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I was thinking about the infinite expression:

$$\sqrt{6+\sqrt{6+\sqrt{6+...}}}$$

It seems to me a bit informal or underspecified. You could state it better by defining an infinite series:

$$a_n = \sqrt{6+a_{n-1}}$$

And then ask about $\lim_{n\rightarrow\infty}{a_n}$. But then you have to wonder what $a_0$ is. Although anyway it's not too hard to figure out that if $a_0 \geq -6$, then the limit is 3. I fully understand this is the typical treatment and has been covered before. But where does this $a_0 \geq -6$ come from? Is there a way to evaluate it without making any assumptions like this?

It also dawned on me that maybe you could also define this sequence going "outside-in" instead of "inside-out". For instance:

$$b_n = b_{n-1}^2-6$$

But this seems a bit more unruly. It diverges for values other than $3$ and $-2$.

So I guess my question is: is that original expression just underdefined, and requires a small "leap" to formalize? Or is there a better way to treat that kind of thing strictly? Are there examples of similar expressions where the "informality" is troublesome - like maybe there's more than one way to formalize it that leads to different answers?

Joe K
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    Why isn't it just straight forward to say $a_0 = \sqrt{6}$? – Wintermute Aug 09 '18 at 21:20
  • Related: https://math.stackexchange.com/questions/115501/sqrtc-sqrtc-sqrtc-cdots-or-the-limit-of-the-sequence-x-n1-sq – Aryabhata Aug 09 '18 at 21:23
  • You do not have to think of this as a sequence and limit case (which will get you in trouble with $a_0$ and whatnot). Instead, think of this number as the solution to $x=\sqrt{6+x}$. – Hamed Aug 09 '18 at 21:23
  • There is no contradiction here. The "outside-in" method you wrote is just not finding the limit of the given expression. – Cave Johnson Aug 09 '18 at 21:24
  • @Hamed doesn't that only work if you first assume the epxression converges or a solution exists? For instance, doing that with $6+(6+(6+(6+...)^3)^3)^3$ gives the answer $-2$, but maybe that expression should just be divergent? – Joe K Aug 09 '18 at 21:31

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Actually it is not ambiguous. The map $f:x\mapsto \sqrt{6+x}$ is a contraction of the metric space $[-1,7]$, due to $f'(x)=\frac{1}{2\sqrt{6+x}}$, so by the Banach fixed point theorem for any $x\in[-1,7]$ the sequence $$ x, f(x), f(f(x)), f(f(f(x))),\ldots $$ converges towards the unique fixed point of $f$ in $[-1,7]$, i.e. $3$. Summarizing $$ \sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=3.$$

Jack D'Aurizio
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  • Why must $x$ be constrained to $[-1,7]$? – Joe K Aug 09 '18 at 21:33
  • @JoeK: it mustn't. You may take any other closed neighbourhood of the origin which is mapped into itself by $f$, provided by $\left|f'\right|<1$ over there. $[-1,7]$ just looked like a convenient choice. – Jack D'Aurizio Aug 09 '18 at 21:36
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The use of "$\cdots$" in this way will always indicate a limit. On the face of it, it indicates repeating some process an infinite number of times -- which is of course impossible. You can only take the limit of a sequence of finite performable operations (or determine that such a limit does not exist). So the expression $$\sqrt{6 + \sqrt{6 + \sqrt{6 + \ldots}}}$$ can only mean the limit of the sequence $s_n$ (if it exists), where

$$s_0 =\sqrt{6}$$ $$s_1 = \sqrt{ 6 + \sqrt{6}} = \sqrt{6 + s_0}$$ $$s_2 = \sqrt{ 6 + \sqrt{ 6 + \sqrt{6}}} = \sqrt{ 6 + s_1}$$ and so $$s_n = \sqrt{6 + s_{n-1}}$$ So in other words, your expression means unambiguously $$\lim_{n\to\infty}s_n$$ if this limit exists.

MPW
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