funció

si f(x) = ( (y+(-x))/(2s) ) ==> [∀s][ s ≠ 0 ==> f(y+(-s)) = 1+f(y+s) ]


f(y+(-s)) = ( (y+(-y)+s)/(2s) ) = ( s/(2s) ) = (1/2)
f(y+s) = ( (y+(-y)+(-s))/(2s) ) = (-1)( s/(2s) ) = (-1)(1/2)


si f(x) = ( (y+(-1)·x^{n})/(2s) ) ==> [∀s][ s ≠ 0 ==> f( (y+(-s))^{(1/n)} ) = 1+f( (y+s)^{(1/n)} ) ]