Cube Root of x Show that for any non-zero starting point x0 , Newton’s method will never ﬁnd the exact value x for which x3 = 0. Solution Given a value x0 close to some x where f (x) = 0, Newton’s method usually produces a series of values x0 , x1 , x2 , ... whose values approach x. However, there is no guarantee that any of those values xi actually equal x. Newton’s method generates a sequence according to the formula: xk+1 = xk − f (xk ) . f � (xk ) In our example, f (x) = x3 and f � (x) = 3x2 , so: xk+1 xk+1 x3k 3x2k xk = xk − 3 2 = xk 3 = xk − Starting with x0 , Newton’s method generates the sequence: x1 = x2 = x3 = x4 = 2 x0 3 � � 2 2 22 2 x1 = · x 0 = 2 x0 3 3 3 3 2 23 x 2 = 3 x0 3 3 2 24 x 3 = 4 x0 3 3 .. . xn = 2n x0 3n If x0 = � 0, the values of the xi get closer and closer to the desired value 0 but never exactly equal zero.

Cube Root of x

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