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I have a two iterative method for approximating the inverse of given square matrix $A$ whose error terms are given as follows

Error estimate of method $1$: $\lVert A^{-1} - X_{k}\rVert \leq q^{2^{k+1}-3}(1-q^{2^{k+1}})^{-1}\|R_{0}\|$, $0<q<1$, $k = 0, 1,\dots$

$R_{0} = I - AX_{0}$ , $X_{k}$ is the sequence of approximations for $A^{-1}$ and $X_{0}$ is the initial approximation

Error estimate of method $2$: $\lVert A^{-1} - X_{k}\rVert \leq q^{2^{k}}(1-q^{2^k})^{-1}\|R_{0}\|$

Clearly method $1$ has correct order $q^{2^{k+2}-3}$ , while second one has order $q^{2^{k}}$. Both the methods are of second order.

My question is could we compare the speed of convergence of two different iteration methods of same order looking at their error estimates?

I beg pardon if this question is stupid.

Thanks for the help.

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  • $\begingroup$ What do you mean "compare the speed of convergence"? Are you looking to validate the estimates above? Are you looking for the value of $q$? Is it the same $q$ in both cases? Do you have access to $A^{-1}$ to compute the errors? If not you could look at ratios of the form $\|X_{k+1} - X_k\|/\|X_k - X_{k-1}\|$ for both methods. $\endgroup$ – Dominique Dec 28 '13 at 0:12
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Normally when one makes error estimates they are not particularly tight bounds, i.e. there could easily be a factor of a constant off. Your two methods differ by approximately a constant, namely $q^{-3}$, so you do not have enough data to conclude that Method 2 has smaller error than Method 1. To conclude this you would need the difference to approach 0 as $k\rightarrow \infty$, and even then you would only know that Method 2 has smaller error for sufficiently large $k$.

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  • $\begingroup$ Dear prof. thanks for the answer. Could you please tell me what differences are you talking about? How could I take? Thanks $\endgroup$ – srijan Jun 7 '13 at 14:36

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