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Recurrence Relation for Iterative Solution of Ax=b

  1. Jun 16, 2011 #1
    [itex] _nx [/itex] represents the nth iteration.

    1. The problem statement, all variables and given/known data

    Show that [itex] {_{n+2}}x = _nx + M^{-1}(b-A{_nx}) [/itex]

    2. Relevant equations

    [itex] {_{n+1}}x = _nx + M_1^{-1}(b-A{_nx}) [/itex]
    [itex] {_{n+2}}x = {_{n+1}}x + M_2^{-1}(b-A{_{n+1}x}) [/itex]
    [itex] M = M_1(M_1+M_2-A)^{-1}M_2 [/itex]

    3. The attempt at a solution

    I tried was adding the two equations, which gives [itex] _{n+2}x = M_2^{-1}(b-A{_{n+1}x}) + _nx + M_1^{-1}(b-A{_nx}) [/itex]. But [itex] M_2^{-1}(b-A{_{n+1}}x) [/itex] devolves into a huge mess once I expand it. Subtraction is less helpful, as you are left without any [itex] _nx [/itex]. Directly substituting for [itex]_{n+1}x [/itex] in the second equation is even less helpful, as you are left with a huge mess of terms that can't really be factored. What am I missing?
     
  2. jcsd
  3. Jun 17, 2011 #2
    Never mind, figured it out. Here's my solution:

    [tex]
    \begin{align*}
    {_{n+1}x} &= {_nx} + M_1^{-1}(b-A_nx)\\
    {_{n+2}x} &= {_{n+1}x} + M_1^{-1}(b-A_{n+1}x) \\
    {_{n+2}x} &= {_nx} + M_1^{-1}(b-A_nx) + M_2^{-1}(b-A[{_nx} + M_1^{-1}(b-A_nx)])\\
    {_{n+2}x} &= {_nx} + M_1^{-1}b - M_1^{-1}A_nx + M_2^{-1}b-M_2^{-1}A_nx - M_2^{-1}AM_1^{-1}b + M_2^{-1}AM_1^{-1}A_nx \\
    {_{n+2}x} &= {_nx} + M_1^{-1}(b-A_nx) + M_2^{-1}(b-A_nx) - M_2^{-1}AM_1^{-1}(b-A_nx) \\
    {_{n+2}x} &= {_nx} + M_1^{-1}(M_1+M_2-A)M_2^{-1}(b-A_nx)
    \end{align*}

    [/tex]
     
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