Другие алгоритмы.

Математика: Быстрое вычисление функций и констант.

Точное вычисление обратного числа, квадратного корня

Простейший классический алгоритм вычисления квадратного корня представлен здесь. Более серьезно вопрос рассмотрен в довольно простой и конкретной статье:

High precision inverse and square roots

Inverting a number of size n can be done in time O(n²) with the classical school division. This time cost can be reduced to O(nlog(n)) when using a fast multiplication, thanks to a Newton process. The same type of techniques apply to the n-th root computation of a number. For those operations we also describe some very efficient algorithms with high order of convergence (that is quartic, quintic, sextic, ... algorithms).

1  High precision inversion

In this section we want to compute the inverse 1/A of the number A to a great accuracy.

1.1  Newton's iteration

Starting from an approximation x₀ of 1/A, we apply Newton iteration [1] to the function f(x) = 1/x-A, which gives the approximating sequence

The iteration (*) have a nice feature : the only needed operations are the product by 2, the difference and the product of high precision numbers. The Newton convergence being quadratic, the precision needed to compute iteration (*) is not the full final precision but only with the precision wanted at step n.

As an example, we compute the inverse of p, starting from the approximation x₀ = 0.31831 (5 digits). The first three iterations become

The first 38 digits of x₃ agree with those of 1/p.

The cost of this process is of the order of the multiplication cost : since each iteration involves two multiplications on high precision numbers, to find 1/A we need to compute 2 multiplications of size n (last step), 2 multiplications of size n/2 (previous step), 2 multiplications of size n/4 , etc. Using for example an FFT multiplication, with cost O(nlogn), the final cost is of order

Note : a better iteration is obtained by writing (*) in the form

A division B/A is performed by multiplying B by the inverse of A.

1.2  A cubic iteration

From the general Householder's iteration [2]

applied on f(x) = 1/x-A, we find

Starting with a good initial point x₀, the convergence is cubical, that is, the number of good digits is multiplied by 3 at each step of the process. (The second form of this iteration is given in [3]). Since h_n tends to zero, the new term h_n² can be computed at a low cost.

As an example, we compute again, the inverse of p, starting from the approximation x₀ = 0.31831 (5 digits). The two first iterations are

and x₃has 174 good digits...

1.3  High order iterations

   Quartic iteration  

It's possible to find a quartic iteration, just apply twice Newton's iteration on f(x) = 1/x-A, we find

If we take a look at our example (inverse of p), starting as usual with x₀ = 0.31831, we find that x₁ has 26 good digits, x₂ is 103 digits exact, x₃ has 413 good digits ... The sequence of correct digits is then for this example

   Quintic iteration  

A quintic iteration founded by the application of a general quintic modified iteration (here, we omit the details of the demonstration which is deduced from the essay on Newton's methods) is given by

the rate of convergence is impressive, again on the previous example, x₁ has 32 correct digits and x₂ has 161 good digits, x₃ has 806 good digits ...

   Higher order  

The inverse of a number can be computed with an iteration at any desired order, it's possible to show that the general algorithm of order r is given by

where P^m(u) is a polynomial of degree m, given by

For example, for m = 3,m = 4 and m = 5 (respectively quartic, quintic, sextic iterations)

In practical cases the number h_n tends to zero, therefore the evaluation of the powers h_n^k is more and more efficient when k increases. A careful implementation of those high order iterations is very efficient.

2  Square roots

In this section we apply the same technique to compute square roots.

2.1  Newton's iteration

Using the function f(x) = x²-A, Newton iteration gives

This iteration has one disadvantage : one should compute the division A/x_n. When one wants to compute 1/ЦA, a better iteration is obtained from the function f(x) = 1/x²-A, yielding the iteration

Note : as for the inverse, the best way to compute the iteration (**) is to write it in the form

2.2  A cubic iteration

Again, from the Householder's iteration, applied on the function f(x) = 1/x²-A, we find after some easy manipulations

This process converges cubically to the inverse of the square root of the number A (This form of the iteration is also given in [3]). Just like for the quadratic iteration, a more efficient way is to write it

2.3  High order iterations

Just like for the inverse, it's possible to find a quartic iteration, just apply twice Newton's iteration on f(x) = 1/x²-A :

This algorithm converges quartically to 1/ЦA.

It is interesting to compare this result to the direct application of the general quartic modified iteration to our function (see the essay on Newton's methods)

This easier relation suggests that it may be more efficient to use the quartic iteration than twice the Newton iteration. The best form of this quartic algorithm is

As an application of the modified iterations, we provide the general expression of some high order iterations to compute square roots. The general pattern of a high order process is given by

with P(u) being a polynomial with rational coefficients.

   Quintic iteration  

   Sextic iteration  

   Septic iteration  

   Octic iteration  

The coefficient of the polynomials P(u) are in fact given by the series expansion of

this allow to define easily an algorithm at any order to compute square roots.

3  m-th roots

The same techniques may be used in order to compute the m-th root (m being an integer) of the number A. Therefore, to find with a great accuracy the number A^-1/m, use Newton's method on the function f(x) = 1/x^m-A and this will produce the process

Householder's iteration also becomes

which converges cubically to A^-1/m.

3.1  High order iteration

The general pattern for any order iteration is given by

and to find P(u) you need to compute the following series expansion

The truncated series at order k will provide an algorithm of order k+1. Observe that for the value m = 1, we retrieve the polynomials used to compute the inverse of a number and for m = 2, it gives back the algorithms to find square roots.

3.1.1  Example

For m = 4 the following algorithm converges quartically to A^-1/4

and A^1/4 is given by at the end of the process by

References

[1]

I. Newton, Methodus fluxionum et serierum infinitarum, (1664-1671)

[2]

A. S. Householder, The Numerical Treatment of a Single Nonlinear Equation, McGraw-Hill, New York, (1970)

[3]

J.M. Borwein and P.B. Borwein, ''Pi and the AGM - A study in Analytic Number Theory and Computational Complexity'', A Wiley-Interscience Publication, New York, (1987)

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