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Hermite 插值
不少實(shí)際問題不但要求在節(jié)點(diǎn)上函數(shù)值相等,而且還要求它的導(dǎo)數(shù)值相等,甚至要求高階導(dǎo)數(shù)值也相等。滿足這種要求的插值多項(xiàng)式就是 Hermite 插值多項(xiàng)式。
下面只討論函數(shù)值與導(dǎo)數(shù)值個數(shù)相等的情況。設(shè)在節(jié)點(diǎn) a ≤ x 0 < x 1 < ? < x n ≤ b a \leq x_0 < x_1 < \cdots < x_n \leq b a≤x0?<x1?<?<xn?≤b 上, y j = f ( x j ) , m j = f ′ ( x j ) ( j = 0 , 1 , ? , n ) y_j = f(x_j), m_j = f'(x_j) (j=0,1,\cdots,n) yj?=f(xj?),mj?=f′(xj?)(j=0,1,?,n),要求插值多項(xiàng)式 H ( x ) H(x) H(x),滿足條件
H ( x j ) = y j , H ′ ( x j ) = m j ( j = 0 , 1 , ? , n ) . H(x_j) = y_j,\quad H'(x_j) = m_j \quad(j = 0,1,\cdots,n). H(xj?)=yj?,H′(xj?)=mj?(j=0,1,?,n).
這里給出的 2 n + 2 2n+2 2n+2 個條件,可唯一確定一個次數(shù)不超過 2 n + 1 2n+1 2n+1 的多項(xiàng)式
H 2 n + 1 ( x ) = H ( x ) , H_{2n+1}(x) = H(x), H2n+1?(x)=H(x),
其形式為
H 2 n + 1 ( x ) = a 0 + a 1 x + ? + a 2 n + 1 x 2 n + 1 . H_{2n+1}(x) = a_0 + a_1x + \cdots + a_{2n+1}x^{2n+1}. H2n+1?(x)=a0?+a1?x+?+a2n+1?x2n+1.