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?

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    $\begingroup$ 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. $\endgroup$ – lulu Jan 1 '19 at 19:18
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    $\begingroup$ 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. $\endgroup$ – KM101 Jan 1 '19 at 19:25
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    $\begingroup$ @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. $\endgroup$ – lulu Jan 1 '19 at 19:25
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    $\begingroup$ 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. $\endgroup$ – KM101 Jan 1 '19 at 19:29
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    $\begingroup$ @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 $\sqrt x = 1-x$. $\endgroup$ – lulu Jan 1 '19 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.

  • $\begingroup$ 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. $\endgroup$ – S.Cramer Jan 1 '19 at 19:37
  • $\begingroup$ @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. $\endgroup$ – Mark Bennet Jan 1 '19 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 $\phi$.

  • $\begingroup$ 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 $\phi$. $\endgroup$ – 雨が好きな人 May 7 '19 at 18:35
  • $\begingroup$ Good point, I'll add that in. Thanks! $\endgroup$ – Eevee Trainer May 7 '19 at 18:37

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