1.数学分析，王昆扬，北京师范大学，http://video.chaoxing.com/play_400001575_18227.shtml

2.高等代数，丘维生，北京大学，http://video.chaoxing.com/serie_400015565.shtml

3.高等代数，李尚志，北京航天航空大学，http://video.chaoxing.com/serie_400001055.shtml

4.实变函数，上海交大，http://video.1kejian.com/university/ggkc/12481/或者http://www.dxzy163.com/view/index6750.html后者包括一些泛函分析

5.抽象代数，邓少强，南开大学http://www.icourses.cn/jpk/viewCharacterDetail.action?courseId=5968§ionId=100992

6.抽象代数，章璞，上海交大，http://video.chaoxing.com/serie_400008049.shtml

7.泛函分析.国立台湾大学，http://video.1kejian.com/video/?68356-0-0.html

8.高等代数,杜现昆,吉林大学,http://www.icourses.cn/coursestatic/course_3399.html

9.常微分方程,史少云,吉林大学,http://www.icourses.cn/coursestatic/course_6531.html

10.数学物理方法,吴崇试,北京大学,http://www.icourses.cn/coursestatic/course_3569.html

11.偏微分方程,朱长江,华中师范大学,http://www.icourses.cn/coursestatic/course_4147.html

12.复变函数,刘太顺,湖州师范学院,http://www.icourses.cn/coursestatic/course_3950.html

13.泛函分析,孙炯,内蒙古大学,http://www.icourses.cn/coursestatic/course_7021.html

14.高等代数,林亚南,厦门大学,http://www.icourses.cn/coursestatic/course_3077.html

国立交通大学开放式课程：http://ocw.nctu.edu.tw/index.php

15.台湾大学陈金次老师高等微积分http://ocw.aca.ntu.edu.tw/ntu-ocw/index.php/ocw/cou/101S130/2/V/1?v=ntu

豆瓣视频推荐http://www.douban.com/group/topic/50463538/

善科数学视频：http://www.mysanco.cn/index.php?class=video&action=video_collection#tab9

爱课程网http://www.icourses.cn/home/

1.Integrals of the Ising class (D.H. Bailey J.M. Borwein R.E.Crandall)

2.Hypergeometric forms for Ising-class integrals (D.H. Bailey, D. Borwein, J.M. Borwein,R.E. Crandall)

3. Finding General Explicit Formulas for Ising Integral Recursions (D.H. Bailey J.M. Borwein)

4.On Recurrences for Ising Integrals (Johannes Kepler University Linz, Austria)

1.THREE TRIPLE INTEGRALS (G. N. WATSON)

2.WATSON'S THIRD INTEGRAL (Hannah Cairns)

3.ON THE EVALUATION OF GENERALIZED WATSON INTEGRALS (G. S. JOYCE AND I. J. ZUCKER)

1.Box integrals (D.H. Bailey J.M. Borwein R.E. Crandall)

2.Higher-dimensional box integrals (Jonathan M. Borwein O-Yeat Chan y R. E. Crandall)

3.ADVANCES IN THE THEORY OF BOX INTEGRALS (D. H. BAILEY, J. M. BORWEIN, AND R. E. CRANDALL)

1.Euler Sums and Contour Integral Representations (Philippe Flajolet and Bruno Salvy)

2.Experimental evaluation of Euler sums (D.H.Bailey J.M.Borwein andR.Girgensohn)

3.Evaluation of triple euler sums (Jonathan M. Borwein)

4.Harmonic sums,Mellin transforms and Integrals (J.A.M.Vermaseren,NIKHEF)

1.EllipticK 范例

在三维立点阵中随机访问并返回原点的概率：

1 - \[Pi]^2/

72 (6 + 2 Sqrt[3] + Sqrt[6]) EllipticK[

35 + 24 Sqrt[2] - 20 Sqrt[3] - 14 Sqrt[6]]^-2 // N

测试程序

BlockRandom[SeedRandom[11]; 
Count[Table[walkerPosition = {0, 0, 0}; steps = 0; 
While[steps == 0 || (steps < 100 && walkerPosition =!= {0, 0, 0}), 
steps++; 
walkerPosition = 
walkerPosition + {{1, 0, 0}, {-1, 0, 0}, {0, 1, 0}, {0, -1, 
0}, {0, 0, 1}, {0, 0, -1}}[[Random[Integer, {1, 6}]]]]; 
steps, {1000}], _?(# < 100 &)]]

2.Beta

贝塔函数倒数的$n\times n$ 矩阵的行列式为$n!$：

9月5号逆神在数学竞赛交流群里给了一份试题，建模结束后自己才真正仔细思考起来，经过各位大神的指教，终于能够把所有试题的答案给补全，难免存在错误，联系2609480070@qq.cm进行纠正.

数学分析练习题

来源：逆蝶

整理：1729

2014年9月5日

下面的习题均来自大学生数学竞赛群群友逆蝶提供的数学分析练习题，难免存在错误，请联系（2609480070）进行纠正！

1.已知$a_1=a_2=1,a_{n+2}=2a_{n+1}+3a_n,n=1,2,\ldots$，求幂级数$\sum\limits_{n=1}^\infty{a_n x^n}$的收敛半径，收敛域以及和函数.

解.\begin{align*}&{a_{n + 2}} + {a_{n + 1}} = 3\left( {{a_{n + 1}} + {a_n}} \right) \\\Rightarrow &{a_{n + 1}} + {a_n} = \left( {{a_2} + {a_1}} \right) \times {3^{n - 1}} = 2 \times {3^{n - 1}}\\&{a_{n + 1}} - \frac{1}{2} \times {3^n} = \left( { - 1} \right)\left( {{a_n} - \frac{1}{2} \times {3^{n - 1}}} \right) \\\Rightarrow &{a_n} - \frac{1}{2} \times {3^{n - 1}} = \frac{1}{2}{\left( { - 1} \right)^{n - 1}}.\end{align*}

即

\begin{align*}\sum\limits_{n = 1}^\infty {{a_n}{x^n}} &= \frac{1}{2}\sum\limits_{n = 1}^\infty {\left( {{3^{n - 1}} + {{\left( { - 1} \right)}^{n - 1}}} \right){x^n}} \\&= \frac{1}{6}\sum\limits_{n = 1}^\infty {{{\left( {3x} \right)}^n}} - \frac{1}{2}\sum\limits_{n = 1}^\infty {{{\left( { - x} \right)}^n}} \\&= \frac{1}{2}\frac{x}{{1 - 3x}} + \frac{1}{2}\frac{x}{{1 + x}} = \frac{{x\left( {1 - x} \right)}}{{\left( {1 + x} \right)\left( {1 - 3x} \right)}}.\end{align*}

由\[\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{3^n} + {{\left( { - 1} \right)}^n}}}{{{3^{n - 1}} + {{\left( { - 1} \right)}^{n - 1}}}} = 3\]

知幂级数收敛半径为$\frac13$，收敛域为$\left(-\frac13,\frac13\right)$.

2.计算级数\[\sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n}} \right)} \]的和.

解.由Stolz公式

\[\mathop {\lim }\limits_{n \to \infty } \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 2}} = 0.\]

我们有

\begin{align*}&\sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n}} \right)} \\&= \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}\left[ {\left( {n + 1} \right)\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n}} \right) - n\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n} + \frac{1}{{n + 1}}} \right) + \frac{n}{{n + 1}}} \right]} \\&= \sum\limits_{n = 1}^\infty {\left( {\frac{{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{n} - \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}}} \right)} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^2}}}} \\&= 1 - \mathop {\lim }\limits_{n \to \infty } \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}} + \frac{{{\pi ^2}}}{6} - 1 = \frac{{{\pi ^2}}}{6}.\end{align*}

3.设$\alpha$是实数，计算\[\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {x^\alpha }} \right)}}} .\]

解.令$x=\tan t$，我们知

\begin{align*}\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {x^\alpha }} \right)}}} &= \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} \\&= \frac{1}{2}\left( {\int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} + \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} } \right)\\&= \frac{\pi }{4}.\end{align*}

4.设$f(x)$在$[0,\pi]$连续，求证：不能同时有\[\int_0^\pi {{{\left| {f\left( x \right) - \sin x} \right|}^2}dx} < \frac{\pi }{4},\int_0^\pi {{{\left| {f\left( x \right) - \cos x} \right|}^2}dx} < \frac{\pi }{4}.\]又问何时上面的两个不等式成为等式？

证.注意到

\[a^2+b^2\geq \frac{(a-b)^2}{2}.\]

我们有

\[\int_0^\pi {{{\left| {f\left( x \right) - \sin x} \right|}^2}dx} + \int_0^\pi {{{\left| {f\left( x \right) - \cos x} \right|}^2}dx} \ge \int_0^\pi {\frac{{{{\left( {\sin x - \cos x} \right)}^2}}}{2}dx} = \frac{\pi }{2}.\]

由抽屉原理知题给两式不能同时成立.由取等条件知当且仅当

\[f\left( x \right) - \sin x = \cos x - f\left( x \right) \Rightarrow f\left( x \right) = \frac{{\sin x + \cos x}}{2}\]

时取等成立.

5.设$f(x)$在$[0,\infty]$上有$n+1$阶连续导函数，且$f(0)\geq 0,f'(0)\geq0,\ldots,f^{(n)}(0)\geq0.$又对任意$x>0$，有$f(x)\leq f^{(n+1)}(x)$.求证：$f(x)\geq0$.

证.1. （gg）若$f(x)$在$x=0$的某个领域$(0,\xi)$内，满足$f(x)>0$.不妨设存在某个$x>0$，有$f(x)<0$，则此时由连续性，存在某个$x_1>0$，使得$f(x_1)=0$.当$x\in(0,x_1)$时，有$f(x)>0$，则$f^{n+1}(x)\geq f(x)>0$.易推得$f(x)$在$(0,x_1)$上为增函数，$f(x_1)>0$，故此时假设不成立；

2. 若$f(x)$在$x=0$的某个邻域$(0,\xi)$内，满足$f(x)<0$，下证矛盾.构造$g\left( x \right) = \sum\limits_{k = 0}^n {{f^{\left( k \right)}}\left( x \right)} /{e^x}$，则$g(x)$为增函数，所以$g\left( x \right) \ge g\left( 0 \right) \ge 0 \Rightarrow \sum\limits_{k = 0}^n {{f^{\left( k \right)}}\left( x \right)} \ge 0$.由于$f(x)$在$x=0$的某个邻域$(0,\xi)$内，满足$f(x)<0$，则必存在某个$\xi_1$，使得当$x\in(0,\xi_1)$时，有$f(x)<0$.对$k=1,2,\ldots,n$，均存在$\xi_k>0$，使得当$x\in(0,\xi_k)$时，使得$f^{(k)}(x)<0$.取$\eta=\min\{\xi_1,\xi_2,\ldots,\xi_n\}$.当$x\in(0,\eta)$时，有$\sum\limits_{k = 0}^n {{f^{\left( k \right)}}\left( x \right)}<0$，矛盾；

3. 若$f(x)$在$x=0$的某个邻域$(0,\xi)$内，满足$f(x)=0$，且$f(x)$不恒为0，易知此时可推得$(0,\xi)$内，有$f^{(k)}(x)=0,k=1,2,3,\ldots,n$.可转化为在$x=\xi$为初始点的情况，这时我们可采用类似1.,2.的讨论；

4. 若$f(x)=0$，则命题得证.

综上，命题成立.

------------------------------------------

（逆蝶）对于$x\in(0,1]$，由Taylor公式，存在$x_1\in(0,1)$使\[f\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} + \frac{{{f^{\left( {n + 1} \right)}}\left( {{x_1}} \right)}}{{\left( {n + 1} \right)!}}{x^{n + 1}}.\]根据条件得\[f\left( x \right) \ge f\left( {{x_1}} \right)\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}}.\]同样将$f(x_1)$展开，可得$x_2\in(0,x_1)$使得\[f\left( {{x_1}} \right) \ge f\left( {{x_2}} \right)\frac{{x_1^{n + 1}}}{{\left( {n + 1} \right)!}}.\]继续这个过程，可得$(0,x)$中严格递减序列$\{x_k\}$使得\[f\left( {{x_k}} \right) \ge f\left( {{x_{k + 1}}} \right)\frac{{x_k^{n + 1}}}{{\left( {n + 1} \right)!}}.\]于是\[f\left( x \right) \ge f\left( {{x_{k + 1}}} \right)\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}}\frac{{x_1^{n + 1}}}{{\left( {n + 1} \right)!}} \cdots \frac{{x_k^{n + 1}}}{{\left( {n + 1} \right)!}}.\]因为$x$及$x_k$都在$[0,1]$中，上式右端当$k\to+\infty$时趋于0，于是对于$x\in[0,1]$有$f(x)\geq0$.由此\[f'\left( x \right) = f'\left( 0 \right) + f''\left( 0 \right)x + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{\left( {n - 1} \right)!}}{x^{n - 1}} + \frac{{{f^{\left( {n + 1} \right)}}\left( \xi \right)}}{{n!}}{x^n} \ge \frac{{f\left( \xi \right)}}{{n!}}{x^n} \ge 0,\]其中$\xi\in(0,x)$.归纳可证$f^{(k)}\geq0,x\in[0,1],k=1,2,\ldots,n+1$.对函数$g(x)=f(x+1)$重复以上过程可知$f(x)\geq0,x\in[1,2]$.用归纳法可证对任意自然数$m$，$f(x)$在$[m,m+1]$上非负.于是结论得证.

6. 设$f(x)$是$[0,+\infty)$上连续函数，满足$0<f(x)<1$，而且无穷积分在$\int_0^{+\infty}{f(x)\, dx}$和$\int_0^{+\infty}{xf(x)\, dx}$都收敛.求证：\[\int_0^{ + \infty } {xf\left( x \right)\, dx} > \frac{1}{2}{\left( {\int_0^{ + \infty } {f\left( x \right)\, dx} } \right)^2}.\]

证.令\[g\left( y \right) = \int_0^y {xf\left( x \right)dx} - \frac{1}{2}{\left( {\int_0^y {f\left( x \right)dx} } \right)^2},y>0.\]

我们得到\[g'\left( y \right) = yf\left( y \right) - f\left( y \right)\int_0^y {f\left( x \right)dx} = f\left( y \right)\left( {y - \int_0^y {f\left( x \right)dx} } \right).\]

又$0<f(x)<1$，我们得到\[0 < \int_0^y {f\left( x \right)dx} < \int_0^y {dx} = y.\]因此我们有$g'(y)>0,g(y)>g(0)=0.$再令$y\to+\infty$即可.

7. 设$0<\alpha\leq1,\beta>0,\alpha+\beta>1$，$f(x)$是$[1,+\infty)$的正函数，且$\int_1^{+\infty}{f(x)\, dx}$收敛.求证：$\int_1^{+\infty}{\frac{{(f(x))}^\alpha}{x^\beta}\, dx}$收敛.

证.当$\alpha=1$时，由\[0 < \int_1^{ + \infty } {\frac{{f\left( x \right)}}{{{x^\beta }}}dx} \le \int_1^{ + \infty } {f\left( x \right)dx} .\]可知积分收敛.

当$0<\alpha<1$时.

（Holder积分不等式）若函数$f(x)$与$g(x)$在区间$[a,b]$上连续非负，且$p>1,\frac1p+\frac1q=1$，则有不等式\[\int_a^b {f\left( x \right)g\left( x \right)dx} \le {\left( {\int_a^b {{{\left[ {f\left( x \right)} \right]}^p}dx} } \right)^{\frac{1}{p}}}{\left( {\int_a^b {{{\left[ {g\left( x \right)} \right]}^q}dx} } \right)^{\frac{1}{q}}}.\]

取$p=\frac1\alpha,q=\frac1{1-\alpha},f(x)={[f(x)]}^\alpha,g(x)=\left(\frac1x\right)^\beta$，对于任给的正数$A$，我们有

\[0 < \int_1^A {\frac{{{{\left( {f\left( x \right)} \right)}^\alpha }}}{{{x^\beta }}}dx} \le {\left( {\int_1^A {f\left( x \right)dx} } \right)^\alpha }{\left( {\int_1^A {\frac{1}{{{x^{\frac{\beta }{{1 - \alpha }}}}}}dx} } \right)^{1 - \alpha }} \le {\left( {\int_1^{ + \infty } {f\left( x \right)dx} } \right)^\alpha }{\left( {\int_1^{ + \infty } {\frac{1}{{{x^{\frac{\beta }{{1 - \alpha }}}}}}dx} } \right)^{1 - \alpha }}.\]

再注意到$\int_1^{ + \infty } {f\left( x \right)dx} ,\int_1^{ + \infty } {\frac{1}{{{x^{\frac{\beta }{{1 - \alpha }}}}}}dx} \left( {\frac{\beta }{{1 - \alpha }} > 1} \right)$均收敛即可得证.

8. 设$\{a_n\}$是正的递增数列.求证：级数$\sum\limits_{n=1}^\infty \left(\frac{a_{n+1}}{a_n}-1\right)$收敛的充分必要条件是$\{a_n\}$有界.

证.1. ($\Rightarrow$)注意到\[\ln(1+x)<x,x>0.\]我们有\[A = \sum\limits_{n = 1}^\infty {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}} - 1} \right)} > \sum\limits_{n = 1}^\infty {\ln \frac{{{a_{n + 1}}}}{{{a_n}}}} = \mathop {\lim }\limits_{n \to \infty } \ln \frac{{{a_n}}}{{{a_1}}}.\]故\[\mathop {\lim }\limits_{n \to \infty } {a_n} = B \le {a_1}{e^A}.\]

2. ($\Leftarrow$)又\[\sum\limits_{n = 1}^\infty {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}} - 1} \right)} = \sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_n}}}} < \sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_1}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n} - {a_1}}}{{{a_1}}} = \frac{1}{{{a_1}}}\mathop {\lim }\limits_{n \to \infty } {a_n} - 1.\]故$\sum\limits_{n=1}^\infty \left(\frac{a_{n+1}}{a_n}-1\right)$收敛.

9. 设$\alpha>0,\{a_n\}$是递增正数列.求证：级数$\sum\limits_{n=1}^\infty \frac{a_{n+1}-a_n}{a_{n+1}a_n^\alpha}$收敛.

证.1. 当$0<\alpha<1$时，有\[\sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_{n + 1}}a_n^\alpha }}} \le \sum\limits_{n = 1}^\infty {\frac{1}{\alpha }\left( {\frac{1}{{a_n^\alpha }} - \frac{1}{{a_{n + 1}^\alpha }}} \right)} = \frac{1}{{\alpha a_1^\alpha }} - \frac{1}{\alpha }\mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^\alpha }}.\]

事实上，由Lagrange中值定理，我们有\[\frac{{a_{n + 1}^\alpha - a_n^\alpha }}{{{a_{n + 1}} - {a_n}}} = \alpha {\xi ^{\alpha - 1}} > \alpha a_{n + 1}^{\alpha - 1},\]其中$\xi\in(a_n,a_{n+1})$.

因此，根据$a_n$单调递增的性质，对于其有界和无界两种情况，不等式右端的级数都是收敛的，由此得证.

2. 当$\alpha\geq1$时，又有

\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_{n + 1}}a_n^\alpha }}} &= \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{a_n^\alpha }} - \frac{{a_n^{1 - \alpha }}}{{{a_{n + 1}}}}} \right)} \le \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{a_n^\alpha }} - \frac{{a_{n + 1}^{1 - \alpha }}}{{{a_{n + 1}}}}} \right)} \\&= \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{a_n^\alpha }} - \frac{1}{{a_{n + 1}^\alpha }}} \right)} = \frac{1}{{a_1^\alpha }} - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^\alpha }}.\end{align*}

同理，不等式右端的级数亦是收敛的，证毕.

10. 设$0<\alpha<1$，证明数列\[{a_n} = \frac{1}{{1 + {n^\alpha }}} + \frac{1}{{2 + {n^\alpha }}} + \cdots + \frac{1}{{n + {n^\alpha }}},n = 1,2, \cdots \]发散.

证.注意到\[x>\ln(x+1).\]我们有\[\frac{1}{{k + {n^\alpha }}} > \ln \frac{{k + 1 + {n^\alpha }}}{{k + {n^\alpha }}}.\]

因此\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{k + {n^\alpha }}}} > \mathop {\lim }\limits_{n \to \infty } \ln \frac{{n + 1 + {n^\alpha }}}{{1 + {n^\alpha }}} \to \infty .\]

所以此数列发散.

11. 计算\[\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{1^2} + \sqrt n + {n^2}}} + \frac{n}{{{2^2} + 2\sqrt n + {n^2}}} + \cdots + \frac{n}{{{n^2} + n\sqrt n + {n^2}}}} \right).\]

解.注意到

\begin{align*}&\mathop {\lim }\limits_{n \to \infty } \left[ {\sum\limits_{k = 1}^n {\frac{n}{{{k^2} + {n^2}}}} - \sum\limits_{k = 1}^n {\frac{n}{{{k^2} + k\sqrt n + {n^2}}}} } \right] = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{{kn\sqrt n }}{{\left( {{k^2} + k\sqrt n + {n^2}} \right)\left( {{k^2} + {n^2}} \right)}}} \\&\le \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{{kn\sqrt n }}{{{n^2} \cdot {n^2}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n\sqrt n }}{{{n^2} \cdot {n^2}}} \cdot \frac{{n\left( {n + 1} \right)}}{2} = 0.\end{align*}

因此

\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{n}{{{k^2} + k\sqrt n + {n^2}}}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{n}{{{k^2} + {n^2}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{{{\left( {\frac{k}{n}} \right)}^2} + 1}}} = \int_0^1 {\frac{1}{{{x^2} + 1}}dx} = \frac{\pi }{4}.\]

12. 设$f(x)$是$[0,2\pi]$上可导的凸函数，$f'(x)$有界.试证\[{a_n} = \frac{1}{\pi }\int_0^{2\pi } {f\left( x \right)\cos nx\, dx} \ge 0.\]

证.由分部积分，我们有

\[{a_n} = \frac{1}{\pi }\int_0^{2\pi } {f\left( x \right)\cos nxdx} = \frac{1}{\pi }\int_0^{2\pi } {f\left( x \right)d\frac{{\sin nx}}{n}} = - \frac{1}{{n\pi }}\int_0^{2\pi } {f'\left( x \right)\sin nxdx} .\]

又由第二积分中值定理，我们有

\begin{align*}\int_0^{2\pi } {f'\left( x \right)\sin nxdx} &= f'\left( {0 + } \right)\int_0^\xi {\sin nxdx} + f'\left( {2\pi - } \right)\int_\xi ^{2\pi } {\sin nxdx} \\&= \frac{{1 - \cos n\xi }}{n}\left[ {f'\left( {0 + } \right) - f'\left( {2\pi - } \right)} \right] \le 0.\end{align*}

故\[{a_n} = - \frac{1}{{n\pi }}\int_0^{2\pi } {f'\left( x \right)\sin nxdx} \ge 0.\]

13. 设$\{a_n\}$是正数列使得$\sum\limits_{n=1}^\infty{\frac{1}{a_n}}$收敛.求证\[\sum\limits_{n = 1}^\infty {\frac{n}{{{a_1} + {a_2} + \cdots + {a_n}}}} \le 2\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} ,\]而且上式右端的系数2是最佳的.

证.由柯西不等式我们得

\[\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} \sum\limits_{m = 1}^n {{a_m}}  \ge {\left( {1 + 2 +  \cdots  + n} \right)^2} = \frac{1}{4}{n^2}{\left( {n + 1} \right)^2},\]

即\[\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}} \le \frac{4}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} .\]

因此

\begin{align*}\sum\limits_{n = 1}^\infty  {\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}}}  &\le 4\sum\limits_{n = 1}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} }  = 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}} } \\&\le 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right]} }  = 2\sum\limits_{m = 1}^\infty  {\frac{1}{{{a_m}}}} .\end{align*}

这里用到了\[\frac{1}{{n{{\left( {n + 1} \right)}^2}}} \le \frac{1}{2}\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}} = \frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{n{{\left( {n + 1} \right)}^2}}}} \right].\]

注意到$a_n=n^\alpha,\alpha>1$时有

\[\mathop {\lim }\limits_{\alpha \to 1} \frac{{\sum\limits_{n = 1}^\infty {\frac{n}{{{a_1} + {a_2} + \cdots + {a_n}}}} }}{{\sum\limits_{j = 1}^\infty {\frac{1}{{{a_j}}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\sum\limits_{n = 1}^N {\frac{n}{{{1^\alpha } + {2^\alpha } + \cdots + {n^\alpha }}}} }}{{\sum\limits_{n = 1}^N {\frac{1}{{{n^\alpha }}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\frac{N}{{\frac{1}{{\alpha + 1}}{N^{\alpha + 1}} + O\left( {{N^\alpha }} \right)}}}}{{\frac{1}{{{N^\alpha }}}}} = 2.\]

14. 设$f(x)$是$[0,+\infty)$上正的连续函数，且$\int_0^{+\infty}{\frac{1}{f(x)}\, dx}$收敛.记$F(x)=\int_0^x{f(t)\, dt}$.求证\[\int_0^{ + \infty } {\frac{x}{{F\left( x \right)}}dx} < 2\int_0^{ + \infty } {\frac{1}{{f\left( x \right)}}dx} ,\]且上式右端的系数2是最佳的.

证.（陈洪葛）由$Cauchy-Schwarz$不等式，得到\[\left( {\int_0^x {f\left( t \right)dt} } \right) \cdot \left( {\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} } \right) \ge {\left( {\int_0^x {tdt} } \right)^2} = \frac{1}{4}{x^4}.\]

所以

\[\int_0^{ + \infty } {\frac{x}{{F\left( x \right)}}dx} \le \int_0^{ + \infty } {\frac{4}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} dx} .\]

注意到\[\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{{x^2}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2f\left( x \right)}} = 0.\]

以及\[\frac{1}{{{x^2}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} = \int_0^\xi {\frac{1}{{f\left( t \right)}}dt} < \int_0^{ + \infty } {\frac{1}{{f\left( t \right)}}dt} .\]

故\begin{align*}&\int_0^A {\frac{4}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} dx} = \int_0^A {\left( {\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} } \right)d\left( { - \frac{2}{{{x^2}}}} \right)} \\&= \int_0^A {\frac{2}{{f\left( x \right)}}dx} - \frac{2}{{{A^2}}}\int_0^A {\frac{{{t^2}}}{{f\left( t \right)}}dt} < \int_0^A {\frac{2}{{f\left( x \right)}}dx} - 2\int_\xi ^A {\frac{{{t^2}}}{{f\left( t \right)}}dt}.\end{align*}

令$A\to+\infty$，得到

\[\int_0^{ + \infty } {\frac{4}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} dx} \le \int_0^{ + \infty } {\frac{2}{{f\left( x \right)}}dx} - 2\int_\xi ^{ + \infty } {\frac{{{t^2}}}{{f\left( t \right)}}dt} < \int_0^{ + \infty } {\frac{2}{{f\left( x \right)}}dx}.\]

另外，当我们取$f(x)=x^a+1(a>1)$时，有$\int_0^{ + \infty } {\frac{{dx}}{{{x^a} + 1}}} = \frac{\pi }{{a\sin \frac{\pi }{a}}}$收敛.此时有

\[\mathop {\lim }\limits_{a \to 1} \frac{{\int_0^{ + \infty } {\frac{x}{{F\left( x \right)}}dx} }}{{\int_0^{ + \infty } {\frac{1}{{f\left( x \right)}}dx} }} = \mathop {\lim }\limits_{a \to 1} \frac{{\int_0^{ + \infty } {\frac{x}{{{x^{a + 1}}/\left( {a + 1} \right) + x}}dx} }}{{\int_0^{ + \infty } {\frac{{dx}}{{{x^a} + 1}}} }} = \mathop {\lim }\limits_{a \to 1} \frac{{\int_0^{ + \infty } {\frac{1}{{{x^a}/\left( {a + 1} \right) + 1}}dx} }}{{\int_0^{ + \infty } {\frac{{dx}}{{{x^a} + 1}}} }} = \mathop {\lim }\limits_{a \to 1} {\left( {a + 1} \right)^{\frac{1}{a}}} = 2.\]

15. 设$f(x)$在$\mathbb{R}$上有二阶导函数，$f(x),f'(x),f''(x)$都大于零，假设存在正数$a,b$使得$f''(x)\leq af(x)+bf'(x)$对一切$x\in\mathbb{R}$成立.

1. 求证：$\mathop {\lim }\limits_{x \to - \infty } f'\left( x \right) = 0$;

2. 求证：存在常数$c$使得$f'(x)\leq cf(x)$;

3. 求使上面不等式成立的最小常数$c$.

证.1. 显然由单调有界定理知$\mathop {\lim }\limits_{x \to - \infty } f\left( x \right),\mathop {\lim }\limits_{x \to - \infty } f'\left( x \right)$均存在，不妨设$\mathop {\lim }\limits_{x \to - \infty } f\left( x \right)=c_1,\mathop {\lim }\limits_{x \to - \infty } f'\left( x \right)=c_2$.因此我们有$f(x)\geq c_1\geq0,f'(x)\geq c_2\geq0$.又\[f\left( 0 \right) = \int_x^0 {f'\left( x \right)dx} + f\left( x \right) \ge - {c_2}x + f\left( x \right) \Rightarrow f\left( x \right) \le {c_2}x + f\left( 0 \right).\]我们知$c_2=0$，否则令$x\to-\infty$，矛盾.

2. 令

\begin{align*}&g\left( x \right) = \left( {\frac{{b + \sqrt {4a + {b^2}} }}{2}f\left( x \right) - f'\left( x \right)} \right){e^{\frac{{ - b + \sqrt {4a + {b^2}} }}{2}x}},\\&g'\left( x \right) = \left( {af\left( x \right) + bf'\left( x \right) - f''\left( x \right)} \right){e^{\frac{{ - b + \sqrt {4a + {b^2}} }}{2}x}} \ge 0.\end{align*}

又$\lim\limits_{n\to-\infty} g(x)=0$.故$g(x)\geq g(-\infty)=0$.即\[f'\left( x \right) \le \frac{{b + \sqrt {4a + {b^2}} }}{2}f\left( x \right).\]故存在常数$c={\frac{{b + \sqrt {4a + {b^2}} }}{2}}$使得$f'(x)\leq cf(x)$成立.

3. 我觉得应该把“求使上面不等式成立的最小常数$c$”改成“求使上面不等式成立的最大常数$c$”.事实上，我们取：

\[\left\{ \begin{array}{l}h\left( x \right) = {e^{cx}}\\h'\left( x \right) = c{e^{cx}}\\h''\left( x \right) = {c^2}{e^{cx}}\end{array} \right.\left( {c > 0} \right).\]

只需保证\[c^2\leq bc+a\Rightarrow 0<c\leq {\frac{{b + \sqrt {4a + {b^2}} }}{2}}.\]

故常数$c$的最大值为${\frac{{b + \sqrt {4a + {b^2}} }}{2}}$.

16. 设$f(x)$是$\mathbb{R}$上有下界或者有上界的连续函数且存在正数$a$使得\[f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)dt} \]为常数.求证：$f(x)$必为常数.

证.（Slade）由题意，$f(x)$必存在上下界，否则，在等式\[f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)dt}=C\]两端同时取$x\to \infty$，等式左端无界，而右端为常数，矛盾.

再之，注意到$\int_{x-1}^x f(t)\,dt$是可微的，故$f$也是可微的.又\[f'\left( x \right) = a\left( {f\left( {x - 1} \right) - f\left( x \right)} \right) \Rightarrow \left| {f'\left( x \right)} \right| \le a\left( {\left| {f\left( {x - 1} \right)} \right| + \left| {f\left( x \right)} \right|} \right).\]由于$f$有界，故$f'$亦有界.记$N=[a]+1$，则$0<\frac aN<1$.令$x=\frac x N$，我们得到\[f'\left( {\frac{x}{N}} \right) = a\left( {f\left( {\frac{x}{N} - 1} \right) - f\left( {\frac{x}{N}} \right)} \right).\]

固定$x$，由Lagrange中值定理，存在${a_1} \in \left( {\frac{x}{N} - 1,\frac{x}{N}} \right)$使得\[f\left( {\frac{x}{N} - 1} \right) - f\left( {\frac{x}{N}} \right) = - \frac{1}{N}f'\left( {{a_1}} \right),\]即$f'\left( {\frac{x}{N}} \right) = - \frac{a}{N}f'\left( {{a_1}} \right)$.类似地，我们有数列$\{a_n\}$使得\[f'\left( {\frac{x}{N}} \right) = - \frac{a}{N}f'\left( {{a_1}} \right) = {\left( { - \frac{a}{N}} \right)^2}f'\left( {{a_2}} \right) = \cdots = {\left( { - \frac{a}{N}} \right)^n}f'\left( {{a_n}} \right).\]令$n\to\infty$，我们有$f'\left( {\frac{x}{N}} \right) = 0 $即$f'\left( x \right) = 0$.故$f(x)$必为常数.

17. 设$f:[0,+\infty)\to [0,+\infty)$且对任意$x\geq0$有$f\circ f(x)=af(x)+bx$，其中$a<0,b>0$.求$f(x).$

证.先证明几个引理：

引理1.设$f\in C^0(\mathbb{R},\mathbb{R})$是方程$f(f(x))=af(x)+bx$的解.若方程中的常数$b\neq0$，则$f:\mathbb{R}\to\mathbb{R}$即单射又是满射，即是一个一一映射.

引理2.设$f\in C^0(\mathbb{R},\mathbb{R})$是方程$f(f(x))=af(x)+bx$的解.若$b\neq0$，且$\lambda$的多项式$\lambda^2-a\lambda-b$的两个根$r$与$s$不相等，则对任意$x\in\mathbb{R}$及任意$n\in\mathbb{Z}$均有

\[{f^n}\left( x \right) = \frac{{{s^n}\left( {f\left( x \right) - rx} \right) + {r^n}\left( {sx - f\left( x \right)} \right)}}{{s - r}}.\]

事实上，当$n$为非正整数时下一式成立：\[{f^n}\left( x \right) = \frac{{{s^n}\left( {{f^{ - 1}}\left( x \right) - \frac{x}{r}} \right) + {r^n}\left( {\frac{x}{s} - {f^{ - 1}}\left( x \right)} \right)}}{{\frac{1}{s} - \frac{1}{r}}}.\]以$f(x)$代替$x$，以$n-1$代替$n$，可推知$n$为非正整数时亦成立.

引理3.设$g_n\in C^0(\mathbb{R},\mathbb{R})$，且${(-1)}^n g_n(x)$对$x$单调递增（$n=0,1,2,\ldots$）.若极限$g(x)=\lim\limits_{n\to\infty}g_n(x)$对任$x\in \mathbb{R}$均存在，则极限函数$g$是个常值函数.

事实上，因$g(x)=\lim\limits_{n\to\infty}g_{2n}(x)$，而$g_{2n}(x)$对$x$递增.又因$g(x)=\lim\limits_{n\to\infty}g_{2n+1}(x)$，而$\lim\limits_{n\to\infty}g_{2n+1}(x)$对$x$递减，故$g(x)$对$x$递减.于是$g$只能是个常值函数.

回到原题，我们有定理：设$r$及$s$是$\lambda$的二次多项式$\lambda^2-a\lambda-b=0$的两个根.若$r<0<s$，且$r\neq -s$，则$f\in C^0(\mathbb{R},\mathbb{R})$是方程$f(f(x))=af(x)+bx$的解的充分必要条件是$f(x)=rx$（对任$x\in\mathbb{R}$），或者$f(x)=sx$（对任$x\in\mathbb{R}$），或者，当$s=1$时还可以是$f(x)=rx+c$（对任$x\in\mathbb{R}$，$c$可以是任一个给定的实数）.

定理的充分性部分是显然的.下面证明定理的必要性部分，设$f$是方程$f(f(x))=af(x)+bx$的一个解，据引理1知$f$严格单调.若$f$严格递增，可推出

\[sx - f\left( x \right) = \left\{ \begin{array}{l}\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{r^n}}},\text{若}\left| r \right| > s,\\\mathop {\lim }\limits_{n \to - \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{r^n}}},\text{若}\left| r \right| < s.\end{array} \right.\]

因${(-1)}^n(s-r)f^n(x)/r^n$是$x$的递增函数（任意$n\in\mathbb{Z}$），由引理3可知极限函数$sx-f(x)$恒取常值，即存在某$c\in\mathbb{R}$使得$f(x)=sx+c$对任意$x\in\mathbb{R}$成立，把$f(x)$的这一表达式代入得到$s^2x+sc+c=(r+s)(sx+c)-rsx$，推出$c=0$.因此，当$f$递增时$f(x)=sx$（对任意$x\in\mathbb{R}$）.

若$f$严格递减，可推出\[f\left( x \right) - rx = \left\{ \begin{array}{l}\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{s^n}}},s > \left| r \right|,\\\mathop {\lim }\limits_{n \to - \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{s^n}}},s < \left| r \right|.\end{array} \right.\]

因${(-1)}^n(s-r)f^n(x)/s^n$是$x$的递增函数（任意$n\in\mathbb{Z}$），由引理3可同样推出$f(x)=rx+c$（对任意$x\in\mathbb{R}$）.将此表达式代入得到$r^2x+rc+c=(r+s)(rx+c)-rsx$，可化简为$(s-1)c=0$.因此，当$s\neq1$时$c=0$，当$s=1$时$c$可为任意给定的实数.

另外：注意到$f:[0,+\infty)\to [0,+\infty)$这一条件，我们知$f(x)=sx(s>0)$.可参阅：

[1]关于迭代函数方程$f^2(x)=af(x)+bx$的通解，麦结华，数学研究与评论第17卷第1期83-90页，1997年2月.

[2]J.Matkowski and Zhang Weinian, Method of characteristics for functional equations in polynomial

form, Acta Math.Sinica, New Series.

18. 设$f(x)$在$(-\infty,+\infty)$上连续，且对任意$x$有$f(2x-f(x))=x$.求证：$f(x)\equiv x+c$，其中$c$为常数.

证.（逆蝶）令$g(x)=2x-f(x)$，则$f[g(x)]=x$，显然$g$是单射且为增函数.若$g$有上界，则$\lim\limits_{x\to+\infty}{g(x)}=a$存在.又$f$连续，$\lim\limits_{x\to+\infty}{f[g(x)]}=f(a)$，矛盾.从而$g$无上界也无下界.由于$g$连续，$\forall x\in R,\exists y$使得$x=g(y)$.从而$f(x)=f[g(y)]=y$，于是$g[f(x)]=g(y)=x$.由$g$递增知$f$递增.

若$\exists x_1,x_2$使得$f(x_1)-x_1\neq f(x_2)-x_2$，则由于$f(x)-x$的连续性知其可以取遍$f(x_1)-x_1$及$f(x_2)-x_2$之间的任何数.

设$d,d'$在它们之间，且$d/d'>1$是正无理数，设$d=f(x)-x,d'=f(x')-x'$.令$x_0=x,x_{n+1}=2x_n-f(x_n)$，则$f(x_{n+1})=x_n$.归纳可得$x_n=x+nd$，同理$x'_n=x'+nd'$.

令$x_m>x'_n,x_{m+1}<x'_n+1$，即$x+md>x'+nd,x+d+md<x'+d'+nd'$.即$(x'-x)/d<(md')/d-n<(x'-x)/d+(d/d'-1)$.由$d/d'$为无理数知上式关于$m,n$有解，这与$f$递增矛盾.

下面是一个推广：对$\forall m\neq0$，若一个连续函数$f:\mathbb{R}\to\mathbb{R}$满足函数方程\[f\left(2x-\frac{f(x)}{m}\right)=mx,\]则有$f(x)=m(x-c)$.

证.（陈洪葛）我们设$g(x)=2x-\frac{f(x)}{m}$，显然$g(x)$是连续函数且有\[g(g(x))=2g(x)-x,\forall x\in\mathbb{R}.\]

若$g(x_1)=g(x_2)$，则有$g(g(x_1))=g(g(x_2))$，我们得到\[x_1=x_2.\]故$g(x)$是一个单射，而我们知道，若$g(x)$是一个连续的单射，则$g(x)$严格单调（关于这点可以用反证法证明）.因此，$g(x)$有2种情况，严格递增或者严格递减.下面证明$g(x)$只能严格递增.

（反证）设$g(x)$严格递减，则对于$x_1<x_2$，我们有$g(x_1)>g(x_2)$，接着又有$g(g(x_1))<g(g(x_2))$，二者等价于\[2g(x_1)-x_1<2g(x_2)-x_2\Leftrightarrow 2[g(x_1)-g(x_2)]<x_1-x_2.\]上面不可能成立，因为左边大于0而右边小于0，故$g(x)$只能严格递增.

改写$g(g(x))=2g(x)-x$为\[g(g(x))-g(x)=g(x)-x.\]递推后得到\[g^n(x)=ng(x)-(n-1)x,(n\geq1),\]这里$g^{(n)}(x)$表示$n$次复合.那么有\[g^n(x)-g^n(0)=n[g(x)-x-g(0)]+x\Leftrightarrow\frac{g^n(x)-g^n(0)}{n}=g(x)-x-g(0)+\frac{x}{n}.\]而$g(x)$严格递增，$g^n(x)$也严格递增，故对上式令$n\to\infty$，由$g(x)$的单调性，我们得到

\begin{align*}g(x)&\leq x+g(0),&&x<0\\g(x)&\geq x+g(0),&&x>0.\end{align*}

这样，我们得到$g(x)$的值域也是$\mathbb{R}$，故$g(x)$是一个一一映射，且$g^{-1}$存在.现在，用$x=g^{-1}(g^{-1}(y))$代入原来的方程，则有\[g^{-1}(g^{-1}(y))=2g^{-1}(y)-y.\]$g^{-1}(y)$同样满足这个方程，则用相同的手段，我们得到

\begin{align*}g^{-1}(y)&\leq y+g(0),&&y<0\\g^{-1}(y)&\geq y+g(0),&&y>0.\end{align*}

现在，用$x=g^{-1}(y)$代入\[g(g(x))-g(x)=g(x)-x\]得到\[g(y)-y=y-g^{-1}(y).\]令$y=0$得到$g^{-1}(0)=-g(0)$.

假设$g(0)\geq0$，则对$x>0$有$g(x)\geq x+g(0)>0$，则对$y=g(x)>0$有$x>g(x)+g^{-1}(0)=g(x)-g(0)$.故得到

\[g(x)=x+g(0),x>0\]同理可得\[g(x)=x+g(0),x<0.\]这样我们得到$f(x)=m(x-g(0))$对$x\in\mathbb{R}$成立.

19. 设$0<a<2$.求证：不存在$(-\infty,+\infty)$上连续的函数$f(x)$，使得对任意$x$有$f(ax-f(x))=x$.

证.我们设$g(x)=ax-f(x)$，显然$g(x)$是连续函数且有\[g(g(x))=ag(x)-x,\forall x\in\mathbb{R}.\]假设此方程有一连续函数解$g:\mathbb{R}\to\mathbb{R}$且多项式方程$r^2-ar+1=0$有一对复数特征根\[{r_1} = a - ib = S\exp \left( { - i\theta } \right),{r_2} = a + ib = S\exp \left( {i\theta } \right)，\]其中$a,b\in \mathbb{R},b>0,S>0$及$\theta\in(0,\pi)$.易知$f$是单调的且$f^2$是严格递增的.并且对于$x\neq0$有$f(x)\neq x$.因此当$f$是严格递增时，数列$\left\{ {{f^{n + 1}}\left( x \right)-{f^n}\left( x \right)} \right\}$对于任意固定的$x\neq0$同样是严格递增的.因此，我们有

\begin{align*}{f^n}\left( x \right) &= \frac{{r_2^n}}{{{r_2} - {r_1}}}\left( {f\left( x \right) - {r_1}x} \right) + \frac{{r_1^n}}{{{r_2} - {r_1}}}\left( {{r_2}x - f\left( x \right)} \right)\\&= \frac{1}{b}{S^n}\sin \theta \cdot f\left( x \right) - \frac{1}{b}{S^{n + 1}}\sin \left( {n - 1} \right)\theta \cdot x.\end{align*}

则

\[{f^{n + 1}}\left( x \right) - {f^n}\left( x \right) = r_2^nU\left( x \right) + r_1^nV\left( x \right),\]

其中$U\left( x \right) = \frac{{{r_2} - 1}}{{{r_2} - {r_1}}}\left( {f\left( x \right) - {r_1}x} \right),V\left( x \right) = \frac{{{r_1} - 1}}{{{r_2} - {r_1}}}\left( {{r_2}x - f\left( x \right)} \right)$.显然$\overline U\left( x \right) = V\left( x \right)$，故对于固定的$x\neq0$我们可令

\[U\left( x \right) = T\exp \left( {it} \right)\text{和}V\left( x \right) = T\exp \left( { - it} \right),\]

其中$T\geq0$且$t\in[0,2\pi]$.因此\[\begin{array}{l}{f^{n + 1}}\left( x \right) - {f^n}\left( x \right) = {S^n}T\left[ {\exp \left( {i\left( {n\theta + t} \right)} \right) + \exp \left( { - i\left( {n\theta + t} \right)} \right)} \right]\\= 2{S^n}T\cos \left( {n\theta + t} \right)\end{array}.\]

因为当$T>0$时$S>0$，与数列$\left\{ {{f^{n + 1}}\left( x \right)-{f^n}\left( x \right)} \right\}$的性质矛盾；当$T=0$时我们有$U(x)=V(x)=0$，因此对所有$x\neq0$我们有$f(x)=r_1x=r_2x$，由此我们得到$r_1=r_2$这一矛盾的结论.证毕.

20. 求证：不存在可微函数$f:(0,+\infty)\to(0,+\infty)$满足方程\[f'(x)=f\circ f(x),x\in(0,+\infty).\]

证.（刘畅）由$f(x)>0$知$f(f(x))>0$，从而有$f'(x)>0$，显然$\lim\limits_{x\to+\infty}f(x)=+\infty$（否则得到$\lim\limits_{x\to+\infty}f'(x)=0$，矛盾），因此$\lim\limits_{x\to+\infty}f'(x)=\lim\limits_{x\to+\infty}f(f(x))=+\infty$.当$x$充分大时，我们有$f'(x)>1$，故存在足够大的$n\in\mathbb{N_+}$使得$f(n)>n+1$，我们有\[f\left( {n + 1} \right) - f\left( n \right) = \int_n^{n + 1} {f'\left( x \right)dx} = \int_n^{n + 1} {f\left( {f\left( x \right)} \right)dx} > f\left( {n + 1} \right),\]矛盾.

21. 设正数列$\{a_n\}$满足$\varliminf\limits_{n\to+\infty}{a_n}=1,\varlimsup\limits_{n\to+\infty}{a_n}<+\infty,\lim\limits_{n\to+\infty}\sqrt[n]{a_1a_2\ldots a_n}=1.$求证：

\[\lim\limits_{n\to+\infty}\frac{a_1+a_2+\ldots+a_n}{n}=1.\]

证.（别人解答）记$x_n=\ln a_n$，由题设条件，

\begin{align*}&\varliminf\limits_{n\to+\infty}x_n=0,\varlimsup\limits_{n\to+\infty}x_n\leq A<+\infty(A>0),\\&\lim\limits_{n\to\infty}\frac1n\sum_{k=1}^n x_k=0.\end{align*}

假设所有的$x_n\geq0$，则当$x\leq\ln2$时，成立不等式$e^x\leq1+2x$.对于固定的$n$，记${S_n} = \left\{ {i \in \mathbb{Z}\left| {1 \le i \le n,{x_i} \le \ln 2} \right.} \right\},{T_n} = \left\{ {i \in \mathbb{Z}\left| {1 \le i \le n,{x_i} > \ln 2} \right.} \right\}$，则

\[\frac{1}{n}\sum\limits_{k = 1}^n {{x_k}} = \frac{1}{n}\sum\limits_{k \in {S_n}} {{x_k}} + \frac{1}{n}\sum\limits_{k \in {T_n}} {{x_k}} \ge \frac{{\left| {{T_n}} \right|}}{n}\ln 2 \ge 0.\]

由此即知\[\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{T_n}} \right|}}{n} = 0.\]

从而\[\frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} = \frac{1}{n}\sum\limits_{k \in {S_n}} {{e^{{x_k}}}} + \frac{1}{n}\sum\limits_{k \in {T_n}} {{e^{{x_k}}}} \le 1 +\frac{{\left| {{T_n}} \right|}}{n}\left( {C+ {e^A}} \right) + \frac{2}{n}\sum\limits_{k = 1}^n {{x_k}} .\]而\[\frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} \ge {e^{\frac{1}{n}\sum\limits_{k = 1}^n {{x_k}} }}.\]由迫敛性即得\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = 1.\]在一般情况下，作序列\[{z_n} = \left\{ \begin{array}{l}- {x_n},{x_n} < 0\\0,{x_n} \ge 0\end{array} \right.,n = 1,2, \cdots \]由题设可知\[\mathop {\lim }\limits_{n \to \infty } {z_n} = 0.\]故$y_n=x_n+z_n\geq0$，得到$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{y_k}} = 0.$从而$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{y_k}}}} = 1$，再由$\frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} \le \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{y_k}}}}$，可得\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = 1.\]

22. 设$f(x)$在$\mathbb{R}$上有二阶连续导数且满足方程\[f^3+{(f')}^3=1.\]求证：$f=1$.

证.（刘畅）显然$f=1$满足题意.当$f\neq1$时，我们有

\[f' = \sqrt[3]{{1 - {f^3}}},f'' = \frac{{ - {f^2}f'}}{{\sqrt[3]{{{{\left( {1 - {f^3}} \right)}^2}}}}} = \frac{{ - {f^2}}}{{\sqrt[3]{{1 - {f^3}}}}}.\]

若$\exists x_0\in\mathbb{R}$使$f(x_0)>1$，则$f'(x_0)<0$，故对$x\leq x_0$都有$f'(x)<0,f''(x)>0$,从而$f(-\infty)=+\infty$.若$\exists x_1>x_0$使得$f(x_1)<1$，则对$x\leq x_1$都有$f'(x)>0,f''(x)<0$，故$f(-\infty)=-\infty$，矛盾.

故对$x>x_0$，均有$f(x)\geq1$.

若$\exists a>x_0$使得$f(a)=1$（其中$a$为使得$f(x)=1$的最小实数），则我们有$x>a$均有$f(x)=1$，这时有$f''(x)$在$x=a$处不连续，矛盾.

若不$\exists b>x_0$使得$f(b)=1$，则对于$x\in\mathbb{R}$均有$f(x)>1$，设$f(+\infty)=1$（否则有$\mathop {\lim }\limits_{x \to + \infty } f'\left( x \right) < 0$，与$\mathop {\lim }\limits_{x \to + \infty } f'\left( x \right) = 0$矛盾），而\[\mathop {\lim }\limits_{x \to + \infty } f'\left( x \right) = 0,\mathop {\lim }\limits_{x \to + \infty } f''\left( x \right) = + \infty .\]亦矛盾.证毕.

PDF下载：http://pan.baidu.com/s/1jGqVEoY

一边参考葛神的日志，一边把这题做出来了，在今年中秋节这算是给我的不错礼物.

前几天在群里被问到一个定长木棍在竖直墙边上运动的轨迹问题，还是挺有趣的，涉及到星形线，解决起来也不简单.

前阵子四叶群里有人问道下面这题，虽感觉此题结论优美但无从下手.

求解\[\displaystyle \int_0^1 \frac{\log^2(1-x)\log(x)}{x}dx=-\frac{\pi^4}{180}.\]

解.$For $|z|<1$ we have that