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\[
V(x) = \begin{cases}
+\infty & (x < 0) \\
-\frac{\hbar^2 \kappa^2}{2m} \delta(x - a) & (x \geq 0)
\end{cases}
\] |
|
\begin{tabular}{cc}
\hline
\textbf{周} & \textbf{人数} & \textbf{犹太人} \\
\hline
-2 & 55 & 141 \\
-1 & 33 & 145 \\
1 & 70 & 139 \\
2 & 49 & 161 \\
\hline
\end{tabular} |
|
\[
m \ddot{x} &= F_x(x, y, z) \quad \textit{coupled} \\
m \ddot{y} &= F_y(x, y, z) \quad \textit{eqns} \\
m \ddot{z} &= F_z(x, y, z) \quad \textit{in general}
\] |
|
\[
(\alpha^2 - c^2 \delta^2) x^2 + y^2 + z^2 - 2 \left( {v\alpha ^2 + \beta \delta c^2} \right) x T = \left( c^2 \beta^2 - \alpha^2 v^2 \right) {t^2}
\]
Comparing 竭? and 竭?:
\begin{align*}
\alpha^2 - c^2 \delta^2 &= 1 \quad \rightarrow \text{(i)} \\
-2 \left( {v\alpha ^2 + \beta \delta c^2} \right) &= 0 \quad\Rightarrow v\alpha ^2 + \beta \delta c^2\rightarrow\text{(ii)} \\
c^2 \beta^2 - \alpha^2 v^2 &= c^2 \quad \rightarrow\text{(iii)}
\end{align*}From (ii) \(\Rightarrow \alpha^2 = -\frac{\beta x c^2}{v}\) \(\rightarrow\) (iv)
\qquad(i) \(\Rightarrow -\frac{\beta x c^2}{v^2} = c^2 \delta^2 = 1\)
\[-c^2 \delta \left( \frac{\beta + v \delta} {v}\right)=1\] \(\Rightarrow c^2 \delta \left( \beta + v \delta \right) = -v \quad\) (v)
(iii ) \(\Rightarrow c^2 \beta^2 - \left(-\frac{\beta x c^2}{v}\right) v^2 - c^2 \Rightarrow \varphi^2 \beta (\beta + \delta v) = \varphi^2\)
\(\Rightarrow\) \(\beta (\beta + \delta v) = 1\) \(\rightarrow\) (vi) |
|
\textbf{证明} 因为
\[
f(z) - A = u(x, y) - u_0 + i [v(x, y) - v_0].
\]
利用不等式
\[
|u(x, y) - u_0| \leq |f(z) - A|, \quad |v(x, y) - v_0| \leq |f(z) - A|, \tag{2-1}
\]
及
\[
|f(z) - A| \leq |u(x, y) - u_0| + |v(x, y) - v_0|. \tag{2-2}
\]
根据极限的定义，由式 (2-1) 可得必要性部分的证明，由式 (2-2) 可得充分性部分的证明。
定义2.1的重要意义在于：它揭示了复变函数的极限与实变函数极限的紧密关系，即将求复变函数的极限问题转化为求两个一元实值函数的极限问题。 |
|
To consider differential operators we restrict further to the inner product space \(C_c^\infty(\mathbb{R})\) of complex-valued, infinitely times differentiable functions with compact support, still with scalar product defined by Eq. (1.59), setting \(a = -\infty\) and \(b = \infty\). What is the adjoint of the differential operator \(D = d/dx\)? The short calculation
\[
(f, D(g)) = \int_{-\infty}^{\infty} dx \, f(x) g'(x) = \left[ f(x) g(x) \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} dx \, f'(x) g(x) = - \langle D(f), g \rangle\tag{1.66}
\] |
|
Using Equation (5.1), but recognizing that \(w\) is a negative number
\[
\alpha_i(y_o) = \tan^{-1} \left( -\frac{w(y_o)}{V_\infty} \right) \tag{5.1b}
\]
for small angles Eq. (5.1b) gives
\[
\alpha_i(y_o) = \frac{1}{4\pi V_\infty} \int_{-b/2}^{b/2} \left( \frac{d\Gamma/dy}{y_o - y} \right) dy \tag{5.20}
\] |
|
因此，\(a_0, \rho_0^k a_k, \rho_0 b_k\) 就是 \(f(\theta)\) 在 \([0, 2\pi]\) 上展开为富里埃级数时的系数，即有
\[
\begin{cases}
a_0 &= \frac{1}{\pi} \int_0^{2\pi} f(\theta) d\theta \\
a_k &= \frac{1}{\rho_0^k \pi} \int_0^{2\pi} f(\theta) \cos k\theta d\theta \quad (k = 1, 2, 3, \ldots) \\
b_k &= \frac{1}{\rho_0^k \pi} \int_0^{2\pi} f(\theta) \sin k\theta d\theta
\end{cases}
\tag{3.2.27}
\] |
|
22. 一小木块以初速度 \(v_0\) 沿斜面匀减速下滑，经过距离 \(s_1\) 停下，以初速度 \(v_0\) 沿斜面向上滑行时，经过距离 \(s_2\) 停下。若在斜面上沿水平方向装一条光滑的板条，当小木块仍以初速度 \(v_0\) 在斜面上沿光滑板条滑行时，通过的距离是多少？(图 3-19) |
|
\begin{align*}
H_o(j\omega) = \frac{1}{j\omega} - e^{-j\omega t} \frac{1}{j\omega}
\end{align*} |
|
\[
\text{Thus}
\quad
\begin{bmatrix}
\cos \theta \, x_1 & -\sin \theta \, x_2 \\
\sin \theta \, x_1 & \cos \theta \, x_2
\end{bmatrix}
\quad
(*)
=
\begin{bmatrix}
\lambda x_1 \\
\lambda x_2
\end{bmatrix}
.
\]
\textit{In particular,} |
|
\[
U = \sum_{\epsilon_i} \epsilon_i \Delta \rho = \frac{2\pi m S}{h^2} \int_0^\infty \frac{d\mu}{e^{\frac{\epsilon}{kT}} - 1} d\epsilon\\
= \frac{2\pi m S}{h^2} \frac{1}{z^2/\mu^2} = \frac{\pi N L^2 h}{m S}\\
\epsilon = \frac{p^2}{2m} \neq \mu B
\] |
|
\[
u_0(z) &= \alpha, \quad v_0(z) = \beta,\\
u_{n+1}(z) &= \alpha + \int_{z_0}^z \left[ a u_n + b v_n \right] d\zeta,\\
v_{n+1}(z) &= \beta + \int_{z_0}^z \left[ c u_n + d v_n \right] d\zeta.
\] |
|
we go back to Eqs. (5.14) and (5.29)
\[
L &= \int_{-b/2}^{b/2} L'(y) dy = \rho_\infty V_\infty \Gamma_0 \int_{-b/2}^{b/2} \sqrt{1 - \left(\frac{2y}{b}\right)^2} dy \\
&= \rho_\infty V_\infty \Gamma_0 \int_{\pi}^{0} \sqrt{1 - \cos^2 \theta} \left(-\frac{b}{2}\sin \theta d\theta\right) \\
&= \rho_\infty V_\infty \Gamma_0 \frac{b}{2} \int_{0}^{\pi} \sin^2 \theta d\theta \tag{5.38} \\
&= \rho_\infty V_\infty \Gamma_0 \frac{b}{2} \int_{0}^{\pi} \frac{1 - \cos 2\theta}{2} d\theta \\
&= \rho_\infty V_\infty \Gamma_0 \frac{b}{2} \left[\frac{\theta}{2} - \frac{\sin 2\theta}{4}\right]_0^{\pi} \\
&= \rho_\infty V_\infty \Gamma_0 \frac{b}{2} \left[\frac{\pi}{2} - 0\right] \\
&= \rho_\infty V_\infty \Gamma_0 \frac{b\pi}{4} \\
L &= \rho_\infty V_\infty \Gamma_0 \frac{b}{4} \pi \tag{5.39}
\] |
|
Two approaches to solve:
\textbf{(a)} \textit{indefinite integration}
\[
\dot{y} &= - \int g \, dt + C_1 \\
&= - gt + C_1
\]
and integrate again:
\[
y(t) = - \frac{1}{2} gt^2 + C_1 t + C_2
\] |
|
定理 1
给定流形 \(N\), 其上一点 \(p \in N\) 处有两个图 \((U, \varphi)\) 和 \((V, \phi)\)。\(p\) 出发的两条道首 \(r_1\) 和 \(r_2\), 它们 \(\varphi \circ r_1\) 和 \(\varphi \circ r_2\) 收敛于同一个切向量。当且仅当 \(\phi \circ r_1\) 和 \(\phi \circ r_2\) 也收敛于同一个切向量。
证明：为了便于，将 \(\phi \circ \varphi^{-1}\) 记为 \(f\)。
由于 \(f : \mathbb{R}^n \rightarrow \mathbb{R}^m\) 是一个双向光滑双射，即 \(f\) 和 \(f^{-1}\) 都是双对局光滑。于是它的 Jacobi 矩阵 \(\partial f / \partial v\) 是非零的；换句话说，如果局部自层值函数 \(f\) 的复合函数被改写为 \(f_j: \mathbb{R}^m \rightarrow \mathbb{R}\)，那么 \(f_j\) 的偏导数 \(\nabla f_j\) 处处非零不同条。类比地，\(f^{-1}\) 的分量函数的梯度也处在不同不平素。在以下证明中，为了方便，我们将省略 Jacobi 矩阵表示方法。 |
|
we may simplify \(\mathbf{J}\) to
\[
\mathbf{J} = \frac{i\hbar}{2} \sum_{\lambda \lambda'} \int d^3k \left( \mathbf{e}_{\mathbf{k}}^{(\lambda)} \times \mathbf{e}_{\mathbf{k}}^{(\lambda')} \right) \left( a_{\mathbf{k}}^{(\lambda) \dagger} a_{\mathbf{k}}^{(\lambda')} - a_{\mathbf{k}}^{(\lambda) \dagger} a_{\mathbf{k}}^{(\lambda')} \right) \tag{22.44}
\]
or
\[
\mathbf{J} = i\hbar \int d^3k \left( \mathbf{e}_{\mathbf{k}}^{(1)} \times \mathbf{e}_{\mathbf{k}}^{(2)} \right) \left( a_{\mathbf{k}}^{(2) \dagger} a_{\mathbf{k}}^{(1)} - a_{\mathbf{k}}^{(1) \dagger} a_{\mathbf{k}}^{(2)} \right) \tag{22.45}
\]
or
\[
\mathbf{J} = i\hbar \int d^3k \, \mathbf{k} \left( a_{\mathbf{k}}^{(2) \dagger} a_{\mathbf{k}}^{(1)} - a_{\mathbf{k}}^{(1) \dagger} a_{\mathbf{k}}^{(2)} \right). \tag{22.46}
\] |
|
\textbf{定理 1}
对任意的仿射集 \(C \subseteq V\), 存在唯一的向量子空间 \(U \subseteq V\), 使得对任意的 \(x \in C\), 我们有
\[
C = x + U = \{x + v \mid v \in U\}.
\] |
|
the \(\zeta\) plane, then
\[
q(x, y) = \kappa(\xi, \eta) \frac{dz}{d\zeta}.
\]
It is, however,
\[
\left| \frac{dz}{d\zeta} \right| = \frac{1}{\sigma^2} \sqrt{ \left( \sigma^2 - \frac{l^2}{4} \right)^2 + l^2 \eta^2}, \quad \sigma^2 = \xi^2 + \eta^2,
\] |
|
常用微分算符运算法式：
\[
&\nabla \times (\nabla \times \mathbf{A}) = \nabla (\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A}\\
&\nabla \cdot (\phi \mathbf{A}) = \phi \nabla \cdot \mathbf{A} + \mathbf{A} \cdot \nabla \phi
\]
积分变换公式：
\[
&\oiint \mathbf{E} \cdot \mathrm{d}\mathbf{S} = \iiint (\nabla \cdot \mathbf{E}) \, \mathrm{d}V \quad \text{Gauss(高斯)定理}\\
&\oint \mathbf{A} \cdot \mathrm{d}\mathbf{l} = \iint (\nabla \times \mathbf{A}) \cdot \mathrm{d}\mathbf{S} \quad \text{Stokes(斯托克斯)定理}
\] |
|
\[
w(y_o) = \frac{\Gamma_0}{\pi b^2} \int_{-b/2}^{b/2} \frac{y \, dy}{\sqrt{1 - \frac{4y^2}{b^2}} (y_o - y)}
\] |
|
Akaike 信息准则 (AIC)。考虑模型的集合 \(\{M_1, M_2, \ldots\}\)。令 \(\hat{f}_j(x)\) 是用模型 \(M_j\) 的极大似然估计得到的估计概率函数。因此，\(\hat{f}_j(x) = \hat{f}(x; \hat{g}_j)\)，这里 \(\hat{\beta}_j\) 是模型 \(M_j\) 的参数 \(\beta_j\) 的极大似然估计。使用损失函数 \(D(f, \hat{f})\)，其中，
\[
D(f, g) = \sum_x f(x) \log \left( \frac{f(x)}{g(x)} \right)
\] |
|
\begin{align*}
\text{iii)} \quad d(x,y) = 0 & \iff |x-y| = 0 \\
& \iff x-y = 0 \\
& \iff x = y \\
\end{align*}
\begin{align*}
\text{iv)} \quad d(x,z) &= |x-z| \\
& = |x-y+y-z| \\
& \leq |x-y| + |y-z| \\
d(x,z) & \leq d(x,y) + d(y,z) \\
\end{align*}
Then \((X,d)\) is a metric space. |
|
1.1 一维理想体，长度 L, N.
U.S. 物态方程？
\[
1) \quad Z &= \sum \omega e^{-\beta \epsilon} \\
\epsilon &= \frac{p^2}{2m} \quad \Rightarrow d\epsilon = \frac{p}{m}dp \\
& \Rightarrow dp = \frac{m}{p} d\epsilon = \sqrt{\frac{m}{2}} \frac{1}{\sqrt{\epsilon}} d\epsilon \\
Z &= \frac{L}{h} \sqrt{\frac{m}{2}} \int_0^\infty \frac{e^{-\beta \epsilon}}{\sqrt{\epsilon}} d\epsilon
\] |
|
\textbf{3. The Riemann-Stieltjes Integral}
Let \(f\) and \(\phi\) be two functions which are defined and finite on a finite interval \([a, b]\). If \(\Gamma = \{ a = x_0 < x_1 < \cdots < x_m = b \}\) is a partition of \([a, b]\), we arbitrarily select intermediate points \(\{\xi_i\}_{i=1}^m\) satisfying \(x_{i-1} \leq \xi_i \leq x_i\), and write
\[
R_\Gamma = \sum_{i=1}^m f(\xi_i) [ \phi(x_i) - \phi(x_{i-1}) ].
\] |
|
\textbf{Lemma 2.5.} \textit{Assume that} \(s \in (0,1]\), \(1 < r, r_1, r_2, q_1, q_2 < +\infty\) and satisfy \(\frac{1}{r} = \frac{1}{r_1} + \frac{1}{q_1} = \frac{1}{r_2} + \frac{1}{q_2}\). \textit{Then}
\[
\|| \nabla|^s (fg) \|_{L^r_x} \lesssim \| f \|_{L^{r_1}_x} \|| \nabla|^s g \|_{L^{q_1}_x} + \|| \nabla|^s f \|_{L^{r_2}_x} \| g \|_{L^{q_2}_x}.
\]
\textbf{Lemma 2.6.} \textit{Assume that} \(s \in (0,1]\), \(1 < q, q_1, q_2 < +\infty\) \textit{and satisfy} \(\frac{1}{q} = \frac{1}{q_1} + \frac{1}{q_2}\), \(G \in C^1 \mathbb{C}\). \textit{Then}
\[
\||\nabla|^s G(u) \|_{L^q_x} \lesssim \|G'(u)\|_{L^{q_1}_x} \|| \nabla|^s u \|_{L^{q_2}_x}.
\]
\textbf{Lemma 2.7.} \textit{Assume that} \(G\) \textit{is a H?lder continuous function of order }\(0 < p < 1\). \textit{Then}
\[
\||\nabla|^s G(u) \|_{L^q_x} \lesssim \||\ u |^{p - \frac{s}{q}} \|_{L^{q_1}_x} \||\nabla|^\sigma u
\|^{\frac{s}{\sigma}}_{L^{\frac{sq2}{\sigma}}_x}.
\] |
|
(iii) To obtain the matrix representation of \(D\) w.r.t. \(B_{p_2}^{2} P_{2}^{3}\) and \(B_{P_{1}}^{1}\) directly, observe that
\[
D(t-1) &= 1 = 1(1) + 0(t) \\
D(t+1) &= 1 = 1(1) + 0(t) \\
D((t-1)(t+1)) &= D(t^{2} - 1) = 0(1) + 2(t) so
\] |
|
\textbf{Definition:}
Let \(X\) be a non-empty set. A function \(d: X \times X \rightarrow \mathbb{R}\) is said to be a metric on \(X\) if for all \(x, y, z \in X\), it satisfies the following axioms:
\(M_1) d(x, y) \geq 0\)
\(M_2) d(x, y) = d(y, x)\)
\(M_3) d(x, y) = 0 \Longleftrightarrow x = y\)
\(M_4) d(x, z) \leq d(x, y) + d(y, z).\) |
|
\textbf{2.18 例} 如下是取值为0, 1的两个随机变量\(X, Y\)的二元分布：很明显，\(f(1, 1) = P(X = 1, Y = 1) = 4/9\).
\begin{tabular}{|c|c|c|c|}
\hline
& \(Y = 0\) & \(Y = 1\) & \\
\hline
\(X = 0\) & 1/9 & 2/9 & 1/3 \\
\hline
\(X = 1\) & 2/9 & 4/9 & 2/3 \\
\hline
& 1/3 & 2/3 & 1 \\
\hline
\end{tabular} |
|
我们对 (1.2.4) 微分，有
\[
A^{-1} \frac{\partial^2 A}{\partial \sigma^a \partial \sigma^b} - A^{-1} \frac{\partial A}{\partial \sigma^a} A^{-1} \frac{\partial A}{\partial \sigma^b} = \sum_{d=1}^r \frac{\partial v^d_b}{\partial \sigma^a} T_d.
\]
把 \(a\) 与 \(b\) 交换而相减，得出
\[
A^{-1} \frac{\partial A}{\partial \sigma^a} A^{-1} \frac{\partial A}{\partial \sigma^b} - A^{-1} \frac{\partial A}{\partial \sigma^b} A^{-1} \frac{\partial A}{\partial \sigma^a}
= \sum_{d=1}^r \left( \frac{\partial v^d_b}{\partial \sigma^a} - \frac{\partial v^d_a}{\partial \sigma^b} \right) T_d.
\] |
|
\[
\psi &= - \frac{\hbar}{2m} \sum_{j=1}^n \left( \frac{\partial^2 \psi}{\partial x_j^2} + \frac{\partial^2 \psi}{\partial y_j^2} + \frac{\partial^2 \psi}{\partial z_j^2} \right) - \frac{ne^2}{\hbar} \sum_{j=1}^n \frac{\psi}{\sqrt{x_j^2 + y_j^2 + z_j^2}} \\
&\quad - \frac{e^2}{\hbar} \sum_{\substack{i,j=1 \\ i \ne j}}^n \frac{\psi}{r_{i,j}}
\] |
|
\[
V_Y(t) &= \mathbb{E} \left[ \exp(t Y_m) \right] = \prod_{i=1}^{m} \mathbb{E} \left[\exp(t X_i)\right] \\
&= \prod_{i=1}^{m} \exp(\mu_i t + \frac{1}{2} \sigma_i^2 t^2) \\
&= \exp \left( \left(\sum \mu_i \right)t + \frac{1}{2} \left(\sum \sigma_i^2 \right)t^2 \right).
\] |
|
玻尔兹曼系统 \(\Omega_{M.B.} = \frac{N!}{\prod l a!} \prod l \omega_{l}^{a_{l}}\)
由 \(\delta \ln \Omega - \alpha \delta N - \beta \delta E = 0 \Rightarrow a_{l} = \omega_{l} e^{-\alpha - \beta \varepsilon_{l}}\)
同理
玻尔兹曼分布 \(a_{l} = \omega_{l} e^{-\alpha - \beta \varepsilon_{l}}\)
玻色分布 \(a_{l} = \frac{\omega_{l}}{e^{\alpha + \beta \varepsilon_{l}} - 1}\)
费米分布 \(a_{l} = \frac{\omega_{l}}{e^{\alpha + \beta \varepsilon_{l}} + 1}\)
关系：非简并条件下，\(e^{\alpha} \gg 1\)，遵循玻尔兹曼分布
弱简并 \(\rightarrow\) 一阶近似 |
|
(ii) 存在实解析函数 \(\psi (\sigma, \tau) = (\psi^1(\sigma, \tau), \cdots, \psi^r(\sigma, \tau))\)，在 \((\sigma, \tau) \in W_{\epsilon_1} \times W_{\epsilon_1}\)，\((0 < \epsilon_1 \leq \epsilon)\) 定义的乘法函数，使得 \(\psi (\sigma, \tau) \in W_{\epsilon}\)，且
\[
A(\sigma) A(\tau) = A(\psi (\sigma, \tau)).
\]
这时 \(G\) 称为 \(r\) 维矩阵李群。
若 \(B_0 \in G\)，但 \(B_0 \notin W_{\epsilon}\)，令
\[
B_0 W_{\epsilon} = \{ B \in G | B = B_0 A, A \in W_{\epsilon} \},
\]
则当 \(B \in B_0 W_{\epsilon}\) 时，能写为
\[
B = B_{\sigma} A(\sigma), \sigma \in W_{\epsilon},
\] |
|
考虑到三维方向 \(V = L^3 \to \infty\) 时
\[
\sum_{\vec{K}} \frac{V}{8 \pi^3} \int \text{d}^3 \vec{K},
\]再对螺度求和
\[
\sum_{\vec{K}, \lambda} \frac{2V}{8 \pi^3} \int \text{d}^3 \vec{K}.
\]由 (1.39) 式可求出能量
\[
E = - \frac{\partial}{\partial \beta} \ln Q = \sum_{\vec{K}, \lambda} \frac{\hbar \omega e^{\hbar \omega / kT}}{1 - e^{-\hbar \omega / kT}}.
\] |
|
利用刚刚证明过的结论：
\[
\oint_{S_0 + S_1} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{S} = \int_{S_0} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{S} - \int_{S_1} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{S} = \frac{q}{\varepsilon_0}
\]
注意，\(\boldsymbol{S}\) 仍用 \(\boldsymbol{n}_1\) 的方向！
\[
\oint_{S_0 + S_2} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{S} = \int_{S_0} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{S} + \int_{S_2} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{S} = \frac{q}{\varepsilon_0}
\]
由第二式第二等号两边 \textcolor{red}{减去} 第一式相应部分可得：
\[
\int_{S_1 + S_2} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{S} = \oiint_{S} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{S} = 0
\]
到此，证明了对单个点电荷的情况下高斯定理成立。 |
|
\textbf{Axiom I:} \(p(A, \alpha, \varnothing) = 0, \, p(A, \alpha, R) = 1\) represents the empty set
\[
p(A, \alpha, E_1 \cup E_2 \cup \ldots) = \sum_{j=1}^{\infty} p(A, \alpha, E_j)
\] |
|
\(2.5.6\ \Rightarrow\)
Suppose \(\lambda\) is an eigenvalue of \(S\), so that \(\exists\) a nonzero vector \(v\) such that \(S(v) = \lambda v\). Let \(w = R^{-1}(v)\), and note that because \(v\) is nonzero and \(R\) is invertible, \(R^{-1}(v)\) is also nonzero. Then
\[
T(w) &= T(R^{-1}(v)) = (R^{-1} \circ S \circ R)(R^{-1}(v)) \\
&= R^{-1} \circ S(v) = R^{-1}(\lambda v) = \lambda R^{-1}(v) = \lambda w,
\] |
|
\[
\frac{d^{l+m}}{dx^{l+m}}(x^2 - 1)^l &= \sum_{r=0}^{l+m} \binom{l+m}{r} \frac{d^r}{dx^r}(x - 1)^l \frac{d^{l+m-r}}{dx^{l+m-r}}(x + 1)^l \\
&= \sum_{r=m}^{l} \binom{l+m}{r} \frac{l!}{(l-r)!} (x - 1)^{l-r} \frac{l!}{(r - m)!} (x + 1)^{r-m} \\
&= \sum_{r=0}^{l-m} \binom{l+m}{r + m} \frac{l!}{(l-m - r)!} (x - 1)^{l-m-r} \frac{l!}{r!} (x + 1)^r \\
&= \frac{(l + m)!}{(l-m)!} (x^2 - 1)^{-m} \sum_{r=0}^{l-m} \binom{l-m}{r}
\] |
|
In the following, let us denote by \(a_k\) and \(a_k^*\) the annihilation and, respectively, creation operators associated with the plane waves \(x \mapsto \varphi_k(x) := e^{ikx} \in L^2(\Lambda)\) of momentum \(k\), for \(k \in \Lambda^* := 2\pi \mathbb{Z}^3\). They satisfy the canonical commutation relations \([a_p, a_q^*] = \delta_{p,q}\) and \([a_p, a_q] = [a_p^* , a_q^*] = 0\), and they can be used to express \(H_N\) as
\[
H_N = \sum_{r \in \Lambda^*} |r|^2 a_r^* a_r + \frac{N \kappa}{2N} \sum_{p,q,r \in \Lambda^*} \widehat{V} \left({r}/{N^{1-k}}\right) a_{p+r}^* a_{q-r}^* a_p a_q.
\] |
|
\[
L' = \frac{1}{2} \rho_\infty V_\infty^2 c(y_o) C_l = \rho_\infty V_\infty \Gamma(y_o)
\] |
|
\begin{tabular}{ccc}
\hline
\textbf{病人数} & \textbf{恶心病例数} \\
\hline
Placebo & 80 & 45 \\
Chlorpromazine & 75 & 26 \\
Dimenhydrinate & 85 & 52 \\
Pentobarbital (100 mg) & 67 & 45 \\
Pentobarbital (150 mg) & 85 & 37 \\
\hline
\end{tabular} |
|
1）定积分的几何意义
① \(y = f(x) \geq 0, \quad x = a, \quad x = b\) 将由所围曲边梯形面积为：
\[
S = \int_a^b f(x) \, dx
\]
② \(y = f(x) \leq 0\) ， \(\ S = -\int_a^b f(x) \, dx\) \quad \(x = a, \quad x = b,\) \(x\) 轴。
③ \(y = f(x)\) 有正有负 \(x = a, \quad x = b,\) \(x\) 轴
\[
S &= \int_a^b f(x) \, dx - \int_c^b f(x) \, dx + \int_a^b f(x) \, dx\\
&= \int_a^b |f(x)| \, dx
\] |
|
\textbf{Theorem 2.4.5 (Lebesgue's Dominated Convergence Theorem).} Let \((X, \mathcal{A}, \mu)\) be a measure space, let \(g\) be a \([0, +\infty]\)-valued integrable function on \(X\), and let \(f\) and \(f_1, f_2, \ldots\) be \([- \infty, +\infty]\)-valued \(\mathcal{A}\)-measurable functions on \(X\) such that
\[
f(x) = \lim_{n \to \infty} f_n(x) \tag{4}
\]
and
\[
|f_n(x)| \leq g(x), \quad n = 1, 2, \ldots \tag{5}
\]
hold at \(\mu\)-almost every \(x\) in \(X\). Then \(f\) and \(f_1, f_2, \ldots\) are integrable, and
\[
\int f \, d\mu = \lim_{n \to \infty} \int f_n \, d\mu.
\] |
|
\begin{align*}
\frac{d}{dt}\left( m \dot{\gamma } \right) - \left( m \gamma \dot{\theta}^2 - \frac{\partial V}{\partial \gamma } \right) = 0 \\
\frac{d}{dt}\left( m \dot{\gamma } \right) + \frac{\partial V}{\partial \gamma } - m \gamma \dot{\theta}^2 = 0 \\
m \ddot{\gamma } - m \gamma \dot{\theta}^2 = -\frac{\partial V}{\partial \gamma } \\
m \ddot{\gamma } = -\frac{\partial V}{\partial \gamma } + m \gamma \dot{\theta}^2
\end{align*} |
|
代入式(16.10) 可得极化状态动作值函数
\[
\begin{cases}
Q_T^k(x, a) = \sum_{x \in X} P_{x \to x^i} (\frac{1}{p} R_{x \to x^i}^a + \frac{T-1}{T} \max_{a \in A} Q_{T-1}^k(\sigma', a')); \\
Q_T^k(x, a) = \sum_{x \in X} P_{x \to x^i}^a (R_{x \to x^i}^a + \gamma \max_{a' \in A} Q_T^k(x', a')).
\end{cases}
\]
上述关于最优值函数的等式, 称为极化 Bellman等式, 其中唯一解是极化值函数。
最优 Bellman等式揭示了非最优策略的改进方式: 将策略选择的动作改变为当前最优的动作, 显然, 这样的改变能使策略更好. 不妨令动作改变后对应的策略为 \(\pi'\), 改变动作的条件为 \(Q^T(x, \pi'(x)) \geq V^T(x)\), 以 \(\gamma\) 折扣累积奖数为例, 由式(16.10) 可计算出递推不等式 |
|
对数运算
\[
\log_a (M \cdot N) & = \log_a M + \log_a N \\
\log_a \left(\frac{M}{N}\right) & = \log_a M - \log_a N \\
\log_a M^n & = n \log_a M \quad (n \in \mathbb{R})
\]
\[
a^b = b \quad \log_a b = \frac{\log_c b}{\log_c a} \quad \log_a b \cdot \log_b a = 1 \quad \log_a b^n = n \log_a b
\] |
|
The minimum sampling rate \(\omega_s = 2 \omega_m\) or \(f_s = 2 f_m\) required to recover the message signal is known as Nyquist rate.
\textit{Nyquist interval}
\begin{align*}
T_s &= \frac{1}{f_s} = \frac{1}{2f_m}
\end{align*} |
|
\textbf{9.33 例} 令 \(X_1, \cdots, X_n \sim \text{Bernoulli}(p)\)，则 \(L(p) = p^S (1 - p)^{n - S}\)，其中，\(S = \sum_i X_i\)，所以 \(S\) 是充分的。
\textbf{9.34 例} 令 \(X_1, \cdots, X_n \sim N(\mu, \sigma)\)，且令 \(T = (\overline{X}, S)\)。则
\[
f(X^n; \mu, \sigma) &= \left( \frac{1}{\sigma \sqrt{2 \pi}} \right)^n \exp \left\{ -\frac{n S^2}{2\sigma^2} \right\} \exp \left\{ -\frac{n (\overline{X} - \mu)^2}{2\sigma^2} \right\},
\] |
|
若得到的打点纸带如图6-10所示，并把测得的各计数点间距标在图上，\(A\)为运动起始的第一点，则应选 \_ 段计算小车\(A\)的碰前速度，应选 \_ 段计算小车\(A\)和\(B\)碰后的共同速度（以上两格填“\(AB\)”“\(BC\)”“\(CD\)”或“\(DE\)”）； |
|
and that \(R(f)\) is a subspace of \(\mathbb{F}\), so \(r(f) = 0\) or 1.
If \(r(f) = 0\) then \(f\) is the 0 map so \(\textit{span} \left( \{ f \} \right) = \{ 0 \}\)
(in \(U^1\)), so \(R(f^1) = Span \left( \{ f \} \right) = \{ 0 \}\), therefore, \(r(f^1) = 0\).
If \(r(f) = 1\) then \(f\) is not the 0 map so
\[
r(f^1) = \dim \left( {Span} \left( \{ f \} \right) \right) = 1. \\
\] |
|
\[
Y =
\begin{cases}
1, & 0 < x < b, \\
0, & \text{其他}.
\end{cases}
\]
\[
Z =
\begin{cases}
1, & a < x < 1, \\
0, & \text{其他}.
\end{cases}
\] |
|
\[
\langle A | \mathbf{1}_{\mathbf{k}}^{(\lambda)} \rangle \left( -\frac{e}{m_e c} \mathbf{A} \cdot \mathbf{p} - \frac{e}{m_e c} \mathbf{s} \cdot (\nabla \times \mathbf{A}) \right) | 0 \rangle | B \rangle.
\]
The relevant photonic part is
\[
\langle \mathbf{1}_{\mathbf{k}}^{(\lambda)} | \mathbf{A}(0) | 0 \rangle &= \hbar^{1/2} c \sum_{\lambda'} \int \frac{d^3 k'}{(2\pi)^{3/2} \sqrt{2 \omega_{\mathbf{k}'}}} \mathbf{e}_{\mathbf{k}'}^{(\lambda')} \langle 0 | a_{\mathbf{k}'}^{(\lambda')} a_{\mathbf{k}}^{(\lambda) \dagger} | 0 \rangle e^{i \omega t} e^{-i \mathbf{k} \cdot \mathbf{r}}
\] |
|
Recalling the definition of a differential, i.e. the differential of \(f\) at \(x\) B the linear operator satisfying
\[
\lim_{\Sigma \to 0} \frac{|f(x+\Sigma h) - f(x) - \Sigma {\nabla}f(x)[h]|}{\Sigma} = 0,
\]
we have that
\[
Df(A)[\epsilon H] &= \epsilon \text{Tr}(C(H)) \\
&= \epsilon \text{Tr}(C A^{-1/2} H A^{-1/2}) = \epsilon \langle \nabla f(A), H \rangle.
\]
And recalling that the inner product on \(S_n^{++}\) is \(\langle A, B \rangle = \text{Tr}(AB)\), we find that
\[
\nabla f(A) = A^{-1}.
\] |
|
\[
\delta \overline {H} &= \sum\limits_i\int \left( \delta \phi_{k_i}^* H_i \phi_{k_i} + \phi_{k_i}^* H_i \delta \phi_{k_i} \right) d^3 x_i \\
&\quad + \frac{1}{2} \sum\sum_{i \neq j} \iint \left( \delta \phi_{k_i}^* \phi_{k_i} + \phi_{k_i}^* \delta \phi_k \right) \frac{1}{r_{ij}} \left| \phi_{k_i} \left( r_i \right) \right|^2 d^3 x_i d^3 x_j \\
&\quad + \frac{1}{2} \sum\sum_{i \neq j} \iint \left| \phi_{k_i} \left( r_i \right) \right|^2 \frac{1}{r_{ij}} \left( \delta \phi_{k_j}^* \phi_{k_j} + \phi_{k_j}^* \delta \phi_{k_j} \right) d^3 x_i d^3 x_j \\
&= \sum_i \int\left( \delta \phi_{k_j}^* H_i \phi_{k_j} + \phi_{k_i}^* H_i \delta \phi_{k_i} \right) d^3 x_i \\
&\quad + \sum\sum_{i \neq j} \iint \left( \delta \phi_{k_i}^* \phi_{k_i} + \phi_{k_i}^* \delta \phi_{k_i} \right) \frac{1}{r_{ij}} \left| \phi_{k_j} \left( r_j \right) \right|^2 d^3 x_i d^3 x_j.
\] |
|
Yes. Since \(Y\) is a subspace, \(0 \in Y\), and \(0 \in T^{-1}(0) \subseteq T^{-1}(Y)\), \(T^{-1}(Y)\) is nonempty. To show that \(T^{-1}(Y)\) is closed under addition and scalar multiplication, let \(u_1, u_2 \in T^{-1}(Y)\) and \(c \in \mathbb{F}\). Then
\[
T(u_1), T(u_2) &\in Y. \\
T(u_1 + u_2) &= T(u_1) + T(u_2) \in Y
\]
since \(Y\) is a subspace (and hence closed under addition). So \(u_1 + u_2 \in T^{-1}(Y)\). |
|
因此，\(c = 21/4\)。现在来计算\(P(X \geq Y)\)，相应的集合为\(A = \{(x, y) : 0 \leq x \leq 1, x^2 \leq y \leq x\}\)（该者可以通过图示来理解），所以
\[
P(X \geq Y) &= \frac{21}{4} \int_{0}^{1} \left( \int_{x^2}^{x} x^2 y dy \right) dx = \frac{21}{4} \int_{0}^{1} x^2 \left( \int_{x^2}^{x} y dy \right) dx \\
&= \frac{21}{4} \int_{0}^{1} x^2 \left( \frac{x^2 - x^4}{2} \right) dx = \frac{21}{4} \int_{0}^{1} \left( \frac{x^4 - x^6}{2} \right) dx \\
&= \frac{21}{4} \left( \frac{1}{2} \int_{0}^{1} (x^4 - x^6) dx \right) = \frac{21}{4} \left( \frac{1}{2} \left( \frac{x^5}{5} - \frac{x^7}{7} \right) \bigg|_{0}^{1} \right) \\
&= \frac{21}{4} \left( \frac{1}{2} \left( \frac{1}{5} - \frac{1}{7} \right) \right) = \frac{21}{4} \left( \frac{1}{2} \left( \frac{7 - 5}{35} \right) \right) = \frac{21}{4} \left( \frac{1}{2} \left( \frac{2}{35} \right) \right) \\
&= \frac{21}{4} \left( \frac{1}{35} \right) = \frac{21}{140} = \frac{3}{20}.
\] |
|
\[
\sigma^2 &= \frac{l^2}{4} - 2 l r \sin \frac{\omega}{2} \sin \left( \frac{\psi}{2} + A \right) + 4 r^2 \sin^2 \frac{\omega}{2} \\
&= \frac{l^2}{4} + 4 r \sin \frac{\omega}{2} \left[ \delta \sin \frac{\omega}{2} - \frac{l}{2} \cos \left( \frac{\omega}{2} + A \right) \right] \tag{4}
\] |
|
Here \(P_{E_1} + P_{E_2} + \cdots\) means the unique projection \(P\) such that \(P_{E_1} \phi + P_{E_2} \phi + \cdots = P \phi\) for all \(\phi\) or equivalently the projection on the closed linear subspace generated by the ranges of \(P_{E_j}\). Whenever we have a function from real Borel sets to projections which has properties (a), (b), and (c) we shall say that we have a \textit{projection-valued measure}. Let \(E \rightarrow P_E\) be any projection-valued measure. Then for each vector \(\phi\) in \(X\), \(E \rightarrow \langle P_E(\phi), \phi \rangle\) will be an ordinary non-negative measure in the real line and we can form integrals \(\int f(x) \, d\alpha(x)\), where \(\alpha(E) = \langle P_E(\phi), \phi \rangle\). We shall write \(\int f(x) \, d(P_x(\phi), \phi)\). Now, if \(P\) comes from \(\lambda P_1 + \cdots + \lambda P_n\) as indicated above, then |
|
4. 如图 5-1 所示，长度相同的三根轻杆构成一个正三角形支架，在 \(A\) 处固定质量为 \(2m\) 的小球，\(B\) 处固定质量为 \(m\) 的小球，支架悬挂在 \(O\) 点，可绕过 \(O\) 点与支架所在平面相垂直的固定轴转动，开始时 \(OB\) 与地面相垂直，放手后开始运动，在不计任何阻力的情况下，下列说法正确的是 （ ）
A. \(A\) 球到达最低点时速度为零。
B. \(A\) 球机械能减少的量等于 \(B\) 球机械能增加量。
C. \(B\) 球向左摆动所能达到的最右位置应高于 \(A\) 球开始运动的高度。
D. 当支架从左向右回摆时，\(A\) 球一定能回到起始高度。 |
|
\[
&= \alpha_1^A \alpha_1^B \left[ \Psi_{11}^{-1} \alpha_C^1 \alpha_D^1 + \Psi_{11}^{-1} (\alpha_C^1 \alpha_D^1 + \alpha_C^2 \alpha_D^1) \right] \\
&\quad + \Psi_{22}^{11} \alpha_C^2 \alpha_D^2] + (\alpha_1^i \alpha_2^j + \alpha_2^i \alpha_1^j) [ \Psi_{12}^{-1} (\alpha_C^1 \alpha_D^1\\
&\quad + \Psi_{22}^{11} (\alpha_C^2 \alpha_D^1 + \alpha_C^2 \alpha_D^1) + \Psi_{22}^{11} \alpha_C^2 \alpha_D^2]\\
&\quad + \alpha_2^A \alpha_2^B \epsilon_B^1 [\Psi_{11}^{22} \alpha_C^2 \alpha_D^2 + \Psi_{12}^{22} (\alpha_C^1 \alpha_D^2 + \alpha_C^2 \alpha_D^2) \\
&\quad + \Psi_{22}^{11} (\alpha_C^2 \alpha_D^2],
\] |
|
\textit{2.4.1} Suppose \([u_1]_V, \ldots, [u_k]_V\) are linearly dependent, so
There exist \(a_1, \ldots, a_k \in \mathbb{F}\), not all zero, such that
\[
a_1[u_1]_V + \ldots + a_k[u_k]_V = [0]_V.
\]
Then
\[
a_1[u_1]_W + \ldots + a_k[u_k]_W &= \tilde{I}(a_1[u_1]_V + \ldots + a_k[u_k]_V) \\
&= \tilde{I}([0]_V) \\
&= [0]_W
\]
so \([u_1]_W, \ldots, [u_k]_W\) are linearly dependent.// |
|
(2.3.2) 可写为
\[
\tilde{\Gamma}_{\alpha \beta i}^{\gamma}(x) = \Gamma_{\mu \nu k}^{\lambda}(x) O_{\alpha}^{\mu^{-1}} O_{\beta}^{\nu^{-1}} O_{i}^{k^{-1}}. \tag{2.3.3}
\]
令 \(\xi^a, \eta^b, \zeta^c\) 为 \(V\) 中的向量, 由上式可知
\[
\tilde{\Gamma}_{\alpha \beta i}^{\gamma}(x) \xi^a \eta^b \zeta^i = \Gamma_{\mu \nu k}^{\lambda}(x) \xi^{\mu} \eta^{\nu} \zeta^k,
\]
即在正交群下是不变的。由 H. Weyl 的关于正交群下不变量的定理可知, \(\Gamma_{\mu \nu k}^{\lambda}(x) \xi^{\mu} \eta^{\nu} \zeta^k\) 必定是向量 \(x, \xi^a, \eta^b, \zeta^i\) 的各种可能内积的函数, 但是它是 \(\xi^a, \eta^b, \zeta^i\) 的线性函数, 因此只可能是下面的形式 |
|
\[
t_0 = \frac{v - v_0}{a} = \frac{0 - 30}{-5} = 6s
\]
由于 \(t_0 < t_1\) , 所以刹车后 \(10s\) 内滑行的距离即为汽车停止
\[
S = \frac{v_0^2}{2}t_0 = \frac{30}{2} \times 6m = 90m.
\]
(2) 设从刹车到滑行 \(50m\) 所经过的时间为 \(t'\), 则有:
\[
x = v_0 t' + \frac{1}{2} a t'^2
\] |
|
11. 证明欧勒多项式的傅里叶展开式
\[
E_{2n}(x) &= (-1)^n 4 (2n)! \sum_{r=0}^{\infty} \frac{\sin \left[ (2r + 1) \pi x \right]}{\left[ (2r + 1) \pi \right]^{2n+1}}, \quad n = 1, 2, \cdots, \quad 0 \le x \le 1,\\
E_{2n+1}(x) &= (-1)^{n+1} 4 (2n + 1)! \sum_{r=0}^{\infty} \frac{\cos \left[ (2r + 1) \pi x \right]}{\left[ (2r + 1) \pi \right]^{2n+2}}, \quad n = 0, 1, 2, \cdots, \quad 0 \le x \le 1.
\] |
|
\begin{tabular}{ccccc}
\hline
$X_1$ & $X_2$ & $V$ & $T$ & $U$ \\
\hline
0 & 0 & 0 & 0 & (0,0) \\
0 & 1 & 0 & 1 & (1,0) \\
1 & 0 & 1 & 1 & (1,1) \\
1 & 1 & 1 & 2 & (2,1) \\
\hline
\end{tabular} |
|
\textbf{Scattering in the Delta Function Potential} \((E > 0)\). The solution of the Schroedinger equation is
\[
\psi_k(x) = \begin{cases}
A e^{ikx} + B e^{-ikx} & (x \leq 0) \\
C e^{ikx} & (x \geq 0)
\end{cases}
\tag{3.88}
\] |
|
\[
\begin{bmatrix}
\cos\varphi & 0 & 0 & \sin\varphi \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-\sin\varphi & 0 & 0 & \cos\varphi
\end{bmatrix}
\] |
|
另外，由矢量运算公式得
\[
\nabla \cdot (M \times x) &= M + x \cdot \nabla M \\
\nabla (x \cdot M) &= x \times (\nabla \times M) + (x \cdot \nabla)M + M \times (\nabla \times x) + (M \cdot \nabla)x \\
&= x \times (\nabla \times M) + (x \cdot \nabla)M + M \\
\nabla \times (x \times M) &= (M \cdot \nabla)x + (\nabla \cdot M)x - M(\nabla \cdot x) - (x \cdot \nabla)M \\
&= (\nabla \cdot M)x - (x \cdot \nabla)M - 2M
\]
综合以上三式，得
\[
2M &= x \times (\nabla \times M) + \nabla (M \times x) - \nabla (x \cdot M) - \nabla \times (x \times M)
\] |
|
\begin{align*}
x(t) &= \cos^2(2\pi t) = \frac{1 + \cos 4\pi t}{2} \\
x(t) &= \frac{1}{2} + \frac{1}{2} \cos 4\pi t \\
x(t) &= 0.5 + 0.5 \cos 4\pi t
\end{align*} |
|
\[
\begin{cases}
u_t &= a^2 u_{zz} \\
u |_{t=0} &= \varphi(x) =
\begin{cases}
\frac{hx}{c} & 0 \le x \le c \\
\frac{h(l-x)}{l-c} & c \le x \le l
\end{cases} \\
u |_{z=0} &= 0 \\
u|_{z=l} &= 0 \quad t \ge 0 \\
\end{cases}
\]
设 \(u(x,t) = X(x)T(t)\)，代入方程得
\[
\frac{T''(t)}{a^2 T(t)} = \frac{X''(x)}{X(x)} = -\lambda
\]
由边界条件得固有值问题
\[
\begin{cases}
X''(x) + \lambda X(x) = 0 \\
X(0) = X(l) = 0
\end{cases}
\] |
|
Optimizing in \(\lambda\) to obtain a tighter bound, we find that \(-t + \frac{\lambda^2}{2} \text{log}^2 k\) is minimized at \(\lambda^*\) satisfying
\[
-t + \lambda^* \text{log}^2 k = 0 \quad \text{i.e.} \quad \lambda^* = \frac{t}{\text{log}^2 k}.
\]
Then
\[
e^{-t + \frac{t^2}{2\text{log}^2 k} \text{log}^2 k} = e^{\frac{t^2}{2\text{log}^2 k}} = e^{\frac{t^2}{2\text{log}^2 k}} \quad \text{which is the desired result.}
\] |
|
如何衡量电场的大小和指向，我们可以在电场中某处放置一电荷\(q\)，设在电场的作用下它所受的力为\(F\)，定义该处的电场强度为
\[
E = \frac{F}{q} \tag{1.2}
\]
于是，在真空中存在有电荷离为\\(Q \\)的点电荷，它产生的电场弥散于全空间。在距原点电荷\\(r \\)处，点电荷所产生的电场为
\[
E = \frac{Q}{4\pi\varepsilon_0 e_r} = \frac{Qr}{4\pi\varepsilon_0} \tag{1.3}
\]
当空间中存在若千点状电荷时，提供电荷的存在与否并不改变当前电场的相互作用形式。因此，...
（后面部分未完整显示，不包括范围内的信息可忽略） |
|
Position vector \(\vec{r}(t)\) defines trajectory. Velocity vector \(\vec{v}(t)\) is tangent. Acceleration vector \(\vec{a}(t)\) gives rate of change of \(\vec{v}(t)\).
\[
\vec{v}(t) &:= \lim_{\Delta t \to 0} \frac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t} \\
&= \frac{d\vec{r}}{dt}
\] |
|
对于 \(\tilde{x}\) 的分量 \(\zeta_{\alpha}(\tilde{x})\) 满足如下的关系：
\[
\zeta_{\alpha}(\tilde{x}) = \sum_{\beta=1}^{N} \zeta_{\beta}(x) [\varphi_{\tilde{V} V}(x)^{-1}]_{\beta}^{\alpha}, \tag{1.1.6}
\]
其中 \(\varphi_{\tilde{V} V}(x)^{-1}\) 表 \(\varphi_{\tilde{V} V}(x)\) 的逆方阵。 \(\zeta\) 称为 \(G\) 型协向量。反之，如在 \(V\) 的点 \(x\) 有一组数 \(\zeta_{\alpha}(x)\) 连续依赖于 \(x\)，对局部坐标变换满足 (1.1.6)，则 \(\zeta_{\alpha}(x)\) 定义一 \(x\) 点的 \(G\) 型协向量。 |
|
\[
&\frac{1}{2\pi i} \oint_C f(\zeta) \frac{\psi'(\zeta)}{\psi(\zeta)} d\zeta = \frac{1}{2\pi i} \oint_C f(\zeta) \frac{1 - t\varphi'(\zeta)}{\zeta - a - t\varphi(\zeta)} d\zeta \\
&= \frac{1}{2\pi i} \oint_C f(\zeta) \left[ 1 - t\varphi(\zeta) \right] \sum_{n=0}^{\infty} \frac{t^n \left[ \varphi(\zeta) \right]^n}{(\zeta - a)^{n+1}} d\zeta \\
&= \sum_{n=0}^{\infty} \frac{t^n}{n!} \frac{d^n}{da^n} \left[ f(a) \left[ \varphi(a) \right]^n \right] \\
&\quad - \sum_{n=0}^{\infty} \frac{t^{n+1}}{(n+1)!} \frac{d^n}{da^n} \left[ f(a) \frac{d}{da} \left[ \varphi(a) \right]^{n+1} \right] \\
&= f(a) + \sum_{n=1}^{\infty} \frac{t^n}{n!} \frac{d^{n-1}}{da^{n-1}} \left[ f'(a) \left[ \varphi(a) \right]^n \right],
\] |
|
\[
\Sigma \omega = \int D \omega d\omega \\
= \frac{S}{\hbar^{2}} \int 2\pi p dp \\
= \frac{S}{\hbar^{2}} \int 2\pi p \frac{m}{h} d\epsilon \\
\] |
|
\begin{align*}
p^3 &=\beta^3 e^{-a^3 x} \\
\alpha ^3 &= A + B s^{2/3} .
\end{align*} |
|
\textbf{Vector Space}
\textit{Def}: Let \(V\) be a non-empty set and \(F\) be a field. Then \(V\) is called a vector space if:
i) \(V\) is an abelian group under addition.
ii) \(a(v + w) = av + aw \quad \forall \, a \in F, \, v, w \in V\).
iii) \((a + b)v = av + bv \quad \forall \, a, b \in F, \, v \in V\).
iv) \(a(bv) = (ab)v \quad \forall \, a, b \in F, \, v \in V\).
v) \(1 \cdot v = v \cdot 1 = v \quad \forall \, 1 \in F, \, v \in V\).
i.e., 1 is the identity under multiplication. |
|
\[
\text{自然数：} & \, N \\
\text{正整数：} & \, N_{+} \\
\text{整数：} & \, Z \\
\text{有理数：} & \, Q \\
\text{实数：} & \, R \\
\text{复数：} & \, C \\
\] |
|
\subsection*{1. 线性组合}
若
\[
f(z) &\sim \sum_{n=0}^{\infty} A_n z^{-n}, \\
g(z) &\sim \sum_{n=0}^{\infty} B_n z^{-n}, \tag{10}
\]
则
\[
a f(z) + \beta g(z) \sim \sum_{n=0}^{\infty} (a A_n + \beta B_n) z^{-n}, \tag{11}
\]
其中 \(a, \beta\) 是任意常数。根据渐近展开的定义 (3)，立刻可以证明。
\textbf{2. 相乘}
若有 (10)，则
\[
f(z) g(z) \sim \sum_{m=0}^{\infty} C_m z^{-m}, \tag{12}
\]
其中
\[
C_m = \sum_{k=0}^m A_k B_{m-k}. \tag{13}
\] |
|
其中函数 \(\psi(y)\) 待定, 以使由 \((1.10)\) 规定的函数 \(\Phi(x, y)\) 适合 \((1.8)\) 的第二式。由 \((1.10)\) 得：
\[
\frac{\partial \Phi}{\partial y} &= \frac{\partial}{\partial y} \int_{x_0}^x P(x, y) \, dx + \psi'(y) \\
&= \int_{x_0}^x \frac{\partial P(x, y)}{\partial y} \, dx + \psi'(y).
\] |
|
We introduce norms
\begin{itemize}
\item vectors \(\|v\|_2 = \sqrt{\sum_{i=1}^{n} v_i^2}\) Note it implies \(\langle x, y \rangle \leq \|x\|_2 \cdot \|y\|_2\)
\item and matrices: \(\|A\|_2 = \max_{x \in \mathbb{R}^n} \frac{\|A \cdot x\|_2}{\|x\|_2}\)
\begin{itemize}
\item This implies \(\|A \cdot x\|_2 \leq \|A\|_2 \cdot \|x\|_2\)
\item This also implies \(\|A\| = \rho(A)\)
\end{itemize}
\end{itemize} |
|
\[
\iiint\limits_{\Omega} f(x, y, z) \, dx \, dy \, dz = \iint\limits_{D} dx \, dy \int_{z_1(x, y)}^{z_2(x, y)} f(x, y, z) \, dz
\]
\[
\iiint\limits_{\Omega} f(x, y, z) \, dx \, dy \, dz = \int_a^b dz\iint\limits_{DZ} f(x, y, z) \, dx \, dy
\]
\[
\iiint\limits_{\Omega} f(x, y, z) \, dV = \iiint_\limits{\Omega'} f(r \cos \theta, r \sin \theta, z) \, r \, dr \, d\theta \, dz
\] |
|
\textbf{Exercise 1.22.} Show the properties of unitary maps in Lemma \textit{1.1.}
For finite-dimensional vector spaces we know that, relative to an ortho-normal basis \(e_1, \ldots, e_n\), a unitary map \(\hat{U}\) is described by a unitary matrix (orthogonal matrix in the real case). Indeed, introducing the matrix \(U\) with matrix elements (in Dirac notation)
\[
U_{ij} = \langle \epsilon_i | \hat{U} | \epsilon_j \rangle, \tag{1.51}
\]
this statement is verified by the following short calculation.
\[
\sum_j (U^{\dagger} U)_{ij} \overline{U}_{jk} = \sum_j \langle \epsilon_i | \hat{U} | \epsilon_j \rangle \langle \epsilon_j | \hat{U} | \epsilon_k \rangle = \langle \epsilon_i | \hat{U} \hat{U}^{\dagger} | \epsilon_k \rangle = \langle \epsilon_i | \epsilon_k \rangle = \delta_{ik}. \tag{1.52}
\] |
|
\[
u(x,t) &= \int_0^t R(x,t,\tau) \, d\tau \\
&= \int_0^t \sum_{k=1}^{\infty} \frac{(-1)^{k+1} 2 l sh l}{a(k^2 \pi^2 + l^2)} \sin \frac{k \pi a}{l} (t - \tau) \sin \frac{k \pi}{l} x \, d\tau \\
&= \sum_{k=1}^{\infty} \int_0^t \frac{(-1)^{k+1} 2 l sh l}{a(k^2 \pi^2 + l^2)} \sin \frac{k \pi a}{l} (t - \tau) \sin \frac{k \pi}{l} x \, d\tau \\
&= \frac{2 l^2 sh l}{a^2 \pi} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(k^2 \pi^2 + l^2)} \left[ 1 - \cos \frac{k \pi a}{l} t \right] \sin \frac{k \pi}{l} x
\] |
|
3.4.5 I claim that the characteristic polynomial of T is \(p_T(t) = t^n\).
There are several ways to see why this is true.
The simplest would be to recall that if \(T\) is nilpotent of degree \(n\), then \(\exists\) a basis \(B\) for \(V\) such that the matrix |
|
\[
= -\frac{1}{2} \begin{pmatrix}\begin{bmatrix} -2 & 2 & 0 \\ 0 & -1 & 0 \\ 4 & -3 & 2 \end{bmatrix}\end{pmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & \frac{1}{2} & 0 \\ -2 & \frac{3}{2} & -1 \end{bmatrix}.
\] |
|
利用 (17), (19), (20), 由 (16) 得
\[
|c_n| &< \frac{1}{n|n + s|} \sum_{k=1}^{n} M r^{-k} (n - k + 1) |c_{n-k}| \\
&\leq \frac{1}{n^2} \sum_{k=1}^{n} M r^{-k} (n - k + 1) M^{n-k} |c_0| \\
&= \frac{|c_0| r^{-n} M^n}{n^2} \sum_{m=1}^{n} m \\
&= \frac{|c_0| r^{-n} M^n}{n^2} \cdot \frac{n(n + 1)}{2} \\
&= \frac{|c_0| M^n r^{-n} n(n + 1)}{2 n^2} < M^n r^{-n} |c_0|, \quad \text{当} \, n \geq 2,
\] |
|
\[
A^{-1} = \frac{1}{-2} \left(A^2 - 2A - I \right) = -\frac{1}{2} \left( \begin{bmatrix} 1 & 6 & 0 \\ 0 & 4 & 0 \\ 0 & -5 & 1 \end{bmatrix} \right)
\] |
|
\[
V(x) = \begin{cases}
V_1 & (x < 0) \\
-V_0 & (0 < x < L) \\
0 & (x > L)
\end{cases}
\] |
|
Given \(A_m = (a_1, \ldots, a_n)\) with \(m\) rows, we transform the decomposition of \(A_m\) into the decomposition of \(A_{m-1}\) with \(m-1\) rows by \(H_1 = I - 2w_1w_1^T\), \(w_1 = \frac{a_1 - b_1}{\| a_1 - b_1 \|}\).
\[
H_1 a_1 &= b_1 = \begin{bmatrix} \| a_1 \| \\ 0 \\ \vdots \\ 0 \end{bmatrix} \Rightarrow H_1 A_m = \begin{bmatrix} \| a_1 \| & \tilde{a}_1^T \\ 0 & \\ \vdots & A_{m-1} \\ 0 & \end{bmatrix} \Rightarrow A_m = H_1 \begin{bmatrix}r_{11} & \tilde{a}_1^T \\ 0 & \\ \vdots & A_{m-1} \\ 0 & \end{bmatrix}, \quad r_{11} = \| a_1 \|.
\] |
|
曲面：\(z = f(x, y)\)
法：\(\vec{\mathbf{n}} = (-f_x, -f_y, 1)\)\quad单法\(\vec{\mathbf{n}} = \frac{(-f_x, -f_y, 1)}{\sqrt{f_x^2 + f_y^2 + 1}}\)\quad\(dS = \sqrt{f_x^2 + f_y^2 + 1} \, dx \, dy\)
\[
\Rightarrow \iint\limits_S P \, dy \, dz + Q \, dx \, dz + R \, dx \, dy = \iint\limits_D \left[ P \cdot (-f_x) + Q \cdot (-f_y) + R \cdot 1 \right] \, dx \, dy
\]
\(P(x, y, f(x, y))\) |
|
可知 \[
U(x_2(t), t )- U(x_B, t_B) &= \int_{t_B}^t \left[ (H_a) U_1 \left( \frac{c-2\tau}{1-a} \right) - U_0' \left( \frac{c-2\tau}{1-a} \right) \right] d\tau , \\
\text{因此：} \\
U(x_2(t),t) &= U(x_B, t_B) + \int_{t_B}^t \left[ (H_a) U_1 \left( \frac{c-2\tau}{1-a} \right) - U_0' \left( \frac{c-2\tau}{1-a} \right) \right] d\tau \\
&= U_0(x_B) +(1+a) \int_{x_B}^x U_1(\xi) \left( \frac{1-a}{-2} \right) d\xi - \int_{x_0}^{x} U_0' (\xi) \frac{(1-a)}{-2} d\xi \\
&= U_0 \left( \frac{c}{1+a} \right) + \frac{1-a^2}{-2} \int_{\frac{c-2t_0}{1-a}}^{\frac{c-2t}{1-a}} u(\xi) d\xi + \frac{(1-a)}{2} \\
\left[ U_t \left( \frac{c-2t}{1-a} \right) - U_0 \left( \frac{c-2t_0}{1-a} \right) \right].
\] |
|
\subsection*{差分关系}
\[
&\varphi_0(x+1) = \varphi_0(x), \\
&\varphi_1(x+1) = \varphi_1(x) + 1, \\
&\varphi_n(x+1) = \varphi_n(x) + nx^{n-1} \quad (n \ge 2). \tag{14}
\]
证明如下：由 (1) 有 |
|
\begin{align*}
F(u(t)) &= \int_{0}^{\infty} u(t) \cdot e^{-j \omega t} \, dt \\
&= \int_{0}^{\infty} 1 \cdot e^{-j \omega t} \, dt \\
F[u(t)] &= \left[ \frac{e^{-j \omega t}}{-j \omega} \right]_{0}^{\infty} = 0 + \frac{1}{j \omega} = \frac{1}{j \omega}
\end{align*} |
|
in agreement with (23).
Linear operators. An operator \(\alpha\) is said to be linear if for arbitrary operands \(u\) and \(v\)
\[
\alpha(u + v) = \alpha u + \alpha v,
\]
and for an arbitrary constant \(c\)
\[
\alpha c = c \alpha.
\] |
|
\[
-N \frac{\partial}{\partial \beta} \left[ \text{lnConst.} - \frac{1}{2} \ln \beta \right] \\
\frac{1}{2} N \frac{1}{\beta} = \frac{1}{2} N K T
\] |
|
Consequently, \(F\) attains its maximum value at the angle \(\psi_0\), which is given by the formula
\[
\tan \psi_0 = \frac{ab - \sin \beta \cos \beta}{a^2 + \sin^2 \beta}. \tag{7}
\] |
|
如果方程 \((8)\) 有两个相异的根, 则方程 \(w'' + pw' + qw = 0\) 在它的奇点 \(z_0\) 的邻域 \(0 < |z - z_0| < R\) 内有两个线性无关的解。其形式为 (为了使用方便, 将上文中的 \(\lambda_0, \lambda_\beta, w_{\alpha}, w_{\beta}\) 改为 \(\lambda_1, \lambda_2, w_1, w_2\) )
\[
w_1(z) &= (z - z_0)^{\lambda_1} \sum_{n=-\infty}^{\infty} c_n (z - z_0)^n, \\
w_2(z) &= (z - z_0)^{\lambda_2} \sum_{n=-\infty}^{\infty} d_n (z - z_0)^n.
\tag{10}
\] |
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