While reading about polylogarithms, I came across the nice polylogarithm ladder,

$$6\operatorname{Li}_2(x^{-1})-3\operatorname{Li}_2(x^{-2})-4\operatorname{Li}_2(x^{-3})+\operatorname{Li}_2(x^{-6}) = \frac{7\pi^2}{30}\tag{1}$$

where $x = \phi = \frac{1+\sqrt{5}}{2}$, the golden ratio, or the root $1<x<2$ of,

$$x^n(2-x) = 1\tag{2}$$

for the case $n=2$. I wondered if there was anything for $n=3$ (the tribonacci constant $T$, or the real root of $x^3-x^2-x-1 = 0$). After a little expt with Mathematica's integer relations command, I found,

$$4\operatorname{Li}_2(x^{-1})-4\operatorname{Li}_2(x^{-3})-3\operatorname{Li}_2(x^{-4})+\operatorname{Li}_2(x^{-8}) = \frac{\pi^2}{6}\tag{3}$$

However, for general $n$ (including the tetranacci constant and higher), it seems they obey,

$$4\operatorname{Li}_2(x^{-1})-\operatorname{Li}_2(x^{-2})+\operatorname{Li}_2(x^{-n+1})-2\operatorname{Li}_2(x^{-n}) = 2\operatorname{Li}_2(-x)+\frac{\pi^2}{2}\tag{4}$$

where $x$ is the root $1<x<2$ of $(2)$.

Q: Anybody knows how to prove if $(4)$ is indeed true for all integer $n\geq2$?

  • 1
    $\begingroup$ For how many values of $n$ did you verify your conjecture (4)? $\endgroup$ – Franklin Pezzuti Dyer Nov 23 '18 at 16:53
  • $\begingroup$ @Frpzzd: I can't remember exactly, but I think it was > 20 $\endgroup$ – Tito Piezas III Nov 23 '18 at 17:34
  • $\begingroup$ I have no idea how to go about proving this (other than scattered attempts that may-or-may not yield anything), but I figured I'd throw a bounty up to see if anyone has any ideas. $\endgroup$ – Brevan Ellefsen Jun 5 at 8:27
  • $\begingroup$ @BrevanEllefsen: Thanks for putting up the bounty. After my recent posts on the Nielsen polylogs, I'm now wondering if those have ladders as well. $\endgroup$ – Tito Piezas III Jun 6 at 15:32

I'll change $x \rightarrow 1+x$ to have \begin{equation} \left(1+x\right)^n\left(1-x\right) = 1,\quad 0 < x < 1 \label{eq1} \tag{1} \end{equation} Consider \begin{align} f_n &= 4\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \operatorname{Li}_2\left(\dfrac{1}{\left(1+x\right)^2}\right) + \operatorname{Li}_2\left(\dfrac{1+x}{\left(1+x\right)^n}\right) - 2\operatorname{Li}_2\left(\dfrac{1}{\left(1+x\right)^n}\right) - \phantom{a} \\ &- 2\operatorname{Li}_2\left(-1-x\right) \end{align} where $x$ is the solution to $(1)$ with parameter $n$.

I restate here main functional equations for the dilogarithm (most of them can be found here)

\begin{align} \operatorname{Li}_2(x)+\operatorname{Li}_2(-x) &= \dfrac{1}{2}\operatorname{Li}_2(x^2) \label{dl2} \tag{2} \\ \operatorname{Li}_2\left(1-x\right)+\operatorname{Li}_2\left(1-\dfrac{1}{x}\right) &= -\dfrac{1}{2}\ln^2 x \label{dl3} \tag{3} \\ \operatorname{Li}_2(x)+\operatorname{Li}_2\left(1-x\right) &= \dfrac{1}{6}\pi^2 - \ln\left(1-x\right)\ln x \label{dl4} \tag{4} \\ \operatorname{Li}_2(-x)-\operatorname{Li}_2\left(1-x\right)+\dfrac{1}{2}\operatorname{Li}_2\left(1-x^2\right) &= -\dfrac{1}{12}\pi^2 - \ln\left(1+x\right)\ln x \label{dl5} \tag{5} \\ \operatorname{Li}_2(-x)+\operatorname{Li}_2\left(-\dfrac{1}{x}\right) &= -\dfrac{1}{6}\pi^2 - \dfrac{1}{2}\ln^2 x \label{dl6} \tag{6} \end{align}

Now use $\left(1+x\right)^n\left(1-x\right) = 1$ to get rid of $n$-th power of $(1+x)$ in $f_n$ and start to use dilogarithm identities \begin{align*} f_n &\overset{(1)}{=} 4\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \color{red}{\operatorname{Li}_2\left(\dfrac{1}{\left(1+x\right)^2}\right)} + \operatorname{Li}_2\left(1-x^2\right) - 2\operatorname{Li}_2\left(1-x\right) - 2\operatorname{Li}_2\left(-1-x\right) \\ &\overset{(2)}{=} \color{red}{2}\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \color{red}{2\operatorname{Li}_2\left(-\dfrac{1}{1+x}\right)} + \color{blue}{\operatorname{Li}_2\left(1-x^2\right)} - \color{blue}{2\operatorname{Li}_2\left(1-x\right)} - 2\operatorname{Li}_2\left(-1-x\right) \\ &\overset{(5)}{=} 2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \color{magenta}{2\operatorname{Li}_2\left(-\dfrac{1}{1+x}\right)} - \color{blue}{\dfrac{1}{6}\pi^2} - \color{blue}{2\ln\left(1+x\right)\ln x} - \color{blue}{2\operatorname{Li}_2(-x)}- \color{magenta}{2\operatorname{Li}_2\left(-1-x\right)} \\ &\overset{(6)}{=} 2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) + \color{magenta}{\dfrac{1}{3}\pi^2} + \color{magenta}{\ln^2\left(1+x\right)} - \dfrac{1}{6}\pi^2 - 2\ln\left(1+x\right)\ln x - \color{green}{2\operatorname{Li}_2(-x)} \\ &\overset{(3)}{=} 2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) + \dfrac{1}{3}\pi^2 + \ln^2\left(1+x\right) - \dfrac{1}{6}\pi^2 - 2\ln\left(1+x\right)\ln x + \color{green}{\ln^2\left(1+x\right)} + \phantom{a} \\ &+ \color{green}{2\operatorname{Li}_2\left(\dfrac{x}{1+x}\right)} \\ &= \color{red}{2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right)} + \color{red}{2\operatorname{Li}_2\left(\dfrac{x}{1+x}\right)} + \dfrac{1}{6}\pi^2 + 2\ln^2\left(1+x\right)- 2\ln\left(1+x\right)\ln x \\ &\overset{(4)}{=} \color{red}{\dfrac{1}{3}\pi^2} - \color{red}{2\ln\left(\dfrac{1}{1+x}\right)\ln\left(\dfrac{x}{1+x}\right)} + \dfrac{1}{6}\pi^2 + 2\ln^2\left(1+x\right)- 2\ln\left(1+x\right)\ln x \\ &= \dfrac{\pi^2}{2} \end{align*}

  • $\begingroup$ Looks great at first glance. I'll take a longer look later and assign the bounty. Thank you so much for answering this question! $\endgroup$ – Brevan Ellefsen Jun 5 at 17:10
  • $\begingroup$ @Knas: Very clever! (I made some minor edits for clarity.) $\endgroup$ – Tito Piezas III Jun 6 at 15:48
  • $\begingroup$ @BrevanEllefsen: I've accepted his answer. By getting rid of the $(1+x)^n$ expression, it simplified matters considerably. $\endgroup$ – Tito Piezas III Jun 6 at 15:52

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