武汉大学2019年数学分析真题
一，
1）求极限$\lim\limits _{x\rightarrow 0}\left( 1+\sin x\right) ^{\tfrac {1}{x}}.$
解:
$$由罗比达法则，\lim _{x\rightarrow 0}{\dfrac {\ln\left( 1+\sin x\right) }{x}}
=\lim _{x\rightarrow 0}{\dfrac {\cos x}{(1+\sin x)}} = 1$$
$$因此\lim _{x\rightarrow 0}\left( 1+\sin x\right) ^{\tfrac {1}{x}}
=e^{\lim _{x\rightarrow 0}{\dfrac {\ln\left( 1+\sin x\right) }{x}}}=e $$
2) $f(x) =\ln \left(\sqrt{1+x^2}-1\right ), $求$f(0)^{(2k+1)} ,k$为自然数.
解：$$f'(x) = {\dfrac{1}{\left( x+\sqrt{x^2+1}\right) }}\cdot(1+\dfrac{1}{\sqrt{x^2+1}} )
= \dfrac{1}{\sqrt{x^2+1}}$$
$$f^{(2)}(x) = ((x^2+1)^{-\tfrac{1}{2}})' = -(x^2+1)^{-\tfrac {3}{2}}\cdot x $$
$$由Leibniz公式，f^{(2k+1)}(x) = [(1+x^2)^{\tfrac{1}{2}} ]^{(2k)} = [-(1+x^2)^{\tfrac{3}{2}}\cdot x ]^{(2k-1)}
=(-1)[(1+x^2)^{-\tfrac{3}{2}}]^{2k-1}\cdot x + C^1_{2k-1}(-1)\cdot[(1+x^2)^{-\tfrac{3}{2}}]$$
$$= P_1\cdot x + (-1)^1\cdot (2k-1)\cdot 1\cdot [(x^2+1)^{-\tfrac{3}{2}}]^{(2k-2)}
= P_1\cdot x + (-1)^1\cdot (2k-1)\cdot 1\cdot[3(x^2+1)^{-\tfrac{5}{2}}\cdot x]^(2k-3)$$
$$\xrightarrow{对第二项再次使用Leibniz公式} = P_2\cdot x + (-1)^2\cdot(2k-1)\cdot(2k-3)\cdot 3\cdot 1
\cdot[(x^2+1)^{-\tfrac{5}{2}}]^{2k-4} $$
$$=\dotsm = P_i\cdot x + (-1)^i\cdot\dfrac{(2k-1)!!}{(2k-2i+1)!!}
\cdot(2i-1)!!\cdot[(x^2+1)^{-\tfrac{2i+1}{2}}]^{(2k-2i)} $$
$$=\dotsm = P_k\cdot x + (-1)^k\cdot [(2k-1)!!]^2\cdot [(x^2+1)^{-\tfrac{2k+1}{2}}]^{(0)} （其中P_i是关于(x^2+1)的多项式）$$
$$因此 f^{2k+1}(0) = (-1)^k\cdot[(2k-1)!!]^2 $$
3)$f(x,y) = x^yy^x$,求$f(x,y)$的全微分.
解：$$因\dfrac{\partial f}{\partial x}=x^{y-1}y^{x+1} + x^yy^x\ln y ;
\dfrac{\partial f}{\partial y}=x^{y+1}y^{x-1} + x^yy^x\ln x $$
$$均在非(0,0)点连续，因此f在非(0,0)点可微，df
= \dfrac{\partial f}{\partial x}dx + \dfrac{\partial f}{\partial y}dy $$
$$= (x^{y-1}y^{x+1} + x^yy^x\ln y)dx + (x^{y+1}y^{x-1} + x^yy^x\ln x)dy $$
$$在(0,0)点处，取任意方向\theta ,求其方向导数 f_{\theta}=\lim_{r\rightarrow0}
{\dfrac{f(r\cos\theta,r\sin\theta)-f(0,0)}{r}} $$
$$ = \lim_{r\rightarrow0}(er\cos\theta)^{\sin\theta}\cdot(r\sin\theta)^{r\cos\theta}
+(er\sin\theta)^{\cos\theta}\cdot(r\cos\theta)^{r\sin\theta}= 0 (当\theta = 0或\dfrac{\pi}{2})或1 (当\theta\neq 0或\dfrac{\pi}{2})$$$$ 因此f(x,y)在(0,0)处不可微$$
二，计算下面积分
1)$\int_{-1}^{1} {\dfrac{1+x^2}{1+x^4}}dx.$
解：$$当x\in (-1,1),\dfrac{1}{1+x^4} = \sum_{n=0}^{\infty}(-x^4)^n
，\dfrac{x^2}{1+x^4}=\sum_{n=0}^{\infty}(-1)^n\cdot x^{4n+2} $$
$$故，\dfrac{1+x^2}{1+x^4} = \sum_{n=0}^{\infty} (-1)^{n}(x^{2n}+x^{2n+2} )
=\sum_{n=0}^{\infty} (-1)^{[\tfrac{n}{2}]}x^{2n} $$
$$取极小正值\delta ，\int_{-1+\delta}^{1+\delta}{\dfrac{1+x^2}{1+x^4}}dx
= \int_{-1+\delta}^{1+\delta} \sum_{n=0}^{\infty} (-1)^{[\tfrac{n}{2}]}x^{2n}dx $$
$$因\sum_{n=0}^{\infty} (-1)^{[\tfrac{n}{2}]}x^{2n}dx 在(-1,1)上一致收敛
，故\int_{-1+\delta}^{1+\delta} \sum_{n=0}^{\infty} (-1)^{[\tfrac{n}{2}]}x^{2n}dx
= \sum_{n=0}^{\infty} (-1)^{[\tfrac{n}{2}]}\int_{-1+\delta}^{1+\delta} x^{2n}dx
= \sum_{n=0}^{\infty} (-1)^{[\tfrac{n}{2}]}\dfrac{2}{2n+1}(1-\delta)^{2n+1}$$
$$于是，I = \int_{-1}^{1} {\dfrac{1+x^2}{1+x^4}}dx
= \lim_{\delta\rightarrow0}\int_{-1+\delta}^{1+\delta}{\dfrac{1+x^2}{1+x^4}}dx
= \lim_{\delta\rightarrow0} \sum_{n=0}^{\infty} (-1)^{[\tfrac{n}{2}]}\dfrac{2}{2n+1}(1-\delta)^{2n+1}$$
$$ = \sum_{n=0}^{\infty} (-1)^{[\tfrac{n}{2}]}\dfrac{2}{2n+1}$$
2)$\iiint _{V} {\dfrac{dxdydz}{(1+x+y+z)^{3}}}$，$V=\{{x+y+z\leq{1}}, x,y,z\geq0 \}$
解：$$\iiint _{V} {\dfrac{dxdydz}{(1+x+y+z)^{3}}} = \int_0^1 dz\int_0^{1-z}dy\int_0^{1-z-y}
\dfrac {1}{(1+x+y+z)^3}=(-\dfrac{1}{2})\int_0^1 dz\int_0^{1-z}(\dfrac{1}{4}-\dfrac{1}{(1+y+z)^2})dy $$
$$=(-\dfrac{1}{2})\int_0^1(\dfrac{1-z}{4}+\dfrac{1}{2}-\dfrac{1}{1+z})dz
=\dfrac{\ln 2}{2}-\dfrac{5}{16} $$
3)$\int_L{\dfrac{xdy-ydx}{x^2+y^2}}$,$L$是不过原点的简单封闭曲线.
解：$$由于(\dfrac{x}{x^2+y^2})_x = \dfrac{y^2 - x^2}{(x^2+y^2)^2},
(\dfrac{-y}{x^2+y^2})_y = \dfrac{y^2 - x^2}{(x^2+y^2)^2} $$
$$ 设D为L所包围区域， 若(0,0)\in D，则由Green公式知所求积分与路径无关
\int_L{\dfrac{xdy-ydx}{x^2+y^2}} = 0；$$
$$ 若(0,0)\notin D，由于\dfrac{xdy-ydx}{x^2+y^2}在D上(0,0)处不连续，故积分不满足Green公式条件$$
$$故以(0,0)为圆心， 某极小正r为半径做圆O，则{L^{+}+O^{-}}上满足Green公式条件$$
$$ 则\int_L{\dfrac{xdy-ydx}{x^2+y^2}} = \int_{L^{+}+O^{-}}{\dfrac{xdy-ydx}{x^2+y^2}}
- \int_{O^{-}}{\dfrac{xdy-ydx}{x^2+y^2}} = \int_{O^{+}}{\dfrac{xdy-ydx}{x^2+y^2}} $$
$$令x=r\cos\theta; y=r\sin\theta， 则上式= \int_0^{2\pi}{\dfrac{r^2\cos^2\theta d\theta
+r^2\sin^2\theta d\theta}{r^2}}= \int_0^{2\pi}d\theta=2\pi $$
三，1）判断$\sum_{n=1}^{\infty}\left({\sqrt[n]{n}-1}\right)^2$的敛散.
解：$$设I_n=\left({\sqrt[n]{n}-1}\right)^2，则n = (\sqrt{I_n}+1)^{n}>1+C^4_{n}\cdot I_n^2
\Rightarrow I_n<(\dfrac {n-1}{C_n^4})^{\tfrac{1}{2}} \sim n^{-\tfrac{3}{2}}
$$
$$由于\sum_{n=0}^{\infty} n^{-\tfrac{3}{2}}收敛，由比较判别知\sum_{n=0}^{\infty}I_n收敛$$
2)若$\sum_1^{\infty}a_n\sin^nx$在[0,$2\pi$]收敛，请问它是否一致收敛?
解：$$因为\sum_1^{\infty}a_n\sin^nx在[0,2\pi]收敛，则 \sum_1^{\infty}a_n\sin^n(\dfrac{\pi}{2})
=\sum_1^{\infty}a_n收敛，即\sum_1^{\infty}a_n关于x\in[0,2\pi]一致收敛 $$
$$又因 \sin^nx 对于任意x\in[0,2\pi],关于n单调，且\left|\sin^nx \right| \le 1$$
$$由abel判别知，级数\sum_1^{\infty}a_n\sin^nx[0,2\pi]上一致收敛$$
四，1）$f(x)$连续可微，$f(0)$不为$0$，其Maclaurin级数（Cauchy余项）：
$$f(x) = f(0)+f^{'}(0)x+\dfrac{f^{(2)}(0)}{2!}x^2+...+\dfrac{f^{(n)}(0)}{n!}x^n+\dfrac{f^{(n+1)}(\theta x)}{n!}\left(1-\theta\right)^nx^{n+1}$$
证明:$\lim_{x\rightarrow0}\theta = 1-\sqrt [n]{\dfrac{1}{n+1}}$\\
证：$$ 令R_n=f(x) - \sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}\cdot x^k，则有 $$
$$R_n^{'} (x) = f'(x) - \sum_{k=1}^{n}\frac{f^(k)(0)}{(k-1)!}\cdot x^{k-1}$$
$$R_n^{(2)}(x) = f^{(2)}(x) - \sum_{k=2}^{n}\frac{f^(k)(0)}{(k-2)!}\cdot x^{k-2}$$
$$\dotsm$$
$$R_n^{(n)}(x) = f^{(n)}(x) - f^{(n)}(0)$$
$$R_n^{(n+1)}(x) = f^{(n+1)}(x)$$
$$故R_n^{'}(0) = R_n^{(2)}(0) = \dotsm = R_n^{(n)}(0)$$
$$则R_n(x)= \int_0^x R_n^{'} (t) dt= \int_0^x R_n^{'} (t)-R_n^{'} (0) d(t-x)
= -\int_0^x (t-x)R_n^{(2)}(t)dt = \dfrac{1}{2!}\int_0^x (x-t)^2 R_n^{(3)}(t)dt$$
$$=\dotsm = \dfrac{1}{n!}\int_0^x (x-t)^n R_n^{(n+1)}(t)dt
= \dfrac{1}{n!} \int_0^x (x-t)^n f^(n+1)(t)dt $$
由中值定理知,$$R_n(x) = \dfrac {f^{n+1}(\theta_1x)}{(n+1)!}\cdot x^{n+1}$$
因此$$\dfrac {f^{n+1}(\theta_1x)}{(n+1)!}\cdot x^{n+1} = \dfrac{f^{(n+1)}(\theta x)}{n!}\left(1-\theta\right)^nx^{n+1}$$
将上等式两边同时取$$x\rightarrow0 $$ 极限，即得题设所需结果。
2)$\{a_n\}严格单调递减，a_n\rightarrow0\left(当n\rightarrow0\right)，证明：
\sum_{n=1}^{\infty}a_n收敛\leftrightarrow\sum_{n=1}^{\infty}n\left(a_n-a_{n+1}\right)收敛$
证：$$必要性，由于0<\sum_{n=1}^{\infty}n\left(a_n-a_{n+1}\right) = \sum_{n=1}^{\infty}a_n - na_{n+1}
<\sum_{n=1}^{\infty}a_n (1)$$
$$显然，\sum_{n=1}^{\infty}n\left(a_n-a_{n+1}\right)收敛.$$
$$充分性，由于\{a_n\}严格单调递减，a_n\rightarrow0，因此a_n>0,且当m>n时，a_m<a_n$$
$$反证法，若\sum_{n=1}{\infty}a_n不收敛，则\exists p\in N^{+}，对任意n\in N^{+}，\sum_{k=n}^{n+p}a_n>\epsilon_0 (i)$$
$$由于\sum_{n=1}^{\infty}n\left(a_n-a_{n+1}\right)收敛，则\exists N_1 \in N^{+}，对N>N_1,
使\sum_{n=N}^{N+p}n\left(a_n-a_{n+1}\right) < \epsilon(\forall \epsilon > 0)$$
$$即，\sum_{n=N+1}{N+p}a_n+Na_N-(N+p)a_{N+p+1}<\epsilon$$
$$将(i)代入，得Na_N-(N+p)a_{N+p+1}<\epsilon-\epsilon_0,由\epsilon的任意性，知Na_N-(N+p)a_{N+p+1}<0
，即\dfrac{a_N}{a_{N+p+1}}<\dfrac{N+p}{N} (ii) $$
$$由于\exists N_2 \in N^{+},当n>N_2, \dfrac{n+p}{n}\rightarrow 1 $$
$$因此 对于N>max{N_1,N_2},由(ii)式知，\dfrac{a_N}{a_{N+p+1}}<1，与\{a_n\}递减矛盾，因此\sum_{n=1}^{\infty}a_n收敛$$
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