Формули MS Word MathType
А це зовсім інший спосіб (не впевнений, що будемо його підтримувати) <wikitex> Let $Q$ be any finite set, and $\mathcal B=2^Q$ be the collection of the subsets of $Q$. Let $f:\mathcal B\rightarrow \mathbb R$ be a function assigning real numbers to the subsets of $Q$ and suppose $f$ satisfies the following conditions:
- (i) $f(A)\ge 0$ for all $A\subseteq Q$, $f(\emptyset)=0$,
- (ii) $f$ is monotone, i.e. if $A\subseteq B\subseteq Q$ then $f(A)\le f(B)$,
- (iii) $f$ is submodular, i.e. if $A$ and $B$ are different subsets of $Q$ then
$$ \eqno{(2)}
f(A)+f(B)\ge f(A\cap B) + f(A\cup B).
$$ </wikitex>
[math]\operatorname{erfc}(x) = \frac{2}{\sqrt{\pi}} \int_x^{\infty} e^{-t^2}\,dt = \frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n)!}{n!(2x)^{2n}}[/math]
[math]\sum\nolimits_{1}^{1sdfsd}{sdfg}\frac{-b\pm \sqrt{{{b}^{2}}-4asdfsdf\frac{1}{2}\sqrt{sdf\prod\limits_{234}^{sdf}{sdf}}c}}{2a}[/math]
[math]\frac{n!}{r!\left( n-r \right)!}\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\sqrt{{{b}^{2}}-4ac}\underset{x\to \infty }{\mathop{\lim }}\,\infty \bigcap\limits_{{}}{\Epsilon \sum\nolimits_{\frac{n!}{r!\left( n-r \right)!}}^{{}}{\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}}}[/math]
\[\fracШаблон:N! {{r!\left( {n - r} \right)!}}\frac{{ - b \pm \sqrt {{b^2} - 4ac} }} Шаблон:2a\sqrt {{b^2} - 4ac} \mathop {\lim }\limits_{x \to \infty } \infty \bigcap\limits_{} {{\rm E}\sum\nolimits_{\fracШаблон:N! {{r!\left( {n - r} \right)!}}}^{} {\frac{{ - b \pm \sqrt {{b^2} - 4ac} }} Шаблон:2a} } \]
