数值分析实验程序C语言

1拉格朗日插值：

#include

#define N 3

void main()

{

double X,Y=0.0,r,x[N],y[N];

int i,j;

printf("请输入xi:\n");

for(i=0;i

scanf("%lf",&x[i]);

printf("请输入相应的yi:\n");

for(i=0;i

scanf("%lf",&y[i]);

printf("请输入X\n");

scanf("%lf",&X);

for(i=0;i

{

r=1.0;

for(j=0;j

{

if(i==j) continue;

r=((X-x[j])/(x[i]-x[j]))*r;

}

Y=Y+r*y[i];}

printf("%lf\n",Y);

}

2牛顿插值：

#include

#define N 6

void main()

{

int i,k,n=N-1;

float X,Y,x[N],y[N],c[N],b[N]; printf("请输入xi:\n");

for(i=0;i

scanf("%f",&x[i]);

printf("请输入相应的yi:\n"); for(i=0;i

scanf("%f",&y[i]);

printf("请输入X\n");

scanf("%f",&X);

for(i=0;i

{c[i]=y[i];b[i]=c[i];}

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

{

for(i=0;i<=n-k;i++)

{

c[i+k]=(c[i+k]-b[i+k-1])/(x[i+k]-x[i]); }

for(i=0;i

b[i]=c[i];}

Y=c[n];

for(k=n-1;k>=0;k--)

Y=(X-x[k])*Y+c[k];

printf("%f\n",Y);

}

3变步长梯形法

#include

#define e 0.0000001

#include

void main()

{

int i=0;

double f(double x);

double a,b,S,x,h,T1,T2;

printf("请输入a,b:\n");

scanf("%lf,%lf",&a,&b);

h=b-a;

T1=(h/2)*(f(a)+f(b));

do

{ i++;

if(i>1){h=h/2;T1=T2;}

S=0.0;

x=a+h/2;

do

{

S=S+f(x);

x=x+h;

} while(x

T2=(T1/2)+(h/2)*S;

printf("T%d= %0.7f\n",i,T2);

}while(fabs(T2-T1)>=e);

}

double f(double x)

{

double y;

if(x==0)

y=1;

else

y=sin(x)/x;

return y;

}

4改进Euler方法：

#include

#define N 10

void main()

{

float f(float x,float y);

int n;

float h,x0,y0,x1,y1,yc,yp;

printf("请输入x0,y0,h\n");

scanf("%f,%f,%f",&x0,&y0,&h);

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

{

if(n>1){x0=x1;y0=y1;}

x1=x0+h;

yp=y0+h*f(x0,y0);

yc=y0+h*f(x1,yp);

y1=(yp+yc)/2;

printf("X%d = %f, Y%d = %f\n",n,x1,n,y1); }

}

float f(float x,float y)

{

float s;

s=y-(2*x)/y;

return s;

}

5牛顿迭代法：

#include

#include

#define e 1e-7

void main()

{

double f(double x);

double fd(double x);

int k,N;

double x0,x1;

printf("请输入x0,N\n");

scanf("%lf,%d",&x0,&N);

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

{

if(fd(x0)==0){printf("奇异标志\n");break;}

x1=x0-(f(x0)/fd(x0));

printf("X%d= %lf\n",k,x1);

if(fabs(x1-x0)

{printf("%lf\n",x1);break;}

else

if(k==N)

{printf("迭代失败\n");break;}

else

x0=x1;

}

}

double f(double x)

{

double y;

y=exp(x)*x-1;

return y;

}

double fd(double x)

{

double y;

y=x*exp(x)+exp(x);

return y;

}