Advanced Engineering Mathematics by erwin kreyszig 9th edition Chapter 1 and 2 Answers PDF

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Answer of Chapter 1 and Chapter 2 of Advanced Engineering Mathematics by erwin kreyszig 9th edition.
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Student Solutions Manual and Study Guide Chapters 1 & 2 Preview

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Numerical Analysis 9th EDITION

Richard L. Burden Youngstown State University

J. Douglas Faires Youngstown State University

Prepared by Richard L. Burden Youngstown State University

J. Douglas Faires Youngstown State University

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∼

x − (ln x)x = 0

[4, 5]. x

f (x) = x − (ln x)x = x − exp(x(ln(ln x))). f

[4, 5]

f (4) ≈ 0.3066 f (5) ≈ −5.799, 0 = f (x) = x − (ln x)x . x (4, 5)

3

2

f (x) = x − 2x − 4x + 3

x3 − 2x2 − 4x + 3 = 0 f

0 = f ′ (x) = 3x2 − 4x − 4 = (3x + 2)(x − 2); x = − 32 f (4) = 19

x = 2.

f (−2) = −5

max |f (x)|

0≤x≤1

f

f (x) = 0   f − 32 ≈ 4.48 f (0) = 3 [−2, −2/3] [0, 1],

f (x) f (1) = −2 [2, 4].

f (2) = −5

f (x) = (2 − ex + 2x) /3 f ′ (x) = (−ex + 2) /3 [0, 1] |f (x)|

f

x = ln 2

max{|f (0)|, |f (ln 2)|, |f (1)|} = max{1/3, (2 ln 2)/3, (4 − e)/3} = (2 ln 2)/3. f (x) = x3 + 2x + k x 2x + k > 0 f (c) = 0

x

k x < − 21 k

f (x) < 2x + k < 0 x > − 12 k

f (c ′ ) = 0 c ′ 6= c f (c) = 0 f ′ (x) = 3x2 + 2 > 0 c′ f ′ (p) = 0 f (c) = 0 f (c ′ ) = 0 c ′ 6= c f (c) = 0

x>0 c p

c

x c

f (x) = ex cos x P 2 (0.5)

x0 = 0

f (0.5)

Z

1

0

|f (0.5) − P 2 (0.5)|

x |f (x) − P 2 (x)| Z 1 P 2 (x) dx f (x) dx

[0, 1]

0

f ′ (x) = ex (cos x − sin x),

f ′′(x) = −2ex (sin x),

f (0) = 1 f ′ (0) = 1

f ′′′(x) = −2ex (sin x + cos x),

f ′′(0) = 0. R2 (x) =

P 2 (x) = 1 + x

−2eξ (sin ξ + cos ξ) 3 x . 3!

P 2 (0.5) = 1 + 0.5 = 1.5    1  −2eξ (sin ξ + cos ξ ) (0.5)2  ≤ (0.5)2 max |eξ (sin ξ + cos ξ)|. |f (0.5) − P 2 (0.5)| ≤ max  3! 3 ξ∈[0.0.5] ξ∈[0,0.5] Dx ex (sin x + cos x) = 2ex cos x > 0 ex (sin x + cos x [0, 0.5]

[0, 0.5]

[0, 0.5]

e0 (sin 0 + cos 0) = 1 < e0.5 (sin 0.5 + cos 0.5) ≈ 2.24. |f (0.5) − P 2 (0.5)| ≤

1 (0.5)3 (2.24) ≈ 0.0932. 3 x ∈ [0, 1]

1 |f (x) − P 2 (x)| ≤ (1.0)3 e1 (sin 1 + cos 1) ≈ 1.252. 3 Z Z Z

0

1

1 0

f (x) dx ≈

1 0

|R2 (x)| dx ≤

ex cos x dx =



Z

0

1

Z

0

1

1  3 x2 = . 1 + x dx = x + 2 2 0

1 1 e (cos 1 + sin 1)x3 dx = 3

Z

1

1.252x3 dx = 0.313.

0

1 1 ex e (cos x + sin x) = (cos 1 + sin 1) − (1 + 0) ≈ 1.378, 2 2 2 0

|1.378 − 1.5| ≈ 0.12.

x

sin x ≈ x

sin 1◦

π 180◦ = π f (x) = sin x f ′ (x) = cos x f ′′(x) = − sin x, 1◦ = 180 f ′′′(x) = − cos x, f ′′(0) = 0. f (0) = 0 f ′ (0) = 1, sin x ≈ x ξ 3 x . R2 (x) = − cos | cos ξ| ≤ 1 f (x) ≈ P 2 (x) 3!     π    π    π   − cos ξ  π 3   ≤ 8.86 × 10−7 . sin − =  = R2 180 180  180 180 3!

f (x) = ex/2 sin 3x P 3 (x) f

(4)

(x) f (x)

f :=

g :=

p3 :=

x  2

·

x 3

f := e(1/2)x sin

(f, x = 0, 4)

(g,

)

1 x 3



23 3 1 1 x + x2 + x 6 3 648

(f, x, x, x, x) f 4 := −

f 5 :=



  23 3 1 1 x + O x4 g := x + x2 + 648 6 3

p3 :=

f 4 :=

[0, 1]

|f (x) − P 3 (x)|

119 (1/2x) e sin 1296



   1 5 (1/2x) 1 x x + e cos 3 3 54

(f 4, x) f 5 := −

    1 61 (1/2x) 1 199 (1/2x) x e cos x + e sin 3 3888 3 2592 [0, 1]

p :=

(f 5 = 0, x, 0..1) p := .6047389076 x=0 1

p

c1 :=

(

(x = p, f 4)) c1 := .09787176213

c2 :=

(

(x = 0, f 4)) c2 := .09259259259

c3 :=

(

(x = 1, f 4)) c3 := .09472344463 f (4)(x)

c1

:= c1/24 := .004077990089 x

| sin x| ≤ |x|

x≥0

sin x ≤ x

f (x) = x − sin x x=0

x

−1 ≤ cos x ≤ 1 | sin x| ≤ 1

f ′ (x) = 1 − cos x ≥ 0 |x| ≥ π

f (x) = x − sin x

f (x)

f (x) > f (0) = 0 x>0 | sin x| = sin x ≤ x = |x| −π < x < 0 sin(−x) ≤ (−x)

x x≥0

x1

| sin x| ≤ |x|

x2

ξ

[a, b] x1

f (ξ) = c1

0≤x≤π

sin x π ≥ −x > 0 | sin x| = − sin x ≤ −x = |x| x

f ∈ C[a, b]

x ≥ sin x

x2

f (x1 ) + f (x2 ) 1 1 = f (x1 ) + f (x2 ). 2 2 2

c2

ξ f (ξ) =

x1

x2

c 1 f (x1 ) + c 2 f (x2 ) . c1 + c2 c1

c 1 6= −c 2

1 (f (x1 ) + f (x2 )) 2

c2

f (x1 )

f (x2 )

f ξ

f (ξ) = m = min{f (x1 ), f (x2 )} m ≤ f (x2 ) ≤ M,

x1

x2

1 1 1 (f (x1 ) + f (x2 )) = f (x1 ) + f (x2 ). 2 2 2 M = max{f (x1 ), f (x2 )}.

c 1 m ≤ c 1 f (x1 ) ≤ c 1 M

m ≤ f (x1 ) ≤ M

c 2 m ≤ c 2 f (x2 ) ≤ c 2 M.

(c 1 + c 2 )m ≤ c 1 f (x1 ) + c 2 f (x2 ) ≤ (c 1 + c 2 )M m≤

c 1 f (x1 ) + c 2 f (x2 ) ≤ M. c1 + c2 x1

ξ

x1

x2

x2 f (ξ) =

c 1 f (x1 ) + c 2 f (x2 ) . c1 + c2

f (x) = x2 + 1 x1 = 0 x2 = 1 c 1 = 2

f (x) > 0

c 2 = −1

x

c 1 f (x1 ) + c 2 f (x2 ) 2(1) − 1(2) = = 0. 2−1 c1 + c2

√ 2

p∗  √  p∗ − 2 √  ≤ 10−4 ,  2 p∗

10−4

 √  √  p∗ − 2 ≤ 2 × 10−4 ;

√ √ √ − 2 × 10−4 ≤ p∗ − 2 ≤ 2 × 10−4 . √ √  2(0.9999), 2(1.0001) . − 67 , 2e − 5.4 13 14

13 14

13 6 − = 0.0720 14 7

= 0.929

6 7

= 0.857

2e − 5.4 = 5.44 − 5.40 = 0.0400.

e = 2.72.

13 14

− 76 0.0720 = = 1.80. 2e − 5.4 0.0400 |1.80 − 1.954| = 0.0788, 1.954

|1.80 − 1.954| = 0.154

13 14

13 6 − = 0.0710 14 7

= 0.928

6 7

= 0.857

e = 2.71.

2e − 5.4 = 5.42 − 5.40 = 0.0200.

− 76 0.0710 = = 3.55. 2e − 5.4 0.0200 13 14

|3.55 − 1.954| = 0.817, 1.954

|3.55 − 1.954| = 1.60,

  π = 4 arctan 21 + arctan 13

P (x) = x − 13 x3 + 51 x5 . 

π = 4 arctan

P

1  2

= 0.464583

P

1 1 + arctan ≈ 3.145576. 2 3

|π − 3.145576| ≈ 3.983 × 10−3

f (x) =



ex − e−x . x

f (0.1) limx→0 ex − e−x = 1 − 1 = 0 x→0

3

= 0.3218107,

|π − 3.145576| ≈ 1.268 × 10−3 . |π|

limx→0 f (x)

lim

 1

limx→0 x = 0

1+1 ex − e−x ex + e−x = = 2. = lim x→0 1 1 x

e0.100 = 1.11 f (0.100) =

1.11 − 0.905 0.205 = = 2.05. 0.100 0.100

1 1 e x ≈ 1 + x + x2 + x3 2 6 f (x) ≈

e−0.100 = 0.905

1 1 e−x ≈ 1 − x + x2 − x3 , 2 6

    1 + x + 12 x2 + 61 x3 − 1 − x + 12 x2 − 61 x3 2x + 13 x3 1 = 2 + x2 . = x x 3

1 f (0.100) ≈ 2 + (0.100)2 = 2 + (0.333)(0.001) = 2.00 + 0.000333 = 2.00. 3

0

01111111111 0101001100000000000000000000000000000000000000000000.

 2  4  7  8 ! 1 1 1 1 +2 1+ + + + 2 2 2 2   1 1 1 83 1 =1+ + + = 1.32421875. = 20 1 + + 128 256 4 16 256 1023−1023

0

01111111111 0101001100000000000000000000000000000000000000000000.

0

01111111111 0101001011111111111111111111111111111111111111111111   =1.32421875 − 21023−1023 2−52 =1.3242187499999997779553950749686919152736663818359375,

0

01111111111 0101001100000000000000000000000000000000000000000001   =1.32421875 + 21023−1023 2−52 =1.3242187500000002220446049250313080847263336181640625.

f (x) = 1.01e4x − 4.62e3x − 3.11e2x + 12.2ex − 1.99.

f (1.53) f (x) n

enx = (ex )

f (x) = ((((1.01)ex − 4.62) ex − 3.11) ex + 12.2) ex − 1.99. e1.53 = 4.62 e3(1.53) = (4.62)2 (4.62) = (21.3)(4.62) = 98.4,

e2(1.53) = (4.62)2 = 21.3 e4(1.53) = (98.4)(4.62) = 455

f (1.53) = 1.01(455) − 4.62(98.4) − 3.11(21.3) + 12.2(4.62) − 1.99 = 460 − 455 − 66.2 + 56.4 − 1.99 = 5.00 − 66.2 + 56.4 − 1.99

= −61.2 + 56.4 − 1.99 = −4.80 − 1.99 = −6.79.

f (1.53) = (((1.01)4.62 − 4.62)4.62 − 3.11)4.62 + 12.2)4.62 − 1.99 = (((4.67 − 4.62)4.62 − 3.11)4.62 + 12.2)4.62 − 1.99 = ((0.231 − 3.11)4.62 + 12.2)4.62 − 1.99

= (−13.3 + 12.2)4.62 − 1.99 = −7.07. 7.61 | − 6.79 + 7.61| = 0.82 0.0710. f l(y)

k

0.108

| − 7.07 + 7.61| = 0.54 y      y − f l(y)  ≤ 0.5 × 10−k+1 .   y

d k+1 > 5.

d k+1 ≤ 5

d k+1 ≤ 5,   n −k   0.5 × 10−k  y − f l(y) = 0.d k+1 . . . × 10 ≤ = 0.5 × 10−k+1 .  n  0.d 1 . . . × 10 y 0.1

d k+1 > 5,   n −k   (1 − 0.5) × 10−k  y − f l(y)  = (1 − 0.d k+1 . . .) × 10 < = 0.5 × 10−k+1 . n   0.d 1 . . . × 10 0.1 y

T

0.995 ≤ P ≤ 1.005,

0.0995 ≤ V ≤ 0.1005,

0.082055 ≤ R ≤ 0.082065,

15

◦

0.004195 ≤ N ≤ 0.004205.

287.61 ≤ T ≤ 293.42. 288.16

P

1.99 ≤ P ≤ 2.01

19

◦

V

0.0497 ≤ V ≤ 0.0503,

286.61 ≤ T ≤ 293.72. 292.16 290.15

= 17◦

.

n arctan x = lim P n (x) = n→∞

∞ X i=1

(−1)i+1

x2i−1 (2i − 1)

|4P n (1) − π| < 10−3 10−10

π = 4 arctan 1 = 4

∞ X i=1

π

(−1)i+1

1 2i − 1 n

4 < 10−3 2(n + 1) − 1

4000 < 2n + 1.

n ≥ 2000. 4 < 10−10 2(n + 1) − 1

n > 20,000,000,000.

π 1 1 π = 4 arctan − arctan . 4 5 239 10−3

π

π=4

∞ X

∞

(−1)i+1

i=1

X 1 1 (−1)i+1 − 2i−1 (2i − 1) 2i−1 239 5 (2i − 1) i=1 i 10−3

i+1 i := 1 :

4 51 (1)

4 = , 5

i=2:

4 53 (3)

4 375

=

4

i=3:

n X i X

55 (5)

=

4 = 2.56 × 10−4 . 15625

ai b j ?

i=1 j=1

Pi

i n X i=1

i=

j=1

ai b j

n(n + 1) 2

i

n X i=1

n

i−1 =

i−1 n(n + 1) −n 2

.

n−1 n(n + 1) 2

(n + 2)(n − 1) 2 i n X X

ai b j =

i=1 j=1

bj

n X i=1

i−1=

bj

i n X X ai bj , i=1

j=1

i

+ 1)

n.

i−1

n(n + 1) −n 2

n

1 n(n 2

.

ai 1 (n 2

+ 2)(n − 1)

n−1

A B C x1 x2 A=0 B=0 x1 = −C/B

x1

D = B 2 − 4AC

D=0

D...