西神的问题

陈洪葛 posted @ May 02, 2014 05:50:19 PM in 数学分析 , 810 阅读

计算无穷级数
\[ \sum_{k=1}^{\infty}\frac{1}{k^2}\cos\left(\frac{9}{k\pi+\sqrt{k^2\pi^2-9}}\right)\]
西神说一般的我们有
\[ \sum_{k=1}^{\infty}\frac{\cos(k\pi-\sqrt{k^2\pi^2-a^2})}{k^2}=\frac{\pi^2}{12}\left(\cosh(a)+\frac{3}{a}\sinh(a)\right)\]
令$a=3$就有
\[ \sum_{k=1}^{\infty}\frac{1}{k^2}\cos\left(\frac{9}{k\pi+\sqrt{k^2\pi^2-9}}\right)=-\frac{\pi^2}{12e^{3}}\]
继续可以推广
\[ \sum_{n=0}^{\infty}\frac{n^{2}\pi^{2}+\phi^{2}}{(n^{2}\pi^{2}-\phi^{2})^{2}}(-1)^{n}\cos\sqrt{n^{2}\pi^{2}+a^{2}-\phi^{2}}=\frac{\cos\sqrt{a^2-\phi^{2}}}{2\phi^{2}}+\frac{a\cos{a}\cot\phi+\phi\sin{a}}{2a\sin\phi}\]


\[ \iiint_{D}\ln{x}\ln{y}\ln{z}\cos(x^2+y^2+z^2)dxdydz\]
其中
\[ D:[0,+\infty)\times[0,+\infty)\times[0,+\infty)\]


设$x_{1},x_{2},\cdots,x_{n}(n\geq 3)$是任意实数,且满足$x_{1}x_{2}\cdots x_{n}=1$,求证
\[ \sum_{k=1}^{n}\frac{x_{k}^{2}}{x_{k}^{2}-2x_{k}\cos\frac{2\pi}{n}+1}\geq 1 \]


设$x_{1},x_{2},\cdots,x_{n}(n\geq 3)$是正实数,且满足
\[ \sum_{1\leq i,j\leq n}|1-x_{i}x_{j}|=\sum_{1\leq i,j\leq n}|x_{i}-x_{j}| \]
求证:对任意的实数$a_{1},a_{2},\cdots,a_{n}$,存在实数$t$,使得
\[ \sum_{i=1}^{n}|\sin(t-a_{i})|\leq \cot\left(\frac{\pi}{2\sum\limits_{i=1}^{n}x_{i}}\right)\]

 


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