O'Neil 7e Errata - Higher Ed eBooks & Digital Learning … · 2013-05-20 · Errata (Includes critical ... Advanced Engineering Mathematics, 7e Peter V. O’Neil ... 134 5th and 6th

Wednesday, May 15, 2013

Errata (Includes critical corrections only for the 1st & 2nd reprint)

Advanced Engineering Mathematics, 7e Peter V. O’Neil

ISBN: 9781111427412

Page # Description

38 Line 14: change "w = v2–a = v2" to "w = v1–a = v2"

46 Line 2: change "y2 = ky2" to "y2 = ky1"

68 3rd line from the bottom: change "…2bcos(ω0t)…" to "…2bω0cos(ω0t)…"

95 Line 7: change … A A

to … A A

103 Theorem 3.6, line 8: change

( )0

( ) ( ).f t d f tτ δ τ τ∞

− =∫

to

( )0

( ) ( ).f t d f tτ δ τ τ∞

− =∫

104 Example 3.17, line 12: change "y(t) = H(t – 3)e(t – 3)sin(t – 3)" to "y(t) = H(t – 3)e–(t – 3)sin(t – 3)"

104 Figure 3.32 should look like this:

108 Line 2: change

1 1 1 2 1 2 2 1( ) ( )m x k k x k x f t′′ = − + + + to

( )1 1 1 2 1 2 2 1( ) ( )m x k k x k f tt x′′ = − + + +


114 Line 14: change "…or ." to "…or / "

114 Line 20: change In order to have lim 0

to In order to have lim 0

115 Line 7: change

( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )2 0 ' 0 0 0 0.d ds Y s sy y sY s y sY s yds ds

− − − + − + − =

to

( ) ( ) ( )( ) ( ) ( ) ( )2 0 ' 0 0 0.d Y sd s Y s sy y sY s ys dsd

− − − + =−−

125 7th line from the bottom: change

73

(7)(6) 0aa

= − =

to

7 31 0

(7)(6)a a= − =

127 Example 4.4, lines 4 & 9: change

2

0 0( )( 1) 5( )n r n r

n nn n

n r n r c x n r c x∞ ∞

+ − +

= =

+ + − + +∑ ∑

to

0 0

( )( 1) 5( )n r n rn n

n n

n r n r c x n r c x∞ ∞

+ +

= =

+ + − + +∑ ∑

132 Lines 10 and 11 should read:

24 1

!2 1

!2 2 3

5 2 4 0

132 2nd line from the bottom: change "(n + r)(n + r – 1)cn + (+r – 1)…" to "(n + r)(n + r – 1)cn + (n + r – 1)…"

134 5th and 6th lines from the bottom: change

( )20 n=0

*1

1 0

1 12!( 1)!!

( 1) 0.

n n

n

n nn n

n n

k x k xn nn

c n n x c x

∞

=

∞ ∞

+= =

− ∞+

+ + − =

∑ ∑

∑ ∑

to


( )20 n=0

* *1

1 0

1 12!( 1)!!

( 1) 0.

n n

n

n nn n

n n

k x k xn nn

c n n x c x

∞ ∞

=

∞ ∞

+= =

−+

+ + − =

∑ ∑

∑ ∑

149 Line 24: change

1 2 1 2 1 2, , .a a b b c c+ = + + +F G to

1 1 2 2 3 3, . ,a b a b a b+ + ++ =F G

163 9th line from the bottom: change .+ ≤ +F G F G

to

.+ ≤ +FF G G

170 3rd line from the bottom: change

21 1 2

1 1 1

1 .kk

ccc c c

= − − − −V G V V

to 2

1 21 1 1

11 .k

kcc

c c c− − −= G VV V

172 Corollary 6.1, line 8: change

1 1 2 21

.k

k k j kj

c c c c=

= + + + =∑X v v v v

to

1 1 2 21

.j

k

k k jj

c c c c=

= + + + =∑X v v v v

176 Line 4: change

3 1 3 1 1 1 0,d⋅ = ⋅ − ⋅ =V V X V V V to

3 1 3 1 1 2 11 0,d h⋅ = ⋅ − ⋅ =− ⋅V V X V V V V V

176 Line 6: change

3 1

1 1

3 112

1

.

d ⋅=

⋅⋅

=

V VV VV V V

V

to


3 1

1 1

3 12

1

.

d =⋅⋅⋅

=

X VV VX V

V

178 Last line: change .S

⊥ ⊥− = −u U u U to

.S⊥ ⊥−− = U uu U


S≠u u to

S≠v u

180 Theorem 6.8, line 12: change 0.S− >u u

to 0.s − >u v

180 Theorem 6.8, line 14: change 2 2

S− > −u v u v to

2 2s−− > u uu v

180 Example 6.21, line 1: change “R5” to “R6”

182 Line 12, change 2

1 2 3 0nnc c x c x c x+ + + + =

to 2

0 1 2 0nnc cc x c x x+ + ++ =

184 Line 11: change

( ) ( ) ( )2

0

8 8sin sin 327

x x x x dxπ

ππ π

⎛ ⎞= − − −⎜ ⎟⎝ ⎠∫

to

( ) ( ) ( )2

0

8 8sin sin 327

x x xx dxπ

π ππ⎛ ⎞= − −⎜

⎝− ⎟

⎠∫

191 6th line from the bottom: change 86

21

to 6

21


229 3rd line from the bottom: change “ ” to “ ”

229 2nd and 3rd lines from the bottom: change “…A and whose second n columns are In.” to “…In and whose second n columns are A.”

230 Example 7.28, line 11: change 1 1/56 1

to 1 1/50 1


12 21 33 13 21 32 13 22 31 .b b b b b b b b b= + − to

12 21 33 13 21 32 13 22 31 .b b b b b b b b b+ −−

250 15th line from the bottom: change "But by (3), |A| = –|B| = |A|." to "But by (3), |A| = –|B| = –|A|."

254 7th line from the bottom: change "Add –1 times…" to "Add –15 times…"

256 Theorem 8.2, line 1: change “1 < i < n” to “1 < k < n”

256 Example 8.4: change 6 3 7

A 12 5 62 4 6

−⎛ ⎞⎜ ⎟= − −⎜ ⎟⎜ ⎟−⎝ ⎠

to 6 3 7

A 12 52 4

96

−⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟−⎝ ⎠

−

284 9.3.1 Orthogonal Matrices, line 2: change .t−1A = A

to .t −1A = A

289 Lemma 9.2, line 5: change

1. If S is n × n skew-hermitian, then t

Z HZ to

1. If S is n × n skew-hermitian, then St

Z Z

289 Lemma 9.2, line 10: change

( ) .t t ttt t t =Z HZ = (Z HZ) Z H Z = Z HZ

to

( ) .tt t tt t t t=Z HZ = (Z HZ) Z = ZZH HZ

292 Example 9.17, line 4: change


1 7/27/2 1

−⎛ ⎞⎜ ⎟−⎝ ⎠

to 1 7/27/2 1

A=−⎛ ⎞

⎜ ⎟−⎝ ⎠

298 4th line from the bottom: change ; (0) .′ = =X AX X O

to

0; ( ) .t′ = =X AX X O

299 Theorem 10.3, line 10: change ( ); (0) j′ = =X AX X E

to

0( ); ( ) jt′ = =X AX X E

301 Lines 3/4: change 3 3

13 3

2

3 3 3 31 2

1 23 3 3 31 2

2 (1 2 )

( 2 ) (1 2 ) 2 (1 2 )

t t

t t

t t t t

t t t t

ce t ece te

c e c t e e t ec c

c e c te e te

⎛ ⎞− + − ⎛ ⎞Ω = ⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − −= = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

C

to 3 3

13 3

2

3 3 3 31 2

1 23 3 3 31 2

2 (1 2 )

( 2 ) (1 2 ) 2 (1 2 )

t t

t t

t t t t

t t t t

ce t ece te

c e c t e e t ec c

c e c te e te

⎛ ⎞− − ⎛ ⎞Ω = ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − −= = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝+ ⎠

+

⎠ ⎝ ⎠⎝

C


5 14 412 11 12 .4 4 5

⎛ ⎞⎜ ⎟′ = −⎜ ⎟⎜ ⎟−⎝ ⎠

X X

to 5 4

12 11 12 .4

4 4 5

⎛ ⎞⎜ ⎟′ = −⎜ ⎟⎜ ⎟−⎝ ⎠

−X X

305 6th line from the bottom: change "The eigenvalues of A are –1/10, –1/5 with corresponding eigenvalues,…" to "The eigenvalues of A are –1/10, –3/5 with corresponding eigenvectors,…"

309 Line 6: change


2 .(1 3 ) / 2

αα

⎛ ⎞= ⎜ ⎟+⎝ ⎠E

to

2 .(1 3 ) / 3

αα

⎛ ⎞= ⎜ ⎟+⎝ ⎠E

310 Line 11: change 2 2

2 1 2

2 2 2

( )

1 1 15 4 4 5 .

1 1 1

t t

t t t

t te e

tte e t e

t

− −

− −

Φ = +

− − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − + − = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠ ⎝ ⎠

E E

to 2 2

2 1 2

2 22

( )

1 1 15 4 4 5 .

1 1 1

t

t tt

tt te e

tte e t e

t

−−

− −

−

Φ = +

− − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − + − = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠ ⎝ ⎠

E E


2 1 32 .− =E E AE to

32 32 .− =E AEE

311 Line 2: change

3

0 1 5 1125 5 0 40 1 5 1

− − −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

E

to

3

0 1 525 5 0 40

1

1 5 1

− −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

−E

311 4th, 7th, 8th, 9th, and 10th lines from the bottom: change "r – k" to "k – r"

318 Line 13: change 3 3 3

0

3 3 3

0

[(1 2 2 ) 4 ( ) ]

[( ) (1 2 2 ) ]

t t s t s

t t s t s

t s e e s t s e e ds

t s e e t s e e ds

− −

− −

⎛ ⎞− + − −⎜ ⎟= ⎜ ⎟⎜ ⎟− + + −⎝ ⎠

∫∫

to 3 3 3

0

3 3 3

0

[(1 2 2 ) 4 ( ) ]

[( ) (1 2 2 ) ]

t t s t s

t t s t s

t s e e s t s e e ds

t s e e t s es e ds

− −

− −

⎛ ⎞− + − −⎜ ⎟= ⎜ ⎟⎜ ⎟− + + −⎝ ⎠

∫∫



( ) 3 3 3 3 1cos sin ,10 10 10 10 10

s s s⎛ ⎞ ⎛ ⎞′ = − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

G i j k

to

( ) 3 3 3 3 1 ,10 10 1

sin cos0 10 10

s s s⎛ ⎞ ⎛ ⎞′ = − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

G i j k

357 Line 2: change 2 2 2 22 cos( ) [ cos( ) sin( )] sin( ) .xy yz x yz x z yz x y yzϕ∇ = + − −i j k

to 2 2 2 22 cos( ) [ cos( ) sin( )] sin( ) .xy yz x yz x z yz x y yzyϕ∇ = + − −i j k

361 Example 11.8, line 1: change 2 2( , , )x y z z x yϕ = − +

to 2 2 0( , , )x y z z x yϕ = + =−


1 1 1 2 2 2( ( ), ) ), ( )) ( ( ), ( ), ( ));x t y t z t x t y t z t≠ to

1 1 1 2 2 2( ( ), ), ( )) ( ( ), (( ), ( ));x t y t z t x t y t z t≠

373 Example 12.7, lines 5, 9, & 13: change 4sin 4cos

to 4sin 4cos

373 3rd line from the bottom: change 16 2 2 2

0

3 3[2 cos( )][2 sin( )] 4 sin ( ) 4 cos16

t t t dt= +∫

to 2 2 2

0

/23 3[2 cos( )][2 sin( )] 4 sin ( ) 4 cos16

( )t t t dttπ

= +∫

380 12.4 Independence of Path and Potential Theory, line 9: change

C Cd dx dy dyx y zϕ ϕ ϕ∂ ∂ ∂= + +∂ ∂ ∂∫ ∫F. R

to

C Cd dxy

zdy dx zϕ ϕ ϕ∂ ∂ ∂= + +∂ ∂ ∂∫ ∫F. R

390 Example 12.16, line 2: change ( ) ( )cos , sin ,x au v y au v z u= = =

to ( ) ( )cos , sin ,x au v y ub v z u= = =


391 11th line from the bottom: change / /( , ) ( , ) ,1 0, ( , )

S x S yz x z x Su v x y y

∂ ∂ ∂ ∂∂ ∂ ∂= = = −

∂ ∂ ∂

to / /( , ) ( , ) ,1 0( ,( ), )

S x S yz x z x Sv yu x y

∂ ∂ ∂ ∂∂ ∂ ∂= = = −

∂ ∂ ∂ 394 Line 9: change

2 5 2

0 1cos( )sin( )sin(2 ) 1v v v dv u u du

π= +∫ ∫

to 2 5 2

0 1cos( )sin( )sin(2 ) 11

2v v v dv u u du

π= +∫ ∫

394 Example 12.21, change

: 0 2,0 1 1.D x≤ ≤ ≤ ≤ to

: 0 2, 1.y0D x≤ ≤ ≤ ≤


1

mass of ( ) ( )n

j jj

P P u vδ=

≈ Δ Δ∑ NΣ

to

1

mass of ( ) ( )jj

j

n

PP u vδ=

≈ Δ Δ∑ NΣ

397 Last line: change ( )n j

n j

V t AV A t

tΔ

= = ⋅ ΔΔ

V n

to ( )n j

n j j

V t AV A

tA

Δ= = ⋅

ΔV n

401 Line 12: change

( ) .g fx y∂ ∂

∇× = −∂ ∂

F

to

( ) .g fx y

k ∂ ∂∇× = −⋅

∂ ∂F

401 Line 14: change

( ) ( )[ , , ] dx dyds f x y g x yds ds

⎛ ⎞⋅ = + ⋅ +⎜ ⎟⎝ ⎠

F T i j i k

to


( ) ( )[ , , ] dx dyds f x y g yds s

dxd

s⎛ ⎞⋅ = + ⋅ +⎜ ⎟⎝ ⎠

F T i j ji

410 Line 5: change 2

0[ 3 sin( )( 3 cos( )) 3 cos( )(3 cos( ))]t t t t dt

π= − − +∫

to 2

0[ 3 sin( )( 3 ( )) 3 cos( )(3 cos( )s )in ]t t t t dt

π= − − +∫

410 Line 15: change 2 3 2 2 2 2

0 0[ cos ( ) sin ( )] r r r dr d

πθ θ θ= −∫ ∫

to 2 3 2 2 2 2

0 0[ cos ( ) sin ( ) ] 2r r r dr d

πθ θ θ− += ∫ ∫

414 Line 11: change "q1 = q2 (x, y, z)…" to "q1 = q1 (x, y, z)…"

428 Line 20: change

( ) ( )30

1 11 4 ( )sin .n

nb x x nx dxn

ππ

π π− −

= − =∫

to

( ) ( )30

1 12 4 ( )sin .n

nb x x nx dxn

ππ

π π− −

= − =∫429 8th line from the bottom: change

( )0 01 2 .2

a L aπ= =

to

( )0 01 2 .2

a L La= =

430 Equation (13.6): change 1 ( )cos( / ) .

L

n La f x n x L dx

Lπ= ∫

to 1 ( ) cos( / ) .

L

n La f x n x L dx

Lπ

−= ∫

430 5th & 6th lines from the bottom: change "[L, L]" to “[–L, L]"

432 Figure 13.2 should look like this:


439 5th line from the bottom: change "PL[–L, L]" to "PC[–L, L]"

442 Theorem 13.2, line 2: change "[0, L]” to "(0, L)”

444 Theorem 13.3, line 2: change "[0, L]” to "(0, L)"

446 Last line: change

1

1

2 ( 1) (cos( ) ( 1) ).n n

nnx

n

∞+

=

= − − −∑

to

21

2 ( 1) (cos( ) ( 1) ).n n

nnx

n

∞

=

= − − −∑

447 Lines 11-14: change

( )

( )

0

0

1 ( ) cos

1 1( ) sin sin ( )

1 1 sin2

1 1sin sin2

.

L

n L

LL

LL

L

L

L L

L L

n

n xA F t dtL L

L n x L n xF t F t dtL n L L n L

n xf t a dtn L

n x n xf t dt a dtn L n LL b

n

π

π ππ π

ππ

π ππ π

π

−

−−

−

− −

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞ ′= −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −

∫

∫

∫

∫ ∫

to


( )

( )

0

0

1 ( ) cos

1 1( ) sin sin ( )

1 1 sin2

1 1sin sin2

.

L

n L

LL

LL

L

L

L L

L L

n

nA F t dtL L

L n L nF t F t dtL n L L n L

nf t a dtn

t

t

Ln nf t dt a dt

n L

t

t

tn L

L bn

t

π

π ππ π

ππ

π ππ π

π

−

−−

−

− −

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞ ′= −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −

∫

∫

∫

∫ ∫


( ) 01

cos( )2 n

n

L LF L a b nππ

∞

=

= − ∑

to

( ) 02LF L a=

447 3rd line from the bottom: 12

to

2


( ) 12

2 21

14 16 cos( / 2)3

n

nx n x

nπ

π

+∞

=

−= + ∑

to

( )22 2

1

14 16 cos( / 2)3 n

n

x n xn

ππ

∞

=

−= + ∑

449 Line 3: change “3. If exists, then the Fourier coefficients of f(x)…” to “3. If exists, then the Fourier coefficients of g(x)…”

449 Line 4: change “(g(x))2” to “(f(x))2”

450 Line 4: change

2 2

01

2 ( ( )) .N L

nn

b f x dxL=

≤∑ ∫

to


2 2

01

2 ( ( )) .N L

nn

gb x dxL=

≤∑ ∫450 Line 6: change

2 2

01

2 ( ( )) .L

nn

b f x dxL

∞

=

≤∑ ∫

to

2 2

01

2 ( ( )) .L

nn

b x xL

g d∞

=

≤∑ ∫458 6th line from the bottom: change

0 01 1 ( )2

L

Ld a f x dx

L −= = ∫

to

0 01 1 ( )2 2

L

Ld a f x dx

L −= = ∫

466 Line 11: change

( ) ( )( ) ( )

( ) ( )( ) ( )

1 cos cos

sin sin .

f d x

d x d

ξ ωξ ξ ωπ

ξ ωξ ξ ω ω

∞

−∞

∞

−∞

⎡⎢⎣

⎤+ ⎥⎦

∫

∫

to

( ) ( )( ) ( )

( ) ( )( ) ( )

1 cos cos

sin sin .f

f d x

d x d

ξ ωξ ξ ωπ

ξ ωξ ξ ω ω

∞

−∞

∞

−∞

⎡⎢⎣

⎤+ ⎥⎦

∫

∫


( ) ( )1 cos 0.eB f dω ξ ωξ ξπ

∞

−∞= =∫

to

( ) ( )s n1 0.ieB f dω ξ ωξ ξπ

∞

−∞= =∫


( ) ( ) ( )

( )

55 5

0 0

5

0

ˆ

1 1 .5 5

i tt i t t i t

i t

f H t e e dt e e dt e dt

ei i

ωω ω

ω

ω

ω ω

∞ ∞ ∞ − +− − − −

−∞

∞− +

= =

⎡ ⎤= − =⎣ ⎦+ +

∫ ∫ ∫

to

( ) ( ) ( )

( )

55 5

0 0

5

0

ˆ

1 1 .5 5

i tt i t t i t

i t

f H t e e dt e e dt e dt

ei i

ωω ω

ω

ω

ω ω

∞ ∞ ∞ − +− − − −

−∞

∞− +

= =

⎡

=

⎤= − =⎣ ⎦+ +

∫ ∫ ∫


476 Proof: change

( )

0 0

00

[ ( )]( ) ( )

ˆ ( ).t

t ti i t

i

e f t e f t e dt

e dt f

ω ω ω

ω ω

ω

ω ω

∞− −

−∞

∞ − −

−∞

=

= = −

∫∫

F

to

( )

0 0

00

[ ( )]

(

( ) (

)

)

ˆ ( ).t

t ti i t

i

ie f t e

f t

f t e dt

e dt f

ω ω ω

ω ω

ω

ω ω

∞ −

−∞

∞ − −

−∞

=

= = −

∫∫

F


( ) ( )

ˆ ( ) [ ( )]( )

[ ( )]( ) .t

f g t

i g t i f d

ω ω

ω ω ω τ τ ω−∞

′=

⎡ ⎤= = ⎢ ⎥⎣ ⎦∫

F

F

to

( ) ( )

ˆ ( ) [ ( )]( )

[ ( )]( ) .t

f g t

i g t i f d

ω ω

ω ω ω τ τ ω−∞

′=

⎡ ⎤= = ⎢ ⎥⎣ ⎦∫F

F

481 14.3.1 Filtering and the Dirac Delta Function, line 6: change “H(t – t0)” to “ δ(t – t0)”

482 Line 12: change

( ) ( )1

1[ ( )] ( )]

sin1 2 .

aa i t i t

a

ia ia

H t a H t a e dt ei

ae e

i

ω ω

ω ω

ωω

ω ω

− −

−−

−

⎤+ − − = = − ⎥⎦

= − =

∫F

to

( ) ( )

1[ ( )] ( )]

sin1 2 .

aa i t i t

a

ia ia

a

H t a H t a e dt ei

ae e

i

ω ω

ω ω

ωω

ω ω

− −

−

−

−

⎤+ − − = = − ⎥⎦

= − =

∫F

486 Line 11: change

/1( ) .2

L n i L i t

L

nf t f e e dL L

π ω ωπ π ωπ

−

−

⎛ ⎞= ⎜ ⎟⎝ ⎠∫

to

/1( ) .2

L n i L i t

Ln

nf t f e e dL L

π ω ωπ π ωπ

∞−

−=−∞

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑∫487 Line 8: change

( )0 0

1 ˆ ) ...2

i tf t f e ωω ω

ωπ

∞

−∞= ∫

to


( )0 0

1 ˆ ) ...2

( i tf t f e ωω ω

ωπ

∞

−∞= ∫

493 2nd line from the bottom: change 1 1

2 / 2 /

0 0

1 12 2

N Nija ijk N ija ijk N

kj k

U e e e ei i

π π− −

− − −

= =

= −∑ ∑

to 1 1

2 / 2 /

0 0

1 12 2

N Nija ijk N ija ijk N

kjj

U e e e ei i

π π− −

− − −

= =

= −∑ ∑

494 Line 2: change

( ) ( )2 / 2 /

2 / 2 /

1 11 12 1 2 1

N Nia ijk N ia ijk N

k ia ijk N ia ijk N

e eU

i e i e

π π

π π

− −

− − −

− −= −

− −

to

( ) ( )2 / 2 /

2 / 2 /

1 11 12 1 2 1

N Nia ik N ia ik N

k ia ik N ia ik N

e eU

i e i e

π π

π π

− − −

− − −

− −= −

− − 494 Line 9: change

5 2 5 2

2 2 /5 2 2 /5

1 1 1 1 .2 21 1

i

k i ik i ik

e eUi ie eπ π

−

− − −

− −= −

− −

to 5 2

2 2 /5

5

2 2 /

2

5

1 1 1 1 .2 21 1

ii

k i ik i ik

e eUi ie eπ π

−

− − −

− −= −

− −

494 Line 14: change 5 2

1 2 /52 2 /5

1 1 5 2 1 12 21 1 2

i

ii i

i eUi ie e i

ππ

−

−−

− −= − −

− − −

to 5 2

1 2 /52 2 /5

5 21 1 1 12 21 1 2

i

ii

i

i

eU ei ie e i

ππ

−

−−

− −= − −

− − −

495 10th line from the bottom: change 2 ( ) 2 ( ) /( ) 1 and 1.r j N r j r j i r j NW e W eπ π− − − − −= = = ≠

to 2 ( )/2( )( ) 1 and 1.r j N r j r ji r j NiW e W e ππ −− − − −−= = = ≠

496 Line 9: change

( )4 14 , for 0, 1, , 1.jj j N

N n+⎡ ⎤

= −⎢ ⎥⎣ ⎦

to

( )4 14 , for 0, 1, , 1.jj jN

NN

+⎡ ⎤= −⎢ ⎥

⎣ ⎦


496 10th line from the bottom: change 4sin .j

jun

⎛ ⎞= ⎜ ⎟⎝ ⎠

to 4sin .j N

ju ⎛ ⎞= ⎜ ⎟⎝ ⎠

497 Lines 6 and 8: change

( )4 4

4 4

4 4 4 42 2

1 1 1 1 12 [1 ( 4 / 2 / )]1 1 4 / 2 /

1 1 14 2 21 1 [4 ( ) 2( )]4 4

i i

k

i i

i i i i

e eUN N i i N ik Ni N ik N

e ek k

i e e e ek

ππ

π π

ππ

−

−

− −

⎡ ⎤− −= −⎢ ⎥

+ − −− + −⎡ ⎤⎢ ⎥⎣ ⎦⎣ ⎦⎡ ⎤− −

= − −⎢ ⎥− + +⎣ ⎦

= − − − − +−

to

( )4 4

4 4

4 4 4 42 2

1 1 1 1 12 [1 ( 4 / 2 / )]1 1 4 / 2 /

1 1 14 2 21 1 [4 ( ) 2( )

4

1

4

2

]

i i

k

i i

i i i i

e eUN N i i N ik Ni N ik N

e ek k

e e e ek

i

k

ππ

π π

ππ

−

−

− −

⎡ ⎤− −= −⎢ ⎥

+ − −− + −⎡ ⎤⎢ ⎥⎣ ⎦⎣ ⎦⎡ ⎤− −

= − −⎢ ⎥− + +⎣ ⎦

= − − − − +−

498 14.6 Sampled Fourier Series, line 7: change

( ) 2 / .M

ikt pN k

k MS t d e π

=−

= ∑

to

( ) 2 / .M

ikt p

kM k

MS t d e π

=−

= ∑

498 Last line: change 1

2 / 2 /

0

1 .M M

ijk N ijk Nk k

k N M k

U e U eN

π π−

= − =

= +∑ ∑

to 1

2 / 2 /

0

11 .M

ijk N ijk Nk k

N

k N M kU e U e

NNπ π

−

= − =

= +∑ ∑


/64

0

.64

ijkk

j

jU e π−= ∑

to


0

127/64.

64ijk

jk

jU e π

=

−= ∑

501 Line 3: change /2

3 / 2 2

5 / 2 3

7 /2 4

9 /2 5

127 ( 1 40.735 ) ( 1 20.355 ) ( 1 13.557 ) ( 1 10.153 ) ( 1 8.1078 ) ( 1 6.7415 ) ( 1 5.7631 ) ( 1 5.0273 ) ( 1 4.4532 ) ( 1 3.9922 ) ( 1 3.9922

i i

i i

i i

i i

i i

i e i ei e i ei e i e

i e i ei e i ei

π π

π π

π π

π π

π π

= + − + + − +

+ − + + − +

+ − + + − +

+ − + + − +

+ − + + − +

+ − − 118 / 2 1119 / 2

120 /2 121 / 2

122 / 2 123 /2

124 / 2 125 /2

126 / 2 127

) ( 1 4.4532 ) ( 1 5.0273 ) ( 1 5.7631 ) ( 1 6.7415 ) ( 1 8.1078 ) ( 1 10.153 ) ( 1 13.557 ) ( 1 20.355 ) ( 1 30.735 )

i i

i

i i

i i

i i

e i ei e i ei e i ei e i ei e i e

π π

π π

π π

π π

π π

+ − −

+ − − + − −

+ − − + − −

+ − − + − −

+ − − + − − / 2

61.04832.=

to /2

3 /2 2

5 /2 3

7 / 2 4

9 /2 5

127 ( 1 40.735 ) ( 1 20.355 ) ( 1 13.557 ) ( 1 10.153 ) ( 1 8.1078 ) ( 1 6.7415 ) ( 1 5.7631 ) ( 1 5.0273 ) ( 1 4.4532 ) ( 1 3.9922 ) ( 1 3.9922

i i

i i

i i

i i

i i

i e i ei e i ei e i e

i e i ei e i ei

π π

π π

π π

π π

π π

= + − + + − +

+ − + + − +

+ − + + − +

+ − + + − +

+ − + + − +

+ − − 118 /2

120 / 2 121 /2

122 /2 123 /2

124 /

1

2 125 /2

126 /

19

7

2

2

/

12 /

) ( 1 4.4532 ) ( 1 5.0273 ) ( 1 5.7631 ) ( 1 6.7415 ) ( 1 8.1078 ) ( 1 10.153 ) ( 1 13.557 ) ( 1 20.355 ) ( 1 0.4 735 )

ii

i

i i

i i

i i

e i ei e i ei e i ei e i ei e i e

ππ

π π

π π

π π

π π

+ − −

+ − − + − −

+ − − + − −

+ − − + − −

+ − − + − − 2

61.04832.= 512 6th line from the bottom: change

( ) 2 2 2

4

6 1 6 34

n

n

n nc

nπ π

π

⎡ ⎤− + − + −⎣ ⎦=

to

( ) 2 2 2

4

6 1 6 344

n

n

n nc

nπ π

π

⎡ ⎤− + − + −⎣ ⎦=



( )( )

2 2 2

21

6 1 6 3cos 2 .

4

n

n

n nnx

nπ π

π

∞

=

⎡ ⎤− + − + −⎣ ⎦+∑

to

( )( )

2 2 2

21

6 1 6 3cos 2 .

44

n

n

n nnx

nπ π

π

∞

=

⎡ ⎤− + − + −⎣ ⎦+∑


1 1 10

N N N

n n n n m mn n n

f c f c f c= = =

⎛ ⎞ ⎛ ⎞≤ − Φ = − Φ ⋅ − Φ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

to 2

1 1 10

N N N

n n n n m mmn n

f c f c f c= = =

⎛ ⎞ ⎛ ⎞≤ − Φ = − Φ ⋅ − Φ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∑ ∑ ∑

522 Lines 10 & 11: change 2 3 4 3 3

2

2 4 2 5 5 6 3 5 2 6

1 1 3 3 512 2 8 21 2

15 15 15 5 35 1054 8 8 16 4 16

xt t xt t x txt t

x t x t xt t x t x t

= + − − + +− +

− + + − + +

to 2 3 4 3 3

2

2 4 2 5 5 6 3 5 2 6

2 2

4 4

1 1 3 3 512 2 8 21 2

15 1

32

5 15 5 35 1054 8 8 1 1686

354

xt t xt t x txt t

x t x t x

x t

t t xx t t x t

= + − − + +− +

− + + − +

+

+ −

539 Line 8: change

1 2( ) ( ) ( )a c a cy x c x J bx c x J bxν ν= + to

1 2( ) ( ) ( )a c a cy x c x J bx c bYx xν ν= +

539 8th line from the bottom: change “xb” to “x6”

539 Example 15.6, line 6: change "Then b = 2. Next," to "Then b = 7. Next,"

545 Equation (15.13): change

01( ) (1 ...

8

xceI xxx

= +

to

01( ) (1 ...1

8

xceI xxx

= +


1( ) ( )J J x J xxν ν νν

−′ + =


to

( ) 1( ) ( )J J x Jx

x xν ν νν

−′ + =

551 Line 16: change

( ) ...uv vu dxα π

α

+′ ′− =∫

to

( ) ...uv vu dxα π

α

+′ ′ ′− =∫

554 Line 8: change

( )( )1 2

0...vx J x dx =∫

to

( )( )1 2

0...nvx J dxj x =∫

554 Line 18: change

( )( ) ( )( )1 22 2

02 .n v n n v nj J J j J j x dx′= − ∫

to

( )( ) ( )( )2 1 22 2

02 .n v n n v nj J J j x J j x dx′= − ∫


( )

( )

2 2

2 2

2 2

2 2

( , )

ˆ( , ) , .

i x

i x

y yF x t e dxt t

y x t e dx f tt t

ω

ω

ω

ω

∞ −

−∞

∞ −

−∞

⎡ ⎤∂ ∂=⎢ ⎥∂ ∂⎣ ⎦∂ ∂

= =∂ ∂

∫

∫

to

( )

( )

2 2

2 2

2 2

2 2

( , )

( , ) , .ˆ

i x

i x

y yF x t e dxt t

y x t e ydx tt t

ω

ω

ω

ω

∞ −

−∞

∞ −

−∞

⎡ ⎤∂ ∂=⎢ ⎥∂ ∂⎣ ⎦∂ ∂

= =∂ ∂

∫

∫

582 Last line: change 2

2 22

ˆ ˆ( , ) ( , ).y t c y x tt

ω ω∂= −

∂

to 2

2 22 ˆ ˆ( , ) ( , ).y t c y t

tω ω ω∂

= −∂

583 Line 2: change

( )2

2 22 ˆ ˆ, ( , ) 0.y t c y x t

tω ω∂

+ =∂

to


( )2

2 22 ˆ ˆ, ( , ) 0.y t c y t

tω ω ω∂

+ =∂

583 Line 9: change

( ) ( )

ˆ ( , 0) ( , 0) ( ,0)

ˆ[ ( )] ,

y yc b F xt t

F g x g

ωω ω ω

ω ω

∂ ∂⎡ ⎤= = ⎢ ⎥∂ ∂⎣ ⎦= =

to

( ) ( )

ˆ ( , 0) ( , 0)

ˆ[ ( ) ,

( )

]

y yc b F xt t

F g x g

ωω

ω ω

ωω∂ ∂⎡ ⎤= = ⎢ ⎥∂ ∂⎣ ⎦= =

590 Line 1: change "Y (s, x)" to "Y (x, s)"

590 Case 1, line 3: change

( ) KF s LKs

= =

to

[ ]( ) KF s L Ks

= =

591 Lines 7 & 8: change

(( (2 1) )/ )2

0

(( (2 1) )/ )2

0

1( 1)

1 .

n n L x c s

n

n L x c s

n

cK eE scK eE s

∞− + −

=

∞− + +

=

= −

−

∑

∑

to

( )

(((2

0

20

2 1) )/ )

(((2 1) )/ )

( 1

1 . 1

1) n L x c s

n

n

n

n

n L x c s

cK eE scK eE s

∞

=

− + −

=

+ +∞

−

= −

− −

∑

∑

592 Line 3: change

( )for 0 2 / ,

4 / for2 / 4 / .t t L c

g tL c L c t L c

≤ ≤⎧= ⎨ ≤ ≤⎩

to

( )for 0 2 / ,

4 / for 2 / 4 / .t t L c

g tL c L ct t L c

≤ ≤⎧= ⎨ ≤ ≤− +⎩


to

1



( )| 2 | | 2 |1 1 sin(4 ) cos(8 ).2 8

x t x te e x t− − + − += +

to

( )| 2 | | 2 |1 1 (4 ) (8 ).2 8

cos sinx t x te e x t− − + − += +

600 Line 2: change

2 2 1 t x tM M Mu dx c u dt u c dt c dx c du

c⎛ ⎞+ = + =⎜ ⎟⎝ ⎠∫ ∫ ∫

to

2 2 1 xt x tM M Muu dx c u dt u c dt c dx c du

c⎛ ⎞+ = + =⎜ ⎟⎝ ⎠∫ ∫ ∫

600 Line 6: change

( )0 0

0 0

( , ) x ct

x ctF x y dA g w dw

+

Δ −− =∫ ∫ ∫ ∫

to

( )0 0

0 0

( , ) x ct

x ctF x y dA g w dw

+

Δ −− = ∫∫ ∫

600 3rd line from the bottom: change 5 2 2

5

1 1 .10 10

x t w

x te dw X T d X dT

+

− Δ+ +∫ ∫ ∫

to 5 2 2

5

1 1 .10 10

wx t

x te dw X T dX dT−+

− Δ+ +∫ ∫ ∫

604 Figure 16.22: change “J0(x)” to “y = J0(x)”

605 Last line: change 2 0T c T′′ + =

to 2 0T c Tλ′′+ =

606 Lines 2 & 5: change (1/ )F r FF

′′ ′+

to 2r F rF

F′′ ′+

606 13th line from the bottom: change 2 2 2( ) ( ) ( ) ( ) 0; ( ) 0.r F r r F r kr n F r F R′′ ′+ + − = =

to 2 2 2( ) ( ) ( ) ( ) 0; ( ) 0.r F r r F r r n F r F Rλ′′ ′+ + − = =



( )

( )

0 01

1 1 1 1

,

cos( ) sin .

nkk

k

nk nknk n nk n

n k n k

jf r a J rR

j ja J r n b J r nR R

θ

θ θ

∞

=

∞ ∞ ∞ ∞

= = = =

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞+ +⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎝ ⎠

∑

∑ ∑ ∑ ∑

to

( )

( )

0

1

00

1 1

1

si

,

cos( ) .n

kk

k

nknk n

n k

nknk n

k

jf r a J rR

ja J r jb JnR

r nR

θ

θ

θ

∞

=

∞ ∞

= =

∞

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎡ ⎤⎛ ⎞+ +⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

∑

∑

∑ ∑


0 0 01

1 ( , ) ( ),2

nkk

k

ja J r f r d rR

π

πθ θ α

π

∞

−=

⎛ ⎞ = =⎜ ⎟⎝ ⎠

∑ ∫

to

00 0 0

1

1 ( , ) ( ),2

kk

k

ja J r f r d rR

π

πθ θ α

π

∞

−=

⎛ ⎞ = =⎜ ⎟⎝ ⎠

∑ ∫


( ) ( ) ( )2 2 22 1 /

1

2 12 1, cos .2 2 1

n kt L

n

n xA Au x t en L

πππ

∞− −

=

−⎛ ⎞= + ⎜ ⎟− ⎝ ⎠

∑

to

( ) ( ) ( ) ( )2 2 22 /

1

112 12, cos

1.

2 2 1

nn kt L

n

n xA Au x t en L

πππ

∞−

+−

=

− −⎛ ⎞= + ⎜ ⎟− ⎝ ⎠

∑

616 Line 3: change ( ) ( ) 0X L AX L c cL′ + = + =

to ( ) ( ) 0X L AX L Ac cL′ + = + =

619 Line 7: change 2 2 2/

1

2( , ) sin n kt Ln

n

n xU x t c eL L

ππ∞−

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑

to 2 2 2/

1( , ) sin n kt L

nn

n xU x t c eL

ππ∞−

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑


0

0

2 3 1 1 sin2

2 1 1 sin2

L

n

L

nc x dL L L

nx dL L L

πξ ξ

πξ ξ

⎡ ⎤ ⎛ ⎞= − − ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫

∫

to


0

0

2 3 1 1 sin2

2 1 1 sin2

L

n

L

nc dL L L

n dL L L

πξ ξ

π

ξ

ξ ξξ

⎡ ⎤ ⎛ ⎞= − − ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫

∫620 Equation (17.6): change

sin to

sin


( ) ( ) ( )20

sin2 2sin .b dπ

ω

πω πωπ ξ ωξ ξ

π ω ω−

= − =∫

to

( ) ( ) ( )20

sin2 2sin .b dπ

ω

πω πωξ

ππ ωξ ξ

π ω−

= − =∫

633 Line 5: change

denominator by / :s k xe− to

denominator by / :Ls ke−

635 Line 2: change “c2 = 0” to “c1 = 0”

635 Line 3: change /

1( , ) .s kx AU x s c es

= +

to /

2( , ) .s k xc e AU x ss

−= +

635 Lines 4, 8, 9 & 10: change “c1” to “c2”

635 Line 6: change

0, to

0,

636 15th line from the bottom: change 10 and 0.T T F F Fr

λ λ′ ′′ ′+ = + + =

to 10 and 0.T T Fk F Fr

λ λ′ ′′ ′+ = + + =


637 Line 7: change 2 2( ) ( ) ( ) 0.r F r rF r F rω′′ ′+ + =

to 2 2 2( ) ( ) ( ) 0.r F r rF r F rrω′′ ′+ + =


( ) ( )01

n nn

f r a J jξ ξ∞

=

=∑

to

( ) ( )01

n nn

f a J jRξ ξ∞

=

=∑

637 2nd line from the bottom: change 2

to 2

638 13th line from the bottom: change 0, 0, ( ) 0,X X Y Y T Tλ μ λ μ′′ ′′ ′+ = + = + + =

to 0, 0, ( ) 0,X X Y Y T Tkλ μ λ μ′′ ′′ ′+ = + = + + =

638 4th line from the bottom: change 2 2 2( ) 0T n m Tπ′+ + =

to 2 2 2( ) 0n m TkT π′+ + =


30

2 4( )sin( ) ( 1) .nx n dn

ππ ξ ξ ξ

π π− = −∫

to

30

2 4( )sin( ) ( 1) .nn dn

ππ ξ ξ ξ

π πξ − = −∫


( ) ( )1 1cos( ) and sin( ) .n n nn

a f n d b f n dR R

π π

π πξ ξ ξ ξ ξ ξ

− −= =∫ ∫

to

( ) ( )1 1cos( ) and sin( ) .n n nn

a f n d b f n dR R

π π

π πξ ξ

πξ ξ ξ

πξ

− −= =∫ ∫

645 2nd line from the bottom: change 1

… to


12 …

647 Line 2: change

( ) ( )2 21 8181 cos sin ,2 4

dπ

πξ ξ ξ π

π −=∫

to

( ) ( )2 2 8181 cos sin ,4

dπ

πξ ξ ξ π

−=∫


1 1

1 2 cos( ) Re 1 2 Re 1 21

n n

n n

zr n zz

θ∞ ∞

= =

⎛ ⎞ ⎛ ⎞+ = + = +⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠∑ ∑

to

1 1

1 2 cos( ) Re 1 2 Re 1 21

n n

n n

zr n zz

ζ∞ ∞

= =

⎛ ⎞ ⎛ ⎞+ = + = +⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠∑ ∑

650 18.5.1 The Upper Half-Plane, line 7: change ( , ) ( ) ( )u x t X x T t=

to ( , ) ( () )u x t X Yx y=

650 18.5.1 The Upper Half-Plane, line 8: change 0, 0.X X T Tλ λ′′ ′′+ = − =

to 0, 0.X X Y Yλ λ′′−′′ + = =


1

to

655 Line 11: change

( )2 ...

sinhnmnm

CAC Bβ

=

to

( )...

sinh4

nmnm

CAC Bβ

=

657 Line 2: change

[ ( )sin( ) ...dd

ϕ ϕϕ

′Φ =

to


[ ( )sin( ]) ...dd

ϕ ϕϕ

′Φ =

659 Proof of Lemma 18.1, line 2: change

( ). .( ) .C C Dgg ds g f ds g f dAn∂

= ∇ = ∇ ∇∂∫ ∫ ∫∫n

to

( ). .( ) .C C Dg ds g f ds g f df An∂

= ∇ = ∇ ∇∂∫ ∫ ∫∫n

659 2nd line from the bottom: change 2 2

2 2

f f g f g fgx y x x y

⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂= + + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠

to 2 2

2 2

f f g f g fgx y x x y y

⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂= + + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠

660 Example 18.7, line 2: change 2 0 for 0 ,0 1,u x y∇ = < < < <

to 2 0 for 10 ,0 1,u x y∇ = < < < <

660 Example 18.7, line 10: change: 2 2

0

1 0.3C

uds y dyn∂

= = ≠∂∫ ∫

to 2

0

1 1 0.3C

uds y dyn∂

= = ≠∂∫ ∫

661 Line 19: change (0) 0,X d′ = =

to (0) 0,X c′ = =

663 Line 8: change 1 ( ) sin( )nb f n d

π

πξ ξ ξ

π −= ∫

to 1 1 ( ) sin( )n

n f dnb R nπ

πξ ξ ξ

π −

− = ∫


0, 0.u vy x∂ ∂

= =∂ ∂

to


0, .yu vy x∂ ∂

= =∂ ∂


( ) ( )1cos( ) and sin( ) .2 2

iz iz z izz e e z e ei

− − −1= + = −

to

( ) ( )1cos( ) and sin( ) .2 2

iz iz iiz zz e e z e ei

− −1= + = −

689 Example 19.13, line 1: change ( /4 2 )2 i n iz e π π+=

to ( /4 2 )2 i nz e π π+=


log(1 ) ln 2 2 . 4

iz i n iπ π⎡ ⎤+ = + +⎢ ⎥⎣ ⎦

to

log(1 ) ln 2 2 . 4

i i nπ π⎡ ⎤+ = + +⎢ ⎥⎣ ⎦

691 Example 19.15, line 3: change 1/8 /16 1/8 ( /4 2 ) 1/8 ( /4 4 )/4 1/8 ( /4 6 )/42 , 2 , 2 , and 2 .i i i ie e e eπ π π π π π π+ + +

to

1/8 /16 1/8 ( /4 2 )/4 1/8 ( /4 4 )/4 1/8 ( /4 6 )/42 , 2 , 2 , and 2 .i i i ie e e eπ π π π π π π+ + +

726 Example 21.7, line 4: change “ 0 | z< <∞ ” to “ 0 z< < ∞”

728 Left-hand side column, line 8: change

( ) ( )1 2

1 1 ( )2 2

f w f wf w dw dwi w z i w zπ πΓ Γ

= +− −∫ ∫

to

( ) ( )1 2

1 1 ( )2 2

f w f wf dw dwi w z i z

zwπ πΓ Γ

= +− −∫ ∫


( ) ( )2 1

1 1 ( )2 2

f w f wf z dz dwi w z i w zγ γπ π

= −− −∫ ∫

to

( ) ( )2 1

1 1 ( )2 2

f w f wf z d dwi w z i

ww zγ γπ π

= −− −∫ ∫

736 Figure 22.1: change “Γ” to “γ” (change uppercase Gamma to lowercase Gamma)

738 Line 8: change


1 1( ) 2 1 1 cos( ) 1 cos( ) 2 ( 1 cos( )).2 2

f z dz i i i i iγ π π⎛ ⎞= − + + − + = − +⎜ ⎟⎝ ⎠∫

to 1 1( ) 2 1 1 cos( ) 1 cos( ) 2 ( 1 cos( )).2 2

f z dz i i i i iγ π π⎛ ⎞= − + + − + = − +⎜ ⎟⎝ ⎠∫


22

sin( )Res( , 2 ) lim( 2 )( 2 )z i

zf iz z i z i→

=− +

to

22

sin( )Res( , 2 ) lim( 2 )z i

zf iz z i→

=+

744 10th line from the bottom: change “2 √3” to “ 2 √3”

752 Example 23.2, line 7: replace a with x to obtain cos and sin

752 Last line: change sin( ) and cos( ).x xu e b v e b= =

to ( ) and ( )cos sin .x xu e b v e b= =

753 Lines 1 & 2: change sin( ) cos( )x xe b ie b+

to ( ) ( )cos sinx xe b ie b+

754 Figure 23.5, left side: change leftmost “π/2” to “–π/2”

754 Blue box, #1: change All “L1” to “C1” (×2) All “L2” to “C2” (×2)

758 Example 23.4, line 2: change “…horizontally by Re(z) and vertically by Im(z).” to “…horizontally by Re(b) and vertically by Im(b).”

758 Example 23.4, line 3: change ( ) 2T z i= −

to ( ) 2 .zT z i+= −


1 1 2 3 1 1 2 1 2( )( )( ) ( )( )( ).w w z z z z z z z z w w− − − = − − − to

31 1 2 3 1 2 1 2( )( )( ) ( )( )( ).w w z z z z z z wz z w− − − = − − −

767 Figure 23.22 caption: change “…|z| onto |w| > 3...” to “…|z| < 1 onto |w| > 3…”


773 8th line from the bottom: change 1 2

1 2( ) ( ) ( ) ( ) nng z a x x xα α αξ ξ ξ− − −= − − −

to 1 2

1 2( ) ( ) ( ) ( ) nnz zg z a x x xzα α α− − −= − − −

775 Right-hand column, last line: change ( )9 arg / 3w π< <

to ( )arg / 30 w π< <

806 2nd line from the bottom: change “Y(t) = y(xt)” to “Y(t) = y(et)”

816 Problem 19, line 1: change “a1,…,ck,” to “a1,…,ak,”

818 Problem 9, line 5: change “(EA)sj = ( row s of E) · column j of A)” to “(EA)sj = (row s of E) · (column j of A)”

840 Problem 1, line 4: change

( ) ( )1

2 14 1 sin .2 1 2N

n

n tS t

nπ

π

∞

=

−⎛ ⎞= ⎜ ⎟− ⎝ ⎠

∑

to

( ) ( )1

2 14 1 sin .2 1 2N

n

N n tS t

nπ

π =

−⎛ ⎞= ⎜ ⎟− ⎝ ⎠

∑


( ) ( ) ( )( ) ( )1

2 cos / 2 1 sinnN

nS t n n t

nπ π

π

∞

=

= − −∑

to

( ) ( ) ( )( ) ( )1

2 cos / 2 1 sinnN

N

nS t n n t

nπ π

π=

= − −∑


( (2 2 2 22 4 , so 2 and 4 .w u iv x y xyi u x y v xy= + = − + = − =

to

( ) ( )2 2 2 22 4 , so 2 and 4 .w u iv x y xyi u x y v xy= + = − + = − =

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O'Neil 7e Errata - Higher Ed eBooks & Digital Learning … · 2013-05-20 · Errata (Includes critical ... Advanced Engineering Mathematics, 7e Peter V. O’Neil ... 134 5th and 6th