Linear Algebra and the C Language/a089

Install and compile this file in your working directory.

/* ------------------------------------ */
/*  Save as :   c00b.c                  */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
int main(void)
{
int     n = 1;

double x1 = +0.50000000;
double x2 = +1.00000000;
double x3 = +0.50000000;
double x4 = +1.00000000;	
    
    do{ 
	   clrscrn();
	   printf(" Continue until integers appear : \n\n\n");	 
	         
	   printf(" n = %d -> n*x1 = %0.4f  \n",n, n*x1);	
	   printf(" n = %d -> n*x2 = %0.4f  \n",n, n*x2);	
	   printf(" n = %d -> n*x3 = %0.4f  \n",n, n*x3);		   
	   printf(" n = %d -> n*x4 = %0.4f  \n",n, n*x4);	   
	   	   
	   n++;
	   
    }while(stop_w());

    return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */

• With n = 2

n = 2 -> n*x1 = 1.0000  
n = 2 -> n*x2 = 2.0000  
n = 2 -> n*x3 = 1.0000  
n = 2 -> n*x4 = 2.0000  

This gives us the coefficients of the equation:

${\displaystyle \mathrm {x_{1}CH_{4}+x_{2}O_{2}\longrightarrow x_{3}CO_{2}+x_{4}H_{2}O} }$

“Simply replace x1, x2,... with their numerical values:

${\displaystyle \mathrm {CH_{4}+2O_{2}\longrightarrow CO_{2}+2H_{2}O} }$

Screen output example:

 Continue until integers appear : 


 n = 1 -> n*x1 = 0.5000  
 n = 1 -> n*x2 = 1.0000  
 n = 1 -> n*x3 = 0.5000  
 n = 1 -> n*x4 = 1.0000  

 Press   return to continue
 Press X return to stop    


 Continue until integers appear : 


 n = 2 -> n*x1 = 1.0000  
 n = 2 -> n*x2 = 2.0000  
 n = 2 -> n*x3 = 1.0000  
 n = 2 -> n*x4 = 2.0000  

 Press   return to continue
 Press X return to stop