Anyone who can help me with this? 1 5. Consider the iterative improvement algori

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Anyone who can help me with this?

1 5. Consider the iterative improvement algorithm discussed in lecture Solve Ax = b for initial approximation io. for i = 0, 1, . . . until convergence compute ri = b-Ari solve Azi = ri update zi+1 = a, + end for We gave an intuitive explanation of why this algorithm could improve the initial approximate solution ro, but we were vague on the conditions required for convergence. In this question, you will attempt to derive the precise conditions required for convergence Page 2 of 3 pages

Explanation / Answer

Iterative improvement
Suppose t – digit where is base arithmetic , then we will calculated result ˆ x assures
(A + A)ˆx = b
||A|| equaling u||A|| , u = 1/2 ()^1t . The solution ˆx is the vector b Axˆ.
As per Gaussian elimination creates a solution ˆx with comparatively small residual
||b Axˆ|| equaling u||A||||xˆ||.
Small residuals don’t entail high exactness.
we can use the formula that computed true theorem
(||xˆ x|| ) / (||x||) uK(A).
If the one round off and condition satisfy at u is nearer to 10d and K(A) nears equals to 10^q
So Gaussian elimination gives a result for ˆx that is d q accurate decimal digits.
Suppose Ax = b has been solved via the partial pivoting factorization

On the off chance that the machine accuracy u and condition number fulfill u 10d and K(A) 10q at that point .after k executions of the algorithm, x has around min(d, k(d q)) revise digits.
On the off chance that uK(A) 1, at that point iterative change can deliver an answer x that has t redress digits. Note that the procedure is moderately shabby in flops. Each improvement costs O(n^2 ) flops
while the original disintegration costs O(n^3 ) flops.
The main drawbacks is memory requirements are two times for the reason that an original copy of A be supposed to be stored in memory. And it is memory dependent.

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