V.A
Newton’s Method in the Special Case
V.2
(i)
P
0
(
x
) has no root in [
a, b
].
(ii)
P
(
a
) and
P
(
b
) are nonzero and have opposite signs, that is,
P
(
a
)
P
(
b
)
<
0.
Then
P
(
x
) has a exactly one root
r
in (
a, b
) and there is an algorithm to approximate
r
to any desired
accuracy.
Proof:
First we prove the assertion that
P
(
x
) has a single root. Since
P
(
a
) and
P
(
b
) have opposite signs,
the IVT tells us that
P
(
x
) must vanish at some point
r
∈
(
a, b
); i.e. there is
at least
one root. On the other
hand, Rolle’s theorem says that two roots of
P
(
x
) are separated by a root of
P
0
(
x
). Since
P
0
(
x
) has no roots
in [
a, b
],
P
(
x
) cannot have two roots in [
a, b
], it has
at most
one root.
The algorithm to approximate the root,
r
, consists of the following steps.
Step 1
(Bounding derivatives):
Find constants
M
1
and
M
2
so that
0
< M
1
≤ 
P
0
(
x
)

for all
x
∈
[
a, b
]
and

P
00
(
x
)
 ≤
M
2
for all
x
∈
[
a, b
]
.
(Important Note:

P
0
(
x
)

has to be bounded
from below
, whereas

P
00
(
x
)

has to be bounded
from above
!)
Let
K
=
M
2
2
M
1
and choose a constant
h >
0 so that
Kh
≤
1
.
Step 2
(Shrinking the interval):
Find inside [
a, b
] an interval [
u, v
] which contains the root, and such that
v

u
≤
h
. This can be done as follows:
Method 1
(Bisection):
•
if
b

a
≤
h
set
u
=
a, v
=
b
. (The original interval is already narrow enough.)
•
otherwise evaluate
P
(
x
) at the midpoint of [
a, b
].
This divides [
a, b
] into two equal subintervals and from the sign of
P
(
a
),
P
(
b
) and
P
(midpoint) you
can tell which subinterval contains the root.