DURAND-KERNER METHOD
In numerical analysis, the 'Durand-Kerner method' (established 1960-66) or 'method of Weierstrass' (established 1859-91) is a root-finding algorithm for solving polynomial equations. In other words, the method can be used to solve numerically the equation
:''f''(''x'') = 0
where ''f'' is a given polynomial, which can be taken to be normed so that the leading coefficient is 1.
The explanation is for equations of degree four. It is easily generalized to other degrees.
Let the normed polynomial ''f'' be defined by
:''f''(''x'') = ''x''4 + ''ax''3 + ''bx''2 + ''cx'' + ''d''
for all ''x''.
The known numbers ''a, b, c, d'' are the coefficients.
Let the (complex) numbers ''P,Q,R,S'' be the roots of this polynomial ''f''.
Then
:''f''(''x'') = (''x'' − ''P'')(''x'' − ''Q'')(''x'' − ''R'')(''x'' − ''S'')
for all ''x''.
One can isolate the value ''P'' from this equation,
: .
The substitution
:
is a strongly stable fixed point iteration in that every initial point ''x'' ≠ ''Q,R,S'' delivers after one iteration the root ''P''.
If one replaces the zeros ''Q'', ''R'' and ''S'' by approximations ''q,r,s'' ≠ ''P'', then ''P''
is still a fixed point of the perturbed fixed point iteration since
:
Note that the denominator is still different from zero. This fixed point iteration is a contraction mapping, around ''P''.
The clue to the method now is to combine the fixed point iteration for ''P'' with similar iterations for ''Q,R,S'' into a simultaneous iteration for all roots.
Initialize ''p, q, r, s'':
:''p''0 := (0.4 + 0.9 i)0 ;
:''q''0 := (0.4 + 0.9 i)1 ;
:''r''0 := (0.4 + 0.9 i)2 ;
:''s''0 := (0.4 + 0.9 i)3 ;
There is nothing special about choosing 0.4 + 0.9 i except that it is neither a real number nor a root of unity.
Make the substitutions for ''n'' = 1,2,3,···
:{|
|-
|
|-
|
|-
|
|-
|
|}
Re-iterate until the numbers ''p, q, r, s'' stop essentially changing. Then they have the values ''P, Q, R, S'' in some order and in the chosen precision. So the problem is solved.
Note that you must use complex number arithmetic, and that the roots are found simultaneously rather than one at a time.
This example is from the reference 1992. The equation solved is ''x''3 − 3''x''2 +3''x'' − 5 = 0 . The first 4 iterations move ''p, q, r'' seemingly chaotically, but then the roots are located to 1 decimal. After iteration number 5 we have 4 correct decimals, and the subsequent iteration number 6 confirms that the computed roots are fixed. This general behaviour is characteristic for the method.
::{|class="wikitable"
|----
!it.-no.
!p
!q
!r
|----
!0
| +1.0000+0.0000i
| +0.4000+0.9000i
| −0.6500+0.7200i
|----
!1
| +1.3608+2.0222i
| −0.3658+2.4838i
| −2.3858−0.0284i
|----
!2
| +2.6597+2.7137i
| +0.5977+0.8225i
| −0.6320−1.6716i
|----
! 3
| +2.2704+0.3880i
| +0.1312+1.3128i
| +0.2821−1.5015i
|----
! 4
| +2.5428−0.0153i
| +0.2044+1.3716i
| +0.2056−1.3721i
|----
! 5
| +2.5874+0.0000i
| +0.2063+1.3747i
| +0.2063−1.3747i
|----
! 6
| +2.5874+0.0000i
| +0.2063+1.3747i
| +0.2063−1.3747i
|----
|}
Note that the equation has one real root and one pair of complex conjugate roots, and that the sum of the roots is 3.
For every ''n''-tuple of complex numbers, there is exactly one monic polynomial of degree ''n'' that has them as its zeros (keeping multiplicities). This polynomial is given by multiplying all the corresponding linear factors, that is
:
This polynomial has coefficients that depend on the prescribed zeros,
:
Those coefficients are, up to a sign, the elementary symmetric polynomials of degrees ''1,...,n''.
To find all the roots of a given polynomial with coefficient vector simultaneously is now the same as to find a solution vector to the system
:
The Durand-Kerner method is obtained as the multidimensional Newton's method applied to this system. It is algebraically more comfortable to treat those identities of coefficients as the identity of the corresponding polynomials, . In the Newton's method one looks, given some initial vector , for an increment vector such that is satisfied up to second and higher order terms in the increment. For this one solves the identity
:
If the numbers are pairwise different, then the polynomials in the terms of the right hand side form a basis of the ''n''-dimensional space of polynomials with maximal degree ''n'' − 1. Thus a solution to the increment equation exists in this case. The coordinates of the increment are simply obtained by evaluating the increment equation
:
at the points , which results in
:, that is
In the quotient ring (algebra) of residue classes modulo ''f(X)'', the multiplication by ''X'' defines an endomorphism that has the zeros of ''f(X)'' as eigenvalues with the corresponding multiplicities. Choosing a basis, the multiplication operator is represented by its coefficient matrix ''A'', the companion matrix of ''f(X)'' for this basis.
Since every polynomial can be reduced modulo ''f(X)'' to a polynomial of degree ''n-1'' or lower, the space of residue classes can be identified with the space of polynomials of degree bounded by ''n-1''.
A problem specific basis can be taken from Lagrange interpolation as the set of ''n'' polynomials
:
where are pairwise different complex numbers. Note that the kernel functions for the Lagrange interpolation are .
For the multiplication operator applied to the basis polynomials one obtains from the Lagrange interpolation
{|
|-
|
|
|-
|
|,
|}
where are again the Weierstrass updates.
The companion matrix of ''f(X)'' is therefore
.
From the transposed matrix case of the Gershgorin circle theorem it follows that all eigenvalues of ''A'', that is, all roots of ''f''(''X''), are contained in the union of the disks with a radius .
Here one has , so the centers are the next iterates of the Weierstrass iteration, and radii that are multiples of the Weierstrass updates. If the roots of ''f(X)'' are all well isolated (relative to the computational precision) and the points are sufficidently close approximations to these roots, then all the disks will become disjoint, so each one contains exactly one zero. The midpoints of the circles will be better approximations of the zeros.
Every conjugate matrix of ''A'' is as well a companion matrix of ''f(X)''. Choosing ''T'' as diagonal matrix leaves the structure of ''A'' invariant. The root close to is contained in any isolated circle with center regardless of ''T''. Choosing the optimal diagonal matrix ''T'' for every index results in better estimates (see ref. Petkovic et al 1995).
The connection between the Taylor series expansion and Newton's method suggests that the distance from to the corresponding root is of the order , if the root is well isolated from nearby roots and the approximation is sufficiently close to the root. So after the approximation is close, Newton's method converges ''quadratically''; that is: the error is squared with every step (which will greatly reduce the error once it is less than 1). In the case of the Durand-Kerner method, convergence is quadratic if the vector is close to some permutation of the vector of the roots of ''f''.
For the conclusion of linear convergence there is a more specific result (see ref. Petkovic et al 1995). If the initial vector and its vector of Weierstrass updates satisfies the inequality
:
:''f''(''x'') = 0
where ''f'' is a given polynomial, which can be taken to be normed so that the leading coefficient is 1.
| Contents |
| Explanation |
| Example |
| Derivation of the method via Newton's method |
| Root inclusion via Gerschgorin's circles |
| Convergence results |
| References |
| External links |
Explanation
The explanation is for equations of degree four. It is easily generalized to other degrees.
Let the normed polynomial ''f'' be defined by
:''f''(''x'') = ''x''4 + ''ax''3 + ''bx''2 + ''cx'' + ''d''
for all ''x''.
The known numbers ''a, b, c, d'' are the coefficients.
Let the (complex) numbers ''P,Q,R,S'' be the roots of this polynomial ''f''.
Then
:''f''(''x'') = (''x'' − ''P'')(''x'' − ''Q'')(''x'' − ''R'')(''x'' − ''S'')
for all ''x''.
One can isolate the value ''P'' from this equation,
: .
The substitution
:
is a strongly stable fixed point iteration in that every initial point ''x'' ≠ ''Q,R,S'' delivers after one iteration the root ''P''.
If one replaces the zeros ''Q'', ''R'' and ''S'' by approximations ''q,r,s'' ≠ ''P'', then ''P''
is still a fixed point of the perturbed fixed point iteration since
:
Note that the denominator is still different from zero. This fixed point iteration is a contraction mapping, around ''P''.
The clue to the method now is to combine the fixed point iteration for ''P'' with similar iterations for ''Q,R,S'' into a simultaneous iteration for all roots.
Initialize ''p, q, r, s'':
:''p''0 := (0.4 + 0.9 i)0 ;
:''q''0 := (0.4 + 0.9 i)1 ;
:''r''0 := (0.4 + 0.9 i)2 ;
:''s''0 := (0.4 + 0.9 i)3 ;
There is nothing special about choosing 0.4 + 0.9 i except that it is neither a real number nor a root of unity.
Make the substitutions for ''n'' = 1,2,3,···
:{|
|-
|
|-
|
|-
|
|-
|
|}
Re-iterate until the numbers ''p, q, r, s'' stop essentially changing. Then they have the values ''P, Q, R, S'' in some order and in the chosen precision. So the problem is solved.
Note that you must use complex number arithmetic, and that the roots are found simultaneously rather than one at a time.
Example
This example is from the reference 1992. The equation solved is ''x''3 − 3''x''2 +3''x'' − 5 = 0 . The first 4 iterations move ''p, q, r'' seemingly chaotically, but then the roots are located to 1 decimal. After iteration number 5 we have 4 correct decimals, and the subsequent iteration number 6 confirms that the computed roots are fixed. This general behaviour is characteristic for the method.
::{|class="wikitable"
|----
!it.-no.
!p
!q
!r
|----
!0
| +1.0000+0.0000i
| +0.4000+0.9000i
| −0.6500+0.7200i
|----
!1
| +1.3608+2.0222i
| −0.3658+2.4838i
| −2.3858−0.0284i
|----
!2
| +2.6597+2.7137i
| +0.5977+0.8225i
| −0.6320−1.6716i
|----
! 3
| +2.2704+0.3880i
| +0.1312+1.3128i
| +0.2821−1.5015i
|----
! 4
| +2.5428−0.0153i
| +0.2044+1.3716i
| +0.2056−1.3721i
|----
! 5
| +2.5874+0.0000i
| +0.2063+1.3747i
| +0.2063−1.3747i
|----
! 6
| +2.5874+0.0000i
| +0.2063+1.3747i
| +0.2063−1.3747i
|----
|}
Note that the equation has one real root and one pair of complex conjugate roots, and that the sum of the roots is 3.
Derivation of the method via Newton's method
For every ''n''-tuple of complex numbers, there is exactly one monic polynomial of degree ''n'' that has them as its zeros (keeping multiplicities). This polynomial is given by multiplying all the corresponding linear factors, that is
:
This polynomial has coefficients that depend on the prescribed zeros,
:
Those coefficients are, up to a sign, the elementary symmetric polynomials of degrees ''1,...,n''.
To find all the roots of a given polynomial with coefficient vector simultaneously is now the same as to find a solution vector to the system
:
The Durand-Kerner method is obtained as the multidimensional Newton's method applied to this system. It is algebraically more comfortable to treat those identities of coefficients as the identity of the corresponding polynomials, . In the Newton's method one looks, given some initial vector , for an increment vector such that is satisfied up to second and higher order terms in the increment. For this one solves the identity
:
If the numbers are pairwise different, then the polynomials in the terms of the right hand side form a basis of the ''n''-dimensional space of polynomials with maximal degree ''n'' − 1. Thus a solution to the increment equation exists in this case. The coordinates of the increment are simply obtained by evaluating the increment equation
:
at the points , which results in
:, that is
Root inclusion via Gerschgorin's circles
In the quotient ring (algebra) of residue classes modulo ''f(X)'', the multiplication by ''X'' defines an endomorphism that has the zeros of ''f(X)'' as eigenvalues with the corresponding multiplicities. Choosing a basis, the multiplication operator is represented by its coefficient matrix ''A'', the companion matrix of ''f(X)'' for this basis.
Since every polynomial can be reduced modulo ''f(X)'' to a polynomial of degree ''n-1'' or lower, the space of residue classes can be identified with the space of polynomials of degree bounded by ''n-1''.
A problem specific basis can be taken from Lagrange interpolation as the set of ''n'' polynomials
:
where are pairwise different complex numbers. Note that the kernel functions for the Lagrange interpolation are .
For the multiplication operator applied to the basis polynomials one obtains from the Lagrange interpolation
{|
|-
|
|
|-
|
|,
|}
where are again the Weierstrass updates.
The companion matrix of ''f(X)'' is therefore
.
From the transposed matrix case of the Gershgorin circle theorem it follows that all eigenvalues of ''A'', that is, all roots of ''f''(''X''), are contained in the union of the disks with a radius .
Here one has , so the centers are the next iterates of the Weierstrass iteration, and radii that are multiples of the Weierstrass updates. If the roots of ''f(X)'' are all well isolated (relative to the computational precision) and the points are sufficidently close approximations to these roots, then all the disks will become disjoint, so each one contains exactly one zero. The midpoints of the circles will be better approximations of the zeros.
Every conjugate matrix of ''A'' is as well a companion matrix of ''f(X)''. Choosing ''T'' as diagonal matrix leaves the structure of ''A'' invariant. The root close to is contained in any isolated circle with center regardless of ''T''. Choosing the optimal diagonal matrix ''T'' for every index results in better estimates (see ref. Petkovic et al 1995).
Convergence results
The connection between the Taylor series expansion and Newton's method suggests that the distance from to the corresponding root is of the order , if the root is well isolated from nearby roots and the approximation is sufficiently close to the root. So after the approximation is close, Newton's method converges ''quadratically''; that is: the error is squared with every step (which will greatly reduce the error once it is less than 1). In the case of the Durand-Kerner method, convergence is quadratic if the vector is close to some permutation of the vector of the roots of ''f''.
For the conclusion of linear convergence there is a more specific result (see ref. Petkovic et al 1995). If the initial vector and its vector of Weierstrass updates satisfies the inequality
:
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