simplify $\sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} }$ -- if $x$ is negative? Was given this recreational problem: simplify $$\sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} }$$
The solution isn't hard. Let $y = \sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} }$, then $y^3 = xy$, $y=\sqrt{x}$
The problem didn't specify, but if $x$ is negative, would this work? 
 A: Note if $x$ is real, with $x \lt 0$, then you have
$$\sqrt[3]{x} \lt 0 \tag{1}\label{eq1A}$$
$$\sqrt[3]{x\sqrt[3]{x}} \gt 0 \tag{2}\label{eq2A}$$
$$\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x}}} \lt 0 \tag{3}\label{eq3A}$$
I trust you see the issue that occurs as you keep expanding the expression to have more cube roots in it.
A: You've tagged your question with "complex analysis" and "complex numbers".  In this context, however, the function $x \mapsto \sqrt[3]{x}$ needs to be defined more clearly.
For instance, if we define $$\sqrt[3]{re^{\theta i}} = \sqrt[3]{r}\cdot e^{\theta i / 3} \\ \sqrt{re^{\theta i}} = \sqrt{r}\cdot e^{\theta i / 2}$$ for $r \geq 0$ and $0 \leq \theta < 2\pi$, then you can show $$\sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} } = \sqrt{x}$$
for any $x \in \mathbb{C}.$  This works both as a fixed point of the function $y \mapsto \sqrt[3]{xy}$ and as a limit of the sequence $x_0 = \sqrt[3]{x}$ and $x_{n+1} = \sqrt[3]{xx_n}$.
A: Let $y=\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\dots}}}$
So,
$y=\sqrt[3]{xy}$
Now raise both side to the power $3$, we get:
$y^3=xy$
Now, $y\ne0$ because $x$ is negative as mentioned in the problem statement, we can divide both sides by $y$, we get:
$y^2=x$
However $y^2=x$ has no real solution since $x$ is negative. Hence the given expression does not have a closed form. [THIS IS THE REQUIRED ANSWER].

When $x$ is positive, then the equation $y^2=x$ reduces to $y=\sqrt{x}$.
