Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Do three iterations (by hand) of the bisection method, applied to f (x) = x³ – 2, using a = 0 and b = 2
Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Do three iterations (by hand) of the bisection method, applied to f (x) = x³ – 2, using a = 0 and b = 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
answer this and I provide some examples and the table of bisection method (pls follow the attachment that i gave)
Transcribed Image Text:Direction. Answer the following: Finds the roots of equations using required relative error
dk < 0.001
1. Do three iterations (by hand) of the bisection method, applied to f (x) = x³ – 2, using
a = 0 and b = 2
![BISECTION METHOD TABLE!!!!!
0.625
0.562
0.593
-0.040
0.593
0.562
0.577
-0.015
0.577
0.562
0.569
-0.002
k
br
f(Ck+1)
dk
Remarks
0.569
0.562
0.565
0.003
0.007
Continue
ak
Ck+1
8
0.569
0.565
0.567
0.0002
0.003
Continue
9
0.569
0.567
0.568
0.001
Stop
ΕΧΑMPLES
therefore, X = 0. 568 = x10
Example 1.) If f (x) = 2 – e*, and take the original interval to be [a, b] = [0,1].
Example 3.) f(x) = 6x-5x2 + 7x - 2
Solution:
Solution:
to
-10
2
40
f(x)
-20
-2
f(a) = 1, f(b) = -0.7183
c = 10+ 1] - 1
f(c) = 0.3513 > 0. Therefore, [a, b] = ;,1|
k
S(Ck+1)
Remarks
Ck+1
0.5
1
1.000
Continue
1
0.5
0.25
-0.040
2
.5
0.375
0.375
2
0.5
0.25
0.375
0.238
3.
0.25
0.312
0.343
0.327
-0.12
%3D
4
0.312
0.054
-
0.343
0.312
-0.035
0.343
0.327
0.335
0.009
0.023
Continue
f(a) = 0.3513, f(b) = -0.7183
%3D
%3D
Continue solution until X = 0.333
3
c =
2
4
Example 4.) Solve 2x3 – 2.5x – 5 = O for the root in the interval [1, 2].
f(c) = -0.1170 < 0. Therefore, [a, b] =
Solution:
k-0; a, = 1; bo = 2
s (do* bo) = s (2) - r(1.5) = -2 < 0
= F(1.5) = -2 < o
Therefore, f(ao)s (be) = 11 > 0 and If(1.5)| > E
(ao +
f(a) = 0.3513, f (b) = –0.1170
C =
8.
k = 1; a, = 1.5; b, = 2
f(c) = 0.1318 > 0. Therefore, [a, b] =
r() - r (15* 2)
s(aD:) = s (15*2) =
f(1.75) = 1.344
%3D
Therefore, f(a)f (b) = -2.688 < 0 and IS(1.75)| >e
k = 2; az = 1.5; b, = 1.75
Details of the remaining steps are provided in the table below:
Example 2.) f(x) = e¬*-x
k
(Ck+1)
Remarks
Solution:
ak
1
Ck+1
1.5
-2.000
Continue
1
15
175
1 344
x2 = 0; f(x?) = 1
xº = 1; f(xº) = -0.632
1.5
1.75
1.625
-0.481
3
1.625
1.75
1.688
0.392
4
1.625
1.688
1.656
-0.054
5
1.656
1.688
1.672
0.167
k
f(ck+1)
dk
Remarks
6.
1.656
1.672
1.664
0.056
Ck+1
0.5
7
1.656
1.664
1.660
0.001
1
0.106
Continue
8
1.656
1.660
1.658
-0.027
1
0.5
0.75
-0.277
9
1.658
1.660
1.659
-0.013
10
1.660
1.660
1.660
-0.006
2
0.75
0.5
0.625
-0.089
11
1.660
1.660
1.660
-0.003
3
0.625
0.5
0.562
0.008
12
1.660
1.660
1.660
-0.001
--](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6216739-1791-4d30-8145-58e01cdd285d%2Fc473f2dc-980c-4099-9aa7-e52bc4ce78ce%2Fglwccse_processed.png&w=3840&q=75)
Transcribed Image Text:BISECTION METHOD TABLE!!!!!
0.625
0.562
0.593
-0.040
0.593
0.562
0.577
-0.015
0.577
0.562
0.569
-0.002
k
br
f(Ck+1)
dk
Remarks
0.569
0.562
0.565
0.003
0.007
Continue
ak
Ck+1
8
0.569
0.565
0.567
0.0002
0.003
Continue
9
0.569
0.567
0.568
0.001
Stop
ΕΧΑMPLES
therefore, X = 0. 568 = x10
Example 1.) If f (x) = 2 – e*, and take the original interval to be [a, b] = [0,1].
Example 3.) f(x) = 6x-5x2 + 7x - 2
Solution:
Solution:
to
-10
2
40
f(x)
-20
-2
f(a) = 1, f(b) = -0.7183
c = 10+ 1] - 1
f(c) = 0.3513 > 0. Therefore, [a, b] = ;,1|
k
S(Ck+1)
Remarks
Ck+1
0.5
1
1.000
Continue
1
0.5
0.25
-0.040
2
.5
0.375
0.375
2
0.5
0.25
0.375
0.238
3.
0.25
0.312
0.343
0.327
-0.12
%3D
4
0.312
0.054
-
0.343
0.312
-0.035
0.343
0.327
0.335
0.009
0.023
Continue
f(a) = 0.3513, f(b) = -0.7183
%3D
%3D
Continue solution until X = 0.333
3
c =
2
4
Example 4.) Solve 2x3 – 2.5x – 5 = O for the root in the interval [1, 2].
f(c) = -0.1170 < 0. Therefore, [a, b] =
Solution:
k-0; a, = 1; bo = 2
s (do* bo) = s (2) - r(1.5) = -2 < 0
= F(1.5) = -2 < o
Therefore, f(ao)s (be) = 11 > 0 and If(1.5)| > E
(ao +
f(a) = 0.3513, f (b) = –0.1170
C =
8.
k = 1; a, = 1.5; b, = 2
f(c) = 0.1318 > 0. Therefore, [a, b] =
r() - r (15* 2)
s(aD:) = s (15*2) =
f(1.75) = 1.344
%3D
Therefore, f(a)f (b) = -2.688 < 0 and IS(1.75)| >e
k = 2; az = 1.5; b, = 1.75
Details of the remaining steps are provided in the table below:
Example 2.) f(x) = e¬*-x
k
(Ck+1)
Remarks
Solution:
ak
1
Ck+1
1.5
-2.000
Continue
1
15
175
1 344
x2 = 0; f(x?) = 1
xº = 1; f(xº) = -0.632
1.5
1.75
1.625
-0.481
3
1.625
1.75
1.688
0.392
4
1.625
1.688
1.656
-0.054
5
1.656
1.688
1.672
0.167
k
f(ck+1)
dk
Remarks
6.
1.656
1.672
1.664
0.056
Ck+1
0.5
7
1.656
1.664
1.660
0.001
1
0.106
Continue
8
1.656
1.660
1.658
-0.027
1
0.5
0.75
-0.277
9
1.658
1.660
1.659
-0.013
10
1.660
1.660
1.660
-0.006
2
0.75
0.5
0.625
-0.089
11
1.660
1.660
1.660
-0.003
3
0.625
0.5
0.562
0.008
12
1.660
1.660
1.660
-0.001
--
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