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I'd really love your help with finding the rational number which the continued fraction $[1;1,2,1,1,2,\ldots]$ represents.

With the recursion for continued fraction $( p_0=a_0, q_0=1, p_{-1}=1, q_{-1}=o), q_s=a_sq_{s-1}+q_{s-2},p_s=a_sp_{s-1}+p_{s-2}$. I found out the $p_k=1,2,5,7,12,32, q_k=1,1,3,4,7,18 $ (anything special about these series? perhaps I did a mistake?), and I know that $r=\lim_{c_k}=\lim\frac{p_k}{q_k}$, but I can't see anything special about $p_k, q_k$ or the realtion between them, Any help?

Thanks a lot!

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    $\begingroup$ Your problem might be that the continued fraction for a rational number terminates. The classical method for evaluating recurrent continued fractions is to set $x=[1;1,2,1,1,2, ...]$ and note that $x=[1;1,2,x]$ (if I have read your notation right) which gives a quadratic in x. $\endgroup$ – Mark Bennet Jun 28 '12 at 22:03
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    $\begingroup$ The number is a quadratic irrationality. $\endgroup$ – André Nicolas Jun 28 '12 at 22:08
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    $\begingroup$ Does the continued fraction continue indefinitely? does it repeat? A number is rational if and only if its continued fraction terminates. A number is the root of a quadratic equation with integer coefficients if and only if its continued fraction repeats. $\endgroup$ – robjohn Jun 28 '12 at 22:09
  • $\begingroup$ That $x$ in the brackets might also be $\frac 1 x$ depending on how the notation works. It still gives a quadratic. $\endgroup$ – Mark Bennet Jun 28 '12 at 22:11
  • $\begingroup$ The 32 in the question should be 31; it's $2\times12+7$. $\endgroup$ – Gerry Myerson Jun 28 '12 at 23:33
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We have that $$\Large x=1+\frac{1}{1+\frac{1}{2\,+\,\frac{1}{1\,+\,\frac{1}{1\,+\,\frac{1}{2\,+\,\cdots}}}}}=1+\frac{1}{1+\frac{1}{2+\frac{1}{x}}}$$ so that $$\frac{1}{x-1}=1+\frac{1}{2+\frac{1}{x}}$$ hence $$\frac{1}{\frac{1}{x-1}-1}=\frac{1}{\frac{1}{x-1}-\frac{x-1}{x-1}}=\frac{1}{\frac{2-x}{x-1}}=\frac{x-1}{2-x}=2+\frac{1}{x}$$ and thus $$x^2-x=2x(2-x)+(2-x)$$ $$x^2-x=-2x^2+4x+2-x$$ $$3x^2-4x-2=0$$ The roots of this equation are $$x=\frac{4\pm\sqrt{40}}{6}=\frac{4\pm 2\sqrt{10}}{6}$$ but we know it can't be $x=\frac{4-2\sqrt{10}}{6}$ since that number is negative, so we can conclude that $x=\frac{4+2\sqrt{10}}{6}$. Note that this number is not rational.

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    $\begingroup$ If you use \cfrac, you can make the continued fraction look like this: $\displaystyle 1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+ \cdots}}}}}$ rather than like this: $\displaystyle 1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{1+\frac{1}{2+ \cdots}}}}}$ $\endgroup$ – Michael Hardy Jun 28 '12 at 22:14
  • $\begingroup$ Nice! Thanks a lot for the answer. $\endgroup$ – Jozef Jun 28 '12 at 22:15
  • $\begingroup$ @Michael: Ah, thanks for the LaTeX tip! $\endgroup$ – Zev Chonoles Jun 28 '12 at 22:15
  • $\begingroup$ just one thing,It is suppose to be $x=\frac{4+/-\sqrt{40}}{6}$ and so on :-) $\endgroup$ – Jozef Jun 29 '12 at 6:53
  • $\begingroup$ @Jozef: Haha, thanks for pointing that out! I'll fix it. That's what I get for typing faster than I was thinking :) $\endgroup$ – Zev Chonoles Jun 29 '12 at 6:57
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It's not rational if its simple continued fraction expansion repeats rather than terminating.

If you mean this: $$ 1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+ \cdots}}}}} $$ with $1,1,2$ repeating forever, then look at it like this: $$ x = 1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{\left( 1+\cfrac{1}{1+\cfrac{1}{2+ \cdots}}\right)}}} = 1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{x}}}. $$

You then have $$ x = 1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{x}}}. $$ Simplify the fraction and you've got a quadratic equation that you can solve for $x$.

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Since this continued fraction is infinite,this can't be a rational number(it is surely an irrational number).For more details, http://en.wikipedia.org/wiki/Continued_fraction

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