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determine
r
(note that
q
is also unknown). It turnsoutthat the result, which isex
act for the specialcase considered here, works well as an iterativeformulawith any
polynomial.
Differentiating Eq. (a) with respect to
x
, we get
P
n
(
x
)
q
)
n
−
1
q
)
n
−
2
=
(
x
−
+
(
n
−
1)(
x
−
r
)(
x
−
P
n
(
x
)
1
n
−
1
=
+
x
−
r
x
−
q
Thus
P
n
(
x
)
P
n
(
x
)
1
n
−
1
=
+
(b)
x
−
r
x
−
q
which upondifferentiationyields
P
n
(
x
)
P
n
(
x
)
2
P
n
(
x
)
P
n
(
x
)
−
1
n
−
1
=−
−
(c)
(
x
−
r
)
2
(
x
−
q
)
2
It isconvenienttointroduce the notation
P
n
(
x
)
P
n
(
x
)
P
n
(
x
)
P
n
(
x
)
G
2
(
x
)
G
(
x
)
=
H
(
x
)
=
−
(4.14)
so that Eqs. (b) and (c) become
1
n
−
1
G
(
x
)
=
+
(4.15a)
x
−
r
x
−
q
−
1
n
1
H
(
x
)
=
+
(4.15b)
(
x
−
r
)
2
(
x
−
q
)
2
If we solve Eq. (4.15a)for
x
−
q
and substitute the result into Eq. (4.15b), weobtain a
quadraticequation for
x
−
r
.
The solution of thisequationis the
Laguerre's formula
n
x
−
r
=
(
n
(4.16)
1)
nH
(
x
)
G
2
(
x
)
G
(
x
)
±
−
−
The procedurefor finding a zeroofageneral polynomial by Laguerre'sformulais:
1.
Let
x
be aguess for the root of
P
n
(
x
)
=
0(any value will do).
P
n
(
x
) and
P
n
(
x
) using the procedureoutlinedinEqs. (4.10)
2. Evaluate
P
n
(
x
)
,
and (4.11).
3. Compute
G
(
x
) and
H
(
x
)fromEqs. (4.14).
4. Determine the improvedroot
r
fromEq. (4.16) choosing the sign that results in the
largermagnitude of thedenominator
(thiscanbe shown to improveconvergence).
←


<ε

−

<ε
ε
5.
Let
x
r
and repeat steps 25until
P
n
(
x
)
or
x
r
, where
is the error
tolerance.
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