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At first, we sample $f(x)$ in the $N$ ( $N$ is odd) equidistant points around $x^*$ :

$f_k = f(x_k),\: x_k = x^*+kh,\: k=-\frac{N-1}{2},\dots,\frac{N-1}{2}$

where $h$ is some step.
Then we interpolate points $\{(x_k,f_k)\}$ by polynomial

(1) $\begin{equation*} P_{N-1}(x)=\sum_{j=0}^{N-1}{a_jx^j}\end{equation*}$

Its coefficients $\{a_j\}$ are found as a solution of system of linear equations:

(2) $\begin{equation*} \left\{ P_{N-1}(x_k) = f_k\right\},\quad k=-\frac{N-1}{2},\dots,\frac{N-1}{2}\end{equation*}$

Here are references to existing equations: (1), (6).
Here is reference to non-existing equation (??).

(3) $\begin{equation*} \left\{ P_{N-1}(x_k) = f_k\right\},\quad k=-\frac{N-1}{2},\dots,\frac{N-1}{2}\end{equation*}$

(4) $\begin{equation*} \left\{ P_{N-1}(x_k) = f_k\right\},\quad k=-\frac{N-1}{2},\dots,\frac{N-1}{2}\end{equation*}$

(5) $\begin{equation*} \left\{ P_{N-1}(x_k) = f_k\right\},\quad k=-\frac{N-1}{2},\dots,\frac{N-1}{2}\end{equation*}$

(6) $\begin{equation*} r = \frac{\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})}{\sqrt[]{\sum_{i=1}^{n}(x_i - \bar{x})^2 \sum_{i=1}^{n}(y_i - \bar{y})^2}}\end{equation*}$

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Matrix

[katex display=true] 
\begin{bmatrix}
1 & 2 & 3\\
a & b & c
\end{bmatrix}
[/katex]

Test Latex

Matrix

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