Editing Newton's method (section)

===Convergence dependent on initialization===
The function {{math|''f''(''x'') {{=}} ''x''(1 + ''x''<sup>2</sup>)<sup>−1/2</sup>}} has a root at 0. The Newton iteration is given by
:<math>x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n(1+x_n^2)^{-1/2}}{(1+x_n^2)^{-3/2}}=-x_n^3.</math>
From this, it can be seen that there are three possible phenomena for a Newton iteration. If initialized strictly between {{math|±1}}, the Newton iteration will converge (super-)quadratically to 0; if initialized exactly at {{math|1}} or {{math|−1}}, the Newton iteration will oscillate endlessly between {{math|±1}}; if initialized anywhere else, the Newton iteration will diverge.<ref>Yuri Nesterov. Lectures on convex optimization, second edition. Springer Optimization and its Applications, Volume 137.</ref> This same trichotomy occurs for {{math|''f''(''x'') {{=}} arctan ''x''}}.<ref name="dennis" />

In cases where the function in question has multiple roots, it can be difficult to control, via choice of initialization, which root (if any) is identified by Newton's method. For example, the function {{math|''f''(''x'') {{=}} ''x''(''x''<sup>2</sup> − 1)(''x'' − 3)e<sup>−(''x'' − 1)<sup>2</sup>/2</sup>}} has roots at −1, 0, 1, and 3.{{sfnm|1a1=Süli|1a2=Mayers|1y=2003}} If initialized at −1.488, the Newton iteration converges to 0; if initialized at −1.487, it diverges to {{math|∞}}; if initialized at −1.486, it converges to −1; if initialized at −1.485, it diverges to {{math|−∞}}; if initialized at −1.4843, it converges to 3; if initialized at −1.484, it converges to {{math|1}}. This kind of subtle dependence on initialization is not uncommon; it is frequently studied in the [[complex plane]] in the form of the [[Newton fractal]].