Transform a recurrence relation into an ODE. What is the relation between the ODE solution and the CF closed form?

Question

Assume a recurrence relation of the following type with $n\ge 0$

\begin{equation} F(n) a_n+P(n) a_{n+1}+ Q(n)a_{n-1}=\eta \delta_{n,0} \end{equation} where $F(n), P(n), Q(n)$ are polynomials of $n$ and $\delta_{n,0}$ is the Kronecker delta. One way to solve such a problem would be to construct the continued fraction (CF) and if you are lucky, you can find a closed form, let's call it $\Gamma$.

Another approach is to transform this recurrence relation into a differential equation:

First, we shift the coefficients by +1, such that \begin{equation} F(n+1) a_{n+1}+P(n+1) a_{n+2}+ Q(n+1)a_{n}=0 \end{equation}

Second, we assume a power series of a generating function $X(z)=\sum_{n\ge 0} a_n z^n$. We use that given a recurrence with a polynomial coefficient $P(n)a_n$, it can be written as $P(z D) X(z)$, where $D$ is the differentiation operator.

\begin{equation} (F(zD)+1)\frac{X(z)-a_0}{z}+(P(zD)+1)\frac{X(z)-a_0-a_1 z}{z^2} a_{n+2}+ (Q(z D)+1)X(z)=0 \end{equation}

Up until here, it is quite general, so let's simplify by assuming that the polynomial coefficients are up to 2nd order, i.e. $F(n)=f_2 n^2$, $P(n)=p_1 n$, and $Q(n)=q_0$.

\begin{equation} f_2 ((zD)^2+1)\frac{X(z)-a_0}{z}+p_1 (zD+1)\frac{X(z)-a_0-a_1 z}{z^2} a_{n+2}+ q_0 X(z)=0 \end{equation}

Without going into the details, after simplifying we obtain a second order ODE with the following structure

\begin{equation} x_2(z^2)X^{\prime \prime}(z)+x_1(z^2,z)x_1X^{\prime}(z)+x_0(z^2,z)x_0X(z)=a_0 b(z^2,z)+a_1 c(z^2,z) \end{equation}

Let's assume that this is an ODE we can solve analytically (and this is a huge assumption). I have three questions:

1. In order to solve this ODE, I need 2 initial conditions $X(0)$ and $X^{\prime}(0)$. Is it correct to assume that they are $X(0)=a_0$ and $X^{\prime}(0)=a_1$ from the recurrence relation?

2. What is the relation between the solution to this ODE $X(z)$ and the closed form of the CF from the recurrence relation $\Gamma$?

3. To derive my ODE, I used a power series generating function. However, I am aware that sometimes people use an exponential generating function. In it clear in which cases one is more useful than the other?

Answer 1

Here, the question "3." has to be addressed first. The given recurrence, with $$F(n)=f_2 n^2,\ P(n)=p_1 n,\ Q(n)=q_0,$$ imply the factorial-like growth of $|a_n|$ as $n\to\infty$. Thus $X(z)=\sum_{n\geq 0}a_n z^n$ doesn't exist as a function of $z$; if we like to work with functions (and not formal power series), we have to consider (something like) the exponential generating function. Another "guide" (for a choice of the sequence whose generating function is to be considered) is the "simplicity" of the resulting ODE; $a_n/n!$ may yield a harder-to-solve thing than, say, $a_n/\Gamma(n+\alpha)$ for some $\alpha\neq 1$. I wonder if there is some machinery for this...

Once the ODE is set up, it's time for the question "1.". The answer is often affirmative, but not always. Say, for the ODE I've got at the end of this answer, $w'(0)=1$ holds automatically (and thus cannot be used as an initial condition).

Finally the question "2.": once the ODE is solved, say, we know $f(z)=\sum_{n\geq 0}a_n z^n$ and $g(z)=\sum_{n\geq 0}b_n z^n$, and wish to compute $\lim\limits_{n\to\infty}a_n/b_n$, preferably indirectly. (Of course there's an option to get $a_n$ and $b_n$ explicitly; sometimes, this is reachable.)

If we know it exists, and $b_n$ are positive real numbers, the following Abelian-type theorem can serve as a tool. Let $r$ be the radius of convergence of $g(z)$; if it is finite, suppose additionally that $\lim\limits_{z\to r}g(z)=\infty$. Then $\lim\limits_{n\to\infty}a_n/b_n=\lim\limits_{z\to r}f(z)/g(z)$. (A proof is easy; here is half-the-matter.)

If we don't know that the limit exists, or work with complex numbers say, another (often useful) tool is the singularity analysis [of $g$].