# Closed form for the infinite product $\prod_{k=0}^\infty \frac{1}{2} (1+x^{1/3^k})$

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:

• A quickie comment ... see MGF on the page en.wikipedia.org/wiki/Cantor_distribution – GEdgar Aug 23 '16 at 18:48
• @GEdgar, thank you. Though I can't get my product into the same form by substituting $x=e^{2t}$, or $x=e^{t}$, it's still not the same. – Yuriy S Aug 23 '16 at 19:13

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 $f$.
• @YuriyS ... Yes, put $u=i\theta$ to switch from $\cosh$ to $\cos$. The case with denominators $2^k$ is easier because in that case we get the moment generating function for Lebesgue measure on an interval. – GEdgar Aug 24 '16 at 0:00