Editing Newton's method (section)

===Undefinedness of Newton's method===
In some cases, it is not even possible to perform the Newton iteration. For example, if {{math|''f''(''x'') {{=}} ''x''<sup>2</sup> − 1}}, then the Newton iteration is defined by
:<math>x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} = x_n-\frac{x_n^2-1}{2x_n} = \frac{x_n^2+1}{2x_n}.</math>
So Newton's method cannot be initialized at 0, since this would make {{math|''x''<sub>1</sub>}} undefined. Geometrically, this is because the tangent line to {{mvar|f}} at 0 is horizontal (i.e. {{math|''f'' ′(0) {{=}} 0}}), never intersecting the {{math|''x''}}-axis.

Even if the initialization is selected so that the Newton iteration can begin, the same phenomenon can block the iteration from being indefinitely continued.

If {{mvar|f}} has an incomplete domain, it is possible for Newton's method to send the iterates outside of the domain, so that it is impossible to continue the iteration.<ref name="judd" /> For example, the [[natural logarithm]] function {{math|''f''(''x'') {{=}} ln ''x''}} has a root at 1, and is defined only for positive {{mvar|x}}. Newton's iteration in this case is given by
:<math>x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n(1- \ln x_n).</math>
So if the iteration is initialized at {{math|e}}, the next iterate is 0; if the iteration is initialized at a value larger than {{math|e}}, then the next iterate is negative. In either case, the method cannot be continued.