$$\begin{aligned}
(c^{\mathrm{ext}})^\alpha_{\beta\gamma}=&\sum_{\mu=1}^N\eta_{D_N}^{\alpha\mu} \frac{\d^3 F_{D_N}}{\d t^\mu\d t^\beta\d t^\gamma},&& 1\le\alpha\le N,&& 1\le\beta,\gamma\le N+1,\label{eq:first property for DN}\\
(c^{\mathrm{ext}})^{N+1}_{\alpha\beta}=&\frac{\d^2 F^o_{D_N}}{\d t^\alpha\d t^\beta},&&1\le\alpha,\beta\le N+1.&&\label{eq:second property for DN}\end{aligned}$$ Since the subspace ${\mathbb C}[v_1,\ldots,v_N,v_{N+1},v_{N+1}^{-1}]{\left<}[w]{\right>}$ is an ideal in the ring ${{\widehat{\mathcal{A}}}}^{{\mathrm{ext}}}_{D_N}$ and the quotient by this ideal coincides with the ring ${{\widehat{\mathcal{A}}}}_{D_N}$, we have $$(c^{\mathrm{ext}}_v)^a_{b,c}=
\begin{cases}
(c_v)^a_{b,c},&\text{if $1\le a,b,c\le N$},\\
0,&\text{if $1\le a\le N$ and one of the indices $b,c$ is equal to $N+1$}.
\end{cases}$$ This implies equation  and it remains to prove .

We have two substantially different cases: the case $\alpha\in\{N,N+1\}$ or $\beta\in\{N,N+1\}$ and the case $\alpha,\beta\in\{1,2,\dots,N-1\}$.

[*Case $\alpha\in\{N,N+1\}$ or $\beta\in\{N,N+1\}$*]{}. If $\alpha,\beta\in\{N,N+1\}$, then $(c^{\mathrm{ext}})^{N+1}_{\alpha,\beta}=(c^{\mathrm{ext}}_v)^{N+1}_{\alpha,\beta}$ and $\frac{\d^2 F^o_{D_N}}{\d t^\alpha\d t^\beta}=\frac{\d^2 F^o_{D_N}}{\d v_\alpha\d v_\beta}$. The equation $(c^{\mathrm{ext}}_v)^{N+1}_{\alpha,\beta}=\frac{\d^2 F^o_{D_N}}{\d v_\alpha\d v_\beta}$ immediately follows from formulas ,  and .

If $1\le\alpha\le N-1$ and $\beta=N+1$, then $$\begin{gathered}
(c^{\mathrm{ext}})^{N+1}_{\alpha,N+1}=\sum_{k=1}^{N-1}\frac{\d v_k}{\d t^\alpha}(c^{\mathrm{ext}}_v)^{N+1}_{k,N+1}\stackrel{\text{eq.\eqref{eq:xy for DN}}}{=}\sum_{k=1}^{N-1}\frac{\d v_k}{\d t^\alpha}\frac{v_{N+1}^{2k-2}}{2^{k-1}}=\frac{\d}{\d t^\alpha}\frac{\d F^o_{D_N}}{\d v_{N+1}}=\frac{\d^2 F^o_{D_N}}{\d t^\alpha\d t^{N+1}}.\end{gathered}$$

If $1\le\alpha\le N-1$ and $\beta=N$, then $$\begin{gathered}
(c^{\mathrm{ext}})^{N+1}_{\alpha,N}=\sum_{k=1}^{N-1}\frac{\d v_k}{\d t^\alpha}(c^{\mathrm{ext}}_v)^{N+1}_{k,N}\stackrel{\text{eq.\eqref{eq:xy for DN}}}{=}0=\frac{\d}{\d t^\alpha}\frac{\d F^o_{D_N}}{\d v_N}=\frac{\d^2 F^o_{D_N}}{\d t^\alpha\d t^N}.\end{gathered}$$

[*Case $\alpha,\beta\in\{1,2,\ldots,N-1\}$*]{}. We have to check that $$\begin{gathered}
(c^{\mathrm{ext}})^{N+1}_{\alpha\beta}=\frac{\d^2 F^o_{D_N}}{\d t^\alpha\d t^\beta} \Leftrightarrow \sum_{1\le a,b\le N-1}\frac{\d v_a}{\d t^\alpha}\frac{\d v_b}{\d t^\beta}(c^{\mathrm{ext}}_v)^{N+1}_{a,b}=\sum_{k=1}^{N-1}\frac{\d^2 v_k}{\d t^\alpha\d t^\beta}\frac{v_{N+1}^{2k-1}}{2^{k-1}(2k-1)},\end{gathered}$$ that is equivalent to the equation $$\begin{gathered}
\label{eq:equation for cab for DN}
(c^{\mathrm{ext}}_v)^{N+1}_{a,b}=-\sum_{k=1}^{N-1}\frac{\d v_k}{\d t^\gamma}\frac{\d^2 t^\gamma}{\d v_a\d v_b}\frac{v_{N+1}^{2k-1}}{2^{k-1}(2k-1)},\quad 