Program for solving equation in numerical techniques by secant method using c++.

#include<iostream.h>

#include<conio.h>

#include<math.h>

float f(float x)

{

float f=pow(x,3)+x-1;

return(f);

  }

void main()

{ clrscr();

cout<<"the given equation is x^3+x-1:"<<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);

x1=x2;

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+x-1:

enter the interval(a,b)

enter x1=0

enter x2=1

the value of f(0):-1

the value of f(1):1

the value of f(9.490434e-41):-1root of the given equation on given tolerance is

0.682326

no. of iteration5