s********r
发帖数: 529 | 1 其实可以用数学归纳法证明
设N*N的对阵矩阵行列式为f_n,则f_n(rho)=(1-\rho)^n*f_{n-1}(rho/(rho+1)),其中f_
n半正定的要求是\rho \in [-1/n,1] |
|
G*****n
发帖数: 7 | 2 correct me if i'm wrong. Think about a simple case: suppose x_i = a * \rho +
\epsilon_i, where \rho is const, a, \epsilon_i are iid. then r is a
function of \rho. if \rho = 0, r = 0; if \rho = \inf, then r = 1. so r can
be anywhere between 0 and 1. applies to arbitrary N case.
just my 2 cents |
|
l**n
发帖数: 67 | 3
irrotational->\vec{v} = \div{phi} which is defined by you
incompressible->\laplace{phi}=0???(no source,sink)
by definition, kinetic energy is
\int_V \rho/2*(\grad{phi})^2 dV
= \rho/2*\int_V{(\grad{phi})^2}dV
= \rho/2*\int_V{\phi*\laplace{phi}+(\grad{phi})^2}dV
using Green's first identity
= \rho/2*\int_s{\phi*\vec{n}.\grad{\phi}dA}
= \rho/2*\int_s{\phi*\partial{\phi}/\partial{n}}dA
The green's first identity is:
for any well-behaved scalar function a,b,
\int_V{a\laplace{b}+\grad{a}.\grad{b})dV |
|
g******i
发帖数: 251 | 4 Using Gauss' Law, we can obtain E1=rho/epslon. when we calculate the force on
the unit area, we should know the field at this point is the sum of the field
of created by all other charges plus the field created by the unit area. We
can prove these two are equal. So the field due to all other charges are
E=rho/2*epslon, so the force is f=Rho*Rho/(2*epslon).
By the way, rho should be surface charge density ba.
导体表面将受到张力,单 |
|
z**********i
发帖数: 12276 | 5 我只保留了一个COVARIATE IN THE MODEL, 如果加上其他的,花的时间更长.
加RANDOM 之前,只要2分钟,但如果加了RANDOM,花几个小时或更久.
CODE:
proc nlmixed data=ami1_04Q1 tech=newrap qpoints=50;
parms b0=-2.53 b24=-0.055 rho=0.007 varRin=0.90 cov=-0.011
varRslp=0.0028 ;
eta=(b0+ (b24+Rslp)*nqt+Rin);
expeta = exp(eta);
p = expeta/(1+expeta);
N=totdenom; r=R_totnum;
A=p*(1-rho)/rho;
B=(1-p)*(1-rho)/rho;
loglike=(lgamma... 阅读全帖 |
|
z**********i
发帖数: 12276 | 6 这个就是BETA BINOMIAL加RANDOM EFFECTS,该怎么变一下呢? 多谢!
proc nlmixed data=ami1_04Q1 tech=newrap ;
parms b0=-2.53 b24=-0.055 rho=0.007 varRin=0.90 cov=-0.011
varRslp=0.0028 ;
eta=(b0+ (b24+Rslp)*nqt+Rin);
expeta = exp(eta);
p = expeta/(1+expeta);
N=totdenom; r=R_totnum;
A=p*(1-rho)/rho;
B=(1-p)*(1-rho)/rho;
loglike=(lgamma(n + 1)-lgamma(r + 1)-lgamma(n-r+1))
+lg... 阅读全帖 |
|
Y****a
发帖数: 243 | 7 Y(t)- rho Y(t-1) = bata (X(t) - rho X(t-1)) + e
where e is iid normal(0,sigma^2)
apply EM algorithm to estimate beta and rho.
1. initial value rho = 0 => beta(hat)
2. plug in beta(hat), transform your Y and X, estimate rho(hat)
3. repeat steps 1 & 2 until converge. |
|
D********i
发帖数: 103 | 8 简单做个量级估算，把目前中国全国风力发电的功率,9000万千瓦，也即10^11瓦，用来
刮风。比如专门吹北京，宽50公里，高10公里吧
大气密度大约 rho=1kg/m^3
风速v
风的动能密度~rho v^2
面积A=50e3*10e3=5e8m^2
动能flux=(rho v^2)v
所以10^11W=(rho v^2)v*A
v=(1e11W/5e8m^2/1kg/m^3)^(1/3)=5.8m/s
也就是说，用全国风力发电的能量给北京50km*10km的面积刮风，刮的风也只有5.8m/s=
20km/h，大约只是三级微风，和人在静止空气中中等速度长跑时感到的风速一样，不知
道能刮跑多少雾霾 |
|
发帖数: 1 | 9 呵呵，有点意思了。
你可以想象把木块提起来无穷小的距离，那么木块收到的浮力就是rho g V. 假设底面
水平，那么底面向上的力就是 rho g h S, h是水深。如果底面和容器底贴住了，这部
分浮力消失，所以收到的实际浮力就是
F=rho g V- rho g h S
所以当 V>hS的时候这个力向上，反之就向下压。当然只有这个力向上且超快木块自重
的话木块才会上浮。这就是说要是在海底，贴紧地面的可能性会更大。
★ 发自iPhone App: ChinaWeb 1.1.5 |
|
y***g
发帖数: 1492 | 10 你看看极坐标的Jacobian就知道了
dV=\rho^2 sin(\phi) d(\rho)d(\theta)d(\phi)
即使\rho \theta \phi 都是uniformly distributed
但是单位球体内的点的数量 明显是\rho 和\phi的函数（而不是常数）
换个元就行了 |
|
w****b
发帖数: 623 | 11 1. Vul vs not. hold
xx Axxxx AJT9x x
RHO opens 1S. The vul made you a coward and, say, you pass. LHO bids a forcing
1NT, RHO bid 2C.
(a) do you double this?
(b) if you pass, LHO takes preference to 2S, and two passes to you. Do you
balance?
2. All white. Hold
x AQJ9xxx Jx Jxx
Partner opens 3D, RHO overcalls 3S, what now?
3. All white. Hold
Kx Kxx AQT9x Jxx, bidding went
LHO CHO RHO YOU
1N - 2D x
- - xx -
- -
1N's 15-17, 2D's transfer.
you lead a club, and dummy comes down to
Axxx Ax K |
|
b***n
发帖数: 13455 | 12 So far, I'd guess LHO holds CK, and RHO holds DK, and SA,SQ split. I'd play S
from the dummy to K in my hand. If SA is in RHO's hand, I will make the deal
easily. If my SK falls to SA of LHO, he has only two options: 1. return S to
SQ of RHO. Then if RHO returns a D, dummy Q. If DK is in LHO, no way to make
it. If not, DQ will take this one, and dummy's SJ/CA will take care of the
rest. 2. return D, also easy for the the dealer. |
|
f*****x
发帖数: 545 | 13
~~~~~~~~~~~~~~~~~~~~~
that is what i did. wait and see policy. after pd bid 4s, i think rho show
his shape and i have good control, so i think i should play a 6s
This is basically my line. however, rho turned out to have sqx and down 1.
I thought about this hand later. rho showed 55 minor, so he has 3 card in
major only, either 2-1 0r 1-2.
lho choosed 3c after rho's 2n, showing his 4 card club spt and his opening
lead showed he has only two d. so lho has either 4 |
|
p***r
发帖数: 20570 | 14 Some thoughts on one hand
This hand was played in the bbo money bridge game. You play with a robot,
your LHO also plays with a robot. The score format is total points.
You hold Ax K9x xx AKQ8xx , both white,
The first question is what to open?
Of course 1NT is a possible choice. However, this hand is probably
too strong for 1NT because of the strong suit in clubs. You don't need
a lot from partner to make 3NT. So you should open 1C, even play with
a robot.
The bidding develops :
1C p 1D 1H ?
Now... 阅读全帖 |
|
C*****9
发帖数: 147 | 15 你基本不会满意直接击落单张HK的概率，希望能偷到SK。DA赢下来后，需要想一会，
给RHO足够的思考时间，第2墩出得太快容易引起警惕，而且如果RHO不得不想一想，
之后结果是只好出A。你也应该想一想是SJ还是Sx，RHO能够估算你S是K或Kx:
出SJ 如果只是解决S猜断，也可以打不扑就是未持有Q，所以可以干脆单防单K
出Sx，一旦LHO持Qx，RHO持Axxx, 上A就整个帮助你建立S套了 |
|
j*******e
发帖数: 2168 | 16 en, it still involves guess.
Now you play DQ, RHO indeed plays DK, you are almost certain that LHO has CA
(otherwise RHO will not pass in 1st seat), but CJ is still a guess right?
LHO can very well have 2452.
Here's what happened at the table (board 3 Italy vs Netherlands):
http://www.bridgebase.com/tools/handviewer.html?linurl=http://w
I don't quite understand the Dutch guy's play.
How do you like the play of a small heart to HK after 3 round of trumps?
I am inclined to place LHO with HA (oth... 阅读全帖 |
|
p***r
发帖数: 20570 | 17 You have to hope RHO to hold 3-2-4-4.
If RHO holds 3-2-4-4 shape, you can play the 3rd C, ruff it, take S finesse
and claim later.
If RHO holds 4-2-4-3 shape. You also need to ruff the 3rd C with low. Then
SJ and hope RHO doesn't cover it. |
|
b***y
发帖数: 2804 | 18 If you keep SA
1) you make the contract whenever LHO has HA
2) when RHO has HA, you make the contract when H are 3-3 or RHO has A8xx or
A10
The only time when playing SA at trick 1 gain, is when RHO has ATxx, AND you
decide to play H9 on second heart. Note that you will lose when RHO has
A8xx. You cannot cater for either ATxx or A8xx. |
|
b***y
发帖数: 2804 | 19 If you keep SA
1) you make the contract whenever LHO has HA
2) when RHO has HA, you make the contract when H are 3-3 or RHO has A8xx or
A10
The only time when playing SA at trick 1 gain, is when RHO has ATxx, AND you
decide to play H9 on second heart. Note that you will lose when RHO has
A8xx. You cannot cater for either ATxx or A8xx. |
|
b***y
发帖数: 2804 | 20 "Try DKQA first. If LHO has 4D, cash SKQ. If LHO also has long S, it is a
simple squeeze."
If you want to be exotic, you can play SK then SA. If RHO has long S, you
have a double-squeeze.
"If RHO has 4 or 5 D, after we have only 3S in dummy and hand, RHO may have
shown 4-3 or 5-2 in minors, then major holding can be 4-2 or 1-5, and guess
is still needed. No guess needed for all other holding."
You also have a guess if RHO has 4-2 in minors, for example. Can be 4-3 or 1
-6 in majors...
guess |
|
k*******1
发帖数: 526 | 21 不知道你算的是什么，你假设10kg水，高度100m，假设为圆柱形，面积自然就算出来了
。m=rho*V=rho*s*h, s=m/(rho*h),so s=0.0001 m^2
rho是水的密度，1000 kg/m^3
s是面积
h是水的高度，100m
m水的质量，10kg
=
^5
at |
|
H********g
发帖数: 43926 | 22 假如雨的密度是rho，竖直下落，下落速度是v，人是个长方体，该长方体顶面积是A 前
面面积是B，那静止的话时间t里脑袋顶上落雨是（rho）Avt，
如果人以速度w跑步的话，单位时间里顶上落雨还是（rho）Avt，但是人的正面开始兜
雨，这个面上的兜雨量是（rho）Bwt，所以跑步的时候两面淋雨，单位时间里的确是要
多沾水。
如果是从一个避雨的地方到另一个地方，路程是s，因为s=wt，那么w越大，t越小，所
以顶面沾水会减小，但是前面沾水的量是一个衡量=rhoBs，也就是B面扫过的体积。所
以跑得越快，顶面沾水越小。所以如果考虑一般情况，还是跑得越快淋水越少。 |
|
S*****T
发帖数: 400 | 23 【 以下文字转载自 Physics 讨论区 】
【 原文由 susygut 所发表 】
发信人: Newton (牛顿), 板面: Science
标 题: 爱因斯坦自述（18）
发信站: 飘渺水云间 (Tue Feb 15 13:19:58 2005), 转信
那时，我认为，冒险尝试把总场b）表示出来，并为它确定场定律，是没有希望的。
因此，我宁愿为表示整个物理实在建立一个初步的形式框架；至少为了能初步研究广义相
对性的基本思想是否有用，这是必要的。这是这样进行的：
在牛顿的理论中，在物资密度 \rho 等于零的那些点上，引力场定律可以写成：
\Delta \varphi = 0 （\varphi=引力势）。一般则写成（泊松方程）
\Delta \varphi = 4\pi k \rho, \ \ \ (\rho = 张量密度 )
在引力场的相对论性理论中，R_{ik} 代替了 \Delta \varphi。于是，我们在等式右边也
必须同样用一个张量来代替 \rho。因为我们从狭义相对论知道，（惯性）质量等于能量，
所以在等式右边应该是能量密度的张量，就其不属于纯粹的引力场而论， |
|
s********k
发帖数: 6180 | 24 【 以下文字转载自 EE 讨论区 】
发信人: silverhawk (silverhawk), 信区: EE
标 题: 求助时间序列correlation的问题
发信站: BBS 未名空间站 (Fri May 7 14:43:51 2010, 美东)
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 |
|
y***d
发帖数: 2330 | 25 有一个泛函(物理上的自由能) F[rho(x)] = some integrations，
想求使 F 最小的 rho(x), 并且要求满足约束条件 \int dx rho(x) = A，
从数学上我知道是要用拉格朗日乘子法，
\delta (F + k(\int dx rho(x) - A)) |
|
y***d
发帖数: 2330 | 26 x 只是坐标，离散化，变成 x_i；然后在每个 x_i 有一个密度 rho(x_i)；
rho(x_i) =def= rho_i 是真正的自变量；
所以 \int dx rho(x) = \sum_i rho(x_i) = \sum_i rho_i ＝ A, 算是线性关系吧 |
|
z*******e
发帖数: 15 | 27 Hi all,
I need to plot such types of graph in two-dimension, for example, it
could be the union of all R_rho, 0=< rho <=1, P=constant,
R_rho = {R1 <= log(1+ rho^2*P), R2 <= log(1+ (1-rho^2)*P),
R1+R2 <= log(1+ P+ 2*rho*P)}
I would like to observe the boundary of the final region.
How could I plot this boundary? I tried matlab, using command "area", but
the whole area is colored.
Are there any better ways or tools to finish this?
Thanks. |
|
s********k
发帖数: 6180 | 28 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 |
|
z*****n
发帖数: 7639 | 29 不能，因为rho是多对一的映射。就是说，从A，B可以到
rho，但是不能从A，rho到B，或者B，rho到A。
时间信息已经在计算correlation的时候丢失了。 |
|
n********r
发帖数: 84 | 30 Let $\rho(u,v)$ denote the correlation of $u$ and $v$, and $t_1, t_2$
increasing functions. Given a bivariate normal distribution (x,y) where $\
rho(x,y)>0$, how does one show
\[
\rho(t_1(x),t_2(y)) \leq \rho(x,y)?
\] |
|
s********k
发帖数: 6180 | 31 【 以下文字转载自 EE 讨论区 】
发信人: silverhawk (silverhawk), 信区: EE
标 题: 求助时间序列correlation的问题
发信站: BBS 未名空间站 (Fri May 7 14:43:51 2010, 美东)
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 |
|
l*****e
发帖数: 65 | 32 包子收到了， 多谢。
令 B=A+kI >=0, B^T 的Perron Vector 为 z>=0。 那么z^T v >0 (严格正). 设 Bv=
av. 那么 z^T B v= (rho(A)+k) z^T v = a z^T v, 故 a= rho(A) +k, 从而
Av= rho(A) v.
这里不需要什么 Irreducibility 或 Primitivity 设定， 主要是 V的元素恒正 是个
非常强的约束。
符号说明， 对非负矩阵，谱半径 rho(A) 属于 s(A)。 |
|
y***d
发帖数: 2330 | 33 【 以下文字转载自 Computation 讨论区 】
发信人: ylsdd (河伯), 信区: Computation
标 题: 带约束条件的 minimization 怎么做？
发信站: BBS 未名空间站 (Wed Mar 10 11:33:49 2010, 美东)
有一个泛函(物理上的自由能) F[rho(x)] = some integrations，
想求使 F 最小的 rho(x), 并且要求满足约束条件 \int dx rho(x) = A，
从数学上我知道是要用拉格朗日乘子法，
\delta (F + k(\int dx rho(x) - A)) |
|
t******n
发帖数: 6242 | 34 今天无聊，看到这篇无聊的文章，就给你无聊地翻译成英文吧。
探索星系的捷径
A Shortcut to Galaxy Exploration
1. 问苍天情为何物？
1. Ask what love is heaven?
有人类的历史以来，人类就不断地探索自然，扩展了解自然的疆界。哥伦布发现新大陆
就是最好的例子。人类探索自然的勇气是从哪儿来的呢？来自人类情感的力量。那么，
情感是什么东西呢？这是千万年来，文学家探索的主题。在有一些文学家的眼中，物理
学家是没有情感的冷血动物。我就是一个物理学家。但是，我一点也不埋怨这些文学家
。虽然物理学家在自然界的探索中做出了伟大的贡献，但是，物理学家从来就没有给出
情感的物理定义是什么，也没有找到自然界蕴含情感的丝毫证据。所以，虽然人类知道
自己的生存发展跟自然环境是密不可分的，但是，很少有人类表达出对自然环境感谢的
情感。在很多人的眼中，自然界是死亡无情的物质世界。中国有一句古语，叫做“天若
有情天亦老”。因为几乎所有人类相信天是不老的，所以，认为苍天是没有情感的，就
不足为奇了。我提出了一个宇宙生长衰老的模型。虽然这个模型及其简单，能解释大多
... 阅读全帖 |
|
g*********r
发帖数: 543 | 35 首先石头被吹起来，可以沿水平方向被吹动 （drag force）， 或者沿垂直方向被抬起
（lift force), 一般drag force 要大很多，所以我们假设风施加drag force大到一
定值（大于石头的来自地面的摩擦力），则石头开始移动。
drag force 主要来源于 （1） 正压力 （2） 以及 风的摩擦力； 哪一个占主要作用
取决于物体的形状。 流线型物体风的摩擦力占主导地位，可以计算风的摩擦力F=C*1/
2*rho*V^2*A 根据物体的形状跟尺寸，以及风速以及风的密度， drag coefficient: C
一般有手册可以查寻到。(2) 如果物体的迎风面很大（比如一面墙），并且不是流线设
计， 则压力差占主导： 迎风面的压力由于风速产生一个动态压力 P = 1/2 *rho*V^2,
则F近似等于 1/2 *rho*V^2*A.
F=C*1/2 *rho*V^2*A = N*mu (drag force = 地面摩擦力）
风速很大，所以雷诺数很大
假设石头是个正方体，则drag coefficient C 应该在1 到2 之间 （包裹正压力跟摩擦
力... 阅读全帖 |
|
i********e
发帖数: 31 | 36 Let z = x+yi, \rho = abs(z) = |z|, \theta = arg(z)
case A:
z -> z^2/|z|^2: (\rho,\theta) -> (1, 2\theta)
case B:
z -> z^2/|z|: (\rho,\theta) -> (\rho, 2\theta)
fact I:
if z ~ complex normal, then
\theta ~ uniform distribution on the unit circle [0, 2\pi)
fact II:
the uniform distribution on the unit circle is
translation and scale invariant |
|
n********r
发帖数: 84 | 37 Let $\rho(u,v)$ denote the correlation of $u$ and $v$, and $t_1, t_2$
increasing functions. Given a bivariate normal distribution of (x,y) where $
\rho(x,y)>0$. How does one show
\[
\rho(t_1(x),t_2(y)) \leq \rho(x,y)?
\] |
|
p*****k
发帖数: 318 | 38 thought this is just a question about the greek: Rho of a straddle. dont have much intuition about rho outside the BS world. but in a BS world, it's not related to either delta or the volatility.
as longhei showed, rho of an european option is proportional to the risk-neutral probability of this option finishing ITM (note call has positive rho while negative for put; also the probabilities of a call and a put with the same strike add up to 1).
when the strike is S*exp(r-sigma^2/2), the prob. i |
|
m*****n
发帖数: 2152 | 39 Z = X when X <=\rho
Z = (1-\rho)Y + \rho when X > \rho
(X , Z) is the answer? |
|
b***k
发帖数: 2673 | 40 One question on the following solution I copied from wilmott.
let Z be an independant uniform variable
and
Y2 = X if Z< rho
Y2 = Y if Z > rho
Cov(X,Y2) = cov(X,X) * rho
and Y2 is uniform
(X,Y2) is our solution.
对于上面构造的Y2，如何计算E(Y2),Var(Y2),
为什么Cov(X,Y2)=cov(X,X)*rho?
是从公式Cov(X,Y2)=E(X*Y2)-E(X)*E(Y2)得到的吗？我推不出来啊。 |
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m******g
发帖数: 12 | 41 Let Z = (X+Y)/sqrt(2), then X, Z is bivariate standard normal with
correlation 1/sqrt(2). So conditioning on value of Z, X has variable 1-
rho^2
= 1/2.
Actually conditioning on any value of X+Y, the conditional variance is
1/2.
By the way, the conditional mean is given by rho Z. In this case,
conditional on Z=1/sqrt(2), the mean of X is 1/sqrt(2)*1/sqrt(2) = 1/2.
In general, if X and Z are marginally standard normal, with correlation
rho,
then conditional on Z=z, X is normal with mean rho*z, and |
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r***a
发帖数: 58 | 42 For i
Step one: generate r_{i}, r2_{i}
Step two: generate x_{i} = \sigma*r_{i}
step three: generate y_{i} = \sigma_{xy}/\sigma^2*x_{i} +
\sigma* sqrt(1-rho^2)*r2_{i}
where rho = sigma_{xy}/sigma^2
The theory behind this is
if x, y ~ bivirate normal (0, 0, sigma_x, sigma_y, rho)
then p(y|x) ~ normal(beta*x, sigma_y^2*(1-rho^2) )
where beta = sigma_{xy}/sigma_x^2 |
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l**n
发帖数: 67 | 43 I am confused here and don't know you are asking or answering.
Doesnot distinguish partial deriv. and deriv. is very bad notation.
But just to remind you, you are dealing with a field equation
here.(mass conservation) Whenever you want to use the field equation,
all the variables should be expressed in terms of field variables
which are coordinates \vec{x} and time. So
\rho=\rho(\vec{x},t)
\vec{u}=\vec{u}(\vec{x},t)
\vec{a}=\vec{a}(\vec{x},t)
and you have the \partial\rho/\partial t + div(\rho u |
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s***e
发帖数: 911 | 44 一般说来,所谓扩散针对分子(或者分子集团)的运动可以遍及整个空间的
情况而言. 这种情况最普遍就是气态和液态物质.性质"良好"的典型气体
和液体分子的热运动被假设为完全随机的, 在这个基础上可以推导出扩散
方程.
想象空间中一小团分子, 比如说, 一个小球形区域内的分子, 分子密度为
rho. 显然单位时间内穿越球面出去的分子正比于密度, 而从球面进入的
分子数正比于外面的分子浓度. 实际上实验指出单位时间内的净穿出数目
正比于浓度的梯度:
J=-D*Grad[rho]
其中D是扩散系数. J是扩散流矢量, 指向浓度减小最快的方向. 如果
假设流体是连续的不可压缩的, 那么小体积元内单位时间内净流出的
分子必然对应密度的减少, 凭这句话就可以推出著名的扩散方程:
(d/dt) rho-D*Laplacian[rho]=0
其中(d/dt)是时间导数, Laplacian=(d/dx)^2+(d/dy)^2+(d/dz)^2是拉普
拉斯算子.
最后在谈谈扩散系数D: 既然扩散过程是由于分子热运动导致, 那么扩散系数
D必然和热力学量有关系. 爱因斯坦推出了这个关系是:
D=(K_B*T) |
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s********k
发帖数: 6180 | 45 【 以下文字转载自 EE 讨论区 】
发信人: silverhawk (silverhawk), 信区: EE
标 题: 求助时间序列correlation的问题
发信站: BBS 未名空间站 (Fri May 7 14:43:51 2010, 美东)
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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r*g
发帖数: 3159 | 46 我来说一个plausible的这些系数的外行理解：
出发点是小范围内,密度随温度变化和密度本身成正比： d rho/d t = k rho， k是常
数。（1）
这样对体膨胀： rho * V = m， 两边对温度求导， dV/dt/V = -k是体膨胀系数,所以
这个定义有意义，要除以V。
如果是一维的，v= Al， A是常数， -k = dV/dt/V = Adl/dt/A/l = dl/dt/l (2)
如果是各向同性三维的， v=l^3, dV/dt/V = 3l^2*dl/dt/l^3 = 3 * dl/dt/l = -3k.
(3)
(1),（2），（3）选一个做出发点就好了。其中 (2)和（3）无所谓哪个更基本， 但
（1）和 （2
）究竟是哪个推出哪个呢？ |
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f*****x
发帖数: 545 | 47 i played this hand with a seeming advanced player. i bid 3h, he raised to 4s,
rho dbled.
his hand:
S: KX
H: JX
D: J10XXX
C: KQXX
rho has aqxxx 5 spades.
AND he blamed my 3h, i think he should bid 3n, right?
if he bid 3n, and if rho lead sx, then 9 tricks are there. |
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j*******e
发帖数: 2168 | 48 I ducked the spade and RHO switched to a low heart. I looked at the H9 on
table for some seconds... and played low from my hand... LHO won with H10 and
continued hearts. turned out RHO had HKQ...5555
I won the third round heart with A, LHO pitched a spade. I Played Club A and
Q. Then DK, all low. Then DQ, RHO played low again (obviously she did not have
DA), and I knew I was dead if LHO had Axx because:
1) If I play low diamond from my hand, LHO could duck and win the third round
D and then play |
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c****u
发帖数: 3277 | 49 LHO 大概是拿了6-4的H和C, RHO大概是双张H, 4张或者强3张C,
LHO 大概有可能把自己的双张花色的垫到RHO可能的边花赢墩上.
看样子, LHO更象是D单张, S双张, 所以可以考虑攻S.
另外同伴没有double过2D, 他拿强D的可能性也相应会低一些.
也许他拿了SAQxxx坐在明手的Kx后面, 或者是SA外加一个将牌赢墩.
可是我不理解为什么RHO一定要叫6C, 他总应该尊重搭档的意见才是.
庄稼也许拿了DAKxx Sxxxx Hxx CAQx
明手也许是SKx HAKQJxx CKJxx Dx
不过我以为LHO叫的也有问题, 在缺一个A的情况下也总应该叫满贯的.
否则就不该问A. |
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c****u
发帖数: 3277 | 50 you can try to cash all your hearts first,
pitch one spade, two clubs. one spade.
it's reasonable to play your RHO for 4 spades.
so after you finished drawing your trumps, RHO couldn't afford to
discard any spades. he has to discard some of his diamonds and clubs.
if you can read the situation well, I think he would discard his "useless"
diamond, you can cash your diamond first, the 3rd diamond might really
squeeze your RHO, he can't afford to pitch one spade, so he'd
discard one club, now you c |
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