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If we define for $0<x<1$ $$f(x):=\prod_{n=1}^\infty\left(1-x^n\right)^{\frac{\lambda(n)}{n}},\tag{1}$$ where $\lambda(n)$ is the Liouvile function (and notice is the similar infinite product than (16) of this MathWorld involving the Möbius function ) then taking the logarithmic derivative of $(1)$, if we presume the uniform convergence we can combine with the Lambert series for the Liouville function to get $$x\frac{f'(x)}{f(x)}=-\frac{1}{2}\left(\vartheta_3(x)-1\right),\tag{2}$$ where $\vartheta_3(x)$ is the Jacobi theta function (see the Lambert series in this Wikipedia).

Question. Is it known or do you know how calculate $f(x)$ from $(2)$? Many thanks.

If it is well known cite the literature.

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  • $\begingroup$ That I want to know is if there is a simple function, as same as occurs for the Möbius function, now for the Liouville function. I hope that my question has mathematical meaning, that is solving the differential equation, if it is possible get such simple function. Many thanks for your attention @JackD'Aurizio $\endgroup$ – user243301 Apr 26 '17 at 22:03
  • $\begingroup$ Yes, now I understand because is direct integration of $\frac{-f'(x)}{f(x)}$, and I see that is more nice than the infnite product and similar than the formula for the Möbius identity (16), because is a negative exponential. I don't know nothing about the graph of your function, but your comment is the answer of my question @JackD'Aurizio $\endgroup$ – user243301 Apr 26 '17 at 22:10
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    $\begingroup$ Formally, $$\log f(x) = \sum_{n\geq 1}\frac{\lambda(n)}{n}\log(1-x^n) = -\sum_{n\geq 1}\sum_{m\geq 1}\frac{\lambda(n) x^{nm}}{nm}=-\sum_{k\geq 1}(\lambda *1)(k)x^k=-\sum_{h\geq 1}x^{h^2} $$ hence $f(x)=\exp\left(\frac{1-\vartheta_3(x)}{2}\right)$. You do not need to consider a logarithmic derivative, but just a logarithm. $\endgroup$ – Jack D'Aurizio Apr 26 '17 at 22:15
  • $\begingroup$ Thanks, now I am going to read your last remarks but feel free to add an answer @JackD'Aurizio , and then from Wolfram Alpha, the plot should be using Plot e^((1-ϑ_3(x))/2), for 0<x<1 $\endgroup$ – user243301 Apr 26 '17 at 22:23
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Formally,

$$\log f(x) = \sum_{n\geq 1}\frac{\lambda(n)}{n}\log(1-x^n) = -\sum_{n\geq 1}\sum_{m\geq 1}\frac{\lambda(n) x^{nm}}{nm}=-\sum_{k\geq 1}(\lambda *1)(k)x^k=-\sum_{h\geq 1}x^{h^2} $$ hence $f(x)=\exp\left(\frac{1-\vartheta_3(x)}{2}\right)$. We do not need to consider a logarithmic derivative, just a logarithm. For any $x\in\mathbb{C}$ such that $|x|<1$ we have that $-\sum_{h\geq 1}x^{h^2}$ is a fast-convergent series, so there are no issues in the numerical computation of $$ f(x)=\prod_{n\geq 1}(1-x^n)^{\frac{\lambda(n)}{n}}.$$ Here it is the graph of $f(x)$ over $(0,1)$:

$\hspace{2cm}$enter image description here

Additionally, by Jacobi triple product we have $$ \vartheta_3(x) = \prod_{m\geq 1}(1-x^{2m})(1+x^{2m-1})^2 $$ hence $\log\log f(x)$ is related with $\sum_{h\geq 1}\frac{\sigma(h)}{h}x^h$, where $\sigma(h)=\sum_{d\mid h}d$.

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  • $\begingroup$ You are incredible, many thanks for your answers in this site. $\endgroup$ – user243301 Apr 26 '17 at 22:38
  • $\begingroup$ Supposing that $nm=k$ in the first equation, differentiation is needed to cancel the $k$ in the denominator. $\endgroup$ – ccorn May 18 '17 at 8:13

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