We have claimed that the lower bound for the radius of convergence by the distance to the nearest singular point is usually the actual radius of convergence. In the next four problems you will justify this claim for the problem $$ \begin{align}(x^2+4)y''+y&=0 \\ y(0)=1,\quad y'(0)=&0 \end{align} $$
Spherical coordinates are defined by the transformations
$$ \begin{align} x&=\rho\sin(\phi)\cos(\theta) \\ y&=\rho\sin(\phi)\sin(\theta) \\ z&=\rho\cos(\phi) \end{align} $$ | $$ \begin{align} \rho&=\sqrt{x^2+y^2+z^2} \\ \tan(\theta)&=\frac{y}{x} \\ \tan(\phi)&=\frac{\sqrt{x^2+y^2}}{z} \end{align} $$ |
Answers to the odd numbered exercises.