WikiSandBox 뜻(meaning)

set noautoscale
set yrange [0:3]
set xrange [0:3]
plot sqrt(x), x*x, log(x)

RC회로
축전기 양 끝의 전위차
$V(t)=\mathcal{E}_0(1-e^{-\frac{t}{RC}})$

discharging:
$q=q_0e^{-\frac{t}{RC}}$
$V=V_0e^{-\frac{t}{RC}}$

limits vs nolimits
$\limits\int_a^bf(x)dx$
$\nolimits\int_a^bf(x)dx$

$|\limits_a^b$
$|\nolimits_a^b$




영인자만들기:
$\begin{pmatrix}a&a\\b&b\end{pmatrix}\begin{pmatrix}x&-y\\-x&y\end{pmatrix},\; \begin{pmatrix}1&1\\-1&-1\end{pmatrix}\begin{pmatrix}1&1\\-1&-1\end{pmatrix}$

TeX의 그리스 문자


단순히

대문자 명칭 나열:
$\Alpha \Beta \Gamma \Delta \Epsilon \Zeta \Eta \Theta \Iota \Kappa \Lambda \Mu \Nu \Xi \Omicron \Pi \Rho \Sigma \Tau \Upsilon \Phi \Chi \Psi \Omega$
소문자 명칭 나열:
$\alpha \beta \gamma \delta \epsilon \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi \omicron \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega$

TeX는 그리스 문자를 표현할 때 로마자와 모양이 같은 것을 재활용한다. 따라서 그리스 대문자 목록을 표현하려면
$A B \Gamma \Delta E Z H \Theta I K \Lambda M N \Xi O \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega$
소문자 목록은
$\alpha \beta \gamma \delta \epsilon \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi o \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega$
var- 소문자 목록은
$\epsilon-\varepsilon/\theta-\vartheta/\pi-\varpi/\rho-\varrho/\sigma-\varsigma/\phi-\varphi$


표준 (italic)
$ABCDEFGHIJKLMNOPQRSTUVWXYZ abcdefghijklmnopqrstuvwxyz 0123456789$
$A B \Gamma \Delta E Z H \Theta I K \Lambda M N \Xi O \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega \alpha \beta \gamma \delta \epsilon \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi o \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega \varepsilon\vartheta\varpi\varrho\varsigma\varphi$
을 다음과 같이 다양하게 표현을 시도, 아래 렌더링에서 어떤 꼴이 표현 지원되거나 지원되지 않는 지 확인 가능

mathcal:
$\mathcal{ABCDEFGHIJKLMNOPQRSTUVWXYZ abcdefghijklmnopqrstuvwxyz 0123456789}$
$\mathcal{A B \Gamma \Delta E Z H \Theta I K \Lambda M N \Xi O \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega \alpha \beta \gamma \delta \epsilon \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi o \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega \varepsilon\vartheta\varpi\varrho\varsigma\varphi}$

mathrm:
$\mathrm{ABCDEFGHIJKLMNOPQRSTUVWXYZ abcdefghijklmnopqrstuvwxyz 0123456789}$
$\mathrm{A B \Gamma \Delta E Z H \Theta I K \Lambda M N \Xi O \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega \alpha \beta \gamma \delta \epsilon \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi o \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega \varepsilon\vartheta\varpi\varrho\varsigma\varphi}$

mathbb:
$\mathbb{ABCDEFGHIJKLMNOPQRSTUVWXYZ abcdefghijklmnopqrstuvwxyz 0123456789}$
$\mathbb{A B \Gamma \Delta E Z H \Theta I K \Lambda M N \Xi O \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega \alpha \beta \gamma \delta \epsilon \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi o \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega \varepsilon\vartheta\varpi\varrho\varsigma\varphi}$

mathbf:
$\mathbf{ABCDEFGHIJKLMNOPQRSTUVWXYZ abcdefghijklmnopqrstuvwxyz 0123456789}$
$\mathbf{A B \Gamma \Delta E Z H \Theta I K \Lambda M N \Xi O \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega \alpha \beta \gamma \delta \epsilon \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi o \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega \varepsilon\vartheta\varpi\varrho\varsigma\varphi}$

mathscr:
$\mathscr{ABCDEFGHIJKLMNOPQRSTUVWXYZ abcdefghijklmnopqrstuvwxyz 0123456789}$
$\mathscr{A B \Gamma \Delta E Z H \Theta I K \Lambda M N \Xi O \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega \alpha \beta \gamma \delta \epsilon \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi o \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega \varepsilon\vartheta\varpi\varrho\varsigma\varphi}$

mathsf, mathtt는 전혀 지원 안하는 듯

delme 1

$\frac{\rm d}{{\rm d}x}\sin x=\cos x$
Pf.
$\frac{\sin(x+h)-\sin x}{h}=\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$
$=\sin x\left(\frac{\cos h-1}{h}\right)+\cos x\left(\frac{\sin h}{h}\right)$
$h\to 0$ 이면
$=\sin x \cdot 0 + \cos x \cdot 1$
$=\cos x$

$\frac{\operatorname{d}}{\operatorname{d}x}\cos x=-\sin x$
Pf.
$\frac{\cos(x+h)-\cos x}{h}=\frac{\cos x\cos h-\sin x\sin h-cos x}{h}$
$=\cos x\left(\frac{\cos h-1}{h}\right)-\sin x\left(\frac{\sin h}{h}\right)$
$h\to 0$ 이면
$=\cos x\cdot 0 -\sin x \cdot 1$
$=-\sin x$

$\frac{\text{d}}{\text{d}x}$