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:$$\begin{array}{rl}y& =\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\cdots }}}}\\ & =\sqrt{2+\sqrt{2-y}}\end{array}$$Now we have to impose limits on the range of values that $y$ can take. A little inspections shows that: $$\sqrt{2}\le y\le 2$$Now we can proceed to find $y$ by squaring both sides twice and then gathering terms together:$$\begin{array}{rl}{y}^2& =2+\sqrt{2-y}\\ y^2-2& =\sqrt{2-y}\\ (y^2-2{)}^2& =2-y\\ y^4-4y^2+4& =2-y\\ y^4-4y^2+y+2& =0\\ y^2(y^2-4)+y+2& =0\\ y^2(y+2)(y-2)+y+2& =0\\ (y+2)(y^2(y-2)+1)& =0\\ (y+2)(y^3-2y^2+1)& =0\end{array}$$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 $y_4$: $$\begin{array}{rl}{y}_1& =-2\\ y_2& =1\\ y_3& =\frac{1+\sqrt{5}}{2}\\ y_4& =\frac{1-\sqrt{5}}{2}\end{array}$$Due to restrictions placed on $y$ however, only $y_3$ is a valid solution and its value of course is $\varphi$. Thus solution is $$\varphi =\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\cdots }}}}$$This is not the only nested radical to produce $\varphi$. An even simpler expression is: $$\varphi =\sqrt{1+\sqrt{1+\sqrt{1+\cdots }}}$$See WOLFRAM Demonstrations Link titled Nested Square Root Representation of the Golden Ratio for more details. Another site at iiTutor shows that:$$\begin{array}{rl}2& =\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}\\ 1& =\sqrt{2-\sqrt{2-\sqrt{2-\cdots }}}\end{array}$$