Page 285 - Numerical Analysis Using MATLAB and Excel

Factorial Polynomials

a = Δp 0() (7.19)
n
1
Differencing again we obtain

2
⋅
⋅
(
)
–
Δ p x() = 21a + 32a x() 1 () + … + nn 1 a x() ( n – 2 ) (7.20)
n
3
n
2
and for x = , 0
2
2
Δ p 0() Δ p 0()
n
n
a = -------------------- = -------------------- (7.21)
2 21 2!
⋅
In general,
j
Δ p 0()
n
,
,,,
a = ------------------- for j = 0 1 2 … n (7.22)
j
j!
Factorial polynomials provide an easier method of constructing a difference table. With this
method we perform the following steps:
1. We divide p x() in (7.16) by to obtain a quotient q x() and a remainder which turns out
r
x
0
n
0
to be the constant term a 0 . Then, we express (7.16) as
p x() = r + xq x() (7.23)
n
0
0
2. We divide q x() in (7.23) by x –( 1 ) to obtain a quotient q x() and a remainder r 1 which
1
0
turns out to be the constant term a 1 . Then,
)
–
q x() = r + ( x1 q x() (7.24)
1
0
1
By substitution of (7.24) into (7.23), and using the form of relation (7.16), we obtain
[
)
)
]
p x() = r + xr + ( x1 q x() = r + r x() 1 () xx – 1 q x() (7.25)
( +
–
0
1
1
n
1
0
1
3. We divide q x() in (7.25) by x –( 2 ) to obtain a quotient q x() and a remainder r 2 which
1
2
turns out to be the constant term a 2 , and thus
)
q x() = r + ( x2 q x() (7.26)
–
1
2
2
By substitution of (7.26) into (7.25), we obtain
)
)
[
p x() = r + r x() 1 () xx – 1 r + ( x2 q x() ] (7.27)
–
( +
2
2
1
n
0
)
)
(
( +
= r + r x() 1 () + r x() 2 () xx – 1 x – 2 q x()
1
2
0
2
Numerical Analysis Using MATLAB® and Excel®, Third Edition 7−7
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