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

BISECTION METHOD-

 if you have an equation like f(x)=x*e^x-1
 then to find it's one root we will use bisection method
choose a & b such that f(a)=-ve & f(b)=+ve   let a=0 &b=1
make a table
n                       a                         b                         x=(a+b)/2                     f(x)
0                       0                         1                            0.5                            -ve
1                     0.5                        1                           0.75                        find f(x)

if f(x) is +ve then x=b  if -ve , x=a .if 0 then x will be a root.
solve further up to 12 iterations means till n=12
and then whatever will be the value of x will be a approximate root.

code is given below

#include<iostream.h>

#include<conio.h>

#include<math.h>

float e=2.718;

float f(float x)

{

   float f=x*pow(e,x)-1;

   return(f);

  }

void main()

{ clrscr();

  cout<<"the given equation is x*e^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+x2)/2;

  c=f(x1)*f(x0);

  if(c<0)

  x2=x0;

  else  if(c>0)

  x1=x0;

  n++;

  if(c==0)

  break;

  xm=(x1+x2)/2;

  }

  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*e^x-1:

enter the interval(a,b)

 enter x1=0

 enter x2=1

 the value of f(0):-1

 the value of f(1):1.718

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

0.567261

no. of iteration13