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很久很久没动了。。。
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2022-2-9 17:03

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2022-2-9 17:03
数学暗恋者,程序员,喜欢古典文学/历史,个人主页: https://zhcosin.coding.me/

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回复 17# hbghlyj
多谢指出,我来一一修改下.

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回复 20# hbghlyj
笔记中绝大部分内容我都是自行推证的,即便一时半会想不出,宁可空着,实在搞不定了,只有先学习一下书上的证明,再用自己的理解叙述出来,绝对不会照搬的,不然就失去了笔记的意义.

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楼主又有空撸数学了

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春节综合征还没过

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本帖最后由 hbghlyj 于 2022-2-10 19:07 编辑

建议楼主也整一个在线版,这样可以被更多人搜索到
例如https://www.jirka.org/ra/html/sec_ift.html
现在有tex4ht,pandoc,latexml,pretext这些工具也比较方便

或者可以直接发到论坛上,"发帖选项"可以勾选"Html代码"
比如这帖
因为页面太长了,我把3个帖子分为1页...这样的话原先的那些链接如果被分到不同的页就失效了,需要手动去找链接位置...但是这个影响不太大,因为大多数的链接都是在章节内引用,所以不会被分到不同的页

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回复 28# hbghlyj
这个我也想过,但是工程量会比较大,还是以后再说,先学习要紧。

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Stolz定理

本帖最后由 hbghlyj 于 2022-3-27 00:40 编辑

回复 19# hbghlyj
这份讲义的60页的证明:
Theorem 2.3.11 (O.Stolz) Suppose $\left(x_{n}\right)$ and $\left(y_{n}\right)$ are two sequences of real numbers such that
(i) $y_{n} \rightarrow \infty$ as $n \rightarrow \infty$,
(ii) $\left(y_{n}\right)$ is a strictly increasing sequence (for large $\left.n\right)$, and
(iii) the limit
$$
\lim _{n \rightarrow \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}
$$
exists or tends to $\infty$ or $-\infty$. Then
$$
\lim _{n \rightarrow \infty} \frac{x_{n}}{y_{n}}=\lim _{n \rightarrow \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}
$$
Proof. The proof is similar to the proof of Theorem 2.3.6. Consider the case that $l=\lim _{n \rightarrow \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}$ is a number. Then for every $\varepsilon>0$ there is $N$ such that for $n>N$ we have
$$
\left|\frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}-l\right|<\frac{\varepsilon}{2} .
$$
Since $\left(y_{n}\right)$ is strictly increasing eventually, so we can choose $N$ big enough so that $y_{k}-$ $y_{k-1}>0$ for all $k>N$ and therefore
$$
-\frac{\varepsilon}{2}\left(y_{k}-y_{k-1}\right)<x_{k}-y_{k-1}-l\left(y_{k}-y_{k-1}\right)<\frac{\varepsilon}{2}\left(y_{k}-y_{k-1}\right)
$$
Adding these inequalities over $k=N+1, \cdots, n$, where $n>N$, we obtain that
$$
-\frac{\varepsilon}{2}\left(y_{n}-y_{N}\right)<x_{n}-y_{N}-l\left(y_{n}-y_{N}\right)<\frac{\varepsilon}{2}\left(y_{n}-y_{N}\right)
$$
which can be written as, since $y_{n}-y_{N}>0$
$$
\left|\frac{x_{n}-x_{N}}{y_{n}-y_{N}}-l\right|<\frac{\varepsilon}{2}
$$
for all $n>N .$ Next we use the identity (similar to that in the proof of Theorem 2.3.6)
$$
\frac{x_{n}}{y_{n}}-l=\frac{x_{N}-l y_{N}}{y_{n}}+\left(1-\frac{y_{N}}{y_{n}}\right)\left(\frac{x_{n}-x_{N}}{y_{n}-y_{N}}-l\right)
$$
so that
$$
\left|\frac{x_{n}}{y_{n}}-l\right|<\left|\frac{x_{N}-l y_{N}}{y_{n}}\right|+\frac{\varepsilon}{2}
$$
for every $n>N$. Since $y_{n} \rightarrow \infty$ so that
$$
\frac{x_{N}-l y_{N}}{y_{n}} \rightarrow 0 \quad \text { as } n \rightarrow \infty
$$
Therefore there is $N_{1}>N$ such that
$$
\left|\frac{x_{N}-l y_{N}}{y_{n}}\right|<\frac{\varepsilon}{2} \quad \text { for } n>N_{1}
$$
and therefore
$$
\left|\frac{x_{n}}{y_{n}}-l\right|<\left|\frac{x_{N}-l y_{N}}{y_{n}}\right|+\frac{\varepsilon}{2}<\varepsilon
$$
for every $n>N_{1}$. By definition
$$
\lim _{n \rightarrow \infty} \frac{x_{n}}{y_{n}}=l=\lim _{n \rightarrow \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}
$$
and the proof is complete.$\blacksquare$
As as example, if $k$ is a positive integer, then we can show (Exercise) by Stolz's theorem that
$$
\lim _{n \rightarrow \infty} \frac{1^{k}+2^{k}+\cdots+n^{k}}{n^{k+1}}=\frac{1}{k+1}
$$

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回复 28# hbghlyj
这个证明貌似在哪看到过,国内的教材

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本帖最后由 zhcosin 于 2022-3-4 21:02 编辑

利用导函数介值性定理证明拉格朗日中值定理
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2022-3-4 21:02
数学暗恋者,程序员,喜欢古典文学/历史,个人主页: https://zhcosin.coding.me/

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分享一个资源Analysis123.pdf

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