Hermite插值是为了保证,插值函数不仅经过所给节点,而且要保证在该点的导数也相等。
H2n+1(x) = Σf(xj)Hn,j(x) + Σf'(xj)H^n,j(x)
Hn,j(x) = [1 - 2(x-xj)L'n,j(xj)]Ln,j2(x)
H^n,j(x) = (x - xj)Ln,j2(x)
Ln,j = [(x-x0)(x-x1)...(x-x(j-1))(x-x(j+1))(x - xn)] / [(xj - x0)(xj - x1)...(x-x(j-1))(x - x(j+1))(x - xn)];
对于小的n值,产生Hermite逼近的一个替代方法是利用差商型Newton插值公式:
Pn(x) = f(x0) + Σ(k = 1, 2n+1)f[z0,z1,...,zk]*(x-z0)(x-z1)...(x-zk-1)
Z2k+1 = Z2k = Xk;