切比雪夫插值节点VIP免费

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|)(|max1xnbxa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(10xfxLxfn11(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(5kxk(k=10,9,8,···,1,0)11(x)=(x–x0)(x–x1)(x–x2)······(x–x10)5/1811(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)(yxf11)(yxf00)(mxf11)(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)()()()()(11001100xmxmxyxyxH20110100))(21()(xxxxxxxxx20100111))(21()(xxxxxxxxx.))(()(201100xxxxxxx201011))(()(xxxxxxxx0x110000100)(0x)(1x)(0x)(1xx)(1x)(0xx0x100100001)(1x)(0xx9/18210)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)())(()()()(xtxtxCtHtftF0)!4)(()()()4()4(xCfF!4)()()4(fxC210)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