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This is a text to show that \(x^*\) and for equations in one line
\begin{equation*} \frac{1}{2} \end{equation*}
test1
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
\begin{equation} \label{eq:poly}
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:
\begin{equation} \label{eq:sys}
\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: (\ref{eq:poly}), (\ref{eq:sys}).
Here is reference to non-existing equation (\ref{eq:unknown}).
\begin{align*} \frac{1}{2} \end{align*}
$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$
\begin{equation*}
1 + 2 = 3
\end{equation*}
\begin{equation*}
1 = 3 – 2
\end{equation*}
\begin{align*}
1 + 2 &= 3\\
1 &= 3 – 2
\end{align*}
\begin{align*}
f(x) &= x^2\\
g(x) &= \frac{1}{x}\\
F(x) &= \int^a_b \frac{1}{3}x^3
\end{align*}
\begin{matrix}
1 & 0\\
0 & 1
\end{matrix}
\begin{align*}
\frac{1}{\sqrt{x}}
\end{align*}
\begin{align*}
\frac{1}{\sqrt{x}}
\end{align*}
The well known Pythagorean theorem \(x^2 + y^2 = z^2\) was
proved to be invalid for other exponents.
Meaning the next equation has no integer solutions:
\[ x^n + y^n = z^n \]