6.2 牛顿插值多项式

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,第 四 章 插 值 法,第一节拉格朗日插值,第二节牛顿插值多项式,第二节牛顿插值多项式,本节主要内容,:,一,.,均差及其性质,二,.Newton,均差插值公式,三,.,小结,一,.,均差及其性质,对于,n,+1,个节点,的插值,问题,将,n,次插值多项式写成如下形式,为待定系数,.,由插值条件,多项式称为,牛顿,(Newton),插值多项式,.,形如上式的插值,当,时,当,时,当,时,依次递推可得,定义,1,记,称,为,关于,x,i,的,零阶均差,.,称,为,关于,x,i,x,i,+1,的,一阶均差,.,称为,二阶均差,.,一般地,k,阶均差,为,均差有如下基本性质：,定理,1,:,(1),均差与函数值的关系为,(2),均差与节点的排列顺序有关,即,(4),若函数,在,上存在,n,阶导数,且节点,则,使得,一阶均差,二阶均差,三阶均差,n,阶均差,计算均差可按下表逐行进行,二,.,Newton,均差插值公式,定理,2,设,是满足插值条件,的插值多项式,而且余项,(2),(1),则,证 明,将,x,看成,上一点,可得,依次将后一式代入前一式,得,上式中,是,(1),式,就是,(2),式,.,由,(2),式有,因此由,(1),定义的,是满足插值条件,的插值多项式,.,证毕,.,例,1,已知,的离散数据如下表：,0.00,0.20,0.30,0.50,0.00000,0.20134,0.30452,0.52110,用,Newton,插值多项式,计算,估计误差,.,的近似值并,解,均差计算的结果如下表,一阶均差,二阶均差,三阶均差,0.00,0.20,0.30,0.50,0.00000,0.20134,0.30452,0.52110,1.0067,1.0318,1.0829,0.08367,0.17033,0.17332,则,Newton,插值多项式为,由此算出,因,余项为,估计误差,若增加一个节点,只需再增加如下一行：,一阶均差,二阶均差,三阶均差,0.00,0.20,0.30,0.50,0.60,0.00000,0.20134,0.30452,0.52110,0.63665,1.0067,1.0318,1.0829,1.1555,0.08367,0.17033,0.24200,0.17332,0.17918,四阶均差,0.00976,插值多项式为,三,.,小 结,1,.,Newton,均差插值多项式,2.,插值余项,