Householder's method

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In numerical analysis, the class of Householder's methods are root-finding algorithms used for functions of one real variable with continuous derivatives up to some order d+1, where d will be the order of the Householder's method.

The algorithm is iterative and it has rate of convergence of d+1.

Contents

[edit] Method

Like any root-finding method, the Householder's method is a numerical algorithm for solving the nonlinear equation f(x) = 0. In this case, the function f has to be a function of one real variable. The method consists of a sequence of iterations:

x_{n+1} = x_n + d\; \frac {\left(1/f\right)^{(d-1)}(x_n)} {\left(1/f\right)^{(d)}(x_n)}

beginning with an initial guess x0.

If f is a (d+1) times continuously differentiable function and a is a zero of f but not of its derivative, then, in a neighborhood of a, the iterates xn satisfy:

| x_{n+1} - a | \le K \cdot {| x_n - a |}^{d+1} , for some K > 0.\!

This means that the iterates converge to the zero if the initial guess is sufficiently close, and that the convergence has rate (d+1).

[edit] Motivation

An approximate idea of the origin of the Householder's method derives from the geometric series. Let the real-valued, continuously differentiable function f(x) have a simple zero at x=a, that is f(a)=0 while f'(a)\ne0. Then 1/f(x) has a simple pole at a, and close to a the behavior of 1/f(x) is dominated by the factor 1/(x-a). Approximatively one gets

\frac1{f(x)}=\frac1{f(x)-f(a)}=\frac{x-a}{f(x)-f(a)}\cdot\frac{-1}{a(1-x/a)} \approx\frac{-1}{af'(a)}\cdot\sum_{k=0}^\infty\left(\frac{x}{a}\right)^k.

Here f'(a)\ne0 because a is a simple zero of f(x). The coefficient of degree d has the value C\,a^{-d}. Thus, one can now reconstruct the zero a by dividing the coefficient of degree d-1 by the coefficient of degree d. Since this geometric series is an approximation to the Taylor expansion of 1/f(x), one can get estimates of the zero of f(x) − now without prior knowledge of the location of this zero − by dividing the corresponding coefficients of the Taylor expansion of 1/f(x) or, more generally, 1/f(b+x). From that one gets, for any integer d, and if the corresponding derivatives exist,

a \approx b+\frac{(1/f)^{(d-1)}(b)}{(d-1)!}\;\frac{d!}{(1/f)^{(d)}(b)} = b+d\;\frac{(1/f)^{(d-1)}(b)}{(1/f)^{(d)}(b)}.

[edit] The methods of lower order

The Householder's method of order 1 is just Newton's method, since:

\begin{array}{rl} x_{n+1} =& x_n + 1\,\frac {\left(1/f\right)(x_n)} {\left(1/f\right)^{(1)}(x_n)}\\[.7em] =& x_n + \frac{1}{f(x_n)}\cdot\left(\frac{-f'(x_n)}{f(x_n)^2}\right)^{-1}\\[.7em] =& x_n - \frac{f(x_n)}{f'(x_n)}. \end{array}

For the Householder's method of order 2 one gets Halley's method, since the identities

\textstyle (1/f)'(x)=-\frac{f'(x)}{f(x)^2}\ and \textstyle\ (1/f)''(x)=-\frac{f''(x)}{f(x)^2}+2\frac{f'(x)^2}{f(x)^3}

result in

\begin{array}{rl} x_{n+1} =& x_n + 2\,\frac {\left(1/f\right)'(x_n)} {\left(1/f\right)''(x_n)}\\[1em] =& x_n + \frac{-2f(x_n)\,f'(x_n)}{-f(x_n)f''(x_n)+2f'(x_n)^2}\\[1em] =& x_n - \frac{f(x_n)f'(x_n)}{f'(x_n)^2-\tfrac12f(x_n)f''(x_n)}\\[1em] =& x_n + h_n\;\frac{1}{1+\frac12(f''/f')(x_n)\,h_n}. \end{array}

In the last line, h_n=-\tfrac{f(x_n)}{f'(x_n)} is the update of the Newton iteration at the point xn. This line was added to demonstrate where the difference to the simple Newton's method lies.

The third order method is obtained from the identity of the third order derivative of 1/f

\textstyle (1/f)'''(x)=-\frac{f'''(x)}{f(x)^2}+6\frac{f'(x)\,f''(x)}{f(x)^3}-6\frac{f'(x)^3}{f(x)^4}

and has the formula

\begin{array}{rl} x_{n+1} =& x_n + 3\,\frac {\left(1/f\right)''(x_n)} {\left(1/f\right)'''(x_n)}\\[1em] =& x_n - \frac{6f(x_n)\,f'(x_n)^2-3f(x_n)^2f''(x_n)}{6f'(x)^3-6f(x_n)f'(x_n)\,f''(x)+f(x_n)^2\,f'''(x_n)}\\[1em] =& x_n + h_n\frac{1+\frac12(f''/f')(x_n)\,h_n}{1+(f''/f')(x_n)\,h_n+\frac16(f'''/f')(x_n)\,h_n^2} \end{array}

and so on...


[edit] Example

The first problem solved by Newton with the Newton-Raphson-Simpson method was the polynomial equation y3 − 2y − 5 = 0. He observed that there should be a solution close to 2. Replacing y=x+2 transforms the equation into

0 = f(x) = − 1 + 10x + 6x2 + x3.

The Taylor series of the reciprocal function starts with

\begin{array}{rl} 1/f(x)=& - 1 - 10\,x - 106 \,x^2 - 1121 \,x^3 - 11856 \,x^4 - 125392 \,x^5\\ & - 1326177 \,x^6 - 14025978 \,x^7 - 148342234 \,x^8 - 1568904385 \,x^9\\ & - 16593123232 \,x^{10} +O(x^{11}) \end{array}

The result of applying Householder's methods of various orders at x=0 is also obtained by dividing neighboring coefficients of the latter power series. For the first orders one gets the following values after just one iteration step:

d x1
1 0.10000000000000000000000000000000
2 0.094339622641509433962264150943396
3 0.094558429973238180196253345227476
4 0.094551282051282051282051282051282
5 0.094551486538216154140615031261963
6 0.094551481438752142436492263099119
7 0.094551481543746895938379484125813
8 0.094551481542336756233561913325371
9 0.094551481542324837086869382419375
10 0.094551481542326678478801765822985

As one can see, there are a little bit more than d correct decimal places for each order d.

[edit] Derivation

An exact derivation of the Householder's methods starts from the Padé approximation of order (d+1) of the function resp. its Taylor development, where the approximant with linear numerator is chosen. If this has been achieved, the update for the next approximation results from computing the unique zero of the numerator.

The Padé approximation has the form

f(x+h)=\frac{a_0+h}{b_0+b_1h+\dots+b_{d-1}h^{d-1}}+O(h^{d+1}).

The rational function has a zero at h = − a0.

Just as the Taylor polynomial of degree d has d+1 coefficients that depend on the function f, also the Padé approximation always has d+1 coefficients dependent on f and its derivatives. More precisely, in any Pade approximant, the degrees of numerator and denominator polynomials have to add to the order of the approximant. Therefore, bd = 0 has to hold.

One could determine the Padé approximant starting from the Taylor polynomial of f using Euclid's algorithm. However, starting from the Taylor polynomial of 1/f is shorter and leads directly to the given formula. Since

(1/f)(x+h) = (1/f)(x)+(1/f)'(x)h+\dots+(1/f)^{(d-1)}(x)\frac{h^{d-1}}{(d-1)!}+(1/f)^{(d)}(x)\frac{h^d}{d!}+O(h^{d+1})

has to be equal to the inverse of the desired rational function, we get after multiplying with a0 + h in the power hd the equation

0=b_d=a_0(1/f)^{(d)}(x)\frac1{d!}+(1/f)^{(d-1)}(x)\frac1{(d-1)!}.

Now, solving the last equation for the zero h = − a0 of the numerator results in

\begin{array}{rl} h=&-a_0= \frac{\frac1{(d-1)!}(1/f)^{(d-1)}(x)}{\frac1{d!}(1/f)^{(d)}(x)}\\[1em] =&d\,\frac{(1/f)^{(d-1)}(x)}{(1/f)^{(d)}(x)} \end{array}.

This implies the iteration formula

x_{n+1} = x_n + d\; \frac { \left(1/f\right)^{(d-1)} (x_n) } { \left(1/f\right)^{(d)} (x_n) } .

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