Questions in category: 分析 (Calculus and Analysis)

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## 1. 设数列 $\{a_n\}$ 满足递推公式 $a_{n+1}=(n+1)a_n+1$, 且 $a_0=1$, 求通项 $a_n$ 的表达式.

Posted by haifeng on 2024-09-06 17:27:39 last update 2024-09-06 17:27:39 | Answers (3) | 收藏

## 2. 平均不等式, 调和平均不等式

Posted by haifeng on 2024-07-17 07:01:10 last update 2024-07-17 07:01:10 | Answers (2) | 收藏

$\sqrt[n]{a_1 a_2\cdots a_n}\leqslant\frac{a_1+a_2+\cdots+a_n}{n},$

$\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}\leqslant\sqrt[n]{a_1 a_2\cdots a_n},$

## 3. 切比雪夫不等式

Posted by haifeng on 2024-07-14 10:41:11 last update 2024-07-14 10:42:58 | Answers (1) | 收藏

$\sum_{i=1}^{n}a_i b_i\ \geqslant\ \frac{1}{n}(\sum_{i=1}^{n}a_i)\cdot(\sum_{i=1}^{n}b_i)\ \geqslant\ \sum_{i=1}^{n}a_i b_{n+1-i}.$

[Hint] 使用排序不等式证明.

## 4. 排序不等式

Posted by haifeng on 2024-07-14 09:27:39 last update 2024-07-14 10:42:20 | Answers (1) | 收藏

$\begin{split} & a_1 b_1+a_2 b_2+\cdots+a_n b_n\quad\text{(顺序和)}\\ \geqslant\ & a_1 b_{j_1}+a_2 b_{j_2}+\cdots a_n b_{j_n}\quad\text{(乱序和)}\\ \geqslant\ & a_1 b_n+a_2 b_{n-1}+\cdots+a_n b_1\quad\text{(反序和)} \end{split}$

## 5. 求解下面的二元三次方程组.

Posted by haifeng on 2024-05-05 12:52:48 last update 2024-05-05 12:52:48 | Answers (1) | 收藏

$\begin{cases} m^3-3mn=2,\\ 3m^2 n-n^3=11. \end{cases}$

## 6. 三倍角公式

Posted by haifeng on 2023-12-25 16:00:44 last update 2023-12-25 16:00:44 | Answers (1) | 收藏

$\sin 3\theta=3\sin\theta-4\sin^3\theta.$

## 7. 求 $\sin(\frac{k}{5}\pi)$ 的值, 这里 $k\in\mathbb{Z}$.

Posted by haifeng on 2023-12-25 12:46:45 last update 2024-01-04 15:53:38 | Answers (1) | 收藏

$\cos(36^{\circ})=\cos(\frac{1}{5}\pi)=\frac{1+\sqrt{5}}{4},$

$\sin(36^{\circ})=\sin(\frac{1}{5}\pi)=\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}.$

$\sin(72^{\circ})=\sin(\frac{2}{5}\pi)=2\sin(\frac{1}{5}\pi)\cos(\frac{1}{5}\pi)=2\cdot\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}\cdot\frac{1+\sqrt{5}}{4}=\frac{1}{2}\sqrt{\frac{5+\sqrt{5}}{2}}.$

$\sin(108^{\circ})=\sin(\frac{3}{5}\pi)=\sin(\pi-\frac{3}{5}\pi)=\sin(\frac{2}{5}\pi)$. 当然, 也可以利用三倍角公式计算:

$\cos(72^{\circ})=\cos(\frac{2\pi}{5})=2\cos^2(\frac{\pi}{5})-1=2\cdot\Bigl(\frac{1+\sqrt{5}}{4}\Bigr)^2-1=\frac{\sqrt{5}-1}{4}.$

## 8. Lipschitz函数的延拓问题

Posted by LioYu on 2023-02-27 19:27:38 last update 2023-02-27 19:27:38 | Answers (0) | 收藏

## 9. 设函数 $f(x)$ 满足下面的关系, 求 $f(1)$.

Posted by haifeng on 2023-02-16 23:02:18 last update 2023-02-17 09:20:07 | Answers (1) | 收藏

$f(f(x))=\begin{cases} (x-1)^2, & x\geqslant 1,\\ f(x), & x < 1. \end{cases}$

## 10. Теорема Лузина-Данжуа(Luzina-Danjoy theorem)

Posted by haifeng on 2021-07-03 10:27:42 last update 2021-07-03 10:27:42 | Answers (1) | 收藏

# Теорема Лузина-Данжуа

## Luzina-Danjoy theorem

Теорема Лузина-Данжуа — Викиконспекты (ifmo.ru)

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