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\[
\frac{\partial A_3}{\partial x^2} - \frac{\partial A_2}{\partial x^3} &= -\eta \left\{ \int_0^{x^1} \frac{2 \, du}{[u^2 + (x^2)^2 + (x^3)^2]^{3/2}}\right. \\
&- \left. \frac{3[(x^2)^2 + (x^3)^2]}{[u^2 + (x^2)^2 + (x^3)^2]^{5/2}} \, du \right\} \\
&= -\eta \left[ \int_0^{x^1} \frac{\partial}{\partial u} \frac{u \, du}{[u^2 + (x^2)^2 + (x^3)^2]^{3/2}} \right] = \eta \frac{x^1}{r^3} = F_{23}.
\] |
|
此外,在电动力学中,以下的运算也是常常要用到的。
恒等式：
\[
&\nabla \times \nabla \varphi = 0 \quad (\text{梯度的旋度恒为零})\\
&\nabla \cdot (\nabla \times \mathbf{A}) = 0 \quad (\text{旋度的散度恒为零})
\]
矢量变换公式：
\[
&\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = (\mathbf{A} \cdot \mathbf{C}) \mathbf{B} - (\mathbf{A} \cdot \mathbf{B}) \mathbf{C}\\
&\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \mathbf{C} \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\mathbf{C} \times \mathbf{A})
\] |
|
\[
\sin \omega_{\text{min}} = \frac{1}{\sqrt{\frac{4 \delta^2}{r^2} + \frac{4 r^2}{l^2}}}
\] |
|
这里应用了 (3.1.8), 定理得证。
用 \(\sigma_a^{EF}\) 乘 (3.1.12) 得出
\[
\sigma_a^{EF} \Gamma_{BC}^b \Gamma_{Ei}^A + \sigma_a^{EF} \Gamma_{BF}^i = \Gamma_{BC}^i \sigma_a^{AB} \epsilon_E^F = \Gamma_{BC}^i \sigma_a^{AB},
\]
这里应用了 (2.4.36)。由 (2.4.30) 可知
\[
\sigma_{BC}^i = l_{ab} \epsilon_C^{AB} \epsilon_D^{CD},
\]
即 \(\sigma_{BC}^i\) 是一张量, 它的协变微分为
\[
\sigma_{BC,a}^{AB} = \frac{\partial}{\partial x^i} \sigma_{BC}^{AB} - \sigma_{BC}^{AB} \Gamma_{Ei}^A + \sigma_a^{EF} \Gamma_{BF}^i + \sigma_a^{BC} \Gamma_{Fi}^B,
\] |
|
\begin{align*}
\vec{A} &= A_x \hat{\imath} + A_y \hat{\jmath} + A_z \hat{k} \\
\vec{B} &= B_x \hat{\imath} + B_y \hat{\jmath} + B_z \hat{k} \\
\vec{A} \cdot \vec{B} &= A_x B_x + A_y B_y + A_z B_z
\end{align*} |
|
\[
\lim_{x \to 0} \left( \frac{a^x + b^x + c^x}{3} \right)^{\frac{1}{x}}, \quad a > 0, b > 0, c > 0
\]
证明:原式
\[
&=\lim_{x \to 0} \left( 1 + \frac{a^x + b^x + c^x - 3}{3} \right)^{\frac{3}{a^x + b^x + c^x - 3} \cdot \frac{a^x + b^x + c^x - 3}{3x}} \\
&= e^ {\lim_{x\to 0} \frac{a^x + b^x + c^x - 3}{3x}}
\]
而
\[
\lim_{x \to 0} \frac{a^x + b^x + c^x - 3}{3x} = \lim_{x \to 0} \frac{a^x \ln a + b^x \ln b + c^x \ln c}{3} = \frac{1}{3} \ln (abc)
\]
故
\[
\lim_{x \to 0} \left( \frac{a^x + b^x + c^x}{3} \right)^{\frac{1}{x}} = e^{ \ln (abc)^\frac{1}{3}} = \sqrt{abc}
\] |
|
\[
N &= \int_{0}^{\mu_{F}} \frac{4\pi mS}{h^3} \, d\epsilon = \frac{2\pi mS}{h^3} \mu_{F}\\
&\Rightarrow \mu_{F} = \frac{Nh^3}{2\pi mS}
\] |
|
\textbf{定义 1 覆盖}
给定集合 \(A \subset X\)，如果 \(X\) 的若干子集的集合 \(\{X_i\} \subseteq X^*\) 满足 \(A \subseteq \bigcup X_i\)，那么我们说集合 \(\{X_i\}\)覆盖了 \(A\)。称 \(\{X_i\}\)为 \(A\) 的一个覆盖 (cover)。 |
|
当数学模型不能得到精确解时，通常要用数值方法求近似解。
截断误差（方法误差）：近似解（数值解）与精确解之间的误差。
例：函数\(f(x)\)用泰勒（Taylor）多项式
\[
P_n(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \cdots + \frac{f^{(n)}(0)}{n!} x^n
\]
近似代替，则数值方法的截断误差为
\[
R_n(x) = f(x) - P_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} x^{n+1}, \quad \text{在 } 0 \leq \xi \leq x \text{ 之间}.
\]
舍入误差：计算机的字长有限导致数值表示和计算过程产生误差。数值分析主要讨论截断误差和舍入误差。 |
|
欧勒公式(6)中的余项 \(R_n\) 还可以简化。为此，引进周期为1的函数 \(P_\lambda(t)\)：
\[
P_\lambda(t) &= \varphi_\lambda(t)/\lambda!, \quad \text{当 } 0 \leq t < 1, \\
P_\lambda(t + 1) &= P_\lambda(t), \quad \lambda \text{为非负整数。} \tag{8}
\]
于是
\[
R_n = h^{2n+1} \int_0^1 P_{2n}(t) \sum_{j=0}^{m-1} F^{(2n)}[a + h(t + s_j)] dt.
\]
把 \(t+s_j\) 换成 \(t\)，利用 \(P_\lambda(t)\) 的周期性，有
\[
R_n = h^{2n+1} \sum_{j=0}^{m-1} \int_0^1 P_{2n}(t) F^{(2n)}(a + ht) dt.
\] |
|
and is therefore situated in the middle of the lower surface. In general, with increasing \(f/\delta\), the minimum value of \(\sigma\) moves from the trailing edge to the middle of the lower surface; the point where \(\sigma = \frac{1}{2}\) again from the trailing edge to the leading edge, while the maximum value of \(\sigma\) moves simultaneously from the leading edge to the middle of the upper surface. |
|
eigenvalues are all real. Now, since \(i \hbar\) is a constant, it follows from 30 that
\[
x \left( -i \hbar \frac{\partial}{\partial x} \right) - \left( -i \hbar \frac{\partial}{\partial x} \right) x = i \hbar, \tag{2}
\]
so that the specific operators \(x\) and \(-i \hbar \frac{\partial}{\partial x}\) have just the com- |
|
\[
\eta^\perp A^2 \eta^\perp \leq \frac{1}{\mu} \eta^\perp V \eta^\perp \leq \frac{1}{\mu} H_0^2 + \frac{2\varepsilon_0}{\mu}.
\] |
|
在这个区域内的任意连续函数\(f(s,t)\)，有完备性关系
\[
\iint \left| f(s,t) \right|^2 dsdt = \sum_{m=1}^{\infty} \left| \iint \overline{\omega_m (s,t)} f(s,t) dsdt \right|^2, \tag{20}
\]
这里设两正交函数组\(\{\varphi_n(s)\}，\{\psi_m(t)\}\)的权都是1。
证明这定理，只要注意由\(\{\varphi_n\}\)的完备性有
\[
\int_a^b \left| f(s,t) \right|^2 ds = \sum_{n=1}^{\infty} \left| g_n(t) \right|^2, \tag{21}
\]
其中
\[
g_n(t) = \int_a^b \overline{\varphi_n(s)} f(s,t) ds. \tag{22}
\] |
|
With \(A_{m-1} = Q_{m-1} R_{m-1}\), we let \(Q_m = H_1 \begin{bmatrix} 1 \\ Q_{m-1} \end{bmatrix}\) and \(R_m = \begin{bmatrix} r_{11} & \tilde{a}_1^T \\ 0 & R_{m-1} \end{bmatrix}\).
\[
\Rightarrow Q_m R_m = H_1 \begin{bmatrix} r_{11} & \tilde{a}_1^T \\ 0 & Q_{m-1} R_{m-1} \end{bmatrix} = H_1 \begin{bmatrix} r_{11} & \tilde{a}_1^T \\ 0 & A_{m-1} \end{bmatrix} = A_m.
\]
\(\Rightarrow\) QR decomposition by Householder transforms (replacing Gram-Schmidt).
\(Q_m\): a product of at most \(m - 1\) Householder matrices.
Note: For \(H = I - 2w_2 w_2^T\), we have \(\begin{bmatrix} 1 & \\ & H \end{bmatrix} = I - 2w_1 w_1^T \text{ with } w_1 = \begin{bmatrix} 0 \\ w_2 \end{bmatrix}.\) |
|
\[
\begin{cases}
\frac{\partial^2 R}{\partial t^2} - a^2 \frac{\partial^2 R}{\partial x^2} &= 0 \quad (t > \tau) \\
R|_{t=\tau} = 0, \quad \frac{\partial R}{\partial t}|_{t=\tau} &= f(x,\tau) \\
R|_{x=0} = R|_{x=l} &= 0
\end{cases}
\tag{3.3.11}
\]
的解（其中 \(\tau \geq 0\) 是参数），则
\[
u(x,t) &= \int_0^t R(x,t,\tau) \, d\tau \tag{3.3.12}
\]
就是混合问题（I）的解。 |
|
can be constructed by describing on the \(z\) plane, with the aid of the formula \(z = \zeta + \frac{\eta^2}{4\zeta}\), the circle \(K\), determined by the camber and radii difference. This circle passes through the point \(\zeta = -\frac{1}{2}\). |
|
\begin{align*}
(1) \quad & \Rightarrow -\left(1 - \frac{B^2}{v^2}\right) = c^2 - c^2\left(\frac{1 - \beta^2}{\beta v}\right)^2 = 1 \\
& \Rightarrow -c^2 \left(\frac{1 - \beta^2}{v^2}\right) \left(1 + \frac{(1 - \beta^2)}{\beta^2}\right) = 1 \\
& \Rightarrow -c^2 \left(\frac{1 - \beta^2}{v^2}\right) -\left(\frac{\beta^2 + 1 - \beta^2}{\beta^2}\right) = 1 \\
& \Rightarrow -c^2(1 - \beta^2) = v^2 \beta^2 \\
& \quad \Rightarrow -c^2 + c^2 \beta^2 = v^2 \beta^2 \\
& \Rightarrow (c^2 - v^2) \beta^2 = c^2 \quad \Rightarrow \beta^2 = \frac{c^2}{c^2 - v^2} \quad \Rightarrow \beta = \frac{c}{\sqrt{c^2 - v^2}} \\
& \Rightarrow \beta = \frac{c}{c\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - v^2/c^2}} \quad
\end{align*} |
|
\[
\therefore \, A^{10} = A \left( 2A^2 + A - 2I \right)^3
\]
\[
= \begin{bmatrix}
1 & 2 & 0 \\
0 & 2 & 0 \\
-2 & -1 & -1
\end{bmatrix}
\begin{bmatrix}
1 & 1022 & 0 \\
0 & 512 & 0 \\
-2 & -851 & -1
\end{bmatrix}
= \begin{bmatrix}
1 & 2046 & 0 \\
0 & 1024 & 0 \\
0 & -1705 & 1
\end{bmatrix}
\] |
|
\textbf{2.4.2} WTS
\[
\dim(U/V) + \dim(V) = \dim(U).
\]
By the rank equation, we have \(r(\tilde{I}) + n(\tilde{I}) = \dim(U)\), so it is sufficient to show that \(R(\tilde{I}) = U/V\) and \(N(\tilde{I}) = V\). To see this, note that \(\forall [u] \in U/V\),
\[
[u] = \tilde{I}(u) \quad \textit{so} \quad R(\tilde{I}) = U/V.
\]
Also note that \(v \in N(\tilde{I})\)
\[
\Leftrightarrow [v] = \tilde{I}(v) = 0 \Leftrightarrow v \in V, \quad \textit{so} \quad N(\tilde{I}) = V.
\] |
|
\[
&\times \frac{l!}{(l-r)!}(x-1)^{l-r} \frac{l!}{(r+m)!}(x+1)^{r+m} \\
&=(-1)^m \frac{(l+m)!}{(l-m)!}(1-x^2)^{-m} \frac{d^{l-m}}{dx^{l-m}}(x^2-1)^l,
\]
故
\[
P_l^{-m}(x) = (-1)^m \frac{(l-m)!}{(l+m)!} P_l^m(x). \tag{8}
\] |
|
我们先来看一个例子。
\textbf{例1} 求双参数函数族
\[
y = C_1 e^x \cos x + C_2 e^x \sin x \tag{1.14}
\]
所满足的微分方程。
解 把 \((1.14)\) 对 \(x\) 先后求导两次, 得出
\[
y' &= C_1 e^x (\cos x - \sin x) + C_2 e^x (\sin x + \cos x), \tag{1.15} \\
y'' &= C_1 e^x (-2 \sin x) + C_2 e^x (2 \cos x). \tag{1.16}
\]
从 \((1.14)\) 和 \((1.15)\) 两式可知 Jacobi 行列式
\[
\frac{D(y, y')}{D(C_1, C_2)} =
\begin{vmatrix}
e^x \cos x & e^x \sin x \\
e^x (\cos x - \sin x) & e^x (\sin x + \cos x)
\end{vmatrix}
= e^{2x} \neq 0.
\]
这说明 \((1.14)\) 中包含的两个任意常数 \(C_1\) 和 \(C_2\) 是独立的。据此, 可 |
|
\[
N &= \sum a_x = \int_0^\infty g(\epsilon) \frac{1}{e^{\frac{\epsilon}{kT}} + 1} \, d\epsilon \\
&= \int_0^\infty g(\epsilon) f_{FD}(\epsilon) \, d\epsilon
\] |
|
\textbf{3.2. Singular Value Decomposition.}
\textbf{Theorem.} Let \(A\) be an \(m \times n\) matrix with non-zero singular values \(s_1, \ldots, s_r\). Then
\[
A = UDV^*, \quad D = \begin{pmatrix}
s_1 & \cdots & 0 & \cdots & 0 \\
\vdots & \ddots & \vdots & & \vdots \\
0 & \cdots & s_r & \cdots & 0 \\
\vdots & & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & \cdots & 0
\end{pmatrix}_{m \times n},
\]
\(U\) and \(V\) are unitary. |
|
\[
T(\omega) &= T(R^{-1}(v)) = (R^{-1} \circ S \circ R)(R^{-1}(v)) \\
&= R^{-1} \circ S(v) = R^{-1}(\lambda v) = \lambda R^{-1}(v) = \lambda \omega,
\]
so \(\lambda\) is an eigenvalue of \(S\). |
|
\[
&\left| \int_{C_R} f(z) \, dz \right| \leq \int_{C_R} |f(z)z| \left| \frac{dz}{z} \right| \to 0 \\
&\text{当} \, R \to \infty \quad \int_{C_R} f(z) \, dz \to 0 \\
&\text{故} \, \left( \int_{-\infty}^{+\infty} f(x) \, dx + \lim_{R \to \infty} \int_{C_R} f(z) \, dz = 2\pi i \sum_{k=1}^{n} f( b_k) \right. \Big| \, I_m b_k > 0 \Big)
\] |
|
如果 \(z_0\) 是方程 \(w'' + pw' + qw = 0\) 的奇点, 则在 \(z_0\) 的邻域 \(0 < |z - z_0| < R\) 内 ( \(R\) 够小使环状域内无方程的奇点 ), 方程的两个线性无关解为
\[
w_1(z) &= (z - z_0)^{\rho_1} \sum_{n=-\infty}^{\infty} c_n (z - z_0)^n, \tag{15} \\
w_2(z) &= (z - z_0)^{\rho_2} \sum_{n=-\infty}^{\infty} d_n (z - z_0)^n. \tag{16}
\] |
|
根据实验中两球碰撞前后的落点位置(图6-8)，能验证两球作弹性正碰的关系式是
A. \(OP + OM = ON.\)
B. \(OP + OM = O'N.\)
C. \(OP^2 + OM^2 = ON^2.\)
D. \(OP^2 + OM^2 = O'N^2.\) |
|
\[
\frac{\partial F}{\partial x} &= \frac{\partial R}{\partial x} + i \frac{\partial T}{\partial x} \\
\frac{\partial F}{\partial (i y)} &= -i \frac{\partial R}{\partial y} + \frac{\partial T}{\partial y}
\] |
|
We have
\[
\begin{pmatrix}
P^* & 0 \\
0 & 1
\end{pmatrix} \begin{pmatrix}
I_{n-1} & b \\
b^* & h_{nn}
\end{pmatrix} \begin{pmatrix}
P & 0 \\
0 & 1
\end{pmatrix} &= \begin{pmatrix}
P^* P & P^* b \\
b^* P & h_{nn}
\end{pmatrix} = H \quad \text{if} \; b = (P^*)^{-1} \begin{pmatrix}
h_{1n} \\
\vdots \\
h_{n-1,n}
\end{pmatrix}.
\]
Thus,
\[
\det P^* \det \begin{pmatrix}
I_{n-1} & b \\
b^* & h_{nn}
\end{pmatrix} \det P = \det H,
\] |
|
\[
V(f,f) &= k \int_{0}^{\ell} \left( \frac{df}{dx} \right)^2 dx = -k \int_{0}^{\ell} \left( \frac{d^2 f}{dx^2} \right) f \, dx = -\frac{k}{\mu} T \left( \frac{d^2 f}{dx^2}, f \right)
\]
We see that \(I^*\) restricted to \(\mathcal{M}\) is \(-\frac{k}{\mu} \left( \frac{d^2 f}{dx^2} \right)\), so the classical "equations of motion" are
\[
\frac{\partial^2 f}{\partial t^2} = -\frac{k}{\mu} \frac{\partial^2 f}{\partial x^2}
\] |
|
\[
\sum_{a=1}^r v^a_b(\sigma) A^{-1}(\tau) T_a A(\tau) = A^{-1}(\tau) A^{-1}(\sigma) \frac{\partial A(\sigma)}{\partial \sigma^b} A(\tau).
\] |
|
若 \(|a| = 0\)，则 \(b \neq 0\)。
原式：
\[
\int_0^{\frac{\pi}{2}} \ln(b^2 \cos^2 x) \, dx = \int_0^{\frac{\pi}{2}} (2 \ln |b| + 2 \ln \cos x) \, dx
\]
可以改写为：
\[
= \pi \ln |b| + 2 \int_0^{\frac{\pi}{2}} \ln (\cos x )\, dx
\]
而
\[
\int_0^{\frac{\pi}{2}} \ln (\cos x) \, dx = \int_0^{\frac{\pi}{2}} \ln (\sin x) \, dx = -\frac{\pi}{2} \ln 2
\] |
|
and simply means that \(\psi_s\) is a common eigenstate of both \(\hat{B}\) and \(\hat{A}\). A more appropriate symbolization would be \(\psi_{\alpha, \beta}\). Summarizing, we have
\[
\left[ \hat{A}, \hat{B} \right] = 0 \implies
\begin{cases}
\hat{A} \psi_{\alpha, \beta} = \alpha \psi_{\alpha, \beta} \\
\hat{B} \psi_{\alpha, \beta} = \beta \psi_{\alpha, \beta}
\end{cases}
\tag{4.57}
\]
The inverse is also true. Assume that the following hold
\[
\begin{cases}
\hat{A} \psi_{\alpha, s} = \alpha \psi_{\alpha, s} \\
\hat{B} \psi_{\alpha, s} = \beta \psi_{\alpha, s}
\end{cases}
\tag{4.58}
\] |
|
\textbf{Interval:} An interval is the set of real numbers, e.g., \(I = [a, b] = \{ x \mid a \leq x \leq b \}\).
The length of an interval is equal to the difference of the end points of the interval, e.g., if \(I = [a, b]\),
\begin{align*}
L(I) = b - a.
\end{align*}
Thus, intervals \([a, b]\) and \((a, b)^c\) have the same length. |
|
7. 甲、乙两名滑冰运动员的质量分别为 \(M_{\text{甲}} = 80 \text{kg}, M_{\text{乙}} = 40 \text{kg}\)，两人面对面沿水平方向拉着弹簧作圆周运动（图4-5），此时两个相距 \(0.9 \text{m}\)，弹簧秤的示数为 \(9.2 \text{N}\)，则下列判断中正确的是 （ ）
A. 两人的线速度相同，约为 \(40 \text{m/s}\)。
B. 两人的角速度相同，约为 \(6 \text{rad/s}\)。
C. 两人的运动半径相同，都是 \(0.45 \text{m}\)。
D. 两人的运动半径不同，甲为 \(0.3 \text{m}\)，乙为 \(0.6 \text{m}\)。 |
|
全称命题的否定是特称命题
\[
\forall x \in M, P(x) \Rightarrow \exists x \in M, \lnot P(x) \\
\exists x \in M, P(x) \Rightarrow \forall x \in M, \lnot P(x)
\] |
|
\[
N &= \frac{1}{2\pi i} \oint_C \frac{1 - t\varphi(\zeta)}{\zeta - a - t\varphi(\zeta)} d\zeta \\
&= \frac{1}{2\pi i} \oint_C \left[ 1 - t\varphi(\zeta) \right] \sum_{n=0}^{\infty} \frac{[t\varphi(\zeta)]^n}{(\zeta - a)^{n+1}} d\zeta \\
&= \sum_{n=0}^{\infty} \frac{t^n}{n!} \frac{d^n}{da^n} \left[ \varphi(a) \right]^n - \sum_{n=0}^{\infty} \frac{t^{n+1}}{(n+1)!} \frac{d^{n-1}}{da^{n+1}} \left[ \varphi(a) \right]^{n+1} \\
&= 1,
\] |
|
To understand further the meaning of \(Q_1 \leq Q_2\) and \(Q_1 \circ Q_2\), consider the case in which \(Q_1 = Q_{E_1}^A\) and \(Q_2 = Q_{E_2}^A\), where \(A\) is some observable. Then \(E_1 \subset E_2\) implies \(Q_{E_1}^A \leq Q_{E_2}^A\) and \(E_1 \cap E_2 = \varnothing\) implies \(Q_{E_1}^A \circ Q_{E_2}^A\).
We pause at this stage to point out that all our axioms so far are satisfied for the \(p\) of classical mechanics and to see what our various |
|
\(x_1 \in E_1\) with \(|\Omega_{x_1}| > \alpha_1 / 2\). Let
\[
E_2 = E_1 - \Omega_{x_1}, \quad \alpha_2 = \sup \{ |\Omega_x| : x \in E_2 \}.
\]
If \(\alpha_2 \neq 0\), choose \(x_2 \in E_2\) with \(|\Omega_{x_2}| > \alpha_2 / 2\). Proceed in this way, obtaining at the \(k\)th stage
\[
E_k &= E_{k-1} - \Omega_{x_{k-1}} = E - \bigcup_{j=1}^{k-1} \Omega_{x_j}, \\
\alpha_k &= \sup \{ |\Omega_x| : x \in E_k \}, \\
x_k &\in E_k, \quad |\Omega_{x_k}| > \alpha_k / 2.
\] |
|
A question closely related to the computation of the ground state energy is whether the ground state exhibits Bose-Einstein condensation (BEC). If \(\psi_N\) denotes the ground state vector, this means that the largest eigenvalue of the associated reduced one particle density matrix \(\gamma_N^{(1)} = \text{tr}_{2, \ldots, N} |\psi_N \rangle \langle \psi_N |\) remains of size one in the limit \(N \to \infty\):
\[\liminf_{N \to \infty} \| \gamma_N^{(1)} \|_{\text{op}} > 0.\] |
|
\[
\text{triplet} \implies
\begin{cases}
|1, 1\rangle &= |\uparrow\rangle^{(1)} |\uparrow\rangle^{(2)} \\
|1, 0\rangle &= \frac{1}{\sqrt{2}} \left( |\uparrow\rangle^{(1)} |\downarrow\rangle^{(2)} + |\downarrow\rangle^{(1)} |\uparrow\rangle^{(2)} \right) \\
|1, -1\rangle &= |\downarrow\rangle^{(1)} |\downarrow\rangle^{(2)}
\end{cases}
\tag{20.162}
\]
and
\[
\text{singlet} \implies |0, 0\rangle &= \frac{1}{\sqrt{2}} \left( |\uparrow\rangle^{(1)} |\downarrow\rangle^{(2)} - |\downarrow\rangle^{(1)} |\uparrow\rangle^{(2)} \right).
\tag{20.163}
\] |
|
弹性钢球 \(A\) 和 \(B\) 的质量分别为 \(M\) 和 \(m\)，且 \(M \ge m\)，使两钢球分别沿光滑斜面从同样高度 \(h\) 处自由下滑，结果在光滑水平面上的 \(O\) 处发生碰撞，碰后小球 \(B\) 反弹，沿原斜面上升的高度为 \(h'\)，则 （ ）
A. \(h' = h\).
B. \(h' = 2h\).
C. \(h' = 4h\).
D. \(h' = 9h\). |
|
Assume now that \(r(S), r(T) \neq 0\). First suppose \(R(S) \cap R(T) \neq \{ 0 \}\). Let \(\{ u_1, \ldots, u_k \}\) be a basis for \(R(S) \cap R(T)\) and extend it to get bases \(\{ u_1, \ldots, u_k, v_1, \ldots, v_l \}\) and \(\{ u_1, \ldots, u_k, w_1, \ldots, w_m \}\) for \(R(S)\) and \(R(T)\) respectively. It is sufficient to show that \(B := \{ u_1, \ldots, u_k, v_1, \ldots, v_l, w_1, \ldots, w_m \}\) spans \(R(S+T)\), because then we have
\[
r(S+T) \leq k + l + m \leq (k + l) + (k + m) = r(S) + r(T).
\] |
|
\[
K_0(x, x'; t - t_0) = \langle x | e^{-\frac{i}{\hbar} \hat{H}_0 (t - t_0)} | x' \rangle = \langle x | e^{-\frac{i}{\hbar} \hat{H}_0 (t - t_0)} \left( \int_{-\infty}^{+\infty} dp | p \rangle \langle p | \right) | x' \rangle, \tag{24.4}
\]
where we have inserted a complete set of energy eigenstates, which for the free particle coincide with the momentum eigenstates. Next, we have
\[
K_0(x, x'; t - t_0) = \int_{-\infty}^{+\infty} dp \langle x | p \rangle e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t - t_0)} \langle p | x' \rangle = \int_{-\infty}^{+\infty} \frac{dp}{2\pi \hbar} e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t - t_0)} e^{\frac{i}{\hbar} p (x - x')}. \tag{24.5}
\] |
|
Guess the form: let \(P = k \, M_0^a r^b G^c\)
(k is a dimensionless constant)
Dimensions of \(G \to (MLT^{-2}) \cdot L^2 M^{-2}\)
\[
T^1 &= M^a L^b (M^{-1} L^3 T^{-2})^c \\
&= M^{(a-c)} L^{b+3c} T^{-2c}
\] |
|
\(f(z) = u(x, y) + iv(x, y)\)，其中
\[
u(x, y) = v(x, y) =
\begin{cases}
\frac{xy}{x^2 + y^2}, & x^2 + y^2 \neq 0, \\
0, & x^2 + y^2 = 0
\end{cases}
\]
令\(f(z) = u(x, y) + iv(x, y)\)，则在点\(z = 0\)满足\(C-R\)方程：
\[
\frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} = 0 \\
\frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} = 0
\]
但\(u(x, y), v(x, y)\)在点\((0, 0)\)不连续，所以复变函数\(f(z)\)在\(z = 0\)不连续，从而不可导。 |
|
\[
\oint_{|z|=1} \frac{dz}{\varepsilon z^2 + 2z + \varepsilon}
\] |
|
17. 某学生做“验证牛顿第二定律”的实验在平衡摩擦力时，把长木板的一端垫得过高，使得倾角偏大。他所得倒的 \(a-F\) 关系可用下列哪根图线表示？图3-14中 \(a\) 是小车的加速度，\(F\) 是细线作用于小车的拉力。答：\_。 |
|
其中 \(\omega\) 为角频率
\[
\omega = c \lvert \mathbf{K} \rvert.
\]
配分函数是
\[
Q &= \sum_{n_k, \lambda} e^{-E}\\
&= \prod_{k, \lambda} (1 + e^{-\beta \hbar \omega} + e^{-2\beta \hbar \omega} + \cdots)\\
&= \prod_{k, \lambda} \frac{1}{1 - e^{-\beta \hbar \omega}}.
\]将 (1.50) 式代入到 (1.37) 式有
\[
-\frac{F}{kT} = \ln Q = -\sum_{k, \lambda} \ln(1 - e^{-\beta \hbar \omega}).
\] |
|
To get \(8\),
\[
\Phi = V\ \frac{3}{4}\ \pi\ (x=E)^{\frac{3}{2}} = R^3 S
\]
\[
R^2 S^{\frac{2}{3}} = \left(\frac{4}{3}\pi V\right)^{\frac{2}{3}} 2\sqrt{E}
\]
\[
E' = (c_1)^{-1} R^{-2} \left(\frac{4}{3}\pi V\right)^{2/3} S^{2/3}
\]
in which \(E'\) is the energy of the energy-surface containing \(S\) cells. |
|
合力与重力大小相等，方向相反，作出绳OB在两个位置时力的合成图，如图1，由图看出，\(F_{OA}\)逐渐减小，\(F_{OB}\)先减小后增大，
当 \(\theta = 90^{\circ}\)时，\(F_{OB}\)最小，选项B正确。 |
|
\[
\sin \alpha \cos \beta &= \frac{1}{2} \left[ \sin(\alpha + \beta) + \sin(\alpha - \beta) \right] \\
\cos \alpha \sin \beta &= \frac{1}{2} \left[ \sin(\alpha + \beta) - \sin(\alpha - \beta) \right] \\
\cos \alpha \cos \beta &= \frac{1}{2} \left[ \cos(\alpha + \beta) + \cos(\alpha - \beta) \right] \\
\sin \alpha \sin \beta &= -\frac{1}{2} \left[ \cos(\alpha + \beta) - \cos(\alpha - \beta) \right]
\] |
|
\textbf{Examples.} \(A = \begin{pmatrix}
1 & 0 \\
0 & 1 \\
1 & 0
\end{pmatrix}\). We have \(A^* A = \begin{pmatrix}
2 & 0 \\
0 & 1
\end{pmatrix}\) and
\[
s_1 = \sqrt{2}, \quad s_2 = 1, \quad v_1 = \begin{pmatrix}
1 \\
0
\end{pmatrix}, \quad v_2 = \begin{pmatrix}
0 \\
1
\end{pmatrix}.
\]
Then,
\[
A v_1 = \begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix} = \sqrt{2} u_1, \quad A v_2 = \begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix} = u_2, \quad u_3 = \begin{pmatrix}
\frac{1}{\sqrt{2}} \\
0 \\
-\frac{1}{\sqrt{2}}
\end{pmatrix}.
\]
We have
\[
U = \begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
0 & 1 & 0 \\
\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}}
\end{pmatrix}, \quad V = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}, \quad A = \sqrt{2} u_1 v_1^* + u_2 v_2^*.
\] |
|
Similarly, \(T(\alpha u) = \alpha T(u) \in Y\) since \(Y\) is closed under scalar multiplication, so \(du \in T^{-1}(Y)\).
First, note that if \(r(S) = r(T) = 0\), then \(S\) and \(T\) are both identically \(0\), and we have:\(r(S+T) = 0 = r(S) + r(T).\) Similarly, if \(r(S) = 0\) then \(r(S+T) = r(T)\) and if \(r(T) = 0\) then:\(r(S+T) = r(S) = 0.\) |
|
\[
A \xi^{-1} + B \xi &= C \xi^{-1} + D \xi \\
i k (C \xi^{-1} - D \xi) - i k (A \xi^{-1} - B \xi) &= g (C \xi^{-1} + D \xi) \\
F \xi &= C \xi + D \xi^{-1} \\
i k F \xi - i k (C \xi^{-1} - D \xi^{-1}) &= g (C \xi + D \xi^{-1})
\]
\[
\Rightarrow
\begin{cases}
A \xi^{-1} + B \xi &= C \xi^{-1} + D \xi \\
A \xi^{-1} - B \xi &= (1 + i \xi^{-1}) C \xi^{-1} - (1 - i \xi) D \xi \\
F \xi &= C \xi + D \xi^{-1} \\
F \xi &= (1 - i \xi) C \xi + (-1 - i \xi) D \xi^{-1}
\end{cases}
\] |
|
Two integration constants \(c_1\) and \(c_2\) are determined by two initial conditions. |
|
and that corresponding to \(A^{2}B'\) must be both
\[
\left[ A \left( \frac{AB + BA}{2} \right) + \left( \frac{AB + BA}{2} \right) A \right] / 2 = \frac{A^2 B + 2ABA + BA^2}{4}
\]
and
\[
\frac{A^2 B + BA^2}{2}
\] |
|
1. 一向右运动的车厢顶上悬挂两单摆 \(M\) 与 \(N\)，它们只能在如图3-1所示平面内摆动，某一瞬时出现图示情景，由此可知车厢的运动及两单摆相对车厢运动的可能情况是 （ ）
A. 车厢作匀速直线运动，\(M\) 在摆动，\(N\) 静止。
B. 车厢作匀速直线运动，\(M\) 在摆动，\(N\) 也在摆动。
C. 车厢作匀速直线运动，\(M\) 静止，\(N\) 在摆动。
D. 车厢作匀加速直线运动，\(M\) 静止，\(N\) 也静止。 |
|
\[
\mathscr{F} = \text{LinSpan} \left( \prod_{i \in [k]} \phi_i : k \in \mathbb{N} \right)
\] |
|
\[
pL &= \frac{N}{\beta} \frac{\partial}{\partial L} \ln Z \\
&= \frac{NkT}{L} \Rightarrow pL = NkT
\] |
|
\[
\sin \alpha + \sin \beta &= 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \\
\sin \alpha - \sin \beta &= 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}
\] |
|
\[
f(x) &= \sum_{k \in \mathbb{Z}} \alpha_k e_k(x) = \frac{1}{\sqrt{2a}} \sum_{k \in \mathbb{Z}} \alpha_k \exp \left( \frac{ik\pi x}{a} \right), \\
\alpha_k &= \langle e_k, f \rangle = \frac{1}{\sqrt{2a}} \int_{-a}^{a} dx \exp \left( \frac{-ik\pi x}{a} \right) f(x).
\] |
|
\textbf{Example.} Let \(A = \begin{pmatrix}
3 & 2 - i & -3i \\
2 + i & 0 & 1 - i \\
3i & 1 + i & 0
\end{pmatrix}\). Find \(A^{100}\).
Matrix \(A\) is Hermitian. We have
\[
\det(A - \lambda I) = -(\lambda + 1)(\lambda + 2)(\lambda - 6), \quad \lambda_1 = -1, \; \lambda_2 = -2, \; \lambda_3 = 6.
\]
One can find three orthogonal eigenvectors:
\[
v_1 &= \begin{pmatrix}
-1 \\
1 + 2i \\
1
\end{pmatrix}, \quad v_2 = \begin{pmatrix}
1 + 3i \\
-2 - i \\
5
\end{pmatrix}, \quad v_3 = \begin{pmatrix}
1 - 21i \\
6 - 9i \\
13
\end{pmatrix}.
\] |
|
关系：给出集合\(A \times A\)中元素的一个性质\(R\)。若\(a, b \in A\)，\((a, b)\)有性质\(R\)，则称\(A\)与\(B\)有关系\(R\)，记为\(aRb\)。
命题：集合\(A\)中关系\(R\)可由\(A \times A\)中子集\(\{(a, b) | a, b \in A, aRb\}\)来刻画；反之，由\(A \times A\)的一个子集\(R\)，也可确定\(A\)的一个关系\(R: aRb,\)若\((a, b) \in R\)。 |
|
11. 如图2-3所示，\(A, B\)两小车相距\(s = 7 \, \text{m}\)时，\(A\)在水平拉力和摩擦力作用下，正以\(v_A = 4 \, \text{m/s}\)的速度向右匀速运动，小车\(B\)此时以速度\(v_B = 10 \, \text{m/s}\)向右匀减速运动，加速度\(a = -2 \, \text{m/s}^2\)，则小车\(A\)追上\(B\)需经历的时间为 \_。 |
|
\[
f(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty} \left( a_k \cos \left( \frac{k \pi x}{a} \right) + b_k \sin \left( \frac{k \pi x}{a} \right) \right).
\] |
|
零点，把 \(f(x)\) 的零点叫做函数 \(y = f(x)\) 的零点（需要知道一点高级方程 ）。
二分法：对于区间 \([a, b]\) 上连续不断且 \(f(a) \cdot f(b) < 0\) 的函数 \(y = f(x)\)，通过不断地选择 \(f(x)\) 的零点所在的区间与第一个，使区间的两个端点逐步逼近零点，进而得到 \(f(x)\) 的近似零点。 |
|
The following formula holds good for the velocity \(q\) at which the air flows by every point on the Joukowski figure.
\[
q = \frac{\kappa (\xi, \eta)}{\left| \frac{dz}{d\xi} \right|}
\]
From \(z = \xi + \frac{b^2}{4\xi}\) it follows that |
|
\[
\begin{cases}
\frac{dy}{dx} = \frac{\partial \varphi}{\partial x}(x, C_1, \cdots, C_n), \\
\cdots \cdots \cdots \\
\frac{d^n y}{dx^n} = \frac{\partial^n \varphi}{\partial x^n}(x, C_1, \cdots, C_n).
\end{cases}
\] |
|
The geometry can be found by going back to the lift coefficient and Eqs. (5.23), (5.21) and (5.2)
\[
C_l &= \frac{2\Gamma(y_o)}{V_\infty c(y_o)} \tag{5.23} \\
C_l &= 2\pi \left[ \alpha_{\text{eff}} - \alpha_{L=0} \right] \tag{5.21} \\
\alpha_{\text{eff}} &= \alpha - \alpha_i \tag{5.2}
\] |
|
\[
\text{Recall the Frenet–Serret formula:} \quad
(2.27)\begin{pmatrix}
T_s \\
N_s \\
B_s
\end{pmatrix}
&=
\begin{pmatrix}
0 & k & 0 \\
-k & 0 & t \\
0 & -t & 0
\end{pmatrix}
\begin{pmatrix}
T \\
N \\
B
\end{pmatrix}.
\]
In particular,\[
(2.28) \gamma_{ss} &= \kappa = kN, \\
(2.29) \gamma_{sss} &= k_s N - k^2 T + ktB.
\]
Computing the normal derivatives, we obtain\[
\nabla_s \kappa &= (\gamma_{sss})^\perp = k_s N + ktB, \\
(2.30) \nabla_s^2 \kappa &= ((k_s N + ktB)_s)^\perp = (k_{ss} - kt^2)N + (2k_s t + kt_s)B.
\] |
|
\[
C_L &= 0.77 \\
C_{D_i} &= \frac{C_L^2}{\pi AR} = \frac{(0.77)^2}{\pi (5.09)} = 0.037
\] |
|
10. 如图3-9所示，\(A, B\) 两木块靠在一起沿斜面下滑，已知 \(A, B\) 与斜面间的动摩擦因数分别为 \(\mu_A\) 与 \(\mu_B\)，设 \(A, B\) 间相互作用力为 \(N\)，则下列判断中正确的是
A. 当 \(\mu_A = \mu_B\) 时，一定一起匀速下滑。
B. 当 \(\mu_A > \mu_B\) 时，不可能一起匀速下滑。
C. 当 \(\mu_B = \mu_B\) 时，一定有 \(N = 0\)。
D. 两者一起加速下滑时，一定有 \(N \neq 0\)。 |
|
3) For any \(z \in [0,1]\):
\[
\int_{X+y} (z)= \int_{[0,1]}\mathbf{1}_{[0,z]}(z-x) dx &= \int_{\max(0,z-1)}^{\min(1,z)} dz = \min(1,z) - \max(0,z-1) = z - \max(0,z-1) = 1 - |z-1|
\]
4) For all \(z > 0\):
\[
\int e^{-\lambda x} \mathbf{1}_{[0,\infty)}(x) \cdot e^{-\lambda(z-x)} \mathbf{1}_{[0,\infty)}(z-x) dx &= \lambda e^{-\lambda z} \int_0^z dx = \lambda^2 e^{-\lambda z} e^{-\lambda z}
\] |
|
这两个不等式的右方是指数函数数 \(m \exp(2M\rho)\) 的幂级数展开的普通项, 因此, 序列
\[
u_n &= u_0 + (u_1 - u_0) + \cdots + (u_n - u_{n-1}), \\
v_n &= v_0 + (v_1 - v_0) + \cdots + (v_n - v_{n-1}),
\]
在 \(|z| \leq R_1\) 中一致收敛. 于是, 按外氏关于解析函数序列的定理, 这两个序列的极限函数 \(u(z)\) 和 \(v(z)\) 是 \(|z| < R\) 内的解析函数. 又
\[
u(z) &= \lim_{n \to \infty} (z) = \alpha + \lim_{n \to \infty} \left[ \int_0^{z} [a u_{n-1} + b v_{n-1}] d \zeta \right] \\
&= \alpha + \left[ \int_0^{z} [a u + b v] d \zeta \right]
\] |
|
\[
\begin{cases}
X''(x) + \lambda X(x) = 0 \\
X(0) = X(L) = 0
\end{cases}
\]
和关于 \(T(t)\) 的常微分方程
\[
T''(t) + a^2 \lambda T(t) = 0
\]
第二步：解固有值问题，求出固有值 \(\lambda_k\) 和固有函数 \(X_k(x)\) 后，对于每一个 \(\lambda_k\) ，求出常微分方程
\[
T''(t) + a^2 \lambda_k T(t) = 0
\]
的通解 \(T_k(t)\)，得到满足方程及边界条件的一系列特解 \(u_k(x,t) = X_k(x) T_k(t)\)。 |
|
\[
\Gamma(n &+ \mu + 1) \int_{-1}^1 (1 - t^2)^{\mu - \frac{1}{2}} H_{2n} \left( \sqrt{x} t \right) dt \\
&= (-1)^n \sqrt{\pi} (2n) ! \Gamma \left( \mu + \frac{1}{2} \right) L_n^{\mu}(x) \ \left( \text{Re}(\mu) > -\frac{1}{2} \right),
\]
其中 \(L_n^{\mu}(z)\) 是广义拉革尔多项式。
31. 证明, 若 \(\rho = 1/\xi\), 有
\[
\frac{d^n}{d\xi^n} f(\xi) = (-)^n \rho^{n+1} \frac{d^n}{d\rho^n} \left\{ \rho^{n-1} f \left( \frac{1}{\rho} \right) \right\}.
\]
利用这公式, 由拉革尔多项式 \(L_n^{\mu}(z)\) 的生成函数 (6.14 节 (8) 式) 推出 \(L_n^{\mu}(z)\) 的微商表示 (6.14 节 (5))。
32. 证明
\[
F\left( \alpha, Y, \frac{xy}{x-1} \right) &= (1 - x)^a \sum_{n=0}^{\infty} \frac{(\alpha)_n}{(Y)_n} L_n^{\gamma-1}(y) x^n \\
&\quad (1 \mid x \mid < 1, y > 0)
\]
(Erdelyi (1953), Vol. 1, p. 276, (5)). 6.14 节 (8) 式是这个展开式的特殊情形: \(a = \gamma = \mu + 1\). |
|
and this value is
\[
F_{\text{max}} = \frac{\sqrt{\left(\delta \cos \left( \lambda + \gamma \right) - \frac{l}{2} \sin \beta \right)^2 + \left( \frac{l}{2} \sin \beta \right)^2}}{\delta \cos A} \tag{71}
\]
We now make the assumption, corresponding to the already |
|
Another interesting example is the sign function \(f: [-\pi, \pi] \rightarrow \mathbb{R}\) defined by
\[
f(x) := \text{sign}(x) =
\begin{cases}
1 & \text{for } x > 0 \\
0 & \text{for } x = 0 \\
-1 & \text{for } x < 0
\end{cases} \tag{3.43}
\]
The periodically continued version of this function is shown in Fig. 7. Since \(f\) is an anti-symmetric function, the Fourier series only contains sine terms. (Alternatively and equivalently, we can think of \(f\) as a function on the \([0, \pi]\) and work out the sine Fourier series.) For the Fourier coefficients we |
|
\[
Thus \ \exists \, a_1, \ldots, a_k, a_{k+1}, \ldots, a_{k+l}, \, b_1, \ldots, b_k, b_{k+1}, \ldots, b_{k+m} \in \mathbb{F} \ s t
\]
\[
S(u) &= a_1 u_1 + \cdots + a_k u_k + a_{k+1} v_1 + \cdots + a_{k+l} v_l \\
T(u) &= b_1 u_1 + \cdots + b_k u_k + b_{k+1} w_1 + \cdots + b_{k+m} w_m
\] |
|
\[
1 - \frac{1}{\frac{\sigma^2}{\sigma_t^2} + 1} = \frac{\frac{\sigma^2}{\sigma_t^2}}{\frac{\sigma^2}{\sigma_t^2} + 1} = \frac{\sigma^2}{\sigma^2 + \sigma_t^2}
\] |
|
\[
\frac{\pi_\theta \left( a_t \mid s_t \right)}{\pi_{\theta \text{old}} \left( a_t \mid s_t \right)}
\] |
|
\textbf{定义 2 切向量的加法}
给定流形 \(N\)，其上一点 \(p \in N\)。\(p\) 出发的两个道零等价类 \([r_1]\) 和 \([r_2]\) 的和定义为，任取 \(p\) 附近一图 \((U, \varphi)\)，令 \([r_1] + [r_2] = [\varphi^{-1}(\varphi(r_1) + \varphi(r_2))]\)。
\textbf{定理 2}
给定流形 \(N\)，其上一点 \(p \in N\) 处有两个图 \((U, \varphi)\) 和 \((V, \phi)\)。对 \(p\) 出发的两条道首 \(r_1\) 和 \(r_2\)，按以上定义得的和（注意，这是 \(N\) 上的一个道零等价类），在两个图中的计算结果一致。 |
|
\[
\log(\det(A + \Sigma H)) - \log(A) &= \log\left(\det(I + \Sigma A^{-1/2} H A^{-1/2})\right) \\
&= \log(\det(I + \Sigma \hat{H})) \quad \text{where} \quad \hat{H} = A^{-1/2} H A^{-1/2}
\]
\(\hat{H}\) is symmetric positive definite, let \(\lambda_i\) be its eigenvalue. Then
\[
\log(\det(I + \Sigma \hat{H})) &= \sum_{\tau=1}^{m} \log(1 + \Sigma \lambda_i)
&= \Sigma \sum_{i=1}^{m} \lambda_i + o( \Sigma) \\
&= \Sigma \operatorname{Tr}(\hat{H}) + o( \Sigma)
\] |
|
\[
d_j := \frac{d}{dr} - \frac{j}{r} + A(r), \quad d_j^* := -\frac{d}{dr} - \frac{(j+1)}{r} + A(r)
\] |
|
\[
&T(f,f) =\iiint\mu(\ell_{1}^{2}+\ell_{2}^{2}+\ell_{3}^{2}) \mathrm{dx~dy~dz} \\
&\mathrm{V}(\mathbf{f},\mathbf{f}) =c_{1}\iiint_{\mathbf{R}}\left[\frac{\partial\ell_{1}}{\partial\mathbf{x}}+\frac{\partial\ell_{2}}{\partial\mathbf{y}}+\frac{\partial\ell_{3}}{\partial\mathbf{z}}\right]^{2} \mathrm{dx~dy~dz} \\
&+ c_{2}\iiint\left[\left(\frac{\partial\ell_{1}}{\partial z}+\frac{\partial\ell_{3}}{\partial x}\right)^{2}+\left(\frac{\partial\ell_{2}}{\partial z}+\frac{\partial\ell_{3}}{\partial y}\right)^{2}\right. \\
&+\left({\frac{\partial\ell_{1}}{\partial\mathbf{y}}}+{\frac{\partial\ell_{2}}{\partial\mathbf{x}}}\right)^{2}+2\biggl({\frac{\partial\ell_{1}}{\partial\mathbf{x}}}\biggr)^{2} \\
&+ 2\biggl(\frac{\partial\ell_{2}}{\partial \mathrm{y}}\biggr)^{2} + 2\biggl(\frac{\partial\ell_{3}}{\partial \mathrm{z}}\biggr)^{2} \biggr] \mathrm{dx~dy~dz}
\] |
|
\textbf{Ring}
\textit{Def}: A non-empty set \(R\) is called a ring if
i) \(R\) is an abelian group under addition.
ii) \(R\) is a semi-group under multiplication.
iii) Distributive law holds.
\begin{align*}
a \cdot (b + c) &= a \cdot b + a \cdot c\\
(a + b) \cdot c &= a \cdot c + b \cdot c
\end{align*} |
|
This comes from \(4 = 2 \times 2\),
\[
\begin{pmatrix}
4 & 1 \\
1 & 2
\end{pmatrix} &= \begin{pmatrix}
2 & 0 \\
a & b
\end{pmatrix} \begin{pmatrix}
2 & a \\
0 & b
\end{pmatrix} \implies a = \frac{1}{2}, \, b = \frac{\sqrt{7}}{2},
\]
and
\[
\begin{pmatrix}
4 & 1 & -1 \\
1 & 2 & 1 \\
-1 & 1 & 2
\end{pmatrix} &= \begin{pmatrix}
2 & 0 & 0 \\
\frac{1}{2} & \frac{\sqrt{7}}{2} & 0 \\
a & b & c
\end{pmatrix} \begin{pmatrix}
2 & \frac{1}{2} & a \\
0 & \frac{\sqrt{7}}{2} & b \\
0 & 0 & c
\end{pmatrix} \implies a = -\frac{1}{2}, \, b = \frac{5\sqrt{7}}{14}, \, c = \frac{\sqrt{42}}{7}.
\] |
|
1) There is equiprobability for the drawing of cards, denoted \(\{RR, RS, RS, BS\}\), and for each card, each side is equiprobable. We can represent the outcome with the sample space\[
\Omega = \{(RR, R), (BB, B), (RB, R), (RS, R)\}
\]with probabilities \(\frac{1}{3}, \frac{1}{7}, \frac{1}{7}, \frac{1}{7}, \frac{1}{6}\).
2) Let \(A\) be the event "The hidden side is white" and \(B\) the event "The revealed side is red." Then Bayes' rule gives
\[
P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(\{RS, RS\})}{P(\{(RR, R), (RB, R)\})} = \frac{1/6}{1/3 + 1/6} = \frac{1}{3}
\] |
|
For the costing part, see the costing part correction.
To diagonalize an entire matrix using power iteration, under the assumption that the eigenvalues are non-degenerate \(|\lambda_1| > |\lambda_2| > \ldots > |\lambda_k|\), we can simply apply power iteration to the matrix
\[
\Pi^* = \Pi - \sum_{j=1}^{p}\sigma_j\sigma_j^T
\]
to obtain the eigenvalue eigenvector pair \((\lambda_{p+1}, \nu_{p+1})\).
Indeed,
\[
\Pi^* = \Pi - \sum_{j=1}^{p}\sigma_j\sigma_j^T = \sum_{j=p+1}^{k}\sigma_j\sigma_j^T
\]
where we used
\[
\Pi = \sum_{j=1}^{k}\sigma_j\sigma_j^T.
\]
Since \(|\lambda_{p+1}| > |\lambda_{p+2}| > \ldots > |\lambda_k|\), the largest eigenvalue of \(\Pi^*\) is \(\lambda_{p+1}\). |
|
\textit{Example 4.} \textbf{Galileo's Law:} A point-like object in free-fall near the surface of the earth obeys the equation:
\[
\ddot{x} = -g
\]
where \(x\) is the coordinate giving the height of the body above the surface of the earth and \(g\) is a universal constant. In other words, \(F = mg\) is the \textit{gravitational force}.
\textit{Example 5.} \textbf{Newton's Law of gravitation:} suppose \(x\) denotes the spatial position of an object near another point-like object of gravitational mass \(M\) at the origin; then Newton's Law of gravitation says that:
\[
\ddot{x} = -\frac{GMm}{|x|^2}
\]
where \(m\) is the gravitational mass of the object, and \(G\) is a universal constant.
\textit{Example 6.} \textbf{Hooke's Law:} suppose \(x\) is the coordinate describing the position from equilibrium of a point-mass attached to a spring. Then Hooke's Law says that:
\[
\ddot{x} = -k^2x
\]
where \(k\) is some constant associated with the spring. This example is sometimes referred to as \textit{simple harmonic motion}, and we shall return to it frequently. |
|
在这里我们引用了热力学公式（1.67）并不表明统计力学依赖于热力学。对统计力学，本身完全可以自成体系的，我们引入热力学公式是说明可以从正则系综导出这些热力学公式来。
假设由体积 \(V\) 相等的 \(M\) 个相同的系统组成的系综，每个系统的粒子都是同类的。系统虫相同，但处于不同的位置，所以是可以区分的。又设每个系统的粒子数是相当大，图1.6 表示 \(M\) 个系统所组成的系综。 |
|
\[
\frac{\partial \varphi \nu(x)}{\partial \tilde{x}^i} \varphi \tilde{\nu}(x) &= -\varphi \nu(x) \frac{\partial \varphi \tilde{\nu}(x)}{\partial \tilde{x}^i} \\
&= -\varphi \nu(x) \frac{\partial \varphi \nu(x)}{\partial \varphi \tilde{\nu}(x)} \frac{\partial \varphi \tilde{\nu}(x)}{\partial \tilde{x}^i} \\
&= -\sum_{a,b=1}^r v^a_b (\varphi \nu(x)) \frac{\partial \varphi \tilde{\nu}(x)}{\partial \tilde{x}^i} T_b,
\] |
|
\textbf{Theorem.} A Hermitian matrix \(H\) is positive definite if and only if all its leading principal minors are positive, that is,
\[
\det H_{1, \ldots, k; 1, \ldots, k} > 0, \quad k = 1, \ldots, n,
\]
or explicitly,
\[
h_{11} > 0, \quad \det \begin{pmatrix}
h_{11} & h_{12} \\
h_{21} & h_{22}
\end{pmatrix} > 0, \quad \ldots, \quad \det \begin{pmatrix}
h_{11} & \cdots & h_{1n} \\
\vdots & \ddots & \vdots \\
h_{n1} & \cdots & h_{nn}
\end{pmatrix} > 0.
\]
\textit{Proof.} Assume that \(H\) is positive definite. Let \(H_k = H_{1, \ldots, k; 1, \ldots, k} = \begin{pmatrix}
h_{11} & \cdots & h_{1k} \\
\vdots & \ddots & \vdots \\
h_{k1} & \cdots & h_{kk}
\end{pmatrix}\). Let \(y \in \mathbb{C}^k\). |
|
The gravitational force law
\[
\vec{\mathbf{F}} = -G \frac{m_1 m_2}{r^2} \frac{\vec{r}}{r}
\]
can be approximated in the vicinity of Earth's surface:
\(M =\) Earth's mass
\(r \approx\) E. radius \(R_0 = 6370 \, \text{km}\)
\[
g = \frac{GM}{R_0^2} \approx 9.81 \, \frac{\text{m}}{\text{s}^2}
\]
\(F = mg\) along vertical towards Earth center |
|
\textbf{Example.} Matrix \(A = \begin{pmatrix}
1 & 1 \\
0 & -1
\end{pmatrix}\) is simple, but not normal. \(A\) has two distinct eigenvalues \(\lambda_1 = 1\), \(\lambda_2 = -1\). Hence, \(A\) is simple. It is easily verified that \(A^* A \neq A A^*\).
\[
A \begin{pmatrix}
1 \\
0
\end{pmatrix} &= \begin{pmatrix}
1 \\
0
\end{pmatrix}, \quad \text{but} \quad A^* \begin{pmatrix}
1 \\
0
\end{pmatrix} = \begin{pmatrix}
1 & 1 \\
0 & -1
\end{pmatrix} \begin{pmatrix}
1 \\
0
\end{pmatrix} = \begin{pmatrix}
1 \\
0
\end{pmatrix} \quad \text{is not an eigenvector of } A^*.
\]
\[
A^* \begin{pmatrix}
2 \\
1
\end{pmatrix} &= \begin{pmatrix}
2 \\
1
\end{pmatrix}, \quad A \begin{pmatrix}
1 \\
-2
\end{pmatrix} = -\begin{pmatrix}
1 \\
-2
\end{pmatrix} \quad \begin{pmatrix}
1 \\
0
\end{pmatrix} \quad \text{and} \quad \begin{pmatrix}
1 \\
-2
\end{pmatrix} \quad \text{are not orthogonal}.
\] |
|
Going back to the expressions for the absorption and emission rates, we have
\[
\frac{\Gamma_{\text{abs}}}{\Gamma_{\text{em}}} &= \frac{|\langle n_{\mathbf{k}}^{(\lambda)} - 1 | \langle B | \mathbf{e}_{\mathbf{k}}^{(\lambda)} \cdot \mathbf{p} e^{i \mathbf{k} \cdot \mathbf{r}} | A \rangle | n_{\mathbf{k}}^{(\lambda)} \rangle |^2}{|\langle n_{\mathbf{k}}^{(\lambda)} + 1 | \langle A | \mathbf{e}_{\mathbf{k}}^{(\lambda)} \cdot \mathbf{p} e^{-i \mathbf{k} \cdot \mathbf{r}} | B \rangle | n_{\mathbf{k}}^{(\lambda)} \rangle |^2} \\
&= \frac{n_{\mathbf{k}}^{(\lambda)} |\langle B | \mathbf{e}_{\mathbf{k}}^{(\lambda)} \cdot \mathbf{p} e^{i \mathbf{k} \cdot \mathbf{r}} | A \rangle |^2}{(n_{\mathbf{k}}^{(\lambda)} + 1) |\langle A | \mathbf{e}_{\mathbf{k}}^{(\lambda)} \cdot \mathbf{p} e^{-i \mathbf{k} \cdot \mathbf{r}} | B \rangle |^2}. \tag{23.56}
\]
However, since\(^2\)
\[
\langle B | \mathbf{e} \cdot \mathbf{p} e^{i \mathbf{k} \cdot \mathbf{r}} | A \rangle &= \langle A | e^{-i \mathbf{k} \cdot \mathbf{r}} (\mathbf{e} \cdot \mathbf{p}) | B \rangle^* = \langle A | (\mathbf{e} \cdot \mathbf{p}) e^{-i \mathbf{k} \cdot \mathbf{r}} | B \rangle^*,
\] |
|
\[
&= (t-1)(-1)^{1+1} \det \begin{pmatrix}\begin{bmatrix} t-2 & 0 \\ 1 & t+1 \end{bmatrix}\end{pmatrix} \\
&\quad + 2(-1)^{1+3} \det \begin{pmatrix}\begin{bmatrix} -2 & 0 \\ t-2 & 0\end{bmatrix} \end{pmatrix}
\] |
|
The corresponding 1-D Schrodinger equation for moving electrons can be written as
\begin{align*}
\frac{d^2 \psi}{dx^2} + \frac{\theta x^2 m}{h^2} \left[ E - V(x) \right] \psi = 0 \tag{2}\\
V(x_1) = V(x + a) \tag{1}\\
\frac{d^2 \psi}{dx^2} + \frac{\theta x^2 m}{h^2} \left[ E - V(x) \right] \psi = 0 \tag{2}
\end{align*} |
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