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One of the problems in Muir's Treatise on determinants in exercise set XXX asks to evaluate the determinant of a continuant matrix $$ f_n(x,y)=\left|\begin{array}{cccc} x & 1 & 0 & 0 & \ldots & 0 & 0\\ y-1 & x & 2 & 0 & \ldots & 0 & 0\\ 0 & y-2 & x & 3 & \ldots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \ldots & y-n+2 & x & n-1\\ 0 & 0 & 0 & \ldots & 0 & y-n+1 & x \end{array} \right|_n. $$ No answer or any hints are given.

I was able to transform this determinant to \begin{equation*} (-1)^{n-1} \left| \begin{array}{cccccc} \ldots & n-y & y-n & n-y & y-n & n+x-1 \\ \ldots & y-n & n-y & y-n & n+x-3 & n-1 \\ \ldots & n-y & y-n & n+x-5 & n-2 & 0 \\ \ldots & y-n & n+x-7 & n-3 & 0 & 0 \\ \ldots & n+x-9 & n-4 & 0 & 0 & 0 \\ & \vdots & \vdots & \vdots & \vdots & \vdots \\ \end{array} \right|_n. \end{equation*} by a series of lengthy calculations. Of course from this I can calculate the determinant when $y=n$, in which case it becomes diagonal. But not sure if this is the right way to proceed if $y\neq n$ and have no clue what to do next. Does anybody have any ideas?

I also established the following recurrence relation (which should not be confused with the standard recurrence for a determinant of a continuant): $$ f_n(x,y)=(x+n-1)f_{n-1}(x-1,y-1)+(1-n)(y-n)f_{n-2}(x-1,y-1). $$ Again, one can see that it can be easily solved when $y=n$. But what to do when $y\neq n$ is not clear.

EDIT: It turns out that when treated as a polynomial in $x$, the the system of functions $f_n(x,y)$ form an orthogonal polynomial system. I found that then $f_n(x,y)$ is related to Meixner polynomials (from Chihara's book on orthogonal polynomials) enter image description here

with $f=d=0$ and $g$ and $h$ suitably chosen. But Muir's book was written in 19-th century, well before the Meixner polynomials were discovered by Meixner in 1934. So this problem in Muir's book is really confusing.

Addendum. The problem exactly how it looks in the 1960 edition of the book: enter image description here

It is equivalent to the formulation I gave above after a simple eqiuivalence transformation and change of notation.

Hans
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    Maybe it's a different edition, but I can't find this problem in exercise set XXX. (I was trying to figure out the methods involved in this chapter, which is about continuants). Like you I was able to find the closed form for $y=n$, but the general case looks more difficult. – Jean-Claude Arbaut Oct 02 '20 at 21:21
  • @Jean-ClaudeArbaut It is the first problem. I gave it in a slightly different form. – Hans Oct 03 '20 at 06:47
  • @Jean-ClaudeArbaut I managed to get an answer from Muir's book. Your comment was helpful and I upvoted several of your answers to other questions, because there is no other way to thank you for your help. – Hans Oct 03 '20 at 12:49
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    Thanks but don't do that. Answers shall only be upvoted if you believe they should be, not because of an "external" reason. – Jean-Claude Arbaut Oct 03 '20 at 13:02
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    As @Jean-ClaudeArbaut said, you should not upvote the answers of another user just because you like them, or because they helped you with one particular question. This is considered targeted voting, and will likely be overturned by the system. Even if it is not, it is an inappropriate use of the voting system. – Xander Henderson Oct 03 '20 at 14:12
  • Hans- @Jean-ClaudeArbaut has an 18k+ reputation. A genuine and sincere comment thanking him for his help is probably a nicer reward than some up-votes on unrelated questions. I know it would be for me. – JonathanZ Oct 03 '20 at 15:29

2 Answers2

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Thanks to a comment by Jean-Claude Arbaut I finally got the answer from the page 559 of Muir's book (after correcting for some typos) $$ f_n(x,y)=\sum _{k=0}^n (-1)^{n-k} \binom{n}{k} \left(\frac{x+y-1}{2}\right)^{(k)} \left(\frac{-x+y-1}{2}\right)^{(n-k)}, $$ where $(a)^{(n)}=a(a-1)\ldots (a-n+1)$ denotes the falling factorial.

Hans
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A Hint:

Denote the determinant by $D_n$, and expand it along the last row, yields \begin{align*} D_n & = (-1)^{n + n - 1}(y - n + 1)[xD_{n - 2} - (n - 1)0] + (-1)^{2n}D_{n - 1} \\ & = -(y - n + 1)xD_{n - 2} + D_{n - 1}, \end{align*} with $D_1 = x, D_2 = x^2 - (y - 1)$.

Now solve this recursion using, say the power series tool.

Zhanxiong
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    It is very unlikely that your method can give the solution even in the simplest case $y=n$, even if you correct the mistake and manage to solve it will lead to something like $(225 - 345 y + 135 y^2 - 15 y^3) + (439 - 300 y + 45 y^2) x^2 + (55 - 15 y) x^4 + x^6$ (this is the determinant when $n=6$), or $(225 + 439 x^2 + 55 x^4 + x^6) + (-345 - 300 x^2 - 15 x^4) y + (135 + 45 x^2) y^2 - 15 y^3$, and it is not clear how this method can give closed form. – Hans Oct 02 '20 at 16:20