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Consider the function $$f(x)=\ln\prod_{k=1}^\infty\frac{1+t^k}{1+t^{k+x}}$$ where $t\in(0,1)$. According to some theorems I found that $f(x)$ is well defined for any $x\in\mathbb{R}$. In order to find the closed form for $f$, I dealt to the following equality as follows $$\prod_{k=1}^\infty\frac{1+(1/2)^k}{1+(1/2)^{k+x}}=\frac{(-1;1/2)_\infty(2^{-x}+1)}{2(-2^{-x};1/2)_\infty}$$ where $t=1/2$ and $(a;q)_n$ gives q-pochhammer symbol.

Is the above equality true? If the answer is yes, how can I compute the exact value of above product, Also can it help to find closed form for $f$.

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  • $\begingroup$ Do you have any reason to believe there exists a closed form for this function? Unless $x\in\mathbb{N}$ (in which case the product easily telescopes), I see no reason to think it would, and ample reason to think otherwise... $\endgroup$ – Steven Stadnicki Oct 27 '17 at 22:07
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Using \begin{align} \prod_{k=0}^{\infty} (1 - a \, q^{k}) &= (a;q)_{\infty} \\ \frac{(a;q)_{\infty}}{(a \, q^{n}; q)_{\infty}} &= (a;q)_{n} \end{align} then $$\prod_{k=1}^{\infty} (1 + t^{k+x}) = (-t^{x+1}; t)_{\infty}$$ which leads to $$f(x) = \ln\left(\frac{\prod_{k=1}^{\infty}(1+t^{x})}{\prod_{k=1}^{\infty} (1 + t^{k+x})}\right) = \ln\left(\frac{(-t;t)_{\infty}}{(-t^{x+1};t)_{\infty}}\right) = \ln((-t;t)_{x}).$$ when $t=1/2$ then $$f(x) = \ln\left(\left(-\frac{1}{2}; \frac{1}{2}\right)_{x}\right).$$ A table of values may be obtained based on the value of $x$.

It is of interest to note that $$\lim_{n \to \infty} f(n) = \ln\left(\left(-\frac{1}{2}; \frac{1}{2}\right)_{\infty}\right) = .86887665\cdots.$$

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  • $\begingroup$ would you lead me that how do you obtain (compute) these result, or $\endgroup$ – soodehMehboodi Nov 21 '17 at 19:34
  • $\begingroup$ would you introduce some references which can help me. $\endgroup$ – soodehMehboodi Nov 21 '17 at 19:36
  • $\begingroup$ @soodehMehboodi See, for example, q-Pochhammer Symbol $\endgroup$ – Leucippus Nov 21 '17 at 20:04

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