Let’s consider the following system:
\[m_k = \sum_{j=1}^N a_j x_j^k, \quad k \in \{0, 1, \ldots, 2N-1\}.\]
Introduce variables \(\{u_j\}\):
\[\begin{cases} u_1 = x_1 + x_2 + \ldots + x_N, \\ u_2 = x_1x_2 + x_1x_3 + \ldots + x_{N-1}x_N, \\ \ldots \\ u_N = x_1x_2\ldots x_N. \end{cases}\]
From the Viet’s theorem, we have:
\[\label{eq3} m_{l} = m_{l-1}u_1 - m_{l-2}u_2 + \ldots + (-1)^{N-1}m_{l-N} u_N.\]
Thus, \(\{u_j\}\) are solution of the following system of equations:
\[\begin{pmatrix} m_{N-1} & -m_{N-2} & \ldots & (-1)^{N-1} m_0 \\ m_{N} & -m_{N-1} & \ldots & (-1)^{N-1} m_1 \\ \vdots & \vdots & \ddots & \vdots \\ m_{2N-2} & -m_{2N-3} & \ldots & (-1)^{N-1} m_{N-1} \end{pmatrix} \times \begin{pmatrix} u_1 \\ u_2 \\ \vdots \\ u_N \end{pmatrix} = \begin{pmatrix} m_N \\ m_{N+1} \\ \vdots \\ m_{2N-1} \\ \end{pmatrix}.\]
We can find \(\{x_j\}\) from the following equation:
\[x^N - u_1 x^{N-1} + \ldots + (-1)^N u_N = 0.\]
We can find \(\{a_j\}\) as a solution of the following system:
\[\begin{pmatrix} 1 & 1 & \ldots & 1 \\ x_1 & x_2 & \ldots & x_d \\ x_1^2 & x_2^2 & \ldots & x_N^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{N-1} & x_2^{N-1} & \ldots & x_N^{N-1} \\ \end{pmatrix} \times \begin{pmatrix} a_1 \\ a_2 \\ a_3\\ \vdots \\ a_N \end{pmatrix} = \begin{pmatrix} m_0 \\ m_1 \\ m_2 \\ \vdots \\ m_{N-1} \end{pmatrix}.\]