Numerical Methods for Engineers 7th Edition - Solution Manual (Chapters 18-21) by Steven Chapra

1


CHAPTER 18
18.1 (a)
1.0791812  0.90309
f (10)  0.90309  (10  8)  0.991136
1
12  8
1 0.991136
  100%  0.886%
t
1

(b)
1.0413927  0.9542425
f (10)  0.9542425  (10  9)  0.997818
1
11 9
1 0.997818
  100%  0.218%
t
1

18.2 First, order the points

x0 = 9 f(x0) = 0.9542425
x1 = 11 f(x1) = 1.0413927
x2 = 8 f(x2) = 0.9030900

Applying Eq. (18.4)

b0 = 0.9542425

Equation (18.5) yields

1.0413927  0.9542425
b1   0.0435751
11 9

Equation (18.6) gives

0.9030900 1.0413927
 0.0435751
8 11 0.0461009  0.0435751
b2    0.0025258
89 89

Substituting these values into Eq. (18.3) yields the quadratic formula

f2 (x)  0.9542425  0.0435751(x  9)  0.0025258(x  9)(x 11)

which can be evaluated at x = 10 for

f2 (10)  0.9542425  0.0435751(10  9)  0.0025258(10  9)(10 11)  1.0003434

18.3 First, order the points

x0 = 9 f(x0) = 0.9542425
x1 = 11 f(x1) = 1.0413927
x2 = 8 f(x2) = 0.9030900
x3 = 12 f(x3) = 1.0791812

The first divided differences can be computed as

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1.0413927  0.9542425
f [x1, x0 ]   0.0435751
11 9
0.9030900 1.0413927
f [x 2 , x1 ]   0.0461009
8  11
1.0791812  0.9030900
f [x 3 , x 2 ]   0.0440228
12  8

The second divided differences are
0.0461009  0.0435751
f [x , x , x ]   0.0025258
89
2 1 0
0.0440228  0.0461009
f [x , x , x ]   0.0020781
12  11
3 2 1



The third divided difference is

0.0020781 (0.0025258)
f [x 3, x 2 , x1, x 0 ]   0.00014924
12  9

Substituting the appropriate values into Eq. (18.7) gives

f3 (x)  0.9542425  0.0435751(x  9)  0.0025258(x  9)(x 11)
 0.00014924(x  9)(x 11)(x  8)

which can be evaluated at x = 10 for

f3 (x)  0.9542425  0.0435751(10  9)  0.0025258(10  9)(10 11)
 0.00014924(10  9)(10 11)(10  8)  1.0000449

18.4
18.1 (a):
x0 = 8 f(x0) = 0.9030900
x1 = 12 f(x1) = 1.0791812
10 12 10  8
f (10)  0.9030900  1.0791812  0.991136
1
8 12 12  8

18.1 (b):
x0 = 9 f(x0) = 0.9542425
x1 = 11 f(x1) = 1.0413927
10 11 10  9
f (10)  0.9542425  1.0413927  0.997818
1
9 11 11 9

18.2 :
x0 = 8 f(x0) = 0.9030900
x1 = 9 f(x1) = 0.9542425
x2 = 11 f(x2) = 1.0413927


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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
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(10  9)(10 11) (10  8)(10 11)
f 2 (10)  0.9030900  0.9542425
(8  9)(8 11) (9  8)(9 11)
(10  8)(10  9)
 1.0413927  1.0003434
(11 8)(11 9)
18.3 :
x0 = 8 f(x0) = 0.9030900
x1 = 9 f(x1) = 0.9542425
x2 = 11 f(x2) = 1.0413927
x3 = 12 f(x3) = 1.0791812

(10  9)(10 11)(10 12) (10  8)(10 11)(10 12)
f3 (10)  0.9030900  0.9542425
(8  9)(8 11)(8 12) (9  8)(9 11)(9 12)
(10  8)(10  9)(10 12) (10  8)(10  9)(10 11)
 1.0413927  1.0791812  1.0000449
(11 8)(11 9)(1112) (12  8)(12  9)(12 11)

18.5 First, order the points so that they are as close to and as centered about the unknown as possible

x0 = 2.5 f(x0) = 14
x1 = 3.2 f(x1) = 15
x2 = 2 f(x2) = 8
x3 = 4 f(x3) = 8
x4 = 1.6 f(x4) = 2

Next, the divided differences can be computed and displayed in the format of Fig. 18.5,

i xi f(xi) f[xi+1,xi] f[xi+2,xi+1,xi] f[xi+3,xi+2,xi+1,xi] f[xi+4,xi+3,xi+2,xi+1,xi]
0 2.5 14 1.428571 -8.809524 1.011905 1.847718
1 3.2 15 5.833333 -7.291667 -0.651042
2 2 8 0 -6.25
3 4 8 2.5
4 1.6 2

The first through third-order interpolations can then be implemented as

f1(2.8)  14 1.428571(2.8  2.5)  14.428571
f2 (2.8)  14 1.428571(2.8  2.5)  8.809524(2.8  2.5)(2.8  3.2)  15.485714
f3 (2.8)  14 1.428571(2.8  2.5)  8.809524(2.8  2.5)(2.8  3.2)
 1.011905(2.8  2.5)(2.8  3.2)(2.8  2.)  15.388571

The error estimates for the first and second-order predictions can be computed with Eq. 18.19 as

R1  15.485714 14.428571  1.057143
R2  15.38857115.485714  0.097143

The error for the third-order prediction can be computed with Eq. 18.18 as

R3  1.847718(2.8  2.5)(2.8  3.2)(2.8  2)(2.8  4)  0.212857

18.6 First, order the points so that they are as close to and as centered about the unknown as possible


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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.

, 4



x0 = 3 f(x0) = 19
x1 = 5 f(x1) = 99
x2 = 2 f(x2) = 6
x3 = 7 f(x3) = 291
x4 = 1 f(x4) = 3

Next, the divided differences can be computed and displayed in the format of Fig. 18.5,

i xi f(xi) f[xi+1,xi] f[xi+2,xi+1,xi] f[xi+3,xi+2,xi+1,xi] f[xi+4,xi+3,xi+2,xi+1,xi]
0 3 19 40 9 1 0
1 5 99 31 13 1
2 2 6 57 9
3 7 291 48
4 1 3

The first through fourth-order interpolations can then be implemented as

f1(4)  19  40(4  3)  59
f2 (4)  59  9(4  3)(4  5)  50
f3 (4)  50 1(4  3)(4  5)(4  2)  48
f4 (4)  48  0(4  3)(4  5)(4  2)(4  7)  48

Clearly this data was generated with a cubic polynomial since the difference between the 4th and the
3rd-order versions is zero.

18.7
First order:
x0 = 3 f(x0) = 19
x1 = 5 f(x1) = 99
45 43
f (10)  19  99  59
1
35 53

Second order:
x0 = 3 f(x0) = 19
x1 = 5 f(x1) = 99
x2 = 2 f(x2) = 6
(4  5)(4  2) (4  3)(4  2) (4  3)(4  5)
f 2 (10)  19  99  6  50
(3  5)(3  2) (5  3)(5  2) (2  3)(2  5)

Third order:
x0 = 3 f(x0) = 19
x1 = 5 f(x1) = 99
x2 = 2 f(x2) = 6
x3 = 7 f(x3) = 291
(4  5)(4  2)(4  7) (4  3)(4  2)(4  7)
f3 (10)  19  99
(3  5)(3  2)(3  7) (5  3)(5  2)(5  7)
(4  3)(4  5)(4  7) (4  3)(4  5)(4  2)
 6 291  48
(2  3)(2  5)(2  7) (7  3)(7  5)(7  2)




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may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the
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