X****r 发帖数: 3557 | 1 O()的结果是一个集合,不能这样加减。
准确地说应该这样写(这里用LaTex的记号表示方法):N^3-N^2+n \in O(N^3)
O(N^3)和O(N^3-N^2+N)是同一个集合,所以没有人用后者。
O只是上界,要精确表示的话还有下界。比如\Theta记号表示上下界:
N^3-N^2+N \in \Theta(N^3) |
|
T*******n 发帖数: 493 | 2 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} |
|
g********m 发帖数: 956 | 3 【 以下文字转载自 Science 讨论区,原文如下 】
发信人: glassdream (dashao), 信区: Science
标 题: 这叫什么方向?
发信站: Unknown Space - 未名空间 (Tue Apr 27 17:02:27 2004) WWW-POST
在极坐标(r, theta)中,r变化的方向称为radial direction,
那么theta变化的方向如何称呼?
多谢赐教。 |
|
w*********r 发帖数: 488 | 4 原题如下:
given function V in terms of S and t,V(S,t)
parameters are:rho,alpha,lambda,theta,mu,sigma
the PDE is:
rho*V(S,t)=alpha*S+lambda*(mu-S)*Vs(S,t)+theta*(Vs(S,t)).^2+0.5*sigma*(S-S.^2)*Vss(S,t)-Vt(S,t)
S is in [0,1],and t is in [0,T],T is a constant.
the boundary condition is V(S,0)=0
Vs is the first-order partial derivative of V wrt S, and Vt is first-order partial derivative of V wrt t.
昨天晚上在MATLAB里面我试了approximation+Broyden method,因为发现Broyden有现成的function(也我们老师自己编的,不是MATLAB自带的),newton那个总说有错误 |
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s*******g 发帖数: 18 | 5 刚刚接触maple, 需要解决实际问题,遇到机器人手臂问题, 需要用角度theta i定义第
n个手臂的位移和速度。
矩阵A,是个i×1矩阵,元素a[i]=l[i]*sin(theta[ i])*omega[ i]
之前用不用定义a[i]是个array之类的?
然后对A求对omega[i]的偏微分,得到的A'=Diff(A,omega[i])。 |
|
v***o 发帖数: 51 | 6 在分析一个VALUE FUNCTION关于STATE VARIABLE的二阶导数时,要看一个函数期望值关
于其分布均值和方差的二阶。稍微想了想,如果分布是正态的,这个函数是连续递增和
CONVEX的,那么这个函数的期望值对于分布均值(和方差)应该也是CONVEX的(因为函
数右端CONVEX地发散至无限Dominate了其他因素)。好像如果是CONCAVE的函数就难说
了。
想更多地了解一下各种函数期望值关于其分布概率参数的性质(如单调和CONVEXITY/
CONCAVITY,即E[f(x)|theta]关于theta(x分布的参数) 的各阶导数。哪位大牛指导
一下哪些文献可以参考? |
|
t****g 发帖数: 715 | 7 If you know what your f(x) is, why not simply compute derivatives of E[f(x)|
theta] wrt theta to see what happens? I can not see any technical difficulty
there. |
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v***o 发帖数: 51 | 8 哈哈。我犯错误了。谢谢指正。当时在想general的分析,就只想着积分号里的东西,
没顾着这个其实是可积的了。
其实我问题没问好。最初源自这么个想法:“要使E[f(x)|theta]关于theta是concave
(or convex),f(x)和x的分布应该符合什么一些假设”。 |
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a*****g 发帖数: 19398 | 9 围棋帮助提高 ADHD 儿童的大脑认知能力zz
Baduk (the Game of Go) Improved Cognitive Function and Brain Activity in Chi
ldren with Attention Deficit Hyperactivity Disorder
Abstract
Objective
Attention deficit hyperactivity disorder (ADHD) symptoms are associated with
the deficit in executive functions. Playing Go involves many aspect of cogn
itive function and we hypothesized that it would be effective for children w
ith ADHD.
Methods
Seventeen drug na?ve children with ADHD and seventeen age and sex matched co
mpa... 阅读全帖 |
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w********h 发帖数: 12367 | 10 no...
there is difference between non-solvent and no-good solvent.
generally no-good solvent means theta-solvent.
as to PS, methylcyclohexane is a theta-solvent while toluene is a good one. |
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d*****w 发帖数: 124 | 11
>>我是自己写, 不超过十行就搞定了.
>>当然可以,你听说过Kramers Theorem吗?自己看书罢
>>一个几何参数,从定义出发当然不考虑什么氢键.如果你要具体讨论氢键与具体分子Rg的
关系,那时你以后的讨论了.
>>我说Rh没有很好定义,是因为它是通过实验测定的扩散系数来等价于一个不可穿流的实
心球的扩散系数来类比的. 举个例子(JCP, 73,5971)温度T> theta, Rg>Rh;温度T<
theta,Rg |
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b*o 发帖数: 47 | 12 I am very interested in this and looking forward to your publication. In
Edwards' derivation, I think the attraction part has the physical origin of
the monomer space packing entropy de Gennes mentioned. The polymer chain in
melt (supposedly polymer good solvent, i.e., kai=0) and in theta solvent both
shrink from that in good monomer solvent, but with different physical driving
force: entropy in the melt case while enthalpy in the theta solvent case, I
think.
off.
=
correlation
of
interaction
i |
|
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s*******g 发帖数: 18 | 14 刚刚接触maple, 需要解决实际问题,遇到机器人手臂问题, 需要用角度theta i定义第
n个手臂的位移和速度。
矩阵A,是个i×1矩阵,元素a[i]=l[i]*sin(theta[ i])*omega[ i]
之前用不用定义a[i]是个array之类的?
然后对A求对omega[i]的偏微分,得到的A'=Diff(A,omega[i])。 |
|
o**a 发帖数: 76 | 15 用Poisson公式可以证明(见最后),但这道题在Ahlfors书上出现在Poisson公式之前,
大家有没有什么好办法?
证明(Poisson公式):
由于
\int_{|z|=r} u(r*exp(i\theta)) d\theta = alpha * log(r) + beta
而 u有界,所以alpha=0
定义u(0)=beta, 则u在|z|<\rou连续,并且满足平均值公式
从而满足极大和极小值原理
根据|z|=r上的u的值以及Poisson公式我们可以得到一个|z|<=r上的调和函数v
u-v在|z|<=r满足极大和极小值原理,而在边界上是0,从而恒等于0
从而u在z=0也调和。 |
|
|
n*******l 发帖数: 2911 | 17 Yes. For any complex number z = r e^(i\theta), we have
e^(-i\theta)z = r. |
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n*******l 发帖数: 2911 | 18 Yes. For any complex number z = r e^(i\theta), we have
e^(-i\theta)z = r. |
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h****f 发帖数: 228 | 19 普通球面坐标用的是(r, theta, phi). theta 是co-latitude, phi 是经度。 我需要
把这个坐标系下的值转换到另外一种坐标,(r, t, p), t 是任意的球面向量,p=(r
x t). 在这个坐标系,p 和 t 是互相垂直的,(p,t) 定义的平面与 r 垂直。 请问
有什么地方可以找到这样的转换矩阵。 |
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s****g 发帖数: 8 | 20 【 以下文字转载自 shopping 讨论区 】
发信人: szhang (sisi), 信区: shopping
标 题: 问个简单的平面几个问题
发信站: BBS 未名空间站 (Thu Apr 26 18:04:55 2007)
有三个点vec1=(x1,y1),vec2=(x2,y2),vec3=(x3,y3)
知道前两个点,vec1,vec2,知道(vec2-vec1)和(vec3-vec2)的夹角theta。 另外点
vec3到点vec2的距离是r。
(vec2-vec1).*(vec3-vec2)=cos(theta);
norm(vec3-vec2)=r;
有没有用matlab算vec3=(x3,y3),比较简单的办法? 我压根没算出来,唉。。。 |
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a***n 发帖数: 40 | 21 【 以下文字转载自 Statistics 讨论区 】
发信人: alvin (al), 信区: Statistics
标 题: a question about least favorable prior
发信站: BBS 未名空间站 (Tue May 29 23:34:29 2007)
Can a least favorable prior be an improper prior? My impression is that it
can, but it seems not true from the proof of the following theorem.
The theorem is: if Delta0 is a Bayes rule w.r.t. proper prior Pi0, and R(
theta, Delta0) <= r(Pi0, Delta0) for any theta. Then Pi0 is least favorable.
One of the steps of the proof is: integral of "R(delta0, the |
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c***r 发帖数: 63 | 22 J(r, \theta) = \frac{\partial(t, u)}{\partial(r, \theta)} |
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C********n 发帖数: 6682 | 23 how about
\int {e^{a*cos\theta + b*sin\theta},x=0..2Pi}
dec
get |
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r******g 发帖数: 112 | 24 见附图。
在xy坐标系内,x轴绕某A点转theta角,得到新的绿色的2条线
现在已知黑色OB,OC长度,以及theta角,请问如何能够反推得出A点的x,y坐标?
因为afa依A点的x,y坐标而变,所以也是未知
身为薄厚的我,实在是没有头绪了,多谢指教 |
|
v***o 发帖数: 51 | 25 想更多地了解一下各种函数期望值关于其分布概率参数的性质(如单调和CONVEXITY/
CONCAVITY,即E[f(x)|theta]关于theta(x分布的参数) 的各阶导数。
在分析一个VALUE FUNCTION关于STATE VARIABLE的二阶导数时,要看一个函数期望值关
于其分布均值和方差的二阶。稍微想了想,如果分布是正态的,这个函数是连续递增和
CONVEX的,那么这个函数的期望值对于分布均值(和方差)应该也是CONVEX的(因为函
数右端CONVEX地发散至无限Dominate了其他因素)。好像如果是CONCAVE的函数就难说
了。
哪位大牛指导
一下哪些文献可以参考? |
|
a***n 发帖数: 3633 | 26 请问如果知道复平面上一个围线地方程z(theta),theta是幅角,
如何可以求出z围成的图像的面积? |
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f*********g 发帖数: 632 | 27 The Siegel theta function is implemented in Mathematica as SiegelTheta[Omega
, s].
This function was investigated by many of the luminaries of nineteenth
century mathematics, Riemann, Weierstrass, Frobenius, Poincaré. Umemura has
expressed the roots of an arbitrary polynomial in terms of Siegel theta
functions (Mumford 1984).
http://mathworld.wolfram.com/SiegelThetaFunction.html
the |
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x***u 发帖数: 6421 | 28 a(d^2/dx^2+d^2/dy^2)f(x,y)-f(x,y)=H(r)exp(i\theta)
求f(x,y)=g(r)exp(i\theta)类型的解,H(r)是已知函数。 |
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w******0 发帖数: 1404 | 29 如果\theta 可写成f1(r)f2(\theta,\phi)的形式显然有。 |
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s****i 发帖数: 216 | 30 对heaviside function theta(x) 做傅里叶变换得到
-i/w+pi*dirac(w)
那么对theta(x-a)做傅里叶变换理应的到
e^(-ia)*[-i/w+pi*dirac(w)]
但是用mathematica公式
FourierTransform[HeavisideTheta[x - a], x, \[Nu],
FourierParameters -> {1, -1}]
得到的是计算
e^(-ia)*[-i/w]+pi*dirac(w)
这是怎么回事呢? 以前没学过傅里叶变换, 但是要用到下, 不是很了解 求板上达人
指教 |
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g****t 发帖数: 31659 | 31 你不是按陀螺绕杆的角速度w列的方程么?
"
陀螺还有一个自由度, 是绕杆的角速度w
"
"
所以拉格朗日量的表达式是这样的: L = A*w^2 + f(r,theta,r',theta')
其中A*w^2表示的是陀螺绕杆自转的转动能.
"
在惯性系下,你写的陀螺绕杆自转的动能是错误的.
A*w^2是对的,但w不是绕杆的角速度.
陀螺绕杆的动能,除了和绕杆的角速度有关,也必然和杆的转动速度有关.
一个postition r(t)的导数,分成两个部分,有一个部分是叉乘,
这个叉乘你给漏了. |
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x********i 发帖数: 905 | 32 http://iccm.mcm.ac.cn/dct/page/1
Invited Lectures
Group 1
Fan Qin: Cluster algebras and monoidal categorification
Fang Li: Positivity of acyclic sign-skew-symmetric cluster algebras via
unfolding method and some related topics
Cheng-Chiang Tsai: An attempt for affine Springer theory
Li Cai: The Gross-Zagier formula: arithmetic applications
Ming-Hsuan Kang: Geometric zeta functions on reductive groups over non-
archimedean local fields
Huanchen Bao: Canonical bases arising... 阅读全帖 |
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s*******g 发帖数: 18 | 33 刚刚接触maple, 需要解决实际问题,遇到机器人手臂问题, 需要用角度theta i定义第
n个手臂的位移和速度。
矩阵A,是个i×1矩阵,元素a[i]=l[i]*sin(theta[ i])*omega[ i]
之前用不用定义a[i]是个array之类的?
然后对A求对omega[i]的偏微分,得到的A'=Diff(A,omega[i])。 |
|
d****s 发帖数: 174 | 34 in the section of Gene Mapping:
LOD=Log10 P theta/P50%
Since the linkage frequency (theta) should be always less than 50%, how
could it be possible to have the LOD >3?
Because linckage have less than 50% reconbination frequency, why it is
linkage rather than nonlinkage when the recombination frequecy increase?
THank! |
|
n*s 发帖数: 752 | 35 我有一个多晶材料,已知[002]面大致平行于材料的一个表面
我用omega-2theta模式测了那个表面,得到了[002][004][014]三个peak
直接套用scherrer's formula,得到的结果比较一致
2 theta plane FWHM grain size (nm)
36.211 [200] 0.028 298.5
38.498 [130] 0.036 233.7
42.958 [220] 0.029 294.4
这样做可信么?
我的FWHM是直接把peak fit成gaussian得到的,3个peak看上去都不太对称,都是左半
边(小角度)比右半边高些。需要用2个gaussian来fit么?
如果我想考虑strain和instument broadening的话,可以直接fit这个公式里的A,B,C
么?
2 A 2 2 2
FWHM = (---------- ) + (B tan(theta) |
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r******g 发帖数: 112 | 36 见附图。
在xy坐标系内,x轴绕某A点转theta角,得到新的绿色的2条线
现在可以测出黑色OB,OC长度,以及知道theta角,请问如何能够反推得出A点的x,y坐
标?
因为afa依A点的x,y坐标而变,所以也是未知
身为薄厚的我,实在是没有头绪了,多谢指教 |
|
r******g 发帖数: 112 | 37 在线等,哪里对不上号?
A点的坐标一变,afa角也变,转动theta角之后的绿线的位置也要变化
现在是可以通过实验测出绿线从x轴或者y轴返回O点的长度和已知theta角
不知道能否倒推出A点的坐标,谢谢 |
|
S*********g 发帖数: 5298 | 38 球坐标?
p(theta) = sin theta
p(phi) = 1/2pi |
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S**********9 发帖数: 217 | 39
They may be different depending on what people think about. Omega scan and
keeing 2 theta fixed is to check the crystal qualities.
theta 2theta scan can be used for both powder and single crystal. However,
for single crystal, one needs to use 4-circle diffractometer to get
appropriate geometry for the scans. |
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z****0 发帖数: 1351 | 40 例如球上有一点坐标(rho,theta,phi),整个球随机转动theta1,phi1的角度,这个点
得到一个新的坐标(rho,theta',phi'),请问有没有类似绕XYZ转动的矩阵存在?谢谢 |
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z****0 发帖数: 1351 | 41 我想生成足够多的点,所以只生成了球面上的一小块patch,然后把这块patch的中心任
意的转(theta,phi),得到新的patch,希望求新patch对应的每一点坐标(rho,theta'
,phi') |
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x******i 发帖数: 3022 | 42
。。
假定运动在theta=pi/2,“赤道平面“内
先求4-速度u^a
再求归一化的theta方向单位矢量,记作v^b
然后求1/2*epsilon_{abcd}u^a v^b F^{cd}
这个应该就是磁场的大小 |
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r*****t 发帖数: 286 | 43 ☆─────────────────────────────────────☆
theta (delta, gamma and vega) 于 (Tue May 1 22:59:25 2007) 提到:
(x1,x2,x3)--multiNorm(u,sigma), and we know u and sigma.
then condition on x3=x, what is the covariance matrix of muitinorm distr
ibution (x1,x2)?
anyone can shed light on this?
☆─────────────────────────────────────☆
quantler (quant) 于 (Tue May 1 23:03:41 2007) 提到:
熬了一个通宵后 我的神智已经不是很清醒了
不能帮您了:)
☆─────────────────────────────────────☆
theta (delta, gamma and vega) 于 (Tue May 1 2 |
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K*****Y 发帖数: 629 | 44 If the marginal distribution does not belong to the family of elliptical distribution, then correlation is not a good measure of dependence, copula is more general here.
Your joint distribution will take the form: F(x,y) = C(F1(x),F2(y),theta). Where F1,F2 are your marginal distributions, while C is the copula function. You need to estimate the unknown parameter theta though.
If you want to stick to the correlation framework, another simpler approach would be like this:
Simulate (Z1,Z2), w |
|
m*e 发帖数: 146 | 45 could you expain why it is s/(1-cos(2*pi/n)) for n side?
Thanks..
starting position]?
trajectory, which is r(theta)=r0*exp(-theta). the distance traveled is then:
's argument still holds: only the toward (i.e., along the relative motion)
component of the relative velocity matters. and one gets the distance
traveled is (side length)/[1-cos(2*pi/n)]. |
|
t******m 发帖数: 255 | 46 完全正确
starting position]?
trajectory, which is r(theta)=r0*exp(-theta). the distance traveled is
then:
apc999's argument still holds: only the toward (i.e., along the relative
motion) component of the relative velocity matters. and one gets the
distance traveled is (side length)/[1-cos(2*pi/n)]. |
|
y*a 发帖数: 7 | 47 It actually has a very elegant ansewer:
E[ |x_n-x_0|^2] = 2*R^2 - 2*R^2*cos(theta)^n
where theta = a/R, a is the distance walked each time (in this case, a=1
meter)
and it converges to 2*R^2 when n --> infinity.
Also if let R --> infinity, the result approaches n*a^2, whihc is the result
for a 2-D random walk on a flat plan.
Simple vector analysis can get you the answer. |
|
n******r 发帖数: 1247 | 48 P{X,Y,Z构成钝角三角形 or 构不成三角形}
=3*积分x:0-1,y:0-sqrt(1-x^2),z:sqrt(x^2+y^2):1
=3*\int\int_{x^2+y^2<1}(1-sqrt(x^2+y^2))dxdy
=3*pi/2\int_0^1(1-r)rdr (极坐标代换x=rcos\theta, y=rsin\theta)
=3*pi/12
=pi/4
P{X,Y,Z构成锐角三角形}=1-pi/4 |
|
T*******t 发帖数: 9274 | 49 还是画图快...
p = \cos\theta, with \theta between 0 and 2/3\pi |
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n****e 发帖数: 629 | 50 你问的是E(X-Y|2X+Y)?
这种题可以化成矢量做。X,Y就是长度为a,b的矢量,夹角\theta满足\cos(\theta)=\
rho
然后你这里要求的 就是X-Y在2X+Y方向的投影
作个内积就出来了 |
|