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12:02 AM
@TedShifrin When you say uniformly distributed, does that mean all placements of $B$ have to end up with the same $\theta$ measure?
 
no, no, I mean try doing $\theta = 2\pi k/n$, $k=1,\dots,n-1$ for the location of $B$. And then for each one, what's the likelihood of a $C$ that works? Try to work it out.
I'll be back later ...
 
Bob
12:18 AM
Could somebody here look at my post: math.stackexchange.com/questions/2955619/…
 
@anakhro What is the statement of the lemma?
 
@MikeMiller the one which symplectic geometry simplifies?
 
yea
 
12:34 AM
If $M\subseteq\mathbb C^n$ is a non-singular affine algebraic variety in complex n-space of dimension $2k$, then $H_i(M;\mathbb Z) = 0$ for any $i>k$.
7.1 in Milnor.
@MikeMiller
 
Bob
Is math taught well in this country?
 
@Bob what country is "this country"?
 
Bob
The United States
 
Never been taught math in the USA, can't say, sorry.
 
12:44 AM
Being a current high school student, I think there is a large spectrum of what is considered "good teaching" of mathematics.
 
generally it sucks
 
Bob
I have a feeling US students are falling behind the rest of the world in math
 
we have bigger problems than sucking at math
 
Some of my teachers make a big effort to explain why things work and why we need them. Others just hand out cheat sheets and say "memorize this for test".
@EricSilva True.
 
Bob
my impression is that high school math lacks rigor today. That is, the concept of a proof is not taught much.
 
12:47 AM
@Bob 100% true. I had to teach myself how to do proofs because the calculus teacher said they were a college concept.
 
Bob
well parents want their kids to get to calculus
 
idt wanting to show high schoolers proofs is the right hill to die on
 
Bob
and therefore things like proofs get dropped
right hill to die on?
 
the actual problems start way earlier and are way more boring and less sexy than "proof based math"
@Bob it's an expression, it means it's just not the right issue to focus on
 
@Bob I didn't know that calculus was a big deal until I reached it.
It's not as hard as people make it out to be.
Unless you're bad at algebra.
 
Bob
12:51 AM
one thing that makes Calculus hard is the integration
often how to perform a particular integration is not clear and you may have to try several methods
 
I'm going to practice like crazy before the AP exam.
 
Bob
one of the things I wonder about is if you pass the AP exam and then take Calc III
will you be ready for Calc III
I know somebody who did that and when he got to Calc III he was not ready for it
 
@Bob the issue is not the lack of proofs.
The issue is that what they are teaching simply isn't math.
And this isn't just an issue for the USA. It's world-wide.
 
@Bob I'm going to self-study the concepts for the class before I get there so that it will be second exposure for things I don't understand. That's what I did for the calculus class I'm in now.
 
And I think it's okay to some extent. Not everyone has to learn mathematics. Just like you can't expect everyone to learn to play an instrument or learn to draw well.
 
Bob
12:56 AM
sounds like a good plan
sounds like a good plan CaptainAmerica16
and with that
I am going to sign off
have a nice evening
 
Peace out.
@anakhro What do you mean by that?
 
By what?
 
When you say what they are teaching "simply isn't math"
 
Somewhat along these lines.
 
Oh, lol. I didn't realize.
 
1:00 AM
What they teach is a bunch of symbol pushing, and largely unmotivated rules.
And I believe that doesn't seem like math at all.
 
I can say I definitely didn't get the point of algebra until I learned proofs. Now I really like math.
 
But it's O.K. they teach it this way because I know if they taught it the way that I'd want them to, then there would likely be serious deficiencies in "symbol pushing skills" in engineers, scientists, social scientists, etc.
 
at least in the US the problem w math education is like way more complicated than this
 
@EricSilva Care to expand?
 
@EricSilva there are many problems with math education. There are many problems with education in general.
This is just one of the key distinctions I like to make when it comes to math education.
 
1:05 AM
i mean the problems with us math education are mostly just instances of the broad political problems w education in the us in general
 
@EricSilva Ah...I see. I still not well-versed in all that stuff because I only just realized it was an issue. lol.
 
why would good teachers stick around if we treat teachers like garbage and dont pay them?
the answer is that they dont really stick around
 
@CaptainAmerica16 well it's only subjectively an issue.
For many people there are no issues.
 
"it wasnt a problem for me" $\neq$ "it isnt a problem"
it isnt "subjectively" an issue
 
No, it really is.
What one person calls a problem, another calls a feature.
 
1:14 AM
you clearly do not know what you are talking about
 
I don't think that's up to you to decide, heh. :P
In your subjective opinion, I don't know what I am talking about. Otherwise, it's quite a benign stance to take.
 
 
2 hours later…
3:29 AM
Let $D \subseteq \Bbb C$ be open. Then $D$ is simply-connected iff $H^1_{dR}(D) = 0$.
 
3:52 AM
Hi guys, taking a real analysis course; is it sufficient to use induction to prove that something holds for all natural numbers in order to prove that it is true as n goes to infinity?
 
checking my memory: Ray Z, right?
that's definitely true but if they ask you to show it as $n \to \infty$, I bet it's not true for all finite $n$
 
1/n>0 for any finite n, but 1/n ->0 as n->infty
so what's true for finite n needn't be true in the limit n->infty
 
@MikeMiller yup :) taking 131AH
I see... I'm trying to prove that $lim_{N \to \infty}sup \{a_n:n\geq N\} \geq lim_{N \to \infty}inf \{a_n:n\geq N\}$ for some sequence $\{a_n\}_{n \geq 0}$
 
in this case you can indeed reduce to the finite case because it is $\le$ instead of $<$. But this is a non-trivial theorem that you should first formulate and then prove.
 
4:11 AM
I'm a bit confused; why is $\leq$ vs. $<$ a significant difference?
I wanted to give an argument that's similar to:
denote $ x_N := sup{a_n : n \geq N} $ and $ y_N := inf{a_n : n \geq N} $, and give a proof like the following:
$x_N \geq y_N$ for all $N$
or, use the property of sets and just claim that the infimum of a set and the supremum of a set will always be the upper/lower bounds and so the $sup \geq inf$ is true in the limit (but ofc in a more formal way than what I just described)
 
4:36 AM
because $[0,\infty)$ is closed and $(0,\infty)$ is not.
 
@MaryStar Sorry I was asleep, I saw that JoeShmo got you covered and you also know you can prove the contrapositive if you want something more than a conterexample
 
@Secret how do hell do you do proof by contrapositive to a false statement? Counterexample is great and their idea for how to construct the counterexample is also great
 
@LeakyNun which $\leq$ were you referring to?
 
38 mins ago, by Leaky Nun
in this case you can indeed reduce to the finite case because it is $\le$ instead of $<$. But this is a non-trivial theorem that you should first formulate and then prove.
do the former first.
what does it mean that "you can reduce to the finite case"? What is the general theorem?
 
@LeakyNun I guess my question wasn't clear; I meant in that quote, "reduce to the finite case because it is ≤ instead of <.", what you meant by "≤ instead of <". As in which inequality you were talking about
$n \geq N$? or $limsup \geq liminf$?
 
4:50 AM
@Holo Ah, clearly, my brain is not functioning, I forgot that the starting implication is false thus its contrapositive is also false. Sorry about the confusion
 
@OneRaynyDay limsup >= liminf
 
(Unrelated)
Soemtimes however, if I want to really understand something, I am not satisfied on just getting one or a few counterexample
Jan 10 '17 at 7:54, by Secret
I also not just satisfied at finding a counterexample, I will try (if it is feasible) to find the whole set of counterexamples, thus finding counterexamples are really finding a set of examples obeying a proposition that is required for a given counterexample
Jan 10 '17 at 7:55, by Secret
That is, I like to understand what makes a pathological thing pathlogical, what makes an unatural structure unatural and so on
 
@Secret lol, it is okay.
You just like weird things, this is why you like pathological things
 
@LeakyNun Gotcha, well, can you prove by induction on finite n? as in, for some finite $N$, $sup \{ a_n:n\geq N\} \geq inf \{ a_n:n\geq N\}$?
I've already done that
 
you mean for all finite $N$.
you still haven't answered my question:
11 mins ago, by Leaky Nun
what does it mean that "you can reduce to the finite case"? What is the general theorem?
 
4:58 AM
It will be interesting how to "fail prove" the false proposition:
> $M\cap N$ is finite $\implies$ $M$ and $N$ are finite sets
So that at the end of this, we get the proposition that describes all the counterexample that make this statement false
 
Well, it's not immediately clear what you wanted, but I could reduce to the finite case for finite $N$, or a finite set $\{a_n\}_{n \in [0,M]}$
 
Here's an example on what I mean by "fail proof"
3
A: Ideas of finding counterexamples?

SecretMy favorite approach in finding counterexamples whenever I am not familiar enough with the mathematical object $M$ is a top down approach. Start with the most general example of $M$, and the find the condition that all counterexamples of M has to violate, then pick one concrete counterexample fr...

I use this technique a lot when constructing highly pathological objects
 
the general theorem that the question asks to prove is $limsup \geq liminf$ for all sequences
 
@OneRaynyDay I mean, why can you reduce to the finite case? What would be the general theorem that allows you to do so?
 
@LeakyNun Not sure, fyi my background is in CS, not in math, so I either don't know which one you meant, or I just don't know it.
 
5:01 AM
then why are you studying real analysis?
 
@LeakyNun ...? I am taking this class because it doesn't have any pre-requisites? This is irrelevant to the question
 
the answer to my question is that the general theorem would be that if $x_n$ and $y_n$ are two convergent sequences such that $x_n \le y_n$ for all $n$, then $\lim_{n\to\infty} x_n \le \lim_{n\to\infty} y_n$.
 
@Secret I see, but it is long and people are generally lazy :\
 
@LeakyNun I didn't go over this theorem in class, but I gotcha
 
...
 
5:03 AM
Btw @Secret, do we have a proof that amorphous sets don't have linear order?
 
I was expecting you to come up with the statement yourself despite not having gone over that theorem
because that's what "you can reduce to the finite case" means
formulating ideas rigorously is an important skill in maths
but since you said your background is in CS, I gave you the answer
 
(I'm still having issues formulating ideas rigorously myself. Half the time when Leaky prompts me to formulate something I just start drawing a blank)
 
@Rithaniel I forgot what you asked me before
could you remind me lol
 
It's not that I didn't think about what you asked me, it's more like I had no idea what you wanted me to think about
but sure, I get it, that's a good intermediate step
that theorem doesn't require induction to prove, but rather just by the definition of convergence using $\epsilon$
 
If it was anything, it would have been a homework problem in topology. (It would have been a while ago, too)
 
5:07 AM
I see
@OneRaynyDay :o you know how to prove the theorem?
by the way are you taking the class for fun or is it for your degree?
(they aren't mutually exclusive)
 
@LeakyNun for fun; and let me give it a try
 
I've not taken any lambda calculus courses, but that's a big topic in CS. Maybe real analysis might help in that topic?
 
5
A: Construction of amorphous subset of $\Bbb R$

Noah SchweberThere is no amorphous set of reals - in fact, no amorphous set can be linearly ordered. (For further discussion of what kind of structure amorphous sets, and more generally Dedekind-finite sets can have, see this paper of Truss or Agatha Walczak-Typke's Ph.D. thesis.) To see this, suppose $A$ i...

 
I think lambda calculus is too discrete for real analysis
 
I am not ready on amorphous sets yet, the chemistry keep pulling me around so I don't have much time recently to deal with set theory stuff which is why I am mostly lurking in the chat
 
5:10 AM
Is that any relation between normal in group theory and normal operator in linear algebra?
 
Suppose $x_n \to x$ and $y_n \to y$, and for the sake of contradiction, $y < x$. Then by definition, there exists $\epsilon > 0$ such that $|x_n - x| < \epsilon$ for all $n \geq N_{\epsilon}$, and similarly for y
 
I planned to do a very thorough discussion on all classes of finite and infinite sets once I stabilise my PhD again
and that will involve going through all the proofs from the most elementary to the most complicated
 
I just found that once the word normal is used then something can commute
 
you can then pick $\epsilon := \frac{x-y}{2} > 0$
and then pick $N_{\epsilon} = max(N^x_{\epsilon}, N^y_{\epsilon})$
where $N^x_{\epsilon}$ denotes the N such that all $x_n$ is within $\epsilon$ of x, and similarly for y
then you can pick any element past $N_{\epsilon}$ in x and y that contradicts the original statement, and so we proved that it must be $\leq$ and not $>$
is that sufficient?
 
brilliant
 
5:12 AM
amorphous sets proofs are interesting because it helps to teach us how to reason about partitions at a very fundamental level without the considerations in nice orderings
 
In group theory the coset commute $aH=Ha$, where H is normal subgroup, and in linear algebra $AA^*=A^*A$, A and its adjoint commute
 
Cool :) thanks
 
just as how d finite sets can teach us how to handle bijections carefully
 
@IsanaYashiro I believe you just answered your own question
@AkivaWeinberger buenos dias, perro comó mi
 
@LeakyNun really?
 
5:14 AM
@IsanaYashiro I think it's just something commuting
 
But why does something commute then it's called normal
I know little more of that on linear algebra but I just on the way of group thoery
 
@Secret I see
 
in linear algebra if an operator is normal then its all eigenspaces are actually normal, i.e. perpendicular, to each other
(hope this is correct)
got this from spectrum theorem
but is this apply to group theory?
 
The terminology "normal" is ubiquitous in mathematics. The only commonality I can find is that the objects involved does not in a way "depends on each other"
 
Cool, sounds so perpendicular
sometimes the silence make me think of whether I were banned by everyone.
 
5:26 AM
I have self-conscious thoughts like that too, sometimes.
(dangit, I'm on the last three quizzes to grade and my pen runs out of ink)
 
I can easily check whether someone has ignored me by probing the continuity of the chat flow
I can also do that check between A and B by checking whether their conversation flows
It's a bit hard for people to apply this onto me because I do have instances where I generate a series of monologues that is directed to no particular people
though this is a lot less common now because the chemistry will often pull my soul elsewhere
 
Well, just watching conversation flow doesn't always paint the whole picture. For example, I have been in chats before where I just decided to change the topic I was talking about at the moment solely because another thought occurred to me.
 
5:43 AM
Why is -1 the residue of 1/z at infinity despite it having a removable singularity there?
 
@LeakyNun Why is the integral of $1/z$ on $[1,\infty]$ infinite despite being defined everywhere on that compact interval?
(Defining the value at $\infty$ to be $0$)
Normally the integral of $f(z)$ on $[a,b]$, where $f$ is finite everywhere on that interval, is finite.
 
wat
what's the intuition of the residue at infinity?
 
You don't have an antiderivative in a neighborhood of infinity.
 
I see
 
$\int_a^b \frac{1}{z}dz = \ln (b) - \ln (a)$
hmm...
 
5:54 AM
If, in a neighborhood of $\infty$, you could integrate it over every interval ending at $\infty$, then you would have an antiderivative
and the residue would be zero
 
the 'invariant' statement of the residue theorem is that the sum of zeroes - poles of a meromorphic 1-form is the euler characteristic.
 
Example: $\int_1^\infty1/z^2\ dz$ is finite, and the residue at infinity is zero
 
@MikeMiller waaaaat
 
figure out why
 
Oh, wow. Topollology
 
5:56 AM
wait, is that the same as $\sum_{p} \operatorname{ord}(f,p)$?
@MikeMiller for any complex manifold?
 
closed riemann surface
maybe poles minus zeroes, signs are dumb
 
does closed mean compact?
 
a closed manifold is a compact manifold without boundary
 
what do you mean by the sum of zeroes - poles?
anyway on $\Bbb C$, $1$ has no zeroes nor poles, but $\frac1z$ has no zeroes and one pole
oh wait, compact
nope, I have no idea how to prove such a statement.
 
sorry, the poles - zeroes counted with multiplicity is equal to the euler characteristic and the sum of the residues at the poles is zero.
assuming they're isolated, which is true as long as it's nonzero. $d(1) = 0$ identically.
 
6:06 AM
@MikeMiller So not $\Bbb C\simeq\Bbb R^2$.
 
but yes $\Bbb{CP}^1$
 
Ah, I'm confusing "closed" with "complete"
 
anyway this should be enough keywords to do the rest of the googling
 
is this related to de Rham cohomology and homology theories and the fact that they have the same rank
 
to get the Google ball rolling
 
6:08 AM
I thought you told us to "figure out why"
 
to get the google boogle roogle
 
6:20 AM
I feel like degree might kick in here
Though we don't know whether zeroes/poles are regular
(Unless multiplicity is what handles that)
 
6:33 AM
from the pov of AG this is because the degree of the canonical line bundle is 2g-2 (which one shows using Riemann-Roch)
 
6:55 AM
Does anyone know of any online writing app (something like pasteOfcode for programming codes) to write Latex math notations to be suitable for reading?
 
Overleaf?
 
Yeah I TeX all my psets in Overleaf
 
Same, especially since we are allowed to do them in small groups so having an online editor is super handy
 
Ah nice
 
Hmm, doesn't that only contain features for like writing books or personal projects? It's not really suitable for sharing on casual platforms like these with just the sending of a link of some sort?
 
7:06 AM
Hmm, I'm kinda wondering, who are some of the more active algebraic topologists floating around these days? (also would be interested in other folk)
I only know Peter May, Jacob Lurie, Mike Hopkins, and Mike Hill
 
I'm trying to help my 12th grader with a math problem. Given a real $a$ such that $a^3+a^2 = 1$ (hence in $(0,1)$), compute the infinite product $(1+a)(1+a^2)(1+a^4)(1+a^8)...$. I've been struggling for a while and am not seeing the light. Any obvious tricks?
 
@Daminark have u not asked peter
 
I mean this more or less has just come to mind now
Like, huh who's actually doing AT nowadays?
 
r u not interested in at for grad
 
It's one option for sure. Peter has told me about some places to look at (though his suggestions were fairly ambitious so I should prob ask for some places that are more safe-ish), though he hasn't mentioned too many specific names of folk except for Mike Hill at UCLA
 
7:17 AM
oh i think it’s normal to like
mention ppl u would be interested in working w when u write ur statements
but i guess ppl don’t always know what they wanna do exactly
 
Yeah I'm not sure exactly how specific I'll be, since I don't really know exactly what I'd like to do later on
 
i think a lot of apps have like a drop down menu where u put in which of their faculty u would be interested in working w
definitely its v flexible since as u said ur not sure exactly
 
AG is real nice so far and the vibe I get from the more advanced bits is that it intersects almost everything, AT (also possibly other types of topology) is something that I think I'd like, then NT/algebra/rep theory, tbh even stuff along the lines of functional analysis wouldn't be out of the picture
 
Hello!!

$ F $ is the set of all points in the plane that are at least as far from the origin as $ P = (3 \mid 0) $.

Do we get the set $\{(x,y)\mid x^2+y^2\leq (x-3)^2+y^2\}$ ? Or have I understood wrong the definition of $F$ ?
 
ah yes nerd math
 
7:29 AM
no u
 
I solved my problem...
 
Hi, sorry for the handwriting if it's not legible, my tutor tried proving part (c) by contrapositive approach and I don't understand what he's doing. Does anyone understand it?
 
8:01 AM
@Daminark did you get your GRE results yet? I realize it's a low priority, but still...
 
Yeah, I got my September score back
 
oh
 
790, which is a bit less than I was gunning for so I'm doing it again in a week
 
I see, best of luck!
 
Thanks fam
 
8:03 AM
:-)
 
8:36 AM
I thought number of zeroes - number of poles is still 0 for modular forms on $\Bbb C/\Bbb Z^2$
I can't come up with any example
 
Meromorphic functions have the same number of zeros as the number of poles on any compact Riemann surface yes
 
what?
 
but yeah I'm not sure if there are modular forms on $\mathbb{C}/\mathbb{Z}^2$
 
3 hours ago, by Mike Miller
the 'invariant' statement of the residue theorem is that the sum of zeroes - poles of a meromorphic 1-form is the euler characteristic.
3 hours ago, by Mike Miller
figure out why
 
1-form
 
8:42 AM
what is a 1-form?
 
something that looks like $fdz$
 
how is that different from just $f$
and do you have any example of that
 
when you take a different chart the transformation behaves differently
 
it's quite interesting how you can one-chart $\Bbb C/\Bbb Z$ but not $\Bbb C/\Bbb Z^2$
and that you can one-chart $C = \{(z,w) \in \Bbb C^2 \mid z^2+w^2 = 1\}$
 
e.g. take $\mathbb{P}^1$, with the usual charts (so the transition map sends $z\mapsto 1/z$ )

So if you consider the function $f(z) = z$ on $\mathbb{P}^1 = \mathbb{C} \cup \{\infty\}$ (so here implicitly I already took a chart) , if you look at this function on the other chart with coordinate $w$ it looks like $f(w) = \frac{1}{w}$

so you see there's one zero and one pole

On the other hand if you look at the meromorphic 1-form $zdz$, this has a zero at $0$. On the other chart, this looks like $(1/w) d(1/w) = -1/w^3 dw$, so there's a pole of order $3$ at $0$ on the other chart (in other wo
 
8:55 AM
@LeakyNun, how does this show that $\sum\frac{\sqrt{a_n}}n$ indeed converges? I think this is not sandwitched.
 
@Silent well but I am sandwiched
 
oh sorry
 
between your messages B)
 
hmm :)
 
@loch I thought the sphere has 0 betti number
oh wait nvm
@Silent sandwiched by 0 of course
 
9:02 AM
@LeakyNun I think right hand side is $\lim_ {n\to \infty}(\sum a_n)^{1/2}\cdot(\frac{\pi^2}{6})^{1/2}$
 
so?
 
so, it may be positive, so how is $\sum\frac{\sqrt{a_n}}n$ sandwitched?
 
you don't know what sandwich theorem is
 
@LeakyNun i see that $0\le \sum\frac{\sqrt{a_n}}n\le\biggl(\sum a_n\biggr)^{1/2}\biggl(\sum\frac1{n^2}\biggr)^{1/2}.$ holds and squeeze theorem says that limits of left and right i'parts' should be equal. But here that may not be satisfied
 
9:09 AM
oh! checkmate :) I will look into that. thank you
But, in your link they are talking about terms! and in my original link, he is talking about partial sums!
 
your original link?
there are two sandwich theorems
one for sequences and one for series
 
@LeakyNun no, no, I m applying series version of squeeze theorem to this, that is the link i was mentioning as original.
 
ah
@Silent it's ok because the partial sums are increasing and bounded
 
oh! thank you very much.
 
this feels like a generalization of the squeeze theorem for series
@Silent can you state the general theorem?
 
9:58 AM
If $X_1$, $X_2$, and $X_3$ are normal distributed with $N(5,2)$, and $X_4$, $X_5$, and $X_6$ are as well but with $N(4,4)$. Then, what is $P(X_1+X_2+X_3>X_4+X_5+X_6)$ if $X_1$, $\dots$, $X_6$ are independent?
Never mind! I solved it. :)
 
 
3 hours later…
1:23 PM
So every general conic is isomorphic to $\Bbb P^1(k)$? @loch
for any field $k$
where $2 \ne 0$
 
1:34 PM
"Without loss of generality we shall often assume that the homomorphism $v : F^\times \to \Bbb Z$ is surjective", the wlog just serves to remind the reader that it is always possible to normalise a discrete valuation right?
 
1:45 PM
@LeakyNun If you don't mind, how you have mastered abstract algebra? I have seen you to reply any kind of problems in abstract algebra very quickly.
 
@ÍgjøgnumMeg I would think so
@taritgoswami I've hardly mastered abstract algebra
and I don't know how to answer your question.
 
@LeakyNun Maybe your understanding is not satisfactory to you, it's pretty good though!
@LeakyNun I mean how you approached to learn Abstract algebra?
 
I first learnt group theory 1.5 years ago from A First Course in Abstract Algebra by John Fraleigh; then I learnt Galois theory 1 year ago from Ian Stewart; and then I learnt some commutative algebra 0.5 year ago from Atiyah-Macdonald; but of course the people here helped me a lot.
 
@LeakyNun Thanks for sharing
Actually I follow [this](https://math.stackexchange.com/a/2895315/579780)(read somewhere) as a definition of understanding, but, in abstract algebra all proofs seems to me as a combination of multiple ideas instead of a single idea. I don't know
if I have fully understood it.
 
is there some specific topic you would like to discuss?
 
1:57 PM
@Mathein hoi, komische frage (die nichts mit der Mathe zu tun hat) aber hast du je "Brükke" statt "Brücke" gesehen? lol
 
@LeakyNun I am reading it for first time. Can you share your understanding about group actions?
 
hmm
do you know modules?
or vector spaces?
 
groups
 
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