微积分与数学分析学习帖

zhcosin

很久很久没动了。。。

zhcosin

回复 17# hbghlyj
多谢指出，我来一一修改下.

zhcosin

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.

kuing

楼主又有空撸数学了

zhcosin

春节综合征还没过

hbghlyj

建议楼主也整一个在线版,这样可以被更多人搜索到
例如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

zhcosin

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。

hbghlyj

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
$$
\lim _{n \rightarrow \infty} \frac{1^{k}+2^{k}+\cdots+n^{k}}{n^{k+1}}=\frac{1}{k+1}
$$

zhcosin

回复 28# hbghlyj
这个证明貌似在哪看到过，国内的教材

zhcosin

利用导函数介值性定理证明拉格朗日中值定理

hbghlyj

分享一个资源Analysis123.pdf
电子版:
bananaspace.org/wiki/%E8%AE%B2%E4%B9%89:%E6%9 … A6%E5%88%86%E6%9E%90

hbghlyj

hbghlyj 发表于 2022-3-2 17:55
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 $\ ...

imomath

Proof of L’Hopital’s theorem

We will first prove a discrete version of the L’Hopital’s theorem: the case when instead of functions we are dealing with sequences. We may see a sequence \((a_n)_{n=1}^{\infty}\) as a special case of a function. The finite difference \(\frac{a_{n+1}-a_n}{(n+1)-n}\) plays a role that derivative plays when studying functions.

Cesaro-Stolz’s theorem

Assume that \((x_n)_{n=1}^{\infty}\) and \((y_n)_{n=1}^{\infty}\) are two sequences of real numbers such that \(y_n\) is increasing and \(\lim_{n\to\infty}y_n=+\infty\).

    (a) The following three inequalities hold: \[ \liminf \frac{x_{n+1}-x_n}{y_{n+1}-y_n}\leq \liminf\frac{x_n}{y_n}\leq \limsup\frac{x_n}{y_n}\leq \limsup \frac{x_{n+1}-x_n}{y_{n+1}-y_n}.\]

    (b) Assume that \(\displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=L\). Then the limit \(\displaystyle \lim_{n\to\infty}\frac{x_n}{y_n}\) exists and is equal to \(L\).

Notice that (b) is a consequence of (a), so we will only prove part (a).

The middle inequality is obvious as \(\limsup\) is always bigger than \(\lim\inf\). The leftmost inequality is analogous to the rightmost, so we will only prove one of them: the right-most.

Let \(\gamma\) be any number greater than \(\displaystyle \frac{x_{n+1}-x_n}{y_{n+1}-y_n}\). Since \(\gamma\) is not \(\limsup\) of the sequence, there exists \(N\) such that for all \(n > N\) the following inequality holds \(\displaystyle \frac{x_{n+1}-x_n}{y_{n+1}-y_n} < \gamma\). Let \(m\in\mathbb N\) be any integer greater than \(N\). According to the previous inequality we have: \begin{eqnarray*} x_{N+1}-x_N&< &\gamma\left(y_{N+1}-y_N\right)\newline x_{N+2}-x_{N+1}&< &\gamma\left(y_{N+2}-y_{N+1}\right)\newline x_{N+3}-x_{N+2}&< &\gamma\left(y_{N+3}-y_{N+2}\right)\newline &\vdots&\newline x_{m}-x_{m-1}&< &\gamma\left(y_{m}-y_{m-1}\right). \end{eqnarray*} Adding all of the previous \(m-N\) inequalities yield: \[x_m-x_N < \gamma\left(y_m-y_N\right) \quad\quad\Rightarrow \quad\quad \frac{x_m}{y_m}-\frac{x_N}{y_m} < \gamma-\gamma\frac{y_N}{y_m}.\] Since \(y_m\to+\infty\) as \(n\to\infty\) we conclude that \[\limsup \frac{x_m}{y_m}\leq\gamma.\] This holds for every \(\displaystyle \gamma > \limsup \frac{x_{n+1}-x_n}{y_{n+1}-y_n}\) thus \[\limsup \frac{x_m}{y_m}\leq\limsup\frac{x_{n+1}-x_n}{y_{n+1}-y_n}.\]

L’Hopital’s theorem

Assume that \(f\) and \(g\) are differentiable functions on an open interval containing real number \(a\). Assume that \(g\neq 0\) and \(g^{\prime}\neq 0\) on that interval. Assume that \(\displaystyle \lim_{x\to a} \frac{f^{\prime}(x)}{g^{\prime}(x)}=L\) for some real number \(L\) and assume that one of the following two conditions is satisfied:

    (a) \(\displaystyle \lim_{x\to a}g(x)=+\infty\).

    (b) \(\displaystyle \lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0\).

Then \(\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=L\).

Proof
    (a) Since we assumed that \(g^{\prime}\neq 0\), according to Darboux’s mean value theorem we may assume that \(g^{\prime}\) is either positive or negative. That means that \(g\) is monotone. It suffices to prove that for each monotone sequence \(x_n\) such that \(\displaystyle \lim_{n\to \infty}x_n=a\) we have \(\displaystyle\lim_{n\to \infty}\frac{f(x_n)}{g(x_n)}=L\).

    The sequence \(g(x_n)\) is increasing and converges to \(+\infty\) hence according to Cesaro-Stolz theorem we have \[\liminf \frac{f(x_{n+1})-f(x_{n})}{g(x_{n+1})-g(x_n)}\leq \liminf \frac{f(x_n)}{g(x_n)}\leq \limsup \frac{f(x_n)}{g(x_n)}\leq \limsup \frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}.\] According to Cauchy mean-value theorem we have that there exists a sequence of numbers \(\xi_n\in(x_n,x_{n+1})\) such that \[\frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}=\frac{f^{\prime}(\xi_n)}{g^{\prime}(\xi_n)}.\] We must have \(\xi_n\to a\) since \(\xi_n\in(x_n,x_{n+1})\). According to our assumption that \(\displaystyle \lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}=L\) we have: \[ L=\liminf \frac{f’(\xi_n)}{g’(\xi_n)}\leq \liminf \frac{f(x_n)}{g(x_n)}\leq \limsup \frac{f(x_n)}{g(x_n)}\leq \limsup \frac{f’(\xi_n)}{g’(\xi_n)}=L. \] This completes the proof in the case (a).

    (b) Let \(\varepsilon > 0\). There exists a neighborhood \(U\) of \(a\) such that \(x\in U\) implies that \(\displaystyle \left|\frac{f^{\prime}(x)}{g^{\prime}(x)}-L\right| < \varepsilon \). Take any \(x\in U\) and let \(y_n\in U\) be a monotone sequence converging to \(a\). According to Cauchy’s mean value theorem we have that there exists a sequence \((\xi_n)_{n=1}^{\infty}\) such that \(\xi_n\in (x,y_n)\) and \[\frac{f(x)-f(y_n)}{g(x)-g(y_n)}=\frac{f^{\prime}(\xi_n)}{g^{\prime}(\xi_n)}\in (L-\varepsilon, L+\varepsilon).\] Therefore the sequence \(\displaystyle \left(\frac{f(x)-f(y_n)}{g(x)-g(y_n)}\right)_{n=1}^{\infty}\) has all its terms in \((L-\varepsilon, L+\varepsilon)\). It remains to notice that \(\displaystyle\lim_{n\to\infty} \frac{f(x)-f(y_n)}{g(x)-g(y_n)}=\frac{f(x)}{g(x)}\) hence \(\displaystyle \frac{f(x)}{g(x)}\in (L-\varepsilon, L+\varepsilon)\) which completes the proof of the part (b) of the theorem.