Mathc initiation/Fichiers c : c24cb
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c16b.c |
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/* --------------------------------- */
/* save as c16b1.c */
/* --------------------------------- */
#include "x_hfile.h"
#include "fb.h"
/* --------------------------------- */
int main(void)
{
double a = -2;
double b = 0;
pt2d p = {a,b}; /* initialize first method */
int n = 5;
clrscrn();
p = i_pt2d(a,b); /* initialize second method */
printf(" Use Newton's method to approximate, \n");
printf(" the solutions of the following system :\n\n\n");
printf(" | %s = 0 \n", feq);
printf(" | %s = 0\n\n\n", geq);
printf(" As a first approximation x = %.1f y = %.1f \n\n", a, b);
stop();
clrscrn();
p_newton_fxy( n, f, g, p);
stop();
clrscrn();
p = newton_fxy( n, f, g, p);
printf(" the solutions of the following system is :\n\n\n");
printf(" x = %f y = %f \n\n\n",p.x,p.y);
printf(" f(%f,%f) = %f \n",p.x,p.y, f(p.x, p.y));
printf(" g(%f,%f) = %f\n\n",p.x,p.y, g(p.x,p.y) );
stop();
return 0;
}
/* --------------------------------- */
/* --------------------------------- */
Voir le fichier x_nwtn.h pour étudier l'algorithme.
Exemple de sortie écran :
Use Newton's method to approximate,
the solutions of the following system :
| (x**2)/4 + (y**2)/9 - 1 = 0
| ((x-1)**2)/10 + ((y+1)**2)/5 - 1 = 0
As a first approximation x = 2.0 y = 1.0
Press return to continue.
Exemple de sortie écran :
x1 =-2.000000 y1 =+0.000000
delta_p.x1 =-0.000000 delta_p.y1 =-0.250000
x2 =-2.000000 y2 =-0.250000
delta_p.x2 =+0.008333 delta_p.y2 =-0.025000
x3 =-1.991667 y3 =-0.275000
delta_p.x3 =+0.000102 delta_p.y3 =-0.000244
x4 =-1.991565 y4 =-0.275244
delta_p.x4 =+0.000000 delta_p.y4 =-0.000000
x5 =-1.991565 y5 =-0.275244
Press return to continue.
Exemple de sortie écran :
the solutions of the following system is :
x = -1.991565 y = -0.275244
f(-1.991565,-0.275244) = -0.000000
g(-1.991565,-0.275244) = 0.000000
Press return to continue.