数学钟的12个问题 - Eufisky - The lost book
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2013年武汉大学数学分析考研真题

数学钟的12个问题

Eufisky posted @ 2014年4月12日 16:47 in 数学分析 , 1364 阅读

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

上面是自己在网上看到的一个数学钟,设计者真是别有用心,于是利用周末闲暇时光,对上面的12个问题,一一给出说明.

数字1:Euler公式
\[e^{i\pi}+1=0.\]
数字2
对于
\[\Gamma \left( {x + 1} \right) = \int_0^\infty  {{e^{ - v}}{v^x}dv} .\]
我们有
\[\Gamma '\left( {x + 1} \right) = \int_0^\infty  {{e^{ - v}}{v^x}\ln vdv} .\]
另外
\[\frac{{\Gamma '\left( {x + 1} \right)}}{{\Gamma \left( {x + 1} \right)}} =  - \gamma  + \sum\limits_{k = 1}^\infty  {\frac{x}{{k\left( {x + k} \right)}}}  \Rightarrow \Gamma '\left( 1 \right) =  - \gamma .\]
所以我们有
\[\Gamma '\left( 1 \right) = \int_0^\infty  {{e^{ - v}}\ln vdv}  =  - \gamma. \]
由此得
\begin{align*}\int_0^\infty  {\frac{{{e^{ - {x^2}}} - {e^{ - x}}}}{x}dx}  &= \left[ {\left( {{e^{ - {x^2}}} - {e^{ - x}}} \right)\ln x} \right]_0^\infty  + \int_0^\infty  {2x{e^{ - {x^2}}}\ln xdx}  - \int_0^\infty  {{e^{ - x}}\ln xdx}  \\&= 0 + \frac{1}{2}\int_0^\infty  {{e^{ - t}}{\mathop{\rm lnt}\nolimits} dt}  - \int_0^\infty  {{e^{ - x}}\ln xdx}  =  - \frac{1}{2}\int_0^\infty  {{e^{ - x}}\ln xdx}  = \frac{\gamma }{2}.\end{align*}
数字3
\[\frac{\pi}{4}=5\arctan{\frac{1}{7}}+2\arctan{\frac{3}{79}}.\]
证明和数字5的大致是一样的,只是计算量偏大了点
\[{\left( {7 - i} \right)^5}\left( {1 + i} \right) = {\left( {48 - 14i} \right)^2}\left( {8 + 6i} \right) = \left( {2108 - 1344i} \right)\left( {8 + 6i} \right) = 8\left( {3116 + 237i} \right) = 4{\left( {79 + 3i} \right)^2}.\]
由此得
\[\arg \left( {1 + i} \right) + \arg {\left( {7 - i} \right)^5} = \arg \left[ {4{{\left( {79 + 3i} \right)}^2}} \right] = \arg {\left( {79 + 3i} \right)^2}.\]
若取$\displaystyle -\pi<\arg\leq\pi$,则由上式可知
\[\frac{\pi }{4} - 5\arctan \frac{1}{7} = 2\arctan \frac{3}{{79}}.\]
\[\frac{\pi }{4} =5\arctan \frac{1}{7} + 2\arctan \frac{3}{{79}}.\]
这一结果属于Leonhard Euler.
数字4:
\[\arctan x = \sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{2n + 1}}} .\]
再令$\displaystyle x=1$即可.
数字5:
应用
\[\frac{\pi}{4}=4\arctan{\frac{1}{5}}-\arctan{\frac{1}{239}}.\]
下面是泉叔书上给出的一个较为简便的证明方法
\[{\left( {5 - i} \right)^4}\left( {1 + i} \right) = {\left( {24 - 10i} \right)^2}\left( {1 + i} \right) = \left( {476 - 480i} \right)\left( {1 + i} \right) = 4\left( {239 - i} \right).\]
于是
\[\arg \left( {1 + i} \right) + \arg {\left( {5 - i} \right)^4} = \arg \left[ {4\left( {239 - i} \right)} \right] = \arg \left( {239 - i} \right).\]
若取$\displaystyle -\pi<\arg\leq\pi$,则由上式可知
\[\frac{\pi }{4} - 4\arctan \frac{1}{5} =  - \arctan \frac{1}{{239}}.\]
\[\frac{\pi }{4} =4\arctan \frac{1}{5}   - \arctan \frac{1}{{239}}.\]
这一结果归功于 John Machin.
数字6:
\[\int_0^1 {\frac{{\ln x}}{{x - 1}}dx}  = \int_1^0 {\frac{{\ln \left( {1 - t} \right)}}{t}dt}  = \mathrm{Li}_2\left( 1 \right) = \frac{{{\pi ^2}}}{6}.\]
数字7:
\[\tan \left( {2\arctan \frac{1}{3}} \right) = \frac{{2 \cdot \frac{1}{3}}}{{1 - {{\left( {\frac{1}{3}} \right)}^2}}} = \frac{3}{4}\]
 \[\Rightarrow \tan \left( {\frac{\pi }{4} - 2\arctan \frac{1}{3}} \right) = \frac{{1 - \frac{3}{4}}}{{1 + \frac{3}{4}}} = \frac{1}{7}.\]
数字8:
\[\sum\limits_{k = 0}^\infty  {\frac{1}{{{{\left( {2k + 1} \right)}^2}}}}  = \sum\limits_{k = 1}^\infty  {\frac{1}{{{k^2}}}}  - \sum\limits_{k = 1}^\infty  {\frac{1}{{{{\left( {2k} \right)}^2}}}}  = \frac{3}{4}\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^2}}}}  = \frac{{{\pi ^2}}}{8}.\]
数字9:注意到
\[\prod\limits_{k = 1}^m {\tan \frac{{k\pi }}{{2m + 1}}}  = \sqrt {2m + 1}.\]
令$\displaystyle m=4$即可得到.
数字10:
\[\sin \frac{\pi }{{10}} = \frac{{\sqrt 5  - 1}}{4}.\]
数字11
首先证明:$\displaystyle\sum\limits_{k = 1}^{n - 1} {{{\tan }^2}\frac{{k\pi }}{{2n}}}  = \frac{{\left( {n - 1} \right)\left( {2n - 1} \right)}}{3}.$
由Euler公式
\[\left(\cos{\frac{k\pi}{2n}} + i\sin{\frac{k\pi}{2n}}\right)^{2n}=(-1)^{k}.\]
由牛顿二项式定理
\[\sum_{t=0}^{2n}\binom{2n}{t}\left(\cos{\frac{k\pi}{2n}}\right)^t \cdot \left(i\sin{\frac{k\pi}{2n}}\right)^{2n-t}=(-1)^{k}.\]
只考虑虚部
\[\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(\cos{\frac{k\pi}{2n}}\right)^{2r+1} \cdot \left(i\sin{\frac{k\pi}{2n}}\right)^{2n-2r-1}=0.\]
两边同除 $\displaystyle \left(\cos{\frac{k\pi}{2n}}\right)^{2n}$
\[\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(i\tan{\frac{k\pi}{2n}}\right)^{2n-2r-1}=0.\]
两边同乘$\displaystyle i\tan{\frac{k\pi}{2n}}$
\[\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(i\tan{\frac{k\pi}{2n}}\right)^{2n-2r}=0.\]
故 $\displaystyle \tan^2{\frac{k\pi}{2n}}$ 是以下多项式方程的根
\[\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(-x\right)^{n-r}=0.\]
根据 Vieta公式我们有
\[\frac{\binom{2n}{3}}{\binom{2n}{1}} =\frac{(2n-1)(n-1)}{3}.\]
进一步,我们有
\[\sum\limits_{k = 1}^{2n - 1} {{{\csc }^4}\frac{{k\pi }}{{2n}}}= \frac{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 11} \right)}}{{45}}.\]
证:
\begin{align*}\sum\limits_{k = 1}^{2n - 1} {{{\csc }^4}\frac{{k\pi }}{{2n}}}  &= \sum\limits_{k = 1}^{2n - 1} {\frac{1}{{{{\sin }^4}\frac{{k\pi }}{{2n}}}}}  = \sum\limits_{k = 1}^{2n - 1} {\frac{1}{{{{\sin }^4}\frac{{k\pi }}{{2n}}}}}  \\&= \sum\limits_{k = 1}^{2n - 1} {\frac{{{{\left( {{{\sin }^2}\frac{{k\pi }}{{2n}} + {{\cos }^2}\frac{{k\pi }}{{2n}}} \right)}^2}}}{{{{\sin }^4}\frac{{k\pi }}{{2n}}}}}  \\&= \sum\limits_{k = 1}^{2n - 1} {\left( {1 + 2{{\cot }^2}\frac{{k\pi }}{{2n}} + {{\cot }^4}\frac{{k\pi }}{{2n}}} \right)}  \\&= 2n - 1 + 2\sum\limits_{k = 1}^{2n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}}  + \sum\limits_{k = 1}^{2n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}}.\end{align*}
先考虑前者
\begin{align*}\sum\limits_{k = 1}^{2n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}}  &= \sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}}  + \sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{\left( {2n - k} \right)\pi }}{{2n}}}  + {\cot ^4}\frac{{n\pi }}{{2n}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}}  = 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{\left( {n - k} \right)\pi }}{{2n}}}  \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\tan }^2}\frac{{k\pi }}{{2n}}}  = \frac{{2\left( {n - 1} \right)\left( {2n - 1} \right)}}{3}.\end{align*}
再考虑后者
\begin{align*}\sum\limits_{k = 1}^{2n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}}  &= \sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}}  + \sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{\left( {2n - k} \right)\pi }}{{2n}}}  + {\cot ^4}\frac{{n\pi }}{{2n}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}}  = 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{\left( {n - k} \right)\pi }}{{2n}}}  \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\tan }^4}\frac{{k\pi }}{{2n}}}  = 2\left[ {{{\left( {\sum\limits_{k = 1}^{n - 1} {{{\tan }^2}\frac{{k\pi }}{{2n}}} } \right)}^2} - 2\sum\limits_{1 \le i < j \le n - 1} {{{\tan }^2}\frac{{i\pi }}{{2n}}{{\tan }^2}\frac{{j\pi }}{{2n}}} } \right]\\&= 2\left[ {\frac{{{{\left( {n - 1} \right)}^2}{{\left( {2n - 1} \right)}^2}}}{9} - 2\frac{{\left( \begin{array}{c}2n\\5\end{array} \right)}}{{\left( \begin{array}{c}2n\\1\end{array} \right)}}} \right] \\&= \frac{{2{{\left( {n - 1} \right)}^2}{{\left( {2n - 1} \right)}^2}}}{9} - \frac{{2\left( {n - 2} \right)\left( {n - 1} \right)\left( {2n - 3} \right)\left( {2n - 1} \right)}}{{15}} \\&= \frac{{\left( {n - 1} \right)\left( {2n - 1} \right)\left( {24{n^2} + 36n - 78} \right)}}{{135}} = \frac{{2\left( {n - 1} \right)\left( {2n - 1} \right)\left( {4{n^2} + 6n - 13} \right)}}{{45}}.\end{align*}
由此
\begin{align*}\Rightarrow \sum\limits_{k = 1}^{2n - 1} {{{\csc }^4}\frac{{k\pi }}{{2n}}}  &= 2n - 1 + \frac{{4\left( {n - 1} \right)\left( {2n - 1} \right)}}{3} + \frac{{2\left( {n - 1} \right)\left( {2n - 1} \right)\left( {4{n^2} + 6n - 13} \right)}}{{45}} \\&= 2n - 1 + \frac{{\left( {n - 1} \right)\left( {2n - 1} \right)\left( {8{n^2} + 12n + 34} \right)}}{{45}} \\&= \frac{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 11} \right)}}{{45}}.\end{align*}
特别地有
\[ \Rightarrow \sum\limits_{k = 1}^{999} {{{\csc }^4}\frac{{k\pi }}{{1000}}}  = 1000009999989 = {10^{12}} + {10^7} - 11.\]
数字12
\begin{align*}\int_0^\infty  {\frac{{xdx}}{{\sinh \left( {\sqrt 3 x} \right)}}} &= \int_0^\infty  {\frac{{xdx}}{{\frac{{{e^{\sqrt 3 x}} - {e^{ - \sqrt 3 x}}}}{2}}}}  = \int_0^\infty  {\frac{{2xdx}}{{{e^{\sqrt 3 x}} - {e^{ - \sqrt 3 x}}}}} \\&\underline{\underline {t = {e^{ - \sqrt 3 x}}}} \frac{2}{3}\int_0^1 {\frac{{\ln t}}{{{t^2} - 1}}dt}  = \frac{1}{3}\int_0^1 {\frac{{\ln t}}{{t - 1}}dt}  - \frac{1}{3}\int_0^1 {\frac{{\ln t}}{{t + 1}}dt}  \\&= \frac{1}{3}\int_0^1 {\frac{{\ln t}}{{t - 1}}dt}  - \frac{1}{3}\left[ {\ln t\ln \left( {t + 1} \right)} \right]_0^1 + \frac{1}{3}\int_0^1 {\frac{{\ln \left( {t + 1} \right)}}{t}dt} \\&= \frac{1}{3}\int_1^0 {\frac{{\ln \left( {1 - t} \right)}}{t}dt}  - \frac{1}{3}\int_{ - 1}^0 {\frac{{\ln \left( {1 - t} \right)}}{t}dt}  = \frac{1}{3}\mathrm{Li}_2\left( 1 \right) - \frac{1}{3}\mathrm{Li}_2\left( { - 1} \right) \\&= \frac{1}{3}\left( {\frac{{{\pi ^2}}}{6} + \frac{{{\pi ^2}}}{{12}}} \right) = \frac{{{\pi ^2}}}{{12}}\end{align*}

 

 


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