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If I have the fraction,

\begin{equation} \cfrac{1}{1-\cfrac{1}{1-1}} \end{equation}

the value of it is undefined because it involves a division by zero.

The same holds for a finite number of iterations as shown below:

\begin{equation} (1) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-1}}} \ \ \ \ \ \ (2) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-1}}}} \ \ \ \ \ \ (3) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-1}}}}} \end{equation}

But for an infinite number of iterations the fraction becomes,

\begin{equation} \cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\space\cdots}}}} \end{equation}

and I can write,

\begin{equation} \begin{split} & \cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\space\cdots}}}}=x \\\\ & \implies \frac{1}{1-x}=x \\\\ & \implies x^2-x+1=0 \\\\ & \implies x = \frac{1\pm\sqrt{3}i}{2} \end{split} \end{equation}

and so it has a complex value.

This does not make intuitive sense to my mathematically illiterate mind : the starting fraction was undefined and so were its finite iterations, so why would the infinite iteration take on a complex value?

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    $\begingroup$ It would not make sense if you assume that this expression converges to any value. When you do assume that it is $x$, you assume that it converges too. $\endgroup$ Commented Aug 22 at 10:04
  • $\begingroup$ So by letting the infinite fraction equal $x$ meant making the assumption that $x$ is a finite value (whether real or complex)? $\endgroup$ Commented Aug 22 at 10:32
  • $\begingroup$ To show there is no convergence of repeated applications of $f(x)=\frac1{1-x}$ to real numbers: if $x \not\in \{0,1\}$ then $f(f(f(x)))=x$ so you have a cycle and one of $f(f(f(x)))$, $f(f(x))$, $f(x)$ is greater than $1$, one is less than $0$ and one is between $0$ and $1$. $\endgroup$ Commented Aug 23 at 19:39

2 Answers 2

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By definition, a continued fraction has a finite value if and only if the sequence of the so-called convergents (the results of the iterations) converges.

The fact that you can get an equation that must be fulfilled in case that the continued fraction has a finite value does not mean that we can assign a finite value to that continued fraction. In other words:

"continued fraction has finite value $\Longrightarrow$ equation holds for that value"

but not the other way round.

Your sequence of convergents does not converge. Therefore there is no finite value associated with the continued fraction.

It is similar with other sequences or series. Famous example, you can perform manipulations on the series $\sum_{n=1}^{\infty} n$ which make it seem that its value is $-\frac{1}{12}$, but this is obviously nonsense when we apply the usual meaning of summation.

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  • $\begingroup$ "The fact that you can get an equation that must be fulfilled in case that the continued fraction has a value does not mean that we can assign a value to that continued fraction." - Could you please elaborate more on that? $\endgroup$ Commented Aug 22 at 10:29
  • $\begingroup$ You derived an equation (specifically $x^2-x+1=0$) from the continued fraction. If the value of the continued fraction were $x$, this equation would obviously have to be fulfilled. However, you cannot "revert" this logic. You cannot say "If the equation is fulfilled, then its solution $x$ is the value of the continued fraction." It only works one way. You do not know if the continued fraction has a value at all in the first place. In order to calculate the value of a continued fraction, you always have to do two things: Get the solution candidates and show the convergence for one of them. $\endgroup$ Commented Aug 22 at 10:50
  • $\begingroup$ Thanks, I get it now. Since the fraction doesn't have a value in the first place any solutions to any equations derived from it are meaningless, correct? $\endgroup$ Commented Aug 22 at 10:56
  • $\begingroup$ Yes, this is correct. $\endgroup$ Commented Aug 22 at 11:08
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(A) Consider $S=1-1+1-1+1-1\cdots$
It alternates between $0$ & $+1$ , hence there is no limiting value.
We can still attempt this : $S=1-1+1-1+1-1\cdots=1-(1-1+1-1+1\cdots)=1-S$
Hence $2S=1$ where we get $S=1/2$ ??!!
We should not make the Equation without checking whether it "Converges" , otherwise the Equation will give invalid Solution , because the "Convergence Assumption" was invalid to start with . . . .

(B) Consider your Case , where we can try making the Equation without checking for Convergence :

\begin{equation} (E1) \space\cfrac{1}{1-\cfrac{1}{1-x}}=x \end{equation}

$(x-1)/x=x$
We get $x^2-x+1=0$ here.

\begin{equation} (E2) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-x}}}=x \end{equation}

We get $x=x$ here ??!!
Every number satisfies this Equation !! More-over , all numbers will work out even with the Current Continued Fraction too . . . .

\begin{equation} (E3) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-x}}}}=x \end{equation}

$1/(1-x)=x$
We get $x^2-x+1=0$ here.

\begin{equation} (E4) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-x}}}}}=x \end{equation}

$(x-1)/x=x$
We get $x^2-x+1=0$ here.

\begin{equation} (E5) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-x}}}}}}=x \end{equation}

We get $x=x$ here too ??!!

\begin{equation} (E6) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-x}}}}}}}=x \end{equation}

$1/(1-x)=x$
We get $x^2-x+1=0$ here.

Consistency is clearly missing , there is no Unique Equation covering how-ever deep we make the Continued Fraction Structure go.

(C) Consider your Case , where we try checking for Convergence :

Let us take the extended real numbers where we have $1/0=\infty$ , while $1/\infty=0$

\begin{equation} (C1) \space\cfrac{1}{1-\cfrac{1}{1-1}}=0 \end{equation}

\begin{equation} (C2) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-1}}}=1 \end{equation}

\begin{equation} (C3) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-1}}}}=\infty \end{equation}

\begin{equation} (C4) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-1}}}}}=0 \end{equation}

\begin{equation} (C5) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-1}}}}}}=1 \end{equation}

\begin{equation} (C6) \space\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-1}}}}}}}=\infty \end{equation}

Consistency is clearly missing here , there is no Unique Convergent Value covering how-ever deep we make the Continued Fraction Structure go.

(D) Try the same with all $-$ changed to $+$ :
We will easily see the Convergence ; more-over , we will get the same Equation , no matter how deep we make the Continued Fraction Structure.
Eventually , we will get something involving $\sqrt{5}$ . . . .

(E) SUMMARY :

Check for Convergence before making the Equation , else the Equation is going to be meaningless towards getting value of the given non-number !

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    $\begingroup$ I like the way you explain, thanks for putting in the effort $\endgroup$ Commented Aug 22 at 16:50
  • $\begingroup$ At the very end you advise checking convergence first “or else” things will be meaningless towards the goal. That is not necessarily true! Playing around with an expression may reveal what its value ought to be, giving an insight leading to a proof of the desired convergence. An example: if I don’t know what $1+r+r^2+r^3+\cdots$ should be and call it $S$, then playing around suggests $rS=S-1$, so $S=1/(1-r)$. Now that we have a goal in mind, let $S_n=1+r+\cdots+r^n$ and consider $S_n-1/(1-r)$. It’s $((1+r+\cdots+r^n)(1-r)-1)/(1-r)=-r^{n+1}/(1-r)$, which tends to $0$ if $|r|<1$. $\endgroup$ Commented Aug 23 at 11:32
  • $\begingroup$ The point of my previous example is that it did not require the trick of knowing ahead of time the formula for summing the terms in a finite geometric progression. Subtracting off the desired limit, once we found it by symbol-pushing without even knowing a criterion under which convergence occurs, leads us to a concise error term and at that point we can see exactly when that error term tends to $0$. $\endgroup$ Commented Aug 23 at 11:38
  • $\begingroup$ The answer by Bill Dubuque in math.stackexchange.com/questions/1078781 is another example of the idea that playing around with something whose meaningfulness you don’t already know can provide an important clue to its existence, which can then be checked by other methods: in a ring where $ab \not= ba$, if $1-ba$ has a left, resp. right, inverse then $1-ab$ has a left, resp. right, inverse. The amusing aspect is that the appropriate inverse of $1-ab$ is discovered by using infinite series even when the ring isn’t assumed to have a topology giving meaning to infinite series. $\endgroup$ Commented Aug 23 at 11:57
  • $\begingroup$ (1) I get what you want to convey , @KCd , with which I agree. (2) My own example would be some limit $L=\infty/\infty$ , whose Existence I would assume and use LH rule to get some new equivalent limit maybe $L=L^2/2+1/2$ , hence I would say that limit is $L=1$ , without even knowing the Existence. I will still have the show the Existence. (3) Likewise , your Examples are good , though we will still have to show the Existence. (4) I was merely restricting to the Case where the Existence is Negative , hence I said "[or] else .... meaningless towards getting value of the given non-number" !! $\endgroup$ Commented Aug 25 at 12:50

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