But I admit that the room is not so active,perhaps there are not many interested in chats or on the topic but i try to search users who are interested in this to discuss with them and explore @nitsua60
If I put a bunch of resistors of resistance $\{R_k\}$ in parallel, then the equivalent resistance will be $$R_{eq}=\frac{1}{R_1^{-1}+R_2^{-1}+\cdots +R_n^{-1}}$$
since the system is symmetric around that one node, the current will have to split up equally between the four directions i.e. 1/4 in each direction
I could do the same for any node, and in particular I can do so for one of the adjacent nodes
and since flipping the direction of current doesn't change things, i can have a configuration where there's 1/4 current coming into that node and an overall current of 1 leaving it
if I now take a superposition of those two systems, I'll have a unit-1 current going into one node and a unit-1 current exiting the other node.
anyways. So I've got a current I=1 in my system, and if I look at the current in the edge connecting the two systems I'll find that the total current there is 1/4+1/4 = 1/2.
from the book " a group action of G on a set A just means that every element g in G acts as a permutation on A in a manner consistent with the group operations in G @LeakyNun
[Some rambles] Juts found out today that the weirdness has a limit. Today's accidental overestimation of the influence will serve as a good lesson to not behave so close to some of my uni friends
@bwDraco I haven't made any non-custom flags in the past 6 months at least. Also, I have been trying several times to tell everyone to leave me alone. I no longer wish to be on this site, and I no longer wish to be contacted through here. I might return next year. Until then, leave me alone.
@Dodsy I've been making no flags toward you. In fact, I'm trying to flat out leave stack exchange and the only reason I'm online is because you insist on rude accusations. Call a moderator and have then deal with it. However, I do not appreciate the rudeness.
@LeakyNun Thanks, it's just that this site has a very bad way of dragging me back into it. I don't want to get caught up talking on here so much anymore so I really shouldn't risk being on here. That's probably extreme but I've learned that it's probably for the best.
Ahhhh hm. Picard's theorem says near essential singularities a holomorphic function takes any value infinitely often, and that'd immediately break injectivity of $f$.
But this seems like a huge machinery to use
Can we use Casorati-Weierstrass? $f(1/z)$ sends a deleted neighborhood around $z= 0$ to a dense subset of $\Bbb C$.
@Dodsy Since I am a nice guy, I will state right now that I am flagging your accusations towards me with a custom message asking for a staff member. I'm not saying you did anything wrong in speculating. However, if there is this issue then it needs to be resolved properly instead of throwing random insults and darts. You're doing the equivalent of putting people's names on a wall and throwing darts to decide who is at fault. For all we know, nobody on the wall is even responsible for this.
Also, I am leaving the site for the next month or so. If someone knows how to disable email notifications for chat that would be great. It's annoying. Otherwise, please star this post so people are aware that I am NOT involved in any of these issues. If I have to be called back here again for an accusation I will flag the accusation. Other than that, I shall not flag short of blatant lewd pornography, illegal links, or spam.
Ah ok but if that was the case for any point $p$ outside the deleted neighborhood with $U$ an open disk around $p$ disjoint from the deleted neighborhood, $f(U)$, being an open set around $p$, would intersect that dense image. That breaks injectivity of $f$
@bwDraco just so you are aware, I was under a chat ban until last night or the night before for falsely admitting to the flag spam at which point shog said "no you didn't" and gave me a one week timeout since apparently falsely admitting to something to calm the chat down isn't appropriate. Alrighties? It was last week and is in the log. So.... yeah I think you both need to back off. When the staff literally bans me for confessing because I didn't do it it should be pretty obvious.
I was making that flag I promised to do and noticed some more posts directed towards me I hadn't noticed. These are serious accusations and I would prefer not to be suspended for no good reason while I'm gone.
Plus, if this account is making flag spam and I'm not aware that is also pretty alarming. However, I'll chalk it up to Dodsy overthinking it.
> En verkan av G på X säges vara [...] Trogen om det för alla par av distinkta element g1 och g2 finns ett element i x så att g1.x inte är lika med g2.x. Ett ekvivalent villkor är att det neutrala elementet i G är det enda element i G som har samtliga punkter i X som fixpunkter under gruppverkan.
@LeakyNun Thanks. My text's exercises call to check for closure under sums and negatives, but the previous seems like a much quicker way to affirm falsehood