<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>矢量微积分 on Haifei's Home</title><link>https://haifei-home.pages.dev/tags/%E7%9F%A2%E9%87%8F%E5%BE%AE%E7%A7%AF%E5%88%86/</link><description>Recent content in 矢量微积分 on Haifei's Home</description><generator>Hugo</generator><language>zh-CN</language><managingEditor>hfwang132@gmail.com (hfwang132)</managingEditor><webMaster>hfwang132@gmail.com (hfwang132)</webMaster><copyright>This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.</copyright><lastBuildDate>Fri, 26 May 2023 21:00:37 +0800</lastBuildDate><atom:link href="https://haifei-home.pages.dev/tags/%E7%9F%A2%E9%87%8F%E5%BE%AE%E7%A7%AF%E5%88%86/index.xml" rel="self" type="application/rss+xml"/><item><title>高维空间中的旋度是什么？</title><link>https://haifei-home.pages.dev/post_20230526_%E9%AB%98%E7%BB%B4%E7%A9%BA%E9%97%B4%E4%B8%AD%E7%9A%84%E6%97%8B%E5%BA%A6/</link><pubDate>Fri, 26 May 2023 21:00:37 +0800</pubDate><author>hfwang132@gmail.com (hfwang132)</author><guid>https://haifei-home.pages.dev/post_20230526_%E9%AB%98%E7%BB%B4%E7%A9%BA%E9%97%B4%E4%B8%AD%E7%9A%84%E6%97%8B%E5%BA%A6/</guid><description>&lt;h3 id="微分形式"&gt;微分形式&lt;/h3&gt;
&lt;p&gt;在介绍旋度之前，我们得先介绍一下微分形式和外微分算子。&lt;/p&gt;
&lt;p&gt;一个 n 阶形式可以定义为一个交替多重线性映射 \(\omega:(T_pM)^n\rightarrow \mathbb{R}\) 。它把多个向量映射成一个实数。另外，它还满足交替性，即交换两个输入向量，输出多一个负号。&lt;/p&gt;</description></item><item><title>矢量乘积法则的简洁证明</title><link>https://haifei-home.pages.dev/post_20210227_%E7%9F%A2%E9%87%8F%E4%B9%98%E7%A7%AF%E6%B3%95%E5%88%99%E7%9A%84%E7%AE%80%E6%B4%81%E8%AF%81%E6%98%8E/</link><pubDate>Sat, 27 Feb 2021 00:46:19 +0800</pubDate><author>hfwang132@gmail.com (hfwang132)</author><guid>https://haifei-home.pages.dev/post_20210227_%E7%9F%A2%E9%87%8F%E4%B9%98%E7%A7%AF%E6%B3%95%E5%88%99%E7%9A%84%E7%AE%80%E6%B4%81%E8%AF%81%E6%98%8E/</guid><description>&lt;p&gt;相信第一次学矢量微积分的小伙伴都对矢量的乘积法则非常头疼：&lt;/p&gt;
&lt;p&gt;简单难度：&lt;/p&gt;
\[\nabla(fg)=f\nabla g+g\nabla f\]\[\nabla\cdot(f\mathrm{A})=f\nabla\cdot\mathrm{A}+\nabla f\cdot \mathrm{A}\]\[\nabla\times(f\mathrm{A})=f\nabla\times\mathrm{A}+\nabla f\times\mathrm{A}\]&lt;p&gt;困难难度：&lt;/p&gt;
\[\nabla\cdot(\mathrm{A}\times\mathrm{B})=\mathrm{B}\cdot(\nabla\times\mathrm{A})-\mathrm{A}\cdot(\nabla\times\mathrm{B})\]&lt;p&gt;地狱难度：&lt;/p&gt;</description></item></channel></rss>