Since it cannot increase unboundedly, this implies that equivalent problem converges and consequently algorithm Fig. 3 also converges.
Hence ({ T^{i}y } _{i=1}^{infty}) is Cauchy and consequently converges to some (zin K).
This implies that and consequently converges uniformly on compact subsets of U. Set.
Thence every subsequence converges to as and consequently, as.
This implies that z converges to 0 and, consequently, that (lim_{ttoinfty} (K-S t))+ R(t)=0), which is a contradiction.
The sequence { ρ k ( x, y ) } being decreasing and lower bounded by 0, consequently it converges to some finite limit, says ρ ( x, y ).
end{aligned} (3.11) We see that the sequence ({s_{n}(x^{ast},y^{ast})}) is decreasing and lower bounded by 0; consequently, it converges to some finite limit which is denoted by (s(x^{ast},y^{ast})).
Since B(1,1) is equivalent to the uniform distribution the denominator is 1 for the score, which is given by Obviously for a→1 the density of the signal component converges to that of the background model and consequently the score converges to 0 for all x.
It is also illustrated that the fluctuation associated with this quantization may enable the network to escape from local minima, to converge to global minima, and consequently to obtain optimal solutions very frequently and much more quickly than pure quantized Hopfield networks (QHN).
and consequently it does converge to a unique minimum point.
