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\[
\quad c_2 = y(0) \qquad \qquad c_1 = \dot{y}(0) = v_y(0)
\]
initial height initial vertical speed |
|
从上面的结果可以推想，如果 \(v(x)\) 是 \(l\) 次勒让德方程的解，则 \(y = (1 - x^2)^{m/2} v^{(m)}(x)\) 满足方程 (1)。事实上可以通过直接计算证明这一点。由此立得方程 (1) 的另一解
\[
Q_i^n(x) = (-1)^n (1 - x^2)^{m/2} \frac{d^n}{dx^n} Q_i(x) \quad (-1 \leq x \leq 1), \tag{6}
\]
\(Q_i(x)\) 是由 5.8 节 (10) 式规定的 \(l\) 次第二类勒让德函数 \(Q_l^m(x)\) 称为 \(m\) 阶 \(l\) 次第二类连带勒让德函数。这也是霍布森的定义,费瑞尔的定义没有前面 \((-)^m\) 这个因子。 |
|
公理1：如果一条直线上有两个点在一个平面内，那么这条直线在此平面内
推论：判定一条直线是否在平面内
判定点在平面内
检验平面
公理2：过不在一条直线上的三点，有且只有一个平面
推论：经过一条直线和这条直线外一点，有且只有一个平面
经过两条相交直线，有且只有一个平面
经过两条平行直线，有且只有一个平面 |
|
20. 图 3-17 是只用天平和米尺测定动摩擦因数的简易装置。在水平桌面上装有一定滑轮, 两侧用细绳连接着两个物体 \(A, B\)，先用手托住 \(A\), 它离开地面的距离 \(h\) 远小于 \(B\) 离开滑轮的距离 \(l\), 放手后, 使物体 \(A\) 带着 \(B\) 一起加速运动, \(A\) 着地后不跳起, \(B\) 继续滑行一段距离后静止。用天平测出 \(A, B\) 的质量分别为 \(m_1, m_2\), 用米尺量出 \(A\) 离地的起始高度 \(h\), \(B\) 开始时离开滑轮的距离 \(l\) 和停止时离开滑轮的距离 \(l'\), 试由 \(m_1, m_2, h, l\) 和 \(l'\) 导出桌面与物体 \(B\) 之间的动摩擦因数表达式为 \(\mu =\) \_ 。 |
|
\[
\Theta(x) = \begin{pmatrix}
\mathcal{L}(x) & \delta^t(x) \\
\delta(x) & \sigma(x)
\end{pmatrix}
\] |
|
将式(3.3.17)及式(3.3.18)代入(3.3.8)及初始条件式(3.3.10)得
\[
\begin{cases}
\sum_{k=1}^{\infty} \left[ T_k(t) + \left( \frac{k \pi a}{l} \right)^2 T_k(t) - f_k(t) \right] \sin \frac{k \pi}{l} x &= 0 \\
\sum_{k=1}^{\infty} T_k(0) \sin \frac{k \pi}{l} x = 0, \quad \sum_{k=1}^{\infty} T'_k(0) \sin \frac{k \pi}{l} x &= 0
\end{cases}
\]
于是确定函数 \(T_k(t)\) 归结为解下列常系数线性非齐次常微分方程的零初值问题
\[
\begin{cases}
T'_k(t) + \left( \frac{k \pi a}{l} \right)^2 T_k(t) &= f_k(t) \\
T_k(0) = T'_k(0) &= 0 \quad (k = 1,2,3,\ldots) \tag{3.3.19}
\end{cases}
\] |
|
\textbf{EXAMPLE 1:}
Let \(X = \mathbb{R}\) then \(d(x, y) = |x-y|\).
\textbf{SOLUTION:}
i) As \(|x-y| \geq 0\)
d(x, y) \(\geq 0\)
ii)
\begin{align*}
d(x, y) &= |x-y| \\
&= |y-x| \\
&= d(y, x)
\end{align*} |
|
In this paper, we are interested in understanding low energy properties of the Bose gas in regimes that interpolate between the GP and thermodynamic limits. Based on 20, 32, it is well-known that the ground state energy \(E_N := \inf \operatorname{spec}(H_N)\) is equal to
\[
E_N = 4\pi aN^{1+k} + o(N^{1+k}),
\]
where \(a\) denotes the scattering length of the potential \(V\) and where \(o(N^{1+k})\) denotes an error of subleading order, that is \(\lim_{N \to \infty} o(N^{1+k})/N^{1+k} = 0\). Recall that under our assumptions the scattering length of \(V\) is characterized by
\[
8\pi a = \inf \left\{ \int_{\mathbb{R}^3} dx \left( 2|\nabla f(x)|^2 + V(x)|f(x)|^2 \right): \lim_{|x| \to \infty} f(x) = 1 \right\}.
\] |
|
由功的定义 \(W = FL \cdot \cos \theta\) 可知结果相同，
即 \(W = qEL \cdot \lambda = qE \cdot L \cdot \cos \theta\)，电场力做功
与沿电场力方向移动的位移有关。 |
|
（2）利用叠加原理来求带电体系对点电荷\(q_0\)的作用力；体电荷密度为\(\rho_e(r)\)的带电体\(V\)，可以看成无数个点电荷\(\Delta q = \rho_e(r') \Delta V'\)组成，于是便有：\[
F(r) &= \frac{q_0}{4\pi\varepsilon_0} \iiint_V \frac{\rho_e(r')}{|r - r'|^3} (r - r') dV', \tag{1.3.5}
\]同理，带电面和带电线对点电荷\(q_0\)的作用力分别为：
\[
F(r) &= \frac{q_0}{4\pi\varepsilon_0} \iint_S \frac{\sigma_e(r')}{|r - r'|^3} (r - r') dS', \tag{1.3.6} \\
F(r) &= \frac{q_0}{4\pi\varepsilon_0} \int_L \frac{\lambda_e(r')}{|r - r'|^3} (r - r') dl', \tag{1.3.7}
\] |
|
\[
\mathbb{E} \left[ \exp(\alpha k;x,t) \right] &= \frac{1}{2} \exp(\alpha;t \lambda) + \frac{1}{2} \exp(-\alpha;t) = \cosh(\alpha;t); \\
\cosh(x) &= \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!} \quad \text{and} \quad \exp \left( \frac{x^2}{2} \right) = \sum_{k=0}^{\infty} \frac{x^{2k}}{2^k k!}.
\] |
|
高斯定理以成立，是由于库仑定律是距离平方反比律的结果。假如我们设想库仑定律是下面形式:
\[
F \propto \frac{1}{r^{2+\Delta}}
\]其中 \(\Delta\) 是任意一小量。则有
\[
E \propto F \propto \frac{1}{r^{2+\Delta}} \tag{1.5.6}
\]
对于点电荷 \(q\)，以它为球心，作半径为 \(r\) 的球面。取该球面为高斯面，有: |
|
计算：\[
\lim_{n \to \infty} \prod_{i=0}^{n-1}{(2 + \cos \frac{i\pi}{n})^\frac{\pi}n}
\]
证：看到连乘符号，首先考虑取对数
\[
&\lim_{n \to \infty} \ln\left[ (2 + \cos \frac{o\pi}{n})^\frac{\pi}{n} (2 + \cos \frac{\pi}{n})^\frac{\pi}{n}\cdots(2 + \cos \frac{(n-1)\pi}{n})^\frac{\pi}{n}\right] \\
&= \lim_{n \to \infty} \sum_{i=0}^{n-1} \ln (2 + \cos \frac{i\pi}{2})^\frac{\pi}{n}
&= \lim_{n \to \infty} \frac{\pi}{n} \sum_{i=0}^{n-1} \ln (2 + \cos \frac{i\pi}{2}) \, \\
&= \int_{0}^{\pi} \ln (2 + \cos x) \, dx = \int_{0}^{\pi} \ln \left( 2 \sin^2 \frac{x}{2} +2\cos^2\frac{x}{2}+\cos^2\frac{x}{2}-\sin^2 \frac{x}{2}
\right) \, dx \\
&= \int_{0}^{\pi} \ln (3 \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} ) \, dx
\]
而结论为：
\[
\int_{0}^\frac{\pi}{2} \ln (a^2 \sin^2 x + b^2 \cos^2 x) \, dx = \pi \ln \frac{|a| + |b|}{2} \quad (a^2 + b^2 \neq 0)
\] |
|
\[
\lim_{n \to \infty} \frac{1}{n} \left( \sqrt{1 + \frac{1}{n}} + \sqrt{1 + \frac{2}{n}} + \ldots + \sqrt{1 + \frac{n}{n}} \right) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \sqrt{1 + \frac{k}{n}} = \int_{0}^{1} \sqrt{1 + x} \, dx
\]
\[
= \frac{2}{3} (1 + x)^{\frac{3}{2}} \bigg|_{0}^{1} = \frac{2}{3} .(2\sqrt{2} - 1)
\] |
|
\textbf{Example.} \(A = \begin{pmatrix}
4 & 1 & -1 \\
1 & 2 & 1 \\
-1 & 1 & 2
\end{pmatrix}\).
\[
L = \begin{pmatrix}
2 & 0 & 0 \\
* & * & 0 \\
* & * & *
\end{pmatrix} = \begin{pmatrix}
2 & 0 & 0 \\
\frac{1}{2} & * & 0 \\
-\frac{1}{2} & * & *
\end{pmatrix} = \begin{pmatrix}
2 & 0 & 0 \\
\frac{1}{2} & \frac{\sqrt{7}}{2} & 0 \\
-\frac{1}{2} & \frac{5\sqrt{7}}{14} & \frac{\sqrt{42}}{7}
\end{pmatrix}.
\] |
|
1) The derivative of the \textbf{log-normal} density reads:
\[
f(x) &= -\frac{1}{x^{2}\sqrt{2\pi}\sigma} \exp\left(-\frac{1}{2\sigma^{2}}(\log(x) - \lambda)^{2}\right) + \frac{1}{x\sqrt{2\pi}\sigma}\left(-\frac{1}{x\sigma^{2}}(\log(x)-\lambda)\right)\exp\left(-\frac{(\log(x)-\lambda)^{2}}{2\sigma^{2}}\right)\\
&= -\frac{1}{x^{2}\sqrt{2\pi}\sigma} \exp\left(-\frac{1}{2\sigma^{2}}(\log(x) - \lambda)^{2}\right)\left(1+\frac{\log(x)-\lambda}{\sigma^{2}}\right)
\] |
|
计算 \(A = f(x_1, x_2, \cdots, x_n)\)。若 \((x_1, x_2, \cdots, x_n)\) 的近似值为 \((x_1^*, x_2^*, \cdots, x_n^*)\)，\(A\) 的近似值 \(A^* = f(x_1^*, x_2^*, \cdots, x_n^*)\)，则由泰勒展开得到
\[
e(A^*) &:= A - A^* = f(x_1, x_2, \cdots, x_n) - f(x_1^*, x_2^*, \cdots, x_n) \\
&\approx \sum_{k=1}^n \left( \frac{\partial f(x_1^*, x_2^*, \cdots, x_n^*)}{\partial x_k} \right) (x_k - x_k^*) = \sum_{k=1}^n \left( \frac{\partial f}{\partial x_k} \right)^* e_k(x_k^*)
\]
则
\[
\varepsilon(A^*) \approx \sum_{k=1}^n \left| \frac{\partial f}{\partial x_k} \right| \varepsilon(x_k^*)
\]
且
\[
\varepsilon_r(A^*) = \frac{e(A^*)}{|A^*|} \approx \sum_{k=1}^n \left| \frac{\partial f}{\partial x_k} \right| \frac{\varepsilon(x_k^*)}{|A^*|}
\] |
|
奇宇称态为
\[
\psi_{n^-}(x) =
\begin{cases}
\psi_{n+}(x), & x > 0 \\
-\psi_{n+}(x), & x < 0
\end{cases}
\]
它们在点 \(x=0\) 的导数则是
\[
\psi_{n+}'(0^+) = &1, \qquad\psi_{n+}'(0^-) = -1, \\
\psi_{n-}'(0^+&) = \psi_{n-}'(0^-) = 1
\] |
|
\textbf{Theorem 1.2.} Let \(s \in (0,1)\), \(\gamma > 0\), \(q > n\), and \(\gamma + 2s \leq q\). Let \(f\) be a solution to the Boltzmann equation in \((0,T) \times \mathbb{R}^n \times \mathbb{R}^n\) with \(n \geq 2\) (see Definition 1.1). Assume that \(f\) satisfies (1.9), (1.13), and (1.14) with \(q > n\).
Then, for any multi-index \(k \in \mathbb{N}^{1+2n}\) and any \(\tau > 0\) and \(p \geq 0\), it holds
\[
\| |v|^p D^k f \|_{L^{\infty}([ \tau, T] \times \mathbb{R}^n \times \mathbb{R}^n)} \leq C_{k,p},
\]
where \(C_{k,p}\) depends only on \(n, s, \gamma, m_0, M_0, p_0, M_q, q, p, \tau, k\). |
|
\[
\begin{cases}
C_1 = g_1 \left( x, y, \frac{dy}{dx}, \cdots, \frac{d^{n-1} y}{dx^{n-1}} \right), \\
\cdots \cdots \cdots \\
C_n = g_n \left( x, y, \frac{dy}{dx}, \cdots, \frac{d^{n-1} y}{dx^{n-1}} \right).
\end{cases}
\] |
|
类似地, 可从
\[
\frac{dz}{dx} &= \lambda \cos(\lambda \varphi) z, \\
z(x_0) &= 1
\]
解得
\[
\frac{\partial \varphi}{\partial y_0} = e^{\int_{x_0}^{x} \lambda \cos(\lambda \varphi(z; x, x_0, y_0, \lambda)) dx}
\]
再注意 \(\varphi(x, 0, 0, \lambda) \equiv 0\), 因此
\[
\left. \frac{\partial \varphi}{\partial y_0} \right|_{x_0 = y_0 = 0} = e^{\int_{0}^{x} \lambda dz} = e^{\frac{\lambda}{2} z^2}.
\] |
|
Since \(S_f + I_f = 1\) and \(S_m + I_m = 1\), the four-dimensional system reduces to the two-dimensional system
\[
I_f &= \frac{\lambda_f}{r} (1 - I_f) I_m - \frac{I_f}{d_f}, \\
I_m &= r \lambda_m (1 - I_m) I_f - \frac{I_m}{d_m},
\]
where \(r = N_f / N_m\). The system has two equilibrium points \((I_f, I_m) = (0, 0)\) and
\[
(I_f, I_m) = \left( \frac{d_f d_m \lambda_f \lambda_m - 1}{d_m \lambda_m (r + d_f \lambda_f)}, \frac{r(d_f d_m \lambda_f \lambda_m - 1)}{d_f \lambda_f (1 + r d_m \lambda_m)} \right).
\] |
|
Normal problem set-up: given the force law, determine the motion \(\vec{r}(t)\).
Using linear momentum \(\vec{p} = m \vec{v}\) we rewrite Newton's eqn as a pair of first-order ODE s:
\[
\frac{d\vec{r}}{dt} = \frac{1}{m} \vec{p}(t)\quad, \quad
\frac{d\vec{p}}{dt} = \vec{F}(\vec{r}(t))
\] |
|
质量为\(M\)的玩具炮，沿水平方向发射一颗质量为\(m\)的炮弹，在炮身固定和可自由反冲两情况下，炮弹的初速度之比为 （ ）
A. \(1:1\)
B. \(\frac{M}{M+m}\)
C. \(\frac{M+m}{m}\)
D. \(\sqrt{\frac{M}{M+m}}\)
E. \(\sqrt{\frac{M+m}{M}}\) |
|
\textbf{定理 1 闭集套}
集合 \(K \subseteq \mathbb{R}\) 是紧致的, 当且仅当下述命题成立: 如果 \(\{C_{\alpha}\}_{\alpha \in A}\) 是一族在 \(K\) 中闭的子集, 而且其中任意有限个的交集都非空, 那么交集 \(\bigcap_{\alpha \in A} C_{\alpha}\) 本身非空。 |
|
\textbf{Theorem.} Let \(H\) be a positive semi-definite matrix. Then there exists a unique positive semi-definite matrix \(H_0\) such that \(H = H_0^2\).
\(H_0\) is called the square root of \(H\) and denoted by \(H^{\frac{1}{2}}\).
\textbf{Proof. Existence.} Let \(H = U \Lambda U^*\), \(\Lambda = \begin{pmatrix}
\lambda_1 & 0 \\
0 & \ddots & 0 \\
0 & \lambda_n
\end{pmatrix}\), \(\lambda_j \geq 0\), and \(U\) is unitary. Let
\[
H_0 = U \Lambda_0 U^*, \quad \Lambda_0 = \begin{pmatrix}
\lambda_1^{\frac{1}{2}} & 0 \\
0 & \ddots & 0 \\
0 & \lambda_n^{\frac{1}{2}}
\end{pmatrix}.
\] |
|
\[
P_h(B) &= \int_0^\infty e^{-x/B} \, d\beta_h(x) \\
&= h + he^{-a/B} + he^{-2a/B} + \cdots \\
&= h \left(1 + e^{-h\nu/B} + e^{-2h\nu/B} + \cdots \right)
\] |
|
6. 一只气球以 \(10 \, \text{m/s}\) 的速度匀速上升，某时刻在气球正下方距气球 \(4 \, \text{m}\) 的地方有一小石子以 \(20 \, \text{m/s}\) 的初速度竖直上抛，下列判断中正确的是(取 \(g = 10 \, \text{m/s}^2\)) （ ）
A. 石子一定不会击中气球。
B. 气球上升速度小于 \(8 \, \text{m/s}\) 时，才能被石子击中。
C. 若气球上升速度大于 \(11 \, \text{m/s}\)，就不会被石子击中。 |
|
Since \(x \neq 0\), this is only possible if \(\lambda = \pm 1\). Now suppose \(\theta \neq k \pi\) for any integer \(k\). Then \(\sin \theta \neq 0\) and \(\vert\cos \theta \vert< 1\), so combining the equations from \((*)\) implies that
\[
\sin \theta \, x_2 = (\cos \theta - \lambda) x_1 = (\cos \theta - \lambda) \left( -\frac{(\cos \theta - \lambda)}{\sin \theta} x_2 \right) = -\frac{(\cos \theta - \lambda)^2}{\sin \theta} x_2.
\] |
|
设 \(f(x)\) 是焦点为土1 的椭圆 \(C\) 上及其内的解析函数，则对于位于这椭圆内部的任意一个共焦椭圆中的点 \(t\)，有
\[
f(t) = \sum_{n=0}^{\infty} a_n P_n(t), \tag{6}
\]其中
\[
a_n = \frac{2n+1}{2\pi i} \oint_C f(x) Q_n(x) dx, \tag{7}
\]
而且级数是一致收敛的。证明如下：按科希公式和 (3) 式
\[
f(t) &= \frac{1}{2\pi i} \int_C \frac{f(x) dx}{x - t} \\
&= \sum_{n=0}^{\infty} (2n + 1) P_n(t) \cdot \frac{1}{2\pi i} \int_C f(x) Q_n(x) dx,
\]
故有 (6)。这个展开称为诺埃曼展开。 |
|
\begin{align*}
(1) \quad p(x | \mu_1, \Sigma) &= \frac{1}{(2\pi)^{m/2}} \frac{1}{(\det(\Sigma)^{1/2}} \exp\left(-\frac{1}{2}(x-\mu)^T \Sigma^{-1} (x-\mu)\right).
\end{align*}
(2) \quad The log likelihood reads
\begin{align*}
\ell(\mu_1, \Sigma) &= \log p(x_1, \ldots, x_m \mid \mu_1, \Sigma) \\
&= -\left(\frac{mm}{2} \log(2\pi) + \frac{m}{2} \log(\det(\Sigma)) + \frac{1}{2} \sum_{r=1}^{m} (x_r-\mu)^T \Sigma^{-1} (x_r-\mu)\right)
\end{align*}
and the gradient w.r.t. \(\mu\) gives
\begin{align*}
\nabla_\mu \ell(\mu, \Sigma^{-1}) &= \sum_{r=1}^{m} \Sigma^{-1} (\mu-x_i) \\
&= \Sigma^{-1}\left(m\mu-\sum_{r=1}^{m} x_i\right).
\end{align*} |
|
PROPOSITION 2. \textit{If a feld} \(F\) \textit{is invariant under all linear motions of} \(\mathbb{R}^3\) \textit{fixing the origin, then} \(L\) \textit{is conserved.}
\textit{Proof.} We differentiate：
\[
\frac{dL}{dt} = m\dot{x} \times \dot{x} + m x \times \ddot{x} = m x \times \ddot{x}
\]
which is zero，since \(x\) and \(\ddot{x}\) are parallel。
We also define the \textbf{torque} of a force \(F\) acting on the body \(x\) by \(\tau = x \times F\)。Then in angular coordinates，Newton's Law takes the form： |
|
法二：令 \[
\Delta x = r\cos\theta, \, \Delta y = r\sin\theta \\
\lim_{(\Delta x, \Delta y) \to (0,0)}\frac{ \Delta x^2 \Delta y}{(\Delta x^2+ \Delta y^2) ^\frac{3} {2} }= \lim_{r \to 0}{\cos^2\theta \sin\theta }
\]
\(\theta\) 不同，极限不同，从而不可微。 |
|
To show that the sum is direct, suppose that \(p(t) \in N(T) \cap R(T)\). Then \(p(t) \in N(T) \Rightarrow p(t) = at\) for some \(a \in F\) and \(p(t) \in R(T) \Rightarrow a = 0\), so \(p(t) = 0\).
\[
\therefore \, P_n = N(T) \oplus R(T).
\] |
|
termed KN Scores, as follows:
\[
\text{FFN}^{(l)}(H^{(l)}) &= W_{2}^{(l)} \text{SiLU}(H^{(l)} W_{1}^{(l)}) \tag{3} \\
\omega_{i}^{l} &= \text{SiLU}(H^{(l)} W_{i}^{(l)}), \quad \forall \omega_{i}^{l} \in \omega \tag{4} \\
\text{KN Scores} &= \frac{1}{|\omega|} \sum \omega_{i}^{l}, \quad \forall \omega_{i}^{l} \in \omega \tag{5}
\]
where \(\omega_{i}^{l}\) denotes the \(i\)-th neuron in the \(l\)-th intermediate layer of FFN and symbol \(\omega\) represents the KNs associated with a specific fact triplet, denoted as \((s, r, o)\). For the first-hop fact and the second-hop fact, we designate their respective sets of KNs as \(\omega_{1}\) and \(\omega_{2}\). Under the context of a single-hop query, we denote KN Scores as \(\{\omega \mid QT1H\}\). Similarly, within the two-hop reasoning context, KN Scores are represented as \(\{\omega \mid QT2H\}\). |
|
其图形见图3.25。不难计算出，在 \(|x|>L\) 区域中找到粒子的概率为
\[
2 \int_{L}^{\infty} |\psi(x)|^2 dx = e^{-2} = 0.1353
\]
2. 奇宇称态
\[
\psi(x) =
\begin{cases}
Ae^{-\beta x}, & x > 0 \\
-Ae^{\beta x}, & x < 0
\end{cases} \tag{3.5.26}
\]
由波函数连续条件，要求奇宇称态 \(\psi(0) = 0\)， |
|
函数及其相应导数的给定的初值。这样, 不失一般性, \(n\) 阶微分方程的初值问题可以提成如下形式
\[
\begin{cases}
y^{(n)} = F(x, y, y', y'', \cdots, y^{(n-1)}), \\
y(x_0) = y_0, y'(x_0) = y_0', \cdots, y^{(n-1)}(x_0) = y_0^{(n-1)}
\end{cases}
\tag{1.13}
\]
自然要问: 当函数 \(F\) 满足什么条件时, 初值问题 \((1.13)\) 的解是存在的, 或者更进一步, 是存在而且唯一的？这是常微分方程理论中的一个基本问题。在第三章中我们将就 \(n = 1\) 的情况进行证明： |
|
\[
&\to\underline{v} = \underline{v}_0 + \underline{a} t \quad \to \underline{v} = \frac{d\underline{r}}{dt} \\
&\int_0^r d\underline{r} = \int_0^t (\underline{v}_0 + \underline{a} t) \, dt \\
&\to\underline{r} = \underline{v}_0 t + \frac{1}{2} \underline{a} t^2
\] |
|
\textbf{证明} 先证必要性：设 \((1.4)\) 是一个恰当方程, 则存在函数 \(\Phi(x, y)\), 满足
\[
\frac{\partial \Phi}{\partial x} = P(x, y), \quad \frac{\partial \Phi}{\partial y} = Q(x, y). \tag{1.8}
\]
然后, 我们在上面的第一式和第二式中, 分别对 \(y\) 和 \(x\) 求偏导数, 就可得到
\[
\frac{\partial P}{\partial y} = \frac{\partial^2 \Phi}{\partial y \partial x}, \quad \frac{\partial Q}{\partial x} = \frac{\partial^2 \Phi}{\partial x \partial y}. \tag{1.9}
\]
由 \(\frac{\partial P}{\partial y}\) 与 \(\frac{\partial Q}{\partial x}\) 的连续性推知 \(\frac{\partial^2 \Phi}{\partial y \partial x}\) 和 \(\frac{\partial^2 \Phi}{\partial x \partial y}\) 是连续的, 从而 \(\frac{\partial^2 \Phi}{\partial y \partial x} = \frac{\partial^2 \Phi}{\partial x \partial y}\)。 |
|
\begin{tabular}{cccc}
\hline
& 不拒绝 \(H_0\) & 拒绝 \(H_0\) & 总计 \\
\hline
\(H_0\) 为真 & \(U\) & \(V\) & \(m_0\) \\
\(H_0\) 为假 & \(T\) & \(S\) & \(m_1\) \\
\hline
总计 & \(m - R\) & \(R\) & \(m\) \\
\hline
\end{tabular} |
|
\[
2i \int_{0}^{\infty} \frac{\sin x}{x} \, dx &= i \pi \\
\int_{0}^{\infty} \frac{\sin x}{x} \, dx &= \frac{\pi}{2}
\] |
|
\[
u(x,t) = \frac{1}{2a} \int_0^tdt \int_{x-a(t-s)}^{0} f(-\xi,t) \, d\xi \, + \frac{1}{2a} \int_0^tdt \int_{0}^{x+at-s} f(\xi,t) \, d\xi \,
\]
为使有关 \(x\) 的偏导所得到在 \(x=0\) 处可积，应要求 \(f(x,0,t) = 0\)。
此时上述给出的积分是原问题的解。 |
|
then
\[
C_l = \frac{2\Gamma(y_o)}{V_\infty c(y_o)}
\]
Combining Eqs. (5.21) \& (5.23) gives
\[
\alpha_{\text{eff}} = \frac{\Gamma(y_o)}{\pi V_\infty c(y_o)} + \alpha_{L=0}
\] |
|
\[
u_1(x) &=
\begin{cases}
0 & \text{if } x < x_0, \\
1 & \text{if } x \geq x_0,
\end{cases} \\
u_2(x) &=
\begin{cases}
0 & \text{if } x \neq x_0, \\
1 & \text{if } x = x_0.
\end{cases}
\] |
|
可以验证上述函数的确是一密度函数
\[
\int_{-\infty}^{\infty} f(x) dx &= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{dx}{1 + x^2} = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{d \arctan(x)}{dx} \\
&= \frac{1}{\pi} [\arctan(\infty) - \arctan(-\infty)] = \frac{1}{\pi} \left[ \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) \right] = 1.
\]
\(\chi^2\)分布 如果
\[
f(x) = \frac{1}{\Gamma(p/2) 2^{p/2}} x^{(p/2) - 1} e^{-x/2}, \quad x > 0,
\]
则\(X\)服从自由度为\(p\)的\(\chi^2\)分布，记为\(X \sim \chi^2_p\)。如果\(Z_1, \cdots, Z_p\)是独立标准正态随机变量，则\(\sum_{i=1}^{p} Z_i^2 \sim \chi^2_p\)。 |
|
\[
V = \frac{\Gamma}{2\pi h} \tag{5.11}
\]
same as the two dimensional theory.
If we have a “semi-infinite” filament we get
\[
V &= \frac{\Gamma}{4\pi} \int_{0}^{\infty} \frac{\sin \theta}{r^2} dl = -\frac{\Gamma}{4\pi h} \int_{\pi/2}^{0} \sin \theta d\theta \\
V &= \frac{\Gamma}{4\pi h} \tag{5.12}
\] |
|
\[
W_b := \{ F \in W \mid \| F \| = \sup_{j \in \mathbb{Z}} \sup_{r > 0} | F_j(r) | < \infty \}.
\] |
|
如图 6-9 所示，是在水平气垫导轨上验证动量守恒的实验装置，小车 \(A\) 的前端粘有橡皮泥，后面连着穿过打点计时器的纸带，小车 \(B\) 静止在导轨上，实验时，轻推小车 \(A\) 使它沿轨道作匀速运动，然后与前方的小车 \(B\) 相碰并粘合在一起，继续作匀速运动。已知打点计时器的电源频率为 \(50 \, \text{Hz}\)。 |
|
弦段\(\Omega_t\) 介质在\(t\)时刻的能量积分为：
\[
E(t) = \int_\Omega (v_0^2 + a^2 v_x^2) \, dx
\] |
|
\[
\partial^a \equiv & \frac{\partial}{\partial x_a} = \left( \frac{\partial}{\partial x^0}, -\nabla \right) \\
\partial_a \equiv & \frac{\partial}{\partial x^a} = \left( \frac{\partial}{\partial x^0}, \nabla \right) \tag{11.76}
\]
四元矢量 \(A\) 的四维散度是一个不变量，
\[
\partial^a A_a = \partial_a A^a = \frac{\partial A^0}{\partial x^0} + \nabla \cdot A \tag{11.77}
\] |
|
令\[
G(t) = \iint_{ki} \left( u_t^2 + cu^2 + a^2 u_x^2 \right) dx, \quad 则有 \quad \frac{dG(\tau)}{d\tau} \leq G(\tau) + F(\tau),
\]
其中
\[
F(\tau) = \int_\Omega \left( \psi^2 + c \psi^2 + a^2 \psi_x^2 \right) dx + \iint_{ki} f^2 dx \, d\tau.
\]
由Gronwall不等式，\(G(\tau) \leq (e^\tau-1)F(\tau)\) \(\implies\) 当 \(\varphi = \psi = f = 0\) 时，\(u \equiv 0\)。
因此\(u=u_1-u_2 = 0\) \(\implies u _1= u_2\)，即解唯一。 |
|
比较上式两端的系数即得
\[
f_k(t) = \begin{cases}
\sin \frac{2\pi a}{l} t & \text{当 } k = 2 \text{ 时} \\
0 & \text{当 } k \geq 1 \text{ 且 } k \neq 2 \text{ 时}
\end{cases}
\]
由式(3. 3. 13)得
\[
T_2(t) = \frac{l}{2\pi a} \int_0^t \sin \frac{2\pi a}{l} \tau \sin \frac{2\pi a(t - \mathrm{r})}{l} d\tau \\&= \frac{l}{4a\pi} \left( \frac{L}{2a\pi} \sin \frac{2\pi a}{l} t - \cos \frac{2\pi}{l} t \right)
\] |
|
This rather involved expression is simplified by the introduction of a new variable, the angle \(\omega\) at the center of the circle \(K\), measured from the radius \(HH\). In this angle, the coordinates \(\xi, \eta\) and the quantities connected with them are expressed as follows: For abbreviation, we designate the radius of the circle \(K\) with \(r = \sqrt{\frac{\xi^2 + \eta^2}{2}} + \delta\) and introduce the angle \(A\) by
\[
\cos A = \frac{l}{\sqrt{\xi^2 + \eta^2}} \quad \text{and} \quad \sin A = \frac{\eta}{\sqrt{\xi^2 + \eta^2}}.
\] |
|
所以问题归结于求下列定解问题
\[
\begin{cases}
\Delta u &= \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial u}{\partial \rho} \right) + \frac{1}{\rho^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \quad 0 < \rho < \rho_0 \\
u(\rho_0, \theta) &= \rho(\theta) \quad 0 \leq \theta \leq 2\pi \\
u(0, \theta) &< \infty \\
u(\rho, 0) &= u(\rho, 2\pi) \\
\frac{\partial}{\partial \theta} u(\rho, 0) &= \frac{\partial}{\partial \theta} u(\rho, 2\pi)
\end{cases}
\]
先令
\[
u(\rho, \theta) = R(\rho) \Phi(\theta) \tag{3.2.17}
\] |
|
故有
\[
P_i(x) = \frac{1}{2k + 1} \left[ P_{i-1}(x) - P_{i+1}(x) \right]
\]
\textbf{例} \(11\) 求证勒让德多项式的递推公式
\[
(k + 1) P_{i-1}(x) - (2k + 1) x P_i(x) + k P_{i-1}(x) = 0
\]
证：先将函数 \(x P_A(x)\) 展成勒让德多项式的级数。设 |
|
(ii) We have
\[
t-1 &= -1(1) + 1(t) + 0(t^2) \\
t+1 &= 1(1) + 1(t) + 0(t^2) \\
(t-1)(t+1) &= t^2 - 1 = -1(1) + 0(t) + 1(t^2)
\]
So the transition matrix is
\[
\begin{bmatrix}
-1 & 1 & -1 \\
1 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\] |
|
\[
T_a = \left[ \frac{\partial A(\tau)}{\partial \tau^a} \right]_{\tau = 0}, \quad a = 1, \ldots, r,
\]
这是 \(N \times N\) 常数方阵，它们是线性独立的，若若不然，有一组数 \(\lambda^1, \ldots, \lambda^r\) 使得 \(\sum_{a=1}^r \lambda^a T_a = 0\)，于是有 |
|
\[
\zeta_i (x) &= \frac{\partial \psi (x)}{\partial x^i}, \\
\delta_{i k} &=
\begin{cases}
1, & \text{当 } i = k; \\
0, & \text{当 } i \ne k.
\end{cases} \tag{1.1.8}
\] |
|
\textbf{D. D'Alembert Ratio Test \textit{(Easy Convergence Test to Apply)}}
1. If \\(\frac{a_{n+1}}{a_n} \leq r < 1 \\) for \\(n > N \\) and \(r\) is independent of \(n\),
\[
\sum_{n=1}^{\infty} a_n \text{ is convergent}
\] |
|
We have a geometric series with ratio \(e^{-h\nu/B}\) whose sum is \(h / (1 - e^{-h\nu/B})\). The corresponding \(E\) is given by
\[
E_h(B) = B^2 \frac{P_h'(B)}{P_h(B)} = B^2 \frac{h\nu e^{-h\nu/B}}{1 - e^{-h\nu/B}} = \frac{h\nu}{e^{h\nu/B} - 1}
\] |
|
\[
J = 1 \otimes \left( -i \partial_{\theta} + \frac{1}{2} \sigma_3 \right) \quad \text{on} \quad L^2(\mathbb{R}_+) \otimes L^2(S)^2
\] |
|
\[
\rho^2 + (A_r - 1) \rho + B_r = 0 \quad (r = 1, 2, \cdots, n). \tag{5}
\]
在 \(\infty\) 点, 由 2.6 节 (1) 有
\[
\lim_{t \to 0} \left\{ \frac{2}{t} - \frac{1}{t^2} p \left( \frac{1}{t} \right) \right\} &= 2 - \lim_{z \to \infty} zp(z) = 2 - \sum_{r=1}^n A_r, \\
\lim_{t \to 0} t^2 \cdot \frac{1}{t^2} q \left( \frac{1}{t} \right) &= \lim_{z \to \infty} z^2 q(z) = \lim_{z \to \infty} z^2 \sum_{r=1}^n \left( \frac{B_r}{(z - a_r)^2} + \frac{C_r}{z - a_r} \right) \\
&= \sum_{r=1}^n B_r + \lim_{z \to \infty} \sum_{r=1}^n C_r \left[ 1 + a_r z^{-1} + O(z^{-2}) \right] \\
&= \sum_{r=1}^n B_r + \sum_{r=1}^n a_r C_r \quad (\text{因 (4)}).
\] |
|
\[
& \bullet \text{玻尔兹曼系统 } \Omega_{M.B.} = \frac{N!}{\prod_{l} a_{l}!} \prod_{l} \omega_{l}^{a_{l}} \quad (n \text{全排列为} n!) \\
& \bullet \text{波色系统 } \Omega_{B.E.} = \prod_{l} \frac{(\omega_{l} + a_{l} - 1)!}{a_{l}!(\omega_{l} - 1)!} \quad (\text{使用} \omega_{l} - 1 \text{个隔板}) \\
& \bullet \text{费米系统 } \Omega_{F.D.} = \prod_{l} \frac{\omega_{l}!}{a_{l}!(\omega_{l} - a_{l})!} \quad (\omega_{l} \text{里面选前} a_{l} \text{个}) \\
& \bullet \text{当 } \frac{a_{l}}{\omega_{l}} \ll 1 \text{时}, \quad \Omega_{B.E.} \approx \frac{\Omega_{M.B.}}{N!}, \quad \Omega_{F.D.} \approx \frac{\Omega_{M.B.}}{N!} \quad (\text{经典极限条件})
\] |
|
公式 (3.2.29) 称为圆域内的泊松 (Poisson) 公式。
\textbf{例 5} 求两端自由的弦的自由振动所对应的混合问题
\[
\frac{\partial^2 u}{\partial t^2} - a^2 \frac{\partial^2 u}{\partial x^2} &= 0 \quad 0 < x < l \quad t > 0 \quad \text{(3.2.30)} \\
u \big|_{t=0} &= \varphi(x), \quad \frac{\partial u}{\partial t} \big|_{t=0} = \psi(x) \quad 0 < x < l \quad \text{(3.2.31)} \\
\frac{\partial u}{\partial x} \big|_{x=0} &= 0, \quad u \big|_{x=l} = 0 \quad t > 0 \quad \text{(3.2.32)}
\]
的解。 |
|
Furthermore, the CLT shows that
\[
\frac{1}{\sqrt{m}} \sum_j X_j \xrightarrow{d} \rightarrow \frac{1}{\sqrt{2}} Z
\]
where \(Z \sim \mathcal{N}(0,1)\). We may conclude using Slutsky's theorem that
\[
\sqrt{m} \overline{Y}_m \xrightarrow{d} \rightarrow \mathcal{N}(0, \frac{2}{3})
\] |
|
\textbf{(10.46) Lemma} Let \(\{\Omega_k\}_{k=1}^{\infty}\) be a sequence of cubes with centers \(\{x_k\}\) such that if \(j < k\), then \(x_k \notin \Omega_j\) and \(|\Omega_k| \leq 2|\Omega_j|\). Then \(\{\Omega_k\}\) has bounded overlaps, and the constant \(c\) for which \(\sum \chi_{\Omega_k} \leq c\) can be chosen to depend only on \(n\). |
|
\[
\frac{d\Gamma}{dy} = -\frac{4\Gamma_0}{b^2} \frac{y}{\sqrt{1 - \frac{4y^2}{b^2}}}
\] |
|
this should be of the form\[
\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) = \frac{\partial L}{\partial x}, \quad \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{y}} \right) = \frac{\partial L}{\partial y}, \quad \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{z}} \right) = \frac{\partial L}{\partial z}
\]
we \(L\) such that
\[
\frac{\partial L}{\partial \dot{x}} = \frac{m_0 x}{\sqrt{1 - v^2/c^2}} = m \dot{x} \quad \frac{\partial L}{\partial \dot{y}} = \frac{m_0 y}{\sqrt{1 - v^2/c^2}} = m \dot{y},
\]
\[
\frac{\partial L}{\partial \dot{z}} = \frac{m_0\dot z}{\sqrt{1 - v^2/c^2}} = m \dot{z}
\]
\[
\frac{\partial L}{\partial\dot x} = -\frac{\partial V}{\partial x}, \quad \frac{\partial L}{\partial \dot y} = -\frac{\partial V}{\partial y}, \quad \frac{\partial L}{\partial \dot z} = -\frac{\partial V}{\partial z}
\]
\((4)\) and \((5)\) are Satisfied
\[
L = -m_0 c^2 \left(1 - v^2/c^2\right)^{1/2} - V
\]
where \(V\) is potential |
|
8. 两木块自左至右运动,现用高速摄影机在同一底片上多次曝光,记录下木块每次曝光时的位置,如图所示,连续两次曝光的时间间隔是相等的,由图2-2可知
A. 在时刻\(t_2\)以及时刻\(t_5\)两木块速度相同。
B. 在时刻\(t_5\)两木块速度相同。
C. 在时刻\(t_3\)以及时刻\(t_4\)之间某瞬时两木块速度相同。
D. 在时刻\(t_4\)以及时刻\(t_6\)之间某瞬时两木块速度相同。 |
|
A、B两球在光滑水平面上沿一直线同方向运动，A球的动量是\(5\text{kg}\cdot\text{m/s}\)，B球的动量是\(7\text{kg}\cdot\text{m/s}\)，当A球追上B球发生碰撞后，A、B两球的动量可能值是（ ）
A. \(P_A = 6\text{kg}\cdot\text{m/s}, \ P_B = 6\text{kg}\cdot\text{m/s}\).
B. \(P_A = 3\text{kg}\cdot\text{m/s}, \ P_B = 9\text{kg}\cdot\text{m/s}\).
C. \(P_A = -2\text{kg}\cdot\text{m/s}, \ P_B = 14\text{kg}\cdot\text{m/s}\).
D. \(P_A = -5\text{kg}\cdot\text{m/s}, \ P_B = 17\text{kg}\cdot\text{m/s}\). |
|
For hydrogen at 1 atm, this is \(6 \times 10^4\), and so very large in comparison with 1. So here the classical theory holds nicely. The error increases with increasing density and decreasing temperature rapidly, and is for helium in the neighborhood of the critical point quite considerable; no longer can one talk of a perfect gas everywhere.
We now calculate from (12) the entropy for our limiting case. Replacing in (21) \(\log(1-e^{-\alpha s})\) by \(-e^{-\alpha s}\) and this by |
|
and by this and \(\det H_{n-1} = \det P^* \det P\),
\[
0 &< \frac{\det H}{\det H_{n-1}} = \det \begin{pmatrix}
I_{n-1} & b \\
b^* & h_{nn}
\end{pmatrix} = \det \begin{pmatrix}
I_{n-1} & b \\
0 & h_{nn} - b^* b
\end{pmatrix} = h_{nn} - b^* b.
\]
For \(x \in \mathbb{C}^n\), \(x \neq 0\), let \(\begin{pmatrix}
z \\
x_n
\end{pmatrix} = \begin{pmatrix}
P & 0 \\
0 & 1
\end{pmatrix} x\). We get
\[
x^* H x &= \begin{pmatrix}
z^* & x_n^*
\end{pmatrix} \begin{pmatrix}
I_{n-1} & b \\
b^* & h_{nn}
\end{pmatrix} \begin{pmatrix}
z \\
x_n
\end{pmatrix} = (z^* + x_n^* b^*) \begin{pmatrix}
z^* & x_n^*
\end{pmatrix} \begin{pmatrix}
z \\
x_n
\end{pmatrix} \\
&= z z^* + x_n^* b^* z + z^* b x_n + h_{nn} x_n^* x_n = (z + b x_n)^* (z + b x_n) - b^* b x_n^* x_n + h_{nn} x_n^* x_n \\
&= (z + b x_n)^* (z + b x_n) + (h_{nn} - b^* b) x_n^* x_n > 0.
\]
Therefore, \(H\) is positive.
\(\square\) |
|
证明：利用定理19.3：若 \(f(x,y)\) 与 \(f_x(x,y)\) 在 \([a,b] \times [c,d]\) 上连续则
\(\int_c^d f(x,y) \, dy\) 在 \([a,b]\) 上可微且 \(\frac{d}{dx} \int_c^d f(x,y) \, dy = \int_c^d f_x(x,y) \, dy
\) |
|
We apply Chernoff bounds. Since the exponential function is an order-preserving bijection, we have for any \(s > 0\)
\[
\mathbb{P}\left( \frac{1}{n} \sum Z_i > t \right) = \mathbb{P}\left( \exp\left( s \sum Z_i \right) > \exp(st) \right) &\leq e^{-st} \mathbb{E} \left[ e^{s \sum Z_i} \right] \quad \text{(Markov)} \\
&= e^{-stn} \prod \mathbb{E} \left[ e^{sZ_i} \right],
\]
where in the last equality we have used the independence of the \(Z_i\).
We therefore need to control the term \(\mathbb{E}\left[ e^{sZ_i} \right]\), known as the moment-generating function of \(Z_i\). If the \(Z_i\) were normally distributed, we could compute the moment-generating function analytically. The following lemma establishes that we can do something similar when the \(Z_i\) are bounded.
\textbf{Lemma (Hoeffding’s Lemma):} If \(Z \in [a, b]\) almost surely and \(\mathbb{E}Z = 0\), then
\[
\mathbb{E}e^{sZ} \leq e^{\frac{s^2(b-a)^2}{8}}.
\]
\textit{Proof of Lemma:} Consider the log-moment generating function \(\psi(s) = \log \mathbb{E}e^{sZ}\), and note that it suffices to show that \(\psi(s) \leq \frac{s^2(b-a)^2}{8}\). We will investigate \(\psi\) by computing the |
|
\textbf{Derivation of Lagrange's Equation of Motion}
\textit{(By Variational Method)}
\textit{Conservative System Without Constraints}
For a single particle:
K.E. = \(T = T(q_1, q_2, \dot{q_1}, \dot{q_2}, t)\)
P.E. = \(V = V(q_1, q_2, t)\)
\[
L = T - V = L(q_1, q_2, \dot{q_1}, \dot{q_2}, t)
\]
The equations of motion are derived by an appeal to the principle of least action \textit{(Hamilton Principle)}. |
|
Consider a square barrier of a height inversely proportional to its width. To be specific we take
\[
V(x) = \begin{cases}
\frac{\lambda}{2\epsilon} & (|x| < \epsilon) \\
0 & (|x| > \epsilon)
\end{cases}
\tag{3.84}
\]
In the limit \(\epsilon \rightarrow 0\) we have a potential that is everywhere zero except the point \(x = 0\) where it has an infinite strength. At the same time the integral of the potential is finite and equal to \(\lambda\)
\[
\int_{-\infty}^{+\infty} dx \, V(x) = \int_{-\epsilon}^{+\epsilon} dx \, \left( \frac{\lambda}{2\epsilon} \right) = \lambda.
\] |
|
\[
f(z) = \frac{\varphi(z)}{(z-b)^n}
\] |
|
定理：\(f(z) = u(x, y) + iv(x, y)\)在\(z_0 = x_0 + iy_0\)处连续的充要条件是\(u(x, y)\)和\(v(x, y)\)在\((x_0, y_0)\)处连续。
即，设\(f(z_0) = u_0 + iv_0\)，则\(\lim_{z \to z_0} f(z) = f(z_0)\)
\[
\iff \lim_{x \to x_0, y \to y_0} u(x, y) = u_0, \quad \lim_{x \to x_0, y \to y_0} v(x, y) = v_0
\] |
|
Consider a clock moving with frame \(S'\) placed at a point \((x_0', y_0', z_0')\) relative to frame \(S\)The time is
\[
t = \frac{{t' + \frac{v x_0'}{c^2} }}{\sqrt{1 - \frac{v^2}{c^2}}}
\]
In sintorval \(t_2 - t_1\) is related to the corresponding S' interval \(t_2' - t_1'\) is given by
\[
t_2 - t_1 = t_2' - t_1' /{\sqrt{1 - {v^2}/{c^2}}}
\] |
|
\[
\sqrt{m}Y_m = \frac{\frac{1}{\sqrt{m}} \sum_j x_j}{\frac{1}{m} \sum_j x_j^2 + \frac{1}{m} \sum_j x_j^3}
\] |
|
用 \(\varepsilon-N\) 方法证明
\[
\lim_{n \to \infty} \frac{3n^2 + n}{2n^2 - 1} = \frac{3}{2}
\] |
|
我们得出
\[
\sum_{a=1}^r v^a_b(\sigma) A^{-1}(\tau) T_a A(\tau) &= A^{-1}(\Psi(\sigma, \tau)) \frac{\partial A(\Psi(\sigma, \tau))}{\partial \Psi^a} \frac{\partial \Psi^a(\sigma, \tau)}{\partial \sigma^b} \\
&= \sum_{c=1}^r v^c_b(\Psi(\sigma, \tau)) \frac{\partial \Psi^a(\sigma, \tau)}{\partial \sigma^b} T_c.
\]
令 \(\sigma = 0\)，于是有 \(A(\Psi(0, \tau)) = I\)，即 \(\Psi(0, \tau) = 0\)，因此得出 |
|
\textbf{定义3} 设 \(n\) 阶微分方程 \((1.1)\) 的解 \(y = \varphi(x, C_1, C_2, \cdots, C_n)\) 包含 \(n\) 个独立的任意常数 \(C_1, C_2, \cdots, C_n\)，则称它为 \textbf{通解}；如果微分方程 \((1.1)\) 的解 \(y = \varphi(x)\) 不包含任意常数，则称它为 \textbf{特解}。
这里说 \(n\) 个任意常数 \(C_1, C_2, \cdots, C_n\) 是独立的，其含义是 \(\varphi, \varphi', \cdots, \varphi^{(n-1)}\) 关于 \(C_1, C_2, \cdots, C_n\) 的 Jacobi 行列式
\[
\frac{D(\varphi, \varphi', \cdots, \varphi^{(n-1)})}{D(C_1, C_2, \cdots, C_n)} \neq 0
\]
其中 \(\varphi', \cdots, \varphi^{(n-1)}\) 分别是 \(\varphi\) 关于自变量 \(x\) 的相应导数。 |
|
\textit{Proof}. Since \(\nu\) and \(\mu\) are mutually singular, there is a set \(Z\) with \(\nu(\mathbb{R}^n - Z) = \mu(Z) = 0\). Let \(E = \mathbb{R}^n - Z\), and consider the sets
\[
T_\alpha &= \left\{ x \in E : \limsup_{h \to 0} \frac{\nu(Q_x(h))}{\mu(Q_x(h))} > \alpha \right\}, \quad (\alpha > 0), \\
T &= \left\{ x \in E : \limsup_{h \to 0} \frac{\nu(Q_x(h))}{\mu(Q_x(h))} > 0 \right\}.
\] |
|
\[
\frac{d^2}{d|\xi|^2} \psi + |\xi| \psi = 0
\]
作变换 \(z = \frac{2}{3} |\xi|^{3/2}\), \(\psi = |\xi|^{1/2} u\), 则上列方程将化为 \(1/3\) 阶的 Bessel 方程
\[
\frac{d^2 u}{d z^2} + \frac{1}{z} \frac{d u}{d z} + \left(1 - \frac{(1/3)^2}{z^2}\right) u = 0
\]
它的两个线性无关解可以取为
\[
J_{1/3} \left( \frac{2}{3} |\xi|^{3/2} \right), \quad J_{-1/3} \left( \frac{2}{3} |\xi|^{3/2} \right)
\]
在 \(\xi = 0\) 点能够与式 (3.7.12) 光滑地连接起来的解是 |
|
7. 如图5-3所示，一小物块以\(2 \, \text{m/s}\)的初速度沿曲面由\(A\)处下滑，到达较低的\(B\)点时速度恰好也是\(2 \, \text{m/s}\)，如果此小物块以\(1 \, \text{m/s}\)的初速度仍旧从\(A\)点下滑，则它到达\(B\)点处的速度（ ）
A. 大于\(1 \, \text{m/s}\).
B. 小于\(1 \, \text{m/s}\).
C. 等于\(1 \, \text{m/s}\).
D. 等于\(2 \, \text{m/s}\). |
|
用证明 (6) 式的同样方法有
\[
\frac{1}{2\pi i} \oint_{(a_r,+)} f(\zeta) \zeta^{-k-1} d\zeta &= \frac{1}{2\pi i} \oint_{(a_r,+)} G_r \left( \frac{1}{\zeta - a_r} \right) \zeta^{-k-1} d\zeta \\
&= -\frac{1}{2\pi i} \oint_{(0,+)} G_r \left( \frac{1}{\zeta - a_r} \right) \zeta^{-k-1} d\zeta \\
&= -\frac{1}{k!} \left[ \frac{d^k}{d\zeta^k} G_r \left( \frac{1}{\zeta - a_r} \right) \right]_{\zeta = 0},
\]
代入前式，并用 (7) 式，得 |
|
如图3-21所示，把一个可看作质点的金属块\(A\)，轻轻放到一块在水平地面上匀速向右运动的薄木板\(B\)上，刚放到木板上时，\(A\)可看成静止，且离木板左端距离\(d=0.64\text{m}\)。已知\(A\)与木板间的动摩擦数\(\mu=0.2\)，\(A\)放上后木板在外力作用下仍以原来的速度做匀速运动。试问：木板的速度\(v_0\)应满足什么条件，才可把木板从金属块\(A\)下抽出？取\(g=10\text{m/s}^2\)。 |
|
\[
C_L &= \frac{2}{V_\infty S} \int_{-b/2}^{b/2} \Gamma(y) \, dy \\
D_i &= \int_{-b/2}^{b/2} L'_i \alpha_i \, dy \\
C_{D_i} &= \frac{2}{V_\infty S} \int_{-b/2}^{b/2} \Gamma(y) \alpha_i(y) \, dy
\] |
|
\(-h/2\) 到 + \(h/2\)，其中 \(h\) 为极的厚度；而沿 \(x\) 和 \(y\) 积分时，积分域是整个板面。最后求出形变板的总自由能 \(F_{\mu \nu} = \int F dV\) 的形式：
\[
F_{\mu 1} = \frac{Eh^3}{24 \left(1 - \sigma^2 \right)} \iint \left\{ \left( \frac{\partial^2 \xi}{\partial x^2} + \frac{\partial^2 \xi}{\partial y^2} \right)^2 + 2 \left(1 - \sigma \right) \left[ \left( \frac{\partial^2 \xi}{\partial x \partial y} \right)^2 - \frac{\partial^2 \xi}{\partial x^2} \frac{\partial^2 \xi}{\partial y^2} \right] \right\} dx dy
\] |
|
Let \(\mathbb{R}^n\) be the dual space of \(\mathbb{R}^n\), then we have to show that \(\mathbb{R}^n\) is isomorphic to \(\mathbb{R}^n\). For this we define a function \(T: \mathbb{R}^n \to \mathbb{R}^n\) as \(T(f) = c_f\).
Where \(f \in \mathbb{R}^n\) and \(B = \{e_1, e_2, \ldots, e_n\}\) be a basis of \(\mathbb{R}^n\) and for each \(x \in \mathbb{R}^n\) we have \(x = \sum_{i=1}^n x_i e_i\) as well as\(f(x) = f\left(\sum_{i=1}^n x_i e_i\right) = \sum_{i=1}^n x_i f(e_i) = \sum_{i=1}^n x_i c_i\) where we take \(c_i = f(e_i)\) and \(c_f = (c_1, c_2, \ldots, c_n) \in \mathbb{R}^n\). |
|
\[
r_H \left( 1 - \frac{H}{K} \right) = sP, \quad P = cH.
\]
If we put
\[
h = \frac{H}{H^*}, \quad p = \frac{P}{P^*}, \quad \rho = \sqrt{\frac{r_H}{r_P}}, \quad k = \frac{K}{H^*}, \quad \tau = \sqrt{r_H r_P} t, \tag{3.8}
\]
Equations (3.7) become
\[
\frac{dh}{d\tau} &= \rho h \left( 1 - \frac{h}{k} \right) - \alpha p h, \\
\frac{dp}{d\tau} &= \frac{1}{\rho} p \left( 1 - \frac{p}{h} \right). \tag{3.9}
\] |
|
1）若\(p, q\) 中至少一个为非整数。
1）含无理数 \(\Rightarrow f(x)\) 非周期函数
2）皆有理数，及分 \(p = b_n = 9 - a_n = 9 - a_n = \left(a, b, a_2, b_2\in Z, a, h, a_3, b_3 \right)\) 至到
则 \(\frac{a_1}{b_1n}, \frac{a_2}{b_2n}\) 是整数 \(\Rightarrow \frac{1}{b_1n}.\frac{1}{a_1n}\) 是整数
最小 \(n =\frac{1}{|\text{cm}|(|b_n|, |b_0|)}\Rightarrow T = 2 |\text{cm}(b_n, b_0) \cdot \pi\)。 |
|
求 \[
\lim_{x \to 0} \frac{1-x^2-e^{-x^2}}{x \sin^3 2x}
\]
证：原式
\[
=\lim_{x \to 0} \frac{1-x^2-(1-x^2+\frac{1}{2}{x^4}+0(x^4))}{x(2x)^3} = \lim_{x \to 0} \frac{-\frac{1}{2}{x^4}+0(x^4)}{8x^4} = -\frac{1}{16}
\] |
|
which is a standard integral form
\[
w(\theta_o) = -\frac{\Gamma_0 \pi \sin n \theta_o}{2 \pi b \sin \theta_o} \quad \text{with } n=1 \tag{5.35}
\]
\[
w(\theta_o) = -\frac{\Gamma_0}{2b} \tag{5.36}
\]
\textbf{Downwash is constant for an Elliptical Lift Distribution} |
|
so \begin{align}
V = S(u) + T(u) = (a_1 + b_1) u_1 + \cdots + (a_k + b_k) u_k \\
+ a_{k+1} v_1 + \cdots + a_{k+l} v_l + b_{k+1} w_1 + \cdots + b_{k+m} w_m \in Span (B).
\end{align}
If \(R(S) \cap R(T) = \{0\}\), then a similar argument shows that
\(\{v_1, \ldots, v_l, w_1, \ldots, w_m\}\) spans \(R(S+T)\), where \(\{v_1, \ldots, v_l\}\) is a basis for \(R(S)\) and \(\{w_1, \ldots, w_m\}\) is a basis for \(R(T)\). 2.2 .11 (i) We have
\[
D(1) &= 0 = 0 \cdot 1 + 0 t \\
D(t) &= 1 = 1 \cdot 1 + 0 t \\
D(t^2) &= 2t = 0 \cdot 1 + 2 t
\] |
|
\[
\iint\limits_D f(x,y) \, d\sigma\qquad\qquad
\iint\limits_D f(x,y) \, dx \, dy\qquad\qquad \iint\limits_D d\sigma = D \, \text{面积}\]
\[
\iint\limits_D f(x, y) \, dx \, dy = \int_a^b \left[ \int_{\varphi_1(x)}^{\varphi_2(x)} f(x, y) \, dy \right] \, dx
= \int_a^b dx \int_{\varphi_1(x)}^{\varphi_2(x)} f(x, y) \, dy
\]
\[
\iint\limits_D f(x,y) \, d\sigma = \iint\limits_{D'} f(r \cos \theta, r \sin \theta) \, r \, dr \, d\theta
\] |
|
\textbf{Lemma 3.1.} \textit{There holds}
\[
\sum_{y \in V, y \ne x} W_s(x, y) \leq \frac{C_x \mu(x)}{s (1-s) \Gamma(1-s)} ,
\]
\textit{where the constant} \(C_x = \max \{ \mu(x) \max_{s \in [0,1]} | \partial_0 p(t, x, x, t) |, 1 \}\).
\textit{Proof}. For any fixed \(x \in V\), since the vertices set \(V\) is countably infinite, then we rewrite it as \(V = \{ x, y_1, y_2, \cdots, y_k, \cdots \}\). It follows from (2.4) that
\[
\sum_{y \in V, y \ne x} W_s(x, y) = \sum_{k=1}^{+\infty} W_s(x, y_k) = \frac{s \mu(x)}{\Gamma(1-s)} \sum_{k=1}^{+\infty} \int_0^{+\infty} p(t, x, y_k) \mu(y_k) t^{-1-s} dt.
\] |
|
The \textit{Wigner–Eckart Theorem} states that
\[
\left( j_A, m_A \left| \hat{V}_q \right| j_B, m_B \right) = \left( j_B, 1, m_B, q \left| j_B, 1, j_A, m_A \right. \right) \frac{\left\langle A \left| \left| \hat{V}_q \right| \right| B \right\rangle}{\sqrt{2j_A + 1}}. \tag{23.22}
\]
The matrix element in the right-hand side vanishes unless
\[
|j_A - j_B| = 1 \quad (E1 \text{ selection rule}). \tag{23.23}
\] |
|
\[
\int_{-\infty}^{\infty} e^{i(\omega_1 - \omega_2) t} \, dt = 2\pi \delta(\omega_1 - \omega_2)
\] |
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