Consider a "sharp loop" to be a sequence of $k$ squares $S_i$ such that $S_i$ is adjacent or diagonal to $S_{i+1}$ and $S_1 = S_k$
Notice that this is the only possible formation that divides the plane into 2 unaccessible parts
Also, notice that the sharp loop must all be the same color
However, if the two squares are divided by a sharp loop (the only case that the path doesn't exist), then it contradicts the fact that "There exists a path avoiding the blue or orange", because this sharp loop is only one color
therefore, these two squares must not be divided by a sharp loop, and thus there must exist a path between them
But I suppose that works. I'm just slightly worried about whether or not we should prove that these are the only structures that divide the plane into disconnected pieces.
@MickLH do me a favour please. The next time you'll see stuff being told here around just ask them what they created in mathemtics, what their original ideas are.
Unique ideas one cannot find elsewhere. Only one single original idea that revolutionized any area of mathematics.
@ZachHauk Here is a different point: when you're dedicated to mathematics in the most profound sense, you don't have time for s*it. I'm too tired of seeing so much envy around.
@ZachHauk I don't do it for me now, I do it more for younger people that think like me and that can be kept unfortunately in a range of very low performance and wrong beliefs about what they actually can do
The time evolution of the position and momentum is then determined by Hamilton's equations: $\frac{dx}{dt}=\frac{\partial H}{\partial p}$ and $\frac{dp}{dt}=-\frac{\partial H}{\partial x}$.
Find how much gallon of paint it takes to paint a room. If the lenght, height and width is given and also we know that it takes a gallon to paint 350 square feet of the room, why do we need height if width is the height?
Moreover, Hamilton's equations determine the time evolution of any function $f(x,p)$ through the chain rule: $$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial p}\frac{dp}{dt}=\frac{\partial f}{\partial x}\frac{\partial H}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial H}{\partial x}$$
One wall is length*height. The next wall is width*height. The next wall is opposite the first one, so it has the same size, so it's length*height again. The last wall is opposite the second one, so it's width*height.
The definition from the book I have is: suppose $X'=AX,X'=BX$ have flow $\Phi^A$ and $\Phi^B$. These two systems are topologically conjugate if there exists a homeomorphism $h:R^2\to R^2$ that satisfies $\Phi^B(t,h(X_0))=h(\Phi^A(t,X_0))$. The homermorphism $h$ is called a conjugacy
@MATHASKER Look at the floor. The floor is a rectangle. One side (edge), the longer side, we'll the length. The other side we'll call the width. The distance between floor and ceiling is the height.
@Simple I suppose, but not necessarily a linear transformation
("Edge" as in, it's one of the twelve edges of the cube-shaped room you're in.
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What's neat about the above formulation is that it has a direct quantum-mechanical version, namely Ehrenfest's theorem: $\dfrac{d}{dt}\langle \hat{A}\rangle = \dfrac{1}{i\hbar}\langle [\hat{A},\hat{H}]\rangle$
is there a way I can like write the questions correctly and send them here like if i want to square something like is there a way like in the main site where i can square instead of doing 2^2
If $\vec{x},\vec{p}$ have canonically conjugate position and momentum components, then their Poisson brackets (appropriately defined for the case of 3 conjugate variables rather than just 1) are $\{x_j,x_k\}=0$, $\{p_j,p_k\}=0$, $\{x_j,p_k\}=\delta_{jk}$.
Moreover: If you work through the (rather tedious) algebra, you find that the Poisson brackets between the angular momentum components are neat: $\{L_j,L_k\}=\epsilon_{jkl}L_l$.
Please don't mind , is there any text you can suggest , so that i can review them ... i know it sounds awkward but i think reading a lucid text is very much helpful in understanding ..
@ZachHauk Here's a quick exercise: find two disjoint closed subsets of $\Bbb R^2$ whose distance is $0$. (The distance between two sets is the infimum of the distance between a point of one set and a point of the other.)
Anyways, to connect this back to what you were interested in earlier: The above algebra is known as the $\mathfrak{so}(3)$ algebra. (It's also the $\mathfrak{su}(2)$ algebra. There's no difference between the two in this case, but with higher dimensions that changes.)
Sure.
What's important is that said structure is relevant in both quantum mechanics and in classical mechanics, though of course the meaning of stuff like position and momentum isn't the same.
For a system of differential equation $X'=AX$, when $A$ has eigenvalues $\lambda_{1,2}=a\pm ib$, the general solution is $X=c_1u(t)+c_2v(t)$ where $u(t),v(t)$ are two linearly independent solution. My question is where is $i$ since $e^{it}=\cos(t)+i\sin(t)$
Lol today in office hours Schlag was all like "Hopefully in the last lecture or 2 of 209, Marianna will teach you about $L^p$ spaces, for which the ones we deal with are a special case by taking $\mathbb{N}$ with the counting measure"
This apparently isn't even a phenomenon that started in the Soug/Schlag days, even Sally's book has chapter 4 on functional analysis and chapter 6 on integration
We're definitely feeling it this quarter, because our problems are basically crunching out stuff with $\ell^p$ spaces, and the only books anyone in our class has so far found that don't just sweep it under the $L^p$ umbrella are K&F and Kreyszig
@Daminark idk anything about this in particular but apparently there was some conflict when trying to figure out foundations of QM about whether we should use an $\ell^{2}$ formulation or an $L^{2}$ formulation but then they ended up being the same lol
at least this is what a grad student told me last year
but that's why newer books just sweep $\ell^{2}$ under the $L^{2}$ umbrella apparently.
