MTH3051 Introduction to Computational Mathematics - User Web ...

School of Mathematical Sciences
Monash University
Example 4.4 Newton-Raphson with f ′ (x) = 0
Find x such that 0 = f(x) = (x − 1) 2 . Clearly f(x) = 0 and f ′ (x) = 0 at x = 1 so x = 1
is our root. Let’s take the initial guess of x = 0.5. Here is what we get from a naive
application of the Newton-Raphson algorithm.
Newton-Raphson iterations x n+1 = x n − f n /f n, ′ f(x) = (x − 1) 2
Iteration n Old guess x n New guess x n+1 (x n+1 − 1) 2 ɛ n+1 /ɛ n
0 0.500000000000 2.500e-01
1 0.500000000000 0.750000000000 6.250e-02 5.000e-01
2 0.750000000000 0.875000000000 1.563e-02 5.000e-01
3 0.875000000000 0.937500000000 3.906e-03 5.000e-01
4 0.937500000000 0.968750000000 9.766e-04 5.000e-01
5 0.968750000000 0.984375000000 2.441e-04 5.000e-01
6 0.984375000000 0.992187500000 6.104e-05 5.000e-01
7 0.992187500000 0.996093750000 1.526e-05 5.000e-01
8 0.996093750000 0.998046875000 3.815e-06 5.000e-01
9 0.998046875000 0.999023437500 9.537e-07 5.000e-01
10 0.999023437500 0.999511718750 2.384e-07 5.000e-01
The iteration do converge, but not as fast as in the previous example. Notice that
ɛ n+1 /ɛ n remains constant. That is, each iteration reduces the error by a constant factor
(in this case 1/2). This is typical of Newton-Raphson iteration when x is a root of both
f(x) = 0 and f ′ (x) = 0.
4.3.1 Some notation.
◮ Simple root: When f(x) = 0 and f ′ (x) ≠ 0.
◮ Double root: When f(x) = 0, f ′ (x) = 0 and f ′′ (x) ≠ 0.
◮ Root of order m:
When f(x) = 0 and all derivatives up to f (m−1) (x) are
zero at x while f (m) (x) ≠ 0.
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