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Calculus and Beyond Homework Help
Proving Equivalence of Iterative Refinement Equations for Linear Systems
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[QUOTE="beth92, post: 4993062, member: 389357"] Still can't seem to get anywhere...in the above equations wouldn't it be x[SUB]1[/SUB] - x = nx[SUB]0[/SUB] - m ? I feel it would be more useful to express the difference between x[SUB]0[/SUB] and x as m = (x[SUB]0[/SUB] - x) as you can then just look for equation (2) as x[SUB]k+1[/SUB] - x = m * n[SUP]k[/SUP]. Either way, in order to iterate the algorithm for k+1 i need to substitute in x[SUB]k[/SUB], not x[SUB]k[/SUB] - x so I end up having to get rid of the m anyway. Maybe I am overcomplicating it but iterating it just seems to give me a long list of terms beginning with x[SUB]0[/SUB] + ... or m + ... Also, one of my classmates brought up that she thinks the exponent in equation 2 should be k+1 rather than k (although I can't get either!) I can't manage to make (1) equal to (2) even when I choose k=0 and write down both expressions. As you say above, using equation (1) I will get x[SUB]1[/SUB] - x = (x[SUB]0[/SUB] - x) + (I - B[SUP]-1[/SUP]A) x[SUB]0[/SUB] Then, using equation 2 for k=0 I will get: x[SUB]1[/SUB] - x = (x[SUB]0[/SUB] - x) * (I - B[SUP]-1[/SUP] A)[SUP]0[/SUP] = x[SUB]0[/SUB] - x If equation (1) and (2) are equivalent then the above two expressions should be equal but as far as I can see they are not, unless x[SUB]0[/SUB]( I - B[SUP]-1[/SUP]A) is zero. Even if my classmate is correct and the exponent should be k+1, I can't see any way to show that x[SUB]1[/SUB] - x = (x[SUB]0[/SUB] - x) + (I - B[SUP]-1[/SUP]A) x[SUB]0[/SUB] = (x[SUB]0[/SUB] - x) * (I - B[SUP]-1[/SUP] A) Sorry for the long reply but I feel I have tried everything! Thanks for your help. [/QUOTE]
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Proving Equivalence of Iterative Refinement Equations for Linear Systems
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