Why is $1 - \frac{1}{1 - \frac{1}{1 - \ldots}}$ not real?

Question

So we all know that the continued fraction containing all $1$s...

$$ x = 1 + \frac{1}{1 + \frac{1}{1 + \ldots}}. $$

yields the golden ratio $x = \phi$, which can easily be proven by rewriting it as $x = 1 + \dfrac{1}{x}$, solving the resulting quadratic equation and assuming that a continued fraction that only contains additions will give a positive number.

Now, a friend asked me what would happen if we replaced all additions with subtractions:

$$ x = 1 - \frac{1}{1 - \frac{1}{1 - \ldots}}. $$

I thought "oh cool, I know how to solve this...":

\begin{align} x &= 1 - \frac{1}{x} \\ x^2 - x + 1 &= 0. \end{align}

And voila, I get...

$$ x \in \{e^{i\pi/3}, e^{-i\pi/3} \} .$$

Ummm... why does a continued fraction containing only $1$s, subtraction and division result in one of two complex (as opposed to real) numbers?

(I have a feeling this is something like the $\sum_i (-1)^i$ thing, that the infinite continued fraction isn't well-defined unless we can express it as the limit of a converging series, because the truncated fractions $1 - \frac{1}{1-1}$ etc. aren't well-defined, but I thought I'd ask for a well-founded answer. Even if this is the case, do the two complex numbers have any "meaning"?)

Answer 1

You're attempting to take a limit.

$$x_{n+1} = 1-\frac{1}{x_n}$$

This recurrence actually never converges, from any real starting point. Indeed, $$x_2 = 1-\frac{1}{x_1}; \\ x_3 = 1-\frac{1}{1-1/x_1} = 1-\frac{x_1}{x_1-1} = \frac{1}{1-x_1}; \\ x_4 = x_1$$

So the sequence is periodic with period 3. Therefore it converges if and only if it is constant; but the only way it could be constant is, as you say, if $x_1$ is one of the two complex numbers you found.

Therefore, what you have is actually basically a proof by contradiction that the sequence doesn't converge when you consider it over the reals.

However, you have found exactly the two values for which the iteration does converge; that is their significance.

Alternatively viewed, the map $$z \mapsto 1-\frac{1}{z}$$ is a certain transformation of the complex plane, which has precisely two fixed points. You might find it an interesting exercise to work out what that map does to the complex plane, and examine in particular what it does to points on the real line.

Answer 2

When you substitute $a_n=a_{n+1}=x$ in $$a_{n+1}=1-\frac{1}{a_n}$$ you assume that the sequence converges to a fixed-point.

If this assumption is true (as in the + case), this method will help you find the fixed-point.

However, since the sequence does not converge, the solution of $$x=1-\frac{1}{x}$$ cannot be the fixed-point (since there is none).

Answer 3

This general method really is used. In the 1981 book Zetafunktionen und Quadratischer Körper by D. B. Zagier, he uses $$ x = n_0 - \frac{1}{n_1 - \frac{1}{n_2 - \frac{1}{n_3 - \ldots}}} $$ with $n_1,n_2,n_3,\ldots \geq 2$ as his basic way to represent quadratic irrationals. In the question above, the OP has all the $n_j = 1,$ which Zagier forbids. Zagier begins with this on page 126. It is necessary for him to do this because he wants to define "reduced" indefinite binary quadratic forms (page 122) as $A x^2 + B xy + C y^2$ with $B^2 - 4AC> 0$ but not a square, and $A>0, C>0, B > A+C.$ Here we go, page 126: the real number $w$ is the larger root of $Ax^2 - Bx + C=0,$ where $\langle A,B,C\rangle$ is reduced, if and only if the continued fraction (with the minus signs) for $w$ is purely periodic. Good, that is exactly how this should work. Meanwhile, just as with ordinary continued fractions, for finite continued fractions we do not want $n_j = 1,$ as that just replaces the integer $n_{j-1}$ by $n_{j-1} - 1.$ Of course, for infinite fractions we need $n_j \geq 2$ for convergence.

Oh, Gauss-Lagrange reduced indefinite forms, integer coefficients, have $AC<0, B > |A+C|.$ These go together with traditional continued fractions. Very similar theorem about purely periodic continued fractions.

Answer 4

I guess that what you're asking for is how the the imaginary unit, i.e. the square root of $-1$ is involved. Indeed, it comes from the known identity between continued fractions and continuous square roots, i.e. $$ \sqrt{a-b\sqrt{a-b\sqrt{a-b\sqrt{a-b\sqrt{\cdots}}}}}= -\cfrac{a}{b-\cfrac{a}{b-\cfrac{a}{b-\cfrac{a}{\ddots}}}} $$ Then you have in your case $a=1$ and $b=1$ you have $$ \sqrt{1-1\sqrt{1-1\sqrt{1-1\sqrt{1-1\sqrt{\cdots}}}}}= -\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\ddots}}}} $$ The solution is the well known solution of the equation $$ x=a/(b+x)$$ which brings to the result you found.

Answer 5

For any continued fraction we have some inequalities, that $[a]< [a,b,c]<[a,b]$ or with big fractions:

$$ a < a + \frac{1}{b + \frac{1}{c}} < a + \frac{1}{b} \tag{$\ast$}$$

What should the values of $a,b,c$ be in your case. Let's try adjusting the signs little bit:

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

Looks like $a = 1, b = -1, c = x$. Do we have that $[1] < [1,-1,x] < [1,x]$ ? We need $a,b,c > 0$. In your case we have the sequence:

$$ 1 \to 1 - \frac{1}{1} = 0 \to 1 - \frac{1}{0}= \infty \to 1 + \frac{1}{\infty}= 1$$

Then $1 \to 0 \to \infty \to 1$ is a cycle of size 3.

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This still seems like a cop-out to me. We should look for an explanation why normal operations like $x \to x+1$ and $x \mapsto \frac{1}{x}$ takes us outside the realm of the fractions $\mathbb{Q}$.

Answer 6

Have you even looked at the approximations, i.e., what happens if you stop filling up the "$\ldots$"? You get things that look like this: $$ 1-\frac1{1-\frac1{1-\frac1{\underbrace{\color{red}{1-\frac1{1}}}_0}}}$$ and that means that there is always some division by zero lurking deep inside. Hence we certainly cannot define this sort-of continued fraction as the limit of its finite approximants.

Answer 7

Note that the answer, $x =1-\frac{1}{1-x}$.

$(x-1)(1-x)=1$

$-x^2+2x^2-1=1$

$-x^2+2x^2-2=0$

Now, take the quadratic formula. You get $x=1\pm i $

Lets look into the quadratic, and take the rest for real within the quadratic formula.

If you have a formula in the form of $ax^2+bx+c=0$, you can take $b^2-4ac$. If that value is positive, it is real. If not, it is imaginary. Since $a=-1, b=2,c=-2$, $b^2-4ac=4-(4\cdot(-1)\cdot(-2))=-4$, which means that the square root of that number will be an imaginary number, as shown above.

Answer 8

what have you done can be seen as a proof of divergence, we assume that the given continued fraction converges to a finite real value $x$, thus once you prove that $x$ is a complex number with $ Im(x) \neq 0$ you get the contradiction.