一些数学题目的解答
函数、极限、连续
求极限 \displaystyle \lim_{n\to\infty} \left(\frac{1}{2}\tan\frac{\pi}{2^2} + \frac{1}{2^2}\tan\frac{\pi}{2^3} + \cdots + \frac{1}{2^n}\tan\frac{\pi}{2^{n+1}}\right)
解： 先积分后求导
\begin{aligned}
\int & \sum_{i=1}^n \frac{1}{2^i}\tan \frac{t}{2^{i}} dt\\
=& \sum_{i=1}^n \ln \sec \frac{t}{2^i} \\
=& \ln \prod_{i=1}^n \sec \frac{t}{2^i} \\
=& \ln \frac{1}{\prod\limits_{i=1}^n\cos \frac{t}{2^i}} \\
\end{aligned}
其中：
\begin{aligned}
& \prod_{i=1}^n\cos \frac{t}{2^i} \\
=& \frac{\sin \frac{t}{2^n} \prod\limits_{i=1}^n\cos \frac{t}{2^i}}{\sin \frac{t}{2^n}} \\
=& \frac{\sin \frac{t}{2^{n-1}} \prod\limits_{i=1}^{n-1}\cos \frac{t}{2^i}}{2\sin \frac{t}{2^n}} \\
=& \frac{\sin t}{2^n\sin \frac{t}{2^n}} \\
\end{aligned}
于是：
\begin{aligned}
& \sum_{i=1}^n \frac{1}{2^i}\tan \frac{t}{2^{i}} \\
& \left(\int \sum_{i=1}^n \frac{1}{2^i}\tan \frac{t}{2^{i}} dt\right)'\\
=& \left(\ln \frac{2^n\sin \frac{t}{2^n}}{\sin t}\right)' \\
=& \frac{\sin t}{2^n\sin \frac{t}{2^n}} \cdot \frac{2^n \cos\frac{t}{2^n} \cdot \frac{1}{2^n} \cdot \sin t - \cos t \cdot 2^n\sin \frac{t}{2^n} }{\sin^2 t} \\
=& \frac{\cos\frac{t}{2^n} \cdot \sin t - \cos t \cdot 2^n\sin \frac{t}{2^n} }{2^n\sin \frac{t}{2^n} \sin t } \\
=& \frac{\cos\frac{t}{2^n}}{2^n\sin \frac{t}{2^n} } - \frac{\cos t}{\sin t } \\
=& \frac{\cos\frac{t}{2^n}}{2^n\sin \frac{t}{2^n} } - \cot t \\
\end{aligned}
故：
\begin{aligned}
&\lim_{n\to\infty} \sum_{i=1}^n \frac{1}{2^i}\tan \frac{t}{2^{i + 1}} \\
=& \lim_{n\to\infty} \frac{\cos\frac{t}{2^n}}{2^n\sin \frac{t}{2^n} } - \cot t\\
=& \lim_{u\to 0} \frac{u\cos ut}{\sin ut } - \cot t\\
=& \frac{1}{t } - \cot t\\
\end{aligned}
即：
\begin{aligned}
I =& \lim_{n\to\infty} \sum_{i=1}^n \frac{1}{2^i}\tan \frac{\pi}{2^{i + 1}} = \frac{2}{\pi} - \cot \frac{\pi}{2} = \frac{2}{\pi}\\
\end{aligned}
微分中值定理
设 f(x) 在 [a, b] 上连续，在 (a, b) 内可导，其中 a, b 同号， f(a) = f(b) = 1；证明：存在 \xi, \eta \in (a, b)，使得 abe^{\eta-\xi} =\eta^2[f(\eta) - f'(\eta)].
证明：
已知 f(a) = f(b) = 1，a,b 同号
设 \displaystyle h(x) = \frac{1}{e^x}，则 \displaystyle h'(x) = -\frac{1}{e^x}，对 h(x) 在 (a,b) 上使用拉格朗日中值定理，则存在 \xi \in (a, b) 使得：
\begin{aligned}
\frac{h(b) - h(a)}{b - a} &= h'(\xi) \\
\frac{{e^{-b}} - e^{-a}}{b - a} &= -\frac{1}{e^\xi} \\
\frac{{e^{-b}} - e^{-a}}{a - b} &= e^{-\xi}\\
\end{aligned}
设 \displaystyle F(x) = \frac{f(x)}{e^x}，则 \displaystyle F'(x) = \frac{f'(x)e^x - e^xf(x)}{e^{2x}} = \frac{f'(x) - f(x)}{e^{x}},
设 \displaystyle G(x) = \frac{1}{x}，a,b 同号，故 G(x) 在 (a,b) 上连续可导，则 \displaystyle G'(x) = -\frac{1}{x^2} \neq 0，
对 F(x)，G(x) 在 (a, b) 上使用柯西中值定理，则存在 \eta \in (a, b) 使得：
\begin{aligned}
\frac{F(b) - F(a)}{G(b) - G(a)} &= \frac{F'(\eta)}{G'(\eta)}\\
\frac{\frac{f(b)}{e^b} - \frac{f(a)}{e^a}}{\frac{1}{b} - \frac{1}{a}} &= \frac{f'(\eta) - f(\eta)}{e^\eta} \cdot (-\eta^2)\\
ab\frac{e^{-b} - e^{-a}}{a - b} &=\eta^2 e^{-\eta}[f(\eta) - f'(\eta)]\\
\end{aligned}
又 \displaystyle\frac{{e^{-b}} - e^{-a}}{a - b} = e^{-\xi}，故：
\begin{aligned}
abe^{-\xi} &=\eta^2 e^{-\eta}[f(\eta) - f'(\eta)]\\
abe^{\eta-\xi} &=\eta^2[f(\eta) - f'(\eta)]\\
\end{aligned}
即存在 \xi, \eta \in (a, b)，使得 abe^{\eta-\xi} =\eta^2[f(\eta) - f'(\eta)]
证毕。
不定积分
麻省理工学院期末考试最难的不定积分题目：
\begin{aligned}
& \int \frac{x^4}{(\sqrt{1-x^2})^5} dx \\
\xlongequal{x=\sin t} & \int \frac{\sin^4 t}{(\sqrt{1-\sin^2 t})^5} \cos t dt \\
= & \int \frac{\sin^4 t}{\cos^5 t} \cos t dt \\
= & \int \tan^4 t dt \\
\xlongequal{u=\tan t} & \int \frac{u^4}{1 + u^2} du \\
= & \int \frac{u^4 + u^2 - u^2 + 1 - 1}{1 + u^2} du \\
= & \int u^2 - 1 + \frac{1}{1 + u^2} du \\
= & \frac{1}{3}u^3 - u + \arctan u + C \\
= & \frac{1}{3}\tan^3 t - \tan t + t + C \\
= & \frac{1}{3}\tan^3 \arcsin x - \tan \arcsin x + \arcsin x + C \\
= & \frac{1}{3}\frac{x^3}{(\sqrt{1 - x^2})^3} - \frac{x}{\sqrt{1-x^2}} + \arcsin x + C
\end{aligned}
空间第二型曲线积分
计算曲线积分 \displaystyle I=\oint_L ydx + zdy + xdz，其中 L 为球面 x^2 + y^2 + z^2 = a^2 与平面 x + y + z = 0 的交线，从 x 轴正项往 x 轴负项看去为逆时针方向。
解法一：
记 F(x, y, z) = x + y + z 则 F(x, y z) 的法向量为：
(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}) = (1, 1, 1)
单位法向量为 \displaystyle\frac{1}{\sqrt{3}}(1, 1, 1)
记 \Sigma 为平面 x + y + z = 0 与曲线 L 围成的区域，则由斯托克斯公式：
\begin{aligned}
& I=\oint_L ydx + zdy + xdz \\
= & \frac{1}{\sqrt{3}}\iint\limits_\Sigma
\begin{vmatrix}
1 & 1 & 1 \\
\frac{\partial}{\partial x} &
\frac{\partial}{\partial y} &
\frac{\partial}{\partial z} \\
y & z & x
\end{vmatrix} dS \\
= &\frac{1}{\sqrt{3}} \iint\limits_\Sigma{
\frac{\partial x}{\partial y} +
\frac{\partial y}{\partial z} +
\frac{\partial z}{\partial x} -
\frac{\partial y}{\partial y} -
\frac{\partial x}{\partial x} -
\frac{\partial z}{\partial z}
}dS \\
= &\frac{1}{\sqrt{3}} \iint\limits_\Sigma -3 dS \\
= & -\sqrt{3} \iint\limits_\Sigma dS\\
= & -\sqrt{3} \pi a^2
\end{aligned}
解法二：
将 z = -x - y 代入球 x^2 + y^2 + z^2 = a^2 得曲线在 xOy 面得投影为椭圆：
L_1:\begin{cases}
x^2 + y^2 + xy = \frac{a^2}{2} \\
z = 0 \\
\end{cases}
记投影曲线围成得区域为 D；
当 x^2 = y^2 时，x^2 + y^2 取得极值，易知椭圆 L_1 的长半轴为 \displaystyle l_1=a，短半轴为 \displaystyle l_2=\frac{a}{\sqrt{3}}，椭圆面积为：\displaystyle S = l_1l_2\pi = \frac{a^2}{\sqrt{3}}\pi
将 z = -x - y 代入积分得：
\begin{aligned}
& I=\oint_L ydx + zdy + xdz \\
=& \oint_L ydx + (-x - y)dy + x(- dx - dy) \\
=& \oint_L (y - x)dx - (2x + y)dy \\
\xlongequal{格林公式}& \iint\limits_D -2-1 dxdy \\
=& -3\iint\limits_D dxdy \\
=& -3 \frac{a^2}{\sqrt{3}} \pi \\
=& -\sqrt{3} a^2 \pi \\
\end{aligned}
随机变量及其分布
设随机变量 X，Y 相互独立，且都服从 G(p)
- 求 Z = \min\{X, Y\} 的数学期望
根据题意，设 q = 1-p，则 P\{X = k\} = P\{Y = k\} = q^{k-1}p
解法一：
\begin{aligned}
Z =& \min \{X, Y\} \\
P\{Z = k\} =& P\{\min \{X, Y\} = k\} \\
= & P\{X=k, Y \geqslant k \} + P\{Y=k, X > k \} \\
\xlongequal{独立} & P\{X=k\}P\{Y \geqslant k \} + P\{Y=k\}P\{X > k \} \\
= & P\{X=k\}(1 - P\{Y < k \}) + P\{Y=k\}(1 - P\{X \leqslant k \}) \\
= & P\{X=k\}(2 - P\{Y < k \} - P\{X \leqslant k \}) \\
= & q^{k-1}p\left(2 - \sum_{i=1}^{k-1} q^{i-1}p - \sum_{i = 1}^{k} q^{i-1}p\right) \\
= & 2q^{k-1}p - q^{k-1}p\left(\sum_{i=1}^{k-1} q^{i-1}p + \sum_{i = 1}^{k} q^{i-1}p\right) \\
= & 2q^{k-1}p - q^{k-1}p^2\left(\sum_{i=1}^{k -1} q^{i-1} + \sum_{i = 1}^{k} q^{i-1}\right) \\
= & 2q^{k-1}p - q^{k-1}p^2\left(\frac{1 - q^{k-1}}{1 - q} + \frac{1 - q^{k}}{1 - q}\right) \\
= & 2q^{k-1}p - q^{k-1}p^2\left(\frac{2 - q^{k-1} - q^k}{1 - q}\right) \\
\xlongequal{q = 1-p} & 2q^{k-1}p - q^{k-1}p(2 - q^{k-1} - q^k) \\
= & q^{k-1}p(q^{k-1} + q^k) \\
= & (q^{2k-1} + q^{2k-2})p \\
\end{aligned}
故 P\{Z = k\} = (q^{2k-1} + q^{2k-2})p
\begin{aligned}
E(Z) =& \sum_{k=1}^{\infty} k(q^{2k-1} + q^{2k-2})p \\
=& p\sum_{k=1}^{\infty} k(q^{2k-1} + q^{2k-2}) \\
=& p\sum_{k=1}^{\infty} kq^{2k-1} + p \sum_{k=1}^{\infty} kq^{2k-2} \\
=& pq\sum_{k=1}^{\infty} k(q^2)^{k-1} + p \sum_{k=1}^{\infty} k(q^2)^{k-1} \\
=& (pq + p)\sum_{k=1}^{\infty} k(q^2)^{k-1} \\
=& (pq + p)\left(\sum_{k=1}^{\infty}x^k\right)'\bigg|_{x=q^2}\\
=& (pq + p)\left(\frac{x}{1-x}\right)'\bigg|_{x=q^2}\\
=& (pq + p)\frac{1}{(1-x)^2}\bigg|_{x=q^2}\\
=& \frac{(pq + p)}{(1-q^2)^2}\\
\xlongequal{q=1-p}& \frac{1}{p(2 - p)}\\
\end{aligned}
解法二：
\begin{aligned}
F_X(k) =& P\{X \leqslant k\} \\
=& \sum_{i=0}^k q^{k-1}p \\
=& p\sum_{i=0}^k q^{k-1} \\
=& p\frac{1-q^k}{1-q} \\
\xlongequal{q=1-p}& 1-q^k \\
=& 1-(1-p)^k \\
\end{aligned}
\begin{aligned}
F_Z(k) =& 1 - [1 - F_X(k)]^2 \\
=& 1 - [1 - (1-q^k)]^2 \\
=& 1 - q^{2k} \\
\end{aligned}
\begin{aligned}
P\{Z = k\} =& P\{Z \leqslant k\} - P\{Z \leqslant k - 1\} \\
=& F_Z(k) - F_Z(k-1) \\
=& (1 - q^{2k}) - [1 - q^{2(k-1)}] \\
=& q^{2(k-1)} - q^{2k} \\
=& q^{2k-2}(1 - q^2) \\
\end{aligned}
\begin{aligned}
E(Z) =& \sum_{k=1}^{\infty} k[q^{2k-2}(1 - q^2)]\\
=& (1 - q^2)\sum_{k=1}^{\infty} kq^{2k-2}\\
=& (1 - q^2)\sum_{k=1}^{\infty} k(q^2)^{k-1}\\
=& (1 - q^2)\left(\sum_{k=1}^{\infty}x^k\right)'\bigg|_{x=q^2}\\
=& (1 - q^2)\left(\frac{x}{1-x}\right)'\bigg|_{x=q^2}\\
=& (1 - q^2)\frac{1}{(1-x)^2}\bigg|_{x=q^2}\\
=& \frac{(1 - q^2)}{(1-q^2)^2}\\
=& \frac{1}{(1-q^2)}\\
\xlongequal{q=1-p}& \frac{1}{p(2 - p)}\\
\end{aligned}
评论