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B. \textbf{Two Important Series}:
1. \textit{Geometric Series} a.
\[
\sum_{n=0}^{\infty} r^n = 1 + r + r^2 + r^3 + \ldots
\]
\textit{NOTE}: Lower limit \(n = 0\). |
|
22. 如图1-18所示，一个质量为\(m\)的物块，与甲、乙两轻弹簧相连接，乙的下端与地面相连，两弹簧的劲度系数分别为\(k_1, k_2\)，现用手拉住甲弹簧的上端\(A\)缓缓上提，要求乙弹簧中的弹力变为原来的\(\frac{2}{3}\)，则\(A\)端上移的距离应为多少? |
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\(f(x)\) 在实轴上无极点，且为有理函数，当 \(z \to \infty\) 时 \(|z^2 f(z)|\) 有界。则取积分围道如图，运用留数定理，关于 \(z^\alpha f(z)\)
\[
\oint f(z) z^\alpha \, dz = \left[ 1 - e^{i 2 \pi \alpha} \right] \int_0^\infty x^\alpha f(x) \, dx = 2 \pi i \sum_k \text{res} (b_k).
\] |
|
利用 Bessel 函数的渐近展开式 [ \(\lambda \to \infty,\) 即 \, \(E \to \infty,\)见附录六, 式(A6.13)] 可得
\[
&J_{1/3} \left( \frac{2}{3} \lambda^{3/2} \right) + J_{-1/3} \left( \frac{2}{3} \lambda^{3/2} \right) \\
&\propto \sqrt{\frac{2}{\pi} \cdot \frac{3}{2} \lambda^{-3/2}} \left[ \cos \left( \frac{2}{3} \lambda^{3/2} - \frac{5\pi}{12} \right) + \cos \left( \frac{2}{3} \lambda^{3/2} - \frac{\pi}{12} \right) \right]\\
&= \sqrt{\frac{3}{\pi}} \cdot \lambda^{-3/4} 2 \cos \left( \frac{2}{3} \lambda^{3/2} - \frac{\pi}{4} \right) \cdot \cos \frac{\pi}{6}\\
\]
对于 \(\lambda \gg 1\)，方程(3.7.16)的根由下式给出
\[
\cos \left( \frac{2}{3} \lambda^{3/2} - \frac{\pi}{4} \right) = 0
\] |
|
12. 证明欧勒多项式的积分表达式
(i) \(E_{2n}(x) = (-1)^n 4 \int_{0}^{\infty} \frac{t^{2n} \sin \pi x \, \mathrm{ch} \pi t}{\mathrm{ch} 2\pi t - \cos 2\pi x} \, dt,\)
\quad \(n = 0, 1, 2, \cdots, \quad 0 < \Re(x) < 1;\)
(ii) \(E_{2n+1}(x) = (-1)^{n+1} 4 \int_{0}^{\infty} \frac{t^{2n+1} \cos \pi x \, \mathrm{sh} \pi t}{\mathrm{ch} 2\pi t - \cos 2\pi x} \, dt,\)
\quad \(n = 0, 1, 2, \cdots, \quad 0 < \Re(x) < 1.\)
[提示：参考第3题。] |
|
\begin{align*}
t \left( \frac{(1 + \omega v)}{c^2} \right) - \frac{x}{c^2} (v + \omega) \\
\sqrt{\left( \frac{(1 + \omega v)}{c} \right)^2 - \frac{1}{c^2} (v + \omega)^2}
\end{align*} |
|
叫作常微分方程 \(\circled{1}\), 其中导数实际出现的最高阶数 \(n\) 叫作常微分方程 \((1.1)\) 的阶.
例如，下面的方程都是常微分方程:
\begin{align}
\frac{dy}{dx} + \frac{1}{x} y &= x^3, \quad (x \neq 0) \tag{1.2} \\
\frac{dy}{dx} &= 1 + y^2 \tag{1.3} \\
y'' + yy' &= x, \tag{1.4} \\
\ddot{\theta} + a^2 \theta &= 0, \quad (\cdot 表示 \frac{d}{dt}, \text{常数 } a > 0). \tag{1.5}
\end{align} |
|
是 \(f(z)\) 在 \(a_n\) 点的主部，
\[
\varphi_{n,p}(z) = \sum_{k=0}^p \left[ \frac{d^k}{dz^k} G_n \left( \frac{1}{\zeta - a_n} \right) \right]_{\zeta = 0} \frac{z^k}{k!}. \tag{4}
\]
\textbf{证} 设 \(m\) 够大使 \(z\) 在 \(C_m\) 内，则
\[
\frac{1}{2\pi i} \oint_{C_m} \frac{f(\zeta) d\zeta}{\zeta - z} = \frac{1}{2\pi i} \oint_{(\zeta_+)} \frac{f(\zeta) d\zeta}{\zeta - z} + \sum_{r=1}^m \frac{1}{2\pi i} \oint_{(a_r,+)} \frac{f(\zeta) d\zeta}{\zeta - z}.
\] |
|
\[
P_{0 \rightarrow \{n\}}(t) = \frac{1}{\hbar^2} \int dE \, \rho(E) \left| \langle E | \hat{V}_0 | 0 \rangle \right|^2 \left( \frac{\sin \left( \frac{(E - E_0)(t - t_0)}{2\hbar} \right)}{\frac{(E - E_0)(t - t_0)}{2\hbar}} \right)^2. \tag{18.228}
\]
At very late times \(t - t_0 \rightarrow +\infty\) we may use the fact that
\[
\lim_{t \rightarrow \infty} \left( \frac{\sin^2 (xt)}{x^2 t} \right) = \pi \delta(x). \tag{18.229}
\]
Then, we have
\[
P_{0 \rightarrow \{n\}}(t) = \frac{2\pi}{\hbar} \int dE \, \rho(E) \left| \langle E | \hat{V}_0 | 0 \rangle \right|^2 \delta(E - E_0)(t - t_0) = \frac{2\pi}{\hbar} (t - t_0) \left| \langle 0 | \hat{V}_0 | 0 \rangle \right|^2 \rho(E_0). \tag{18.230}
\]
Dividing this probability by the time, we obtain the total rate of transitions: |
|
\[
\frac{\sin\left(\pi \sqrt{l^2 + x^2}\right)}{\pi \sqrt{l^2 + x^2}} = \sum_{n=1}^{\infty} \frac{(-\pi x^2)^n}{n! 2^n} \cdot \frac{1}{l^n} \sqrt{\frac{1}{2l}} J_{n+1/2}(\pi l)
\] |
|
(1.1.2)，有
\[
\xi^\alpha(x) = \xi^\beta(x) [\varphi \nu(x)]^\alpha_\beta,
\]
微分后有
\[
\frac{\partial \xi^\alpha}{\partial x^\beta} = \frac{\partial \xi^\lambda}{\partial x^k} [\varphi \nu(x)]^\alpha_\lambda \frac{\partial x^k}{\partial x^\beta} + \xi^\lambda \frac{\partial [\varphi \nu(x)]^\alpha_\lambda}{\partial x^\beta}. \tag{1.2.7}
\]
从上述式中可看出，\(\xi^\alpha\) 的微分不再成为 \(G\) 型张量，为了使之成为另一矩阵李群 \(G_1\) 型的张量，我们引进一个联络的概念。 |
|
\[
C_l = a_0 \left[ \alpha_{\text{eff}} - \alpha_{L=0} \right] = 2\pi \left[ \alpha_{\text{eff}} - \alpha_{L=0} \right]
\] |
|
\[
0 \leq& 1 + \mu = \rho + \delta(r) - \rho_1 - \delta(r_1)\\
\leq& \frac{1}{2}[\gamma(r) - \gamma(r_1)][1 + \frac{1}{\gamma(r)}] \leq 1 + \gamma(r).\\
\] |
|
\textit{Exercise 3}
1) For all \(x, y \in \mathbb{Z}\), \(P(X = x, Y = y) = P(X = x)P(Y = y)\) (by independence via the laws, for all \(g \in \mathbb{Z}\)),
\[
P(X + Y = g) = \sum_{x+y = g} P(X = x, Y = y) = \sum_{x} P(X = x) P(Y = g-x)
\] (discrete convolution). |
|
随机事件的概率（发生可能性的大小）：当观测次数\(N\)趋于无穷时，某一事件\(A\)发生的次数\(N_A\)与总观测次数的比值将趋于稳定的极限值—概率\(P(A)\):
\[
P(A) = \lim_{N \to \infty} \frac{N_A}{N}
\] |
|
6. \textbf{Def. Oscillatory: Ex:} \(\sum_{n=1}^{\infty} U_n = 1 - 1 + 1 - 1 + \ldots (-1)^{n} + \ldots\)
often classed with divergence (not convergence) |
|
\[
\oint (xP) \cdot d\Sigma = \iiint (\nabla \cdot xP) \, dV = \iiint P \, dV - \iiint \rho_p x dV = \iiint \rho_p x dV
\]极化强度 \(P\) 满足
\[
& P = \lim_{V \to 0} \frac{1}{V} \iiint P dV = \lim_{V \to 0} \frac{1}{V} \left[ \iiint \rho_p x dV + \oint (xP) \cdot d\Sigma \right] \\
& \quad = \lim_{V \to 0} \frac{1}{V} \left[ \left( \sum P_i \right) + \iiint \sigma_p \cdot d\Sigma \right]
\]
其中 \(\sigma_p = P \cdot e_n\) , 忽略后一项，得 \(P \approx \frac{\sum P_i}{\Delta V}\)。 |
|
则原式 \[
&= 2 \int_0^{\pi} \ln (3 \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) d(\frac{x}{2}) \\
&= 2 \int_0^{\frac{\pi}{2}} \ln (\sin^2 t + 3 \cos^2 t) \, dt = 2 \pi \ln \frac{1+\sqrt{3}}{2}
\]
从而
\[
\lim_{n \to \infty} \prod_{i=0}^{n-1} (2 + \cos \frac{i \pi}{n})^{\frac{\pi}{n}} = e^{2 \pi \cdot \ln \frac{1+\sqrt{3}}{2}} = \left( \frac{1+\sqrt{3}}{2} \right)^{2\pi}
\] |
|
(上式中常数因子都略去未记,这对以下讨论无影响)。我们试用上列波函数来计算粒子流密度\(j\)及粒子密度\(\rho\)。
\[
j &= \frac{\hbar}{2im} \left( \psi^* \frac{d\psi}{dx} - \psi \frac{d\psi^*}{dx} \right)
\]
用自然单位（注意\(dx = 2^{-1/3}d\xi\)），得
\[
j &= \frac{1}{2i}2^{1/3} \left( \psi^* \frac{d\psi}{d\xi} - \psi \frac{d\psi^*}{d\xi} \right) = 2^{1/3}
\]
\(\rho = \psi^* \psi = \xi^{1/2}\)
因此 |
|
\[
\frac{\partial \tilde{x}}{\partial x} = \begin{pmatrix}
\frac{\partial \tilde{x}^1}{\partial x^1} & \cdots & \frac{\partial \tilde{x}^1}{\partial x^m} \\
\vdots & \ddots & \vdots \\
\frac{\partial \tilde{x}^m}{\partial x^1} & \cdots & \frac{\partial \tilde{x}^m}{\partial x^m}
\end{pmatrix} \tag{1.1.3}
\]
是非异的情形。此时，变换 (1.1.2) 称为局部坐标变换或简称坐标变换。
令 \(\mathbb{K}\) 代表实数域或复数域， \(GL(N, \mathbb{K})\) 表示矩阵元素属于 \(\mathbb{K}\) 的 \(N \times N\) 非异方阵所成的群。由于 \(GL(N, \mathbb{K})\) 是欧氏空间 \(\mathbb{K}^{N^2}\) 的开集，其幺元素，即单位方阵 \(1\)，必有 \(\mathbb{K}^{N^2}\) 中的邻域仍然包含于 \(GL(N, \mathbb{K})\)。 |
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2) By independence and the Fubini-Tonelli Theorem, for aP \(I \times J \subseteq K\),
\[
P((X, Y) \in I \times J) = P(X \in I)P(Y \in J) = \left( \int_I f_X(x) dx \right) \left( \int_J f_Y (y) dy \right) = \int_{I \times J} f(x) f(y) dx dy
\]
so that
\[
f_{(X, Y)} = f_X f_Y
\] |
|
in its first variable and intolerable in its second variable. Assume furthermore that the Lipschitz constant of \(f\) its frst variable is \(\nu\) -integrable. Then
\[
\sup_{t \in [0,T]} \left| X_t(x_0, \omega) - x_0 - \int_0^t \bar{f}(x_0, \varphi_{s/t}(\omega)) \, ds \right| = o \left( \sup_{[0,T]} \int_0^t \bar{f}(x_0, \varphi_{s/t}(\omega)) \, ds \right),
\]
for \(\nu\)-almost every \(\omega \in \mathcal{M}\).
If moreover \((\mathcal{M}, \varphi_t, \nu)\) satisfies a law of large numbers of the following form
\[
\forall g \in L^1(\nu), \quad bt \int_0^t g \circ \varphi_s \, ds \overset{t \to +\infty}{\Longrightarrow} L_r \int_\mathcal{M} g \, d\nu Y, \tag{13}
\] |
|
设在 \(x \in V\) 有一组量 \(T_{\alpha_1 \cdots \alpha_r}^{\beta_1 \cdots \beta_s}(x) (\alpha_1, \cdots, \alpha_r, \beta_1, \cdots, \beta_s = 1, \cdots, N)\)，它经坐标变换 (1.1.2) 有如下的变换关系：
\[
\tilde{T}_{\alpha_1' \cdots \alpha_r'}^{\beta_1' \cdots \beta_s'}(\tilde{x}) = [\det \varphi_{\tilde{V} V}(x)]^{-1} T_{\lambda_1 \cdots \lambda_r}^{\gamma_1 \cdots \gamma_s}(x) [\varphi_{\tilde{V} V}(x)]_{\alpha_1'}^{\lambda_1} \cdots [\varphi_{\tilde{V} V}(x)]_{\alpha_r'}^{\lambda_r} \times [\varphi_{\tilde{V} V}(x)^{-1}]_{\gamma_1}^{\beta_1'} \cdots [\varphi_{\tilde{V} V}(x)^{-1}]_{\gamma_s}^{\beta_s'}. \tag{1.1.7}
\] |
|
Exercise 1 \(m^{\circ}1\)(\(m^{\circ}2\))
\(X \in \mathbb{R}^p\) is a random vector and \(x\) denotes a sample of \(X\), \(w \in \mathbb{R}^p\) is a weight vector.
\(\Sigma\) is a scalar random variable with distribution \(\mathcal{N}(0, \sigma^2)\).
1) Conditionally on \(X\) and \(w\), the quantity \(w^T X\) can be considered deterministic.
Thus, the random variable \(w^T X + \Sigma\) is a Gaussian random variable with mean
\(w^T X\) and variance Var\([\Sigma] = \sigma^2\), meaning \(Y|X \sim \mathcal{N}(w^T X, \sigma^2)\) |
|
\[
P'_{n+1}(x) &- P_{n-1}'(x) = \frac{1}{2^{n-1} (n-1)!} \frac{d'}{dx^n} \left( (x^2-1)^n P_n(x) \right) \\
&= \frac{\alpha n}{\alpha^n (n!)} \frac{d^n}{dx^n} (x^2-1)^n + P_n(x) \\
&= -2n P_n(x) + P_n(x) \\
&= (2n+1) P_n(x)
\] |
|
\[
A &= \frac{S}{2} \left(1 + \frac{ik}{\kappa}\right) e^{i\kappa a-na} \\
B &= \frac{S}{2} \left(1 - \frac{ik}{\kappa}\right) e^{i\kappa a+na}
\]
比较式(3.3.11)和(3.3.12)，消去\(A, B\)，得
\[
\left(1 + \frac{ik}{\kappa}\right) + R \left(1 - \frac{ik}{\kappa}\right) &= S \left(1 + \frac{ik}{\kappa}\right) e^{i\kappa a-na} \\
\left(1 - \frac{ik}{\kappa}\right) + R \left(1 + \frac{ik}{\kappa}\right) &= S \left(1 - \frac{ik}{\kappa}\right) e^{i\kappa a+na}
\]
再从上式消去\(R\)，得
\[
\frac{S e^{i\kappa a-na} - 1}{S e^{i\kappa a+na} - 1} = \left(\frac{1 - ik/\kappa}{1 + ik/\kappa}\right)^2
\] |
|
\textbf{DEFINITION 6.} Suppose \(F : X \to Y\) is a smooth map between manifolds: then the \textit{derivative} of \(F\) at \(x\) is the linear map \(D_x F : T_x X \to T_{F(x)} Y\) which takes a path \(\gamma : (-\varepsilon, \varepsilon) \to X\) with \(\gamma(0) = x\) in \(T_x X\) to the path \(F \circ \gamma \in T_{F(x)} Y\).
The meaning of this should be intuitively clear, but let's check that this actually recovers the original derivative when \(X, Y\) are open subsets of \(\mathbb{R}^n, \mathbb{R}^m\) respectively. In this case, given \(v \in \mathbb{R}^n\), the map \(\mathbb{R}^n \cong T_x X \to T_{F(x)} Y \cong \mathbb{R}^m\) is given explicitly by:
\[
v \mapsto \left. \frac{d}{dt} \right|_{t=0} F(x + tv) = D_v F(x)
\]
which is exactly the directional derivative of \(F\) in the direction \(v\). In fact, we can always reduce to this case by using the local identifications with open subsets of \(\mathbb{R}^n\) given by local coordinates: if \(G = \psi \circ F \circ \phi^{-1}\) is |
|
\[
&\begin{cases}
& 0 \leq \delta(r) - \delta(s) \leq \min\{1 + \gamma(r), d(1 - \gamma(r))\}, \\
& 1 < s, r < \infty \quad \text{if} \quad d = 2,
&\end{cases}\\
\] |
|
从 \((1.14)\) 和 \((1.15)\) 两式解出 \(C_1\) 和 \(C_2\) 作为 \(x, y\) 和 \(y'\) 的函数, 即
\[
\begin{cases}
C_1 = e^{-x} [y (\sin x + \cos x) - y' \sin x], \\
C_2 = e^{-x} [y (\sin x - \cos x) + y' \cos x].
\end{cases}
\]
然后把它们代入 \((1.16)\) 式, 得到一个二阶微分方程
\[
y'' - 2y' + 2y = 0, \tag{1.17}
\] |
|
在水平桌面上固定着一个光滑圆轨道，在轨道的 \(B\) 点静止着一个质量为 \(m_2\) 的弹性小球，另一个质量为 \(m_1\) 的弹性小球以初速度 \(v_0\) 运动，与质量为 \(m_2\) 的小球发生第一次碰撞后，恰好 \(C\) 点发生第二次碰撞（图6-2），则两球质量之比为 \(m_1:m_2\) 等于
A. \(5:3\)
B. \(9:1\)
C. \(1:7\)
D. \(2:3\) |
|
The solution that, for \(t = 0\), satisfies \(d(x_i, 0) = 0\) (the initial network consists of \(N\) vertices and no edges) is
\[
d(x_i, t) = \frac{2(N-1)}{N(N-2)} t + Ct^{N/2(N-1)},
\]
where \(C\) is a constant. Since \(N \gg 1\), the expression above can be written
\[
d(x_i, t) \approx \frac{2}{N} t + Ct^{1/2}.
\]
Taking into account the relation \(\sum_{j=1}^N d(x_j) = 2t\), the constant \(C\) is equal to zero, and we finally obtain
\[
d(x_i, t) \approx \frac{2}{N} t.
\] |
|
\[
\kappa(\xi, \eta) &= \frac{1}{\sqrt{\frac{\xi^2 + \eta^2}{2}} + \delta} \left| 2 \left(\xi + \frac{l}{2}\right) \sin \beta + \eta \cos \beta \right|, \\
\frac{q}{V} &= \frac{\sigma^2}{\sqrt{\frac{\xi^2 + \eta^2}{2}} + \delta} \frac{2 \left(\xi + \frac{l}{2}\right) \sin \beta + \eta \cos \beta}{\sqrt{ \left( \sigma^2 - \frac{l^2}{4} \right)^2 + l^2 \eta^2}}, \\
\sigma^2 &= \xi^2 + \eta^2.
\] |
|
结果：
\[
\int_{0}^{\frac{\pi}{2}} \ln(a^2\sin^2x + b^2\cos^2x)\, dx = \pi \ln\frac{|a| + |b|}{2} \quad (a^2 + b^2 \neq 0)
\] |
|
\begin{align*}
h^2 + \chi E^2 \chi &= T + V + \chi E^2 \chi \\
&= T + \chi V \chi + \chi^\perp (V - E^2) \chi^\perp + E^2 \\
&\geq \chi h^2 \chi + \chi^\perp T \chi^\perp + \chi^\perp (V - E'^2) \chi^\perp + E^2 - 2 \varepsilon_0.
\end{align*} |
|
\begin{itemize}
\item 从近独立粒子到作用粒子
\item 刘维尔定理：随着一个代表点沿正则方程所确定的轨道在相空间中运动，其邻域的代表点密度是不随时间改变的常数
\[
\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t} + \sum_\alpha \left( \frac{\partial \rho}{\partial q_\alpha} \frac{\partial H}{\partial p_\alpha} - \frac{\partial \rho}{\partial p_\alpha} \frac{\partial H}{\partial q_\alpha} \right) = 0
\]
\item 分布函数
\begin{itemize}
\item 含义 \(P = \rho(q, p, t) d\Omega\)
\item 归一化 \(\int \rho(q, p, t) d\Omega = 1\)
\item 宏观量 \(\bar{B}(t) = \int B(q, p) \rho(q, p, t) d\Omega\)
\end{itemize}
\end{itemize} |
|
Then, for any \(z \in \mathbb{R}\),
\[
f_{X_1 + X_2}(z) = \frac{1}{\sqrt{2\pi\sigma_1^2}} \cdot \frac{1}{\sqrt{2\pi\sigma_2^2}} \int e^{-\frac{1}{2\sigma_1^2}(x-\lambda_1)^2} e^{-\frac{1}{2\sigma_2^2}(z-x-\lambda_2)^2} dx.
\] |
|
\textbf{9.47 定理} 令 \(X\) 的密度是指数族的，则
\[
\mathbb{E}(T(X)) &= A'(\eta), \\
\text{Var}(T(X)) &= A''(\eta).
\]
如果 \(\theta = (\theta_1, \cdots, \theta_k)\) 是一个向量，且
\[
f(x; \theta) = h(x) \exp \left\{ \sum_{j=1}^k \eta_j(\theta) T_j(x) - B(\theta) \right\},
\] |
|
where now \(\delta\) is real. The velocity function is
\[
F' = \frac{dF}{d\delta^2} - \frac{dF}{d\delta} \frac{d\delta}{d\delta^2} = -\frac{n e^{-n\delta}}{\sin \delta} - \frac{n}{\sin \delta} ( \cos n\delta - i \sin n\delta)
\]
giving at points along the line the transverse component |
|
于是，可取以\(O\)为球心，以\(OA = r\)为半径的\textbf{球面作为高斯面}，如右下图所示。根据高斯定理可得：\[
\iint E(r) \cdot dS = 4\pi r^2 E(r) = \frac{1}{\varepsilon_0} \sum \limits_{\text{(S内)}} q
\](1) 高斯面位于带电球外（\(r > R\)）。对这种情况，S内的电量等于带电球的总电量\(Q = 4\pi R^3 \rho_e / 3\)，代入上式可得：\[
E(r) = \frac{Q}{4\pi \varepsilon_0 r^2} = \frac{R^3 \rho_e}{3 \varepsilon_0 r^2}, \quad (r > R)
\] |
|
\begin{align*}
&\sum_{n=0}^{2} x(n) z^{-n} \\
&x(0)z^{0} + x(1)z^{-1} + x(2)z^{-2} \\
&1 \cdot z^{-0} + 2 \cdot z^{-1} + 3 \cdot z^{-2} \\
&1 + \frac{2}{z} + \frac{3}{z^2}
\end{align*} |
|
\[
Z &= \sum e^{-\beta E} \\
&= \frac{2 \pi m S}{h^2} \frac{1}{\beta} e^{\beta \varepsilon_0}
\] |
|
法：\(\vec{n}_2 = (f_x, f_y, -1)\) 以上下侧而定
\[
\Rightarrow \iint\limits_S P \, dy \, dz + Q \, dx \, dz + R \, dx \, dy = \iint\limits_P (P f_x + Q f_y - R) \, dx \, dy
\]
注：\(\iint\limits_S R \, dx \, dy\) 并非二重积分，而是二型曲面积分的第三项。 |
|
如图4-2所示，\(AB\) 是竖直平面内的光滑圆弧面，一小物块从 \(A\) 点静止释放，它滑上静止不动的水平皮带后，从 \(C\) 点离开皮带作平抛运动，落在水平地面的 \(D\) 点。现在皮带轮转动，皮带的上表面以某一速率向左或向右匀速运动，小物块仍从 \(A\) 点静止释放，则小物块将可能落在地面上的位置是 （ ）
A. \(D\) 点右边的 \(M\) 点；
B. \(D\) 点；
C. \(D\) 点右边的 \(N\) 点；
D. 右皮带轮正下方的 \(O\) 点。 |
|
\[
\delta_{k_1 \cdots k_r}^{i_1 \cdots i_r} =
\begin{vmatrix}
\delta_{k_1}^{i_1} & \cdots & \delta_{k_r}^{i_1} \\
\vdots & \ddots & \vdots \\
\delta_{k_1}^{i_r} & \cdots & \delta_{k_r}^{i_r}
\end{vmatrix},
\] |
|
求 \(f(x, y) = x \sin\frac{1} {y} + y \sin\frac {1} {X}\) 在 \((0,0)\) 的 \textbf{重极限} 和 \textbf{累次极限}。
证：
\[
0 \leq |x \sin \frac{1} {y} + y \sin\frac{1} {x} | \leq |x \sin \frac{1} {y} | + |y \sin \frac{1} {x} | \leq |x| + |y|
\]
而
\[
\lim_{(x, y) \to (0, 0)} |x| + |y| = 0 \textbf{由迫敛性知得重极限为}，0。
\]
而
\[
\lim_{x \to 0} \sin \frac{1}{x} \quad \lim_{y \to 0} \sin \frac{1}{y}
\]
不存在，故累次极限不存在。 |
|
因为 \(\left( \frac{\partial p}{\partial T} \right)\) 在相Ⅰ和相Ⅱ是连续的，故由相Ⅰ过渡到相Ⅱ在 \(T = T_t\) 处，定容比热是连续的。现在问 \(\frac{dC_V}{dT}\) 是否连续呢？
可以证明，由相Ⅰ过渡到相Ⅱ在 \(T = T_t\) 处 \(\frac{dC_V}{dT}\) 是不连续的，其差为
\[
\Delta \left( \frac{\partial C_V}{\partial T} \right) &= \left( \frac{\partial C_V}{\partial T} \right)_{T_t^+} - \left( \frac{\partial C_V}{\partial T} \right)_{T_t^-} \\
& = -\frac{27}{16 \pi} \frac{Nk}{T_t^2} \quad (2.612)^2. \tag{1.245}
\] |
|
\[
\lim_{\epsilon \to 0^+} [\psi'(\epsilon) - \psi'(-\epsilon)] = \lim_{\epsilon \to 0^+} [\kappa A (2\kappa\epsilon) + \kappa B (2\kappa\epsilon)] \\
= \lim_{\epsilon \to 0^+} 2\kappa^2 \epsilon (A + B) = \frac{2\mu \gamma}{\hbar^2} \psi(0)
\] |
|
where
\[
a_0 = \frac{1}{a} \int_{-a}^{a} dx \, f(x), \quad a_k = \frac{1}{a} \int_{-a}^{a} dx \, \cos \left(\frac{k \pi x}{a}\right) f(x), \quad b_k = \frac{1}{a} \int_{-a}^{a} dx \, \sin \left(\frac{k \pi x}{a}\right) f(x)
\]
for \(k = 1, 2, \ldots\) Parseval's equation now takes the form
\[
\frac{1}{a} \int_{-a}^{a} dx \, |f(x)|^2 = \frac{1}{a} |f|^2 = \frac{1}{a} \left(\sum_{k=0}^{\infty} |(c_k, f)|^2 + \sum_{k=1}^{\infty} |(s_k, f)|^2 \right) = \frac{|a_0|^2}{2} + \sum_{k=1}^{\infty} \left( |a_k|^2 + |b_k|^2 \right).
\] |
|
为 \(0\) , 也势能的变化量等于静电力所做的功。
即 \(W = Ep_{\infty} - E_{pA} = 0 - E_{pA}\)
所以 \(E_{pA} = -W = 1.2 \times 10^{-4} J\)
\(V_A = \frac{E_{pA}}{q} = \frac{1.2 \times 10^{-4}}{1.0 \times 10^{-8}}\)
\(V = 1.2 \times 10^4 V\) |
|
So \(R(T) = \{ a_0 + a_2 t^2 + \ldots + a_n t^n \mid a_0, a_2, \ldots, a_n \in \mathbb{F} \}\).
Similarly, \[ p(t) \in N(T)\iff [T]_B[p(t)]_B \iff 0\]
\[
\iff
\begin{bmatrix}
p(t)
\end{bmatrix}_B
=
\begin{bmatrix}
0 \\
\vdots \\
0
\end{bmatrix}
\text{ for some } a_1 \in \mathbb{F}
\]
\[
\iff
p(t) = a_1 t \text{ for some } a_1 \in \mathbb{F}
\]
so \(N(T) = \{ a, t \mid a_1 \in \mathbb{F} \}\). To see that \(P_n = N(T) \oplus R(T)\), first observe that if \(p(t) \in P_n\) then \(\exists a_0, a_1, \ldots, a_n \in \mathbb{F}\) such that \(p(t) = a_0 + a_1 t + \ldots + a_n t^n\),
and \(a_1 t \in N(T), a_0 + a_2 t^2 + \ldots + a_n t^n \in R(T)\) so \(p(t) \in N(T) + R(T)\)
\[
\Rightarrow P_n = N(T) + R(T)
\] |
|
条件概率 \(P(B \mid A)\) 是在事件 \(B\) 已出现，而又出现事件 \(A\) 的概率（按雷克书中的定义，但很多书与此相反）。
我们有：
\[
P(B \mid A) &= \frac{P(A \cap B)}{P(B)}
\]
由 \(P(A \cap B) = P(B \cap A)\) 我们有（Bayes' theorem）:
\[
P(A)P(B \mid A) &= P(B)P(A \mid B)
\]
若 A 和 B 独立，则 \(P(B \mid A) = P(A)\)。 |
|
increases further, up to the angle \(\omega_{\max} = \omega_{\min} + \pi\). The value of \(\omega_{\min}\) increases with \(f/\delta\). For \(\frac{\delta}{f} = 0\), \(\omega_{\min} = 0\), hence the value of \(\sigma\) is smallest on the trailing edge and greatest on the leading edge. Conversely, for \(\frac{\delta}{f} = 0\), \(\omega_{\min} = (\pi/2) - A\) |
|
Let \(v \in R(S+T)\), so \(v = (S+T)(u) = S(u) + T(u)\) for some \(u \in U\).
Then \(\exists\, a_1, \ldots, a_k, a_{k+1}, \ldots, a_{k+l}, b_1, \ldots, b_{k}, b_{k+1}, \ldots, b_{k+m} \in \mathbb{F}\) s.t.
\[
S(u) &= a_1 u_1 + \cdots + a_k u_k + a_{k+1} v_1 + \cdots + a_{k+l} v_l \\
T(u) &= b_1 u_1 + \cdots + b_k u_k + b_{k+1} w_1 + \cdots + b_{k+m} w_m
\] |
|
由于 \(x > 0\)，有 \(1 + x < e^x\)，即 \(\ln(1+x) < x\) 从而 \(\ln(1+\frac{1}{2n}) < \frac{1}{2^n}\)。
\[
\ln(x_n) &= \ln(1+\frac{1}{2}) + \ln(1+\frac{1}{2^2}) + \ldots + \ln(1+\frac{1}{2^n}) \\
&< \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^n} = \frac{\frac{1}{2} (1 - (\frac{1}{2})^n)}{1-\frac{1}{2} }= 1 - \frac{1}{2^n} < 1
\]
故 \(x_n < e\)，从而 {\(x_n\)} 有上界。
\[
\frac{x_{n+1}}{x_n} &= 1 + \frac{1}{2^{n+1}} > 1 \quad \text{且} \quad x_n > 0
\]
故 \(\{x_n\}\) 递增。
从而 \(\{x_n\}\) 有敛散。即 \(x_n = (1 + \frac{1}{2})(1 + \frac{1}{2^2}) \ldots (1 + \frac{1}{2^n})\)极限存在。 |
|
\[
\mathcal{F}_{\theta}(x_t) &= \int p(\epsilon) \min \left\{ 0, \log \frac{\pi(T_{\theta}(x_t, \epsilon))}{\pi(x_t)} + \log \frac{q_{\theta}(x_t | T_{\theta}(x_t, \epsilon))}{q_{\theta}(T_{\theta}(x_t, \epsilon) | x_t)} \right\} d\epsilon + \beta \mathcal{H}_{q_{\theta}(y | x_t)}.
\]
Since MCMC at the \(t\)-th iteration proposes a specific state \(y_t\) constructed as \(\epsilon_t \sim p(\epsilon_t)\), \(y_t = T_{\theta}(x_t, \epsilon_t)\), an unbiased estimate of the exact gradient \(\nabla_{\theta} \mathcal{F}_{\theta}(x_t)\) can be obtained by
\[
\nabla_{\theta} \mathcal{F}_{\theta}(x_t, \epsilon_t) &= \nabla_{\theta} \min \left\{ 0, \log \frac{\pi(T_{\theta}(x_t, \epsilon_t))}{\pi(x_t)} + \log \frac{q_{\theta}(x_t | T_{\theta}(x_t, \epsilon_t))}{q_{\theta}(T_{\theta}(x_t, \epsilon_t) | x_t)} \right\} + \beta \nabla_{\theta} \mathcal{H}_{q_{\theta}(y | x_t)},
\] |
|
此外, 通常把反称张量写为
\[
(F_{ik}) &= \begin{pmatrix}
0 & F_{01} & F_{02} & F_{03} \\
F_{10} & 0 & F_{12} & F_{13} \\
F_{20} & F_{21} & 0 & F_{23} \\
F_{30} & F_{31} & F_{32} & 0
\end{pmatrix} = \begin{pmatrix}
0 & E_1 & E_2 & E_3 \\
-E_1 & 0 & H_3 & -H_2 \\
-E_2 & -H_3 & 0 & H_1 \\
-E_3 & H_2 & -H_1 & 0
\end{pmatrix},
\]
其中 \(\mathbf{E} = (E_1, E_2, E_3)\) 称为电场向量, \(\mathbf{H} = (H_1, H_2, H_3)\) 称为磁场向量. |
|
The result for \(n\) independent \(X_i \sim \mathcal{N}(\mu, \sigma^2)\) follows from an immediate induction.
Since the \(X_i\) are independent, denoting \(\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i\), we have that, for any \(t \in \mathbb{R}\): |
|
故
\[
B_k = 0 \quad (k = 1, 2, 3, \ldots)
\]
\[
u|_{t=0} = \sum_{k=1}^{\infty} A_k \sin \frac{k \pi x}{l} =
\begin{cases}
\frac{hx}{c} & 0 \leq x \leq c \\
\frac{h(l-x)}{l-c} & c < x \leq l
\end{cases}
\]
故
\[
A_k = \frac{2}{l} \left[ \int_0^c \frac{h}{c} x \sin \frac{k \pi x}{l} \, dx + \int_c^l \frac{h(l-x)}{l-c} \sin \frac{k \pi x}{l} \, dx \right]
\]
\[
\quad = \frac{2hl^2}{k^2 \pi^2 c (l-c)} \sin \frac{k \pi c}{l}
\]
于是
\[
u(x,t) = \sum_{k=1}^{\infty} \frac{2hl^2}{k^2 \pi^2 c (l-c)} \sin \frac{k \pi c}{l} \cos \frac{k \pi a t}{l} \sin \frac{k \pi x}{l}
\] |
|
对数函数： \(y = \log_a x \ (a > 0 \ 且 \ a \neq 1)\)
(\(a = 4\)依底表，亦叫依数表)
\(0 < a < 1\)时，底数表为减函数
\(a > 1\)时，底数表为增函数 |
|
向量\(\nabla u = grad \, u = \frac{\partial u}{\partial x} \, \vec{i} + \frac{\partial u}{\partial y} \, \vec{j} + \frac{\partial u}{\partial z} \, \vec{k}\)
1) \(\nabla C = 0\)
2) \(\nabla (u+v) = \nabla u + \nabla v\)
3) \(\nabla (uv) = v \nabla u + u \nabla v\)
4) \(\nabla \left(\frac{u}{v}\right) = \frac{1}{v} (v \nabla u - u \nabla v)\)
5) \(\nabla \varphi(u) = \varphi'(u) \nabla u\)
6) \(\nabla \psi(u,v) = \psi_u \nabla u + \psi_v \nabla v\) |
|
\textit{Theorem} Let \(C\) be a curve defined by (2.11). Then \(C\) is rectifiable if and only if both \(\phi\) and \(\psi\) are of bounded variation. Moreover,
\[
V(\phi), V(\psi) \leq L \leq V(\phi) + V(\psi).
\]
\textit{Proof}. We will use the simple inequalities
\[
|a|, |b| \leq (a^2 + b^2)^{1/2} \leq |a| + |b|
\]
for real \(a\) and \(b\). Thus, if \(C\) is rectifiable and \(\Gamma = \{t_i\}\) is any partition of \([a,b]\), the inequality
\[
l(\Gamma) = \sum (\left(\phi(t_i) - \phi(t_{i-1})\right)^2 + \left(\psi(t_i) - \psi(t_{i-1})\right)^2)^{1/2} \leq L
\] |
|
为了使误差最小化，采用梯度下降法，对误差求负导数
\[
-\frac{\partial E_{\theta}}{\partial \theta} &= \mathbb{E}_{x \sim \pi} \left[ 2 \left( V^{\pi}(x) - V_{\theta}(x) \right) \frac{\partial V_{\theta}(x)}{\partial \theta} \right] \\
&= \mathbb{E}_{x \sim \pi} \left[ 2 \left( V^{\pi}(x) - V_{\theta}(x) \right) x \right], \tag{16.34}
\]
于是可得到对于单个样本的更新规则
\[
\theta = \theta + \alpha \left( V^{\pi}(x) - V_{\theta}(x) \right) x \cdot \tag{16.35}
\]
我们并不知道策略的真实值函数 \(V^{\pi}\)，但可借助时序差分学习，基于 \(V^{\pi}(x) = r + \gamma V^{\pi}(x)\) 用当前估计的值函数代替真实值函数，即 |
|
\[
\alpha(\theta_o) = \frac{2b}{\pi c(\theta_o)} \sum_{n=1}^{N} A_n \sin n\theta_o + \alpha_{L=0}(\theta_o) \frac{d\theta}{dy} d\theta + \frac{1}{\pi} \sum_{n=1}^{N} \int_{0}^{\pi} \frac{nA_n \cos n\theta}{\cos \theta - \cos \theta_o} d\theta
\] |
|
我们知道，对于自由粒子，其定态为非束缚态，能量是连续变化的，
\[
E = {\hbar^2 k^2}/{2m} \quad (-\infty < k < +\infty)
\]
对于一维自由粒子，能级为二重简并。而一维谐振子场中运动的粒子，定态为束缚态(不简并)，能量是不连续的，
\[
E = E_n = \left(n + \frac{1}{2}\right) \hbar \omega_0, \quad n = 0, 1, 2, \cdots
\] |
|
then the infinite series \textit{converges} \(\sum_{n=1}^{\infty} U_n = S\)
a. \textit{Necessary condition} \(\lim_{{n \to \infty}} U_n = 0\) (but not sufficient) for convergence
5. \textit{Def. Divergence} \(\lim_{{i \to \infty}} S_i = \pm \infty\) |
|
\begin{align*}
\frac{dE}{dt} &= \vec{v} \cdot \vec{f} + (c^2 - v^2) \frac{dm_0}{dT}
\end{align*}
Hence the rate of increase of energy due to external source
\begin{align*}
\frac{dE}{dT} &= (c^2 - v^2) \frac{dm_0}{dT}
\end{align*}
\textbf{LANGRANGES EQUATION OF MOTION}\\
Lagrange's equation for a single particle in the classical form
\begin{align*}
\frac{d}{dt} (m \dot{x}) - \frac{\partial V}{\partial x}
\end{align*}
\begin{align*}
\frac{d}{dt} (m \dot{y}) &= -\frac{\partial V}{\partial y}, \quad \frac{d}{dt} (m \dot{z}) = -\frac{\partial V}{\partial z}
\end{align*}
Hence
\begin{align*}
\frac{d}{dt} \left(\frac{m_0}{\sqrt{1 - v^2/c^2}} \dot{x}\right) - \frac{\partial V}{\partial x} = 0 \quad \frac{d}{dt} \left(\frac{m_0}{\sqrt{1 - v^1/c^1}} \dot{y}\right) - \frac{\partial V}{\partial y} = 0 \quad \frac{d}{dt} \left(\frac{m_0}{\sqrt{1 - v^2/c^2}} \dot{z}\right) - \frac{\partial V}{\partial z} = 0
\end{align*} |
|
6. 物体 \(A\) 静止在粗糙斜面上，\(B\) 放在粗糙水平地面上，在物体 \(A\) 上作用一水平推力 \(F\)，如图 1–5 所示。当 \(F\) 逐渐增大，物体 \(A\) 仍保持静止时，则（ ）
A. 斜面 \(B\) 对物体 \(A\) 的支持力增大。
B. 物体 \(A\) 所受的静摩擦力可能为零。
C. 物体 \(A\) 所受的静摩擦力可能沿斜面向下。
D. 物体 \(A\) 所受的合力增大。 |
|
Taking the gradient w.r.t \(A = \Sigma^{-1}\) of the log-likelihood, we obtain
\begin{align*}
\nabla_A \ell(\mu, A) = -\frac{m}{2} A^{-1} + \frac{1}{2} \sum_{\tau =1}^{m} (x_\tau - \mu)(x_\tau - \mu)^\tau .
\end{align*}
so that
\begin{align*}
\hat{ \sum }= \frac{1}{m} \sum_{\tau=1}^{m} (x_i - \mu)(x_i - \mu)^\tau.
\end{align*} |
|
Note that if \(a \in \mathbb{R}\) and \(f \in L^p(X, \mathcal{A}, \mu, \mathbb{R})\), then \(\alpha f \in L^p(X, \mathcal{A}, \mu, \mathbb{R})\), and that if \(a \in \mathbb{C}\) and \(f \in L^p(X, \mathcal{A}, \mu, \mathbb{C})\), then \(\alpha f \in L^p(X, \mathcal{A}, \mu, \mathbb{C})\). Furthermore, if \(f\) and \(g\) belong to \(L^p(X, \mathcal{A}, \mu, \mathbb{R})\) or to \(L^p(X, \mathcal{A}, \mu, \mathbb{C})\), then
\[
|f(x) + g(x)|^p &\leq (|f(x)| + |g(x)|)^p \leq (2 \max\{|f(x)|, |g(x)|\})^p \\
&\leq 2^p |f(x)|^p + 2^p |g(x)|^p.
\] |
|
them according to Equation 6.
\[
\begin{cases}
\text{Enhance: } \omega_{i}^{l} = n \times \omega_{i}^{l}, n > 1, \forall \omega_{i}^{l} \in \omega_{(s,r,o)} \\
\text{Suppress: } \omega_{i}^{l} = 0, \quad \omega_{i}^{l} \in \omega_{(s,r,o)}
\end{cases}
\] |
|
\begin{itemize}
\item How does \(Z(t)\) evolve?
\[
\frac{d}{dt} Z_u(t) = \sum_v \beta A_{u,v} Z_v(t) - Z_u(t)
\]
- This is a linear evolution!
\item In expectation, it is
\[
\frac{d}{dt} \mathbb{E}[Z_u(t)] = \sum_v \beta A_{u,v} \mathbb{E}[Z_v(t)] - \mathbb{E}[Z_u(t)]
\]
- This is a linear deterministic evolution (N dimension)
\item Which is
\[
\mathbb{E}[Z(t)] = e^{t(\beta A - I_d)} Z(0) = e^{t(\beta A - I)} X(0)
\]
\item So that
\[
\mathbb{P}[Z(t) \ne 0] \leq ||e_1|| ||\exp(t \cdot (\beta A - I)) X(0)||
\]
\end{itemize} |
|
13. \(n\)阶\(\nu\)次欧勒多项式\(E_{\nu}^{(n)}(x)\)的定义是
\[
\frac{2 e^{\nu x}}{(e^t + 1)^n} = \sum_{\nu=0}^{\infty} \frac{t^{\nu}}{\nu!} E_{\nu}^{(n)}(x),
\]
证明加法公式
\[
E_{\nu}^{(m+n)}(x + y) = \sum_{k=0}^{\nu} \binom{\nu}{k} E_{k}^{(m)}(x) E_{\nu-k}^{(n)}(y).
\]
14. 用达布公式（1.3节）或其他方法证明下列展开式
\[
f(z) - f(a) = \sum_{n=1}^{\infty} \frac{(z - a)^n}{n! (1 - r)^n} \left[f^{(n)}(z) - r^n f^{(n)}(a)\right],
\]
求出\(n\)项以后的余项，并讨论级数的收敛性。 |
|
\[
\frac{-t}{e^{-t} - 1} &= 1 + \frac{t}{2} + \sum_{r=1}^{n} \frac{(-r)^{r-1} B_r}{(2r)!} t^{2r} \\
&\quad + \frac{t^{2n+2}}{e^{-t} - 1} \int_{0}^{1} P_{2n+1}(x) e^{-t x} dx. \tag{1}
\]
代入上节 \((22)\) 式的积分中，级数 \(\sum\) 这一项的贡献为
\[
\sum_{r=1}^{n} \frac{(-r)^{r-1} B_r}{(2r)!} \int_{0}^{\infty} t^{2r-2} e^{-s t} dt = \sum_{r=1}^{s} \frac{(-r)^{r-1} B_r}{(2r)!} \frac{(2r-2)!}{z^{2r-1}},
\]
积分项的贡献计算如下：用 \(1.3\) 节 (15) 式，有
\[
\left| \int_{0}^{1} P_{2n+1}(x) e^{-t x} dx \right| \leq \frac{4}{(2\pi)^{2n+1}} \int_{0}^{1} e^{-t x} dx = \frac{4}{(2\pi)^{2n+1}} \frac{e^{-t} - 1}{e^{-t}},
\] |
|
考虑对自变量 \(x\) 有 \(n\) 次连续导数的函数族
\[
y = \varphi(x, C_1, C_2, \cdots, C_n), \tag{1.18}
\]
其中 \(C_1, C_2, \cdots, C_n\) 是独立的, 即: \(\varphi, \frac{\partial \varphi}{\partial x}, \cdots, \frac{\partial^{n-1} \varphi}{\partial x^{n-1}}\) 对 \(C_1, C_2, \cdots, C_n\) 存在连续偏导数, 而且 Jacobi 行列式
\[
\frac{D \left( \varphi, \frac{\partial \varphi}{\partial x}, \cdots, \frac{\partial^{n-1} \varphi}{\partial x^{n-1}} \right)}{D(C_1, C_2, \cdots, C_n)} \neq 0 \tag{1.19}
\] |
|
\[
-(ad-bc)I &= A^2 - (a+d)A = A(A - (a+d)I) \implies \\
A^{-1} &= \frac{1}{-(ad-bc)} \left( A - (a+d)I \right),
\] |
|
we can again invoke geometry and use
\[
y &= \frac{b}{2} \cos \theta, \\
dy &= -\frac{b}{2} \sin \theta d\theta
\]
where \(\theta: \pi \rightarrow 0\) as \(y: -\frac{b}{2} \rightarrow \frac{b}{2}\) |
|
4. 设 \(W_1, W_2\) 为 \(V\) 的子空间, \(u,v \in V\) , 使得 \(u \in W_1\) 且 \(u+v \in W_2\) , 则得 \(W_1 = W_2\)
证明: \(v+W_1 = v+W_2\) \(\Rightarrow v-u+W_2 = v-u+0 \oplus W_2\) 同理 \(0 \in W_1 \subseteq W_2\)
\(\forall u, v \in V,\) \(v-u+W_1 = W_2\) \(\Rightarrow W \subseteq W_2\) 同理 \(W_2 \subseteq W_1 \Rightarrow W_1 = W_2\) |
|
So that if one integrates from wingtip to wingtip
\[
w(y_o) = -\frac{1}{4\pi} \int_{-b/2}^{b/2} \left( \frac{d\Gamma/dy}{y_o - y} \right) dy \tag{5.19}
\]
Relationship between \(\Gamma\) distribution and downwash at \(y_o\) |
|
\[
\frac{1}{F} &= (a^2 + \sin^2 \beta) \tan^2 \psi - 2 (ab - \sin \beta \cos \beta) \tan \psi + (b^2 + \cos^2 \beta), \\
a &= \frac{2}{l} \left( \delta \cos (A + \beta) - \frac{l}{2} \sin \beta \right), \\
b &= \frac{2}{l} \left( \delta \sin (A + \beta) + \frac{l}{2} \cos \beta \right). \tag{6}
\] |
|
事件\(A\)或事件 \(B\) 或者两者同时出现的概率为：\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)。互斥事件：\(P(A \cup B) = P(A) + P(B)\)。
若事件 \(A1, A2,..., Am\) 是互斥和完备的，则 \(\sum P(A_i) = 1\)。事件A和事件B二者同时出现的概率为 \(P(A \cap B)\)。当且仅当 \(P(A \cap B) = P(A)P(B)\) 时，事件A和B是独立的。事件A和B互斥时 \(P(A \cap B) = 0\)。 |
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\[
&* \text{注意这里是电子的自旋简并2.} \\
N &= \Sigma a_{+} = \int_{0}^{\infty} D(\epsilon) e^{\frac{-\epsilon}{kT}} d\epsilon \\
f &= \frac{1}{e^{\frac{\epsilon}{kT}} + 1} = \frac{?}{?} \\
N &= \int_{0}^{k} \frac{4\pi ms}{h^{2}} d\epsilon = \frac{2\pi ms}{h^{2}} \mu_{F}
\] |
|
Combining Eqs. (5.47) and (5.23) gives
\[
\text{const.} = \frac{2\Gamma(y_o)}{V_\infty c(y_o)} \implies c(y_o) = \text{const.}_2 \Gamma(y_o) \tag{5.48}
\] |
|
Differentiating these equations with respect to \(dx\) and \(dy\)
\[
\frac{\partial^2 R}{\partial x \partial y} + \frac{\partial^3 T}{\partial y^3} - \frac{\partial^2 T}{\partial x^2} - \frac{\partial^2 T}{\partial y^2} &= 0, \quad i. \ e. \\
\frac{\partial^2 T}{\partial x^2} - \frac{\partial^2 R}{\partial y^2} - \frac{\partial R}{\partial x^2} + \frac{\partial R}{\partial y^2} &= 0, \quad i. \ e.
\]
or again
\[
\frac{\partial^2 T}{\partial x \partial y} - \frac{\partial R}{\partial y^2} - \frac{\partial R}{\partial x^2} + \frac{\partial^2 R}{\partial y^2} = 0
\]
Hence, it appears that the real part as well as the imaginary part of any analytical complex |
|
\[
\forall g, h \in L^1(\nu), & \quad \frac{\int_0^{t/\epsilon} g \circ \varphi_s ds}{\int_0^{t/\epsilon} h \circ \varphi_s ds} \overset{\nu-a.e.}{\underset{\epsilon \to 0}{\longrightarrow}} \frac{\int_M g \, d\nu}{\int_M h \, d\nu}, \tag{9}
\]
provided \(\int_M h \, d\nu \neq 0\), and
\[
\forall T > 0, \forall h \in L^1(\nu), & \quad \left( \epsilon \int_0^{t/\epsilon} h \circ \varphi_s ds \right)_{t \in [0,T]} \overset{L_{\mu}||\cdot||_\infty}{\underset{\epsilon \to 0}{\longrightarrow}} \epsilon \to 0 \int_M h \, d\nu \, (\tilde{L}_t(0))_{t \in [0,T]},\tag{10}
\] |
|
So that
\[
\alpha_i = \frac{C_L}{\pi AR} \tag{5.44}
\]
We can then get induced drag from Eq. (5.27)
\[
C_{D_i} = \frac{2}{V_\infty S} \int_{-b/2}^{b/2} \Gamma(y) \alpha_i(y) dy \tag{5.27}
\] |
|
\[
S &= Nk \left( \ln Z - \frac{\partial}{\partial \beta} \ln Z \right) \\
&= Nk \left( \ln Z - \frac{E_0}{kT} + 1 \right)
\] |
|
\[
\text{res} \, f(i) = \frac{1}{2\pi i} \oint\limits_{L_1} \frac{\frac{1}{z + i} }{z - i}\, dz = \left(\frac{1}{z +i}\right) _i = \frac{1}{2i}
\] |
|
\begin{align*}
F \chi &= 0 \\
(F')^2 &\leq \left( \frac{1-a}{2} \right)^2 \chi^\perp (V - E^2) \chi^\perp \\
\sup_{r > 0} |F_j - F_k| &\leq (\beta/2) |j-k|, \quad \forall j, k \in \mathbb{Z}.
\end{align*} |
|
\({E}(x) =\sum\limits_{n=1}^{N} \frac{(n-1)!}{(n-1)!(N-n)!} \, p^n(1-p)^{N-n}\) |
|
\[
\vec{r}(t + \Delta t) &= \vec{r}(t) + \Delta t \, \vec{v}(t) \\
\vec{r}_{i+1} &= \vec{r}_i + \Delta t \, \vec{v}_i
\] |
|
\[
u(x, t) = & \int_0^L dx_0 \int_0^t dt_0 G(x, t; x_0, t_0) f(x_0, t_0) + \int_0^L dx_0 (u G) \big|_{t_0=0} \\
& + a^2 \int_0^t dt_0 \left( G \frac{\partial u}{\partial x_0} - u \frac{\partial G}{\partial x_0} \right) \bigg|_{x_0=L} \\
& - a^2 \int_0^t dt_0 \left( G \frac{\partial u}{\partial x_0} - u \frac{\partial G}{\partial x_0} \right) \bigg|_{x_0=0}
\]
其中Green函数满足
\[
\begin{cases}
G_t - a^2 G_{xx} = \delta(t-t_0)\delta(x-x_0) & (0 < x < L, \, t > 0) \\
\left(\alpha_1 G_x + \beta_1 G\right) \big|_{x=0} = 0, & \left(\alpha_2 G_x + \beta_2 G\right) \big|_{x=L} = 0 \\
G \big|_{t=0} = 0
\end{cases}
\] |
|
\textbf{(7.43) Theorem} If \(\phi\) is convex in \((a, b)\), then it satisfies a Lipschitz condition on every closed subinterval of \((a, b)\). In particular, if \(a < x_1 < x_2 < b\), we have
\[
\phi(x_2) - \phi(x_1) = \int_{x_1}^{x_2} \phi' .
\]
\textit{Proof}. Let \([x_1, x_2]\) be a closed subinterval of \((a, b)\), and let \(x_1 \leq y < x \leq x_2\). Then as before
\[
D^+ \phi(y) \leq \frac{\phi(x) - \phi(y)}{x - y} \leq D^- \phi(x),
\] |
|
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 \(U\) such that the matrix representation of \(T\) with respect to \(B\) is a \(n \times n\) shift matrix,
\[
S =
\begin{pmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & 1 & 0
\end{pmatrix}
\]
(see (2.5.43), or Theorem 2.21).
So \(p_T(t) = p_S(t) = \det(tI - S) = t^n\). |
|
再利用条件 \((1.5)\) 得到
\[
\frac{\partial \Phi}{\partial y} &= \int_{x_0}^x \frac{\partial Q(x, y)}{\partial x} \, dx + \psi'(y) \\
&= Q(x, y) - Q(x_0, y) + \psi'(y).
\]
由此可见, 为了使 \((1.8)\) 的第二式成立, 即
\[
\frac{\partial \Phi}{\partial y} = Q(x, y),
\]
只要令
\[
\psi'(y) = Q(x_0, y),
\] |
|
Owing to the established result:
\[
P \left( \sum_{i} a_{i} x_{i} \geq t \right) &\leq e^{-\frac{\lambda t}{2}} \prod_{i=1}^{m} \exp \left( \frac{\lambda^{2} \sigma_{i}^{2}}{2} \right)
\] |
|
\[
\int_{0}^{a} \left| \psi_n(x) \right|^2 dx = 1
\]
可以求出 \(|A_n|^2 = 2/a\)，取 \(A_n\) 为实数，
\[
A_n = \sqrt{2/a}
\]
则得到归一化的实波函数为
\[
\psi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi }{a}x\right)
\] |
|
9. 证明乘法公式
\[
E_n(m x) &= m^n \sum_{r=0}^{m-1} (-1)^r E_n \left( x + \frac{r}{m} \right) \quad (m \text{ 奇}),\\
E_n(m x) &= \frac{2 m^n}{n + 1} \sum_{r=0}^{m-1} (-1)^{r+1} \varphi_{n+1} \left( x + \frac{r}{m} \right) \quad (m \text{ 偶}).
\] |
|
\textbf{2.31 例} 令\(X\)和\(Y\)具有如下分布：
\begin{tabular}{|c|c|c|c|}
\hline
& \(Y = 0\) & \(Y = 1\) & \\
\hline
\(X = 0\) & 1/4 & 1/4 & 1/2 \\
\hline
\(X = 1\) & 1/4 & 1/4 & 1/2 \\
\hline
& 1/2 & 1/2 & 1 \\
\hline
\end{tabular}
则\(f_X(0) = f_X(1) = 1/2\)，\(f_Y(0) = f_Y(1) = 1/2\)，因\(f_X(0) f_Y(0) = f(0,0)\)，\(f_X(0) f_Y(1) = f(0,1)\)，\(f_X(1) f_Y(0) = f(1,0)\)，\(f_X(1) f_Y(1) = f(1,1)\)，所以\(X\)和\(Y\)是独立的，假设\(X\)和\(Y\)具有如下联合分布函数： |
|
\[
\psi_{n+}(x) =
\begin{cases}
x e^{-x/n} F\left(1 - n, 2, \frac{2x}{n}\right), & x > 0 \\
-x e^{x/n} F\left(1 - n, 2, -\frac{2x}{n}\right), & x < 0
\end{cases}
\] |
|
\textbf{Example 20.4} Consider the scattering of particles of mass \(m\) and energy \(E\) by the potential
\[
V(r) = \begin{cases}
0 & (0 \leq r < a) \\
\frac{U_0}{r} & (a < r < 2a) \\
0 & (r > 2a)
\end{cases}
\tag{20.109}
\]
Calculate the scattering amplitude in the Born approximation. Are there special values of the energy for which the Born approximate scattering amplitude vanishes? We have: |
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