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α i
α 1 α i
α i
α n + 1 α i
= U diag
, ... ,
0
, ... ,
i th position
η
.
1
ζ
r
2
2
u i
η
.
η
α
i
α i α 1
0
.
η
= u 1
u n + 1
···
α i
η
α
α
i
n
+
1
So the following result has been proved:
d u i
d ζ
α i
α i α 1 u 1 +···+
α i
α i α i 1 u i 1 +
α i
α i α i + 1 u i + 1 +···
= η
α i α n + 1 u n + 1
α i
+
1
ζ ( 1 ζ )
α i
α i α j u j
=
(6.53)
j
=
i
Defining τ f , e as
#
$
α g
α g α f
for g = f
τ f , g =
(6.54)
%
0
otherwise
eq. (6.53) can be written in a more compact form as
τ 1, i
0
.
τ n + 1, i
d u i
d ζ
= η U
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