@Obliv What on earth is that supposed to mean? When you write down a formula, always explain your notation! What are $E,L$, what is $T$, where did $f$ go?
@ACuriousMind Hey, can you explain $$\int_{B_y(3\rho_j)\cap B_p(\rho)}|\nabla f|^2\phi^2\,dV\ge C_1^2C^{-8}_{\mathcal{VD},\phi}C_{\mathcal P,\phi}\rho^{-2\epsilon}|f_{B_0}-f(y)|^2\rho_j^{2\epsilon-2}V_\phi(B_y(\rho_j))?$$
@Obliv So, what you're trying to say is that $E,L\subset\mathbb{R}^2$ and you have a function $t : \mathbb{R}^2\to\mathbb{R}^2, (x,y)\mapsto (u(x,y),v(x,y))$ such that $t(E) = L$? I don't know what "f transforms into a function T(u(x,y),v(x,y))" is supposed to mean, though.
@ACuriousMind the function $T$ is defined such that the domain is still a subset of $\mathbb{R}^2$ but the image is transformed by $(x,y) \to (u(x,y),v(x,y))$ as you stated above. I think you knew what i was talking about but you are teaching me to be specific anyway, right?
@acuriousmind how would I phrase it correctly? $\int \int_E f(x,y) dA$ where $E$ is the domain of $f(x,y): \mathbb{R}^2 \to \mathbb{R}$ becomes $\int \int_L T(u(x,y),v(x,y))\mid J \mid ~dv~du$ where $T(u,v): \mathbb{R}^2 \to \mathbb{R}$ and $E,L \subset \mathbb{R^2}$ and $\mid J \mid = \frac{\partial(u,v)}{\partial(x,y)}$
no mention of the word transformation anywhere. Strict definitions. Beat that
@Obliv I have no idea what $T$ is supposed to be! I would write the change of variables formula as $\int_L f(u,v)\mathrm{d}u\mathrm{d}v = \int_E f(u(x,y),v(x,y)) \det(D \phi) \mathrm{d}x\mathrm{d}y$ where $\det(D\phi)$ your Jacobian determinant.
@0celo7 Only for the Lebesgue integral, and it was not elegant.
@Obliv Try: Let $U,V\subset\Bbb R^n$ be open, $\phi:U\to V$ a diffeomorphism. A function $f:V\to\bar{\Bbb R}$ is integrable precisely when $(f\circ\phi)|\det D\phi|$ is integrable over $U$, and in this case one has $$\int_{\phi(U)}f=\int_U (f\circ\phi)|\det D\phi|.$$
@ACuriousMind It's apparently worse for the Riemann integral.
In fact, when Fleming was writing his analysis book, he just wanted to do Riemann because it's an undergrad book. But he found proving change of variables for Riemann was too hard, so he decided to just do Lebesgue integration
The extra effort for that was less than coming up with the necessary epsilons and deltas for the Riemannian proof
@0celo7 No. Take the three spin-1 matrices $J_1,J_2,J_3$ and show that for any vector $v$, the span of $v,J_1v,J_2v,J_3v$ is three-dimensional, i.e. the entire space. Therefore, this representation doesn't have invariant proper subspaces.
1. A representation of an algebra $\mathfrak{g}$ is irreducible iff it doesn't contain proper invariant subspaces. 2. If $(V,\rho)$ contains an invariant subspace $W$, then for $v\in W$, $\rho(\mathfrak{g})v$ is also an invariant subspace of $W$. 3. If you can show that $\rho(\mathfrak{g})v = V$ for all $v\in V$, then there are no invariant proper subspaces.
@0celo7 Write it in terms of the basis vectors upon which you know the action of the $J_i$ (or linear combinations of them) and construct three linearly independent vectors of the form $v + \alpha J_1 v + \beta J_2 v + \gamma J_3 v$.
@0celo7 A leaf electroscope does that. There is actually a tiny current in the device, but it is motion of the conduction electrons in the metal of the device.
There are no drivers for the hardware on it though
Although, theoretically, once a jailbreak is possible you could port something like Bochs to it's architecture (AARCH64 I'm guessing?) And then emulate Linux on Bochs
That will be slow though
Does AARCH64 imply Hypervisor support? If so you could minimize the iOS kernel and run Linux on Ring -1 as a VM
He arrived as me and Michelle were opening the door, drove on the driveway, then backed up, then drove up, then backed again, then drove up and left the car
