西西11年送的几个CMC培训题整理

Eufisky posted @ 2014年8月24日 04:27 in 数学分析 with tags 大学生数学竞赛 , 1424 阅读

1.若存在正整数$a$，使得连续函数$f:[0,+\infty)\to[0,+\infty)$满足$f[f(x)]=x^a$，对任意的$x\in[0,+\infty)$，求证：$\int_0^1{[f(x)]}^2 dx\geq\frac{2a-1}{a^2+6a-3}$.

2.$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{{\left( {2n - 1} \right)}^{2011}}}}\sum\limits_{k = 0}^{n - 1} {\int_{2k\pi }^{\left( {2k + 1} \right)\pi } {{x^{2010}}{{\sin }^3}x{{\cos }^2}xdx} } .$

3.$\int_0^\infty {\frac{{\left[ {\frac{x}{\pi }} \right]}}{{{e^x} + 1}}dx} ,$其中$[a]$为取整函数.

4.$\int_1^\infty {\frac{{dx}}{{2\left[ x \right] + 3{{\left[ x \right]}^2} + {{\left[ x \right]}^3}}}} .$

5.$\sum\limits_{n = 2}^\infty {\ln \left( {1 - \frac{1}{{n\left( {n - 1} \right)}}} \right)} .$

6.求证：$\mathop {\lim }\limits_{n \to \infty } \frac{{\ln C_{np}^0 + \ln C_{np}^p + \ln C_{np}^{2p} + \cdots + \ln C_{np}^{np}}}{{{n^2}}} = \frac{p}{2},\frac{p}{2} \in {N^ * }.$

7.求极限$\mathop {\lim }\limits_{x \to 0} \int_0^{ + \infty } {\frac{{{y^3}\left( {y - 3} \right){e^{ - y}}}}{{1 - {e^{ - xy}}}}dy} .$

8.设$f_1(x)=x,f_2(x)=x^x,\ldots,f(x)={{x^x}^{\scalebox{-1}[1]{$\ddots$}}}^x$，求$\mathop {\lim }\limits_{x \to 1}\frac{{{f_n}\left( x \right) - {f_{n - 1}}\left( x \right)}}{{{{\left( {1 - x} \right)}^n}}}.$

9.求$\int_0^1 {\left( {\frac{{{x^{A - 1}}\cos \left( {B\ln x} \right) - {x^{CA - 1}}\cos \left( {BC\ln x} \right)}}{{\ln x}}\sum\limits_{k = 0}^\infty {{x^{D{C^k}}}} } \right)dx}$的值，其中$A,B,D>0$，且$C>1$（后面的$DC^k$单指$C$的$k$次方）.

1.（_伽罗华_）易得$f$为单射且严格单调递增，$f(0)=0,f(1)=1$，从而$\int_0^1 {f\left( x \right)dx} + \int_0^1 {{f^{ - 1}}\left( x \right)dx} = 1.$

2.

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5.（tian27546西西）因为$\cos x = \prod\limits_{n = 1}^\infty {\left( {1 - \frac{{4{x^2}}}{{{{\left( {2n - 1} \right)}^2}{\pi ^2}}}} \right)} .$令$x=\frac{\sqrt{5}}{2}\pi$得到$\cos \frac{{\sqrt 5 }}{2}\pi = \prod\limits_{n = 1}^\infty {\frac{{4{n^2} - 4n - 4}}{{{{\left( {2n - 1} \right)}^2}}}} .$又$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{2n!!}}{{\left( {2n - 1} \right)!!}}} \right)^2}\frac{1}{{2n + 1}} = \frac{\pi }{2}.$两式作除整理得：$\mathop {\lim }\limits_{n \to \infty } \prod\limits_{i = 2}^n {\frac{{{i^2} - i - 1}}{{i\left( {i - 1} \right)}}} = \mathop {\lim }\limits_{n \to \infty } \prod\limits_{i = 2}^n {\left( {1 - \frac{1}{{i\left( {i - 1} \right)}}} \right)} = - \frac{{\cos \left( {\frac{{\sqrt 5 }}{2}\pi } \right)}}{\pi }.$故$\sum\limits_{n = 2}^\infty {\ln \left( {1 - \frac{1}{{n\left( {n - 1} \right)}}} \right)} = \ln \left( { - \frac{{\cos \left( {\frac{{\sqrt 5 }}{2}\pi } \right)}}{\pi }} \right).$

\begin{align*}\prod\limits_{n = 2}^\infty  {\frac{{{n^2} - n - 1}}{{n\left( {n - 1} \right)}}}  = \prod\limits_{n = 2}^\infty  {\frac{{\left( {n - \alpha } \right)\left( {n - \beta } \right)}}{{n\left( {n - 1} \right)}}} \\= \frac{1}{{\Gamma \left( { - \alpha } \right)}} \cdot \frac{1}{{\Gamma \left( { - \beta } \right)}} \cdot \frac{{{n^{ - \alpha }}}}{{ - \alpha \left( {1 - \alpha } \right)}} \cdot \frac{{{n^{ - \beta }}}}{{ - \beta \left( {1 - \beta } \right)}}\\= \frac{1}{{\Gamma \left( { - \alpha } \right)}} \cdot \frac{1}{{\Gamma \left( { - \beta } \right)}} = \frac{{ - 1}}{{\Gamma \left( {1 - \alpha } \right)\Gamma \left( {1 - \beta } \right)}}\\= \frac{{ - 1}}{{\frac{\pi }{{\sin \left[ {\left( {1 - \frac{{1 + \sqrt 5 }}{2}} \right)\pi } \right]}}}} =  - \frac{{\cos \frac{{\sqrt 5 }}{2}\pi }}{\pi },\end{align*}其中$\alpha = \frac{{1 + \sqrt 5 }}{2},\beta = \frac{{1 - \sqrt 5 }}{2},\alpha + \beta = 1,\alpha \beta = - 1.$

6.

\begin{align*}\sum\limits_{k = 0}^n {\ln C_{np}^{kp}}  &= \sum\limits_{k = 0}^n {\ln \frac{{\left( {np} \right)!}}{{\left( {kp} \right)!\left( {np - kp} \right)!}}}  = \left( {n + 1} \right)\ln \left( {np} \right)! - \sum\limits_{k = 0}^n {\ln \left( {\left( {kp} \right)!\left( {np - kp} \right)!} \right)} \\&= \left( {n + 1} \right)\ln \Gamma \left( {np + 1} \right) - 2\sum\limits_{k = 0}^n {\ln \Gamma \left( {kp + 1} \right)} \\&= \left( {n + 1} \right)\left\{ {\left( {np + \frac{1}{2}} \right)\ln \left( {np + 1} \right) - \left( {np + 1} \right) + \frac{1}{2}\ln \left( {2\pi } \right) + 2\int_0^\infty  {\frac{{\arctan \frac{t}{{np + 1}}}}{{{e^{2m}} - 1}}dt} } \right\}\\&- 2\sum\limits_{k = 0}^n {\left\{ {\left( {kp + \frac{1}{2}} \right)\ln \left( {kp + 1} \right) - \left( {kp + 1} \right) + \frac{1}{2}\ln \left( {2\pi } \right) + 2\int_0^\infty  {\frac{{\arctan \frac{t}{{kp + 1}}}}{{{e^{2m}} - 1}}dt} } \right\}} .\end{align*}因为$\int_0^\infty {\frac{{\arctan \frac{t}{{kp + 1}}}}{{{e^{2m}} - 1}}dt} \le \int_0^\infty {\frac{{\arctan t}}{{{e^{2m}} - 1}}dt} \approx 0.0405,$所以上式的主要部分是$\left( {n + 1} \right)\left( {np + \frac{1}{2}} \right)\ln \left( {np + 1} \right) - 2\sum\limits_{k = 0}^n {\left( {kp + \frac{1}{2}} \right)\ln \left( {kp + 1} \right)} .$再利用下Euler-Maclaurin求和公式\begin{align*}&2\sum\limits_{k = 0}^n {\left( {kp + \frac{1}{2}} \right)\ln \left( {kp + 1} \right)}  \approx \frac{2}{p}\int_0^{np} {\left( {x + \frac{1}{2}} \right)\ln \left( {x + 1} \right)dx}  + \left( {np + \frac{1}{2}} \right)\ln \left( {np + 1} \right)\\&\approx  - \frac{{{n^2}p}}{2} + n\ln \left( {np + 1} \right) + {n^2}p\ln \left( {np + 1} \right) + \left( {np + \frac{1}{2}} \right)\ln \left( {np + 1} \right).\end{align*}所以原极限等于$\frac p2.$
（tian27546西西）由Stirling公式得：存在常数$C_1,C_2$使得$\frac{{{C_1}}}{n}\frac{{{n^{pn}}}}{{{k^{pk}}{{\left( {n - k} \right)}^{p\left( {n - k} \right)}}}}\le \left( \begin{array}{l}np\\kp\end{array} \right) \le {C_2}\frac{{{n^{pn}}}}{{{k^{pk}}{{\left( {n - k} \right)}^{p\left( {n - k} \right)}}}}.$

（ygc136441788）求$\mathop {\lim }\limits_{n \to \infty } \frac{{\ln C_{np}^0 + \ln C_{np}^p + \ln C_{np}^{2p} + \cdots \ln C_{np}^{np}}}{{{n^2}}}.$

\begin{align*}&\frac{{C_{\left( {n + 1} \right)p}^{kp}}}{{C_{np}^{kp}}} = \frac{{\left( {np + p} \right)!}}{{\left( {kp} \right)!\left( {np + p - kp} \right)!}} \cdot \frac{{\left( {kp} \right)!\left( {np - kp} \right)!}}{{\left( {np} \right)!}}\\&= \frac{{\left( {np + p} \right)\left( {np + p - 1} \right)\left( {np + p - 2} \right) \cdots \left( {np + 1} \right)}}{{\left( {np + p - kp} \right)\left( {np + p - 1 - kp} \right)\left( {np + p - 2 - kp} \right) \cdots \left( {np - kp + 1} \right)}}.\end{align*}
\begin{align*}&\prod\limits_{k = 0}^n {\left( {np + p - kp} \right)\left( {np + p - 1 - kp} \right)\left( {np + p - 2 - kp} \right) \cdots \left( {np - kp + 1} \right)} \\&= \left[ {p\left( {p - 1} \right)\left( {p - 2} \right) \cdots \left( 1 \right)} \right]\left[ {\left( {2p} \right)\left( {2p - 1} \right) \cdots \left( {p + 1} \right)} \right] \cdots \left[ {\left( {\left( {n + 1} \right)p} \right)\left( {\left( {n + 1} \right)p - 1} \right) \cdots \left( {np + 1} \right)} \right]\\&= \left( {np + p} \right)!.\end{align*}
\begin{align*}&{X_{n + 1}} - {X_n} = \ln \prod\limits_{k = 0}^n {\frac{{C_{\left( {n + 1} \right)p}^{kp}}}{{C_{np}^{kp}}}}  + \ln C_{\left( {n + 1} \right)p}^{np + 1} + \ln C_{\left( {n + 1} \right)p}^{np + 2} +  \cdots \ln C_{\left( {n + 1} \right)p}^{\left( {n + 1} \right)p}\\&= \ln \frac{{{{\left( {np + p} \right)}^{n + 1}}{{\left( {np + p - 1} \right)}^{n + 1}}{{\left( {np + p - 2} \right)}^{n + 1}} \cdots {{\left( {np + 1} \right)}^{n + 1}}}}{{\left( {np + p} \right)!}}.\end{align*}

\begin{align*}&\mathop {\lim }\limits_{n \to \infty } \frac{{\ln C_{np}^0 + \ln C_{np}^p + \ln C_{np}^{2p} +  \cdots \ln C_{np}^{np}}}{{{n^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\ln \frac{{{{\left[ {\left( {np + p} \right)\left( {np + p - 1} \right)\left( {np + p - 2} \right) \cdots \left( {np + 1} \right)} \right]}^n}}}{{\left( {np + p} \right)!}}}}{{2n + 1}}\\&= \mathop {\lim }\limits_{n \to \infty } \frac{{\ln \frac{{{{\left( {np} \right)}^n}{{\left[ {\left( {1 + \frac{p}{{np}}} \right)\left( {1 + \frac{{p - 1}}{{np}}} \right)\left( {1 + \frac{{p - 2}}{{np}}} \right) \cdots \left( {1 + \frac{1}{{np}}} \right)} \right]}^n}}}{{\left( {np + p} \right)!}}}}{{2n + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\ln \frac{{{{\left( {np} \right)}^{np}}{{\left[ {\left( {1 + \frac{p}{{np}}} \right)\left( {1 + \frac{{p - 1}}{{np}}} \right)\left( {1 + \frac{{p - 2}}{{np}}} \right) \cdots \left( {1 + \frac{1}{{np}}} \right)} \right]}^n}}}{{{{\left( {\frac{{np + p}}{e}} \right)}^{np + p}}\sqrt {2\pi \left( {np + p} \right)} }}}}{{2n + 1}}\\&= \mathop {\lim }\limits_{n \to \infty } \frac{{np\ln \left( {np} \right) + n\sum\limits_{k = 1}^p {\ln \left( {1 + \frac{k}{{np}}} \right)}  - \left( {np + p} \right)\ln \left( {np + p} \right) + np + p - \ln \sqrt {2\pi \left( {np + p} \right)} }}{{2n + 1}} = \frac{p}{2}.\end{align*}

7.（博士哥）\begin{align*}&\int_0^\infty  {\frac{{{y^3}\left( {y - 3} \right){e^{ - y}}}}{{1 - {e^{ - xy}}}}dy}  = \int_0^\infty  {{y^3}\left( {y - 3} \right){e^{ - y}}\sum\limits_{n = 0}^\infty  {{e^{ - ny}}} dy} \\&= \sum\limits_{n = 0}^\infty  {\frac{{6 - 18nx}}{{{{\left( {1 + nx} \right)}^5}}}}  =  - \frac{{3x\mathrm{PolyGamma}\left[ {3,\frac{1}{x}} \right] + \mathrm{PolyGamma}\left[ {4,\frac{1}{x}} \right]}}{{{x^5}}}.\end{align*}可以利用$\mathrm{PolyGamma\left[0,\frac1x\right]}$的渐近表示式求出该极限3.

(tian27546西西)我们先证明几个引理：

$\sum\limits_{k = 0}^\infty {\frac{{f\left( k \right) + f\left( {k + 1} \right)}}{2}} = - \frac{{f\left( 0 \right)}}{2} + \sum\limits_{k = 0}^\infty {f\left( k \right)} = - \frac{n}{2} + \sum\limits_{k = 0}^\infty {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}} .$因为$y=f(x)$曲线下凹，所以折线下的梯形面积大于$[0,\infty)$上曲线$y=f(x)$下的曲边梯形面积，有

\begin{align*}&- \frac{n}{2} + \sum\limits_{k = 0}^\infty  {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}}  > \int_0^\infty  {f\left( x \right)dx}  = \int_0^\infty  {\frac{n}{{{{\left( {1 + xt} \right)}^{n + 1}}}}dx}  =  - \frac{1}{{t{{\left( {1 + xt} \right)}^n}}}\left| {_0^\infty } \right. = \frac{1}{t}\\&\Rightarrow \frac{1}{t} + \frac{n}{2} < \sum\limits_{k = 0}^\infty  {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}} .\end{align*}

$- n + \sum\limits_{k = 0}^\infty {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}} < \int_{\frac{1}{2}}^\infty {f\left( x \right)dx} = \int_{\frac{1}{2}}^\infty {\frac{n}{{{{\left( {1 + xt} \right)}^{n + 1}}}}dx} = - \frac{1}{{t{{\left( {1 + xt} \right)}^n}}}\left| {_{\frac{1}{2}}^\infty } \right. = \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^n}}}$

$\sum\limits_{k = 0}^\infty {\frac{{1 - 3kt}}{{{{\left( {1 + kt} \right)}^5}}}} = \sum\limits_{k = 0}^\infty {\frac{{4 - 3\left( {1 + kt} \right)}}{{{{\left( {1 + kt} \right)}^5}}}} = \sum\limits_{k = 0}^\infty {\frac{4}{{{{\left( {1 + kt} \right)}^5}}}} - \sum\limits_{k = 0}^\infty {\frac{3}{{{{\left( {1 + kt} \right)}^4}}}} .$

$\frac{1}{t} + 2 < \sum\limits_{k = 0}^\infty {\frac{4}{{{{\left( {1 + kt} \right)}^5}}}} < \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^4}}} + 4,\frac{1}{t} + \frac{3}{2} < \sum\limits_{k = 0}^\infty {\frac{3}{{{{\left( {1 + kt} \right)}^4}}}} < \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^3}}} + 3.$

$\frac{1}{t} + 2 - \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^3}}} - 3 < \sum\limits_{k = 0}^\infty {\frac{4}{{{{\left( {1 + kt} \right)}^5}}}} - \sum\limits_{k = 0}^\infty {\frac{3}{{{{\left( {1 + kt} \right)}^4}}}} < \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^4}}} + 4 - \frac{1}{t} - \frac{3}{2}.$

\begin{align*}&\mathop {\lim }\limits_{t \to 0} \frac{{{{\left( {1 + \frac{t}{2}} \right)}^3} - 1}}{{t{{\left( {1 + \frac{t}{2}} \right)}^3}}} = \mathop {\lim }\limits_{t \to 0} \frac{{\frac{{3t}}{2} + \frac{{3{t^2}}}{4} + \frac{{{t^3}}}{8}}}{{t{{\left( {1 + \frac{t}{2}} \right)}^3}}} = \mathop {\lim }\limits_{t \to 0} \frac{{\frac{3}{2} + \frac{{3t}}{4} + \frac{{{t^2}}}{8}}}{{{{\left( {1 + \frac{t}{2}} \right)}^3}}} = \frac{3}{2}\\&\mathop {\lim }\limits_{t \to 0} \frac{{{{\left( {1 + \frac{t}{2}} \right)}^4} - 1}}{{t{{\left( {1 + \frac{t}{2}} \right)}^4}}} = \mathop {\lim }\limits_{t \to 0} \frac{{2t + \frac{{3{t^2}}}{2} + \frac{{{t^3}}}{2} + \frac{{{t^4}}}{{16}}}}{{t{{\left( {1 + \frac{t}{2}} \right)}^4}}} = \mathop {\lim }\limits_{t \to 0} \frac{{2 + \frac{{3t}}{2} + \frac{{{t^2}}}{2} + \frac{{{t^3}}}{{16}}}}{{{{\left( {1 + \frac{t}{2}} \right)}^4}}} = 2.\end{align*}

8.（博士哥）$x\to1$时，可以得到$f_n(x)=1-(1-x)f_{n-1}(x)+O((1-x)^2)$，所以$x\to1$时${f_n}\left( x \right) \approx {\left( {x - 1} \right)^{n - 1}}\left( {x - \frac{1}{{2 - x}}} \right) + \frac{1}{{2 - x}}.$因此$\mathop {\lim }\limits_{x \to 1} \frac{{{f_n}\left( x \right) - {f_{n - 1}}\left( x \right)}}{{{{\left( {1 - x} \right)}^n}}} = \mathop {\lim }\limits_{x \to 1} \frac{{{{\left( {x - 1} \right)}^{n - 1}}\left( {x - \frac{1}{{2 - x}}} \right) - {{\left( {x - 1} \right)}^{n - 2}}\left( {x - \frac{1}{{2 - x}}} \right)}}{{{{\left( {1 - x} \right)}^n}}} = 1.$

9.\begin{align*}\int_0^1 {\left( {\frac{{{x^{A - 1}}\cos \left( {B\ln x} \right) - {x^{CA - 1}}\cos \left( {BC\ln x} \right)}}{{\ln x}}\sum\limits_{k = 0}^\infty  {{x^{D{C^k}}}} } \right)dx} \\= \int_0^\infty  {\left( {\frac{{{e^{ - Ax}}\cos \left( {Bx} \right) - {e^{ - ACx}}\cos \left( {BCx} \right)}}{x}\sum\limits_{k = 0}^\infty  {{e^{ - D{C^k}x}}} } \right)dx} .\end{align*}

\begin{align*}&\mathcal{L}\left( {\left( {{e^{ - Ax}}\cos \left( {Bx} \right) - {e^{ - ACx}}\cos \left( {BCx} \right)} \right)\sum\limits_{k = 0}^\infty  {{e^{ - D{C^k}x}}} } \right)\\&= \sum\limits_{k = 0}^\infty  {\frac{{A + D{C^k} + p}}{{{{\left( {A + D{C^k} + p} \right)}^2} + {B^2}}} - \frac{{AC + D{C^k} + p}}{{{{\left( {AC + D{C^k} + p} \right)}^2} + {{\left( {BC} \right)}^2}}}}. \end{align*}

$\mathcal{L}^{-1}\left(\frac 1p\right)=1.$所以原积分

\begin{align*}I& = \int_0^\infty  {\sum\limits_{k = 0}^\infty  {\frac{{A + D{C^k} + x}}{{{{\left( {A + D{C^k} + x} \right)}^2} + {B^2}}} - \frac{{AC + D{C^k} + x}}{{{{\left( {AC + D{C^k} + x} \right)}^2} + {{\left( {BC} \right)}^2}}}} dx} \\&= \frac{1}{2}\sum\limits_{k = 0}^\infty  {\ln \frac{{{{\left( {AC + D{C^k}} \right)}^2} + {{\left( {BC} \right)}^2}}}{{{{\left( {A + D{C^k}} \right)}^2} + {B^2}}}}  = \mathop {\lim }\limits_{n \to \infty } \frac{1}{2}\ln \prod\limits_{k = 0}^n {\frac{{{{\left( {AC + D{C^k}} \right)}^2} + {{\left( {BC} \right)}^2}}}{{{{\left( {A + D{C^k}} \right)}^2} + {B^2}}}} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{2}\ln \left( {\frac{{\prod\limits_{k = 0}^{n - 1} {{{\left( {AC + D{C^{k + 1}}} \right)}^2} + {{\left( {BC} \right)}^2}} }}{{\prod\limits_{k = 0}^{n - 1} {{{\left( {A + D{C^k}} \right)}^2} + {B^2}} }}\frac{{{{\left( {AC + D} \right)}^2} + {{\left( {BC} \right)}^2}}}{{{{\left( {A + D{C^n}} \right)}^2} + {B^2}}}} \right)\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{2}\ln \left( {{C^{2n}}\frac{{{{\left( {AC + D} \right)}^2} + {{\left( {BC} \right)}^2}}}{{{{\left( {A + D{C^n}} \right)}^2} + {B^2}}}} \right) = \frac{1}{2}\ln \left( {\frac{{{{\left( {AC + D} \right)}^2} + {{\left( {BC} \right)}^2}}}{{{D^2}}}} \right).\end{align*}

Kar Board Model Pape 说:
2022年9月29日 18:41