Closed-form expressions for the zeros of $\text{Li}_{-n}(x)$?

Question

Consider the first few polylogarithm functions $\text{Li}_{-n}(x)$, where $-n$ is a negative integer and $x\in\mathbb R$ (plotted below). Observation suggests that $\text{Li}_{-1}(x)$ has one zero (at $x=0$), $\text{Li}_{-2}(x)$ has two zeros (one at $x=0$, and one at $x=-1$), and more generally, $\text{Li}_{-n}(x)$ has $n$ zeroes (one at $x=0$, and the rest satisfying $x<0$).

Are there closed-form expressions for these zeroes? In other words, is there a general formula giving the $m$th zero of $\text{Li}_{-n}(x)$?

Answer 1

If you consider $$P_n(x)=\frac{(1-x)^{n+1} }{x}\,\text{Li}_{-n}(x)$$ they are palindromic polynomials of degree $(n-1)$. $$\left( \begin{array}{cc} n & P_n(x) \\ 1 & 1 \\ 2 & x+1 \\ 3 & x^2+4 x+1 \\ 4 & x^3+11 x^2+11 x+1 \\ 5 & x^4+26 x^3+66 x^2+26 x+1 \\ 6 & x^5+57 x^4+302 x^3+302 x^2+57 x+1 \\ 7 & x^6+120 x^5+1191 x^4+2416 x^3+1191 x^2+120 x+1 \end{array} \right)$$ Then, as usual, you can have their analytical solutions up to $n=5$ remembering that is $a$ is a root $\frac 1 a$ is another one and that, if $n$ is even, $-1$ is a root of $P_n(x)$.

Beside this very limited information, I do not know anything else but I would be surprised to see a closed form solution.

Edit

However, we can make rather good approximation using firt Taylor series built around $x=0$ $$P_n(x)=1+\sum_{k=1}^p c_k x^k$$ where the very first coefficients are given by $$\left( \begin{array}{cc} k & c_k \\ 1 & 2^n-(n+1) \\ 2 & 3^n-2^n (n+1)+\frac{ (n+1)n}{2} \\ 3 & 4^n-3^n (n+1)+2^n \frac{ (n+1)n}2-\frac{(n+1)n(n-1)}{6} \\ 4 & 5^n-4^n (n+1)+3^n\frac{ (n+1)n}{2}-2^{n}\frac{(n+1)n(n-1)}{6} +\frac{(n+1)n(n-1) (n-2)}{24} \end{array} \right)$$ which reveals clear and simple patterns.

From this, we could make series reversion to get $$x=-\frac {1}{c_1}-\frac {c_2}{c_1^3}-\frac {2c_2^2-c_1c_3}{c_1^5}-\frac {5c_2^3-5c_1c_2c_3+c_1^2c_4}{c_1^7}\tag1$$ or, more than likely better, use one level above Householder method $$x=-\frac{c_1^3-2 c_1 c_2+c_3}{c_1^4-3 c_1^2c_2 +2 c_1 c_3+c_2^2-c_4}\tag2$$

Some results

$$\left( \begin{array}{cccc} n & \text{from }(1) & \text{from }(2) & \text{exact } \\ 3 & -0.267883300781 & -0.267942583732 & -0.267949192431 \\ 4 & -0.100918342637 & -0.101010101010 & -0.101020514434 \\ 5 & -0.043044294676 & -0.043091520063 & -0.043096288203 \\ 6 & -0.019506385927 & -0.019522790309 & -0.019524242690 \\ 7 & -0.009143733121 & -0.009148332808 & -0.009148694810 \\ 8 & -0.004374958453 & -0.004376077576 & -0.004376157646 \\ 9 & -0.002121044838 & -0.002121290533 & -0.002121306903 \\ 10 & -0.001037447933 & -0.001037497866 & -0.001037501034 \end{array} \right)$$ For sure, all of the above can be improved pushing the expansion or increasing the order of the iterative method.

Another thing which could be done (in particular if accurate asymptotics are looked for) is to perform a first order Taylor expansion around $x=a=-\frac 1 {2^n-(n+1)}$ and obtain $$x=\frac{a ((a-1) \text{Li}_{-n-1}(a)+(a (n-1)+2) \text{Li}_{-n}(a))}{(a-1) \text{Li}_{-n-1}(a)+(a n+1) \text{Li}_{-n}(a)}$$

For $n=20$, the above would give $x=-9.56717000\times 10^{-7}$ while the exact solution is $-9.56717029\times 10^{-7}$; for the same $n=20$, $(1)$ and $2$ would give exactly the result.

Answer 2

Extending a bit off of Claude Leibovici's answer, the factorization of palindromic polynomials can be done algebraically in general up to the 9th degree. In particular, if the degree is odd ($n$ is even) then $x+1$ can be factored out, giving an even degree palindromic polynomial. Then the degree is even and the substitution $t=x+\frac1x$ may be used to halve the degree of the polynomial.

For example with $n=7$ we compute $t^2$ and $t^3$ to get

$$\begin{cases}t^2=x^2+2+\frac1{x^2}\\t^3=x^3+3x+\frac3x+\frac1{x^3}\end{cases}\implies\begin{cases}x^2+\frac1{x^2}=t^2-2\\x^3+\frac1{x^3}=t^3-3t\end{cases}$$

and thus see that

\begin{align}&x^6+120x^5+1191x^4+2416x^3+1191x^2+120x+1\\&=x^3\left(x^3+120x^2+1191x+2416+\frac{1191}x+\frac{120}{x^2}+\frac1{x^3}\right)\\&=x^3[(t^3-3t)+120(t^2-2)+1191t+2416]\\&=x^3(t^3+120t^2+1188t+2176)\end{align}

which may then be algebraically solved.

Interestingly, it is shown in this paper that these Eulerian polynomials of degree $k$ have at least one irreducible factor over $\mathbb Q$ of degree at greater than $p_L(k)$, the largest prime less than $k$, and for $k=10$ ($n=11$) it is irreducible over $\mathbb Q$.