Discussion in 'Business Directory' started by Formations Factory, Jul 24, 2015.
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text:
February 13, 2019 April 9, 2019 Beth Young 0 Comments
• There's a 20 percent chance of rain before 8 a.m. today, with skies gradually clearing and a steady temperature around 41. The wind will be out of the west at 15 to 22 miles per hour, gusting as high as 38. It will be partly cloudy overnight, with a low around 27. Thursday will be mostly sunny, with a high near 43. We're expecting rain likely after noon on Friday, with mostly cloudy skies and a high near 49.
• Polystyrene packaging bans are going into effect in both Southampton and East Hampton towns, after the Southampton Town Board voted unanimously to approve a ban Tuesday afternoon. East Hampton's ban will go into effect April 22 and Southampton's will go into effect May 8.
• Riverhead Town Supervisor Laura Jens-Smith's listening tour of the town will continue this evening with a session at the Riverhead Free Library at 6 p.m. Last night's meeting in Wading River was cancelled due to inclement weather. We'll have the details when it's rescheduled.
• Our Peconic Dish columnist, Alison Boyd, has some delicious recipes for us this week, incorporating blood oranges into a delicious salad and upside down cake that are bound to wow your Valentine.
• The centerfold of the print edition of the East End Beacon has always been a love letter to the Peconic Estuary. This week, the National Estuary Program embarks on its annual Valentine to estuaries, the #iheartestuaries campaign. Help us join in! Get the details in The Peconic Bathtub.
• The Southold Town Anti-Bias Task Force has announced that its annual Synergy conversation between the community and the police will be held on Thursday, March 28 at 6 p.m. at the Peconic Lane Recreation Center. All are welcome to attend.
The high tides on the East End for the next two days are as follows:
Plum Gut Harbor: 3:44 a.m., 4:15 p.m.
Montauk Harbor: 2:52 a.m., 3:23 p.m.
Greenport: 4:21 a.m., 4:52 p.m.
Mattituck Inlet: 4:56 a.m., 5:38 p.m.
Sag Harbor: 4:16 a.m., 4:47 p.m.
New Suffolk: 5:43 a.m., 6:14 p.m.
South Jamesport: 5:50 a.m., 6:21 p.m.
Shinn. Bay Entrance: 2:21 a.m., 2:56 p.m.
Shinn. Inlet: 12:30 a.m., 1:05 p.m.
Shinn. Inlet: 1:30 a.m., 2:11 p.m.
Good morning! • The New York State DEC has hatched a plan to kill or…
Good Morning! • With the coldest temperatures of the winter expected to hit Long Island…
Good Morning! • The National Weather Service is predicting we'll see four to eight inches…
← South Fork Towns Adopt Polystyrene Bans
Winterfest Tunes Up for 12th Season →
February 18, 2015 Beth Young 0
Please prove you're human: seven + seven =

redpajama_set_name: RedPajamaCommonCrawl


Homeopathic medicine world's best for kidney affection. The most important part of a body different get defective then we might see a lot of changes in our body.
Kidney failure and kidney problems can take you to the end of your life. We need to cure kidney problems. Let's Discuss some Homeopathic medicines for renal colic and calculi.
Here are some best medicines in homeopathic treatment which helps to cure the renal colic pain and calculi problem.
If there is any kidney Disease or kidney stone then you can use these mother tinctures.
You can take 15-20 drops in a cup of fresh water four times in a day. With the help of this treatment, INSHALLAH stone will be torn down into the pieces and excreted with the urine.
If there is severe pain you can use these medicines in hot water after 20-30 minutes again and again.
But wait once you get rid from the kidney stone. Take it to the laboratory and test this stone.
If the test comes that the stone is made with the phosphate stone. Then you can use the below mentioned treatment.
For the treatment of the phosphate stone, you can use the following h* mother tinctures for the stone excretion.
Take these two remedies equal amount and use 10 drops of in 1/4th cup of water, three times in a day.
But if the stone is of oxalic acid which is hard and difficult to break. Then for the treatment of oxalic acid kidney stone following h* remedies can help you a lot.
You can use these both medicines. You can use 4 tablets from each for 4 times in a day.


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text:
Pfizer's COVID vaccine causes miscarriages, stillbirths
After initially being denied, the Public Health and Medical Professionals for Transparency finally gained access to documents containing Pfizer's data on their COVID-19 vaccine, revealing that the big pharma company received over 150,000 reports of serious adverse events, including miscarriages and stillbirths, within three months of vaccine rollout.
What is especially concerning about the data found are the number of pregnant and lactating women who received the shots in the first few months of the rollout. The data states that of 270 "unique pregnancies" exposed to the vaccine, "nu outcome was provided for 238 pregnancies," leaving only 32 with known outcomes. Out of these, 32 known outcomes, 23 resulted in spontaneous abortions (miscarriages), two premature births with neonatal deaths, two spontaneous abortions with intrauterine death, one spontaneous abortion with neonatal death and one pregnancy with "normal outcome."
This means that of 32 pregnancies with known outcomes, 28 resulted in fetal death. If results are to be based only on the known outcomes of the vaccines in pregnant and lactating women, there is an 87.5 percent pregnancy loss rate.
Public health agencies remain adamant that pregnant women and their children are more likely to die of the virus than from the vaccine, but there is little evidence that proves such claims, as the study they usually refer to comes from the Centers for Disease Control and Prevention (CDC) itself.
A comparison of stillbirth rates in 1,249,634 deliveries at 736 hospitals from March 2020 to September 2021 among women with and without COVID infections established that there was indeed a surge in stillbirths, but not during the height of the first deadly wave of the virus: only during the period when the delta variant was dominant, or after pregnant women were pressured into getting vaccinated. (Related: UK now pushing pregnant women to take covid vaccines despite miscarriage, infertility risks.)
Despite the numbers (only 0.98 percent of COVID-19 affected deliveries pre-delta compared to 2.70 percent after vaccines were introduced) the CDC still won't consider that the experimental injections could be the reason for the increasing stillbirth rates.
Because the COVID-19 vaccines are said to be highly effective and vaccination among pregnant women was approximately 30 percent as of July 2021, most women with COVID-19 at delivery were likely to be unvaccinated, according to the CDC. This information does not sound very scientific because science isn't usually an assumption based on a slogan added to the guesstimates.
Another study that experts heavily relied on was a study that was later corrected when analysts recognized the error in their data collections, adding the possibility of an 82 percent miscarriage rate in early pregnancy despite concluding that COVID shots were safe and effective.
Too much unknown information to definitively say vaccines are "for the good of all"
The Pfizer vaccine has already been approved by the Food and Drug Administration (FDA) and these COVID vaccines have already caused 1.8 million adverse events that included deaths, miscarriages, bells palsy, heart attacks and various permanent disabilities, to name a few.
While research continues regarding how natural immunity from COVID is more robust than vaccines, 79 percent of fully vaccinated individuals had been infected with any one of the variants of the virus.
This means that injecting bodies with foreign substances that have been rushed through development causes harm or even death, negating any "duty to seek to preserve our bodies."
If previous studies are any indication, it shows that vaccines are anything but safe and effective.
In mid-2021, Dr. Simone Gold, founder of America's Frontline Doctors previously explained that there is a known potential mid-pregnancy fertility risk with the use of the vaccine. He acknowledged that while getting COVID-19 has been associated with a high risk of mid-pregnancy miscarriage because "the placenta fails," he noted that the vaccine may do the exact same thing.
"It's lunacy to get this experimental vaccine if you're a young female," she said. "It's that simple … I would flat out forbid any young female from getting this vaccine, and I think it's very unethical for any physician to offer this to a young female," he said.
Watch the video below to learn more about vaccine injuries:
This video is from The Resistance Chicks on Brighteon.TV.
Read more about COVID-19 vaccines at Vaccines.news.
Sent-Trib.com
LifeSiteNews.com
Tagged Under: abortions, Big Pharma, biological weapon, covid vaccines, depopulation, genocide, infanticide, miscarriage, pandemic, pharmaceutical fraud, spike protein, stillbirth, Vaccine deaths, vaccines, women's health


This spicy shrimp egg roll skillet is the quick and easy weeknight meal you need! It's packed full of napa cabbage, carrots, garlic, ginger, and spicy shrimp. It's ready in just 30 minutes and packed full of flavor!
My husband could eat Chinese food every single day. I'm not even kidding. When I travel for work I'm fairly certain he orders take out every single day.
However, I'm not a huge fan of takeout. Mostly because everything is so damn salty I just can't.
I don't need to step on that scale in the morning and be annoyed with myself that I decided to eat all the sodium the night before.
Well, with this recipe I get all the flavor of takeout without all that extra sodium.
Plus I get to use high-quality ingredients which makes me happy.
I made this skillet recipe extra spicy but if you have kids (or you don't like spice) just tame down the addition of chili garlic sauce.
Which, if you don't own a bottle you need to buy ASAP! It's so good and I put it in EVERYTHING.
Plus it's super cheap and that always makes me happy.
When I decided to make this dish I just happened to have a big head of napa cabbage sitting in my fridge.
However, I'm pretty sure if you only had green cabbage on hand that would work just as well.
OR if you wanted to make your life super duper simple buy a couple bags of coleslaw mix and you don't even have to cut anything.
I also decided to stir in cilantro right at the end because nothing beats that fresh cilantro taste. I mean, unless you're one of those people who thinks cilantro tastes like soap.
In that case, I'm sorry. Cilantro is so dang good.
Since I'm allergic to sesame oil I used olive oil to cook up my spicy shrimp egg roll skillet but if you're not allergic to sesame you could 100% use sesame oil to cook your cabbage.
Oh, and I know you see sesame seeds hanging out in that bowl. Don't worry, raw sesame seeds in small doses don't trigger my allergy.
Love this spicy shrimp egg roll skillet?
Try my savory Asian oat bran, Instant Pot Asian chicken and rice soup, spicy cashew chicken, 20 minute healthier orange chicken, spicy beef with peanuts, spicy garlic stir fry, or my spicy pork noodle bowl.
This spicy shrimp egg roll skillet is the quick and easy weeknight meal you need! It's packed full of napa cabbage, carrots, garlic, ginger, and spicy shrimp. It's ready in just 30 minutes and packed full of flavor!
In a large skillet add oil and set over medium-high heat.
When the oil is hot add the cabbage and carrots, cook for 5 minutes.
Add the garlic, ginger, lemongrass, chili garlic paste, soy sauce, rice wine vinegar, fish sauce and brown sugar. Cook for about 5 minutes or until the cabbage is slightly cooked down and most of the liquid is cooked down.
Add the shrimp and stir to combine. Cook for about 5 minutes or until the shrimp is firm, pink and fully cooked.
Stir in the cilantro and garnish with sesame seeds and sliced green onions.
This is all I can think about having right now. Looks SO good!
I love a bit of spice but appreciate the suggestion on taming it a bit for my kids.
Ooh, that looks absolutely mouthwatering, YUM!
WOW! This looks inanely delicious! Must try!
These look so delish! A must try recipe!
I'm not a fan of take out either, so this is perfect! Love it!
This looks like a new favorite for weeknights!


text:
\section{Introduction}\label{intro}
Many integrable systems first appeared in mathematical physics and mathematical  biology for real-life problems.
Time discretizations of such integrable systems can be useful for understanding physical and biological phenomena numerically.
Discrete-time evolutions in integrable discrete systems also contribute to scientific computing as key components of numerical algorithms.
\par
The Toda equation is one of the most famous integrable systems, and describes the motion governed by a nonlinear spring \cite{Toda1981},
but studies have branched into, for example, nonlinear electric circuits \cite{Hirota1973}, explicit soliton solutions \cite{Nakamura1997}, and the connection with simple Lie algebra \cite{Bogoyavlensky1976}.
A time-discretization of the Toda equation is
\begin{align}\label{adToda}
\left\{
\begin{array}{l}
\displaystyle q_{k}^{(s+1)}+e_{k-1}^{(s+1)}=q_{k}^{(s)}+e_{k}^{(s)},\quad k=1,2,\dots,m,\\[2pt]
\displaystyle q_{k}^{(s+1)}e_{k}^{(s+1)}=q_{k+1}^{(s)}e_{k}^{(s)},\quad k=1,2,\dots,m-1,\\[2pt]
\displaystyle e_{0}^{(s)}:=0,\quad e_{m}^{(s)}:=0,\\[2pt]
s=0,1,\dots,
\end{array}
\right.
\end{align}
where the subscripts $k$ and superscripts $s$ with parentheses are discrete-space and discrete-time variables, respectively.
Interestingly, the Toda equation and its time discretization both have relationships to well-known algorithms for computing eigenvalues of tridiagonal matrices.
The time evolution in the Toda equation corresponds to the $1$-step of the $QR$ algorithm for tridiagonal matrix exponentials \cite{Symes1982}.
The discrete Toda (dToda) equation \eqref{adToda} is equivalent to the recursion formula that generates the similarity $LR$ transformations of tridiagonal matrices in the quotient-difference (qd) algorithm \cite{Rutishauser1990}.
The qd algorithm is also used to compute singular values of bidiagonal matrices.
\par
In the theory of orthogonal polynomials, the compatibility condition of a transformation for discrete-time evolution and its inverse yields the  following dToda equation with arbitrary constants $\mu^{(t)}$ and $\mu^{(t+1)}$ \cite{Hirota1997}:
\begin{align}\label{nadToda}
\left\{
\begin{array}{l}
\displaystyle Q_{k}^{(t+1)}+E_{k-1}^{(t+1)}+\mu^{(t+1)}=Q_{k}^{(t)}+E_{k}^{(t)}+\mu^{(t)},
\quad k=1,2,\dots,m,\\[2pt]
\displaystyle Q_{k}^{(t+1)}E_{k}^{(t+1)}=Q_{k+1}^{(t)}E_{k}^{(t)},\quad k=1,2,\dots,m-1,\\[2pt]
\displaystyle E_{0}^{(t)}:=0,\quad E_{m}^{(t)}:=0,\\[2pt]
 t=0,1,\dots.
\end{array}
\right.
\end{align}
We consider $t$ appearing in the superscripts in \eqref{nadToda} to be a different discrete-time variable from $s$ in the dToda equation \eqref{adToda}.
Since $\mu^{(t)}$ and $\mu^{(t+1)}$ depend on the independent variable $t$, we refer to \eqref{nadToda} as the non-autonomous dToda equation, and refer to \eqref{adToda} as the autonomous dToda equation.
If we set $s=t$ and $\mu^{(t)}=\mu^{(t+1)}=0$, then the non-autonomous dToda equation \eqref{nadToda} is equivalent to the autonomous dToda equation \eqref{adToda}.
The non-autonomous dToda equation \eqref{nadToda} also gives the $LR$ transformations with implicit shifts of tridiagonal matrices.
In other words, \eqref{nadToda} can be considered a recursion formula of the shifted qd algorithm that may converge faster than the original algorithm.
The constants $\mu^{(t)}$ and $\mu^{(t+1)}$ then correspond to the shifts that bring about the convergence acceleration.
\par
The integrable Lotka-Volterra (LV) system is a simple biological model describing a predator-prey interaction of several species.
Aside from the discretization of the Toda equation, the discrete LV (dLV) system has discretization parameters $\delta^{(t)}$ and $\delta^{(t+1)}$ as follows:
\begin{align}\label{dLV}
\left\{
\begin{array}{l}
\displaystyle
u_k^{(t+1)}(1+\delta^{(t+1)}u_{k-1}^{(t+1)})
=u_k^{(t)}(1+\delta^{(t)}u_{k+1}^{(t)}),\quad
k=1,2,\dots,2m-1,\\[2pt]
\displaystyle
u_0^{(t)}:=0,\quad u_{2m}^{(t)}:=0,\quad\\[2pt]
 t=0,1,\dots.
\end{array}
\right.
\end{align}
Here, $u_k^{(t)}$ indicates density of the $k$th species at the discrete-time $t$.
The dLV system \eqref{dLV} also generates the $LR$ and shifted $LR$ transformations of
positive-definite tridiagonal matrices for computing singular values of bidiagonal matrices \cite{Iwasaki2002,Iwasaki2004,Tsujimoto2001}.
\par
Several observations have been made regarding the asymptotic behavior of solutions
to the autonomous dToda equation \eqref{adToda} and the dLV system \eqref{dLV}.
The latter, however, has only been analyzed with positive initial values.
These asymptotic behaviors have been individually shown using distinct frameworks in separate books \cite{Henrici,Rutishauser1990} and separate papers \cite{Iwasaki2002,Iwasaki2004,Tsujimoto2001}.
Moreover, to the best of our knowledge, asymptotic analysis for the non-autonomous dToda equation \eqref{nadToda} has not been reported yet.
In this paper, based on only an infinite sequence with respect to two types of discrete-time variables $s$ and $t$,
we report on the asymptotic behavior of all three systems.
\par
The remainder of this paper is organized as follows.
In Section~\ref{sec:2}, we first introduce an infinite sequence with respect to two types of discrete-time variables $s$ and $t$.
The infinite sequence is also associated with matrix eigenvalues.
We then show properties of the Hankel determinants associated with the infinite sequence.
In Sections~\ref{sec:3} and~\ref{sec:31}, we examine relationships between the Hankel determinants,
their Hankel polynomials, and Hadamard polynomials with respect to $s$ and $t$.
Referring to \cite{Maeda2013},
we simultaneously derive the autonomous dToda equation \eqref{adToda} and the non-autonomous dToda equation \eqref{nadToda}.
In Section~\ref{sec:4}, we observe eigenvalue problems of square matrices involving the dToda variables,
and then express the determinant solutions with sufficient degrees of freedom to the dToda equations using the Hankel determinants.
In Section~\ref{sec:5}, we present asymptotic analysis of the dToda equations by considering asymptotic expansion of the Hankel determinants as $s\rightarrow\infty$ or $t\rightarrow\infty$.
In Section~\ref{sec:6}, by relating the non-autonomous dToda equation \eqref{nadToda} to the dLV system \eqref{dLV},
we describe the determinant solution to the dLV system \eqref{dLV} and its asymptotic behavior as $t\rightarrow\infty$.
Finally, we conclude in Section~\ref{sec:7}.

\section{Hankel determinants with two discrete-time variables}\label{sec:2}
In this section, we first introduce an infinite sequence $\{f_s^{(t)}\}_{s,t=0}^{\infty}$ with respect to two types of discrete-time variables $s$ and $t$, and then associate it with the Hankel determinants.
Next, we derive identities concerning the Hankel determinants
under certain restrictions on the infinite sequence $\{f_s^{(t)}\}_{s,t=0}^{\infty}$.
\par
For arbitrary complex $\lambda_1,\lambda_2,\dots,\lambda_m$,
let us introduce an $m$-degree polynomial with respect to complex $z$:
\begin{align}\label{pz}
p(z)=(z-\lambda_1)(z-\lambda_2)\cdots(z-\lambda_m).
\end{align}
The polynomial \eqref{pz} can be regarded as the characteristic polynomial of $m$-by-$m$ matrices
with eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_m$.
Clearly, the polynomial $p(z)$ can be expanded as
\begin{align}\label{pza}
p(z)=z^m+a_1z^{m-1}+\dots+a_{m-1}z+a_{m},
\end{align}
where $a_1,a_2,\dots,a_n$ are complex constants.
\par
At each $t=0,1,\dots$, let $\{f_{s}^{(t)}\}_{s=0}^{\infty}$ be an infinite sequence
whose entries satisfy the linear equation with coefficients $a_1,a_2,\dots,a_m$ appearing in \eqref{pza},
\begin{align}\label{fnmt}
f_{s+m}^{(t)}+a_1f_{s+m-1}^{(t)}+\dots+a_mf_{s}^{(t)}=0,\quad s=0,1,\dots,
\end{align}
where $f_0^{(t)},f_{1}^{(t)},\dots,f_{m-1}^{(t)}$ are arbitrary.
Moreover,
let the infinite sequence $\{f_s^{(t)}\}_{s,t=0}^{\infty}$ satisfy
\begin{align}\label{key}
f_{s}^{(t+1)}=f_{s+1}^{(t)}-\mu^{(t)}f_{s}^{(t)}, \quad s,t=0,1,\dots,
\end{align}
where $\{\mu^{(t)}\}_{t=0}^{\infty}$ is an arbitrary constant sequence. \par
We next consider the determinants of symmetric square matrices of order $k$:
\begin{align}\label{Hankel}
H_k^{(s,t)}:=
\left|\begin{array}{cccc}
f_{s}^{(t)}	&f_{s+1}^{(t)}	&\cdots&	f_{s+k-1}^{(t)}\\
f_{s+1}^{(t)}	&f_{s+2}^{(t)}	&\cdots&	f_{s+k}^{(t)}\\
\vdots    &\vdots      	&\vdots&\vdots\\
f_{s+k-1}^{(t)}	&f_{s+k}^{(t)}	&\cdots&	f_{s+2k-2}^{(t)}
\end{array}\right|,
\quad s,t=0,1,\dots,
\end{align}
where $H_{-1}^{(s,t)}:=0$ and $H_{0}^{(s,t)}:=1$.
The determinants $H_k^{(s,t)}$ are known as the Hankel determinants and
can be used for expressing solutions to several integrable systems.\par
We now derive a proposition concerning the case $k=m+1$ in the Hankel determinants $H_k^{(s,t)}$.
\begin{proposition}\label{prop1}
It holds that
\begin{align}
H_{m+1}^{(s,t)}=0,\quad s,t=0,1,\dots.\label{taubc}
\end{align}
\end{proposition}
\begin{proof}
By multiplying the $1$st, $2$nd, $\dots$, $m$th columns in the Hankel determinant $H_{m+1}^{(s,t)}$
by $a_m$, $a_{m-1}$, $\dots$, $a_1$, respectively, and by adding these results to the $(m+1)$th column,
we obtain
\begin{align}\label{taum+1st}
H_{m+1}^{(s,t)}=
\left|
\begin{array}{ccccc}
\bvec{f}_{m+1}^{(s,t)}&\bvec{f}_{m+1}^{(s+1,t)}&\cdots&\bvec{f}_{m+1}^{(s+m-1,t)}&
\displaystyle\sum_{\ell=0}^{m}a_{\ell}\bvec{f}_{m+1}^{(s+m-\ell,t)}\\
\end{array}
\right|,
\end{align}
where $a_0:=1$ and  $\bvec{f}_k^{(s,t)}:=(f_{s}^{(t)}, f_{s+1}^{(t)},\dots,f_{s+k-1}^{(t)})^{\top}$.
It is obvious from \eqref{fnmt} that
\begin{align}\label{fbc}
\sum_{\ell=0}^m a_\ell\bvec{f}_{m+1}^{(s+m-\ell,t)}=0.
\end{align}
Thus, combining \eqref{taum+1st} with \eqref{fbc} gives \eqref{taubc}.
\end{proof}
\par
The following proposition gives evolutions with respect to $s$ and $t$ in the Hankel determinants $H_k^{(s,t)}$ associated
with the restricted infinite sequence $\{f_s^{(t)}\}_{s=0}^{\infty}$.
\begin{proposition}\label{prop2}
For $k=0,1,\dots,m$, it holds that
\begin{align}
&H_k^{(s+2,t)}H_k^{(s,t)}=H_{k}^{(s+1,t)}H_{k}^{(s+1,t)}+H_{k+1}^{(s,t)}H_{k-1}^{(s+2,t)},
\quad s,t=0,1,\dots,\label{tauJacobin}\\
&H_{k}^{(s,t+2)}H_{k}^{(s,t)}=H_{k}^{(s,t+1)}H_{k}^{(s,t+1)}+H_{k+1}^{(s,t)}H_{k-1}^{(s,t+2)},
\quad s,t=0,1,\dots.\label{tauJacobit}
\end{align}
\end{proposition}
\begin{proof}
For any determinant $D$, it is shown in \cite{Hirota2003} that
\begin{align}\label{Jacobi}
D\left[\begin{array}{c} i_1\\j_1\end{array}\right]
D\left[\begin{array}{c} i_2\\j_2\end{array}\right]
=
D\left[\begin{array}{c} i_1\\j_2\end{array}\right]
D\left[\begin{array}{c} i_2\\j_1\end{array}\right]
+
DD\left[\begin{array}{cc}i_1&i_2\\j_1&j_2\end{array}\right],
\end{align}
where $D\left[\begin{array}{cccc}i_1&i_2&\dots&i_n\\j_1&j_2&\dots&j_n\end{array}\right]$
denotes the cofactor obtained from $D$ by deleting the
$i_1$th, $i_2$th, $\dots$, $i_n$th rows and $j_1$th, $j_2$th, $\dots$, $j_n$th columns.
Equation \eqref{Jacobi} is well known as the Jacobi identity for determinants.
Then, by letting $i_1=j_1=1$, $i_2=j_2=k+1$ and $D=H_{k+1}^{(s,t)}$ in \eqref{Jacobi},
we obtain
\begin{align}\label{temp}
H_{k+1}^{(s,t)}\left[\begin{array}{c} 1\\1\end{array}\right]
H_{k+1}^{(s,t)}\left[\begin{array}{c} k+1\\ k+1\end{array}\right
=
H_{k+1}^{(s,t)}\left[\begin{array}{c} 1\\ k+1\end{array}\right]
H_{k+1}^{(s,t)}\left[\begin{array}{c} k+1\\ 1\end{array}\right]
+
H_{k+1}^{(s,t)}H_{k+1}^{(n,t)}\left[\begin{array}{cc}1&k+1\\1&k+1\end{array}\right].
\end{align}
It follows that
\begin{align*}
&\left|
\begin{array}{cccc}
\bvec{f}_k^{(s+2,t)}&\bvec{f}_{k}^{(s+3,t)}&\dots&\bvec{f}_{k}^{(s+k+1,t)}
\end{array}
\right|
\left|
\begin{array}{cccc}
\bvec{f}_k^{(s,t)}&\bvec{f}_{k}^{(s+1,t)}&\dots&\bvec{f}_{k}^{(s+k-1,t)}
\end{array}
\right|\\
&\quad=
\left|
\begin{array}{cccc}
\bvec{f}_k^{(s+1,t)}&\bvec{f}_{k}^{(s+2,t)}&\dots&\bvec{f}_{k}^{(s+k,t)}
\end{array}
\right|\left|
\begin{array}{cccc}
\bvec{f}_k^{(s+1,t)}&\bvec{f}_{k}^{(s+2,t)}&\dots&\bvec{f}_{k}^{(s+k,t)}
\end{array}
\right|\\
&\qquad
+\left|
\begin{array}{cccc}
\bvec{f}_{k+1}^{(s,t)}&\bvec{f}_{k+1}^{(s+1,t)}&\dots&\bvec{f}_{k+1}^{(s+k-1,t)}
\end{array}
\right|\left|
\begin{array}{cccc}
\bvec{f}_{k-1}^{(s+2,t)}&\bvec{f}_{k-1}^{(s+3,t)}&\dots&\bvec{f}_{k-1}^{(s+k,t)}
\end{array}
\right|,
\end{align*}
which immediately leads to \eqref{tauJacobin}.\par
By multiplying the $k$th, $(k-1)$th, \dots, $1$st rows by $-\mu^{(t)}$, and then adding these to the $(k+1)$th, $k$th, \dots, $2$nd rows in the entries of the Hankel determinants $H_{k+1}^{(s,t)}$ in \eqref{Hankel}, respectively, and then by considering \eqref{key},
we obtain
\begin{align}\label{temp16}
H_{k+1}^{(s,t)}=\left|
\begin{array}{cccc}
f_{s}^{(t)}&f_{s+1}^{(t)}&\cdots&f_{s+k}^{(t)}\\[2pt]
f_{s}^{(t+1)}&f_{s+1}^{(t+1)}&\cdots&f_{s+k}^{(t+1)}\\
\vdots&\vdots&\vdots&\vdots\\
f_{s+k-1}^{(t+1)}&f_{s+k}^{(t+1)}&\cdots&f_{s+2k-1}^{(t+1)}
\end{array}
\right|.
\end{align}
Performing a similar procedure the columns in the right-hand side of \eqref{temp16},
we can express the Hankel determinants $H_{k+1}^{(s,t)}$ as
\begin{align}\label{temp1}
H_{k+1}^{(s,t)}=\left|
\begin{array}{ccccc}
f_{s}^{(t)}&f_{s}^{(t+1)}&f_{s+1}^{(t+1)}&\cdots&f_{s+k-1}^{(t+1)}\\[2pt]
f_{s}^{(t+1)}&f_{s}^{(t+2)}&f_{s+1}^{(t+2)}&\cdots&f_{s+k-1}^{(t+2)}\\[2pt]
f_{s+1}^{(t+1)}&f_{s+1}^{(t+2)}&f_{s+2}^{(t+2)}&\cdots&f_{s+k}^{(t+2)}\\
\vdots&\vdots&\vdots&\vdots&\vdots\\
f_{s+k-1}^{(t+1)}&f_{s+k-1}^{(t+2)}&f_{s+k}^{(t+2)}&\cdots&f_{s+2k-2}^{(t+2)}
\end{array}
\right|.
\end{align}
It is obvious from \eqref{temp1} that
$H_{k+1}^{(s,t)}\left[\begin{array}{c}1\\1\end{array}\right]=H_{k}^{(s,t+2)}$ for $k=1,2,\dots,m$.
By considering the other cofactors obtained from $H_{k+1}^{(s,t)}$ in \eqref{temp1},
we can see that
\begin{align}
&H_{k+1}^{(s,t)}\left[\begin{array}{c} k+1\\ k+1\end{array}\right]=
\left|\begin{array}{ccccc}
f_{s}^{(t)}&f_{s}^{(t+1)}&f_{s+1}^{(t+1)}&\cdots&f_{s+k-2}^{(t+1)}\\[2pt]
\bvec{f}_{k-1}^{(s,t+1)}&\bvec{f}_{k-1}^{(s,t+2)}&\bvec{f}_{k-1}^{(s+1,t+2)}&\cdots&\bvec{f}_{k-1}^{(s+k-2,t+2)}
\end{array}
\right|,\label{temp2-17}\\
&H_{k+1}^{(s,t)}\left[\begin{array}{c} 1\\ k+1\end{array}\right]=
\left|\begin{array}{ccccc}
\bvec{f}_{k}^{(s,t+1)}&\bvec{f}_{k}^{(s,t+2)}&\bvec{f}_{k}^{(s+1,t+2)}&\cdots&\bvec{f}_{k}^{(s+k-2,t+2)}
\end{array}
\right|,\\
&H_{k+1}^{(s,t)}\left[\begin{array}{c} k+1\\ 1\end{array}\right]=
\left|\begin{array}{cccc}
f_{s}^{(t+1)}&f_{s+1}^{(t+1)}&\cdots&f_{s+k-1}^{(t+1)}\\[2pt]
\bvec{f}_{k-1}^{(s,t+2)}&\bvec{f}_{k-1}^{(s+1,t+2)}&\cdots&\bvec{f}_{k-1}^{(s+k-1,t+2)}
\end{array}
\right|,\\
&H_{k+1}^{(n,t)}\left[\begin{array}{cc}1&k+1\\1&k+1\end{array}\right]
=
\left|\begin{array}{cccc}
\bvec{f}_{k-1}^{(s,t+2)}&\bvec{f}_{k-1}^{(s+1,t+2)}&\cdots&\bvec{f}_{k-1}^{(s+k-2,t+2)}
\end{array}
\right|.
\end{align}
By multiplying the $1$st row in the right-hand side of \eqref{temp2-17} by $\mu^{(t)}$, then adding it to the $2$nd row and by using \eqref{key},
we obtain
\begin{align}
H_{k+1}^{(s,t)}\left[\begin{array}{c} k+1\\ k+1\end{array}\right]
=\left|
\begin{array}{ccccc}
f_{s}^{(t)}&f_{s}^{(t+1)}&f_{s+1}^{(t+1)}&\cdots&f_{s+k-2}^{(t+1)}\\[2pt]
f_{s+1}^{(t)}&f_{s+1}^{(t+1)}&f_{s+2}^{(t+1)}&\cdots&f_{s+k-1}^{(t+1)}\\[2pt]
\bvec{f}_{k-2}^{(s+1,t+1)}&\bvec{f}_{k-2}^{(s+1,t+2)}&\bvec{f}_{k-2}^{(s+2,t+2)}&\cdots&\bvec{f}_{k-2}^{(s+k-1,t+2)}
\end{array}
\right|.\label{temp2-21}
\end{align}
Similarly, for the $2$nd, $3$rd, $\dots$, $k$th rows in \eqref{temp2-21},
we derive
\begin{align}
H_{k+1}^{(s,t)}\left[\begin{array}{c} k+1\\ k+1\end{array}\right]
=\left|
\begin{array}{ccccc}
f_{s}^{(t)}&f_{s}^{(t+1)}&f_{s+1}^{(t+1)}&\cdots&f_{s+k-2}^{(t+1)}\\[2pt]
f_{s+1}^{(t)}&f_{s+1}^{(t+1)}&f_{s+2}^{(t+1)}&\cdots&f_{s+k-1}^{(t+1)}\\[2pt]
\vdots&\vdots&\vdots&\ddots&\vdots\\
f_{s+k-1}^{(t)}&f_{s+k-1}^{(t+1)}&f_{s+k}^{(t+1)}&\cdots&f_{s+2k-3}^{(t+1)}
\end{array}
\right|.\label{temp2-22}
\end{align}
Performing a similar procedure for the $1$st, $2$nd, $\dots$, $(k-1)$th columns in \eqref{temp2-22},
it follows that
$H_{k+1}^{(s,t)}\left[\begin{array}{c} k+1\\ k+1\end{array}\right]=H_{k}^{(s,t)}$.
Moreover, by rewriting the other cofactors as
$H_{k+1}^{(s,t)}\left[\begin{array}{c} 1\\ k+1\end{array}\right]=
H_{k+1}^{(s,t)}\left[\begin{array}{c} k+1\\ 1\end{array}\right]=H_{k}^{(s,t+1)}$
and
$H_{k+1}^{(n,t)}\left[\begin{array}{cc}1&k+1\\1&k+1\end{array}\right]=H_{k-1}^{(s,t+2)}$,
we obtain \eqref{tauJacobit}.
\end{proof}

\section{Hadamard polynomials and autonomous discrete Toda}\label{sec:3}
In this section, we consider the Hankel polynomials associated with the Hankel determinants $H_k^{(s,t)}$
and define the Hadamard polynomials by rates of the Hankel determinants $H_k^{(s,t)}$
and their corresponding Hankel polynomials.
We then present relationships between the Hankel determinants $H_k^{(s,t)}$,
the Hankel polynomials, and the Hadamard polynomials with respect to $s$.
From these relationships, we derive the autonomous dToda equation \eqref{adToda}.
\par
For $k=1,2,\dots,m$, as the Hankel polynomials associated with the Hankel determinants $H_k^{(s,t)}$,
let us introduce $k$-degree polynomials with respect to $z$ associated with the infinite sequence $\{f_{s}^{(t)}\}_{s,t=0}^{\infty}$:
\begin{align}
H_{k}^{(s,t)}(z):=\left|
\begin{array}{ccccc}
f_{s}^{(t)}&f_{s+1}^{(t)}&\cdots&f_{s+k-1}^{(t)}&1\\
f_{s+1}^{(t)}&f_{s+2}^{(t)}&\cdots&f_{s+k}^{(t)}&z\\
\vdots&\vdots&\vdots&\vdots&\vdots\\
f_{s+k}^{(t)}&f_{s+k+1}^{(t)}&\cdots&f_{s+2k-1}^{(t)}&z^k\
\end{array}
\right|,\quad s,t=0,1,\dots,\label{Tkz}
\end{align}
where $H_{-1}^{(s,t)}(z):=0$ and $H_{0}^{(s,t)}(z):=1$.\par
The following lemma gives the relationships of the Hankel determinants $H_k^{(s,t)}$ and the Hankel polynomials $H_k^{(s,t)}(z)$ at each $t$.
\begin{lemma}\label{lem1}
For $k=0,1,\dots,m$ and $s,t=0,1,\dots$, the Hankel determinants $H_k^{(s,t)}$ and the Hankel polynomials $H_{k}^{(s,t)}(z)$
associated with the infinite sequence $\{f_{s}^{(t)}\}_{s,t=0}^{\infty}$ satisfy
\begin{align}
&zH_k^{(s,t)}H_{k-1}^{(s+1,t)}(z)
=H_k^{(s+1,t)}H_{k-1}^{(s,t)}(z)+H_{k-1}^{(s+1,t)}H_k^{(s,t)}(z),
\label{TkzJacobi}\\
&H_{k}^{(s+1,t)}H_{k}^{(s,t)}(z)
=H_{k+1}^{(s,t)}H_{k-1}^{(s+1,t)}(z)+H_{k}^{(s,t)}H_{k}^{(s+1,t)}(z).\label{TkzPlucker}
\end{align}
\end{lemma}
\begin{proof}
By letting $i_1=j_1=k+1$, $i_2=1$, $j_2=k$ and $D=H_k^{(s,t)}(z)$ in \eqref{Jacobi},
we obtain
\begin{align*}
&H_k^{(s,t)}(z)\left[\begin{array}{c}k+1\\k+1\end{array}\right]
H_k^{(s,t)}(z)\left[\begin{array}{c}1\\k\end{array}\right]\\
&\quad=
H_k^{(s,t)}(z)\left[\begin{array}{c}k+1\\k\end{array}\right]
H_k^{(s,t)}(z)\left[\begin{array}{c}1\\k+1\end{array}\right]
+
H_k^{(s,t)}(z)
H_k^{(s,t)}(z)\left[\begin{array}{cc}k+1&1\\k+1&k\end{array}\right].
\end{align*}
Thus, it follows that
\begin{align*}
&\left|\begin{array}{cccc}
\bvec{f}_k^{(s,t)}&\bvec{f}_{k}^{(s+1,t)}&\cdots&\bvec{f}_k^{(s+k-1,t)}
\end{array}\right|\left|\begin{array}{ccccc}
\bvec{f}_k^{(s+1,t)}&\bvec{f}_{k}^{(s+2,t)}&\cdots&\bvec{f}_k^{(s+k-1,t)}&z\bvec{z}_{k}
\end{array}\right|\\
&=
\left|\begin{array}{ccccc}
\bvec{f}_k^{(s,t)}&\bvec{f}_{k}^{(s+1,t)}&\cdots&\bvec{f}_k^{(s+k-2,t)}&\bvec{z}_{k}
\end{array}\right|\left|\begin{array}{cccc}
\bvec{f}_k^{(s+1,t)}&\bvec{f}_{k}^{(s+2,t)}&\cdots&\bvec{f}_k^{(s+k,t)}
\end{array}\right|\\
&\qquad+\left|\begin{array}{ccccc}
\bvec{f}_{k+1}^{(s,t)}&\bvec{f}_{k+1}^{(s+1,t)}&\cdots&\bvec{f}_{k+1}^{(s+k-1,t)}&\bvec{z}_{k+1}
\end{array}\right|\left|\begin{array}{cccc}
\bvec{f}_{k-1}^{(s+1,t)}&\bvec{f}_{k-1}^{(s+2,t)}&\cdots&\bvec{f}_{k-1}^{(s+k-1,t)}
\end{array}\right|,
\end{align*}
where $\bvec{z}_{k}:=(1,z,\dots,z^{k-1})^{\top}$.
This identity is easily checked to be \eqref{TkzJacobi}.\par
For any $(k+1)$-dimensional vectors $\bvec{x}_1,\bvec{x}_2,\dots,\bvec{x}_{k+3}$,
the determinants of the submatrices in the $(k+1)$-by-$(k+3)$ matrix
$X_k:=\left(\begin{array}{cccc}\bvec{x}_1&\bvec{x}_2&\cdots&\bvec{x}_{k+3}\end{array}\right)$ satisfy
\begin{align}\label{Plucker}
&\left|\begin{array}{cccccc}
\bvec{x}_1&\bvec{x}_2&\cdots&\bvec{x}_{k-1}&\bvec{x}_{k}&\bvec{x}_{k+1}
\end{array}\right|\left|\begin{array}{cccccc}
\bvec{x}_1&\bvec{x}_2&\cdots&\bvec{x}_{k-1}&\bvec{x}_{k+2}&\bvec{x}_{k+3}
\end{array}\right|\nonumber\\
&-\left|\begin{array}{cccccc}
\bvec{x}_1&\bvec{x}_2&\cdots&\bvec{x}_{k-1}&\bvec{x}_{k}&\bvec{x}_{k+2}
\end{array}\right|\left|\begin{array}{cccccc}
\bvec{x}_1&\bvec{x}_2&\cdots&\bvec{x}_{k-1}&\bvec{x}_{k+1}&\bvec{x}_{k+3}
\end{array}\right|\nonumber\\
&+\left|\begin{array}{cccccc}
\bvec{x}_1&\bvec{x}_2&\cdots&\bvec{x}_{k-1}&\bvec{x}_{k}&\bvec{x}_{k+3}
\end{array}\right|\left|\begin{array}{cccccc}
\bvec{x}_1&\bvec{x}_2&\cdots&\bvec{x}_{k-1}&\bvec{x}_{k+1}&\bvec{x}_{k+2}
\end{array}\right|=0.
\end{align}
This is known as the Pl\"ucker relation.
By letting $\bvec{x}_1=\bvec{f}_{k+1}^{(s+1,t)}$, $\bvec{x}_2=\bvec{f}_{k+1}^{(s+2,t)}$,
$\dots$, $\bvec{x}_{k}=\bvec{f}_{k+1}^{(s+k,t)}$, $\bvec{x}_{k+1}=\bvec{f}_{k+1}^{(s,t)}$,
$\bvec{x}_{k+2}=\bvec{z}_{k+1}$, $\bvec{x}_{k+3}=\bvec{e}_{k+1}:=(0,0,\dots,0,1)^{\top}$ in \eqref{Plucker},
we obtain
\begin{align*}
&\left|\begin{array}{cccccc}
\bvec{f}_{k+1}^{(s+1,t)}&\bvec{f}_{k+1}^{(s+2,t)}&\cdots&\bvec{f}_{k+1}^{(s+k,t)}&\bvec{f}_{k+1}^{(s,t)}
\end{array}\right|\nonumber\\
&\quad\times\left|\begin{array}{cccccc}
\bvec{f}_{k+1}^{(s+1,t)}&\bvec{f}_{k+1}^{(s+2,t)}&\cdots&\bvec{f}_{k+1}^{(s+k-1,t)}&\bvec{z}_{k+1}&\bvec{e}_{k+1}
\end{array}\right|\nonumber\\
&-\left|\begin{array}{cccccc}
\bvec{f}_{k+1}^{(s+1,t)}&\bvec{f}_{k+1}^{(s+2,t)}&\cdots&\bvec{f}_{k+1}^{(s+k-1,t)}&\bvec{f}_{k+1}^{(s+k,t)}&\bvec{z}_{k+1}
\end{array}\right|\nonumber\\
&\quad\times\left|
\begin{array}{cccccc}
\bvec{f}_{k+1}^{(s+1,t)}&\bvec{f}_{k+1}^{(s+2,t)}&\cdots&\bvec{f}_{k+1}^{(s+k-1,t)}&\bvec{f}_{k+1}^{(s,t)}&\bvec{e}_{k+1}
\end{array}\right|\nonumber\\
&+\left|\begin{array}{cccccc}
\bvec{f}_{k+1}^{(s+1,t)}&\bvec{f}_{k+1}^{(s+2,t)}&\cdots&\bvec{f}_{k+1}^{(s+k-1,t)}&\bvec{f}_{k+1}^{(s+k,t)}&\bvec{e}_{k+1}
\end{array}\right|\nonumber\\
&\quad\times\left|
\begin{array}{cccccc}
\bvec{f}_{k+1}^{(s+1,t)}&\bvec{f}_{k+1}^{(s+2,t)}&\cdots&\bvec{f}_{k+1}^{(s+k-1,t)}&\bvec{f}_{k+1}^{(s,t)}&\bvec{z}_{k+1}
\end{array}\right|=0,
\end{align*}
which implies \eqref{TkzPlucker}.

\end{proof}
\par
Moreover, for $k=1,2,\dots,m$, let us define another set of polynomials ${\cal H}_{k}^{(s,t)}(z)$ of degree $k$ as
\begin{align}
{\cal H}_{k}^{(s,t)}(z):=\frac{H_{k}^{(s,t)}(z)}{H_{k}^{(s,t)}},\quad s,t=0,1,\dots,\label{HadamardHankel}
\end{align}
where ${\cal H}_{-1}^{(s,t)}(z):= 0$ and ${\cal H}_{0}^{(s,t)}(z):=1$.
The polynomials ${\cal H}_k^{(s,t)}(z)$ are the Hadamard polynomials.
Now, we derive a proposition for the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ with $k=m$.
\begin{proposition}\label{prop3}
The Hadamard polynomial ${\cal H}_m^{(s,t)}(z)$ coincides with
the characteristic polynomial $p(z)$ of matrices with eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_m,$
\begin{align}
{\cal H}_{m}^{(s,t)}(z)=p(z).\label{phizpz}
\end{align}
\end{proposition}
\begin{proof}
The $m$-degree polynomial $H_m^{(s,t)}(z)$ can be written as
\begin{align}\label{tmp6}
H_m^{(s,t)}(z)=
\left|\begin{array}{ccccc}
\bvec{f}_{m+1}^{(s,t)}&\bvec{f}_{m+1}^{(s+1,t)}&\cdots&\bvec{f}_{m+1}^{(s+m-1,t)}&\bvec{z}_{m+1}
\end{array}
\right|.\end{align}
By multiplying the $1$st, $2$nd, $\dots$, $m$th rows on the right-hand side of \eqref{tmp6} by $a_m$, $a_{m-1}$, $\dots$, $a_1$, respectively,
and adding these to the $(m+1)$th row,
we obtain
\begin{align*}
H_m^{(s,t)}(z
=\left|\begin{array}{ccccc}
\bvec{f}_{m}^{(s,t)}&\bvec{f}_{m}^{(s+1,t)}&\cdots&\bvec{f}_{m}^{(s+m-1,t)}&\bvec{z}_{m}\\
\displaystyle \sum_{\ell=0}^{m}a_{\ell}f_{s+m-\ell}^{(t)}
&\displaystyle \sum_{\ell=0}^{m}a_{\ell}f_{s+m+1-\ell}^{(t)}
&\cdots
&\displaystyle \sum_{\ell=0}^{m}a_{\ell}f_{s+2m-1-\ell}^{(t)}
&\displaystyle \sum_{\ell=0}^{m}a_{\ell}z^{m-\ell}
\end{array}\right|.
\end{align*}
From \eqref{fnmt}, it is obvious that the ($m+1,1$), ($m+1,2$), $\dots$, ($m+1,m$) entries are $0$.
The ($m+1,m+1$) entry is equal to the characteristic polynomial $p(z)$.
Thus, we can express the characteristic polynomial $p(z)$
using the Hankel determinant $H_{m}^{(s,t)}$ and the Hankel polynomial $H_{m}^{(s,t)}(z)$ as $H_m^{(s,t)}(z)=p(z)H_m^{(s,t)}$.
By recalling the definition of the Hadamard polynomial ${\cal H}_m^{(s,t)}(z)$ in \eqref{HadamardHankel},
we obtain \eqref{phizpz}.

\end{proof}
\par
The following lemma gives identities of the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$.
\begin{lemma}\label{lem3}
Let us assume that for the infinite sequence $\{f_s^{(t)}\}_{s,t=0}^{\infty}$,
the Hankel determinants $H_k^{(s,t)}$ are all nonzero.
Then, for $k=1,2,\dots,m$,
the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ satisfy
\begin{align}
&z{\cal H}_{k-1}^{(s+1,t)}(z)={\cal H}_{k}^{(s,t)}(z)+q_k^{(s,t)}{\cal H}_{k-1}^{(s,t)}(z),
\quad s,t=0,1,\dots,\label{Christoffeln}\\
&{\cal H}_k^{(s,t)}(z)={\cal H}_{k}^{(s+1,t)}(z)+e_{k}^{(s,t)}{\cal H}_{k-1}^{(s+1,t)}(z),
\quad s,t=0,1,\dots,
\label{Geronimusn}
\end{align}
where $q_k^{(s,t)}$ and $e_k^{(s,t)}$ are given using the Hankel determinants $H_k^{(s,t)}$ as
\begin{align}
&q_{k}^{(s,t)}:=\frac{H_{k-1}^{(s,t)}H_{k}^{(s+1,t)}}{H_{k}^{(s,t)}H_{k-1}^{(s+1,t)}},\label{qkst}\\
&e_k^{(s,t)}:=\frac{H_{k+1}^{(s,t)}H_{k-1}^{(s+1,t)}}{H_{k}^{(s,t)}H_{k}^{(s+1,t)}}.\label{ekst}
\end{align}
\end{lemma}
\begin{proof}
By dividing both sides of \eqref{TkzJacobi} by $H_{k-1}^{(s+1,t)}H_{k}^{(s,t)}$,
we obtain
\begin{align*}
z\frac{H_{k-1}^{(s+1,t)}(z)}{H_{k-1}^{(s+1,t)}}=
\left(\frac{H_{k-1}^{(s,t)}H_{k}^{(s+1,t)}}{H_{k}^{(s,t)}H_{k-1}^{(s+1,t)}}\right)
\frac{H_{k-1}^{(s,t)}(z)}{H_{k-1}^{(s,t)}}+\frac{H_{k}^{(s,t)}(z)}{H_{k}^{(s,t)}}.
\end{align*}
By considering the definition of the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ in \eqref{HadamardHankel},
we thus derive \eqref{Christoffeln} with \eqref{qkst}.
Similarly, it follows from \eqref{TkzPlucker} that
\begin{align}\label{tmp3}
\frac{H_{k+1}^{(s,t)}H_{k-1}^{(s+1,t)}(z)}{H_{k}^{(s,t)}H_{k}^{(s+1,t)}}
={\cal H}_k^{(s,t)}(z)-{\cal H}_{k}^{(s+1,t)}(z).
\end{align}
The left-hand side of \eqref{tmp3} can be rewritten as
\begin{align*}
\frac{H_{k+1}^{(s,t)}H_{k-1}^{(s+1,t)}(z)}{H_{k}^{(s,t)}H_{k}^{(s+1,t)}}
=\left(\frac{H_{k+1}^{(s,t)}H_{k-1}^{(s+1,t)}}{H_{k}^{(s,t)}H_{k}^{(s+1,t)}}\right)
{\cal H}_{k-1}^{(s+1,t)},\end{align*}
and so we have \eqref{Geronimusn} with \eqref{ekst}.

\end{proof}

The identities \eqref{Christoffeln} in Lemma~\ref{lem3} can be regarded as the recursion formulas for generating evolutions of discrete-time variable $s$ of the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$.
In the theory of orthogonal polynomials, \eqref{Christoffeln} correspond to
the Christoffel transformations for the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ in \cite{Chihara}.
Similarly, \eqref{Geronimusn}, which is the inverse transformation of \eqref{Christoffeln},
is the Geronimus transformation for the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$.
The Christoffel and Geronimus transformations are useful in the study of integrable discrete systems \cite{Zhedanov}.
\par
From Lemma~\ref{lem3}, we derive a theorem for relationships involving $q_k^{(s,t)}$ and $e_k^{(s,t)}$.
\begin{theorem}\label{lem5}
Let us assume that for the infinite sequence $\{f_{s}^{(t)}\}_{s,t=0}^{\infty}$,
the Hankel determinants $H_k^{(s,t)}$ are all nonzero.
Then, the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ for $s,t=0,1,\dots$ satisfy the three-term recursion formulas:
\begin{align}\label{3termqe}
{\cal H}_{k+1}^{(s,t)}(z)=(z-q_{k+1}^{(s,t)}-e_{k}^{(s,t)}){\cal H}_{k}^{(s,t)}(z)
-q_{k}^{(s,t)}e_{k}^{(s,t)}{\cal H}_{k-1}^{(s,t)}(z),
\quad k=0,1,\dots,m-1.
\end{align}
More importantly, $q_k^{(s,t)}$ and $e_k^{(s,t)}$ satisfy
\begin{align}\label{q+en}
\left\{
\begin{array}{l}
q_k^{(s+1,t)}+e_{k-1}^{(s+1,t)}=q_k^{(s,t)}+e_k^{(s,t)},\quad k=1,2,\dots,m,\\[2pt]
q_k^{(s+1,t)}e_{k}^{(s+1,t)}=q_{k+1}^{(s,t)}e_{k}^{(s,t)},\quad k=1,2,\dots,m-1,\\[2pt]
s,t=0,1,\dots.
\end{array}
\right.
\end{align}
\end{theorem}
\begin{proof}
Substituting \eqref{Christoffeln} for \eqref{Geronimusn},
we obtain
\begin{align}\label{CG1}
z{\cal H}_k^{(s,t)}(z)={\cal H}_{k+1}^{(s,t)}(z)+(q_{k+1}^{(s,t)}+e_{k}^{(s,t)}){\cal H}_{k}^{(s,t)}(z)+q_{k}^{(s,t)}e_k^{(s,t)}{\cal H}_{k-1}^{(s,t)}(z).
\end{align}
By replacing $s$ and $k$ with $s+1$ and $k-1$ in \eqref{CG1}, respectively,
we derive
\begin{align}\label{tmp7}
z{\cal H}_{k-1}^{(s+1,t)}(z)
={\cal H}_{k}^{(s+1,t)}(z)+(q_{k}^{(s+1,t)}+e_{k-1}^{(s+1,t)}){\cal H}_{k-1}^{(s+1,t)}(z)
+q_{k-1}^{(s+1,t)}e_{k-1}^{(s+1,t)}{\cal H}_{k-2}^{(s+1,t)}(z).
\end{align}
Moreover, by substituting \eqref{Geronimusn} for \eqref{Christoffeln},
we obtain
\begin{align}\label{CG2}
z{\cal H}_{k-1}^{(s+1,t)}(z)
={\cal H}_{k}^{(s+1,t)}(z)+(q_{k}^{(s,t)}+e_{k}^{(s,t)}){\cal H}_{k-1}^{(s+1,t)}(z)+q_{k}^{(s,t)}e_{k-1}^{(s,t)}{\cal H}_{k-2}^{(s+1,t)}(z).
\end{align}
By observing both sides of \eqref{tmp7} and \eqref{CG2},
we have \eqref{q+en}.
\end{proof}
\par
Here, since \eqref{q+en} does not give evolution with respect to $t$,
we may simplify $q_k^{(s,t)}$ and $e_k^{(s,t)}$ as $q_k^{(s)}$ and $e_k^{(s)}$, respectively,
in the case where $t$ is fixed in Theorem~\ref{lem5}.
Thus, \eqref{q+en} with fixed $t$ in Theorem~\ref{lem5} is equivalent to the autonomous dToda equation \eqref{adToda}.
In other words, the autonomous dToda equation \eqref{adToda} is given by a special case of \eqref{q+en}.

\section{Hadamard polynomials and non-autonomous discrete Toda}\label{sec:31}
In this section, with respect to $t$, but not $s$,
we present relationships between the Hankel determinants $H_k^{(s,t)}$,
the Hankel polynomials  $H_k^{(s,t)}(z)$, and the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ defined in Section~\ref{sec:3}.
From these relationships, we also derive the non-autonomous dToda equation \eqref{nadToda}.\par
We derive a lemma for the Hankel determinants $H_k^{(s,t)}$ and the Hankel polynomials $H_{k}^{(s,t)}(z)$ at each $s$.
\begin{lemma}\label{lem2}
For $k=1,2,\dots,m$ and $s,t=0,1,\dots$, the Hankel polynomials $H_{k}^{(s,t)}(z)$ associated with the infinite sequence $\{f_{s}^{(t)}\}_{s,t=0}^{\infty}$ satisfy
\begin{align}
&(z-\mu^{(t)})H_{k}^{(s,t)}H_{k-1}^{(s,t+1)}(z)=H_{k}^{(s,t+1)}H_{k-1}^{(s,t)}(z)+H_{k-1}^{(s,t+1)}H_{k}^{(s,t)}(z),\label{TkzJacobit}\\
&H_{k}^{(s,t+1)}H_k^{(s,t)}(z)=H_{k+1}^{(s,t)}H_{k-1}^{(s,t+1)}(z)+H_{k}^{(s,t)}H_{k}^{(s,t+1)}(z).\label{TkzPluckert}
\end{align}
\end{lemma}
\begin{proof}
By multiplying the $k$th row in the Hankel polynomials $H_k^{(s,t)}(z)$ by $-\mu^{(t)}$,
then adding it to the $(k+1)$th row and by using \eqref{key},
we derive
\begin{align*}
H_k^{(s,t)}(z)=
\left|\begin{array}{ccccc}
f_s^{(t)}&f_{s+1}^{(t)}&\dots&f_{s+k-1}^{(t)}&1\\
f_{s+1}^{(t)}&f_{s+2}^{(t)}&\dots&f_{s+k}^{(t)}&z\\
\vdots&\vdots&\ddots&\vdots&\vdots\\
f_{s+k-1}^{(t)}&f_{s+k}^{(t)}&\dots&f_{s+2k-2}^{(t)}&z^{k-1}\\
f_{s+k-1}^{(t+1)}&f_{s+k}^{(t+1)}
&\dots&f_{s+2k-2}^{(t+1)}&(z-\mu^{(t)})z^{k-1}
\end{array}\right|.
\end{align*}
Similarly, for the $k$th, $(k-1)$th, $(k-2)$th, $\dots$, $2$nd rows in the Hankel polynomials $H_k^{(s,t)}(z)$,
it follows that
\begin{align}
H_{k}^{(s,t)}(z)
=\left|\begin{array}{ccccc}
f_{s}^{(t)}&f_{s+1}^{(t)}&\cdots&f_{s+k-1}^{(t)}&1\\
\bvec{f}_{k}^{(s,t+1)}&\bvec{f}_{k}^{(s+1,t+1)}&\cdots&\bvec{f}_{k}^{(s+k-1,t+1)}&
\left(z-\mu^{(t)}\right)\bvec{z}_k
\end{array}\right|.\label{Tkzjacobitproof}
\end{align}
By letting $i_1=j_1=k+1$, $i_2=1$, $j_2=k$ and $D=H_{k}^{(s,t)}(z)$
in the Jacobi identity \eqref{Jacobi} and considering \eqref{Tkzjacobitproof},
we obtain
\begin{align}
&\left|\begin{array}{ccccc}
f_s^{(t)} & f_{s+1}^{(t)}&\cdots&f_{s+k-1}^{(t)}\\
\bvec{f}_{k-1}^{(s,t+1)}&\bvec{f}_{k-1}^{(s+1,t+1)}&\cdots&\bvec{f}_{k-1}^{(s+k-1,t+1)}
\end{array}\right|\nonumber\\
&\quad\times\left|
\begin{array}{ccccc}
\bvec{f}_{k}^{(s,t+1)}&\bvec{f}_{k}^{(s+1,t+1)}&\cdots&\bvec{f}_{k}^{(s+k-2,t+1)}&(z-\mu^{(t)})\bvec{z}_k
\end{array}\right|\nonumber\\
&=\left|\begin{array}{ccccc}
f_s^{(t)} & f_{s+1}^{(t)}&\cdots&f_{s+k-2}^{(t)}&1\\
\bvec{f}_{k-1}^{(s,t+1)}&\bvec{f}_{k-1}^{(s+1,t+1)}&\cdots&\bvec{f}_{k-1}^{(s+k-2,t+1)}&
(z-\mu^{(t)})\bvec{z}_{k-1}
\end{array}\right|\nonumber\\
&\quad\times\left|\begin{array}{ccccc}
\bvec{f}_{k}^{(s,t+1)}&\bvec{f}_{k}^{(s+1,t+1)}&\cdots&\bvec{f}_{k}^{(s+k-1,t+1)}
\end{array}\right|\nonumber\\
&+\left|\begin{array}{ccccc}
f_s^{(t)} & f_{s+1}^{(t)}&\cdots&f_{s+k-1}^{(t)}&1\\
\bvec{f}_{k-1}^{(s,t+1)}&\bvec{f}_{k-1}^{(s+1,t+1)}&\cdots&\bvec{f}_{k-1}^{(s+k-1,t+1)}
&(z-\mu^{(t)})\bvec{z}_k
\end{array}\right|\nonumber\\
&\quad\times\left|\begin{array}{ccccc}
\bvec{f}_{k-1}^{(s,t+1)}&\bvec{f}_{k-1}^{(s+1,t+1)}&\cdots&\bvec{f}_{k-1}^{(s+k-2,t+1)}
\end{array}\right|.\label{tmp4}
\end{align}
Equation \eqref{tmp4} immediately leads to \eqref{TkzJacobit}.\par
By multiplying the $k$th, $(k-1)$th, $\dots$, $1$st columns in the Hankel polynomials $H_k^{(s,t)}(z)$ by $-\mu^{(t)}$
and adding these to the $(k+1)$th, $k$th, $\dots$, $2$nd columns, respectively,
we derive
\begin{align}\label{tmp5}
H_k^{(s,t)}(z)=\left|\begin{array}{cccccc}
\bvec{f}_{k+1}^{(s,t)}&\bvec{f}_{k+1}^{(s,t+1)}&\bvec{f}_{k+1}^{(s+1,t+1)}&\cdots&
\bvec{f}_{k+1}^{(s+k-2,t+1)}&\bvec{z}_{k+1}
\end{array}\right|.
\end{align}
By letting
$\bvec{x}_1=\bvec{f}_{k+1}^{(s,t+1)}$, $\bvec{x}_2=\bvec{f}_{k+1}^{(s+1,t+1)}$,
$\dots$, $\bvec{x}_k=\bvec{f}_{k+1}^{(s+k-1,t+1)}$, $\bvec{x}_{k+1}=\bvec{f}_{k+1}^{(s,t)}$,
$\bvec{x}_{k+2}=\bvec{z}_{k+1}$, $\bvec{x}_{k+3}=\bvec{e}_{k+1}$
in the Pl\"uker relation \eqref{Plucker}, we obtain
\begin{align}
&\left|\begin{array}{cccccc}
\bvec{f}_{k+1}^{(s,t+1)}&\bvec{f}_{k+1}^{(s+1,t+1)}&
\cdots&\bvec{f}_{k+1}^{(s+k-1,t+1)}&\bvec{f}_{k+1}^{(s,t)}
\end{array}\right|\nonumber\\
&\quad \times \left|\begin{array}{cccccc}
\bvec{f}_{k+1}^{(s,t+1)}&\bvec{f}_{k+1}^{(s+1,t+1)}&
\cdots&\bvec{f}_{k+1}^{(s+k-2,t+1)}&\bvec{z}_{k+1}&
\bvec{e}_{k+1}
\end{array}\right|\nonumber\\
&-\left|\begin{array}{cccccccc}
\bvec{f}_{k+1}^{(s,t+1)}&\bvec{f}_{k+1}^{(s+1,t+1)}&
\cdots&\bvec{f}_{k+1}^{(s+k-1,t+1)}&
\bvec{z}_{k+1}
\end{array}\right|\nonumber\\
&\quad \times\left|\begin{array}{cccccccc}
\bvec{f}_{k+1}^{(s,t+1)}&\bvec{f}_{k+1}^{(s+1,t+1)}&
\cdots&\bvec{f}_{k+1}^{(s+k-2,t+1)}&
\bvec{f}_{k+1}^{(s,t)}&\bvec{e}_{k+1}
\end{array}\right|\nonumber\\
&+\left|\begin{array}{cccccccc}
\bvec{f}_{k+1}^{(s,t+1)}&\bvec{f}_{k+1}^{(s+1,t+1)}&
\cdots&\bvec{f}_{k+1}^{(s+k-1,t+1)}&\bvec{e}_{k+1}
\end{array}\right|\nonumber\\
&\quad \times\left|\begin{array}{cccccccc}
\bvec{f}_{k+1}^{(s,t+1)}&\bvec{f}_{k+1}^{(s+1,t+1)}&
\cdots&\bvec{f}_{k+1}^{(s+k-2,t+1)}&
\bvec{f}_{k+1}^{(s,t)}&\bvec{z}_{k+1}
\end{array}\right|=0.\label{Tkzprooftem4}
\end{align}
Thus, combining \eqref{tmp5} with \eqref{Tkzprooftem4} gives \eqref{TkzPluckert}.

\end{proof}
\par
We give a lemma concerning identities of the Hadamard polynomials ${\cal H}_k^{(s,t)}$.
\begin{lemma}\label{lem4}
Let us assume that for the infinite sequence $\{f_{s}^{(t)}\}_{s,t=0}^{\infty}$,
the Hankel determinants $H_k^{(s,t)}$ are all nonzero.
Then, for $k=1,2,\dots,m$,
the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ satisfy
\begin{align}
&(z-\mu^{(t)}){\cal H}_{k-1}^{(s,t+1)}(z)={\cal H}_{k}^{(s,t)}(z)+Q_k^{(s,t)}{\cal H}_{k-1}^{(s,t)}(z),
\quad s,t=0,1,\dots,\label{Christoffelt}\\
&{\cal H}_{k}^{(s,t)}(z)
={\cal H}_{k}^{(s,t+1)}(z)+E_{k}^{(s,t)}{\cal H}_{k-1}^{(s,t+1)}(z),
\quad s,t=0,1,\dots,\label{Geronimust}
\end{align}
where $Q_k^{(s,t)}$ and $E_k^{(s,t)}$ are given using the Hankel determinants $H_k^{(s,t)}$ as
\begin{align}
&Q_k^{(s,t)}:=\frac{H_{k-1}^{(s,t)}H_{k}^{(s,t+1)}}{H_{k}^{(s,t)}H_{k-1}^{(s,t+1)}},\label{Qkst}\\
&E_k^{(s,t)}:=
\frac{H_{k+1}^{(s,t)}H_{k-1}^{(s,t+1)}}{H_{k}^{(s,t)}H_{k}^{(s,t+1)}}.\label{Ekst}
\end{align}
\end{lemma}
\begin{proof}
By dividing both sides of \eqref{TkzJacobit} by $H_{k-1}^{(s,t+1)}H_{k}^{(s,t)}$
and considering the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ in \eqref{HadamardHankel},
we obtain
\begin{align*}
(z-\mu^{(t)}){\cal H}_{k-1}^{(s,t+1)}(z)=
\left(\frac{H_{k-1}^{(s,t)}H_{k}^{(s,t+1)}}{H_{k}^{(s,t)}H_{k-1}^{(s,t+1)}}\right)
{\cal H}_{k-1}^{(s,t)}(z)+{\cal H}_{k}^{(s,t)}(z),
\end{align*}
which immediately leads to \eqref{Christoffelt} with \eqref{Qkst}.
Similarly, it follows from \eqref{TkzPluckert} that
\begin{align*}
\left(\frac{H_{k+1}^{(s,t)}H_{k-1}^{(s,t+1)}}{H_{k}^{(s,t)}H_{k}^{(s,t+1)}}\right)
{\cal H}_{k-1}^{(s,t+1)}(z)={\cal H}_{k}^{(s,t)}(z)-{\cal H}_{k}^{(s,t+1)}(z),
\end{align*}
which is equivalent to \eqref{Geronimust} with \eqref{Ekst}.

\end{proof}
\par
With the help of Lemma~\ref{lem4},
we give the following theorem concerning $Q_k^{(s,t)}$ and $E_k^{(s,t)}$.
\par
\begin{theorem}\label{lem6}
Let us assume that for the infinite sequence $\{f_{s}^{(t)}\}_{s,t=0}^{\infty}$,
the Hankel determinants $H_k^{(s,t)}$ are all nonzero.
Then, the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ for $s,t=0,1,\dots$ satisfy the three-term recursion formulas:
\begin{align}\label{3termQE}
{\cal H}_{k+1}^{(s,t)}(z)=(z-Q_{k+1}^{(s,t)}-E_{k}^{(s,t)}-\mu^{(t)}){\cal H}_{k}^{(s,t)}(z)
-Q_{k}^{(s,t)}E_{k}^{(s,t)}{\cal H}_{k-1}^{(s,t)}(z),
\quad k=0,1,\dots,m-1.
\end{align}
More importantly, $Q_k^{(s,t)}$ and $E_k^{(s,t)}$ satisfy
\begin{align}\label{Q+Et}
\left\{
\begin{array}{l}
Q_{k}^{(s,t+1)}+E_{k-1}^{(s,t+1)}+\mu^{(t+1)}
=Q_{k}^{(s,t)}+E_{k}^{(s,t)}+\mu^{(t)},\quad k=1,2,\dots,m,\\[2pt]
Q_{k}^{(s,t+1)}E_{k}^{(s,t+1)}
=Q_{k+1}^{(s,t)}E_{k-1}^{(s,t)},\quad k=1,2,\dots,m-1,\\[2pt]
s,t=0,1,\dots.
\end{array}
\right.
\end{align}
\end{theorem}
\begin{proof}
Similar to the proof for Lemma~\ref{lem5},
it follows from \eqref{Christoffelt} and \eqref{Geronimust} that
\begin{align}
(z-\mu^{(t)}){\cal H}_{k}^{(s,t)}(z)
={\cal H}_{k+1}^{(s,t)}(z)+(Q_{k+1}^{(s,t)}+E_{k}^{(s,t)}){\cal H}_{k}^{(s,t)}(z)
+Q_{k}^{(s,t)}E_{k}^{(s,t)}{\cal H}_{k-1}^{(s,t)}(z),\label{lem6-1}\\
(z-\mu^{(t)}){\cal H}_{k-1}^{(s,t+1)}(z)
={\cal H}_{k}^{(s,t+1)}(z)+(Q_{k}^{(s,t)}+E_{k}^{(s,t)}){\cal H}_{k-1}^{(s,t+1)}(z)
+Q_{k}^{(s,t)}E_{k-1}^{(s,t)}{\cal H}_{k-2}^{(s,t)}(z).\label{gct}
\end{align}
Comparing \eqref{lem6-1} with \eqref{gct} gives \eqref{Q+Et}.

\end{proof}
\par
By considering $Q_k^{(s,t)}, E_k^{(s,t)}$ as $Q_k^{(t)}, E_k^{(t)}$ at each $s$,
we immediately simplify \eqref{Q+Et} as the non-autonomous dToda equation \eqref{nadToda}.\par
Theorems~\ref{lem5} and~\ref{lem6} suggest that the two types of discrete-time variables play a key role in simultaneously deriving
both the autonomous dToda equation \eqref{adToda} and the non-autonomous dToda equation \eqref{nadToda}.

\section{Determinant solutions}\label{sec:4}
In this section, we clarify eigenpairs of tridiagonal matrices involving $q_k^{(s,t)},e_k^{(s,t)}$ in Theorem~\ref{lem5} and $Q_k^{(s,t)},E_k^{(s,t)}$ in Theorem~\ref{lem6}.
We then show the determinant solutions with sufficient degrees of freedom to the autonomous dToda equation \eqref{adToda}
and the non-autonomous dToda equation \eqref{nadToda}. \par
Let us introduce $m$-by-$m$ bidiagonal matrices with $e_1^{(s,t)}$, $e_2^{(s,t)},\dots,e_{m-1}^{(s,t)}$ and $q_1^{(s,t)},q_2^{(s,t)},\dots,q_m^{(s,t)}$:
\begin{align*}
L^{(s,t)}:=\left(
\begin{array}{cccc}
1&\\
e_1^{(s,t)}&1&\\
&\ddots&\ddots&\\
&&e_{m-1}^{(s,t)}&1
\end{array}
\right),\quad
R^{(s,t)}:=\left(
\begin{array}{cccc}
q_1^{(s,t)}&1\\
&q_2^{(s,t)}&\ddots\\
&&\ddots&1\\
&&&q_m^{(s,t)}
\end{array}
\right).
\end{align*}
Then, the following proposition shows eigenpairs of $A^{(s,t)}:=L^{(s,t)}R^{(s,t)}$.
\begin{proposition}\label{prop4}
Eigenvalues of $A^{(s,t)}$ coincide with the roots $\lambda_1,\lambda_2,\dots,\lambda_m$ of the polynomial
$p(z)=(z-\lambda_1)(z-\lambda_2)\dots(z-\lambda_m)$.
Moreover, the eigenvectors corresponding to $\lambda_1,\lambda_2,\dots,\lambda_m$
are ${\cal H}_1^{(s,t)},{\cal H}_2^{(s,t)},\dots,{\cal H}_m^{(s,t)}$, respectively,
where ${\cal H}_k^{(s,t)}:=({\cal H}_0^{(s,t)}(\lambda_k),{\cal H}_1^{(s,t)}(\lambda_k),\dots,{\cal H}_{m-1}^{(s,t)}(\lambda_k))^\top$.
\end{proposition}
\begin{proof}
Let us prepare the $k$th principal submatrix of $A^{(s,t)}$, namely,
\begin{align*}
A_k^{(s,t)}
:=\left(
\begin{array}{cccc}
q_1^{(s,t)}&1\\
q_1^{(s,t)}e_{1}^{(s,t)}&q_2^{(s,t)}+e_1^{(s,t)}&\ddots\\
&\ddots&\ddots&1\\
&&q_{k-1}^{(s,t)}e_{k-1}^{(s,t)}&q_k^{(s,t)}+e_{k-1}^{(s,t)}
\end{array}
\right).
\end{align*}
Moreover, let $I_k$ be the $k$-by-$k$ identity matrix.
Then, by considering cofactor expansion along the $(k+1)$th row of $\det(zI_{k+1}-A_{k+1}^{(s,t)})$, we derive
\begin{align}\label{phizI-A}
\det(zI_{k+1}-A_{k+1}^{(s,t)})
=(z-q_{k+1}^{(s,t)}-e_k^{(s,t)})\det(zI_k-A_{k}^{(s,t)})
-q_k^{(s,t)}e_k^{(s,t)}\det(zI_{k-1}-A_{k-1}^{(s,t)}).
\end{align}
By comparing \eqref{3termqe} in Theorem~\ref{lem5} with \eqref{phizI-A},
we obtain
\begin{align}
{\cal H}_{k}^{(s,t)}(z)=\det(zI_k-A_k^{(s,t)}),\quad k=1,2,\dots,m.\label{phikstdet}
\end{align}
Thus, by combining Proposition~\ref{prop3} with \eqref{phikstdet},
we have
\begin{align}
p(z)=\det(zI_m-A^{(s,t)}).
\end{align}
Consequently, by taking into account that $p(z)=(z-\lambda_1)(z-\lambda_2)\cdots(z-\lambda_m)$,
we see that
$\lambda_1,\lambda_2,\dots,\lambda_m$ are the eigenvalues of $A^{(s,t)}$.\par
Equation \eqref{3termqe} with $z=\lambda_k$ leads to
\begin{align}\label{qephisystem}
\left\{
\begin{array}{l}
(q_{1}^{(s,t)}+e_{0}^{(s,t)}){\cal H}_{0}^{(s,t)}(\lambda_k)+{\cal H}_{1}^{(s,t)}(\lambda_k)=\lambda_k{\cal H}_{0}^{(s,t)}(\lambda_k),\\
q_{1}^{(s,t)}e_{1}^{(s,t)}{\cal H}_{0}^{(s,t)}(\lambda_k)+
(q_{2}^{(s,t)}+e_{1}^{(s,t)}){\cal H}_{1}^{(s,t)}(\lambda_k)+{\cal H}_{2}^{(s,t)}(\lambda_k)=\lambda_k{\cal H}_{1}^{(s,t)}(\lambda_k),\\
\quad\vdots\\
q_{m-1}^{(s,t)}e_{m-1}^{(s,t)}{\cal H}_{m-2}^{(s,t)}(\lambda_k)+
(q_{m}^{(s,t)}+e_{m-1}^{(s,t)}){\cal H}_{m-1}^{(s,t)}(\lambda_k)=\lambda_k{\cal H}_{m-1}^{(s,t)}(\lambda_k).
\end{array}
\right.
\end{align}
Noting that $A^{(s,t)}=A_m^{(s,t)}$, we can rewrite \eqref{qephisystem} as
\begin{align}\label{eigenpqe}
A^{(s,t)}{\cal H}_k^{(s,t)}=\lambda_k{\cal H}_k^{(s,t)},\quad k=1,2,\dots,m.
\end{align}
Equation \eqref{eigenpqe} shows that ${\cal H}_{1}^{(s,t)}$, ${\cal H}_{2}^{(s,t)}$, $\dots$, ${\cal H}_{m}^{(s,t)}$ are eigenvectors of $A^{(s,t)}$ corresponding to $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$, respectively.\par

\end{proof}
\par
It is emphasized here that the discrete-time variable $t$ is meaningless for investigating the autonomous dToda equation \eqref{adToda}.
Without loss of generality, we can remove the discrete-time variable $t$ from the superscript in the case of the autonomous dToda equation \eqref{adToda}.
Thus, we henceforth regard $q_k^{(s,t)}, e_k^{(s,t)}$ in \eqref{q+en} with $t=0$ as $q_k^{(s)}, e_k^{(s)}$ in \eqref{adToda}, that is, $q_k^{(s)}=q_k^{(s,0)}$ and  $e_k^{(s)}=e_k^{(s,0)}$.
With the help of Proposition~\ref{prop4},
we derive the following theorem with respect to the solution to the autonomous dToda equation \eqref{adToda}.
\begin{theorem}\label{thm1}
The solution to the autonomous dToda equation \eqref{adToda} can be expressed as
\begin{align}
&q_k^{(s)}=\frac{H_{k-1}^{(s,0)}H_{k}^{(s+1,0)}}{H_{k}^{(s,0)}H_{k-1}^{(s+1,0)}},
\quad k=1,2,\dots,m,\label{qkstau}\\
&e_k^{(s)}=\frac{H_{k+1}^{(s,0)}H_{k-1}^{(s+1,0)}}{H_{k}^{(s,0)}H_{k}^{(s+1,0)}},\label{ekstau}
\quad k=1,2,\dots,m-1,
\end{align}
where $f_0^{(0)},f_{1}^{(0)},\dots, f_{m-1}^{(0)}$ appearing in the Hankel determinants $H_k^{(0,0)}$
are uniquely given from $q_1^{(0)}$, $q_2^{(0)}$, $\dots$, $q_m^{(0)}$
and $e_1^{(0)}$, $e_2^{(0)}$, $\dots$, $e_{m-1}^{(0)}$ in $A^{(0,0)}$.
\end{theorem}
\begin{proof}
By ignoring the discrete-time variable $t$ in Lemma~\ref{lem3} and Theorem~\ref{lem5},
we derive \eqref{qkstau} and \eqref{ekstau} as the solution to the autonomous dToda equation \eqref{adToda}.
\par
Using a cofactor expansion along the $(k+1)$th column for the Hankel polynomials $H_k^{(s)}(z)$ without the discrete-time variable $t$ in \eqref{Tkz},
we obtain
\begin{align}\label{thm1-1}
H_k^{(s)}(z)=H_k^{(s)}z^k+\tilde{H}_{k-1}^{(s)}z^{k-1}+\tilde{H}_{k-2}^{(s)}z^{k-2}+\dots+\tilde{H}_0^{(s)},
\end{align}
where $\tilde{H}_{k-i}^{(s)}$ denote determinants given by replacing the $(k-i+1)$th column with $-\bvec{f}_{k}^{(s+k)}$
in the Hankel determinants $H_k^{(s)}$, that is,
\begin{align*}
\tilde{H}_{k-i}^{(s)}:=\left|
\begin{array}{cccccccccccc}
\bvec{f}_{k}^{(s)}&\bvec{f}_k^{(s+1)}&
\cdots&\bvec{f}_k^{(s+k-i-1)}&
-\bvec{f}_k^{(s+k)}&\bvec{f}_k^{(s+k-i+1)}&
\cdots&\bvec{f}_k^{(s+k-1)}&
\end{array}\right|.
\end{align*}
Thus, by comparing \eqref{thm1-1} with ${\cal H}_{k}^{(s,t)}(z)=z^k+b_{k,1}^{(s,t)}z^{k-1}+b_{k,2}^{(s,t)}z^{k-2}+\cdots+b_{k,k}^{(s,t)}$,
we derive $b_{k,1}^{(s)}=\tilde{H}_{k-1}^{(s)}/H_k^{(s)},b_{k,2}^{(s)}=\tilde{H}_{k-2}^{(s)}/H_k^{(s)},\dots,
b_{k,k}^{(s)}=\tilde{H}_{0}^{(s)}/H_k^{(s)}$.
Applying the inverse of Cramer's rule to these with $s=0$, we obtain
\begin{align}
\left(\begin{array}{cccc}
\bvec{f}_{k}^{(0)}&\bvec{f}_{k}^{(1)}&\cdots&\bvec{f}_{k}^{(k-1)}
\end{array}
\right)\left(
\begin{array}{c}
b_{k,k}^{(0)}\\b_{k,k-1}^{(0)}\\\vdots\\b_{k,1}^{(0)}
\end{array}\right)=
-\bvec{f}_{k}^{(k)},\quad
k=1,2,\dots,m.\label{beqs}
\end{align}
Therefore it follows that
\begin{align}
\left\{
\begin{array}{l}
f_0^{(0)}=1,\\
f_1^{(0)}=-b_{1,1}f_0^{(0)},\\
f_2^{(0)}=-b_{2,1}f_{1}^{(0)}-b_{2,2}f_0^{(0)},\\
\quad\vdots\\
f_{m-1}^{(0)}=
-b_{m-1,1}f_{m-2}^{(0)}-b_{m-1,2}f_{m-3}^{(0)}-\cdots-b_{m-1,m-1}f_{0}^{(0)},
\end{array}
\right.\label{fbdet}
\end{align}
which implies that $f_0^{(0)},f_{1}^{(0)},\dots, f_{m-1}^{(0)}$ are uniquely determined
using $b_{k,1}^{(0)}$, $b_{k,2}^{(0)}$, $\dots$, $b_{k,k}^{(0)}$.\par
Moreover, it is obvious from \eqref{phikstdet} with $s=0$ and without the discrete-time $t$ that
$q_k^{(0)}$ and $e_k^{(0)}$ generate the Hadamard polynomial ${\cal H}_k^{(0)}(z)$.
Thus, we see that $b_{k,1}^{(0)}$, $b_{k,2}^{(0)}$,
$\dots$, $b_{k,k}^{(0)}$ are uniquely given.

\end{proof}
\par
Similar to $L^{(s,t)}$ and $R^{(s,t)}$ involving $e_{k}^{(s,t)}$ and $q_k^{(s,t)}$, respectively,
let us define $m$-by-$m$ matrices involving $E_k^{(s,t)}$ and $Q_k^{(s,t)}$ appearing in Lemma~\ref{lem4} and Theorem~\ref{lem6}:
\begin{align*}
{\cal L}^{(s,t)}:=\left(
\begin{array}{cccc}
1&\\
E_1^{(s,t)}&1&\\
&\ddots&\ddots&\\
&&E_{m-1}^{(s,t)}&1
\end{array}
\right),\quad
{\cal R}^{(s,t)}:=\left(
\begin{array}{cccc}
Q_1^{(s,t)}&1\\
&Q_2^{(s,t)}&\ddots\\
&&\ddots&1\\
&&&Q_m^{(s,t)}
\end{array}
\right).
\end{align*}
Then, we have the following proposition concerning eigenpairs of ${\cal A}^{(s,t)}:={\cal L}^{(s,t)}{\cal R}^{(s,t)}$.
\begin{proposition}\label{prop5}
Eigenvalues of ${\cal A}^{(s,t)}+\mu^{(t)}I_m$ coincide with the roots $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$ of the polynomial
$p(z)=(z-\lambda_1)(z-\lambda_2)\dots(z-\lambda_m)$.
Moreover, the eigenvectors corresponding to $\lambda_1,\lambda_2,\dots,\lambda_m$
are ${\cal H}_1^{(s,t)},{\cal H}_2^{(s,t)},\dots,{\cal H}_m^{(s,t)}$, respectively.
\end{proposition}
\begin{proof}
Let us prepare the $k$th principle submatrix of ${\cal A}^{(s,t)}$, namely,
\begin{align*}
{\cal A}_k^{(s,t)}:=
\left(
\begin{array}{cccc}
Q_1^{(s,t)}&1\\
Q_1^{(s,t)}E_1^{(s,t)}&Q_2^{(s,t)}+E_1^{(s,t)}&\ddots\\
&\ddots&\ddots&1\\
&&Q_{k-1}^{(s,t)}E_{k-1}^{(s,t)}&Q_k^{(s,t)}+E_{k-1}^{(s,t)}
\end{array}
\right).
\end{align*}
Then, by considering cofactor expansion along the $(k+1)$th row for
$\det[(z-\mu^{(t)})I_{k+1}-{\cal A}_{k+1}^{(s,t)}]$,
we obtain
\begin{align}\label{detzmuicalA}
\det[(z-\mu^{(t)})I_{k+1}-{\cal A}_{k+1}^{(s,t)}

=(z-Q_{k+1}^{(s,t)}-E_{k}^{(s,t)}-\mu^{(t)})
\det[(z-\mu^{(t)})I_{k}-{\cal A}_{k}^{(s,t)}]\nonumber\\

\quad\quad-Q_{k}^{(s,t)}E_{k}^{(s,t)}
\det[(z-\mu^{(t)})I_{k-1}-{\cal A}_{k-1}^{(s,t)}].
\end{align}
Comparing this with \eqref{3termQE},
we derive
\begin{align*}
{\cal H}_{k}^{(s,t)}=\det[(z-\mu^{(t)})I_k-{\cal A}_{k}^{(s,t)}],
\end{align*}
which implies $p(z)=\det[(z-\mu^{(t)})I_m-{\cal A}^{(s,t)}]$.
Thus, by recalling $p(z)=(z-\lambda_1)(z-\lambda_2)\cdots(z-\lambda_m)$,
we see that $\lambda_1,\lambda_2,\dots,\lambda_m$ are eigenvalues of ${\cal A}^{(s,t)}+\mu^{(t)}I_m$.
Similarly to the case of $q_k^{(s,t)}$ and $e_k^{(s,t)}$,
by reconsidering \eqref{3termQE} with $z=\lambda_k$,
we have
\begin{align}\label{eigenpairsofcalA}
({\cal A}^{(s,t)}+\mu^{(t)}I_m){\cal H}_k^{(s,t)}=\lambda_k{\cal H}_k^{(s,t)},\quad k=1,2,\dots,m.
\end{align}
Thus, ${\cal H}_{k}^{(s,t)}$ are eigenvectors of  ${\cal A}^{(s,t)}+\mu^{(t)}I_m$ corresponding to the eigenvalues $\lambda_k$.

\end{proof}
\par
With the help of Proposition~\ref{prop5},
we derive the following theorem concerning the solution to the non-autonomous dToda equation \eqref{nadToda}.
\begin{theorem}\label{thm2}
The solution to the non-autonomous dToda equation \eqref{nadToda} can be expressed as
\begin{align}
&Q_k^{(t)}=\frac{H_{k-1}^{(0,t)}H_{k}^{(0,t+1)}}{H_{k}^{(0,t)}H_{k-1}^{(0,t+1)}},
\quad k=1,2,\dots,m,\label{Qkttau}\\
&E_k^{(t)}=\frac{H_{k+1}^{(0,t)}H_{k-1}^{(0,t+1)}}{H_{k}^{(0,t)}H_{k}^{(0,t+1)}},
\quad k=1,2,\dots,m-1,\label{Ekttau}
\end{align}
where $f_0^{(0)},f_{1}^{(0)},\dots, f_{m-1}^{(0)}$ appearing in the Hankel determinants $H_k^{(0,0)}$
are uniquely given by $Q_1^{(0)}$, $Q_2^{(0)}$, $\dots$, $Q_m^{(0)}$
and $E_1^{(0)}$, $E_2^{(0)}$, $\dots$, $E_{m-1}^{(0)}$ in ${\cal A}^{(0,0)}$.
\end{theorem}
\begin{proof}
The proof is essentially given using Lemma~\ref{lem4} and Theorem~\ref{lem6} instead of Lemma~\ref{lem3} and Theorem~\ref{lem5}, and replacing $s$ with $t$.

\end{proof}

\par
It is obvious that the solutions $q_k^{(s)}, e_k^{(s)}$ in Theorem~\ref{thm1} or $Q_k^{(t)},E_k^{(t)}$ in Theorem~\ref{thm2} have sufficient arbitrariness from the constants $a_1,a_2,\dots,a_m$ and the initial values $f_0^{(0)}, f_1^{(0)},\dots,f_{m-1}^{(0)}$ of the infinite sequence $\{f_s^{(t)}\}_{s,t=0}^{\infty}$.
From the perspective of the eigenvalue problem,
we also discuss the arbitrariness of $q_k^{(s)}, e_k^{(s)}$ or $Q_k^{(t)},E_k^{(t)}$.
If $q_k^{(0,0)}, e_k^{(0,0)}$ or $Q_k^{(0,0)},E_k^{(0,0)}$ are given,
we uniquely obtain $A_k^{(0,0)}$ or ${\cal A}_k^{(0,0)}$.
In other words, the eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_m$ of both $A^{(0,0)}$ and ${\cal A}^{(0,0)}+\mu^{(0)}I_m$
are determined by $q_k^{(0,0)}, e_k^{(0,0)}$ or $Q_k^{(0,0)},E_k^{(0,0)}$.
Thus, the eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_m$ have sufficient arbitrariness.
\section{Asymptotic behavior of discrete Toda equations}\label{sec:5}
In this section, we find a general term to the infinite sequence $\{f_{s}^{(t)}\}_{s,t=0}^{\infty}$
associated with distinct $\lambda_1,\lambda_2,\dots,\lambda_m$,
and then derive asymptotic expansions of the Hankel determinants $H_k^{(s,t)}$
involving $f_s^{(t)}$ as $s\rightarrow\infty$ or $t\rightarrow\infty$.
Using the asymptotic expansions,
we clarify asymptotic behavior of the determinantal solutions to the autonomous Toda equation \eqref{adToda}
and the non-autonomous dToda equation \eqref{nadToda}.
\par
The following lemma gives a general term to  the infinite sequence $\{f_{s}^{(t)}\}_{s=0}^{\infty}$ at fixed $t$.
\begin{lemma}\label{lem7}
Let us assume that the elements of the infinite sequence $\{f_{s}^{(t)}\}_{s=0}^{\infty}$
satisfy the linear dependence \eqref{fnmt} at fixed $t$.
Then, $f_{s}^{(t)}$ can be expressed using distinct $\lambda_1,\lambda_2,\dots,\lambda_m$ as
\begin{align}\label{gsfst}
f_{s}^{(t)}=\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s},\quad s=0,1,\dots,
\end{align}
where $c_1^{(t)},c_2^{(t)},\dots,c_m^{(t)}$ are constants determined by
\begin{align}\label{cdetermine}
\left(
\begin{array}{c}
c_{1}^{(t)}\\c_{2}^{(t)}\\\vdots\\c_{m}^{(t)}\\
\end{array}
\right)=
\left(
\begin{array}{cccc}
1&1&\cdots&1\\
\lambda_1&\lambda_2&\cdots&\lambda_m\\
\vdots&\vdots&\ddots&\vdots\\
\lambda_1^{m-1}&\lambda_{2}^{m-1}&\cdots&\lambda_{m}^{m-1}
\end{array}
\right)^{\hspace{-2mm}-1}
\left(
\begin{array}{c}
f_{0}^{(t)}\\f_{1}^{(t)}\\\vdots\\f_{m-1}^{(t)}\\
\end{array}
\right),
\end{align}
and satisfying
\begin{align}
c_i^{(t)}=c_i^{(0)}\prod_{\ell=0}^{t-1}(\lambda_{i}-\mu^{(\ell)}),\quad i=1,2,\dots,m.\label{citoi0}
\end{align}
\end{lemma}
\begin{proof}
Using \eqref{gsfst}, we can rewrite the left-hand side of the linear dependence \eqref{fnmt} as
\begin{align*}
\sum_{i=0}^{m}a_{i}f_{s+m-i}^{(t)}
&=
\sum_{i=0}^{m}a_{i}\left(
\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s+m-i}\right)\\
&=\sum_{\ell=1}^m
c_{\ell}^{(t)}\lambda_\ell^{s}\left(
\sum_{i=1}^{m}a_{i}\lambda_{\ell}^{m-i}
\right).
\end{align*}
From \eqref{pza}, it is obvious that
$\sum_{i=1}^{m}a_{i}\lambda_{\ell}^{m-i}=p(\lambda_\ell)$.
By recalling that $\lambda_\ell$ are zeros of $p(z)$,
we derive $\sum_{i=1}^{m}a_{i}\lambda_{\ell}^{m-i}=p(\lambda_\ell)=0$.
Thus, we see that $f_s^{(t)}$ in \eqref{gsfst} satisfy \eqref{fnmt}.
From \eqref{gsfst}, it follows that
\begin{align}\label{coefmatrix}
\left(\begin{array}{c}
f_0^{(t)}\\f_1^{(t)}\\\vdots\\f_{m-1}^{(t)}
\end{array}\right)=
\left(\begin{array}{cccc}
1&1&\cdots&1\\
\lambda_1&\lambda_2&\cdots&\lambda_m\\
\vdots&\vdots&\ddots&\vdots\\
\lambda_1^{m-1}&\lambda_{2}^{m-1}&\cdots&\lambda_{m}^{m-1}
\end{array}
\right)\left(
\begin{array}{c}
c_{1}^{(t)}\\c_{2}^{(t)}\\\vdots\\c_{m}^{(t)}\\
\end{array}
\right).
\end{align}
Then, by considering that the $m$-by-$m$ matrix involving $\lambda_1,\lambda_2,\dots,\lambda_m$
is the Vandermonde matrix with determinant $\prod_{i<j}(\lambda_j-\lambda_i)$, we obtain \eqref{cdetermine}.
Moreover, by combining \eqref{key} in Proposition~\ref{prop2} with \eqref{gsfst},
we derive
\begin{align}\label{cit+1}
c_i^{(t+1)}=(\lambda_i-\mu^{(t)})c_i^{(t)},\quad i=1,2,\dots,m,
\end{align}
which immediately leads to \eqref{citoi0}.

\end{proof}
\par
It is of significance that the infinite sequence $\{f_s^{(t)}\}_{s=0}^{\infty}$ at any $t$ in Lemma~\ref{lem7}
involves arbitrary constants $\lambda_1,\lambda_2,\dots,\lambda_m$ and initial values $f_0^{(t)}, f_1^{(t)},\dots,f_{m-1}^{(t)}$.
Thus, the infinite sequence $\{f_s^{(t)}\}_{s=0}^{\infty}$ has $2m$ degrees of freedom,
which is greater than in the initial setting of the autonomous dToda equation \eqref{adToda}.
In other words, the freedom in the infinite sequence $\{f_s^{(t)}\}_{s=\infty}^{\infty}$ sufficiently covers
that in the setting of the autonomous dToda equation \eqref{adToda}.
\par
The following lemma gives an expansion of the Hankel determinants $H_k^{(s,t)}$.
\begin{lemma}\label{lem8}
At any $t$, the Hankel determinants $H_k^{(s,t)}$ can be expanded as
\begin{align}\label{aeHankels}
H_k^{(s,t)}=\sum_{1\le \kappa_1<\kappa_2<\dots<\kappa_k\le m}
c_{\kappa_1}^{(t)}c_{\kappa_2}^{(t)}\cdots c_{\kappa_k}^{(t)}
(\lambda_{\kappa_1}\lambda_{\kappa_2}\cdots \lambda_{\kappa_k})^s
\prod_{\ell=2}^{k}\lambda_{\kappa_\ell}^{\ell-1}
\prod_{i<j}(\lambda_{\kappa_j}-\lambda_{\kappa_i}).
\end{align}
\end{lemma}
\begin{proof}
Since the entries of the Hankel determinants $H_k^{(s,t)}$ satisfy \eqref{gsfst} in Lemma~\ref{lem7},
we derive
\begin{align*}
H_k^{(s,t)}=\left|
\begin{array}{cccc}
\displaystyle\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s}
&\displaystyle\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s+1}
&\cdots
&\displaystyle\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s+k-1}\\
\displaystyle\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s+1}
&\displaystyle\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s+2}
&\cdots
&\displaystyle\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s+k}\\
\vdots&\vdots&\ddots&\vdots\\
\displaystyle\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s+k-1}
&\displaystyle\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s+k}
&\cdots
&\displaystyle\sum_{\ell=1}^{m}c_{\ell}^{(t)}\lambda_{\ell}^{s+2k-2}\\
\end{array}
\right|,\quad k=1,2,\dots,m.
\end{align*}
By reorganizing these determinants, we obtain
\begin{align}\label{taukccclllv}
&H_k^{(s,t)}=
\sum_{1\le \kappa_1<\kappa_2<\dots<\kappa_k\le m}
c_{\kappa_1}^{(t)}c_{\kappa_2}^{(t)}\cdots c_{\kappa_k}^{(t)}
(\lambda_{\kappa_1}\lambda_{\kappa_2}\cdots \lambda_{\kappa_k})^s
\left|
\begin{array}{cccc}
1&\lambda_{\kappa_2}&\cdots&\lambda_{\kappa_k}^{k-1}\\
\lambda_{\kappa_1}&\lambda_{\kappa_2}^2&\cdots&\lambda_{\kappa_k}^k\\
\vdots&\vdots&\ddots&\vdots\\
\lambda_{\kappa_1}^{k-1}&\lambda_{\kappa_2}^k&\cdots&\lambda_{\kappa_k}^{2k-2}
\end{array}
\right|,\nonumber\\
&\quad
k=1,2,\dots,m.
\end{align}
Noting that the Vandermonde determinants appear on the right hand side of \eqref{taukccclllv},
we thus have \eqref{aeHankels}.

\end{proof}
\par
Lemma~\ref{lem8} immediately leads to asymptotic expansions of the Hankel determinants $H_k^{(s,t)}$ as $s\rightarrow\infty$.
\begin{lemma}\label{lem9}
Let us assume that $\lambda_1,\lambda_2,\dots,\lambda_m$ are constants
such that $|\lambda_1|>|\lambda_2|>\dots>|\lambda_m|$.
Then, the Hankel determinants $H_k^{(s,t)}$ can be expanded as
\begin{align}\label{astauks}
H_k^{(s,t)}=\hat{c}_{k}^{(t)} (\lambda_1\lambda_2\cdots\lambda_k)^s\left(1+O\left(\rho_k^s\right)\right),
\quad s\rightarrow\infty,
\end{align}
where $\hat{c}_k^{(t)}:=c_1^{(t)}c_2^{(t)}\cdots c_k^{(t)}\prod_{\ell=2}^{k}\lambda_{\kappa_\ell}^{\ell-1}
\prod_{i<j}(\lambda_{\kappa_j}-\lambda_{\kappa_i})$
and $\rho_k$ are some constants such that $\rho_k>|\lambda_{k+1}|/|\lambda_{k}|$.
\end{lemma}
\begin{proof}
Under the assumption $|\lambda_1|>|\lambda_2|>\dots>|\lambda_m|$,
the dominant terms of the Hankel determinants $H_k^{(s,t)}$ as $s\rightarrow\infty$
are the terms with  $\kappa_1=1,\kappa_2=2,\dots,\kappa_k=k$ appearing in Lemma~\ref{lem8}.
\end{proof}
\par
With the help of Lemma~\ref{lem9},
we present an asymptotic analysis of the autonomous dToda equation \eqref{adToda}
as $s\rightarrow\infty$.
\begin{theorem}\label{thm3}
Let us assume that for the infinite sequence $\{f_s\}_{s=0}^{\infty}$, the Hankel determinants $H_k^{(s)}$ are all nonzero.
Moreover, let $\lambda_1,\lambda_2,\dots,\lambda_m$
be constants such that $|\lambda_1|>|\lambda_2|>\dots>|\lambda_m|$.
Then, the asymptotic behavior as $s\rightarrow\infty$ of the autonomous dToda variables $q_k^{(s)}$ and $e_k^{(s)}$ are given as
 \begin{align}
&\lim_{s\rightarrow\infty}
q_{k}^{(s)}=\lambda_k,\quad k=1,2,\dots,m,\label{qkstsinfty}\\
&\lim_{s\rightarrow\infty}
e_k^{(s)}=0,\quad k=1,2,\dots, m-1.\label{ekstsinfty}
\end{align}
\end{theorem}
\begin{proof}
By applying Lemma~\ref{lem9} to the expressions of $q_{k}^{(s,t)}$ and $e_k^{(s,t)}$ in Lemma~\ref{lem3},
we obtain, as $s\rightarrow\infty$,
\begin{align}
&q_k^{(s)}=\lambda_k \frac{\left(1+O\left(\rho_{k-1}^s\right)\right)\left(1+O\left(\rho_k^{s+1}\right)\right)}
{\left(1+O\left(\rho_k^s\right)\right)\left(1+O\left(\rho_{k-1}^{s+1}\right)\right)},
\quad k=1,2,\dots,m,\label{qkstOs}\\
&e_k^{(s)}=\frac{\hat{c}_{k+1}^{(t)}\hat{c}_{k-1}^{(t)}}{(\hat{c}_k^{(t)})^2}
\frac{1}{\lambda_k}
\left(\frac{\lambda_{k+1}}{\lambda_k}
\right)^{\hspace{-1mm}s}
\frac{\left(1+O\left(\rho_{k+1}^s\right)\right)\left(1+O\left(\rho_{k-1}^{s+1}\right)\right)}
{\left(1+O\left(\rho_k^s\right)\right)\left(1+O\left(\rho_{k}^{s+1}\right)\right)},
\quad k=1,2,\dots,m-1.\label{ekstOs}
\end{align}
Thus, by taking the limit $s\rightarrow\infty$ in \eqref{qkstOs} and \eqref{ekstOs},
we see that $q_k^{(s,t)}\rightarrow\lambda_k$ and $e_k^{(s,t)}\rightarrow 0$ as $s\rightarrow\infty$.
We may remove the discrete-time variable $t$ from the superscripts in $q_k^{(s,t)}$ and $e_k^{(s,t)}$
in the case of analysis for the autonomous dToda equation \eqref{adToda}.
Therefore, we have \eqref{qkstsinfty} and \eqref{ekstsinfty}

\end{proof}
\par
The discrete-time evolution from $s$ to $s+1$ in $Q_k^{(s,t)}$ and $E_k^{(s,t)}$
is not considered in either the autonomous dToda equation \eqref{adToda} and the non-autonomous dToda equation \eqref{nadToda}.
Although the explicit formula for generating such a discrete-time evolution has not been shown,
we can describe an asymptotic behavior of $Q_k^{(s,t)}$ and $E_k^{(s,t)}$ as $s\rightarrow\infty$.
\begin{theorem}\label{thm4}
Let us assume that for the  infinite sequence $\{f_s^{(t)}\}_{s,t=0}^{\infty}$, the Hankel determinants $H_k^{(s,t)}$ are all nonzero.
Moreover, let $\lambda_1,\lambda_2,\dots,\lambda_m$ be constants such that $|\lambda_1|>|\lambda_2|>\dots>|\lambda_m|$.
Then, it holds that, as $s\rightarrow\infty$,
\begin{align}
&\lim_{s\rightarrow\infty}Q_k^{(s,t)}=\lambda_{k}-\mu^{(t)},\quad k=1,2,\dots,m,\\
&\lim_{s\rightarrow\infty}E_k^{(s,t)}=0,\quad k=1,2,\dots,m-1.
\end{align}
\end{theorem}
\begin{proof}
For the expressions of $Q_k^{(s,t)}$ and $E_k^{(s,t)}$ in Lemma~\ref{lem4} as $s\rightarrow\infty$,
it follows that, as $s\rightarrow\infty$,
\begin{align}
&Q_k^{(s,t)}=\frac{\hat{c}_{k-1}^{(t)}\hat{c}_{k}^{(t+1)}}{\hat{c}_k^{(t)}\hat{c}_{k-1}^{(t+1)}}
\frac{\left(1+O\left(\rho_{k-1}^s\right)\right)\left(1+O\left(\rho_k^{s}\right)\right)}
{\left(1+O\left(\rho_k^s\right)\right)\left(1+O\left(\rho_{k-1}^{s}\right)\right)},
\quad k=1,2,\dots,m,\label{QkstOs}\\
&E_k^{(s,t)}=
\frac{\hat{c}_{k+1}^{(t)}\hat{c}_{k-1}^{(t+1)}}{\hat{c}_{k}^{(t)}\hat{c}_k^{(t+1)}}
\left(\frac{\lambda_{k+1}}{\lambda_k}
\right)^{\hspace{-1mm}s}
\frac{\left(1+O\left(\rho_{k+1}^s\right)\right)\left(1+O\left(\rho_{k-1}^{s}\right)\right)}
{\left(1+O\left(\rho_k^s\right)\right)\left(1+O\left(\rho_{k}^{s}\right)\right)}
\quad k=1,2,\dots,m-1.\label{EkstOs}
\end{align}
Equations \eqref{QkstOs} and \eqref{EkstOs}
imply that, as $s\rightarrow\infty$, $Q_k^{(s,t)}\rightarrow\hat{c}_{k-1}^{(t)}\hat{c}_{k}^{(t+1)}/(\hat{c}_{k}^{(t)}\hat{c}_{k-1}^{(t+1)})$
and $E_k^{(s,t)}\rightarrow0$, respectively.
Noting that $\hat{c}_{k}^{(t+1)}/\hat{c}_{k}^{(t)}
=\prod_{\ell=1}^{k}c_{\ell}^{(t+1)}/c_{\ell}^{(t)}
=\prod_{\ell=1}^{k}(\lambda_{\ell}-\mu^{(t)})$,
we also derive $Q_k^{(s,t)}\rightarrow\lambda_{k}-\mu^{(t)}$ as $s\rightarrow\infty$.

\end{proof}
\par
Theorems~\ref{thm3} and~\ref{thm4} with
Propositions~\ref{prop4} and~\ref{prop5} claim that, for fixed $t$, $q_k^{(s)}=q_k^{(s,t)}$ and $Q_k^{(s)}=Q_k^{(s,t)}$ converge to
the eigenvalues of $A^{(s)}=A^{(s,t)}$ and ${\cal A}^{(s)}={\cal A}^{(s,t)}$ as $s\rightarrow\infty$, respectively.
The asymptotic convergence of $q_k^{(s)}$ and $e_k^{(s)}$ as $s\rightarrow\infty$ obviously coincides with that shown in \cite{Henrici,Rutishauser1990}.
\par
Let us turn to the asymptotic analysis of $q_k^{(s,t)}, e_k^{(s,t)}$ and $Q_k^{(s,t)}, E_k^{(s,t)}$ as $t\rightarrow\infty$.
The following lemma gives asymptotic expansions of the Hankel determinants $H_k^{(s,t)}$ as $t\rightarrow\infty$.
\begin{lemma}\label{lem10}
Let us assume that $\lambda_1,\lambda_2,\dots,\lambda_m$ are constants such that
$|\lambda_1-\mu^{(t)}|>|\lambda_2-\mu^{(t)}|>\dots>|\lambda_m-\mu^{(t)}|$.
Then, as $t\rightarrow\infty$,
\begin{align}\label{asympexpantaukt}
&H_k^{(s,t)}=
\check{c}_k^{(s)}
\prod_{i=0}^{t}
(\lambda_{1}-\mu^{(i)})(\lambda_{2}-\mu^{(i)})\cdots(\lambda_{k}-\mu^{(i)})
\left(
1+O\left(\varrho_k^t\right)
\right),
\quad k=1,2,\dots,m,
\end{align}
where $\check{c}_{k}^{(s)}$ are constants given by
$\check{c}_k^{(s)}:=c_1^{(0)}c_2^{(0)}\cdots c_k^{(0)}(\lambda_1\lambda_2\cdots \lambda_k)^s$
$\prod_{\ell=2}^{k}\lambda_{\kappa_\ell}^{\ell-1}$
$\prod_{i<j}(\lambda_{\kappa_j}-\lambda_{\kappa_i})$
and $\varrho_k$ are constants satisfying $\varrho_{k}>|\lambda_{k+1}-\mu^{(t)}|/|\lambda_{k}-\mu^{(t)}|$.
\end{lemma}
\begin{proof}
Equation \eqref{asympexpantaukt} is easily proved by noting that the terms with
$\kappa_1=1,\kappa_2=2,\dots,\kappa_k=k$ appearing in Lemma~\ref{lem8}
are dominant terms of the Hankel determinants $H_k^{(s,t)}$ as $t\rightarrow\infty$
under the assumption $|\lambda_1-\mu^{(t)}|>|\lambda_2-\mu^{(t)}|>\dots>|\lambda_m-\mu^{(t)}|$.

\end{proof}
\par
Neither the autonomous dToda equation \eqref{adToda} nor the non-autonomous dToda equation \eqref{nadToda}
generates the discrete-time evolution of $q_k^{(s,t)}$ and $e_k^{(s,t)}$ with respect to $t$.
However, Lemma~\ref{lem10} immediately yields asymptotic behavior of $q_k^{(t)}$ and $e_k^{(t)}$ obtained by removing the discrete-time variable $s$ from the superscripts in $q_k^{(s,t)}$ and $e_k^{(s,t)}$, respectively, as $t\rightarrow\infty$.
\begin{theorem}\label{thm5}
Let us assume that for the infinite sequence $\{f^{(t)}\}_{t=0}^{\infty}$,
the Hankel determinants  $H_k^{(s,t)}$ are all nonzero.
Moreover, let $\lambda_1,\lambda_2,\dots,\lambda_m$ be constants
such that $|\lambda_1-\mu^{(t)}|>|\lambda_2-\mu^{(t)}|>\dots>|\lambda_m-\mu^{(t)}|$.
Then, it holds that
\begin{align}
&\lim_{t\rightarrow\infty} q_k^{(t)}
=\lambda_k,\quad k=1,2,\dots,m,\\
&\lim_{t\rightarrow\infty} e_k^{(t)}
=0,\quad k=1,2,\dots,m-1.
\end{align}
\end{theorem}
\begin{proof}
By combining Lemma~\ref{lem10} with the expressions of $q_k^{(s,t)}$ and $e_{k}^{(s,t)}$ in Lemma~\ref{lem3},
we obtain, as $t\rightarrow\infty$,
\begin{align}
&q_k^{(s,t)}=\frac{\check{c}_{k-1}^{(s)}\check{c}_k^{(s+1)}}{\check{c}_k^{(s)}\check{c}_{k-1}^{(s+1)}}
\frac{\left(1+O\left(\varrho_{k-1}^{t}\right)\right)
\left(1+O\left(\varrho_{k}^{t}\right)\right)}
{\left(1+O\left(\varrho_{k}^{t}\right)\right)
\left(1+O\left(\varrho_{k-1}^{t}\right)\right)},\quad
k=1,2,\dots,m,\label{qkstOt}\\
&e_k^{(s,t)}=\frac{\check{c}_{k+1}^{(s)}\check{c}_{k-1}^{(s+1)}}{\check{c}_k^{(s)}\check{c}_k^{(s+1)}}
\prod_{i=1}^{t}\frac{\lambda_{k+1}-\mu^{(i)}}{\lambda_k-\mu^{(i)}}
\frac{\left(1+O\left(\varrho_{k}^{t}\right)\right)
\left(1+O\left(\varrho_{k-1}^{t}\right)\right)}
{\left(1+O\left(\varrho_{k}^{t}\right)\right)
\left(1+O\left(\varrho_{k}^{t}\right)\right)},
\quad k=1,2,\dots,m-1.\label{ekstOt}
\end{align}
From the limit $t\rightarrow\infty$ in \eqref{qkstOt} and \eqref{ekstOt},
we find that $q_k^{(s,t)}
\rightarrow
\check{c}_{k-1}^{(s)}\check{c}_k^{(s+1)}/$
$(\check{c}_k^{(s)}\check{c}_{k-1}^{(s+1)})
=\lambda_k$ and  $e_k^{(s,t)}\rightarrow 0$ as $t\rightarrow\infty$.

\end{proof}
\par
Lemma~\ref{lem10} also enables us to clarify the asymptotic analysis of the non-autonomous dToda equation \eqref{nadToda}.
\begin{theorem}\label{thm6}
Let us assume that for the infinite sequence $\{f^{(t)}\}_{t=0}^{\infty}$,
the Hankel determinants  $H_k^{(s,t)}$ are all nonzero.
Moreover, let $\lambda_1,\lambda_2,\dots,\lambda_m$ be constants
such that $|\lambda_1-\mu^{(t)}|>|\lambda_2-\mu^{(t)}|>\dots>|\lambda_m-\mu^{(t)}|$.
Then, it holds that
\begin{align}
&\lim_{t\rightarrow\infty} Q_k^{(t)}=\lambda_k-\mu^\ast,\quad k=1,2,\dots,m,\\
&\lim_{t\rightarrow\infty} E_k^{(t)}=0,\quad k=1,2,\dots,m-1,
\end{align}
where $\mu^{\ast}:=\lim_{t\rightarrow\infty}\mu^{(t)}$.
\end{theorem}
\begin{proof}
From the expressions of $Q_k^{(s,t)}$ and $E_k^{(s,t)}$ in Lemma~\ref{lem4} and Lemma~\ref{lem10},
it follows that, as $t\rightarrow\infty$,
\begin{align*}
&Q_k^{(s,t)}=(\lambda_k-\mu^{(t+1)})
\frac{\left(1+O\left(\varrho_{k-1}^{t}\right)\right)
\left(1+O\left(\varrho_{k}^{t+1}\right)\right)}
{\left(1+O\left(\varrho_{k}^{t}\right)\right)
\left(1+O\left(\varrho_{k-1}^{t+1}\right)\right)},
\quad k=1,2,\dots,m,\\
&E_k^{(s,t)}=\frac{\check{c}_{k+1}^{(s)}\check{c}_{k-1}^{(s)}}{(\check{c}_k^{(s)})^2}
\frac{1}{\lambda_k-\mu^{(t+1)}}\prod_{i=1}^{t}\frac{\lambda_{k+1}-\mu^{(i)}}{\lambda_k-\mu^{(i)}}
\frac{\left(1+O\left(\varrho_{k+1}^{t}\right)\right)
\left(1+O\left(\varrho_{k-1}^{t+1}\right)\right)}
{\left(1+O\left(\varrho_{k}^{t}\right)\right)
\left(1+O\left(\varrho_{k}^{t+1}\right)\right)},
\quad k=1,2,\dots,m-1,
\end{align*}
which imply that $Q_k^{(s,t)}\rightarrow \lambda_k-\mu^{\ast}$ and $E_k^{(s,t)}\rightarrow 0$ as $t\rightarrow\infty$.

\end{proof}
\par
From Theorems~\ref{thm5} and~\ref{thm6} with Propositions~\ref{prop4} and~\ref{prop5},
it can be concluded that $q_k^{(t)}$ and  $Q_k^{(t)}$ converge to the eigenvalues of $A^{(t)}$
and their shifted values of ${\cal A}^{(t)}$, as $t\rightarrow\infty$, respectively.
In particular,
Theorem~\ref{thm6} with Proposition~\ref{prop5} also describes asymptotic convergence to matrix eigenvalues
in the so-called qd with implicit shift algorithm.

\section{Discrete Lotka-Volterra system, its determinant solution and asymptotic behavior}\label{sec:6}
In this section, we first derive the dLV system \eqref{dLV} by observing properties concerning the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$ involving the infinite sequence $\{f_{s}^{(t)}\}_{s,t=0}^{\infty}$ again.
Then, we show the determinant solution to the dLV system \eqref{dLV}, and give an asymptotic analysis for the dLV system  \eqref{dLV}.\par
For the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$,
let us prepare symmetrical polynomials $\tilde{\cal H}_k^{(s,t)}(z)$ given by
\begin{align}\label{psiphi}
\left\{
\begin{array}{l}
\tilde{\cal H}_{-1}^{(s,t)}(z):=0,\\
\tilde{\cal H}_{2k}^{(s,t)}(z):={\cal H}_k^{(s,t)}(z^2),\quad k=0,1,\dots,m,\\
\tilde{\cal H}_{2k+1}^{(s,t)}(z):=z{\cal H}_{k}^{(s+1,t)}(z^2),\quad k=0,1,\dots,m.
\end{array}
\right.
\end{align}
Hereinafter, we refer to the symmetric polynomials $\tilde{\cal H}_k^{(s,t)}(z)$ as the symmetric Hadamard polynomials.
\par
Similar to the case of the Hadamard polynomials ${\cal H}_k^{(s,t)}(z)$,
we easily derive three-term recurrence relations with respect to
the symmetric Hadamard polynomials $\tilde{\cal H}_{k}^{(s,t)}(z)$.
\begin{lemma}\label{lem11}
The symmetric Hadamard polynomials $\tilde{\cal H}_k^{(s,t)}(z)$ satisfy
\begin{align}\label{psi3term}
z\tilde{\cal H}_k^{(s,t)}(z)=\tilde{\cal H}_{k+1}^{(s,t)}(z)+v_k^{(s,t)}\tilde{\cal H}_{k-1}^{(s,t)}(z),\quad
k=0,1,\dots,2m,
\end{align}
where $v_{2k}^{(s,t)}:=e_k^{(s,t)}$ and $v_{2k-1}^{(s,t)}:=q_k^{(s,t)}$.
\end{lemma}
\begin{proof}
By replacing $z$ with $z^2$ in Lemma~\ref{lem3},
we obtain
\begin{align*}
&z^2{\cal H}_{k-1}^{(s+1,t)}(z^2)={\cal H}_{k}^{(s,t)}(z^2)+q_{k}^{(s,t)}{\cal H}_{k-1}^{(s,t)}(z^2),\quad k=1,2,\dots,m,\\
&{\cal H}_k^{(s,t)}(z^2)={\cal H}_{k}^{(s+1,t)}+e_k^{(s,t)}{\cal H}_{k-1}^{(s+1,t)}(z^2),\quad k=0,1,\dots,m.
\end{align*}
Thus, by considering \eqref{psiphi}, we have \eqref{psi3term}.

\end{proof}
\par
For the parameters $\mu^{(t)}$ in the non-autonomous dToda equation \eqref{nadToda},
let us introduce an infinite sequence $\{\kappa^{(t)}\}_{t=0}^{\infty}$ satisfying
\begin{align}\label{kappamu}
(\kappa^{(t)})^2:=\mu^{(t)},\quad t=0,1,\dots.
\end{align}
Then, we derive another three-term recurrence relations of the symmetric Hadamard polynomials
$\tilde{\cal H}_k^{(s,t)}(z)$ involving the infinte sequence $\{\kappa^{(t)}\}_{t=0}^{\infty}$.
\begin{lemma}\label{lem12}
The symmetric Hadamard polynomials $\tilde{\cal H}_{k}^{(s,t)}(z)$ satisfy
\begin{align}\label{psichrist}
[z^2-(\kappa^{(t)})^2]
\tilde{\cal H}_{k-1}^{(s,t+1)}(z)
=\tilde{\cal H}_{k+1}^{(s,t)}(z)+V_{k}^{(s,t)}\tilde{\cal H}_{k-1}^{(s,t)}(z),\quad k=1,2,\dots,2m,
\end{align}
where $V_{2k-1}^{(s,t)}:=Q_{k}^{(s,t)}$ and $V_{2k}^{(s,t)}:=Q_{k}^{(s+1,t)}$.
\end{lemma}
\begin{proof}
From \eqref{Christoffelt} in Lemma~\ref{lem4},
it is obvious that
\begin{align*}
&[z^2-(\kappa^{(t)})^2]\tilde{\cal H}_{2k-2}^{(s,t+1)}(z)
=\tilde{\cal H}_{2k}^{(s,t)}(z)+Q_{k}^{(s,t)}\tilde{\cal H}_{2k-2}^{(s,t)}(z),\\
&[z^2-(\kappa^{(t)})^2]\tilde{\cal H}_{2k-1}^{(s,t+1)}(z)
=\tilde{\cal H}_{2k+1}^{(s,t)}(z)+Q_{k}^{(s+1,t)}\tilde{\cal H}_{2k-1}^{(s,t)}(z),
\end{align*}
which are equivalent to \eqref{psichrist}.

\end{proof}
Lemmas~\ref{lem11} and~\ref{lem12} lead to the following lemma with respect to $v_k^{(s,t)}$ and $V_k^{(s,t)}$.
\begin{lemma}
The variables $v_k^{(s,t)}$ and $V_k^{(s,t)}$ satisfy
\begin{align}
&v_k^{(s,t+1)}+V_{k+2}^{(s,t)}=v_{k+2}^{(s,t)}+V_{k+1}^{(s,t)},
\quad k=0,1,\dots,2m-2,\label{v+V}\\
&v_{k}^{(s,t+1)}V_{k}^{(s,t)}=v_{k}^{(s,t)}V_{k+1}^{(s,t)},
\quad k=1,2,\dots,2m-1. \label{vV}
\end{align}
\end{lemma}
\begin{proof}
From Lemmas~\ref{lem11} and~\ref{lem12},
it follows that
\begin{align*}
(v_{k}^{(s,t+1)}+V_{k+2}^{(s,t)}-v_{k+2}^{(s,t)}-V_{k+1}^{(s,t)})\tilde{\cal H}_{k+1}^{(s,t)}(z)
+(v_{k}^{(s,t+1)}V_{k}^{(s,t)}-v_{k}^{(s,t)}V_{k+1}^{(s,t)})\tilde{\cal H}_{k-1}^{(s,t)}(z)=0.
\end{align*}
It is obvious that the symmetric Hadamard polynomials $\tilde{\cal H}_{k+1}^{(s,t)}(z)$ and $\tilde{\cal H}_{k-1}^{(s,t)}(z)$
are linear independent.
Thus, we have \eqref{v+V} and \eqref{vV}.
\end{proof}
\begin{lemma}\label{lem14}
Let $u_k^{(s,t)}:=v_k^{(s,t)}(\tilde{\cal H}_{k-1}^{(s,t)}(\kappa^{(t)})/\tilde{\cal H}_{k}^{(s,t)}(\kappa^{(t)}))$ for $k=0,1,\dots,2m$.
Then, $v_k^{(s,t)}$ and $V_k^{(s,t)}$ can be expressed in terms of $u_k^{(s,t)}$ as
\begin{align}
&v_k^{(s,t)}=u_k^{(s,t)}(\kappa^{(t)}-u_{k-1}^{(s,t)}),\quad k=1,2,\dots,2m,\label{vu}\\
&V_k^{(s,t)}=-(\kappa^{(t)}-u_{k-1}^{(s,t)})
(\kappa^{(t)}-u_{k}^{(s,t)}),\quad k=1,2,\dots,2m.\label{Vu}
\end{align}
Moreover, it holds that
\begin{align}
u_k^{(s,t+1)}(\kappa^{(t+1)}-u_{k-1}^{(s,t+1)})
=
u_{k}^{(s,t)}(\kappa^{(t)}-u_{k+1}^{(s,t)}),\quad
k=1,2,\dots,2m-1.\label{odLV}
\end{align}
\end{lemma}
\begin{proof}
By letting $z=\kappa^{(t)}$ in Lemma~\ref{lem11}
and by replacing $\tilde{\cal H}_{k-1}^{(s,t)}(\kappa^{(t)})/\tilde{\cal H}_{k}^{(s,t)}(\kappa^{(t)})$ with $u_k^{(s,t)}/v_k^{(s,t)}$,
we obtain
\begin{align*}
\kappa^{(t)}=\frac{v_{k+1}^{(s,t)}}{u_{k+1}^{(s,t)}}+u_{k}^{(s,t)},\quad k=0,1,\dots,2m-1,
\end{align*}
which leads to \eqref{vu}.
Moreover, Lemma~\ref{lem12} with $z=\kappa^{(t)}$ yields
\begin{align*}
V_{k}^{(s,t)}=-\frac{\tilde{\cal H}_{k+1}^{(s,t)}(\kappa^{(t)})}{\tilde{\cal H}_{k-1}^{(s,t)}(\kappa^{(t)})}.
\end{align*}
Noting that
\begin{align*}
\frac{\tilde{\cal H}_{k+1}^{(s,t)}(\kappa^{(t)})}{\tilde{\cal H}_{k-1}^{(s,t)}(\kappa^{(t)})}=\frac{\tilde{\cal H}_{k+1}^{(s,t)}(\kappa^{(t)})}{\tilde{\cal H}_{k}^{(s,t)}(\kappa^{(t)})}\frac{\tilde{\cal H}_{k}^{(s,t)}(\kappa^{(t)})}{\tilde{\cal H}_{k-1}^{(s,t)}(\kappa^{(t)})},
\end{align*}
we derive
\begin{align}
V_k^{(s,t)}=-\frac{v_{k}^{(s,t)}v_{k+1}^{(s,t)}}{u_{k}^{(s,t)}u_{k+1}^{(s,t)}}.\label{Vvvuu}
\end{align}
Combining \eqref{vu} with \eqref{Vvvuu} gives \eqref{Vu}.\par
Equation \eqref{vV} with \eqref{vu} and \eqref{Vu} immediately leads to \eqref{odLV}.
It is easy to check that $v_k^{(s,t)}$ in \eqref{vu} and $V_k^{(s,t)}$ in \eqref{Vu} also satisfy \eqref{v+V}.

\end{proof}
\par
Without loss of generality, we can fix $s=0$ for examining the dLV system \eqref{dLV}
because the system does not give the evolution from $s$ to $s+1$.
Thus, the replacements $u_k^{(0,t)}=[1/(\kappa^{(t)})]u_k^{(t)}$
and $(\kappa^{(t)})^2=-1/\delta^{(t)}$ in \eqref{odLV} of Lemma~\ref{lem14}
simply generate the dLV system \eqref{dLV}.
From $e_0^{(s,t)}=0$ and $e_{m}^{(s,t)}=0$,
we also derive the boundary condition $u_0^{(t)}=0$ and $u_{2m}^{(t)}=0$
in the dLV system \eqref{dLV}.
\par
Lemma~\ref{lem14} with $v_{2k-1}^{(s,t)}=q_k^{(s,t)}$ and $v_{2k}^{(s,t)}=e_k^{(s,t)}$ suggests
that the dLV variables $u_k^{(t)}$ can be expressed using $q_k^{(s,t)}$ and $e_k^{(s,t)}$.
Recalling that $q_k^{(s,t)}$ and $e_k^{(s,t)}$ are expressed using the Hankel determinants $H_k^{(s,t)}$,
we obtain a theorem concerning the solution to the dLV system \eqref{dLV}.
\begin{theorem}\label{thm7}
Let us assume that for the infinite sequence $\{f_s^{(t)}\}_{s,t=0}^{\infty}$,
 $H_k^{(s,t)}$ are all nonzero.
Then, the dLV variables $u_k^{(t)}$ can be expressed using the Hankel determinants $H_k^{(s,t)}$ as
\begin{align}
&u_{2k-1}^{(t)}=\frac{H_k^{(1,t)}H_{k-1}^{(0,t+1)}}{H_k^{(0,t)}H_{k-1}^{(1,t+1)}},\quad k=1,2,\dots,m,\label{u2k-1tau}\\
&u_{2k}^{(t)}=\frac{1}{\delta^{(t)}}
\frac{H_{k+1}^{(0,t)}H_{k-1}^{(1,t+1)}}{H_k^{(1,t)}H_{k}^{(0,t+1)}},\quad k=0,1,\dots,m.
\label{u2ktau}
\end{align}
\end{theorem}
\begin{proof}
By taking into account that
$u_{2k-1}^{(s,t)}=v_{2k-1}^{(s,t)}(\tilde{\cal H}_{2k-2}^{(s,t)}(\kappa^{(t)})/\tilde{\cal H}_{2k-1}^{(s,t)}(\kappa^{(t)}))$
in Lemma~\ref{lem14} and by using
\eqref{psiphi},
we derive
\begin{align}\label{u2k-1q}
u_{2k-1}^{(s,t)}=q_k^{(s,t)}\frac{{\cal H}_{k-1}^{(s,t)}(\mu^{(t)})}{\kappa^{(t)}{\cal H}_{k-1}^{(s+1,t)}(\mu^{(t)})}.
\end{align}
Since \eqref{Tkzjacobitproof} with $z=\mu^{(t)}$ becomes $H_k^{(s,t)}(\mu^{(t)})=(-1)^kH_k^{(s,t+1)}$,
we see that ${\cal H}_{k-1}^{(s,t)}(\mu^{(t)})=H_{k-1}^{(s,t)}(\mu^{(t)})/H_{k-1}^{(s,t)}=(-1)^kH_{k-1}^{(s,t+1)}/H_{k-1}^{(s,t)}$.
Thus, by combining it with \eqref{u2k-1q}, we obtain
\begin{align}\label{temp128}
u_{2k-1}^{(s,t)}=\frac{q_k^{(s,t)}}{\kappa^{(t)}}
\frac{H_{k-1}^{(s,t+1)}H_{k-1}^{(s+1,t)}}{H_{k-1}^{(s,t)}H_{k-1}^{(s+1,t+1)}}.
\end{align}
Considering $q_k^{(s,t)}$ in \eqref{qkst}, we can rewrite \eqref{temp128} as
\begin{align*}
&u_{2k-1}^{(s,t)}=\frac{1}{\kappa^{(t)}}
\frac{H_k^{(s+1,t)}H_{k-1}^{(s,t+1)}}{H_k^{(s,t)}H_{k-1}^{(s+1,t+1)}},\quad k=1,2,\dots,m.
\end{align*}
Noting that $u_{2k-1}^{(t)}=\kappa^{(t)}u_{2k-1}^{(0,t)}$, we have \eqref{u2k-1tau}.
\par
Similarly, it follows that
\begin{align*}
u_{2k}^{(s,t)}
&=e_k^{(s,t)}\frac{\kappa^{(t)}{\cal H}_{k-1}^{(s+1,t)}(\mu^{(t)})}{{\cal H}_{k}^{(s,t)}(\mu^{(t)})}\nonumber\\
&=-e_k^{(s,t)}\kappa^{(t)}
\frac{H_{k-1}^{(s+1,t+1)}H_k^{(s,t)}}
{H_{k-1}^{(s+1,t)}H_k^{(s,t+1)}}.
\end{align*}
By combining $e_k^{(s,t)}$ in \eqref{ekst} with this,
we obtain
\begin{align*}
u_{2k}^{(s,t)}=-\kappa^{(t)}
\frac{H_{k+1}^{(s,t)}H_{k-1}^{(s+1,t+1)}}{H_k^{(s+1,t)}H_{k}^{(s,t+1)}},\quad k=0,1,\dots,m.
\end{align*}
Therefore, the specialization $u_{2k}^{(t)}=\kappa^{(t)}u_{2k}^{(0,t)}$ yields \eqref{u2ktau}.

\end{proof}
\par
Using the substitution $\tilde{H}_k^{(s,t)}:=(\delta^{(0)}\delta^{(1)}\cdots\delta^{(t-1)})^{k}H_k^{(s,t)}$,
the resulting solution is equivalent to that shown in \cite{Iwasaki2002,Spiridonov1997}.
Although the discrete-time variable $s$ does not appear in the dLV system \eqref{dLV},
it plays a key role in expressing the determinant solution to the dLV system \eqref{dLV}.
It is emphasized that the determinant solution to the dLV system \eqref{dLV} is given similar to the cases of the autonomous dToda equation \eqref{adToda} and the non-autonomous dToda equation \eqref{nadToda} from the viewpoint of the infinite sequence $\{f_s^{(t)}\}_{s,t=0}^{\infty}$ involving two types of discrete-time variables $s$ and $t$.
\par
Theorem~\ref{thm7} with Lemma~\ref{lem10} enables us to asymptotically analyze the dLV system \eqref{dLV}.
\begin{theorem}\label{thm8}
Let us assume that for the infinite sequence $\{f_s^{(t)}\}_{s,t=0}^{\infty}$,  $H_k^{(s,t)}$ are all nonzero.
Moreover, let $\lambda_1,\lambda_2,\dots,\lambda_m$ be constants such that
$|\lambda_1-\mu^{(t)}|>|\lambda_2-\mu^{(t)}|>\dots>|\lambda_m-\mu^{(t)}|$
and the limit $\delta^{(t)}$ exists.
Then, it holds that
\begin{align}
&\lim_{t\rightarrow\infty}
u_{2k-1}^{(t)}=\lambda_k,\quad k=1,2,\dots,m,\label{limu2k-1tau}
\\
&\lim_{t\rightarrow\infty}
u_{2k}^{(t)}=0,\quad k=1,2,\dots,m-1.\label{limu2ktau}
\end{align}
\end{theorem}
\begin{proof}
From Lemma~\ref{lem10} and Theorem~\ref{thm7},
we derive
\begin{align*}
&\lim_{t\rightarrow\infty}
u_{2k-1}^{(t)}=
\frac{\check{c}_k^{(1)}\check{c}_{k-1}^{(0)}}{\check{c}_k^{(0)}\check{c}_{k-1}^{(1)}},\\
&\lim_{t\rightarrow\infty}
u_{2k}^{(t)}=0.
\end{align*}
Noting that
$\check{c}_k^{(1)}\check{c}_{k-1}^{(0)}/(\check{c}_k^{(0)}\check{c}_{k-1}^{(1)})=\lambda_k$,
we immediately have \eqref{limu2k-1tau} and \eqref{limu2ktau}.
\end{proof}
\par

The convergence theorem concerning the dLV system \eqref{dLV} is restricted in \cite{Iwasaki2002,Tsujimoto2001} to the case where $\delta^{(t)}$ is positive at every $t$.
In addition, in \cite{Iwasaki2002,Tsujimoto2001},
the dLV system \eqref{dLV} is associated with matrix similarity transformations for analyzing the asymptotic behavior and the positivity of the parameter $\delta^{(t)}$ is shown to be a sufficient condition for convergence.
Without discussing the associated similarity transformations,
Theorem~\ref{thm8}, however, claims that the convergence theorem similar to in \cite{Iwasaki2002,Tsujimoto2001} holds
even if $\delta^{(t)}$ is negative at each $t$.
Suitable negative $\delta^{(t)}$ in the dLV system \eqref{dLV} realize the introduction of the effective shifts into the similarity transformations for accelerating the convergence \cite{Iwasaki2004,Yamamoto2010}.
Theorem~\ref{thm8} thus shows the convergence of a shifted algorithm for computing singular values of bidiagonal matrices,
which is designed based on the dLV system \eqref{dLV}.

\section{Conclusion}\label{sec:7}
The main objective of this paper is to describe and understand the well-known integrable discrete systems, the discrete Toda equations (dToda) and the discrete Lotka-Volterra (dLV) system, in terms of their determinant solutions and asymptotic behavior.
The key point is to introduce an infinite sequence with respect to two types of discrete-time variables.
\par
We first examined properties of the Hankel determinants and the Hadamard polynomials associated with the infinite sequence.
Then, we showed a Hankel determinant expression of solutions with sufficient degrees of freedom for the autonomous dToda equation with no parameter and the non-autonomous dToda equation with parameters.
Next, by using asymptotic expansions of the Hankel determinants, we presented asymptotic analysis of the autonomous and non-autonomous dToda equations as the discrete-time variables go to infinity.
We finally clarified the determinant solution to the dLV system with parameters, and observed its asymptotic behavior as the discrete-time variable goes to infinity.
In particular, asymptotic analysis of the non-autonomous dToda equation and the dLV system concluded the asymptotic convergence of the shifted qd and dLV algorithms for computing singular values of bidiagonal matrices providing a regularity condition on the Hankel determinants.

redpajama_set_name: RedPajamaArXiv


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redpajama_set_name: RedPajamaStackExchange


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