处理虚幻的数字，跳起来抓住

Question

我有这段代码应该使用正割法计算 f(x) = ln(x+1)+1 的根。
输入 xold1 = 0; xold2 = 1;

do {
        try {
            iteration++;
            fxold1 = Math.log(xold1+1)+1;
            fxold2 = Math.log(xold2+1)+1;
            xnew = xold2 - (fxold2 * (xold2 - xold1))/(fxold2 - fxold1);

            //Show iterations and results       
            System.out.println("Iteration: " + iteration + "; x = " + xnew);

            diff = Math.abs(xnew-xold1);

            //Replace old variables with new ones
            xold2 = xold1;
            xold1 = xnew; 
        } catch(Exception e) {
            System.out.println("No solution for this starting point.");                
        }
    } while(diff > 0.00001);

输出：

  Iteration: 1; x = -1.4426950408889634
  Iteration: 2; x = NaN

在纸上做数学运算，第二次迭代给出一个虚数：0.185125859 + 3.14159265 i。所以，这个想法是程序应该跳起来捕捉。为什么它没有这样做，我应该怎么做呢？谢谢！

Answer 1

通过查看 the docs

很容易回答这个问题

If the argument is NaN or less than zero, then the result is NaN.

如果您给它一个否定参数，

Math.log 不会抛出异常。它returnsNaN。您应该 check for that 而不是试图捕获异常。

xnew = xold2 - (fxold2 * (xold2 - xold1))/(fxold2 - fxold1);
if(Double.isNaN(xnew)){
    System.out.println("No solution for this starting point.");
    break;
}

处理虚幻的数字，跳起来抓住

handle unreal numbers, jump to catch

java

math

try-catch