Calculus and Mathmatics Analysis
Example 1
$\large \begin{array}{rl} Newton-Leibniz\ formula \\
\int_{a}^{b}{f'(x) dx} =& f(b) - f(a) \\
=& \underset{n \rightarrow \infty}{\lim} \overset{ n }{\underset{k=1}{\sum}} { ( f'(x_k) \cdot \Delta{x_k} ) }, 黎曼和形式\\
=& \underset{n \rightarrow \infty}{\lim} \overset{ n }{\underset{k=1}{\sum}} { \Delta{f(x_k)} },\ 無窮微分劃分形 \\
\end{array}$
\(\large \begin{array}{rl} Newton-Leibniz\ formula \\
\int_{a}^{b}{f'(x) dx} =& f(b) - f(a) \\
=& \underset{n \rightarrow \infty}{\lim} \overset{ n }{\underset{k=1}{\sum}} { ( f'(x_k) \cdot \Delta{x_k} ) }, 黎曼和形式\\
=& \underset{n \rightarrow \infty}{\lim} \overset{ n }{\underset{k=1}{\sum}} { \Delta{f(x_k)} },\ 無窮微分劃分形 \\
\end{array}\)
Example 2
$$\large e = \lim_{\Delta X \to 0} \frac{ (\frac{\Delta Y}{Y}) } { (\frac{\Delta X}{X}) } = \frac{ (\frac{dY}{Y}) }{ (\frac{dX}{X}) } = (\frac{dY}{dX}) * (\frac{X}{Y})$$
hypothesis $ \large Y = f(X) $:
- 弧彈性係數: 兩點\((X_0, Y_0)\) 與 \((X_1, Y_1)\) 之間的彈性係數:\[\large e = \frac{ (\frac{\Delta Y}{Y}) } { (\frac{\Delta X}{X}) } = \frac{ (\frac{\Delta Y}{\Delta X}) }{ (\frac{Y}{X}) } \]通常\(\large (X, Y)\)取\(\large (X=\frac{X_0+X_1}{2}, Y=\frac{Y_0+Y_1}{2})\)
- 點彈性係數, 點\((X, Y)\)處的彈性係數, 即當 \(\Delta X \to 0\) 並且 \(\Delta Y \to 0\)時:\[\large e = \lim_{\Delta X \to 0} \frac{ (\frac{\Delta Y}{Y}) } { (\frac{\Delta X}{X}) } = \frac{ (\frac{dY}{Y}) }{ (\frac{dX}{X}) } = (\frac{dY}{dX}) * (\frac{X}{Y}) \]
Matrix Analysis and Advanved Linear Algebra
矩陣可以用bmatrix Environment和pmatrix Environment，分別為方括號和圓括號，例如
$$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$$
效果是：
如果要輸入行列式的話，可以使用vmatrix Environment，用法同上。
Probability and Statistics
\sim: \(\large \sim\)
$\large X_{\theta_1} \sim {\mathcal {N}}(\mu_1 ,\sigma_1 ^{2})$, \(\large X_{\theta_1} \sim {\mathcal {N}}(\mu_1 ,\sigma_1 ^{2})\)
$\large X_{\theta_2} \sim {\mathcal {N}}(\mu_2 ,\sigma_2 ^{2})$, \(\large X_{\theta_2} \sim {\mathcal {N}}(\mu_2 ,\sigma_2 ^{2})\)
parameter vector is$\large \theta = [\mu, \sigma^{2}]$, \(\large \theta = [\mu, \sigma^{2}]\)
Example Population Parameters
設總體 $\large X$ 容量為 $\large N$, 個體取值為$\large \{x _i \},\ i \in [1, N]$, 定義：
$\large \begin{array}{ll} \\
\text{ population }mean : & \mu = \dfrac{1}{N} \overset{N}{\underset{i =1}{\sum}}{ x_i } \\
\text{ population }total : & \tau = \dfrac{1}{N} \overset{N}{\underset{i =1}{\sum}}{ x_i } = N \cdot \mu \\
\text{ population }variance : & \sigma^2 = \dfrac{1}{N} \overset{N}{\underset{i =1}{\sum}}{ (x_i - \mu)^2 } \\
\text{ population }standard\ deviation : & s = \sqrt{ \sigma^2 } \\
\end{array}$
設總體 \(\large X\) 容量為 \(\large N\), 個體取值為\(\large \{x _i \},\ i \in [1, N]\), 定義：
\(\large \begin{array}{ll} \\
\text{ population }mean : & \mu = \dfrac{1}{N} \overset{N}{\underset{i =1}{\sum}}{ x_i } \\
\text{ population }total : & \tau = \dfrac{1}{N} \overset{N}{\underset{i =1}{\sum}}{ x_i } = N \cdot \mu \\
\text{ population }variance : & \sigma^2 = \dfrac{1}{N} \overset{N}{\underset{i =1}{\sum}}{ (x_i - \mu)^2 } \\
\text{ population }standard\ deviation : & s = \sqrt{ \sigma^2 } \\
\end{array}\)
Example 1
$\large \begin{array}{rrll} \\
\bm{ Population } & \bm{ Parameters } & \bm{ Statistics } & \bm{ Sample } \\
quantity(count)\ =& \bm{ N } & \bm{ n } &=\ quantity (count) \\
mean\ =& \bm{ \mu } & \bm{ \overset{-}{x} } &=\ mean \\
variance \ =& \bm{ \sigma^2 } & \bm{ s^2 } &=\ variance \\
standard\ deviation \ =& \bm{ \sigma } & \bm{ s } &=\ standard\ deviation \\
\end{array}$
$\large \begin{array}{rrll} \\ \bm{ Population } & \bm{ Parameters } & \bm{ Statistics } & \bm{ Sample } \\ quantity(count)\ =& \bm{ N } & \bm{ n } &=\ quantity (count) \\ mean\ =& \bm{ \mu } & \bm{ \overset{-}{x} } &=\ mean \\ variance \ =& \bm{ \sigma^2 } & \bm{ s^2 } &=\ variance \\ standard\ deviation \ =& \bm{ \sigma } & \bm{ s } &=\ standard\ deviation \\ \end{array}$