陈洪葛的博客

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
\[ f'(\eta)-\frac{f(\eta)-f(a)}{\eta-a}=\frac{f(b)-f(a)}{g(b)-g(a)}\cdot\left[g'(\eta)-\frac{g(\eta)-g(a)}{\eta-a} \right] \]

_________________________________________________________________________________________________
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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计算

\[ \sum_{n=1}^{+\infty}(-1)^{n-1}\frac{\ln{n}}{n} \]

解:
首先，显然看到上面的级数是条件收敛的，所以设
\[ S_{n}=\sum_{k=1}^{n}(-1)^{k-1}\frac{\ln{k}}{k} \]
于是
\begin{align*}
S_{2n}&=\sum_{k=1}^{n}\frac{\ln(2k-1)}{2k-1}-\sum_{k=1}^{n}\frac{\ln(2k)}{2k}\\
&=\sum_{k=1}^{2n}\frac{\ln{k}}{k}-\sum_{k=1}^{n}\frac{\ln{2k}}{k}\\
&=\sum_{k=1}^{2n}\frac{\ln{k}}{k}-\sum_{k=1}^{n}\frac{\ln{k}}{k}-\ln{2}\cdot\sum_{k=1}^{n}\frac{1}{k}\\
&=\sum_{k=n+1}^{2n}\frac{\ln{k}}{k}-\ln{2}\sum_{k=1}^{n}\frac{1}{k}\\
&=\sum_{k=1}^{n}\frac{\ln(n+k)}{n+k}-\ln{2}\sum_{k=1}^{n}\frac{1}{k}\\
&=\sum_{k=1}^{n}\frac{1}{n+k}\cdot\ln\left(1+\frac{k}{n}\right)+\ln{n}\sum_{k=1}^{n}\frac{1}{n+k}-\ln{2}\sum_{k=1}^{n}\frac{1}{k}\\
\end{align*}
这时注意到
\[ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n+k}\cdot\ln\left(1+\frac{k}{n}\right)=\int_{0}^{1}\frac{\ln(1+x)}{1+x}dx=\frac{1}{2}\ln^{2}{2}\]
以及熟悉的
\[ \sum_{k=1}^{n}\frac{1}{k}=\ln{n}+\gamma+\varepsilon_{n} \]
于是
\[ \ln{n}\sum_{k=1}^{n}\frac{1}{n+k}-\ln{2}\sum_{k=1}^{n}\frac{1}{k}=\ln{n}\left(\sum_{k=1}^{n}\frac{1}{n+k}-\ln{2}\right)-\gamma\ln{2}\]

事实上又有
\[  \int_{\frac{1}{n}}^{1+\frac{1}{n}}\frac{1}{1+x}dx\leq\sum_{k=1}^{n}\frac{1}{n+k}\leq \ln{2}\]
不难算到
\[ \sum_{k=1}^{n}\frac{1}{n+k}-\ln{2}\sim o\left(\frac{1}{2n}\right) \]
于是
\[ \lim_{n\to\infty}\left[\ln{n}\sum_{k=1}^{n}\frac{1}{n+k}-\ln{2}\sum_{k=1}^{n}\frac{1}{k}\right]=-\gamma\ln{2} \]
最后
\[ \sum_{n=1}^{+\infty}(-1)^{n-1}\frac{\ln{n}}{n}=\frac{1}{2}\ln^{2}{2}-\gamma\ln{2} \]