\alpha_k \mathbf{C}(k), \qquad
\mathbf{C}(k):=\left((r\beta_k)^{i+j-2}\right)_{i,j=1}^n,$$ where ${\operatorname{rank}}\mathbf{C}(k)=1$ (each next row is the previous one multiplied by $r\beta_k$). From this by the property (\[rank\_inequality\]) we obtain the required bound for ${\operatorname{rank}}\mathbf{R}$.

[**.2. Parameters of the regularization.**]{} In the problems under consideration the ranks of matrices $\left(s_{i+j-2}\right)_{i,j=1}^n$ are small. Consequently, taking into account Lemma \[ranks\], we will consider only the overall case ${\nu}=n$ to solve the regularization problem. Then the formula (\[f\_reg\]) has $2n$ summands (if $\sigma\ne 0$, $r\ne 0$), and in this sense the amplitude and frequency sums obtained have no advantage over the $h$-sums of order $2n$. However, an appropriate choice of $\alpha_k$, $\beta_k$, $\sigma$ and $r$ can essentially simplify the latter sum in (\[f\_reg\]). Indeed, let $p$ and $q$ be fixed non-zero complex numbers and $$\label{alpha-beta} \alpha_k=\beta_k=\exp \left(\frac{2\pi (k-1)
i}{n}\right),\qquad r=\left(\frac{q}{p}\right)^{1/n},\qquad
\sigma=\frac{p^2}{nq} r,\qquad k=\overline{1,n},$$ where the number $r$ is any of the $n$ values of the root. Then, as it can be easily seen, in (\[Var\_disc\_mom\_syst\]) we obtain $$\label{TAU} \tau_{n-1}=p,\quad \tau_{2n-1}=q;\qquad \tau_{m}=0 \quad
\hbox{   for other   }\quad  m.$$ Indeed, $$\tau_m=\frac{p^2}{nq} r^{m+1}\sum_{k=1}^n \exp\left({\frac{2\pi
(k-1) i}{n}(m+1)}\right),$$ where the sum of the exponents equals $n$ if $m+1$ is divisible by $n$ and zero if not. Consequently, $$\label{SVORACH} \sigma H_n(\{\alpha_k\},\{r\beta_k\},h;z)= p\,
h_{n-1}z^{n-1}+q\,h_{2n-1}z^{2n-1}+O(z^{2n}).$$ Thus, assuming 