Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...

4.2. NEWTON’S METHOD 51
4.2.2 Problems
As mentioned above, convergence is dependent on f(x), and the initial estimate x 0 . A number of
conceivable problems might come up. We illustrate them here.
Example 4.4. Consider Newton’s method applied to the function f(x) = x j with j > 1, and with
initial estimate x 0 ≠ 0.
Note that f(x) = 0 has the single root x = 0. Now note that
x k+1 = x k −
xj k
jx j−1
k
=
(
1 − 1 )
x k .
j
Since the equation has the single root x = 0, we find that x k is converging to the root. However,
it is converging at a rate slower than we expect from Newton’s method: at each step we have a
constant decrease of 1−(1/j) , which is a larger number (and thus worse decrease) when j is larger.
Example 4.5. Consider Newton’s method applied to the function f(x) = ln x
x
, with initial estimate
x 0 = 3.
Note that f(x) is continuous on R + . It has a single root at x = 1. Our initial guess is not too far
from this root. However, consider the derivative:
f ′ (x) = x 1 x − ln x
x 2
= 1 − ln x
x 2
If x > e 1 , then 1 − ln x < 0, and so f ′ (x) < 0. However, for x > 1, we know f(x) > 0. Thus taking
x k+1 = x k − f(x k)
f ′ (x k ) > x k.
The estimates will “run away” from the root x = 1.
Example 4.6. Consider Newton’s method applied to the function f(x) = sin (x) for the initial
estimate x 0 ≠ 0, where x 0 has the odious property 2x 0 = tan x 0 .
You should verify that there are an infinite number of such x 0 . Consider the identity of x 1 :
Now consider x 2 :
x 1 = x 0 − f(x 0)
f ′ (x 0 ) = x 0 − sin(x 0)
cos(x 0 ) = x 0 − tan x 0 = x 0 − 2x 0 = −x 0 .
x 2 = x 1 − f(x 1)
f ′ (x 1 ) = −x 0 − sin(−x 0)
cos(−x 0 ) − x 0 + sin(x 0)
cos(x 0 ) = −x 0 + tan x 0 = −x 0 + 2x 0 = x 0 .
Thus Newton’s method “cycles” between the two values x 0 , −x 0 .
Of course, Newton’s method may find some iterate x k for which f ′ (x k ) = 0, in which case, there
is no well-defined x k+1 .
4.2.3 Convergence
When Newton’s Method converges, it actually displays quadratic convergence. That is, if e k = x k −r,
where r is the root that the x k are converging to, that |e k+1 | ≤ C |e k | 2 . If, for example, it were
the case that C = 1, then we would double the accuracy of our root estimate with each iterate.
That is, if e 0 were 0.001, we would expect e 1 to be on the order of 0.000001. The following theorem
formalizes our claim: