lims

1. $ \lim\limits_{x -> 0} \frac{sinx} {x} = $

2. $ \lim\limits_{x -> \infty} (1 + \frac{1} {x})^x = $

Derivative

1. $ (x^\alpha)^{(n)} = $

2. $ (x^n)^{(n)} = $

3. $ (\frac{1}{x})^{(n)} = $

4. $ (sinx)^{(n)} $

5. $ (cosx)^{(n)} $

6. $ (\frac{1}{x + a})^{(n)} = $

7. $ (a^x)^{(n)} = $

   series

   Fourier

8. $ a_n = $

   $ b_n = $

   $ f(x) ~ $

9. Dirichlet
   $ S(x) = $

10. ans

    lims

11. 1

12. e

Derivative

1. $ \alpha(\alpha - 1)(\alpha - 2) … (\alpha - n + 1)x^{\alpha - n} $

2. n!

3. $\frac{(-1)^n n!}{x^{n+1}} $

4. $ sin(x + \frac{n}{2} \pi) $

5. $ cos(x + \frac{n}{2} \pi) $

6. $ \frac{(-1)^n n!}{(x + a)^{n+1}} $

7. $ a^x ln^n a ,(a > 0)$

series

Fourier

1. $ \frac{1}{l} \int^l_{-l} f(x) cos \frac{n \pi x}{l} dx , n = 0,1,2,3…$

   $ \frac{1}{l} \int^l_{-l} f(x) sin \frac{n \pi x}{l} dx, n = 1,2,3…$

   $ \frac{a0}{2} + \sum\limits{n = 1}^{\infty}(a_n cos\frac{n \pi x }{l} + b_n sin\frac{n \pi x}{l}) $

2. 当x是f(x)的连续点时，f(x)
   当x是f(x)的第一类间断点时, $ \frac{f(x-0) + f(x + 0)}{2} $

English

1. what if 如果。。。将会。。。有没有这样一种可能……

2. how 如何，多么

2023_Eng_301(Powered By LQR)

Here is a thrilling scene, featuring two teams racing on dragon boats with crowds of people cheering on the bridge. At the corner of the picture, an old woman is holding her husband’s hand, smiling with satisfaction, expressing her happiness seeing the dragon boat race of their village is getting more and more popular.

The idea of the artist lurking behind the caricature can be perceived as an appreciation of the boosting tourism based on traditional Chinese festival in villages. A majority of relevant departments tend to develop blinking forms of activities while overlooking the very essence of them, leading to the result that commercial interest is overstated and the beauty of conventions are forgotten. This tendency may seem innocuous in the short term, risks erasing the sense of belonging in dwellers’ hearts and jeopardizing the real interactions between people and the past in a larger sense. To address the obstacles met in advancing a county requires a shift in mindset-focusing on the daily lives of the public and the thriving of cultural activities. The responsibility urges city authorities to fulfill their role through leading the trend, establishing proper environment for citizens to enjoy such festivals.

With collective commitment in promoting traditional affairs, we can aspire to see more villages or cities to have significant changes and win both ecologically and admirably.

通过查阅资料，发现RTX30系对cuda9有支持问题

ERROR Graphviz 的 dot 工具未被正确找到或配置

assertionerror 3221225477

网站内下载Graphviz并配置环境变量

failed to run cuBLAS routine: CUBLAS_STATUS_EXECUTION_FAILED

解决方案conda install blas

已安装pydot库但是仍然报错ImportError: Failed to import pydot. Please install pydot

解决方案pip install pydot

问题背景

一体机从购入（2017年中）至2025年1月使用，打印，扫描一切正常。2025年2月出现打印速度很慢，打印测试页一张大约需要5min左右。扫描文件正常，包括传输速度

一台HP-M126nw一体机。一台小米路由器4a千兆版

连接方式

节点：打印机、路由器、终端设备（PC/手机）

打印机通过2.4Ghz Wi-Fi连接路由器，其他终端设备（PC，手机）通过2.4Ghz Wi-Fi和5Ghz Wi-Fi连接路由器

路由器购入于2018年中，打印机购入于2017年中

解决方案

路由器换新

尝试过的解决方案【以下于本例中均无效】

1. 重启Print Spooler

2. 重新安装驱动

3. 将驱动替换为PCL

4. 将驱动替换为aserjet p1108驱动

5. 打印机重置

6. 因为电脑是通过网卡启动，所以切换系统比较自由。一开始是以为Windows 11的问题，而且HP社区也有相关的情况，所以当时就退回了当时安装的Windows 10, 发现Windows 10下也有问题，然后又退回Windows 7，Windows 7下问题依旧。OS X测试了：OS X10.9、macOS16。手机无线打印测试了AirPrint

7. 关于有线打印，由于打印机距离电脑过于远，没有尝试

数学运算

代入排除法

题目选项信息充分，题目有几个量则选项有几个量与之对应，优先考虑排除，甚至可以只算部分，比如只算最后一位

可以考虑边界

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江苏2020-B类

题目大意：参加某活动，第一次汇总1742人，第二次汇总1796，发现第一次计算有误，将某单位参加人数的个位数字与十位数字颠倒，已知这个单位参加人数的个位与十位数字之和10，求可能参加人数.

A: 164  B:173  C:182

ANS：C

解析：由这个单位参加人数的个位与十位数字之和10，代入验证，再由第一次汇总与复核结果相差1796 - 1742 == 54。依次验证即可. 如C：182 - 128 == 54 OK

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2024江苏
售卖A，B两类物品，A类¥80/件，B类¥50/件，每买20A则送一个B。如果花了¥12100买了200个（包含赠送的B），求其中的A有多少。
A:75  B:79

ANS: A

解析：和差倍比。代入求出赠送的B，然后根据总数量求出A，验证是不是12100即可

代入A：如果A == 75；则B = 75 / 20 == 3。所以设买了x个B，则75 + x + 3 == 200; x == 122;
total_price == 75 * 80 + 122 * 50 == 12100 OK

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2024浙江选调
公园正六边形灯塔，底边12m。A，B分别从对角处沿边顺时针走，V_A = 3 m/s V_B = 2 m/s.经过每个顶点都停下一秒，求A出发后多久能看到B【不在同一边看不到】
A: 29  B: 34

解析：注意到：假设A在A点，B在D点。A走一条边需 A_time = 12 / 3 + 1 == 5s; B_time = 12 / 2 + 1 == 7s。
代入A：A：29 / 5 == 5, 29 % 5 == 4, 4s又走了一条边. A_x = 5 + 1 == 6;所处A点; B_x = 29 / 7 + 1，所处BC，所以排除