Order Of Convergence Of An Iterative Scheme
Let the sequence of iterative values { xn }
¥n = 0 converges
to 's'. Also let e
n = s-xn and e
n+1= s-xn+1 for n > 0 are the errors at nth and
(n+1)th iterations respectively. If two positive constants A
¹ 0 and R > 0 exist, and
lim
n
®¥ |
|s-xn+1| |
|
lim
n ®¥ |
|en+1
| |
|
= |
|
= A |
| |s-xn|R |
|
|e
n|R |
|
then the sequence is said to converge to 's'
with order of convergence R. The number A is called the asymptotic error
constant.
This can be derived from Taylor series as follows
Let xi+1= g(xi) define an iterative
method, and let 's' and xn respectively are the exact and
approximate solutions of x = g(x).
Then xn= s + e
n , where
en is the error in xn. Suppose
that g is differentiable any number of times, then from Taylor's formula
xn+1 = g(xn) = g(s +
en)
| = g(s) +
en g'(s) + |
1 |
e
n2 g''(s) + ... |
| 2 |
The exponent of
en in the first non-vanishing term after g(s) is
called the order of the iteration process defined by g. Since xn+1
- s = en+1 (and
g(s) = s) the above equation can now written as
| e
n+1 = en
g'(s) + |
1 |
e
n2 g''(s) + ... |
| 2 |
that is the error at (n+1)th iteration can be
written as a function of error at the previous iteration. In the case of
convergence en
is small for large n and hence the order is a measure for the speed of convergence.
For example if a scheme is second order, that is
then the number of significant digits are approximately
doubled in each step.
Order of Newton's
Method: Since for Newton's method
| g'(x) = |
f(x) * f ''(x) |
|
|
| (f '(x))2 |
|
at x = s, g' = 0 since f(s) = 0 and
at x = s, g''(s) need not be zero, hence
Newton-Raphson method is of order two. That is for each iteration the scheme
converges approximately to two significant digits.
Order of
Fixed Point Iteration method : Since the convergence of this
scheme depends on the choice of g(x) and the only information available about
g'(x) is |g'(x)| must be lessthan 1 in some interval which brackets
the root. Hence g'(x) at x = s may or may not be zero. That is the order
of fixed point iterative scheme is only one.
Order
of Secant method : The g(x) for secant method is
| xi+1= xi - |
(xi - xi-1) * f(xi) |
|
| f(xi) - f(xi-1) |
If xi+1 = s
+ ei+1, xi = s +
ei and xi-1 = s +
ei-1 then
| s + ei+1 = (s +
ei
) - |
(ei -
ei-1) * f(s + ei
) |
|
| f(s + ei ) - f(s +
ei-1) |
| ® e
i+1 = ei - |
(ei -
ei-1) * f(s + ei
) |
|
|
| f(s + ei ) - f(s +
ei-1) |
|
=
ei - |
(ei - e
i-1) * (eif '(s) + (
½)ei
2f ''(s) + . . .) |
|
|
| (eif '(s) + (
½)ei
2f ''(s) + . . .) - (ei-1
f '(s) + (½)e
i-12 f ''(s) + . . .) |
|
=
ei - |
(eif '(s) + (
½)ei
2f ''(s) + . . .) |
|
|
| (f '(s) + (½)(e
i+ ei-1
)f ''(s) + . . .) |
|
= ei - (
ei + |
1 |
ei2 |
f ''(s) |
+ . . .) (1 + |
1 |
(ei-1 +
ei) |
f ''(s) |
+ . . .)-1 |
| 2 |
f '(s) |
2 |
f '(s) |
| e
i+1 = |
1 |
ei
ei-1 |
f ''(s) |
+ 0(ei
2ei-1 + e
iei-12) |
| 2 |
f '(s) |
| ei+1
= C
eiei-1
where C = |
1 |
f ''(s) |
| 2 |
f '(s) |
and higher order terms are neglected.
since
ei
= Aepi-1
ei-1
= A-1/pe
i1/p
=>
ei+1
= Aei(A
-1/pei
1/p) =
A1-1/pe
i1+1/p
=> order of the scheme
p = 1 + 1/p
|
p = |
1 |
(1 ± 51/2)
= 1.618 |
| 2 |
and A =C
p/(p+1)
|
