$$
x=1+\cfrac{1}{1+\cfrac{1}{1+...}}\implies x=1+\frac{1}{x}\implies x=\frac{1\pm \sqrt{5}}{2}
$$
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.
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