Homework Assignment #3

Homework Assignment #3
3-1 Use appropriate Lagrange interpolating polynomials of degrees one, two and three to approximate
f(0.25):
f(0.1) = 0.62049958, f(0.2) = −0.28398668, f(0.3) = 0.00660095, f(0.4) = 0.24842440.
3-2 Use Neville’s method to approximate f(0.25) for above data.
3-3 Neville’s method is used to approximate f(0.4), giving the following table.
x 0 = 0 P 0 = 1
x 1 = 0.25 P 1 = 2 P 0,1 = 2.6
x 2 = 0.5 P 2 P 1,2 P 0,1,2
x 3 = 0.75 P 3 = 8 P 2,3 = 2.4 P 1,2,3 = 2.96 P 0123 = 3.016
3-4 Suppose x j = j, for j = 0, 1, 2, 3 and it is known that P 0,1 (x) = x + 1, P 1,2 (x) = 3x − 1, P 1,2,3 (1.5) = 4.
Find P 0,1,2,3 (1.5).
3-5 For a function f, the forward divided differences are given by
x 0 = 0.0 f[x 0 ]
f[x 0 , x 1 ]
x 1 = 0.4 f[x 1 ] f[x 0 , x 1 , x 2 ] = 50 7
f[x 1 , x 2 ] = 10
x 2 = 0.7 f[x 2 ] = 6
Determine the missing entries in the table.
3-6 Given P n (x) = f[x 0 ] + f[x 0 , x 1 ](x − x 0 ) + a 2 (x − x 0 )(x − x 1 ) + a 3 (x − x 0 )(x − x 1 )(x − x 2 ) + ... + a n (x −
x 0 )(x − x 1 )...(x − x n−1 ), use P n (x 2 ) to show that a 2 = f[x 0 , x 1 , x 2 ].
3-7 Show that f[x 0 , x 1 , ..., x n , x] = f (n+1) (ξ(x))
for some ξ(x). [Hint: Considering the interpolation polynomial
(n+1)!
of degree n + 1 on x 0 , x 1 , ..., x n , x, we have f(x) = P n+1 (x) = P n (x) + f[x 0 , x 1 , ..., x n , x](x − x 0 )(x −
x 1 )...(x − x n ).]
3-8 Construct the Hermite interpolation polynomial f(x) using the following data:
(1) f(0) = 1, f(1) = 0, f ′ (0) = 0, f ′ (1) = 0.
(2) f(0) = 0, f(1) = 1, f ′ (0) = 0, f ′ (1) = 0.
(3) f(0) = 0, f(1) = 0, f ′ (0) = 1, f ′ (1) = 0.
(4) f(0) = 0, f(1) = 0, f ′ (0) = 0, f ′ (1) = 1.
3-9 Let z 0 = x 0 , z 1 = x 0 , z 2 = x 1 , z 3 = x 1 . From the following divided-difference table
z 0 = x 0 f[z 0 ] = f(x 0 )
f[z 0 , z 1 ] = f ′ (x 0 )
z 1 = x 0 f[z 1 ] = f(x 0 ) f[z 0 , z 1 , z 2 ]
f[z 1 , z 2 ] f[z 0 , z 1 , z 2 , z 3 ]
z 2 = x 1 f[z 2 ] = f(x 1 ) f[z 1 , z 2 , z 3 ]
f[z 2 , z 3 ] = f ′ (x 1 )
z 3 = x 1 f[z 3 ] = f(x 1 )
Show that the cubic Hermite polynomial H 3 (x) can also be written as
f[z 0 ] + f[z 0 , z 1 ](x − x 0 ) + f[z 0 , z 1 , z 2 ](x − x 0 ) 2 + f[z 0 , z 1 , z 2 , z 3 ](x − x 0 ) 2 (x − x 1 ).
1

Solution of Homework Assignment #3
3-1
3-2
3-3
n x 0 , x 1 ,..., x n P n (0.25)
1 0.2, 0.3 −0.13869287
2 0.2, 0.3, 0.4 −0.13259734
3 0.2, 0.3, 0.4, 0.1 −0.21033722
x 0 = 0.1 P 0 = 0.62049958
x 1 = 0.2 P 1 = −0.28398668 P 0,1 = −0.73622981
x 2 = 0.3 P 2 = 0.00660095 P 1,2 = −0.13869287 P 0,1,2 = −0.28807711
x 3 = 0.4 P 3 = 0.24842440 P 2,3 = −0.11431078 P 1,2,3 = −0.13259734 P 0123 = −0.21033722
x 0 = 0 P 0 = 1
x 1 = 0.25 P 1 = 2 P 0,1 = 2.6
x 2 = 0.5 P 2 = 4 P 1,2 = 3.2 P 0,1,2 = 3.08
x 3 = 0.75 P 3 = 8 P 2,3 = 2.4 P 1,2,3 = 2.96 P 0123 = 3.016
3-4
P 0,1,2,3 (1.5) = P 1,2,3(1.5)(x − x 0 ) − P 0,1,2 (1.5)(x − x 3 )
and
x 3 − x 0
P 0,1,2 (1.5) = P 1,2(1.5)(x − x 0 ) − P 0,1 (1.5)(x − x 2 ) 3.5 × (1.5 − 0) − 2.5 × (1.5 − 2)
=
x 2 − x 0 2 − 0
Then P 0,1,2,3 (1.5) = 3.625.
= 3.25
3-5
x 0 = 0.0 f[x 0 ] = 1
f[x 0 , x 1 ] = 5
x 1 = 0.4 f[x 1 ] = 3 f[x 0 , x 1 , x 2 ] = 50 7
f[x 1 , x 2 ] = 10
x 2 = 0.7 f[x 2 ] = 6
3-8 (1) 2x 3 − 3x 2 + 1. (2) −2x 3 + 3x 2 . (3) x 3 − 2x 2 + x. (4) x 3 − x 2 .
3-9 H(x) = f[z 0 ] + f[z 0 , z 1 ](x − x 0 ) + f[z 0 , z 1 , z 2 ](x − x 0 ) 2 + f[z 0 , z 1 , z 2 , z 3 ](x − x 0 ) 2 (x − x 1 ),
where f[z 0 ] = f(x 0 ), f[z 0 , z 1 ] = f ′ (x 0 ), f[z 0 , z 1 , z 2 ] = f(x 1) − f(x 0 ) − f ′ (x 0 )(x 1 − x 0 )
x 1 − x 0
,
f[z 0 , z 1 , z 2 , z 3 ] = f ′ (x 1 )(x 1 − x 0 ) − 2f(x 1 ) + 2f(x 0 ) + f ′ (x 0 )(x 1 − x 0 )
(x 1 − x 0 ) 3 . Then H(x 0 ) = f[z 0 ] = f(x 0 ),
H(x 1 ) = f(x 1 ), H ′ (x 0 ) = f ′ (x 0 ) and H ′ (x 1 ) = f ′ (x 1 ). Thus H satisfies the requirements of the cubic
Hermite polynomial H 3 , and the uniqueness of H 3 implies H 3 = H.
2