Questions in category: 分析 (Calculus and Analysis)

## 1. 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

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

## 2. 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

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

## 3. 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:

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

## 4. 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

## 5. 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

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

## 6. 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)

## 7. 证明 $\cos 3\theta=4\cos^3\theta-3\cos\theta$.

Posted by haifeng on 2019-12-31 20:15:07 last update 2019-12-31 20:15:27 | Answers (3) | 收藏

## 8. 等额本息贷款的计算公式

Posted by haifeng on 2019-02-23 09:40:49 last update 2019-02-24 11:51:46 | Answers (1) | 收藏

$A$: 贷款本金

$x$: 每月还款金额（简称【每月本息】）

$R$: 年利率

$r$: 月利率（为年利率的$1/12$, 即 $R=12r$.）

$N$: 还款月数

$x=A\times\frac{r(1+r)^N}{(1+r)^N-1}$

$Q_i$ 为第 $i$ 期还款后的欠款总金额。$n$ 为当前还款的期数. $n=1,2,\ldots,N$.

$Q_1=A(1+r)-x$

$n=2$, （指还完第二期贷款）

$\begin{split} Q_2&=Q_1(1+r)-x\\ &=[A(1+r)-x](1+r)-x\\ &=A(1+r)^2-[1+(1+r)]x \end{split}$

$n=3$, （指还完第三期贷款）

$\begin{split} Q_3&=Q_2(1+r)-x\\ &=\bigl[A(1+r)^2-[1+(1+r)]x\bigr](1+r)-x\\ &=A(1+r)^3-[1+(1+r)+(1+r)^2]x \end{split}$

$n=k$, （指还完第 $k$ 期贷款）

$Q_k=A(1+r)^k-[1+(1+r)+(1+r)^2+\cdots+(1+r)^{k-1}]x\tag{*},$

$\begin{split} Q_{k+1}&=Q_k(1+r)-x\\ &=\bigl[A(1+r)^k-[1+(1+r)+(1+r)^2+\cdots+(1+r)^{k-1}]x\bigr](1+r)-x\\ &=A(1+r)^{k+1}-[(1+r)+(1+r)^2+\cdots+(1+r)^k]x-x\\ &=A(1+r)^{k+1}-[1+(1+r)+(1+r)^2+\cdots+(1+r)^k]x.\\ \end{split}$

$\begin{split} Q_k&=A(1+r)^k-[1+(1+r)+(1+r)^2+\cdots+(1+r)^{k-1}]x\\ &=A(1+r)^k-\frac{1-(1+r)^k}{1-(1+r)}\cdot x\\ &=A(1+r)^k-\frac{(1+r)^k-1}{r}\cdot x \end{split}$

$A(1+r)^N=x\cdot\frac{(1+r)^N-1}{r}$

$x=\frac{Ar(1+r)^N}{(1+r)^N-1}.$

## 9. Jan 1, 2000 A.D. 对应的 Julian Day Number

Posted by haifeng on 2019-02-15 10:36:52 last update 2019-02-15 10:36:52 | Answers (1) | 收藏

Jan 1, 2000 A.D. 对应的 Julian Day Number 是 2451545.