物理 & 數學

--- title: "物理 & 數學" path: "physics" ---  Physics === 動量 --- $P = mv\quad$ $v = a * \Delta T$ $P = mv$ , $F = ma \implies m = \frac{F}{a}$ $\implies P = \frac{F}{a} * v$ $\implies P = F * \Delta T$ ## 衝量 $\vec{J} = \vec{F} * \Delta t$ $\vec{J} = \Delta \vec{p} = m\Delta \vec{v}$ <center>在沒有受外力的情況下動量守恆，比能量守恆好用一點</center> (有外力就不能用嗎，有外力不是一樣動量守恆) $\implies$ 要加上衝量    ## 質心 **質心可以代表系統** $兩物體重m_1, m_2, 位置x_1, x_2 \\\implies 質心位置 = \displaystyle \frac{m_1 *　x_1 +　m_2 *　x_2}{m_1 + m_2}$ $兩物體重m_1, m_2　速度v_1, v_2 \\\implies 質心速度 =\displaystyle \frac{m_1 *　v_1 +　m_2 *　v_2}{m_1 + m_2}$ $兩物體重m_1, m_2　加速度a_1, a_2 \\\implies 質心加速度 =\displaystyle \frac{m_1 *　a_1 +　m_2 *　a_2}{m_1 + m_2}$ :::spoiler pf $\displaystyle v = \frac{\Delta x}{\Delta T} = \frac{m_1 * \displaystyle \frac{\Delta x_1}{\Delta T} + m_2 * \frac{\Delta x_2}{\Delta T}}{m_1 + m_2} \\\implies \displaystyle \frac{m_1 *　v_1 +　m_2 *　v_2}{m_1 + m_2}$ :::  ## 單純好玩 :::success $\displaystyle E = \frac{1}{2} mv^2$ $P = mv$ $\displaystyle E = P * \frac{1}{2} v$ $\displaystyle E = \frac{1}{2} * F * \Delta T * v$ ::: :::info $\displaystyle E = \frac{p^2}{2m}$ ::: > [color=purple][name=???][time=May, 8, 2023] ## 能量 $為什麼E= \frac{1}{2} mv^2$ $E =F * S \quad F = m * a$  $\implies E = m * a * S$ $v_1^2 = v_0^2 + 2aS$ 動能應該是相對於靜止狀態所以$v_0 = 0$   $\displaystyle \implies a*S = \frac{v^2}{2}$ $E = m * a * S$ $\displaystyle \implies E = \frac{1}{2} mv^2$ $E = mc^2$ ## 向心加速度（單純我知道的） $F = ma$ $\displaystyle a = \frac{v^2}{r}$ $\displaystyle E = \frac{1}{2}mv^2 \implies v^2 = \frac{2E}{m}$ $\displaystyle a = \frac{\displaystyle \frac{2E}{m}}{r} \implies \frac{2E}{mr}$ #### 角速率 $T : 週期$ $\omega : 角速率$ $\displaystyle T = \frac{2\pi r}{v}$ $\displaystyle \omega = \frac{d\theta}{dt} \\\implies\displaystyle \frac{2\pi}{T} \\\implies \displaystyle \frac{2\pi}{\displaystyle \frac{2\pi r}{v}} \implies \frac{v}{r}$ #### 向心力 $F = ma \\\implies\displaystyle m\frac{v^2}{r} = mw^2r = mvw$ 總結: 1. $\displaystyle w = \frac{2\pi}{T} = \frac{v}{r}$ 2. $\displaystyle a_{c} = \frac{v^2}{r} = vw = w^2r = \frac{4\pi^2 r}{T^2}$ ## 理想氣體方程式 $PV = nRT$ $P : 壓力$ $V : 體積 (L)$ $n：莫耳數$ $R : 理想氣體常數 \approx 0.082 atm.L/mole.K$  $T : 絕對溫度 \quad 0k \approx -273.15^{\circ}C$ 沸騰時溫度$100^{\circ}(373K)$ $一大氣壓$ $請算1mol 水蒸氣的體積?$ ::: spoiler answer $PV = nRT$ $\quad \quad n =1, P =1$ $V(L) = RT \implies V = 0.082 * (373)(L)$　 ::: ## 都普勒效應 $v = f * \lambda$ $f_0: 原本波的平率$ $v: 波在介質的移動速度$ $v_1: 波源的移動速度(往觀測者方向為正)$ $v_2: 觀測者的移動速度(往波源反方向為正)$ $f_1: 觀測者觀測到的頻率$ $波源 \implies 頻率不變$ > 介質相對於波源的移動速度為$v - v_1$ > 波長為$\displaystyle \frac{v - v_1}{f_0}$ $觀測者 \implies 波長不變$ >介質相對於觀測者的移動速度為$v - v_2$ >$\displaystyle (v - v_2) = f_1 * \frac{v - v_1}{f_0}$ $\displaystyle f_1 = f_0(\frac{v - v_2}{v - v_1})$ ## 角動量  ![](https://i.imgur.com/fIkBTm4.png) $L:角動量$ $r:半徑$ $p:動量$ $\theta:半徑和動量的夾角$ $\vec{L} = \vec{r} \times \vec{p}$  :::info $|\vec{L}| = |\vec{r}||\vec{p}|sin\theta$ > 修改[color=purple][name=???][time=May, 8, 2023] ::: >$令|\vec{r}| = r, |\vec{p}| = p, |\vec{L}| = L$ $當\theta = 90^{\circ} (圓周運動）\implies$ `L = r p` > $L = r * p$ > $L = rmv$ > $\displaystyle L = mr^2w \quad w = \frac{v}{r}$　 > $\displaystyle L = \frac{mv^2}{w} = \frac{2E}{w}$  ## 轉動慣量比較： > $p = mv$ > $L = mr^2w$ $v可以跟ｗ對應，所以讓m跟mr^2對應$ 轉動慣量$I = mr^2$ >$L = Iw$ ## 力矩 $\vec{M} = \vec{r} \times \vec{F} = |\vec{r}||\vec{F}|sin\theta$ $M = r * F$ > $\displaystyle F = ma = m\frac{\Delta v}{\Delta t} = \frac{m\Delta v}{\Delta t} = \frac{\Delta p}{\Delta t}$ > $\displaystyle M = r * \frac{\Delta p}{\Delta t} = \frac{r\Delta p}{\Delta t} = \frac{\Delta L}{\Delta t}$ $\displaystyle M = \frac{\Delta L}{\Delta t}$ $方向一樣自行判斷$ ## 簡諧運動 :::info $$F = -kx$$ $$F \propto x \\ m\cdot a\propto x \\ a\propto x$$ ::: ![](https://i.imgur.com/MxxUi90.png) https://www.geogebra.org/calculator/eyec3xsj 等速圓周運動在直線上的投影速率:v 半徑:r 夾角:$\theta$ 加速度 $\displaystyle \frac{v^2}{r}$ $P'$加速度: >$\displaystyle \frac{v^2}{r} cos\theta \implies rw^2cos(wt)$ $P'$位置 >$\displaystyle rcos\theta \implies rcos(wt)$ $P'$速度 >$\displaystyle -vsin\theta \implies -vsin(wt)$ $\displaystyle \frac{rw^2cos\theta}{rcos\theta} = w^2$ $加速度 \propto 位置 \implies簡協運動$ ### 延伸(擺錘週期) ![](https://i.imgur.com/cMfnfnY.png) https://www.geogebra.org/calculator/hjk4np74 設最大角度為$\beta$ 目前的角度為$\alpha$ 加速度 $\implies gsin\alpha$ 當$\alpha$ 很小時 $sin\alpha \approx \alpha$ 詳請見泰勒展開 >$加速度 \implies g\alpha$ 位置 $\implies l\alpha$ $加速度\propto 位置 \implies 簡協運動$ 簡協運動 >$r = l\beta$ >$x = r cos\theta = l\alpha$ >$\displaystyle a_x = \frac{v^2}{r} cos\theta = \frac{4\pi^2r}{T^2}cos\theta = g\alpha$ $\displaystyle \frac{l\alpha}{g\alpha} = \frac{l}{g} = \frac{rcos\theta}{\displaystyle \frac {4\pi^2r}{T^2} cos\theta} = \frac{T^2}{4\pi^2}$ $\\\implies \displaystyle \frac{l}{g} = \frac{T^2}{4\pi^2} \\\implies \displaystyle T = \sqrt{4\pi^2\frac{l}{g}} = 2\pi\sqrt{\frac{l}{g}}$  # 數學 ## 向量   ### 內積(點積) #### 幾何表示 $|\vec{v}| = \vec{v} 的長度$ $\vec{v}=(a,b)$ $|\vec{v}|=\sqrt{a^2+b^2}$ $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|cos\theta \quad$ 物理意義:能量 #### 矩陣表示 $a\begin{bmatrix} x_1, y_1\end{bmatrix} \quad b\begin{bmatrix} x_2, y_2\end{bmatrix}$ $\vec{a} \cdot \vec{b} = \vec{a} \times \vec{b}^T = \begin{bmatrix} x_1, y_1 \end{bmatrix} \times \begin{bmatrix} x_2 \\ y_2\end{bmatrix} = \begin{bmatrix} x_1*x_2 +y_1 * y _2\end{bmatrix}$ :::spoiler ***pf 兩個相同*** 設$\angle a = \alpha, \angle b = \beta$ >$a[x_1, y_1], b[x_2, y_2]$ >$|a| = \sqrt{x_1^2 +y_1^2} \quad |b| = \sqrt{x_2^2 + y_2^2}$ >$tan\alpha = \frac{y_1}{x_1} \quad tan \beta = \frac{y_2}{x_2} \implies cos\alpha = \frac{x_1}{\sqrt{x_1^2 +y_1^2}} \quad cos\beta = \frac{x_2}{\sqrt{x_2^2 + y_2^2}}$ $cos\theta = cos(\alpha - \beta) = cos\alpha cos\beta + sin\alpha sin\beta$ $=\frac{x_1 * x_2 + y_1 * y_2}{\sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2 )}}$ $|\vec{a}||\vec{b}|cos\theta =\sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2 )} * \frac{x_1 * x_2 + y_1 * y_2}{\sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2 )}} = x_1 * x_2 + y_1 * y_2$ ::: --- ### 外積(向量積)   #### 幾何表示有方向性 $\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|sin\theta\quad 0^{\circ} \leq \theta \leq 180^{\circ}$ 物理意義: 力矩幾何意義: 面積(大小為面積) --- #### 行列式 $a[x_1, y_1], b[x_2, y_2]$ $\begin{vmatrix} x_1 && x_2 \\ y_1 && y_2 \end{vmatrix} = |x_1 y_2 - x_2y_1|$ ---  --- ## 相消級數  $\frac{1}{\displaystyle x * \displaystyle(x+1)} \implies \frac{\displaystyle 1}{\displaystyle x} - \frac{\displaystyle 1}{\displaystyle x + 1}$ $\displaystyle \frac{\displaystyle 1}{\displaystyle x * (x+ 1) * (x+ 2)} \implies \frac{1}{2} \left[ \frac{1}{x *(x + 1)} - \frac{1}{(x + 1)* (x + 2)}\right]$ $\displaystyle \frac{1}{1 * 2 * 3} *\frac{1}{2 * 3 * 4} + \frac{1}{3 * 4 * 5} ... +\frac{1}{99 * 100 * 101}$ $\displaystyle \frac{1}{2} \begin{Bmatrix} \displaystyle \frac{1}{1 * 2} - \frac{1}{2 * 3} + \frac{1}{2 * 3} \dots - \frac{1}{100 * 101}\end{Bmatrix} \\\implies \displaystyle \frac{5049}{20200}$ --- 另一種解法 $\displaystyle \frac{1}{x*(x+1)*(x+2)}+\frac{1}{(x+1)*(x+2)*(x+3)}\implies \frac{1}{2}*(\frac{1}{x}-\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+3})$  就用$\displaystyle \frac{1}{(x+1)(x+2)}*(\frac{1}{x}+\frac{1}{x+3})$推 ---  ## 乘法公式 1. $a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2} \left[ (a-b)^2 + (b - c)^2 + (c - a)^2\right]$ 2. $(a \pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3$ 3. $a^3 \pm b^3 = (a\pm b)(a^2 + b^2 \mp ab)$ 4. $a^3 + b^3 + c^3 - 3abc= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ 5. $a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)$ :::spoiler ***pf 3*** $a^3 + b^3 = (a + b)^3 - 3ab(a + b) = (a + b)\left[ (a + b)^2 - 3ab \right] \\\implies (a + b)(a^2 + b^2 - ab)$ 負號同理 ::: :::spoiler ***pf 4*** $K為多項式$ $a^3 + b^3 + c^3 = (a + b + c)^3 - K$ $K = 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b\quad$ 想法:強行湊出$(a + b + c)$ $\implies K = 3(a + b + c)(ab + bc + ca) - 3abc \implies -K = -3(a + b + c)(ab + bc + ca) + 3abc$ $a^3 +b^3 + c^3 - 3abc = (a + b + c)\left[ (a + b + c)^2 - 3ab - 3bc - 3ca \right] \\\implies a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ ::: :::spoiler ***pf 5*** $a^4 + a^2b^2 + b^4 = (a^2 + b^2)^{2} - a^2b^2 \\\implies (a^2 + ab + b^2)(a^2 - ab + b^2)$ ::: ## 三角函數 ### 特別的直角三角形 ![](https://i.imgur.com/990c0g9.png) ### 和角公式 - $sin(\alpha\pm\beta) = sin\alpha cos\beta \pm cos\alpha sin\beta$ - $cos(\alpha \pm \beta) = cos\alpha cos\beta \mp sin\alpha sin\beta$ - $\displaystyle tan(\alpha \pm \beta) = \frac{tan\alpha \pm tan\beta}{1 \mp tan\alpha tan\beta}$ ### 延伸 #### 積化合差和角公式延伸 - $sin\alpha cos\beta = \frac{1}{2}\left[ sin(\alpha + \beta) +sin(\alpha - \beta)\right]$ - $cos\alpha sin\beta = \frac{1}{2} \left[ sin(\alpha + \beta) - sin(\alpha - \beta) \right]$ - $cos\alpha cos\beta = \frac{1}{2} \left[ cos(\alpha + \beta) + cos(\alpha - \beta)\right]$ - $sin\alpha sin\beta = -\frac{1}{2} \left[cos(\alpha + \beta) - cos(\alpha - \beta)\right]$ #### 二倍角公式 $sin(2\theta) = sin(\theta + \theta) = 2sin\theta cos\theta$ $cos(2\theta) = cos(\theta + \theta) = cos^2\theta - sin^2\theta \\\implies 1 - 2sin^2\theta \\\implies 2cos^2\theta - 1$ #### 半角公式 - $\displaystyle cos(\theta) = cos^2(\frac{\theta}{2}) - sin^2(\frac{\theta}{2}) \\\implies 1 - 2sin^2(\frac{\theta}{2}) \\\implies 2cos^2(\frac{\theta}{2}) - 1$ - $\displaystyle sin^2(\frac{\theta}{2}) = \frac{1 - cos\theta}{2} \implies sin(\frac{\theta}{2}) = \pm\sqrt{\frac{1 - cos\theta}{2}}$ - $\displaystyle cos^2(\frac{\theta}{2}) = \frac{1 + cos\theta}{2} \implies cos(\frac{\theta}{2}) = \pm\sqrt{\frac{1 + cos\theta}{2}}$ - $\displaystyle tan(\frac{\theta}{2}) = \frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})} \\\implies \displaystyle \pm\sqrt {\frac{1 - cos\theta}{1 + cos\theta}}\\\implies\displaystyle \pm\frac{\sqrt{1 - cos^2\theta}}{1 + cos\theta} \\\implies\displaystyle \frac{sin\theta}{{1+cos\theta}}$ #### 正切半角公式(萬能公式)  - $\displaystyle sin(\theta)=\frac{2tan\frac{\theta}{2}}{1+tan^2\frac{\theta}{2}}$ - $\displaystyle cos\theta=\frac{1-tan^2\theta}{1+tan^2\frac{\theta}{2}}$ $⋯以此類推$ ![](http://i.imgur.com/Idgpum6.png) ### 其他令 $a = sin\theta,\quad b = cos\theta \implies a^2 + b^2 = 1$ $sin^4\theta + cos^4\theta = 1 - 2sin^2\theta cos^2\theta$ $sin^6\theta + cos^6\theta = 1 - 3sin^2\theta cos^2\theta$ :::spoiler ***pf*** $(a^2 + b^2)(a^2 + b^2) = (a^2 + b^2) = 1 \implies a^4 + b^4 + 2a^2b^2 = 1$ $\implies a^4 + b^4 = 1 - 2a^2b^2$ ::: :::spoiler ***pf2*** $(a^4 + b^4)(a^2 + b^2) = a^4 + b^4 = 1 - 2a^2 b^2 \implies a^6 + b^6 + a^2b^2(a^2 + b^2)^{=1} = 1 - 2a^2b^2 \\\implies a^6 + b^6 = 1 - 3a^2b^2$ ::: ## 矩陣 ### 零矩陣 $O = \begin{bmatrix} 0 && 0\\0 && 0\end{bmatrix}$ ### 單位矩陣 $I = \begin{bmatrix} 1 && 0 \\ 0 && 1\end{bmatrix}$ ### 矩陣相加 :::info $矩陣大小必須相等$ $\implies \quad (2 \times 2) + (2 \times 2)$ ::: $\begin{bmatrix} a && b \\ c && d\end{bmatrix}+\begin{bmatrix} w && x \\ y && z\end{bmatrix}=\begin{bmatrix} a+w && b+x \\ c+y && d+z\end{bmatrix}$ ### 矩陣相乘 :::danger 先決條件:第一個矩陣行數 = 第二個矩陣的列數  ::: $\begin{bmatrix} b_1 && b_2\end{bmatrix} *\begin{bmatrix} a_1 \\ a_2\end{bmatrix} = \begin{bmatrix} a_1*b_1 + a_2 * b_2\end{bmatrix}$  向量 -> 行列式 | | | $a_1$ | | -------- | -------- | -------- | | | | $a_2$ | | $b_1$ | $b_2$ | $a_1 \times b_1 + a_2 \times b_2$ | #### 實例費氏數列 $\begin{cases}a_n = a_{n - 1} + a_{n -2} \end{cases}$ 矩陣$A =\begin{bmatrix} a_{n - 1} && a_{n - 2}\end{bmatrix}$ 轉移矩陣$B = \begin{bmatrix} 1 && 1 \\ 1 && 0\end{bmatrix}$  $A * B = \begin{bmatrix} a_{n - 1} + a_{n - 2} && a_{n - 1}\end{bmatrix} = \begin{bmatrix} a_n && a_{n - 1}\end{bmatrix}$ 可以做快速冪啊這有什麼意義快速冪是什麼 $a^(2^n) 最少需要幾次運算$ 這什麼 $123^4 如果用手算的話你會怎麼算$ $((100+20+3)^2)^2$ $((123)^2)^2 就是快速冪$還是不懂轉成二次方? 就是把次方想成二進位 $a^4, 4 = 100_{(2)}, 然後從1次開始,$ $a^n \,\, n\in\mathbb{Z}$ $任何整數都可以表示成2的次方相加, 5 = 2^0 + 2^2$ $如果知道a^1, 即可以在k次運算中得到a^{2^k}$ 如果$n = 11, n = 1011_{(2)}$ $a^{11} = a^1 * a^2 * a^8$ $ans = 1$ $開始運算$ >第一次運算 $ans = ans * a$ >第二次運算 $a * a = a^2$ >第三次運算 $ans = ans * a^2$ >第四次運算 $a^2 * a^2 = a^4$ >第五次運算 $a^4 * a^4 = a^8$ >第六次運算 $ans = ans * a^8$   ## 微分  ### 極限為了處理一些分母可能為0的式子，做出一個接近於０但不為０的值代替  $$\lim_{x \to a} f(x) \implies x無限接近於 a時f(x)的值$$ #### 例題 $$\lim_{x \to 3}\frac{x^2 - 2x - 3}{x - 3} =$$ 作法 $$\lim_{x \to 3}\frac{(x+1)(x-3)}{x-3}, x - 3$$ x - 3不為0 (無限接近), 所以約掉 $$\lim_{x\to 3} = x + 1 = 4$$ ### 斜率那現再來嘗試用$lim$來表示斜率吧設斜率為$f^{'}(0)$ $$\displaystyle f^{'}(x) =\frac{\Delta x}{\Delta y}$$ $$\displaystyle f^{'}(x)=\lim_{dx \to 0}\frac{f(x +　dx) - f(x)}{dx} $$ 來嘗試算$x^2$得斜率吧 $$\displaystyle f^{'}(x) = \lim_{dx \to 0}\frac{(x + dx)^2 - x^2}{dx}$$ $$\displaystyle f^{'}(x) = \lim_{dx \to 0}\frac{2x(dx) + dx^2}{dx}$$ $$\displaystyle f^{'}(x) = \lim_{dx\to 0}2x + dx$$ `dx = 0`, 帶入 $$m = 2x$$ ### 微分公式1 $x^n 很在f(x)中是最常見的，先來嘗試微分吧$ $先備知識：二項式定理$ $\displaystyle(x + a)^n = x^n + n\cdot ax^{n - 1} + k\cdot x^{n - 2}\dots$  $\implies(x+a)^n=\displaystyle\sum_{k=0}^{n}C_{k}^{n}\cdot x^n\cdot a^{n-k}$ #### n為正整數 $$f(x) = x^n$$ $$\displaystyle f^{'}(x) = \lim_{dx \to 0}\frac{f(x+dx) - f(x)}{dx}$$ $$\displaystyle f^{'}(x) = \lim_{dx \to 0}\frac{(x + dx)^n - (x)^n}{dx}$$ $$\displaystyle f^{'}(x) = \lim_{dx \to 0}\frac{n \cdot dx \cdot x^{n - 1} + dx^2(\dots)} {dx}$$ $$f^{'}(x) = \lim_{dx \to 0}n \cdot x^{n - 1} + dx(\dots)$$ $$f^{'}(x) = n \cdot x^{n - 1}$$ --- #### n為負整數 $$f(x) = x^n$$ $$f^{'}(x) = \lim_{dx \to 0}\frac{f(x + dx) - f(x)}{dx}$$ $$f^{'}(x) = \lim_{dx \to 0} \frac{\displaystyle \frac{1}{(x + dx)^n} - \frac{1}{x^n}}{dx}$$ $$f^{'}(x) = \lim_{dx \to 0}\frac{-(n \cdot dx \cdot x^{n - 1}) - dx^2(\cdots)}{dx \cdot x^n \cdot (x + dx)^n}$$ $$f^{'}(x) = \lim_{dx \to 0} \frac{-n \cdot x^{n - 1} - (\dots)}{x^n(x + dx)^n}$$ `dx = 0`帶入 $$f^{'}(x) = -n \cdot \frac{x^{n - 1}}{x^{2n}} = -n \cdot x^{-n-1}$$ --- ### 微分公式2　$\implies 加減法$  $$f(x) = g(x) \pm q(x)$$ $$f^{'}(x) = g^{'}(x) \pm q^{'}(x)$$ 想法：斜率代表著變動率，在`dx'相同時，變動率是可以相加的 --- ### 泰勒係數 :::info $$f^{(i)}(x) = f(x) 微分i次$$ ::: 一多項式`f(x)` $$f(x) 必定可以表示為 k_0 + k_1(x - a) + \cdots k_n(x - a)^n$$ $x = a$帶入 $$k_0 = f(a)$$ 微分 $$f^{'}(x) = k_1 + 2*k_2(x - a)+... +n(x - a)^{n - 1}$$ $x = a$帶入 $$k_1 = f^{'}(a)$$ 以此類推 $$k_i = \frac{1}{i!} \cdot f^{(i)}(a)$$ $$f(x)=\displaystyle\sum_{i=1}^{\infty}\frac{1}{i!}\cdot f^{(i)}(a)\cdot(x - a)^i$$ ---  ### 微分公式3　$\implies 乘法$  $$f(x) = p(x) *q(x)$$ $$f^{'}(x) = p^{'}(x)\cdot q(x) + p(x) \cdot q^{'}(x)$$ ***pf:*** $$p(x) = p(a) + p^{'}(a)(x - a) +　\cdots + \frac{1}{i!} \cdot p^{(i)}(a) \cdot (x - a)^i$$ $$q(x) = q(a) + q^{'}(a) \cdot (x - a) +　\cdots + \frac{1}{i!} \cdot q^{(i)}(a) \cdot (x - a)^i$$ $$f(x) = f(a) + f^{'}(a)(x - a) +　\cdots + \frac{1}{i!} \cdot f^{(i)}(a) \cdot (x - a)^i = p(x) \cdot q(x)$$ $$(1), (2)帶入$$ $$f(x) = p(x) \cdot q(x) = p(a)\cdot q(a) +\left[ p(a) \cdot q^{'}(a) + p^{'}(a)\cdot q(a) \right] (x - a) ......$$ $$和f(x)的泰勒係數比對$$ $$f^{'}(a) = p(a) \cdot q^{'}(a) + p^{'}(a) \cdot q(a)$$  --- ### 連鎖律(learning) :::info $[u]^{'} = 對u作微分$ ::: $a為多項式$ $$\left[ a^{\displaystyle r} \right] ^{'} = r \; \cdot a^{r - 1} \cdot a^{'}$$ ***pf*** : https://www.youtube.com/watch?v=K0j99cEtAr8 嘗試解釋.ing $$f(g(x)) $$ ```graphviz digraph main{ x u y x -> u[label = "g(x)"] u -> y[label = "f(u)"] {rank = same; x, u, y} } ``` :::info $$u = g(x),\quad b = g(a)$$ ::: $$目標 :f'(a)\cdot (x - a)$$ $$f(u) = f(b) + f^{'}(b)\cdot (u - b) \cdots \quad -(1)$$ $$g(x) = g(a) + g^{'}(a) \cdot (x - a) \cdots \quad-(2)$$ $$u - b = g(x) - g(a) = g^{'}(a) \cdot (x - a) + g^{"}(a) \cdot (x - a)^2 \cdots\quad帶入(1)$$ $$f(u) = f(b) + f^{'}(b) \cdot [g^{'}(a) +(x - a)\cdot(...)](x - a)\:+\: (x - a)^2[\cdots] +\cdots$$ $$\begin{aligned}f(u) = f(b) + f^{'}(b)\cdot g^{'}(a)\cdot &(x - a) + \cdots \\ f'(a)\cdot \quad\: &(x - a)\end{aligned}$$ $$f^{'}(a) = f'(b)\cdot g'(a)$$ $$a = x, b = g(a) =g(x)帶入$$ :::success $$[f(g(x))]' = f'(g(x)) \cdot g'(x) \quad or$$ $$[f(u)]' = f'(u) \cdot [u]'$$ ::: --- ### 自然常數 :::info $$[e^x]' = e^x$$ $$\lim_{dt\to 0} \frac{e^{dt} - 1}{dt} = e$$ ::: $$[Inx]' = \frac{1}{x}$$ ::: spoiler ***pf*** $$f(x) = In x, 令x = e^k$$ $$f'(x) = \frac{\Delta y}{\Delta x}= \lim_{dt \to 0} \frac{In e^k * e^{dt} - In e^k}{e^k \cdot e^{dt} - e^k}$$ $$f'(x) = \frac{dt}{e^k(e^dt - 1)} = \frac{1}{e^k} \times (\frac{e^{dt} - 1}{dt})^{-1} = \frac{1}{e^k} = \frac{1}{x}$$ ::: 處理指數微分特別好用 #### 廣義微分公式 :::success $$[x^r]' = rx^{r - 1}$$ ::: $$f(x) = y = x^n, 同時取對數$$ $$In y = In x^n = n In x$$ $$微分, In y 用連鎖律解決$$ ::: spoiler 如果不知道怎麼做點我 ::: success $$令g(k) = In k, g'(k) = \frac{1}{k}$$ $$[g(y)]' = g'(y) * y' = \frac{1}{y} *y'$$ ::: $$\frac{1}{y} * y' = \frac{n}{x}$$ $$y' = y * x^{-1} * n,\ y = x^n代回$$ #### 指數微分 $$f(x) = y = 2^x$$ $$In y = In2^x = xIn2$$ $$\frac{1}{y} \cdot y' = In 2$$ $$y' = yIn2$$ $$f(x) = 2^x In2$$ $$

Read more

向量

分治

Treap 暑訓題目

一些小小的思考方式