/*
    newton1_3.c: 関数ポインタ使用バージョン
 */
#include <stdio.h> // fprintf, printf, fgets, sscanf
#include <math.h>  // fabs
double a = 2;

double f(double x)
{
    return x * x - a; // f(x) = x^2 - a
}

double df(double x)
{
    return 2 * x; // f'(x) = 2x
}

double newton(double x, double (*f)(double), double (*df)(double), double eps)
/*
    Solving f(x) = 0 by Newton Method

    x: initial value of x
    f(x): left term of equation f(x)=0
    df(x): derivative of f(x)
    eps: error
*/
{
    int n = 0;
    double x0, err;

    printf("# n, x, err\n");
    printf("%4d, % .15e\n", n, x);
    do {
        n++;
        x0 = x;
        x = x0 - f(x0) / df(x0);
        err = fabs(x - x0);
        printf("%4d, % .15e, % .15e\n", n, x, err);
    } while (err >= eps);
    return x;
}

int main(void)
{
    double x, eps = 1.0e-10;
    char s[128];

    fprintf(stderr, " a = "); fgets(s,128,stdin); sscanf(s, "%lf", &a);
    while (a <= 0.0) {
        fprintf(stderr, "'a' には正の数を入れてください。\n");
        fprintf(stderr, " a = "); fgets(s,128,stdin); sscanf(s, "%lf", &a);
    }

    x = (a + 1.0) / 2.0;

    printf("\n# sqrt(%e) = % .15e\n", a, newton(x, f, df, eps));
    return 0;
}
/*  ---   実行例   ---
Z:\proen1\kadai4>newton1_3
 a = 2
# n, x, err
   0,  1.500000000000000e+000
   1,  1.416666666666667e+000,  8.333333333333326e-002
   2,  1.414215686274510e+000,  2.450980392156854e-003
   3,  1.414213562374690e+000,  2.123899820016817e-006
   4,  1.414213562373095e+000,  1.594724352571575e-012

# sqrt(2.000000e+000) = 1.414213562373095e+000
*/