609 reputation
412
bio website componentowl.com
location Brno, Czech Republic
age 28
visits member for 3 years
seen Aug 8 at 19:50

Founder of two companies focusing on digital imaging, computer vision and .NET components.

Currently focuses on Image alignment and stitching library for .NET, software consulting and .NET component development.

He loves digital photography, programming, yoga, math, comics drawing, healthy lifestyle and green tea.


1d
awarded  Yearling
Jul
26
comment Rotating one 3-vector to another
Thanks. Could you please point me to the theory? The matrix seems to be simply $R=ab^{T}$ where $a$ and $b$ are normalized vectors. It is not clear to me why $R^{T}=R^{-1}$ (i.e. $R^{T}R=I$) in that case. The book is unfortunately hard to obtain and too much hassle just to get this information...
Jul
5
awarded  Taxonomist
Jul
2
awarded  Curious
Jan
7
awarded  Popular Question
Nov
7
comment Differentiating second order term of Taylor polynomial (multivariable)
I have expanded the question - I am trying to derive Newton iteration step (formula for $\Delta x$) derived from second-order Taylor expansion. I can't find appropriate answer on how to get to that formula...
Nov
7
revised Differentiating second order term of Taylor polynomial (multivariable)
clarified application of the problem
Nov
7
comment Differentiating second order term of Taylor polynomial (multivariable)
Thanks for clarifying the expansion, yet I need to know how to compute derivative of the second-order term with respect to $[ \Delta x \; \Delta y \; \Delta z]$...
Nov
7
comment Differentiating second order term of Taylor polynomial (multivariable)
And $g^{T}+H(\Delta x)$ is a 3-vector since $H(\Delta x)$ is a matrix multiplied by a vector.
Nov
7
comment Differentiating second order term of Taylor polynomial (multivariable)
$g^{T}\Delta x$ is a $3\times 3$ matrix so is the second term of the Taylor polynomial. This is copied from a book, but the differentiation was missing and I don't know how they arrive at the results.
Nov
7
asked Differentiating second order term of Taylor polynomial (multivariable)
Aug
26
awarded  Yearling
Jun
27
awarded  Popular Question
Jun
9
comment Levenberg-Marquardt - Is forcing Hessian to be positive definite OK?
Okay I have solved it - the model fitted just perfectly on the points and hence the system was singular for $\lambda = 0$...
Jun
9
accepted Levenberg-Marquardt - Is forcing Hessian to be positive definite OK?
Jun
9
comment Levenberg-Marquardt - Is forcing Hessian to be positive definite OK?
The problem actually appeared because I have implemented the method from "Numerical Recipes" book. They suggested to set lambda=0 after several improvements and go for "final solution"...
Jun
9
comment Levenberg-Marquardt - Is forcing Hessian to be positive definite OK?
Well the problem is actually that the model fits the data exactly with initial parameters, hence Jacobian is zero and so does Hessian. It was a programming bug because lambda happened to be zero as well and this made the system singular. Thanks for pointing out the effect of any positive lambda - that helped.
Jun
9
asked Levenberg-Marquardt - Is forcing Hessian to be positive definite OK?
May
19
comment Optimal sampling for an arbitratry area
Thanks. It is not extremely important to have the points distributed the best way possible, just to extend the idea of uniform sampling from rectangular images to the arbitrary shaped areas.
May
19
comment Optimal sampling for an arbitratry area
As for the segmentation, I have a clue it may have to do something with Voronoi diagram but I am not sure.