Numerical Analysis Solution Edition 9th PDF

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Mathematics, the science of structure, order, and relation that has evolved from elemental practices of counting, measuring, and describing the shapes of objects. It deals with logical reasoning and quantitative calculation, and its development has involved an increasing degree of idealization and a...

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Student Solutions Manual and Study Guide Chapters 1 & 2 Preview

for

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...