Other solutions?
- Thread starter Techtonic
- Start date
People who work with continued fractions may have some neat tricks to show you, but I do not. I'd basically do what you did but might try a series of substitutions because I can see a lot of opportunity for sign errors and other fussy details.
[MATH]f(x) = \dfrac{x}{1 - \dfrac{1}{1 - \dfrac{1}{1 - \dfrac{1}{x}}}}.[/MATH]
[MATH]\text {Let } p = \dfrac{1}{x} \implies x = \dfrac{1}{p} \text { and}\\ f(x) = \dfrac{x}{1 - \dfrac{1}{1 - \dfrac{1}{1 - p}}}[/MATH][MATH]\text {Let } q = \dfrac{1}{1 - p} \implies q - pq = 1 \implies p = \dfrac{q - 1 }{q} \implies\\ f(x) = \dfrac{x}{1 - \dfrac{1}{1 - \dfrac{1}{1 - p}}} = \dfrac{x}{1 - \dfrac{1}{1 - q}} = \\ x \div \left ( 1 - \dfrac{1}{1 - q} \right ) = x \div \dfrac{-q}{1 - q} = x * \dfrac{q - 1}{q}.\\ \text {But } p = \dfrac{q - 1 }{q}.\\ \therefore f(x) = px.\\ \text {But } x = \dfrac{1}{p} \implies f(x) = \dfrac{1}{p} * p = 1.[/MATH]Substitutions make it easy to avoid potential errors and make very clear why you get the answer you get.
[MATH]f(x) = \dfrac{x}{1 - \dfrac{1}{1 - \dfrac{1}{1 - \dfrac{1}{x}}}}.[/MATH]
[MATH]\text {Let } p = \dfrac{1}{x} \implies x = \dfrac{1}{p} \text { and}\\ f(x) = \dfrac{x}{1 - \dfrac{1}{1 - \dfrac{1}{1 - p}}}[/MATH][MATH]\text {Let } q = \dfrac{1}{1 - p} \implies q - pq = 1 \implies p = \dfrac{q - 1 }{q} \implies\\ f(x) = \dfrac{x}{1 - \dfrac{1}{1 - \dfrac{1}{1 - p}}} = \dfrac{x}{1 - \dfrac{1}{1 - q}} = \\ x \div \left ( 1 - \dfrac{1}{1 - q} \right ) = x \div \dfrac{-q}{1 - q} = x * \dfrac{q - 1}{q}.\\ \text {But } p = \dfrac{q - 1 }{q}.\\ \therefore f(x) = px.\\ \text {But } x = \dfrac{1}{p} \implies f(x) = \dfrac{1}{p} * p = 1.[/MATH]Substitutions make it easy to avoid potential errors and make very clear why you get the answer you get.