一些数学分析考研题（2024年）的解析

证明\[\because \lim_{n\to\infty }a_n=a.\]\[\therefore \forall \varepsilon > 0,\exists N_1\in \mathbb{N},n> N_1,s.t.a-\varepsilon < a_n< a+\varepsilon .\]同理，\[\forall \varepsilon > 0,\exists N_2\in \mathbb{N},n> N_2,s.t.b-\varepsilon < b_n< b+\varepsilon .\]令\[N=\max \{N_1.N_2\},\]由\c_n=\frac{a_1b_N+a_2b_{N-1}+\cdots +a_Nb_1}{n}+\frac{a_{N+1}b_{Nn}+\cdots +a_{n}b_{N+1}}{n},\]而\[\frac{a_1b_N+a_2b_{N-1}+\cdots +a_Nb_1}{n}+\frac{n(a-\varepsilon )(b-\varepsilon )}{n}< c_n< \frac{a_1b_N+a_2b_{N-1}+\cdots +a_Nb_1}{n}+\frac{n(a+\varepsilon )(b+\varepsilon )}{n}.\]\[\Rightarrow \lim_{n\to\infty }c_n=ab.\]

证明 由泰勒公式：\[f(x+h)=f(x)+f'(x)h+\frac{1}{2}f''(\xi )h^2,\xi \in (x,x+h)\]\[f(x-h)f(x)-f'(x)h+\frac{1}{2}f''(\eta )h^2,\eta \in (x-h,x)\]两式相减，得\[f(x+h)-f(x-h)=2M_1h+\frac{1}{2}f''(\xi )h^2-\frac{1}{2}f''(\eta )h^2,\]\[|2f'(x)h|\le |f(x+h)-f(x-h)|+\frac{1}{2}|f''(\xi )-f''(\eta )|h^2\le 2M_0+M_2h^2.\]\[M_1\le \frac{2M_0}{2h}+\frac{M_2h^2}{2h}=\frac{M_0}{h}+\frac{M_2h}{2},\]令\[h=2\sqrt{\frac{M_0}{M_2}},\]即有\[M_1^2\le 2M_0M_2.\]

证明  $ f(x)在(a,b)上一致收敛\Rightarrow 。$\[\forall \varepsilon > 0,\exists n> \mathrm{N}\in \mathbb{N},p> 0\delta > 0,0< |x_{n+p}-x_n|< \delta ,\]\[|f(x_{n+p})-f(x_{n})|< \varepsilon .(\{f(x_n)\}\subseteq \{f(x)\})\]即函数值列$\{f(x_n)\}也是cauchy列.$
       $ f(x)在(a,b)上一致收敛\Leftarrow 。$\[\forall \varepsilon > 0,\forall x_n.x_m\in (a,b),\exists n,m> \mathrm{N_1}\in \mathbb{N},\delta > 0,\]\[\forall x_n,x_m\in (a,b),0< |x_n-x_m|< \delta ,s.t.\]\[|f(x_n)-f(x_m)|< \varepsilon .\]
       $对于点x_n,x_m的邻域\cup ^0(x_n,\frac{\delta }{2}),\cup ^0(x_m,\frac{\delta }{2}),分别存在收敛于x_n和x_m的子列\{x_{n_k}\} 和\{x_{m_l}\},$\[\exists N_2\in \mathbb{N},n_k,m_l> N_2,|x_n-x_{n_k}|<\frac{\delta }{3} ,|x_m-x_{m_l}|<\frac{\delta }{3},|x_{n_k}-x_{m_l}|<\frac{\delta }{3} .\]\[由于\{f(x_n)\}为cauchy列，故必有|f(x_{n_k})-f(x_{m_l})|< \frac{\varepsilon }{3}.\]\[而子列\{x_{n_k}\} 和\{x_{m_l}\}分别收敛于x_n和x_m，\]\[所以有|f(x_n)-f(x_{n_k})|< \frac{\varepsilon }{3}，|f(x_m)-f(x_{m_l})|< \frac{\varepsilon }{3}.\]
        $令N=\max \{N_1,N_2\}.f(x_n),f(x_m)\in \{f(x_n)\}此时$\[|f(x_n)-f(x_m)|< |f(x_n)-f(x_{n_k})|+|f(x_{n_k})-f(x_{m_l})|+|f(x_m)-f(x_{m_l})|< \frac{\varepsilon }{3}+\frac{\varepsilon }{3}+\frac{\varepsilon }{3}=\varepsilon .\]也即\[f(x)在(a,b)上存在cauchy列，f(x)在(a,b)上一致收敛.\]

证明  令\[F(x)=\frac{1}{2}[\int_{0}^{x}f(t)dt]^2.\]则\[F'(x)=\varphi (x)=f(x)\int_{0}^{x}f(t)dt.\]显然有\[F'(0)=\varphi (0)=0.\]又\[\because \varphi (x)\downarrow ，\]\[\therefore x\in (-\infty ,+\infty ）,0\le F'(x)\le F'(0)=0.\]从而必有\[\int_{0}^{x}f(t)dt=0，\Rightarrow f(x)\equiv 0.(可以证明)\]

解  \[\because |\displaystyle\sum_{n=1}^{\infty }\frac{\sgn(x-x_n)}{2^n}|\le \displaystyle\sum_{n=1}^{\infty }|\frac{\sgn(x-x_n)}{2^n}|\le \displaystyle\sum_{n=1}^{\infty }\frac{1}{2^n}< \infty .\]\[f(x)在（0，1）上一致收敛.\]令\[u_n(x)=\frac{\sgn(x-x_n)}{2^n}.\]\[\because \lim_{x\to x_n^+}=\frac{|x-x_n|}{2^n}=0=\lim_{x\to x_n^-}=-\frac{|x-x_n|}{2^n}.\]\[\therefore u_n(x)在x=x_n时，连续.\]因此\[f(x)\in C(-\infty,+\infty).\]

解（1）、\[\because I(r)=\int_{0}^{+\infty }re^{-rx}dx=e^{-rx}|_0^{+\infty }=-1.收敛.\]\[\lim_{x\to+\infty }e^{-rx}=0.\]\[\limsup_{x\to+\infty } |e^{-ax}-0|=0.\]\[\therefore I(r)在[a,b]上一致收敛.（0<a<b）\]
   （2）、由于\[r\in [0,b],\because I(0)=\int_{0}^{+\infty }re^{-rx}dx=\int_{0}^{+\infty }e^{-rx}d(rx)=\infty.\]\[\therefore I(r)在[0,b]上非一致收敛.\]

解（1）、Abel公式：\[\displaystyle\sum_{k=1}^{n}a_nb_n=\displaystyle\sum_{k=1}^{n-1}A_k(b_k-b_{lk+1})+A_nb_n.\]其中\[A_k=\textstyle\sum_{k=1}^{n}a_n.\]
（2）、由\[\displaystyle\sum_{n=1}^{\infty }(b_{n+1}-b_n)\le |\displaystyle\sum_{n=1}^{\infty }(b_{n+1}-b_n)|< \infty .\]所以\[\displaystyle\sum_{n=1}^{\infty }(b_{n+1}-b_n)=\lim_{n\to\infty }\displaystyle\sum_{k=1}^{n }(b_{n+1}-b_n)=\lim_{n\to\infty }(b_{n+1}-b_1)=\displaystyle\sum_{n=1}^{\infty }b_{n+1}-b_1< \infty .\]
  （3）、设\[\displaystyle\sum_{n=1}^{\infty }a_n=A,\displaystyle\sum_{n=1}^{\infty }b_n=B.\]\[\begin{align*}\therefore  \lim_{n\to\infty }\displaystyle\sum_{k=1}^{n}a_nb_n &=\lim_{n\to\infty }[\displaystyle\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})+A_nb_n \\
&\le M\lim_{n\to\infty }\displaystyle\sum_{k=1}^{n-1}(b_k-b_{k+1})+Ac,(A_k\le M有界,c=\lim_{n\to\infty }b_n有界) \\
&\le Mb_1-Mc+Ac \\
&< \infty .
\end{align*}\]

证明
           致密性定理也称威尔斯特拉斯(Weierstrass)聚点定理：有界数列必含有收敛子列。
           由条件知：\[R_f=\{f(x)|x\in [a,b]\}有界,f(x)连续.\]\[\because R_f=\{f(x)|x\in [a,b]\}有界,有上下确界.\]记\[\alpha =\inf R_f，\beta =\sup R_f.\]上下确界也是$R_f$的最大、小值.
           由上下确界的定义\[\forall x\in [a,b],f(x)\geqslant \alpha .\forall \varepsilon > 0,\exists x\in [a,b],f(x)< \alpha +\varepsilon .\] \[对于[a,b]存在有界\{x_n\}.由致密性定理，\{x_n\}一定存在收敛子列\{x_｛n_k｝\},\lim_{n_k\to\infty }x_{n_k}=c.\]\[对于x_{n_k}\in [a,b],\alpha \le f(x_{n_k})< \alpha +\varepsilon .\]\[取\varepsilon =\frac{1}{n_k}，则当n_k\to\infty 时，有f(c)=\alpha .即f(x)能取到最小值.\]\[同样的方法，可证f(x)能取到最大值.\]

解 对$x+y+z=e^z,$隐函数求偏导.\[1+\frac{\partial z}{\partial x}=e^z\frac{\partial z}{\partial x}.\frac{\partial z}{\partial x}=\frac{1}{e^z-1}.\]\[1+\frac{\partial z}{\partial y}=e^z\frac{\partial z}{\partial y}.\frac{\partial z}{\partial y}=\frac{1}{e^z-1}.\]\[\frac{\partial^2z}{\partial x \partial y}=e^z\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}+e^z\frac{\partial^2z}{\partial x \partial y}.\]\[\frac{\partial^2z}{\partial x \partial y}=\frac{e^z}{1-e^z}\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}=\frac{e^z}{(1-e^z)^3}.\]

解 \[\lim_{n\to\infty }\displaystyle\sum_{k=1}^{n}\frac{n}{n^2+k^2}\sin \frac{k}{n}=\lim_{n\to\infty }\frac{1}{n}\displaystyle\sum_{k=1}^{n}\frac{1}{1+(\frac{k}{n})^2}\sin \frac{k}{n}.\]\[\because \frac{1}{1+\frac{1}{n}}< \frac{1}{1+(\frac{k}{n})^2}< \frac{1}{1+\frac{1}{n^2}}.\]\[\therefore \lim_{n\to\infty }\frac{1}{n}\displaystyle\sum_{k=1}^{n}\frac{1}{1+(\frac{k}{n})^2}\sin \frac{k}{n}=\lim_{n\to\infty }\frac{1}{n}\displaystyle\sum_{k=1}^{n}\sin \frac{k}{n}=\int_{0}^{1}\sin xdx=1-\cos 1.\]