设$f$是定义在$R$上的实函数，对任意绝对收敛的级数$\displaystyle\sum_{n=1}^{\infty}a_{n}$,若$\displaystyle\sum_{n=1}^{+\infty}f(a_{n})$收敛，则
\[ f(x)=O(x) (x\to 0) \]
证明：若$f(x)$并不是$O(x)$,(当$x\to 0$)时，必要时我们能找到一个单调的序列$x_{n}\to 0$,且对任意的$n$,都有$x_{n}\neq 0$,同时$f(x_{n})$保持着一致的符号，且
\[ \lim_{k\to\infty}\left|\frac{f(x_{k})}{x_{k}}\right|=+\infty \]
对任意的$n>0$,存在这样的$k_{n}\in \mathbf{N}^{+}$,当$k\geq k_{n}$时，有$|x_{k}|\leq \frac{1}{2n^2}$且
\[ \left|\frac{f(x_{k})}{x_{k}}\right|\geq n \]
令$ j_{n}=\left[\frac{1}{n^2|x_{k_{n}}|}\right]$,则
\[ \frac{1}{2n^2}\leq \overbrace{|x_{k_{n}}+x_{k_{n}}+\cdots+x_{k_{n}}|}^{j_{n}\text{个}}\leq \frac{1}{n^2} \]
对$n$求和，就有
\[ \left|\sum_{n=1}^{\infty}\overbrace{x_{k_{n}}+x_{k_{n}}+\cdots+x_{k_{n}}}^{j_{n}\text{个}} \right|\leq \sum_{n=1}^{\infty}\frac{1}{n^2} \]
是收敛的，但这时
\[ \left|\sum_{n=1}^{\infty}\overbrace{f(x_{k_{n}})+f(x_{k_{n}})+\cdots+f(x_{k_{n}})}^{j_{n}\text{个}}\right|\geq \sum_{n=1}^{\infty}\frac{1}{2n}\]
发散。这一矛盾说明了
\[ f(x)=O(x) (x\to 0) \]
计算
\[ \arctan\left(\frac{r\sin{\theta}}{1+r\cos{\theta}}\right)\]
的Fourier series.
解:
考虑$z=r(\cos{\theta}+i\sin{\theta})$,则
\[ \frac{1}{1+z}=\frac{1+\overline{z}}{(1+z)(1+\overline{z})}=\frac{1}{1+r\cos{\theta}+ir\sin{\theta}}=\frac{1+r\cos{\theta}}{1+2r\cos{\theta}+r^2}-i\cdot\frac{r\sin{\theta}}{1+2r\cos{\theta}+r^2}\]
于是，记
\[ a=\frac{1+r\cos{\theta}}{1+2r\cos{\theta}+r^2},b=-\frac{r\sin{\theta}}{1+2r\cos{\theta}+r^2}\]
我们得到
\[ \frac{1}{1+z}=a+ib \]
那么它的辐角
\[ \varphi=\arctan\frac{b}{a}=-\arctan\left(\frac{r\sin{\theta}}{1+r\cos{\theta}}\right)\]
\[ a+ib=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}+i\cdot\frac{b}{\sqrt{a^2+b^2}}\right)=\sqrt{a^2+b^2}(\cos\varphi+i\sin\varphi)=\sqrt{a^2+b^2}e^{i\varphi}\]
又注意到
\[ \sqrt{a^2+b^2}=|a+bi|=\frac{1}{|1+z|}=\frac{1}{\sqrt{(1+z)(1+\overline{z})}}\]
所以
\[ \frac{1}{1+z}=\frac{1}{\sqrt{(1+z)(1+\overline{z})}}e^{i\varphi} \]
就是
\[ -i\varphi=i\arctan\left(\frac{r\sin{\theta}}{1+r\cos{\theta}}\right)=\frac{1}{2}\left(\ln(1+z)-\ln(1+\overline{z})\right)\]
这时，利用
\[ \ln(1+z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}z^{n}\]
及$z=r\cos\theta+ir\sin{\theta}$,得到
\[ \ln(1+z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(r^{n}\cos n\theta+ir^{n}\sin n\theta)\]
\[ \ln(1+\overline{z})=\overline{\ln(1+z)}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(r^{n}\cos n\theta-ir^{n}\sin n\theta)\]
于是，自然就得到
\[ \arctan\left(\frac{r\sin{\theta}}{1+r\cos{\theta}}\right)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}r^{n}\sin{n\theta} \]
这个不是别的，就是它的Fourier Series.
另外，若令$r=-1$,则
\[ \sum_{n=1}^{\infty}\frac{\sin{n\theta}}{n}=\arctan\left(\frac{\sin{\theta}}{1-\cos\theta}\right)=\frac{\pi-\theta}{2}\]
求证：
\[ \lim_{n\to\infty}n\left[\left(\frac{1}{\pi}\left(\sin\left(\frac{\pi}{\sqrt{n^2+1}}\right)+\sin\left(\frac{\pi}{\sqrt{n^2+2}}\right)+\cdots+\sin\left(\frac{\pi}{\sqrt{n^2+n}}\right) \right)\right)^{n}-\frac{1}{\sqrt[4]{e}}\right]=-\frac{1}{\sqrt[4]{e}}\left(\frac{15}{96}+\frac{\pi^2}{6} \right) \]
————————————————————————————————————————
证明
记
\[ I=n\left[\left(\frac{1}{\pi}\left(\sin\left(\frac{\pi}{\sqrt{n^2+1}}\right)+\sin\left(\frac{\pi}{\sqrt{n^2+2}}\right)+\cdots+\sin\left(\frac{\pi}{\sqrt{n^2+n}}\right) \right)\right)^{n}-\frac{1}{\sqrt[4]{e}}\right]\]
则
\[ I=\frac{n}{\sqrt[4]{e}}\left(\exp\left(n\ln\frac{\sin\frac{\pi}{\sqrt{n^2+1}}+\sin\frac{\pi}{\sqrt{n^2+2}}+\cdots+\sin\frac{\pi}{\sqrt{n^2+n}}}{\pi}+\frac{1}{4}\right)-1 \right)\]
注意到
\[ \sin\frac{\pi}{\sqrt{n^2+k}}=\frac{\pi}{\sqrt{n^2+k}}-\frac{1}{6}\left(\frac{\pi}{\sqrt{n^2+k}}\right)+o\left(\frac{1}{n^3}\right)\qquad (n\to+\infty) \]
所以
\[ \frac{1}{\pi}\sum_{k=1}^{n}\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]+o\left(\frac{1}{n^2}\right)\qquad (n\to+\infty) \]
而
\begin{align*}
\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}&=\frac{1}{n}\sum_{k=1}^{n}\left(1+\frac{k}{n^2}\right)^{-\frac{1}{2}}\\
&=\frac{1}{n}\sum_{k=1}^{n}\left(1-\frac{k}{2n^2}+\frac{3}{8}\left(\frac{k}{n^2}\right)^{2}+o\left(\frac{1}{n^2}\right)\right)\\
&=1-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}+o\left(\frac{1}{n^2}\right)
\end{align*}
所以
\[ \frac{1}{\pi}\sum_{k=1}^{n}\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)=1-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]+o\left(\frac{1}{n^2}\right) \]
\begin{align*}
&\ln\left[\frac{1}{\pi}\sum_{k=1}^{n}\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)\right]\\
&=-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]-\frac{1}{2}\left[ \frac{(n+1)}{4n^2}-\frac{(n+1)(2n+1)}{16n^4}+\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]\right]^{2}+o\left(\frac{1}{n^2}\right)\\
&=-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]-\frac{1}{2}\left[\frac{n+1}{4n^2}+o\left(\frac{1}{n}\right)\right]^2+o\left(\frac{1}{n^2}\right)\qquad (n\to+\infty)
\end{align*}
\begin{align*}
&n\ln\left[\frac{1}{\pi}\sum_{k=1}^{n}\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)\right]+\frac{1}{4}\\
&=\frac{1}{4}+n\left[-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]-\frac{1}{2}\left[\frac{n+1}{4n^2}+o\left(\frac{1}{n}\right)\right]^2+o\left(\frac{1}{n^2}\right)\right]\\
&=-\frac{15}{96n}-\frac{\pi^2}{6n}+o\left(\frac{1}{n}\right)\quad (n\to+\infty)
\end{align*}
这里得注意到事实
\[ \left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]\sim \frac{\pi^2}{n^2} \]
所以就有
\[ \lim_{n\to\infty}I=\lim_{n\to\infty}\frac{n}{\sqrt[4]{e}}\left(e^{-\frac{15}{96n}-\frac{\pi^2}{6n}+o\left(\frac{1}{n}\right)}-1\right)= -\frac{1}{\sqrt[4]{e}}\left(\frac{15}{96}+\frac{\pi^2}{6} \right) \]
幸神的问题：设$\displaystyle p(x)=\sum_{i=0}^{n}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}$,若$\displaystyle a_{0}+\sum_{a_{i}<0}\left(1-\frac{i}{n}\right)C_{n}^{i}a_{i}>0$, 且$\displaystyle a_{n}+\sum_{a_{i}<0}C_{n}^{i}\cdot\frac{i}{n}\cdot a_{i}>0 $,求证：$\forall x\in[0,1]$,有$p(x)>0$.
证明：由Weight-AM-GM,有
\[ \left(1-\frac{i}{n}\right)(1-x)^{n}+\frac{i}{n}x^{n}\geq (1-x)^{n-i}x^{i} \]
这时对$p(x)$,有
\begin{align*}
p(x)&=\sum_{i=0}^{n}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}\\
&=a_{0}(1-x)^{n}+a_{n}x^{n}+\sum_{i=1}^{n-1}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}\\
&\geq a_{0}(1-x)^{n}+a_{n}x^{n}+\sum_{a_{i}<0}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}\\
&\geq a_{0}(1-x)^{n}+ a_{n}x^{n}+\sum_{a_{i}<0}C_{n}^{i}a_{i}\left[\left(1-\frac{i}{n}\right)(1-x)^{n}+\frac{i}{n}x^{n} \right]\\
&=(1-x)^{n}\left[a_{0}+\sum_{a_{i}<0}\left(1-\frac{i}{n}\right)C_{n}^{i}a_{i}\right]+x^{n}\left[a_{n}+\sum_{a_{i}<0}C_{n}^{i}\cdot\frac{i}{n}\cdot a_{i}\right]\\
&>0
\end{align*}
设$f,g$在$[a,b]$上可导，且$f,g$在$a$处二阶可导，满足$g(a)\neq g(b)$且满足
\[ \left[f'(a)-\frac{f(b)-f(a)}{g(b)-g(a)}\cdot g'(a)\right]\cdot\left[f''(a)- \frac{f(b)-f(a)}{g(b)-g(a)}\cdot g''(a)\right]>0 \]
(西西提供)
证明：存在$\eta\in(a,b)$使得
\[ f'(\eta)-\frac{f(\eta)-f(a)}{\eta-a}=\frac{f(b)-f(a)}{g(b)-g(a)}\cdot\left[g'(\eta)-\frac{g(\eta)-g(a)}{\eta-a} \right] \]
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证明：设
\[ K=\frac{f(b)-f(a)}{b-a} \]
\[ F(x)=f(x)-f(a)-K(g(x)-g(a)) \]
我们有
\[ F(a)=F(b)=0 \]
\[ F'(a)\cdot F''(a)>0 \]
于是
\[ F(x)=F(a)+F'(a)(x-a)+\frac{1}{2}F''(a)(x-a)^2+o((x-a)^2)\qquad (x\to a^{+}) \]
定义
\[ H(x)=\left\{
\begin{array}{ll}
\dfrac{F(x)}{x-a}, & \hbox{$x\in(a,b]$;} \\
F'(a), & \hbox{$x=a$.}
\end{array}\right.\]
$H(x)$是$[a,b]$上的可导函数，又有
\[ H(b)=0 \]
\[ H'(x)=F'(x)\cdot \frac{1}{x-a}-\frac{F(x)}{(x-a)^2}=\frac{F'(x)-F'(a)}{x-a}-\frac{1}{2}F''(a)+o(1) \]
所以就有
\[ \lim_{x\to a^{+}}H'(x)=\frac{1}{2}F''(a) \]
若不存在这样的$\eta$,就是对任意的$x\in(a,b)$有$H'(x)\neq 0 $,不妨设$H'(x)>0.(\forall x\in(a,b))$,则$H(x)$在$[a,b]$上严格递增，又由$H(b)=0$知必有$H(a)=F'(a)<0$,推得$F''(a)<0$,这时就有
\[ \lim_{x\to a^{+}}H'(x)=\frac{1}{2}F''(a)<0 \]
也就是说存在$\delta>0$,对任意的$x\in(a,a+\delta)$有
\[ H'(x)<0 \]
这与假设矛盾。所以必然存在一点$\eta\in(a,b)$使得$H'(\eta)=0$,这时，不难算得
\[ H'(\eta)=\frac{1}{\eta-a}\left[ f'(\eta)-\frac{f(\eta)-f(a)}{\eta-a}-\frac{f(b)-f(a)}{g(b)-g(a)}\cdot\left[g'(\eta)-\frac{g(\eta)-g(a)}{\eta-a} \right]\right]=0 \]
即
\[ f'(\eta)-\frac{f(\eta)-f(a)}{\eta-a}=\frac{f(b)-f(a)}{g(b)-g(a)}\cdot\left[g'(\eta)-\frac{g(\eta)-g(a)}{\eta-a} \right] \]
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最近有点忙，可能暂时不会更新博客了。
这几天得回去去一趟学校，办各种杂事各种忙。希望能早点回来学习。大家有问题可以发到我邮箱:pxchg1200@sina.com。
