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CHE/COSC/MATH 4340-01 Numerical Analysis
Solution of Writing Assignment 06 Due Date: Thursday, 03/31/
Problem 29, pp. 150 We will show that divided differences are linear maps by induction. Let α and β be two constants. For n = 1, we have
(α f + β g)[x 0 , x 1 ] = (α f + β g)(x 1 ) − (α f + β g)(x 0 ) x 1 − x 0
(by definition of divided difference)
=
α f (x 1 ) + β g(x 1 ) − α f (x 0 ) − β g(x 0 ) x 1 − x 0 (by linearity of the combination)
= α f (x 1 ) − f (x 0 ) x 1 − x 0
- β g(x 1 ) − g(x 0 ) x 1 − x 0 = α f [x 0 , x 1 ] + β g[x 0 , x 1 ].
This shows that divided differences are linear maps for n = 1. Next, assume that it is true for n − 1, i.e.,
(α f + β g)[xi, xi+ 1 , · · · , xi+n− 1 ] = α f [xi, xi+ 1 , · · · , xi+n− 1 ] + β g[xi, xi+ 1 , · · · , xi+n− 1 ]. (0.2)
Using this assumption, we will show it is true for n. To this end, we write
(α f + β g)[x 0 , x 1 , · · · , xn]
= (α f + β g)[x 1 , x 2 , · · · , xn] − (α f + β g)[x 0 , x 1 , · · · , xn− 1 ] xn − x 0 = α f [x 1 , x 2 , · · · , xn] + β g[x 1 , x 2 , · · · , xn] − α f [x 0 , x 1 , · · · , xn− 1 ] − β g[x 0 , x 1 , · · · , xn− 1 ] xn − x 0
where we have used the assumption for n − 1 on the last line. Continuing on by rearranging terms, we see that
(α f + β g)[x 0 , x 1 , · · · , xn]
= α f [x 1 , x 2 , · · · , xn] − f [x 0 , x 1 , · · · , xn− 1 ] xn − x 0
- β g[x 1 , x 2 , · · · , xn] − g[x 0 , x 1 , · · · , xn− 1 ] xn − x 0 = α f [x 0 , x 1 , · · · , xn] + β g[x 0 , x 1 , · · · , xn],
which confirms that divided differences are linear maps for n.
Problem 35 pp. 150 Newton form for interpolating polynomial p 2 (x) is
p 2 (x) = f (x 0 ) + f [x 0 , x 1 ](x − x 0 ) + f [x 0 , x 1 , x 2 ](x − x 0 )(x − x 1 ). (0.5)
Using definition of divided differences, we write
p 2 (x) = f (x 0 ) + f [x 0 , x 1 ](x − x 0 ) + f [x 0 , x 1 , x 2 ](x − x 0 )(x − x 1 )
= f (x 0 ) + ( f (x 1 ) − f (x 0 )) x − x 0 x 1 − x 0
- ( f [x 1 , x 2 ] − f [x 0 , x 1 ]) (x − x 0 )(x − x 1 ) x 2 − x 0 = f (x 0 ) + ( f (x 1 ) − f (x 0 )) x − x 0 x 1 − x 0
( (^) f (x 2 ) − f (x 1 ) x 2 − x 1
f (x 1 ) − f (x 0 ) x 1 − x 0
)(x − x 0 )(x − x 1 ) x 2 − x 0 = f (x 0 )m 0 (x) + m 1 (x) f (x 1 ) + m 2 (x) f (x 2 ),
where
m 0 (x) = 1 − x − x 0 x 1 − x 0
(x − x 0 )(x − x 1 ) (x 1 − x 0 )(x 2 − x 0 ) m 1 (x) = x − x 0 x 1 − x 0
(x − x 0 )(x − x 1 ) (x 2 − x 1 )(x 2 − x 0 )
(x − x 0 )(x − x 1 ) (x 1 − x 0 )(x 2 − x 0 ) m 2 (x) = (x − x 0 )(x − x 1 ) (x 2 − x 0 )(x 2 − x 1 )
If we can show that mi(x) = li(x), then we have confirmed that Newton form is the same as Lagrange form. Obviously m 2 (x) = l 2 (x). Furthermore,
m 0 (x) = 1 − x − x 0 x 1 − x 0
(x − x 0 )(x − x 1 ) (x 1 − x 0 )(x 2 − x 0 ) = x 1 − x 0 − x + x 0 x 1 − x 0
(x − x 0 )(x − x 1 ) (x 0 − x 1 )(x 0 − x 2 ) = x − x 1 x 0 − x 1
(x − x 0 )(x − x 1 ) (x 0 − x 1 )(x 0 − x 2 ) = x − x 1 x 0 − x 1
x − x 0 x 0 − x 2
x − x 1 x 0 − x 1
(x 0 −^ x 2 +^ x^ −^ x 0 x 0 − x 2
(x − x 1 )(x − x 2 ) (x 0 − x 1 )(x 0 − x 2 ) = l 0 (x).
Similar derivation shows that m 1 (x) = l 1 (x).
Problem 10, pp. 176
a. Suppose f (x) is interpolated by a polynomial of degree 2, i.e., p 2 (x) = c 0 + c 1 (x − x 0 ) + c 2 (x − x 0 )(x − x 1 ) with
f (x 0 ) = p 2 (x 0 ) = c 0 f (x 1 ) = p 2 (x 1 ) = c 0 + c 1 (x 1 − x 0 ) = c 0 + hc 1 f (x 2 ) = p 2 (x 2 ) = c 0 + c 1 (x 2 − x 0 ) + c 2 (x 2 − x 0 )(x 2 − x 1 ) = c 0 + c 1 (x 2 − x 1 + x 1 − x 0 ) + c 2 (x 2 − x 1 + x 1 − x 0 )(x 2 − x 1 ) = c 0 + c 1 ( 1 + α)h + c 2 ( 1 + α)αh^2.
These equations give
c 0 = f (x 0 )
c 1 = f (x 1 ) − f (x 0 ) h c 2 = f (x 2 ) − f (x 0 ) − ( f (x 1 ) − f (x 0 ))( 1 + α) ( 1 + α)αh^2
h^2
[ (^) f (x 2 ) ( 1 + α)α
f (x 1 ) α
f (x 0 ) α
( 1 + α)
)]
h^2
[ (^) f (x 0 ) 1 + α
f (x 1 ) α
f (x 2 ) ( 1 + α)α
]
Now we set f ′′(x) ≈ p′′ 2 (x) = 2 c 2.
b. Assume f ′′(x) ≈ A f (x 0 ) + B f (x 1 ) +C f (x 2 ), and assume that this approximation is exact for poly- nomial of degree 2. Thus it is exact for 1, (x − x 1 ) and (x − x 1 )^2. To this end we get
A + B +C = 0 −hA + 0 + αhC = 0 h^2 A + 0 + α^2 h^2 C = 2.
From the second equation we get A = αC, substitute to third equation to get h^2 (α + α^2 )C = 2, from which we get C =
h^2 α( 1 + α)
, and thus A =
h^2 ( 1 + α)
. Substitute the resulting A and C to
the first equation to give B =
h^2 α
