@Daminark while finite graphs are well-understood...

oops, you plugged in x=2 ?

yes.

eh, it happens

but I marked that at x=1 y = 5

and marked that at x = 2 y = 14

but then just wrote it wrong.

very silly mistake.

I spose'

yeah, the silly mistakes are small but frustrating

@Semiclassical imgur.com/a/82NPA

yeah, that shouldn't be a big deal

hopefully not. :(

it's not for marks though.

@Daminark every connected graph has a spanning tree (axiom of choice), and the spanning tree is acyclic?

The kinds of mistakes which bother me more (in the context of physics) are ones which give answers that are obviously wrong

lmfao

@Leaky I think this question isn't even well-formed otherwise

Like

You don't need to worry about infinite graphs

for instance, if someone gave an answer of the form $a+b^2$ where $a,b$ are supposed to have units of length

well, then I know right off the bat that they did something wrong. the units don't work

@Daminark I'm just messing around lol

I know nothing about graphs

$\aleph_6 \le \frac{1}{2}\aleph_6$ so actually it works in that context anyway

dimensional analysis @Semiclassical

yep

@Semiclassical that makes sense.

@Daminark $\dfrac12 \aleph_0$ is not well-defined

@Daminark started class yet?

Is there not such a thing as cardinal arithmetic?

It's a simple dictum, but one which people forget all the time

IIRC Saddle behaviour is observed when $\triangle < 0$ of the system $\dot{X} = AX$@nitsua60

especially on a quiz, presumably because of the time constraints

Like, $\frac{1}{2}\aleph_0$ would be the cardinality of a set such that there exists a 2-1 surjection from $\mathbb{N}$ to it, perhaps

to my mind, though, one should always always always check units at the end of a problem

if it passes a units check, then at least the answer isn't obviously wrong

haha, right.

@Dodsy nope, starting a week from tomorrow

if it doesn't, then you know something's up; if you find the error, great, but if not you can at least point out why you know your answer isn't right

But I think I finally know exactly which classes I intend to do

Sort of

Not really

and to me that kind of evaluation step is a Good Thing (tm)

But I know 3/4 for sure and the last has 3 options

It shows the student wasn't just blindly plugging-and-chugging but was actually thinking about whether their answer made sense

That's not the only check possible, of course.

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

@Semiclassical Good Thing$^{TM}$ ?

Rip @Dodsy

yep

@Daminark :o rip dodsy!? !?!??! :(

$\text{Good Thing}^\tm$

aww.

:)

rip semic

$\circ$ _ $\circ$

I feel like I'm missing something

hm...

Good Thing$^{TM}$

My favorite use of this was when I was explaining tangent bundles to someone

The Tangent Bundle${}^{TM}$

That's what the eunuch said. "I feel like I'm missing something".

Dodsy would've been said to snipe me there but he had to edit it the first time so no snipe for you!

Come back one year!

hey mr. meanhead.

I have a friend who has a very... interesting, first name, and his last two initials are T and M

arithmetic mean or the geometric mean?

Geometric, of course!

So he always writes his name as *insert first name here*$^{TM}$

HAH

psh, harmonic mean all the way

that's awesome.

Wait even that's a thing? @Semi

I have a BS name

yeah.

even that's a thing!?

$H=\left(\frac{1}{n}\sum_{k=1}^n \frac{1}{k}\right)^{-1}$

harmonnic mean is useful when determining average velocities eg

more generally, the $p$-th power mean is given by $(\frac1n \sum_{k=1}^n k^p)^{1/p}$

oh someone told me this amazing thing

okay so the inverse sine function has a range of $[-\frac{\pi}{2},\frac{\pi}{2}]$ and a domain of $[-1,1]$ ? @Semiclassical

that RMS > AM > GM > HM can be proved by using p-th power mean, defined just now by Semi

$p=1$ is the arithmetic mean, $p=2$ is the quadratic mean, $p=-1$ is the harmonic mean, and $p\to 0$ gives the geometric mean

@Dodsy ya

and we keep continuing each other's sentence

@Semiclassical ty, trying to do this stuff from memory so it solidifies :)

my favorite proof is still Lagrange extremizing the shit out

yeah, hard to beat that

my favourite proof is the geometrical proof

in the case of just two numbers, though, there's a cool geometric argument

...loool

yep

I used that image in one of my questions, lol

for 2 numbers it's trivial so

when did I pass 40 k lol

oh, sure. two numbers is pretty boring

but it's cute to have a proof without words

that's probably true

I like geometry but I know absolutely nothing about it. :(

just pythagoreans theorem.

oh I know something about a cone

if you know pythagoras' theorem then you know the Riemannian metric on the Euclidean space

That's right!

how'd you know that?

in particular you know the Riemannian metric

in particular you know Riemannian geometry

the fanciest metric I know is the special relativity one

I like the taxicab metric :c

actually why don't we use -1 norm

$dx_\mu dx^\mu = dx^2+dy^2+dz^2-c^2 dt^2$

A cone is some percentage of a cylinder or something?

is it a third?

$$\| a \|_{-1} = \dfrac{1}{a_1^{-1} + a_2^{-1} + \cdots + a_n^{-1}}$$

a cylinder is a cone with the apex removed :)

@BalarkaSen

:(

@Dodsy yes

a cone is a hat with the head spot filled in.

why are you pinging me

@BalarkaSen because it's topology

why are you pinging me with topology

@LeakyNun amusingly, that expression comes up in physics a lot (the rhs, I mean)

@Semiclassical :O

@BalarkaSen because I thought you like topology

@Dodsy I said yes to a third, so no to a half

or else ZFC would be inconsistent and both would be provable

why are you pinging someone who likes topology

yeah you're right it's a third.

I knew it

@BalarkaSen because I just mentioned topology

why are you pinging me mentioning topology

because you're topologically a meanhead blob

If I put a bunch of resistors of resistance $\{R_k\}$ in parallel, then the equivalent resistance will be $$R_{eq}=\frac{1}{R_1^{-1}+R_2^{-1}+\cdots +R_n^{-1}}$$

:{

@Semiclassical oh, that

now, while the case with finitely many resistors is well-understood...

draws an infinite grid

must I cite the obligatory xkcd?

places a 1-ohm resistor at each edge of the lattice

@Semiclassical yes, you must.

Why do people get Eulers equation as a tattoo on their bodies?

@Dodsy because it's beautiful?

oh that explains it.

I don't actually know how one solves the version in that XKCD, tbh

I know I've seen the solution for the equivalent resistance between two adjacent points

I should go to bed

@Semiclassical how?

7 am alarm tomorrow

-____________-

I also have my first tutorial for physics.

It goes something like this. Suppose you connect one node of that lattice up to a current source, and pump in 1 unit of current

lol

interesting.

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.

> take a superposition of those two systems

@_@

(((superposition)))

Sure. Resistors are linear elements.

superpostion sounds familiar.

probably not related

@Semiclassical wait, when can you not take superposition?

When stuff isn't linear.

night lads!

@Semiclassical eg?

$y''+y^2=x$

oh

ok

challenge accepted

lol

that's the riccatti equation, if memory serves

fails miserably

it doesn't have nice solutions.

main point is that, if you had solutions $y_1,y_2$, you couldn't add them together to get a third solution $y_1+y_2$

right

by comparison, $y''+y=0$ would permit this kind of adding of solutions

so it isn't a vector space

right

nonlinear solution spaces kinda suck

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.

linear systems ftw

so by ohms law the voltage drop across that edge will be V=IR=(1/2)(1)=1/2

@LeakyNun 美好的一天

So I need to apply a voltage of 1/2 to those two nodes to get a current of 1 through the system

@KasmirKhaan god dag

:)

hence, equivalent resistance is V/I = (1/2)/(1)=1/2

@LeakyNun busy with something or can we finish group action :D

@KasmirKhaan sure

All righty

problem with this approach is that I could easily enough say that the currents exiting the first node would be all be equal i.e. 1/4

I showed that sigma_G : A--> A is a permutation of A

but I don't know what the currents will be in the next closest set of edges

and the map G to S_G is a homomorphism

Hm, I guess it's not entirely obvious that a biholomorphic automorphism of $\Bbb C$ is indeed linear

shouldn't an automorphism preserve addition? though i guess 'automorphism' doesn't always assume that

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

@Semiclassical By biholomorphic automorphism I mean a biholomorphism from C to itself

ah, okay

@LeakyNun am trying to understand the examples now on page 43 =p

Maybe I see how to do it

ok

@LeakyNun the avious homomorphsm between two groups , that sends all elemtns to e

@LeakyNun how does one prove that it is a homomorphism

maybe I ishould not call it homomorphsim yet, but just a map

you prove

for all x and y

f(x) f(y) = ?

what does it mean for it to be a homomorphism?

it obeys this law f(xy) = f(x) f(y)

right. but what's f(x)?

an element of H

which one?

thats what is not clear to me

what's the definition of $f$?

think.

what i know is that f(e_G) = e_Hh

@Semi shut up

this shouldn't need any hint

let me think all righty

I'm pretty sure that such a biholomorphism can't have essential singularities at infinity but I am having trouble seeing this.

@LeakyNun oh I got it :D f(x) = e_H =f (y)

Hm, say $f : \Bbb C \to\Bbb C$ is the thing. $f(1/z)$ then needs to have an essential singularity at $z = 0$.

e_H * e_H = e_H

@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.

@KasmirKhaan good

@LeakyNun Dodsy did. you were in the convo so I linked you.

that was pretty avious but the notation scares me many times :D

@Typhon we apologize for your experience.

@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.

regardless

throwing accusations is not appropriate

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 $f$

apologies for the profanity

@LeakyNun how long you gonna be here? ?=p

@KasmirKhaan idk

Okay =p what is a vector space

hey chat

like i saw the definition but , can we do group actions on it?

I have not been abusing flags.

now farewell chat

For n >3 , D_2n and S_n are not ismomorphic, but are they homomorphic ?

hopefully I don't have to come back till 2018

ugh

yet you are still talking

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.

just call a fricken staff member

@KasmirKhaan any groups are homomorphic

@LeakyNun yes but if we forget about the trivial case

because D_2n acts on the set of n vertices

@LeakyNun like why is it the case that all groups have some structure in commen

there is an injective homomorphism from D_2n to S_n

hmm

faithfull action =p

yes, faithful like my love towards mathematics

No really its called that

I know :P

:D

I thught that was funny name

if ga = a for all g in G its called the trivial action

but if ga are all disticnt it is faithfull

well leaky thanks alot for the help :D I think I need to get more confident on reading and thinking by myself =p

> 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.

:D

it sounds funny in swedish =p

lol

haha

but its easiar to understand

all righty last Q !

Am gonna start doing quotient groups ( chapter 3 )

I like to have an idea of it

i know that Z / Zn is a quotient group

btw on another note , are factor groups and quotient the same thing ?

yes

why they use diff names

like its super comfusing ><

no idea

okay just opened up Paint 3D on windows and it's honestly pretty cool

When verifying $H<(\Bbb R,+)$, would $0\notin H$ be an automatic failure? It seems every subgroup by def. requires inclusion of the neutral element

@Brody right

@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

@Brody the former is required to prove the latter

Experiment: $\langle a,b \mid a^{10} = b^{10} = 1, ab=b^3a \rangle$

@BalarkaSen Hard for a function with an essential singularity to be one-to-one :P

hi Brody, Leaky, Kasmir

@TedShifrin do you have any idea what that is?

What what what is?

Good grief, no.

It's a semi-direct product of $\Bbb Z_{10}$ with itself.

@TedShifrin oh!

@LeakyNun I don't understand what this denotes

@Brody group presentation

it isn't related to your question

You look at all words in $a$, $b$, and their inverses, @Brody, subject to the rules he listed.

oh. well, not there yet anyway :p

Hi @Ted

@Kasmir: I presume you got my email?