# Evaluating $\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}$.

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.

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.

• @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? Feb 10 '20 at 0:53
• Thanks for the edit. I redacted my vote to close. Feb 10 '20 at 4:33
• @JyrkiLahtonen no worries :) Feb 10 '20 at 4:40
• 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. Feb 10 '20 at 4:52
• (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. Feb 10 '20 at 5:04

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.

• Actually, I don't know. Who am I supposed to believe? lol Feb 9 '20 at 9:04
• @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. Feb 9 '20 at 9:07
• @JyrkiLahtonen The three dots just means "and so on", with $(9-5, 3), (9-5, 3)$ and etc. Feb 9 '20 at 11:59
• @JyrkiLahtonen I may need to work on my rigour. Thanks for letting me know. Feb 10 '20 at 4:41

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 $$3$$. $\sqrt{9-5\sqrt{3-x}}$">

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:

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

• Yes! I needed it to be $2$. Thank you so much! Feb 9 '20 at 8:56
• 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 :) Feb 9 '20 at 9:01