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the question is to find the value of this ugly non-stopping fraction $$\frac{1}{2+\frac{1}{4+\frac{1}{4+\frac{1}{\ldots}}}}$$.

I have totally no clue; thanks for the help! How am I suppose to solve this thing? This thing certainly looks ugly.

Thank you for all the help.

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  • $\begingroup$ Are the terms in $\ldots$ repeating like $4 + 1/(\ldots)$? $\endgroup$ – Tunococ Aug 3 '13 at 2:55
  • $\begingroup$ Can you at least do a search here on MSE first before ... copy and pasting your question? I'm sure (similar) questions has been asked numerous times already. $\endgroup$ – user67258 Aug 3 '13 at 2:57
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    $\begingroup$ @user88786: Did you create five different accounts to ask five questions in a row? One account would work. $\endgroup$ – Amzoti Aug 3 '13 at 2:57
  • $\begingroup$ It is usually a good idea to give some idea of what the $\dots$ represent. I can't tell. Is it fours all the way down? $\endgroup$ – Thomas Andrews Aug 3 '13 at 3:10
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Let $x$ be your continued fraction so that:

$$x = \cfrac{1}{2+\cfrac{1}{4+\cfrac{1}{4+\cdots}}}.$$

Notice the repeating nature after a while in the continued fraction (this is very important). We want to make use of this. To do so let's invert both sides:

$$x^{-1} = 2 + \cfrac{1}{4+\cfrac{1}{4+\cfrac{1}{4+\cdots}}}.$$

Or written a slightly different way..

$$x^{-1} - 2 = \cfrac{1}{4+\cfrac{1}{4+\cfrac{1}{4+\cdots}}}.$$

We are now left with evaluating the continued fraction on the right. This is actually simpler than it looks because of the self-similar behavior.

Define the following:

$$y = \cfrac{1}{4+\cfrac{1}{4+\cfrac{1}{4+\cdots}}}.$$

Then:

$$y^{-1} = 4 + \cfrac{1}{4+\cfrac{1}{4+\cfrac{1}{4+\cdots}}}.$$

Or..

$$y^{-1} = 4 + y.$$

Can you solve this for $y$? Do you see how this solves the problem at hand?

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  • $\begingroup$ @MJD Wow thanks for that. I didn't know that command existed. $\endgroup$ – Cameron Williams Aug 3 '13 at 3:01
  • $\begingroup$ Glad to be of service. See also… $\endgroup$ – MJD Aug 3 '13 at 3:03
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    $\begingroup$ Also, $y=2+\frac1{2+y}$ work as well. $\endgroup$ – user67258 Aug 3 '13 at 3:03

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