一些数学题目的解答

CreateTime 2020-11-27
UpdateTime 2020-12-04

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

f(x)[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) = 1a,b 同号

\displaystyle h(x) = \frac{1}{e^x},则 \displaystyle h'(x) = -\frac{1}{e^x},对 h(x)(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}

\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}},

\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

F(x)G(x)(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}

\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:

F(x, y, z) = x + y + zF(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)

\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:

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

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

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 XY 相互独立,且都服从 G(p)

  • 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}

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}