polynomials

This is a discussion on polynomials within the C Programming forums, part of the General Programming Boards category; I'm new to C programming and I have an assignment that's really been bugging me. I have to write a ...

  1. #1
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    Question polynomials

    I'm new to C programming and I have an assignment that's really been bugging me. I have to write a function
    double eval(double p[], double x, int n)
    {
    }
    that returns the value of polynomial p evaluated at x using Horner's Rule which is p(x) = a0 + x(a1 + (a2 + x(a3 + x(a4 + x(a5 )))))

    Here's what I've written so far:

    #include <stdio.h>
    #include <math.h>
    #define N 5
    double eval(double p[], double x, int n);

    int main() {

    double polynomial; /* Used for return value from eval function */

    double p[N + 1];

    /* Print results of polynomial evaluation */
    printf("Polynomial evaluated at x: %f\n\n", polynomial);

    return 0;
    }

    /* Polynomial evaluation function */
    double eval(double p[], double x, int n) {
    double result;
    int i;
    if (n <= 0)
    {
    return 0.0;
    }
    result = p[0];

    for (i=1; i <= n; i++)
    {
    result = result + p[i] * pow(x, i);
    }

    return result;
    }
    Am I heading in the right direction? Comments???

  2. #2
    eh ya hoser, got a beer? stumon's Avatar
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    >>result = result + p[i] * pow(x, i);

    Below i have added a picture taken from google for that guys rule. I would say, yes you are in the right direction, are there any problems?

    --edit--
    Question: What is x? What do you pass? and where is the function call in main?
    Attached Images Attached Images  
    Last edited by stumon; 04-09-2003 at 06:04 PM.
    The keyboard is the standard device used to cause computer errors!

  3. #3
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    ... but notice that you are not using Homer's Rule (D'oh!). You have correctly described it at the top of your post. Think about a for-loop that evaluates each of the brackets in turn. The resulting code will be more efficient than repeatedly using pow(a,b).

    actually you almost got the algorithm right (probably just a typo) it should be:

    a0 + a1 * x + a2 * x^2 + a3 * x^3 + a4 * x^4 + a5 * x^5 =
    a0 + x * (a1 + x * (a2 + x * (a3 + x * (a4 + x * (a5)))))

    Good luck,
    Last edited by DavT; 04-10-2003 at 07:30 AM.
    DavT
    -----------------------------------------------

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