JavaScript: Calculate the nth root of a number

Question

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 Math object. Am I overlooking something?
If not...

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

What's the best algorithm to do this myself?

Answer 1

Image: calculate the nth root of a number

To calculate the nth root of a number in JavaScript, you can use exponentiation. To find the nth root of a number x, you can use the following formula: nth root of x = x^(1/n)

power function (review):

Math.pow(base, exponent);

Now, to calculate the nth root of a number x, you can use this function as follows:

function nthRoot(x, n){
    return Math.pow(x, 1 / n);
}
    
let number = 27;
let root = 3;
let result = nthRoot(number, root);
console.log(result);
// result = 27^(1/3) = 3

Another example:

console.log(nthRoot(1024, 5));
// 1024^(1/5) = 4

Answer 2

JavaScript Math object doesn't have a built-in method specifically for calculating the nth root of a number. However, you can still calculate it using the Math.pow() method by taking advantage of the property that the nth root of a number x is equivalent to raising x to the power of 1/n.

For example, to find the cube root (3rd root) of a number x, you can do:

const x = 27;
const n = 3;
const result = Math.pow(x, 1 / n);
console.log(result); // Output: 3

If you are looking for a library that provides more advanced mathematical functions, you can use a popular JavaScript math library called "math.js". This library extends the capabilities of the built-in Math object and provides additional functions like nthRoot().

First, you need to include the library in your HTML file:

<script src="https://cdnjs.cloudflare.com/ajax/libs/mathjs/11.7.0/math.min.js"></script>

Then you can use the nthRoot() function:

const math = require("mathjs");
const x = 125;
const n = 3;
const result = math.nthRoot(x, n);
console.log(result); // Output: 5

Hope this helps!

Answer 3

The ultra short version: const nthroot=(x,root)=>x**(1/root);

Answer 4

const original = 16;

const rootOf = 0.33;
const root = Math.pow(original, 1 / rootOf);
console.log(`${rootOf} Root of ${original} is ${root.toFixed(2)}`);

Answer 5

I have written an algorithm but it is slow when you need many numbers after the point:

https://github.com/am-trouzine/Arithmetic-algorithms-in-different-numeral-systems

NRoot(orginal, nthRoot, base, numbersAfterPoint);

The function returns a string.

E.g.

var original = 1000;
var fourthRoot = NRoot(original, 4, 10, 32);
console.log(fourthRoot); 
//5.62341325190349080394951039776481

Answer 6

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:

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:

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

Example (Imaginary number result):

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

Answer 7

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

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;
    }

Answer 8

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

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

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.

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

Answer 9

The nth root of x is the same as x to the power of 1/n. You can simply use Math.pow:

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

Answer 10

You could use

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

Answer 11

The n-th root of x is a number r such that r to the power of 1/n is x.

In real numbers, there are some subcases:

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

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

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:

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)

Answer 12

Use Math.pow()

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

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){}
}

Answer 13

Can you use something like this?

Math.pow(n, 1/root);

eg.

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