135 reputation
9
bio website
location Italy
age
visits member for 3 years, 3 months
seen yesterday

Currently I'm scientific software developer with proficiency in C/C++ with their related technologies Boost, STL, Qt, Python, computer graphics, OpenGL, Mathematica, MatLab, Bash scripting, NI Labview, LATEX, CMake, CUDA.


Sep
27
awarded  Curious
Sep
26
asked Is $d(i,j) = 1-\textrm{corr}(i,j)$ a metric?
Sep
24
awarded  Autobiographer
Jul
23
comment Multiple Hypergeometric Distributions
If the draws are independent then you can multiply the probabilities.
Jul
10
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 $\binom{a}{b} = \frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}$
Jul
10
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 $\sum_{i=mi}^m \binom{pi}{i} \binom{p-pi}{m-i}/\binom{p}{m}$
Jul
9
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?
Jun
17
awarded  Peer Pressure
Apr
29
awarded  Tumbleweed
Apr
22
revised Proving a disequality involving hypergeometric distribution
added 259 characters in body
Apr
22
asked Proving a disequality involving hypergeometric distribution
Jun
19
accepted Texture mapping and conformal transformation
Jun
19
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
Jun
19
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;
Jun
18
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?
Jun
17
asked Texture mapping and conformal transformation
Jun
14
awarded  Supporter
Jun
14
awarded  Commentator
Jun
14
comment Is nonlinear conjugate gradient a quasi-newton optimization technique?
Thank you, this question is what I was looking for.
Jun
7
asked Is nonlinear conjugate gradient a quasi-newton optimization technique?