华中科技大学2017年数学分析考研试题参考解答

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华中科技大学2017年数学分析考研试题参考解答

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1、 计算 $\displaystyle \lim_{x\to\infty}\left[x-x^2\ln \left(1+\frac{1}{x}\right)\right]$ . fl: 函数极限  

纸质资料/答疑/pdf1/pdf2 / $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} \mbox{原式}&=\lim_{t\to 0}\left[\frac{1}{t}-\frac{1}{t^2}\ln(1+t)\right] =\lim_{t\to 0}\frac{t-\ln(1+t)}{t^2} \overset{\tiny\mbox{Taylor 展开}}{=} \frac{1}{2}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

跟锦数学微信公众号. 在线资料/公众号/

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2、 计算 $\displaystyle \iint_S x\mathrm{ d} y\mathrm{ d} z+y\mathrm{ d} x\mathrm{ d} z+z\mathrm{ d} x\mathrm{ d} y$ , 其中 $\displaystyle S$ 为 $\displaystyle z=\sqrt{1-x^2-y^2}$ , 取上侧. fl:曲线曲面积分  

纸质资料/答疑/pdf1/pdf2 / 取 $\displaystyle \varSigma: x^2+y^2\leq 1, z=0$ , 取上侧, 则 $\displaystyle \iint_\varSigma\cdots=0$ , 而 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} \mbox{原式}=&\mbox{原式}-\iint_\varSigma \cdots \stackrel{\tiny\mbox{Gauss}}{=} \iiint_{x^2+y^2+z^2\leq 1\atop z\geq 0}3\mathrm{ d} x\mathrm{ d} y\mathrm{ d} z =3\cdot \frac{1}{2}\cdot \frac{4\pi}{3}=2\pi. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

跟锦数学微信公众号. 在线资料/公众号/

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3、 计算 $\displaystyle I=\int_0^\infty \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x}\mathrm{ d} x$ . fl:广义积分  

纸质资料/答疑/pdf1/pdf2 / $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} \int_0^\infty \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x}\mathrm{ d} x &=\int_0^\infty \int_a^b \mathrm{e}^{-xy}\mathrm{ d} y\mathrm{ d} x =\int_a^b \int_0^\infty \mathrm{e}^{-xy}\mathrm{ d} x\mathrm{ d} y\\ & =\int_a^b \frac{1}{y}\mathrm{ d} y =\ln \frac{b}{a}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

积分可以交换次序是因为 $\displaystyle \sup_{y\in [a,b]}\mathrm{e}^{-xy}=\mathrm{e}^{-ax}, \int_0^\infty \mathrm{e}^{-ax}\mathrm{ d} x$ 收敛及 Weierstrass 判别法蕴含 $\displaystyle \int_0^\infty \mathrm{e}^{-xy}\mathrm{ d} x$ 关于 $\displaystyle y\in [a,b]$ 一致收敛.  跟锦数学微信公众号. 在线资料/公众号/

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4、 求 $\displaystyle \sum_{n=0}^\infty \frac{x^{2n}}{2n+1}$ 的收敛域, 并求其和. fl:幂级数  

纸质资料/答疑/pdf1/pdf2 / 易知收敛域为 $\displaystyle [-1,1)$ . 当 $\displaystyle x\in [-1,0)\cup (0,1)$ 时, $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} \mbox{原式}&=\frac{1}{x}\sum_{n=0}^\infty \int_0^x t^{2n}\mathrm{ d} t =\frac{1}{x}\int_0^x \frac{1}{1-t^2}\mathrm{ d} t =\frac{1}{2x}\ln\frac{1+x}{1-x}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

故原式 $\displaystyle =\left\{\begin{array}{llllllllllll}\frac{1}{2x}\ln\frac{1+x}{1-x},&x\in [-1,0)\cup (0,1),\\ 0,&x=0.\end{array}\right.$ 跟锦数学微信公众号. 在线资料/公众号/

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5、 在 $\displaystyle [-\pi,\pi]$ 上, 将 $\displaystyle \arcsin (\cos x)$ 用 Fourier 级数展开, 并计算 $\displaystyle \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$ 的值. fl:Fourier级数  

纸质资料/答疑/pdf1/pdf2 /

(1)、 设 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} x\sim \frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx,\quad  x\in [0,\pi]. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

则 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} a_0=&\frac{2}{\pi}\int_0^\pi x\mathrm{ d} x=\pi,\\ a_n=&\frac{2}{\pi} \int_0^\pi x\cos nx\mathrm{ d} x =\frac{2}{\pi} \int_0^\pi x\mathrm{ d} \frac{\sin nx}{n}\\ =&-\frac{2}{n\pi}\int_0^\pi \sin nx\mathrm{ d} x =\frac{2[(-1)^n-1]}{\pi n^2}\quad (n\geq 1). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

于是 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} x=\frac{\pi}{2}-\frac{4}{\pi}\sum_{k=0}^\infty \frac{\cos(2k+1)\pi}{(2k+1)^2},\quad  x\in [0,\pi]. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

(2)、 $\displaystyle f(x)=\arcsin (\cos x)$ 是偶函数, 且 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} 0\leq x\leq \pi&\Rightarrow t\equiv \cos x=\sin\left(\frac{\pi}{2}-x\right) \Rightarrow f(x)=\arcsin t=\frac{\pi}{2}-x. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

故由第 1 步知 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} f(x)\sim\frac{4}{\pi}\sum_{k=0}^\infty \frac{\cos(2k+1)\pi}{(2k+1)^2}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

令 $\displaystyle x=0$ 得 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} \frac{\pi}{2}=\frac{4}{\pi}\sum_{k=0}^\infty \frac{1}{(2k+1)^2} \Rightarrow \sum_{k=1}^\infty \frac{1}{(2k+1)^2}=\frac{\pi^2}{8}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

跟锦数学微信公众号. 在线资料/公众号/

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6、 用确界定理证明单调有界数列必有极限. fl:实数理论  

纸质资料/答疑/pdf1/pdf2 / 设 $\displaystyle a_n$ 递增有上界, 则 $\displaystyle a=\sup a_n$ 存在. 由上确界定义, $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} \forall\ \varepsilon\gt 0,\exists\ N,\mathrm{ s.t.} a_N\gt a-\varepsilon. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

于是当 $\displaystyle n\geq N$ 时, $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} a-\varepsilon\lt a_N\leq a_n\lt a\lt a+\varepsilon\Rightarrow |a_n-a|\lt \varepsilon. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

这就证明了 $\displaystyle \lim_{n\to\infty}a_n$ 存在且等于 $\displaystyle a$ .  跟锦数学微信公众号. 在线资料/公众号/

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7、 已知 $\displaystyle \lim_{x\to a^+}f'(x)$ 存在, $\displaystyle f(x)$ 在 $\displaystyle (a,b)$ 内可微. 求证: $\displaystyle f(x)$ 在 $\displaystyle x=a$ 处右可导. fl: 微分  

纸质资料/答疑/pdf1/pdf2 / $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} f'_+(a)=\lim_{x\to a^+}\frac{f(x)-f(a)}{x-a} \overset{\tiny\mbox{Lagrange 中值}}{=} \lim_{x\to a^+}f'(\xi_x) \overset{\tiny\mbox{归结原理}}{=} \lim_{x\to a^+}f'(x). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

故 $\displaystyle f(x)$ 在 $\displaystyle x=a$ 处右可导.  跟锦数学微信公众号. 在线资料/公众号/

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8、 已知 $\displaystyle \sum_{n=0}^\infty a_n$ 条件收敛, 设 $\displaystyle a_n^+=\max\left\{a_n,0\right\}, a_n^-=\max\left\{-a_n,0\right\}$ . 证明: $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} \lim_{n\to\infty}\frac{\sum_{k=0}^n a_k^+}{\sum_{k=0}^n a_n^-}=1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

fl: 数项级数  

纸质资料/答疑/pdf1/pdf2 / 由题设, $\displaystyle \sum_{n=1}^\infty |a_n|=+\infty, \sum_{n=1}^\infty a_n$ 收敛. 于是 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} b_n\equiv \frac{\sum_{k=1}^n a_k}{\sum_{k=1}^n |a_k|} \Rightarrow \lim_{n\to\infty}b_n\overset{\tiny\mbox{有界乘以无穷小}}{=} 0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

进一步, $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} &\lim_{n\to\infty}\frac{\sum_{k=0}^n a_k^+}{\sum_{k=0}^n a_n^-} =\lim_{n\to\infty}\frac{2\sum_{k=0}^n \max\left\{a_k,0\right\}}{2\sum_{k=0}^n \max\left\{-a_k,0\right\}} =\lim_{n\to\infty}\frac{\sum_{k=0}^n (|a_k|+a_k)}{\sum_{k=0}^n (|a_k|-a_k)}\\ =&\lim_{n\to\infty}\frac{\sum_{k=0}^n |a_k|+\sum_{k=0}^n a_k}{\sum_{k=0}^n|a_k|-\sum_{k=0}^n a_k} =\lim_{n\to\infty}\frac{1+b_n}{1-b_n}=1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

跟锦数学微信公众号. 在线资料/公众号/

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9、 已知 $\displaystyle f(x,t)$ 在 $\displaystyle \mathbb{R}^2$ 上二阶可微, 且满足 $\displaystyle f_{xx}(x,t)=f_{tt}(x,t)$ . 设 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} E(t)=\frac{1}{2}\int_{t-1}^{1-t} \left[|f_x(x,t)|^2+|f_t(x,t)|^2\right]\mathrm{ d} x. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

证明: $\displaystyle E(t)$ 单调递减. fl: 微分  

纸质资料/答疑/pdf1/pdf2 / 由 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} &\int_{t-1}^{1-t}\left[f_x(x,t)f_{xt}(x,t) +f_t(x,t)f_{tt}(x,t)\right]\mathrm{ d} x\\ =&\int_{t-1}^{1-t}\left[f_x(x,t)f_{xt}(x,t) +f_t(x,t)f_{xx}(x,t)\right]\mathrm{ d} x\\ =&\int_{t-1}^{1-t}\left[f_x(x,t)f_t(x,t)\right]_x\mathrm{ d} x =f_x(x,t)f_t(x,t)|_{x=t-1}^{x=1-t} \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

知 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} E'(t)=&\frac{1}{2}\left[|f_x(1-t,t)|^2+|f_t(1-t,t)|^2\right]\cdot (-1)\\ &-\frac{1}{2}\left[|f_x(t-1,t)|^2+|f_t(t-1,t)|^2\right]\\ &+\int_{t-1}^{1-t}\left[f_x(x,t)f_{xt}(x,t) +f_t(x,t)f_{tt}(x,t)\right]\mathrm{ d} x\\ =&-\frac{1}{2}|f_x(1-t,t)-f_t(1-t,t)|^2 -\frac{1}{2}|f_x(t-1,t)+f_t(t-1,t)|^2|\leq 0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

而 $\displaystyle E(t)$ 递减.  跟锦数学微信公众号. 在线资料/公众号/

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10、 设 $\displaystyle D=\left\{(x,y)\in\mathbb{R}^2;x^2+y^2\leq 1\right\}$ , $\displaystyle f\in C^1(D)$ , 试证: $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} \iint_D |f(x,y)-f(0,0)|\mathrm{ d} x\mathrm{ d} y \leq \iint_D \frac{\sqrt{f_x^2(x,y)+f_y^2(x,y)}}{2\sqrt{x^2+y^2}}\mathrm{ d} x\mathrm{ d} y. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

fl: 积分法与不等式  

纸质资料/答疑/pdf1/pdf2 / 作极坐标变换 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} x=r\cos\theta,\quad  y=r\sin\theta, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

则由 Newton-Leibniz 公式及 Cauchy-Schwarz 不等式知 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} &\quad |f(r\cos\theta,r\sin\theta)-f(0,0)|\\ &=\left|\int_0^r \frac{\partial}{\partial t}f(t\cos\theta,t\sin\theta)\mathrm{ d} t\right|\\ & =\left|\int_0^r f_x(t\cos\theta,t\sin\theta)\cos\theta +f_y(t\cos\theta,t\sin\theta)\sin\theta\mathrm{ d} t\right|\\ &\leq \int_0^r \sqrt{f_x^2(t\cos\theta,t\sin\theta)+f_y^2(t\cos\theta,t\sin\theta)} \cdot \sqrt{\cos^2\theta+\sin^2\theta}\mathrm{ d} t\\ &=\int_0^r \sqrt{f_x^2(t\cos\theta,t\sin\theta)+f_y^2(t\cos\theta,t\sin\theta)} \mathrm{ d} t \equiv F_\theta(r). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

进一步, $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} &\quad \int_0^1 |f(r\cos\theta,r\sin\theta)-f(0,0)|r\mathrm{ d} r\\ &\leq \int_0^1 F_\theta(r)r\mathrm{ d} r =\int_0^1F_\theta(r)\mathrm{ d}\frac{r^2}{r} =F_\theta(r)\frac{r^2}{2}|_0^1 -\int_0^1 F_\theta'(r)\frac{r^2}{r}\mathrm{ d} r\\ &\leq \frac{F_\theta(1)}{2}\quad \left(F_\theta'(r) =\sqrt{f_x^2(r\cos\theta,r\sin\theta)+f_y^2(r\cos\theta,r\sin\theta)}\geq 0\right), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

$\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} &\quad \iint_D |f(x,y)-f(0,0)|\mathrm{ d} x\mathrm{ d} y\\ &=\int_0^{2\pi}\mathrm{ d} \theta \int_0^1 |f(r\cos\theta,r\sin\theta)-f(0,0)|r\mathrm{ d} r \leq \int_0^{2\pi}\frac{F_\theta(1)}{2}\mathrm{ d} \theta\\ &=\int_0^{2\pi} \frac{1}{2}\int_0^1 \sqrt{ f_x^2(t\cos\theta,t\sin\theta) +f_y^2(t\cos\theta,t\sin\theta) } \mathrm{ d} t \mathrm{ d} \theta\\ &=\int_0^{2\pi} \frac{1}{2}\int_0^1 \frac{\sqrt{ f_x^2(t\cos\theta,t\sin\theta) +f_y^2(t\cos\theta,t\sin\theta) }}{t}\cdot t \mathrm{ d} t \mathrm{ d} \theta\\ &=\iint_D \frac{\sqrt{f_x^2(x,y)+f_y^2(x,y)}}{2\sqrt{x^2+y^2}}\mathrm{ d} x\mathrm{ d} y. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

跟锦数学微信公众号. 在线资料/公众号/

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