Newton Iteration
Consider solution to the equation
z
2
=
2
on the real line.
The solution is clearly
z
= ±
.
Now consider approximating the solutions of this equation using Newton’s
Method.
We first set up the function to be set to zero,
f
(
x
)
=
x
2
!
2
=
0
.
Then our
Newton Function is the following.
g
(
x
)
=
x
!
x
2
!
2
2
x
The derivative of
g
(
x
)
is the following.
!
g
(
x
)
=
1
2
"
1
x
2
Clearly
!
g
(
x
)
=
0
if and only if
x
= ±
.
The graph of
!
g
(
x
)
is given below.
Based on the graph of
!
g
(
x
)
and using the Mean Value Theorem, we can analyze exactly
what will happen when we apply the Newton Function to any starting value.
If we have
any positive number
0
<
x
<
2
, then the Mean Value Theorem says that
g
(
x
)
>
2
.
For
x
>
2
, we have that
g
(
x
)
>
2
as well.
We also have that
g
(
x
)
!
2
<
1
2
x
!
2
.
So, for
x
>
2
we have that
g
n
(
x
)
!
<
1
2
"
#
$
%
&
'
n
x
!
2
.
This guarantees that
g
n
(
x
)
!
2
as
n
!"
whether
x
is less than
2
or greater than
2
as long as
x
is
positive.
If
x
is negative, then
g
n
(
x
)
! "
by a similar argument.
So, Newton’s
!
g
(
x
)
=
1
2
"
1
x
2
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View Full DocumentMethod converges for all
x
except
x
=
0
.
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 Spring '08
 BLOCK
 Calculus, newton function

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