Newton’s Method for Solving Equations
3.5
Intuition of Newton’s method:
Suppose we want to find the solution of
f x  0. Let
f  xn 1   f xn   f ' xn  xn 1  xn ,
where
xn 1 and xn are some values. If xn 1 is the solution, then
f  xn 1   f  xn   0  f  xn   f '  xn  xn 1  xn 
 xn 1  xn 
f xn 
f '  xn 
f x 
x

x
That is, n 1 can be obtained by evaluating the function
f ' x 
at
xn . Starting from n  0 , i.e., x0 , the following sequence can be
obtained,
x1  x0 
x2  x1 
f  x0 
,
f '  x0 
f  x1 
,
f '  x1 

xn 1  xn 
f  xn 
f '  xn 
Example 9:
Let
f x   x 2  2 . Use x0  1
as the initial guess and find
1
x1 , x2 , x3 .
[solutions:]
Since
f ' x   2 x ,
xn 1
x
f  xn 
xn2  2
1 

 xn  '
 xn 
 xn   n 
f  xn 
2 xn
2
x
n 

x
1
 n 
2
xn

n
0
1
2
1
2 
 xn 

2
xn 
xn
1
1.5
1.41666
2
xn
xn  2
2
1.33333
1.41176
xn
xn1  1  xn  2 
2
xn 
3
2.83333
2.82843
1.5
1.41666
1.41421
f xn 
-1
0.25
0.006945
◆
Several questions need to be answered:
1. Is there a unique solution to
f x  0.
2. Converge or not??
3. How fast is the convergence??
Theorem 7 (unique solution):
f ' is continuous on a, b and
(i) f a  and f b have different signs,
'
(ii) f  x   0 on
then
a, b ,
f x   0 has a unique solution, s, on a, b .
2
(b,f(b))
f(x)
f(b)
0
(a,f(a))
f(a)
a
b
x
[Intuition:]
'
Since f  x   0 on
on
a, b
a, b
a, b
'
or f x   0 on
or decreasing on
and f
'
a, b .
'
is continuous, then f  x   0
That is, f
a, b . The figure above is the plot for
being increasing.
Theorem 8 (convergence):
f , f ' , f '' are continuous on a, b . Suppose that
(i) f a  and f b have different signs,
(ii)
f ' x   m  0
(iii)
f ''  x   M
(iv)
(v)
is increasing on
M
4
m
x0  a, b .
3
f
Then,
(a) The sequence
xn 1  xn 
f  xn 
, n  0, 1, 2,
f '  xn 
f x   0 in a, b .
converge to the unique solution, s, of
(b) The convergence is quadratic. That is, each iterate is accurate to twice
as many decimal places as the one that precedes it.
Example 9 (continue):
Let
a, b  1,2. Then,
(i)
f 1  1, f 2  2 .
(ii)
f ' x   2 x  2  m
(iii)
f '' x   2  2  M
(iv)
(v)
M 2
 1 4
m 2
x0  1  1,2  a, b
Then, by theorem 8, the sequence based on Newton’s method will converge to the true
solution and the convergence is quadratic.
◆
Note:
Another criterion for convergence is
f x  f
f
'
''
x 
 x 2
 1.
As the inequality is true, the sequence based on Newton’s method
converges to the solution of
f x   0 .
4
Example 9 (continue):
For
x  1,2  a, b,
f x  f
f
'
''
x 
x 2

x
2

 2 2
1
1


 1,
2
4x 2
x2
the sequence based on Newton’s method converges to the true solution.
◆
Note:
It is not necessary to check the conditions of theorem 7 and theorem 8.
However, the following problem might happen:
(i)
The sequence based on Newton’s method could grow without
bound or oscillate.
(ii) The sequence based on Newton’s method could converge to a
solution, but there could be other solutions that would be
missed.
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