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Monday, 31 March 2014

Program to solve equations in numerical techniques by regula falsi method using c++.


#include<iostream.h>
#include<conio.h>
#include<math.h>
float f(float x)
{
float f=pow(x,3)-18;
return(f);
  }
void main()
{ clrscr();
cout<<"the given equation is x^3-18:"<<endl;
  float x1,x2,x0,c,xm,n;
cout<<"enter the interval(a,b)"<<endl;
cout<<"\n enter x1=";
cin>>x1;
cout<<"\n enter x2=";
cin>>x2;
cout<<"\n the value of f("<<x1<<"):"<<f(x1);
cout<<"\n the value of f("<<x2<<"):"<<f(x2);
cout<<"\n the value of f("<<x0<<"):"<<f(x0);
if (f(x1)*f(x2)<0)
  {do
  {x0=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1));
  c=f(x1)*f(x0);
if(c<0)
x1=x0;
else  if(c>0)
x2=x0;
n++;
if(c==0)
break;
xm=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1));
  }
while(fabs(xm-x0)>=0.0001);
cout<<"root of the given equation on given tolerance is"<<x0<<endl;
cout<<"no. of iteration"<<n<<endl;
  }
else
cout<<"can not found in the given inter val";
getch();
  }
OUTPUT-
the given equation is x^3-18:
enter the interval(a,b)

enter x1=2

enter x2=3

the value of f(2):-10
the value of f(3):9
the value of f(9.490434e-41):-18root of the given equation on given tolerance is
2.620708


no. of iteration4

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