微积分A(2)期末复习笔记，仅供参考。 PDF版本链接

网页较大预警。如果公式解析不正确请刷新页面。长时间加载不出来建议打开PDF用Ctrl+S键下载到本地。流量访问请谨慎。

多元函数定义

内积: $x,y \in \mathbb R^n$, $\langle x,y \rangle = \sum_{i=1}^n x_iy_i$.

范数: $\forall N: \mathbb R^n \to \mathbb R$ s.t.: $\forall \lambda \in \mathbb R,x,y \in \mathbb R^n$,

1. $N(x) \ge 0$且$N(x)=0 \Leftrightarrow x=0$.
2. $N(\lambda x) = \vert \lambda \vert N(x)$.
3. $N(x+y) \le N(x)+N(y)$.

$n$维空间范数等价性:
$N,M$为$\mathbb R^n$中任意两个范数, $\exists \alpha, \beta \in \mathbb R$ s.t.: $\forall x \in \mathbb R^n$,

$\alpha M(x) \le N(x) \le \beta M(x)$

因此通常只考虑2-范数: $\| x \| = \sqrt{\langle x,x \rangle}$.

三角不等式:

$\| x+y \| \le \| x \| + \| y \|$

Cauchy-Schwarz不等式:

$\vert \langle x,y \rangle \vert \le \| x \| \cdot \| y \|$

距离: $x,y \in \mathbb R^n, \Omega, \Lambda \subseteq \mathbb R^n$,

1. $\text{dis}(x,y) = \| x-y \|$
2. $\text{dis}(\Omega,y) = \inf \left\lbrace \| x-y \| \Big \vert x \in \Omega \right \rbrace$
3. $\text{dis}(\Omega,\Lambda) = \inf \left\lbrace \| x-y \| \Big \vert x \in \Omega, y \in \Lambda \right \rbrace$

直径: $\text{diam}(\Omega) = \text{sup} \left\lbrace \| x-y \| \Big \vert x, y \in \Omega \right \rbrace$.

邻域: $x_0 \in \mathbb R^n, \delta \in \mathbb R$,

1. $B(x_0,\delta) = \lbrace x \vert \text{dis}(x,x_0) \lt \delta \rbrace$.
2. $B_0(x_0,\delta) = \lbrace x \vert 0 \lt \text{dis}(x,x_0) \lt \delta \rbrace$.

补集: $\Omega^c = \mathbb R^n \setminus \Omega$.

内点: $\exists \delta \gt 0, B(x_0,\delta) \subseteq \Omega$, 则$x_0 \in \Omega_0$.

外点: $\exists \delta \gt 0, B(x_0,\delta) \bigcap \Omega = \varnothing$, 则$x_0 \in \Omega_0^c = \overline \Omega$.

孤立点: $x_0 \in \Omega, \exists \delta \gt 0, B_0(x_0,\delta) = \varnothing$, 则$x_0$为孤立点.

边界点: $\forall \delta \gt 0, B(x_0,\delta) \bigcap \Omega \ne \varnothing \text{ and } B(x_0,\delta) \bigcap \Omega \ne B(x_0,\delta)$, 则$x_0$为边界点.

聚点: $\forall \delta \gt 0, B(x_0,\delta) \bigcap \Omega \ne \varnothing$, 则$x_0 \in \Omega'$.

闭包: $\text{Cl}(\Omega) = \Omega' \bigcup \Omega$.

$x$为非聚点, $x \not \in \Omega$, 则$x$为外点.
$x$为非聚点, $x \in \Omega$, 则$x$为孤立点.

开集: $\Omega$ s.t.: $\Omega = \Omega_0$.

闭集: $\Omega$ s.t.: $\Omega = \overline \Omega$.

有界集: $\Omega$ s.t.: $\exists r \gt 0, \Omega \subseteq B(0,r)$.

$\mathbb R^n$与$\varnothing$既开又闭.
开集的并是开集, 有限个开集的交是开集.
闭集的交是闭集, 有限个闭集的并是闭集.

点列收敛: $\lbrace x_i \rbrace$为$\mathbb R^n$中点列, 若$\lim_{i\to\infty} \| x_i-x_0 \| = 0$, 则$\lim_{i\to\infty} x_i = x_0$.

收敛点列极限唯一.

Cauchy列(基本列): $\forall \varepsilon \gt 0, \exists N \gt 0, \forall l,k \in \mathbb N_+, l,k \gt 0$, $\| x_k-x_l \| \lt \varepsilon$.

基本列 $\Leftrightarrow$ 收敛.

$\Omega$为闭集, $\lbrace x_i \rbrace \subseteq \Omega$而$\lim_{i\to\infty} x_i$存在, 则$\lim_{i\to\infty} x_i \in \Omega$. 特别地, 因为$\mathbb R^n$为闭集, 所以$\mathbb R^n$具有完备性.

连通集: $\forall x,y \in \Omega$, $\exists \varphi_i \in \mathscr C[a,b]$, $\varphi_i(a) = x^{(i)}, \varphi_i(b) = y^{(i)}$. 连通非空开集为开区域, 开区域闭包为闭区域.

二元函数: $f:\Omega \to \mathbb R$ s.t.: $\Omega \subseteq \mathbb R^2, \forall (x,y) \in \Omega, \exists! z \in \mathbb R, z = f(x,y)$.

二元函数极限与连续

二重极限: $p_0 \in \Omega \subseteq \mathbb R^2$为聚点, 二元函数$f$定义在$B_0(p_0,\delta_0) \bigcap \Omega$上, 若$\exists A \in \mathbb R, \forall \varepsilon \gt 0, \exists 0 \lt \delta \lt \delta_0, p \in B_0(p_0,\delta_0) \bigcap \Omega$, 都有: $\vert f(p)-A \vert \lt \varepsilon$, 则$\lim_{p\to p_0 \atop p\in\Omega} = A$. 若$p_0$为$\Omega$内点, 则可简记为$\lim_{p \to p_0} f(p) = \lim_{\| p-p_0 \| \to 0} f(p) = \lim_{x\to x_0 \atop y\to y_0} f(x,y) = A$.

若二重极限存在, 则任意路径趋近于$p_0$, 极限值相同.

累次极限: $\lim_{x \to x_0} \lim_{y \to y_0} f(x,y)$与$\lim_{y \to y_0} \lim_{x \to x_0} f(x,y)$.

若二重极限和累次极限均存在, 则相等.
若累次极限不相等, 则二重极限不存在.

连续: $\forall \varepsilon \gt 0, \exists \delta \gt 0, \forall p \in B(p_0,\delta) \bigcap D(f)$, 都有$\vert f(p)-f(p_0) \vert \lt \varepsilon$, 则$f$在$p_0$处连续.
否则$f$在$p_0$处间断, $p_0$为$f$间断点. $\lim_{p \to p_0}f(p)$存在但不等于$f(p_0)$则为可去间断点, 不存在则为本性间断点.

$f$在$p_0$二重极限存在, 则$f$在$p_0$处连续.
若$p_0$为$D(f)$聚点, $f$在$p_0$处连续 $\Leftrightarrow$ $f$在$p_0$二重极限为$f(p_0)$.
若$p_0$为$D(f)$孤立点, $f$在$p_0$处必连续.
若$p_0$为$f$间断点, $p_0$必为$D(f)$聚点.
若$p_0$处$f$连续, 则一元函数$f(x,y_0)$与$f(x_0,y)$在$(x_0,y_0)$处连续.

若$f$在开集$D$上连续或在闭集$D$内部和边界点上连续, 则$f \in \mathscr C(D)$.

$f$在连通集$D$上连续, 则$f$在$D$上最值存在.

二元函数导数

偏导数:

$\begin{aligned} f'_x(p_0) &= \left.\frac{\partial f}{\partial x}\right\vert_{p_0} = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x} \\ f'_y(p_0) &= \left.\frac{\partial f}{\partial y}\right\vert_{p_0} = \lim_{\Delta y \to 0} \frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y} \\ \end{aligned}$

全微分:

$\begin{aligned} \Delta f(p_0) &= f'_x(p_0)\Delta x + f'_y(p_0)\Delta y + \underbrace{o\left(\sqrt{\Delta x^2+\Delta y^2}\right)}_{o(\rho)} \\ \text{d}f(p_0) &= f'_x(p_0)\;\text{d}x + f'_y(p_0)\;\text{d}y \\ \end{aligned}$

可微 $\Rightarrow$ 连续, 偏导存在
偏导连续 $\Rightarrow$ 可微

可微的充要条件:

$\lim_{p \to p_0} \frac{f(x+\Delta x,y+\Delta y)-f(x,y)-f'_x(p_0)\Delta x-f'_y(p_0)\Delta y}{\sqrt{\Delta x^2 + \Delta y^2}} = 0$

复合函数求导:
若$z = f(x,y) = f(x(t,s),y(t,s))$:

$\begin{aligned} \frac{\partial z}{\partial t} &= \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} \\ \frac{\partial z}{\partial s} &= \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial s} \\ \end{aligned}$

高阶偏导数:

$\begin{aligned} f''_{xx}(p_0) &= \left.\frac{\partial^2 f}{\partial x^2}\right\vert_{p_0} = \lim_{\Delta x \to 0} \frac{f'_x(x_0+\Delta x,y_0)-f'_x(x_0,y_0)}{\Delta x} \\ f''_{xy}(p_0) &= \left.\frac{\partial^2 f}{\partial y \partial x}\right\vert_{p_0} = \lim_{\Delta y \to 0} \frac{f'_x(x_0,y_0+\Delta y)-f'_x(x_0,y_0)}{\Delta y} \\ f''_{yx}(p_0) &= \left.\frac{\partial^2 f}{\partial x \partial y}\right\vert_{p_0} = \lim_{\Delta x \to 0} \frac{f'_y(x_0+\Delta x,y_0)-f'_y(x_0,y_0)}{\Delta x} \\ f''_{yy}(p_0) &= \left.\frac{\partial^2 f}{\partial y^2}\right\vert_{p_0} = \lim_{\Delta y \to 0} \frac{f'_y(x_0,y_0+\Delta y)-f'_y(x_0,y_0)}{\Delta y} \\ \end{aligned}$

若$f''_{xy},f''_{yx}$在$p_0$处连续, 则$f''_{xy}(p_0)=f''_{yx}(p_0)$.

方向导数: 对于$e \in \mathbb R^2, \| e \| = 1$, 可定义:

$f'_e(p_0) = \left.\frac{\partial f}{\partial e}\right\vert_{p_0} = \lim_{t\to0^+} \frac{f(p_0+te)-f(p_0)}{t} = \frac{\text{d} [f(p_0+te)]}{\text{d} t}$

注意到上式中$e$仅指代射线方向, 即$\forall \lambda \gt 0$:

$\left.\frac{\partial f}{\partial(\lambda e)}\right\vert_{p_0} = \left.\frac{\partial f}{\partial e}\right\vert_{p_0}$

可微 $\Rightarrow$ 方向导数存在.
方向导数存在且同一点处反方向导数为正方向导数相反数 $\Rightarrow$ 偏导数存在.

$\begin{aligned} \left.\frac{\partial f}{\partial(-e)}\right\vert_{p_0} &= - \left.\frac{\partial f}{\partial e}\right\vert_{p_0} \\ \left.\frac{\partial f}{\partial \vec i}\right\vert_{p_0} &= \left.\frac{\partial f}{\partial x}\right\vert_{p_0} \;\;\;\;\;\;\;\; \text{ if } \left.\frac{\partial f}{\partial \vec i}\right\vert_{p_0} = -\left.\frac{\partial f}{\partial(-\vec i)}\right\vert_{p_0} \\ \left.\frac{\partial f}{\partial \vec j}\right\vert_{p_0} &= \left.\frac{\partial f}{\partial y}\right\vert_{p_0} \;\;\;\;\;\;\;\; \text{ if } \left.\frac{\partial f}{\partial \vec j}\right\vert_{p_0} = -\left.\frac{\partial f}{\partial(-\vec j)}\right\vert_{p_0} \\ \end{aligned}$

梯度:

$\nabla f(p_0) = \text{grad} f(p_0) = (f'_x(p_0),f'_y(p_0))$

$\left.\frac{\partial f}{\partial e}\right\vert_{p_0} = \nabla f(p_0) \cdot e = \| \nabla f(p_0) \| \cos \langle \nabla f(p_0), e \rangle$

向量值函数与隐函数

向量值函数: $f:\Omega \to \mathbb R^m$ s.t.: $\Omega \subseteq \mathbb R^n, \forall \vec x \in \Omega, \exists! \vec y \in \mathbb R^m, \vec y = f(\vec x)$.

连续: $f \in \mathscr C(\Omega) \Leftrightarrow \forall 1 \le i \le n, f_i \in \mathscr C(\Omega_i)$.

映射微分:

$\begin{aligned} \Delta f(\vec x) &= A\vec{\Delta x} + \underbrace{o\left(\|\vec{\Delta x} \|\right)}_{o(\rho)} \\ \text{d}f(\vec x) &= A\text{d} \vec x \\ \end{aligned}$

其中$A \in M_{m \times n}(\mathbb R)$, 满足:

$A_{ij} = \left.\frac{\partial y_i}{\partial x_j}\right\vert_{p_0}$

称$A$为$f$的Jacobi矩阵, 记作$J f(\vec x_0) = \left.\frac{\partial (y_1,\cdots,y_m)}{\partial (x_1,\cdots,x_n)}\right\vert_{\vec x_0} = A$.
$\Phi(\vec x_0)(\vec x) = J f(\vec x_0) \times \vec x$为$f$在$\vec x_0$的微分映射.
$\text{d} f(\vec x) = J f(\vec x_0) \times \text{d} \vec x$为$f$在$\vec x_0$的微分.

$\begin{bmatrix} \text{d}y_1 \\ \text{d}y_2 \\ \cdots \\ \text{d}y_m \\ \end{bmatrix} = \frac{\partial (y_1,\cdots,y_m)}{\partial (x_1,\cdots,x_n)} \times \begin{bmatrix} \text{d}x_1 \\ \text{d}x_2 \\ \cdots \\ \text{d}x_n \\ \end{bmatrix}$

复合映射微分:
若向量值函数$f$在$\vec x_0$处可微, $g$在$\vec u_0 = f(\vec x_0)$处可微, $\text{Im}(f) \subseteq D(g)$. 则:

$J(g \circ f)(\vec x_0) = J g(\vec u_0) \times J f(\vec x_0)$

逆映射微分:

$\begin{aligned} J (f \circ f^{-1}) &= I_{n \times n} \\ J f^{-1}(\vec y) &= \left( J f(\vec x) \right)^{-1} \\ \end{aligned}$

隐函数: $D(F) = W \times E \subseteq \mathbb R^n \times \mathbb R^m$, $\forall \vec x \in W, \exists! y \in E$ s.t.: $F(\vec x,y) = 0$. 则$F(\vec x,y)=0$确定隐函数$y = f(\vec x)$.

隐函数存在性:
若$F$在$W$内有定义, 且:

1. $F \in \mathscr C^{(q)}$, $q \ge 1$.
2. $\exists \vec p_0 \in W \times E, F(\vec p_0) = 0$.
3. $F'_y(\vec p_0) \ne 0$.

则: $\exists I \times J \subseteq W \times E$ s.t.: $\vec x_0 \in I, y_0 \in J$ (i.e.: $\vec p_0$的邻域):

1. $\forall \vec x \in I$, $\exists ! y = f(\vec x) \in J$. (隐函数存在唯一性)
2. $y_0 = f(\vec x_0)$. (初值条件)
3. $f \in \mathscr C^{(q)}(I)$. (隐函数连续性)
4. $\forall \vec x \in I$,

f'_j(\vec x) = -\frac{F'_j(\vec p)}{F'_y(\vec p)} $$ (隐函数导数)

若二元函数$F(x,y)=0$确定隐函数$f(x)$, $f^{-1} = g$存在 $\Leftrightarrow$ $F(x,y)=0$确定隐函数$g(y)$.

隐函数方程组: $\forall 1 \le i \le m, F_i(x_1,\cdots,x_n,y_1,\cdots,y_m) = 0$, 其中$D(F_i) = W_x \times W_y$.

隐函数方程组解存在性:
若$F$在$W_x$内有定义, 且: $\forall 1 \le i \le m$,

1. $F \in \mathscr C^{(q)}$, $q \ge 1$.
2. $\exists p_0 \in W_x \times W_y, F_i(\vec p_0) = 0$.
3. $\left.\frac{\partial (F_1,\cdots,F_m)}{\partial (y_1,\cdots,y_m)}\right\vert_{\vec p_0}$可逆.

则: $\exists \vec p_0$的邻域$I_x \times I_y$: $\forall 1 \le i \le m$,

1. $\forall \vec x \in I_x, \exists! \vec y$ s.t.: $F_i(\vec x,\vec y) = 0$. 可以相应定义定义$y_i = (\vec y)_i = f_i(\vec x)$.
2. $(\vec y_0)_i = f_i(\vec x_0)$.
3. $f_i \in \mathscr C^{(q)}(I_x)$.
4. $\forall \vec x \in I_x$,

$\left.\frac{\partial(y_1,\cdots,y_m)}{\partial(x_1,\cdots,x_n)}\right\vert_{\vec x} = -\left( \left.\frac{\partial (F_1,\cdots,F_m)}{\partial (y_1,\cdots,y_m)}\right\vert_{\vec p} \right)^{-1} \times \left.\frac{\partial(F_1,\cdots,F_m)}{\partial(x_1,\cdots,x_n)}\right\vert_{\vec x}$

空间曲线和曲面

空间曲线切向量: $\vec p = t \vec \tau$, 其中切向量$\vec \tau = (x'_t,y'_t,z'_t)(\vec p_0)$. 若空间曲线处处有非零、连续的切向量, 则称空间曲线光滑.

空间曲面切向量与法向量: 曲面$S:F(\vec p) = 0$在$\vec p_0$处可微, 且$\nabla F(\vec p_0) \ne 0$, $F$在邻域内连续, 则:

$\forall \vec \tau, F'(t_0) = \vec n \cdot \vec \tau = 0$

其中法向量$\vec n = \nabla F(\vec p_0) = (F'_x,F'_y,F'_z)(\vec p_0)$, 切向量$\vec \tau = (x'_t,y'_t,z'_t)(\vec p_0)$对于任意曲线$l:x=x(t),y=y(t),z=z(t)$ s.t: $\vec p_0 \in l \subseteq S$.

切向量求切线方程:

$\frac{x-x_0}{\tau_x} = \frac{y-y_0}{\tau_y} = \frac{z-z_0}{\tau_z}$

法向量求法线方程:

$\frac{x-x_0}{n_x} = \frac{y-y_0}{n_y} = \frac{z-z_0}{n_z}$

切向量求法平面方程:

$\tau_x(x-x_0) + \tau_y(y-y_0) + \tau_z(z-z_0) = 0$

法向量求切平面方程:

$n_x(x-x_0) + n_y(y-y_0) + n_z(z-z_0) = 0$

切向量求法向量(法向量求切向量类似):

$\vec n = \vec \tau_1 \times \vec \tau_2 = \left \vert \begin{matrix} \vec i & \vec j & \vec k \\ \tau_{1x} & \tau_{1y} & \tau_{1z} \\ \tau_{2x} & \tau_{2y} & \tau_{2z} \\ \end{matrix} \right \vert = \left( \det \frac{\partial (y,z)}{\partial (u,v)}, \det \frac{\partial (z,x)}{\partial (u,v)}, \det \frac{\partial (x,y)}{\partial (u,v)} \right)$

显函数法向量: $\vec n = (f'_x,f'_y,-1)(p_0)$.

二元函数泰勒展开与极值

高阶全微分: 若$f \in \mathscr C^{(n)}$,

$\text{d}^n f = \left( \text{d}x \cdot \frac{\partial}{\partial x} + \text{d}y \cdot \frac{\partial}{\partial y} \right)^nf(x,y) = \sum_{i=0}^n C_n^i \frac{\partial^n f}{\partial x^i \partial y^{n-i}}(x,y) \cdot \text{d}x^i \cdot \text{d}y^{n-i}$

二元函数泰勒展开: 若$f \in \mathscr C^{(n)}$,

$\begin{aligned} T_f(x,y) &= \sum_{k=0}^n \frac{1}{k!} \left( (x-x_0) \cdot \frac{\partial}{\partial x} + (y-y_0) \cdot \frac{\partial}{\partial y} \right)^kf(x_0,y_0) \\ f(x,y) &= T_f(x,y) + R_n \\ \end{aligned}$

皮亚诺余项: $f \in \mathscr C^{(n)}$, $R_n = o \left( \sqrt{(x-x_0)^2+(y-y_0)^2}^n \right)$.
拉格朗日余项: $f \in \mathscr C^{(n+1)}$, $R_n = \frac{1}{(n+1)!}\left( \Delta x \cdot \frac{\partial}{\partial x} + \Delta y \cdot \frac{\partial}{\partial y} \right)^{n+1}f(x_0+\theta \Delta x,y_0+\theta \Delta y)$, $\theta \in [0,1]$.

e.g.: $n=2$

$\begin{aligned} f(x,y) &= f(x_0,y_0) \\ &+ (x-x_0)f'_x(x_0,y_0) + (y-y_0)f'_y(x_0,y_0) \\ &+ \frac{1}{2}(x-x_0)^2f''_{xx}(x_0,y_0) + (x-x_0)(y-y_0)f''_{xy}(x_0,y_0) + \frac{1}{2}(y-y_0)^2f''_{yy}(x_0,y_0) \\ &+ R_2 \end{aligned}$

二元函数微分中值定理: $f \in \mathscr C$, $\theta \in [0,1]$,

$f(x,y)-f(x_0,y_0) = \Big(\Delta xf'_x+\Delta yf'_y\Big)(x_0+\theta\Delta x,y_0+\theta\Delta y)$

二元函数极值: $B(p_0,\delta) \subseteq D(f)$, 若$\forall p \in B(p_0,\delta), f(p) \le f(p_0)$, 则$p_0$为极大值点, $f(p_0)$为极大值. 极小值点和极小值同理.

极值点 $\Rightarrow$ 内点.
极值点, 偏导存在 $\Rightarrow$ 驻点(临界点).

$p_0$为$f$驻点, $f \in \mathscr C^{(2)}(B(p_0,\delta_0))$:

• $f''_{xx}f''_{yy} - {f''_{xy}}^2 \gt 0$, $p_0$为极值点
  • $f''_{xx} \gt 0, f''_{yy} \gt 0$, $p_0$为极小值点.
  • $f''_{xx} \lt 0, f''_{yy} \lt 0$, $p_0$为极大值点.
• $f''_{xx}f''_{yy} - {f''_{xy}}^2 \lt 0$, $p_0$不为极值点
• $f''_{xx}f''_{yy} - {f''_{xy}}^2 = 0$, 无法确定
  • 若$\exists 0 \lt \delta \lt \delta_0, \forall p \in B(p_0,\delta)$, $f''_{xx}f''_{yy} - {f''_{xy}}^2 \ge 0$, 则为极值点, 按照第一条方式判断.

一般地, 对于$n$原函数, 判断Hessian矩阵$H_f(p_0)$:

$\begin{bmatrix} f''_{x_1x_1} & f''_{x_1x_2} & f''_{x_1x_3} & \cdots & f''_{x_1x_n} \\ f''_{x_2x_1} & f''_{x_2x_2} & f''_{x_2x_3} & \cdots & f''_{x_2x_n} \\ f''_{x_3x_1} & f''_{x_3x_2} & f''_{x_3x_3} & \cdots & f''_{x_3x_n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ f''_{x_nx_1} & f''_{x_nx_2} & f''_{x_nx_3} & \cdots & f''_{x_nx_n} \\ \end{bmatrix}$

若正定或在邻域内连续半正定, 则为极小值点; 若负定或在邻域内连续半负定, 则为极大值点.

最小二乘法求回归直线:

$\begin{bmatrix} k \\ b \\ \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^n x_i^2 & \sum_{i=1}^n x_i \\ \sum_{i=1}^n x_i & n \\ \end{bmatrix}^{-1} \begin{bmatrix} \sum_{i=1}^n x_iy_i \\ \sum_{i=1}^n y_i \\ \end{bmatrix} = \frac{1}{\text{Var}[x]} \begin{bmatrix} 1 & -\mathbb E[x] \\ -\mathbb E[x] & \mathbb E[x^2] \\ \end{bmatrix} \begin{bmatrix} \sum_{i=1}^n x_iy_i \\ \sum_{i=1}^n y_i \\ \end{bmatrix}$

拉格朗日乘子法求条件极值: 求$f(x,y)$在$\varphi(x,y)=0$条件下的极值, 可构造拉格朗日函数并求解其驻点:

$L(x,y,\lambda) = f(x,y) + \lambda \varphi(x,y)$

含参定积分

一致连续: $\forall \varepsilon \gt 0, \exists \delta \gt 0, \forall p_1,p_2 \in \Omega, \| p_1-p_2 \| \lt \delta$, 都有$\vert f(p_1)-f(p_2) \vert \lt \varepsilon$, 则$f$在$\Omega$上一致连续.

有界闭集上连续则一致连续.

含参定积分: $I$为任意区间, $D = [a,b] \times I \subseteq D(f)$, 若$\forall u \in I$, $\varphi(u) = \int_a^b f(x,u)\;\text{d}x$存在, 则称其为$f$含参$u$的定积分.

连续性: $f \in \mathscr C(D)$ $\Rightarrow$ $\varphi \in \mathscr C(I)$, 即:

$\lim_{u\to u_0}\varphi(u) = \lim_{u\to u_0} \int_a^b f(x,u) \;\text{d}x = \int_a^b \lim_{u\to u_0} f(x,u) \;\text{d}x = \varphi(u_0)$

可导性: $f'_u \in \mathscr C(D)$ $\Rightarrow$ $\varphi'$存在且:

$\varphi'(u) = \int_a^b f'_u(x,u) \;\text{d}x$

可积性: $f \in \mathscr C(D)$ $\Rightarrow$ $\varphi \in \mathscr R(I)$且:

$\int_\alpha^\beta\varphi(u)\;\text{d}u = \int_\alpha^\beta \int_a^b f(x,u) \;\text{d}x \;\text{d}u = \int_a^b \int_\alpha^\beta f(x,u) \;\text{d}u \;\text{d}x$

变限积分: $f'_u \in \mathscr C(D)$, $a(u),b(u) \in \mathscr C[a,b]$在$[\alpha,\beta]$上可导 $\Rightarrow$ $\varphi'$在$[\alpha,\beta]$上存在且:

$\varphi'(u) = \int_{a(u)}^{b(u)} f'_u(x,u) \;\text{d}x + b'(u) \cdot f(b(u),u) - a'(u) \cdot f(a(u),u)$

含参广义积分: $I$为任意区间, $D = [a,+\infty) \times I \subseteq D(f)$, 若$\forall u \in I$, $\varphi(u) = \int_a^{+\infty} f(x,u)\;\text{d}x$存在, 则称其为$f$含参$u$的无穷积分.

一致收敛: $\forall \varepsilon \gt 0, \exists A_0 \gt a, \forall A \gt A_0, u \in I$, 都有$\vert \int_A^{+\infty}f(x,u)\;\text{d}x \vert \lt \varepsilon$, 则$\int_a^{+\infty}f(x,u)\;\text{d}x$在$I$上一致收敛.

一致收敛Cauchy原理:
$\int_a^{+\infty}f(x,u)\;\text{d}x$在$I$上一致收敛 $\Leftrightarrow$ $\forall \varepsilon \gt 0, \exists A_0 \gt a, \forall A_1,A_2 \gt A_0, u \in I$, 都有$\vert \int_{A_1}^{A_2} f(x)\;\text{d}x \vert \lt \varepsilon$.

Weierstrass判别法:
$D = [a,+\infty) \times I \subseteq D(f)$, $f \in \mathscr C[a,+\infty)$. 若$\exists F \in \mathscr C[a,+\infty)$ s.t: $\forall (x,u) \in D, f(x) \le F(x)$且$\int_a^{+\infty} F(x)\;\text{d}x$收敛, 则$\int_a^{+\infty} f(x,u)\;\text{d}x$在$I$上一致收敛.

积分第二中值定理:
$f,g \in \mathscr C([a,b] \times I), g'_x \in \mathscr C([a,b] \times I)$, 若$g$关于$x$单调, 则$\exists \xi \in (a,b)$ s.t.:

$\int_a^b f(x,u)g(x,u) \;\text{d}x = g(a,u)\int_a^\xi f(x,u) \;\text{d}x + g(b,u)\int_\xi^b f(x,u) \;\text{d}x$

Dirichlet判别法:
$f,g \in \mathscr C([a,b] \times I)$, 若:

1. $\int_a^{+\infty} f(x,u)\;\text{d}x$关于$u \in I$一致有界.
2. $g$关于$x$单调且关于$u \in I$一致趋于$0$.

则$\int_a^b f(x,u)g(x,u) \;\text{d}x$在$I$上一致收敛.

Abel判别法:
$f,g \in \mathscr C([a,b] \times I)$, 若:

1. $\int_a^{+\infty} f(x,u)\;\text{d}x$关于$u \in I$一致收敛.
2. $g$关于$x$单调且关于$u \in I$一致有界.

则$\int_a^b f(x,u)g(x,u) \;\text{d}x$在$I$上一致收敛.

局部一致收敛: $\forall u_0 \in I, \delta \gt 0$, 若都有$\int_a^{+\infty} f(x,u)\;\text{d}x$在$(u_0-\delta,u_0+\delta) \bigcap I$上一致收敛, 则$\int_a^{+\infty} f(x,u)\;\text{d}x$在$I$上局部一致收敛.

连续性: $f \in \mathscr C(D)$, $\varphi(u)$在$I$上局部一致收敛 $\Rightarrow$ $\varphi \in \mathscr C(I)$, 即:

$\lim_{u\to u_0}\varphi(u) = \lim_{u\to u_0} \int_a^{+\infty} f(x,u) \;\text{d}x = \int_a^{+\infty} \lim_{u\to u_0} f(x,u) \;\text{d}x = \varphi(u_0)$

可导性: $f'_u \in \mathscr C(D)$, $\varphi(u)$在$I$上一致收敛, $\int_a^{+\infty} f'_u(x,u) \;\text{d}x$在$I$上局部一致收敛 $\Rightarrow$ $\varphi'$存在且:

$\varphi'(u) = \int_a^{+\infty} f'_u(x,u) \;\text{d}x$

可积性: $f \in \mathscr C(D)$, $\varphi(u)$在$I$上一致收敛 $\Rightarrow$ $\varphi \in \mathscr R(I)$且:

$\int_\alpha^\beta\varphi(u)\;\text{d}u = \int_\alpha^\beta \int_a^{+\infty} f(x,u) \;\text{d}x \;\text{d}u = \int_a^{+\infty} \int_\alpha^\beta f(x,u) \;\text{d}u \;\text{d}x$

Gamma函数: $\Gamma(\alpha) \in \mathscr C^{(\infty)}(0,+\infty)$,

$\Gamma(\alpha) = \int_0^{+\infty} x^{\alpha-1}e^{-x} \;\text{d}x$

$\Gamma^{(n)}(\alpha) = \int_0^{+\infty} x^{\alpha-1}e^{-x} \ln^n x \;\text{d}x$

$\Gamma(x+1) = x\Gamma(x)$

$\Gamma(n+1) = n\Gamma(n) = n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$

$\Gamma(\alpha)\Gamma(1-\alpha) = \frac{\pi}{\sin(\alpha\pi)}$

$\Gamma\left(\frac{1}{2}\right) = \int_0^{+\infty} \frac{1}{\sqrt x e^x} \;\text{d}x = \sqrt \pi$

Beta函数: $\Beta(p,q) \in \mathscr C((0,+\infty)\times(0,+\infty))$

$\Beta(p,q) = \int_0^1 x^{p-1}(1-x)^{q-1} \;\text{d}x$

$\Beta(p,q) = \Beta(q,p)$

$\Beta(p+1,q) = \frac{p}{p+q}\Beta(p,q)$

$\Beta(p+1,q+1) = \frac{pq}{(p+q)(p+q+1)}\Beta(p,q)$

$\Beta(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} = \frac{1}{q C_{p+q-1}^{p-1}}$

多重积分

二重积分: $D \subseteq R^2$, $T$是$D$的一个分割, $\vert T \vert = \max_{1 \le i \le n} \lbrace \text{diam}(\Delta D_i) \rbrace$, 其中$\Delta D_i$面积为$S(\Delta D_i) = \Delta a_i$, $(x_i,y_i) \in \Delta D_i$, 则:

$V = \lim_{\vert T \vert\to 0} \sum_{i=1}^n f(x_i,y_i) \cdot \Delta a_i = \iint_D f(x,y) \;\text{d}S$

若该极限值存在且对于不同的$T$一致, 则$f$在$D$上可积, $f \in \mathscr R(D)$.

$\iint_D 1 \;\text{d}a = S(D)$.
$f \in \mathscr R(D)$ $\Rightarrow$ $f$在$D$上有界.
$f \in \mathscr R(D)$ $\Leftrightarrow$ 达布上和=达布下和 $\Leftrightarrow$ 振幅刻画$\lim_{\vert T \vert\to 0} \sum_{i=1}^n \omega_i \cdot \Delta a_i$.
$D$为有界闭区域(或间断点集零面积且有界), 则$f \in \mathscr C(D)$ $\Rightarrow$ $f \in \mathscr R(D)$.

若$D = \bigcup_{i=1}^m D_i$, $\forall i \ne j, (D_i)_0 \bigcap (D_j)_0 = \varnothing$, 则:

$\iint_D f \;\text{d}S = \sum_{i=1}^m \iint_{D_i} f \;\text{d}S$

且$f \in \mathscr R(D)$ $\Leftrightarrow$ $\forall i, f \in \mathscr R(D_i)$.

$f \ge g$ $\Rightarrow$ $\iint_D f \;\text{d}S \ge \iint_D g \;\text{d}S$.
$\left\vert \iint_D f \;\text{d}S \right\vert \le \iint_D \vert f \vert \;\text{d}S$.

二重积分中值定理: $f \in \mathscr C(D), g \in \mathscr R(D)$, $g$在$D$上不变号, 则$\exists (\xi,\eta) \in D$ s.t.:

$\iint_D fg \;\text{d}S = f(\xi,\eta) \cdot \iint_D g \;\text{d}S$

二重积分的计算: 若$D = \lbrace (x,y) \vert x \in [a,b], y_1(x) \le y \le y_2(x) \rbrace$, 其中$y_1,y_2 \in \mathscr C[a,b]$且$y_1 \le y_2$恒成立. 则:

$\iint_D f \;\text{d}S = \int_a^b \int_{y_1(x)}^{y_2(x)} f \;\text{d}y \;\text{d}x$

$D = \lbrace (x,y) \vert y \in [a,b], x_1(y) \le x \le x_2(y) \rbrace$情况类似.
注意若$\varphi(x,y) = (u,v)$为连续可微双射, 即$\forall (x,y) \in D, \frac{\partial (u,v)}{\partial (x,y)}$可逆, 则:

$\;\text{d}x\text{d}y = \left\vert \det \frac{\partial (x,y)}{\partial (u,v)} \right\vert \;\text{d}u\text{d}v = \left\vert \det \frac{\partial (u,v)}{\partial (x,y)} \right\vert^{-1} \;\text{d}u\text{d}v$

特别地, 对于极坐标有:

$\;\text{d}x\text{d}y = \left\vert \det \begin{bmatrix} \cos\theta & -\rho\sin\theta \\ \sin\theta & \rho\cos\theta \end{bmatrix} \right\vert \;\text{d}\rho\text{d}\theta = \rho \;\text{d}\rho\text{d}\theta$

三维二重积分: 面积微元$\text{d}S$为:

$\text{d}S = \sqrt{1+(f'_x)^2 + (f'_y)^2} \;\text{d}x\text{d}y$

若曲面单位法向量为$\vec n = \frac{\vec \tau_1 \times \vec \tau_2}{\| \vec \tau_1 \times \vec \tau_2 \|} = (\cos \alpha, \cos \beta, \cos \gamma)$, 则也可写作:

$\text{d}S = \frac{1}{\vert \cos\gamma \vert} \;\text{d}x\text{d}y$

若切向量具体为$\vec \tau_1 = (x'_u,y'_u,z'_u), \vec \tau_2 = (x'_v,y'_v,z'_v)$, 则:

$\begin{aligned} \vec n &= \det \begin{bmatrix} \vec i & \vec j & \vec k \\ x'_u & y'_u & z'_u \\ x'_v & y'_v & z'_v \\ \end{bmatrix} = \left( \det \frac{\partial (y,z)}{\partial(u,v)}, \det \frac{\partial (z,x)}{\partial(u,v)}, \det \frac{\partial (x,y)}{\partial(u,v)} \right) = (A,B,C) \\ \text{d}S &= \frac{\sqrt{A^2+B^2+C^2}}{\vert C \vert} \;\text{d}x\text{d}y \\ &= \frac{\sqrt{A^2+B^2+C^2}}{\left \vert \det \frac{\partial (x,y)}{\partial(u,v)} \right \vert} \cdot \left \vert \det \frac{\partial (x,y)}{\partial(u,v)} \right \vert\;\text{d}u\text{d}v \\ &= \sqrt{A^2+B^2+C^2} \;\text{d}u\text{d}v \\ &= \| \vec \tau_1 \times \vec \tau_2 \| \;\text{d}u\text{d}v \end{aligned}$

Gauss系数:

$\begin{aligned} E &= \| \vec \tau_1 \|^2 = (x'_u)^2 + (y'_u)^2 + (z'_u)^2 \\ F &= \| \vec \tau_2 \|^2 = (x'_v)^2 + (y'_v)^2 + (z'_v)^2 \\ G &= \vec \tau_1 \cdot \vec \tau_2 = x'_ux'_v + y'_uy'_v + z'_uz'_v \\ \text{d}S &= \sqrt{EF-G^2}\;\text{d}u\text{d}v \\ \end{aligned}$

三重积分的计算: 三重积分与二重积分定义类似. 特别地, 三重积分可以转化为"先二后一"或"先一后二"的积分.

$\iiint_V f(x,y) \;\text{d}x\text{d}y\text{d}z = \iint_D \int_{z_1(x,y)}^{z_2(x,y)} f(x,y) \;\text{d}z \;\text{d}x\text{d}y$

$\iiint_V f(x,y) \;\text{d}x\text{d}y\text{d}z = \int_{z_1}^{z_2} \iint_{D_z} f(x,y) \;\text{d}x\text{d}y \;\text{d}z$

与二重积分类似有:

$\;\text{d}x\text{d}y\text{d}z = \left\vert \det \frac{\partial (x,y,z)}{\partial (u,v,w)} \right\vert \;\text{d}u\text{d}v\text{d}w = \left\vert \det \frac{\partial (u,v,w)}{\partial (x,y,z)} \right\vert^{-1} \;\text{d}u\text{d}v\text{d}w$

特别地, 对于柱坐标系有:

$\;\text{d}x\text{d}y\text{d}z = \left\vert \det \begin{bmatrix} \cos\theta & -\rho\sin\theta & 0 \\ \sin\theta & \rho\cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \right\vert \;\text{d}\rho\text{d}\theta\text{d}z = \rho \;\text{d}\rho\text{d}\theta\text{d}z$

对于球坐标系有($x = \rho\sin\varphi\cos\theta, y = \rho\sin\varphi\sin\theta, z = \rho\cos\varphi$):

$\;\text{d}x\text{d}y\text{d}z = \left\vert \det \begin{bmatrix} \sin\varphi\cos\theta & \rho\cos\varphi\cos\theta & -\rho\sin\varphi\sin\theta \\ \sin\varphi\sin\theta & \rho\cos\varphi\sin\theta & \rho\sin\varphi\cos\theta \\ \cos\varphi & -\rho\sin\varphi & 0 \end{bmatrix} \right\vert \;\text{d}\rho\text{d}\varphi\text{d}\theta = \rho^2 \sin \varphi \;\text{d}\rho\text{d}\varphi\text{d}\theta$

物理量计算:
质量:

$M = \iiint_V \rho(x,y,z) \;\text{d}x\text{d}y\text{d}z$

质心(静力矩$/M$):

$\left\lbrace \begin{aligned} \widetilde x &= \frac{1}{M}\iiint_V x\rho(x,y,z) \;\text{d}x\text{d}y\text{d}z \\ \widetilde y &= \frac{1}{M}\iiint_V y\rho(x,y,z) \;\text{d}x\text{d}y\text{d}z \\ \widetilde z &= \frac{1}{M}\iiint_V z\rho(x,y,z) \;\text{d}x\text{d}y\text{d}z \\ \end{aligned} \right.$

转动惯量:

$I = \iiint_V \rho(x,y,z) \cdot r^2(x,y,z) \;\text{d}x\text{d}y\text{d}z$

引力:

$\left\lbrace \begin{aligned} F_x &= Gm_0\iiint_V \frac{(x-x_0)\rho(x,y,z)}{r^3} \;\text{d}x\text{d}y\text{d}z \\ F_y &= Gm_0\iiint_V \frac{(y-y_0)\rho(x,y,z)}{r^3} \;\text{d}x\text{d}y\text{d}z \\ F_z &= Gm_0\iiint_V \frac{(z-z_0)\rho(x,y,z)}{r^3} \;\text{d}x\text{d}y\text{d}z \\ \end{aligned} \right.$

曲线积分

第I型曲线积分:

$\int_L f(x,y,z) \;\text{d}\ell$

其中:

$\;\text{d}\ell = \sqrt{(x'_t)^2 + (y'_t)^2 + (z'_t)^2}\;\text{d}t$

特别地, 对于封闭曲线记为:

$\oint_L f(x,y,z) \;\text{d}\ell$

第I型曲线积分不具有方向性:

$\int_{\overgroup{AB}} f(x,y,z) \;\text{d}\ell = \int_{\overgroup{BA}} f(x,y,z) \;\text{d}\ell$

$f \ge g$ $\Rightarrow$ $\int_L f \;\text{d}\ell \ge \iint_L g \;\text{d}\ell$.
$\left\vert \int_L f \;\text{d}\ell \right\vert \le \iint_L \vert f \vert \;\text{d}\ell$.

第I型曲线积分中值定理: $f \in \mathscr C(L)$, $\exists (\xi,\eta,\zeta) \in L$ s.t.:

$\iint_L f(x,y,z) \;\text{d}\ell = f(\xi,\eta,\zeta) \cdot \ell$

第II型曲线积分: $\vec F(x,y,z) = (X(x,y,z),Y(x,y,z),Z(x,y,z))$, $\vec r = \vec \tau \text{d}\ell = (\text{d}x,\text{d}y,\text{d}z)$, 有时$\vec r$也写做$\vec \ell$.

$\begin{aligned} \int_L \vec F \cdot \text{d}\vec r &= \int_L \vec F \cdot \vec \tau \text{d}\ell = \int_L X\text{d}x + Y\text{d}y + Z\text{d}z \\ \end{aligned}$

注意$\int_L X\text{d}x + Y\text{d}y + Z\text{d}z = \int_L X\;\text{d}x + \int_L Y \;\text{d}y + \int_L Z \;\text{d}z$, 其中的曲线积分$\int_L$不能直接等价于一重积分$\int$, 因为另一变量在改变而非常数. $\int_L f(x,y) \;\text{d}x$这种曲线积分在封闭情况下可以求解, 参见下方"Green公式"与"Stokes公式"(另外, 如果被积函数和另一变量无关也可以求解).

对于闭区域$D$的边界曲线$\partial D = L$, 一般将其正方向定义为$D$内部在$L$左侧, 记为$L^+$. 默认封闭曲线积分为沿正方向的积分.

若$L$为多条有向曲线段$L_i$首尾相接而成, 则:

$\iint_L \vec F \cdot \vec \tau \;\text{d}\ell = \sum_{i=1}^m \iint_{L_i} \vec F \cdot \vec \tau \;\text{d}\ell$

且$\vec F \cdot \vec \tau \in \mathscr R(L)$ $\Leftrightarrow$ $\forall i, \vec F \cdot \vec \tau \in \mathscr R(L_i)$.

第II型曲线积分具有方向性:

$\int_{\overgroup{AB}} \vec F(x,y,z) \cdot \text{d}\vec r = -\int_{\overgroup{BA}} \vec F(x,y,z) \cdot \text{d}\vec r$

事实上, $\int F \cdot \vec \tau \;\text{d}\ell$也可看做是第I型曲线积分不具有方向性, 但$\vec F \cdot (-\vec \tau) = - \vec F \cdot \vec r$, 故第II型曲线积分具有方向性.

Green公式:
$D \subseteq \mathbb R^2$, $P,Q \in \mathscr C^{(1)}(D)$, $\partial D = L$光滑或分段光滑, 则:

$\oint_{L^+} P\text{d}x + Q\text{d}y = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \;\text{d}x\text{d}y = \iint_D \det \begin{bmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ P & Q \\ \end{bmatrix} \;\text{d}x\text{d}y$

更进一步, 事实上有:

$\begin{aligned} \oint_{L^+} P\text{d}x &= - \iint_D \frac{\partial P}{\partial y}\;\text{d}x\text{d}y \\ \oint_{L^+} Q\text{d}y &= \iint_D \frac{\partial Q}{\partial x}\;\text{d}x\text{d}y \\ \end{aligned}$

注意到如果构造$P,Q$使得$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = C$为非零常数, 则可用于求出$D$区域的面积. 例如:

$S(D) = \frac{1}{2}\oint_{\partial D^+} x\text{d}y-y\text{d}x$

另外, 分别考虑曲线在平面内的切向量和$\vec \tau$和朝向区域外侧的单位法向量$\vec n$, Green公式也可写作:

$\begin{aligned} \oint_{L^+} (P,Q) \cdot \vec \tau\text{d}\ell &= \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \;\text{d}x\text{d}y \\ \oint_{L^+} (P,Q) \cdot \vec n\text{d}\ell &= \iint_D \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \;\text{d}x\text{d}y \\ \end{aligned}$

路径无关积分:
$D \subseteq \mathbb R^2$, $P,Q \in \mathscr C^{(1)}(D)$, $L$光滑或分段光滑, 则以下命题等价:

1. $\int_{L(A,B)} P\text{d}x+Q\text{d}y$路径无关.
2. $\exists U$ s.t.: $\forall (x,y) \in D, \text{d}U = P\text{d}x + Q\text{d}y$. 即:

   $U(x,y) = \int_{(x_0,y_0)}^{(x,y)} P\text{d}x + Q\text{d}y$

3. $\forall (x,y) \in D$, $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = \frac{\partial^2 U}{\partial x \partial y}$.
4. $\forall L \subseteq D$, 若$L$为光滑或分段光滑封闭曲线, 则$\oint_L P\text{d}x + Q\text{d}y = 0$.

全微分方程: 若$\exists U, \text{d}U(x,y) = P\text{d}x+Q\text{d}y$, 则方程$P\text{d}x+Q\text{d}y = 0$为全微分方程. 由上述路径无关积分结论可得:

1. $P\text{d}x+Q\text{d}y = 0$是全微分方程 $\Leftrightarrow$ $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$.
2. $P = \frac{\partial U}{\partial x}, Q = \frac{\partial U}{\partial y}$.
3. 全微分方程的通解为$U(x,y) = C$.

求全微分方程的通解函数$U$的方法:

1. 任选一条路径$L$进行曲线积分.
2. 定义$U_x(x,y) = \int P\;\text{d}x$, 则$\frac{\partial U_x}{\partial x} = P$. 若$U(x,y) = U_x(x,y) + U_y(y)$, 则$Q = \frac{\partial U}{\partial y} = \frac{\partial U_x}{\partial y} + \frac{\text{d} U_y}{\text{d} y}$, 即$U_y = \int Q - \frac{\partial U_x}{\partial y} \;\text{d}y$. 即:

   $U = \int P\;\text{d}x + \int Q - \frac{\partial\left(\int P\;\text{d}x\right)}{\partial y} \;\text{d}y + C$

3. 直接凑微分$P\text{d}x + Q\text{d}y = \text{d}U$.

常见全微分形式:

$y\text{d}x + x\text{d}y = \text{d}(xy)$

$\frac{y\text{d}x - x\text{d}y}{y^2} = \text{d}\left(\frac{x}{y}\right)$

$\frac{y\text{d}x - x\text{d}y}{xy} = \text{d}\left(\ln \left\vert \frac{x}{y} \right\vert \right)$

$\frac{y\text{d}x - x\text{d}y}{x^2+y^2} = \text{d}\left(\arctan \frac{x}{y} \right)$

$\frac{x\text{d}x + y\text{d}y}{x^2+y^2} = \text{d}\left(\frac{1}{2}\ln(x^2+y^2) \right)$

$\frac{y\text{d}x - x\text{d}y}{x^2-y^2} = \text{d}\left(\frac{1}{2}\ln\left\vert \frac{x-y}{x+y} \right\vert \right)$

积分因子: 若$P\text{d}x + Q\text{d}y = 0$不是全微分方程, 但$\mu P\text{d}x + \mu Q\text{d}y = 0$是全微分方程(即$\frac{\partial(\mu \cdot P)}{\partial y} = \frac{\partial(\mu \cdot Q)}{\partial x}$, 也即$\mu'_yP + \mu P'_y = \mu'_xQ + \mu Q'_x$), 则$\mu(x,y)$为$P\text{d}x + Q\text{d}y = 0$的积分因子.

若$\mu(x,y) = \mu(x)$(即$\mu'_y = 0$), 则:

$\mu'_x = \frac{P'_y - Q'_x}{Q}\mu \Rightarrow \mu = \exp\left(\int \frac{P'_y - Q'_x}{Q} \;\text{d}x \right)$

若$\mu(x,y) = \mu(y)$(即$\mu'_x = 0$), 则:

$\mu'_y = \frac{P'_y - Q'_x}{-P}\mu \Rightarrow \mu = \exp\left(\int \frac{P'_y - Q'_x}{-P} \;\text{d}y \right)$

若求出积分因子, 则全微分方程$\mu P\text{d}x + \mu Q\text{d}y = 0$通解为$U(x,y) = C$也就是原方程的通解.

曲面积分

第I型曲面积分:

$\iint_\Sigma f(x,y,z) \;\text{d}S$

其中(与三维二重积分类似):

$\begin{aligned} \text{d}S &= \sqrt{1+(f'_x)^2 + (f'_y)^2} \;\text{d}x\text{d}y \\ \text{d}S &= \frac{1}{\vert \cos\gamma \vert} \;\text{d}x\text{d}y \\ \text{d}S &= \frac{\sqrt{A^2+B^2+C^2}}{\vert C \vert} \;\text{d}x\text{d}y \\ \text{d}S &= \sqrt{A^2+B^2+C^2} \;\text{d}u\text{d}v \\ \text{d}S &= \| \vec \tau_1 \times \vec \tau_2 \| \;\text{d}u\text{d}v \\ \text{d}S &= \sqrt{EF-G^2}\;\text{d}u\text{d}v \\ \end{aligned}$

事实上, 由行列式的物理意义即可得到, 对于任意$\vec \tau_u$和$\vec \tau_v$确定的$\text{d}S$, 以及$\vec \tau_p$和$\vec \tau_q$确定的$\text{d}S'$, 都有:

$\text{d}S = \left\vert \det \frac{\partial (u,v)}{\partial (p,q)} \right\vert \;\text{d}S'$

特别地, 对于封闭曲线记为:

$\oiint_\Sigma f(x,y,z) \;\text{d}S$

第I型曲面积分中值定理: $f \in \mathscr C(\Sigma)$, $\Sigma$光滑, $\exists (\xi,\eta,\zeta) \in \Sigma$ s.t.:

$\iint_\Sigma f(x,y,z)\;\text{d}S = f(\xi,\eta,\zeta) \cdot S(\Sigma)$

Poisson公式:

$\begin{aligned} \iint_{x^2+y^2+z^2=1} f(ax+by+cz)\;\text{d}S &= 2\pi\int_{-1}^1 f\left(u\sqrt{a^2+b^2+c^2}\right)\;\text{d}u \\ \iint_{x^2+y^2+z^2=r^2} f(ax+by+cz)\;\text{d}S &= 2\pi r\int_{-r}^r f\left(u\sqrt{a^2+b^2+c^2}\right)\;\text{d}u \\ \end{aligned}$

双侧曲面: 曲面$S$上任意一点$p_0$处存在两个共线反向法向量, 选定其中一个为正方向, 当动点$p$从$p_0$出发沿$S$上任意封闭曲线回到$p_0$时, 法向量正方向与出发时正方向相同, 则$S$为双侧曲面. (单侧曲面例如莫比乌斯环). 规定方向的双侧曲面为有向曲面.

第II型曲面积分:$\vec F(x,y,z) = (X(x,y,z),Y(x,y,z),Z(x,y,z))$, $\vec S = \vec n \text{d}S = (\cos\alpha,\cos\beta,\cos\gamma)\text{d}S = (\text{d}y \wedge \text{d}z, \text{d}z \wedge \text{d}x, \text{d}x \wedge \text{d}y)$, 其中$\vec n$是有向曲面正方向一侧的单位法向量:

$\begin{aligned} \iint_{\Sigma^+} \vec F \cdot \text{d}\vec S &= \iint_{\Sigma^+} \vec F \cdot \vec n \text{d}S = \iint_{\Sigma^+} X \text{d}y \wedge \text{d}z + Y \text{d}z \wedge \text{d}x + Z \text{d}x \wedge \text{d}y \\ \end{aligned}$

其中: $\text{d}y\wedge\text{d}z = \cos\alpha\;\text{d}S = \frac{A}{\sqrt{A^2+B^2+C^2}}\;\text{d}S = \frac{A}{\sqrt{A^2+B^2+C^2}} \times \frac{\sqrt{A^2+B^2+C^2}}{\vert A \vert}\;\text{d}y\text{d}z = \text{sgn}(A)\;\text{d}y\text{d}z$. 另外两项同理.

注意$\iint_{\Sigma^+} X \text{d}y \wedge \text{d}z + Y \text{d}z \wedge \text{d}x + Z \text{d}x \wedge \text{d}y = \iint_{\Sigma^+} X \text{d}y \wedge \text{d}z + \iint_{\Sigma^+} Y \text{d}z \wedge \text{d}x + \iint_{\Sigma^+} Z \text{d}x \wedge \text{d}y$, 其中的曲面积分$\iint_{\Sigma^+}$不能直接等价于二重积分$\iint$, 因为第三变量在改变而非常数. 不同于曲线积分, $\iint_{\Sigma^+} f(x,y,z) \;\text{d}y\wedge\text{d}z$这种曲面积分在封闭情况下可以求解, 参见下方"Gauss公式"(另外, 类似于曲线积分如果被积函数和第三变量无关也可以求解).

对于闭区域$V$的边界曲线$\partial V = \Sigma$, 一般将其正方向定义为$\Sigma$的外侧, 记为$\Sigma^+$. 默认封闭曲线积分为沿正方向的积分.

第II型曲面积分具有方向性:

$\iint_{\Sigma^+} \vec F(x,y,z) \cdot \text{d}\vec S = -\iint_{\Sigma^-} \vec F(x,y,z) \cdot \text{d}\vec S$

事实上, $\int F \cdot \vec n \;\text{d}S$也可看做是第I型曲面积分不具有方向性, 但$\vec F \cdot (-\vec n) = - \vec F \cdot \vec n$, 故第II型曲面积分具有方向性.

$\iint_{\Sigma^+} \vec F \cdot \text{d}\vec S = \pm \iint_D XA + YB + ZC \;\text{d}u\text{d}v$

其中:

$A = \det \frac{\partial (y,z)}{\partial(u,v)}, B = \det \frac{\partial (z,x)}{\partial(u,v)}, C = \det \frac{\partial (x,y)}{\partial(u,v)}$

Gauss公式:
$V \subseteq \mathbb R^3$, $P,Q,R \in \mathscr C^{(1)}(V)$, $\partial V = \Sigma$光滑或分片光滑, 则:

$\oiint_{\Sigma^+} P \text{d}y \wedge \text{d}z + Q \text{d}z \wedge \text{d}x + R \text{d}x \wedge \text{d}y = \iiint_V \frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z} \;\text{d}x\text{d}y\text{d}z$

更进一步, 事实上有:

$\begin{aligned} \oiint_{\Sigma^+} P\text{d}y\wedge\text{d}z &= \iiint_V \frac{\partial P}{\partial x}\;\text{d}V \\ \oiint_{\Sigma^+} Q\text{d}z\wedge\text{d}x &= \iiint_V \frac{\partial Q}{\partial y}\;\text{d}V \\ \oiint_{\Sigma^+} R\text{d}x\wedge\text{d}y &= \iiint_V \frac{\partial R}{\partial z}\;\text{d}V \\ \end{aligned}$

注意到如果构造$P,Q,R$使得$\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z} = C$为非零常数, 则可用于求出$V$区域的体积.

Stokes公式:
$S$为光滑双侧曲面, $P,Q,R \in \mathscr C^{(1)}(S)$, $\partial S = L$光滑或分段光滑, 则:

$\begin{aligned} \oint_{L^+} P\text{d}x + Q\text{d}y + R\text{d}z = \iint_{S^+} &\left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right) \text{d}y\wedge\text{d}z \\ +& \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right) \text{d}z\wedge\text{d}x \\ +& \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \text{d}x\wedge\text{d}y \\ =\iint_{S^+} &\det \begin{bmatrix} \text{d}y\wedge\text{d}z & \text{d}z\wedge\text{d}x & \text{d}x\wedge\text{d}y \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \\ \end{bmatrix} \\ =\iint_{S^+} &\det \begin{bmatrix} \cos\alpha & \cos\beta & \cos\gamma \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \\ \end{bmatrix} \;\text{d}S \end{aligned}$

其中$L^+$和$S^+$方向满足右手法则(大拇指指向$S^+$正法向量方向, 四指为$L^+$正方向).

更进一步, 事实上有:

$\begin{aligned} \oint_{L^+} P\text{d}x &= \iint_{S^+} \frac{\partial P}{\partial z}\text{d}z\wedge\text{d}x - \frac{\partial P}{\partial y}\text{d}x\wedge\text{d}y \\ \oint_{L^+} Q\text{d}y &= \iint_{S^+} \frac{\partial Q}{\partial x}\text{d}x\wedge\text{d}y - \frac{\partial Q}{\partial z}\text{d}y\wedge\text{d}z \\ \oint_{L^+} R\text{d}z &= \iint_{S^+} \frac{\partial R}{\partial y}\text{d}y\wedge\text{d}z - \frac{\partial R}{\partial x}\text{d}z\wedge\text{d}x \\ \end{aligned}$

特别地, 在二维特殊情况下有:

$\begin{aligned} \oint_{L^+} P\text{d}x &= - \iint_{S^+} \frac{\partial P}{\partial y}\text{d}x\wedge\text{d}y \\ \oint_{L^+} Q\text{d}y &= \iint_{S^+} \frac{\partial Q}{\partial x}\text{d}x\wedge\text{d}y \\ \end{aligned}$

与Green公式相吻合.

空间路径无关积分:
若$\Omega \subseteq \mathbb R^3$中任意一条简单封闭区县$L$可以通过连续形变收缩为一点, 则称$L$为零伦的. $L$为零伦的当且仅当$L$为某分片光滑曲面$S \subseteq \Omega$的边界曲线.
若区域$\Omega$内任意简单闭曲线都为零伦的，则$\Omega$为单连通体.
若$\Omega \subseteq \mathbb R^3$为单连通体, $P,Q \in \mathscr C^{(1)}(\Omega)$, $L$光滑或分段光滑, 则以下命题等价:

1. $\int_{L(A,B)} P\text{d}x+Q\text{d}y+R\text{d}z$路径无关.
2. $\exists U$ s.t.: $\forall (x,y,z) \in \Omega, \text{d}U = P\text{d}x + Q\text{d}y + R\text{d}z$. 即:

   $U(x,y) = \int_{(x_0,y_0)}^{(x,y)} P\text{d}x + Q\text{d}y + R\text{d}z$

3. $\forall (x,y,z) \in \Omega, \frac{\partial R}{\partial y} = \frac{\partial Q}{\partial z}$, $\frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}$, $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$.
4. $\forall L \subseteq D$, 若$L$为光滑或分段光滑封闭曲线, 则$\oint_L P\text{d}x + Q\text{d}y + R\text{d}z = 0$.

场论

场: $\Omega \subseteq \mathbb R^3$, $\forall M \in \Omega$: 若$\exists! u = f(M)$, 则$f$为$\Omega$上的数量场; 若$\exists \vec u = \vec F(M)$, 则$\vec F$为$\Omega$上的向量场.

梯度场: 若数量场$U$在$\Omega$上连续可微, 则$\nabla U(M) = (U'_x(M),U'_y(M),U'_z(M))$为$\Omega$上的向量场, 即为$u$的梯度场. $\nabla$又称哈密顿算子:

$\nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$

注意$U$为数量场而$\nabla U$为向量场, 不能写作$\nabla \cdot U$, 后者中$U$为向量场而$\nabla \cdot U$为数量场(散度场).

任意连续可导数量场存在对应向量场(梯度场), 但并非所有连续向量场存在对应数量场, 其存在的充要条件是势函数存在(积分路径无关).

保守场: 若向量场$\vec F$在$\Omega$上积分路径无关, 即$\int_{L(A,B)} \vec F \cdot \vec \tau = U(B)-U(A)$, 则$\vec F$为$\Omega$上的保守场.

有势场: 若向量场$\vec F$在$\Omega$上存在可微函数$U$, 使得$\vec F = \nabla U$, 则$\vec F$为$\Omega$上的有势场.

$\vec F$为有势场 $\Leftrightarrow$ $\vec F$为保守场.

散度场: 若向量场$\vec F(M) = (P(M),Q(M),R(M))$在$\Omega$上连续可微, 则:

$\text{div} \vec F = P'_x+Q'_y+R'_z = \nabla \cdot \vec F$

为$\Omega$上的数量场, 即$\vec F$的散度场.

通量/流量: $S^+$为连续可微向量场$\vec F$中的定向曲面, $\vec n$为$S^+$单位法向量, 则$\vec F$通过曲面$S$向$\vec n$方向的通量/流量为:

$\iint_{S^+} \vec F \cdot \vec n \text{d}S$

注意到根据Gauss公式, $\vec F$在$\Omega$上连续可微, 故:

$\oiint_{\partial\Omega^+} \vec F \cdot \vec n \;\text{d}S = \iiint_\Omega \text{div}\vec F \;\text{d}V$

且根据积分中值定理, $\exists \xi \in \Omega$:

$\oiint_{\partial\Omega^+} \vec F \cdot \vec n \;\text{d}S = \iiint_\Omega \text{div}\vec F \;\text{d}V = \text{div}\vec F(\xi) \cdot V(\Omega)$

故散度场可以定义为, 对于以$M$为球心的球$D$:

$\text{div}\vec F(M) = \lim_{V(D)\to0}\frac{1}{V(D)}\oiint_{\partial D^+} \vec F \cdot \vec n\;\text{d}S$

无源场: 若$\text{div}\vec F(M) \ne 0$, 称$\vec F$在$M$有流源; 若$\text{div}\vec F(M) \gt 0$, 称$\vec F$在$M$有正流源; 若$\text{div}\vec F(M) \lt 0$, 称$\vec F$在$M$有负流源. 若$\forall M \in \Omega$, $\text{div}\vec F(M) = 0$, 则$\vec F$为$\Omega$上的无源场.

旋度场: 若向量场$\vec F(M) = (P(M),Q(M),R(M))$在$\Omega$上连续可微, 则:

$\text{rot}\vec F = \left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial z},\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x},\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) = \det\begin{bmatrix}\vec i & \vec j & \vec k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \\ \end{bmatrix} = \nabla \times \vec F$

为$\Omega$上的向量场, 即为$\vec F$的旋度场.

环量/环流量: $L^+$为向量场$\vec F$中光滑或分段光滑的有向闭合曲线, $\vec \tau$为$L^+$单位切向量, 则$\vec F$沿$L^+$的环量/环流量为:

$\oint_{L^+} \vec F \cdot \vec \tau \text{d}\ell$

注意到根据Stokes公式:

$\oint_{L^+} \vec F \cdot \vec \tau \text{d}\ell = \iint_{\Sigma^+} \text{rot}\vec F \cdot \vec n\;\text{d}\Sigma$

且根据积分中值定理, $\exists \xi \in \Sigma^+$:

$\oint_{L^+} \vec F \cdot \vec \tau \text{d}\ell = \iint_{\Sigma^+} \text{rot}\vec F \cdot \vec n\;\text{d}S = \text{rot}\vec F(\xi) \cdot \vec n \cdot S(\Sigma)$

故旋度场可以定义为, 对于以$M$为圆心的圆$D$:

$\text{rot}\vec F(M) \cdot \vec n = \lim_{S(D)\to0}\frac{1}{S(D)}\oint_{\partial D^+} \vec F \cdot \vec \tau\;\text{d}\ell$

无旋场: 故$\text{rot}\vec F(M) \cdot \vec n$表示$\vec F$在$M$处环绕$\vec n$的方向旋量, 则$\text{rot}\vec F(M)$的三个方向分量分别表示$\vec F$在$M$处环绕$x,y,z$三个坐标的方向旋量. 若$\| \text{rot}\vec F(M) \| \ne 0$, 则$M$为$\vec F$的旋涡, $\| \text{rot}\vec F(M) \|$越大旋转越快. 若$\forall M \in \Omega$, $\| \text{rot}\vec F(M) \| = 0$, 则$\vec F$为$\Omega$上的无旋场.

调和场: 若向量场$\vec F$在$\Omega$上同时为有势场和无源场, 则$\vec F$为$\Omega$上的调和场.
$\vec F$在$\Omega$上同时为有势场和无源场, 故:

$0 = \text{div}\vec F = \nabla \cdot \vec F = \nabla \cdot (\nabla U) = (\nabla \cdot \nabla) U = \Delta U$

其中$\Delta$为拉普拉斯算子:

$\Delta = \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$

注意若$U$为数量场则$\Delta U$也为数量场, 若$U$为向量场则$\Delta U$也为向量场, 不能写作$\Delta \cdot U$, 后者无意义.

拉普拉斯方程: $\Delta U = 0$为拉普拉斯方程, 满足次方程的解函数$U$称为调和函数.

场的联合运算: 若数量场$U$在$\Omega$上二阶连续可微, 向量场$\vec F$在$\Omega$上二阶连续可微, 则:

$\text{div}(\text{rot}\vec F) = \nabla \cdot (\nabla \times \vec F) = \vec F \cdot \underbrace{(\nabla \times \nabla)}_{=\vec 0} = 0 \tag{1}$

$\text{rot}(\nabla U) = \nabla \times (\nabla U) = \underbrace{(\nabla \times \nabla)}_{=\vec 0} U = \vec 0 \tag{2}$

$\text{div}(\nabla U) = \nabla \cdot (\nabla U) = (\nabla \cdot \nabla) U = \Delta U \tag{3}$

$\begin{aligned} \nabla(\text{div}\vec F) - \text{rot}(\text{rot}\vec F) &= (\nabla \cdot \vec F)\nabla - \underbrace{\nabla \times (\nabla \times \vec F)}_{\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\vec b - (\vec a \cdot \vec b)\vec c} \\ &= (\nabla \cdot \vec F)\nabla - \Big( (\nabla \cdot \vec F)\nabla - (\nabla \cdot \nabla) \vec F \Big) \\ &= (\nabla \cdot \nabla) \vec F \\ &= \Delta \vec F \\ \end{aligned} \tag{4}$

由(1)., 旋度场均为无源场.
由(2)., 梯度场均为无旋场.
由(2)., $\vec F$为有势场/保守场 $\Leftrightarrow$ $\vec F$为无旋场, 即$\| \text{rot}\vec F \| = 0$.

平面通量: $L^+$为连续可微向量场$\vec F$中的定向曲线, $\vec n$为$L^+$在平面内的单位法向量, 则$\vec F$穿过曲线$L$的通量为:

$\int_{L^+} \vec F \cdot \vec n \;\text{d}\ell$

平面环量: 与空间环量一致.

平面散度与旋度:

$\oint_{\partial D^+} \vec F \cdot \vec n \;\text{d}\ell = \iint_D \text{div}\vec F\;\text{d}x\text{d}y$

$\oint_{\partial D^+} \vec F \cdot \vec \tau \;\text{d}\ell = \iint_D \text{rot}\vec F\;\text{d}x\text{d}y$

级数

级数: 无穷项数列求和:

$S = u_1+u_2+u_3+\cdots+u_n+\cdots = \sum_{n=1}^{\infty} u_n$

其中$u_n$为通项/一般项, 若$\forall i, u_i$不含有参数变量, 则该级数为**(常数项)级数**. 正项级数、负项级数、交错级数、任意项级数等都为常数项级数的特殊情况.

部分和: 级数的前$n$项部分和为:

$S_n = u_1+u_2+u_3+\cdots+u_n = \sum_{i=1}^n u_i$

级数收敛当且仅当部分和数列收敛, 且此时两者相等.

$S = \sum_{n=1}^{\infty} u_n = \lim_{n\to\infty} S_n$

级数收敛柯西原理: 又数列收敛柯西原理和级数收敛的部分和等价条件可得, 级数收敛当且仅当: $\forall \varepsilon \gt 0, \exists N \gt 0, \forall n \gt N, p \gt 0$, 都有:

$\vert u_{n+1}+\cdots+u_{n+p} \vert \lt \varepsilon$

$S$收敛 $\Rightarrow$ 通项$u_n$收敛至$0$.
改变$S$的有限项不影响$S$的敛散性(但收敛时会影响级数值, 这与数列极限不同).

余合: $r_n = S-S_n$为级数$S$的第$n$项余和.

级数收敛当且仅当余和收敛至$0$.

$S=\lim_{n\to\infty} r_n = \lim_{n\to\infty} S-S_n = S-\lim_{n\to\infty}S_n = 0$

收敛级数的运算: 若级数$\sum_{n=1}^{\infty} u_n$收敛到$A$, 级数$\sum_{n=1}^{\infty} v_n$收敛到$B$, $\forall \alpha,\beta \in \mathbb R$, $0 = n_0 \lt n_1 \lt n_2 \lt \cdots$, 则:

$\begin{aligned} \sum_{n=1}^{\infty}(\alpha u_n + \beta v_n) &= \alpha A + \beta B \\ \sum_{k=1}^{\infty}(u_{n_{k-1}+1}+\cdots+u_{n_k}) &= A \end{aligned}$

注意新级数$\sum_{k=1}^{\infty}(u_{n_{k-1}+1}+\cdots+u_{n_k})$收敛不保证原级数收敛(e.g.: $1,-1$交错数列). 若收敛的新级数中每一项内符号相同，则原级数收敛.

同号级数: $\forall n \ge 1, u_n \ge 0$, 则$\sum_{n=1}^\infty u_n$为正项级数. $\forall n \ge 1, u_n \le 0$, 则$\sum_{n=1}^\infty u_n$为负项级数. 不失一般性, 通常仅考虑正项级数即可.

正项级数的级数部分和数列$S_n$单调递增, 故正项级数收敛当且仅当$S_n$.

正项级数$\sum_{n=1}^\infty u_n$与$\sum_{n=1}^\infty v_n$满足$\exists N \in \mathbb N_+$ s.t.: $\forall n \gt N$, $u_n \le cv_n$, 其中$c \gt 0$. 则:

1. $\sum_{n=1}^\infty v_n$收敛 $\Rightarrow$ $\sum_{n=1}^\infty u_n$收敛.
2. $\sum_{n=1}^\infty u_n$发散 $\Rightarrow$ $\sum_{n=1}^\infty v_n$发散.

e.g.:
广义调和级数/$p$-级数:

$\zeta(p) = \sum_{n=1}^\infty \frac{1}{n^p}$

因为有:

$\frac{1}{p-1} = \int_1^{+\infty} \frac{1}{x^p} \;\text{d}x \lt \zeta(p) \lt 1+\int_1^{+\infty} \frac{1}{x^p} \;\text{d}x = \frac{p}{p-1}$

故由于$\int_1^{+\infty} \frac{1}{x^p} \;\text{d}x$收敛当且仅当$p \gt 1$, 进而$p$-级数收敛当且仅当$p \gt 1$.

比较判别法:
正项级数$\sum_{n=1}^\infty u_n$与$\sum_{n=1}^\infty v_n$满足$\lim_{n\to\infty}\frac{u_n}{v_n} = k$, 则:

1. $\sum_{n=1}^\infty v_n$收敛, $0 \le k \lt +\infty$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$收敛.
2. $\sum_{n=1}^\infty v_n$发散, $0 \lt k \le +\infty$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$发散.

比阶判别法:
对于正项级数$\sum_{n=1}^\infty u_n$, 若存在$p$使得$\lim_{n\to\infty} n^p u_n = k$.

1. 若$p \gt 1$, $0 \le k \lt +\infty$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$收敛.
2. 若$p \le 1$, $0 \lt k \le +\infty$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$发散.

比值判别法:

$\sum_{n=1}^\infty q^n$

几何级数收敛当且仅当$\vert q \vert \lt 1$. 故对于正项级数$\sum_{n=1}^\infty u_n$, 若$\lim_{n\to\infty}\frac{u_{n+1}}{u_n} = c$, 其中$0 \le c \le +\infty$, 则:

1. $c \lt 1$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$收敛.
2. $c \gt 1$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$发散.
3. $c = 1$, 无法判断.

特别地, 即使$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}$不存在, 若存在对应上界和下界满足上述条件, 也可用于判别.

根式判别法(柯西判别法):
对于正项级数$\sum_{n=1}^\infty u_n$, 若$\lim_{n\to\infty}\sqrt[n]{u_n} = c$, 其中$0 \le c \le +\infty$, 则:

1. $c \lt 1$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$收敛.
2. $c \gt 1$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$发散.
3. $c = 1$, 无法判断.

拉贝判别法:
对于正项级数$\sum_{n=1}^\infty u_n$, 若存在$\mu$使得$n\to\infty$时有$\frac{u_n}{u_{n+1}} = 1 + \frac{\mu}{n} + o(\frac{1}{n})$, 则:

1. $\mu \gt 1$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$收敛.
2. $\mu \lt 1$ $\Rightarrow$ $\sum_{n=1}^\infty u_n$发散.
3. $\mu = 1$, 无法判断.

积分判别法:
对于正项级数$\sum_{n=1}^\infty u_n$, 若$\exists f(x) \in \mathscr C[1,+\infty)$ s.t.: $\forall n \in \mathbb N_+, f(n) = u_n$且$f(x)$为单调递减非负函数, 则: $\sum_{n=1}^\infty u_n$收敛当且仅当$\int_1^{+\infty} f(x) \;\text{d}x$收敛. 特别地,

$\int_1^{+\infty} f(x)\;\text{d}x \le \sum_{n=1}^\infty u_n \le u_1 + \int_1^{+\infty} f(x)\;\text{d}x$

$u_n$非负单调递减, 则$\sum_{n=1}^\infty u_n$收敛当且仅当$\sum_{n=1}^\infty 2^n u_{2^n}$收敛.

Proof:
记$S_n = \sum_{k=1}^n u_k$, $P_n = \sum_{k=1}^n 2^k u_{2^k}$, 则:

$S_{2^{n+1}-1} = \sum_{k=0}^n \sum_{j=2^k}^{2^{k+1}-1} u_j \le \sum_{k=0}^n 2^ku_{2^k} = u_1 + P_n$

$S_{2^{n+1}} = \sum_{k=0}^n \sum_{j=2^k+1}^{2^{k+1}} u_j \ge \sum_{k=0}^n 2^ku_{2^{k+1}} = u_1 + \frac{1}{2}P_{n+1}$

莱布尼茨判别法:
$u_n$非负单调递减趋于$0$, 则$S = \sum_{n=1}^\infty (-1)^{n-1}u_n$必定收敛: $u_1-u_2 \le S \le u_1$. 且部分和数列$S_n = \sum_{k=1}^n (-1)^{k-1}u_k$满足$u_{n+1}-u_{n+2} \le \vert S-S_n \vert \le u_{n+1}$. 注意莱布尼茨判别法仅仅是充分条件.

任意项级数: 任意一项都可以为正项或负项的级数. 若$\sum_{n=1}^\infty \vert u_n \vert$收敛, 则原任意项级数$\sum_{n=1}^\infty u_n$绝对收敛; 若原任意项级数收敛但$\sum_{n=1}^\infty \vert u_n \vert$不收敛, 则原任意项级数条件收敛.

定义任意项级数的正项部分$u_n^+ = \max\lbrace u_n,0 \rbrace$和负项部分$u_n^- = \max\lbrace -u_n,0 \rbrace$. 则:

1. $\sum_{n=1}^\infty u_n$绝对收敛当且仅当$\sum_{n=1}^\infty u_n^+$和$\sum_{n=1}^\infty u_n^-$均收敛.
2. $\sum_{n=1}^\infty u_n$条件收敛则$\sum_{n=1}^\infty u_n^+$和$\sum_{n=1}^\infty u_n^-$均发散(单调递增无上界).

Proof:
(1) $\Rightarrow$显然: $0 \le u_n^+ \le \vert u_n \vert$, $0 \le u_n^- \le \vert u_n \vert$.
(1) $\Leftarrow$显然: $\sum_{n=1}^\infty \vert u_n \vert = \sum_{n=1}^\infty u_n^++u_n^-$.
(2), 根据(1), 条件收敛则至少一个发散, 不妨假设$\sum_{n=1}^\infty u_n^-$发散. 若$\sum_{n=1}^\infty u_n^+$收敛, 则$\sum_{n=1}^\infty u_n$必定发散, 则不条件收敛.

绝对收敛级数满足交换律.

Proof:
绝对收敛级数的正项部分级数和负项部分级数均收敛, 收敛同号级数满足交换律.

黎曼定理:
条件收敛级数必定不满足交换律: 必定存在一个重排列方法使其发散; 且$\forall A \in \mathbb R$, 存在一个重排列方法使其收敛到$A$.

Proof:
注意到条件收敛级数的正项部分级数和负项部分级数均发散.
$\forall M \in \mathbb R$, 可以取$\min k_1$ s.t.: $\sum_{j=1}^{k_1} u_j^+ \ge M$, 再取$\max k_2$ s.t.: $\sum_{j=1}^{k_1} u_j^+ - \sum_{j=1}^{k_2} u_j^- \ge M$, 以此类推, 故级数发散.
$\forall A \in \mathbb R$, 可以取$\min k_1$ s.t.: $\sum_{j=1}^{k_1} u_j^+ \ge A$, 再取$\min k_2$ s.t.: $\sum_{j=1}^{k_1} u_j^+ - \sum_{j=1}^{k_2} u_j^- \le A$, 以此类推, 故级数收敛到$A$.

级数乘法: 级数乘法有两种定义:
级数对角线加法(卷积):

$\begin{aligned} c_n &= \sum_{i=1}^n a_ib_{n+1-j} \\ \sum_{k=1}^n c_k &= \sum_{i+j \le n+1} a_ib_j \\ \end{aligned}$

方块加法(多项式乘法):

$\begin{aligned} c_n &= \sum_{i=1}^n (a_ib_n + a_nb_i) \\ \sum_{k=1}^n c_k &= \left(\sum_{i=1}^n a_i\right) \cdot \left(\sum_{j=1}^n b_j\right) \\ \end{aligned}$

柯西定理:
$\sum_{n=1}^\infty a_n$与$\sum_{n=1}^\infty b_n$绝对收敛, 则任意顺序相加的级数$\sum_{i,j}^\infty a_ib_j$均绝对收敛到级数乘法结果.

Proof:
考虑绝对值方块相加部分和$P_n' = \left(\sum_{k=1}^n \vert a_k \vert \right) \cdot \left(\sum_{k=1}^n \vert b_k \vert \right)$, 故其对应原级数$P_n = \left(\sum_{k=1}^n a_k \right) \cdot \left(\sum_{k=1}^n b_k \right)$绝对收敛. 而绝对收敛数列满足交换律: $\left(\sum_{k=1}^n a_k \right) \cdot \left(\sum_{k=1}^n b_k \right) = \sum_{i,j}^\infty a_ib_j$.

Dirichlet判别法:
$u_n$单调递减趋于$0$, $\sum_{n=1}^\infty v_n$部分和有界, 则$\sum_{n=1}^\infty u_nv_n$收敛.
Abel判别法:
$u_n$单调有界, $\sum_{n=1}^\infty v_n$收敛, 则$\sum_{n=1}^\infty u_nv_n$收敛.

函数项级数: 函数$u_n(x)$定义在$I$上, 则:

$\sum_{n=1}^\infty u_n(x)$

$\forall \alpha \in I$, 若$\sum_{n=1}^\infty u_n(\alpha)$收敛, 则$\alpha$为收敛点. $\sum_{n=1}^{\infty} u_n(x)$的收敛点集合为其收敛域$D$. 若$x \in D$, 则有和函数:

$S(x) = \sum_{n=1}^\infty u_n(x)$

与常数项级数类似对于$x \in D$, 部分和函数有:

$\begin{aligned} S_n(x) = \sum_{k=1}^n u_k(x) \\ \lim_{n\to\infty} S_n(x) = S(x) \\ \end{aligned}$

一致收敛: 若$\sum_{n=1}^\infty u_n(\alpha)$在区间$I$上处处收敛, 部分和函数列$\lbrace S_n(x) \rbrace$在区间$I$上一致收敛到和函数, 则该函数项级数在区间$I$上一致收敛. 具体地, 对于部分和函数列的一致收敛: $\forall \varepsilon \gt 0$, $\exists N \gt 0$, $\forall n \gt N, x \in I$, $\vert S_n(x) - S(x) \vert \lt \varepsilon$, 记为$S_n(x) \stackrel{I}{\rightrightarrows} S(x)\;\;\;\;(n \to \infty)$.

$\lbrace f_n(x) \rbrace$一致收敛当且仅当$\forall \varepsilon \gt 0$, $\exists N \gt 0$, $\forall n \gt N, x \in I$, $\forall p \ge 1$, $\vert f_{n+p}(x) - f_n(x) \vert \lt \varepsilon$.

$f_n(x) \stackrel{I}{\rightrightarrows} f(x)$当且仅当$\lim_{n\to\infty} \sup_{x \in I}\vert f_n(x)-f(x) \vert = 0$.

若$\lbrace f_n(x) \rbrace$在$[a,b]$上有定义, $\forall n \ge 1$, $f_n(x)$在$a$处右连续, 但$\lbrace f_n(a) \rbrace$发散, 则$\forall 0 \lt \delta \lt b-a$, $\lbrace f_n(x) \rbrace$在$(a,a+\delta)$内非一致收敛.

函数项级数一致收敛柯西原理: $\sum_{n=1}^\infty u_n(x)$在区间$I$上一致收敛当且仅当$\forall \varepsilon \gt 0$, $\exists N \gt 0$, $\forall n \gt N, x \in I$, $\forall p \in \mathbb N^\ast$, $\left\vert \sum_{k=n+1}^{n+p} u_k(x) \right\vert \lt \varepsilon$.

若$\lbrace u_n(x) \rbrace$在$I$上非一致收敛到$0$, 则$\sum_{n=1}^\infty u_n(x)$在$I$上非一致收敛.

若$\sum_{n=1}^\infty u_n(x)$在$[a,b]$上有定义, $\forall n \ge 1$, $u_n(x)$在$a$处右连续, 但$\sum_{n=1}^\infty u_n(x)$发散, 则$\forall 0 \lt \delta \lt b-a$, $\sum_{n=1}^\infty u_n(x)$在$(a,a+\delta)$内非一致收敛.

Majorant判别法:
$\forall n \ge 1$, $\vert u_n(x) \vert$在区间$I$上有定义且有上界$c_n$. 若$\sum_{n=1}^\infty c_n$收敛, 则$\sum_{n=1}^\infty u_n(x)$在$I$上绝对收敛且一致收敛.

Dirichlet判别法:
若$\forall x \in I$, $\lbrace u_n(x) \rbrace$单调且$u_n(x) \stackrel{I}{\rightrightarrows} 0$, 且$\sum_{n=1}^\infty v_n(x)$部分和函数列一致有界(i.e.: $\exists M \gt 0$ s.t.: $\left\vert \sum_{k=1}^n v_k(x) \right\vert \le M$), 则$\sum_{n=1}^\infty u_n(x)v_n(x)$在$I$上一致收敛.

Abel判别法:
若$\forall x \in I$, $\lbrace u_n(x) \rbrace$单调且函数列$\lbrace u_n(x) \rbrace$在$I$上一致有界, 且$\sum_{n=1}^\infty v_n(x)$在$I$上一致收敛, 则$\sum_{n=1}^\infty u_n(x)v_n(x)$在$I$上一致收敛.

若$\sum_{n=1}^\infty u_n(x)$在$I$上一致收敛到其和函数$S(x)$, 且$\forall n \ge 1$, $u_n(x) \in \mathscr C(I)$, 则$S(x) \in \mathscr C(I)$.

若$f_n(x) \stackrel{I}{\rightrightarrows} f(x)$且$\forall n \ge 1$, $f_n(x) \in \mathscr C(I)$, 则$f(x) \in \mathscr C(I)$.

若$\forall n \ge 1$, $u_n(x) \in \mathscr C(I)$但$S(x) \not \in \mathscr C(I)$, 则$\sum_{n=1}^\infty u_n(x)$在$I$上非一致收敛.

若$\sum_{n=1}^\infty u_n(x)$在$[a,b]$上一致收敛到其和函数$S(x)$, 且$\forall n \ge 1$, $u_n(x) \in \mathscr C[a,b]$, 则$S(x) \in \mathscr R[a,b]$且:

$\int_a^b S(x)\;\text{d}x = \sum_{n=1}^\infty \int_a^b u_n(x)\;\text{d}x = \lim_{n\to\infty} \int_a^b S_n(x)\;\text{d}x$

一般地, 若$f_n(x) \stackrel{[a,b]}{\rightrightarrows} f(x)$, 且$\forall n \ge 1$, $f_n(x) \in \mathscr C[a,b]$, 则$f(x) \in \mathscr R[a,b]$且:

$\int_a^b f(x)\;\text{d}x = \lim_{n\to\infty} \int_a^b f_n(x)\;\text{d}x$

若$\sum_{n=1}^\infty u_n(x) = S(x)$, $\sum_{n=1}^\infty u_n'(x)$在$[a,b]$上一致收敛, 且$\forall n \ge 1$, $u_n(x) \in \mathscr C^{(1)}[a,b]$, 则$S(x) \in \mathscr C^{(1)}[a,b]$且:

$S'(x) = \sum_{n=1}^\infty u'_n(x)$

一般地, 若$f_n(x) \stackrel{[a,b]}{\rightrightarrows} f(x)$, 且$\forall n \ge 1$, $f_n(x) \in \mathscr C^{(1)}[a,b]$, 则$f^{(1)}(x) \in \mathscr C[a,b]$且:

$\int_a^b f'(x)\;\text{d}x = \lim_{n\to\infty} \int_a^b f'_n(x)\;\text{d}x$

幂级数:

$\sum_{n=0}^\infty a_n(x-a)^n$

Abel第一定理:
$\sum_{n=0}^\infty a_nx^n$在$x_0 \ne 0$处收敛, 则$\forall \vert h \vert \lt \vert x_0 \vert$, $\sum_{n=0}^\infty a_nx^n$在$[-h,h]$上绝对收敛且一致收敛.

Proof:
$\vert a_nx^n \vert = \vert a_nx_0^n \vert \cdot \left\vert \left(\frac{x}{x_0}\right)^n \right\vert \le \vert a_nx_0^n \vert \cdot \left\vert \frac{h}{x_0} \right\vert^n$. 注意到$a_nx_0^n$趋于$0$故有上界$M$, 根据Majorant判别法, 该式$\le M \cdot \left\vert \frac{h}{x_0} \right\vert^n$, 后者为公比小于$1$的等比数列, 故收敛. 故原级数绝对收敛且一致收敛.

$\sum_{n=0}^\infty a_nx^n$在$x_1 \ne 0$处发散, 则$\forall \vert x_2 \vert \gt \vert x_1 \vert$, $\sum_{n=1}^\infty a_nx_2^n$发散.

$\sum_{n=0}^\infty a_nx^n$在$x_0 \ne 0$处收敛, 在$x_1 \ne 0$处发散, 则$\exists ! r \gt 0$, 使得$\sum_{n=0}^\infty a_nx^n$在$(-r,r)$内绝对收敛, 在$[-r,r]$外处处发散.

收敛半径: 若$\exists! r \in [0,\infty]$ s.t.: $\forall \vert x \vert \lt r$, $\sum_{n=0}^\infty a_nx^n$绝对收敛, $\forall \vert x \vert \gt r$, $\sum_{n=0}^\infty a_nx^n$发散. 则$r$为$\sum_{n=0}^\infty a_nx^n$的收敛半径; $(-r,r)$为$\sum_{n=0}^\infty a_nx^n$的收敛区间(注意收敛区间一定是开区间).

幂级数的收敛域为收敛区间并上收敛的区间端点. 收敛域是以原点为中心的区间.

对于$\sum_{n=0}^\infty a_nx^n$的收敛半径$r$, $\forall [a,b] \subseteq (-r,r)$, $\sum_{n=0}^\infty a_nx^n$在$[a,b]$上绝对收敛且一致收敛.

若$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = l$或$\lim_{n\to\infty}\sqrt[n]{\vert a_n \vert} = l$, 则$\sum_{n=0}^\infty a_nx^n$的收敛半径为$r = 1/l$. 特别地, $l = 0$时$r = +\infty$; $l = +\infty$时$r = 0$.

Abel第二定理:
若$\sum_{n=0}^\infty a_nx^n$收敛半径为$R$且在$R$(或$-R$)处收敛, 则$\forall 0 \lt r \lt R$, $\sum_{n=0}^\infty a_nx^n$在$[-r,R]$(或$[-R,r]$)上一致收敛.

Proof:
$[-r,R] = [-r,0] \bigcup [0,R]$. $[-r,0] \subseteq (-R,R)$显然一致收敛, 对于$[0,R]$: $\sum_{n=0}^\infty a_nx^n = \sum_{n=0}^\infty a_nR^n \cdot \frac{x^n}{R^n}$, 根据Abel判别法, 前者与$x$无关在$[0,R]$一致收敛, 后者在$[0,R]$单调递减且一致收敛, 则原级数在$[0,R]$一致收敛.

若$\sum_{n=0}^\infty a_nx^n$收敛半径为$r$, 则:

1. $S(x) \in \mathscr C(-r,r)$.
2. 若级数在$r$处收敛, 则$S(x) \in \mathscr C(-r,r]$.
3. 若级数在$-r$处收敛, 则$S(x) \in \mathscr C[-r,r)$.
4. $\forall [a,b] \subseteq (-r,r)$, $S(x) \in \mathscr R[a,b]$, 且:

   $\int_a^b S(x)\;\text{d}x = \sum_{n=0}^\infty \int_a^b a_nx^n \;\text{d}x = \sum_{n=0}^\infty \frac{1}{n+1}a_n(b^{n+1}-a^{n+1})$

$\sum_{n=0}^\infty \frac{1}{n+1}a_nx^{n+1}$, $\sum_{n=0}^\infty a_nx^n$, $\sum_{n=1}^\infty na_nx^{n-1}$收敛半径相同.

若$\sum_{n=0}^\infty a_nx^n$收敛半径为$r$, 则$S(x) \in \mathscr C^{(\infty)}(-r,r)$且:

$S^{(k)}(x) = \sum_{n=k}^\infty n^{\underline k} a_n x^{n-k}$

(注意$S(x)$仅在收敛域上有意义)

泰勒级数: 若$f(x)$在$(a-r,a+r)$内能展开成幂级数$f(x) = \sum_{n=0}^\infty a_n(x-a)^n$, 则$r$为该级数的收敛半径, $f(x)$为该级数的和函数, $f(x)$在$(a-r,a+r)$内存在任意阶导数且$f^{(k)}(x) = S^{(k)}(x)$. 特别地, $f^{(k)}(a) = k!a_k$, 故$a_k = \frac{f^{(k)}(a)}{k!}$唯一确定, 则:

$f(x) \sim \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$

称右侧级数为$f(x)$在$a$点的泰勒级数, $f(x)$在$a=0$处的泰勒级数为$f(x)$的麦克劳林级数.

$f(x)$在点$a$处可以展开为泰勒级数当且仅当$f(x)$在$(a-r,a+r)$内任意阶导数存在, 且在$\forall x \in (a-r,a+r)$, $n\to\infty$时拉格朗日余项$R_n(x)$趋近于$0$.

$\lim_{n\to\infty} R_n = \lim_{n\to\infty}\frac{f^{(n+1)}(\xi)(x-a)^{n+1}}{(n+1)!}$

注意到一个充分条件为: $f(x)$在$(a-r,a+r)$内各阶导数一致有界, 则:

$\lim_{n\to\infty} \vert R_n \vert \le \lim_{n\to\infty}\frac{Mr^{n+1}}{(n+1)!} = M\lim_{n\to\infty}\frac{r^{n+1}}{(n+1)!} = 0$

e.g.:

$\begin{aligned} e^x &\sim \sum_{n=0}^\infty \frac{1}{n!}x^n \\ \sin x &\sim \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1} \\ \cos x &\sim \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n} \\ \ln(1+x) &\sim \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n &x \in (-1,1] \\ (1+x)^\alpha &\sim \sum_{n=0}^\infty \frac{\alpha^{\underline n}}{n!}x^n &x \in \begin{cases}(-1,1) & \alpha \le -1 \\ (-1,1] & -1 \lt \alpha \lt 0 \\ [-1,1] & \alpha \gt 0 \\\end{cases} \\ \arcsin x &\sim \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n+1)(2n)!!}x^{2n+1} &x \in [-1,1] \\ \arctan x &\sim \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} &x \in [-1,1] \\ \end{aligned}$

特别地,

$\begin{aligned} \frac{1}{\sqrt{1+x}} &= (1+x)^{-1/2} = \sum_{n=1}^\infty \frac{(-1)^n(2n-1)!!}{(2n)!!} x^{2n} \\ \sqrt{1+x} &= (1+x)^{1/2} = 1+\frac{1}{2}x+\sum_{n=2}^\infty (-1)^{n+1}\frac{(2n-3)!!}{(2n)!!}x^n \\ \frac{\pi}{2} &= \lim_{n\to\infty} \left( \frac{(2n)!!}{(2n-1)!!}\right)^2\frac{1}{2n+1} \end{aligned}$

三角函数系: $\forall m,n \in \mathbb N^+$:

$\int_{-\pi}^\pi \sin nx \cos mx \;\text{d}x = \frac{1}{2}\int_{-\pi}^\pi \sin((n+m)x) + \sin((n-m)x)\;\text{d}x = 0$

$\int_{-\pi}^\pi \sin nx \sin mx \;\text{d}x = \pi [m=n]$

$\int_{-\pi}^\pi \cos nx \cos mx \;\text{d}x = \pi [m=n]$

故$\lbrace \frac{1}{\sqrt{2\pi}},\frac{1}{\sqrt \pi}\cos (nx),\frac{1}{\sqrt \pi}\sin(nx) \rbrace$为标准正交三角函数系: 即任意不同的二者内积为$0$, 相同的内积为$1$.

Fourier级数: 设$f(x) \in \mathscr R[-\pi,\pi]$, 则存在Fourier系数$\lbrace a_n \rbrace_{n=0}^\infty, \lbrace b_n \rbrace_{n=1}^\infty$ s.t.: $f(x)$的Fourier级数为:

$f(x) \sim \frac{a_0}{2} +\sum_{n=1}^\infty a_n \cos (nx) + b_n \sin (nx) \qquad \qquad (x \in [-\pi,\pi])$

若Fourier级数收敛则可记为:

$f(x) = \frac{a_0}{2} +\sum_{n=1}^\infty a_n \cos (nx) + b_n \sin (nx) \qquad \qquad (x \in [-\pi,\pi])$

对等式两侧同时乘$\cos nx$并积分, 可得:

$a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos(nx) \;\text{d}x$

对等式两侧同时乘$\sin nx$并积分, 可得:

$b_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin(nx) \;\text{d}x$

Dirichlet收敛定理:
若$f(x)$在$[a,b]$上有界, 且存在分割$a = x_0 \lt x_1 \lt \cdots \lt x_{n-1} \lt x_n = b$使得在任意区间$(x_{k-1},x_k)$上$f(x)$单调, 则称$f(x)$分段单调. 若$f(x)$以$2\pi$为周期, 在$[-\pi,\pi]$分段单调有界, 则$f(x)$的Fourier级数在$\mathbb R$上收敛, 其在$x$处的展开收敛到:

$\frac{1}{2}\Big(f_+(x) + f_-(x)\Big)$

若$f(x)$为奇函数, 则$a_n = 0$; 若$f(x)$为偶函数, 则$b_n = 0$.

若$f(x)$周期为$2l$, 则可将$\varphi(y) = f(\frac{l}{\pi}y) = f(x)$展开为Fourier级数. 故:

$f(x) \sim \frac{a_0}{2} +\sum_{n=1}^\infty a_n \cos \left(\frac{n\pi}{l}x\right) + b_n \sin \left(\frac{n\pi}{l}x\right)$

$a_n = \frac{1}{l}\int_{-l}^l f(x)\cos\left(\frac{n\pi}{l}x\right) \;\text{d}x$

$b_n = \frac{1}{l}\int_{-l}^l f(x)\sin\left(\frac{n\pi}{l}x\right) \;\text{d}x$

三角多项式: 对于$\mathbb R$上的$\lbrace c_k \rbrace_{k=0}^n, \lbrace d_k \rbrace_{k=1}^n$, 若$c_nd_n \ne 0$, 则有$n$阶三角多项式:

$T_n(x) = \frac{1}{2}c_0 + \sum_{k=1}^n c_k \cos(kx) + d_k \sin(kx)$

若$f(x) \in \mathscr R[-\pi,\pi]$以$2\pi$为周期, 则其Fourier级数的$n$阶部分和$S_n(f,x)$为一个阶数$\le n$的三角多项式, 而且该三角多项式最小化方均误差:

$S_n(f,x) = \argmin_{T_k(x), k \le n} \int_{-\pi}^\pi (f(x)-T_k(x))^2\;\text{d}x$

Proof:
$\forall T_k(x)$ s.t.: $k \le n$:

$\begin{aligned} \frac{1}{2\pi}\int_{-\pi}^\pi (f(x)-T_k(x))^2\;\text{d}x &= \frac{1}{2\pi}\int_{-\pi}^\pi f^2(x) + T^2_k(x) \;\text{d}x - \frac{1}{\pi} \int_{-\pi}^\pi f(x)T_k(x)\;\text{d}x \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi f^2(x) \;\text{d}x + \frac{1}{2}\left(\frac{1}{2}c_0^2 + \sum_{k=1}^n c_k^2+d_k^2 \right) - \left(\frac{1}{2}a_0c_0 + \sum_{k=1}^n c_ka_k+d_kb_k\right) \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi f^2(x) \;\text{d}x + \underbrace{\frac{1}{2}\left(\frac{1}{2}(a_0-c_0)^2+\sum_{k=1}^n (a_k-c_k)^2+(d_k-b_k)^2 \right)}_{\ge 0} - \frac{1}{2}\left(\frac{1}{2}a_0^2 + \sum_{k=1}^n a_k^2+b_k^2 \right) \\ &\ge \frac{1}{2\pi}\int_{-\pi}^\pi f^2(x) \;\text{d}x - \frac{1}{2}\left(\frac{1}{2}a_0^2 + \sum_{k=1}^n a_k^2+b_k^2 \right) \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi (f(x)-S_n(f,x))^2\;\text{d}x \end{aligned}$

第一逼近定理:
$\forall f(x) \in \mathscr C[a,b]$, $\forall \varepsilon \gt 0$, $\exists$ 多项式 $p(x)$ s.t.: $\forall x \in [a,b], \vert f(x)-p(x) \vert \lt \varepsilon$.

第二逼近定理:
$\forall f(x)$连续且周期为$2\pi$, $\forall \varepsilon \gt 0$, $\exists$ 三角多项式 $q(x)$ s.t.: $\forall x \in \mathbb R, \vert f(x)-q(x) \vert \lt \varepsilon$.

Lebesgue可积函数逼近定理:
$\forall f(x) \in \mathscr R[a,b]$, $\forall \varepsilon \gt 0$, $\exists g(x) \in \mathscr C[a,b]$ s.t.: $\int_a^b (f(x)-g(x))^2\;\text{d}x \lt \varepsilon$.

由上述定理可进一步得到推论:
$\forall f(x)$可积且周期为$2\pi$, $\forall \varepsilon \gt 0$, $\exists$ 三角多项式 $q(x)$ s.t.: $\int_{-\pi}^{\pi} (f(x)-q(x))^2\;\text{d}x \lt \varepsilon$.

Bessel不等式: 若$f(x) \in \mathscr R[-\pi,\pi]$以$2\pi$为周期, 则$\forall n \in \mathbb N^+$:

$\frac{1}{\pi}\int_{-\pi}^\pi f^2(x) \;\text{d}x \ge \frac{1}{2}a_0^2 + \sum_{k=1}^n a_k^2+b_k^2$

Parseval等式: 若$f(x) \in \mathscr R[-\pi,\pi]$以$2\pi$为周期, 则:

$\frac{1}{\pi}\int_{-\pi}^\pi f^2(x) \;\text{d}x = \frac{1}{2}a_0^2 + \sum_{k=1}^\infty a_k^2+b_k^2$

更一般地, 若$f(x),g(x) \in \mathscr R[-\pi,\pi]$以$2\pi$为周期, 则:

$\frac{1}{\pi}\int_{-\pi}^\pi f(x)g(x) \;\text{d}x = \frac{1}{2}a_0c_0 + \sum_{k=1}^\infty a_kc_k+b_kd_k$

Riemann-Lebesgue引理:
$f(x) \in \mathscr R[a,b]$, 则:

$\begin{aligned} \lim_{\vert \lambda \vert \to \infty} \int_a^b f(x) \cos(\lambda x)\;\text{d}x = 0 \\ \lim_{\vert \lambda \vert \to \infty} \int_a^b f(x) \sin(\lambda x)\;\text{d}x = 0 \\ \end{aligned}$

故若$f(x) \in \mathscr R[a,b]$, 则$\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = 0$.
更进一步, 若$f'(x) \in \mathscr R[a,b]$且$f(-\pi)=f(\pi)$, 则$a_n=o(\frac{1}{n}), b_n=o(\frac{1}{n})$. 这是因为$a_n = \frac{1}{n}b_n', b_n = -\frac{1}{n}a'_n$.

复谐振动:

$x = Ce^{i\omega t} = re^{i(\theta + \omega t)} = r(\cos(\theta+\omega t)+ i \sin(\theta+\omega t))$

其中$C = re^{i\theta}$为复振幅, $\vert C \vert$为振幅, $\theta$为初相, $\omega \in \mathbb R$为圆频率.

复数形式Fourier级数:
设$f(t) \in \mathscr R[-l,l]$周期为$2l$, 令$\omega = \pi/l$, 则其Fourier级数可以写成:

$\begin{aligned} f(t) &\sim \frac{a_0}{2} + \sum_{n=1}^\infty a_n\cos(n \omega t) +b_n\sin( n\omega t) \\ &\sim \frac{a_0}{2} + \frac{1}{2}\sum_{n=1}^\infty a_n(e^{in\omega t}+e^{-in\omega t})+b_n(ie^{-in\omega t}-ie^{in\omega t}) \\ &\sim \frac{a_0}{2} + \frac{1}{2}\sum_{n=1}^\infty (a_n-ib_n)e^{in\omega t} + (a_n+ib_n)e^{-in\omega t} \\ &\sim \sum_{k=-\infty}^\infty c_ke^{ik\omega t} \end{aligned}$

其中$c_0 = a_0/2$, $c_k = \frac{a_k-ib_k}{2}$, $c_{-k} = \frac{a_k+ib_k}{2} = \overline{c_k}$ ($k \gt 0$), 则有:

$c_k = \frac{1}{2l}\int_{-l}^l f(t)e^{-ik\omega t}\;\text{d}t \qquad \qquad (k \in \mathbb Z)$

复值函数内积: 若$f(x),g(x)$为$[-l,l]$上可积实变复值函数, 其内积为:

$\langle f,g \rangle = \frac{1}{2l}\int_{-l}^l f(t)\overline{g(t)}\;\text{d}t$

$\lbrace e^{ik\omega t} \vert k \in \mathbb Z\rbrace$为标准正交函数系.

$f(t) \sim \sum_{k=-\infty}^\infty c_ke^{ik\omega t} \sim \sum_{k=-\infty}^\infty \langle f,e^{ik\omega t} \rangle e^{ik\omega t}$

Fourier积分: $f(t) \in\mathscr C(\mathbb R)$为非周期函数, $f_l(t)$为$f(t)$定义在$[-l,l]$上的部分延拓产生的周期为$2l$的周期函数(i.e.: $f(t+2l) = f(t)$). 若$f$在任意有限区间上分段单调, 则$f$可以展开成Fourier级数, 则:

$\begin{aligned} f_l(t) &= \sum_{k=-\infty}^\infty c_k e^{ik\omega t} \\ &= \sum_{k=-\infty}^\infty \langle f,e^{ik\omega t} \rangle e^{i\omega_k t} \\ &= \frac{1}{2l}\sum_{k=-\infty}^\infty \int_{-l}^l f(u)e^{ik\omega(t-u)}\;\text{d}u \\ \end{aligned}$

考虑构造连续实变量$\omega_k = k\omega$. 则$\frac{1}{2l} = \frac{1}{2\pi}(\omega_k-\omega_{k-1}) = \frac{1}{2\pi} \Delta \omega_k$. 那么:

$\begin{aligned} f(t) &= \lim_{l\to+\infty} f_l(t) \\ &= \lim_{l\to+\infty} \frac{1}{2\pi}\sum_{k=-\infty}^\infty \Delta \omega_k \cdot \int_{-l}^l f(u)e^{ik\omega(t-u)}\;\text{d}u \\ &= \lim_{\Delta\omega_k=\frac{\pi}{l} \to 0} \frac{1}{2\pi}\sum_{k=-\infty}^\infty \Delta \omega_k \cdot \int_{-l}^l f(u)e^{i\omega_k(t-u)}\;\text{d}u \\ &= \frac{1}{2\pi}\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(u)e^{i\omega_{\omega}(t-u)}\;\text{d}u\;\text{d}\omega \\ f(t) &= \frac{1}{2\pi}\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(u)e^{i\omega(t-u)}\;\text{d}u\;\text{d}\omega \\ \end{aligned}$

若$f$在$\mathbb R$上绝对可积且在$t$处连续, 则上述积分式为$f(t)$的Fourier积分.

Fourier变换: 根据Fourier积分可以定义含参积分:

$\begin{aligned} \hat f(\omega) &= \frac{1}{2\pi} \int_{-\infty}^{+\infty} f(u)e^{-i\omega u}\;\text{d}u \\ f(t) &= \int_{-\infty}^{+\infty} \hat f(\omega)e^{i\omega t}\;\text{d}\omega \end{aligned}$

$\hat f$为$f$的Fourier变换, $f$为$\hat f$的Fourier逆变换.

特别地, 由于$e^{i\omega u} = \cos(\omega u) + i \sin(\omega u)$, 而$\sin(\omega u)$为关于原点的奇函数, 积分为$0$, 故也可写作:

$\begin{aligned} \hat f(\omega) &= \frac{1}{2\pi} \int_{-\infty}^{+\infty} f(u)\cos(\omega u)\;\text{d}u \\ f(t) &= \int_{-\infty}^{+\infty} \hat f(\omega)\cos(\omega t)\;\text{d}\omega \end{aligned}$

离散Fourier变换:
若$f$以$2\pi$为周期, 则:

$\begin{aligned} \hat f(n) &= \frac{1}{2\pi}\int_{-\infty}^\infty f(u)e^{-inu}\;\text{d}u \\ f(t) &= \sum_{n=-\infty}^{+\infty} \hat f(n)e^{int} \\ \end{aligned}$

$f,g$在$\mathbb R$上绝对可积, 则:

$\widehat{\alpha f + \beta g} = \alpha \hat f(t) + \beta \hat g$

$f(x),f'(x)$在$\mathbb R$上绝对可积且$\lim_{t\to\infty} f(t) = 0$, 则:

$\hat f'(x) = (ix)\hat f(x)$

更进一步, 若$\forall k \in \mathbb N$, $f^{(k)}$在$\mathbb R$绝对可积且$\lim_{t\to\infty} f^{(k)}(t) = 0$, 则:

$\hat f^{(k)}(x) = (ix)^k \hat f(x)$

$f,g$满足常系数微分方程$g(t) = \sum_{k=0}^n a_k f^{(k)}(t)$, 则:

$\hat f = \frac{\hat g}{\sum_{k=0}^n a_k(it)^k}$

卷积:

$(f \ast g)(t) = \frac{1}{2\pi} \int_{-\infty}^\infty f(t-u)g(u)\;\text{d}u$

$f,g$在$\mathbb R$上绝对可积, 则:

$\widehat{f \ast g} = \hat f \cdot \hat g$

$f,g,h$满足卷积方程$f = g + h \ast f$, 则:

$\hat f = \frac{\hat g}{1-\hat h}$

扫描二维码即可在手机上查看这篇文章，或者转发二维码来分享这篇文章: