Power series solution of $f'' + (z^{-1})f'+f = 0$ I am trying to do what the title says, and I get to the point where I have a recursive relationship to even numbered terms of the expansion, but I'm stuck on two things.
EDIT:  we are given that f(0) = 1
EDIT 2:  We just have gone over complex power series, radius of convergence.  Thus we do not yet have advanced tools to solve this problem.  My work again so far is in the attached image link.


*

*How do I get the odd numbered terms? 

*And how do I turn both even and odd term series into series notation (i.e. how to get $a_k$)?
Thanks!
my work so far
 A: Find the power series solution of
\begin{equation}zf^{\prime\prime}+f^\prime+zy=0\tag{1}\end{equation}
Since $z=0$ is a regular singular point of the equation, will attempt the Method of Frobenius.
Let
\begin{eqnarray}
f&=&\sum_{n=0}^{\infty}a_nz^{n+r}\\
f^\prime&=&\sum_{n=0}^{\infty}(n+r)a_nz^{n+r-1}\\
f^{\prime\prime}&=&\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nz^{n+r-2}\\
\end{eqnarray}
Then
\begin{eqnarray}
zf^{\prime\prime}+f^\prime+zf&=&z^r\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nz^{n-1}\\
&+&z^r\sum_{n=0}^{\infty}(n+r)a_nz^{n-1}+z^r\sum_{n=0}^{\infty}a_nz^{n+1}\\
&=&z^r\sum_{n=0}^{\infty}(n+r)^2a_nz^{n-1}+z^r\sum_{n=0}^{\infty}a_nz^{n+1}\\
&=&0
\end{eqnarray}
For $z\ne0$ this implies that
\begin{equation}
\sum_{n=0}^{\infty}(n+r)^2a_nz^{n-1}+\sum_{n=0}^{\infty}a_nz^{n+1}=0
\end{equation}
Separating out the $z$ terms with negative exponents results in the following:
\begin{equation}
r^2a_0z^{-1}+\sum_{n=1}^{\infty}(n+r)^2a_nz^{n-1}+\sum_{n=0}^{\infty}a_nz^{n+1}=0
\end{equation}
Thus, the indicial equation is $r=0$. So, although the equation is not analytic at $z=0$, the equation could have been solved without using the Frobenius Method.
\begin{equation}
\sum_{n=1}^{\infty}n^2a_nz^{n-1}+\sum_{n=0}^{\infty}a_nz^{n+1}=0
\end{equation}
After calibrating the sums we get
\begin{equation}
a_1+\sum_{n=1}^{\infty}[(n+1)^2a_{n+1}-a_{n-1}]=0
\end{equation}
So $a_1=0$ and we get a recursive relation
\begin{equation}
a_{n+1}=\frac{a_{n-1}}{(n+1)^2}\text{ for }n\ge1
\end{equation}
Thus, all the odd power terms are $0$ and the even powered terms satisfy
$$ a_{2n}=\frac{a_0}{2^{2n}(n!)^2} $$
giving one solution
$$ f=a_0\sum_{n=0}^{\infty}\frac{x^{2n}}{2^{2n}(n!)^2} $$
As is frequently the case when expanding about a singular point, this gives us just one of two linearly independent solutions. To find a second linearly independent solution it will be necessary to use reduction of order.
But I will give you a hint: If you multiply equation $(1)$ by $z$ you get
$$ z^2f^{\prime\prime}+zf^\prime+z^2y=0\tag{2} $$
Recall that
$$ z^2f^{\prime\prime}+zf^\prime+(z^2-\nu^2)y=0\tag{3} $$
is Bessel's equation. So what we have is Bessel's equation with $\nu=0$.
