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?
29 Answers
Can you use something like this?
Math.pow(n, 1/root);eg.
Math.pow(25, 1/2) == 5The 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)Use Math.pow()
Note that it does not handle negative nicely - here is a discussion and some code that does
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){}
}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();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
xis positive andris even. - There is one positive solution when
xis positive andris odd. - There is one negative solution when
xis negative andris odd. - There is no solution when
xis negative andris 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)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 81This doesn't work with negative bases, though.
let root3 = (-32) ** 5; // NaNWell, 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 -4Example (Imaginary number result):
Math.numberRoot(-729, 6); // Returns a string containing "3i".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; }I have written an algorithm but it is slow when you need many numbers after the point:
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
