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kontsev_
Asked:2022-04-24 00:55:31 +0000 UTC2022-04-24 00:55:31 +0000 UTC

切线法

  • 772

程序给出-nan，错误是什么？

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
double f(double x) {
    return (x * x - 2);
}
double f1(double x) {
    double h = 0.01;
    return ((f(x+h) - f(x-h))/2*h);
}
double f2(double x) {
    double h = 0.01, s;
    s = (f(x+h) - 2*f(x) + f(x-h))/h*h;
    return s;
}
double newton (double a, double b, double eps) {
    double xold, xnew, c;
    c = (a + b)/2;
        if(f1(c)*f2(c) > 0)
            xnew = a;
        else
            xnew = b;
        do {
            xold = xnew;
            xnew = xold - f(xold)/f1(xold);
        } while(fabs(xold - xnew) > eps);
        return xnew;
}
int main () {
    double a, b, eps;
    puts("eps a b");
    scanf ("%lf %lf %lf", &eps, &a, &b);
    printf ("%lf\n", newton(a, b, eps));
}
c

1 个回答

  1. Best Answer

    从学习语言的基础开始。

    ((f(x+h) - f(x-h))/2*h)

    表示 - 将差值除以 2并乘以h。

    而这个表达

    (f(x+h) - 2*f(x) + f(x-h))/h*h

    基本一样

    (f(x+h) - 2*f(x) + f(x-h))

    由于不正确的导数（顺便说一句，为什么不直接采用2*x？:)）方法发散......

    • 5

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