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 -
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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