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Week 12
 Iterative solutions to
systems of linear equations
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View Full Document Iterative solutions
3x
z =2
x +2y
=1
4x+4y+8z=28
Solve for x,y,z:
x= (2+z)/3
y= (1+x)/2
z= (28+4x4y)/8
Iterative solutions
x= (2+z)/3
y= (1+x)/2
z= (28+4x4y)/8
Use these from an initial guess to
estimate the solution.
x
n
= (2+z
n1
)/3
y
n
= (1+x
n1
)/2
z
n
= (28+4x
n1
4y
n1
)/8
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View Full Document So where do we start?
Iterative solutions
Answer: Anywhere, closer
just converges faster.
I shall pick b
0
= [0, 0, 0]
Iterative solutions
x
n
= (2+z
n1
)/3
y
n
= (1+x
n1
)/2
z
n
= (28+4x
n1
4y
n1
)/8
b
0
= [0, 0, 0]
b
1
:
b
2
:
x
2
= (2+z
1
)/3
y
2
= (1+x
1
)/2
z
2
= (28+4x
1
4y
1
)/8
x
1
= (2+z
0
)/3
y
1
= (1+x
0
)/2
z
1
= (28+4x
0
4y
0
)/8
b
1
= [2/3, 1/2, 28/8]
b
2
= [1.83, 0.166, 4.083]
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View Full Document Iterative solutions
x
n
= (2+z
n1
)/3
y
n
= (1+x
n1
)/2
z
n
= (28+4x
n1
4y
n1
)/8
b
0
= [0, 0, 0]
b
1
:
x
1
= (2+z
0
)/3
y
1
= (1+x
0
)/2
z
1
= (28+4x
0
4y
0
)/8
b
1
= [2/3, 1/2, 28/8]
x
n
= (2+z
n1
)/3
y
n
= (1+x
n1
)/2
z
n
= (28+4x
n1
4y
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This note was uploaded on 10/21/2011 for the course CSCI 2031 taught by Professor Meyer during the Spring '08 term at Minnesota.
 Spring '08
 MEYER

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