GENERALIZATION OF LINEAR REGRESSION PROBLEMS - Neural-based Orthogonal Data Fitting: The EXIN Neural Networks - page 145

Theorem 96 (Validity) The ( n + 1 ) th -dimensional MCA EXIN ODE con-
strained on the TLS hyperplane is the ( n + 1 ) th -dimensional TLS EXIN ODE
and its first n components are eq. ( 5.13 ) , while its ( n + 1 ) th component is
asymptotically an identity. This justifies the use of eq. ( 5.13 ) as the ODE and the
respect of the hyperplane constraint by the discrete TLS EXIN learning law.

Proof. From the MCA EXIN ODE,

R −

T

d ( t )
dt

1 ( t ) R ( t )

=− ( t )

T

( t ) ( t )

T

( t ) ( t )

n + 1 .Replacing ( t ) with x T

( t ) ;− 1 T

where R is given by eq. (5.14) and ∈

yields

dx / dt
0

− Rx − r
− r T x

1
1 + x T x

=

x T Rx − 2 x T r + x
− x T Rx + 2 x T r −

1

+ (5.15)

1 + x T x

Its first n components are eq. (5.13). Define as the ( n + 1 ) th component of
the right-hand side and as the expression

= m 1 + x T x = x T x b T b − x T x b T Ax − x T A T Ax + b T Ax

= x T x b T b b T Ax − x T A T Ax A T b − − (5.16)

At the TLS solution (minimum), eq. (1.18) holds:

[ A ; b ] T [ A ; b ] x
− 1

A T AA T b
b T Ab T b

x
− 1

1 x

2
n

= = σ (5.17)

+

− 1

Hence,

→ x T x σ x T σ

n + 1 x =

t →∞

2
n + 1

2

− 0 (5.18)

showing that the ( n + 1 ) th component of the MCA EXIN ODE, in the limit,
carries no information.