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Durand–Kerner method : ウィキペディア英語版
Durand–Kerner method
In numerical analysis, the Durand–Kerner method, established 1960–66 and named after E. Durand and Immo Kerner, also called the method of Weierstrass, established 1859–91 and named after Karl Weierstrass, is a root-finding algorithm for solving polynomial equations. In other words, the method can be used to solve numerically the equation
: ƒ(''x'') = 0
where ƒ is a given polynomial, which can be taken to be scaled so that the leading coefficient is 1.
==Explanation==
The explanation is for equations of degree four. It is easily generalized to other degrees.
Let the polynomial ƒ be defined by
:ƒ(''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 ƒ.
Then
:ƒ(''x'') = (''x'' − ''P'')(''x'' − ''Q'')(''x'' − ''R'')(''x'' − ''S'')
for all ''x''. One can isolate the value ''P'' from this equation,
:P=x-\frac.
So if used as a fixed point iteration
:x_1:=x_0-\frac,
it is strongly stable in that every initial point ''x0'' ≠ ''Q,R,S''
delivers after one iteration the root ''P=x1''.
Furthermore, if one replaces the zeros ''Q'', ''R'' and ''S''
by approximations ''q'' ≈ ''Q'', ''r'' ≈ ''R'', ''s'' ≈ ''S'',
such that ''q,r,s'' are not equal to ''P'', then ''P''
is still a fixed point of the perturbed fixed point iteration
:x_:=x_k-\frac,
since
:P-\frac = P - 0 = P.
Note that the denominator is still different from zero.
This fixed point iteration is a contraction mapping
for ''x'' 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,···
:)})(p_-r_)(p_-s_) };
|-
| q_n = q_ - \frac-r_)(q_-s_) };
|-
| r_n = r_ - \frac-q_n)(r_-s_) };
|-
| s_n = s_ - \frac-q_n)(s_-r_n) }.
|}
Re-iterate until the numbers ''p, q, r, s''
stop essentially changing in relative to the desired precision.
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.

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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