Exercises
[0DN]Note:Babylonian method for square root. Let \(S{\gt}0\) and consider the sequence defined by recurrence as
\[ x_{n+1} = \frac 1 2 \left( x_ n + \frac{S}{x_ n} \right) \quad ; \]show that \(x_ n→ \sqrt S\) and that, for \(S∈ [1/4,1]\) and \(x_ 0=1\), convergence is superquadratic, i.e.
\[ \left| x_ n-\sqrt{S} \right| ≤ 2^{1-2^ n }\quad . \]Find a function \(f(x)\) (dependent on \(S\) ) such that the previous iteration can be seen as a Newton’s method, i.e.
\[ x - \frac{f(x)}{f'(x)} = \frac 1 2 \left( x + \frac{S}{x} \right). \]Generalize the Babylonian method to find a root \(\sqrt[k]{S}\).
Solution 1