Does a given infinite nested radical have infinitely many solutions?

Given a standard infinitely nested radical such as:

$$x = \sqrt{1 + \sqrt{1 + \sqrt{ 1 + ...}}}$$

depending on where you choose to first substitute $$x$$ in the nest, aren't there infinitely many solutions when solving for $$x$$?

For example, you could substitute $$x$$ as follows:

$$x = \sqrt{1 + x}$$

in which case, you get the equation $$x^2 - x - 1 = 0$$.

However, couldn't I just as well substitute $$x$$ in the following way:

$$x = \sqrt{ 1 +\sqrt{1 + x}}$$

in which case, you get the equation $$x^4 - 2x^2 - x = 0$$

Obviously, you could keep this up and generate infinitely different equations depending on where in the infinite nested loop you decided to substitute $$x$$. So, my questions are:

1. Is this an actual phenomenon or did I violate some sort of rule?

2. What are the implications of this?

• Why do you think these give different answers? Both seem to give $\frac {1+\sqrt 5}2$. Note that, if the value exists at all, it is clearly $>0$ so we can discard non-positive roots. – lulu Jan 1 at 19:18
• But those solutions are extraneous and therefore discarded, regardless of how many there are. The only correct solution is the golden ratio, as pointed out. – KM101 Jan 1 at 19:25
• @S.Cramer True but each time you square you pick up an "extra" solution which must be discarded. Here the fact that any value for this expression would have to exceed $1$ lets us discard the irrelevant values. – lulu Jan 1 at 19:25
• Because $\sqrt{1+{\sqrt{1+…}}}$ is clearly positive. Any non-positive solution obtained is the solution to the new polynomial constructed, not the original nested radicals. – KM101 Jan 1 at 19:29
• @S.Cramer The problem is that squaring "loses" signs. Study my example of $\sqrt x = x-1$. That only has a single real solution (easy to see) but squaring also picks up the real solution to $\sqrt x = 1-x$. – lulu Jan 1 at 19:30

Every time you square this you add extraneous solutions which arise from taking the negative value of some of the square roots. The use of $$x$$ implies that the pattern is recurrent, but the quartic arising from squaring twice corresponds to the four possible choices for the square roots you have removed.

• Does this logic apply if, instead of square roots, I used cube roots to construct this nesting? Then, I would have odd powered solutions half of the time. – S.Cramer Jan 1 at 19:37
• @S.Cramer There are three (complex) cube roots of any non-zero real number - so you have three solutions of the resulting cubic - but quite often only one of these is a real number, corresponding to the original expression. – Mark Bennet Jan 1 at 19:40

No, there aren't infinite solutions, because $$f(x)=\sqrt{x+1}$$ is a contraction of $$[1,2]$$ ($$\frac{1}{2\sqrt{3}}\leq f'(x)\leq \frac{1}{2\sqrt{2}}$$ for any $$x\in\left[1,2\right]=I$$), hence the Banach fixed point theorem ensures that the iteration $$x_{n+1} = \sqrt{1+x_n}$$ with starting point $$x_0=0$$ (such that both $$x_1$$ and $$x_2$$ belong to $$I$$) converges to the unique solution of $$f(x)=x$$ in $$I$$, i.e. $$\frac{1+\sqrt{5}}{2}$$. In fancy terms,

$$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\ldots}}}}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}$$ In general, when squaring both sides of an equation involving one or many square roots you may introduce spurious solutions.

The issue is the convergence of the nested radical.

Let $$(x_n)$$ be the sequence of iterated radicals. We'll have

$$x_n = \left\{\begin{matrix} 1 & n = 1\\ \sqrt{1+x_{n-1}} & n > 1 \end{matrix}\right.$$

A sequence can logically only converge to one unique value, i.e. $$(x_n) \to x$$. Take $$n \to \infty$$ and we have the infinite nested radical described,

$$x = \sqrt{1 + \sqrt{1+\sqrt{...}}}$$

Suppose we have some number $$\alpha$$ we want to express as an infinite root. A video by blackpenredpen on YouTube goes over generating such expressions, and touches on why some of these roots may seem to be equal to multiple values or have multiple solutions - namely, the implicit introduction of extraneous solutions.

We can start by saying the following: let $$x = \alpha$$ be a solution to our hypothetical infinite root. Then, throwing another arbitrary constant $$\beta$$ in, where $$\beta \neq \alpha$$,

\begin{align} x = \alpha &\implies x - \alpha = 0 \\ &\implies (x - \alpha)(x - \beta) =0 \\ &\implies x^2 -(\beta+\alpha)x + \beta\alpha = 0 \\ &\implies x^2 = (\beta+\alpha)x - \beta \alpha \\ &\implies x = \sqrt{(\beta+\alpha)x - \beta \alpha} \\ \end{align}

And from here, we can use this sort of recursive definition to generate our infinitely nested radical:

\begin{align} x &= \sqrt{ -\beta \alpha + ( \beta + \alpha ) x } \\ &= \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) x } } \\ &= \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) x } } } \\ &= \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) ... } } } \\ \end{align}

Notice how through all this, the other root $$\beta$$ still was introduced as an extraneous solution. And of course our chose to multiply by $$(x - \beta)$$ was arbitrary: we could've gone

\begin{align} x = \alpha &\implies x - \alpha &= 0 \\ &\implies (x - \alpha)(x - \beta_1) &=0 \\ &\implies (x - \alpha)(x - \beta_1)(x - \beta_2) &=0 \\ &\implies (x - \alpha)(x - \beta_1)(x - \beta_2)(x - \beta_3) &=0 \\ &\implies (x - \alpha)(x - \beta_1)(x - \beta_2)(x - \beta_3)...(x - \beta_n) &=0 \\ \end{align}

and thus introduced $$n$$ arbitrary constants $$\beta_1 ... \beta_n$$, all not equal to $$\alpha$$, as these "fake solutions." (Fake in that we begin with the presumption that $$\alpha$$ is the only solution, so to say $$\beta_k$$ for any $$k$$ is a solution when $$\beta_k \neq \alpha$$ is an immediate contradiction.)

The algebra would be a pain in the butt in constructing these radicals for $$n>1$$, but it's doable. Our situation when presented with solving such a root, instead of constructing it, is essentially the same, but in reverse: we have no inherent idea of how many arbitrary solutions might have been generated, and indeed in a sense infinitely many were depending on your interpretation of the root and its recursion. So we always have to hearken back to what would make the root a proper solution:

That being, convergence of the iterated root to that value.

For example, for the sequence $$(x_n)$$ introduced initially, you can show that:

• It is monotone increasing
• It is bounded above by $$\phi = (1 + \sqrt 5) / 2$$
• We have $$\sup(x_n) = \phi$$ as well
• As a consequence of the previous statements, $$(x_n) \to \phi$$

Thus, $$\phi$$ is the only actual solution to this nested radical. Sure, you can have manipulations and such that supply other solutions - but take $$x_1, x_2, x_3, x_4, ...$$, take the limit as $$n \to \infty$$ of $$(x_n)$$: it's not going to approach any of them but $$\phi$$.

• Nice answer. I’m going to nitpick purely because I can! $(x_n) \to \phi$ because the $(x_n)$ is monotonically increasing and $\sup (x_n) = \phi$, not simply because it is increasing and bounded above by $\phi$. – 雨が好きな人 May 7 at 18:35
• Good point, I'll add that in. Thanks! – Eevee Trainer May 7 at 18:37