Bahasa Pemrograman C++ - Metode Terbuka #1
#include <stdio.h>
#include <conio.h>
#include <math.h>
main()
{
float konvergen, f, ft1, ft2, xold, xnew, e, error;
int kondisi=1,i;
mulai:
puts ("\nPersamaan yang akan diselesaikan adalah e^x-3x^3 ");
printf ("\nMasukkan error : ");scanf("%f",&e);
printf ("\nMasukkan Taksiran Nilai Akar awal : ");scanf("%f",&xold);
puts
("=============================================================================================");
f=exp(xold) - (3*(xold * xold * xold));
ft1 = exp(xold)-(9*(xold));
ft2 = exp(xold)-9;
printf ("\nnilai fungsi awal f(%f) = %.4f", xold, f);
printf ("\nnilai fungsi turunan satu awal f(%f) = %.4f", xold, ft1);
printf ("\nnilai fungsi turunan dua awal f(%f) = %.4f", xold, ft2);
konvergen = (f*ft2)/((ft1*ft1));
printf ("\nnilai konvergen = %.4f", konvergen);
if ((f!=0) && (ft1 !=0) && (fabs(konvergen)< 1)) {
puts("Konvergen"); }
else {
puts ("\n Taksiran Akar awal salah, silahkan diulangi");
goto mulai; }
kondisi = 1;
i=1;
puts
("=============================================================================================");
puts ("i xold f(x) f'(x) xnew abs(fx)");
while (kondisi)
{
f=exp(xold)-(3*(xold * xold * xold));
ft1 = exp(xold)-(9*(xold));
xnew = xold-(f/ft1);
error=fabs(f);
printf("%3i %5.4f %9.4f %9.4f %9.4f %9.4f\n", i, xold, f, ft1, xnew, error);
if (error>=e)
{
xold=xnew;
}
else{
kondisi=0;
break; }
i++;
}
puts
("=============================================================================================");
printf ("\nHasil Akar = %.4f", xnew);
getch();
}
