CODEWITHSUNDEEP
c
Alex Podworny
Alex Podworny

Reputation: 1019

Approximating e using loops based on user input

I'm working on an assignment in a first year C class and we are in the loop section of a textbook. I've already learned a few languages but I believe I am doing this wrong somehow since I am not getting the correct output. I believe I need to do this question using loops (so no additional math libraries). Normally I'd use a debugger, but I'm programming C using sublime text and the command prompt so I don't think that's possible. We also have not gone through methods/functions/whatever C uses yet, so my solution can't use that stuff.

Using only C89 is preferred.

Here is the question:

The value of the mathematical constant e can be expressed as an infinite series: e = 1 + 1/1! + 1/2! + 1/3! + ... Write a program that approximates e by computing the value of 1 + 1/1! + 1/2! + 1/3! + ... + 1/n! where n is an integer entered by the user.

Note that I believe ! in this case means factorial.

I'm checking my output against that of this sigma calculator, and just adding 1 to the calculator's output to check if my result is correct or not.

http://www.mathsisfun.com/numbers/sigma-calculator.html

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    float e = 1;/* Start off as 1 since the equation adds 1 anyways */
    int input, i, j;

    /* Ask the user for the value that they want to use, and store it in a variable */
    printf("Enter an integer to use to approximate: ");
    scanf("%d", &input);

    for (i = 1; i < input; i++)
    {
        /* This inside loop is for the factorial, where j goes through all of the factorials for each i */
        for(j = i; j > 0; j--)
        {
            e += (1/(float)j);
        }

    }

    printf("e equals %f\n", e);
    return 0;
}

Upvotes: 0

Views: 4140

Answers (4)

omerfarukdogan
omerfarukdogan

Reputation: 869

The loop should be this way:

for(i=1; i<=input; i++)
{
    int result = 1;
    for(int j=1; j<=i; j++)
    {
        result = result * j;
    }
    //now the "result" is the factorial of i
    e += 1 / (float)result; // 1/factorial(1) + 1/factorial(2) + ...
}

Upvotes: 4

a_pradhan
a_pradhan

Reputation: 3295

The loop can be simple enough like this :

int fact = 1 ;
for(int i = 1; i < input; ++i)
{
    fact *= i ;
    e += (1.0f/(float)fact);
}

No nested loop is required. Here is a working version.

Upvotes: 2

Dimitri Mockelyn
Dimitri Mockelyn

Reputation: 1545

You are not doing the right thing in your factorial calculation. Your are summing when you should be multiplying. Your inner loop should probably look like this :

    /* This inside loop is for the factorial, where j goes through all of the factorials for each i */
    float inverse_factorial_i = 1;
    for(j = i; j > 0; j--)
    {
        inverse_factorial_i *= (1/(float)j);
    }

And then

e += inverse_factorial_i

Upvotes: 2

Jean-Baptiste Yun&#232;s
Jean-Baptiste Yun&#232;s

Reputation: 36431

No your code just makes sums over i of sums over j in [1,i] of 1/j. So you calculates 1/1 + (1/1+1/2) + (1/1+1/2+1/3) + ... and not 1/1 + (1/1*1/2) + (1/1*1/2*1/3) + ...

That should be something like:

for (i = 1; i < input; i++)
{
    float inversefact = 1;
    for(j = i; j > 0; j--)
    {
        inversefact *= (1/(float)j);
    }
    e += inversefact;

}

Upvotes: 1

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