Unformatted text preview: ives and the following
Lemma shows that this is the case.
8 Lemma 5.2.1. Let f (t) 2 C k a b] and let t0 t1 : : : tk be k + 1 distinct points on a b],
then f (k)( )
f t0 t1 : : : tk ] = k!
2 (a b):
Proof. Consider the function
g(t) = f (t) ; Pk (t): (5.2.10) Since f (tj ) = Pk (tj ), j = 0 1 : : : k, the function g(t) has k + 1 distinct zeros in a b].
Thus, according to Rolle's theorem, g0(t) vanishes at k distinct points on (a b) (Figure
5.2.1). Similarly, g00 (t) vanishes at k ; 1 points on (a b) and, continuing in this manner,
g(k) will vanish at one point on (a b). Let us call this point . Thus, g(k)( ) = f (k)( ) ; Pk(k)( ) = 0
According to (5.2.8a) or (5.2.9), the k th derivative of Pk (t) is Pk(k)(t) = k!f t0 t1 : : : tk ]
Combining the above two results establishes the result.
With Lemma 5.2.1, we can obtain formulas for interpolation errors. Theorem 5.2.2. Let f (t) 2 C k+1 a b] and let t0 t1 : : : tk be k + 1 distinct points on
a b], then there exists a point = (t) 2 (a b) such that
Ek (t) = f (t) ; Pk (t) = f(k + 1)!) (t ; tj ):
j =0 (5.2.11) Proof. Construct a polynomial Pk+1(t) that interpolates f (t) at t0 t1 : : : tk and the
additional point . According to (5.2.8a), Pk+1(t) = Pk (t) + f t0 t1 : : : tk ] k
j =0 (t ; tj ): Since Pk+1( ) = f ( ), we have f ( ) = Pk ( ) + f t0 t1 : : : tk ]
j =0 ( ; tj ): g(t) t
t0 t1 tk g’(t) t g’’(t) t (k) g (t)
ξ Figure 5.2.1: Zeros and extrema of the function g(t) = f (t) ; Pk (t).
Thus, Ek ( ) = f ( ) ; Pk ( ) = f t0 t1 : : : tk ]
j =0 ( ; tj ): Using (5.2.10) k
f (k+1)( ) Y( ; t ):
Ek ( ) = (k + 1)!
Since is arbitrary we have established the result. The points in the divided-di erence polynomial have to be distinct but do not have
to be uniform or ordered. However, the computation simpli es greatly when the interpolation points are ordered and uniformly spaced. This will be our main interest when
developing formulas for ODEs, so we'll restrict ti = t0 + ih i = 0 1 : : : k: (5.2.12) With t...
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