Diagonal element of the resolvent of bi-infinite tridiagonal Laurent operator For $\alpha, \beta, \gamma \in \mathbb{C}$ consider the bi-infinite tridiagonal Laurent operator $T$  with $\beta $ on the diagonal given by.
\begin{pmatrix}
\dots &  \dots & \dots & \dots  & \dots & \dots & \dots & \dots \\ 
 \dots & \alpha & \beta & \gamma & 0 & 0 & 0 & \dots \\ 
\dots & 0 & \alpha & \beta & \gamma & 0 & 0 & \dots\\ 
\dots &  0 & 0 & \alpha & \beta  & \gamma & 0 & \dots\\ 
\dots & 0 & 0 & 0 & \alpha & \beta & \gamma &  \dots \\ 
\dots &  \dots & \dots & \dots  & \dots & \dots & \dots & \dots \\ 
\end{pmatrix}
General theory tells us that $T$ is invertible if and only the symbol curve (which is in this case is an ellipsis) given by
$
\{ z \in \mathbb{T} \mid \frac{\alpha}{z} + \beta + z \gamma \} 
$
does not enclose $0$.
Suppose that this is the case and let $e_0$ denote the a unit vector in the standard basis. Then what is the value
$
\langle e_0 , T^{-1} e_0 \rangle ? 
$
 A: As we explain in the Appendix here: https://arxiv.org/abs/2206.09879
this is a standard calculation:
"First, $\sigma(T)$ is the image of the symbol curve
$
 a(z) =  \alpha z^{-1} + \beta +  \gamma z $
for $z \in \mathbb{T}$. Since $T$ is invertible it holds that $ a(z) \neq 0$ for all $z \in \mathbb{T}$  and therefore, the symbol curve of the inverse is given by
\begin{align*}
 \frac{1}{a(z)} =  \frac{1}{ \frac{\alpha}{ z}  + \beta +  \gamma z }  =   \frac{z}{  \alpha  + \beta z  +  \gamma z^2} = \frac{z}{ \gamma(   \frac{\alpha}{\gamma}   +  \frac{\beta}{\gamma}  z  +   z^2) }.
 \end{align*}
We can rewrite the denominator $ \gamma ( z - \lambda_+) (z- \lambda_-) $ with
\begin{align*}
 \lambda_{\pm} = \frac{ - \beta}{2 \gamma} \pm \sqrt{  \left( \frac{\beta}{2 \gamma} \right)^2 -  \frac{\alpha }{\gamma} }.
 \end{align*}
Notice that
\begin{align*}
\lambda_+ \lambda_- = \frac{\alpha}{\gamma} \text{ ,  }
\lambda_+ + \lambda_- = - \frac{\beta}{\gamma},
\text{ and }
\lambda_+ -  \lambda_- =  2 \sqrt{  \left( \frac{\beta}{2 \gamma} \right)^2 -   \frac{\alpha }{\gamma} }.
\end{align*}
Now, assuming that $\vert \lambda_2 \vert < 1< \vert \lambda_1 \vert $ (where $\{\lambda_2, \lambda_1 \} = \{\lambda_+, \lambda_- \}$ has implications on how we write this up as a geometric series):
\begin{align*}
\frac{1}{a(z)} & =  \frac{z}{ \gamma ( z - \lambda_+) (z- \lambda_-) } = \frac{z}{ \gamma ( \lambda_1 - \lambda_2)} \left( \frac{1}{z-\lambda_1} -  \frac{1}{z-\lambda_2}  \right) \\ 
& = \frac{z}{ \gamma ( \lambda_1 - \lambda_2)} \left( - \frac{1}{\lambda_1} \frac{1}{1- \frac{z}{\lambda} } - \frac{1}{z}  \frac{1}{1-\frac{\lambda_2}{z} }  \right) \\
& = \frac{z}{ \gamma ( \lambda_1 - \lambda_2)} \left( - \frac{1}{\lambda_1} \sum_{n=0}^\infty (\frac{z}{\lambda_1})^n   - \frac{1}{z}  \sum_{n=0}^\infty (\frac{\lambda_2}{z})^n  \right) \\ 
& =  \frac{1}{ \gamma ( \lambda_2 - \lambda_1)} \left( \sum_{n=1}^\infty (\frac{z}{\lambda_1})^n   +   \sum_{n=0}^\infty (\frac{\lambda_2}{z})^n  \right).
\end{align*}"
Then one can read of coefficients to obtain the answer.
