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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
5.
Let x
r and repeat steps 2-5until
P n ( x )
or
x
r
, where
is the error
tolerance.
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