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Recurrence Relation for Iterative Solution of Ax=b
Never mind, figured it out. Here's my solution: \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 -...- Kidnapster
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- Forum: Calculus and Beyond Homework Help
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K
Recurrence Relation for Iterative Solution of Ax=b
_nx represents the nth iteration. Homework Statement Show that {_{n+2}}x = _nx + M^{-1}(b-A{_nx}) Homework Equations {_{n+1}}x = _nx + M_1^{-1}(b-A{_nx}) {_{n+2}}x = {_{n+1}}x + M_2^{-1}(b-A{_{n+1}x}) M = M_1(M_1+M_2-A)^{-1}M_2 The Attempt at a Solution I tried was...- Kidnapster
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- Iterative Recurrence Relation
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- Forum: Calculus and Beyond Homework Help