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$$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$

My answer is:

$$\begin{align} &A = 3 - \frac {2}{A}\\ \implies &\frac {A^2-3A+2}{A}=0\\ \implies &A^2-3A+2=0\\ \implies &(A-1)\cdot(A-2)=0\\ \implies &A=1\;\text{ or }\; A=2 \end{align}$$

I should note that I'm not sure if the above answer is true. Because I expected just one answer for A (A is a numeric expression), but I found two, $1$ and $2$. This seems to be a paradox.

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    $\begingroup$ It remains to show that one of them is not possible. $\endgroup$ – 1-___- Aug 6 '15 at 18:23
  • $\begingroup$ Alternatively, compute the first continued fraction then the second. Remember this sequence must converge to "something". $\endgroup$ – 1-___- Aug 6 '15 at 18:26
  • $\begingroup$ @user2770287 I have no idea how to write such a computer program. $\endgroup$ – mike Aug 6 '15 at 18:59
  • $\begingroup$ I meant the partial continued fraction. So you can do it by hand. $\endgroup$ – 1-___- Aug 6 '15 at 19:03
  • $\begingroup$ @user2770287 Have you tried? $\endgroup$ – mike Aug 6 '15 at 19:04
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Let us define two series. The first is \begin{align} a_1 &= 3 \\ a_2 &= 3 - \frac{2}{3} \\ a_3 &= 3 - \frac{2}{3 - \frac{2}{3}} \\ a_4 &= 3- \frac{2}{3 - \frac{2}{3 - \frac{2}{3}}} \\ &\vdots \\ a_{n+1} &= 3 - \frac{2}{a_n} \quad (*) \end{align} and \begin{align} b_1 &= 3 - 2 \\ b_2 &= 3 - \frac{2}{3-2} \\ b_3 &= 3 - \frac{2}{3 - \frac{2}{3-2}} \\ b_4 &= 3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{3-2}}} \\ &\vdots \\ b_{n+1} &= 3 - \frac{2}{b_n} \quad (**) \\ \end{align}

Note: This is the same recurrence relation $(*)$ or $(**)$ but with different start value $a_1 = 3$ and $b_1 = 1$.

For convergence we need $a_{n+1} - a_{n} \to 0$ or $a_n \to a$.

This way (in case of convergence) equations $(*)$ and $(**)$ have a limit $$ a = 3 - \frac{2}{a} \quad (\#) $$ which has indeed the solutions $a = 1$ and $a = 2$.

However that means we could also try $$ c_n = 3 - \frac{2}{c_{n+1}} $$ or $$ c_{n+1} = \frac{2}{3 - c_n} \quad (\#\#) $$ because it has the limit form $(\#)$.

Note that equation $(\#\#)$ is quite different from equation $(*)$ (see image below).

And indeed this recurrence relation $(\#\#)$ works too. Using $c_1 = 1$ will give $c_n \to 1$, Using $c_1 = 2$ will give $c_n \to 2$. Using $c_1 = 1000$ will give $c_n \to 1$.

So why is this? Still two solutions and the start value decides the limit.

Here is an image:

fixed points

The green graph is related to $(*)$: $$ f(x) = 3-\frac{2}{x} $$ the blue graph is related to $(\#\#)$: $$ g(x) = \frac{2}{3-x} $$ and the red graph is the identity function: $$ \mbox{id}(x) = x $$

We see that both $f$ and $g$ hit the identity at $x=1$ and $x=2$. Those points are fixed points of $f$ and $g$: \begin{align} x^* &= f(x^*) \\ x^* &= g(x^*) \end{align} And one could now try to apply the theory of fixed points, esp. properties of fixed point iterations. \begin{align} x_{n+1} &= f(x_n) \quad (\$) \\ x_{n+1} &= g(x_n) \end{align}

The fixed point iteration of $f$ is like the the iteration of original continued fractions (compare $(\$)$ with $(*)$ or $(**)$).

The theory behind can now help with statements about convergence and the dependency of start values.

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  • $\begingroup$ nice and complete solution! $\endgroup$ – mike Aug 6 '15 at 19:18
  • $\begingroup$ One of my favorite solutions so far on this site. $\endgroup$ – Chinny84 Aug 7 '15 at 10:49
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This continued fraction is the limit of the sequence $a_n=3-2/a_{n-1}$. Computing the first few terms shows that $2$ is the correct limit; if our initial term $a_1=3$ were different then the limit could be $1$.

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  • $\begingroup$ Good idea! This continued fraction could also be the limit of the sequence$a_{n-1}=3-2/a_{n}$. The limit might then be 2. $\endgroup$ – mike Aug 6 '15 at 19:11
  • $\begingroup$ can we use an initial start at 1? :-) $\endgroup$ – Math-fun Aug 6 '15 at 19:15
  • $\begingroup$ I guess yes because it IS the second solution (1) $\endgroup$ – mike Aug 6 '15 at 19:24
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Note that as we add more terms to the continued fraction, it oscillates between $1$ and slightly higher than $2$. $$ \begin{align} n&=1& 3&=3& 3-2&=1\\\\ n&=2& 3-\cfrac23&=\frac73& 3-\cfrac{2}{3-2}&=1\\\\ n&=3& 3-\cfrac{2}{3-\cfrac23}&=\frac{15}7& 3-\cfrac2{3-\cfrac2{3-2}}&=1\\\\ n&=4& 3-\cfrac2{3-\cfrac2{3-\cfrac23}}&=\frac{31}{15}& 3-\cfrac2{3-\cfrac2{3-\cfrac2{3-2}}}&=1\\\\ && 3-\cfrac2{\cfrac{2^n-1}{2^{n-1}-1}}&=\frac{2^{n+1}-1}{2^n-1}& 3-\cfrac21&=1 \end{align} $$ Therefore, the limit of the continued fractions with $2n-1$ twos and threes is $$ \lim_{n\to\infty}\frac{2^{n+1}-1}{2^n-1}=2 $$ and the limit of the continued fractions with $2n$ twos and threes is $$ \lim_{n\to\infty}1=1 $$ Therefore, one value represents the limit of an odd number of twos and threes and the other value represents the limit of an even number of twos and threes.

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  • $\begingroup$ I have clarified what I was trying to say before to make it more explicit. If the reason for the downvote was something else, please let me know. $\endgroup$ – robjohn Aug 7 '15 at 7:03
  • $\begingroup$ Brilliant answer! $\endgroup$ – Harish Chandra Rajpoot Aug 7 '15 at 12:59
  • $\begingroup$ @HarishChandraRajpoot: Thanks. I just wish the downvoter felt the same way ;-) $\endgroup$ – robjohn Aug 7 '15 at 17:23
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Your two solutions are correct. You can verify them as follows.

$$\color{blue}{1} = 3 - 2=3-\frac{2}{\color{\red}{1}}\tag{1}$$ Now replace the red 1 with the blue 1, which equals to the right hand side in (1).

$$\color{blue}{2} = 3 - 1=3-\frac{\color{cyan}{2}}{\color{\red}{2}}\tag{2}$$ Now replace the red 2 with the blue 2, which equals to the right hand side in (2).

I heard from the internet that this is one of the ways Ramanujan created some of his equalities.

Now 2 questions for you, (1) what if you replace the cyan 2 with blue 2 and so on so forth? (2) what if you replace the cyan 2 with blue 2 and then replace the red 2 with blue 2 and so on so forth?

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  • $\begingroup$ If one write a computer program to solve it, at the end which result the program will output one or two? $\endgroup$ – user153245 Aug 6 '15 at 18:40

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