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Cauchy–Bunyakovsky–Schwarz

For all sequences of real numbers \(a_i\) and \(b_i\), it holds that

\[\left(\sum_{i=1}^na_ib_i\right)^2\leq\left(\sum_{i=1}^na_i^2\right)\cdot\left(\sum_{i=1}^nb_i^2\right).\]

Equality holds if and only if \(a_i=kb_i\) for a non-zero constant \(k\in\mathbb{R}\).

This expression uses \$ to display a dollar sign: $\sqrt{$4}$

here is some more

\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

again?

\[\begin{bmatrix} a & b \\ c & d \end{bmatrix}\] \[\begin{definition}[Group] \label{def:group} \normalfont A group $\GG$ consist of a set and an operator $\circ$, and satisfies the following axioms: \begin{enumerate} \item[]\textbf{Closure:} $\forall a,b\in\GG$ \[a\circ b\in\GG.\] \item[]\textbf{Associativity:} $\forall a,b,c\in\GG$ \[a\circ(b\circ c)=(a\circ b)\circ c.\] \item[]\textbf{Neutral Element:} $\exists e\in\GG \text{ such that } \forall a \in\GG$ \[a\circ e=a=e\circ a.\] In German $e$ is often referred to as \emph{``Einselement"}, \emph{``Eins"} or simply $1$. \item[]\textbf{Inverse Element:} $\forall a\in\GG, \exists a^{-1}\in\GG$ \[a\circ a^{-1}=e=a^{-1}\circ a.\] \end{enumerate} \end{definition}\]

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