# Repeated nested roots

Quite some years ago, I remember being asked the following question:

Suppose $\alpha = \sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}$, what is $\alpha$.

The solution was given by squaring $\alpha$ and solving the arising equation $\alpha^2-\alpha-2 = 0$, to give $\alpha=2$ under the reasonable assumption that $\alpha$ has to be positive.

In the years after this, I passed on the question to new young and aspiring mathematicians, without giving it much thought. Recently, however, I spent some time thinking about it again and noticed some peculiar things.

If we generalise this question to the case where $\alpha = \sqrt{a+\sqrt{a+\sqrt{a+\ldots}}}$ for some $a\in\mathbb{R}$, I wondered how this would behave. We can still square this equation to obtain $\alpha^2-\alpha-a=0$ and thus $\alpha = \frac{1}{2}(1\pm\sqrt{4a+1})$. This is where the questions I have arise. First remark that the solution only makes sense for $a\geq -\frac{1}{4}$, if we want to remain in $\mathbb{R}$.

We saw before that for $a=2$, the choice for $\pm$ would be $+$. If we take $a=0$, we want $\alpha$ to be $0$ as well and so the choice there suddenly switches to $-$. I wonder if there's any intuitive explanation why the choice of solution for $\alpha$ would suddenly switch at $a=0$. (for $a>0$ the root is $>1$, so we will certainly want to choose $+$ for $\pm$)

The other strange thing that arises from the solution of $\alpha$ is that if $a=-\frac{1}{4}$ we would find $\sqrt{a+\sqrt{a+\sqrt{a+\ldots}}}=0$, which completely baffles me. What is the point that I'm missing? Is there something about convergence of $\sqrt{a+\sqrt{a+\sqrt{a+\ldots}}}$ that causes troubles that I so far fail to see?

(if you consider this question to be wrongly tagged, please alter the tas and remove this bracketed statement)

• – Aryabhata Apr 17 '13 at 0:46

I think the resolution of these strange observations is through remembering that the nested radical is a limit: $\sqrt{a+\sqrt{a+\sqrt{a+\ldots}}} = \lim_{n\to\infty} b_n$ where $b_1=\sqrt a$, $b_2=\sqrt{a+\sqrt a}$, $b_3=\sqrt{a+\sqrt{a+\sqrt a}}$, and so on. So:

• when $a=-1/4$ (indeed, when $a<0$), no element of the sequence is well-defined, so it doesn't make sense to assign $\sqrt{a+\sqrt{a+\sqrt{a+\ldots}}}$ a value.
• even though each of $b_1,b_2,b_3,\dots$ depend continuously upon the parameter $a$, that's no guarantee that $\lim_{n\to\infty} b_n$ should depend continuously upon $a$. Indeed, here's a case where it doesn't. The relevant cause here is basically the fact that $\sqrt0=0$ but $\sqrt\epsilon$ is much larger than $\epsilon$ when $\epsilon$ is small.
• That more or less makes sense. It still leaves me with two similar questions. 1) Is there an intuitive way to see why this sequence doesn't continuously depend on $a$? (just for clarification: when I mentioned convergence of the nested roots, I meant your sequence of $b_i$) 2) For negative $a$, is it reasonable to expect the series to converge within $\mathbb{C}$? For $a=-\frac{1}{4}$, why would it make sense that it goes to 0 then? – HSN Apr 16 '13 at 22:48
• (1) I would say the burden of proof is on the other side. Is there any reason to expect that something that doesn't have to be continuous should be continuous? (2) I calculated some numbers with $a=-1/4$, and in fact it seems to converge to $1/2$, not $0$. – Greg Martin Apr 17 '13 at 4:16
• Thanks. The question that remains then is why the sequence for $a=-1/4$ seems to converge to something like $1/2$ in $\mathbb{C}$. I'll accept this answer, but anything sensible on this remaining question would still be welcome. – HSN Apr 18 '13 at 20:11

Here are a few comments on Lubin's observation that this problem can be thought of in the context of dynamical systems.

First, as he correctly points out, we are essentially iterating the function $F_{\alpha}(x) = \sqrt{\alpha+x}$ starting from the point $x=0$ or, put another way, we are iterating one branch of the inverse of $f_{a}(x) = x^2 - a$ starting from the point $x_0=0$. In this context, we are basically looking for a fixed point of $f_{a}$. We would like this fixed point to be repulsive under iteration of $f_{a}$ so that it is attractive under iteration of $F_{a}$.

Now from the standpoint of dynamics, an extremely concept is that a fixed point $x_0$ of a function $g$ will be attractive under iteration if and only if $|g'(x_0)|<1$. It will be repulsive when $|g'(x_0)|>1$. If $|g'(x_0)|=1$, then $x_0$ is called neutral and a variety of things can happen.

Returning to the specific case of interest, let's suppose for now that $a >0$, as the OP appeared to be interested in sequences of real numbers. There is one positive fixed point of $f_a$, namely $x_0 = (1+\sqrt{1+4a})/2$. Furthermore, $f_{a}'(x_0) = 1+\sqrt{1+4a} > 1$ for $a>0$. Thus, this fixed point is always repulsive for $f_a$ and we'd expect convergence under iteration of $F_a$.

As far as the Julia set goes, Lubin is correct that the sequence of full inverse images converges to the Julia set. While $F_{a}$ is just a single branch of the inverse, we'd still expect iteration of $F_{a}$ to generate a sequence of points that converges to some point on the Julia set. This is illustrated in the animation below. The origin is the green point and the fixed limit is the red. The blue points represent the sequence of iterates of $F_a$ starting from the green and ending at the red.

• I was hoping that someone who actually knows the field would step in! Thanks, this is very helpful. – Lubin Apr 20 '13 at 19:21

This question falls into the domain of dynamics, maybe even algebraic dynamics, and it makes sense to discuss what’s happening when the values are complex numbers.

Now, I’ve barely dipped the littlest of my little toes into these waters, and I may get way over my head before I’m finished here, so I hope some specialists can step in. Let’s name the members of your sequence: \begin{align} s_1&=\sqrt a\\ s_2&=\sqrt{a+s_1}\\ s_3&=\sqrt{a+s_2}\,, \end{align} in other words, $s_{i-1}=F(s_i)$, where $F(z)=z^2-a$. Look familiar? In this situation, we’re not asking about the (forward) orbit of the complex number $0$ under $F$, as is done in defining the Mandelbrot set, but the backward orbit, and at the $n$-th stage, we properly want to look not merely at a single number $s_n$ such that $F^{\circ n}(s_n)=0$, but at the complete inverse image under $F^{\circ n}$ of $0$. There will typically be $2^n$ points in this set $B_n=\{s\in\mathbb C\colon F^{\circ n}(s)=0\}$. My vague recollection is that when you look at the sets $B_n$, they look more and more like the Julia set of $F$, but here I have to defer to people who know what they’re talking about.

So what? Well, at least you shouldn’t be surprised that when you change $a$ a little, you get radically different behavior. That’s what chaotic behavior just is.

• Well ... except that you've changed the question. It is interesting to look at the backwards orbit. However, the notation $\sqrt{a+s_j}$ (when $a$ and $s_j$ are real numbers) implicitly means the nonnegative inverse image. So there really is only one point each time. – Greg Martin Apr 17 '13 at 4:10
• Lubin and @GregMartin I've posted an answer that places your observation in the context of dynamical systems. – Mark McClure Apr 17 '13 at 20:07