切比雪夫插值节点带导数条件的插值函数分段插值函数二元函数插值简介《数值分析》15取插值结点:a≤x0<x1<······<xn≤b满足Ln(xk)=f(xk)的n次多项式插值余项)()!1()()()()(1)1(xnfxLxfxRnnnnn)())(()(101nnxxxxxxx其中,选取:x0,x1,······,xn,使min|)(|max1xnbxa结论:切比雪夫多项式Tn+1(x)的全部零点。拉格朗日插值余项2/18n+1阶切比雪夫多项式:Tn+1=cos(n+1)cos=x代入得Tn+1(x)=cos((n+1)arccosx)))1(2)12(cos(nkxk即2)12(arccos)1(kxn(k=0,1,···,n)取f(x)∈C[–1,1],令x=cos,则有[–1,1][0,]将g()=f(cos)展开成余弦级数10cos21)(nnnaag——切比雪夫结点3/18211)(xxf例1.函数取等距插值结点:-5,-4,-3,-2,-1,0,1,2,3,4,5x[-5,5]∈11(x)=(x+5)(x+4)(x+3)(x+2)(x+1)x(x-1)(x-2)(x-3)(x-4)(x-5))(!11)()()(11)11(10xfxLxfn11(x)4/18-4.9491-4.5482-3.7787-2.7032-1.40870.00001.40872.70323.77874.54824.9491在[-5,5]区间上,取11个切比雪夫结点)22)12(cos(5kxk(k=10,9,8,···,1,0)11(x)=(x–x0)(x–x1)(x–x2)······(x–x10)5/1811(x)-5-4-3-2-1012345-0.200.20.40.60.811.2-5-4-3-2-1012345-0.500.511.52插值函数L10(x)取切比雪夫结点插值插值函数L10(x)取等距结点插值6/18已知节点x0和x1处的函数值及导数值00)(yxf11)(yxf00)(mxf11)(mxf求三次插值函数H(x)=a0+a1x+a2x2+a3x3满足插值条件jjyxH)(jjmxH)((j=0,1)三次Hermite插值问题xx0x1H(x)y0y1H’(x)m0m17/18例2.已知插值条件:求3次插值函数.332210)(xaxaxaaxH解:设得a0=0,a1=0,列出方程组03213232aaaa求解,得a2=3,a3=–2所以,有H(x)=3x2–2x3=(3–2x)x2-0.4-0.200.20.40.60.811.21.41.6-0.6-0.4-0.200.20.40.60.81x01H(x)01H’(x)008/18利用基函数表示Hermite插值)()()()()(11001100xmxmxyxyxH20110100))(21()(xxxxxxxxx20100111))(21()(xxxxxxxxx.))(()(201100xxxxxxx201011))(()(xxxxxxxx0x110000100)(0x)(1x)(0x)(1xx)(1x)(0xx0x100100001)(1x)(0xx9/18210)4(3)])([(!4)()()()(xxxxfxHxfxR两点Hermite插值的误差估计式证明:由插值条件知R(x0)=R’(x0)=0,R(x1)=R’(x1)=0构造辅助函数2120)())(()()()(xtxtxCtHtftF利用f(x)–H(x)=C(x)(x–x0)2(x–x1)22120)())(()(xxxxxCxR取x异于x0和x1,设10/18反复应用Roll定理,得F(4)(t)有一个零点设为ξ2120)())(()()()(xtxtxCtHtftF0)!4)(()()()4()4(xCfF!4)()()4(fxC210)4(2120)])([(!4)()())(()(xxxxfxxxxxCxR显然,F(t)有三个零点x0,x,x1,由Roll定理知,存在F’(t)的两个零点t0,t1满足x0