JavaScript: compute the nth root of a number

Question

I am trying to get the nth root from a number using JavaScript, but I see no way to do this using the built-in Math object. Am I not noticing something?
If not...

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

What is the best algorithm for this?

Answer 1

Can you use something like this?

 Math.pow(n, 1/root); 

eg.

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

Answer 2

The root n th of x coincides with x degree 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 3

Use Math.pow ()

Note that it does not handle a negative result - 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 4

you can 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 5

n 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 even.
• There is one positive solution when x positive and r is odd.
• There is one negative solution when x negative and r is odd.
• There is no solution when x negative and r is even.

Since Math.pow does not like a negative base with a non-integer metric, 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 6

For special cases of square and cubic roots, it is better to use the eigenfunctions Math.sqrt and Math.cbrt respectively.

In 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 does not work with negative bases.

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

Answer 7

Here is a function that tries to return an imaginary number. At first, he also checks a few common things, for example: getting the square root of 0 or 1 or getting the 0th root of 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

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 the ES6 function, the arrow function, pattern strings and exponent. Thus, it may 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 result result):

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