n-th Root Algorithm

问题

What is the fastest way to calculate the n-th root of a number?

I'm aware of the Try and Fail method, but I need a faster algorithm.

回答1:

The canonical way to do this is Newton's Method. In case you don't know, the derivative of xⁿ is nx^n-1. This will come in handy. 1 is a good first guess. You want to apply it to the function a - xⁿ

IIRC, it's superconvergent on functions of the form a - xⁿ, but either way, it's quite fast. Also, IIRC, the warning in the wiki about it failing to converge would apply to more complex functions that have properties that the 'nice' functions you are interested in lack.

回答2:

Not the fastest, but it works. Substitute your chosen type:

    private static decimal NthRoot(decimal baseValue, int N)
    {
        if (N == 1)
            return baseValue;
        decimal deltaX;
        decimal x = 0.1M;
        do
        {
            deltaX = (baseValue / Pow(x, N - 1) - x) / N;
            x = x + deltaX;
        } while (Math.Abs(deltaX) > 0);
        return x;
    }

    private static decimal Pow(decimal baseValue, int N)
    {
        for (int i = 0; i < N - 1; i++)
            baseValue *= baseValue;
        return baseValue;
    }

回答3:

Are you referring to the nth root algorithm ? This is not a try-and-fail method, but an iterative algorithm which is repeated until the required precision is reached.

回答4:

Extracting n'th root using vedic mathematics is fastest method. For more detail about algorithm refer http://www.vedicmaths.org/Journal.asp

来源：https://stackoverflow.com/questions/3848640/n-th-root-algorithm