Using c to solve equations

[zeb1d1ah]

Hello,

I am trying to solve an equation using c programming. Basically I need to write a program that calulates solutions to the equation for a range of values of x, where x >0. I am new to c programming and would appreciate some help. I am trying to use arrays for the x values and a for loop so that it cycles through all the possible scenarios. I then want to print this to a file so that it can be imported to excel.

Thanks for any help,

Z

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
double y_value[100];
double x_value[100];
int n1;
int a;
a=0.4+(8039/25000);
for(n1=0;n1<100;n1++)
{
x_value[n1]=0 + (double)n1*0.02;
y_value[n1]=function(x_value[n1];
}
for(n1=0;n1<10;n1++)
{
printf("%g\t %g\n",x_value[n1],y_value[n1]);
}
return 0;
}


Is what I have so far, am unsure of how to define the function so it gets the result I want

[quzah]

Without knowing what your function 'function' is actually supposed to do, it's hard to really tell you where to go.

double function( double x )
{
    ...do stuff...

    return somedouble;
}

You get to fill in the 'do stuff' part.


Quzah.

[slingerland3g]

Please indent. Also you are missing a paren for your function call. So what are you expecting this function to do against your x_value[] argument you are passing in? What are your requirments. Can you explain a bit more on what it is you are trying to achieve with this function?

[zeb1d1ah]

insert

int main()
{
    double y_value[100];
    double x_value[100];
    int n1;
    int a;
    File*excel data;
    
    a=0.4+(8039/25000);
    for(n1=0;n1<100;n1++)
    {
        x_value[n1]=0 + (double)n1*0.02;
        y_value[n1]=function(x_value[n1];
    }
    for(n1=0;n1<10;n1++)
    {
        printf("%g\t %g\n",x_value[n1],y_value[n1]);
    }
    return 0;
}

Thanks for getting back to beThe function is an equation based on the constant 'a' which I defined and various values of x. I think it should be something like this:

function=sin(pow(x_value[n1],a)-pow(x_value[n1],(1/a))+(a*x_value[n1]));

Any more help appreciated, have tried to indent this time, sorry!

[slingerland3g]

You will need to include the math.h file. Then read what quzah mentioned about how to implement your function. Since you have your algorithm in place assign its value to a variable and just return its value, which happens to be a double.

[zeb1d1ah]

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double result(double );
int main()
{
    double y_value[100];
    double x_value[100];
    int n1;
    int a;

    a=0.4+(8039/25000);

    for(n1=0;n1<100;n1++)
    {
        x_value[n1]=0+(double)n1*0.02;
        y_value[n1]=result(double(x_value[n1]));
    }

    for(n1=0;n1<10;n1++)
    {
        printf("%g\t %g\n",x_value[n1],y_value[n1]);
    }

    double result(double x_value[n1])
    {
        double result;
        result = sin(pow(x_value[n1],a)-pow(x_value[n1],(1/a))+(a*x_value[n1]));
    }   return result;
    return 0;
}

Thanks for the nudge in the right direction guys, still don't have it working, am I going along the right tracks?

[slingerland3g]

Place your result() function outside the main() block of code. Also prior to main() you will need to declare your result() and define its argument types.

#includes....

/*Function declarations */
double result( double x);

int main()
{
   ...
}

/*Your custom functions outside of main() */

double result(double x)
{
   ...
  /* throw in a few printf()s here and there to test your results, good for debugging */
}

[Adak]

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

double result(double );

int main()
{
    double y_value[100];
    double x_value[100];
    int n1;
    int a;

    a=0.4+(8039/25000);

    for(n1=0;n1<100;n1++)
    {
        x_value[n1]=0+(double)n1*0.02;
        y_value[n1]=result(double(x_value[n1]));
    }

    for(n1=0;n1<10;n1++)
    {
        printf("%g\t %g\n",x_value[n1],y_value[n1]);
    }

    return 0;
}

double result(double x_value[n1])
{
    double result;
    result = sin(pow(x_value[n1],a)-pow(x_value[n1],(1/a))+(a*x_value[n1]));
    return result;
}

You can't have a function inside another function. I've moved the result function out from inside the main function, in this code.

Also, the close brace - } needed to be moved to the line *after* the return, instead of being before at the start of that line of code.

You can't return anything after the closing brace.

Personally, I don't let variable names be the same as function names. You should be OK since you have the prototype for the function result(), above main(). Just watch your naming conventions.

[zeb1d1ah]

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double test(double );
int a;
a=0.4+(8039/25000);
int main()
{
    double y_value[100];
    double x_value[100];
    int n1;


    for(n1=0;n1<100;n1++)
    {
        x_value[n1]=0+(double)n1*0.02;
        y_value[n1]=test(x_value[n1]);
    }

    for(n1=0;n1<100;n1++)
    {
        printf("%g\t %g\n",x_value[n1],y_value[n1]);
    }

    return 0;
}

double test(double argument1)
    {
        double result;
        result = sin(pow(argument1,a)-pow(argument1,(1/a))+(a*argument1));
        return result;
    }

Thanks again, This works without any errors but when the printf runs it crashes the programme and doesn't display the results, sorry for being a pain but have only been doing c for a few days! This advice is helping though!

[zalezog]

Are you sure 'a' is supposed to be an int ?

int a;
a=0.4+(8039/25000);

I changed it to double(and 8039.0 / 25000) and it ran without any compile-time or runtime errors.

[BEN10]

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double test(double );
int a;
a=0.4+(8039/25000);
int main()
{
    double y_value[100];
    double x_value[100];
    int n1;


    for(n1=0;n1<100;n1++)
    {
        x_value[n1]=0+(double)n1*0.02;
        y_value[n1]=test(x_value[n1]);
    }

    for(n1=0;n1<100;n1++)
    {
        printf("%g\t %g\n",x_value[n1],y_value[n1]);
    }

    return 0;
}

double test(double argument1)
    {
        double result;
        result = sin(pow(argument1,a)-pow(argument1,(1/a))+(a*argument1));
        return result;
    }

Thanks again, This works without any errors but when the printf runs it crashes the programme and doesn't display the results, sorry for being a pain but have only been doing c for a few days! This advice is helping though!

Look what you're doing is

int a;
a=0.4+(8039/25000);

Here 'a' will be zero coz 8039/25000 is zero(since both of them are int) and 0.4 added to 0 will be 0.4 but since you're declaring a as an int thus a=integer part of 0.4=0. Now in test function you're doing 1/a which is division by zero.
The solution would be to declare a as float and do 8039.0/25000.

[zeb1d1ah]

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double test(double );
double a;
a=0.4+(8039.0/25000);
int main()
{
    double y_value[100];
    double x_value[100];
    int n1;


    for(n1=0;n1<100;n1++)
    {
        x_value[n1]=0+(double)n1*0.02;
        y_value[n1]=test(x_value[n1]);
    }

    for(n1=0;n1<100;n1++)
    {
        printf("%g\t %g\n",x_value[n1],y_value[n1]);
    }

    return 0;
}

double test(double argument1)
    {
        double result;
        result = sin(pow(argument1,a)-pow(argument1,(1/a))+(a*argument1));
        return result;
    }

Guys, thanks for the feedback and advice, I still can't get this to do what I want. I am trying to produce a list of 'x_values' and the corresponding answer from 'result = ' in a table, so that I can produce a text file and transfer to excel.

Any more guidance greatly appreciated as I am a bit stumped!

[Adak]

Why not drop the size of the arrays (x_value & y_value), down to like 2, and then step through it with several printf() statements added if needed, and see where the numbers go flooey?

As small a program as this is, the problem can barely run, let alone hide.

[iMalc]

You can't put the assignment to 'a' outside of a function like that.
If you write it as:

const double a = ....;

it will then be evaulated during compilation and is allowed to be anywhere.

[zeb1d1ah]

I finally got this working, many thanks to those who contributed! My understanding of c is starting to increase!

Thanks,

Z