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What does the below expression converge to and why? $$ \cfrac{2}{3 -\cfrac{2}{3-\cfrac{2}{3-\cfrac2\ddots}}}$$

Setting it equal to $ x $, you can rewrite the above as $ x = \dfrac{2}{3-x} $, which gives the quadratic equation $x^2 - 3x + 2 = 0 $, and the roots are $ 1 $ and $2 $, both positive. How do we know which to reject?

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    Hint - try looking at this as a recursively defined sequence - is it bounded? From what direction? Does is increase or decrease? Some induction might help. – GSofer Mar 02 '18 at 21:31
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    Ambiguity in the definition of a continued fraction ... see this question https://math.stackexchange.com/questions/417280/continued-fraction-fallacy-1-2 – Donald Splutterwit Mar 02 '18 at 21:34
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    @Hobart Pao You can get some insight from this video, https://www.youtube.com/watch?v=leFep9yt3JY&feature=youtu.be&t=2m59s –  Mar 02 '18 at 21:55
  • I was given this problem last year :) – Mr Pie Mar 03 '18 at 23:56

6 Answers6

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When we ask "What does this expression converge to?" we usually mean "If you cut off this infinite expression after a certain point, you get a number. As you cut off the expression later and later, that number gets closer and closer to some limiting value. What is that limiting value?"

However, with this continued fraction, we could truncate it in two ways: after a $2$, getting something like $$ \cfrac{2}{3 - \cfrac2{3 - \cfrac2{3- \cfrac2{3 - 2}}}}, $$ or after a $3$, getting something like $$ \cfrac{2}{3 - \cfrac2{3 - \cfrac2{3- \cfrac2{3 - \cfrac23}}}}. $$ With many continued fractions, both of these result in the same answer. Here, however, cutting the expression off when the number "at the end" is a $2$ always means that the whole thing simplifies to $2$; cutting it off when the number at the end is a $3$ gives a value approximately equal to $1$ (more precisely, it gives $\frac{2^{n+1}-2}{2^{n+1}-1}$, where $n$ is the number of $3$'s used, and this converges to $1$ as $n \to \infty$).

You could decide that the first kind of truncation is the kind you want, in which case the continued fraction converges to $2$. Or you can decide that the second kind of truncation is the kind you want, in which case the continued fraction converges to $1$. (Usually, the second kind of truncation gets used by convention for continued fractions.)

You can even decide that both kinds of truncations are fine, in which case the continued fraction does not converge: it alternates between values equal to $2$ and values close to $1$. This is the option that we usually pick for the sum $1 - 1 + 1 - 1 + 1 - 1 + \dots$, for instance: this has values of $1$ or $0$ depending on if we cut it off at an even or odd position, but there's no reason to pick one over the other, so we say it does not converge.

gen-ℤ ready to perish
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Misha Lavrov
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I think you need to question the assumption that you need to reject a solution.

You're probably thinking:

But this problem clearly only has one solution!

But why? Didn't you just prove $x = \dfrac{2}{3-x}$ has two solutions?

But that's an implicit equation. This is an explicit expression. An expression can't equal 2 things!

Sure it can. $\sqrt{-1}$ is an expression, and only equal to $i$ because we arbitrarily decided to ignore $-i$.
But if you want to ignore things arbitrarily, you can do that here too—that's just a preference.

But, like, come on. If you chop off this expression, it'll clearly only have one value. So if you keep chopping it off later and later, clearly it can't converge to two values?!

Oh, so you want convergence? In that case, there's no reason to believe there's exactly 1 solution. In fact, you're lucky if you get any solution. But if that's what you want, see the other answers here.


The fundamental problem here is that you have not defined the "$=$" operator between a finite value and expressions that have an infinite number of terms.

In fact, there are multiple definitions, and depending on which you choose, it may make sense to reject none, some, or all of the solutions.

user541686
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    I'd say that the problem is not the definition of "=" (which is just the usual definition of equality of real numbers) but rather the definition of how we take an expression with infinitely many terms and evaluate it to a real number. But this is a very minor nitpick. – Misha Lavrov Mar 03 '18 at 20:26
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    Almost upvoted, but '=' is not an operator that acts on expressions (or an operator at all, unless you want to be pedantic and call it a function from the product of some set $S$ with itself to $\{\text{true},\text{false}\}$), and calling it one is bound to cause a severe misunderstanding. – Deusovi Mar 03 '18 at 22:34
  • @Misha: what does "evaluate" mean if not "give an 'equal' 'value' to"? The definition of equality is the issue here. – user541686 Mar 03 '18 at 22:41
  • @Deusovi: Yeah but I didn't know what else to call it. – user541686 Mar 03 '18 at 22:41
  • It's not the definition of equality, but the definition of an infinite continued fraction. "Evaluating" is not something that happens inside mathematics, but outside it: when we say 2+3 = 5, we *mean* "the expression 2+3 and the expression 5 mean the exact same thing", not "when both sides are evaluated, the results are the same". The equals sign is a symbol *in* mathematics, not outside of it, where the computation is done. – Deusovi Mar 03 '18 at 22:44
  • (Put another way: once we know what the infinite continued fraction means, deciding equality is easy! It's the continued fraction that needs to be defined, not the equals sign.) – Deusovi Mar 03 '18 at 22:44
  • The problem is absolutely not with the $=$ operator. The problem is the $\ddots$ notation. Unlike the other notations used in the expression, it does not in designate a number, so there is no reason why the expression as a whole would designate a number. – Gilles 'SO- stop being evil' Mar 03 '18 at 23:25
  • @Deusovi: So you also have a problem with claiming that $\cfrac{2}{3 -\cfrac{2}{3-\cfrac{2}{3-\cfrac2\ddots}}} = \cfrac{2}{3 -\cfrac{2}{3-\cfrac{2}{3-\cfrac2\ddots}}}$ is true? Because presumably you would find yourself having to evaluate each side and then fail in doing that? – user541686 Mar 04 '18 at 03:33
  • @Mehrdad Are you saying you *don't* have a problem with claiming that $\frac{!^x}{(((((??} = \frac{!^x}{(((((??}$ is true? – Misha Lavrov Mar 04 '18 at 04:33
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    To illustrate that $=$ is not the problem, observe that any *other* question asked of the expression in question is also problematic. Is it even? Is it bigger than $3/2$? These questions don't involve $=$ but still require you to resolve some confusion before you can answer them. – Ben Millwood Mar 04 '18 at 08:42
  • @BenMillwood: What is the logic behind you assuming/deducing that the fact that $=$ is a problem means $>$ or $=\pmod{2}$ cannot also be problems? – user541686 Mar 04 '18 at 08:47
  • If I ask any question at all of an object and end up confused, it seems better to take issue with the object, than with all possible questions :) – Ben Millwood Mar 04 '18 at 08:53
  • @BenMillwood: I [literally just gave you an example to the contrary](/questions/2674205/what-does-this-converge-to-and-why/2674583?noredirect=1#comment5526272_2674583). There is no confusion involved when asking if this expression equals itself. The answer is true. And heck, given that fractions are understood to mean division, if you divided $2$ by the expression, there is again no confusion -- the result is equal to $3 -\cfrac{2}{3-\cfrac{2}{3-\cfrac2\ddots}}$. Problems only come up if you try to declare that to be equal to a number, or to another expression where equality is ill-defined. – user541686 Mar 04 '18 at 09:01
  • @Mehrdad Yes, I *would* have that problem. Once you define the object, then it's obviously true. But if an expression is undefined, then I'd say the question "Is it true that $\frac{**}{!?!+}=\frac{**}{!?!+}$?" is neither true nor false - it's *not a thing that can be answered*, because it doesn't mean anything. You say "there is no confusion involved when asking if this expression equals itself" -- but it's *not* an expression until it's defined. It doesn't represent anything - it's just a meaningless sequence of symbols. [cont. ...] – Deusovi Mar 04 '18 at 13:33
  • [...] Just because it *looks* similar to something you're familiar with doesn't mean you can automatically assume it *is* an object with those properties. (If you haven't defined *sin* yet, you would think you can cancel $n$ from $\frac{\sin x}n$. But you can't, since $\sin$ is not a multiplication of variables. Just because it *looks* like it is, doesn't mean you can assume it has that property.) – Deusovi Mar 04 '18 at 13:36
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You can look at the stability of the solutions. You start with some $x_0$ close to one of your solutions, and check if $$x_1=\frac{2}{2-x_0}$$ is closer to that solution. So if $x_0=1+\epsilon$, where $|\epsilon|\ll 1$, we have $$x_1=\frac{2}{3-(1+\epsilon)}=\frac{2}{2-\epsilon}\approx\left(1+\frac{\epsilon}{2}\right)$$ This is closer to $1$ than $x_0$, so the solution is stable.

We repeat the procedure for $x_0=2+\epsilon$. $$x_1=\frac{2}{3-(2+\epsilon)}\approx2(1+\epsilon)=2+2\epsilon$$ This $x_1$ is further from $2$ than $x_0$, so the solution is not stable.

Andrei
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    I think you need to discuss why stability of solutions is related to the question at hand, for this to be a suitable answer. – Federico Poloni Mar 03 '18 at 09:09
  • If an expression **converges** to a certain limit, one can use numerical approximations to find it. I showed that one of the solutions cannot be found using such procedure – Andrei Mar 03 '18 at 17:10
  • Sure you can approximate both solutions. The sequence $(2, 2/(3-2), 2/(3-2/(3-2)), \ldots)$ converges to $2$. The fact that some other sequences don't converge to $2$ is irrelevant: all it shows is that they aren't valid approximations of the limit of that particular sequence. – Gilles 'SO- stop being evil' Mar 03 '18 at 23:22
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Notations such as $2$, $3$, $\mathord{\bullet} - \mathord{\bullet}$, $\dfrac{\mathord{\bullet}}{\mathord{\bullet}}$ (where $\mathord{\bullet}$ stands for an expression) have a universally-accepted meaning which has a precise mathematical definition: the number two, the number three, subtraction, division. I'm not going to state those precise definitions here, the important thing is that these definitions exist. (There are in fact many possible definitions, and mathematicians might disagree on which one is the “best” one, but the important point is that they all agree that whichever definition they've picked, the other mathematicians' definitions are equivalent.)

On the other hand, the notation $\ddots$ (called ellipsis) does not have a precise mathematical meaning. When you use this definition, it's up to you to make sure that there is only one possible interpretation.

The ellipsis is often a convenient shortcut, but you need to be careful when using it. If you write $$ 1^2 + 2^2 + 3^2 + \ldots + n^2 $$ then everybody will figure out that you mean the sum of the squares of the $n$ first integers, which could be written formally as $\sum_{i=1}^n i^2$. If you write $$ \color{darkred}{1 + 4 + \ldots + n^2} $$ then it's still a good guess, but it's a guess which is not straightforward, so you should avoid it. And if you write $$ \color{red}{1 + \ldots + n^2} $$ then there's more than one sensible interpretation ($\sum_{i=1}^n i^2$ or $\sum_{i=1}^{n^2} i$?), so this is not a good notation at all.

With a nested expression like yours, the notation means “the limit of the sequence with increasing levels of nesting”. This definition only works if, for all the ways to split the nesting, the expression converges to the same limit. Otherwise the definition is ambiguous, so the expression does not denote a number at all.

Let's use one particular way of parsing the expression as nested pieces: take the sequence $$ 2, \quad \cfrac{2}{3 - 2}, \quad \cfrac{2}{3 - \cfrac{2}{3 - 2}}, \quad \ldots $$ which can be defined more formally as $$ \begin{cases} x_0 &= 2 \\ x_{i+1} &= \dfrac{2}{3-x_i} \\ \end{cases} $$ This sequence is constant: all terms have the value $2$, and in particular it converges to $2$.

Now let's we cut differently: $$ \cfrac{2}{3}, \quad \cfrac{2}{3 - \cfrac{2}{3}}, \quad \cfrac{2}{3 - \cfrac{2}{3 - \cfrac{2}{3}}} $$ which is more formally defined by $$ \begin{cases} y_0 &= \dfrac{2}{3} \\ y_{i+1} &= \dfrac{2}{3-y_i} \\ \end{cases} $$ Here we have a sequence that converges to $1$.

Since cutting the expression in two different ways results in different limits, the expression $$ \cfrac{2}{3 - \cfrac{2}{3 - \cfrac{2}{3 - \cfrac{2}{\ddots}}}} $$ is ambiguous. It does not define a number.

How do we know which limit to reject? There are two possible answers.

  • The notation is ambiguous, therefore the expression is invalid. We must reject both.
  • The notation is ambiguous, therefore it's up to the author to clarify. You tell us which one to accept.
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In fact this fraction is the limit of the following sequence$$x_{n+1}=\dfrac{2}{3-x_n}\\x_0=\dfrac{2}{3}$$now if $0<x_n<1$ we have $$2<3-x_n<3\to\\\dfrac{2}{3}<x_{n+1}=\dfrac{2}{3-x_n}<1$$therefore for all $n$ we have $$0.66<x_n<1$$ ant then the limit is $1$.

Mostafa Ayaz
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$x $ is the limit of the recurrent sequence defined by $$x_{n+1}=\frac {2}{3-x_n} =f (x_n) $$

$$f'(x)=\frac {2}{(3-x)^2} $$

$$|f'(1)|=\frac {1}{2}<1$$

$1$ is an attractive point. $$|f'(2)|=2>1$$ $2$ is a repulsive point.

The limit is $x=1$.

hamam_Abdallah
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