MTH3051 Introduction to Computational Mathematics - User Web ...

School of Mathematical Sciences
Monash University
which we re-write as
x n+1 − x = g ′ (x)(x n − x) + O ( (x n − x) 2)
Now notice that x n − x is the error in our approximation at iteration n. So let’s define
the error at each iteration by
ɛ n = |x n − x|
then we have
ɛ n+1 = |g ′ (x)|ɛ n + O ( )
ɛ 2 n
Notice that if |g ′ (x)| < 1 then successive errors will be smaller than the previous errors,
that is the iterations will converge. On the other hand, if |g ′ (x)| > 1 the errors will grow
and the sequence diverges. The only other case is when |g ′ (x)| = 1 and in this case the
errors neither grow nor decay.
Note also that the above arguments all hinge on the assumption that we are near the
fixed point. All bets are off if we have a bad initial guess (i.e. x 1 is far from the fixed
point). However once the sequence gets close to the fixed point (if we should be so lucky)
then the above arguments do apply.
Fixed Point Iteration Convergence
Given an x 1 close to the fixed point x = g(x), then the sequence
x n+1 = g(x n ) n = 1, 2, 3, · · ·
Converges when |g ′ (x)| < 1
Diverges when |g ′ (x)| > 1
Useless when |g ′ (x)| = 1
Example 4.1
Verify, using the above conditions, that the scheme x n+1 = e −xn
scheme x n+1 = − log(x n ) should diverge.
will converge, while the
4.2.5 Programming notes
Since we know that the fixed point iterations might fail we must take care when we write
our computer programs. We need to ask ourselves what can go wrong in the calculations
(too many iterations? not converging? stalled? etc.) and put appropriate tests in our
programs. Here is a very rough sketch of a program that could be used for any algorithm
to solve 0 = f(x) for x.
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