Nth root algorithm

$x^n\; =\; A$

(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:

$x\_\{k+1\}\; =\; frac\{1\}\{2\}left(x\_k\; +\; frac\{A\}\{x\_k\}right)$

In the case of n = 1/2 the rule also simplifies and becomes:

$x\_\{k+1\}\; =\; frac\{2\; x\_k\}\; \{1\; +\; A\; x^2\_k\}$

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.

Derivation from Newton's method

Newton's method is a method for finding a zero of a function f(x). The general iteration scheme is:

1. Make an initial guess $x\_0$
2. Set $x\_\{k+1\}\; =\; x\_k\; -\; frac\{f(x\_k)\}\{f\text{'}(x\_k)\}$
3. Repeat step 2 until the desired precision is reached.

The n^th root problem can be viewed as searching for a zero of the function

$f(x)\; =\; x^n\; -\; A$

So the derivative is

$f^prime(x)\; =\; n\; x^\{n-1\}$

and the iteration rule is

$x\_\{k+1\}\; =\; x\_k\; -\; frac\{f(x\_k)\}\{f\text{'}(x\_k)\}$

$=\; x\_k\; -\; frac\{x\_k^n\; -\; A\}\{n\; x\_k^\{n-1\}\}$

$=\; x\_k\; -\; frac\{x\_k\}\{n\}+frac\{A\}\{n\; x\_k^\{n-1\}\}$

$=\; frac\{1\}\{n\}\; left[\{(n-1)x\_k\; +frac\{A\}\{x\_k^\{n-1\}\}\}right]$

leading to the general n^th root algorithm.

References

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