Saturday, September 12, 2015

Negative solution for a positive continued fraction



$$
x=1+\cfrac{1}{1+\cfrac{1}{1+...}}\implies x=1+\frac{1}{x}\implies x=\frac{1\pm \sqrt{5}}{2}
$$
Can the negative solution be considered as a solution? If yes, how is it possible to have a negative solution for a positive continued fraction? If no, how do we prove that it can't be a solution?




Edit 1: I want to understand the assumption we are considering while forming the equation which results in the "extraneous solution".


Answer



No, the negative number is not a solution. You showed that if $x$ is equal to that fraction, then it is either $\frac{1+\sqrt 5}{2}$ or $\frac{1-\sqrt5}{2}$. You calculated possible candidates for solutions, not the solution itself.



You can prove that $x$ must be positive by simply arguing that $x$ is a limit of a sequence with only positive elements, so the limit (if it exists, which should also be proven) must be positive.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...