u*****y
发帖数: 882 | 1 小弟收到一封邀请信：
Dear Dr. XXXXXXX;
On behalf of the Rho Chi National Honor Society, I would like to
congratulate you on your excellent academic achievements and would like to
extend to you a formal invitation to join the ZZZZZZ Chaper of the Rho Chi
Honor Society at YYYYYYYY University.
Rho Chi is a national pharmacy honor society that recognizes students and
faculty with excellent academic records and personal qualities.
......
Sincerely,
MMMMMMM
President, ZZZZ Chapter
问了系里的senior faculty，这个好像是邀请了才能... 阅读全帖 |
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y***e
发帖数: 6082 | 2 你作了主席都很难claim。。。。
小弟收到一封邀请信：
Dear Dr. XXXXXXX;
On behalf of the Rho Chi National Honor Society, I would like to
congratulate you on your excellent academic achievements and would like to
extend to you a formal invitation to join the ZZZZZZ Chaper of the Rho Chi
Honor Society at YYYYYYYY University.
Rho Chi is a national pharmacy honor society that recognizes students and
faculty with excellent academic records and personal qualities.
......
Sincerely,
MMMMMMM
President, ZZZZ Chapter
问了系里的senior... 阅读全帖 |
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r********r
发帖数: 352 | 3 Being a Rho-Chi member makes you feel good about yourself, and also makes your resume looks good. It might not help you getting a good job or getting a green card faster. In my opinion, being a Rho-chi pharmacist does inspire you to pursue intellectual excellence all through your career, and it is
worthwhile to become a member.
I think for pharmacy students, you need a top 10% GPA to be invited to the
society, but I do know PhD students who joined Rho-Chi just by nomination of
his/her PI. |
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c***k
发帖数: 4349 | 4 ρ.Ρ.rho σ.Σ.sigma τ.Τ.tau υ.Υ.upsil |
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k*****y
发帖数: 86 | 5 我是一个在读研究生 Rho - Chi Pharmacy Honor Society值得加入吗 对今后的找工作
或是绿卡申请有作用吗？ 谢谢 |
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c*******n
发帖数: 1214 | 6 The Rho Chi members are top 10% of the pharmacy students. I do not think it
cost a lot to join, so if you can join, why not. |
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c******h
发帖数: 355 | 7 This is a honor society. you cannot become a Rho Chi member by your own
application. |
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k*****y
发帖数: 86 | 8 谢谢 我是PhD学生 是我老板推荐我的
your resume looks good. It might not help you getting a good job or getting
a green card faster. In my opinion, being a Rho-chi pharmacist does inspire
you to pursue intellectual excellence all through your career, and it is
of |
|
|
k*****y
发帖数: 86 | 9 谢谢 我是PhD学生 是我老板推荐我的
your resume looks good. It might not help you getting a good job or getting
a green card faster. In my opinion, being a Rho-chi pharmacist does inspire
you to pursue intellectual excellence all through your career, and it is
of |
|
y****i
发帖数: 431 | 10 I want to be Rho-Chi member, but my grade is not good enough... |
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k*****n
发帖数: 108 | 11 a 3.75 gpa upon time of graduation is not bad at all... i worked really hard
throughout pharmacy school and only landed a 3.77 gpa by graduation, from
one of the top pharmacy schools in country
also a gpa can only get you that far in your career, the rest is your
capabilities as a person/pharmacist, i was in rho chi for 4 years, and
honestly no one cared to ask me about it during any of my interviews |
|
w********y
发帖数: 35 | 12 那我来厚颜无耻的show off一下吧。我现在P3,GPA 3.968.唯一一门2学分的课拿了B其
他都是A. 我是rho-chi不过真的觉得除了让自己增加几分自信之外，没什么用。另外我
不觉得美国同学成绩比中国人差，我们班很多美国同学成绩也非常好，能力也很强，
更主要的是很刻苦。要不输给他们，我也很刻苦。读pharmD比从前读PhD专心多了。
我们学校不是烂校，top 3之一。 |
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d********t
发帖数: 9628 | 13 把其中一个a*X+b，把另一个c*Y+d，新的distributions的correlation还是rho吗？
谢谢！ |
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d********t
发帖数: 9628 | 14
x,y的sigma分别大了sqrt(a)和sqrt(c)，rho就刚好不变了对吧？ |
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f*******a
发帖数: 663 | 15 开始忘贴代码了，有朋友要求，就把修改后的代码贴在这里。改动不多，可以部分提升
效率。原来的也没删，注释掉了。供参考。
=========================================================================
clear;
close all
disp('The only input needed is a distance matrix file')
disp('The format of this file should be: ')
disp('Column 1: id of element i')
disp('Column 2: id of element j')
disp('Column 3: dist(i,j)')
if(0)
% mdist=input('name of the distance matrix file (with single quotes)?\n'
);
mdist = 'example_distances.dat';
disp('Reading input dis... 阅读全帖 |
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x******n
发帖数: 24 | 16 First you may use F=f/\rho^k, G=g/\rho^k, you can obtain the new equation as
dF/d\rho=(a/\rho+a\lambda/\rho^2)G
dG/d\rho=-(a/\rho+a\lambda/\rho^2)F
which you may get F^2+G^2=C, then, assume
F(\rho)=\sqrt{C}\sin\theta(\rho),
G(\rho)=\sqrt{C}\cos\theta(\rho)
substitute into above equations, you reduce to solve the ODE on \theta(\rho)
d\theta/d\rho=a/\rho+a\lambda/\rho^2
then you can get the solutions... |
|
l*3
发帖数: 2279 | 17 呵呵, 我觉得你可能是不会证明这个显而易见的事实:
"刚体动能可以分解为随质心平动的动能+相对质心转动的动能" 这两部分.
我还是给你证明一下吧:
设初始时刻, 质心位置是r0, 刚体上任意一点位置r, 对应的密度是rho(r)
(注意, 这里r是指矢量, 是一个坐标组, 其实就相当于n维向量, 用r 是为了表示其状
态是 "初始状态", 和下面要提到的x做区分)
x0=x0(r0,t) 表示质心的运动轨迹 (依赖于初始值r0, 时间t)
x=x(r,t) 表示刚体上初始时刻位置为r的点的运动轨迹 (依赖于r, t, t=0时x=r)
则质心的定义意味着下式成立:
int (x-x0) rho(r) dr = 0
两端对 t求微分得到:
int (v-v0) rho(r) dr = 0
其中v=v(r,t) 表示陀螺上初始位置在r的点在t时刻的速度. v0=(r0,t) 表示质心在t时
刻的速度.
而陀螺总动能的表达式是什么呢?
是1/2( int v^2 rho(r) dr )
上式 =1/2( int 1/2 (v-v0 + v0)^2 rho(r) dr)
= 1/2( int ... 阅读全帖 |
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t******n
发帖数: 2939 | 18 ☆─────────────────────────────────────☆
l63 (l63) 于 (Sun Jun 30 11:44:29 2013, 美东) 提到:
具体是这样: 如图, 陀螺理想化模型为一个零质量的杆顶着一个均匀质量的圆盘, 圆盘
重心为C, 陀螺支撑点为O, 固定O点 (就是只是固定O点的位置, 陀螺杆本身是可以往任
意方向摆动的), 初始状态时, 陀螺绕杆的角速度为w0, 杆与竖直方向有一个小的夹角
a0, 初始状态时陀螺重心速度为0. 杆长为D.
计算方法: 建立坐标系, 以柱坐标描述重心的位置C=C(r,theta,z), 由于杆长固定, 故
z=sqrt(D^2-r^2), 相当于重心是两个自由度: r, theta (也就是可以用极坐标描述),
t=0时, theta=0, theta'=0, r=D*cos a0, r'=0; 陀螺还有一个自由度, 是绕杆的角速
度w, 这个w在t=0时的值为w0.
考虑陀螺的拉格朗日量, 即动能-势能.
具体表达式很长, 我会略去一些不必要的部分:
首先考虑陀螺的重心速度对应的平动动能, 重心速度由重... 阅读全帖 |
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x******n
发帖数: 24 | 19 For these eqs, same ideas may apply to, althouth it is unlikely to get a
nalytic solution:
first let F=f*\rho^{-a}, G=g*\rho^a, from eqs, you may get
\rho^{2a}(F^2)'+\rho^{-2a}(G^2)'=0 (*)
which may give you two eqs as:
(G^2)'+(\rho^{4a}F^2)'=4a\rho^{4a-1}F^2
(F^2)'+(\rho^{-4a}G^2)'=-4a\rho^{-4a-1}G^2
integrate these two eqs, the second one will give you
f^2+g^2
the first one will give you an integral ineq related to
\rho^{4a-1}F^2 after get rid of positive term G^2, by Gromell's ineq, y |
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x******n
发帖数: 24 | 20 For these eqs, same ideas may apply to, althouth it is unlikely to get
analytic solution:
first let F=f*\rho^{-a}, G=g*\rho^a, from eqs, you may get
\rho^{2a}(F^2)'+\rho^{-2a}(G^2)'=0 (*)
which may give you two eqs as:
(G^2)'+(\rho^{4a}F^2)'=4a\rho^{4a-1}F^2
(F^2)'+(\rho^{-4a}G^2)'=-4a\rho^{-4a-1}G^2
integrate these two eqs, and notice the definations of F and g, you will get
f^2+g^2 is bounded by two functions [f^2(1)+g^2(1)]\rho^{2a} and
[f^2(1)+g^2(1)]\rho^{-2a} (1)
Secondly, if you multipl |
|
M****e
发帖数: 3715 | 21 If x and y are iid standard normal, then
E[X|aX+bY]= (aX+bY)/(a+b)
E[X|aX+bY+c]= (aX+bY)/(a+b)
You can change two correlated normal distribution into two iid std normal
and use the equation above to get then answer.
X ~ N(ux, ox^2)
Y ~ N(uy, oy^2)
Cov(X,Y) = rho
Let Z1,Z2 ~N(0,1), iid
Then
X = ux +ox*Z1
Y = uy + rho*oy*Z1 + sqrt(1-rho^2)*oy*Z2
Z == X+Y
E[X|X+Y] = E[ux +ox*Z1 | (ox+rho*oy)*Z1 + oy*sqrt(1-rho^2)*Z2 + ux + uy ]
= ux + E[ox*Z1 | (ox+rho*oy)*Z1 + oy*sqrt(1-rho^2)*Z2 + ux + uy ]
= ux ... 阅读全帖 |
|
M****e
发帖数: 3715 | 22 If x and y are iid standard normal, then
E[X|aX+bY]= (aX+bY)/(a+b)
E[X|aX+bY+c]= (aX+bY)/(a+b)
You can change two correlated normal distribution into two iid std normal
and use the equation above to get then answer.
X ~ N(ux, ox^2)
Y ~ N(uy, oy^2)
Cov(X,Y) = rho
Let Z1,Z2 ~N(0,1), iid
Then
X = ux +ox*Z1
Y = uy + rho*oy*Z1 + sqrt(1-rho^2)*oy*Z2
Z == X+Y
E[X|X+Y] = E[ux +ox*Z1 | (ox+rho*oy)*Z1 + oy*sqrt(1-rho^2)*Z2 + ux + uy ]
= ux + E[ox*Z1 | (ox+rho*oy)*Z1 + oy*sqrt(1-rho^2)*Z2 + ux + uy ]
= ux ... 阅读全帖 |
|
p***r
发帖数: 20570 | 23 There could be another stronger line if you take HK on the first trick. When
you cash CA and opp drops C9, you can suspect that it could be a singleton.
So you just play back H right away to cut opp's connections.
If opps cash 4 hearts, you just come back to the squeeze position.
If LHO wins HT and shoot back D, you just play DA, back to CK to check the C
position. When the bad news comes (assume LHO discards a D), you just play
another H and play LHO for 4-3-5-1. If RHO cashes all his hearts, y... 阅读全帖 |
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w*****x
发帖数: 116 | 24 谢谢, 你的解答我看懂了,上面那个也可以用 X=f-g代换.
此外, 我计算对正电子作用,得到相似的方程组,
原来第二式中变为正号, 你的方法就不能用了.
df/d\rho-a*f/\rho=(a/\rho+a\lambda/\rho^2)g
dg/d\rho+a*g/\rho=-(a/\rho+a\lambda/\rho^2)f
而这里第二式的+a*g/\rho为正号.
请教中,谢谢. |
|
l******i
发帖数: 1404 | 25 Transformation Method and Accept-Reject Method
are two basic ways to generate random variables.
To summarize:
1. To generate iid uniform firstly:
Most programming languages have the ability to generate pseudo-random
numbers which are effectively distributed according to the standard
uniform distribution.
For instance, In C++, use srand().
2.
Given \rho to generate two N(0,1) with correlation \rho:
From 1, we can easily generate two iid U(0,1)
random variables P and Q;
Use Box-Muller transformati... 阅读全帖 |
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u*****a
发帖数: 6276 | 26 宾州州立大学警方星期二公布，兄弟会Kappa Delta Rho宾州州立大学分会的专用脸书
（Facebook）网页上流传多名裸体、无知觉、像破玩具娃娃一样的女子照片。警方相信
，那些女子根本不知道她们被拍裸照，更不知道她们的裸照被上传社交媒体。
据多家网站报道，宾州州立大学警方星期二举行记者会，公布对兄弟会Kappa Delta
Rho 该校分会的调查结果。Kappa Delta Rho那些网页显示欺辱和毒品交易证据和裸体
并失去知觉的女子照片。
有线电视新闻网（CNN）说，Kappa Delta Rho宾州州立大学分会已经被禁止活动。那
个分会有144名成员使用那个脸书网页。
哥伦比亚广播公司（CBS News）说，尽管那个兄弟会全国组织已决定暂停宾州州立大
学分会活动一年，但警方的调查仍在继续并有可能导致刑事起诉，包括侵犯隐私和性骚
扰。
Kappa Delta Rho一名前成员今年1月告诉警方说，分会有两个专用脸书网页，名字为“
Covert Business Operations” 和 “2.0”。后面那个网页是在兄弟会的另一网页"
Covert Business Trans... 阅读全帖 |
|
b***y
发帖数: 2804 | 27 OK, reconstruct the play, say LHO started with 4351 shape:
Trick 1, H2 lead, RHO plays HJ, you duck
Trick 2: RHO plays low H, you win HK
Trick 3-4: you cash CAK (LHO pitches a D)
Trick 5: you play a H, RHO wins
Trick 6: RHO cashes 1 more H (LHO pitches a D), leaving 1 H un-cashed
Trick 7: RHO plays a spade, you win SA
Trick 8: you play CQ, LHO pitches 3rd D
At this point, LHO has 3 spades and 2 D, all you need to do is to play D to
DA. The "best" defense you suggested (to cash 4th H but not last... 阅读全帖 |
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p***r
发帖数: 20570 | 28 If H is 3-3, you can always make it if you only lose one trump in C.
if H is 4-2, you can make it if you can ruff two H or HQx appears and you
also lose only one C later.
so it appears that you want to play LHO to hold 2 H and Cxx, CJxx, CKxx, CJx
CKx or CKJ. Or RHO holds 2H and can't overruff you, or after RHO's overruff
with CJ, you just take the success finesse in C and later play a squeeze in
S and H or S is 3-3 with SQ on side.
so it's good to just ruff H right away and see what happens.
If... 阅读全帖 |
|
b***y
发帖数: 2804 | 29 I think we can summarize the best line so far.
First, test diamonds. If 3-3 or DJ drop then claim. If DJ is outstanding in
LHO, claim (there are multiple ways to ensure 13 tricks).
If DJ is outstanding in RHO, next you play two more clubs (dummy pitch 1
heart, then D10).
1) If RHO has only 6 minor cards, you next cash 2 hearts, if LHO can follow
then RHO has at least 2 spades, you can pick up spade suit with confidence;
if LHO shows out then you know exact spade distribution too.
2) If RHO has a... 阅读全帖 |
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r*******s
发帖数: 303 | 30 可以这样想这个问题。
假如你有一个伞，只有三根支撑的棍子.
伞一开始是收起来的。这三根棍子是平行的，所以\rho=1，夹角的cos(0).
慢慢打开伞，直到三根棍子都在一个平面上了。这时候夹角是120，\rho=-1/2
所以这个可能的范围就是 ［-1/2，1］。
当然也可以算矩阵
1 \rho \rho
\rho 1 \rho
\rho \rho 1
det >0 |
|
r*f
发帖数: 731 | 31 Here is a problem:
Consider a Lagrangian description of a fluid motion, i.e.,
x=x(a,t) (x and a are vectors, but I don't know how to type
vectors here,sorry) is the trajectory of a particle identified
by initial spatial coordinate a, rho=rho(a,t),etc.
Compute the time rate of change of rho at a fixed point. The
answer must be expressed in terms of x=x(a,t) and rho=rho(a,t).
Consider a two-D flow.
Actually, in this problem, we have known x=x(a,t) and rho=rho(a,t).
what we need to solve is d(rho)/ |
|
t**s
发帖数: 4026 | 32 能不能自动自动换行啊，一个很长的方程，比如下面这一个，有包子好伐
\noindent\(\frac{1}{2 \gamma \sigma _1 \sqrt{1-\rho _{2,1}^2}}\left(2 (r-\m
athbf{v}) (-1+\gamma ) \sqrt{1-\rho _{2,1}^2} \left(\sigma _4 \rho _{4,1}
g_{\gamma }{}^{(0,0,0,0,1)}[t,S,\mathbf{v},r,\mathbb{D}]+\sigma _3 \rho _{3,
1} g_{\gamma }{}^{(0,0,0,1,0)}[t,S,\mathbf{v},r,\mathbb{D}]+\sigma _2
\rho _{2,1} g_{\gamma }{}^{(0,0,1,0,0)}[t,S,\mathbf{v},r,\mathbb{D}]\right)+
2 S \sigma _1^2 \sqrt{1-\rho _{2,1}^2} \left(\sigma _4 \rho _{4,1} \left(g_{
\gamma
}{}^{(0, |
|
N***r
发帖数: 2539 | 33 是，严格来讲你这个确实不是exact的伯努力方程。
严格的伯努力方程的推导有两种，无粘流体机械能守恒，或者欧拉方程里的动量方程沿
流线的积分。两种情况得到的最终方程是一样的，对不可压缩流体。
P_static + 1/2 rho u^2 + rho*g*h = constant
注意这里头一个非常重要的假定就是 “无粘”。
你的方程一般称为Bernoulli's equation with head loss，也就是考虑能量损失的伯
努力方程，一般很少被称为energy conservation equation，或许称之为mechnical
energy conservation是可能的（我不是很清楚）。在民用设计，计算里很常用，比如
说消防队计算水头什么的。湍流和层流并不是判断伯努力方程能否适用的准则，比如在
一段直圆管内，只要有摩擦，不管是层流还是湍流，严格的伯努力方程都失效，因为你
不能假定 “无粘流体”。这个时候要考虑沿程损失，而且要使用Bernoulli's
equation with head loss，就是这样的：
P_s1 + 1/2 rho u1^2 + rho*g... 阅读全帖 |
|
r*f
发帖数: 731 | 34 谢了.我知道你给我的方程是mass conservation equation.
but the \rho in this equation is the function of x and t or of a and t?
You know, the answer must be expressed in terms of x=x(a,t) and rho=rho(a,t).
Actually, the second question of this problem is :
write the conservation of mass equation.
And I have written it as d{\rho|/dt+div(\rho*u)=0.
But I don't know whether \rho must be the function of x and t or it can be the
function of a and t.
In our general problems, \rho always is the function of x and t, t |
|
s***e
发帖数: 911 | 35 我考虑了一下直接计算几率的可能性. 从n=2开始:
\rho(x1,x2)=\rho(x2)*\rho(x1|x2),
其中\rho(x2)=1/L, \rho(x1|x2)=1/x2 for 0
distribution is normalized, and =1/(4L), =1/(2L), =1/(4L).
Everything seems reasonable so far.
The distribution of d=(x2-x1) is calculated as:
\rho(d)=Integrate[\rho(x1,x2)*\Delta(d-(x2-x1)),{x1,0,L},{x2,0,L}]
I did a rough calculation on this. Again you need to to the transformation
for the delta function. But it's very simple. You can use this calcl |
|
f*******a
发帖数: 663 | 36 原帖见
http://www.mitbbs.com/article_t/DataSciences/6761.html
一点分析心得，与大家共享，以期抛砖引玉。
感谢zhaoce的总结一文让我看到这篇文章；也非常感谢f0008朋友在我始终无法下载附
件的情况下把附件发给了我。
===========================================================================
聚类算法能利用的一般是局部特性，如邻域点距离、基于核函数的密度估计。Mean-
shift算法就是一个非常经典的算法，以梯度方法迭代至局部密度峰值点。
这个算法的思路其实与Mean-shift很类似，虽然作者要在文章中反复说和Mean-shift不
一样，但本质上非常相近。MS以梯度寻找峰值点，而这个算法则是直接在点群中搜索峰
值点。这样做是基于一个近似假设：峰值点和点群中的某点距离不远。举个极端的例子
：只有一类，20个点均匀分布在一个圆上。MS算法可以准确聚到圆心，只要核大小足够
。而这个算法，只能聚在这20个点中的某个点上。这个假设在一般情况下可以接受，也
能... 阅读全帖 |
|
m*******s
发帖数: 3142 | 37 最近遇到一个比较困难的数值问题，不知道大家有没有碰到类似的问题。
有一个算符（或者说矩阵）积分微分方程，结构大概如下
\frac{\partial \hat{\rho} \left(t \right) }{\partial t}+ \hat{H}\hat{\rho}\
left(t \right)-\hat{\rho }\left(t \right)\hat{H}=\int_{0}^{t} d\tau F \left[\hat{\rho}\left(\tau \right),t \right]
右边是关于\hat{\rho} \left(t \right)的某种积分变换，但是没有解析结果。右边积
分的上限是随着t变换的，这就要求在给定的t以前的所有 \hat{\rho} 都要知道。
我知道这样的积分微分方程很难求解，不知道大家大家能否提供一些数值求解的思路？
特别是右边的积分如何数值处理？ |
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l*****e
发帖数: 65 | 38 rho(A)不是矩阵的模, 比如反例
A=(1 1) B=(0 0) A+B= (1 1)
(0 1) (1 0) (1 1)
则 rho(A+B)=2, rho(A)=1, rho(B)=0, 这里A, B是Jordan 标准型或转置。
至于Normal matrix N, 他们是 UD(U^*), 其中U是酉矩阵， D是对角阵，自然的就有
max_{|x|=1} |Nx| = max_{|x|=1} |Dx| =max_i |d_i |= rho(D)=rho(N)
总之，Normal是最接近对角阵的，性质好点很正常的。 |
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c**********e
发帖数: 2007 | 39 Can construct it in the following way:
X, Z ~ uniform on [0,1]. W ~ Bin(1, rho).
All three are independent. Let Y=WX + (1-W)Z
Then X and Y have corr rho.
The density of X+Y is
rho/2 + (1-rho)t for t<=1
rho/2 + (1-rho)(1-t) for 1<=t<=2. |
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S*******s
发帖数: 13043 | 40 from PD, the way to calcuate conditional EL is :
LGD * N( (N-1(PD) + rho*N-1(1-0.1%))/(1-rho^2)^0.5 )
where N is cdf of standard normal, N-1 is inverse cdf of standard normal.
I can not figure out the reasoning.
let a=N-1(PD), b=N-1(1-0.1%), c = (N-1(PD) + rho*N-1(1-0.1%))/(1-rho^2)^0.5,
we can see (1-rho^2)^0.5*c-rho*b =a.
what does it try to capture? |
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f*******d
发帖数: 339 | 41 一般说来，如果总电荷不为零，通过选择坐标有可能使有限系统的电偶极矩为零， 因为
\vec p = \int \vec x \rho(x)
如果不为零， 可以取
\vec x' = \vec x - \vec x0
\vec p' = \int \vec x \rho (x) - \vec x0 \int rho (x)
所以只要选
\vec x0 = (\int rho(x) \vec x)/\int \rho(x) = （\int rho(x) \vec x） / Q_tot
就足以令电偶极矩为零。这里的条件是总电荷不为零， 如果总电荷为零就没有办法了。
电四极矩有五个独立分量，独立的坐标变换有六个（三个平移三个旋转）， 因此也可以
选座标
使之为零， 具体公式我就不写了，总之是写出其变换方程，然后求解。
但是不能同时使电偶极距为零。 如果总电荷和电偶极矩都恰为零， 那么上述平移变换不
起作用，
也无法使电四极矩为零。 总之， 最低一级不为零的多极矩是与坐标无关的，不可能消去
。 |
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h****l
发帖数: 7290 | 42 狗尾续貂，给你加一些信息
P=rho*R*T
其中rho为密度，R为气体常熟，对于空气R=287.06，T为温度。
专业的物理量应该是P1，P2，S，x，rho，u，t
P1和P2为作用在前后截面上的压力（压强），u为这段气流的速度，t为时间
F=(P1-P2)*S
m=rho*S*x
a=du/dt
F=ma====>(P1-P2)*S=rho*S*x*du/dt====>dP/dx=rho*du/dt
接下去要得到压力随空间的分布，而不是密度的分布，因为一般
的速度范围内空气密度变化甚微，而压力变化比较大。当然
这里的压力是流体静压，而静压是与流体速度有关的..... |
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w****b
发帖数: 623 | 43 Our teammates' problem -- we knew those were problem as the results were all
disastrous.
1. Non vul. with KQ KQJ QJTxxxx x, you are first to speak with 1D. The bidding
went
You LHO CHO RHO
1D 1H x P
2D 3C x P
?
2. vul vs not. hold x Qx KQJxxxx xxx
RHO CHO LHO YOU
2S 3C 4S ?
3. All white. Hold AQxxxxx xx Ax Qx, you are 3rd to speak. Bidding went
CHO RHO YOU LHO
1C 1H 1S 3H(weak)
3S P ?
4. All red. Hold ATxxx KTxx AT Kx, RHO opened a 1st seat 1D, bidding went
RHO YOU LHO CHO
1D 1 |
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C*****9
发帖数: 147 | 44 比赛就要结束了，你面临二副高空作业考验。
1. IMP Pair，双有。你拿:
SJ2
HA9432
DJ987
C92
4c(RHO) - P (you) - 4s (LHO) - X (pd)
5c(RHO) - ?
2. IMP Pair，双无。你拿:
SJ987
H4
DJT984
CT87
p(RHO) - P (you) - 1h (LHO) - X (pd)
4h(RHO) - P (you) - p (LHO) - X (pd)
p (RHO) - ? |
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i****e
发帖数: 642 | 45 Team, good opps, red vs white
LHO Pd RHO YOU
P P 1S X
P 2H P 2NT
P 3NT AP
97
7654
QJ2
K652
KQ8642
AT
KT
AQ9
Lead (4th) D5, J, 6, T. S7, and RHO wins with SA, while LHO discards C8.
H3 to LHO's K, and H9 to RHO's J and your A.
Your DK goes to LHO's DA, and RHO follows D8.
LHO returns CJ to you, while RHO follows C3. Now where is the 9th trick? |
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g***a
发帖数: 114 | 46 LZ，我不太确定你的意思。
最后一种情况，我的理解是：
上部分的冲量变化的贡献：-\rho*v^2*s
中间水柱的部分：-\rho*((v+\delta v)^2 - v^2)/2g * g
= -\rho*(2v\delta v + \delta v^2)/2
下部分的冲量变化的贡献：\rho*(v + \delta v)^2 * s
最后的结果是\rho*(v\delta v) + O(\delta v^2).
所以应该是X3 > X1.
为什么你提到 “\delta v和液柱高度有关，新增的动量刚好和重量抵消” |
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g****t
发帖数: 31659 | 47 不可能.
因为不同的A1,A2,
A1,B
A2,B
可能生成同样一个rho
除非你事先知道A的范围或者分布.
或者知道更高阶的或者更多的统计量.
例如A(k-1),B(k)的cor,等等.
简单的说,这就是个几个未知数几个方程的问题.
例如A是高斯分布,那么如果要知道A的分布,则需要两个方程.
那么一个cor是反推不回去的.
假如有两个时间序列A,B. 之间的correlation为rho（假设rho比较大）,有没有什么办
法将A表
示成为一个B和rho的函数？A=f(B,rho).不要求数学意义上相等，数值意义上近似即可
。谢谢 |
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d**e
发帖数: 2420 | 48 great, that's right.
given A>B>0, we can choose sufficiently small d>0 such that
A-dI_n>B>0, then \rho(A-dI_n)>=\rho(B)
Note that \rho(A)=d+\rho(A-dI_n)>\rho(B).
The proof is complete.
非常感谢诸位的帮忙，实在让人敬佩。 |
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V****n
发帖数: 651 | 49 我先介绍一下大致估算方法及结果，如果你有兴趣，推导过程列在后面。
我这个方法需要一个correction factor，这个correction factor可以用一个测量
数据，也可以用多个测量数据导出，后者准确些。
因为你列出了一个测量数据(平躺150mm水柱，坐立350mm水柱，大脑至腰椎距离为0.6米
)，我就暂时用这些数值推导一下correction factor：
350mmH2O = 0.5*1000*w2^2 - 600mmH2O
w2=4.32m/s
w2(reference)=(2*9.8*0.6)^0.5=3.43m/s
correction factor = 4.32/3.43=1.26
那么，假设你有一个新的病人，大脑至腰椎距离为0.4m,那么
新的压力差 = (0.4*1.26^2)*1000mmH2O - 400mmH2O=235mmH2O
再比如，距离为0.5米，压力差=294... 阅读全帖 |
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b***k
发帖数: 2673 | 50 Thanks,miles. From this equation I know
E(Y2)=E(X)*rho + E(Y)*(1-rho),and
Cov(X,Y2)=Cov(X,X)*rho+Cov(X,Y)*(1-rho)=Cov(X,X)*rho
But how to evaluate Var(Y2), since I need to know it because
Corr(X,Y2)=Cov(X,Y2)/[sqrt(Var(X))*sqrt(Var(Y2))] |
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