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Latex数学笔记

米尔嘉
2022-09-02 / 0 评论 / 0 点赞 / 457 阅读 / 1,425 字 / 正在检测是否收录...
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用于记录数学方面遇到的难题

1. 计算积分

计算积分limnnln(n+1)(nn1)\lim_{n\rightarrow \infty} \frac{n}{\ln \left( n+1 \right)}\left( \sqrt[n]{n}-1 \right)

解:

limnnln(n+1)(nn1)=limnnln(n+1)(e1nlnn1)=limnnln(n+1)1nlnn=limnlnnln(n+1)=1limnnln(n+1)(nn1)=t=1nlimt0(1t)t1tln(1+1t)=limt0etln1t1tln(1+1t)=limt0ln1tln(1+1t)=1 \begin{aligned} &\lim_{n\rightarrow \infty} \frac{n}{\ln \left( n+1 \right)}\left( \sqrt[n]{n}-1 \right) =\lim_{n\rightarrow \infty} \frac{n}{\ln \left( n+1 \right)}\left( e^{\frac{1}{n}\ln n}-1 \right) =\lim_{n\rightarrow \infty} \frac{n}{\ln \left( n+1 \right)}\cdot \frac{1}{n}\ln n =\lim_{n\rightarrow \infty} \frac{\ln n}{\ln \left( n+1 \right)}=1\\ &\lim_{n\rightarrow \infty} \frac{n}{\ln \left( n+1 \right)}\left( \sqrt[n]{n}-1 \right) \xlongequal{t=\frac{1}{n}}\lim_{t\rightarrow 0} \frac{\left( \frac{1}{t} \right) ^t-1}{t\ln \left( 1+\frac{1}{t} \right)}=\lim_{t\rightarrow 0} \frac{e^{t\ln \frac{1}{t}}-1}{t\ln \left( 1+\frac{1}{t} \right)}=\lim_{t\rightarrow 0} \frac{\ln \frac{1}{t}}{\ln \left( 1+\frac{1}{t} \right)}=1 \\ \end{aligned}

2. 计算积分

计算积分0π2xcos2x2dxdx\int_0^{\frac{\pi}{2}}{\frac{x}{\cos ^2\frac{x}{2}}}dxdx

解:

0π2xcos2x2dx=0π2xsec2x2dx=2xtanπ20π240π2tanx2dx2=π0+4lncosx20π2=π2ln2 \begin{aligned} \int_0^{\frac{\pi}{2}}{\frac{x}{\cos ^2\frac{x}{2}}}dx&=\int_0^{\frac{\pi}{2}}{x\sec ^2\frac{x}{2}}dx\\&=2\cdot x\cdot \tan \frac{\pi}{2}\mid_{0}^{\frac{\pi}{2}}-4\int_0^{\frac{\pi}{2}}{\tan \frac{x}{2}}d\frac{x}{2}\\&=\pi -0+4\ln |\cos \frac{x}{2}|\mid_{0}^{\frac{\pi}{2}}\\&=\pi -2\ln 2 \end{aligned}

28.求证:limn(2n1)!!(2n)!!=0证明:方法1:记xn=(2n1)!!(2n)!!=1234562n12n1xn=k=1n2k2k1=k=1n(1+12k1)=ek=1nln(1+12k1)limn1xn=elimnk=1nln(1+12k1)因为limnk=1nln(1+12k1)=n=1ln(1+12n1),级数发散于+所以limn1xn=+,从而limn(2n1)!!(2n)!!=limnxn=0方法2:由于2k12k<(2k1)2(2k)21=2k12k+1(k=1,2,)所以0<(2n1)!!(2n)!!=1234562n12n<1335572n12n+1=12n+1则由夹逼原理知所证极限式成立。 28.\text{求证:}\lim_{n\rightarrow \infty} \frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!}=0 \\ \text{证明:} \\ \text{方法}1\text{:记}x_n=\frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!}=\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdots \frac{2n-1}{2n} \\ \text{则}\frac{1}{x_n}=\prod_{k=1}^n{\frac{2k}{2k-1}}=\prod_{k=1}^n{\begin{array}{c} \left( 1+\frac{1}{2k-1} \right)\\ \end{array}}=e^{\sum_{k=1}^n{\ln \left( 1+\frac{1}{2k-1} \right)}} \\ \lim_{n\rightarrow \infty} \frac{1}{x_n}=e^{\lim_{n\rightarrow \infty} \sum_{k=1}^n{\ln \left( 1+\frac{1}{2k-1} \right)}} \\ \text{因为}\lim_{n\rightarrow \infty} \sum_{k=1}^n{\ln \left( 1+\frac{1}{2k-1} \right)}=\sum_{n=1}^{\infty}{\ln \left( 1+\frac{1}{2n-1} \right)}\text{,级数发散于}+\infty \text{,} \\ \text{所以}\lim_{n\rightarrow \infty} \frac{1}{x_n}=+\infty \text{,从而}\lim_{n\rightarrow \infty} \frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!}=\lim_{n\rightarrow \infty} x_n=0 \\ \text{方法}2\text{:由于}\frac{2k-1}{2k}<\sqrt{\frac{\left( 2k-1 \right) ^2}{\left( 2k \right) ^2-1}}=\sqrt{\frac{2k-1}{2k+1}}\left( k=1,2,\cdots \right) \text{,} \\ \text{所以}0<\frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!}=\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdots \frac{2n-1}{2n}<\sqrt{\frac{1}{3}\cdot \frac{3}{5}\cdot \frac{5}{7}\cdots \frac{2n-1}{2n+1}}=\sqrt{\frac{1}{2n+1}} \\ \text{则由夹逼原理知所证极限式成立。}

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