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12:01 AM
hi, @topologicalmagician
howdy, @Leaky and @Mathein
 
Hi @Ted
 
@MatheinBoulomenos and algebras come with adjoints
 
What is the quotient space constructed by identifying opposite points on $S^1$ called?
 
In this page: mathoverflow.net/questions/23478/… , there is the phrase "The closure of open ball of radius $r$ in a metric space, is the closed ball of radius $r$ in that metric space." given as a false belief. And there is the comment giving a simple counter-example: the closure of a ball of radius $r$ is not the closed ball of radius $r$. I would say that this is the trivial example, right? Is there any others?
 
Is it $\Bbb RP^1$?
 
12:14 AM
I am trying to say that is this idea really fundamentally pathetic? Or there are just exceptions?
 
@Ultradark it's mostly called $S^1$
 
@MatheinBoulomenos I mean identifying opposite pairs of points
 
you can call it $\Bbb R P^1$, but mostly people don't bother because it's just homeomorphic to $S^1$ again
 
I am stuck on a really simple step...I cannot determine if an element is $0$ by the universal property of the inverse limit in CRing.
If I have each downward projection maps an element $x\in \lim A_n$ to $0$ how can I say $x$ is $0$ in $\lim A_n$?
 
12:25 AM
0
Q: Prime twins $ (3^n - 2, 3^n - 4) $ conjecture

mickLet $n$ be a positive integer. Conjecture There are infinitely many prime twins of the form $$ ( 3^n - 2, 3^n - 4) $$ Examples include $$(3^2 - 2,3^2 - 4) = ( 7,5 ) $$ $$ ( 3^{37} - 2 , 3^{37} - 4 ) $$ $$ ( 3^{41} - 2 , 3^{41} - 4 ) $$ Notice $2,37,41$ are primes themselves. Coincidence ? ...

 
-1
Q: How can I solve the functional equation to find all involutions whose reciprocals are also involutions?

UltradarkHow can I solve the functional equation to find all involutions whose reciprocals are non trivial involutions? I think this functional equation can be expressed as: $$ f(x)=\frac{1}{f(x)}, $$ with the conditions that $f(f(x))=x$ and $\frac{1}{f(f(x))}=\frac{1}{x}.$ $f(x)=\frac{1}{x}$ does not ...

Solve
 
12:55 AM
Hello guys, I have another quick question about mathematical statements. Is the following statement false? "There exists a x in the reals, for all y in the reals, such that x*y=0"?
 
I don't like the syntax. It should be "There exists x such that for all y we have xy=0."
Or "There exists x such that xy=0 for all y ..."
Anyway, why do you think it's false?
 
Oh, I'm sorry. I wasn't aware of this. Thanks for the pointer. I think it's false because if, for instance, we pick x = 3, then it wouldn't work if y = 1 (since the statement is claiming that it should work for all y in the reals).
 
Well, obviously you picked a bad x.
 
But isn't this statement "lets" us pick x that would make this statement true for all possible numbers, not only 1? Or is just picking one is sufficient to make this statement true? I am quite confused about this concept.
 
When one says "there exists x" that probably means that not very many will work.
There's no "for all x" !! So it's up to you to find the one (or two) that work.
 
1:04 AM
Oh, so this statement structure (more precisely, the 'there is' quantifier) allows us to be given a number for which we could make this statement true?
since the other part states that y can be any real number?
 
Well, to show it's a true statement, you find it. To show it's a false statement, you show that no value of x will work.
 
$\lnot\exists\equiv\forall\lnot$
 
No symbols, DogAteMy.
 
$\ddot\frown$
 
but aren't we allowed to show how this statement is false by choosing at least one counterexample that would make it false?
Sorry, if my questions appear silly.
 
1:06 AM
Let $V$ be an $n$-dimensional inner product space, where $n>0$, and let $F$ be the linear transformation on $V$ defined by $F(\textbf{u})=\langle \textbf{u},\textbf{c} \rangle \textbf{b}-\langle \textbf{b},\textbf{c} \rangle \textbf{u} $, where $\textbf{b},\textbf{c} \in V$ and $ \langle \textbf{b},\textbf{c} \rangle \neq 0 $. Show that $V$ has a basis consisting of eigenvectors of $F$ and find the matrix of $F$ with respect to some such basis.
 
No, because it's not a for all statement. So one counterexample is irrelevant.
 
The strategy here would probably be to find the matrix of $F$. How would one go about doing that?
 
No, @schn. The strategy is first to figure out the eigenvectors.
 
@TedShifrin But those are derived from $(A-\lambda I)$, where $A$ is the matrix of $F$, no?
 
But that's not the definition, is it?
 
1:09 AM
Oh, I see. So just to clarify - basically because the statement is stating that there exists a number (which we're not allowed to choose since it's given), then by finding at least one number that would make this statement work (which is 0 in this case), then we could make this statement work?
 
For the eigenvectors, you mean?
 
Abwatts, I wouldn't say it's given, but it's up to you to find the right one that makes it work ... or to show nothing will work.
3
Yes, @schn.
 
@TedShifrin I see. So it would be sufficient to choose the right x and y to make this statement work?
 
@TedShifrin Okay, by definition of the eigenvectors, do you mean $\text{ker} \ (A-\lambda I)$?
 
The right x. It has to work for all y. Everything depends on the form of your particular statement. You have to pay a lot of attention to "there exists" and "such that" and "for all."
No, @schn. Because I don't know who $\lambda$ is.
 
1:12 AM
@TedShifrin What would be your definition then?
 
@TedShifrin Got it, thanks a lot for the clarification!
 
Have you found me the definition in your book/course, schn?
Cool, @Abwatts.
 
@TedShifrin It's all entangled with the expression $(A-\lambda I)$...
 
I don't believe you. That comes later in every book I've seen.
An eigenvector of $F$ is a nonzero vector $v$ so that ...
 
$F(\textbf{v})=\lambda \textbf{v}$
 
1:15 AM
finish
 
for some $\lambda\in\textbf{R}$.
 
There you go.
OK, so go back to your definition of $F$ and see if you can find some vectors $u$ for which that will work.
 
Okay.
 
Any thoughts?
 
One would probably want to eliminate $\textbf{b}$ in $F(\textbf{u})=\langle \textbf{u},\textbf{c} \rangle \textbf{b}-\langle \textbf{b},\textbf{c} \rangle \textbf{u}$, no?
Or rather $\langle \textbf{u},\textbf{c} \rangle \textbf{b}$.
 
1:21 AM
So, OK, what vectors u will eliminate that?
 
Only $\textbf{0}$, correct?
 
NOOOOO.
 
How to solve $ (2D^2 -3D + 2) =x e^{-\dfrac{x}{2}}$ by operator method, where D is differential operator. It is not the case that I am solving by this method for the first time, but in this problem I get stuck.
 
@Ajay: The indicial equation has complex roots, so you have to use those.
WB symbolic DogAteMy.
 
@TedShifrin Since the inner product is not defined (it could be the dot product or not), it is unclear what would make $\langle \textbf{u},\textbf{c} \rangle =0$, isn't it?
 
1:27 AM
No, it's clear. Just work the problem assuming it's the inner product you know.
 
Okay, then $F(\textbf{c}^\perp)=-\langle \textbf{b},\textbf{c} \rangle \textbf{u}$.
 
What does $\mathbf c^\perp$ mean here?
 
@TedShifrin What do you mean by indicial? The characteristic equation?
 
Right, @Ajay.
 
@TedShifrin Any vector perpendicular to $\textbf{c}$.
 
1:30 AM
That's not what that means, of course. It means the entire subspace orthogonal to $\mathbf c$.
So is any vector orthogonal to $\mathbf c$ an eigenvector of $F$?
 
It depends on if $\textbf{u} \in \mathbf c^\perp$.
 
I guess you are confused with the method of variation of parameters with the operator method, I usually have to use the solution of auxiliary equation(hence characteristics equation) in the method of variation of parameters. But in this case I have to $y_p = (F(D))^{-1} (x e^{-\tfrac{x}{2}} )$
 
Well, my question is asking about vectors in $\mathbf c^\perp$, right?
@Ajay: Don't you usually factor $F(D)$ to do that?
Truthfully, I haven't thought about this in close to 50 years, but ...
@schn: So you have found a lot of eigenvectors now. How many linearly independent ones?
 
@TedShifrin Yes, any vector in $\mathbf c^\perp$ is an eigenvector, since $v \in \mathbf c^\perp$ would give $F(\textbf{v})=-\langle \textbf{b},\textbf{c} \rangle \textbf{v}$
 
Right, good.
 
1:35 AM
Yes, I do that, but I have never used the roots of indicial equation in finding the $y_p$ while using this method.
 
How do you find $(F(D))^{-1}$, @Ajay?
 
@TedShifrin $n-1$?
 
Good, @schn. So we only need one more!!! Can you guess what it'll be?
 
Could it be the zero vector?
 
No, remember eigenvectors are always nonzero.
Change strategies on the right-hand side now.
 
1:38 AM
For the simplest case, say the on the right side of the differential equation, if there is a function of the form $e^{ \alpha x}$, and let the equation is $F(D) y = e^{ \alpha x}$, and if $F( \alpha ) \ne 0$, then for $ (F(D))^{-1} = (F(\alpha))^{-1} $
 
@TedShifrin You mean there's a way to manipulate $F(\textbf{u})=\langle \textbf{u},\textbf{c} \rangle \textbf{b}-\langle \textbf{b},\textbf{c} \rangle \textbf{u}$ to find the last eigenvector?
 
I don't believe that formula makes sense.
 
@TedShifrin Can you please tell, for whom that message was?
 
The "inverse" can't depend on what's on the right-hand side like that. What if $\alpha$ turns into $\beta$?
You. :)
@schn: Your first strategy was to get rid of the $\mathbf b$ term. What is another strategy to get an eigenvector?
@Ajay: I think you need to find something more general. You're being confused by having pure exponentials, since any power of $D$ turns them back into themselves. You need to read about the general method.
 
Assuming, you have pointed out the last formula, $(F(D))^{-1} = (F(\alpha))^{-1}$, My prof. convinced us that for the case when $\alpha$ is constant, $F(D) e ^ {\alpha x} = F( \alpha ) e^ { \alpha x } $.
 
1:44 AM
That's because of what I just finished typing.
But that does not generalize to other functions besides pure exponentials.
 
@TedShifrin $F(\textbf{b})$?
 
Aha, @schn.
 
Yeah, I know the general method ~ by using wronskian, or by the method of variation of paramters. Okay, I see.
 
And note that they told you that $\mathbf b\notin\mathbf c^\perp$, so you win!!! :)
 
@TedShifrin Yes. So 0 is an eigenvalue here?
 
1:45 AM
@Ajay: There are ways to find the formal inverse of $F(D)$ by factoring and using partial fractions.
Aha, @schn.
Yes, the eigenvalue associated to $\mathbf b$ is indeed $0$.
OK, you have almost finished the problem now, @schn. Now I'll go cook dinner :)
 
But how can $\mathbf b\notin\mathbf c^\perp$? Is it because the inner product may not be the dot product? Thanks for your time.
 
No, no, they told you that $\langle b,c\rangle \ne 0$ !!!
 
I tried that, but at the end, I am getting something like $ (D - 2.5)^{-1} x$, which normally doesn't happens witht he problem that I have solved yet. The problem is -2.5 part.
 
@Ajay: I think it's a lot weirder than that, because the characteristic equation has complex roots.
But you should go ask your prof.
 
Okay.
 
1:48 AM
Night, all.
 
Good night!
 
2:24 AM
Hey guys, to prove the following statement, "The number $a^2$ is even if an only if $a$ is even", can I use a direct proof strategy to prove one way, and then the contrapositive of this statement to prove the other way?
 
2:57 AM
The contrapositive of “if P, then Q” is “if not Q, then not P”. It is not equivalent to “P only if Q” @Abwatts
“If a is even then a^2 is even” is equivalent to “If a^2 is not even then a is not even.” “If a^2 is even then a is even” is equivalent to “if a is not even then a^2 is not even.” But the first two statements are not equivalent to the latter two. For the iff , all four need to be true.
 
3:41 AM
Is there a uncountable bounded closed subset of irrational numbers?
 
 
2 hours later…
5:22 AM
0
Q: Tuples with 4 consecutive integers from 1 to 50

maths studentHow many ways are there to pick an ordered tuple of 5 elements from the set of positive integers between 1 and 50 inclusive, provided that there must be at least one consecutive sequence of 4 consecutive elements? So what I think is somehow we need to use inclusion exclusion principle ; So I pro...

 
 
2 hours later…
6:58 AM
@SayanChattopadhyay yes
 
7:28 AM
0
Q: Gcd of two number divides each other

maths student(a) Prove that $\operatorname{gcd}(a, b) | \operatorname{gcd}\left(3 a+b, a^{3}\right)$ Since say $d = gcd(a,b)$ then $d|a$ and $d|b$ this will imply $d|(3a+b)$ and $d|a^3$ and therfore $d$ is divisor of those two.But how to show it is greatest common divisor.

 
 
1 hour later…
8:45 AM
Morning guys, when defining "maximal" Hilbert space my textbook says, "the idea behind maximality is that we can obtain $\mathfrak H$ as a limit of a finite-dimensional projections, i.e. $$\mathfrak H= \bigoplus_{k\in\mathbb N} \{e_k\}$$"
I am not familiar with this notation
what does $\{\cdot\}$ mean in this context?
could it be the span?
 
9:03 AM
0
Q: Divisibility if a is odd.

maths studentSuppose that $a |(4 b+5 c)$ and $a |(2 b+2 c) .$ Prove that if $a$ is odd, then $a | b$ and $a | c$ So since $a |(2 b+2 c)$ this imply $a |(4 b+4 c)$ so $a |(4 b+5 c)$ after subtraction of two dividend we directly get $a | c$ but I am unable to get why it should divide $b$ if $a$ is odd.

 
 
2 hours later…
10:40 AM
Hey! So I have a question but I'm not sure if it's more computer sciencey or mathematical... it has to do with a moving circle on a grid, and the question is about how many points are in that circle
 
@Mathein just saw an absolutely vile proof of quadratic reciprocity
 
@ÍgjøgnumMeg oh wow
 
hahaha
 
what was the proof?
btw, it seems that Rösner got a room for the exercise session tomorrow
 
I‘ll write the idea when I get to my laptop, it was just.. technical af and very boring hahaha
oh nice :)
 
10:52 AM
I don't see the point of bothering with a non-satisfying elementary proof of quadratic reciprocity when there are very satisfying conceptual proofs with ANT
 
Agreed
Agreed
 
@ÍgjøgnumMeg actually we saw that $\theta(\frac{1}{x})=\sqrt{x}\theta(x)$ in the L-functions lecture. That can actually be used as a basis for a proof of QR, too
 
orly, that's nice
I actually really enjoyed the L-functions lecture but he was super fast and my analysis is kinda weak so I'll have to work super hard for it lol
 
you'll be fine
it's a good exercise
 
anyway the point of the proof Vogel just went through was as follows; use Gauß Lemma that says that $\left(\frac{a}{p} \right)= \varepsilon_1 \cdots \varepsilon_{\frac{p-1}{2}}$ where the $\varepsilon_i \in \{\pm 1\}$
ang on
test $\lfloor \rfloor$
okay
so one shows that $\left( \frac{a}{p} \right) = (-1)^{\sum_{i=1}^{p-1/2}\lfloor \frac{2ai}{p}\rfloor}$
this leads you to a proof of the 2nd Ergänzungssatz
then because $\left(\frac{2a}{p}\right) = \left( \frac{2}{p} \right)\left(\frac{a}{p}\right) = \left(\frac{2}{p}\right)(-1)^{\sum_{i=1}^{(p-1)/2}\lfloor \frac{ai}{p}\rfloor}$
and then cancel the 2/p and get some junk for a/p idk
wrote it wrong I think but anyway
Take p,q not equal primes
 
11:08 AM
the nice thing about the proof by the theta inversion formula is that it generalizes to arbitrary number fields!, I haven't looked into this, but I saw it mentioned somewhere
 
and define $M:=\lbrace 1, \cdots, \frac{p-1}{2 }\rbrace \times \lbrace 1, \cdots, \frac{q-1}{2}\rbrace$ and then $\ell_1 := \lvert\lbrace (i,j) \in M : qi > pj\rbrace\rvert$ and $\ell_2 := \lvert\lbrace (i,j) \in M : qi < pj\rbrace\rvert$. Then $qi > pj \iff j < qi/p \iff j \leq \lfloor \frac{qi}{p}\rfloor$
ergh
do it like that lol
and then also something analog for i
and then you get $\ell_1$ is the sum over these floors and $\ell_2$ is the sum over the other floors and then you get that $\ell_1 + \ell_2$ is exactly the number of elements in $M$, which is $\frac{p-1}{2}\frac{q-1}{2}$ so that $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}$
and then move the q/p to the other side...
hahahaha
I was just sitting there thinking "why are we bothering with this"
der hot an Vogel!
hehe
@Mathein I'll try to fill my notes from yesterday with some actual information about the various kinds of L-functions Rösner talked about, I didn't get down all the relevant information about all the different kinds because he speaks so fast hahahaha
 
11:44 AM
@MatheinBoulomenos I like the proof using cyclotomic field
 
12:24 PM
I like the proof using the Albert–Brauer–Hasse–Noether theorem
The Albert-Brauer-Hasse-Noether theorem says that for any global field $K$ with sets of places $\Sigma$, there's a short exact sequence $0 \to \mathrm{Br}(K) \to \bigoplus_{v \in \Sigma} \mathrm{Br}(K_v) \to \Bbb Q/\Bbb Z \to 0$. That splits non-canonically.
By LCFT we have a canonical isomorphism, called invariant map $\mathrm{Br}(K_v)=\Bbb Q/\Bbb Z$ if $v$ is non-archimedean and $\mathrm{Br}(K_v)=(\frac{1}{2}\Bbb Z)/\Bbb Z$ if $v$ is real and obviously $\mathrm{Br}(K_v)=0$ if $v$ is complex. The map $\bigoplus_{v \in \Sigma} K_v \to \Bbb Q/\Bbb Z$ is given by summing up those local invari
 
12:47 PM
cool
 
monka
 
1:10 PM
 
Can someone help me
 
How to quit mathematics successfully if you are a male:
1. Raise your testosterone levels.
 
How can I compute $$/sum \binomial{14}{k}$$
 
1:31 PM
Hey @Mathei @ÍgjøgnumMeg
Do you know a good reference to read about derived functors and Kan extensions?
 
1:47 PM
When you compactify $\Bbb R^n,$ is there a larger ambient space it sits in?
for example, if the map is $f:\Bbb R^n \to S^n$ what larger space does $S^n$ embed into
 
@Alessandro hm, for Kan extensions, you could look at Riehl's "Category in Context" http://www.math.jhu.edu/~eriehl/context.pdf
for derived functors, any text on homological algebra will talk about those
 
@Ultradark $\Bbb RP^n$ is the compactification of $\Bbb R^n$. You can take a colimit over inclusions of $\Bbb RP^n$ into $\Bbb RP^{n+1}$ and obtain $\Bbb RP^\infty$, which is the classifying space of $O(1)$-principal bundles
 
so for example Rotman "An introduction to homological algebra" or the book of the same name bei Weibel
 
@GaloisintheField okay, and if you want the compactification to be conformal, is it still $\Bbb RP^n?$
 
1:54 PM
@Ultradark I don't know what a conformal compactification is, so no idea
 
@GaloisintheField oh okay, do you know how a stereographic projection relates to a compactification?
I know that stereographic projections are always conformal, and that any closed connected manifold is a compactification of $\Bbb R^n$
 
hi, chat! it’s been a long time!
came here with a question i think is not worth for math.se
 
@Ultradark Stereographic projection is used to define charts on the Riemann sphere for example
@Ultradark It's just used for giving the topological spaces you get from compactification nice charts, and you can verify they chart transitions are smooth
 
2:10 PM
let $V$ be a real vector space, $||v|| = d(v,v)$ be a norm with the parallelogram rule, i.e., $||u-v||^2 + ||u+v||^2 = ||u||^2 + ||v||^2$. prove that $<u,v> := \frac{1}{4}( ||u+v||^2 - ||u-v||^2 )$ is an inner product over V.
(hermitian) symmetry and positivity are obvious. linearity on the first component is not.
 
If you took the stereographic projection of $S^2$, but only the portion that maps to the unit disk, and looked at its chart, and then identified the points on the boundary of the disk, would the resulting shape, $S^2$ be an isometry of the sphere you started with
 
2:57 PM
What's the unit for Gaussian curvature? 1/meter^2?
I guess, yeah, 'cause it's the product of two planar curvatures, and those are each one over the radius of an oscillating circle
(and also when we integrate it over the surface the units cancel)
Thinking about units for stuff
Like, did you know that both "kinematic viscosity" and "thermal diffusivity" are measured in units of m^2/s
(whatever those are)
We all know that m^2 is area
Is it possible m^2 is something else as well?
I guess you can think of it as "volume over length"
like if object A has a volume and object B has a length, what's the ratio of their sizes
 
3:13 PM
@TedShifrin Hi.Can you help me with a question
 
What sort of answer would one possibly expect to "Let $D$ be a divisor with $deg(D)=-2$. Describe all meromorphic 1-forms $\omega$ such that $(\omega)=D$."
Is there a nice description given by the one relation we have (coming from the change of charts) for arbitrary rational functions?
Oh, I forgot to say, that was explicitly for CP^1
Perhaps I can deal with a local description where $\omega$ is $f$ on $U_0$ and $g$ on $U_1$, and deduce that
$$a\ne 0, ord_{[a:b]}\omega = ord_{b/a}f,\qquad a=0, ord_{[0:1]}\omega = ord_{\infty}f - 2$$

although I think I am being unrigourous in some sense, since $f:\Bbb C\to \Bbb CP^1$ rather than $f:\Bbb CP^1\to\Bbb CP^1$, meaning the order at $\infty$ may make no sense
 
3:54 PM
@Aladdin What's up?
@Galois: What is $\deg(K)$ on $\Bbb P^1$?
 
Right, so $D$ is a canonical divisor.
What's the dimension of the space of all such meromorphic $1$-forms?
 
$$\iiint_D \z^2 dxdyz$$
 
Huh? @Aladdin ... You saw the answer to that ridiculous spherical coordinates question?
 
yes
 
3:59 PM
I would never in a million years assign such a yucky question. Anyhow, what is your question?
 
but shouldn't you have put z=$rcos(/theta)$ there?
 
Where?
What are we talking about?
 
1
Q: Triple integral in different coordinate systems.

HedgehogMy task is to write volume integral in 3 coordinate systems : Cartesian, cylindrical and spherical. This integral shows volume of intersection of 2 spheres, first with center at $(0, 0,-3)$ and radius $5$ and second with center at $(0,0, 3)$ and radius $\sqrt{13}$.I am able to do it for Cartesian...

 
So that OP was using an unusual angle $\psi$, not the usual $\theta$ or $\phi$.
But you just wrote down $\iiint_D z^2\,dx\,dy\,dz$???
 
ah it's another question
 
4:02 PM
All right. Focus. Let's be clear what we're talking about/asking.
 
Well, apparently $\Omega(D)=\{\omega\in M\Omega^1(\Bbb CP^1)| (\omega)\geq D, \text{ or }\omega = 0\}$ is finite dimensional, and perhaps this is equal to my space, since any differential $1$-form must have degree $-2$, so this perhaps this already equals
$\Omega(D)=\{\omega\in M\Omega^1(\Bbb CP^1)| (\omega)= D, \text{ or }\omega = 0\}$
 
for that integral,I am supposed to find volume between $x^2$+$y^2$=$a^2$ and praboloid $x^2$+$y^2$=z and plane z=0
so i tried doing this in cylindrical coordinates
 
OK, that's a standard question.
 
In any case I can partially deduce the dimension by Riemann Roch as $dim(\Omega(D)) = dim(L(D-K))$
 
@Galois, aha, if you know RR this is totally straightforward.
 
4:04 PM
Well then I'm missing it haha
 
what's $\dim(L(D))$?
Yes, @Aladdin, cylindrical coordinates is correct.
 
i took $\theta$ from (0,$\pi),r from (0,a) and z from (0,a)
 
No, you're making the most basic mistakes.
 
@TedShifrin As a $\Bbb C$-vector space?
 
yes, @Galois
@Aladdin: Why $\pi$?
 
4:06 PM
it should be $\2pi$
i think
 
$2\pi$, of course.
Now if you fix a point inside the circle $x^2+y^2\le a^2$, where does $z$ start and where does $z$ end? Reread the problem.
 
z is from (0,$a^2$)?
 
No.
 
why...they intersect at z=$a^2$?
 
Go back and review how you set up double/triple integrals. You fix the outer variable at some random value and ask how the inner variable varies, depending on the value of the outer variable.
You should watch some of my YouTube lectures on multiple integrals.
 
4:08 PM
ok
 
Hi demonic @Alessandro.
 
Evening all
and/or morning
Hi @Ted and @Alessandro
 
Hi @ÍgjøgnumMeg
 
Hi @ÍgjøgnumMeg
 
4:12 PM
Neat thing
 
Oh, hi, DogAteMy.
 
Take a square of area one
Add layers like in the picture so that each layer has area 1
The shape approaches a circle
 
I'm not sure I understand what a layer is.
Oh, the union of the things of the same color.
 
So like the brown is two rectangles of area 1/2, the yellow is 3/8+1/4+3/8, etc
Yeah
 
There are some rules about how you can form the next layer, I guess.
 
4:15 PM
All but the other two rectangles are determined by the shape of the last one
and the area constraint gives you the outer two layers
 
Right, so you're required to use "shapes" from before.
 
Note that $\dfrac38=\dfrac12\dfrac34$
This ends up being related to the Wallis product
 
@TedShifrin Is it 3?
 
You can prove Wallis from this I mean
 
Nope. @Galois
What do you know, in general, about $L(D)$ when $\deg(D)<0$?
 
4:17 PM
Sanity check: I have $\int_0^1 u^{s/2}\omega(u)\frac{du}{u}$ and I substitute $u := \frac{1}{x}$. Then I get $-\int_{1}^\infty x^{-s/2}\omega(\frac{1}{x})\frac{dx}{x}$... correct? (the $\omega(\frac{1}{x})$ factor doesn't generate a $-1$ so the sign should be correct)
 
Yeah, @ÍgjøgnumMeg, that's right.
Oh, wait.
 
@TedShifrin Well maybe the convention is swapped. In my convention $L(D)=0$ when $deg(D)>0$ and $L(D)=\Bbb C$ when $deg(D)=0$
 
The sign is wrong.
 
Hmm why is the sign wrong
 
4:18 PM
$dx/x = - du/u$.
 
hi @TedShifrin @AkivaWeinberger @everyone
 
@Galois, huh? How do you write RR, then?
 
The next layer works out to $\dfrac5{16}+\dfrac3{16}+\dfrac3{16}+\dfrac5{16}$
 
hi @Lucas
 
@LucasHenrique Hi
and $\dfrac5{16}=\dfrac12\dfrac34\dfrac56$
 
4:19 PM
@TedShifrin $l(-D)=deg(D)-g+1+dim(\Omega(D))=deg(D)-g+1+dim(L(D-K))$
 
oh, crazy convention. So what's $l(-D)$?
 
Also, $1+\dfrac12+\dfrac38+\dfrac5{16}=\dfrac{35}{16}$
 
@TedShifrin 0
 
${}=\dfrac32\dfrac54\dfrac76$
 
In all the times I've studied and taught Riemann surfaces, I've never seen this backwards convention, @Galois.
OK, so what does RR tell you, then?
 
4:21 PM
@Ted ah yeah, thanks
 
which is the radius after four layers
 
Yeah, it's somewhat disturbing :').

RR tells me that dim $dim\Omega(D)=1$
So I guess I just have to find one $\Bbb C$-generator
 
Right, it's a $1$-dimensional vector space.
Think about taking a well-known meromorphic $1$-form on $\Bbb P^1$ and "dividing" yours by it. You'll get a well-defined meromorphic function.
 
@TedShifrin So the most well-known meromorphic $1$-form is probably the one corresponding to $dz = -1/w^2 dw$, i.e. with $1$ on $\phi_0(U_0)$ and $-1/w^2$ on $\phi_1(U_1)$
 
OK, so the meromorphic $1$-form $dz$.
 
4:34 PM
And I take an arbitrary divisor $D$ with $deg(D)=-2$, and we've deduced that it has $dim(\Omega(D))=1$. So maybe I pick a concrete form for $D$ like:
$$D= \sum_{i=1}^n m_i\cdot p_i, \sum m_i = -2$$
 
So I'm suggesting you look at your $\omega$ and take $\omega/dz$. Why is this a well-defined meromorphic function?
 
@TedShifrin It should be well defined because the 'transition relation' will cancel out, moving between the two charts
 
Yes.
 
Since you guys are discussing differential-things: I’ve learned that the Riemann integral over a surface D. The formal definition uses a limit quantifying over arbitrary points of of all partitions of D. I can’t really see how the choice is irrelevant, but I can see this probably relates to analysis (in $\mathh{R}^n), so I skipped it. Should I be worried?
 
No, actually, surface integrals are very subtle. If you don't approximate correctly, you can get very wrong answers. There's a very famous example. It appears in Spivak's Calculus on Manifolds and (stolen) also in my multivariable math book.
The example is originally due to Schwarz.
It's giving a (finite) right circular cylinder with apparently infinite surface area.
 
4:44 PM
@AkivaWeinberger Where does this "square-circle" construction come from? Or is that your own invention?
 
So $D - (dz) = (f)$ for some meromorphic $f$ and hence $D=f dz$ in which case $D=(dfz)$? Well no that's not right...
 
@MartinSleziak Donald Knuth I think
2
 
It seems that everything about derivatives is easier than those about integrals... the whole idea of approximating things in calculus makes me anxious, lol. Studying calculus without analysis is kinda harrowing.
 
Differentiation is a science, integration an art
 
Er no I got it from him but he got it from somewhere else
 
4:46 PM
@Galois: Your meromorphic $1$-form looks like $f(z)\,dz$ (in the main chart), so I'm asking you to give an explicit formula for $f(z)$ given that $(\omega) = D$. You already picked a specific representative for $D$. Let's just suppose $D=-a-b$ ($a,b\in\Bbb C$).
 
@LucasHenrique @ÍgjøgnumMeg Algebraically, differentiation is easier than integration. Numerically, it's the opposite
Also
 
@Lucas: If you're doing multivariable, now might be the time to "enjoy" some of my videos (more proofs than your course, perhaps).
a few things missing on the right, DogAteMy :D
 
@TedShifrin I’ll take a look, thanks.
Oh. I have an accomplishment I’d like to share: I’ve started a scientific research!
I’ll study Galois theory.
 
@TedShifrin $f(z)=z^2/(z-a)(z-b)$?
 
@AkivaWeinberger After you suggested Knuth, I tried to Google a bit. In this blog post there are links to the paper An Elementary Proof of the Wallis Product Formula for pi by Johan Wästlund. It also links to video of Stanford Lecture: Donald Knuth—"Why Pi?"(2010).
 
4:53 PM
@Galois: No, why the $z^2$?
 
@TedShifrin Where I think the $z^2$ is necessary, else otherwise we would have $D= -a-b - 2\cdot\infty$ coming from the other chart
 
@Lucas: Studying a well-established field is "scientific research"?
 
Where unless I'm wrong, $ord_{[z_0:z_1]}\omega = ord_{z_1/z_0}f$ for $z_0\ne 0$ and $ord_{[0:1]}\omega = ord_\infty f -2$ (which may not make sense, since $f$ is not defined at $\infty$ hmm)
 
@Galois: Better double-check that.
Just write down $dz/(z-a)(z-b)$ in the chart at infinity and look at it.
 
4:56 PM
You're too fond of formulas for me :P
 

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