\label{eq:v0set}\\
= \begin{cases} 0, & \text{ if } j + k \text{ even} \\
2 a_j a_k, & \text{ if } j + k \text{ odd} \end{cases} \label{eq:v0case} \\
\intertext{where}
a_k = h_k(b), \quad k = 0, 1, \dotsc \label{eq:akdef}\end{gathered}$$ With $b > 0$ fixed, from now on we will use $(a_k)$ as in . By Lemmas \[lem:tjodd\] and \[lem:jgbehave\] for $n \geq N^*$ $$\begin{aligned}
T_j(n; v^0) {\equiv}T_j(n)& = 0 \, \, \, \, \text{ if } j \text{ odd;} \label{eq:orthooddevenext} \\
\intertext{in particular,}
T_1(n) &= 0 , \quad T_3(n) = 0. \label{eq:orthospec} \end{aligned}$$ To evaluate $T_2(n)$ we’ll sum up (we did it in in general setting) Cauchy integrals of functions $$(z - z_n) \cdot \frac{1}{z - z_m} \cdot v_{mk}^0 \cdot \frac{1}{z - z_k} \cdot v_{km}^{0} \cdot \frac{1}{z - z_m} \label{eq:vgdef}$$ If $m \neq n$ it is analytic for any $k$ so Cauchy integral is zero. If $m = n$ $$v_{mk}^0 = 0 \quad \text{ if } n - k \text{ is even.} \label{eq:orthodiff}$$ Therefore, by Lemma \[lem:jgbehave\], $j = 2$, with $z_n - z_k = 2(n-k)$, $$\begin{aligned}
T_2(n; v^0) {\equiv}T_2(n) & = \sum_{\substack{ k = 0 \\ k - n \text{ odd} }}^{\infty} \frac{v_{nk}^0 v_{kn}^0}{z_n - z_k} =  \sum_{\substack{ 