LaTeX - zakladne prikazy

Pomoc - v prípade problémov s používaním fóra píšte sem

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Martin Sleziak
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Joined: Mon Jan 02, 2012 5:25 pm

LaTeX - zakladne prikazy

Post by Martin Sleziak »

Indexy a exponenty

Pokial exponent ci index obsahuje viac znakov, dam ho do kuceravych zatvoriek: {}

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$p=x_1^2+x_2^2$
$|A|=\sum_{\pi\in S_n} (-1)^{i(\pi)} a_{1,\pi(1)} a_{2,\pi(2)} \dots a_{n,\pi(n)}$
$a_nx^n+\dots+a_1x+a_0$
$p=x_1^2+x_2^2$
$|A|=\sum_{\pi\in S_n} (-1)^{i(\pi)} a_{1,\pi(1)} a_{2,\pi(2)} \dots a_{n,\pi(n)}$
$a_nx^n+\dots+a_1x+a_0$

Mnozinove zatvorky

Kedze {} maju v TeXu specialny vyznam, ak ich chcem pouzit ako mnozinove zatvorky, musim pridat backslash.

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\{x\in\mathbb R; x^2\le\sqrt2\}
$\{x\in\mathbb R; x^2\le\sqrt2\}$



Sumy
Niekedy vyzera lepsie, ked sumacny rozsah pisem nad/pod sumu

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$\sum\limits_{k=1}^n \binom nk = 2^n$
$\prod\limits_{k=1}^\infty e^{-1/k} \ge \prod\limits_{k=1}^\infty \left(1-\frac1k\right)$
$\sum\limits_{k=1}^n \binom nk = 2^n$
$\prod\limits_{k=1}^\infty e^{-1/k} \ge \prod\limits_{k=1}^\infty \left(1-\frac1k\right)$

Nerovnosti

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$a<b\le c$
$a>b\ge c$

$a<b\le c$
$a>b\ge c$

Zlomky

$\frac12 < \frac23$
$\frac{a^3-b^3}{a-b} = a^2+ab+b^2$

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$\frac12 < \frac23$
$\frac{a^3-b^3}{a-b} = a^2+ab+b^2$
Odmocniny

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$\sqrt{16}=4$
$\sqrt[3]8=2$
$\sqrt{16}=4$
$\sqrt[3]8=2$
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