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 2d comment Integral identity involving sin(x)/x What is there to prove? A more proper question would be to ask how to integrate the right hand side if that is what you are trying to figure out. Apr16 comment Integral identity involving sin(x)/x It is not terribly suprising that two definite integrals are related to each other by a constant scale factor. I would expect nothing less. Apr11 comment Can some one help me parametrize $\frac{x^4}{a^4}+\frac{y^4}{b^4}+\frac{z^4}{c^4}=1$ Your'e right, it is a sphere. Apr11 comment 3D vector perpendicular calculation What methods do you understand? Vector math (dot & cross product)? Linear algebra (matrices, transformations, rotations)? Apr3 comment SVM and quadratic programming SVM does reduce to a QP problem in the dual form. There are plenty of resources out there to explain it. To handle b you would add a 1 line, not a zero line. Apr3 comment Matrix properties invariant under scalar multiplication Column space, row space, null space, left null space, eigen vectors, rank, to name a few. Mar31 comment Finding the area of the triangle with vertices at $(ct, c/t)$, $(-ct, -c/t)$, $(ct^{2}, 2ct)$ The reference that you found is an interesting approach. It uses the same principle except it works in 3D space. The thinking goes like this. You are constructing your triangle in 3D space by adding $z=1$ as a third coordinate. So your triangle is floating in space in the horizontal plane $z=1$. You then compute the volume of the parallelepiped (denote as $V_P = D_3$). In the same construction you can create a tetrahedron (i.e. pyramid) by connecting your vertices to the origin. The volume of this is $V_T=1/3(1)A_T$. It is known that $V_P=6V_T =6(1/3A_T)$, so we get $A_T = 1/2D_3$. Mar31 comment Finding the area of the triangle with vertices at $(ct, c/t)$, $(-ct, -c/t)$, $(ct^{2}, 2ct)$ Regarding the sign, that just depends on the order of the points that you chose to put into the determinant. If you changed the order from CW to CCW your sign and mine would match. Mar31 comment Finding the area of the triangle with vertices at $(ct, c/t)$, $(-ct, -c/t)$, $(ct^{2}, 2ct)$ It is because by subtracting $B$, I effectively translated the triangle to the origin (i.e. point B is now at 0,0). Then it is known that the area of the parallelogram spanned by $A-B$ and $A-C$ is equal to the determinant shown above. The area of the triangle is 1/2 the area of the parallelogram, thus the 1/2. Mar30 comment Finding the area of the triangle with vertices at $(ct, c/t)$, $(-ct, -c/t)$, $(ct^{2}, 2ct)$ Also you could use Herron's formula which can be used to calculate the area from the length of the three sides. You can find the formula here: en.wikipedia.org/wiki/Heron%27s_formula Mar27 comment proving Pythagoras Theorem in the third dimension using orthogonal projection from a parallelogram What does $p$ represent? Is it the area of the original parallelogram created by $u$ and $v$? Also, what makes you think that it should be true. Mar23 comment Factoring derivatives It is the product rule. Mar2 comment Integrating a step function using antiderivatives $F(x)$ is not continuous. It should be piece-wise continuous. Feb25 comment Transformations between coordinate frames Your transformations dont look right. In your notation $T_{AB}, is that from A to B or the other way around. Feb23 comment Transformations between coordinate frames Luigi, they are Homogeneous coordinats. Feb22 comment Derivative of angle between two vectors singularity! I think that you might be running into trouble with your assumption that A and B are unit vectors. By that assumption you cannot arbitrarily change one of the vectors. It has to remain on the unit circle. Thus$\bf \dot{\hat A}\dot \bf \hat B$and$\bf {\hat A}\dot \bf \dot{\hat B}$are both zero. So you have a situation where you have$0/0$and you would have to use L'Hopital's rule to solve it. I recommend that you solve for the general case. Feb22 comment Solving a transformation equation involving vectors and quaternions Because quaternions are not commutative, you cannot divide. You have to pre or post multiply to manipulate both sides of the equation. Feb16 comment Quick question about ln(0) You know about L'Hôpital's rule right? Just put the indeterminate form into$\infty/\infty$form as follows$x \ln x = \frac{\ln x}{1/x}$. Then take derivatives. It is pretty straight forward. Feb16 comment Quick question about ln(0) You could also side step the indeterminate form by solving the equivalent integral down the$y$axis.$\int_{0}^{1}\ln x\,\mathrm dx=\int_{-\infty}^{0}e^y\,\mathrm dy = -1\$ Feb8 comment Searching a function for data Seems like your polynomial is already in closed form. Not sure what else you are looking for.