JavaScript：计算数字的第 n 个根

JavaScript: Calculate the nth root of a number

提问by Nathan

I'm trying to get the nth root of a number using JavaScript, but I don't see a way to do it using the built in Mathobject. Am I overlooking something?
If not...

我正在尝试使用 JavaScript 获取数字的第 n 个根，但我看不到使用内置Math对象的方法。我是否忽略了什么？
如果不...

Is there a math library I can use that has this functionality?
If not...

我可以使用具有此功能的数学库吗？
如果不...

What's the best algorithm to do this myself?

自己做这件事的最佳算法是什么？

回答by Digital Plane

Can you use something like this?

你能用这样的东西吗？

Math.pow(n, 1/root);

eg.

例如。

Math.pow(25, 1/2) == 5

回答by Facebook Staff are Complicit

The nth root of xis the same as xto the power of 1/n. You can simply use Math.pow:

的nth 根与 的幂x相同。您可以简单地使用：x1/nMath.pow

var original = 1000;
var fourthRoot = Math.pow(original, 1/4);
original == Math.pow(fourthRoot, 4); // (ignoring floating-point error)

回答by mplungjan

Use Math.pow()

使用 Math.pow()

Note that it does not handle negative nicely - here is a discussion and some code that does

请注意，它不能很好地处理负面影响 - 这是一个讨论和一些代码

http://cwestblog.com/2011/05/06/cube-root-an-beyond/

http://cwestblog.com/2011/05/06/cube-root-an-beyond/

function nthroot(x, n) {
  try {
    var negate = n % 2 == 1 && x < 0;
    if(negate)
      x = -x;
    var possible = Math.pow(x, 1 / n);
    n = Math.pow(possible, n);
    if(Math.abs(x - n) < 1 && (x > 0 == n > 0))
      return negate ? -possible : possible;
  } catch(e){}
}

回答by Somebody

You could use

你可以用

Math.nthroot = function(x,n) {
    //if x is negative function returns NaN
    return this.exp((1/n)*this.log(x));
}
//call using Math.nthroot();

回答by GOTO 0

For the special cases of square and cubic root, it's best to use the native functions Math.sqrtand Math.cbrtrespectively.

对于平方根和立方根的特殊情况，最好分别使用原生函数Math.sqrt和Math.cbrt。

As of ES7, the exponentiation operator **can be used to calculate the nth root as the ¹/_nth power of a non-negative base:

从 ES7 开始，幂运算符**可用于计算作为非负底数的¹/ _n次幂的n次方根：

let root1 = Math.PI ** (1 / 3); // cube root of π

let root2 = 81 ** 0.25;         // 4th root of 81

This doesn't work with negative bases, though.

但是，这不适用于负基数。

let root3 = (-32) ** 5;         // NaN

回答by Oriol

The n-th root of xis a number rsuch that rto the power of 1/nis x.

的n-th 根x是一个数r，其次方r为1/nis x。

In real numbers, there are some subcases:

在实数中，有一些子情况：

• There are two solutions (same value with opposite sign) when xis positive and ris even.
• There is one positive solution when xis positive and ris odd.
• There is one negative solution when xis negative and ris odd.
• There is no solution when xis negative and ris even.

• 当x为正数和r偶数时，有两种解（同值反号）。
• 当x为正且r为奇数时，有一个正解。
• 当x为负且r为奇数时，有一个负解。
• 当x为负且r为偶数时无解。

Since Math.powdoesn't like a negative base with a non-integer exponent, you can use

由于Math.pow不喜欢具有非整数指数的负底，您可以使用

function nthRoot(x, n) {
  if(x < 0 && n%2 != 1) return NaN; // Not well defined
  return (x < 0 ? -1 : 1) * Math.pow(Math.abs(x), 1/n);
}

Examples:

例子：

nthRoot(+4, 2); // 2 (the positive is chosen, but -2 is a solution too)
nthRoot(+8, 3); // 2 (this is the only solution)
nthRoot(-8, 3); // -2 (this is the only solution)
nthRoot(-4, 2); // NaN (there is no solution)

回答by SwiftNinjaPro

Here's a function that tries to return the imaginary number. It also checks for a few common things first, ex: if getting square root of 0 or 1, or getting 0th root of number x

这是一个尝试返回虚数的函数。它还首先检查一些常见的事情，例如：如果得到 0 或 1 的平方根，或者得到数字 x 的第 0 个根

function root(x, n){
        if(x == 1){
          return 1;
        }else if(x == 0 && n > 0){
          return 0;
        }else if(x == 0 && n < 0){
          return Infinity;
        }else if(n == 1){
          return x;
        }else if(n == 0 && x > 1){
          return Infinity;
        }else if(n == 0 && x == 1){
          return 1;
        }else if(n == 0 && x < 1 && x > -1){
          return 0;
        }else if(n == 0){
          return NaN;
        }
        var result = false;
        var num = x;
        var neg = false;
        if(num < 0){
            //not using Math.abs because I need the function to remember if the number was positive or negative
            num = num*-1;
            neg = true;
        }
        if(n == 2){
            //better to use square root if we can
            result = Math.sqrt(num);
        }else if(n == 3){
            //better to use cube root if we can
            result = Math.cbrt(num);
        }else if(n > 3){
            //the method Digital Plane suggested
            result = Math.pow(num, 1/n);
        }else if(n < 0){
            //the method Digital Plane suggested
            result = Math.pow(num, 1/n);
        }
        if(neg && n == 2){
            //if square root, you can just add the imaginary number "i=√-1" to a string answer
            //you should check if the functions return value contains i, before continuing any calculations
            result += 'i';
        }else if(neg && n % 2 !== 0 && n > 0){
            //if the nth root is an odd number, you don't get an imaginary number
            //neg*neg=pos, but neg*neg*neg=neg
            //so you can simply make an odd nth root of a negative number, a negative number
            result = result*-1;
        }else if(neg){
            //if the nth root is an even number that is not 2, things get more complex
            //if someone wants to calculate this further, they can
            //i'm just going to stop at *n√-1 (times the nth root of -1)
            //you should also check if the functions return value contains * or √, before continuing any calculations
            result += '*'+n+√+'-1';
        }
        return result;
    }

回答by Zamy Arkre Nendmed

Well, I know this is an old question. But, based on SwiftNinjaPro's answer, I simplified the function and fixed some NaN issues. Note: This function used ES6 feature, arrow function and template strings, and exponentation. So, it might not work in older browsers:

好吧，我知道这是一个老问题。但是，根据 SwiftNinjaPro 的回答，我简化了函数并修复了一些 NaN 问题。注意：该函数使用了 ES6 特性、箭头函数和模板字符串以及指数。因此，它可能不适用于旧浏览器：

Math.numberRoot = (x, n) => {
  return (((x > 1 || x < -1) && n == 0) ? Infinity : ((x > 0 || x < 0) && n == 0) ? 1 : (x < 0 && n % 2 == 0) ? `${((x < 0 ? -x : x) ** (1 / n))}${"i"}` : (n == 3 && x < 0) ? -Math.cbrt(-x) : (x < 0) ? -((x < 0 ? -x : x) ** (1 / n)) : (n == 3 && x > 0 ? Math.cbrt(x) : (x < 0 ? -x : x) ** (1 / n)));
};

Example:

例子：

Math.numberRoot(-64, 3); // Returns -4

Example (Imaginary number result):

示例（虚数结果）：

Math.numberRoot(-729, 6); // Returns a string containing "3i".