So I'm trying to solve the following ODE which have three ordinary singular points: $$z(z-1)(z-\tau)R''(z)+(a+bz+cz^2+dz^3)R'(z)+(\alpha+\beta z+\gamma z^2)R(z)=0;$$ Using the Frobenius method, $R(z)=\sum a_kz^{k+b}$, finding the roots $b_1=0,b_2=1-\frac{a}{\tau}$, leads me to the patternless recurrence relation: $$a_{k-4}[P_1(k)]+a_{k-3}[P_2(k)]+a_{k-2}[P_2(k)]+a_{k-1}[P_2(k)]=0.$$ Where $P_i(k)$ means a polynomial of degree ($i$) in the variable $k$. Also I supposed $a_0=1$
- How can I check for the radius of convergence of such messy thing? All methods I looked up where meant to well behaved recurrence relations.
2)IF this solution is convergent $\forall$ z. I know that the solution is valid for at most the radius for the nearest singular point. But can the solution be valid for a $z$ greatear than the fartest singular point, like, suppose $0<\tau<1$, can the Frobenius solution at $z=0$ be valid for $z>\tau$???