Saturday, January 26, 2013

Write a C++ program to calculates the following equation for entered numbers (n, x). 1+ (nx/1!) - (n(n-1)x^2/2!)......................................

Exercise 3)
Write a C++ program and call it series.cpp which receives two numbers from input (n, x) and calculates the following equation for entered numbers (n, x).
1+  (nx/1!) - (n(n-1)x^2/2!)......................................

Solution -

#include<iostream>
#include<cmath>
// the user define function is declaired here to get the factorial
unsigned int fac(unsigned int n);

using namespace std;

int main()
{
//-------defining variables and initializing them-------------
double n,x,i,ans=0,ans1=0,ans2=0,ans3=0,ans4=0,ans5=0,power;
char redo;
//--------Printing my name on screen----------------
cout<<"Welcome to the series program  written by Your Name"<<endl;
cout<<"***************************************************************"<<endl;
cout<<endl<<endl<<endl;
//--here do loop is used so that the program can be used more then one time
//without exiting the run screen---------------------------
do
{
//----receiving the variables from input--------------

cout<<" Please enter two numbers to apply your requested operation"<<endl;
cout<<"enter the value of n:";
cin>>n;
cout<<"enter the value of x:" ;
cin>>x;
cout<<endl;
/*If the series 1+((n*x)/fac(1))-(n*(n-1)*pow(x,2)/fac(2))+(n*(n-1)*(n-2)*pow(x,3)/fac(3))
........is  considered as binomial series then the result equal to (2-pow((1-x),n)
therefore below formulae is used*/

ans=2-pow((1-x),n);
cout<<"-------------------------------------------------------------------------"<<endl;
cout<<"The result of the n term if the series is considered as Binomial series\n";
cout<<"(1+((n*x)/fac(1))-(n*(n-1)*pow(x,2)/fac(2))+(n*(n-1)*(n-2)*pow(x,3)/fac(3))....\n";
cout<<"is ="<<ans<<endl;
cout<<"-------------------------------------------------------------------------"<<endl;

/* 1+((n*x)/fac(1))-(n*(n-1)*pow(x,2)/fac(2))+(n*((n-2)*pow(x,3)/fac(3))......
i considered then the reault is given below*/

ans3=1+((n*x)/fac(1));
//the ans2 and ans5 has to be done zero every time it enters the loop
ans2=0;
ans5=0;
// initialising for loop
for( i=1;i<=n;i++){
//------calculating the requested equation for inputs-------------
ans2=ans2+ans5;
ans5=-(n*(n-i)*pow((-x),(i+1))/fac(i+1));
//the below expressions are to check the status of the loop so thet we can moniter the loop at everzý point
/*cout<<"value of i after each loop="<<i<<endl;
cout<<"valu of n after each loop="<<n<<endl;
power=pow((-x),(i+1));
cout<<"valu of power term after each loop="<< power<<endl;
cout<<"valu of ans2 after each loop="<<ans2<<endl;
cout<<"valu of ans5 after each loop="<<ans5<<endl;
*/

}
//------- printing the results on screen-----------
cout<<endl;
cout<<endl;
cout<<"-------------------------------------------------------------------------"<<endl;
cout<<"The result of the n term if the series is not considered as Binomial\n ";
cout<<"series is given below"<<endl;
cout<<endl;
cout<<"The sum of  1st two termsof n termseries is ="<<ans3<<endl;
cout<<"The sum of n terms except 1st two term ="<<ans2<<endl;
cout<<endl;

ans4=ans2+ans3;
cout<<"The final result of the n term  series is  ="<<ans4;
cout<<endl;
cout<<"-------------------------------------------------------------------------"<<endl;

//----now once again the program will ask the user if want to continue or not
cout<<"enter y or Y to continue:";
cin>>redo;
cout<<endl<<endl;

}
while(redo=='y'||redo=='Y');

system("pause");
return 0;

}
// the user body of function is given below
unsigned int fac(unsigned int n)
{
if (n == 0)
{
return 1;
}
else {
return n * fac(n-1);
}
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