Anal 21-22(H)
2021-2022数学分析(甲)I(H)期末练习题
\(1.叙述数列收敛的柯西收敛原理，并用柯西收敛原理证明a_n = \sum\limits^{n}_{k=1}\dfrac{sink}{k(k+1)}收敛\)
\(柯西收敛原理：\forall\epsilon>0,\exists N,\forall 对n,m>N都有|a_n-a_m|<\epsilon\)
\(\forall\epsilon>0,取N > \dfrac{1}{\epsilon},则m+1>N >\dfrac{1}{\epsilon}\)
\(不妨n>m>N,|a_n - a_m| = |\dfrac{sin(m+1)}{(m+1)(m+2)}+...+\dfrac{sinn}{n(n+1)}|<|\dfrac{1}{m+1}-\dfrac{1}{n+1}|<\epsilon\)
\(所以a_n收敛\)
\(2.计算题\)
\((1)\lim\limits_{x\to+\infty}\dfrac{\int_a^x(1+u^4)^{\frac{
1}{4}}du}{x^3}\)
\(由洛必达法则：原式 = \lim\limits_{x\to+\infty}\dfrac{(1+x^4)^{\frac{1}{4}}}{3x^2} = 0\)
\((2)\int\dfrac{3x^2-4x-1}{x^3-2x^2-x+2}dx\)
\(原式 = ln|x^3-2x^2-x+2|+C\)
\(注：观察，凑微分\)
\((3)f(x) = \begin{cases} \dfrac{sinx}{x},x\neq0 \\ 1,x = 0 \end{cases}求f'(0),f''(0).\)
\(f'(0) = \lim\limits_{x\to 0}\dfrac{f(x)-f(0)}{x-0} = \lim\limits_{x\to 0}\dfrac{sinx - x}{x^2} = 0\)
\(x\neq 0时，f'(x) = \dfrac{xcosx - sinx}{x^2}\)
\(f''(0) = \dfrac{f'(x) - f'(0)}{x-0} = \lim\limits_{x\to 0}\dfrac{xcosx - sinx}{x^3} = \lim\limits_{x\to 0}\dfrac{x(cosx-1)+x-sinx}{x^3} = -\dfrac{1}{2} + \dfrac{1}{6} = -\dfrac{1}{3}\)
\(注：导数定义，常见泰勒展开式\)
\((4)设曲线C:y = e^x,过(0,0)向C作切线L，求L，C，y轴所围成图形绕x轴旋转一周形成旋转体的体积\)
\(易知切点(1,e)，作图得V = \int^1_0\pi e^{2x}dx-\dfrac{1}{3}\pi e^2 = \dfrac{\pi e^2}{6} - \dfrac{\pi}{2}\)
\(3.证明：若\{x_n\}无上界但并非正无穷大，求证：\{x_n\}存在一个发散到正无穷大的子列\)
\(无上界：对\forall M,均\exists n,使得x_n>M\)
\(并非正无穷大：\exists G>0,对于\forall N,存在n_0>N,使得x_{n_0}\le G\)
\(取M_1 = 1,则存在n_1,满足x_{n_1}>1\)
\(取M_2 = 2,存在n_2满足x_{n_2}>2\)
\(并且，一定存在n_2>n_1满足上述要求。否则如果找不到n_2>n_1,那么对于\forall n>n_1,都有x_n\le M_2则\{x_n\}在(n_1,+\infty)上有界，矛盾！\)
\(类似地，取M_i = i,存在n_i>n_{i-1}满足x_i>i\)
\(由此得到子列\{x_{n_k}\}满足x_{n_k}>k显然发散到正无穷.\)
\(\bm{注：课本P59T10}\)
\(4.设f(x)在[0,1]上有界，证明:\sup\limits_{x\in[0,1]}f(x)-\inf\limits_{x\in[0,1]}f(x) = \sup\limits_{x',x''\in[0,1]}|f(x')-f(x'')|\)
\(令\sup\limits_{x\in[0,1]}f(x) = M，\inf\limits_{x\in[0,1]}f(x) = m,由上下界的定义：\forall x_1,x_2\in(0,1),M\ge f(x_1),f(x_2)\ge m\)
\(从而得到M-m\ge|f(x_1)-f(x_2)|,这就证明了M-m是上界\)
\(又由确界的定义，对\forall\epsilon>0,存在x_1满足f(x_1)>M-\epsilon,存在x_2满足f(x_2)<m+\epsilon\)
\(f(x_1)-f(x_2) < M-m-2\epsilon，这就证明了上确界\)
\(注：利用上确界的两点定义即可。\)
\(5.f(x)\in D[0,1],且\lim\limits_{x\to 0^+}f'(x)存在，求证\lim\limits_{x\to 0^+}f(x)存在\)
\(极限存在的性质：局部有界性.存在\delta,M满足：0<x<\delta时，|f'(x)|\le M\)
\(由柯西收敛准则：\forall\epsilon>0,\exists\delta = \dfrac{\epsilon}{M},\forall x_1,x_2\in(0,\delta)都有|f(x_1)-f(x_2)| = |f'(\xi)(x_1-x_2)|\le M|x_1-x_2|\le M\delta = \epsilon\)
\(注：联想极限存在的性质，证明极限存在的方法\)
\(当笼统证明极限存在而没说多少的时候，多用柯西收敛原理\)
\(6.f(x)在(0,+\infty)上一致连续，g(x)在[0,+\infty)上连续，且\lim\limits_{x\to+\infty}[f(x)-g(x)] = 0,求证：g(x)在(0,+\infty)上一致连续\)
\(由f(x)在(0,+\infty)上一致连续\Rightarrow\forall\epsilon>0,\exists\delta>0,对于任意的x_1,x_2\in(0,+\infty),只要|x_1-x_2|<\delta,就有|f(x_1)-f(x_2)|<\epsilon\)
\(\lim\limits_{x\to+\infty}[f(x)-g(x)] = 0\Rightarrow\forall\epsilon>0,\exists X，\forall x_1,x_2>X,都有|f(x_1)-g(x_1)-[f(x_2)-g(x_2)]|<\epsilon\)
\(先考虑g(x)在[X,+\infty)的一致连续性，\forall\epsilon>0,\exists\delta>0,对于任意的x_1,x_2\in[X,+\infty),只要|x_1-x_2|<\delta,就有|g(x_1)-g(x_2)| < |f(x_1)-g(x_1)-[f(x_2)-g(x_2)]|+|f(x_1)-f(x_2)|<2\epsilon\)
\(又由Cantor定理，g(x)在[0,X+1]上连续且一致连续，综上g(x)在[0,+\infty)上一致连续\)
\(\bm{注：课本P99T8，T15}\)
\(7.f(x) = \begin{cases} \dfrac{sinx}{x},x>0 \\ 1,x = 0 \end{cases},求证\int_0^{+\infty}f(x)dx,\int_0^{+\infty}f^2(x)dx均收敛，并证明二者相等.\)
\(y = sinx的变上限积分有界，y = \dfrac{1}{x}单调递减且在无穷处的极限为0，由狄利克雷判别法，\int^{+\infty}_0f(x)dx收敛\)
\(f^2(x)\le\dfrac{1}{x^2},由比较判别法，\int_0^{+\infty}f^2(x)dx收敛。\)
\(\int_0^{+\infty}\dfrac{sin^2x}{x^2}dx = \int_0^{+\infty}sin^2xd(\dfrac{-1}{x}) = \dfrac{-sin^2x}{x}|^{+\infty}_0-\int_0^{+\infty}\dfrac{-1}{x}d(sin^2x) = \int_0^{+\infty}\dfrac{sin2x}{x}dx = \int_0^{+\infty}\dfrac{sin2x}{2x}d(2x) = \int_0^{+\infty}\dfrac{sinx}{x}dx.\)
\(8.f(x)三阶可导，f(0) = f'(0) = f''(0) = 0,f'''(0)>0,0<f(x)<1,选取x_1\in(0,1),按递推关系x_{n+1} = x_n(1-f(x_n))得到数列\{x_n\}\)
\((1)求证：\lim\limits_{n\to+\infty}x_n = 0.\)
\((2)求证:存在非零常数c和正数\alpha,满足\lim\limits_{n\to+\infty}cn^\alpha x_n = 1\)
\((1)用数学归纳法证明x_n>0:x_1>0,假设x_k>0,x_{k+1} = x_k(1-f(x_k))>0，成立\)
\(由递推关系：x_{n+1}-x_n = -x_nf(x_n) <0，所以x_n递减\)
\(由单调有界准则，x_n的极限存在，设为A，在递推式两边令n\to+\infty\)
\(A = A(1-f(A))\Rightarrow Af(A) = 0\)
\(易知0 \le A <1,若A = 0，则上式显然成立。\)
\(若0<A<1,则Af(A)不可能为0，综上：\lim\limits_{n\to+\infty}x_n = 0\)
\((2)先考虑求\lim\limits_{n\to+\infty}nx_n^k = \lim\limits_{n\to+\infty}\dfrac{n}{\frac{1}{x_n^k}} = \lim\limits_{n\to+\infty}\dfrac{1}{\frac{1}{x_{n+1}^k}-\frac{1}{x_{n}^k}} = \lim\limits_{n\to+\infty}\dfrac{x_n^k[x_n(1-f(x_n))]^k}{x_n^k-[x_n(1-f(x_n))]^k} = \lim\limits_{n\to+\infty}\dfrac{x_n^k(1-f(x_n))^k}{1-(1-f(x_n))^k}\)
\(由于n\to+\infty,x_n\to 0,f(x_n)\to 0.利用(1+x)^\alpha在0处展开式有：\)
\(\lim\limits_{n\to+\infty}\dfrac{x_n^k(1-f(x_n))^k}{1-(1-f(x_n))^k} = \lim\limits_{t\to 0}\dfrac{t^k(1-kf(t))}{kf(t)} = \lim\limits_{t\to 0}(\dfrac{t^k}{kf(t)}-t^k)\)
\(\lim\limits_{t\to 0}\dfrac{t^k}{kf(t)} = \lim\limits_{t\to 0}\dfrac{(k-1)(k-2)t^{k-3}}{f'''(t)}\)
\(要满足题意，必有k = 3,此时nx_n^3\sim \dfrac{2}{f'''(0)}即x_n\sim (\dfrac{2}{f'''(0)})^\frac{1}{3}n^{-\frac{1}{3}}\)
\(所以\alpha = \dfrac{1}{3},c = (\dfrac{f'''(0)}{2})^{-\frac{1}{3}}\)
\(\bm{注：同类题课本P184T9(5.4)}\)
