How to separate self-defining values from sigma? $$\sum_{k=1}^{m} \sum_{j=1}^{n} a_kx_j^{b_k+b_i} = \sum_{j=1}^{n} y_jx_j^{b_i}$$
What I need to do is solve $a$ for every $i$ given ($i$ is between 1 and $m$), so their result won't be composed of any other $a_i$. (They can contain any $b$ $x$ $y$.)
I am currently stuck here:
$$a_i = \frac{ \sum_{j=1}^{n} y_jx_j^{b_i} - \sum_{k=1}^{i-1} a_k\sum_{j=1}^{n} x_j^{b_k+b_i} - \sum_{k=i+1}^{m} a_k\sum_{j=1}^{n} x_j^{b_k+b_i} }{\sum_{j=1}^{n} x_j^{b_i+b_i}}$$
Notice that $a_i$ equal to expression(?) that contains any $a$ (Find $a_k$) except $a_i$, that means that it's indirectly containing itself. Can I get a hint?
I thought of somehow folding out only $- \sum_{k=i+1}^{m} a_k\sum_{j=1}^{n} x_j^{b_k+b_i}$ to something that doesn't contain any $a$, and then it's easily solvable from here, but how to start?
 A: The sigma notation is somewhat confusing. From the comments, you seem to be looking for a linear combination of $M$ powers, $a_1x^{b_1}+\ldots+a_Mx^{b_M}$, that is "closest" to $N$ points, $(x_1,y_1),\ldots,(x_N,y_N).$
$$
\left[ \begin{array}{lllll}
x_1^{b_1} & \ldots & x_1^{b_k} & \ldots & x_1^{b_M} \\
\vdots    & & & & \vdots \\
x_j^{b_1} & \ldots & x_j^{b_k} & \ldots & x_j^{b_M} \\
\vdots    & & & & \vdots \\
x_N^{b_1} & \ldots & x_N^{b_k} & \ldots & x_N^{b_M} 
\end{array} \right]
\left[ \begin{array}{l}
a_1 \\ \vdots \\ a_k \\ \vdots \\ a_M
\end{array} \right]\  \approx\  \left[\begin{array}{l}
y_1 \\ \vdots \\ y_j \\ \vdots \\ y_N
\end{array} \right]
$$
We can call the $N\times M$ matrix of powers $\mathbf{B};\  $ call the vector of $N$ unknowns $\mathbf{a}\  $ and the $M$ y values we are trying to approximate $\mathbf{y}.\  $ For example, row $j$ of $\mathbf{Ba}\approx\mathbf{y}$ is $\  y_j \approx a_1x_j^{b_1} + \ldots + a_kx_j^{b_K} + \ldots + a_Mx_j^{b_M}.\  $ If we take $\approx$ as a goal to minimize the $\ell_2$ distance from $\mathbf{Ba}$ to $\mathbf{y}$, then the linear least squares solution is given by:
$$
0 < |\mathbf{B^TB}| \implies\  \mathbf{a} = 
\left(\mathbf{B^TB}\right)^{-1}\mathbf{B^Ty}
$$
There are (many) "solutions" for $\mathbf{a}$ when the determinant, $|\mathbf{B^TB}|,$ is 0. We can avoid this difficulty by removing redundant collumns from $\mathbf{B}$ and providing enough points, $M\le N.\  $ Notice that when $N=M$, our solution reduces to an exact "approximation", $\  \mathbf{a} = \mathbf{B^{-1}y}.$
Computing the least squares regression in practice might use orthogonal projections for better numerical stability, while we used the so called "normal" equations. Some sort of least square, aka linear regression, support is provided in every major math toolkit.
