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Sometimes, when getting some numerical results when investigating number theory sequences with a computer, I find myself suspecting that a decimal value ($a$) I have found might be a quadratic irrational. My usual next step is to look at the continued fraction for $a$ and hopefully find that it is periodic, and then I can find a closed expression for $a$.

My question stems from my inability (so far) to do anything along similar lines when I find (by calculating examples) a value $b$ that I suspect might be a cubic irrational such as $\frac{2^{1/3} + 5.2^{2/3}}{7}$. I believe that it is unknown whether there is any pattern for such irrationals when expressed as continued fractions (and maybe not even known if the terms are bounded?).

Does anyone have any ideas about how to investigate such a decimal with a view to identifying exactly which cubic irrational expression might be the an appropriate closed form? Are there any other techniques better than continued fractions?

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You might look it up in a table, e.g. pi.lacim.uqam.ca or oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html. It doesn't look like this will get your example, but lookup for 0.5977055 returns "5977055411582484 = 1/7*2^(1/3)+1/7*5^(2/3)". This isn't really a "technique," but could be a starting point. –  Zander Jun 2 '12 at 10:43
    
Many thanks for pointing these out. I knew about EOIS, but never dreamt anything similar for non-integers existed. –  Old John Jun 2 '12 at 11:36
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Or try something like Brun's algorithm, LLL or PSLQ on $(1, x, x^2, x^3)$ to find a polynomial relation. Brun gives already good results in many cases and is very simple to implement. It is similar to Euclid's algorithm: reduce the largest number in the sequence modulo the second largest. –  WimC Jun 2 '12 at 12:23
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Concerning the parenthetical question, it is believed that every real algebraic irrational of degree 3 or more has unbounded partial quotients, but there isn't even one such irrational for which this has been proved. –  Gerry Myerson Jun 2 '12 at 12:48
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Brun on the example $(2^{1/3} + 5 \cdot 2^{2/3})/7$ gives the minimum polynomial $343 x^3 - 210 x - 502$ after 55 iterations. (Using plain IEEE double precision.) –  WimC Jun 2 '12 at 12:51

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