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Regula Falsi or False Position Method Using C++



f(x)=X3-2X-5
f(2) = -1
f(3) = 16
x=(af(b) - bf(a))/((f(b)-f(a));
///Bismillahir Rahmanir Rahim
///Author: Tanvir Ahmmad
///CSE, Islamic University, Bangladesh
#include<iostream>
using namespace std;
typedef long long ll;
double func(double num)
{
    return ((3*num*num)-(6*num)+2);
}
int main()
{
    char ch=253;
    double a,b,out,tol;
    cout<<"Given equation : f(x) = 3X"<<ch<<"-6X+2"<<endl;
    cout<<"Given two input of x such that f(x) is negative and positive and there difference is just one"<<endl;
    while(1)
    {
        cin>>a>>b;
        if(a==1 && b==2)
        {
            cout<<"Given correct output"<<endl;
            break;
        }
        else
        {
            cout<<"f(a)="<<func(a)<<endl;
            cout<<"f(b)="<<func(b)<<endl;
            cout<<"Incorrect Output"<<endl;
        }
    }
    int step,stp=0;
    cout<<"How much step you want to work : "<<endl;
    cin>>step;
    cout<<"Enter tolerable fraction : "<<endl;
    cin>>tol;
    while(step--)
    {
        out=((a*func(b))-(b*func(a)))/(func(b)-func(a));
        cout<<++stp<<" - iteration "<<out<<" and f(x) = "<<func(out)<<endl;
        if(abs(func(out))<=tol)
        {
            cout<<"After "<<stp<<" step using tolerable fraction root of 3X"<<ch<<"-6X+2 = "<<out<<endl;
            break;
        }
        else if(func(out)<0) a=out;
        else if(func(out)>0) b=out;
        else
        {
            cout<<"After "<<stp<<" step root of 3X"<<ch<<"-6X+2 = "<<out<<endl;
        }
    }
    if(step<=0) cout<<"After using all step this equation's root will be "<<out<<endl;
    return 0;
}
///Alhamdulillah


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