Binomial Series
Before I start I want to explain that I have searched the forums and haven't found my solution yet.
My teacher gave us a problem to solve he wants a program to solve binomials of the form (ax+b)^n where a,b,n are integers given by the user.
I have solved 90% of it but there remains a problem.
usually (a+b)^2 = a^2+2ab+b^2 and so on and so forth
my problem is that my program is giving a^2+ab+b^2 thus calculating wrong can someone help?
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// Project 2a.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <limits>
#include <string>
using namespace std;
unsigned long factorial(unsigned long);
int main()
{
int r = 1;
while (r == 1)
{
int n = 0;
int a = 0;
int b = 0;
int k = 0;
int y = 0;
cout << "This is a program to solve (ax+b)^n given a,b,n \n" << endl;
cout << "Give me a: \t";
while (!(cin >> a))
{
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cout << "\nPlease type a number a:\t";
}
cout << "Give me b: \t";
while (!(cin >> b))
{
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cout << "Error \n \nPlease type a number b:\t";
}
cout << "Give me n: \t";
while (!(cin >> n))
{
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cout << "Error \n \nPlease type a number:\t";
}
n = abs(n);
cout << "(";
cout << a;
cout << "x";
cout << " + ";
cout << b;
cout << ")^";
cout << n;
cout << endl << endl;
while (n >= k)
{
y = y + (factorial(n) / (factorial(k)*factorial(n - k)))*((a ^ (n - k))*(b^k));
cout << (pow(a, (n - k)))*pow(b, k);
if ((n - k) > 0)
{
cout << "x";
cout << "^";
cout << n - k;
cout << " + ";
}
k = k + 1;
}
cout << "\n \nTo repeat press 1, to exit press any other number:\t";
cin >> r;
system("CLS");
}
cout << endl << endl;
return 0;
}
unsigned long factorial(unsigned long number)
{
if (number <= 1)
return 1;
else
return number * factorial(number - 1);
}
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The problem is on line 70. b^k is not b raised to the k'th power, it's the exclusive-or of b and k. You want pow(b,k) instead.
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