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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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S.C.
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    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 01 '19 at 19:18
  • @lulu To be fair, the quartic also has the solutions -1, 0 – PrincessEev Jan 01 '19 at 19:22
  • well, from what I know about algebra, I thought for a given polynomial equation, the highest degree of the polynomial represents the number of roots for that polynomial. So, each of these equations has more and more roots as you delay the substitution of x. I figured this then fundamentally changed solutions to the equation.

    In which case, when someone asks "What is the solution to this infinitely nested radical", my feeling is that there are infinitely many. Which is confusing.

     – S.C. Jan 01 '19 at 19:24
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    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 01 '19 at 19:25
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    @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 01 '19 at 19:25
  • This is already the case for much simpler expressions. If we have $\sqrt x = x-1$ then squaring yields $x=x^2+1$ which has two roots (only one of which is positive of course). But if we write it as $\sqrt x +1 =x$ then square we get $x^2=x+1-2\sqrt x$ which we can rearrange to get $x^2-x-1=-2\sqrt x$ and squaring again would now give a quartic. – lulu Jan 01 '19 at 19:29
  • I guess I do not quite understand why the "extra solution" must be discarded. Why are some solutions 'extraneous' while others are not? – S.C. Jan 01 '19 at 19:29
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    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 01 '19 at 19:29
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    @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 01 '19 at 19:30
  • ohhhh. So all of the extra solutions I am generating by delaying the substitution are solely negative numbers? If this is the case, is there a way of proving that? (i.e. how do I know that I will produce infinitely many negative solutions as I infinitely delay the substitution) – S.C. Jan 01 '19 at 19:31
  • Oh, I'm not even sure that is true (though it might be). In my example, say, squaring gives two positive solutions (despite what I incorrectly said earlier) but only one is a solution. For your particular problem it seems to me that the hard point is to show that the iterated radical actually converges to a value $x$. If you can show that then, as you noted, we have $x^2=x+1$ and $x>1$ so it is easy to get the Golden Ratio. – lulu Jan 01 '19 at 19:33
  • I feel like the argument would have to change if, instead of making a nest equation of square roots, I, instead, used cubic roots. Because then I will have delayed substitutions that are raised to odd powers half of the time. – S.C. Jan 01 '19 at 19:35
  • Convergence questions can be subtle. In this case, I'd think the best approach would be to let $x_n$ be the iterated radical with only $n$ $1's$, then show that $x_n$ is increasing and bounded by the Golden Ratio. That would suffice. – lulu Jan 01 '19 at 19:37
  • Because $\sqrt{x}$ is defined to be positive, each expression has a unique value. This value changes as the expressions get longer, so the issue is whether it converges to a limit. – timtfj Jan 01 '19 at 23:55

3 Answers3

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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.

Mark Bennet
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  • 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.C. Jan 01 '19 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 01 '19 at 19:40
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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 $\phi$.

PrincessEev
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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.

Jack D'Aurizio
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