Math

The Not So Short Introduction to LATEX2Eより

Add a squared and b squared to get c squared. or, using a mathematical approach: a^2 + b^2 + c^2

\TeX{} is pronounced as \tau\epsilon\chi
\times 100~m^3 of water
This comes from my \heartsuit

\[ a^2 + b^2 = c^2 \]

\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{k^2}= \frac{\pi^2}{6}

A d_{e_{e_p}} mathematical expression followed by a h^{i^{g^h}} expression. As opposed to a smashed \smash{d_{e_{e_p}}} expression followed by a \smash{h^{i^{g^h}}} expression.

\forall x \in \mathbf{R}:\qquad x^{2} \geq 0
x^{2} \geq 0\qquad\text{for all }x\in\mathbf{R}
x^{2} \geq 0\qquad \text{for all } x \in \mathbb{R}

\lambda,\xi,\pi,\theta,\mu,\Phi,\Omega,\Delta

p^3_{ij} \qquad m_\text{Knuth}
a^x+y \neq a^{x+y}\qquad e^{x^2} \neq {e^x}^2
\sqrt{x} \Leftrightarrow x^{1/2} \quad \sqrt[3]{2} \quad \sqrt{x^{2} + \sqrt{y}} \quad \surd[x^2 + y^2]
\Psi = v_1 \cdot v_2 \cdot \ldots \qquad n! = 1 \cdot 2\cdots (n-1) \cdot n
0.\overline{3} = \underline{\underline{1/3}}

\underbrace{\overbrace{a+b+c}^6 \cdot \overbrace{d+e+f}^9} _\text{meaning of life} = 42

f(x) = x^2 \qquad f'(x) = 2x \qquad f''(x) = 2
\hat{XY} \quad \widehat{XY} \quad \bar{x_0} \quad \bar{x}_0
\vec{a} \qquad \vec{AB} \qquad \overrightarrow{AB}

\[\lim_{x \rightarrow 0} \frac{\sin x}{x}=1\]
× %\DeclareMathOperator{\argh}{argh} %\DeclareMathOperator*{\nut}{Nut} \[3\argh = 2\nut_{x=1}\]
a\bmod b
 x\equiv a \pmod{b}
\[3/8 \qquad \frac{3}{8} \qquad \tfrac{3}{8} \]
In text style:
× 1\frac{1}{2}~hours \qquad 1\dfrac{1}{2}~hours

\[\sqrt{\frac{x^2}{k+1}}\qquad x^\frac{2}{k+1}\qquad \frac{\partial^2f} {\partial x^2} \]

Pascal’s rule is
\binom{n}{k} =\binom{n-1}{k} + \binom{n-1}{k-1}
f_n(x) \stackrel{*}{\approx} 1
数式の中央寄せはだめみたい
\sum_{i=1}^n \qquad \int_0^{\frac{\pi}{2}} \qquad \prod_\epsilon
× \sum^n_{\substack{0<i<n j\subseteq i}} P(i,j) = Q(i,j)
\sum^n_{\substack{0 \lt; i\text{<}n j\subseteq i}} P(i,j) = Q(i,j)
{a,b,c} \neq \{a,b,c\}
1 + \left(\frac{1}{1-x^{2}} \right)^3 \qquad \left. \ddagger \frac{~}{~}\right)
\Big((x+1)(x-1)\Big)^{2}
\big( \Big( \bigg( \Bigg( \quad \big\} \Big\} \bigg\} \Bigg\} \quad \big\| \Big\| \bigg\| \Bigg\| \quad \big\Downarrow \Big\Downarrow \bigg\Downarrow \Bigg\Downarrow

align と label はだめみたい。
$$\begin{align} f(x) &= (a+b)(a-b) \label{1} $= a^2-ab+ba-b^2 \tag{3.4} $= a^2+b^2 \tag{wrong} \end{align}$$

This is a reference to \eqref{1}.


\mathbf{X} = \left( \begin{array}{ccc} x_1 & x_2 & \ldots \\ x_3 & x_4 & \ldots \\ \vdots & \vdots & \ddots \end{array} \right)

$$\|x| = \left\{ \begin{array}{rl} -x & \text{if } x < 0\\ 0 & \text{if } x = 0\\ x & \text{if } x > 0
\end{array} \right. $$


\begin{equation*}
|x| = \left\{
\begin{array}{rl}
-x & \text{if } x < 0\\
0 & \text{if } x = 0\\
x & \text{if } x > 0
\end{array} \right.
\end{equation*}

\begin{matrix}
1 & 2 \\
3 & 4
\end{matrix} \qquad
\begin{bmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{bmatrix}

\int_1^2 \ln x \mathrm{d}x \qquad \int_1^2 \ln x \,\mathrm{d}x

\newcommand{\ud}{\,\mathrm{d
\int_a^b f(x)\ud x
}}


{}^{14}_{6}\text{C}
\qquad \text{versus} \qquad
{}^{14}_{\phantom{1}6}\text{C}

\Re \qquad \mathcal{R} \qquad \mathfrak{R} \qquad \mathbb{R} \qquad

R = \frac{
\sum_{i=1}^n (x_i-\bar{x})
(y_i- \bar{y})}
{\left[
\sum_{i=1}^n(x_i-\bar{x})^2
\sum_{i=1}^n(y_i-\bar{y})^2
}
\right]^{1/2


$\mu, M \qquad
\mathbf{\mu}, \mathbf{M}$
\qquad \boldmath{$\mu, M$}

$\mu, M \qquad
\boldsymbol{\mu}, \boldsymbol{M}$


\begin{proof}
Trivial, use
\[E=mc^2 \qedhere \]
\end{proof}
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