@Semiclassical If $M=(a~b)^T(c~d)$ you have $a\cdot c+a\cdot d+b\cdot c-b\cdot d$ which factors as $a\cdot(c+d)+b\cdot(c-d)$ which is maximized when $a\|(c+d)$ and $b\|(c-d)$ are parallel which yields a max value of $\|c+d\|+\|c-d\|$. If you draw a unit-side rhombus and notice the perpendicular diagonals split it into four right triangles, you see this is $2[\cos(\theta/2)+\sin(\theta/2)]$ (where $0\le\theta\le\pi/2$ wlog). Maximizing that is simple.
also, what kind of operations on a matrix would yield an expression like $M_{11}+M_{12}+M_{21}-M_{22}$? Maybe like ${\rm tr}([\begin{smallmatrix} 1&1\\1&-1\end{smallmatrix}]M)$
also, where $\theta$ is the angle between $c$ and $d$ ^
@0celo7 oh I forgot to give you the non-Fredholm answer earlier, whoops
In case it's still something you want to know, for matrices rank-nullity is all you need
Let's say you have an $m\times n$ matrix, it's surjective iff the rank is $m$. Well, its transpose is $n\times m$ and also has rank $m$. But by rank-nullity the kernel is $0$
“What is the definition of a circle?” asked the examiner. “It is the set of points in a plane, equidistant from a fixed point,” Edik replied. “Wrong,” said the examiner. “It is the set of *all* points in a plane, equidistant from a fixed point.”
Let $K$ a connected compact subset of the Euclidean plane which has an infinite set of reflection symmetries.
Does this imply that $K$ is an annulus ?
Source : les dattes à Dattier
"A talented Jewish mathematician sometimes found an education at the Institute of Metallurgy or the Pedagogical Institute. Others would enroll in the Institute of Railway Engineers, whose Russian abbreviation sounded like MEED. This led to the saying, “Esli zheed, idi v MEED”—“If you’re a Jew [the rhyme scheme requires a pejorative term here], then go to MEED.”"
I think you define an end of $X$ to be a decreasing sequence $U_1 \supset U_2 \supset U_3 \supset \cdots$ such that $X \setminus U_i = K_i$ is a compact set.
Say I have a differentiable function
$$
f(x) = f(g(x),h(x)) = g(x)h(x)
$$
If I compute $df/dx$ I can apply the product rule, but what about
$$
\frac{df}{dg} = ?
$$
what can I do in this case?
The only attempt is this one
$$
\frac{df}{dg} = \frac{df}{dx} \frac{dx}{dg} = \left(g'(x)h(x) + g(x...
Oh, fait alors Anna toujours a volé à droite à gauche des résultats que tu n'auras jamais trouvé toutes seules, et cela pour avoir ta médaille...lol
Vraiment, cela fait pitié tu insultes les auteur des résultats qui t'intéressent tu les décourage et t'approprie leur travaille
cela porte un nom : VAMPIRE !
Mais tu n'es pas la seule dans ce cas tes collégues aussi suive ta pente
Cela fait vraiment pitié
Heureusement que les Gauss et compagnies ne sont pas tombé de se travers...
Mais va savoir, en louchant sur des livres non publié, parce que les auteurs étaient honnis...lol
PLAGIAIRES !
vous manquez singulièrement d'imagination, et vous croyez qu'en faisant des cérémonies barbares, cela va vous rendre plus intelligent...vous êtes vraiment des idiots...
Hi guys I have a matrix in $\mathbb{R}^{m \times n}$ where each entry $a_{ij}$ depends from a different vector variable, namely
$$
a_{ij} = a_{ij}(x_1^i,x_2^j)
$$
So in case $m = n = 2$ it would be something like
$$
A(\vec{x})=\begin{pmatrix}
a_{11}(x_1^1,x_2^1) & a_{12}(x_1^1,x_2^2) \\
a_{21}...
vous manquez singulièrement d'imagination, et vous croyez qu'en faisant des cérémonies barbares, cela va vous rendre plus intelligent...vous êtes vraiment des idiots... Tchuss.
"But for higher d, these ad hoc methods no longer work. Nevertheless, there is an elegant proof of Theorem 1, due to Mazur, and known as Mazur’s swindle. "
What is the definition of a linear space, when used for the lie group $GL(n)$, it's not a vector space, but then what does it mean to say that it is a linear space?
I also ramble a lot of weird maths I do on this chat and other places, but I never force anyone to accept my ideas, in fact, I am happy whenever they found something that is limiting or makes no sense, because the collaboration process will allow us all to refine our knowledge on the field in question
That's why the literature exists, it is there to give us a bearing on what we knew so far as a species
Say I have a manifold defined by the gluing of a manifold triad $(M, \partial M_1, \partial M_2)$ with a homeomorphism $h : \partial M_1 \to \partial M_2$
Yes. The choice of axis cannot make a difference in whether the problem is solvable or not.
All it can do, for better or worse, is make solving easier.
Moreover, there's at least one benefit to picking this point over one of the support points: The displacement from the midpoint to the center of mass stays the same, regardless of the angle of deflection.
Another choice, I should note, would be to take the center of the disk.
I'm not sure which of these two is easier.
Hmm. Maybe it is just as easy to pick one of the support points, though.
I'm being a bit indecisive, I know.
I think the center of the disk is easiest, in fact, because then the weight of the disk doesn't produce a torque with respect to this axis.
Let
$$
f(x_1,x_2) = \frac{1}{\sqrt{x_1^2+x_2^2+1}} \left(x_1,x_2,1 \right)
$$
For $i = 1,2$ we have
$$
\partial_{x_i} f = \frac{1}{\sqrt{x_1^2+x_2^2+1}}\left(\partial_{x_i}x_1,\partial_{x_i}x_2,0 \right) +\frac{-x_i}{\left(x_1^2+x_2^2+1\right)^{3/2}}(x_1,x_2,1) =
\frac{1}{\left(x_1^2+x_2^2+1\ri...
Let $K$ a connected compact subset of the Euclidean plane which has an infinite set of reflection symmetries.
Does this imply that $K$ is an annulus ?
Source : les dattes à Dattier
Say I have a differentiable function
$$
f(x) = f(g(x),h(x)) = g(x)h(x)
$$
If I compute $df/dx$ I can apply the product rule, but what about
$$
\frac{df}{dg} = ?
$$
what can I do in this case?
The only attempt is this one
$$
\frac{df}{dg} = \frac{df}{dx} \frac{dx}{dg} = \left(g'(x)h(x) + g(x...
