x is a scalar.
$\vec{w} = \begin{bmatrix}w_0 & w_1 & w_2& ... & w_M\end{bmatrix}^T$
$\begin{matrix}A_{ij} = \sum \limits_{n=1}^{N} (x_n)^{i+j}&&0\le i \le M &&0\le j \le M \end{matrix}$
$\begin{matrix}T_i = \sum \limits_{n=1}^{N} (x_n)^i~ t_n&&0\le i \le M \end{matrix}$
given:
$\sum \limits_{n=1}^{N} \sum \limits_{j=0}^{M} \bigg(w_j ~x_n^{i+j}\bigg) - \sum \limits_{n=1}^{N} t_n~x^i=0$
prove that this equation is equal to matrix equation:
$A\vec{w} = \vec{T}$
the index i indicates the row or equation number in the system of equations.
I can kind of see that the second term ($\sum \limits_{n=1}^{N} t_n~x^i$) is equivalent to the vector $\vec{T}$, and the first term ($w_j ~x_n^{i+j}$) is some type of product of matrix A and vector $\vec{w}$. I'm not really sure how to factor first term to make this more obvious that its a product of A and $\vec{w}$
hints:
$z_i=\sum \limits_{k=1}^{n}A_{ik}~x_{k}$
is equivalent to:
$\vec{z}=\mathbf{A}\vec{x}$
we can also drop the summations of j and use Einstein notation, in that case:
$z_i=A_{ik}~x_k$
is equivalent to:
$\vec{z}=\mathbf{A}\vec{x}$
$\sum \limits_{n=1}^{N} \sum \limits_{j=0}^{M} \bigg(w_j ~x_n^{i+j}\bigg) - \sum \limits_{n=1}^{N} t_n~x^i=0$
$\sum \limits_{n=1}^{N} \sum \limits_{j=0}^{M} \bigg(w_j ~x_n^{i+j}\bigg) = \sum \limits_{n=1}^{N} t_n~x^i$
$\sum \limits_{n=1}^{N} \sum \limits_{j=0}^{M} \bigg(w_j ~x_n^{i+j}\bigg) = T_i$
$\sum \limits_{j=0}^{M} \sum \limits_{n=1}^{N} \bigg(w_j ~x_n^{i+j}\bigg) = T_i$
$\sum \limits_{j=0}^{M} \bigg(w_j ~\sum \limits_{n=1}^{N} x_n^{i+j}\bigg) = T_i$
$\sum \limits_{j=0}^{M} \bigg(w_j ~A_{ij}\bigg) = T_i$
$\sum \limits_{j=0}^{M} \bigg(A_{ij}~ w_j \bigg) = T_i$
Converting to Einstein notation:
$A_{ij}~ w_j = T_i$
$A \vec{w} = \vec{T}$