please help stuck on a slight problem, basic, easy to read, getting wrong answer

This is a discussion on please help stuck on a slight problem, basic, easy to read, getting wrong answer within the C Programming forums, part of the General Programming Boards category; hey guys. trying to figure this out... when i run my code my a2 and a3 values are 0 on ...

  1. #1
    jdi
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    please help stuck on a slight problem, basic, easy to read, getting wrong answer

    hey guys.

    trying to figure this out... when i run my code my a2 and a3 values are 0

    on the calculator when i calculate by hand they aren't 0, i dont understand why. changing everything from float to double doesn't work either.

    how can i get this to work??

    im using predefined values for everything as this is a specific case of an application.

    the van der waals equation is rearranged into cubic form, then solved for v. its fairly straight forward but i cant get it working. my general cubic equation solver (which this is based of) works beautifully.

    please help

    Code:
    
    
    
    
    
    
    #include <stdio.h>
    #include <math.h>
    #define PI (3.141592653589793238462643)
    #define a 0.448
    #define b 4.29e-5
    #define R 8.318
    #define n 1.0
    #define P 4.0e7
    #define T 400.0
    
    
    float  Q, S, a1, a2, a3, A11, x1, x2, x3, theta, test1, test2, test3, Qcubed, Ssquared;
    
    //Sign function prototype
    float sign(float S);
    
    
    int main()
    {
    
    
    
    	//Code that handles coefficents
    	a1 = -((P*n*b + n*R*T)/P);
    	a2 = (a*n*n/P);
    	a3 = -(a*b*n*n*n/P);
    
    
    
    	//Doing some calculations to simplify our life
    	Q = (a1*a1 - 3.*a2)/(9.0);
    	S = (2.*a1*a1*a1 - 9.*a1*a2 + 27.*a3)/(54.0);
    	Qcubed = Q*Q*Q;
    	Ssquared = S*S;
    
    	//Printing out all the variables used so far to make sure no inf,NaN or other errors have occured
    	printf("%f = Q, %f = S, %f = a1, %f = a2 %f = a3\n", Q, S, a1, a2, a3);
        printf("%f = sign(S) %f = A11\n", sign(S), A11);
    	printf("%f = Qcubed - Ssquared\n", Qcubed - Ssquared);
    
    
    	//The first case, if our Q^3 - S^2 > 0
    	//The cubic equation has 3 real, easy to find roots, found as below
    	if (Qcubed - Ssquared > 0)
    	{
    		theta = acos(S/sqrt(Qcubed));
    		x1 = -2*sqrt(Q)*cos(theta/3.) - b/3.;
    		x2 = -2*sqrt(Q)*cos((theta + 2*PI)/3.) - a1/3.;
    		x3 = -2*sqrt(Q)*cos((theta + 4*PI)/3.) - a1/3.;
    
    		printf("The cubic has three real roots, %f = x1, %f = x2, %f = x3\n", x1, x2, x3);
    	}
    
    
    	//The second case, if Q^3 - S^2 <= 0, first we calculate A11
    	//After A11 has been calculated we can decide which equation to use to solve the cubic
    	//The cubic has only one real root as expressed as below
    	if (Qcubed - Ssquared <= 0)
    	{
    
    	//Calculating A11 to find out which equation to use to solve the cubic
    		A11 = -sign(S)*pow((sqrt(Ssquared - Qcubed) + fabs(S)),1/3.);
    
    	//If A11 is zero, then we use this to find the root
    		if (A11 = 0)
    		{
    			x1 = A11 - a1/3.;
    			printf("The cubic has only one real root, %f = x1\n", x1);
    		}
    
    	//If A11 is anything APART from 0 then we use this to find the root
    		else
    		{
    			x1 = A11 + Q/A11 - a1/3.;
    			printf("The cubic has only one real root, %f = x1\n", x1);
    		}
    
    	}
    
    
    
    
    	//Accuracy testing, subbing x1, x2, x3 values into equation
    	//If these equal 0, then we know our solutions are genuine
    	test1 = x1*x1*x1 + a1*x1*x1 + a2*x1 + a3;
    	test2 = x2*x2*x2 + a1*x2*x2 + a2*x2 + a3;
    	test3 = x3*x3*x3 + a1*x3*x3 + a2*x3 + a3;
    
    	printf("Subbing x1, x2, and x3 into the equation yields %f = test1, %f = test2, %f = test3\n", test1, test2, test3);
    	printf("If these are equal to 0, our solutions are genuine\n");
    
    
    
    	return 0;
    
    }
    
    
    //The sign function
    float sign(float S)
    {
    	if (S > 0)
    		return 1;
    	else if (S < 0)
    		return -1;
    	else
    		return 0;
    }

  2. #2
    and the Hat of Guessing tabstop's Avatar
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    If you use %f you will only get six decimal places. Since a2, rounded to six decimal places, is 0.000000 that's what you get. Use %g instead.

  3. #3
    jdi
    jdi is offline
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    thank you

    but sadly it still doesnt work!

    for some reason which i cannot pinpoint im getting my root as "inf" and i dont know why

    i can't figure this out!! its killing me :x

  4. #4
    Registered User
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    Before you check into the morticians:

    Add another = to this:

    Code:
    //If A11 is zero, then we use this to find the root
    	if (A11 = 0)  //should be if(A11==0)

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