Lista 1 - Cálculo Numérico

Ex1.c

/*
 * Cálculo do zero da função pelo método da bissecção
 *
 * This program is free software: you can redistribute it and/or modify
 * it under the terms of the GNU General Public License as published by
 * the Free Software Foundation, either version 3 of the License, or
 * (at your option) any later version.

 * This program is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU General Public License for more details.
 *
 * You should have received a copy of the GNU General Public License
 * along with this program.  If not, see <http://www.gnu.org/licenses/>.
 *
 * Copyright 2012 - Christian <[email protected]> 
 *
 */
#include <stdio.h>
#include <math.h>

double f ( double );

int main (void) {
    
    double epsilon = 0.000001; // Resíduo aceito
    
    double a = 0.01, b = 0.1;  // Extremos do intervalo
    
    double x1, x2, x3; // Pontos no intervalo
    double f1, f2, f3; // Valores da função no pontos
    double imax = 100; // Limite de iterações
    double delta_x;    // Resíduo a cada tentativa
    int i;
    
    x1 = a;
    x3 = b;
    
    f1 = f(x1);
    f3 = f(x3);

    for (i = 1; i <= imax; i++) {

        x2 = (x1 + x3) / 2.0;
        f2 = f(x2);
         
        if (f1 * f2 <= 0.0) { // A raíz está na primeira metade do intervalo
            
            delta_x = fabs(x2 - x1) / 2.0;
            f3 = f2;
            x3 = x2;
            
        } else { // A raíz está na segunda metade
            
            delta_x = fabs(x3 - x2) / 2.0;
            f1 = f2;
            x1 = x2;
            
        }
        
        if (delta_x < epsilon) {
            printf("A raíz é: %.10f\n", x2);
            printf("Foi encontrada em %d iterações \n", i);
            
            return 0;
        }
    }
    
    return 0;
}

double f (double r) {

    float M = 6000; // Montante após 5 anos
    float n = 5;    // Anos
    float v = 1000; // Investimento anual
    
    return M - v * ( ( 1 + r ) / r ) * (pow( 1 + r, n ) - 1);
}

Ex2.c

/*
 * Cálculo do zero da função pelo método de Newton-Raphson
 *
 * This program is free software: you can redistribute it and/or modify
 * it under the terms of the GNU General Public License as published by
 * the Free Software Foundation, either version 3 of the License, or
 * (at your option) any later version.

 * This program is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU General Public License for more details.
 *
 * You should have received a copy of the GNU General Public License
 * along with this program.  If not, see <http://www.gnu.org/licenses/>.
 *
 * Copyright 2012 - Christian <[email protected]> 
 *
 */
#include <stdio.h>
#include <math.h>

double f ( double );
double f_linha ( double );

int main(void) {
   
    double epsilon = 0.000000000001; // Resíduo aceito
 
    double x_i, x_i1; // Pontos no intervalo
    double delta_x;    // Resíduo a cada tentativa
    
    int imax = 20; // Limite de iterações
    int i = 0;
    
    double x_0 = 0.03; // Chute inicial
    
    x_i = x_0;
    
    for (i = 1; i <= imax; i++) {
        
        x_i1 = x_i - f( x_i ) / f_linha( x_i );

        if( fabs(x_i1 - x_i) < epsilon ) {
        
            printf("A raíz é: %.10f\n", x_i1);
            printf("Foi encontrada em %d iterações \n", i);
            
            return 0;
        }
        
        x_i = x_i1;
        
    }
    
    return 0;
    
}


double f (double r) {

    float M = 6000; // Montante após 5 anos
    float n = 5;    // Anos
    float v = 1000; // Investimento anual
    
    return M - v * ( ( 1 + r ) / r ) * (pow( 1 + r, n ) - 1);
}

double f_linha (double r) { // Derivada analítica de f
    
    return -1000 * (6 * pow( 1+r, 5 ) / r - pow(1 + r, 6) / pow (r, 2) + 1 / pow(r, 2) );
    
}