In order to proove the converge, proove the cardinality:

ñf^T.Yi>0 

ñf^T.Zi>0

?=min(ñf^t.zi)

a.b=||a||.||b||.cos?;cos?€[-1,1]

||a||.||b||>=a.b

||ñf||.||ñk||>=ñf^t.ñ(k)>k.?

||ñf||².||ñ(k)||²>=k².?²

inecuality1:||ñf||².||ñ(k)||²>=k².?²  ?=min(ñf².zi)

ñ(k) =ñ(k-1)+yk  y={z1,z2,z3}={y1,y2,y3}(we put yk because yk is badly clasified by ñ(k-1).yk<=0)

...ñ(k)^t.ñ(k)=[ñ²(k-1)+yk]^t.[ñ².(k-1)+yk]^t.[ñ(k-1)+yk]=...inecualiti2:||ñ(k)||²=<?^k_i=1yi^tyi<=k?   ?=max(zi^t.zi)

k<=||ñf||².?/?^{2-->it converges}

linear separator

po->(media) n->(final-principio)

n^t(p-p0)->(0,26 -43)(p1;p2)-(0,26 -43)(1,91;131)

0,26p1-43p2+5654=0-->linear ecuation

Euclidear distance

r1= (r1;-1/2r1^tr1)