Anal 22-23Final
2022-2023数学分析(甲)I(H)期末练习题
\(一、计算题\)
\(1.\lim\limits_{x\to 1}(\dfrac{1}{x-1}-\dfrac{1}{lnx})\)
\(令t+1 = x,则t\to 0\)
\(原式 = \lim\limits_{t\to 0}(\dfrac{1}{t}-\dfrac{1}{ln(1+t)}) = \lim\limits_{t\to 0}\dfrac{ln(t+1)-t}{tln(1+t)} = \lim\limits_{t\to 0}\dfrac{-\dfrac{1}{2}t^2+o(t^2)}{t^2+o(t^2)} = -\dfrac{1}{2}.\)
\(2.\lim\limits_{n\to +\infty}\dfrac{1}{n}\sum\limits_{k = 1}^n\dfrac{cos(\frac{k}{n})}{1+sin(\frac{k}{n})}\)
\(由定积分的定义可得：原式 = \int_0^1\dfrac{cosx}{1+sinx}dx = ln(1+sinx)|^1_0 = ln(1+sin1).\)
\(3.\int\dfrac{ln(x+1)}{(x+2)^2}dx\)
\(\int\dfrac{ln(x+1)}{(x+2)^2}dx = \int ln(x+1)d(\dfrac{-1}{x+2})\)
\(= -\dfrac{ln(x+1)}{x+2}+\int\dfrac{1}{(x+2)(x+1)}dx\)
\(= -\dfrac{ln(x+1)}{x+2} +ln\dfrac{x+1}{x+2}+C (x>-1)\)
\(4.求y = \int^x_{-\sqrt{3}}\sqrt{3-t^2}dt在x\in[-\sqrt{3},\sqrt{3}]上的弧长\)
\(y' = \sqrt{3-x^2},弧微分ds = \sqrt{1+(y')^2} = \sqrt{4-x^2}dx\)
\(s = \int_{-\sqrt{3}}^{\sqrt{3}}\sqrt{4-x^2}dx = (x\sqrt{4-x^2}+4arcsin\dfrac{x}{2})|_0^{\sqrt{3}} = \sqrt{3}+\dfrac{4\pi}{3}\)
注：可以画图求面积求这个定积分，更容易一点
\(5.\int_0^{+\infty}e^{-x}cosxdx\)
\(先求\int e^{-x}cosxdx = e^{-x}\dfrac{1}{2}(sinx-cosx)(略去C)\)
\(所以原式 = e^{-x}\dfrac{1}{2}(sinx-cosx)|^{+\infty}_0 = -\dfrac{1}{2}\)
\(二、叙述确界原理，并利用确界原理证明：\)
\(若有界函数f(x)在(0,1)上单调递增，则极限\lim\limits_{x\to 1^-}f(x)存在\)
\(确界原理：非空有界数集必存在确界\)
\(单侧极限存在的证明：设E = \{f(x)|x\in(0,1)\}\)
\(易知E有上界f(1),且E非空，所以E存在上确界A.\)
\(由上确界的性质，\forall \epsilon>0,存在x_0\in(0,1),f(x_0)>A-\epsilon\)
\(则对\forall\epsilon>0,\exists\delta = 1-x_0>0,当1-\delta<x<1即x_0<x<1时，由f(x)的单调性有：\)
\(\Rightarrow |f(x)-A|<\epsilon，所以单侧极限存在\)
注：2020期末第一题.
\(三、已知g(x)有二阶连续导数，g(0) = 1,g'(0) = 0,且f(x) = \begin{cases} \dfrac{g(x)-cosx}{x}(x\neq 0)\\a(x = 0) \end{cases}\)
\(1.若f(x)在x = 0处连续，求a\)
\(\lim\limits_{x\to 0}f(x) = \lim\limits_{x\to 0}\dfrac{g'(x)+sinx}{1} = 0 = a\)
\(2.已知f(x)在x = 0处连续，讨论f'(x)在x = 0处的连续性\)
\(f'(0) = \lim\limits_{x\to 0}\dfrac{f(x)-f(0)}{x-0} = \lim\limits_{x\to 0}\dfrac{g(x)-cosx}{x^2} = \dfrac{g''(0)+1}{2}\)
\(x \neq 0时，f'(x) = \dfrac{x(g'(x)+sinx)-g(x)+cosx}{x^2}\)
\(\lim\limits_{x\to 0}f'(x) = \lim\limits_{x\to 0}\dfrac{x(g''(x)+cosx)}{2x} = \dfrac{g''(0)+1}{2} = f'(0)\)
\(因此f'(x)在x = 0处连续\)
\(四、叙述函数f(x)在区间I上一致连续的定义，并证明f(x) = x^{\frac{1}{2023}}在[0,+\infty)上一致连续\)
\(一致连续的定义是：\forall\epsilon>0,\exists\delta>0,\forall x_1,x_2\in I,只要|x_1-x_2|<\delta,就有|f(x_1)-f(x_2)|<\epsilon\)
\(先证：\forall x,y\ge 0,x^{\frac{1}{2023}}+y^{\frac{1}{2023}}>(x+y)^{\frac{1}{2023}}\)
\(这是显然的，因为两边同时2023次方就可以得到。\)
\(所以可以得到：x^{\frac{1}{2023}}-y^{\frac{1}{2023}}<|x-y|^{\frac{1}{2023}}\)
\(\forall \epsilon>0,取\delta = \epsilon^{2023},|x_1,x_2|<\delta时，|f(x_1)-f(x_2)|<|x_1-x_2|^{\frac{1}{2023}}<\epsilon\)
注：此题也可以用Cantor定理和Lagrange中值定理证出。
\(五、已知连续的非常值函数f(x)满足:\lim\limits_{x\to+\infty}f(x)=f(0),证明：f(x)在[0,+\infty)上有最大值或者最小值。\)
\(因为f(x)非常值，不妨f(1)>f(0),由\lim\limits_{x\to+\infty}f(x)=f(0),得：\forall\epsilon>0,\exists X>1,\forall x>X,有|f(x)-f(0)|<\epsilon\)
\(特别地，取\epsilon = f(1)-f(0),则f(x)-f(0)<f(1)-f(0)\)
\(\Rightarrow f(x)<f(1)(x>X)\)
\(考虑f(x)在闭区间[0,X]上连续，所以一定存在最大值M = f(x_0)\)
\(x\in[0,X],M\ge f(x)\)
\(x\in[X,+\infty),M\ge f(1)\ge f(x)\)
\(显然M为f(x)在[0,+\infty)上的最大值。\)
\(注：若f(1)<f(0),类似可证f(x)存在最小值，所以可以"不妨".\)
注：这种用极限定义来约束无穷处函数状态的手法还是很常见的。
\(六、叙述闭区间套定理，并利用闭区间套定理证明闭区间上的连续函数的零点存在性定理：\)
\(f(x)在[a,b]上连续，且f(a)f(b)<0,则\exists x\in(a,b)满足f(c) = 0.\)
\(闭区间套定理:\)
\(如果一系列闭区间满足：[a_{n+1},b_{n+1}]\subset [a_n,b_n],且\lim\limits_{n\to +\infty}(b_n-a_n) = 0,则存在唯一的实数\xi属于所有闭区间[a_n,b_n],且\xi = \lim\limits_{n\to +\infty}a_n = \lim\limits_{n\to +\infty}b_n\)
\(对f(x)进行如下操作，设a_0 = a,b_0 = b，不妨f(a)<0,f(b)>0.\)
\(设f(a_i)f(b_i)<0,对于\dfrac{a_i+b_i}{2},有以下三种情况：\)
\(①f(\dfrac{a_i+b_i}{2}) = 0,则找到c\)
\(②f(\dfrac{a_i+b_i}{2})>0,则令b_{i+1} = \dfrac{a_i+b_i}{2},a_{i+1} = a_i\)
\(③f(\dfrac{a_i+b_i}{2})<0,则令a_{i+1} = \dfrac{a_i+b_i}{2},b_{i+1} = b_i\)
\(对i=0,1,...,n进行如上操作，如果某一次找到c，则得证，如果均没有找到\)
\(则得到一个闭区间套，b_n-a_n = \dfrac{b-a}{2^n}\to 0(n\to\infty)\)
\(且f(a_n)f(b_n)<0,令n\to \infty,由闭区间套定理和f(x)的连续性可以知道：f'(\xi)\le 0\)
\(则只能f(\xi) = 0,\xi\in (a,b)即所寻找的c\).
注：书本\(\bm{P99T5}\)
\(七、已知f(x)在(-1,2)上二阶连续可导，且f'(\dfrac{1}{2}) = 0,证明：\)
\(在x = \dfrac{1}{2}处应用带Lagrange余项的Taylor展开式将f(x)展开到2阶\)
\(f(x) = f(\dfrac{1}{2})+\dfrac{f''(\xi)}{2}(x-\dfrac{1}{2})^2\)
\(代入x = 0，x = 1\)
\(f(0) = f(\dfrac{1}{2})+\dfrac{f''(\xi_1)}{8}\)
\(f(1) = f(\dfrac{1}{2})+\dfrac{f''(\xi_2)}{8}\)
\(两式作差：f''(\xi_2)-f''(\xi_1) = 8{f(1)-f(0)}\)
\(\Rightarrow 8|f(1)-f(0)| = |f''(\xi_1)-f''(\xi_2)|\le|f''(\xi_1)|+|f''{(\xi_2)}|\le 2|f''(\xi)|\)
\(其中f''(\xi) = max\{f''(\xi_1),f''(\xi_2)\}\)
\(即证得：|f''(\xi)|\ge 4|f(1)-f(0)|.\)
\(八、已知f(x)在[0,1]上二阶连续可导，证明：\)
\(\int_0^1x^nf(x)dx = \dfrac{1}{n+1}\int^1_0f(x)d(x^{n+1})\)
\(=\dfrac{f(1)}{n+1}-\dfrac{1}{n+1}\int_0^1x^{n+1}f'(x)dx\)
\(=\dfrac{f(1)}{n+1}-\dfrac{f'(1)}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)}\int_0^1x^{n+2}f''(x)dx\)
\(=\dfrac{f(1)}{n+1}-\dfrac{f'(1)}{(n+1)(n+2)}+\dfrac{f''(\xi)}{(n+1)(n+2)(n+3)},\xi\in(0,1).(此处用积分第一中值定理)\)
\(注意此时得到的式子只是形式上和结果有差异，故整理成题目中结论所要求的形式\)
\(记\int^1_0x^nf(x)dx = I，当n\to \infty时：\)
\(nI\to f(1)\)
\((n^2I-nf(1))\to -f(1)-f'(1)\)
\(且I最后一项分母最高次数为n^3\)
\(因此：\int^1_0x^nf(x)dx = \dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o(\dfrac{1}{n^2})\)
注：题目中的"二阶连续可导"提示用泰勒展开式也是可以的。
\(f(x) = f(1)+f'(1)(x-1)+\dfrac{f''(\xi)}{2}(x-1)^2\)
