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逆神的数学分析题答案总算补全了

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.

2014年9月5日

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.

\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*}

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.

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

\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$\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {x^\alpha }} \right)}}} .$

\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？

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

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

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

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：$f(x)\geq0$.

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.

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（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：$\int_0^{ + \infty } {xf\left( x \right)\, dx} > \frac{1}{2}{\left( {\int_0^{ + \infty } {f\left( x \right)\, dx} } \right)^2}.$

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.

（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$\int_a^b {f\left( x \right)g\left( x \right)dx} \le {\left( {\int_a^b {{{\left[ {f\left( x \right)} \right]}^p}dx} } \right)^{\frac{1}{p}}}{\left( {\int_a^b {{{\left[ {g\left( x \right)} \right]}^q}dx} } \right)^{\frac{1}{q}}}.$

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

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.

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.

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*}

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.

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

\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*}

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

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

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

\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*}

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.

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

\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*}

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

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.

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

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

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

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$c$.

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*}

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：

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

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.

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$f(x).$

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

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

[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.

（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.

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

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

$g(x)=x+g(0),x>0$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$f(ax-f(x))=x$.

\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*}

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

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

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

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：

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

\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*}

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

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：$f=1$.

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