D**o
发帖数: 2653 | 1 注意作者 \author{YHBKJ}
Atiyah-Bott Localization 1
2012-09-05 09:24:19
\documentclass[a4paper,12pt]{article}
\usepackage{amsfonts}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{CJK,graphicx}
\usepackage{amscd}
\usepackage{amssymb}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{lemma}{Lemma}[section]
\begin{document}
\title{\textbf{\Huge{Atiyah-Bott Localization 1}}}\author{YHBKJ}\date{}\
maketitle
\begin{ab... 阅读全帖 |
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g****g
发帖数: 1828 | 2 In probability theory, the normal (or Gaussian) distribution, is a
continuous probability distribution that is often used as a first
approximation to describe real-valued random variables that tend to cluster
around a single mean value. The graph of the associated probability density
function is “bell”-shaped, and is known as the Gaussian function or bell
curve:[nb 1]
f(x) = \tfrac{1}{\sqrt{2\pi\sigma^2}}\; e^{ -\frac{(x-\mu)^2}{2\sigma^2}
},
where parameter μ is the mean (location of the pe... 阅读全帖 |
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L******r
发帖数: 199 | 3 Tex如下，每次标号都另起一行，很讨厌，如何把公式和标号放进同一行中？
\begin{equation}\label{example}
\textit{\textbf{X}} =
\begin{pmatrix}
&c_1 & c_2 &c_3& \cdots &c_m\\
d_1 & 8 & 0 & 1& \cdots & 7\\
d_2 & 13 & 6 & 0& \cdots &6\\
d_3 & 1 & 0 & 0& \cdots &4\\
\cdots & \cdots & \cdots & \cdots & \ddots &\cdots\\
d_n & 3 & 10 & 0 & \cdots &1\\
\end{pmatrix}
\end{equation} |
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b*m
发帖数: 124 | 4 your matrix is too wide and that is wide the label goes into another line.
A quick fix is
\begin{equation}\label{example}
\textit{\textbf{X}} =
\left(
\begin{matrix}
&c_1 & c_2 &c_3& \cdots &c_m\\
d_1 & 8 & 0 & 1& \cdots & 7\\
d_2 & 13 & 6 & 0& \cdots &6\\
d_3 & 1 & 0 & 0& \cdots &4\\
\cdots & \cdots & \cdots & \cdots & \ddots &\cdots\\
d_n & 3 & 10 & 0 & \cdots &1
\end{matrix}
\right)
\end{equation} |
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g****g
发帖数: 1828 | 5 标准差（Standard Deviation），在概率统计中最常使用作为统计分布程度（
statistical dispersion）上的测量。标准差定义为方差的算术平方根，反映组内个体
间的离散程度。测量到分布程度的结果，原则上具有两种性质：
1. 为非负数值，
2. 与测量资料具有相同单位。
一个总量的标准差或一个随机变量的标准差，及一个子集合样品数的标准差之间，有所
差别。其公式如下所列。
标准差的观念是由卡尔·皮尔逊 (Karl Pearson)引入到统计中。
目录
[隐藏]
* 1 阐述及应用
* 2 标准差的定义及简易计算公式
o 2.1 标准计算公式
o 2.2 简化计算公式
o 2.3 随机变量的标准差计算公式
o 2.4 样本标准差
o 2.5 连续随机变量的标准差计算公式
o 2.6 标准差的性质
* 3 范例
* 4 正态分布的规则
* 5 标准差与平均值之间的关系
* 6 几何学解释
... 阅读全帖 |
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a***s
发帖数: 616 | 6 这个看上去像小学数奥题，用因数分解.
For any $N = \prod_{i=1}^k p_i^r_i$ where $p_i$'s are prime numbers, the to
tal number of factorization $N$ has is $\prod_{i=1}^k (r_i + 1)$. For exampl
e, $24 = 2^3 \cdot 3^1$ and it has $(3+1)\cdot(1+1) = 8$ factorization. For
this case, we divide it by 2 to get the number of tiling.
To get 10 different rectangles, we need at least 20 factorization. 20 can be
achieved by $2 \cdot 2 \cdot 5$ or $4 \cdot 5$. By enumeration, one can fig
ure out the smallest number $2^4 \cdot 3 ... 阅读全帖 |
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N********n
发帖数: 13236 | 7 唐人街能造的房子都造完了，park北面一大块空地我一直以为是Metra场站，没想到是
留着给residential用的，靠近公园、靠近唐人街、靠近downtown，小资最爱、国内兔
毫最喜欢啊：
Mayor proposes new roadway between downtown, Chinatown
By Jon Hilkevitch
Tribune reporter
7:37 p.m. CDT, April 30, 2014
The Emanuel administration presented plans Wednesday for a new road that by
late 2016 would provide smoother connections from downtown to Chinatown and
other areas along the south branch of the Chicago River.
The $62 million project, called the Wells-Wentworth Connector, would also... 阅读全帖 |
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o**a
发帖数: 76 | 8 这里是tex文件，可以直接latex编译:
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\begin{align*}
&\int_0^{\infty} \frac{2\eta}{\eta^2+z^2} \cdot \frac{d\eta}{e^{2\pi\eta}-1} \
\
&=\frac{1}{\pi} \int_0^{\infty} \frac{\eta}{\eta^2+z^2} \cdot \frac{2\pi}{e^{2
\pi\eta}-1} d\eta \\
&=\frac{1}{\pi} \int_0^{\infty} \frac{\eta}{\eta^2+z^2} \cdot \frac{2\pi e^{-2
\pi\eta}}{1-e^{-2\pi\eta}} d\eta\\
&=\frac{1}{\pi} \int_0^{\infty} \frac{\eta}{\eta^2+z^2} \cdot d(\log(1-e^{-2\
pi\eta} |
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c******m
发帖数: 599 | 9 比如A和B都是SPD(symmetric positive definite)
对于一般的R^n上的内积, BA不是SPD,
但是, 重新定义R^n上的内积<\cdot,\cdot>= (A\cdot, \cdot)
那么BA在这个<,> 内积下就还是SPD的
推导自己根据定义自己推一下
ref.
Xu, J. Iterative Methods by Space Decomposition and Subspace Correction
SIAM Review, Vol. 34, No. 4, (Dec., 1992), pp. 581-613. |
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e*******e
发帖数: 248 | 10 I have the following code to write a large matrix. But the bracket enclosing
the matrix looks very ugly. Does any one know how to solve this problem?
The code is:
\begin{equation}
\widehat{\mathbf{Z}^2(t)}=\left[\begin{array}{ccccc}
W^+_1& W^-_1& \cdots& W^+_N&W^-_N
\end{array}
\right] \cdot \left[\begin{array}{c} f^+_1(t) \\
f^-_1(t) \\
\cdots \\
|
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s***e
发帖数: 911 | 11
他肯定有个概念错误在他问题里面. D不可能是能量密度, 如果E是电场的话. D多半
是电位移矢量.
场强使得介质受极化, 得到极化矢量P. 在线性情况下P=X \cdot E,其中X是个3维方阵.
在各向同性介质里面X是对角矩阵; 在均匀各向同性介质里面X的对角元都相等, 于是
可以表达为一个常数标量的单位矩阵的数乘.
D定义为E和P的线性组合. 所以在均匀各向同价值里面最终可以表达为:
D=a1*I \cdot E=a1*E
I是单位阵, a1是某标量;
普遍的线性情况下
D=(c1*I+X) \cdot E
非线性普遍情况下, X的每个分量都是(E_1,E_2,E_3)的函数, 分别作Taylor展开到需要
的阶, 然后按照D的定义可以得到解吸表达;
大多数情况是均匀各向同性的, 这时候特别好, X_{i}=f(E_{i}), 所以
D_{i}=c1*E_{i}+c2*f(E_{i})
i是分量. f就是你的非线性函数了. |
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f**********d
发帖数: 4960 | 12 【 以下文字转载自 Mathematics 讨论区 】
发信人: freelikewind (像风一样自由), 信区: Mathematics
标 题: 范数问题
发信站: BBS 未名空间站 (Wed Oct 23 15:16:13 2019, 美东)
Suppose f(x) in L^2(Omega), then we know that simply
||f(x)||^{2} = int_{Omega} f^2(x)dx
Now consider the situation that the analytic form of the f(x) is unknow,
but its values at a set of sampling position x_{1}, x_{2}, cdots, x_{n},
i.e., what we have are f(x_{1}), f(x_{2}), cdots, f(x_{n}).
The question is, does there exist a better way to approximate the ||f(x)||^{... 阅读全帖 |
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K*****y
发帖数: 1793 | 13 在这里编辑公式，
http://thornahawk.unitedti.org/equationeditor/equationeditor.php
\frac{\prod_{pword\in Pwords}\mathrm{hits}(word\,\, \mathrm{NEAR}\,\, pword)
\cdot \prod_{nword\in Nwords}\mathrm{hits}(nword)}{\prod_{nword\in Nwords}\
mathrm{hits}(word\,\, \mathrm{NEAR}\,\, nword)\cdot \prod_{pword\in Pwords}\
mathrm{hits}(pword)}
出来的正是我要的结果
但是放进tex文档，pword\in Pwords就都变成旁下标了。
谁给点提示？ |
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L******r
发帖数: 199 | 14 \begin{equation}\label{SOLocationWeight}
\[
Weight = \left\{\begin{array}{ll}
0.2, &\mbox{if $l \leq \frac{1}{3}$} \\
2.4\cdot l-0.6, & \mbox{if $\frac{1}{3}\leq l \leq \frac{2}
{3}$} \\
-1.8 \cdot l+2.2, & \mbox{if $\frac{2}{3}\leq l \leq 1
$}}
\end{array} \right.
\]
\end{equation}
前面用同样的都没问题，这个公式应该是2.3，但是没有显示出来
系统也把这个公式给默认标号了，因为后面
的公式是2.4都出来了。
谢谢指教 |
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T*******n
发帖数: 493 | 15 Here is the proper way of doing this:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\label{SOLocationWeight}
\text{Weight} =
\begin{cases}
0.2, & \text{if $l \leq \frac{1}{3}$}; \\
2.4\cdot l-0.6, & \text{if $\frac{1}{3}\leq l \leq \frac{2}{3}$}; \\
-1.8\cdot l+2.2, & \text{if $\frac{2}{3}\leq l \leq 1 $}.
\end{cases}
\end{equation}
\end{document}
2}
1 |
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T*******n
发帖数: 493 | 16 Try this
\documentclass[draft]{article}
\begin{document}
Text text text text text text text text text text text text text
$\theta: \{\theta_0, \theta_1, \cdots, \theta_n\}$.
text text text text text text text text text text text text
\begin{sloppypar}
Text text text text text text text text text text text text text
$\theta: \{\theta_0, \theta_1, \cdots, \theta_n\}$.
text text text text text text text text text text text text
\end{sloppypar}
\end{document} |
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B********e
发帖数: 10014 | 17 a bounded nice domain \Omega in R^N, n is the normal unit vector at boundary;
a vector function q with N components satisfying q\cdot n=0;
a scalor function f on \Omega satisfying f=0 at boundary;
show that q\cdot grad(f)=0 at boundary.
thanks! |
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s***i
发帖数: 2 | 18 First, because x_i\le C_i for all index i, then the k-th largest (smallest)
element among x_i is smaller than the corresponding k-th largest (smallest)
element in C_i. Therefore if you reorder x_i into decreasing order (x_n\le x
_{n-1}\le \cdots \le x_1) it does not change the objective function, and the
constraints remains feasible. So you can add extra constraint x_n\le x_{n-1
}\le \cdots \le x_1 to the original problem without changing objective value.
Now by first order condition, it is easy |
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D**u
发帖数: 204 | 19 Here is a solution by using vector calculations.
Let a = |BC|, c = |AB|, then easy to calculate that:
(1) |AM|^2 = |MB|^2 + |AB|^2 = a^2 -2ac + 2c^2
(2) |CN|^2 = |CB|^2 + \NB|^2 = 2a^2 - 4ac + 4c^2.
We also have
vec(MA)\cdot vec(CN) = (vec(BM)-vec(BA))\cdot (vec(BC)-vec(BN))
= a^2 - 2ac + 2c^2 = cos(pi/4) * |AM| * |CN|.
so 角MPC=45度 |
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n***p
发帖数: 7668 | 20 OK, if $f$ depends on the position on the boundary, then
in fact it should be written as $f(x,u)$. It becomes a slightly
more interesting variational problem. Suppose the minimizer u
exists, which is not always true, then it satisfies
-\Delta u = 0 in the domain
2 \nabla u\cdot n + f_u (x,u) = 0 on the boundary.
Here \nabla u\cdot n is the normal derivative of u. |
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a****a
发帖数: 5763 | 21 一般的解hermitian matrix, numerical recipes上有一小节专门讲
这个问题,可以直接用jacobian transformation，或者用Householder
转换成三角阵然后用QL迭代
或者干脆写成 2nx2n的实matrix 求解
if C=A+iB is hermitian
(A+iB)\cdot (u+iv) =\lambda(u+iv)
then it is equivalent to write as
|A -B | \cdot |u| = \lambda |u |
|B A | |v| |v |
不太清楚解得性质如何，不过后者应该能保证eigenvalue是实数吧
具体参见Numberical Recipes third edtion page 590
I
including |
|
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Y***0
发帖数: 1804 | 23 在数学中，幂级数是一类形式简单而应用广泛的函数级数，变量可以是一个或多个（见
“多元幂级数”一节）。单变量的幂级数形式为：
f(x) = \sum_{n=0}^\infty a_n \left( x-c \right)^n
= a_0 + a_1 (x-c)^1 + a_2 (x-c)^2 + a_3 (x-c)^3 + \cdots |
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n**d
发帖数: 9764 | 25 Pedway--Downtown Pedestrian Walkway System
Chicago’s downtown pedestrian way system, the Pedway, lies in the heart of
the city. This system of underground tunnels and overhead bridges links more
than 40 blocks in the Central Business District, covering roughly five
miles.
Used by tens of thousands of pedestrians each day, the Pedway connects to
public and private buildings, CTA stations and commuter rail facilities.
Development of the Pedway began in 1951, when the City of Chicago built one-
blo... 阅读全帖 |
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发帖数: 1 | 28 http://www.ridebustang.com/fares-and-tickets
3:10 PM - 5:35 PM
3:10 PM from Union Station
$17.00 Walking duration 0 min
3:10 PM Union Station
(Denver, Colorado)
CDOT West Line (DAILY) towards Westbound
5:35 PM Vail Transportation Center
(Vail, Colorado)
Tickets and information
Bustang - Ticket information - 1 (800) 900-3011
These directions are for planning purposes only. You may find that
construction projects, traffic, weather, or other events may cause
conditions to differ from the map re... 阅读全帖 |
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m*********o
发帖数: 440 | 29
village
8
"太刺激我们农村人了\cdots" |
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m*********o
发帖数: 440 | 31 现在经济形势不好, 美男计也起不了作用\cdots |
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l*n
发帖数: 2590 | 34 几分钟前看了CDOT的update，说85% of US34 is total loss；40% of US36 is total
loss; 50% of Highway 7 is total loss.
按照美国人的速度，看来今年年底之前能够修好一条简易公路就不错了。 |
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b*******y
发帖数: 268 | 35 每次email里要写“thanks for your reply” 的时候总是会写成“thanks for your
replay”
tex里打“cdot”的时候总会打成“cdota”
然后每次再回去改，
btw，怎么忽然之间版上的人都不认识了。。。 |
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z****e
发帖数: 54598 | 37 Mathematics
While he is often regarded as a designer of mechanical devices, Archimedes
also made contributions to the field of mathematics. Plutarch wrote: "He
placed his whole affection and ambition in those purer speculations where
there can be no reference to the vulgar needs of life."[42]
Archimedes used the method of exhaustion to approximate the value of pi.
Archimedes was able to use infinitesimals in a way that is similar to modern
integral calculus. Through proof by contradiction (reduc... 阅读全帖 |
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a*****y
发帖数: 33185 | 38 http://en.wikipedia.org/wiki/Pi
Second millennium AD
Until the second millennium AD, estimations of π were accurate to fewer
than 10 decimal digits. The next major advances in the study of π came with
the development of infinite series and subsequently with the discovery of
calculus, which permit the estimation of π to any desired accuracy by
considering sufficiently many terms of a relevant series. Around 1400,
Madhava of Sangamagrama found the first known such series:
{\pi} = 4\sum^\infty_... 阅读全帖 |
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A****h
发帖数: 385 | 39 “如果你打算從二十歲到四十歲，每年都可考慮結婚。假如你每年只考慮一位對 象，
那麼最好的方法大約是M8，也就是說，自你二十八歲起，才開始要 「認真」起來。(年
青的時侯多玩玩無妨)。 ”
可见真是学好数理化，走遍天下全不怕。要想求得好姻缘，不靠摇签靠概率。
==================================================================
什么是秘书问题
在机率及博弈论上，秘书问题（类似名称有相亲问题、止步问题、见好就收问题、
苏丹的嫁妆问题、挑剔的求婚者问题等)内容是这样的：要聘请一名秘书，有n人来面试
。每次面试一人，面试过后便要即时决定聘不聘他，如果当时决定不聘他，他便不会回
来。面试时总能清楚了解求职者的适合程度，并能和之前的每个人作比较。问凭什么策
略，才使选得到最适合担任秘书的人的机率最大？
秘书问题的策略
基本解決策略如下：对于某些整数r，其中1 le r < n。先面试首r人，都不聘请他
们，在之后的n − r人中，如果任何一人比之前面试的人都更佳，便聘请他。
r的最佳值應該是rapprox frac{n}{... 阅读全帖 |
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e**n
发帖数: 478 | 40 Is this you want?
\begin{eqnarray*}
a & = & b+c+d+e+\sqrt{f^{2}+\frac{1}{\sqrt{g^{\frac{h^{2}+i^{2}}{2}}}}}\\
& & +\left(\sqrt{ff^{2}+\frac{1}{\sqrt{gg^{\frac{h^{2}+i^{2}}{2}}}}}+\sqrt
{ff^{2}+\frac{1}{\sqrt{gg^{\frac{h^{2}+i^{2}}{2}}}}}\right)^{3}\\
& & +\sqrt{ff^{2}+\frac{1}{\sqrt{gg^{\frac{h^{2}+i^{2}}{2}}}}}+\cdots\end{
eqnarray*} |
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b******y
发帖数: 2729 | 41 【 以下文字转载自 EE 讨论区 】
【 原文由 bjyyj 所发表 】
the pdf of the time interval between two buses is exponential
f(t)=exp(-t/15)/15
the avg number of passengers who arrives between time interval t is 2t
the avg number per bus:
\int_0^{\infty}2t\cdot f(t) dt
answer of b should be the same as |
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t**********h
发帖数: 100 | 42 不，我不能。如果能放在eqnarray我肯定放，我不应该说公式，应该说是一串\theta:
\{\theta_0, \theta_1, \cdots, \theta_n\}. 仅此而已。 |
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n******d
发帖数: 18 | 43 我不喜欢用word,现在用latex.
就是那种表示price quantity关系的图。
我想知道怎么才能easily 把一些数学符号标注在图上
如 \bar{x}_1 , \alpha\cdot(1-\beta) 一类的。
xfigure 好像不错，不过我不能输入数学符号。 |
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s*******1
发帖数: 146 | 44 Job Title: GIS Analyst
Job ID: 100289
Date Closed: 03/07/2010
Business Unit: Charlotte Dept Transportation
Full/Part Time: Full-Time
Regular/Temporary: Regular
City Job Details
Major Duties and Responsibilities:
The CDOT GIS Analyst is responsible for providing data analysis and support
through the reporting and mapping of work activities within the Street
Maintenance Division of the Charlotte Department of Transportation. Position
is primarily responsible for the deve |
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S******g
发帖数: 365 | 45 最简单的方法，对(n+1)^k展开，然后全加起来，可以在已知1\cdots (k-2)情况下求出
(k-1)的情况。 |
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l******n
发帖数: 9344 | 46 一个distribution function defined as:
phi: [0,T]*Omega*S^2 ----> Omega
Omega is a domain in R^n.(you can take n=3)
phi satisfies the following equation
dphi/dt + (v*\nabla)phi=\Delta phi + \nabla \cdot (g*phi)
v is H^1 in Omega, the first \nabla is direvatives to variables in Omega
, the second \nabla is direvatves to variables in S^2, g is a smooth fun
ction on S^2.
initially phi is smooth, has anybody any idea how to prove that phi is H
^1(Omega) or L^2(Omega)?
Or if you know any papers related, |
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s*x
发帖数: 3328 | 47 这个问题如果不要求整数的话很简单，但是如果要按离散问题来算就比较麻烦，但是如
果不要求最优解，只要求一个k-approximation解的话应该不难，弄个greedy算法，估
计就是个k-approximation，我猜想这个k\leq 2. 算法大概如下：
把 p_1\cdots p_n 按价格由大到小排序，然后从p_1开始加一直到p_n，如果其中加入
某件商品超过了100块，就不加那件商品。最后加剩下的商品价值最小的。这些买下来
。然后第二轮开始，还是类似的做法。直到最后没有剩下的了。
比如如下7件商品 100,40,40,40,30,20,10
第一次 100+10 送 30, 剩下 40,40,40,30,20
第二次 40+40+20 送 30, 剩下 30
第三次 30 |
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m*********s
发帖数: 368 | 48 pu/pn = gradient u \cdot n, 估計你是忘了定義了
你會算那一點的梯度 和 法線方向即可..
数 |
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