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Conditioning and Stability
A problem is
well conditioned
if a small change in the input creates a small change in the
output (solution).
A computation is
backward stable
if it produces the exact solution to a nearby problem.
There is room for debate in this deﬁnition, but it does succinctly capture the idea. I like
better: A computation is
backward stable
if the result is
close
to the exact solution to a nearby
problem. In any case, we say that a method is backward stable for a set of problems if it is
backward stable for each problem in the set.
If the problem we are trying to solve has a unique solution, then we can formulate it as
“evaluate
f
(
x
)”, where
x
represents the input and
f
(
x
) represents the solution (output). Let’s
represent our input space by
D
and our output space by
R
. Then our computed result can be
represented by
¯
f
:
D → R
, where the exact result is represented by
f
:
D → R
. That is,
¯
f
(
x
) is our computed approximation to
f
(
x
).
A problem is
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 Fall '10
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