interpolating polynomial.c

高次方程求解的c代码实现

//	Write a computer program using a Lagrange interpolating polynomial of the 10th order to evaluate f(x)=lnx where [a, b]=[1,2], h=0.1, xi = 1+ih, i=0, 1, …, 10, and find the approximations of ln1.54 and ln1.98.
//
#include "stdio.h"
#include "math.h"
#define N 10

 
float f(float x)
{
 float y;
 y=log(x);
 return(y);
}

void lag(float a[],float b[],int n,float x)
{
 int i,j;
 for(i=0;i<=n;i++)
  for(j=0;j<=n;j++)
   if(i!=j)
    b[i]=b[i]*(x-a[j])/(a[i]-a[j]);
}

float sum(float a[],float b[],int n)
{
 float y=0;
 int i;
 for(i=0;i<=n;i++)
  y=y+f(a[i])*b[i];
 return(y);
}

main()
{
 float x,y;
 float m[N+1];
 float l[N+1];
 int i;
 for(i=0;i<=N;i++)
 {
  m[i]=1+0.1*i;
  l[i]=1;
 }
 do
 {
  printf("input x belong to [1,2]:\n");
  scanf("%f",&x);
 }while((x<1)||(x>2));
 lag(m,l,N,x);
 y=sum(m,l,N); 
 printf("f(%f)=%8.4f\n",x,y);
 getchar();
 getchar();
}
 

//	Write and test a computer program to implement the Bisection algorithm. Test program on the function and interval: 2-x+ex+2cosx-6 on [1, 3].

 
#include "stdio.h"
#include "math.h"

float f(float x)
{
 float y;
 y=pow(0.5,x)+exp(x)+2*cos(x)-6;
 return(y);
}

float mid_point(float x1,float x2)
{
 float y;
 y=(x1+x2)/2;
 return(y);
}

float root(float x1,float x2)
{
 float x,y;
 do
 {
  x=mid_point(x1,x2);
  y=f(x);
  if(y*f(x1)>0)
    {
     x1=x;
    }
  else
    {
     x2=x;
    }
 }while(fabs(y)>=0.001);
 return(x);
}

main()
{
 float x1,x2,f1,f2,x;
 do
 {
  printf("input x1,x2:\n");
  scanf("%f,%f",&x1,&x2);
  f1=f(x1);
  f2=f(x2);
 }while(f1*f2>=0);
 x=root(x1,x2);
 printf("A root of equation is %8.4f\n",x);
 getchar();
 getchar();
}
 
//	Write a brief computer program to solve the equation x3+3x=5x2+7 by Newton’s method. Take ten steps starting at x0=5.

#include "stdio.h"
#include "math.h"
#define N 10

 
float f(float x)
{
 float y;
 y=x*x*x-5*x*x+3*x-7;
 return(y);
}

float f1(float x)
{
 float y;
 y=3*x*x-10*x+3;
 return(y);
}

float root(float x0,int n)
{
 float x;
 int i;
 for(x=x0,i=0;i<n;i++)
  {
   x=x-f(x)/f1(x);
  }
 return(x);
}

main()
{
 float x0,x;
 printf("input initial x0:\n");
 scanf("%f",&x0);
 x=root(x0,N);
 printf("A root of equation is %8.4f\n",x);
 getchar();
 getchar();
} 
 
//	Write a computer program to carry out the Secant method on a function f, assuming that two starting points are given. Test the routine on the function f = x3-12x2+3x+1.

#include "stdio.h"
#include "math.h"

 
float f(float x)
{
 float y;
 y=x*x*x-12*x*x+3*x+1;
 return(y);
}

float f1(float x1,float x2)
{
 float y;
 y=(f(x2)-f(x1))/(x2-x1);
 return(y);
}

float root(float x1,float x2)
{
 float x;
 x=x2;
 while(fabs(f(x))>=0.0001)
  {
   x=x-f(x)/f1(x1,x2);
   x1=x2;
   x2=x;
  }
 return(x);
}

main()
{
 float x1,x2,x;
 printf("input initial x1,x2:\n");
 scanf("%f,%f",&x1,&x2);
 x=root(x1,x2);
 printf("A root of equation is %8.4f\n",x);
 getchar();
 getchar();
}