chapter 4 - iiethus, the third-order result is perfect because the original function is a third...

1

PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

CHAPTER 4

4.1 (a) For this case xi = 0 and h = x. Thus,

2

)0(")0(')0()(2x

fxffxf

1)0(")0(')0( 0 efff

2

1)(2x

xxf

(b)

62

)(32

1

he

heheexf iiii xxxx

i

for xi = 0.2, xi+1 = 1 and h = 0.8. True value = e–1

= 0.367879.

zero order:

818731.0)1( 2.0 ef

%55.122%100367879.0

818731.0367879.0

t

first order:

163746.0)8.0(818731.0818731.0)1( f

%49.55%100367879.0

163746.0367879.0

t

second order:

42574.02

8.0818731.0)8.0(818731.0818731.0)1(

2

f

%73.15%100367879.0

42574.0367879.0

t

third order:

355875.06

8.0818731.0

2

8.0818731.0)8.0(818731.0818731.0)1(

32

f

%26.3%100367879.0

355875.0367879.0

t

2


4.2 Use the stopping criterion

%5.0%105.0 22 s

True value: cos(/3) = 0.5

zero order:

13

cos

%100%1000.5

10.5

t

first order:

451689.0

2

3/1

3cos

2

%66.9t %4.121%1000.451689

10.451689

a

second order:

501796.0

24

3/451689.0

3cos

4

%359.0t %986.9%100501796.0

451689.0501796.0

a

third order:

499965.0

720

3/501796.0

3cos

6

%00709.0t %366.0%100499965.0

501796.0499965.0

a

Since the approximate error is below 0.5%, the computation can be terminated.

4.3 Use the stopping criterion: %5.0%105.0 22 s

True value: sin(/3) = 0.866025…

3


zero order:

047198.133

sin

%92.20%1000.866025

047198.10.866025

t

first order:

855801.0

6

3/047198.1

3sin

3

%18.1t %36.22%100855801.0

047198.1855801.0

a

second order:

866295.0

120

3/855801.0

3sin

5

%031.0t %211.1%100866295.0

855801.0866295.0

a

third order:

866021.0

5040

3/866295.0

3sin

7

%000477.0t %0316.0%100866021.0

866295.0866021.0

a


4.4 True value: f(3) = 554.

zero order:

62)1()3( ff

%191.111%100554

62)(554

t

4


first order:

78)2(7062)13)(1('62)3( ff %921.85t

second order:

35442

13878)13(

2

)1("78)3( 2

ff %101.36t

third order:

55486

150354)13(

6

)1(354)3( 3

)3(

f

f %0t

Thus, the third-order result is perfect because the original function is a third-order

polynomial.

4.5 True value: f(2.5) = ln(2.5) = 0.916291...

zero order:

0)1()5.2( ff

%100%1000.916291

00.916291

t

first order:

5.1)5.1(10)15.2)(1(')1()5.2( fff

%704.63%1000.916291

5.10.916291

t

second order:

375.05.12

15.1)15.2(

2

)1("5.1)5.2( 22

ff

%074.59%1000.916291

375.00.916291

t

third order:

5.15.16

2375.0)15.2(

6

)1(375.0)5.2( 33

)3(

f

f

5


%704.63%1000.916291

5.10.916291

t

fourth order:

234375.05.124

65.1)15.2(

24

)1(5.1)5.2( 44

)4(

f

f

%421.74%1000.916291

234375.00.916291

t

Thus, the process seems to be diverging suggesting that a smaller step would be required for

convergence.

4.6 True value:

2837)2(12)2(75)2('

71275)('

2

2

f

xxxf

function values:

xi–1 = 1.8 f(xi–1) = 50.96 xi = 2 f(xi) = 102

xi+1 = 2.2 f(xi+1) = 164.56

forward:

8.3122.0

10256.164)2('

f

%53.10%100283

312.8283

t

backward:

2.2552.0

96.50102)2('

f

%823.9%100283

255.2283

t

centered:

6


284)2.0(2

96.5056.164)2('

f

%353.0%100283

284283

t

Both the forward and backward have errors that can be approximated by (recall Eq. 4.15),

hxf

E it

2

)("

28812)2(15012150)2(" xf

8.282.02

288tE

This is very close to the actual error that occurred in the approximations

forward: 8.298.312283 tE

backward: 8.272.255283 tE

The centered approximation has an error that can be approximated by,

12.06

150

6

)( 22)3(

hxf

E it

which is exact: Et = 283 – 284 = –1. This result occurs because the original function is a cubic equation which has zero fourth and higher derivatives.

4.7 True value:

28812)2(150)2("

12150)("

f

xxf

h = 0.25:

28825.0

85938.39)102(21406.182

25.0

)75.1()2(2)25.2()2("

22

ffff

h = 0.125:

288125.0

82617.68)102(26738.139

125.0

)875.1()2(2)125.2()2("

22

ffff

7


Both results are exact because the errors are a function of 4th and higher derivatives which are

zero for a 3rd

-order polynomial.

4.8

38666.1

12

)/()/(

c

egmcgte

c

v tmctmc

079989.2)5.1(38666.1~)~(

c

c

vcv

4533.3015.12

)50(8.9)5.12( 50/)6(5.12 ev

079989.24533.30 v

Thus, the bounds computed with the first-order analysis range from 28.3733 to 32.5333. This result can be verified by computing the exact values as

6458.32111

)50(8.9)( 6)50/11( eccv

4769.28114

)50(8.9)( 6)50/14( eccv

Thus, the range of 2.0844 is close to the first-order estimate.

4.9

4533.3015.12

)50(8.9)5.12( 50/)6(5.12 ev

mm

vc

c

vmcv ~~)~,~(

38666.1

12

)/()/(

c

egmcgte

c

v tmctmc

871467.01 )/()/(

tmctmc ec

ge

m

gt

m

v

822923.3742934.1079989.2)2(871467.0)5.1(38666.1)~,~( mcv

822923.34533.30 v

8


4.10 For T~

=20,

TT

HTH

~)

~(

408.8650)1067.5(9.0)15.0(44 383

TAeT

H

169.168)20(408.8)~

( TH

Exact error:

3286.1682

81.1205468.1542

2

)630()670(true

HHH

Thus, the first-order approximation is close to the exact result.

For T~

=40,

3387.336)40(408.8)~

( TH

Exact error:

6124.3372

83.1059055.1735

2

)610()690(true

HHH

Again, the first-order approximation is close to the exact result. The results are good because H(T) is nearly linear over the ranges we are examining. This is illustrated by the following

plot.

800

1200

1600

2000

550 600 650 700 750

4.11 For a sphere, A = 4r2. Therefore,

424 TerH

At the mean values of the parameters,

9


288.1320)550)(1067.5(90.0)15.0(4)550,9.0,15.0( 482 H

TT

He

e

Hr

r

HH

~~~

84.603,178 4

Tre

r

H

987.14664 42

Tr

e

H

6021.916 32

Ter

T

H

4297.441)20(6021.9)05.0(987.1466)01.0(84.17603 H

To check this result, we can compute

6372.936)530)(1067.5(85.0)14.0(4)530,85.0,14.0( 482 H

178.1829)570)(1067.5(95.0)16.0(4)570,95.0,16.0( 482 H

2703.4462

6372.936178.1829true

H

4.12 The condition number is computed as

)~(

)~('~

xf

xfxCN

(a) 617.157003162.1

)1139.158(00001.1

1100001.1

100001.12

100001.1

CN

The result is ill-conditioned because the derivative is large near x = 1.

(b) 101054.4

)1054.4(10)(105

5

10

10

e

eCN

The result is ill-conditioned because x is large.

10


(c) 99999444.00016667.0

)10555556.5(300

3001300

11300

300300

6

2

2

CN

The result is well-conditioned.

(d) 0005.09995.0

)499667.0(001.0

1

12

x

e

x

exex

CNx

xx

The result is well-conditioned.

(e) 001,102.6366

)237,264,20(141907.3

cos1

sin

)cos1(

)(sinsincos)cos1(2

x

x

x

xxxxx

CN

The result is ill-conditioned because, as in the following plot, the function has a singularity at

x = .

-2000

-1000

0

1000

2000

3.08 3.1 3.12 3.14 3.16 3.18 3.2

4.13 Addition and subtraction:

vuvuf ),(

vv

fu

u

ff ~~

1 1

v

f

u

f

vuvuf ~~)~,~(

Multiplication:

11


vuvuf ),(

uv

fv

u

f

vuuvvuf ~~~~)~,~(

Division:

vuvuf /),(

2

1

v

u

v

f

vu

f

vv

uu

vvuf ~~1

)~,~(2

2

~~

)~,~(v

vuuvvuf

4.14

axf

baxxf

cbxaxxf

2)("

2)('

)( 2

Substitute these relationships into Eq. (4.4),

)2(!2

2))(2( 2

12

112

12

1 iiiiiiiiiii xxxxa

xxbaxcbxaxcbxax

Collect terms

cxxbbxxxxxaxxaxaxcbxax iiiiiiiiiiiii )()2()(2 12

12

112

12

1

cbxbxbxaxxaxaxaxxaxaxcbxax iiiiiiiiiiiii 12

12

12

12

12

1 222

cbxbxbxxaxxaxaxaxaxaxcbxax iiiiiiiiiiiii 1112

1222

12

1 )()22()2(

cbxaxcbxax iiii 12

112

1

12


4.15 The first-order error analysis can be written as

SS

Qn

n

QQ

74.50)2(

)(1 5.0

3/2

3/5

2

S

HB

BH

nn

Q 9.2536

2

1

)2(

)(15.03/2

3/5

SHB

BH

nS

Q

228.0076.0152.000003.09.2536003.074.50 Q

The error from the roughness is about 2 times the error caused by the uncertainty in the slope.

Thus, improving the precision of the roughness measurement would be the best strategy.

4.16 Use the stopping criterion

%5.0%105.0 22 s

True value: 1/(1 – 0.1) = 1.111111….

zero order:

11

1

x

%10%1001.11111

11.11111

t

first order:

1.11.011

1

x

%1t %0909.9%1001.1

11.1

a

second order:

11.101.01.011

1

x

%1.0t %9009009.0%1001.11

1.111.1

a

third order:

13


111.1001.001.01.011

1

x

%01.0t %090009.0%1001.111

11.1111.1

a


4.17

d

d 00

sinsin

11

)1)(1(2

sin 0

d

d

where = (ve/v0)2 = 4 and = 0.25 to give,

1305.325.01

)4(25.01

))4(25.025.01)(25.01(2

4sin 0

d

d

1305.3sin 0

For = 0.25(0.02) = 0.005,

015652.0)005.0(1305.3sin 0

559017.0425.01

25.01)25.01(sin 0

Therefore,

574669.0015652.0559017.0sin max 0

543365.0015652.0559017.0sin min 0

076.35180

)574669.0arcsin(max 0

913.32180

)543365.0arcsin(min 0

4.18 The derivatives can be computed as

14


xxxf sin5.01)(

xxf cos5.01)('

xxf sin5.0)("

xxf cos5.0)()3(

xxf sin5.0)()4(

The first through fourth-order Taylor series expansions can be computed based on Eq. 4.5 as

First-order:

))((')()(1 axafafxf

5.122

cos5.012

sin5.012

)(1

xxxf

Second-order:

212 )(

2

)(")()( ax

afxfxf

2

2 )2/)(2/sin(25.05.1)( xxxf

Third-order:

3)3(

23 )(6

)()()( ax

afxfxf

323 )(

6

)2/cos(5.0)2/)(2/sin(25.05.1)( axxxxf

2

3 )2/)(2/sin(25.05.1)( xxxf

Fourth-order:

4)4(

34 )(24

)()()( ax

afxfxf

424 )2/(

24

)2/sin(5.0)2/)(2/sin(25.05.1)(

xxxxf

15


424 )2/(

48

1)2/)(2/sin(25.05.1)( xxxxf

Note the 2nd

and 3rd

Order Taylor Series functions are the same. The following MATLAB script file which implements and plots each of the functions indicates that the 4

th-order

expansion satisfies the problem requirements.

x=0:0.001:3.2;

f=x-1-0.5*sin(x);

subplot(2,2,1);

plot(x,f);grid;title('f(x)=x-1-0.5*sin(x)');hold on

f1=x-1.5;

e1=abs(f-f1); %Calculates the absolute value of the

difference/error

subplot(2,2,2);

plot(x,e1);grid;title('1st Order Taylor Series Error');

f2=x-1.5+0.25.*((x-0.5*pi).^2);

e2=abs(f-f2);

subplot(2,2,3);

plot(x,e2);grid;title('2nd/3rd Order Taylor Series Error');

f4=x-1.5+0.25.*((x-0.5*pi).^2)-(1/48)*((x-0.5*pi).^4);

e4=abs(f4-f);

subplot(2,2,4);

plot(x,e4);grid;title('4th Order Taylor Series Error');hold off

4.19 Here are Excel worksheets and charts that have been set up to solve this problem:

16


-4

0

4

8

12

-2 0 2

exact

backward

centered

forward

17


-4

0

4

8

12

-2 0 2

exact

backward

centered

forward

chapter 4 - iiethus, the third-order result is perfect because the original function is a third...

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