Page 292 - Numerical Analysis Using MATLAB and Excel

Chapter 7 Finite Differences and Interpolation

r0 =
0 0 0 -1
q1 =
8 -4
r1 =
0 0 2
q2 =
8
r2 =
0 12
q3 =
0
r3 =
8
Therefore,

p m() = 8m() 3 () + 12 m() 2 () + 2m() 1 () – 1 (7.43)
n
Taking the antidifference of (7.43) we obtain

–
1
------------------ –
--------------------- +
Δ p m() = 8m() 4 () 12 m() 3 () 2m() 2 () () 1 () (7.44)
m
------------------ +
n
4
2
3
m
= 2m() 4 () + 4m() 3 () + () 2 () – () 1 ()
m
and with (7.40)
∑ cubes = 2m() 4 () + 4m() 3 () + () 2 () – () 1 () n + 1 1
m
m
m =
)
(
)
(
)
)
(
)
)
(
–
2n + 1 nn – 1 n2 + 4n + 1 nn – 1 + ( n + 1 n – ( n + 1 ) (7.45)
( =
– 21() 4 () – 4 1() 3 () – 1 () 2 () + 1 () 1 ()
Since
)
)
(
1 () 4 () 11 1 12 13 = 0
(
)
–
–
( =
–
1 () 3 () 11 1 12 = 0 (7.46)
)
(
)
–
–
( =
1 () 2 () 11 1 = 0
)
–
( =
1 () 1 () = 1
relation (7.45) reduces to
∑ cubes = 2n + 1 nn – 1 n – 2 + 4n + 1 nn – 1 + ( n + 1 n – ( n + 1 + 1 (7.47)
)
(
)
)
(
)
)
(
)
)
(
(
7−14 Numerical Analysis Using MATLAB® and Excel®, Third Edition
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