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Nth Root Calculation

The principal nth root ${\sqrt[{n}]{A}}$ of a positive real number A, is the positive real solution of the equation

$x^{n}=A$

(for integer n there are n distinct complex solutions to this equation if $A>0$, but only one is positive and real).

There is a very fast-converging nth root algorithm for finding ${\sqrt[{n}]{A}}$:

1. Make an initial guess $x_{0}$
2. Set $x_{k+1}={\frac {1}{n}}\left[{(n-1)x_{k}+{\frac {A}{x_{k}^{n-1}}}}\right]$
3. Repeat step 2 until the desired precision is reached.

A special case is the familiar square-root algorithm. By setting n = 2, the iteration rule in step 2 becomes the square root iteration rule:

$x_{k+1}={\frac {1}{2}}\left(x_{k}+{\frac {A}{x_{k}}}\right)$

Several different derivations of this algorithm are possible. One derivation shows it is a special case of Newton's method (also called the Newton-Raphson method) for finding zeros of a function $f(x)$ beginning with an initial guess. Although Newton's method is iterative, meaning it approaches the solution through a series of increasingly accurate guesses, it converges very quickly. The rate of convergence is quadratic, meaning roughly that the number of bits of accuracy doubles on each iteration (so improving a guess from 1 bit to 64 bits of precision requires only 6 iterations). For this reason, this algorithm is often used in computers as a very fast method to calculate square roots.

For large n, the n^th root algorithm is somewhat less efficient since it requires the computation of $x_{k}^{n-1}$ at each step, but can be efficiently implemented with a good exponentiation algorithm.

Rough estimation

Many of the methods for calculating square roots of a positive real number S require an initial seed value. If the initial value is too far from the actual square root, the calculation will be slowed down. It is therefore useful to have a rough estimate, which may be very inaccurate but easy to calculate. If S ≥ 1, let D be the number of digits to the left of the decimal point. If S < 1, let D be the negative of the number of zeros to the immediate right of the decimal point. Then the rough estimation is this:

If D is odd, D = 2n + 1, then use ${\sqrt {S}}\approx 2\cdot 10^{n}.$

If D is even, D = 2n + 2, then use ${\sqrt {S}}\approx 6\cdot 10^{n}.$

Two and six are used because they approximate the geometric means of the lowest and highest possible values with the given number of digits: ${\sqrt {\sqrt {1\cdot 10}}}={\sqrt[{4}]{10}}\approx 2\,$ and ${\sqrt {\sqrt {10\cdot 100}}}={\sqrt[{4}]{1000}}\approx 6\,.$

When working in the binary numeral system (as computers do internally), an alternative method is to use $2^{\left\lfloor D/2\right\rfloor }$ (here D is the number of binary digits).