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In this reference (pg. 4) and few others a specific parameter in a recursion formula is solved by setting the determinant of an infinite matrix to $0$. In this precise case we have

$c_{n-1} - D_n c_n + c_{n+1} = 0 $,

where

$D_n = \frac{(\mu + i n)^2+a}{q}$.

In order to find $\mu$, people assume that

$$ \begin{vmatrix} & & \ddots & & & & & \\ \cdots & 0 & 1 & - D_{n+1} & 1 & 0 & \cdots & \\ & \cdots & 0& 1 & - D_n & 1 & 0 & \cdots \\ & & & & & \ddots & & \\ \end{vmatrix} = 0 $$

as if this were a trivial consequence from the finite-dimensional case. In fact I don't know why this has to be true, but my guess is that $\lim_{n \to \pm \infty } c_n = 0$ is supposed. Is this correct? Still I don't follow how from this condition you get that the determinant has to vanish, I would appreciate any insight into that.

Besides that, I would like to know how to calculate determinants as the one given here. In the article I linked before, it is expected some knowledge about the definition and properties of determinants of infinite matrices. I hope that someone would be able to show me how to calculate this determinant, at least in principle, as it looks fairly easy, since only one (the "main") diagonal contains values different from $0$ and $1$.

EDIT:

I tried to understand what is going on by saying that the determinant of the infinite matrix given by the Fibonacci recurrence should be also $0$, as it has a non-trivial solution:

$$ \begin{vmatrix} & & \ddots & & & & & \\ \cdots & 0 & 1 & 1 & -1 & 0 & \cdots & \\ & \cdots & 0& 1 & 1 & -1 & 0 & \cdots \\ & & & & & \ddots & & \\ \end{vmatrix} = 0 $$

but again I'm stuck, because by truncating this to an $n \times n$ matrix, the determinant is $F_{n+1}$, which goes to infinity for increasing $n$.

Moreover, any infinite matrix representation of the recursion $c_{n+1} = a c_n + b c_{n-1}$ should have determinant $0$ by using the same argument, which brings me back to my original concern: Why does the given recursion (at the top of this post) have solutions only for specific values of $\mu$?

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  • $\begingroup$ This is a very nice problem, although I don't have an answer. But some observations might be useful. First, the fact that the determinant is zero follows from the finite-dimensional case. For an $n \times n$ matrix $\mathbf A$ and an $n$-dimensional vector $\mathbf c$, $\mathbf A \mathbf c = 0$ has a nontrivial solution only if the determinant $|\mathbf A|$ is 0. In our case, the $\mathbf A$ is tridiagonal, with the diagonal elements being $-D_n$, and the off-diagonal ones being 1, and that leads to the determinant equation. BTW, I think there is a sign error in the determinant equation. $\endgroup$ – hbp Oct 14 '15 at 19:47
  • $\begingroup$ Second, the determinant $\Delta_n$ if truncated at the top for some large $n$, satisfies the recurrence relation, $\Delta_{n} = -D_n \, \Delta_{n-1} - \Delta_{n-2}$. We can see that equation satisfied by the determinants are structurally very similar to the equation satisfied by $c_n$. Now for the first few ones, $\Delta_1=-D_1$, $\Delta_2 = D_1 D_2 - 1$, $\Delta_3 = -D_1 \, D_2 \, D_3 + D_1 + D_3$, etc. So $\Delta_n$ is a polynomial of $\mu$. I suppose it asymptotically is dominated by $(-1)^n D_1 \cdots D_n$. This is why the authors divided the recurrence equation of $c_n$ by $-D_n$. $\endgroup$ – hbp Oct 14 '15 at 20:00
  • $\begingroup$ Once we have divided the $\Delta_n(\mu)$ by $(-1)^n D_1 \cdots D_n$, we can guess the form of the function. To do this we seek the zeros or poles of the function. An analogue is this. If we know that a function $f(\mu)$ has poles at $x = \pm\pi, \pm2\pi,\dots$ and it approaches $1$ at $\mu = 0$. Then we can guess that $f(\mu) = \prod_{n \ne 0}(1-\frac{\mu}{n\,\pi})^{-1}=\frac{\mu}{\sin\mu}$. I think that the authors are trying to do something similar, and try to show $\frac{\Delta(\mu)}{(-1)^nD_1\cdots D_n}$ and $1/[\cos(i\pi\mu) - \cos(\pi\sqrt{a})]$ are the same by Liouville's theorem. $\endgroup$ – hbp Oct 14 '15 at 20:12
  • $\begingroup$ @hbp Thanks, I corrected the sign. So basically, in general one has to divide by the asymptotically dominating factor, and this will give rise to poles? $\endgroup$ – Rol Oct 18 '15 at 18:55
  • $\begingroup$ Thanks for the updates. Your observation is correct. In your Fibonacci example, we indeed have an infinite determinant. This is undesirable. But observe that if the diagonal elements are zeros, then the recurrence will be $\Delta_n = \Delta_{n-2}$, and the determinant will be finite (unfortunately, it will oscillate between 0 and 1). So at least for the odd $n$, we have a solution. This feature is general. By setting the parameter $\mu$ to some special value, the diagonal would assume a set lucky numbers to make the determinant finite, hence delivering a solution. $\endgroup$ – hbp Oct 19 '15 at 22:21

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