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dlaed6 (3)
  • >> dlaed6 (3) ( Solaris man: Библиотечные вызовы )
  • 
    NAME
         dlaed6 - compute the positive or negative root  (closest  to
         the  origin)  of  f(x)  =  rho + (z(1) / (d(1)-x)) + (z(2) /
         (d(2)-x)) + (z(3) /  (d(3)-x))   It  is  assumed  that    if
         ORGATI = .true
    
    SYNOPSIS
         SUBROUTINE DLAED6( KNITER, ORGATI, RHO, D,  Z,  FINIT,  TAU,
                   INFO )
    
         LOGICAL ORGATI
    
         INTEGER INFO, KNITER
    
         DOUBLE PRECISION FINIT, RHO, TAU
    
         DOUBLE PRECISION D( 3 ), Z( 3 )
    
    
    
         #include <sunperf.h>
    
         void dlaed6(int kniter, int orgati, double drho, double  *d,
                   double *dz, double finit, double *tau,
                    int *info) ;
    
    PURPOSE
         DLAED6 computes the positive or negative  root  (closest  to
         the origin) of
                          z(1)        z(2)        z(3)
         f(x) =   rho + --------- + ---------- + ---------
                         d(1)-x      d(2)-x      d(3)-x
    
         otherwise it is between d(1) and d(2)
    
         This routine will be called by  DLAED4  when  necessary.  In
         most  cases,  the  root sought is the smallest in magnitude,
         though it might not be in some extremely rare situations.
    
    
    ARGUMENTS
         KNITER    (input) INTEGER
                   Refer to DLAED4 for its significance.
    
         ORGATI    (input) LOGICAL
                   If ORGATI is true, the needed root is between d(2)
                   and  d(3);  otherwise it is between d(1) and d(2).
                   See DLAED4 for further details.
    
         RHO       (input) DOUBLE PRECISION
                   Refer to the equation f(x) above.
    
         D         (input) DOUBLE PRECISION array, dimension (3)
                   D satisfies d(1) < d(2) < d(3).
    
         Z         (input) DOUBLE PRECISION array, dimension (3)
                   Each of the elements in z must be positive.
    
         FINIT     (input) DOUBLE PRECISION
                   The value of f at 0. It is more accurate than  the
                   one  evaluated  inside  this  routine  (if someone
                   wants to do so).
    
         TAU       (output) DOUBLE PRECISION
                   The root of the equation f(x).
    
         INFO      (output) INTEGER
                   = 0: successful exit
                   > 0: if INFO = 1, failure to converge
    
    
    
    


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