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I found this question somewhere and have been unable to solve it. It is a modification of a very common algebra question.

$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the nearest value of $y^2 - y$?

$\textbf {(1) } 1 \qquad \textbf {(2) } \sqrt5 \qquad \textbf {(3) } 4 \qquad \textbf {(4) } 2\sqrt5$

I worked the problem and got $$y^2 - y = 5+ \sqrt{5-\sqrt{5+\sqrt{5-...}}} -\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} \\ = 5+ \sqrt{5-y} - y = 5-y + \sqrt{5-y}$$ What do I do next?

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the $\infty$ is unnecessary. – ILoveMath Dec 1 '13 at 18:36
The $\infty$ even looks wrong. – heinrich5991 Dec 1 '13 at 22:21
up vote 11 down vote accepted

Let's let




and assume that both limits exist. Note that

$$y^2=5+z\quad\text{and}\quad z^2=5-y$$


$$\begin{align} y^2-y&=(5+z)-y\\ &=5-(y-z)\\ &=5-{y^2-z^2\over y+z}\\ &=5-{(5+z)-(5-y)\over y+z}\\ &=5-{z+y\over y+z}\\ &=4 \end{align}$$

Added 5/6/14: I thought I'd go ahead and add a proof that the two limits actually do exist. I'm not sure this is the simplest proof -- I'll be happy to see something slicker -- but here goes.

Define the sequence of $y_n$'s by $y_n=\sqrt{5+\sqrt{5-y_{n-1}}}$ with $y_1=\sqrt{5+\sqrt{5}}$, and let $z_n=\sqrt{5-y_n}$. It's easy to show (by induction) that $2\lt y_n\lt4$ for all $n$, so it suffices to show that the sequence of $y_n$'s has a limit.

I'll do so by showing that it's a Cauchy sequence, and I'll do that by showing that $\sum|y_n-y_{n-1}|$ converges.

We have $y_n^2-5=\sqrt{5-y_{n-1}}$, hence




where we've used the inequalities $2\lt y_n\lt 4$ (for all $n$) in the denominator. It follows that $\sum|y_n-y_{n-1}|$ is bounded by a geometric series with ratio $1/8$, hence converges.

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You were on the right track. Put

$$x=\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}\implies x^2=5+\sqrt{5-\sqrt{5+\ldots}}\implies$$



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,I think the sign '+,-' toggles. – lab bhattacharjee Dec 1 '13 at 18:36
Indeed. Editing on its way and thanks. – DonAntonio Dec 1 '13 at 18:36
what is the answer? The approximation should come around 4. – shaurya gupta Dec 7 '13 at 16:43

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