Practical set 4 (Newton Raspon)

#include <stdio.h>

#include<math.h>

float value(float x)

{

    float f;

//  f=x-(1.5*sin(x))-2.5;

f=pow(x,5)-3*(pow(x,2))-100;

//f=x*tan(x)-1;

//f=(x*x)-(5*x)+6;

    return f;

}

  

 float value1(float y)

 {

     float f;

//    f=1-(1.5*cos(y));

f=5*(pow(y,4))-(6*y);

//f=(y*(1/cos(y)*cos(y)))+tan(y);

//f=(2*y)-5;

     return f;

 }

  

 float fun(float y,float x)

 {

     float p;

     p=x-(value(y)/value1(y));

    return p;

}

    int main()

  {

      int i,n;

      float p[20],a,b;

       

       

      printf("\nQuestion is : x^2-5x+6 =0 \n");

      printf("\nEnter how many iteration : ");

      scanf("%d",&n);

        for(i=0;i<n;i++)

   {

       p[i]=value(i);

       printf("\n f[%d] : %f ",i,p[i]);

   }

    

   for(i=0;i<n;i++)

    {

        if(p[i]*p[i+1]<0)

         {

            a=i;

            b=i+1;

            printf("\n\nThe root lies between: [%2.4f %2.4f]",a,b);

            break;

            }      

        }

      p[0]=3.5;

      printf("\n\n f[0] : %f ",p[0]);

      for(i=1;i<n;i++)

   {

       p[i]=fun(p[i-1],p[i-1]);

       printf("\n f[%d] : %f ",i,p[i]);

   }

    return 0;

}  



By - milan zinzuvadiya