2021-2022春夏学期数学分析甲II(H)期末练习题
\(1.叙述函数列一致收敛的定义，并用定义证明函数列\{\dfrac{sinnx}{n^2}\}在R上一致收敛\)
\((课本P53)设\{f_n(x)\}是定义在I上的一系列函数，f(x)是I上的一个函数，若\forall\varepsilon>0,\exists N,使得\forall n>N,\forall x\in I,|f_n(x)-f(x)|<\varepsilon,则称函数列\{f_n(x)\}一致收敛到f(x)\)
\(\forall\varepsilon>0,\exists N = [\dfrac{1}{\varepsilon}]+1,\forall n>N,\forall x\in R,均有：\)
\(故\{\dfrac{sinnx}{n^2}\}一致收敛到0.\)
\(2.证明函数f(x) = \begin{cases} \dfrac{|xy|}{\sqrt{x^2+y^2}},(x,y)\neq 0 \\ 0,(x,y)=(0,0)\end{cases}在(0,0)处连续，存在偏导数，但不可微.\)
\(证明：先证连续性\)
\(|\dfrac{|xy|}{\sqrt{x^2+y^2}}|\le\dfrac{\dfrac{x^2+y^2}{2}}{\sqrt{x^2+y^2}}\to 0 = f(0,0),(x,y)\to (0,0)\)
\(偏导数：f'_x(0,0) = \lim\limits_{\Delta x\to0}\dfrac{f(\Delta x,0)-f(0,0)}{\Delta x} = 0,同理f'_y(0,0) = 0\)
\(不可微：\lim\limits_{(x,y)\to(0,0)}\dfrac{f(x,y)-0-0-0}{\sqrt{x^2+y^2}} = \lim\limits_{(x,y)\to(0,0)}\dfrac{|xy|}{x^2+y^2}极限不存在\)
\(3.说明e^{x+y+1}+x^2y=e在(0,0)的某邻域内唯一确定y关于x的函数y = f(x)，并求f'(0),f''(0).\)
\(记F(x,y) = e^{x+y+1}+x^2y-e,有：F(0,0) = 0,F(x,y),F'_y(x,y)在B_r(0,0)内连续\)
\(F'_y(x,y) = e^{x+y+1}+x^2\Rightarrow F'_y(0,0) = e\neq 0\)
\(因此F(x,y) = 0在B_r(0,0)内唯一确定隐函数y = f(x)\)
\(e^{x+y+1}+x^2y=e对x求导：\)
\(有y'|_{x=0} = -1.上述方程对x再求一次导：\)
\(f''(0) = 0\)
\(4.计算以下积分：\)
\((1)I= \iiint_Vz^2\sqrt{x^2+y^2+z^2}dxdydz,V为x^2+y^2+z^2\le R^2区域\)
\(换元x = rsin\varphi cos\theta,y = rsin\varphi sin\theta,z = rcos\varphi,\varphi\in[0,\pi],\theta\in[0,2\pi],r\in[0,R]\)
\(I = \iiint_\Omega r^2cos^2\varphi·r·r^2sin\varphi drd\varphi d\theta = \int_0^{2\pi}d\theta\int_0^\pi cos^2\varphi sin\varphi d\varphi\int_0^Rr^5dr = \dfrac{2\pi R^6}{9}\)
\((2).I = \int_L(z-y)dx+(x-z)dy+(y-x)dz,L为x^2+y^2 = 1与x-y+z = 2的交线.方向从z正方向看逆时针\)
\(由Stokes公式：I = \iint_{S}2dydz+2dzdx+2dxdy,S即截得的椭圆面(上侧)\)
\(参数化:z = 2-x+y,(x,y)\in D,\vec{n} = (1,-1,1)\)
\(I = \iint_D2dxdy = 2·\pi·1^2 = 2\pi\)
\(注：参数化x = cost,y=sint,z=2+sint-cost,t\in[0,2\pi]\)
\((3)I = \int_L e^x(1-cosy)dx-e^x(1-siny)dy,L即沿着y = sinx从(0,0)到(\pi,0)\)
\(记A(0,0),B(\pi,0),设\partial D = L \cup\vec{BA}.\)
\(由Green公式：\int_{\partial D}e^x(1-cosy)dx+e^x(siny-1)dy = \iint_D(e^x(siny-1)-e^xsiny)dxdy = \iint_D-e^xdxdy = -\int_0^\pi e^xdx\int_0^{sinx}dy = -\dfrac{e^\pi+1}{2}\)
\(I = (\int_L+\int_{\vec{BA}})e^x(1-cosy)dx-e^x(1-siny)dy\)
\(BA:y = 0,x\in[0,\pi]\)
\(\int_{\vec{BA}}e^x(1-cosy)dx-e^x(1-siny)dy = 0\)
\(因此I = -\dfrac{e^\pi+1}{2}\)
\((4)I = \iint_S2xydydz+2yzdzdx+(z-2yz-z^2+1)dxdy,S为上半球面z = \sqrt{1-x^2-y^2}\)
\(设圆盘D = \{(x,y)|x^2+y^2\le 1\},\partial\Omega = S(上侧)\cup D(下侧)
\)
\(由Gauss公式：\iint_{\partial\Omega}2xydydz+2yzdzdx+(z-2yz-z^2+1)dxdy = \iiint_{\Omega}(2y+2z+1-2y-2z)dxdydz = \dfrac{2}{3}\pi\)
\(\iint_{D下}2xydydz+2yzdzdx+(z-2yz-z^2+1)dxdy = -\iint_D(z-2yz-z^2+1)dxdy = -\iint_Ddxdy = -\pi\)
\(\therefore I = \dfrac{5}{3}\pi\)
\(5.求f(x,y) = xy+x-y在圆盘x^2+y^2\le5上的最大值和最小值\)
\(x^2+y^2<5时，令\dfrac{\partial f}{\partial x} = 0,\dfrac{\partial f}{\partial y} = 0\Rightarrow(x,y) = (1,-1)为唯一驻点，f(1,-1) = 1\)
\(x^2+y^2 = 5时，令x = \sqrt{5}cos\theta,y = \sqrt{5}sin\theta,\theta\in[0,2\pi]\)
\(f = 5sin\theta cos\theta+\sqrt{5}(cos\theta-sin\theta) = g(\theta)\)
\(一方面，g = \dfrac{5}{2}sin2\theta-\sqrt{10}sin(\theta-\dfrac{\pi}{4})\ge-\dfrac{5}{2}-\sqrt{10} = g(\dfrac{3\pi}{4})\)
\(另一方面，令g'(\theta) = 0\Rightarrow(sin\theta+cos\theta)(sin\theta-cos\theta-\dfrac{1}{\sqrt{5}}) = 0\)
\(容易验证sin\theta-cos\theta = \dfrac{1}{\sqrt{5}}时g有最大值3，此时(x,y) = (-1,-2)\)
\(法二：在x^2+y^2=5上运用拉格朗日乘子法，略\)
\(法三：最大值亦可这样求：注意到(x-y)^2 = x^2+y^2-2xy,令x-y = a,xy = b,问题转化为：\)
\(当a^2+2b = 5时，求a+b的最值。。。二次函数\)
\(6.求幂级数\sum\limits_{n=0}^{\infty}\dfrac{x^n}{3^n(n+1)}的收敛半径，收敛域以及和函数\)
\(易求得R = 3，收敛域[-3,3)，令f(x) = \sum\limits_{n=0}^{\infty}\dfrac{x^n}{3^n(n+1)},[xf(x)]' = \sum\limits_{n=0}^{\infty}(\dfrac{x}{3})^n = \dfrac{1}{1-\frac{x}{3}}\)
\(f(0) = 1,x\neq0时，f(x) = \dfrac{-3ln(1-\frac{x}{3})}{x}\)
\(7.f(x) = \dfrac{1}{4}x(2\pi-x),T_f = 2\pi,求f(x)的傅里叶级数展开式，并证明\sum\limits_{n=0}^\infty\dfrac{1}{n^2} = \dfrac{\pi^2}{6}\)
\(a_0 = \dfrac{1}{\pi}\int_0^{2\pi}f(x)dx = \dfrac{1}{3}\pi^2\)
\(a_n = \dfrac{1}{\pi}\int_0^{2\pi}f(x)cosnxdx = -\dfrac{1}{n^2}\)
\(b_n = \dfrac{1}{\pi}\int_0^{2\pi}f(x)cosnxdx = 0\)
\(故f(x) = \dfrac{\pi^2}{6}-\sum\limits_{n=0}^\infty\dfrac{cosnx}{n^2},令x = 0即证明\sum\limits_{n=0}^\infty\dfrac{1}{n^2} = \dfrac{\pi^2}{6}\)
\(8.已知f(x)\in C(-\infty,+\infty),证明函数列f_n(x) = \dfrac{1}{n}\sum\limits_{k=1}^nf(x+\dfrac{k}{n})在任意闭区间[a,b]上一致收敛\)
\(证：极限函数f(x) = \int_0^1f(x+t)dt\)
\(|f_n(x)-f(x)| = |\sum_{k=1}^n\int_{\frac{k-1}{n}}^{\frac{k}{n}}[f(x+\dfrac{k}{n})-f(x+t)]dt|\le\sum_{k=1}^n\int_{\frac{k-1}{n}}^{\frac{k}{n}}|f(x+\dfrac{k}{n})-f(x+t)|dt\)
\(因为f(x)连续，所以在闭区间[a,b+1]上一致连续\)
\(所以\forall\varepsilon>0,\exists\delta>0,\forall x_1,x_2:|x_1-x_2|<\delta,都有|f(x_1)-f(x_2)|<\varepsilon\)
\(所以，\forall\varepsilon>0,\exists N = [\dfrac{1}{\delta}],\forall n>N,\forall x\in R,有：\)
\(|x+\dfrac{k}{n}-(x+t)|<\delta,故|f_n(x)-f(x)|<n·\dfrac{\varepsilon}{n} = \varepsilon\)
\(即证f_n(x)一致收敛到f(x)\)