approx value of (e)/

This is a discussion on approx value of (e)/ within the C++ Programming forums, part of the General Programming Boards category; Value of e is to be found by the formula e=1+1/1!+1/2!+1/3!+.................. The following code give value only upto 5!.Although it ...

  1. #1
    o0o
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    Thumbs down approx value of (e)/

    Value of e is to be found by the formula

    e=1+1/1!+1/2!+1/3!+..................

    The following code give value only upto 5!.Although it should increment num and use the new value (ie 5!*6!*7!*..........)again.What's wrong?

    Code:
    #include<iostream>
    #include<conio.h>
    
    using std::cout;
    using std::cin;
    using std::endl;
    
    int main()
    {
        // declare and initialize variable/s
        
        int num=5,factorial=1;
        double result;
        
        // display the approx value of e
        
        while(num>0)
        {
            factorial=factorial*num;
            num-=1;
            result=1+factorial;
        }
        
        cout<<"\n\n\t\t\te = "<<1/result;
        num+=1;
        
    getch();
    }

  2. #2
    Epo
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    You're setting:
    num=5

    Then, in your While Loop, the condition is:
    while (num > 0)

    And inside your loop, you are decreasing num by 1 each time.
    Meaning the first time around, num = 5.
    Then 4,
    then 3,
    then 2,
    then 1,
    then, it equals 0, and since 0 (Num) is not greater than 0 (Your Condition), it exits the loop. Hence, only doing it 5 times.

  3. #3
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    Does this help :
    Code:
    #include<iostream>
    #include<conio.h>
    
    using namespace std;
    
    int func(int number)
    {
        if(number > 0)
            return number * func(number-1);
        
        return 1;
    }
    
    int main()
    {
        cout << "5! = " << func(5) << endl;
        getch();
    }
    Last edited by laasunde; 12-25-2003 at 02:31 PM.

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