<s>[INST] deriving $y=\sqrt{x+\sqrt{x+\sqrt{x}}\cdots} $ How to derive $y=\sqrt{x+\sqrt{x+\sqrt{x}}\cdots}$ at $x=6$ ?
 [/INST] Assuming that
$$
y = \sqrt{ x + \sqrt{ x + \sqrt{ x + \sqrt{ x + \cdots } } } },
$$
means to "repeat up to infinity".

So we write
$$
\begin{eqnarray}
y_0 &=& 0,\\
y_{n+1} &=& \sqrt{ x + y_n },\\
\end{eqnarray}
$$
and we define
$$
y = \lim_{n \rightarrow \infty} y_n.
$$
We must first show that $y$ exists.


If $y$ exists we obtain
$$
y = \lim_{n \rightarrow \infty} y_n = \lim_{n \rightarrow \infty} y_{n+1}
\Rightarrow y = \sqrt{x+y} \Rightarrow y^2 - y - x = 0 $$
At this point the quadratic formula tells us that $$y = \frac{1 \pm \sqrt{1 + 4x}}{2}$$
and we take the positive solution.
The case $x=6$ yields $y = \tfrac{1}{2} + \tfrac{1}{2} \times 5 = 3$.
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