Question

(Computing e in c++ programming) You can approximate e by using the following series: e = 1 +1/1! +1/2! +1/3! +1/4! + ... + 1/ i! Write a program that displays the e value for i = 10000, 20000, ..., and 100000. (Hint: Since i!= i × (i − 1) × ...× 2 × 1 ,then 1/i! is 1/i(i-1)! . Initialize e and item to be 1 and keep adding a new item to e. The new item is the previous item divided by i for i = 2, 3, 4, ...)

Answer 1

#include <iostream>
using namespace std;

void f(long double& q)
{
& long double& a,s,i;

a=1;
s=1;


& for ( i=1;i<q;i++){
a=a+1/s;
s=s*i;
}

cout.precision(15);


cout<<"e& "<<a;
&

cout<<"\n\n";
&
&
&
& }

int main()
{

cout <<"for i = 10000& \n";
f(10000);

cout <<"for i = 20000& \n";
f(20000);
&
& cout <<"for i = 30000& \n";
f(30000);

cout <<"for i = 40000& \n";
f(40000);

cout <<"for i = 50000& \n";
f(50000);

cout <<"for i = 60000& \n";
f(60000);

cout <<"for i = 70000& \n";
f(70000);

cout <<"for i = 80000& \n";
f(80000);

cout <<"for i = 90000& \n";
f(90000);

cout <<"for i = 100000& \n";
f(100000);



system("PAUSE");
}