It's not a surprise I skimmed over this section on my first read of Griffiths....

@Astrum Yes, matching boundary conditions is somewhat tedious

But its the essence of solving the problem. Its all about boundary conditions.

so for this problem, I need to write it all out as $V(x,y)$

oooookkkk

@Astrum Also, your method while clever is probably not how Griffiths intends for you to solve this problem

@DanielFischer I will prove $V_4C_4\not\simeq Q_8$ by counting elts of order $2$.

@KevinDriscoll to be honest, I don't think it's very clever, because if I think it up, it's probably not the best solution =p

@Astrum No, actually your method is much more general than what Griffiths wants you to do

my method of learning physics and math is to stumble my way through, usualyl stubbing my toes on the way

@Astrum splitting up a diff eq on a weird boundary into a more normal one on smaller boundary is a common trick

@PedroTamaroff Count elements of order $4$, those are fewer.

AGH! OK. =) My ideas are always bettered. =P

@PedroTamaroff Of course, we could do even better.

@Astrum The advanatge is you should get a solution which is exact everywhere in closed form. The way Griffiths intended you to do it will involve an infinite sum, so with finite terms its only approximate

There's a natural inclusion $D_4 \subset S_4$.

@Astrum Your solution will also avoid the so-called 'Gibbs phenomenon'

never heard of it...

I need to study more @.@

@KevinDriscoll Thanks. I'll let him know what you said.

@TedShifrin We're switching books to Farkas-Kra for RS.

Also, hi.

@Astrum its only important in signal analysis, a very technical point about fourier series

LOL, probably good, @Mike.

@DanielFischer ORLY?

@TedShifrin 'ello.

Not true, @Kevin: Important lots of ways. Non- uniform convergence when $f$ has discontinuities.

@PedroTamaroff Yes, $D_4 \subset S_4$, done.

@TedShifrin Well, there was nothing left to do with surfaces in Conway... he defines sheaves and complex manifolds and says "WELP WE'RE DONE HERE"

@TedShifrin Hi.

@DanielFischer So four elements of order $4$ at least?

@Ted Of course you're right, but I've never had the difference between pointwise convergence and absolute convergence ever DEFINED in a physics class, much less leveraged

@Mike: Conway doesn't know from sheaves. Hi @Pedro.

@PedroTamaroff No, $D_4$ has only two elements of order $4$.

So we'll do the uniformization chapter of Farkas, @TedShifrin. What do you mean by "Conway doesn't know from sheaves"?

He's a functional analyst. They don't do sheaf theory.

@Pedro @Daniel @Ted I have a question involving asymptotic expansion of an integral using sums of residues, but its somewhat tedious in that doing it involves knowing specific facts about some of the unknown functions involved in the integral. The problem is giving me FITS, but I'm hesitant to ask it here because I would ahve to provide a number of additional facts that make it pretty specific. Do you guys think its worth it to ask?

So when I have the equation $Be^{-kx}\sin ky$, I just use the BC that at $x=0$ $V= V_0$? or do I need to specify that the interval needs to be $[0,a/2)$? Or maybe that's taken care of by the $V=V_1 = V_2$ at $(0,a/2)$ @KevinDriscoll

Yeah, he defines them and then goes away.

@Astrum Ya you just use $V_1(0,y)=V_0$ and the interval is taken care of byt he fact that $V_1$ is onyl defined on $y \in [0,a/2)$

ok, got it, I'm gonna be offline now while I try to finish this problem, I think I understand. If not, I'll be back later. Thanks for the help!

It looks like Farkas does stuff analytically, too. At least based on the fact that uniformization starts with subharmonic functions, @TedShifrin

@KevinDriscoll Dunno. What kind of integral?

hopefully I don't pop back in in 10 minutes...

@Astrum and $V_1$ $V_2$ are NOT the same at $x=0 , y=a/2$, they are only the same for $y=a/2, x\neq 0$

see ya

erhm

ok

you can't avoid analysis, @Mike, so far as I know.

@Daniel I have a relation which relates one unknown function to the integral of another unknown function times a product of gamma functions along a vertical line int he complex plane

so in other words $V_1 (x,a/2)= V_2 (x, a/2)$ iff $x \ne 0$ ?

@Astrum Exactly

@Astrum at x=0 we have the boundary condition

alright, thanks

see ya!

Urk, @KevinDriscoll. Could be something for Ron Gordon or robjohn.

@Danny what's up?

@Daniel Possibly

@robjohn do u have time 5 minutes , i need to ask smith i dont understand

I guess I might as well ask. I'm not getting anywhere stewing about it. And the worst that can happen is it goes overlooked

@Danny oksy

Yup, @KevinDriscoll. At worst, you don't get an answer.

write it clearly and succinctly on main, @Kevin.

@TedShifrin That's true, but I imagine Gunning does things more straight-up cohomologically, or something. (Haven't had a chance to finish it yet, I'll do that over break.)

@Ted Ya I will attempt it. The succinctly is sort of the problem. I am trying to cobble many disparate facts about several unknown functions into one place to hopefully get an asymptotic expansion for one of the functions in terms of a large parameter

I never considered asking itin chat, its far too technical

@Mike: I don't think Gunning does uniformization, but I might be wrong.

@TedShifrin FUUUUUUUU

My fountain pen just broke.

My fingers are now partially black.

if $Pr(X > a, Y > a, Z >a )$ and $X,Y,Z$ are independent random variables , how can i argue that it is $Pr(a)Pr(b)Pr()$ the definition of independence is $Pr(A\cap B) = Pr(A)Pr(B) $ but how can i transform $Pr(X > a, Y > a, Z >a )$ to the defintion. @robjohn

@Ted He does. After Jacobi varieties, before branched covers.

@Pedro: I still have my Parker I bought in 1963.

@robjohn did u see what i wrote...

@TedShifrin Fixed it!

Also a Parker this one.

The rubber in the cylinder moved an the void was broken.

@Danny I am looking at it

Ah, @Mike, I'll have to look tomorrow, if we aren't canceled

I love that pen, Pedro, 50 years and counting ...

@Danny $P((A\cap B)\cap C)=P(A\cap B)P(C)=P(A)P(B)P(C)$

@Ted I certainly can't comment on what he does, but I know he does something!

@TedShifrin Ah, that's wonderful.

I have a certain love of pens too.

Not farm-pens, though.

=)

@robjohn yes that thing i figured out

I wonder if I can just get the library to let me but their copy of Gunning. I'm the first person to check it ojt since 1990.

That's comforting, @Pedro.

@Danny sorry. Was there something else?

since $(A\cap B) \subset A$

@Mike: You should steal some of my library when I retire :)

@TedShifrin I did something nice today.

Oh?

@robjohn but what is $A \cap B$ in terms of my "question"

lol, My pens last less than a week

@Ted I'd love to. I love building up my library :)

@Danny $A$ is the occurrence that $X>a$

@Danny $B$ is the occurrence that $Y>a$

@Danny $C$ is the occurrence that $Z>a$

is it $\{(x,y) x\in X , y \in Y : x> a, y >a \} $

providing we we only had A and B

@Danny $A\cap B=\{(x,y)\in X\times Y:x\gt a\land y\gt a\}$

@Danny Does that make sense?

@Ted Here's a question I bet you'd love to answer.

No thanks.

@Pedro: How was u nice?

@robjohn i think it is clear now

@Danny :-)

:O

@TedShifrin Sorry, I had to go.

I made a tet.

And showed its motions are iso to $A_4$.

@TedShifrin here

@Pedro Now find the group of isometries of a torus.

@Mike I'm thinking rotations.

mmm, oscillating indian skin burns

@Pedro Well, you can rotate it about more than one axis. But I can't think of the answer myself immediately. So I'll take it back and say symmetry group of the circle. :D

@seaturtles OUCH!

@Mike Aha.

My answer was overly technical, so I added colours and aligns to make light bulbs appear over the OP's head.

That's too wide to read on my phone.

Turn it sideways? Scroll?

barrel roll

@seaturtles I have to prove $A_n$ contains a subgroup iso to $S_{n-2}$ for $n>2$.

seen that at least twice on main

OK.

walk of shame towards main

@TedShifrin Are there Riemann surfaces with anabelian fundamental group?

@PedroTamaroff e.g. here

Dumb question.

FUUU

hiccups.

Indeed @Mike

@Pedro: Thomas was right. You confounded row and column spaces.

Indeed it was a dumb question @Ted? :D

Yes @Mike

and the answer to the question was indeed, indeed.

@TedShifrin Yes, I know. Did you see my tet?

Yes. I have models of all the regular polyhedra for my alg courses.

@TedShifrin I have an icosahedron.

And now a tet.

All paper made.

Three more to go. :)

I also have a cylinder with the lattice of topologies of a set with 3 elements.

How bout the lattice of subgroups/subfields for Galois group of $x^3-2$?

I don't know Galois theory!

Don't poke the wound!

Poking is fun ...

@Pedro You should find the symmetry group of an infinite helix.

I think that'd be fun!

Hm... no. That's I'll pose.

ill posed

@seaturtles I will try to prove the extra part that $S_{n-1}$ cannot be embedded in $A_n$.

Should it be easy?

@Pedro I can prove that for odd $n$ :D

@Mike Right, I see.

Even is less trivial, of course.

@TedShifrin

Did you see this?

@KevinDriscoll uh.... can we start another room?

(if you're not busy right now)

@JasperLoy Hey :)

Guess I missed you.

@Astrum ya just invite me if oy ulike

@KevinDriscoll I have a last question for you: did you do research with anyone at your home university? How did you approach them?

@Mike Ya that was my only real research. Mostly I just read their websites/papers and picked something that I found the most interesting. Then I sent them a straightforward e-mail saying I'd like to gain a flavor of their research and asking if they're accepting independent study students.

Alright, neat. I'll forward on your advice.

@Mike

@Pedro

I figured it what it was, but I still have to prove stuff. =P

What was?

I am looking at the butterfly lemma.

That's an ugly-ass diagram

For example, I have to show $ \{(V\cap u)v\}\cap\{ u(U\cap v)\}=\langle u\cap V,U\cap v\rangle$ where $v\lhd V,u\lhd U$.

@TedShifrin

In particular, the important stuff is that the diagram is correct, which Lang doesn't prove. =P

omg bunny

crickets

@usukidoll by the inductive hypothesis $2^k \ge k +1$

yeah I got that step

my goal is $k+2$ so I have to start with the left with is $2^{k+1} $ which is $2^k$$2$

so you want to show that $2^{k+1} \ge k+2$ then $2^{k+1} = 2^k 2 \ge 2(k+1)$

I need to get $k+2$

so you have $2^{k+1} \ge 2k + 2 \ge k+2$

I got as far as by inductive hypothesis ... $(2)(k+1)$ and some strange stuff appears

since $k\ge 0$

at that step you're done

basically

@Mike Is it relevant where we place $v$ and $u$, i.e. $u(U\cap v)$ versus $(U\cap v)u$?

because the next two steps are $(k+1)+(k+1)$ how did the book get that

followed by $(k+1)+1$

@PedroTamaroff What ar you trying to prove? Also, do you mean $U \cap \langle v \rangle$?

@usukidoll $2(k+1) = (k+1)+(k+1)$

If your group isn't abelian, certainly it matters

@usukidoll and $(k+1)+(k+1) \ge (k+1)+1$ for $k\ge 0$

why is $2(k+1) = (k+1)+(k+1)$

@usukidoll $2(k+1) = 2k +2 = k+k+1+1 = (k+1)+(k+1)$

@Mike No, Lang denote by $v$ a subgroup $v\lhd V$.

And $u\lhd U$.

gOD.

........ does that mean 2(k+1) as write K+1 two times?!

Yeah.

Well, $UV$ and $VU$ may not generate the same subgroup in general, I don't think.

Wait, no.

@usukidoll Yes, thats sorta the definition of multiplication

If those are both normal they certainly will. If any one of them is normal it will.

@usukidoll $2x=x+x$

If neither of your subgroups is normal then in general $UV \neq VU$ p sure

and what next factor out k so it would be K+[1+1] k+2

oh man I'm so not used to bending the rules

@usukidoll You're not bending any rules

@usukidoll once you have $2^{k+1} \ge 2(k+1)$ its just algebra to get the required result

@Pedro

like distributing 2 with the k+1 2k+2 k +1+k+1

also$k+(1+1)k+2$ would be $2k+3$ not $2k+2$

@Mike Horrible physics are horrible.

Doge denies physics.

(k+1)+(k+1) is k+2

huh?

afterwards but ....

$(k+1)+(k+1)=2(k+1)=2k+2$

@usukidoll I don't think you're thinking clearly about this. $(k+1)+(k+1) = k+k+1+1=2(k+1)$

yeah I got that part now

but after that step it's (k+1)+1

so how do I get from (k+1)+k+1 to this step -> (k+1)+1

ummmm @usukidoll $2(k+1) - k+2 = k \ge 0$ for $k\ge 0$

it seems entirely trivial to me

:///

@usukidoll Is it not obvious to you that $2k+2 \ge k+2$?

(for k positive)

there's a whole extra $k$!!! :-P

errr nooo

@usukidoll Perhaps I am wrong here and your brain is just having one of those days, but if that inequality isn't obvious then I think these problems are above the level that you need to be at.

Its just an

'algebra 1' kind of inequality

plug in some values for k and its obviously true

oh that I could just plug in k =1 I could see that 4 is greater than or equal to 3

but you need to know it for more than just $k=1$

Indeed, you could even prove (BY INDUCTION)! that its true for all k

or you could go

@Mike It suffices I show $u(U\cap v)=\langle u\cap V,U\cap v\rangle=(u\cap V)v$

$k \geq 0 \implies 2k \geq k \implies 2k+2 \geq k+2$

@usukidoll Or you could just do what I did and subtract...... and see that k is left over

@Mike Does one really need to prove such a trivial thing though in this type of problem?

I buy that shit.

@KevinDriscoll No.

But I'm convincing her.

ah okay, I'm just being a bothersome outsider, thne

I welcome your comments :D

@Mike OH WELL IN THAT CASE......

the book didn't really explain that part clearly

which part?

after I use the inductive hypothesis

it went all over the place to (k+1)+(k+1)

and then k+1+1

which is k+2 but I want to know why it did that

Ya I imagine the author thought that the algebra was 'elementary' and shouldnt be included. All he doing is simplifying the result to get it into the desired form

k let's try another one Show: $n ≥ 4, n^2 ≥ 3n + 4.$

that is he is trying to show that $2^{k+1} \ge (k+1)+1$ if we alreayd have that $2^k \ge k+1$

OK, I give up on flappy doge

now inductive step...for $P(k)$ $k^2 \geq 3k+4$

for $P(k+1) = (k+1)^2 \geq 3(k+1) +4$

I have a nice induction problem

my goal is to retrieve $3k+7$

let f be a function from reals to reals with property $f\frac{x_1+x_2}{2}=\frac{f(x_1)+f(x_2)}{2}$ Prove $f (\frac{(x_1+x_2+\dots+x_n}{n})=\frac{f(x_1)+f(x_2)\dots f(x_n)}{n}$

so this next step is $3k+7$ so how was this previous step jumped from $3k+2k+5$ to $3k+7$ @KevinDriscoll

can ayone prove it?

@usukidoll Well you have to make an argument. Is $3k+2k+5 \ge 3k+7$??

@usukidoll??

$25 \geq 19$

@usukidoll Certainly. Can you show that its true in general?

that's the thing ._.

Hey can someone help me with showing something is a bijection?

@Mike?

@usukidoll If I gave you two numbers, how would you tell which is bigger?

@Anthony I don't believe bijections exist

@Anthony what is teh function?

the numerical value

@Mike Careful or @Ted is gonna create a 1-to-1 mapping of his fist to your face

Soooo I need to show that a function from N x N to N is bijective

Best joke I've heard in here in ages.

everyone knows that negative numbers are super small if they're big like -1000000000000000000

@usukidoll So suppose I gave you something like $\pi^5$ and $e^4+10$, what then?

$\binom{i+j-1}{2}+j$??

The function is (i,j) to j + 1 + 2 + 3 + 4 + ... (i+j-2)

dang we have to find the real value of pi and the real value of e before we go further

pi is 3.14 e is 2.71

yes, that's the same as what I wrote

@usukidoll Okay suppose you do then, then how do yo uknow which expression is bigger?

of couse now we have exponenials to deal with

Look up triangular numbers

solve it

solve what?

$(3.14)^5$

and $e^4+10$

@user4140 me?

okay and then do what? @usukidoll

@Anthony yes

highest value is the bigger one

yes but how do you know which one has the higher value?

solve it mannn like (3.14)(3.14)(3.14)(3.14)(3.14)

@user4140 How is that useful in showing the bijection?

@usukidoll For example: an ever simpler case. How can we show that $5>2$?

did you already ask this question on stack exchange?

@Anthony Do you know the Cantor-Bernstein-Schroeder theorem?

5 is bigger than 2

No, I didn't.

yes but how do you know that?

prove it to me!

And I know of it, I don't think I can use it.

Sure you can.

Oh, sorry. Misread.

Haha.

alright it's better to have 5 apples than 2 apples

Thats not a proof (especially if you hate apples!)

mathematically give an argument for why $5>2$

That's what you asked right?

:/

@KevinDriscoll o_O

5-2>0

@user4140 Probably. Probably someone from my class.

Yes, I thought that guy is an asshole.

@Mike I don't mean something silly like go to ZFC etc etc etc. Just something using basic operations. The key question is, in general, if we have 2 numbers how do we tell which one is larger?

5-3>0 while 3-5<0

@Mike (its a didactic point, ofc)

does that work for you kevin?

@KevinDriscoll I was pretty sure you were just trolling for a bit there.

@user4140 The guy who asked?

@user4140 Indeed it does!

@Anthony yes

well good night guys

@user4140 Night dawg.

@Mike Yeah I know it seems like it, but the question goes ot the heart of @usukidoll s problem

@usukidoll Do you see @user4140's point? $5>2$ because $5-2=3>0$ and if the difference of 2 numbers if greater than zero then the one which we subtracted from is larger

yeahhhhh...

@usukidoll So based on that same reasoning, given $5k+5$ and $3k+7$ how do we tell which is larger?

5k+5 >0 and 3k+7 >0

Well thats certianly true

but I don't think it helps answer the question

Oh, I see what the intent was behind the question now.

That's a cute way to show what you mean. I'm gonna steal it.

@Mike I'm afraid it might be slightly too cute fo rits own good, but feel free

it has something to do with the inequality

@usukidoll go back to the case of $5>2$ and follow the same reasoning

@usukidoll But now we've reduced the problem all we have to show now is that $5k+5 \ge 3k+7$

let k = 4....

Why are you letting $k$ equal anything?

well it's not going to work if k = 0 for that inequality

Well, just like with $5>2$, you can move things around to find out EXACTLY when it will work.

Just to start, $5k+5 \geq 3k+7 \iff 5k \geq 3k+2$...

Keep going until one side is $k$ and the other side is a number.

oooooooo

Back to you, @KevinDriscoll!

@usukidoll AH HA!! and so for $k \ge 1$ we proved that $5k+5 \ge 3k+7$ and since in our induction problem we have that $k\ge4$.....

:) mhm

blah flat out demorgan's laws since $(A \cup B)^c = (A^c \cap B^c)$

Can somebody tell me how the hint here math.stackexchange.com/questions/205076/… completes the solution.

nah

oh @Ted is here

Off topic: If $a$ is in the upper half plane, why is $e^{2\pi i a}$ in the unit disk? For example, I plugged in the value $a=3-i$ in wolfram and the answer is larger than 535, so it is not within the unit disk.

@TheSubstitute $|e^{2\pi i(x+iy)}| = |e^{2\pi ix}e^{-2 \pi y}| = |e^{-2\pi y}| < 1 \iff y > 0$

@KevinDriscoll hey

Yooooo

did you get the invite?

hi aai

again

aw man my leg is asleep

I don't wanna be alone forever but I can be tonight

I don't wanna be alone forever but I love gypsy life

cause I'm I'm I'm I'm I'm I'm a gypsy gypsy gypsy I'm

da fuq did I just read?

Greetings

@robjohn do you see a nice way to prove that $$\sum_{k=1}^{2N} \frac{(-1)^{k+1}}{k} H_{k-1}-\sum_{k=1}^{2N}\frac{(-1)^{k+1}}{k} H_{2N}+\sum_{k=1}^{2N}\frac{(-1)^{k+1}}{i}H_{k-1}$$ tends to $0$?

@robjohn without using that extension formula of harmonic number.

@robjohn ups, there is a mistake above.

@robjohn it's $$\sum_{k=1}^{2N} \frac{(-1)^{k+1}}{k} H_{N}+\sum_{k=1}^{2N} \frac{(-1)^{k+1}}{k} H_{k-1}-\sum_{k=1}^{2N}\sum_{i=1}^{k}\frac{(-1)^{i+1}}{i}\frac{1}{k}$$

(previously I copied a wrong row)

My god, 55 users in 41 rooms. That's impressive.

Anc Chris's sis - not really sure, and I may come about rater stupid but I think if you look at the odds and the even at each of these sums it might get it where you want

(when k is odd, and when it's even)

@robjohn I think I'll try a different approach.