西电计算方法第三次作业

第四章习题

1利用函数y 二..x 在X 1 =100,X 2 =121处的值，计算-..115的近似值，并估计误差。

1

n

x — X ・

解：LMx) = ' y 」k (x),l k (x) -

7 X k - X i

y °(x) =10, %(x) =11

估计误差：卜帀—^(115) = 10.724-10.714 =0.01

于是 N(x)=16x 7x(x-1) 一 2.5x(x-1)(x-2)

x(x-1)(x-2)(x-4)

6

Ljx)二 y °l °(x) %l 1(x) =10 x -121 100-121 11

x -100

121-100

.LM115) =10

115-121 100-121 11 115-100 121-100 75

7 10.714 一 115

N°(0.125 0.125t) =0.79618— 0.02284t— 0.003395t(t -1) —0.000527t(t — 1)(t - 2) 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求x=0.5635时的函数近似值

C语言程序设计：

# in clude

float Lagrange(float x[],float y[],float xx,int n)

{

int i,j;

float *a,yy=0;

a=new float[ n];

for(i=0;i<=n _1;i++)

{

a[i]=y[i];

for(j=0;j<=n _1;j++)

if(j!=i) a[i]*=(xx-x[j])/(x[i]-x[j]);

yy+=a[i];

}

delete a;

return yy;

}

void mai n()

{

float x[4]={0.56160,0.56280,0.56401,0.56521};

float y[4]={0.82741,0.82659,0.82577,0.82495};

float xx=0.5635,yy;

yy=Lagra nge(x,y,xx,4);

prin tf("x=%f,y=%f\n",xx,yy);

}

运行结果：

x=0.563500

y=0.826116

g *C:\DocuMents and Settings\Ad>inistrator\Debug\Cpp1.exe 1^0.563500/9=0.826116

Press any key to cont in Lie

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M(0.596)和2 (0.895 )。C语言程序设计：

# in clude

# define N 4

void Differen ce(float x[],float y[],i nt n)

{

float *f=new float[ n+1];

int k,i;

for(k=1;k<=n ;k++)

{

f[0]=y[k];

for(i=0;i

f[i+1]=(f[i]-y[i])/(x[k]-x[i]);

y[k]=f[k];

}

delete f;

return;

}

void mai n()

{

int i;

float b,varx=0.596;

float x[N+1]={0.4,0.55,0.65,0.8,0.9};

float y[N+1]={0.41075,0.57815,0.69675,0.88811,1.02652};

Differe nce(x,y,N);

b=y[N];

for(i=N-1;i>=0;i--)

b=b*(varx-x[i])+y[i];

prin tf("N n( %f)=%f',varx,b);

}

运行结果：

N n(0.596)=0.631918

N n(0.895)=1.019368