One root as a continued fraction
- Ryan Hirst
Let where [list=1]
Then f(x) has a root r > b given by the following recursion
- Why not ask for the root directly? If we choose , then as well, and . The equality is The resulting cubic in ε lacks a first-order term, but this simplification is not enough to allow algebraic solution.
- I have chosen the positive side for simplicity, but the single condition real; a \ne b}[/math] guarantees a root in this form. Why?