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I'm attempting to find the continued fraction for the cube root of 2 (no calculator usage)

So I wrote this as 2^(1/3) = 1/(1/2^(1/3))), and since $1^3$ < $2$ < $2^3$, so the cube root is between 1, and 2, so I can write that

2^(1/3) = 1/(1/2^(1/3))) = 1/(1/(2^(1/3)- 1))), and multiplying by the conjugate gives

= 1/((2^(1/3) + 1)/(4^(1/3) - 1)), and a calculator shows that this has an integer part of 3, but I want to continue without a calculator.

Is there an easier way? Do I need a calculator? Is this the only method?

I'm trying to show that at some point, the pattern repeats, so I only need an example that it repeats.

Knocker379
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    the continued fraction is not periodic. Any continued fraction that is, eventually, periodic, represents a real $\frac{A + \sqrt B}{C}$ with integer $A,B,C$ and $B>0$ – Will Jagy Jul 28 '21 at 00:13
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    Although I agree with the comment of @WillJagy, the procedure for manually deriving the (simple) continued fraction representation of $\alpha = \alpha_0 = [a_0; a_1, a_2, \cdots]$ is still unchanged. That is $$a_0 = \lfloor \alpha_0\rfloor$$ and for $n \in \Bbb{Z_{\geq 0}}$ you have that $$\alpha_n - a_n = \frac{1}{\alpha_{(n+1)}} ~~\text{and} ~~a_{(n+1)} = \lfloor \alpha_{(n+1)}\rfloor.$$ – user2661923 Jul 28 '21 at 00:39
  • @user2661923 I can see an algorithm (no calculator, no approximation) to find arbitrarily many correct $a_n$ in the simple continued fraction for $2^{1/3}.$ The cubic $x^3 + 2 y^3 + 4 z^3 - 6xyz$ factors, one of the factors is $x + y \sqrt[3] 2 + z \sqrt[3]4 .$ The result is that we can find the reciprocal of some needed $x + y \sqrt[3] 2 + z \sqrt[3]4 $ as a similar expression with rational coefficients. Nothing easy about this, though. – Will Jagy Jul 28 '21 at 01:08
  • @user2661923 a similar algorithm, all symbols, no decimals, for $\sqrt n$ brought to my attention by Lubin, https://math.stackexchange.com/questions/2215918/continued-fraction-11/2216011#2216011 – Will Jagy Jul 28 '21 at 01:14
  • My recollection is that I have tried just this calculation, always involving $\lfloor \frac1{c_0+c_1\sqrt2+c_2\sqrt4}\rfloor$, with the coefficients ${c_i}$ always integers. But at each step, the coefficients were larger, so that actually calculating the floor function required an estimate best done with a calculator. Very difficult. – Lubin Jul 28 '21 at 03:29
  • The continued fraction very likely has no pattern, so calculation by hand will probably be hopeless. – Peter Jul 29 '21 at 07:32
  • @Lubin, one aspect makes this workable by computer: given $ q=c_0 + c_1 \sqrt[3] 2 + c_2 \sqrt[3] 4,$ with rational $c_j,$ we can tell whether $q$ is positive because $c_0^3 + 2 c_1^3 + 4 c_2^3 - 6 c_0 c_1 c_2,$ divided by $q,$ is positive (the quotient is the magnitude of a complex number). So, no decimal estimation is required. – Will Jagy Jul 30 '21 at 17:40
  • @Lubin so we may subtract integers off $c_0 + c_1 \sqrt[3]2 + c_2 \sqrt[3]4$ until we get a negative result, thus finding the floor of the original. – Will Jagy Jul 30 '21 at 17:50
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    @WillJagy, I never thought of it all in this way. The sign of $\theta$ Is revealed in the sign of its Norm. No matter what, this calculation involves a hella lotta computation. – Lubin Jul 31 '21 at 14:43

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