A Question about continued fraction representations

japplepie
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How powerful are continued fraction representations?

From what I understand, they could be used to exactly represent some irrational numbers

So, could they represent any root of an nth degree polynomial equation?

Specially where n>4, since 5th degree roots are not guaranteed to have an algebraic representation.
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And if they do, will there be a possibility to have a quintic formula which isn't an algebraic solution, but rather an continued fraction solution
 
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An irrational \xi admits a finite or periodic continued fraction representation if and only if there exist integers a, b, c, d with c \geq 1 and d nonzero such that

\displaystyle \xi = \frac{a + b \sqrt{c}}{d}

So in short the answer to your question is no; the only classes of numbers which admit finite or periodic continued fraction representations are integers, rationals, and roots to quadratic polynomials (also known as "quadratic surds"). Any higher degree algebraic number does not admit a periodic continued fraction representation.

Of course, it is possible to have infinite continued fractions with a finitely representable series of coefficients (e.g. some aperiodic pattern, for instance the CF for e) however this does not involve finitely many operations and so cannot be considered a closed-form solution.
 
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japplepie said:
How powerful are continued fraction representations?
They are very similar to power series.

From what I understand, they could be used to exactly represent some irrational numbers
No finite length continued fraction with rational numbers can represent an irrational number. The term "continued fractions" includes infinitely long continued fractions. They can represent irrational numbers.
 
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I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
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