In the Middle of the Tower 2

Given that $\left(\dfrac 12\right)^{x^\frac 13} = x$ , $\implies - x^\frac 13 \ln 2 = \ln x$ or $\ln x + x^\frac 13 \ln 2$ .

Let $f(x) = \ln x + x^\frac 13 \ln 2$ . By Newton-Raphton method , we have:

$x_{n+1} = x_n - \frac {f(x_n)}{f'(x_n)} = x_n - \frac {\ln x + x^\frac 13 \ln 2}{\frac 1{x_n}+\frac {\ln 2}{3x^\frac 23}}$

Assuming $x_0 = 1$ and using an Excel spreadsheet to implement, we have:

where the columns from left to right are $x_n$ , $f(x_n)$ , $f'(x_n)$ , and $x_{n+1}$ . Therefore $x \approx \boxed{0.564}$ .

Let $f(x)=2^{-\sqrt[3]{x}}$ (ie the left-hand side of the given equation).

We see that $f(0)=1$ and $f(1)=\frac12$ ; so the root must lie in this interval.

Let $x_0=\frac12$ and $x_{n+1}=f \left(x_n \right)$ .

Iterating, we find

so the root is $\boxed{0.564}$ .

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The more rigorous version of this is that the form given in the problem leads to a convergent iterated function system (IFS).

It's easy to check that $f'(x)$ is negative, so there is a unique root. As above, it lies in the interval $0<x<1$ . The condition for convergence of an IFS to a root $x=a$ is that $|f'(a)|<1$ ; this is also easy to verify. Since $f(x)$ is pretty well-behaved, any starting value $x_0$ will work.