Is there any known closed form for the infinite product:
$$f(x)=\prod_{k=0}^\infty \frac{1}{2} (1+x^{1/3^k})$$
We can assume $0<x<1$.
I know of a similar product:
$$\prod_{k=0}^\infty \frac{1}{2} (1+x^{1/2^k})=\frac{x^2-1}{2 \ln x}$$
But we can't apply the same method (i.e. using the formula $1-q^2=(1-q)(1+q)$) for the product above.
The product arose when considering the iterative sequence:
$$a_{n+1}=\sqrt[3]{a_n \frac{(a_n+b_n)^2}{4}}, \qquad b_{n+1}=\sqrt[3]{b_n \frac{(a_n+b_n)^2}{4}}$$
If we set:
$$x=\frac{b_0}{a_0},\qquad a_0>b_0$$
Then is follows that:
$$a_n^3=\prod_{k=0}^{n-1} \frac{1}{4} (1+x^{1/3^k})^2 a_0^3$$
Thus:
$$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=\left(f \left(\frac{b_0}{a_0} \right) \right)^{2/3} a_0$$
Comparing the convergence of the product and the sequence I get comparable errors at each step. The product is slightly better initially.
Due to a comment by GEdgar, quite a useful substitution might be:
$$g(t)=\prod_{k=0}^\infty \frac{1}{2} (1+e^{t/3^k})$$
Edit
For all practical purposes this infinite product can be represented by a linear function for $0.2<x<1$ with very high accuracy:
$$f(x) \approx 0.7248447 x+0.2731338$$
The parameters of the linear fit were obtained by least squares for $9$ points from $0.2$ to $1$. Here is the error. I'm sure something a cubic fit will get even better results:
For $x>0$, $$ f(x) = \prod_{k=0}^\infty\;\frac{1+x^{1/3^k}}{2} $$ Consider the partial product $$ f_N(x) = \prod_{k=0}^N\;\frac{1+x^{1/3^k}}{2} $$ Now write $x=e^{2u}$, so that $$ \frac{1+x^{1/3^k}}{2}=\frac{1+e^{2u/3^k}}{2}= e^{u/3^k}\;\frac{e^{-u/3^k}+e^{u/3^k}}{2} = e^{u/3^k}\cosh\frac{u}{3^k} $$ The partial product is $$ f_N(x) = \left(\prod_{k=0}^N\;e^{u/3^k}\right)\left(\prod_{k=0}^N\cosh\frac{u}{3^k}\right) = \exp\left(\frac{3u}{2}-\frac{3u}{2\cdot3^{N+1}}\right)\prod_{k=0}^N\cosh\frac{u}{3^k} $$ and the limit is $$ f(x) = \exp\left(\frac{3u}{2}\right)\prod_{k=0}^\infty\cosh\frac{u}{3^k} =m(3u) $$ where $m$ is the moment generating function for the Cantor distribution, as shown HERE. The reason for $3u$ and not $u$ is that the index $k$ starts at $1$ for $m$, but at $0$ for $f$.