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 Mar 6 comment Linear transformation of $5$ points in a graph. Yes, you have to remove the mean from the data and then project the data onto your eigenvectors (i.e. basis). If you put them in columns as $R = [v1 v2]$ then the formula is something like this: $data2 = R'*(data-mean)'$. You will have to work out the right Matlab, but that is the jist of what you need to do. Mar 6 comment Linear transformation of $5$ points in a graph. Do you know how to form a basis from the eigenvectors? Do you know how to do a general coordinate transformation? Feb 29 comment Coordinate transformations and interpreting what the Jacobian determinant describes Yes or just $dx^{1}dx^{2}\cdots dx^{n}=Jdy^{1}dy^{2}\cdots dy^{n}$ because the Jacobian just relates volumes between one coordinate system and another. The way that you wrote it, both would tie back to a third coordinate system (i.e. $Jdx^{1}dx^{2}\cdots dx^{n}=\tilde{J}dy^{1}dy^{2}\cdots dy^{n} = dz^{1}dz^{2}\cdots dz^{n}$). Feb 28 comment Coordinate transformations and interpreting what the Jacobian determinant describes Yes, you could think about it that way. Just keep in mind that the assumption that one of the coordinate systems in Cartesian is not necessary. They could both be curvilinear coordinate systems. Your last statement is correct Feb 27 comment Coordinate transformations and interpreting what the Jacobian determinant describes "coordinate lines form a rectilinear grid and so the coordinate volume element is simple a cube" -- Yes, that too. But keep in mind that the coordinate system could be stretched in one direction and still be rectangular. Feb 27 comment Coordinate transformations and interpreting what the Jacobian determinant describes "So is the reason why.. is independent of the location?" yes. Aug 21 comment Why are Runge Kutta's method and Euler's so different? Did you mean $\underline {\dot x}=\underline A\cdot \underline x$? Apr 28 comment Drawing an arc within two defined points in C graphics How are a, b, c, d related to the location of the arc? Are the two lines always vertical? How long are the lines? Apr 25 comment Simplifying a Complex Number You should be familiar with Euler's Formula: $e^{ix}=\cos x + i \sin x$ which is periodic. So then $e^{ix} = e^{i(x\pm 2n\pi)}$ where $n \in \mathbb Z$. So then, $e^{ i \frac{2014\pi}{12}} = e^{ i(166\pi+ \frac{22\pi}{12})} = e^{ i \frac{22\pi}{12}}= e^{ i \frac{11\pi}{6}} = e^{- i \frac{\pi}{6}}$ Apr 17 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. Apr 16 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. Apr 11 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. Apr 3 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. Apr 3 comment Matrix properties invariant under scalar multiplication Column space, row space, null space, left null space, eigen vectors, rank, to name a few. Mar 31 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 $A_T = 1/2D_3$. Mar 31 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. Mar 31 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. Mar 30 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 Mar 27 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. Mar 23 comment Factoring derivatives It is the product rule.