工作递归巴比伦平方根，需要合并错误。

Question

我已经创建了一个有效的递归巴比伦平方根方法，但我想加入一个错误。我希望程序在测试编号为 + 或 - 实数平方根误差时停止。在这种情况下，程序将在 5.015... 处停止，因为 5.015 在实数平方根 (5) 的 0.1 范围内。

public class BabyRoot {
private static double testnum = 0;
private static double targetnum = 0;
private static double error = 0;

public static void babyRoot(double num) {
    testnum = 0.5 * (testnum + (targetnum / testnum));
    System.out.println("Test number equals: " + testnum);
    if ((testnum * testnum) == targetnum)
        System.out.println("The correct square root is: " + testnum);
    else
        babyRoot(testnum);
}

public static void main(String[] args) {
    error = 0.1;
    testnum = 25;
    targetnum = testnum;
    babyRoot(testnum);
}

}

输出：

Test number equals: 13.0

Test number equals: 7.461538461538462

Test number equals: 5.406026962727994

Test number equals: 5.015247601944898

Test number equals: 5.000023178253949

Test number equals: 5.000000000053722

Test number equals: 5.0

The correct square root is: 5.0

Answer 1

您需要更改 if 语句以检查数字是否在 targetnum-error 和 targetnum+error 范围内：

public static void babyRoot(double num , double err)
{
    testnum = 0.5 * (testnum + (targetnum / testnum));
    System.out.println("Test number equals: " + testnum);

    if ((testnum >= (Math.sqrt(targetnum) - err)) &&
        (testnum <= (Math.sqrt(targetnum) + err)))
        System.out.println("The correct square root is: " + testnum);
    else
        babyRoot(testnum);
}

工作递归巴比伦平方根，需要合并错误。

Working recursive babylonian square root, need to incorporate an error.

java

recursion

square-root