can be equivalently written as $$\begin{aligned}
&\sum_{k=0}^{a+b-N-1} \omega_k \frac{v_{N+1}^{2(a+b)-2N-1-2k}}{2^{a+b-N-k}}=-\sum_{k=1}^{N-1}\frac{\d v_k}{\d t^\gamma}\frac{\d^2 t^\gamma}{\d v_a\d v_b}\frac{v_{N+1}^{2k-1}}{2^{k-1}(2k-1)}\Leftrightarrow\\
\Leftrightarrow&\sum_{k=1}^{N-1} \omega_{a+b-N-k} \frac{v_{N+1}^{2k-1}}{2^k}=-\sum_{k=1}^{N-1}\frac{\d v_k}{\d t^\gamma}\frac{\d^2 t^\gamma}{\d v_a\d v_b}\frac{v_{N+1}^{2k-1}}{2^{k-1}(2k-1)}.\end{aligned}$$ So we have to prove that $$\begin{gathered}
\label{eq:identity for DN}
\frac{\d v_k}{\d t^\gamma}\frac{\d^2 t^\gamma}{\d v_a\d v_b}=-\frac{2k-1}{2}\omega_{a+b-N-k}\Leftrightarrow \frac{\d^2 t^\gamma}{\d v_a\d v_b}=-\sum_{k=1}^{N-1}\frac{2k-1}{2}\omega_{a+b-N-k}\frac{\d t^\gamma}{\d v_k}.\end{gathered}$$ Recall that $t^\gamma(v_1,\ldots,v_{N-1})$ is a quasi-homogeneous polynomial of degree $N-\gamma$, if we put $\deg v_a=N-a$. This implies that both sides of the last equation in  are zero if $a+b\le N$. Let us assume now that $a+b\ge N+1$. The last equation in  is equivalent to $$\frac{\d^2 t^\gamma}{\d v_a\d v_b}-\sum_{i=1}^{a-1}s_{N-i}\frac{\d^2 t^\gamma}{\d v_{a-i}\d v_b}=-\sum_{k=1}^{N-1}\frac{2k-1}{2}\frac{\d t^\gamma}{\d v_k}\left(\omega_{a+b-N-k}-\sum_{i=1}^{a-1}s_{N-i}\omega_{a-i+b-N-k}\right).$$ Note that for $i\ge a$ we have $a-i+b-N-k<0$ and, therefore, by , the expression in the brackets is equal to zero unless $k=a+b-N$. So we come to the following equivalent identity: $$\begin{gathered}
\label{eq:main identity for DN}
\frac{\d^2 t^\gamma}{\d v_a\d v_b}-\sum_{i=1}^{a-1}s_{N-i}\frac{\d^2 t^\gamma}{\d v_{a-i}\d v_b}=-\frac{2(a+b-N)-1}{2}\frac{\d t^\gamma}{\d v_{a+b-N}},\quad 
\begin{array}{@{}c@{}} 1\le a,b,\gamma\le N-1, \\  a+b\ge N+1. \end{array}\end{gathered}$$

Note that both sides of are quasi-homogeneous polynomials of degree $a+b-\gamma-N$. Differentiating both sides by $\frac{\d^{\sum\alpha_i}}{\d v_1^{\alpha_1}\ldots \d v_{N-1}^{\alpha_{N-1}}}$, putting $v_j=0$ and using formula , wee see that equation  is equivalent to the following family of identities: $$\begin{aligned}
&-\frac{1}{2}A\left(2\gamma-1+2(N-1)\sum\alpha_i\right)+A\sum_{i=1}^{a-1}(N-i-1)\alpha_{N-i}=-A\frac{2(a+b-N)-1}{2}\Leftrightarrow\notag\\
\Leftrightarrow & A\left(-\gamma-(N-1)\sum\alpha_i+\sum_{i=1}^{a-1}(N-i-1)\alpha_{N-i}+a+b-N\right)=0,\label{eq:final identity for DN}\end{aligned}$$ that should be true for 