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Tuesday, 1 April 2014

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

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