A friend of mine in Athens has complained multiple times about Amazon and USPS. I've never had trouble. Did UPS leave you a slip of paper giving you choices for redelivery?
@MikeMiller interesting geometric interpretation of free resolutions : take an acyclic CW-complex $X$, and let $\tilde{X}$ be it's universal cover. lift cell structure of $X$ to $\tilde{X}$. let $C_n$ be the collection of $n$-cells of $\tilde{X}$. $C_n$ is naturally a $\Bbb Z[\pi_1]$ module, coming from the action by deck transformation. so we have a $\Bbb Z[\pi_1]$ resolution of $\Bbb Z$ : $\Bbb Z \leftarrow C_0 \leftarrow C_1 \leftarrow \cdots$, where the maps are the bd maps.
@BalarkaSen: What you're mentioning now is the easy direction of some exciting work of Wall. Indeed, a group $\pi_1$ has a finite CW-complex model for $B\pi_1$ iff $\mathbb Z$ admits a free $\mathbb Z[\pi_1]$-resolution.
The problem is
$$e^{\ln(2x+1)} =5x$$
I've tried using natural logs to both sides like..
$2x+1= \ln 5x $
But I'm not sure if $\ln$ and $e^{\ln}$ cancel out.
@Balarka I am having difficulties understanding components of a connected set.. I cannot think of a picture for it, I tried with $\Bbb{R}_l$ but was of no use
Because the discs are connected and disjoint ? or is there some other reason
Okay the union of these discs is the whole space and they are disjoint and connected .. that is why they are an example of connected components of a set I guess.. Am i right @Balarka
@AlexClark no i dont downvote people usually, i just upvote unless something bugs about the question, i just leave a constructive non-offending critical comment
i dont leave unjustifiable dvt, no
i have posted a peer leveled question which was been closed as unclear, where u still see these (solve x+1=2) questions in an outstanding community like se
Okay I got this so far.. x~t if there is a connected subspace of X which contains both x,t. The collection of all these sets are called the components of a set
Now in our case a disc $D$ is a connected component of the set $R^2$ because a disc is a connected subspace of the $R^2$ and for a given $x,t$ we can find a disc which contains both x and t. .. to make it better let $\epsilon$ and $\epsilon-\dfrac{1}{2}$ be two numbers in $\Bbb{R^2}$ such that $0<\epsilon<1$ then we can find a disc centered at the origin which contains both $\epsilon,\epsilon-1$ . therefore a disc is a connected component of $\Bbb{R^2}$.@Balarka
What are you studying @Huy?
@Balarka The earlier message was too old to edit so writing a new one .. it should be $\epsilon-\dfrac{1}{2}$
I understood that there are 6 letters if we consider "AUE" as a single letter and answer would be 6!. Again for AUE it is 3!, but I didn't get why to do 6! * 3!.
Can't we just add (6! + 3!) to get final result??
PLEASE HELP ME.
@Rememberme connected component $\neq$ connected subset. That's precisely why I told you to read the definition.
@Rememberme No, no, no. The equivalence class of an element is called a connected component. But even before that, you have to verify that the relation is an equivalence relation.
algebraic geometry is a lot more than study of locus of polynomials! but yeah, that's the classical-point of view. You can treat zeros of polynomials over affine/projective spaces as "curves"
I'm reading Nano: The Essentials by T. Pradeep and I came upon this statement in the section explaining the basics of scanning electron microscopy.
However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. At such velocities, one needs to do...
one other thing to be careful about: one can talk about electrons moving at the speed of light in the sense of QFT---i.e. virtual electrons---but you can't measure them as such
@Agawa001 that's way too fancy. near-empty a glass jar, put a cathode and an anode in it and let electricity flow through. you'll get a green florescence inside. this is a standard fact.
that's really how electrons interact: the elementary event is the three-body interaction of two electrons and a photon
now, that photon is a virtual particle: it's not one of the input or output particles. as a consequence, it doesn't have to strictly obey energy-momentum conservation (though there are still constraints)
however, that's not the only possible electron-electron collision. in quantum field theory, there are really an infinity of such diagrams.
for example, i could modify that exchange photon to not just transfer from one electron to the other, but additionally: the virtual photon annihilates into a virtual electron-positron pair, which subsequently collide together again to create a new virtual photon.
in that case, the electron-positron pair so produced don't have to satisfy the usual energy-momentum constraint, and in that sense i think they could move faster than light. but it's a part of the description of the event, not something that one actually can measure
now, re: cathode ray tubes---i haven't actually worked with the theory of those in a while. but i recall my solid state prof saying that, despite there being quantum aspects of it, what one is seeing is more of a classical phenomenon than one might think
because i know only the rudimentary elements of general relativity; i never had to take a course in it, since i ended up going down the condensed matter route in grad school rather than astrophysics
shrug different people like different things. for me, quantum stuff is more interesting than general relativity
So except the first few lines I guess .. The rest of the proof is fine right? After reading the definition again a connected component of a set is a connected subspace if I am not mistaken.@Balarka
Well for the proof that it is an equivalence relation : We sat x~y if there is a connected subspace of X containing both x and y ... y~x just means that there is a connected subspace X containing both y and x that is similar to containing both x and y .
Reflexivity is very obvious ... X contains x, that is just same thing as x~x which would also mean X contains both x and x which is just that X contains x.
Transitivity : Let X be a connected set containing x,y and Y be a connected set containing y,z . Then $X\cup Y$ contains x and z . This will connected because X and Y will have an eleme…
Well what my teacher told me today was that the zeroes of the schrodinger wave equation lead to the orbitals .. As in when you plot the zeroes they lead you to the various orbitals . Is this correct ? @Semiclassical
not sure precisely what you're saying there. but i think you're getting at the fact that the wavefunctions of, say, a hydrogen atom can be labelled by how many zeroes they have
with those being the relevant quantum numbers of the system
Okay so wont the schrodinger wave equation give me many solutions (since it is a pde) and will lead to different quantum numbers . But why is that we only have four of them that is n,l,m,m_s
where m is the magnetic quantum number, m_s is the spin quantum number
I am talking generally about orbitals which(accordance to my teacher) are kind of shapes which you get when you plot the zeroes of the SWE . These are the places where the probability of finding an electron is very high
Though I know one more I guess .... (I found this out from a bit of search) the partial differentiation of x,y,z axis equals some h/4\pi I dont know it might be wrong @Semiclassical
to see the connection, we work in spherical coordinates. the laplacian is then a bit tedious to write out, so i'll refer to Mathworld
what one can notice then is that, while $\phi$ and $\theta$ are a bit tangled up, the $r$-dependence of the Laplacian separates out more nicely. namely, $r^2 \nabla^2\Psi = \partial_r(r^2 \partial_r \Psi)+Q(\phi,\theta)\Psi$ where $Q$ contains the two angular derivatives
as a consequence of that, one is able to do separation of variables in order to split the 3D SWE into a 1D radial part and a 2D angular part
$$\csc^2+\sec^2=1?$$
I thought I could just use reciprocal from the other formula $\sin^2+\cos^2=1$, can you explain what's wrong?
I understood that there are 6 letters if we consider "AUE" as a single letter and answer would be 6!. Again for AUE it is 3!, but I didn't get why to do 6! * 3!.
Can't we just add (6! + 3!) to get final result??
PLEASE HELP ME.
I'm reading Nano: The Essentials by T. Pradeep and I came upon this statement in the section explaining the basics of scanning electron microscopy.
However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. At such velocities, one needs to do...
