for

This is a discussion on for within the C Programming forums, part of the General Programming Boards category; if limit is 4 how to use for calculat this program? #include<stdio.h> int main(void) { double tpt =0; int i, ...

  1. #1
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    Question for

    if limit is 4

    how to use for calculat this program?


    #include<stdio.h>

    int main(void)
    {
    double tpt =0;
    int i, limit;

    for(i=0; i<=limit; i++)
    {
    1 1 1 1 1
    tot += - + - + - + - + -
    1*1 1*2 1*2*3 1*2*3*4 (limit-1)
    (4-1)
    1*2*3
    }

  2. #2
    Code Goddess Prelude's Avatar
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    >1 1 1 1 1 tot += - + - + - + - + - 1*1 1*2 1*2*3 1*2*3*4 (limit-1) (4-1) 1*2*3
    What is this? This isn't even *close* to a valid expression.

    -Prelude
    My best code is written with the delete key.

  3. #3
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    Hmmm, saw this post too late. There was another posting by you with this question. Can you tell us in words what you try to do?

  4. #4
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    Perhaps you should set limit to a value before comparing it.

  5. #5
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    Question for

    Euler's equation

  6. #6
    Mayor of Awesometown Govtcheez's Avatar
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    > Euler's equation

    Um... okay.... Still doesn't explain your problem... Almost nothing in your code is valid C.

  7. #7
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    "Bad programming = Compiler errors" -- Euler's equation.

  8. #8
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    What's wrong in this program?
    How to fix it?

    use loop to terminates when the difference between
    two succesive values of e differ by less than
    0.0000001

    #include<stdio.h>

    void getLimit(int *);
    void calculation(int);

    int main(void)
    {
    int limit;

    getLimit(&limit);
    calculation(limit);

    return 0;
    }

    /**************************GET LIMIT********************/

    void getLimit(int *limit)
    {
    printf("Enter limit: ");
    scanf("%d", limit);

    return;
    }

    /**************************CALCULATION************* *******/

    void calculation(int limit)
    {
    int i, j, p, flag=0;
    double sum, diff, psum=1;

    for(i=1; i<limit; i++)
    {
    p=1;

    for(j=1; j<=i; j++)
    p=p*j;

    sum=psum + (1/p);
    diff=psum-sum;

    if(diff < 0.0000001)
    {
    flag=1;
    break;
    }

    if(flag==1)
    printf("Prev-next:%.2lf %.2lf %.2lf\n", psum, sum, diff);
    else
    printf("Num of iteration: %d, e=%.5lf\n", limit, sum);
    }
    return;
    }

  9. #9
    ....
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    Your program looks a bit unstructured. Take some more effort in designing your algorithm.

    Second, there are many formula's of Euler. But I think your pointing at the formula to calculate e.

    e = sigma (x=0->inf) (1 / x!)

    Which you need to implement as

    e = sigma (x=0->limit) (1 / x!)

    Which is

    e = 1 / 0! + 1 / 1! + 1 / 2! + ... + 1 / limit!

    As you can see there are two main calculations to be made

    1. Sum
    2. Faculty

    A sum can be calculated by using a for-loop and faculty can be a function:

    Code:
    e = 0
    for (x = 0; x < limit; x++)
    {
        e += 1 / fac (x);
    }
    The faculty function is a recursive one

    x! = x * (x-1) * (x-2) * ... * 3 * 2 * 1

    So you see the end-step is reached if x is 1 and the recursion step is x * (x - 1). The function is now easily created:

    Code:
    int fac (int i)
    {
        if (i == 1 || i == 0)
        {
            return 1;
        }
        else
        {
            return (i * fac (i - 1));
        }
    }

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