equations.cpp

/*
    Методы решения нелинейных уравнений:
        1.  Ньютона
        2.  Бисекций
        3.  Ппростых итераций
*/

#include <iostream>
#include <cmath>
#include <cstdio>

using namespace std;

//  ln(x+1)-2x^2+1=0
//  ln(x+1)-2x^2+x+1=x
//  ln(x+1)-(x-1)(2x+1) = x

// 2x^2-x-1=0
// D=1+8 = 9 = 3^2
// x1=(1+3)/4 = 1
// x2=(1-3)/4 = -1/2
// 2x^2-x-1=2(x-1)(x+1/2)=(x-1)(2x+1)

//  0.5*(log(x+1)+1)/x=x

inline double f(double x) {
//  f(x) = 0
    return log(x+1)-2*x*x+1;
}

inline double df(double x) {
//  f(x) = 0
    return 1/fabs(x+1)-4*x;
}

inline double g(double x) {
//  x = g(x)
    return 0.5*(log(x+1)+1)/x;
}

double solve_bisection(double (*f)(double x), double a, double b, double eps, int limit) {
    cout << "\tMethod of bisections:" << endl;
    double x = 0.5*(a+b);
    int count = 0;
    while ( fabs( (*f)(x) ) > eps && count < limit) {
        printf("%d:\t x = %0.9lf \t f(x) = %0.9lf\n", count, x, (*f)(x));
        if ( (*f)(a)*(*f)(x) < 0 )
            b = x;
        else
            a = x;
        x = 0.5*(a+b);
        ++count;
    }
    cout << endl;
    return x;
}

double solve_iteration(double (*f)(double x), double eps, int limit) {
    cout << "\tMethod of simple iterations:" << endl;
    double x = 1, y = (*f)(x);
    int count = 0;
    while (fabs(x - y) >= eps && count < limit) {
        printf("%d:\t x = %0.9lf \t f(x) = %0.9lf\n", count, x, x-(*f)(x));
        x = y;
        y = (*f)(x);
        count++;
    }
    cout << endl;
    return x;
}

double solve_neuton(double (*f)(double x), double (*df)(double x), double eps, double limit) {
    cout << "\tMethod of Neuton:" << endl;
    double x = 10, y = (*f)(x)/(*df)(x);
    int count = 0;
    while (fabs((*f)(x)) >= eps && count < limit) {
        printf("%d:\t x = %0.9lf \t f(x) = %0.9lf\n", count, x, (*f)(x));
        x -= y;
        y = (*f)(x)/(*df)(x);
        count++;
    }
    cout << endl;
    return x;
}

int main() {
    double eps = 10e-9;
    solve_bisection(f, 0.1, 10, eps, 1000);
    solve_iteration(g, eps, 1000);
    solve_neuton(f, df, eps, 1000);
    return 0;
}