also wäre die positive Nullstelle der Funktion f(x) = x2 - 3 gesucht.
Beginne mit Intervall [ 1 ; 2 ] denn f(1) <0 und f(2) > 0
Mitte ist 1,5
Berechne f(1,5) = -0,75 Also neues Intervall [1,5 ; 2 ]
Mitte 1,75
Berechne f(1,75 ) = 0,0625 Also neues Intervall [1,5 ; 1,75 ]
Mitte 1,625 f(1,625) =0,359... Also Intervall [1,625 ; 1,75 ]
Mitte 1,6875 f(1,6875) = -0,152..... Also Intervall [1,6875 ; 1,75 ]
Mitte 1,71875 f(1,71875 ) = -0,045.... Also Intervall [1,71875 ; 1,75 ]
Mitte 1,734375 f(..) =0,0080... Also Intervall [1,71875 ; 1,734375 ]
Mitte 1,7265625 f(..) = -0,018... Also Intervall [ 1,7265625 ; 1,734375 ]
Mitte 1,73046875 f(..)= - 0,0054... Also Intervall [ 1,73046875 ; 1,734375 ]
also 1,73 gesichert.