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I was wondering if it was possible to evaluate

$$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}$$

I let the expression equal $x>0$ and wrote $$x=\sqrt{9-5\sqrt{3-x}}$$ However, there is not just one value $x$ can take; $x=2$ or $x=3$.

How do I find out which one it is, or does this infinite-nested radical converge at all? Perhaps it merely oscillates between $2$ and $3$, but I am not entirely sure. Any help or hints would be much appreciated.

Thank you in advance.


The ellipsis means "and so on". It measures the following: $$\sqrt{9-5}$$ $$\sqrt{9-5\sqrt{3-\sqrt{9-5}}}$$ $$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5}}}}}$$ $$\vdots$$

Incidentally, I did not refuse to clarify the meaning. I am only active on Math.SE for so long. Whatever requests that occur can only be followed up the moment I am active, can see them and have time to act.

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  • $\begingroup$ @JyrkiLahtonen now that I have actually seen your comments, I think the real question is about why there is ambiguity. I mean, both the sequences you have defined recursively are the same, but tend toward different limits, which is not entirely clear to me. Unless, they are not the same? $\endgroup$
    – Mr Pie
    Feb 10 '20 at 0:53
  • $\begingroup$ Thanks for the edit. I redacted my vote to close. $\endgroup$ Feb 10 '20 at 4:33
  • $\begingroup$ @JyrkiLahtonen no worries :) $\endgroup$
    – Mr Pie
    Feb 10 '20 at 4:40
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    $\begingroup$ No, they aren't. One might hope that it wouldn't matter. Like with the sequences of decimals when $$0.27,\ 0.2727,\ 0.272727,\ 0.27272727,\ldots$$ and $$0.2,\ 0.272,\ 0.27272,\ 0.2727272,\ldots,$$ where it doesn't matter. Both sequences converge to $3/11$. The key being that their difference rapidly tends to zero. What we see here is more like the difference between $$1,\ 1-1+1,\ 1-1+1-1+1,\ldots$$ and $$1-1,\ 1-1+1-1,\ 1-1+1-1+1-1,\ldots,$$ with the former sequence being constant $1$ and the latter constant $0$, leaving the meaning of $$1-1+1-1+1-1+1-1\cdots$$ undefined. $\endgroup$ Feb 10 '20 at 4:52
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    $\begingroup$ (reposting a part of a deleted comment of mine). The sequence $$\sqrt{9},\sqrt{9-5\sqrt{3-\sqrt9}}, \sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt9}}}},\ldots$$ with the added stuff ending at a $9$, is a constant sequence of $3$s. On the other hand, if we always end at a $5$, the sequence consists of $2$s only. And if we end at a $3$, the sequence converges towards $2$. This is in line with both the answers. This interpretation is the only way to get $3$ as the answer, but it does highlight the ambiguity. $\endgroup$ Feb 10 '20 at 5:04
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Infinitely nested radicals may not make sense. The usual way to define this expression is as $\lim_{n\to \infty} a_n$, where $a_{n+1} = \sqrt{9 - 5\sqrt{3 - a_n}}$. The problem here is that we have no initial point specified. Choosing $a_0 = 2$ or $a_0 = 3$ will produce two different limits, so the nested radical is not well-defined.

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  • $\begingroup$ @AnthonyLahmann's answer contradicts yours. Unfortunately for you, I am leaning towards their answer instead. :/ $\endgroup$
    – Mr Pie
    Feb 9 '20 at 8:59
  • $\begingroup$ Actually, I don't know. Who am I supposed to believe? lol $\endgroup$
    – Mr Pie
    Feb 9 '20 at 9:04
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    $\begingroup$ @MrPie Ultimately, it's a difference in convention. It's like pointing out that $0^0$ is undefined; it is true, but in certain situations, it makes sense to invent your own convention and say it's $1$. You can ignore unstable fixed points of $f(x) = \sqrt{9 - 5\sqrt{3 - x}}$ if it helps you get where you need to go, but just don't lose sight of the fact that you're using a non-standard convention for nested radicals. $\endgroup$
    – user744868
    Feb 9 '20 at 9:07
  • $\begingroup$ @JyrkiLahtonen The three dots just means "and so on", with $(9-5, 3), (9-5, 3)$ and etc. $\endgroup$
    – Mr Pie
    Feb 9 '20 at 11:59
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    $\begingroup$ @JyrkiLahtonen I may need to work on my rigour. Thanks for letting me know. $\endgroup$
    – Mr Pie
    Feb 10 '20 at 4:41
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The answer is $2$. While $3$ is also a fixed point, it is unstable because if we let $x=3-\epsilon$ for some small $\epsilon$, and iterate $x\leftarrow \sqrt{9-5\sqrt{3-x}}$, it will diverge away from $3$.

Graph of <span class=$\sqrt{9-5\sqrt{3-x}}$">

If you look at the graph, you will find that the slope approaches $\infty$ as $x\to 3$. The derivative of $\sqrt{9-5\sqrt{3-x}}$ is $\frac5{4\sqrt{9-5\sqrt{3-x}}\sqrt{3-x}}$. When $x\to 3$, the $\sqrt{3-x}$ in the denominator will approach $0$, which means the derivative approaches $\infty$ as $x\to 3$. Therefore, the fixed point is unstable and will very quickly diverge away from $3$. Plugging $2$ into the equation gives $\frac58$, which is less than $1$. Therefore, the fixed point is stable.

In conclusion:

$$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}=2$$

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  • $\begingroup$ Yes! I needed it to be $2$. Thank you so much! $\endgroup$
    – Mr Pie
    Feb 9 '20 at 8:56
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    $\begingroup$ If this be true, it means the following formula is true: $$\begin{align}\sqrt 5 - \sqrt 3 &= \sqrt{9+4\sqrt{1+\sqrt{7+3\sqrt{5+\sqrt{7+3\sqrt{5+\sqrt{7+\cdots}}}}}}} \\ &- \sqrt{7+4\sqrt{1+\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}\end{align}$$ The symmetries allign :) $\endgroup$
    – Mr Pie
    Feb 9 '20 at 9:01

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