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In this question we increased solution domain by squaring both sides of equation but what about this one ?

Here the question is to evaluate $1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots }}}}}$


$$x-1=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots }}}}}$$

$$\cfrac{1}{x-1}=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots }}}}}$$


$$x^2-x-1 = 0$$

$$x=\frac{1\pm \sqrt{5}}{2}$$

Now, it is obvious that answer can't be negative so :


share|cite|improve this question – Gahawar Aug 23 '14 at 18:07
@Nick It's just $\frac{1}{\frac{1}{a}}=a$ – Paul Sundheim Aug 23 '14 at 18:26
Maybe this says something about the ambiguity of continued fractions. – Paul Sundheim Aug 23 '14 at 18:28
up vote 11 down vote accepted

I think it's obviously a negative number of magnitude less than 1. After all, $1 / x$ is clearly a negative number of magnitude greater than $1$!

Okay not really. But the reason you think it's obviously positive is that there is a hidden condition in your question: you intend $x$ to be the limit of the sequence $$1, 1 + \frac{1}{1}, 1 + \frac{1}{1 + \frac{1}{1}}, \ldots $$ rather than, say, just a number that satisfies the self-referential relationship $x = 1 + 1/x$. It is this hidden condition that differentiates between the two solutions of this equation.

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+1 for the hidden condition, which is pretty standard when writing continued fractions. – Lee Mosher Aug 23 '14 at 18:45
@LeeMosher Thanks for your answer. one little question : if we were asked what is $1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots }}}}}$ is it wrong to say $x=\frac{1-\sqrt{5}}{2}$ ? – Shabbeh Aug 24 '14 at 2:42
@Shabbeh: The equation that you solve, namely $x^2 - x - 1 = 0$, is a quadratic equation and so has two roots. There are many situations in mathematics where you are asked to solve an equation with multiple roots but with some kind of additional constraint which rules out one or more of the roots. This is one of those situations, an unfamiliar one for you. More familiar to you is the situation where the quadratic equation is obtained by squaring something. Another is maximizing a function by setting its derivative equal to zero and solving. You'll run into more such situations. – Lee Mosher Aug 24 '14 at 12:21
@Shabbeh: The usual interpretation of your symbols is that you have some initial value $x_0$ and recursively define $x_n = 1 + 1/x_{n-1}$, and the problem is to compute $$x = \lim_{n \to \infty} x_n$$ (assuming it exists) Your solution method is justified as $$1 + 1/x = \lim_{n \to \infty} (1 + 1/x_n) = \lim_{n \to \infty} x_{n+1} = x $$ so you get the equation, but you've forgotten that you're trying to compute the limit. Maybe if you really wanted to you could rationalize something about the transformation, but it's not clear to me how to do so. – Hurkyl Aug 24 '14 at 18:56
Here's another example: $x = 2(2(2(2(2(\ldots)-1)-1)-1)-1)-1$. This satisfies the very simple equation $x = 2x - 1$. But it doesn't tell you whether the correct solution is $x = 1$, $x = +\infty$, or $x = -\infty$ (or if the limit exists at all). You can't know the value of $x$ unless you know what initial value you are supposed to start with. – Hurkyl Aug 24 '14 at 18:58

Let $f(z)=1+\frac1z$. The function $f$ has two fixed points, which are $\frac{1\pm\sqrt{5}}{2}$.

At first, when you write $$x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}}$$ you're defining $x$ to be the particular fixed point of $f$ that attracts the orbit of $1$, if it exists. It so happens that $x$ does exist, and it equals $\frac{1+\sqrt{5}}{2}$.

Later, when you write $$x=1+\frac1x$$ you're asserting that $x$ is some fixed point of $f$, but this equation doesn't specify which fixed point that is.

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If we set $x = \dfrac{1}{-1 +\underbrace{\dfrac{1}{-1+\dfrac{1}{-1+\dfrac{1}{-1+\cdots}}}}_{x}}$, then we have $x = \frac{1}{-1 + x}$, which gives also $x^2 - x - 1=0$

That's a similar presentation of the negative answer.

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