Mathc initiation/Fichiers c : c22cd1
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c3d1.c |
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/* --------------------------------- */
/* save as c3d1.c */
/* --------------------------------- */
#include "x_hfile.h"
#include "fd.h"
/* --------------------------------- */
int main(void)
{
int n = 6;
double FirstApproximation = 1.5;
clrscrn();
printf(" The Newton's method : \n"
" f(x_n) \n"
" x_n+1 = x_n - ------- \n"
" f'(x_n) \n"
"\n\n\n\n\n");
printf(" Use Newton's method to approximate \n"
" the largest positive real root of :\n\n"
" f : x-> %s\n\n\n"
" On a graph of f, you can see that, the largest\n"
" positive real root of f is between 1.0 and 2.0.\n\n"
" Choose x = %.1f as a first approximation.\n\n",
feq,FirstApproximation);
stop();
clrscrn();
printf(" The Newton's method : \n"
" f(x_n) \n"
" x_n+1 = x_n - ------- \n"
" f'(x_n) \n"
"\n\n\n");
printf(" As a first approximation x = %.1f \n\n"
" The Newton's method give : \n\n",FirstApproximation);
p_Newton_s_Method(FirstApproximation, n, f, Df);
printf(" f(%.15f) = %.15f\n\n",
Newton_s_Method(FirstApproximation, n, f, Df)
,f(Newton_s_Method(FirstApproximation, n, f, Df)));
stop();
return 0;
}
/* ---------------------------------- */
/* ---------------------------------- */
Fichier de commande gnuplot :
# ---------------------
# Copy and past this file into the screen of gnuplot
#
#
set zeroaxis lt 3 lw 1
plot [-2.5:2.5] [-6.:6.]\
x**3 - 3.0*x + 1.0
reset
# ---------------------
Nous pouvons observer que la plus grande racine est comprise entre 1 et 2. Il faut calculer la racine de cette équation x**3 - 3.0*x + 1.0 = 0. On va donc choisir comme première approximation 1.5. On arrêtera les calculs lorsque l'on aura deux valeurs identiques.
Exemple de sortie écran :
The Newton's method :
f(x_n)
x_n+1 = x_n - -------
f'(x_n)
Use Newton's method to approximate
the largest positive real root of :
f : x-> x**3 - 3.0*x + 1.0
On a graph of f, you can see that, the largest
positive real root of f is between 1.0 and 2.0.
Choose x = 1.5 as a first approximation.
Press return to continue.
*********************
The Newton's method :
f(x_n)
x_n+1 = x_n - -------
f'(x_n)
As a first approximation x = 1.5
The Newton's method give :
x[1] = 1.500000000000000
x[2] = 1.533333333333333
x[3] = 1.532090643274854
x[4] = 1.532088886241467
x[5] = 1.532088886237956
x[6] = 1.532088886237956
f(1.532088886237956) = -0.000000000000000
Press return to continue.