Unformatted text preview: ρ = max ij | u ij | max ij | a ij | . The backward error result for G.E.P.P. is ˜ L ˜ U = ˜ P ( A + δA ) , where k δA k k A k = ρ O( μ ) . This implies G.E.P.P. is backward stable if ρ = O(1). We have a beautiful delimma here that makes further discussion diﬃcult, so we will ever so simply say that G.E.P.P. is backward stable, but not really, . ..but eﬀectively is. Backward and forward substitution, on the other hand, are clearly backward stable. The result for back substitution is that the computed ˜ x satisﬁes ( R + δR )˜ x = b, where k δR k k R k = O( μ ) . Combining the results above, we can say the that the computed solution, ˜ x to Ax = b , using G.E.P.P with forward and backward substitution, satisﬁes ( A + δA )˜ x = b, where k δA k k A k = ρn 3 O( μ ) . The n 3 term above appears quite pessimistic, for in practice we see k δA k k A k = ρn O( μ ) ....
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This note was uploaded on 12/18/2010 for the course PHYS 5073 taught by Professor Mark during the Fall '10 term at Arkansas.
- Fall '10