parallelogram
The parallelogram law in inner product spaces
In a normed space, the statement of the parallelogram law is an equation relating norms:
In an inner product space, the norm is determined using the inner product:
As a consequence of this definition, in an inner product space the parallelogram law is an algebraic identity, readily established using the properties of the inner product:
Adding these two expressions:
as required.
If x is orthogonal to y, then
and the above equation for the norm of a sum becomes:
which is Pythagoras' theorem.
parallelogram的更多相关文章
- Parallelogram Counting(平行四边形个数,思维转化)
1058 - Parallelogram Counting PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit ...
- POJ 1971 Parallelogram Counting (Hash)
Parallelogram Counting Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 6895 Acc ...
- 计算几何 + 统计 --- Parallelogram Counting
Parallelogram Counting Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 5749 Accepted: ...
- 【codeforces 749B】Parallelogram is Back
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- 1058 - Parallelogram Counting 计算几何
1058 - Parallelogram Counting There are n distinct points in the plane, given by their integer coord ...
- Codeforces 749B:Parallelogram is Back(计算几何)
http://codeforces.com/problemset/problem/749/B 题意:已知平行四边形三个顶点,求另外一个顶点可能的位置. 思路:用向量来做. #include <c ...
- Light OJ - 1058 Parallelogram Counting(判定平行四边形)
Description There are n distinct points in the plane, given by their integer coordinates. Find the n ...
- POJ 1971 Parallelogram Counting
题目链接: http://poj.org/problem?id=1971 题意: 二维空间给n个任意三点不共线的坐标,问这些点能够组成多少个不同的平行四边形. 题解: 使用的平行四边形的判断条件:对角 ...
- LightOJ 1058 - Parallelogram Counting 几何思维
http://www.lightoj.com/volume_showproblem.php?problem=1058 题意:给你顶点,问能够成多少个平行四边形. 思路:开始想使用长度来扫描有多少根,但 ...
随机推荐
- React Native 'config.h' file not found 问题、 'glog/logging.h' file not found 问题、configure: error: C compiler cannot create executables问题解决过程记录
1.在github 上面 git clone 一个RN 项目代码,npm install (yarn)后,准备运行iOS工程,发现'config.h' file not found ,恶心!!! 百度 ...
- PAT甲题题解-1058. A+B in Hogwarts (20)-大水题
无语,这种水题还出,浪费时间,但又不得不A... #include <iostream> #include <cstdio> #include <algorithm> ...
- [Latex] Travis-CI与Latex构建开源中文PDF
博主有一本开源书籍,用 latex 排版,托管在Github上.但用 latex 不像是 Markdown,当tex文本更新时,用于最终浏览的PDF文件很难得到及时的更新, 所以博主一直想找到一套工具 ...
- ns3的输入输出奥秘(三) Tracing系统
Tracing系统 (1)ASCII Tracing 还是以myfirst.cc为例子 可以在Simulator::Run()前面加上 AsciiTraceHelper ascii; pointToP ...
- Beta阶段综合报告
一. 敏捷冲刺每日报告 http://www.cnblogs.com/mia0502/p/7767628.html http://www.cnblogs.com/mia0502/p/7773872.h ...
- WPF和js交互 WebBrowser数据交互
this.webBrowser1.ObjectForScripting = new OprateBasic(); this.webBrowser1.Source = new Uri(Environme ...
- Office处理
1.NPOI:一个开源项目,不需要安装Microsoft Office,支持对Office 97-2003,2007文件格式,功能比较强大. http://npoi.codeplex.com/ 2.a ...
- Linux命令(九)比较文件差异 diff
diff 命令介绍 diff 命令的功能为逐行比较两个文本文件,列出其不同之处.对给出的文件进行系统的检查,并显示出两个文件中所有不同的行.如果 diff 命令后跟的是目录,则会对该目录中的同名文件进 ...
- [CB]Intel 2018架构日详解:新CPU&新GPU齐公布 牙膏时代有望明年结束
Intel 2018架构日详解:新CPU&新GPU齐公布 牙膏时代有望明年结束 北京时间12月12日晚,Intel在圣克拉拉举办了架构日活动.在五个小时的演讲中,Intel揭开了2021年CP ...
- Laravel Service Provider 中 boot 方法和 register 方法的区别
register 方法用于绑定服务到容器,框架会先调用所有 provider 的 register 方法,等所有服务都注册完毕再去调用每一个服务的 boot 方法. 所以不能在 register 方法 ...





