degree $n$ and all its roots are pairwise distinct. Moreover, the regular system has a unique solution.

This theorem immediately implies the following proposition about regular solvability of the interpolation problem (\[2n-interpolation\]) for a function $f$, being analytic in a neighbourhood of the origin.

\[th1\] Suppose that the generating polynomial $G_n$, constructed using the numbers $s_m=s_m(h,f)$, $m=\overline{0,2n-1}$, is of degree $n$ and all its roots are pairwise distinct. Then the amplitude and frequency operator $H_n$ is uniquely determined from the system $(\ref{SRS})$ and realizes the $2n$-multiple interpolation $(\ref{2n-interpolation})$ of the function $f$ at the node ${z=0}$.

\[remark\_even\_odd\] If the function $f$ is even or odd, then the local precision of the interpolation can be increased. Indeed, if $f$ is even, then $f(z)=\tilde{f}(t)$, $t=z^2$, for some function $\tilde{f}$, analytic at the point $t=0$, and the interpolation (\[2n-interpolation\]) under the assumptions of Theorem \[th1\] with the basic function $h$ gives $$\label{interpolaton_even}
\tilde{f}(t)=\sum_{k=1}^n\mu_kh(\lambda_kt)+O(t^{2n})\quad
\Leftrightarrow \quad
f(z)=\sum_{k=1}^n\mu_kh(\lambda_kz^2)+O(z^{4n}),$$ where it is necessary that $f_{2m}\neq 0 \Rightarrow h_m\neq 0$, see (\[ssmm\]). If $f$ is odd, then $f(z)=z\tilde{f}(t)$, $t=z^2$, and analogously $$f(z)=\sum_{k=1}^n\mu_kzh(\lambda_kz^2)+O(z^{4n+1}).$$

Note that, in contrast to similar discussions in [@YF Corollary  2], here we do not require the functions $f$ and $h$ to be of the 