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9:00 PM
for example, my linear algebra book uses $\{a_1,a_2,\ldots,a_n\} = \{a_i\}_{i=1}^n$
which makes sense
so i'm complaining to you guys because the book's publisher isnt here
 
@Daminark "Your gminushimon is evolving"
 
@GFauxPas I mean the ordering is just furnished by the index, so you don't need to worry about that too much. I've seen both and neither causes a problem
 
but the ordering of terms changes the sum of a series in general
aannyway
 
@orbit-stabilizer Well, let's be particular. What does g - h vanish on, from definition?
 
How's the Biology going? @Daminark
 
9:02 PM
Ah I mean series is another story, but like, that's where you determine the order of the sum just by the index, so the notation for the sequence still seems to me as being somewhat immaterial
 
the proportion of ambiguous notation just increases the more hard math you do
6
 
Over $E$ @BalarkaSen
 
@skullpatrol I am now done with that class. We had 3 tests during the quarter and no final in final's week
 
So, basically the rationals in [0,1]
 
@orbit-stabilizer Right. So do you know a characteristic property of $E$, as a subset of $[0, 1]$?
Right, rationals in [0, 1]
 
9:03 PM
@EricSilva Which brings to mind the joke we had a day or so ago
 
But yeah I dunno, some mix of my having other things to do and not being particularly interested in bio made this probably the class that I was least engaged with, at least in college and possibly ever
 
It's not connected for one. It's also not complete. It's also not closed. It's not open.
 
Let $X$ be a subset of $\Bbb R$ so that every continuous function $X \to \Bbb R$ can be extended to a continuous function $\Bbb R \to \Bbb R$. What can we say about $X$?
 
@orbit-stabilizer Keep listing :D
 
9:04 PM
@Semiclassical which joke?
 
"What domain is the study of abuse of notation?" "Physics."
5
 
lol
 
Not compact.
 
lol
 
You're getting ahead of yourself, though. Let's stop. Why is it not closed?
 
9:04 PM
@Vrouvrou $\frac{a-b}2$ need not be in $K$. What's important is that $\frac{a+b}2$ is in $K$.
 
It doesn't contain its limit points.
 
What are it's limit points?
 
All points in [0,1]
 
Ah.
That has a name
Do you know the name?
 
Martha?
 
9:05 PM
LMAO
 
That was beautiful
I lol'd IRL
 
that hit me at a time i was not expecting it
 
me too hahaha
 
I am become dead
 
9:06 PM
At least BvS gave us some memes
 
Martha is kind of dense though.
 
What did Avengers ever do
@BalarkaSen ohhh righttt
 
I hope I'm not being annoying or something..
 
Semi I was feeling rushed in my homework for analysis so for one of the problems i extended the function to $\mathbb C \to \mathbb C$ so I could use the identity theorem on it
 
9:07 PM
@FuzzyPixelz I hope that too
 
Okay, so E is dense in [0,1]
 
hope the teacher will be okay with that
 
But if you have time for this,
 
What domain is the study of the abuse of terminology? @Semiclassical "Philosophy" :P
 
9:08 PM
@BalarkaSen i was actually thinking of that scene haha. So pumped for Infinity War
 
I really love MCU
Oh you should see that meme
 
was deriving the power series for arctan by integrating a geometric series, but the geometric series was only valid for $r < 1$, but I made a complex argument that it should hold for $r = 1$ as well
 
not sure how to do it otherwise
 
Aced my geometry final woot woot
 
9:09 PM
@Faust which type of geometry?
 
garts
 
Congrats
 
Euclidean and Axiom geometry
 
@orbit-stabilizer But yeah E is dense in [0, 1]. Or, as you say, that means every element of [0, 1] is a limit point of E. What does that say about g - h?
 
ancient greek geometry ofc, what other kind of geometry is there?
using NUMBERS? pfft
 
9:10 PM
@BalarkaSen hah just watched it.
 
straightedge and compass or gtfo
 
Sick @Faust, congrats!
 
:-)
 
@orbit-stabilizer I think this is my favorite one so far
 
thanks meant that
joke ruined
 
9:11 PM
Still got 3 more though =(
 
@BalarkaSen hmm. I don't know, I mean g - h is still defined over [0,1]
 
hows everyone elses exams going?
 
starting this week, algebra tuesday followed by analysis wednesday
so bleh
 
I've got two exams Friday morning, which is gonna be less than fun
 
and school just ended this friday - so shitty
got the very first two days of exams
 
9:12 PM
@orbit-stabilizer Here's an idea. Suppose $x \to x_n$, and $f$ is continuous. Then $f(x) \to f(x_n)$.
 
But what is Friday? [Vsauce music]
 
@BalarkaSen okay ill think about that
 
It's Friday, friday... ugh I can't even do that
 
Ouch! @Daminark
 
9:13 PM
i got algebra real analysis and number theory left, but at least geometry and graph theory are done ^^
 
Let $f\left(x\right)=\arctan\left(x\right)-\frac{n}{x}+1$ and let $a_n$ be the solution of the equation $f(x)=\frac{\pi}{2}$ on $(0,\infty)$ Prove that $\arctan\left(\frac{1}{a_n}\right)=\frac{n}{a_n}-1$
 
having exams the first day sucks
 
Number theory woot
 
@FuzzyPixelz $\arctan x + \arctan x^{-1} = \frac \pi 2$?
 
Yes, I tried using that, but I end up getting $\arctan\left(\frac{1}{a_n}\right)=1-\frac{n}{a_n}$ @LeakyNun
 
9:14 PM
you're big enough to deal with sign errors on your own
 
$\arctan x+\arctan\frac1x=\frac\pi2$ if $x$ is positive and $-\frac\pi2$ if $x$ is negative
 
FIVE math finals? @Faust
 
though I don't know if that's what happened
 
@AkivaWeinberger $(0,\infty)$
 
@skullpatrol one was an on the last day of class open book final
So sort of only 4
 
9:16 PM
Still...
 
Oh we know that $a_n \gt n$ for all $n$ so there couldn't be a problem with that, I just don't see what's the issue.
 
what is this gfoty album
 
my next two are the 12th and 13th the last is the 16th so at leas ti got some time between my geometry one was the first day of finals :'(
 
this is like music but syrup of ipecac
 
No doubt.
 
9:19 PM
@FuzzyPixelz I think you're right, I think the problem is wrong
Numerically, $a_2\approx2.964$
 
No way, this was on my exam this morning...
 
$\arctan(\frac1{2.964})\approx0.3254$
 
How funny, I din't think about this carefully back then.
 
and $1-\frac2{2.964}\approx0.3252$
 
9:20 PM
are all compact connected subsets of $\Bbb R$ homeomorphic?
 
And 2.964 is a solution of the equation ?
 
@LeakyNun Yes
@FuzzyPixelz It's $a_2$
 
@AkivaWeinberger and if we change "compact" to "open"?
 
Oh wait I'm wrong
Duh, $\{0\}$ and $[0,1]$
 
lol
 
9:21 PM
Point and interval
 
@BalarkaSen: So, $h,g$ cts implies that $\lim_{x \to a} g(x) = g(a)$ and $\lim_{x \to a} h(x) = h(a)$. This happens iff $\lim_{n \to \inf} g(p_n) = g(a)$ and $\lim_{n \to \inf} h(q_n) = h(a)$ for every sequence $\{p_n\}$ such that $\lim_{n \to \inf} p_n = g(a)$ and $\lim_{n \to \inf} q_n = h(a)$.
 
The only connected subsets of $\Bbb R$ are intervals (where we count $\{c\}$ and $[c,c]$ I guess)
@LeakyNun
(Strange how points are both connected and totally disconnected)
 
@AkivaWeinberger so it's true when we change compact to open?
 
Should be yeah
 
you forgot $\Bbb R$
but alright it's an interval
 
9:23 PM
$(-\infty,\infty)$
 
Are point sets connected?
 
@orbit-stabilizer yes
 
Hm? What do you mean
(Also I don't know if the empty set counts as connected)
(so I'mma assume "neither")
 
because you can't divide it into two non-empty sets, period (you don't even need open)
 
@orbit-stabilizer That's true
 
9:25 PM
So if it's connected then, all nonempty connected open subsets of $\Bbb R$ are homeomorphic.
 
Okay, then I'm stuck. I don't know what to do from here
 
What are you trying to show?
 
@orbit-stabilizer What does $E$ being dense in $[0, 1]$ mean?
 
If something's zero on the rationals it's zero everywhere?
 
In sequence lingo?
 
9:25 PM
Every point of [0,1] is a limit point of E or contained in E.
 
@orbit-stabilizer "sequence"
 
But what does being a limit point of E mean?
 
Say $f(x)\ne0$. Then choose an open set around $f(x)$ that doesn't hit zero
and look at its preimage
 
Uh, for every point in [0,1] there exists a rational sequence that converges to it.
 
If all the rationals map to zero, the preimage will contain no rational
 
9:26 PM
@orbit-stabilizer Bingo
 
and thus not be open
and so that's an open set whose preimage is not open so it's not continuous
 
So let your $a$ be any thing in $[0, 1]$
Pick a sequence $\{p_n\}$ in rationals converging to $a$
Now use what you said
 
@AkivaWeinberger I want to show there's only 1 continuous extension of $f$ on [0,1]
 
Namely, $g(p_n) \to g(a)$ and $h(p_n) \to h(a)$
 
@orbit-stabilizer If $g$ and $h$ are different thingies than look at $g-h$
 
9:29 PM
@orbit-stabilizer should one do real analysis before topology?
 
That's what we're doing right now haha
 
It's continuous so preimages of open sets should be open
Say $g(x)-h(x)\ne0$
 
@AkivaWeinberger We're trying to get him to do a much easier argument
 
@orbit-stabilizer so it isn't topology?
 
@LeakyNun uh idk... ive never done topology, but im in a real analysis class rn
 
9:29 PM
Using sequences
 
@orbit-stabilizer then why are you using words like "dense" and "limit point" and "compact"?
 
Ah, I thought this was a topology thingy
Fine, sequences then
 
@LeakyNun Because that's chapter 2 in Rudin!
 
^can confirm
 
9:30 PM
real analysis use those words? :o
 
He has a weird definition of connectedness though
 
I mean, compact is acceptable, but "limit point"?
 
Chapter 2 of Rudin is honest topology, I believe
 
Yeah it's metric topology
 
@LeakyNun I mean, you'll also find limits in real analysis
$\lim$
 
9:30 PM
but not limit point
 
Like it's not just the analyst's take on topology even, it's topology topology
Analysts need a good bit of topology
 
in the sense that a set has many limit points but a sequence has only one limit
 
@BalarkaSen okay looking at the rational sequences, the functions must agree on each point in the sequence, and since they're cts, they must agree on their limits as well?
 
@orbit-stabilizer exactly
 
@orbit-stabilizer Bingo
 
9:31 PM
@Daminark I don't remember, how does it define compactness?
 
Every open cover has a finite subcover
 
@AkivaWeinberger how does he define connectedness?
 
9:31 PM
Oh, OK, it doesn't butcher it then
:P
 
There's nothing wrong with sequential compactness
That's how normal people think about compactness
 
@Leaky$A\cap $ closure($B$) is empty set and $B \cap $ closure($A$) is also empty set
 
@Balarka the thing is, in analysis you often think about the weak topology on a space
 
Idk how to do the bar
 
sequential compactness is compactness in analysis, until you get to the dual of Banach spaces
 
9:33 PM
\overline{whatever}
 
Fair, fair, fair
Functional analysts aren't normal people anyway
 
@BalarkaSen thanks! I'll have to think a bit more to make it rigorous, but I get the idea
 
But yeah the weak topology on a space is basically the smallest one that makes all bounded linear functionals continuous
 
No problem, you have the idea
 
And I think those topologies are not metrizable
(For an infinite dimensional space ofc)
 
9:34 PM
Sounds true
 
The funny part is, in $\ell^1(\mathbb{N})$, it turns out that weak and strong convergence are the same
 
@orbit-stabilizer so a set $S$ is connected if there is no $A$ and $B$ such that $A \cup B \supseteq S$, $A \cap \overline B = \varnothing$, and $B \cap \overline A = \varnothing$?
 
It takes blood to prove it, you have to use this thing called the gliding hump (or at least that's how I've seen it)
Actually wait @Alessandro do you know if Banach-Alaoglu gives you just sequential compactness of the unit ball? Or is it legit compactness
 
@Daminark There are also other spaces with this property (which is called Schur's property) apparently even though I don't know any
 
Oh. There's a Christmas remix of Everyday Bro
I wanna die
 
9:37 PM
@Daminark Doesn't it give weak* compactness?
 
Well yeah it's def weak*
 
@BalarkaSen No, I mean
 
There's another theorem, I think by Banach, that gives sequential compactness, let me look it up
 
I was worried he'd define it as "closed and bounded"
 
@LeakyNun: Two subsets $A$ and $B$ of a metric space $X$ are said to be seperated if both $A\cap \overline{B} = \emptyset$ and $\overline{A} \cap B = \emptyset$.
 
9:37 PM
I'm just wondering like, because weak* is not metrizable, sequential may not be the same as compactness anymore
So I'm wondering which one it gives
 
@Akiva oh
 
A set $E \subset X$ is said to be connected if $E$ is not the union of two nonempty seperated sets.
 
@orbit-stabilizer aha
 
In particular they don't need to be open
and $A\cup B$ equals the set
 
Wiki says that the if $X$ is separable the weak* topology on the closed unit ball of its dual is metrizable, that can't be!
 
9:39 PM
Wait what? Hold on the weak* topology on a Hilbert space is gonna be the weak topology
Oh wait maybe it's only on the unit ball and not on the whole space
 
It is on the ball
 
Oh wait weak* is not weak?
 
still very weird
 
Like that I could buy somehow
 
This is a horrible offense on the English language
 
9:40 PM
en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem look at the "sequential Banach-Alaoglu" paragraph
 
@Akiva so if you have a Banach space $X$, you can embed it naturally into its double dual
By evaluation maps
 
@Daminark After doing so you hope real strong that the embedding is actually an isomorphism
 
So, if you look at $X^*$, you have the normal weak convergence, which is a condition about every sequence in the double dual
But weak* convergence just requires that all the evaluation maps in the double dual (the embedded copy of $X$) satisfy the condition
@Alessandro true
 
which is why reflexive spaces, that is spaces canonically isomorphic to their double dual, are super nice
 
I mean the only Banach spaces I've ever interacted with are the $L^p$ spaces and they're all good so woo
 
9:44 PM
how do i link and show something i said earlier?
 
$1<p<+\infty$ @Dami
 
Fair
 
Anyway $L^2$ and $\ell^2$ are the only Banach spaces
So we're good :P
 
I said this earlier:
So, $h,g$ cts implies that $\lim_{x \to a} g(x) = g(a)$ and $\lim_{x \to a} h(x) = h(a)$. This happens iff $\lim_{n \to \inf} g(p_n) = g(a)$ and $\lim_{n \to \inf} h(q_n) = h(a)$ for every sequence $\{p_n\}$ such that $\lim_{n \to \inf} p_n = g(a)$ and $\lim_{n \to \inf} q_n = h(a)$.
 
@Alessandro kek
 
9:47 PM
That's wrong right. The last two equalities should be $a$, right>
 
I mean what about $L^2(L^2([0,1]))$?
 
@Daminark There's also $c$ and $c_0$ (and $c_{00}$ but that one isn't Banach) which have messy duals I think
 
Both their duals are $\ell^1$
 
@Daminark You have to define a measure on $L^2([0,1])$ for that to make sense, if you really want to...
@Daminark wtf
 
Oh yeah, and that's near hopeless
Like it'll be a garbage measure, not translation invariant (and possibly you'll lose a lot more)
Yeah I can pull up the proof, I had that as a pset problem
 
9:49 PM
Yeah, I don't think there's really a point in doing so
 
Thanks, I can't read it right now, but I saved it for later
 
Aight
 
What's the source by the way? I'm looking for a good book or notes on functional analysis
 
Well that was just my solution to a pset problem
The book we technically used last year was a kind of baby book
 
9:52 PM
The more introductive the better
 
Kolmogorov & Fomin: Elements of the Theory of Functions and Functional Analysis
But it does all the Banach space theory immediately after sets/metric spaces so it doesn't work as much with L^p, and generally feels less comprehensive
The book we're using next quarter in functional analysis is this: people.math.ethz.ch/~salamon/PREPRINTS/funcana.pdf
Traditionally the class used Brezis but our prof felt like it didn't do stuff like spectral theory enough, and treated functional analysis as a bunch of tools that you had to slog through in order to do PDEs, especially those related to fluid dynamics
 
Ah, yeah, I looked at Kolmogorov and Fomin already. The first $3$ chapter of the second one look exactly like the stuff we covered so far, thanks!
 
No problem! But yeah he also said that Lax Functional Analysis isn't too bad, though I'm liking this one a lot so I'll just roll with it
 
Brezis seemed a bit too terse and steep especially in the first part
 
Lol yeah
 
9:56 PM
Dumb question: Suppose $x \in [k,k+1)$ for some $k \in \Bbb{N}$. I need to find $N \in \Bbb{N}$ such that $(x - \frac{1}{n}, x + \frac{1}{n}) \subseteq [k,k+1)$ for every $n \ge N$.
 
I mean he does Hahn-Banach on like page 3
 
For some reason I can't find such an $N$.
 
@user193319 Try drawing it
 
@AlessandroCodenotti champion
 
@user193319 You want $k<x-\frac1n$ and $x+\frac1n\le k+1$
You can solve for $n$ in each case
 
10:05 PM
Ok, gotta go, I'll be back later, bye everyone!
 
Those are both $\le$ actually
 
See you @Alessandro!
 
later
 
10:18 PM
@Daminark
I''m looking at this:
3
Q: Classify all groups of order 182

Bill KayIn studying for an upcoming prelim, I came across this problem: Classify all groups of order $182 = 2*7*13$. Now, the standard tricks here are to look at Sylow's theorems or semi-direct products. Let $n_p$ denote the number of Sylow $p$-subgroups. We can conclude that $n_7 =1$ since $n_7 |...

Since $G/P$ has order 26, why does that tell us that our 13-sylow subgroup has index 2?
 
@orbit-stabilizer Lagrange theorem?
 
How?
$P$ is a 7-sylow subgroup
 
no, the 13-sylow subgroup inside G/P
 
That subgroup can't be contained within the quotient.
But I guess, you're saying look at aP, where a is in the 13-sylow subgroup?
 
G/P is a group of order 26
its 13-subgroup has index 2
 
10:23 PM
Oh. Yes. Then how can we transfer that information back into our original group G?
 
by taking the preimage?
 
Hm okay.
 
Let's try Sylowing the 13-subgroup
The only options are 1 or 14
 
AH, correspondence theorem apparently applies
@Daminark yeah
 
If there were 14, then you'd have 168 elements of order 13. This leaves 12 elements, 6 of which have order 7
Wait no
14 elements left. 6 have order 7
So there are 8 left
 
10:28 PM
8 left. One has order 0.
7 left.
 
I call the identity order 1
 
8 little monkeys, jumping on the bed.
 
But yeah so the remaining ones all have to be order 2
 
Sorry, yeah
 
So then n_2 = 7
 
10:30 PM
But, that's a contradiction, no?
Wait, that's fine.
 
@orbit-stabilizer kek
 
Okay, say I have this group. How do I classify it?
 
Hey! I need help. I'm trying to find a stackexchange site that will help me figure out a clever way to teach solving system of linear equations by substitution because I'm somehow failing with my explanation. I have done so many examples with them. I'm beginning to think that their cognitive ability is not up to the task or I'm also thinking I might suck at reaching them in this topic. :(
 
Where $n_2 = 7, n_7 = 1, n_{13} = 14$.
@randomgirl matheducation?
 
thank you @orbit-stabilizer
 
10:33 PM
Askaway right here
 
Oh wait here's an idea
Take a Sylow 2-subgroup, its normalizer has size 26
That'll intersect trivially with the normal subgroup of order 7
 
How are you approaching the topic? @randomgirl
 
How do we know its normalizer has size 26?
 
Meaning the whole group is a semidirect product of $C_7$ and some group of order 26, so if we can figure out all of those...
It has index 7 by the third Sylow theorem
 
So it is an 8th grade class just so you know. In a way I do think the topic is advanced but you know state standards. I teach them we need to rearrange one equation of the two for one variable and insert it into the other and boom it is just like solving a regular linear equation from there.
 
10:37 PM
Or wait maybe there's something easier we can do
 
I think they are not used to math being so long
 
You take a group of order 13 and then take its semidirect product with the group of order 7, that'll be normal because Index 2
So then take a semidirect product of that and you're good
And the best part is
Groups of order 91 are cyclic
 
Okay, I need to go over semidirect products bcz I can't follow this. I still don't understand the part where you said the normalizer has size 26
 
So I'm abandoning that approach
 
Oh, okay
But, how did you know it had size 26?
 
10:40 PM
But the idea was that n_2=7
So by the third Sylow theorem, the index of the normalizer is 7
 
Oh, I didn't learn that part of the theorems...
 
Rip
But yeah here we've got something
Our group necessarily has a subgroup of order 91
But groups of order 91 are cyclic by Sylow counting
So a group of order 182 is a semidirect product of $C_2$ and $C_{91}$
 
@Daminark thanks, I'll read this over
 
So we want to look at the automorphism group of $C_{91}$, which is $C_6\times C_{12}$
So either the homomorphism from $C_2$ into it is trivial, so we're cyclic, or we have to look at a subgroup of order 2
And I think there are just 3 of those? Thinking about it in additive notation, the elements of order 2 should just be (3,0), (3,6), and (0,6)
So we've narrowed ourselves to at most 4 possibilities
@randomgirl sorry I was doing something else but I can try to help out here
So this system is just 2 linear equations, right?
 
yep.
I also posted the question on math educators
 
10:52 PM
Okay so, this may be a bit strange, but have you tried explaining it in a sort of, "what can we guarantee from each" way?
Keep in mind that I haven't done any pedagogical research so anything backed up by cognitive studies should immediately take precedence over what I'm saying
 
I'm worried it might be too advanced and that is why there is so much torture but it could also mean I just suck at getting the explanation across. I feel like I have explained it myself real well but to 8th graders and how they perceive things they most likely don't feel the same way.
And are you talking about the three differ ways?
 
But when I first learned it, I found it helpful to think about it by looking at the equation and saying well, if we know this equation should be true, then we can rearrange it to guarantee something else is true
 
we haven't even made it to elimination
 
Which textbook are you using? @randomgirl
 
So you have some $ax+by=c$, and then if that has to be satisfied, so much $y = \frac{c-ax}{b}$
 
10:56 PM
prentice hall course 3 mathematics... i probably made a mistake and am teaching them too much at one time. The book jumps from slope-intercept form without doing any rearranging to solving systems of equations that do require rearranging.
 
Now this and another equation both have to be true, this allows us to plug one thing in another. If students have a conceptual problem, that might be a way to approach it
I wouldn't expect this to be too advanced, like I think this is basically standard fare for 8th grade, but maybe
 
I'm sorry I mean the book doesn't have them rearrange equations and then figure out slope and y-intercept.
 
Oh hmm
 
like we jump from things like y=mx+b to solve the system 3x+y=3 and 2x-y=4
And I probably shouldn't have made that jump with the book
 
Originally I was thinking it was conceptually unclear, but to be fair, the algorithm doesn't require much thought
And if they haven't had practice with generically rearranging equations, I'd guess that's more likely the problem
 

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