- 115 -


          5
1=C    +     d. + C, +
     e   i1l  1


1 3            5
     S   C  +     d,   C
i=2           i=l


(6.22)


        13K X   + I   KC
        i=2 C-


Solving for C1 from (6.21) and (6.22) and equating the resulting expressions gives an equation for Ce.


      5
Ce -      oi d
     i=l           C1
 13    K'
   X z.i-i

i=2    C
        e


          5
C-c    -     d.
       e i=l  1

     13 K!
     i=2 Ce


By algebraic manipulation, equation (6.23) can be written as

Hdi + Hd2 + (H-l)d3 + (H+l)d4 + (H-l)d5 + H(C e-1) + Ce = 0


                                                   (6.24)


     1 3   K'
       I z . i-1
    i=2    Ce
H=
         13 K!
      1+i
        i=2 C~


(6.25)


From equations (6.7) through (6.10),

                C 2                C
                e        2          e
             d = ---- -2 d '   2 = K- d
             K01 (K02]             K02


(6.23)


where