Skip to main content

newton raphson method


  ///Bismillahir Rahmanir Rahim

///Author:Tanvir Ahmmad
///CSE,Islamic University,Bangladesh

#include<iostream>
using namespace std;

double fun_1(double num)
{
    return ((num*num)-(4*num)-7);
}

double deri_fun_1(double num)
{
    return ((2*num)-4);
}

int main()
{
    char ch = 253;
    double num,brk_limit;
    int step,cas=1;
    cout<<"Given equation : "<<"X"<<ch<<"-4X-7"<<endl;
    cout<<"       f(x)=X"<<ch<<"-4X-7"<<endl;
    cout<<"      f'(x)=2X-4"<<endl;
    cout<<"Please enter an inter initial guess : ";
    while(1)
    {
        cin>>num;
        if(num>5) cout<<"Please enter a lower number : ";
        else if(num==5)
        {
            cout<<"You picked a right number.Let x0="<<num<<endl;
            break;
        }
        else cout<<"Please enter a upper number : ";

    }
    cout<<"Please enter minimum number of step if you want to do : ";
    cin>>step;
    int j=step;
    cout<<"Please enter a break down limit : ";
    cin>>brk_limit;

    while(step--)
    {
        num=(num-(fun_1(num)/deri_fun_1(num)));
        cout<<"X"<<cas++<<" : "<<num<<endl;
        if(abs(fun_1(num))<=brk_limit)
        {
            cout<<"Answer of "<<"X"<<ch<<"-4X-7 using The Newton-Raphson Method : "<<num<<endl;
            break;
        }
    }

    if(step==(-1))
        cout<<"Answer of "<<"X"<<ch<<"-4X-7 after "<<j<<" step using The Newton-Raphson Method : "<<num<<endl;
    return 0;
}



///Alhamdulillah


Comments

Popular posts from this blog

Codeforces round 1676(A. Lucky?)

Just count the  first three  &  last three  number ///Bismillahir Rahmanir Rahim ///Author: Tanvir Ahmmad ///CSE,Islamic University,Bangladesh #include< iostream > #include< cstdio > #include< algorithm > #include< string > #include< cstring > #include< sstream > #include< cmath > #include< cstring > #include< vector > #include< queue > #include< map > #include< set > #include< stack > #include< vector > #include< iterator > #include   < functional >   ///sort(arr,arr+n,greater<int>()) for decrement of array /*every external angle sum=360 degree   angle find using polygon hand(n) ((n-2)*180)/n*/ ///Floor[Log(b)  N] + 1 = the number of digits when any number is repres...

Maximum Number of Coins You Can Get (LeetCode)

Problem Name: 1561.  Maximum Number of Coins You Can Get Problem Link: https://leetcode.com/problems/maximum-number-of-coins-you-can-get/description/ Difficulty: Medium Tag: Sorting | Array | Math | Greedy | Sorting | Game theory Language: C# | C++ OJ: LeetCode Algorithm: Sort the Piles: Sort the array of piles in descending order, so that the piles with the maximum number of coins are at the beginning. Calculate the Total Number of Piles to Collect From: Determine the total number of piles you can collect coins from based on the rule that for every three consecutive piles, you can only pick coins from two of them. Set this value to 2/3 of the total number of piles. Iterate Through Piles and Collect Coins: Start iterating through the sorted piles from the second pile (index 1) and continue up to the calculated total number of piles to collect from. In each iteration, collect the coins from the current pile. Since you can only collect coins from every second pile, use a step of ...

Codeforces round 1661(B. Getting Zero)

using   second operation  the answer is maximum is  15  because  2 15 =32768 so we need to  pre-calculate all the illegible input 0 to 32768   using  second operation Then  just increment 1(using first operation)  from input check is it  minimum is not ! ///Bismillahir Rahmanir Rahim ///Author: Tanvir Ahmmad ///CSE,Islamic University,Bangladesh #include< iostream > #include< cstdio > #include< algorithm > #include< string > #include< cstring > #include< sstream > #include< cmath > #include< cstring > #include< vector > #include< queue > #include< map > #include< set > #include< stack > #include< vector > #include< iterator > #include   < functional >   ///sort(arr,arr+n,greater<int...