Write a simple c++ program that computes the value of ex by using the.docx

D

Write a simple c++ program that computes the value of ex by using the formula: ex =1+x/1!+x2/2!+x3/3!+...+xN/N! try for N large N = 10000. using <iostream> and cmath of needed only. Solution #include <iostream> #include <cmath> using namespace std; int main(){ double n = 10000; double ex = 1; double x; cout << \"Enter the value of x: \"; cin >> x; double num = 1, den = 1, term = 1; for(double i = 1; i <= n; i++){ term *= (x / i); ex += term; } cout << \"e^\" << x << \" = \" << ex << \"\ \"; } .

Write a simple c++ program that computes the value of ex by using the formula: ex
=1+x/1!+x2/2!+x3/3!+...+xN/N! try for N large N = 10000. using <iostream> and cmath of
needed only.
Solution
#include <iostream>
#include <cmath>
using namespace std;
int main(){
double n = 10000;
double ex = 1;
double x;
cout << "Enter the value of x: ";
cin >> x;
double num = 1, den = 1, term = 1;
for(double i = 1; i <= n; i++){
term *= (x / i);
ex += term;
}
cout << "e^" << x << " = " << ex << " ";
}

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Write a simple c++ program that computes the value of ex by using the.docx

  • 1. Write a simple c++ program that computes the value of ex by using the formula: ex =1+x/1!+x2/2!+x3/3!+...+xN/N! try for N large N = 10000. using <iostream> and cmath of needed only. Solution #include <iostream> #include <cmath> using namespace std; int main(){ double n = 10000; double ex = 1; double x; cout << "Enter the value of x: "; cin >> x; double num = 1, den = 1, term = 1; for(double i = 1; i <= n; i++){ term *= (x / i); ex += term; } cout << "e^" << x << " = " << ex << " "; }