How can I use Euler's Continued Fraction Theorem to find the continued fraction expansion for a (ordinary, finite) fraction via its (terminating or recurring) decimal expansion, rather than via the more obvious Euclid's Algorithm?
The examples I'm thinking of are $5/7$ and $3/8$ each of which can be thought of as a series in a power of $10$: $$ 5/7=0.\overline{571428}=\sum_{k=1}^\infty 571428 (10^{-6})^k $$ and $$ 3/8=0.375=3* 10^{-1}+7* 10^{-2} +5* 10^{-3}. $$ I'd like to use these coefficients and expansion points together with Euler's Theorem to show the continued fraction expansions $$ 5/7=[0;1,2,2] $$ and $$ 3/8=[0;2,1,2]. $$
You could use the repeating digits to come up with a non-standard continued fraction in that the terms will not be integers. With sufficient applications of equivalence transforms these can be converted to integers.
in your example:
$$\ a_0 = .571428$$ $$\ a_i = 10^{-6} $$ $$\ x = 5/7 = a_0 + a_0a_1 + a_0a_1a_2+... = \frac {a_0} {1-\frac{a_1} {1+a_1- \frac{a_2}{1+a_2- \ddots}}} $$ $$\ x = \frac{.571428} {1-\frac{10^{-6}}{1+10^{-6}-\frac{10^{-6}}{1+10^{-6}- \ddots }}} $$ $$\ = \frac {571428}{1000000-\frac{1000000}{1000001-\frac{1000000}{1000001-\ddots}}}$$