ 2, x2 ϭ 1x1 ϭ 7.3, x2 ϭ Ϫ3.2


7.

9.

11.

13. No, since

15.

17.

19.


5. Exact 7.

9. Exact

11. (a)

(b)

13. steps; spectral radius approximately

15. (Jacobi, Step 5);

(Gauss–Seidel)

19.


1.

3.

5. 7. ab ϩ bc ϩ ca ϭ 05, 15, 1, [1 1 1 1 1]

5.9, 113.81 ϭ 3.716, 3, 13 [0.2 0.6 Ϫ2.1 3.0]

18, 1110 ϭ 10.49, 8, [0.125 Ϫ0.375 1 0 Ϫ0.75 0]

1306 ϭ 17.49, 12, 12

[2.00004 0.998059 4.00072]T[1.99934 1.00043 3.99684]T

0.09, 0.35, 0.72, 0.85,8, Ϫ16, 43, 86

x(3)T ϭ [0.50333 0.49985 0.49968]

x(3)T ϭ [0.49983 0.50001 0.500017],

2, 1, 4

x1 ϭ 2, x2 ϭ Ϫ4, x3 ϭ 80.5, 0.5, 0.5

1

16E 21 Ϫ6 Ϫ14 6

Ϫ6 36 Ϫ12 Ϫ4

Ϫ14 Ϫ12 20 Ϫ4

6 Ϫ4 Ϫ4 4

U1

36D 584 104 Ϫ66

104 20 Ϫ12

Ϫ66 Ϫ12 9

TcϪ3.5 1.25

3.0 Ϫ1.0d

xT (ϪA)x ϭ ϪxTAx Ͻ 0; yes; yes; no

E 1 0 0 0

Ϫ1 2 0 0

3 Ϫ1 3 0

2 0 Ϫ1 4

U E1 Ϫ1 3 2

0 2 Ϫ1 0

0 0 3 Ϫ1

0 0 0 4

U , x1 ϭ 2

x2 ϭ Ϫ3

x3 ϭ 4

x4 ϭ Ϫ1

D0.1 0 0

0 0.4 0

0.3 0.2 0.1

T D0.1 0 0.3

0 0.4 0.2

0 0 0.1

T , x1 ϭ 2

x2 ϭ Ϫ1

x3 ϭ 4

1

D 3 0 0

2 3 0

4 1 3

T D3 2 4

0 3 1

0 0 3

T , x1 ϭ 0.6

x2 ϭ 1.2

x3 ϭ 0.4


9. 11.

13.

15.

17.

19. extremely ill-conditioned

21. Small residual but large deviation of .

23.


1. 3.

5. 9.

11.

13.


1. Spectrum

3. Centers Skew-symmetric, hence

5.

7.

9. They lie in the intervals with endpoints Why?

11.

13.

15.

17. Show that

19. 0 lies in no Gerschgorin disk, by (3) with hence


1.

3.

5. Same answer as in Prob. 3, possibly except for small roundoff errors.

7. eigenvalues (4S) 1.697,

3.382, 5.303, 5.618

9.

11.

(Step 8)


1. D 0.98 Ϫ0.4418 0

Ϫ0.4418 0.8702 0.3718

0 0.3718 0.4898

Tƒ P ƒ Ϲ 1.633, Á , 0.7024

q ϭ 1, Á , Ϫ2.8993 approximates Ϫ3 (0 of the given matrix),

P2 Ϲ yTy>xTx Ϫ (yTx>xTx)2 ϭ l2 Ϫ l2 ϭ 0

x ϭ lxTx, yTy ϭ l2xTx,y ϭ Ax ϭ lx, yT

q ϭ 5.5, 5.5738, 5.6018; ƒ P ƒ Ϲ 0.5, 0.3115, 0.1899;

q Ϯ d ϭ 4 Ϯ 1.633, 4.786 Ϯ 0.619, 4.917 Ϯ 0.398

q ϭ 10, 10.9908, 10.9999; ƒ P ƒ Ϲ 3, 0.3028, 0.0275

det A ϭ l1Á ln � 0.Ͼ;

AAT

ϭ ATA.

10.52 ϭ 0.7211

1122 ϭ 11.05

r (A) Ϲ Row sum norm �A �ϱ ϭ maxjak

ƒajk ƒ ϭ maxj

( ƒajj ƒ ϩ Gerschgorin radius)

ajj Ϯ (n Ϫ 1) # 10�5.

t11 ϭ 100, t22 ϭ t33 ϭ 1

2, 3, 8; radii 1 ϩ 12, 1, 12; actually (4S) 1.163, 3.511, 8.326

l ϭ i�, Ϫ0.7 Ϲ � Ϲ 0.7.0; radii 0.5, 0.7, 0.4.

{Ϫ1, 4, 9}5, 0, 7; radii 6, 4, 6.

Ϫ 1.778x2 ϩ 2.852x32.552 ϩ 16.23x, Ϫ4.114 ϩ 13.73x ϩ 2.500x2, 2.730 ϩ 1.466x

1.89 Ϫ 0.739x ϩ 0.207x2

Ϫ11.36 ϩ 5.45x Ϫ 0.589x2s ϭ 90t Ϫ 675, vav ϭ 90 km>hr

1.48 ϩ 0.09x1.846 Ϫ 1.038x

27, 748, 28,375, 943,656, 29,070,279

x~[0.145 0.120],

[Ϫ2 4]T, [Ϫ144.0 184.0]T, � ϭ 25,921,

167 Ϲ 21 # 15 ϭ 315

� ϭ 20 # 20 ϭ 400; ill-conditioned

� ϭ 19 # 13 ϭ 247; ill-conditioned

� ϭ (5 ϩ 15)(1 ϩ 1>15) ϭ 6 ϩ 215� ϭ 5 # 12 ϭ 2.5


3.

5.

7. Eigenvalues 16, 6, 2

9. Eigenvalues (4S)


15.

17.

19.

21.

Exact:

23.

Exact:

25. 27. 30

29. 5 31.

33. 35.

37. Centers respectively. Eigenvalues (3S)

39. Centers respectively; eigenvalues 0, 4.446, Ϫ9.4460, Ϫ1, Ϫ4; radii 9, 6, 7,

2.63, 40.8, 96.615, 35, 90; radii 30, 35, 25,

1.514 ϩ 1.129x Ϫ 0.214x25 # 2163 ϭ 5

3

115 # 0.4458 ϭ 51.27

42, 1674 ϭ 25.96, 21

[2 1 4]T

D1.700

1.180

4.043

T , D1.986

0.999

4.002

T , D2.000

1.000

4.000

T[6.4 3.6 1.0]T

D5.750

3.600

0.838

T , D6.400

3.559

1.000

T , D6.390

3.600

0.997

TD 0.28193 Ϫ0.15904 Ϫ0.00482

Ϫ0.15904 0.12048 Ϫ0.00241

Ϫ0.00482 Ϫ0.00241 0.01205

T[Ϫ2 0 5]T

[3.9 4.3 1.8]T

D141.4 1.166 0

1.166 68.66 0.1661

0 0.1661 Ϫ30.04

TD141.1 4.926 0

4.926 68.97 0.8691

0 0.8691 Ϫ30.03

T , D141.3 2.400 0

2.400 68.72 0.3797

0 0.3797 Ϫ30.04

T ,141.4, 68.64, Ϫ30.04

D 15.8299 Ϫ1.2932 0

Ϫ1.2932 6.1692 0.0625

0 0.0625 2.0010

TD 11.2903 Ϫ5.0173 0

Ϫ5.0173 10.6144 0.7499

0 0.7499 2.0952

T , D 14.9028 Ϫ3.1265 0

Ϫ3.1265 7.0883 0.1966

0 0.1966 2.0089

T ,

E 3 Ϫ67.59 0 0

Ϫ67.59 143.5 45.35 0

0 45.35 23.34 3.126

0 0 3.126 Ϫ33.87

UD 7 Ϫ3.6056 0

Ϫ3.6056 13.462 3.6923

0 3.6923 3.5385



1. (errors of

3. (set (errors of

5. 0.0013, 0.0042 (errors of

7. 0.00029, 0.01187 (errors of

9. Errors 0.03547 and 0.28715 of and much larger

11. error

(use RK with

13. error

15. 0.18, 0.74, 1.73, 3.28, 5.59, 9.04, 14.3, 22.8,

36.8, 61.4

17. 0.2, 3.1, 10.7, 23.2, 28.5,

19. Errors for Euler–Cauchy 0.02002, 0.06286, 0.05074; for improved Euler–Cauchy

0.012086, 0.009601; for Runge–Kutta. 0.0000011, 0.000016, 0.000536


1.

3.

5. RK error smaller in absolute value,

(for

7. 0.232490 (0.34), 0.236787 (0.44),

0.240075 (0.42), 0.242570 (0.35), 0.244453 (0.25), 0.245867 (0.16), 0.246926 (0.09)

9.

13. from to

15. (a) 0, 0.02, 0.0884, 0.215848, (poor)

(b) By 30–


1. errors of (of from 0.002 to 0.5

(from to 0.1), monotone

3. error

exact

5.

(error 0.005), 0.61 (0.01), 0.429 (0.012), 0.2561 (0.0142), 0.0905 (0.0160)

7. By about a factor

9. Errors of (of from to (from to

11.

continuation will give an “ellipse.”(Ϫ0.81);(Ϫ0.59),

Ϫ0.91), (Ϫ0.42,Ϫ0.97), (Ϫ0.23,Á , 0.92), (0.39, 0.98), (0.20, 1),y2) ϭ (0,(y1,

0.6 # 10�5)0.3 # 10�51.3 # 10�50.3 # 10�5y2)y1

Pn(y2) # 106 ϭ 0.08, Á , 0.27

105. Pn (y1) # 106 ϭ Ϫ0.082, Á , Ϫ0.27,

y1r ϭ y2, y2r ϭ y1 ϩ x, y1 (0) ϭ 1, y2 (0) ϭ Ϫ2, y ϭ y1 ϭ e�x Ϫ x, y ϭ 0.8

y ϭ cos 12 xϪ0.005, Ϫ0.01, Ϫ0.015, Ϫ0.02, Ϫ0.0229;

y1r ϭ y2, y2r ϭ Ϫ14 y1, y ϭ y1 ϭ 1, 0.99, 0.97, 0.94, 0.9005,

Ϫ0.01

y2)y1y1 ϭ Ϫe�2x ϩ 4ex, y2 ϭ Ϫe�2x ϩ ex;

50%

y4 ϭ 0.417818, y5 ϭ 0.708887

0.7: Ϫ5, Ϫ11, Ϫ19, Ϫ31, Ϫ41x ϭ 0.3y ϭ exp (x2). Errors # 105

0.133156 (Ϫ74)

0.095411 (Ϫ54),0.066096 (Ϫ39),0.043810 (Ϫ26),0.027370 (Ϫ17),

0.015749 (Ϫ10),y10 (error # 107) 0.008032 (Ϫ4),Á ,y4,y ϭ exp (x3) Ϫ 1,

y ϭ 1>(4 ϩ e�3x), y4, Á , y10 (error # 105)

x ϭ 0.4, 0.6, 0.8, 1.0)

error # 105 ϭ 0.4, 0.3, 0.2, 5.6

1.557626 (Ϫ22)

1.260288 (Ϫ13),1.029714 (Ϫ7.5), 0.842332 (Ϫ4.4),0.684161 (Ϫ2.4),

0.546315 (Ϫ1.2),y10 (error # 105) 0.422798 (Ϫ0.49),Á ,y4,y ϭ tan x,

P10 ϭ Ϫ1.8 # 10�6y10 ϭ 2.718284,y10* ϭ 2.718276,

P5 ϭ Ϫ3.8 # 10�8,y5 ϭ 1.648722,y5* ϭ 1.648717,y ϭ ex,

Ϫ0.000455,

Ϫ3489, ϩ80444

Ϫ1656,Ϫ376,Ϫ32.3,yr ϭ 1>(2 Ϫ x4); error # 109:

y ϭ 3 cos x Ϫ 2 cos2 x; error # 107:

0.83 # 10�7, 0.16 # 10�6, Á , Ϫ0.56 # 10�6, ϩ0.13 # 10�5y ϭ tan x;

h ϭ 0.2)P ϭ 0.0002>15 ϭ 1.3 # 10�5ϩ9 # 10�6;Ϫ4 # 10�8, Á , Ϫ6 # 10�7,Ϫ10�8,y ϭ 1>(1 Ϫ x2>2);

y10y5

y5, y10)y ϭ 1>(1 Ϫ x2>2),

y5, y10)y ϭ ex,

y5, y10)y Ϫ x ϭ u), 0.00929, 0.01885y ϭ x Ϫ tanh x

y5, y10)y ϭ 5e�0.2x, 0.00458, 0.00830



3.

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) . (b) Reduce to 4 equations by symmetry.

13. at the others

15. otherwise

17. (0.1083, 0.3248 are 4S-values

of the solution of the linear system of the problem.)


5.

7. A, as in Example 1, right sides

Solution

13.

Here with from the stencil.

15. (4S)


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 0.100, 0.163

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 0.06538, 0.10603, 0.10565, 0.6543

15.


1.

3. For we obtain

etc.

5.

7. 0.190, 0.308, 0.308, 0.190, (3S-exact: 0.178, 0.288, 0.288, 0.178)

Á (t ϭ 0.2)1.357,

1.296,1.135,0.935,Á (t ϭ 0.1); 0, 0.575,1.834,1.679,1.271,0.766,0.354,0,

Ϫ0.08, Ϫ0.16 (t ϭ 0.6),

0.24, 0.40 (t ϭ 0.2), 0.08, 0.16 (t ϭ 0.4),x ϭ 0.2, 0.4

u (x, 1) ϭ 0, Ϫ0.05, Ϫ0.10, Ϫ0.15, Ϫ0.20, 0

0.1018, 0.1673, 0.1673, 0.1018 (t ϭ 0.04), 0.0219, 0.0355, Á (t ϭ 0.20)

(t ϭ 0.20)

(t ϭ 0.04),

(t ϭ 0.08)(t ϭ 0.04),

b ϭ [Ϫ200, Ϫ100, Ϫ100, 0]T; u11 ϭ 73.68, u21 ϭ u12 ϭ 47.37, u22 ϭ 15.79

43Ϫ14

3 ϭ Ϫ43 (1 ϩ 2.5)

u12 ϭ 1.u21 ϭ 4,u11 ϭ u22 ϭ 2,2u21 ϩ 2u12 Ϫ 12u22 ϭ Ϫ14,

u11 Ϫ 4u12 ϩ u22 ϭ 0,u11 Ϫ 4u21 ϩ u22 ϭ Ϫ12,Ϫ4u11 ϩ u21 ϩ u12 ϭ Ϫ3,

u11 ϭ u21 ϭ 125.7, u21 ϭ u22 ϭ 157.1

Ϫ220, Ϫ220, Ϫ220, Ϫ220.

u11 ϭ 0.766, u21 ϭ 1.109, u12 ϭ 1.957, u22 ϭ 3.293

13, u11 ϭ u21 ϭ 0.0849, u12 ϭ u22 ϭ 0.3170.

u21 ϭ u23 ϭ 0.25, u12 ϭ u32 ϭ Ϫ0.25, ujk ϭ 0

u12 ϭ u32 ϭ 31.25, u21 ϭ u23 ϭ 18.75, ujk ϭ 25

u13 ϭ u23 ϭ u33 ϭ 0

u12 ϭ u32 ϭ Ϫu14 ϭ Ϫu34 ϭ Ϫ64.22, u22 ϭ Ϫu24 ϭ Ϫ53.98,

u11 ϭ u31 ϭ Ϫu15 ϭ Ϫu35 ϭ Ϫ92.92, u21 ϭ Ϫu25 ϭ Ϫ87.45,

u11 ϭ Ϫu12 ϭ Ϫ66

Ϫ3u11 ϩ u12 ϭ Ϫ200, u11 Ϫ 3u12 ϭ Ϫ100



17. 0.038, 0.125 (errors of and

19.

21.

23.

25.

27.

errors between and

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 3 time steps)

33.

35. 0.043330, 0.077321, 0.089952, 0.058488 0.010956, 0.017720, 0.017747,

0.010964


3.

9. Step 5: value 0.000016


7. No

9. is the unused time on respectively.

11.

13.

15.

17. (copper),

19.

21.


3.

5. Eliminate in Column 3, so that 20 goes.

7.

9. on the segment from (3, 0, 0) to (0, 0, 2)

11. We minimize! The augmented matrix is

T0 ϭ D1 1.8 2.1 0 0 0

0 15 30 1 0 150

0 600 500 0 1 3900

T .fmax ϭ 6

fmax ϭ f (6021 , 0, 1500

105 , 0) ϭ 22007

fmin ϭ f (0, 12 ) ϭ Ϫ10.

f (120>11, 60>11) ϭ 480>11

37,500

fmax ϭ f (210, 60) ϭx1>3 ϩ x2>2 Ϲ 100, x1>3 ϩ x2>6 Ϲ 80, f ϭ 150x1 ϩ 100x2,

f ϭ x1 ϩ x2, 2x1 ϩ 3x2 Ϲ 1200, 4x1 ϩ 2x2 Ϲ 1600, fmax ϭ f (300, 200) ϭ 500

fmax ϭ f (45, 30) ϭ 8400

0.5x1 ϩ 0.25x2 Ϲ 30, f ϭ 120x1 ϩ 100x2,0.5x1 ϩ 0.75x2 Ϲ 45

f (9, 6) ϭ 360

f (Ϫ113 , 26

3 ) ϭ 198 13

f (2.5, 2.5) ϭ 100

M1, M2,x3, x4

(0.11247, Ϫ0.00012),

f (x) ϭ 2(x1 Ϫ 1)2 ϩ (x2 ϩ 2)2 Ϫ 6; Step 3: (1.037, Ϫ1.926), value Ϫ5.992

(t ϭ 0.20)

(t ϭ 0.04),

u (P22) ϭ 60u (P12) ϭ u (P32) ϭ 90,

u (P21) ϭ u (P13) ϭ u (P23) ϭ u (P33) ϭ 30,u (P11) ϭ u (P31) ϭ 270,

(t ϭ 0.3.

10�510�6

y1r ϭ y2, y2r ϭ 2ex Ϫ y1, y ϭ ex Ϫ cos x, y ϭ y1 ϭ 0, 0.241, 0.571, Á ;

y1r ϭ y2, y2r ϭ x2y1, y ϭ y1 ϭ 1, 1, 1, 1.0001, 1.0006, 1.002

Ϫ2.5 # 10�5)

y ϭ sin x, y0.8 ϭ 0.717366, y1.0 ϭ 0.841496 (errors Ϫ1.0 # 10�5,

0.5463023 (1.8 # 10�7)0.4227930 (2.3 # 10�7),

0.3093360 (2.1 # 10�7),0.1003346 (0.8 # 10�7) 0.2027099 (1.6 # 10�7),

1.5538 (0.0036)1.2593 (0.0009),1.0299 (Ϫ0.0002),0.84295 (Ϫ0.00066),

0.68490 (Ϫ0.00076),0.54702 (Ϫ0.00072),0.42341 (Ϫ0.00062),

0.30981 (Ϫ0.00048),0.20304 (Ϫ0.00033),0.10050 (Ϫ0.00017),y ϭ tan x; 0 (0),

y10)y5y ϭ ex,


The pivot is 600. The calculation gives

The next pivot is The calculation gives

Hence has the maximum value so that f has the minimum value 13.5, at

the point

13.


1.

3.

5.

7.

9.


9. Step 5: Slower. Why?

11. Of course! Step 5:

17.

19.


9. 11.

13. D0 1 1

0 0 1

1 1 0

TE0 1 1 1

0 0 0 0

1 0 0 0

0 0 0 0

UD0 1 0

0 0 1

1 0 0

Tf (3, 6) ϭ Ϫ54

f (2, 4) ϭ 100

[Ϫ1.003 1.897]T

[0.353 Ϫ0.028]T.

f (4, 0, 12) ϭ 9

f (1, 1, 0) ϭ 13

f (10, 5) ϭ 5500

f (20, 20) ϭ 40

f (6, 3) ϭ 84

fmax ϭ f (5, 4, 6) ϭ 478

(x1, x2) ϭ a2400

600,

105>235>2 b ϭ (4, 3).

Ϫ13.5,Ϫf

T2 ϭ D1 0 0 Ϫ 6175 Ϫ 3

1400 Ϫ272

0 0 352 1 Ϫ 1

40105

2

0 600 0 Ϫ2007

127 2400

T Row 1 Ϫ 1.235 Row 2

Row 2

Row 3 Ϫ 100035 Row 2

352 .

T1 ϭ D1 0 610 0 Ϫ 3

1000 Ϫ11710

0 0 352 1 Ϫ 1

401052

0 600 500 0 1 3900

T Row 1 Ϫ 1.8600 Row 3

Row 2 Ϫ 15600 Row 3

Row 3


15. 1 2

3 4

17. If G is complete.

Edge

19.


1. 5 3. 4

5. The idea is to go backward. There is a adjacent to and labeled etc.

Now the only vertex labeled 0 is s. Hence implies so that

is a path that has length k.

15. Delete the edge

17. No


1.

5.

7.

9.


21. 4 Ϫ 3 Ϫ 5 L ϭ 10

1

13. 5 Ϫ 3 Ϫ 6

ì

êL ϭ 17

2 Ϫ 4

2

5. 1 ì

ê 3 L ϭ 124 ì

ê

5

9. Yes

2

11. 1 Ϫ 3 Ϫ 4 ì

êL ϭ 38

5 Ϫ 6

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.

(1, 5), (2, 3), (2, 6), (3, 4), (3, 5); L2 ϭ 9, L3 ϭ 7, L4 ϭ 8, L5 ϭ 4, L6 ϭ 14

(1, 2), (2, 4), (3, 4); L2 ϭ 10, L3 ϭ 15, L4 ϭ 13

(1, 2), (2, 4), (3, 4), (3, 5); L2 ϭ 2, L3 ϭ 4, L4 ϭ 3, L5 ϭ 6

(1, 2), (2, 4), (4, 3); L2 ϭ 12, L3 ϭ 36, L4 ϭ 28

(2, 4).

s: vkv0 Ϫ v1 Ϫ Á Ϫ vk�1 Ϫ vk

v0 ϭ s,l(v0) ϭ 0

k Ϫ 1,vkvk�1

EϪ1 Ϫ1 1 Ϫ1

1 0 0 0

0 1 Ϫ1 0

0 0 0 1

U1

2

3

4

e1

e2

e3

e4


Ver

tex

ì

ê

19. If we add an edge to T, then since T is connected, there is a path in T

which, together with forms a cycle.


1. If G is a tree.

3. A shortest spanning tree of the largest connected graph that contains vertex 1.

7.

9.

11.


1.

3.

5.

7.

9. One is interested in flows from s to t, not in the opposite direction.

13.

15.

17.

is unique.

19. For instance,

is unique.


3.

5. By considering only edges with one labeled end and one unlabeled end

7. where 6 is

the given flow

9. where 4

is the given flow

15.


1. No 3. No

5. Yes,

7. Yes, 11.

13. is augmenting and gives and

is of maximum cardinality.

15. is augmenting and gives and

is of maximum cardinality.

19. 3 21. 2

23. 3 25. K4

(1, 4), (3, 6), (7, 8)

1 � 4 Ϫ 3 � 6 Ϫ 7 � 81 Ϫ 4 � 3 Ϫ 6 � 7 Ϫ 8

(3, 7), (5, 4)

(1, 2),1 � 2 Ϫ 3 � 7 Ϫ 5 � 41 Ϫ 2 � 3 Ϫ 7 � 5 Ϫ 4

1 Ϫ 2 � 3 Ϫ 7 � 5 Ϫ 4S ϭ {1, 3, 5}

S ϭ {1, 4, 5, 8}

S ϭ {1, 2, 4, 5}, T ϭ {3, 6}, cap (S, T) ϭ 14

1 Ϫ 2 Ϫ 4 Ϫ 6, ¢t ϭ 2; 1 Ϫ 3 Ϫ 5 Ϫ 6, ¢t ϭ 1; f ϭ 4 ϩ 2 ϩ 1 ϭ 7,

1 Ϫ 2 Ϫ 5, ¢t ϭ 2; 1 Ϫ 4 Ϫ 2 Ϫ 5, ¢t ϭ 1; f ϭ 6 ϩ 2 ϩ 1 ϭ 9,

(2, 3) and (5, 6)

f ϭ 3 ϩ 5 ϩ 7 ϭ 15, f ϭ 15

f12 ϭ 10, f24 ϭ f45 ϭ 7, f13 ϭ f25 ϭ 5, f35 ϭ 3, f32 ϭ 2,

f ϭ 17

f13 ϭ f35 ϭ 8, f14 ϭ f45 ϭ 5, f12 ϭ f24 ϭ f46 ϭ 4, f56 ϭ 13, f ϭ 4 ϩ 13 ϭ 17,

1 Ϫ 2 Ϫ 5, ¢f ϭ 2; 1 Ϫ 4 Ϫ 2 Ϫ 5, ¢f ϭ 2, etc.

P1: 1 Ϫ 2 Ϫ 4 Ϫ 5, ¢f ϭ 2; P2: 1 Ϫ 2 Ϫ 5, ¢f ϭ 3; P3: 1 Ϫ 3 Ϫ 5, ¢f ϭ 4

¢12 ϭ 5, ¢24 ϭ 8, ¢45 ϭ 2; ¢12 ϭ 5, ¢25 ϭ 3; ¢13 ϭ 4, ¢35 ϭ 9

S ϭ {1, 4}, 8 ϩ 6 ϭ 14

{3, 6, 7}, 8 ϩ 4 ϩ 4 ϭ 16

{4, 5, 6}, 10 ϩ 5 ϩ 13 ϭ 28

{3, 6}, 11 ϩ 3 ϭ 14

(1, 4), (4, 3), (4, 5), (1, 2); L ϭ 12

(1, 4), (4, 3), (4, 2), (3, 5); L ϭ 20

(1, 4), (1, 3), (1, 2), (2, 6), (3, 5); L ϭ 32

(u, v),

u: v(u, v)



11.

13. To vertex

From vertex

15.

17.

Vertex Incident Edges

1 (1, 2), (1, 4)

2 (2, 1), (2, 4)

3 (3, 4)

4 (4, 1), (4, 2), (4, 3)

19.

23.


1. 3.

5. 7.

9. 11.

13. 15.

17. 3.54, 1.29


1. outcomes: RRR, RRL, RLR, LRR, RLL, LRL, LLR, LLL

3. outcomes first number (second number) referring

to the first die (second die)

5. Infinitely many outcomes

7. The space of ordered pairs of numbers

9. 10 outcomes:

11. Yes

17. implies by the definition of union. Conversely. implies

that because always and if we must have equality

in the previous relation.

A � B,B � A �B,A �B ϭ B

A � BA � BA �B ϭ B

D ND NND Á NNNNNNNNND

H TH TTH TTTH Á (H ϭ Head, T ϭ Tail)

(1, 1), (1, 2), Á , (6, 6),62 ϭ 36

23

x ϭ 1.355, s ϭ 0.136, IQR ϭ 0.15x ϭ 144.67, s ϭ 8.9735, IQR ϭ 16

x ϭ 19.875, s ϭ 0.835, IQR ϭ 1.5qL ϭ 89.9, qM ϭ 91.0, qU ϭ 91.8

qL ϭ 1.3, qM ϭ 1.4, qU ϭ 1.45qL ϭ 199, qM ϭ 201, qU ϭ 201

qL ϭ 138, qM ϭ 144, qU ϭ 154qL ϭ 19, qM ϭ 20, qU ϭ 20.5

(1, 6), (4, 5), (2, 3), (7, 8)

(1, 2), (1, 4), (2, 3); L2 ϭ 2, L3 ϭ 5, L4 ϭ 5

1 2

4 3

E0 1 0 1

1 0 1 0

0 1 0 1

1 0 1 0

U1

2

3

4

1 2 3 4

E0 0 1 1

0 0 1 1

1 1 0 0

1 1 0 0

UA58 App. 2 Answers to Odd-Numbered Problems


1. 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. (a) (c) same as (a).

Why?

13.

15.

17. hence by disjointedness of B

and


1. In ways

3.

5. 7.

9. In ways 11.

13. (b)

15. P (No two people have a birthday in common)

Answer: which is surprisingly large.


1. by (6)

3. by (10),

5. No, because of (6)

7. because of (6) and

9.

11.

13. if if

if if

Answer: 500 cans,

15. etc.


1. 3. cf. Example 2

5. 7.

9. 11.

13. 15.

17. Product of the 2 numbers. cents

19. (0 ϩ 1 � 3 ϩ 3 � 8 ϩ 1 � 27)>8 ϭ 54>8 ϭ 6 � 75

E (X ) ϭ 12.25, 12X ϭ

12 , 1

20 , (X Ϫ 12 )120$643.50

c ϭ 0.073750, 1, 0.002

C ϭ 12 , � ϭ 2, s2 ϭ 4� ϭ 1

4 , s2 ϭ 116

� ϭ p, s2 ϭ p2>3;k ϭ 12 , � ϭ 4

3 , s2 ϭ 29

X Ͼ b, X м b, X Ͻ c, X Ϲ c,

P ϭ 0.125, 0

x Ϲ 10 Ϲ x Ͻ 1, F(x) ϭ 1F(x) ϭ 1 Ϫ 12 (x Ϫ 1)2

Ϫ1 Ϲ x Ͻ 0x Ͻ Ϫ1, F(x) ϭ 12(x ϩ 1)2F(x) ϭ 0

0.53 ϭ 12.5%

k ϭ 5; 50%

1 ϩ 8 ϩ 27 ϩ 64 ϭ 100k ϭ 1100

P(0 Ϲ X Ϲ 2) ϭ 12k ϭ 1

4

k ϭ 155

41%,

ϭ 365 � 364 Á 346>36520 ϭ 0.59.

1>(12n)

9 � 8 ϭ 726!>6 ϭ 120

210, 70, 112, 28A103 B A52 B A62 B ϭ 18,000

26 �

15 �

44 �

33 �

22 �

11 ϭ 4

6 �35 �

24 �

13 �

22 �

11 ϭ 4!2!

6! ϭ 26 �

15 ϭ 1

15

10! ϭ 3,628,800

A�BcP(A) ϭ P(B) ϩ P(A�Bc) м P (B)A ϭ B � (A�Bc),

1 Ϫ 0.8754 ϭ 0.4138 Ͻ 1 Ϫ 0.752 ϭ 0.4375 Ͻ 0.5 (c Ͻ b Ͻ a)

1 Ϫ 0.963 ϭ 11.5%

(a) ϩ (b) ϩ (c) ϭ 1.

100200 �

99199 ϭ 24.874%, (b)

100200 �

100199 ϩ 100

200 �100199 ϭ 50.25%,

498500 �

497499 �

496498 �

495497 �

494496 � 0.98008

89

(a) 0.93 ϭ 72.9%, (b)90

100 ⅐8999 ⅐

8898 ϭ 72.65%

1 Ϫ 4>216 ϭ 98.15%,



3.

5.

7.

9. Answer:

11.

13.

15.


1. 3.

5. 7.

9. About 11. hours

13. About 683 (Fig. 521a)


1. 3.

5. if

7. 27.45 mm, 0.38 mm

11. 25.26 cm, 0.0078 cm 13.

15. The distributions in Prob. 17 and Example 1

17. No


11.

13.

21. Sum over j from 1.

17.

19.

21. 23.

25.


1. In Example 1, so and is as before.

3.

5.

7.

9.

11.

13.

15. Variability larger than perhaps expected

u ϭ 1

u ϭ n>S x j ϭ 1>xl ϭ f ϭ p(1 Ϫ p)x�1, etc., p ϭ 1>x7>12

l ϭ pk(1 Ϫ p)n�k, p ϭ k>n, k ϭ number of successes in n trails

� ϭ x ϭ 15.3n� ϭ nx,

/ ϭ e�n��(x1ϩÁϩxn)>(x1! Á xn!), 0 ln />0� ϭ Ϫn ϩ (x1 ϩ Á ϩ xn)>� ϭ 0,

s� 2an

jϭ1

x j ϭ 0. 0 ln />0/ ϭ 0� ϭ 0

0.1587, 0.6306, 0.5, 0.4950

1, 12f (x) ϭ 2�x, x ϭ 1, 2, Á

f (x) ϭ A50x B0.03x0.9750�x � 1.5xe�1.5>x!

x ϭ 6, s ϭ 3.65

xmin Ϲ x j Ϲ xmax.

x ϭ 111.9, s ϭ 4.0125, s2 ϭ 16.1

QL ϭ 110, QM ϭ 112, QU ϭ 115

50%

a2 Ͻ y Ͻ b2f2(y) ϭ 1>(b2 Ϫ a2)

29 , 19 , 12

18 , 3

16 , 38

t ϭ 108458%

31.1%, 95.4%15.9%

45.065, 56.978, 2.0220.1587, 0.5, 0.6915, 0.6247

1 Ϫ e�0.2 ϭ 18%

42%, 47.2%, 10.5%, 0.3%

1314 %

9%f (x) ϭ 0.5xe�0.5>x!, f (0) ϩ f (1) ϭ e�0.5(1.0 ϩ 0.5) ϭ 0.91.

0.265

A5x B 0.55, 0.03125, 0.15625, 1 Ϫ f (0) ϭ 0.96875, 0.96875

38%



3. Shorter by a factor 5.

7.

9.

11.

13.

15. degrees of freedom.

Hence

17.

19. is normal with mean 105 and variance 1.25.

Answer:


3. accept the hypothesis.

5. do not reject the hypothesis.

7. accept the hypothesis.

9.

11. Alternative

(Table A9, Appendix 5). Reject the hypothesis

13. Two-sided. (Table A9, Appendix 5),

no difference

15. (Table A10. Appendix 5), accept the

hypothesis

17. By (12), Assert that B is better.


1.

3. 27

5. Choose 4 times the original sample size

9.

11.

13. In about of the cases

15. is negative in (b) and we set


1. 3.

5. 7.

9. 11.

13. (by the normal approximation)

15. (1 Ϫ u)5, 3u(1 Ϫ u)5�14r ϭ 0, u ϭ 16, AOQL ϭ 6.7%

a9

xϭ0

a100

xb 0.12x 0.88100�x ϭ 22%

(1 Ϫ 12 )3 ϩ 3 # 1

2 (1 Ϫ 12 )2 ϭ 1

2(1 Ϫ u)n ϩ nu(1 Ϫ u)n�1

19.5%, 14.7%e�25u(1 ϩ 25u), P(A; 1.5) ϭ 94.5, a ϭ 5.5%

0.8187, 0.6703, 0.13530.9825, 0.9384, 0.4060

UCL ϭ � ϩ 31� ϭ 9.3.

LCL ϭ 0, CL ϭ � ϭ 3.6,LCL ϭ � Ϫ 31�

30% (5%)

LCL ϭ np Ϫ 31np(1 Ϫ p), CL ϭ np, UCL ϭ np ϩ 31np(1 Ϫ p)

2.5810.0004>12 ϭ 0.036, LCL ϭ 3.464, UCL ϭ 3.536

LCL ϭ 1 Ϫ 2.58 # 0.02>2 ϭ 0.974, UCL ϭ 1.026

t0 ϭ 116(20.2 Ϫ 19.6)>10.16 ϩ 0.36 Ͼ c ϭ 1.70.

19 � 1.02>0.82 ϭ 29.69 Ͻ c ϭ 30.14

t ϭ (0.55 Ϫ 0)>10.546>8 ϭ 2.11 Ͻ c ϭ 2.37

� ϭ 5000 g.

� � 5000, t ϭ (4990 Ϫ 5000)>(20>150) ϭ Ϫ3.54 Ͻ c ϭ Ϫ2.01

� Ͻ 58.69 or � Ͼ 61.31

s2>n ϭ 1.8, c ϭ 57.8,

c ϭ 6090 Ͼ 6019:

t ϭ (0.286 Ϫ 0)>(4.31>17) ϭ 0.18 Ͻ c ϭ 1.94;

P (104 Ϲ Z Ϲ 106) ϭ 63%

Z ϭ X ϩ Y

CONF0.95{0.74 Ϲ s2 Ϲ 5.19}

k1 ϭ 12.41, k2 ϭ 7.10. CONF0.95 {7.10 Ϲ s2 Ϲ 12.41}.c2 ϭ 129.6.

F(c2) ϭ 0.975,c1 ϭ 74.2,F (c1) ϭ 0.025,n Ϫ 1 ϭ 99

CONF0.95{0.023 Ϲ s2 Ϲ 0.085}

k ϭ 366.66 (Table 25.2), CONF0.99{9166.7 Ϲ � Ϲ 9900}

n Ϫ 1 ϭ 5, F(c) ϭ 0.995, c ϭ 4.03, x ϭ 9533.33, s2 ϭ 49,666.67,

CONF0.99{63.72 Ϲ � Ϲ 66.28}

CONF0.95{125.3 Ϲ � Ϲ 126.7}, CONF0.95{0.1566 Ϲ p Ϲ 0.1583}

c ϭ 1.96, x ϭ 126, s2 ϭ 126 # 674>800 ϭ 106.155, k ϭ cs>1n ϭ 0.714,

4, 1612



3.

5.

7.

9. 42 even digits, accept.

13. (1 degree of

freedom, 95%)

15. Combining the last three nonzero values, we have 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. Hypothesis Alternative

Hypothesis rejected.

11. Consider

13. transpositions, Assert that fertilizing increases yield.

15. Assert that there is an increase.


1. 3.

5. 7.

9.

13.

15.


15. 17.

19.

21.

23. when For we obtain

If n increases, so does whereas decreases.

25. y ϭ 3.4 Ϫ 1.85x

ba,b ϭ (1 Ϫ u)6 ϭ 37.7%.

u ϭ 15%u ϭ 0.01.a ϭ 1 Ϫ (1 Ϫ u)6 ϭ 5.85%,

2.58 � 10.00024>12 ϭ 0.028, LCL ϭ 2.722, UCL ϭ 2.778

c ϭ 14.74 Ͼ 14.5, reject �0; £((14.74 Ϫ 14.50)>10.025) ϭ 0.9353

CONF0.99{27.94 Ϲ � Ϲ 34.81}� ϭ 20.325, s2 ϭ (78)s2 ϭ 3.982

CONF0.95{0.046 Ϲ �1 Ϲ 0.088}

y Ϫ 1.875 ϭ 0.067(x Ϫ 25), 3sx2 ϭ 500, q0 ϭ 0.023, K ϭ 0.021,

CONF0.95{41.7 Ϲ �1 Ϲ 44.7}.

c ϭ 3.18 (Table A9), k1 ϭ 43.2, q0 ϭ 54,878, K ϭ 1.502,

y ϭ 0.32923 ϩ 0.00032x, y(66) ϭ 0.35035

y ϭ 0.5932 ϩ 0.1138x, R ϭ 1>0.1138y ϭ Ϫ10 ϩ 0.55x

y ϭ Ϫ11,457.9 ϩ 43.2xy ϭ 0.98 ϩ 0.495x

P(T Ϲ 2) ϭ 2.8%.

P(T Ϲ 4) ϭ 0.007.n ϭ 8; 4

yj ϭ x j Ϫ �0� .

t ϭ 110 � 1.58>1.23 ϭ 4.06 Ͼ c ϭ 1.83 (a ϭ 5%).

�� Ͼ 0, x ϭ 1.58,�� ϭ 0.

x ϭ 9.67, s ϭ 11.87. t0 ϭ 9.67>(11.87>115) ϭ 3.16 Ͼ c ϭ 1.76 (a ϭ 5%).

(12)18(1 ϩ 18 ϩ 153 ϩ 816) ϭ 0.0038

(12)8 ϩ 8 # (1

2)8 ϭ 3.5%

10,0942608 � 3.87). 0

2 ϭ 12.8 Ͻ c ϭ 16.92.

K Ϫ r Ϫ 1 ϭ 9 (r ϭ 1

02 ϭ

(355 Ϫ 358.5)2

358.5ϩ

(123 Ϫ 119.5)2

119.5ϭ 0.137 Ͻ c ϭ 3.84

02 ϭ 10.264 Ͻ 11.07; yes

02 ϭ 16

10 Ͼ 11.07; yes

02 ϭ (40 Ϫ 50)2>50 ϩ (60 Ϫ 50)2>50 ϭ 4 Ͼ c ϭ 3.84; no


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Answers to Odd-Numbered Problems - Purdue Universitycaiz/MA527-cai/Appendix2.pdf · 2012-09-19 · A6 App. 2 Answers to Odd-Numbered Problems Problem Set 1.7, page 42 1. ; hence is