\Bigg]
\nonumber\\
&\times&
\Bigg[U_{N-1}(\tilde{z}_{\alpha}/2)T_{N+1}(\tilde{z}_{\alpha}/2)-N
\nonumber\\
&-&
i\eta\exp[i\tilde{k}_{\alpha}]\sin(\tilde{k}_{\alpha})[U_{N-1}(\tilde{z}_{\alpha}/2)]^2
\Bigg]^{-1},\end{aligned}$$ where $T_{m}(z)=\cos[m\cos^{-1}(z)]$ is the Chebyshev polynomial of the first kind. Further simplifications occur when these polynomials are expressed in their trigonometric forms. For instance, let’s calculate $|(\Delta \tilde{\mathcal{Z}}_{\alpha})|^2$, with $ \tilde{z}_{\alpha}/2=\cos( \tilde{\theta}_{\alpha})$ where $\tilde{\theta}_{\alpha}=\pi- \tilde{k}_{\alpha}$. We obtain $$\begin{aligned}
\label{abdz2}
\dfrac{|(\Delta \tilde{\mathcal{Z}}_{\alpha})|^2}{4}
=
\dfrac{\sum_{n,m=1}^{N}\epsilon_{n'}\epsilon_{m'}\sin^{2}(n' \tilde{\theta}_{\alpha})\sin^{2}(m'\tilde{\theta}^{*}_{\alpha})}
{|D(\tilde{z_{\alpha}})|^{2}}.\end{aligned}$$ Here $n'$ and $m'$ are respectively $N+1-n$ and $N+1-m$, and $D(\tilde{z_{\alpha}})$ is simply the quantity in the second bracket of Eq. (\[FPT2\]). Averaging releases one of the summation as the $\epsilon_{j}'$s are statistically independent-identically-distributed (i.i.d.) random variables. We simply have $$\begin{aligned}
\label{abdz3}
\langle|(\Delta \tilde{\mathcal{Z}}_{\alpha})|^2\rangle
=
\sigma^{2}\dfrac{\sum_{n=1}^{N}4\sin^{2}(n\tilde{\theta}_{\alpha})\sin^{2}(n\tilde{\theta}^{*}_{\alpha})}
{|D|^{2}},\end{aligned}$$ where $\sigma^{2}$ is variance of the $\epsilon_{j}'$s.

Summation in the above equality can be performed by using trigonometric identities. For instance, we first write $$\begin{aligned}
\label{abdz4}
4\sin^{2}(n\theta)\sin^{2}(n\theta^{*})
&=&
1-\cos(2n\theta)-\cos(2n\theta^{*})
\nonumber\\
&+&
\dfrac{\cos(4\,n\,\Re\{\theta\})+\cos(4\,i\,n\,\Im\{\theta\})}{2},\end{aligned}$$ and we use the summation formula $$\begin{aligned}
\label{abdz5}
\sum_{n=1}^{N}\cos(n\theta)
&=&
\dfrac{1}{2}
\left[
\dfrac{\sin[(N+1/2)\theta]}{\sin(\theta/2)}-1
\right].\end{aligned}$$ It turns out after some trigonometry that one can write the summation in a closed form. We find $$\begin{aligned}
\sum_{n=1}^{N}4\sin^{2}(n\theta)\sin^{2}(n\theta^{*})
&=&N+\dfrac{1}{2}-\dfrac{U_{2N}+U_{2N}^{*}}{2}+
\dfrac{T^{*}_{2N+2}T_{2N}-T_{2N+2}T^{*}_{2N}}{2[T^{*}_{2}-T_{2}]}.
\nonumber
\\\end{aligned}$$ Here the argument of the polynomials is $\tilde{z}_{\alpha}/2$ and for their complex conjugate it is $\tilde{z}^{*}_{\alpha}/2$. Finally, we write down finite-$N$ result for average of the absolute square of the 