What is the iterative method in system of linear equations? 2. Give atleast two examples of each iterative method. Answer question number 1 only. These is the reference that might help to answer the question.
What is the iterative method in system of linear equations? 2. Give atleast two examples of each iterative method. Answer question number 1 only. These is the reference that might help to answer the question.
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
1. What is the iterative method in system of linear equations?
2. Give atleast two examples of each iterative method.
Answer question number 1 only.
These is the reference that might help to answer the question.
Transcribed Image Text:The solution is X1 = 1.0013, X2 = 0.9945, X3 = 0.9983.
The exact solution is x = 1, X2 = 1, X3 = 1.
Example
3x1 + x2 - x3 = 3
2x1 + 4x2 + x3 = 7
X1 - x2 + 4x3 = 4
Consider the initial approximation, x1(0) = 0, x2(0) = 0, x3) = 0. Apply Gauss-
Seidel method until the last iteration has a difference that is less than 0.005.
References:
Kharab, A. & Guenther R. (2019). An introduction t
(4 ed.). Boca Raton: CRC Press.
Gupta, R. (2019). Numerical methods: Fundamentals and applications. India: Cambridge
University Press,
numerical methods: A MATLAB approach
Step 1. Rewrite the system.
X1 =,(3 – x2 + x3)
x2 = (7 – 2x1 – x3)
x3 = (4 - x1 + x2)
Step 2. Get the first approximation.
x,(1)
=(3 – x2(0) + x3(0)) = 1
X2(1)
=-(7 – 2x,
- x3(0)) = 1.25
x3(1) = (4 – x,(1) + x2(1)) = 1.0625
%3D
Step 3. Perform approximation until desired accuracy is obtained.
1
=-(3-x," + x,"
3
(2)
(") 0.9375
(7–2x,2) – x,"))=1.0156
4
(2) =
(2) -
(2)
(4-x,4 +x,2))= 1.0195
4
As required, the difference of the last iteration should be less than 0.005,
which is after the third iteration (faster than Jacobi).
1.0000
| 1.0625
1.0195
0.9983
Iteration 1
1.2500
Iteration 2
0.9375
1.0156
Iteration 3
1.0013
0.9945
![Systems of Linear Equations (Iterative Methods)
Iterative methods are based on successive improvement of initial guesses
for the solution.
The first approximation [x1(1), x2(1), ..., Xn(1)] is used to solve for the second
iteration. The process is repeated until desired accuracy is obtained. The
(k+1)th iteration can be obtained from kth iteration by the following formula:
x,
(k+1) =b,-(a,x,4) + a,zx,«) +•.+a„x„")]
%3D
Iterative Method 1:
Jacobi Method/Method of
Simultaneous
Displacement
1
x, (k).
(k)
(k) +...+azX
(k+1)
x,
The linear system of equations can be rewritten as:
-[b, - (a,,x, +a3X, +…·+a„X,
(k+1)
%3D
(k).
k = 0,1,2,...
a
1
x, =
-b, - (a,x, +a,x, + +a„X,
The Jacobi iteration formula can be written as:
1
(k+1)
%3D
(k)
1sisn k=0,1,2,...
j=1
X =
-b, - (a,x, +aX2 ++a,n-1\,-1
a
Example
Initial approximation is required to get the next approximation. Let the initial
approximation be [x1©, x2), .., Xn]. These values are used to get the next
approximation [x,("), x2(1), ..., xn")] of the Jacobi method.
3x1 + x2 – x3 = 3
2x1 + 4x2 + x3 = 7
X1 - x2 + 4x3 = 4
(1)
%3D
- b, - (a,,x2 + a,3x, +.+a,„X„")]
(0)
(0)
(0)
Consider the initial approximation, x,0) = 0, x20) = 0, x30) = 0. Apply Jacobi
method until the last iteration has a difference that is less than 0.005.
(1)
(0)
,- (a,x +azx
Step 1. Rewrite the system.
1
X1 = (3 – x2 + x3)
(1)
%3D
x2 = (7 – 2x1 – x3)
(0)
b,-(a,x,® +a„zx," +…+a,n-*n-")|
ann
X3 =7(4 – x1 + x2)
06 Handout 1
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*Property of STI
Page 1 of 3
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IT1903
Step 2. Get the first approximation.
Solve for the approximation of variable x1:
1
(1)
(0)
(1)
(0)
(0)
(0)
(3-x, +x,'
3
)=1
Co"x"lv+..+ o*xlv)– 'q ]
x,"
(1)
=-(7-2x,0 – x,)=1.75
(0)
To solve for x2(1), use x1(1) instead of x1(0) (as used in Jacobi):
4
1
b,-(a,x," +a,x,
(1)
(0)
(0)
X,
+..+a,„x,") |
x, ),
-(4-x, + x,")=1
4
%3D
(0
Solve for the first approximation to the ith variable:
Step 3. Perform approximation until desired accuracy is obtained.
+a* +.+ax,
%3D
1
(2)
(1)
(3-x,"+x,")=0.75
3
X" =.
The last variable is calculated as:
6, - (a,,x," +a„X
a
(1)
(1)
(1)
+...+a,n-1Xn-1
(2)
(1)
(7-2x," - x,)=1
4
X, =.
The first approximation [x1("), x2"), ..., x,(")] is used to solve for the second
iteration. The (k+1)th iteration can be obtained from kth iteration by the
following formula:
(2)
(1)
x, =.
(4-x,
+x,")=1.1875
As required, the difference of the last iteration should be less than 0.005,
which is after the sixth iteration.
(k) +...+ aX,
%3D
a
Iteration 1
Iteration 2
1.0000 1.7500| 1.0000
1.1875
1.0625
1.0039
0.9896
0.9962
1.0000
1.0625 1.0781
0.7500
-(
Iteration 3
Iteration 4
Iteration 5
0.9948 0.9531
1.0169 1.0016
0.9960 | 0.9941
Iteration 6
(k+1) + a,X2
(k+1) =
*+1) +.+a,x**)) k= 0,1,2,...
(k+1)
The solution is x1 = 0.9960, x2 = 0.9941, X3 = 0.9962.
The exact solution is x1 = 1, X2 = 1, X3 = 1.
a
The Gauss-Seidel iteration formula can be written as:
Iterative Method 2: Gauss-Seidel Method/Method of Successive
1
(k+1)
(k)
1sisn k=0,1,2,..
(k+1)
Displacement/Liebmann Method
a,
j=l
Let the initial approximation be [x,(0), x20), ..., x,]. The latest available
values are used to get the next approximation. Let the next approximation
be (x"), x2"), ..., Xn()].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe92e190-21eb-4bbd-ab9b-abe2398bf339%2Fa925ab12-b539-4bc6-83ef-921247754266%2F4ne5aqn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Systems of Linear Equations (Iterative Methods)
Iterative methods are based on successive improvement of initial guesses
for the solution.
The first approximation [x1(1), x2(1), ..., Xn(1)] is used to solve for the second
iteration. The process is repeated until desired accuracy is obtained. The
(k+1)th iteration can be obtained from kth iteration by the following formula:
x,
(k+1) =b,-(a,x,4) + a,zx,«) +•.+a„x„")]
%3D
Iterative Method 1:
Jacobi Method/Method of
Simultaneous
Displacement
1
x, (k).
(k)
(k) +...+azX
(k+1)
x,
The linear system of equations can be rewritten as:
-[b, - (a,,x, +a3X, +…·+a„X,
(k+1)
%3D
(k).
k = 0,1,2,...
a
1
x, =
-b, - (a,x, +a,x, + +a„X,
The Jacobi iteration formula can be written as:
1
(k+1)
%3D
(k)
1sisn k=0,1,2,...
j=1
X =
-b, - (a,x, +aX2 ++a,n-1\,-1
a
Example
Initial approximation is required to get the next approximation. Let the initial
approximation be [x1©, x2), .., Xn]. These values are used to get the next
approximation [x,("), x2(1), ..., xn")] of the Jacobi method.
3x1 + x2 – x3 = 3
2x1 + 4x2 + x3 = 7
X1 - x2 + 4x3 = 4
(1)
%3D
- b, - (a,,x2 + a,3x, +.+a,„X„")]
(0)
(0)
(0)
Consider the initial approximation, x,0) = 0, x20) = 0, x30) = 0. Apply Jacobi
method until the last iteration has a difference that is less than 0.005.
(1)
(0)
,- (a,x +azx
Step 1. Rewrite the system.
1
X1 = (3 – x2 + x3)
(1)
%3D
x2 = (7 – 2x1 – x3)
(0)
b,-(a,x,® +a„zx," +…+a,n-*n-")|
ann
X3 =7(4 – x1 + x2)
06 Handout 1
E student.feedback@sti.edu
*Property of STI
Page 1 of 3
ASTI
IT1903
Step 2. Get the first approximation.
Solve for the approximation of variable x1:
1
(1)
(0)
(1)
(0)
(0)
(0)
(3-x, +x,'
3
)=1
Co"x"lv+..+ o*xlv)– 'q ]
x,"
(1)
=-(7-2x,0 – x,)=1.75
(0)
To solve for x2(1), use x1(1) instead of x1(0) (as used in Jacobi):
4
1
b,-(a,x," +a,x,
(1)
(0)
(0)
X,
+..+a,„x,") |
x, ),
-(4-x, + x,")=1
4
%3D
(0
Solve for the first approximation to the ith variable:
Step 3. Perform approximation until desired accuracy is obtained.
+a* +.+ax,
%3D
1
(2)
(1)
(3-x,"+x,")=0.75
3
X" =.
The last variable is calculated as:
6, - (a,,x," +a„X
a
(1)
(1)
(1)
+...+a,n-1Xn-1
(2)
(1)
(7-2x," - x,)=1
4
X, =.
The first approximation [x1("), x2"), ..., x,(")] is used to solve for the second
iteration. The (k+1)th iteration can be obtained from kth iteration by the
following formula:
(2)
(1)
x, =.
(4-x,
+x,")=1.1875
As required, the difference of the last iteration should be less than 0.005,
which is after the sixth iteration.
(k) +...+ aX,
%3D
a
Iteration 1
Iteration 2
1.0000 1.7500| 1.0000
1.1875
1.0625
1.0039
0.9896
0.9962
1.0000
1.0625 1.0781
0.7500
-(
Iteration 3
Iteration 4
Iteration 5
0.9948 0.9531
1.0169 1.0016
0.9960 | 0.9941
Iteration 6
(k+1) + a,X2
(k+1) =
*+1) +.+a,x**)) k= 0,1,2,...
(k+1)
The solution is x1 = 0.9960, x2 = 0.9941, X3 = 0.9962.
The exact solution is x1 = 1, X2 = 1, X3 = 1.
a
The Gauss-Seidel iteration formula can be written as:
Iterative Method 2: Gauss-Seidel Method/Method of Successive
1
(k+1)
(k)
1sisn k=0,1,2,..
(k+1)
Displacement/Liebmann Method
a,
j=l
Let the initial approximation be [x,(0), x20), ..., x,]. The latest available
values are used to get the next approximation. Let the next approximation
be (x"), x2"), ..., Xn()].
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