Householder's method

Method

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

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 x_n satisfy:

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

Despite their order of convergence, these methods are not widely used because the gain in precision is not commensurate with the rise in effort for large d. The Ostrowski index expresses the error reduction in the number of function evaluations instead of the iteration count.

Motivation

An approximate idea of 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 a root f(a)=0 of multiplicity one, which is equivalent to f ′ ( a ) ≠ 0 . Then 1/f(x) has a singularity at a, specifically a simple pole (also of multiplicity one), and close to a the behavior of 1/f(x) is dominated by 1/(x-a). Approximately, one gets

Here f ′ ( a ) ≠ 0 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,

Alternative motivation

Suppose x=a is a simple root. Then near x=a, (1/f)(x) is a meromorphic function. Suppose we have the Taylor expansion:

By König's theorem, we have:

This suggests that Householder's iteration might be a good convergent iteration. The actual proof of the convergence is also based on this idea.

The methods of lower order

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

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

and

result in

In the last line, h n = − f ( x n ) f ′ ( x n ) is the update of the Newton iteration at the point x n . 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

and has the formula

and so on...

Example

The first problem solved by Newton with the Newton-Raphson-Simpson method was the polynomial equation y 3 − 2 y − 5 = 0 . He observed that there should be a solution close to 2. Replacing y=x+2 transforms the equation into

The Taylor series of the reciprocal function starts with

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: For an example, in the case of the 3rd order, x 1 = 0.0 + 106 / 1121 = 0.09455842997324 .

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

Let's calculate the x 2 , x 3 , x 4 values for some lowest order,

And using following relations,

Derivation

An exact derivation of Householder's methods starts from the Padé approximation of order (d+1) of the function, where the approximant with linear numerator is chosen. Once 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

The rational function has a zero at h = − a 0 .

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

has to be equal to the inverse of the desired rational function, we get after multiplying with a 0 + h in the power h d the equation

Now, solving the last equation for the zero h = − a 0 of the numerator results in

This implies the iteration formula