Matrix and Taylor Expansion of a Finite Continued Function

Question

I noticed a pattern between matrix and Taylor series of a finite continued fraction function. However, I don't know how to prove it or why they are related.

Let $$ f_{1}(z)=\frac{1}{-z-1} $$ $$ f_{2}(z)=\frac{1}{-z-\frac{1}{-z-1}} $$ Similarly, we define $f_{3}(z),f_{4}(z)$, etc. That is, $f_{l}(z)$ has $l$ layers.

Let's consider the Taylor expansion of $f_{l}(z)$ at $z=0$, and write $$ f_{l}(z)=a_{l}(0)+a_{l}(1)z+a_{l}(2)z^{2}+\cdots $$

Now, let's consider the following $l \times l$ matrix $$ A_{l}=\begin{pmatrix} 1 & 1 & \cdots & 1 & 1 \\ 1 & 1 & \cdots & 1 & 0\\ &\vdots \\ 1 & 0 & \cdots & 0 & 0 \end{pmatrix} $$

Here is the amazing thing: $$ (A_{l}^{n})_{1,1}= \mid a_{l}(n) \mid $$ where $(A_{l}^{n})_{1,1}$ stands for the $(1,1)$-entry of the matrix $A_{l}^{n}$.

I don't have a proof, but I checked with many $l$ and $n$ and they all hold. My questions are the following:

1. How to prove this equality? (I tried to prove it directly, but taking $n$-th derivative of a continued fraction is a nightmare)
2. What is the intuition behind it? Is there any deeper connection between this matrix and this sequence of finite continued fractions?
3. Is it a special case of some deeper result?

Thank you very much!

Answer 1

Heres my go at a proof: First off, there seems to be a slight mistake in your statement, namely, I've found that $|a_l(n)| = (A_l^{n+1})_{11}$. Regardless, setting $b_i = (A_l^{i+1})_{11}$, see that the power series $b_0 + b_1z + b_2 z^2 + ...$ is just the $(1,1)$ term in the matrix $$A_l + A_l^2 z + A_l^3 z^2 + ... = A_l(1 + A_lz + A_l^2 z^2 + ...) = \frac{A_l}{1 - A_l z} = \frac{1}{A_l^{-1} - z},$$ where the von Neumann series holds as long as $z \approx 0$. It's good to note that $A_l^{-1}$ exists (since $\text{det}(A_l) = \pm 1$) and actually $$A_l^{-1} = \begin{pmatrix} 0 & 0 & \cdots && 1 \\ 0 & && 1 & -1 \\ \vdots && ⋰& ⋰ &0 \\ 0 & 1 & -1 & \\ 1 & -1 & 0 & \cdots & 0 \end{pmatrix}.$$

Lucky for us, working with the matrix $A_l^{-1}$ is actually easier than working with $A_l$ because of all the 0's. From here, it suffices to compute the $(1,1)$ term in $(A_l^{-1} - z)^{-1}$ which is just $\frac{1}{\det(A_l^{-1} - z)}$ times the determinant of the $l-1 \times l-1$ bottom right minor. I'll let $B_l$ denote $A_l^{-1} - z$ and $C_l$ denote the $l-1 \times l-1$ minor.

By the symmetry of the matrices, note that the central $l-2 \times l-2$ submatrix of $B_l$ is exactly $B_{l-2}$, i.e., $$B_l = A_l^{-1} - z = \begin{pmatrix} -z & 0 & \cdots && 1 \\ 0 & -z&& 1 & -1 \\ \vdots && & &0 \\ 0 & 1 & -1\\ 1 & -1 & 0 & \cdots & -z \end{pmatrix} = \begin{pmatrix} -z & 0 & \cdots & 0 & 1 \\ 0 &&&& -1 \\ \vdots && B_{l-2}&& 0 \\ 0 &&&& \vdots \\ 1 & -1 & 0 & \cdots & -z \end{pmatrix}.$$

Because of all the $0$'s, expanding down the first column, $$\det B_l = -z \det C_l + (-1)^{l+1} \det(\text{upper right minor}) = -z \det C_l - \det B_{l-2}$$ (note how the $-1$'s cancel regardless of $l$'s parity). Similarly, expanding down the rightmost column of $C_l$, most terms are 0, so $$\det C_l = -z \det B_{l-2} - \det C_{l-2}.$$ Hopefully all my signs are correct, but you can check for small $l$.

Remember, we want the $(1,1)$ term in $A_l + A_l^2 z + ... = (A_l^{-1} - z)^{-1} = B_l^{-1}$. Plugging in, \begin{align*} (B_l^{-1})_{11} = \frac{\det C_l}{\det B_l} &= \frac{\det C_l }{-z \det C_l - \det B_{l-2}} = \frac{1}{-z - \frac{\det B_{l-2}}{\det C_l}} \\ &= \frac{1}{-z - \frac{\det B_{l-2}}{-z \det B_{l-2} - \det C_{l-2}}} = \frac{1}{-z - \frac{1}{-z - \frac{\det C_{l-2}}{\det B_{l-2}}}}. \end{align*} We are almost done but there is a slight problem because what happened to the $||$ signs around the $a_i(l)$? And also when $l = 1$, note that $(1-z)^{-1} = \frac{1}{-z + 1}$ which is different than $\frac{1}{-z - 1}$. At this point, it is good to note that the $l$ even case is slightly different from the $l$ odd case since for example, $1 - z$ appears in the matrix in the odd case while $-1 - z$ appears in the even case and also the coefficients of your $f_l$ seem to be always positive if $l$ is even but alternate signs when $l$ is odd.

Just change the induction hypothesis to something that is true. For even $l$, I claim that $\frac{\det C_l}{\det B_l }$ indeed equals your $f_l$. Note this holds for $(A_2^{-1} - z)^{-1} = \begin{pmatrix} -z & 1 \\ z & -1 - z \end{pmatrix}$. So then since $\frac{\det C_l}{\det B_l} = \frac{1}{-z - \frac{1}{-z - \frac{\det C_{l-2}}{\det B_{l-2}}}}$, this holds for all even $l$ by induction. For $l$ odd, instead define $g_l(z) = \frac{1}{-z - \frac{1}{\ddots -z - \frac{1}{-z + 1}} }$, i.e., similar to the $f_l$ but where there is $+1$ instead of $-1$ in the most nested fraction. Note that $(A_l^{-1} - z)^{-1}_{11} = g_l(z)$ for $l=1$ and $l=3$ so the same formula above implies it holds for all odd $l$. By playing around with the signs in the continued fractions, it seems there are many different ways to represent the $f_l$ and its cousins. In particular, check that $-g_l(-z) = f_l(z)$. So then the coefficients of $g_l$ are $\pm$ those of $f_l$. In total, this shows how to relate the coefficients of $(A_l)_{11} + (A_l^2)_{11} z + ... $ to those of $f_l(z)$ in such a way that shows your claim.