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 Sep27 awarded Curious Sep26 asked Is $d(i,j) = 1-\textrm{corr}(i,j)$ a metric? Sep24 awarded Autobiographer Jul23 comment Multiple Hypergeometric Distributions If the draws are independent then you can multiply the probabilities. Jul10 comment Is There a Continuous Analogue of the Hypergeometric Distribution? Thanks for the good reference. What I'm still missing is how is possible to compute the cumulative hypergeometric distribution given the Wiener-Askey chaos expansion. Why isn't the sum approximable with an integral as follows? $\sum_{i=m_i}^m \binom{p_i}{i}\binom{p-p_i}{m-i}/\binom{p}{m} \approx \int_{m_i-1/2}^{m+1/2} \binom{p_i}{x}\binom{p-p_i}{m-x}/\binom{p}{m} dx$ where the binomial coefficients are evaluated with the help of $\Gamma$ function as $\binom{a}{b} = \frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}$ Jul10 comment Is There a Continuous Analogue of the Hypergeometric Distribution? can you please point me to some good explanation of the continuous Hahn polynomials? I'm interested in understanding if it is possible to compute the hahn polynomial from the following $_3F_2({1,mi-m,mi-pi},{mi+1,mi+p-pi-m+1},1)$ which comes a simplification of the hypergeometric cumulative distribution $\sum_{i=mi}^m \binom{pi}{i} \binom{p-pi}{m-i}/\binom{p}{m}$ Jul9 comment Is There a Continuous Analogue of the Hypergeometric Distribution? Is it also possible to use the Gauss-Hermite polynomial chaos for the cumulative hypergeometric distribution? Jun17 awarded Peer Pressure Apr29 awarded Tumbleweed Jun19 accepted Texture mapping and conformal transformation Jun19 comment Texture mapping and conformal transformation Oh yes! I've been wrong putting $(r,\theta)$ instead of $(r \cos(\theta), r \sin(\theta) )$ this in fact makes much more sense! Stupid me, seeing the same thing for 2 days hides the evident things! This is the result of conformal requirements: and the resulting shader code float rho = sqrt(v.x*v.x+v.y*v.y); float theta = atan(v.y,v.x); float a=1.0; float r = (pow(4*rho*rho+1,1.5) -1)/(12*a) ; texture_coordinate = vec2(r*cos(theta),r*sin(theta)); gl_Position = gl_ModelViewProjectionMatrix*v the effect is the following: imgur.com/jqusOyl Jun19 comment Texture mapping and conformal transformation Hey, thanks for the ideas! This is the result of conformal and constant area requirements. imgur.com/ygrMoFw My biggest problem is that, i think, the theta is wrong. This is basically the GLSL code to do this texture mapping if your are interested (the constant area case) float rho = sqrt(v.x*v.x+v.y*v.y); float theta = (atan(v.y,v.x)+pi/2)/(pi); float s= sqrt(4*rho*rho+1); float a=1.0; float r = (pow(4*rho*rho+1,1.5) -1)/(12*a) ; texture_coordinate = vec2(r,theta); gl_Position = gl_ModelViewProjectionMatrix*v; Jun18 comment Texture mapping and conformal transformation Thanks for the accurate response. In fact I would also like to understand how to get the constant area element. Excuse me for the double question. You are suggesting that my new coordinates should be $\theta = \arctan(y/x)$ and the $r(\rho)$ in your last equation? Jun17 asked Texture mapping and conformal transformation Jun14 awarded Supporter Jun14 awarded Commentator Jun14 comment Is nonlinear conjugate gradient a quasi-newton optimization technique? Thank you, this question is what I was looking for. Jun7 asked Is nonlinear conjugate gradient a quasi-newton optimization technique? Sep27 comment Extracting angular velocity tensor from orthogonal matrices Yes, this is not correct. I meant I would like to compute $\mathbf{Y}$ in a more robust manner, but I think that given its definition, it's obvious it approximate the identity when $\Delta t \rightarrow 0$, so for high sampling rate, no chances... It seems to me so strange that there are no methods to extract angular velocity from discrete time systems... Sep26 comment Extracting angular velocity tensor from orthogonal matrices Thanks, I understand, when the sampling rate increases $\mathbf{Y}$ approaches the identity, but maybe can be useful to derive a better discrete approximation to Y. Is this the case? How to deal (if possible) with better derivative approximation scheme supposing that I'm able to have $R(t),R(t-1),R(t-2),...,R(t-k)$?