factorial program

[boy3005]

The factorial of a non-negative integer n (written n! and read n factorial) is defined by
n! = 1 ,if n = 0 or 1
= 1*2*3*....n, if n > 1
So for instance, 3! = 1*
2 *
3 = 6 and 5! = 1
* 2 *
3 *
4 *
5 = 120 etc.
The number e the natural base of the exponential function is defined as
e = 1/0! + 1/1! + 1/2! + 1/3! + ....1/n!+ .....
Write a program that estimates the value of e using the first few terms up to 1/n!, that is
e = 1/0! + 1/1! + 1/2! + 1/3! + ....1/n!
The user is asked for the value of n, then compute the above sum.

#include <iostream>
using namespace std;

int main() {

	int n, i;
	long nfact=1;
	cout << "n=";
	cin >>n;
	if ( n >0 )
	for ( i = 1;i <=n;i++)
	nfact += 1.0/i // im confused here.
	cout << "n!= "<< nfact;
	

this is all i got so far or is my code entirely wrong ?

[giblit]

Well you have the basic ideas for a factorial

if( n > 0 )
{
    factorial = 1;
    for( int i = 0; i < n; ++i )
    {
        factorial *= i + 1;
    }
}

How ever you are not understanding the formula for e

it is e = 1/0! + 1/1! .... 1/n!

so say for example we have n = 5 this means:
e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5!
e = 1/0 + 1/1 + 1/2 + 1/6 + 1/24 + 1/120

Basically when they are in order like that you can just multiply the denom by the next value


Basically it would then we could say this

double e = 0; //same as 1/0!
int denominator = 1;

for( int i = 0; i < n; ++i )
{
    e += 1.0 / ( denominator *= i + 1 );
}

I think that is correct I haven't tested I am busy right now sorry but if it's not exactly correct I'll fix it when I have more time.

*edit had a typo with my 5! it was 120 not 60

[robertpires]

I'm not focusing on the logic of the problem but your syntax is off

if/else conditions that have more than one statement require braces
your for loop should also have braces just for good practice

also you should indent after the initial if-conditional and for-loop statement to make it clearer which statements belong to whatever your doing

[Parasin]

You need braces for your if statement and your for loop. otherwise, this won't work.

[boy3005]

#include <iostream>
using namespace std;

int main() {
	
	int n,i;
	cout << " Enter n=";
	cin >> n;
	
	double e=0;
	int denominator=1;

		for (int i=0; i < n;i++)
	{
	e+=1.0/ (denominator *= i+1 );
	cout <<"The sum is " <<e;
	}
	return 0;
}

i dont think the code is right cuz i keep getting numbers that are bigger than what it's supposed to be ( type in n=2, turns out to be 11.5)
and i keep getting "The sum is" multiple times.

[giblit]

Nope I just checked my formula is correct ( according to what he said ) except for one thing. He gave me the wrong formula he said the first value was 1/0! and idk for some reason I was thinking it was = to 0 but thats undefined..the real formula for find e is 1 + 1/1 , + 1/2 + 1/3 + 1/n.. The reason you got 11.5 for 2 idk you must have outputted a one first on accident.


I fixed the formula by changing intial value of e to 1 and I moved the denominator *= outside of the e += to make it look neater.

#include <iostream>

int main()
{
	const int n = 20;
	double e = 1.0;
	int denominator = 1;

	for( int i = 0; i < n; ++i )
	{
		denominator *= i + 1;
    	        e += 1.0 / denominator;
	}
	std::cout << e << std::endl;
}

Cheers.

Oh and you are getting the sum is multiple times because you have it inside of the for loop.

[Chervil]

Purely for reasons of style, I'd prefer to avoid the i + 1 - can't explain why, it is just not aesthetically pleasing.

#include <iostream>

int main()
{
    const int n = 20;
    long double e = 1.0L;
    long double denominator = 1.0L;
    
    for (int i = 1; i <= n; ++i )
    {
        denominator *= i;
        e += 1.0L / denominator;
    }
    std::cout.precision(20);
    std::cout << e << std::endl;
}

[boy3005]

when i enter 2 for n, is it suppose to print 2 and 2.5?? isnt it suppose to just print 2.5?

[giblit]

Are you sure the first 2 isn't the input?

*edit
It should output 2.5 since 1 + 1/1 + 1/2 = 1 + 1 + .5 = 2 + .5 = 2.5

I see your problem maybe it's because you have the n = ... output and never put an endl;

[boy3005]

when i put endl after it it adds another 2....
i think its easier if u just look at my code.

#include <iostream>
using namespace std;

int main() {
	
	int n,i;
	cout << "Enter n=";
	cin >> n;
	double e=1.0;
	int denominator=1;
		for (int i=0; i < n;i++)
	{
	e+=1.0/ (denominator *= i+1 );
	cout << e<<endl;
	}
	return 0;
}

ok it worked out fine somehow when i put the cout out of the loop and deleted all the endl;
Do you know why it works now without all the endl?

[giblit]

because you have the output inside of the for loop switch line 14 and 15