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\begin{enumerate}
\item If \(\phi(x) = x\), \(\int_a^b f \, d\phi\) is clearly just the Riemann integral \(\int_a^b f \, dx\). In this case, Theorem 5.54 in Chapter 5 gives a necessary and sufficient condition for the existence of the integral.
\item If \(f\) is continuous on \([a, b]\) and \(\phi\) is continuously dif... | |
\[
u(x, y, t) &= \frac{1}{2\pi a} [ \frac{\partial}{\partial t} \int_0^{\omega} \int_0^{2\pi} \frac{\varphi(x + r \cos \theta, y + r \sin \theta)}{\sqrt{(at)^2 - r^2}} r dr d\theta \\
&\quad + \int_0^{\omega} \int_0^{2\pi} \frac{\psi(x + r \cos \theta, y + r \sin \theta)}{\sqrt{(at)^2 - r^2}} r dr d\theta ]
\] | |
\[
S &= Nk \left( \ln Z - \beta \frac{\partial}{\partial \beta} \ln Z \right) \\
&= Nk \left\{ \ln \left( \frac{1}{h} \sqrt{\frac{2\pi}{m\beta}} \right) - \frac{3}{2} \ln \beta + \frac{3}{2} \right\} \\
&= Nk \left[ \ln Z + \frac{3}{2} \right]
\] | |
\[
Ax^2 + By^2 + Cxy + Dx + Ey + F &= 0 \quad \text{若该方程表示圆} \\
\text{则 }
\begin{cases}
A = B \neq 0 \\
C = 0 \\
(D/2)^2 + (E/2)^2 - F &> 0
\end{cases}
\] | |
的光子处于圆偏振态，等1. 试问: 从光的量子性观点来看, 应怎样理解经典检偏实验的观测结果? 对于平常宏观观测来讲, 这是容易回答的. 因为实际上是入射的大量的光子(光子数\(N \gg 1\)) 处于同一个偏振态, 总能量为 \(N \hbar \omega\). \(\alpha = 45^{\circ}\) 线偏振光经过晶片时, 半数光子被吸收, 而半数光子通过晶片, 并且通过的光子都处于 \(x\) 方向线偏振态. 但如果只有一个光子入射, 情况将如何? 对于图 2.8(a) 的情况, 光子将通过晶片, 能量及偏振态均不改变. 对于图 2.8(b) 的情况, 光子将被吸收, 因而在晶片后就观不到光子. 对于图 2.8(c)... | |
The following theorem shows that, although \(\hat{h}\) cannot hope to do better than \(\hat{h}\) in general, the difference should not be too large as long as the sample size is not too small compared to \(M\).
\textbf{Theorem:} The estimator \(\hat{h}\) satisfies
\[
R(\hat{h}) \leq R(\hat{h}) + \sqrt{\frac{2 \log (2M/... | |
由 \(v = at\) 得：
\[
0 \sim 45: \quad a = \frac{v_1}{t_1} = \frac{3}{6} \text{ m/s}^2 = 0.5 \text{ m/s}^2,
\]
\[
0 \sim 105: \quad v_2 = at_2 = 0.5 \times 10 \text{ m/s} = 5 \text{ m/s}.
\] | |
\[
L'(y_o) &= \rho_\infty V_\infty \Gamma(y_o) \\
L &= \int_{-b/2}^{b/2} L'(y) \, dy
\] | |
由 Poisson 公式，全球面面积元素 \(dS\) 的投影元素为 \(dw = d\xi dy\)。
\[
V(x, y, t) = \frac{1}{2 \pi a} \int_0^c \left[ \int_{x_0}^{x_1} \frac{\varphi(s) e^{-\frac{a}{b}t}}{[a^2 t^2 - (s-x)^2 - (y-y)^2]} ds dy \right] &+ \frac{1}{2 \pi a} \int_0^c \frac{\psi(s) e^{-\frac{a}{b}t}}{[a^2 t^2 - (s-x)^2 - (y-y)^2]} ds dy \\
&+ \frac{1}{2 \p... | |
So \(\alpha\) is an eigenvalue of \(A\) since \(V \neq 0\).
If \(\alpha = 0\) then \(AV = 0\) but \(V \neq 0\) so \(A\) could not be invertible. Thus \(\alpha \neq 0\). | |
留数：设 \(z_0\) 是 \(f(z)\) 的孤立奇点，\(f(z)\) 在 \(z_0\) 去心邻域内的洛朗级数中 \((z-z_0)^{-1}\) 的系数 \(C_1\) 称为 \(f(z)\) 在 \(z_0\) 的留数，记
\[
\operatorname{Res} [f(z), z_0] = C_1 = \frac{1}{2\pi i} \oint_C f(z) dz
\]
留数定理：\(\oint f(z)_c dz = 2\pi i \sum \operatorname{Res}[f(z), z_k]\)
(1) 可去奇点，\(C_1 = 0\)
(2) 本性奇点，展开洛朗级数求解
(3) \textit{i.} ... | |
\[
\langle A | \mathbf{1}_{\mathbf{k}}^{(\lambda)} \hat{H}_{\text{int}}(t) | B \rangle | 0 \rangle &= \frac{i e \hbar^{1/2}}{m_e (2\pi)^{3/2} \sqrt{2\omega}} \langle A | (\mathbf{k} \cdot \mathbf{r}) (\mathbf{e}_{\mathbf{k}}^{(\lambda)} \cdot \mathbf{p}) | B \rangle e^{i\omega t}. \tag{23.29}
\]
The spin part does not ... | |
For an \(m \times n\) matrix, denoted by \(A_{m \times n}\), we consider \(A_{m \times n} x = b\) for \(m \gg n\), i.e., a system with more equations than unknowns (often overdetermined).
With \(A_{m \times n} = Q_{m \times m} R_{m \times n}\), where \(R_{m \times n} = \begin{bmatrix} R_{n \times n} \\ 0 \end{bmatrix}\... | |
Note: Let \(y= \frac{x-\mu}{\sigma} = \frac{x-\mu}{\sigma}, \quad \dot{y}= \frac{-1}{\sigma} \Rightarrow x = \sigma y + \mu, \quad dx = \sigma dy\)
\[
E[X^n] &= e^{\mu n + \frac{1}{2} \sigma^2 n^2} \frac{1}{\sqrt{2 \pi} \sigma} \int_{-\infty}^{\infty} e^{-\frac{1}{2} (y - \sigma n)^2} dy \\
&= e^{\mu n + \frac{1}{2} \... | |
对该常微分方程初值问题进行求解： \(\quad P(x) = -1, \quad q(x) = 0.\)
\[
u &= e^{-fdt} \left( d + \int 0 e^{fdt} \right) = d \cdot e^{t}
\]
由\(u(0) = arctan c_2, \quad 有：d e^0 = arctan c \quad \Longrightarrow \quad d = \arctan c\)
故 \(u = \arctan c \cdot e^{t}\) | |
Let \(r_t(\theta)\) denote the probability ratio \(r_t(\theta) = \frac{\pi_\theta \left( a_t \mid s_t \right)}{\pi_{\theta \text{old}} \left( a_t \mid s_t \right)},\) so \(r(\theta_{\text{old}}) = 1\). TRPO maximizes a "surrogate" objective
\[
L^{CPI}(\theta) = \hat{\mathbb{E}}_t \left[ \frac{\pi_\theta(a_1 \mid s_1)}{... | |
\[
\mathbb{E}[|Y(\infty)|] = \sum_{u \in V} P[Y_u(\infty) = 1] \leq \sum_{u \in V} \sum_{v \in V} X_v(0) \cdot \sum_{t \geq 0} \beta^t A^t_{v,u}
\]
Rewrite as
\[
\mathbb{E}[|Y(\infty)|] \leq \left\langle e_1, \sum_{t \geq 0} (\beta A)^t X(0) \right\rangle
\]
\begin{itemize}
\item Same using vector/matrix notation
... | |
\[
\vec{F} = (P(x,y,z), Q(x,y,z), R(x,y,z))
\]
散度
\[
\text{div} \ \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}
\]
\(\nabla \cdot \vec{F}\) | |
\textbf{法二：拟合法}
\[
A_n = (A_n -\frac{ A_1 + \cdots + A_{n}}{n} )+ \frac{A_1 + \cdots + A_n}{n}
\]
即证
\[
\lim_{n \to \infty} (A_n -\frac{ A_1 + \cdots + A_n}{n}) = 0
\]
令 \(a _1 = A_1, \ a_2 = A_2 - A_1 \ldots, \ a_n = A_n - A_{n-1}\)。由 \(\lim_{n \to \infty} n(A_n - A_{n-1}) = 0\) 则 \(\lim_{n \to \infty} na_n = 0\),
\[... | |
\begin{align*}
x_1' &= x_1 \cos \varphi + x_4 \sin \varphi \\
x_2' &= x_2 \\
x_3' &= x_3 \\
x_4' &= - x_1 \sin \varphi + x_4 \cos \varphi
\end{align*} | |
一列简谐横波沿 \(x\) 轴正方向传播，到达坐标原点的波形如图 7-3 所示，当波传到 \(N\) 点时，处于 \(O\) 点处的质点通过的路程和该时刻的位移是
A. \(40.5 \text{cm}，1 \text{cm}.\)
B. \(40.5 \text{cm}，-1 \text{cm}.\)
C. \(81 \text{cm}，1 \text{cm}.\)
D. \(81 \text{cm}，-1 \text{cm}.\) | |
\[
\mathcal{F}_L(x_t) &= \int \mathcal{N}(\epsilon | 0, I) \min \left\{ 0, \log \pi \left( x_t + \frac{1}{2} LL^{\top} \nabla \log \pi(x_t) + L \epsilon \right) - \log \pi(x_t) \right. \\
&\quad \left. - \frac{1}{2} \left( \| \frac{1}{2} L^{\top} [\nabla \log \pi(x_t) + \nabla \log \pi(y)] + \epsilon \|^2 - \|\epsilon ... | |
\textbf{Lecture \#1: Infinite Series, Series of Functions, Binomial Theorem}
I. \textit{Infinite Series}
A. \textit{Fundamentals:}
1. Infinite series are an extremely powerful way to represent functions, enabling solution by relatively straightforward means.
a. Determine whether a series is convergent is often critical... | |
当 \(m \to \infty\) 时，根据条件 (1)，在 \(|z| < R\) 中 (R 为任意正数) 有
\[
\left| \frac{z^{p+1}}{2\pi i} \oint_{C_m} \frac{f(\zeta) d\zeta}{\zeta^{p+1} (\zeta - z)} \right| \leq \frac{R^{p+1} M_m}{R_m (R_m - R)} \to 0,
\]
因此有 (2) 式，其中的级数在 \(|z| < R\) (R 任意) 中是一致收敛的 (设 \(z \neq a_n, n = 1, 2, \cdots\)) 。 | |
\[
h(t) = \text{sec}\left(\frac{t-T/2}{T_s}\right)
\]
\(h(t)\) is a rectangular function delayed by \(T/2\) where \(T_s = \text{sampling period}\). | |
\textbf{Examples.}
\begin{enumerate}
\item \(H = \begin{pmatrix}
2 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 2
\end{pmatrix}\) is positive definite.
\[
2 &> 0, \quad \det \begin{pmatrix}
2 & 1 \\
1 & 3
\end{pmatrix} = 5 > 0, \quad \det H = 3 > 0.
\]
\item \(H = \begin{pmatrix}
... | |
Let us now illustrate the symbolic methods of operator algebra by deriving (23) from (20) without further reference to operands. Multiplication of (20) by \(x\) from the left yields
\[
x \frac{d}{dx} x - x^2 \frac{d}{dx} = x,
\]
while multiplication of (20) by \(x\) from the right gives
\[
\frac{d}{dx} x^2 - x ... | |
(11), (12) 这两个方程现在代替了 (3) (\(a_{11} = \lambda_1, a_{12} = 0\)), 与 (8) 式相应的方程是
\[
\begin{vmatrix}
\lambda_1 - \lambda & a_{21} \\
0 & a_{22} - \lambda
\end{vmatrix}
= 0. \tag{13}
\]
已设这方程有重根, 故 \(a_{22} = \lambda_1\), 而
\[
w_2' = a_{21}w_1 + \lambda_1 w_2. \tag{14}
\]
于是, 由 (11) 和 (14) 得
\[
\left( \frac{w_2}{w_1} \right... | |
考虑在无穷远处不发生形变的无限介质，我们把第一个积分的积分曲面推向无穷远处，在那里 \(\sigma_a = 0\)，所以积分值为零。利用张量 \(\sigma_a\) 的对称性，第二个积分可以重新写为：
\[
\int \mathcal{B} R dV &= -\frac{1}{2} \int \sigma_a \left( \frac{\partial \delta u_i}{\partial x_k} + \frac{\partial \delta u_k}{\partial x_i} \right) dV = \\
&= -\frac{1}{2} \int \sigma_a \delta \left( \frac{\part... | |
15. 在粗糙的水平地面上放一个倾角为 \(\alpha = 30^\circ\) 的斜面体，一个质量 \(m = 5 \, \text{kg}\) 的木块静止于斜面上。现用平行于斜面、大小为 \(F = 30 \, \text{N}\) 的力推木块，使木块沿斜面匀速上滑 (图 1-14)。在这个过程中，斜面体与地面保持相静止，则地面对斜面体的摩擦力 \(f = \) \_，方向 \_ 。( \(g = 10 \, \text{m/s}^2\) ) | |
\[P = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x^2(t) \, dt\] | |
假设 \(\psi(x)\) 是方程 (3.1.3) 的一个解。如它是实解，则把它归入实解的集合中去，如它是复解，
则按定理 1，\(\psi^*(x)\) 也是方程 (3.1.3) 的一个解，并且与 \(\psi(x)\) 一样，同属于能量本征值 \(E\)。再根据线性方程解的叠加性定理,
\[
\varphi(x) = \psi(x) + \psi^*(x) \\
\chi(x) = \frac{1}{i} [\psi(x) - \psi^*(x)]
\]
也是方程 (3.1.3) 的解，同属于能量 \(E\)，并彼此独立。不难看出, \(\varphi(x)\) 与 \(\chi(x)\) 均为实解，
而 \(\p... | |
只有沿\textbf{z}轴的分量。于是
\[
dE_z = \frac{\lambda_e \, dl (\mathbf{r} - \mathbf{r'}) \cdot \hat{\mathbf{z}}}{4 \pi \varepsilon_0 |\mathbf{r} - \mathbf{r'}|^3} = \frac{\lambda_e \, z \, R \, d\varphi}{4 \pi \varepsilon_0 (R^2 + z^2)^{3/2}}
\]积分求得\textbf{A}点的电场强度：
\[
E = E_z = \frac{\lambda_e \, R \, z}{4 \pi \varepsi... | |
解：将初值问题化为三个初值问题求未定：
\[
\begin{cases}
-y_1'' + y_1 = 0 \\
y_1(0) = a \\
y_1'(0) = 0
\end{cases}
\begin{cases}
-y_2'' + y_2 = 0 \\
y_2(0) = 0 \\
y_2'(0) = b
\end{cases}
\begin{cases}
-y_3'' + y_3 = f(x) \\
y_3(0) = 0 \\
y_3'(0) = 0
\end{cases}
\]
根据已知条件，
\begin{cases}
-y' + y = 0 \\
y(0) = 0 \\
y'(0) = 1
\end{cases}
的解... | |
\[
\psi(x) = \begin{cases}
0 & (x \leq 0) \\
A \sinh(\kappa x) & (0 \leq x \leq a) \\
B e^{-\kappa x} & (x \geq a)
\end{cases}
\] | |
\[
\min_{P_G} \max_{P_D} \mathbb{E}_{(x, y) \sim p_{\text{data}}} [\log D(P_D, x, y) + \\
\log (1 - D(x, F(P_G, x)))],
\] | |
若 \(B \in B_0 \mathcal{W}_S\)，便能写成 \(B = B_0 A(\sigma)\)，此时有
\[
B^{-1} \frac{\partial B}{\partial \sigma^b} = A^{-1} \frac{\partial A(\sigma)}{\partial \sigma^b} = \sum_{a=1}^r v^a_b(\sigma) T_a, \quad b = 1, \ldots, r,
\]
这证明对 \(G\) 中任一方阵 \(B\)，\(B^{-1} \frac{\partial B}{\partial \sigma^b}\) 能够用常数方阵 \(T_a\) 的线性组合来表示，其... | |
\[
\text{res} \, f(a) &= \frac{1}{(3-1)!} \left. \frac{d^{(3-1)}}{dz^{(3-1)}} \left[ (z-a)^3\frac{ z e^{z}}{(z-a)^3} \right] \right|_{z=a} \\
&= \frac{1}{2} \frac{d^2}{dz^2} (z e^{z}) = \frac{1}{2} e^a (a+2)
\] | |
\[
f'(z) &= \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} + i \frac{\partial u}{\partial y} \\
&= \frac{\partial u}{\partial x} - i \frac{\partial u}{\partial y} = \frac{\partial v}{\partial y} - i \frac{\partial u}{\partial x}
\] | |
\begin{align*}
P_{mj} = 1 \otimes
\begin{pmatrix}
p_j & 0 \\
0 & p_{j+1}
\end{pmatrix}
\end{align*} | |
一列简谐横波的波形如图7-5所示，此时刻质点\(P\)的速度大小为\(v\)，经过0.2s它的速度大小、方向第一次与\(v\)相同，再经过1.0s它的速度大小、方向第二次与\(v\)相同，则下列判断中正确的有 （ ）
A. 波沿\(-x\)方向传播，波速为 \(7.5 \text{m/s}\)。
B. 波沿\(+x\)方向传播，波速为 \(5 \text{m/s}\)。
C. 图中与质点\(P\)的位移大小总是相等、方向总是相反的质点是\(M\)。
D. 从开始计时起，质点经过 \(1.0\text{s}\)，它的位移和回复力均增为原来的\(2\)倍。 | |
\textbf{THEOREM:} Let \((X, a)\) be a metric space. Then:
i) \(\emptyset\) and \(X\) are open sets.
ii) Union of any number of open sets is open.
iii) Intersection of finite number of open sets is open. | |
\begin{tabular}{|c|*{5}{c}|*{5}{c}|}
\hline
& \multicolumn{5}{c|}{类型 I} & \multicolumn{5}{c|}{类型 II} \\
\hline
病人 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
基因 1 & 230 & -1350 & -1580 & -400 & -760 & 970 & 110 & -50 & -190 & -200 \\
基因 2 & 470 & -850 & -0.8 & -280 & 120 & 390 & -1730 & -1360 & -1 & -330 \\
\vd... | |
\[
\langle f, g \rangle &= \int_{-a}^{a} dx \, f(x)^* g(x). \\
e_k &:= \frac{1}{\sqrt{2a}} \exp\left(\frac{ik\pi x}{a}\right), \quad k \in \mathbb{Z}.
\] | |
\begin{align*}
\hat{\mathcal{H}} \equiv \bigoplus_{j \in \mathbb{Z}} L^2(\mathbb{R}_+, \mathbb{C}^2)
\end{align*} | |
用落体法验证机械能守恒定律的实验中，运动系统获得的动能往往 \_（填大于、等于、小于）它所减少的势能，其主要原因是 \_ 。为了减少实验误差，对悬挂在纸带下的重锤的质量要求是 \_。 | |
\[
\underline{\mathbf{F}}_{12} = m_1 \underline{\mathbf{a}}_1 &= \frac{d \underline{\mathbf{P}}_1}{dt} \quad and \quad \underline{\mathbf{F}}_{21} = m_2 \underline{\mathbf{a}}_2 = \frac{d \underline{\mathbf{P}}_2}{dt}\ \\
\underline{\mathbf{F}}_{12} + \underline{\mathbf{F}}_{21} &= \frac{d}{dt} (\underline{\mathbf{P}}_... | |
and along the circle \(K\)
\[
\kappa(\xi, \eta) = \frac{1}{\sqrt{\frac{\xi^2 + \eta^2}{2}} + \delta} \left| 2 v (\xi \sin \theta + \eta \cos \beta) + c \right|,
\] | |
\[
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dy e^{-ixy} = \sqrt{2\pi} \, \delta(x) \\
\int_{0}^{\infty} dx \, e^{-iqx} = \frac{1}{iq} + \pi \, \delta(q)
\] | |
As stated above, Eq. (5.51) must be differentiated to be used in the Prandtl Lifting Line Theory. We get
\[
\frac{d\Gamma}{dy} = \frac{d\Gamma}{d\theta} \frac{d\theta}{dy} = 2bV_\infty \sum_{n=1}^{N} nA_n \cos n\theta \frac{d\theta}{dy} \tag{5.52}
\] | |
电场强度是空间坐标的\textbf{矢量函数}，\(E = E_{(x, y, z)}\)，即\textbf{矢量场}。为与其它矢量场，如速度场、引力场等相区别，我们称它为\textbf{电场}。简言之，电场就是带电体周围的一个具有特定性质的空间。
在此空间中，处一点电荷\(q\)，会受到作用力\(F\)，由(1.4.1)知，应有：
\[
F = qE(r)
\] | |
\textbf{Proof:} Let \(V = \{a_1, a_2, a_3, \ldots, a_n\}\) be a totally ordered set. Now, we arrange its elements according to their order. Then
\[
V = \{a_{i_1}, a_{i_2}, a_{i_3}, \ldots, a_{i_n}\}
\]
As \(V\) is finite, totally ordered, therefore if we take any subset of \(V\) it remains with a least element, and he... | |
\[
V = \int_{-\infty}^{\infty} \frac{\Gamma \sin \theta}{4\pi r^2} dl \tag{5.8}
\]
where \(\theta\) is the angle formed by \(r\) and the filament. The geometry gives
\[
r = \frac{h}{\sin \theta}, \quad l = \frac{h}{\tan \theta}, \quad dl = \frac{h}{\sin^2 \theta} d\theta \tag{5.9}
\]
which gives
\[
V = -\frac{\Gamma}{4... | |
\[
\frac{dz}{dx} &= \lambda \cos(\lambda \varphi) z, \\
z(x_0) &= -\sin(\lambda x_0 y_0)
\]
[ 其中 \(\varphi = \varphi(x; x_0, y_0, \lambda)\) 是 (4.17) 的解 ] 可以解得
\[
\frac{\partial \varphi}{\partial x_0} = -\sin(\lambda x_0 y_0) e^{\int_{x_0}^{x} \lambda \cos(\lambda \varphi) dx}
\]
因此
\[
\left. \frac{\par... | |
\[
2. \, & A, B, E, F^{nm} \quad r(A+B) \le r(A)+r(B) \\
& \text{法一：证明：} r(A+B) \le \dim(im(A,B)) \\
& \forall \alpha \in im(B), \beta \gamma \in F^{nm}, s.t. \, d = (A^{t}B)\beta = \beta t A B \\
& \Rightarrow im(A+B) \subset ma A t im B \\
& dim(im(A+B)) \le dim(A at im B) \le dim(A at im B)
\] | |
\[
&\underline{a} = \frac{dv}{dt} = \text{constant} \\
&\underline{r} = 0 \quad \text{at} \quad t = 0 \\
&\int_{v_0}^{v} d\underline{v} = \int_{0}^{t} \underline{a} \, dt
\] | |
\(\therefore \, T'(F)\) is the zero map from \(U \to F\). | |
\textbf{2.1 Azuma-Hoffding inequality}
Given a filtration \(\{ \mathcal{F}_1 \}_1\) of our underlying space \(\mathcal{X},\) recall that \(\{ \Delta_i \}_1\) are called \textit{martingale differences} if, for every \(i,\) it holds that \(\Delta_i \in \mathcal{F}_n\) and \(\mathbb{E} \left[ \Delta_i | \mathcal{F}_i \rig... | |
\(v = a_r x^r + a_{r+1} x^{r+1} + a_{r+2} x^{r+2} + \cdots,\) with \(a_r \neq 0,\)
that is,
\[
v = \sum_i a_i x^i, \quad i = r, r+1, r+2, \cdots,
\] | |
和奖赏的近似，但它在求平均值 “批处理式” 进行的，即在一个完整的采样轨迹完成后再对所有的状态动作对进行更新。实际上这个更新过程能增大且进行。对于状态动作对 \((x, a)\)，不妨假定基于 \(t\) 个采样已估计出值函数 \(Q_t^\pi(x, a) = \frac{1}{t} \sum_{i=1}^t r_i\) 则在得到第 \(t + 1\) 个采样 \(r_{t+1}\) 时，类似式(16.3)，有
\[
Q_{t+1}^\pi(x, a) = Q_t^\pi(x, a) + \frac{1}{t+1} \left( r_{t+1} - Q_t^\pi(x, a) \right) \tag{16.29}
\]
显然，... | |
Find orthogonal \(Q\) and upper triangular \(R\) such that \(A = QR \Rightarrow QR\) decomposition.
We find \(Q\) through a number of simple Householder transforms.
If \(A_2 = [a_1, \ldots, a_n]\) has only two rows, \(a_1 = \begin{bmatrix} a_{11} \\ a_{21} \end{bmatrix}\), we find \(H\) such that
\[
H a_1 = b_1 = \begi... | |
1. \textbf{Convergence}: If \(0 \leq u_n \leq a_n\) for all \(n\), and \(\sum_{n=1}^{\infty} a_n\) is convergent, then \(\sum_{n=1}^{\infty} u_n\) is convergent.
2. \textbf{Divergence}: If \(0 \leq b_n \leq v_n\) for all \(n\), and \(\sum_{n=1}^{\infty} b_n\) is divergent, then \(\sum_{n=1}^{\infty} v_n\) diverges. | |
\frac{d^2 u}{dz^2} + \left\{ \sum_{r=1}^{4} \frac{1 - \alpha_r - \beta_r}{z - a_r} \right\} \frac{du}{dz}
+ \left\{ \sum_{r=1}^{4} \frac{\alpha_r \beta_r}{(z - a_r)^2} + \frac{A z^2 + 2B z + C}{\prod_{r=1}^{4} (z - a_r)} \right\} u = 0, | |
\[
C_l &= \frac{2\Gamma_0}{V_\infty c_r} = 2\pi [\alpha_{\text{eff}} - \alpha_{L=0}] \\
&= \frac{2\Gamma_0}{V_\infty c_r} = 2\pi [\alpha - \alpha_i - \alpha_{L=0}] \\
&= 2\pi \left[ \alpha - \frac{\Gamma_0}{2bV_\infty} - \alpha_{L=0} \right]
\] | |
\[
\vec{F} = (P(x, y, z), Q(x, y, z), R(x, y, z))
\]
梯度 \(\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\)
散度 \(\nabla \cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\)
旋度 \(\nabla \times... | |
Rearranging, we get
\[
0 = (\sin^2 \theta + (\cos \theta - \lambda)^2) x_2 = (1 + \lambda^2 - 2 \cos \theta \lambda) x_2.
\]
Since \(\lambda^2 = 1\) and \(|\cos \theta| < 1\), \((1 + \lambda^2 - 2 \cos \theta \lambda) > 0\) so we must have \(x_2 = 0\), so \((\ast)\) implies that \(x_1 = 0\) as well \(\Longrightarrow \... | |
3. 甲、乙两物体分别在恒力 \(F_1\) 和 \(F_2\) 的作用下运动，其动量 \(P\) 与时间 \(t\) 的关系如图6-1所示。设甲在时间 \(t_1\) 内受到的冲量为 \(I_1\)，乙在时间 \(t_2\) 内受到的冲量为 \(I_2\)，则 \(F_1, F_2, I_1, I_2\) 的大小关系是
A. \(F_1 > F_2, I_1 = I_2\).
B. \(F_1 < F_2, I_1 < I_2\).
C. \(F_1 > F_2, I_1 > I_2\).
D. \(F_1 = F_2, I_1 = I_2\). | |
\begin{enumerate}[(i)]
\item \(\mu(Q_x(h)) > 0\) for \(x \in \mathbb{R}^n, h > 0\);
\item sets of the form
\[
\left\{ x : \sup_{h > 0} \frac{\nu(Q_x(h))}{\mu(Q_x(h))} > \alpha \right\}, \quad \left\{ x : \limsup_{h \to 0} \frac{\nu(Q_x(h))}{\mu(Q_x(h))} > \alpha \right\}, \text{ etc.,}
\]
are measurable.
\end{enumerate... | |
由 D'Alembert 公式，可分别得到上面三个问题的解。
\[
V(x, t) &= \frac{1}{2} [f(x+at) + f(x-at)] \\
W(x, t) &= \frac{1}{2} [g(y+at) + g(y-at)] + \frac{1}{2a} \int_{y-at}^{y+at} \varphi(\xi) d\xi \\
h(x, t) &= \frac{1}{2a} \int_{x-at}^{x+at} \psi(\xi) d\xi \\
\]
由叠加原理，解的表达式为：
\[
u = \frac{1}{2} [f(x+at) + f(x-at)] + \frac{1}{2} [g(y+at) + ... | |
By the Cayley-Hamilton Theorem, we have
\[
0 &= p_A(A) = A^3 - 2A^2 - A + 2I \implies \\
-2I &= A^3 - 2A^2 - A = A \left( A^2 - 2A - I \right) \implies
\] | |
Recall that if \(p^{(t)} \in P_n\),
\[
[T(p^{(t)})]_B = [T]_B [p^{(t)}]_B,
\]
where \([p^{(t)}]_B\) and \([T(p^{(t)})]_B\) denote the coordinate vectors of \(p^{(t)}\) and \(T(p^{(t)})\) with respect to \(B\), respectively. We use this to find \(R(T)\) and \(N(T)\), but this problem can also be done without using matri... | |
\(\text{(ii)} \quad \frac{d}{dx} B_{\nu}^{(n)}(x) = \nu B_{\nu-1}^{(n)}(x),\)
\(\text{(iii)} \quad \int_a^x B_{\nu}^{(n)}(t) dt = \frac{1}{\nu + 1} \left[ B_{\nu + 1}^{(n)}(x) - B_{\nu + 1}^{(n)}(a) \right],\)
\(\text{(iv)} \quad B_{\nu}^{(n)}(x + 1) = B_{\nu}^{(n)}(x) + \nu B_{\nu - 1}^{(n-1)}(x),\)
\(\text{(v)} \quad... | |
一根长为 \(2l\) 的细线 \(CD\)，其 \(D\) 端挂着一个质量为 \(M\) 的小球，另一端 \(C\) 用两根长为 \(l\) 的细线悬挂在天花板上相距 \(2l\) 的 \(A, B\) 两点 (图 6-13)，在跟 \(ABC\) 同一竖直平面内有一颗质量为 \(m\)、水平速度为 \(v_0\) 的子弹从左方很快击中小球，并留在球内与球一起运动，恰能使 \(CD\) 线摆到竖直方向成 \(\alpha = 60^\circ\) 角的地方；求：
(1) 此时 \(AC\) 段中的张力；
(2) 子弹的人射速度应满足什么条件；
(3) 小球刚开始摆动时 \(CD\) 线中的张力。 | |
\[
\frac{1}{4\pi} \iiint \left( E_X^2 + E_Y^2 + E_Z^2 \right) \, dx \, dy \, dz &< \infty
\]
and \(\text{div} \, \mathbf{E} = 0\). Let
\[
V(\mathbf{E}, \mathbf{E}) = \frac{1}{4\pi} \iiint \left( E_X^2 + E_Y^2 + E_Z^2 \right) \, dx \, dy \, dz
\]
Let
\[
T(\mathbf{H}, \mathbf{H}) = \frac{1}{4\pi} \iiint \left( H_X^2 + H_... | |
exercise in higher dimensional integration to compute that
\[
\beta([0,x]) = \sqrt{n} \sqrt{M_1 \cdots M_{3n}} \, C_n x^{3n/2}
\]
where \(C_n\) depends only on \(n\). Let \(D_n = C_n \sqrt{M_1 \cdots M_{3n}}\). Then
\[
P(B) = 3n/2 \, D_n \sqrt{n} \int_0^\infty e^{-x/B} x^{3n/2 - 1} \, dx
\] | |
\[
H_{i_1, \ldots, i_k; j_1, \ldots, j_k} &= \begin{pmatrix}
h_{i_1 i_1} & \cdots & h_{i_1 i_k} \\
\vdots & \ddots & \vdots \\
h_{i_k i_1} & \cdots & h_{i_k i_k}
\end{pmatrix}, \quad
y = \begin{pmatrix}
y_{i_1} \\
\vdots \\
y_{i_k}
\end{pmatrix}, \quad i_1 < \cdots < i_k.
\]
\[
y^* H_{i_1, \ldot... | |
\textit{(10.49) Theorem} Let \(\mu\) satisfy the stated conditions, and let \(f\) be a Borel measurable function which is integrable (\(d\mu\)) over every bounded Borel set in \(\mathbb{R}^n\). Then
\[
\lim_{h \to 0} \frac{1}{\mu(Q_x(h))} \int_{Q_x(h)} f \, d\mu = f(x) \quad \text{a.e.} (\mu).
\]
\textit{Proof}. Assume... | |
判断函数的奇偶性（定义域关于原点对称）
当 \(f(-x) = f(x)\) 时 为偶函数
当 \(f(-x) = -f(x)\) 时 为奇函数
注：\(f(0) = 0\) 是奇函数也是偶函数 | |
其中 \(n_i\) 和 \(p_j\) 分别是零点 \(a_i\) 和极点 \(b_j\) 的阶，积分是沿 \(C\) 的正向一周(逆时针)。
\textbf{证} 按科希(Cauchy)定理
\[
\frac{1}{2\pi i} \oint_C \frac{\varphi(z) \psi'(z)}{\psi(z)} dz = \frac{1}{2\pi i} \left[ \sum_{(a_i)} \int_{a_i} \frac{\varphi(z) \psi'(z)}{\psi(z)} dz + \sum_{(b_j)} \int_{b_j} \frac{\varphi(z) \psi'(z)}{\psi(z)} dz \... | |
\[
\forall G, H \in L^1 (\mu), \quad \frac{\sum_{k=0}^{n-1} \int_M G \circ T^k \, d\mu}{\sum_{k=0}^{n-1} \int_M H \circ T^k \, d\mu} \overset{\mu-\text{a.e.}}{\underset{ n \to \infty}{\longrightarrow}}\frac{\int_M G \, d\mu}{\int_M H \, d\mu}, \tag{7}
\]
provided \(\int_M H \, d\mu \neq 0\). It follows from (a dir... | |
\[
\Rightarrow \gamma = -\frac{v^2}{(c2-\nu)^2} \frac{1}{\frac{1}{\sqrt{1-{v^2}/{c^2}}}} =\frac{-v}{c\sqrt{(1-\frac{v^2}{c^2})}} = -\frac{v}{c\sqrt{1-v^2/c^2}}
\]
\[
\Rightarrow \alpha^2 = -\left(\frac{-v^2}{c^2\nu^2}\right) \frac{c^2}{v^2} = \frac{1c^2}{c^2(1-v^2/c^2)} = \frac{1}{1-v^2/c^2}
\]
\[
\Rightarrow \alpha = ... | |
In terms of this notion of sum for disjoint questions we can write down certain important properties of the functions \(E \rightarrow Q_E^A\) from \(\Omega\) to \(\mathcal{Q}\) and \(Q \rightarrow m_{\alpha}(Q)\) from \(\mathcal{Q}\) to the reals. For each \(A\), \(Q^A\) has the properties:
\begin{itemize}
\item[(a... | |
21. 如图3-18所示，跨过定滑轮的细绳一端固定在木箱上，另一端被站在箱内的人拉住。已知箱子质量\(M = 40 \, \text{kg}\)，人的质量\(m = 60 \, \text{kg}\)，当人以\(F = 550 \, \text{N}\)的力拉绳时，人对箱子的压力多大？取\(g = 10 \, \text{m/s}^2\)。 | |
证明 由引理 2 可知, \(\mathcal{S}\) 有一个基 (1.6), 它的线性组合生成整个线性空间 \(\mathcal{S}\). 这就是说, (1.7) 式表示齐次线性微分方程组 (1.2) 的通解.
通常称齐次线性微分方程组 (1.2) 的 \(n\) 个线性无关的解为一个**基本解组**. 因此, 求 (1.2) 的通解只须求它的一个基本解组.
假设已知
\[
y_1(x), \cdots, y_n(x) \tag{1.8}
\] | |
\(X\)可看成是抛一枚硬币直到出现一次正面为止所需要抛的次数。泊松分布 如果
\[
f(x) = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x \geq 0,
\]
则\(X\)服从参数为\(\lambda\)的泊松分布，记为\(X \sim \text{Poisson}(\lambda)\)。易见
\[
\sum_{x=0}^{\infty} f(x) = e^{-\lambda} \sum_{x=0}^{\infty} \frac{\lambda^x}{x!} = e^{-\lambda} e^{\lambda} = 1.
\] | |
于是，我们可取以细棒为轴，长度为\(l\)，半径为\(r\)的圆柱面作为高斯面。由高斯定理可得
\[
2\pi r \cdot l \cdot E = \frac{1}{\varepsilon_0} \cdot l \cdot \eta_e,
\quad
E = \frac{\eta_e}{2\pi r \varepsilon_0},
\]
即
\[
E = \frac{\eta_e}{2\pi \varepsilon_0 r^2} \cdot r.
\]
说明！对于\textbf{有限长}的带电细棒，问题将不再是一维的，而是二维的。必须根据电场叠加原理通过积分去计算。 | |
自然, 这里所有的矢量运算都是在二维坐标系xy 内进行的, 将上式右边的第一个积分变换为环绕板面的封闭周线的积分中,
\[
\int \nabla \cdot (\Delta \xi \nabla \delta \xi) \, dV = \oint \Delta \xi (n \cdot \nabla \delta \xi) \, dV = \oint \Delta \xi \, \frac{\partial \delta \xi}{\partial n} \, dV,
\]
式中 \(\alpha \beta \alpha\) 表示沿边界外法线方向的导数。对第二个积分作用样的变换, 我们得到
\[
\int \nabla \del... | |
\[
\alpha_{\text{eff}} &= \alpha_{\text{eff}}(y_o) \quad \text{because of downwash} \\
\alpha_{L=0} &= \alpha_{L=0}(y_o) \quad \text{because of aerodynamic twist}
\] | |
\[
J = \frac{D(x,y,z)}{D(u,v,w)} =
\begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\
\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\par... | |
\[
V_j^+ &= \left( \frac{j+1}{r} \right)^2 \left[ \left( 1 - \frac{rA}{j+1} \right)^2 + \frac{r^2 B}{j+1} \right], \\
&\geq \frac{|j|^2}{4r^2} \left[ \left( 1 - 2\delta^{1+\alpha} \right)^2 - 4(\alpha + 1)\delta^{1+\alpha} \right], \\
&\geq \frac{|j|^{2\alpha\delta}}{4\delta^2} \left[ \left( 1 - 2\delta^{1+\alpha} \rig... | |
\[
\psi'(\epsilon) &= \kappa (Ae^{re} - Be^{-re}) \\
\psi'(-\epsilon) &= \kappa (Ae^{-re} - Be^{re}) \\
\psi'(\epsilon) &- \psi'(-\epsilon) = \kappa A (e^{re} - e^{-re}) - \kappa B (e^{-re} - e^{re})
\] | |
the last step follows from arguing as in part (i).
\(\therefore\) For \(i = 1, \ldots, n\),
\[
1 = a \sum_{j=1}^{n} b_{ij} \implies \sum_{j=1}^{n} b_{ij} = \frac{1}{a}.
\] | |
\[
\alpha \sum_{n=0}^{\infty}P_{x}(\alpha) t^n - \sum_{n=0}^{\infty} P_{n}(x) t^{n+1} &= \sum_{n=0}^{\infty} P_{n}(x) n t^{n-1} \\
&- \alpha, \sum_{n=0}^{\infty} P_{n}(x) n t^n \\
&+ \, \sum_{n=0}^{\infty} P_{n}(x) n t^{n+1}
\] | |
\[
\text{则有} \int_{t_0} v^2 dx dt \leq \iint_{k\tau} {v}^2 dx dt.
\]
再令 \(G(\tau) = \iint_{k\tau} v^2 dx dt,\) 故有 \(\frac{d G(\tau)}{dt} \leq G(\tau) \leq 0 \implies G(\tau) = \int_{t_0}^t v^2 dx dt = 0 \implies v = 0.\)
因而 \(v = u - \bar{u} = 0 \implies u = \bar{u},\) 即解唯一。 | |
电流的形式穿过曲面流走。把这个思想翻译成数学语言可表达为，流出闭合曲面的电流
\[
I = \frac{dq}{dt} = -\frac{d\left( \iiint \rho(\mathbf{x}', t) \, dV \right)}{dt} = -\iiint \frac{\partial \rho(\mathbf{x}', t)}{\partial t} \, dV \tag{1.20}
\]
其中 \(\rho(\mathbf{x}', t)\) 是闭合曲面内的电荷密度。式\((1.20)\)中对函数 \(\rho(\mathbf{x}', t)\) 的两个不同的操作(对变空间的积分和对时间的微分)顺序的交... | |
互斥事件：在一次观测中不可能同时发生的事件。
加法原理：对两互斥事件A和B，它们中任意一个出现的概率为：
\[
P(A + B) = P(A) + P(B)
\]
完备性：完备的互斥事件出现的总概率为1：
\[
P(\sum A_i) = 1
\] | |
真空中静电场的\textbf{高斯定理}：通过任意闭合曲面（或称高斯面）\(S\) 的电通量等于该面内全部电荷的代数和除以 \(\varepsilon _0\)，与面外的电荷无关。
高斯定理的\textbf{数学表述}是：
\[
\Phi _{E} = \iint_{S} \mathbf{E} \cdot d\mathbf{S} = \frac{1}{\varepsilon _0} \sum_{(S内)} q \tag{1.5.4}
\]\textbf{说明}：
（1）静电场有源，电荷是它的源；
（2）\(\mathbf{E}\) 是 \(S\) 面上的电场，E 由全空间电荷的分布决定；
（3）\(d\mathbf{S}\) 的方向... | |
\alpha x - x \alpha = 0,
\[
\alpha \left( -i \hbar \frac{\partial}{\partial x} \right) - \left( -i \hbar \frac{\partial}{\partial x} \right) \alpha = i \hbar 2x. \tag{5}
\]
Equations (5) are just the equations 37, and their general solution is
\[
\alpha = x^2 + c, \tag{6}
\] | |
若 \(\lim_{z \to z_0} f(z)\), \(\lim_{z \to z_0} g(z)\) 存在，则
\[
(1) \lim_{z \to z_0} (f(z) \pm g(z)) &= \lim_{z \to z_0} f(z) \pm \lim_{z \to z_0} g(z) \\
(2) \lim_{z \to z_0} f(z) g(z) &= \lim_{z \to z_0} f(z) \lim_{z \to z_0} g(z) \\
(3) \lim_{z \to z_0} \frac{f(z)}{g(z)} &= \frac{\lim_{z \to z_0} f(z)}{\lim_{z \to z_... | |
\[
C_{D_i} = 2 AR \left\{ \int_{0}^{\pi} \left( \sum_{n=1}^{N} A_n \sin n\theta \right) \left( \sum_{m=1}^{N} m A_m \sin m\theta \right) d\theta \right\} \tag{5.59}
\]
Which is easily evaluated since
\[
\int_{0}^{\pi} \sin m\theta \sin k\theta d\theta &= \begin{cases}
\frac{\pi}{2} & \text{if } m = k \\
0 & \text{if } ... |
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