有没有比 java.lang.Math.sqrt(double) 更准确的平方根函数

Question

I am looking for square root functions (to use in java) that can give square roots upto atleast to 5-6 decimal places accurately. If there is a way to control accuracy of java.lang.Math.sqrt(double) or any external math library, please mention it.

Answer 1

What problem are you trying to solve? I believe java.lang.Math's sqrt is supposed to be accurate to the full double width . Do you need more than this?

Answer 2

Try out the code at this link, which uses Newton's method to compute square roots for large numbers (and gets BigDecimal output) to see if it fits your needs. If not, you can modify it :) .

http://www.merriampark.com/bigsqrt.htm

Answer 3

You can use this formula to get it to an arbitrary accuracy:

http://en.wikipedia.org/wiki/Newton's_method#Square_root_of_a_number

I doubt that it will be particularly fast though. You'd also need a way to store numbers with larger than type(double) accuracy.

Answer 4

You can always implement your own using Newton's Method . Then you can make it as accurate as you want (at the cost of CPU of course).

Numerical Recipes in C has code and in-depth discussion. However make sure you check the license before using their code in your product.

Answer 5

You could roll your own using BigDecimal and a method like Newton-Raphson. That would allow you to specify a precision.

Have a look at Numerical Recipes if you need any inspiration.

Answer 6

Math.sqrt

public static double sqrt(double a)

Returns the correctly rounded positive square root of a double value . Special cases:

• If the argument is NaN or less than zero, then the result is NaN.
• If the argument is positive infinity, then the result is positive infinity.
• If the argument is positive zero or negative zero, then the result is the same as the argument.

Otherwise, the result is the double value closest to the true mathematical square root of the argument value.

Parameters:

• a - a value.

Returns:

• the positive square root of a. If the argument is NaN or less than zero, the result is NaN.

Answer 7

This question has been answered thoroughly enough already, but here is an experimental way to verify that Math.sqrt is accurate:

import static java.lang.Math.*;
public class Test {
  public static void main(String[] args) {
    double max = 0;
    for (int i = 0; i < 100; i++) {
      double r = random();
      double err = abs(pow(sqrt(r), 2) - r) / ulp(r);
      if (err > max) max = err;
    }
    System.out.println(max);
  }
}

This prints out 1.0 , confirming what the documentation says — that the value returned from sqrt will be within one unit of precision to the exact answer.

Answer 8

public class SquareRoot {

    public void sqRoot_Efficient(int n){    //Time Complexity: O(log(n))
        for(int i=1;i<=n;i++){
            if((n/i)*(n/i) < n){
                System.out.println(n/i);
                break;
            }
        }
    }
    
    public static void main(String[] args) {
        SquareRoot sr=new SquareRoot();
        sr.sqRoot_Efficient(27);
    }
}

Answer 9

我通常建议的Apache公地作为第一个地方去寻找大多数的任务。但它出现在commons.math不包括开方（）。