切比雪夫插值節(jié)點(diǎn)

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1、切比雪夫插值節(jié)點(diǎn)帶導(dǎo)數(shù)條件的插值函數(shù)分段插值函數(shù)二元函數(shù)插值簡(jiǎn)介《數(shù)值分析》15取插值結(jié)點(diǎn):a≤x0＜x1＜······＜xn≤b滿(mǎn)足Ln(xk)=f(xk)的n次多項(xiàng)式插值余項(xiàng)其中,選取:x0,x1,······,xn,使結(jié)論:切比雪夫多項(xiàng)式Tn+1(x)的全部零點(diǎn)。拉格朗日插值余項(xiàng)2/18n+1階切比雪夫多項(xiàng)式:Tn+1=cos(n+1)?cos?=x代入得Tn+1(x)=cos((n+1)arccosx)即(k=0,1,···,n)取f(x)∈C[–1,1],令x=cos?,則有[–1,1]??[0,?]將g(?)=f(cos?)展開(kāi)成余弦級(jí)數(shù)——切比雪夫結(jié)點(diǎn)3/18例1

2、.函數(shù)取等距插值結(jié)點(diǎn):-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(x)?4/18-4.9491-4.5482-3.7787-2.7032-1.40870.00001.40872.70323.77874.54824.9491在[-5,5]區(qū)間上,取11個(gè)切比雪夫結(jié)點(diǎn)(k=10,9,8,···,1,0)?11(x)=(x–x0)(x–x1)(x–x2)······(x–x10)5/18?11(x)?插值函數(shù)L10(x)取切比雪夫結(jié)點(diǎn)插值

3、插值函數(shù)L10(x)取等距結(jié)點(diǎn)插值6/18已知節(jié)點(diǎn)x0和x1處的函數(shù)值及導(dǎo)數(shù)值求三次插值函數(shù)H(x)=a0+a1x+a2x2+a3x3滿(mǎn)足插值條件(j=0,1)三次Hermite插值問(wèn)題xx0x1H(x)y0y1H’(x)m0m17/18例2.已知插值條件:求3次插值函數(shù).解:設(shè)得a0=0,a1=0,列出方程組求解,得a2=3,a3=–2所以,有H(x)=3x2–2x3=(3–2x)x2x01H(x)01H’(x)008/18利用基函數(shù)表示Hermite插值x0x110000100xx0x100100001x9/18兩點(diǎn)Hermite插值的誤差估計(jì)式證明:由插值條件知R(x0)

4、=R’(x0)=0,R(x1)=R’(x1)=0構(gòu)造輔助函數(shù)利用f(x)–H(x)=C(x)(x–x0)2(x–x1)2取x異于x0和x1,設(shè)10/18反復(fù)應(yīng)用Roll定理,得F(4)(t)有一個(gè)零點(diǎn)設(shè)為ξ??顯然,F(t)有三個(gè)零點(diǎn)x0,x,x1,由Roll定理知,存在F’(t)的兩個(gè)零點(diǎn)t0,t1滿(mǎn)足x0

5、線(xiàn)性插值函數(shù)12/18分段三次Hermite插值取a≤x0

6、,y)=z1(1–u)(1–v)+z2u(1–v)+z3uv+z4(1–u)v16/18[u,v]=meshgrid(0:0.1:1)；L1=(1-u).*(1-v);surf(u,v,L1)figureL2=u.*(1-v);surf(u,v,L2)figureL3=u.*v;surf(u,v,Lu3)figureL4=(1-u).*v;surf(u,v,L4)x=asinφcosθy=asinφsinθz=bcosφ三角形區(qū)域?線(xiàn)性插值插值條件:z1=P(x1,y1)z2=P(x2,y2)z3=P(x3,y3)(x1,y1)(x3,y3)(x2,y2)拉格朗日方法P(x,y

7、)=l1(x,y)z1+l2(x,y)z2+l3(x,y)z3P(x,y)=ax+by+c(x，y)(x1，y1)(x2，y2)(x3，y3)l1(x,y)100l2(x,y)010l3(x,y)00117/18?l1(x,y)的圖形是空間三角形分片線(xiàn)性插值18/18P(x,y)=l1(x,y)z1+l2(x,y)z2+l3(x,y)z3圖形是空間三角形