西电计算方法第三次作业

1、班级：021113 学号：02111283 姓名：范亚男 第四章习题5.解:x01245y01646880先求差商表：kf()f(,)f()f()f()000167-2.5111630-32246213488-88450于是x(x-1)(x-2)(x-4)9.解：x0.1250.2500.3750.5000.6250.750f(x)0.796180.773340.743710.704130.656320.60228先求差分表h=0.125xff0.1250.79618-0.02284-0.00679-0.003160.2500.77334-0.02963-0.00995-0.001720.37

2、50.74371-0.03958-0.00823-0.002000.5000.70413-0.04781-0.006230.6250.65632-0.054040.7500.60228故三次牛顿差分插值公式当0.125+0.125t=0.1581时，t=0.2648代入上式得f(0.1581)=0.790615当0.125t+0.125=0.636时，t=4.088代入上式得f(0.636)=0.646062上机作业例1：已知函数表：xi 0.56160 0.56280 0.56401 0.56521 yi 0.82741 0.82659 0.82577 0.82495用三次拉格朗日插值多项式

3、求x=0.5635时的函数近似值C语言程序设计：# include<stdio.h>float Lagrange(float x,float y,float xx,int n)int i,j;float *a,yy=0;a=new floatn;for(i=0;i<=n-1;i+)ai=yi;for(j=0;j<=n-1;j+)if(j!=i) ai*=(xx-xj)/(xi-xj);yy+=ai;delete a;return yy;void main()float x4=0.56160,0.56280,0.56401,0.56521;float y4=0.82741

4、,0.82659,0.82577,0.82495;float xx=0.5635,yy;yy=Lagrange(x,y,xx,4);printf("x=%f,y=%fn",xx,yy);运行结果：x=0.563500y=0.826116 例2：已知函数表 xi 0.4 0.55 0.65 0.8 0.9 yi 0.41075 0.57815 0.69675 0.88811 1.02652用牛顿插值多项式求Nn(0.596)和Nn（0.895）。C语言程序设计：# include<stdio.h># define N 4void Difference(float

5、x,float y,int n)float *f=new floatn+1;int k,i;for(k=1;k<=n;k+)f0=yk;for(i=0;i<k;i+)fi+1=(fi-yi)/(xk-xi);yk=fk;delete f;return;void main()int i;float b,varx=0.596;float xN+1=0.4,0.55,0.65,0.8,0.9;float yN+1=0.41075,0.57815,0.69675,0.88811,1.02652;Difference(x,y,N);b=yN;for(i=N-1;i>=0;i-)b=b*(varx-xi)+yi;p