p*****k
发帖数: 318 | 1
should not it be: sqrt(2)*[the distance between the center and the starting position]?
if you don't believe apc999's clever argument, you could solve the trajectory, which is r(theta)=r0*exp(-theta). the distance traveled is then:
sqrt(2)* int{from 0 to infty} [dr/d(theta)] d(theta)
and you got the same answer. |
|
y*a
发帖数: 7 | 2
Sure. Here is my derivation:
if x_i is the location vector originated from the center of the sphere, then
we have
x_i = R * p_i, where p_i is the unit vector in the same direction as x_i
then
R*p_i+1 = R*cos(theta)*p_i + R*sin(theta)*t_i --- (1)
where t_i is an random unit vector with a uniform distribution
in all directions that are perpendicular to p_i. and
theta = a/R, where a is the distance walked each time ( in this question, a
= 1 meter).
from (1) we have
p_n = p_0 * (cos(theta))^n +
|
|
D**u
发帖数: 204 | 3 Here is the counterexample (such that F(P) != 0 for some P).
...
Suppose the vertices of polygon P are (in counterclockwise position) z1, ...
, zn, (z_{n+1} := z1). Let theta_i be the angle of vector
z_{i+1} - z_i.
Define g(theta) = sin(3*theta), then we have g(theta) = - g(theta + pi).
We define F(p) = \sum |z_{i+1} - zi| * g(theta_i).
Ease to check the following:
(1) if P is the union of 2 non-intersecting P1 and P2, because the "line integral" on the common boundary of P1 and P2 vanishes, so |
|
z****i
发帖数: 406 | 4 the second problem, if the samples are from the uniform distribution [0,
theta], then the estimated theta would be the maximum of the samples, if we
are using MLE. so theta =100 if we just pull out one ticket, and theta =200
for the second time. but this estimation doesn't seem quite reasonable...
any ideas? |
|
z****g
发帖数: 1978 | 5 lz面的是trading-nature的position?感觉题目都和trading有关
其实就是让你从trading intuition来说
1. vega和theta没说吧，in-money和out-moeny option的vega和theta不一样，一般做
trading的时候对theta的考虑多一些，很多时候option trading赚的是theta，而不是
volatility
2. 第二个直观来看就是如果USD/EUR, EUR/USD可以分别交易的话，那这两种asset之间
是有套利价值的。因为按Ito推导的话，就假设 p * q = 1这个限制了，但是实际上，
如果两种汇率都是单独交易的话，一定是存在套利价值的（sig) |
|
i**w
发帖数: 71 | 6 Is the following argument natural?
From the perspective of a market maker, whenever he sells a call, he needs
to set up a hedging portfolio, which contains delta shares of stocks and
certain amount of funding balance. The cost to set up such a portfolio is
the price that he quotes for the buyer. (What he actually quotes will of
course includes a premium on top, to make money. but let's forget about the
making money part.)
-(Perfect) Black Scholes world
If he lives in a (perfect) Black-Scholes wo... 阅读全帖 |
|
s******h
发帖数: 539 | 7 I'll give you an example that says x, y, z are marginally normal and
pairwise independent but jointly can have infinitely many possible
situations.
Consider (X, Y, Z) with the following density function (don't mess those '{'
,'}'):
f(x, y, z;theta)
= 1/sqrt{2*pi}^3*exp{-(x^2 + y^2 + z^2)/2}*
{1 + theta*xyz*exp{(x^2 + y^2 + z^2)/2}*I{-1<= x, y, z <= 1}},
theta can be any real number such that theta < exp{-3/2}.
In this case, f(x,y), f(x, z), f(y,z) are all normal densities with x, y, z
pairwise i |
|
S******y
发帖数: 1123 | 8 I am wondering what data structure is "{ } " in R?
for example -
theta ={}
...
theta = rbind(theta, result)
Is theta a list or array?
Thanks. |
|
a****m
发帖数: 693 | 9 you want get the density of f(b|y;theta')
depending on f(y,b|theta')/f(y;theta')， which the top part is the joint
distribution and bottom one is marginal distribution. the theta' is the
initiated value of parameter. |
|
c*****l
发帖数: 1493 | 10 simulating a very basic model: Y|b=X*\beta+Z*b +\sigma^2* diag(ni);
b~N(0,\psi) #bivariate normal
where b is the latent variable, Z and X are ni*2 design matrices, sigma is
the error variance,
Y are longitudinal data, i.e. there are ni measurements for object i.
Parameters are \beta, \sigma, \psi; call them \theta.
I wrote a EM, the M step is to maximize the log(f(Y,b;\theta)) as the
regular way,
the E step involves the evaluation of E step, using Gau... 阅读全帖 |
|
w****o
发帖数: 367 | 11 Yes, Here is the statement:
X~f(x|theta), U_0 is unbiased estimator of 0. (i.e., E(U_0) = 0, for every
theta). Delta(X) is UE of g(theta).
Delta(X) is UMVUE of g(theta) iff Cov(U_0, Delta(X) = 0.
Cov(U_0, Delta(X) = 0 is a very strong condition: 如果你在你的Estimator里面
加入白噪音(white noise),也就是U_0, 不会是你的估计变得更好.
换句话说,如果你的Estimator可以被白噪音improved的话,你的estimator就是存在缺陷
的. |
|
p*****y
发帖数: 799 | 12 据新娘网（www.sima.com.cn）报道，
今年高考阅卷还有一名比古文奇才更奇才的物理奇才，最后一题的解答阅卷老师竟然完
全看不懂，请来专家组的孔子理工大学的物理系王跋诞教授来，王教授阅后称，这道题
考生的解答用到了中学生难以掌握的高深的物理知识，涉及到量子力学，群论，非对易
何，广义相对论，所用语言设计古希腊语，希伯来文，拉丁文，中文，英文，一般大学
教授都难以通晓。尽管王教授这个学期是教古典理论物理的，都是翻阅了三十多本教科
书才刚刚弄明白，刚注释就写满了300多页草纸。王教授认为该生实在是个不可多得难
能可贵的奇才，完全可以直接上理论物理的研究生，甚至当教授博导也不过分。
今年高考物理的最后一道题是：
一个质量为m，半径为r的小球，但是不知道其为空心的还是密度均匀的实心的。将其从
一个斜角为30度摩擦系数为1.2长度为3米的斜面顶部以初速度为零滚下，请问如何判
断球是空心还是实心的？
考试的解答是：
Let斜面的夹角be \theta.
根据被牛顿、雷布尼茨et al研究过的微积分原理，
\theta=\int^\infty_0\exp(-r/\theta)dr，
此处r可 |
|
h**6
发帖数: 4160 | 13 计算几何中，我一般用这个式子表示直线：
xcos(theta) + ysin(theta) = r
其中r是直线到原点的距离，theta是从x轴正半轴逆时针转向原点到该直线的垂线的夹
角。 |
|
g**e
发帖数: 6127 | 14 8. let rand() generates uniform distributed value between [0,1)
r = sqrt( rand() );
theta = rand() * 2 * pi;
x = r*sin(theta);
y = r*cos(theta);
return (x,y);
ob) |
|
v*****a
发帖数: 104 | 15 let x = r*cos(theta) and y = r*sin(theta) and let theta go from 0 to 2pi? |
|
u*****o
发帖数: 1224 | 16 设计题好难。。除了最后一题lz应该答的都不错啊，一定没问题的，bless!
想问一下那道partition document的题你咋答的？我想的是每次用内存处理一小部分，
把结果（hash-table)移到硬盘里，最后处理完再在硬盘里merge。。 或者用类似map/
reduce的办法？预处理怎么答啊？用unix吗？
那个customer personalization的题不知怎么答，如果我答可能从collaborative
filtering上答，设计一个regression model(linear/logistic) y = theta * x, y =
user 喜好， theta = parameter for each user (stay duration, visit frequency
etc), x = feature vector for each page (content...), 每步循环estimate theta
和 x 的值，直到达到optimization...这个你看make sense咩？
partition
理等 |
|
g********t
发帖数: 39 | 17 贡献刚做的online test，职位是Machine Learning related。
Question 1 / 2 (LaserMaze)
You are standing in a rectangular room and are about to fire a laser toward
the east wall. Inside the room a certain number of prisms have been placed.
They will alter the direction of the laser beam if it hits them. There
are north-facing, east-facing, west-facing, and south-facing prisms. If the
laser beam strikes an east-facing prism, its course will be altered to be
East, regardless of what direction it had been goi... 阅读全帖 |
|
M****g
发帖数: 162 | 18 Question 1 / 2 (LaserMaze)
You are standing in a rectangular room and are about to fire a laser toward
the east wall. Inside the room a certain number of prisms have been placed.
They will alter the direction of the laser beam if it hits them. There
are north-facing, east-facing, west-facing, and south-facing prisms. If the
laser beam strikes an east-facing prism, its course will be altered to be
East, regardless of what direction it had been going in before. If it hits
a south-facing prism... 阅读全帖 |
|
B********e
发帖数: 10014 | 19 好吧，我再说一遍我的意思
希望你不要再用‘那么不简单的limit’来羞辱我，我的学生会笑抽的。
pi代表单位圆的半周长。当然它本身是个无维度的数字。
单位圆的半个圆弧以弧度（rad）计长（即半周长），就是pi－－定义的；
对应的就是180度－－显然的。
可以精确用弧度值来描述的半周长，非得切称x份再搅合个sin函数，
再隐含使用sin(\theta)和\theta相差的是\theta的高阶小量来求一个
天真无暇的baby极限来建立degree和rad的联系，纯粹是脱裤子放屁多此一举。
我或许不应该这样驳aphh的面子。但是真心讲，
把小孩子一些似是而非的概念和想象过于拔高对他并不一定有好处。
说实话，很多民科都带有这种影子。
Pi |
|
g****u
发帖数: 695 | 20 You are asked to build a trading strategy with 60% correct ratio. This means
whatever method you give, it will be excuted a large number of times in
market in order to make sure profit with statistic significancy.
^VIX is an index so not tradable directly. There are derivatives using ^VIX
as the underlying, e.g. VIX futures and options. That would be the first
products you should look at. However, due to their lack of liquidity,
neither is suitable for a trading strategy to bet through a large n... 阅读全帖 |
|
B*W
发帖数: 46 | 21 原文发在
http://www.thetainvest.com/2014/07/%E6%9C%9F%E6%9D%83%E9%A3%8E%
看到坛子里大家对期权这么感兴趣，早就想把我一些教训和浅见和大家交流交流。今天
终于把给合作方的胶片写完，注册了域名，争取能一星期写一篇博客。如果能帮助大家
切磋技艺，少走弯路，就善莫大焉了。
读懂期权报价
期待的回报越高，可能性越低，反之亦然。 概率（Probability）和投资回报率 （
Return on Investment ROI）总是此消彼长，高回报需要冒高风险，太稳妥是赚不了大
钱滴。Out of The Money （OTM）期权可以在瞬间带来几倍，甚至几十倍的回报。但这
种小概率事件是普通投资者很难捕捉到，或很难反复捕捉到（有内幕另当别论)。想反
， MM们看透了散户一口想吃成一个大胖子，在ER前豪赌一把的心理，肆无忌惮地抬高
期权的价格，抵消了卖期权者的风险。关于ER再专文讨论。券商可能会提供 期权或期
权组合的Profit Probability, Max Return, Chance of Max Return， Max Lo... 阅读全帖 |
|
D*****T
发帖数: 126 | 22 原文发在
http://tinyurl.com/p8yk3zv
看到坛子里大家对期权这么感兴趣，早就想把我一些教训和浅见和大家交流交流。今天
终于把给合作方的胶片写完，注册了域名，争取能一星期写一篇博客。如果能帮助大家
切磋技艺，少走弯路，就善莫大焉了。
读懂期权报价
期待的回报越高，可能性越低，反之亦然。 概率（Probability）和投资回报率 （
Return on Investment ROI）总是此消彼长，高回报需要冒高风险，太稳妥是赚不了大
钱滴。Out of The Money （OTM）期权可以在瞬间带来几倍，甚至几十倍的回报。但这
种小概率事件是普通投资者很难捕捉到，或很难反复捕捉到（有内幕另当别论)。想反
， MM们看透了散户一口想吃成一个大胖子，在ER前豪赌一把的心理，肆无忌惮地抬高
期权的价格，抵消了卖期权者的风险。关于ER再专文讨论。券商可能会提供 期权或期
权组合的Profit Probability, Max Return, Chance of Max Return， Max Loss,
Change of Max Loss。是这个样子滴：
SPXW p... 阅读全帖 |
|
E***r
发帖数: 1037 | 23
option
2nd 花脸猫~
我补充一下
楼主大致是买了一周到期的 ATM (slightly OTM) Put
Theta，也就是单位时间的 extrinsic value 的损耗，临近过期是加速的
见左上图示意图，从绿线渐变到蓝线
今天 SPY 244 Jun 28 Put 的 Theta 是 -5.54
再过几天就会番两三倍
作为参考，SPY 244 Jun 21 Put （两天到期） 今天 Theta 已经是 -12.37 了 |
|
c*****o
发帖数: 1702 | 24 说的很有道理，你的意思是应该买远期的option?
: option
: 2nd 花脸猫~
: 我补充一下
: 楼主大致是买了一周到期的 ATM (slightly OTM) Put
: Theta，也就是单位时间的 extrinsic value 的损耗，临近过期是加速的
: 见左上图示意图，从绿线渐变到蓝线
: http://www.billiontrader.com/tim/100_6.png
: 今天 SPY 244 Jun 28 Put 的 Theta 是 -5.54
: 再过几天就会番两三倍
: 作为参考，SPY 244 Jun 21 Put （两天到期） 今天 Theta 已经是 -12.37 了
|
|
c*****o
发帖数: 1702 | 25 哥们帮我算算delta跟Vega是多少？
: option
: 2nd 花脸猫~
: 我补充一下
: 楼主大致是买了一周到期的 ATM (slightly OTM) Put
: Theta，也就是单位时间的 extrinsic value 的损耗，临近过期是加速的
: 见左上图示意图，从绿线渐变到蓝线
: http://www.billiontrader.com/tim/100_6.png
: 今天 SPY 244 Jun 28 Put 的 Theta 是 -5.54
: 再过几天就会番两三倍
: 作为参考，SPY 244 Jun 21 Put （两天到期） 今天 Theta 已经是 -12.37 了
|
|
b**l
发帖数: 33123 | 26 ☆─────────────────────────────────────☆
lovesunny123 (lovesunny123) 于 (Mon Dec 25 08:25:02 2017, 美东) 提到:
把大菠萝硬留了三年，瓜熟蒂落的时候被9人捡了个便宜。汤废材都40多了，随时可能
报销退役，错过了大菠萝，这下该怎么办。
☆─────────────────────────────────────☆
ccnpc (FatWolf) 于 (Mon Dec 25 09:31:10 2017, 美东) 提到:
今年再拿个碗就只剩别人郁闷了
☆─────────────────────────────────────☆
ThetaWave (Theta) 于 (Mon Dec 25 09:54:29 2017, 美东) 提到:
Having Jimmy G would secure another ten more years of thriving for Pats. It
makes more business sense. 但是sticking wi... 阅读全帖 |
|
b**l
发帖数: 33123 | 27 ☆─────────────────────────────────────☆
Dominique (曾经木木难为水 除却金色不是云) 于 (Tue Dec 26 15:39:49 2017, 美东) 提到:
竟然说是粉丝不grateful...
:(
didn't we show enough support and gracefulness during this difficult time!!!
"1 win away from another 10 win season! The fans that have turned on
players and coaches. You should be ashamed of yourself. This team has
overcome Tons of adversity and shown incredible resolve. There are teams
out there that haven’t had a winning season in years. Be grateful."
☆───────... 阅读全帖 |
|
s*****h
发帖数: 44903 | 28 ☆─────────────────────────────────────☆
HattoriHeiji (服部平次) 于 (Mon Jan 22 10:58:35 2018, 美东) 提到:
大家还记得以前版上总有人津津乐道的 那个前一秒还在超市搬砖后一秒拿碗的传奇么
Of course, that was not Foles, but he's close
2016赛季开始之前 佛寺一度曾经想要退出联盟
“我很挣扎，我有8个月的时间都拿不到球，我对比赛一点热情都没有了，我想退出。”
佛寺回忆起来那个时候 如此说
这时候 另一个传奇人物出手了
安胖想起来以前在老鹰一起度过的美好时光 一纸合同甩出
于是佛寺又有了打球的机会
或者更准确的说 是在铁匠屁股后面坐板凳的机会
佛寺回家跟老婆进行了一次长谈
但意料之中地 发现盖棉被纯聊天并解决不了问题
于是只好像无数的卢瑟们一样 求助于伟大的主
祷告之后 他听见脑海里一个声音
“我要为安西（划掉）安迪教练打球！”
然后便签了那一份一百多万的替补两毛五合同
一年多以后 当这纸合同已经消失在后视镜的时候
他的前面 是超级碗的大门
而且背景... 阅读全帖 |
|
x5
发帖数: 27871 | 29 ☆─────────────────────────────────────☆
redskeleton (红骨架) 于 (Wed Sep 26 16:04:50 2012, 美东) 提到:
聊了一个小时了。。
☆─────────────────────────────────────☆
tba (清华棒球联盟) 于 (Wed Sep 26 16:05:27 2012, 美东) 提到:
捷克，来TT?
☆─────────────────────────────────────☆
poise (Go Hokies) 于 (Wed Sep 26 16:05:35 2012, 美东) 提到:
你旦的?
☆─────────────────────────────────────☆
xjack (xjack) 于 (Wed Sep 26 16:06:51 2012, 美东) 提到:
哪个啊？
☆─────────────────────────────────────☆
flashingup (闪爷) 于 (Wed Sep 26 16:07:... 阅读全帖 |
|
p***e
发帖数: 29053 | 30 普通青年 : 1+3=4
文艺青年 : 1+3=B
二逼青年: like 獭獭 prove that cosh(z+1/z) = a0 + \sum^\infty_1 a_n(z^n+1/z^
n), where a_n = 1/2pi \int^\infty_0 cos(n\theta)cosh(2cos\theta)d\theta |
|
p**s
发帖数: 2707 | 31 O点只要偏离（0,0,-1），你的三个旋转轴就不能完全沿a’或者theta’方向，必然有w
分量，w=0就不成立，最后结果，肯定是a,theta不变，W变化pi/2。
你要是理解不了，可以继续画图，自己拿笔在纸上画即可，没必要贴出来，提示一你点
，三次旋转的方向，在实验室坐标里是不动的，但是a',w是随着陀螺转动的，所以旋转
角在a,theta，W上的分解，在任意时刻都是不断变化的。
最后，我上面都是胡说八道的，你要是看不懂，或者觉得是错的，就不必劳神反驳了。
组(
角) |
|
i******l
发帖数: 268 | 32 从力矩=角加速度开始，设人爬杆长度r=r0+vt,杆初始倾斜角度theta0=0,
mgr sin(theta)=mr^2 dtheta/dt
g sin(theta) = r dtheta/dt
g(r0+vt)dt = dtheta/sin(theta)
左右积分 |
|
b*****n
发帖数: 3774 | 33 你是问lognorma的expectation和variance哇，随便一本stat的书都有，
E是e^(mu+1/2*theta)
V是（e^theta^2-1)e^(2mu+theta^2)
太难打了。你google一哈嘛。 |
|
X****r
发帖数: 3557 | 34 看看这个\Theta(m^2 n)的对不
br />
/* find the max size of zero-filled 1-D array of integers
in \Theta(n) time
*/
int findZeroArraySize1D( int *x, int n ) {
int i, mx, mxp ;
mx = mxp = 0 ;
for( i = 0 ; i < n ; i ++ ) {
if( x[i] ) {
mxp = 0 ;
}else if( ++mxp > mx ) {
mx = mxp ;
}
}
return mx ;
}
/* find the max size of zero-filled 2-D array of non-negative integers
in \Theta(m^2 n) time
*/
int findZeroArraySize( int **x, int m, int n ) {
int **s, *d, i, j, ia, ib |
|
n********r
发帖数: 65 | 35 /*------------- code begin --------------- */
#include
const double theta = 0.5 * atan(2);
int main(void)
{
double a = 3.14159;
.....
.....
return 0;
}
/*------------- code end --------------- */
some books say the global variables are initilized
at the compile time. Is this true in the above case?
If in this case theta is initialized at the run time,
then when does the code " theta = code 0.5 * atan(2) " been executed?
before " double a = 3.14159 " or after that? |
|
d*********a
发帖数: 255 | 36 M=36;
theta=zeros(1,M);
pho=[1 0.5;0.5 1];
syms u;
syms v;
s=[norminv(u);norminv(v)];
for i=1:M
theta(i)=int(int(10*(exp(-0.5*s'*(inv(pho)-eye(2))*s)/sqrt(det(pho))),u,0,1)
,v,0,1);
end
theta
===============================================
??? Error using ==> <
Function '<' not defined for variables of class 'sym'.
Error in ==> C:\MATLAB6p1\toolbox\stats\norminv.m
On line 36 ==> k = find(sigma <= 0 | p < 0 | p > 1 | isnan(p));
如何修改才可以正确运行？或者有没有什么简单函数可以近似norminv函数？谢谢 |
|
d*******2
发帖数: 340 | 37 一个NxN矩阵，矩阵中心点为(N/2,N/2)。知道一个角度函数 theta(r)，r为各矩阵元素
到中心点的距离，即r(ite1,ite2)=sqrt((ite1-N/2)^2+(ite2-N/2)^2)。
现在需要构造一个新的矩阵(N,N)，如果这个矩阵中某个元素的位置为（r*cos(M*theta
),r*sin(M*theta))，则该元素为1，否则为0（M为整数）。请问怎么构造这个心的矩阵
？先谢了！ |
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w********h
发帖数: 12367 | 38 偶的意思是说“溶剂”分为三种：
(1)非溶剂（呵呵，non-solvent）;
(2)不良溶剂
(3)良溶剂
theta solvent属于（2），特定在theta温度下的稀溶液的情形，
但非溶剂不可以是theta solvent。
one. |
|
w********h
发帖数: 12367 | 39 刚才在写完后就意识到这个问题，一懒就没再改。
对的，若常温是theta温度的话，那么这时候使用就是theta溶剂，
若theta温度高于常温，在常温下该溶剂就只是非良溶剂了. |
|
w********h
发帖数: 12367 | 40 thanks for your further clarification.
as to monomer slovents (here it just means the small molecular weight solvents
generally), they can be good, theta, poor, or even non-solvents.
if it means the exactly the monomer of the solute polymer chain,
generally we think it is good solvent.
the thing is:
when the solvent chain (or matrix chain) changes from a monomer to a polymer
chain as the solute chain, what happens there?
Good---->theta---->all theta to polymer in itself?
take
a
a
was
this
as
ch |
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r*t
发帖数: 34 | 41 I just write a program of DOPRI5(4), but have a problem with automatic step-
size selection. after calculation of K1~K7, the error is estmated by:
Error = (71/57600*k1 - 71/16695*k3 + 71/1920*k4 - 17253/339200*k5 + 22/525*
k6 - 1/40*k7)*DeltaT (eq. 1)
Then I adjust DeltaT by:
Theta = 0.9*(Tolerance/Error)^(1/5) (eq.2)
(Here I select Tolerance = 1e-7)
and if Theta<1, then DeltaT = DeltaT*Theta
However, I found that the simulation is exemely slow (much slower than
without step-size control), |
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a***n
发帖数: 40 | 42 Thanks. I guess then, for theta1
w.r.t. f(x|theta), but f(x|theta) is not absolutely continuous w.r.t f(x|
theta1), is it true? In general, if the support of X depends on parameter
space, then the pdf's cannot be absolutely continuous w.r.t. each other? |
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G***p
发帖数: 59 | 43 记得中学时
划个矩形的内接园
然后划个角度出来
可以直观的得到
\sin\theta < \theta <\tan\theta |
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d*********a
发帖数: 255 | 44 【 以下文字转载自 Computation 讨论区 】
发信人: dragoninsea (龙哥), 信区: Computation
标 题: 帮看看一小段matlab程序
发信站: BBS 未名空间站 (Mon Oct 29 23:24:04 2007)
M=36;
theta=zeros(1,M);
pho=[1 0.5;0.5 1];
syms u;
syms v;
s=[norminv(u);norminv(v)];
for i=1:M
theta(i)=int(int(10*(exp(-0.5*s'*(inv(pho)-eye(2))*s)/sqrt(det(pho))),u,0,1)
,v,0,1);
end
theta
===============================================
??? Error using ==> <
Function '<' not defined for variables of class 'sym'.
Error in ==> C:\MATLAB6p1\toolbox\stats\norminv.m
On |
|
s**c
发帖数: 1247 | 45 \sum [ \theta^r * p^(\theta - 1) ]
\theta takes value 1,2,3,...
0
r is any positive integer
要得到最后的结果
3x |
|
z*******x
发帖数: 11 | 46 大家好,我找到了一个流函数的表达式:
psi=[r-(1+2K1(epsilon)/(epsilon*K0(epsilon)))/r+2K1(r*epsilon)/(epsilon*K0(
epsilon))]*sin(theta)
这里epsilon是常数,K1和K0是modified Bessel functions. r=sqrt(x^2+y^2);sin(
theta)=y/sqrt(x^2+y^2).
因为我对Bessel函数不熟,所以我无法根据定义:
Vr=partial(psi)/partial(theta)/r; Vtheta=-partial(psi)/partial(r);
或者在Cartisian坐标:
U=partial(psi)/partial(y); V=-partial(psi)/partial(x)
得到速度的具体表达式
我自学了相关的Bessel函数后,还是不能得到上面速度场的表达式,想请大家给帮帮忙,
谢谢了.
祝
好
bin |
|
h**********c
发帖数: 4120 | 47 you problem is not mathematical description adequately.
but I make the following assumptions.
the camera move from (a,b,c) to (m,n,q)
the disk is at (r,s,t) up vector vd=(u,v,w)
v1 = (a-r,b-s,c-t)
v2 = ((m-r,n-s,q-t)
v3 = cross(v1,v2)
v4 = cross(vd,v3)
theta = dot (v3,vd)/(norm(v3)*norm(vd))
Rotate disk about v4 with angle theta. calculate theta by the dot product
arccos or sth else.
This is a kind suggestion, no reliability. |
|
o**a
发帖数: 86 | 48 【 以下文字转载自 Physics 讨论区 】
发信人: ouba (comfortably numb), 信区: Physics
标 题: a question on probability density
发信站: BBS 未名空间站 (Thu Feb 25 12:26:21 2010, 美东)
Folks, thanks for your time first!
function f = x.cos(a)**2 + y.cos(b)**2 + z.cos(c)**2
I know x > y > z. a, b, c are the angles between a vector and x, y, z axes
of a cartesian coordinate respectively.
If written in nutation and azimuthal angles,
f = x.sin(theta)**2.cos(phi)**2 + y.sin(theta)**2.sin(phi)**2 + z.cos(theta)
**2.
This vector is und |
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m*******s
发帖数: 3142 | 49 遇到对数函数ln(1+x)的Chebyshev polynomial展开的系数问题。
http://en.wikipedia.org/wiki/Chebyshev_polynomials 中的Example 1
就是要求如下定积分
\int_{-\pi}^{\pi}\ln\left(1+\cos\theta \right)\cos\left(n\theta \right) d\
theta
其中-\pi应该是是积分的奇点，不知道该如何处理。
再变换成一个复变函数沿单位圆的积分，
\int \left[2\ln\left(1+z \right)-\ln z -\ln2 \right]\frac{z^n+z^\left(-n\
right)}{2z}dz
-1是\ln\left(1+z \right)的branch point，正好在积分路径上，可以要用小圆弧绕开
吗？
原点是\ln z的branch point，需要用branch cut把复平面割开，同时还要用小圆绕开
把原点绕开，我发现正是沿这个小圆的积分是发散的，不知道怎么办。
但是最后答案却是能求出的，我觉得很奇怪。
请高手支招。 |
|
a****a
发帖数: 5763 | 50 第一个不等式应该是用 z=1+re^{i\theta} 带入之后，分子取最大模，分母取最小模
这样利用 |\int f(z)dz| < \int |f(z)| dz , 得到一个 最大模对\theta的积分，显然
\theta 不会超过 \pi, 就得出不等式 |
|
