摘要: 以后在这里记下写的代码了。祝我以坚持。牛顿迭代法求非线性方程的根,迭代很简单:xn+1=xn-f(xn)/f'(xn),实现代码也很简单,如下:#include #include#includedouble f(double x);double df(double x);void main(void... 阅读全文
posted @ 2015-08-23 18:16 scian102 阅读(614) 评论(1) 推荐(0)