Perturbation series for $x^5+\varepsilon x-1=0$

Question

I want to find a closed form for the perturbation coefficients $a_n$ defined by the perturbative solution $$ x(\varepsilon)=1+\sum_{n=1}^\infty a_n \varepsilon^n $$ to the quintic equation $$ x^5+\varepsilon x-1=0. $$ By computing the determinant, as was suggested for this related question regarding the cubic case, we can argue that the radius of convergence of the above series must be $$ \rho=\frac{5}{4^{4/5}}=1.64938\dots $$ Furthermore, the Lagrange-Bürmann theorem allows us to formally write down a closed form for the coefficients, namely $$ a_n=\frac{(-1)^n}{n!}\frac{d^{n-1}}{dx^{n-1}}\left(\frac{x}{1+x+x^2+x^3+x^4}\right)^n_{x=1} $$ but this doesn't look very illuminating. (For the case of the cubic equation a slightly more explicit but still cumbersone rewriting was made possible by the simpler form of $a_n$).

What I would like to achieve is obtaining $\rho$ from the explicit closed expression for the $a_n$. Therefore I set out to compute some of them. Here they are: $$ a_1=-\frac{1}{5},\ a_2=-\frac{1}{5^2},\ a_3=-\frac{1}{5^3},\ a_4=0,\\ a_5=\frac{21}{5^6},\ a_6=\frac{78}{5^7},\ a_7=\frac{187}{5^8},\ a_8=\frac{286}{5^9},\ a_{9}=0,\\ a_{10}=-\frac{9367}{5^{12}},\ a_{11}=-\frac{39767}{5^{13}},\ a_{12}=-\frac{105672}{5^{14}},\ a_{13}=-\frac{175398}{5^{15}},\ a_{14}=0. $$ The behavior of the $a_n$ for $n=1,\ldots,30$ supports the following conjecture: $$\boxed{ a_n = -(-1)^{\lfloor n/5\rfloor}\frac{c_n}{5^{\alpha_n}} } $$ where $$ \alpha_n=\sum_{k=0}^\infty \left\lfloor \frac{n}{5^k} \right\rfloor $$ and the $c_n$ are nonnegative integer coefficients which are not divisible by $5$ and vanish for $n=5m-1$. I think that $c_n$ should be something of the form $$ \frac{1}{4n+1}\binom{5n}{n} $$ which for $n=4$ goes very near to reproducing $c_8=286$ and has a scaling similar to that exhibited by the $c_n$. This problem is also motivated by this video where it is suggested that the answer can be guessed with some effort by staring at the coefficients hard enough (27.08). I think I need some help, however!

Answer 1

For convenience let $\varepsilon = 5s$. Maple gives me the following:

$$\sum _{k=0}^\infty \frac{(-16 s^5)^k}{k!} \left(\frac{\left(\frac 2 5\right)_{2k} \left(-\frac 1 {10}\right)_{2k} }{\left(\frac 4 5\right)_k \left(\frac 3 5\right)_k \left(\frac 2 5\right)_k} - \frac{\left(\frac 4 5\right)_{2k} \left(\frac 3 {10}\right)_{2k} s }{ \left(\frac 6 5\right)_k \left(\frac 4 5\right)_k \left( \frac 3 5\right)_k } - \frac{\left(\frac{6}{5}\right)_{2k} \left(\frac{7}{10}\right)_{2k} s^2 }{ \left(\frac{7}{5}\right)_k \left(\frac{6}{5}\right)_k \left( \frac{4}{5}\right)_k } - \frac{\left(\frac{8}{5}\right)_{2k} \left(\frac{11}{10}\right)_{2k} s^3 }{\left(\frac{8}{5}\right)_{k} \left(\frac{7}{5}\right)_k \left( \frac{6}{5}\right)_k} \right) $$ using the Pochhammer symbols $$(a)_m = \frac{\Gamma(a+m)}{\Gamma(a)}$$

Answer 2

Making Robert Israel's formula a bit more compact we can write $$\boxed{ x(\varepsilon)= \sum_{k=0}^\infty (-1)^{k+1}(2)^{4k}\sum_{j=0}^3(-1)^{\delta_{j0}}\frac{\left(\frac{2(j+1)}{5}\right)_{2k}\left(\frac{2(j+1)}{5}-\frac{1}{2}\right)_{2k}}{\left(\frac{j+5}{5}\right)_{k}\left(\frac{j+4}{5}\right)_{k}\left(\frac{j+3}{5}\right)_{k}\left(\frac{j+2}{5}\right)_{k}}\frac{\varepsilon^{5k+j}}{5^{5k+j}} } $$ where $(a)_m$ is the Pochhammer symbol and $$ (a)_m = \frac{\Gamma(a+m)}{\Gamma(a)}=(a+m-1)(a+m-2)\cdots(a+1)a $$ for positive integer $m$. Applying Stirling's asymptotic expansion, $$\begin{aligned} &\lim_{k\to\infty}\left[\frac{2^{4k}}{5^{5k+j}}\frac{\left(\frac{2(j+1)}{5}\right)_{2k}\left|\left(\frac{2(j+1)}{5}-\frac{1}{2}\right)_{2k}\right|}{\left(\frac{j+5}{5}\right)_{k}\left(\frac{j+4}{5}\right)_{k}\left(\frac{j+3}{5}\right)_{k}\left(\frac{j+2}{5}\right)_{k}}\right]^{1/{5k}}\\ =&\ \lim_{k\to\infty}\frac{2^{4/5}}{5}\frac{\left(2+\frac{2(j+1)}{5k}\right)^{2/5}\left(2+\frac{2(j+1)}{5k}-\frac{1}{2k}\right)^{2/5}}{\left(1+\frac{j+5}{k}\right)^{1/5}\left(1+\frac{j+4}{k}\right)^{1/5}\left(1+\frac{j+3}{k}\right)^{1/5}\left(1+\frac{j+2}{k}\right)^{1/5}}\\ =&\ \frac{2^{4/5}}{5}\cdot 2^{2/5}\cdot 2^{2/5}\\ =&\ \frac{4^{4/5}}{5} \end{aligned}$$ independently of $j=0,1,2,3$. Therefore by Hadamard's formula, the radius of convergence is $$\boxed{ \rho=\frac{5}{4^{4/5}}. } $$