Given $a_0=1$, $b_0=1$, and the recurrence relations for $m\geq1$:
$$b_m = \frac{1}{m} \sum_{k=1}^m (6k-m) a_k b_{m-k}$$
$$a_{m-1}=-b_m$$
The first few terms of $a_m$ can be calculated using Mathematica (code below),
$$a_1=-\frac{1}{5},\quad
a_2=-\frac{1}{25},\quad
a_3=-\frac{1}{125},\quad
a_4=0,\quad
a_5=\frac{21}{15625},\quad\cdots
$$
m=5;
a[0]=1;b[0]=1;
For[n=1,n<=m,n++,
b[n]=1/(n*a[0])Sum[(6k-n)a[k]b[n-k],{k,1,n}];
a[n]=(a[Evaluate[n]]/.Solve[b[n]+a[n-1]==0,a[Evaluate[n]]][[1]]);
]
For[i=1,i<=m,i++,Print["a[",i,"]= ",a[i]]]
My questoins are:
- Is it possible to solve the recurrance relations for $a_m$?
- Without explicitly solving the recurrance relations for $a_m$, is it possible to calculate the radius of convergence for the series $$x(\epsilon) = \sum_{m=0}^\infty a_m \epsilon^m$$
I prvoide some context of the problem. To use the perturbation method to solve the algebraic equation: $$x^5+\epsilon x = 1\tag{1}$$ Let us assume that the solution takes the form $$x(\epsilon)=a_0+a_1\epsilon+a_2\epsilon^2+\cdots\tag{2}$$ Substituting (2) into (1), expanding the multinomial power using this formula, and matching the coefficient of same power terms of $\epsilon$, we can obtain the recurrence relations as in the main question.
The radius of convergence for $\epsilon$ is my main interest. By observation, I suspect $$|a_m|\sim\frac{1}{5^{m+s}}\quad\text{as}\quad m\to\infty$$ for some $s>0$ but I couldn't get the right pattern. The denominator $5$ must have something to do with the quintic equation (1).
