Nested Radicals

I came across a problem in Cliff Pickover's Twitter feed. It is depicted in Figure 1.

Figure 1

No solution was offered so I did a search of Google Images and came up with a link to MathWorld. It is there that a solution is offered:$$ \sqrt{x}=\sqrt[3]{x  \, \sqrt[3]{x  \, \sqrt[3]{x \cdots}}}$$This is an instance of a more general formula:$$ x^{1/(n-1)}=\sqrt[n]{x \, \sqrt[n]{x \, \sqrt[n]{x \, \cdots}}}$$When \(n=3\), we get the original formula that Pickover was referencing. There are other interesting results in the MathWorld article. The following is particularly striking:$$ x^{e-2}=\sqrt{x  \, \sqrt[3]{x \, \sqrt[4] {x \, \sqrt[5]{x  \cdots}}}}$$Presh Talwalkar has a very helpful article on this topic that explains how this last result is obtained. See Figure 2.

Figure 2

A few days later, I came across another nested radical problem in a YouTube video. This is the problem: $$ ?=\sqrt{2+\sqrt{2-\sqrt{2 + \sqrt{2 - \cdots}}}}$$The solution is quite different to the previous approach and begins by replacing the ? with a \(y\) and making use of the fact that the nested radical is infinite:$$ \begin{align} y &= \sqrt{2+\sqrt{2-\sqrt{2 + \sqrt{2 - \cdots}}}} \\ &=\sqrt{2+ \sqrt{2 - y}}\end{align}$$Now we have to impose limits on the range of values that \(y\) can take. A little inspections shows that: $$ \sqrt{2} \leq y \leq 2$$Now we can proceed to find \(y\) by squaring both sides twice and then gathering terms together:$$ \begin{align} y^2 &= 2+\sqrt{2-y} \\ y^2-2 &= \sqrt{2-y} \\ (y^2-2)^2 &= 2-y \\ y^4 -4y^2+4 &= 2-y \\ y^4 - 4y^2 +y +2 &= 0 \\ y^2(y^2-4)+y+2 &=0 \\y^2(y+2)(y-2)+y+2 &= 0 \\(y+2)(y^2(y-2)+1) &= 0 \\ (y+2)(y^3-2y^2+1) &= 0     \end{align}$$Now \(y-1\) divides the cubic expression and so the LHS of the quartic equation becomes: $$ (y+2)(y-1)(y^2-y-1) = 0 $$ There are four solutions \(y_1, y_2, y_3\) and \(y_4\): $$ \begin{align} y_1 &= -2 \\ y_2 &= 1 \\ y_3 &= \frac{1+\sqrt{5}}{2} \\ y_4 &= \frac{1-\sqrt{5}}{2} \end{align} $$Due to restrictions placed on \(y\) however, only \(y_3\) is a valid solution and its value of course is \( \phi \). Thus solution is $$ \phi =\sqrt{2+\sqrt{2-\sqrt{2 + \sqrt{2 - \cdots}}}}$$This is not the only nested radical to produce \( \phi \). An even simpler expression is: $$ \phi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$$See WOLFRAM Demonstrations Link titled Nested Square Root Representation of the Golden Ratio for more details. Another site at iiTutor shows that:$$ \begin{align} 2 &= \sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}\\1 &= \sqrt{2-\sqrt{2-\sqrt{2-\cdots}}} \end{align}$$