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00:00
Ah, that was his thesis.
In my paper, I worked with the second-order one in terms of osculating spaces and adapted frames.
The whole problem, in general, is that you have to use sheaves because (for embedded submanifolds) the osculating spaces may drop rank. But look at Pohl.
@leslie That looks like a horrid reference.
With a general (positive-definite) quadratic form, it's just the spectral theorem and the one-dimensional result.
that's done in the first bit of one of the solutions. i was selecting for paste-ability, not ease of reading.
i'm a low quality chatterbot interface to google.
Well, we all know I set a high bar :)
I want to see a picture or three of the scary strawberry with paws.
we're still working out just how exactly this is going to work, but if it comes to fruition, ha ha ha, i will document it.
Maybe a lingonberry garnish would be good.
00:28
@TedShifrin that one, but in $\mathbb{R}^n$. the 1D integral is by squaring, fubini, and then polar coordinates, I know.
@leslietownes, looks like it. Thanks!
look at the section beginning 'wick's theorem' where you diagonalize and it becomes a product of one-dimensional gaussians. i don't know what kind of physics sorcery is going on once you add in what they call a 'source term.'
oh no no I don't want any sorcery
physicists invented nuclear weapons, that's the kind of road you head down if you keep listening to them.
yeah.
I'm a peacenik
when it comes to physics
There’s nothing $n$-dimensional about these questions. They do show up in harmonic analysis, though, integrating spherical harmonics.
00:59
I hate that feeling when I'm trying to show something that my brain tells me should be easy, but I just can't find the right track to follow. It's like trying to climb something and just slipping down on the first step each time
I've got a couple of sequences in a Hilbert space, $\{x_n\}$, which converges weakly to $x$, and $\{y_n\}$, which converges to $y$. I'm trying to show $\langle x_n,y_n\rangle\to\langle x,y\rangle$. I know this has to rely on the fact that $\{y_n\}$ strongly converges, because this isn't true for two weakly convergent subsequences.
My search-fu has failed me again, though
(I found a thousand and one duplicate questions about showing that every bounded sequence in a Hilbert space has a weakly convergent subsequence, though)
Real newbie query: Here's a simple proof from Spivak: $\frac{\left | x \right |}{\left | y \right |} = \left | x \right | \cdot \left | y \right |^{-1} = \left | x \right | \cdot \left | y ^{-1} \right | = \left | x y ^{-1} \right| = \left | \frac{ x }{ y } \right |$ I think I understand how this is derived, but what I do not understand is why $\left| y \right |^{-1}$ is different from $\left| y ^{-1} \right |$ Thanks.
It's the difference between $\frac{1}{\vert y\vert}$ and $\vert\frac{1}{y}\vert$, which are honestly the same thing. It's why Spivak says they're equal
So, he's just messing with me?? :-)
No, didn’t he prove those equal in a previous part?
@TedShifrin Correct...indeed he did. But was just wondering why he would go out of his way to show the two.
01:13
That”s the point :)
Because that’s what’s going on at this point in the book. Proving everything that you always thought was obvious. Just using the properties.
@TedShifrin OK... I think I get it. The logic, which I love mind you, takes a bit of getting used to. OK... I like that.
For example, you’ve always said it was obvious that $0\cdot x=0$ for any $x$. Why is this?
the notation is just so familiar. people are trained in algebra class to perform those manipulations in their sleep, and not to notice the potential differences between them.
Rithaniel: You need an “add a clever zero” proof.
Yes...funnily enough, just did that last night. As you say, not at all "obvious"
01:17
But we all assume it to be gospel truth. Yet, it’s not a standard property that we get to assume.
What I like about it, to be honest, is the rigor that needs to be applied. As you say, it's not rote by any stretch of the imagination.
Spivak (or, indeed, proof math) isn’t the right choice if you want just to assume that everything “obvious” is true.
if someone said here are functions f and g, show that f(g(x)) = g(f(x)), you'd think hmm, that's not a general property that all pairs of functions have, so this might have to do something with these specific f and g, and what do we know about them, etc. but if instead of f and g you write | | and ^(-1) suddenly it just looks like "duh."
Agree... that's why I am trying to go through his book.
Nice comment, @Leslie.
01:19
we need to unlearn the 'duh' so we can relearn the higher 'duh' and step closer to enlightenment.
@Leslie Was my advice to Rith right?
@TedShifrin I was thinking that might be the case. Something like $$\vert\langle x_n,y_n\rangle-\langle x,y_n\rangle+\langle x,y_n\rangle-\langle x_n,y\rangle+\langle x_n,y\rangle-\langle x,y\rangle\vert$$ was what I was originally considering, but I dislike the middle
Thank you Ted and Leslie. That's a really nice summary of what I was actually trying to ask. :-)
Can you do just one zero?
Is that for me?
01:22
@user1115542 Through Spivak (and my books too) you are expected to use what’s been proved in previous parts. Always look for that.
No, the one zero was for Rith.
Yes...that I have figured out...hence my reluctance just to "plow" on and hope for the best.
@TedShifrin I apologize if I do not realize who you are.
I don't see a "one zero" idea immediately. Are we trying to build it into the form $\langle x-x_n,y_n-y\rangle$ or summat?
@Rithaniel try the version indicated here: math.stackexchange.com/questions/2952507/… note that the hypotheses there are not your hypotheses, but you have stronger hypotheses. you can control <u - u_n, v_n> using the strong convergence of u_n to u and the fact that the weakly convergent sequence is norm bounded.
Again, leslie has found a relevant question almost immediately
2
I wanted $x_n,y_n$, $x_n,y$, $x,y$.
01:26
as ted suggested, just one zero is enough. that answer shows why the result can fail if in the absence of the strong convergence assumption.
Oh darn. I was too too lazy to type inner products. I like to think instead of searching. Ahem.
@TedShifrin Oh yeah, weak convergence
I like to think, too, but there is more than one way to bake an apple pie from scratch
One is called "research," and the other is called "reasoning." I do want to understand where and how the strong convergence assumption comes into play, though
You certainly need it for my first term!
Thinking is ultimately research.
2
I’m serious here.
Very true
The skills for becoming a researcher.
2
I’m not a fancy functional analyst. I use my basic Spivak skills.
01:30
you cheated by writing half of his exercises.
Not half
Also, clicked over to that link that leslie had. I did find that one already, but I dismissed it because it was trying to prove this to be the case with both sequences being weakly convergent
Ha ha
FB may go to hell. Zuckerberg deserves it.
Even though I’m an addicted subscriber.
But, I suppose the argument would work if one of the two were strongly convergent, then?
Also, I heard about the whole FB thing
So work out what I wrote, dammit.
If I’m wrong, show me.
01:33
I will, Ted, don't worry
But I was gonna share my ambivalence to FB going down
rith: yes, if one is strongly convergent you can control the term he couldn't control using the schwarz inequality.
that argument breaks down if only weak convergence is assumed because x_n going to x only weakly doesn't force ||x_n - x|| to go to zero.
Oh, I see. I suppose that would have been the step that I would have figured out with the thinking step. (Sorry Ted. Next time you're helping me I'll do it purely myself)
i really should use mathjax.
LOL … Or you can follow the advice not to ask here and find it on main. I prefer making people think.
@leslietownes Duh.
01:38
To further my "climbing up a slippery surface" metaphor, this is like a person pointing out to me that there's a staircase
Read Camus’s The Myth of Sisyphus!
Or, hell, google.
Sisyphus was the guy rolling the boulder up the mountain, right?
@Rithaniel And when he was up, he was up. And when he was down, he was down. And when he was half-way up, he was neither up nor down.
Oh, wait... that's the Duke of York, not Sisyphus.
I know that one! Also, Tantalus is a guy that might have a parallel to the experience of research
01:44
My metaphor comes to life!
@XanderHenderson No, he kept sliding down.
02:03
I've been off FB for years. and I never had IG either
but they have all my data anyway because I use whatsapp
i've never used FB, but my cat and daughter have IG accounts. FB thinks i am interested in cat toys. the algorithm basically advertises to me as if i am a cat.
02:20
its creepy that "the" algorithm had become an institution in our society
02:48
i've got a facebook account but like
i haven't used it in RL years
so i'm not so aware of that algorithm
@JoeShmo 1984
by contrast, i'm very familiar with the shenanigans of the Youtube algorithm
@LeakyNun, Yes. An Orwellian institution, indeed.
03:15
an artificially intelligent Orwellian institution has emerged from the apocalyptic pandemic
Greetings, Leaky.
The year is 19^84. Long ago, we started using exponential notation to keep track of the year because it got too long otherwise
The stars went dark many eons ago, but still people are posting dumb stuff on FB
3
03:31
and my cat is still cute on instagram
In a strawberry costume?
possibly. last year the daughter did go as a black cat. we forgot to get a pic of the two of them.
04:04
\o @AudenYoung
04:38
@user178758 o/
how are you?
04:56
@leslietownes still wanting to be a strawberry ghost?
yes. with paws.
Ted popping up because he sensed I was calling to him through ESP.....
05:25
and claws?🧐
gotta have claws
professor Ted, while writing an answer, I wasn't aware that you were answering that question of dc3rd. Please let me know if I should delete it.
too late since I've read all of it. :), I'm still working it out on my own though...
@dc3rd ahh, I see. I was excited to answer it as soon as I saw the question as I've been studying linear algebra these days :)
no worries....it is all helpful.
still using Axler's book for linear algebra Koro?
05:33
yes, alongwith Hoffman &Kunze's
I've been working through Friedberg, Inse, et al'
"Lang's Algebra changed the way graduate algebra is taught..."
Last year, I'd read D.Poole's and Gilbert Strang's linear algebra.
I'm just now stumbling into the fact that linear algebra is a lot more vast than an undergraduate course.....
@dc3rd Do they use matrices or linear maps or both?
05:44
Insel does both. as should all undergrad linear texts. since they are one in the same.
In Axler's I have noted that, linear maps are highlighted more but matrices are discussed less in Axler's book.
mainly linear maps though because the idea is that everything can be converted to a matrix so make's it less clunky to write
yeah, axler covers the bare minimum of matrix stuff.
Insel has tossed it in here and there and I'm almost done the book. It is more so along the lines of "you should be aware that the linear maps can be turned into matrices", so for ease of typesetting I'll mainly use linear maps unless necessary
Would you agree with the quote above @leslietownes about Lang's Algebra changing the way graduate algebra is taught?
05:53
i dunno, i did not live in a pre-lang world. many of lang's books are somewhat divisive. they have devotees and people who hate them. if you look at early-mid 20th century algebra books they are sometimes a little crusty and obsessed with special cases, to my mind. and they predate the widespread adoption of basic frameworks and terminology.
is lang responsible for bringing that in, or did he just happen to write a book that became popular while that was happening generally in algebra, i dunno.
is graduate algebra built off of linear algebra?..........I have a feeling this is a rhetorical question...
sort of, if you interpret 'built off' very loosely. in a formal sense, not really. a lot of notions that are very interesting and useful in algebra become trivial in linear algebra, so linear algebra does not have analogues of, or stepping stones to, a lot of algebraic topics.
hmmm........one or two of my profs before had called linear algebra a "terminal" course as in once you learn it there is nothing more to learn about it and only can be applied.....
that is probably more reflective of the course offerings at wherever that is than some objective reality. although maybe it depends on what 'linear algebra' means. there's a sense in which even calculus classes are "terminal." you can learn more about the theory, the generalizations, etc. but they are probably not going to make you any better at calculus.
linear algebra over finite fields can have a very algebraic flavor that you don't see too much of if you only work over R and C. and linear algebra + topologies + no finite dimensionality enters into PDE, functional analysis, etc.
today I learned 3 laws of newton in a fist fight
06:12
Did you use the third one your advantage? 👊🏼🤼
it is disadvantage since if I hit someone I get hit too
that's when speed comes in...
p=mv
talkin' about fights, there's a good heavyweight fight on this Saturday
It could go either way
06:44
are you talking about ufc or wwe?
07:17
🥊 boxing
Tyson Fury vs. Deontay Wilder III, billed as Once and For All, is an upcoming professional boxing trilogy fight between WBC and The Ring heavyweight champion, Tyson Fury, and former WBC heavyweight champion, Deontay Wilder. The bout is set to take place at T-Mobile Arena in Paradise, Nevada on October 9, 2021. == Background == Fury and Wilder first fought in December 2018, with the bout ending in a controversial draw. The pair had a rematch in February 2020, with Fury emerging victorious via seventh-round technical knockout to capture the WBC and vacant Ring magazine heavyweight titles. Wilder...
07:33
\o @Slate Welcome 🙏
08:23
Let $X$ be a finite set and $\mathcal{F}\subset\mathcal{P}(X)$ i.e., subcollection of subsets of $X$, such that any $F_1,F_2\in\mathcal{F}$, $F_1\cap F_2$ is nonempty i.e., pairwise intersects. I want to show that by deleting a point $x\in X$ (so deleting any $F\in \mathcal{F}$ with $x\in F$), we can color $X-x$ by two colors such that for any $F'\in\mathcal{F}'$, $F'$ contains two different colors. Here, $\mathcal{F}' = \{F\in\mathcal{F}\mid x\notin F\}$.
Any thoughts? I think the choice of $x$ is important so I need to delete a point $x$ with some special property like a point that contained $F$'s the most.
09:00
Hi @BalarkaSen, no never seen a Siefert surface of the trefoil knot?
09:41
@SMCnotSvenMagnusCarlsen Then you should not be studying Haken manifolds, and so on.
 
1 hour later…
10:49
@BalarkaSen I agree
 
3 hours later…
13:31
dark hadou
13:58
@user178758 'lo :)
 
2 hours later…
15:50
@Xander Probably not the best place for you to have your Mod hat on, but when you put it on, be sure to make it clear to others, "I am now Moderator" before you humiliate them for not treating you like a Mod.
16:38
I don't know if I am the only one who experiences this , but the SE chat notification bell always gives me a heart attack.
yeah it's a bit aggressive
I always keep the browser tab mute so that I don't hear the notification. But when I accidentally forget to mute then....
you can turn off sound notifications by clicking the speaker at the top left corner of the right sidebar and setting it to "none"
@Koro Nah, of course not. But both you and @dc3rd should consider whether my approach is different or better. By that I mean, less guess-and-check.
woah cool
also, hollywood is about to go on strike, what timeline is this
16:49
@hyper-neutrino I wish I knew that setting. =(
well now you do :p
Now SE notification bell will never haunt me again. Thanks hyper-neutrino for enlightening me with your wisdom.
if you'd like to torment yourself again, someone on codegolf made a digital piano that uses SE's ping sounds :p here
i find slapping the laptop hard stops the notifications.
derivative of average velocity is 0 for N particle when they are having fist fight in limited space
hal 9000 isolated me in deep space and now I need to burst my airlock to go toward my spaceship
This is a fast, cookieless domain intended for static content delivered to the Stack Exchange family of websites
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can't run
17:01
did you paste the code into the console?
Hi
what does $f:\mathbb{Z}/r \rightarrow \mathbb{Z}/2r$, given by $f(n)=2n$ mean?
where is the console
is $r$ here supposed to be the ideal genrated by r?
@monoid What do you think? It should be $f([n]_r) = [2n]_{2r}$, to be explicit.
By $[n]_k$ I mean the equivalence class of $n$ mod $k$.
@BannedUser you can open it with ctrl-shift-i and look for the "console" tab
17:06
oh, $\mathbb{Z}/r$ is $\mathbb{Z}_r$
Right. They're writing $\Bbb Z/r$ for $\Bbb Z/\langle r\rangle$.
But it is disturbing not to have some mention of the fact that elements are equivalence classes, not integers.
@hyper-neutrino next step
nevermind just curious
yeah, @TedShifrin. I agree
$\mathbb{Z}/r$ is horrible, I hate it
shrugs
Morning, @robjohn
17:12
I liked your approach Ted. It actually made me apply all the fundamental linear algebra framework, at least in terms in questioning if what I was building was correct. for example: I could have a basis vector for my subspace, but the basis vector representing that subspace doesn't necessarily have to serve as a basis vector for my larger vector space.
maybe "all" is an exaggeration....but you know what I mean.....apply what I've been learning
@TedShifrin morning
Right. But it's forward-moving without guesswork. If you take a basis for the column space to be $(1,0,0)$ (as a column vector, of course), then you just have to make sure that vector is in the nullspace. Then you're done.
And once you understand that, you could do it with an arbitrary vector $(a,b,c)$ if you really wanted to.
@monoidaltransform suppose $r=7$, then $\left\{\begin{array}{}0+7\mathbb{Z}\mapsto0+14\mathbb{Z}\\1+7\mathbb{Z}\mapsto2+14\mathbb{Z}\\2+7\mathbb{Z}\mapsto4+14\mathbb{Z}\\3+7\mathbb{Z}\mapsto6+14\mathbb{Z}\\4+7\mathbb{Z}\mapsto8+14\mathbb{Z}\\5+7\mathbb{Z}\mapsto10+14\mathbb{Z}\\6+7\mathbb{Z}\mapsto12+14\mathbb{Z}\end{array}\right.$
@Ted: is Saturday spay day? I figure a vet would rather do stuff on Saturday than Sunday.
17:31
Now that "neuter day" was canceled for inappropriate gender, yes, it's Saturday :)
Given the map $f: \mathbb{Z}/r\rightarrow \mathbb{Z}/2r$, by $f(n)=2n$, its cokernel is just $\mathbb{Z}_{2r}/\overline{2}\mathbb{Z}_{2r}$?
this notation is confusing lol
@monoidaltransform what does the bar over the $2$ mean?
the equivalence class of 2
17:34
The notation sucks because now they're using $\bar 2$ for $[2]_{2r}$ and they haven't bothered to notate equivalence classes elsewhere.
I would fire the book.
Although, admittedly, I eventually dropped the bar notation in my own algebra book once I got to quotient rings and there would be too many bars if I took a quotient of $\Bbb Z_n$.
@robjohn I should have commented, however, that this vet clinic is open 7 days a week.
@TedShifrin ah, our vet is not open Sunday. We have an after hours vet that we can go to at night or on weekends. Not too far away in Agoura Hills (15 minutes door to door)
Yes, I knew you made a visit to them a few weeks ago. Should I inquire about the pup?
She is hanging in there, trying hard. I am disappointed in the vets around here. Our vet referred me to some oncologists for a diagnosis, because she thinks it might be cancer, but she has no definite evidence other than an inflamed vertebra in an x-ray. The oncologist wants a diagnosis before they will make an appointment. What is it that they do: just care for them afterwards? who can make the diagnosis?
It is highly frustrating.
You have my sympathy. It sounds insane, but even human medical practice is often insane. Maybe ask the after-hours vet who knows the pup for an assessment?
@TedShifrin They have even less evidence, but it may be all we have.
17:43
Well, interpretation of the X-ray is a starting point.
They were treating her for pancreatitis
@TedShifrin Our vet was going to talk to a radiologist, but as I haven't heard back about that, I don't know if she has yet.
Who does the oncologist expect to make the diagnosis? I guess "thinks it might be" is an acting diagnosis. Proceed?
I tend to try not to be annoying, but here I think you are warranted in calling and asking for a progress report.
I called a few times yesterday and talked to staff, but the doctor probably doesn't want to talk to me because she has nothing new to say.
When I had my soft tissue sarcoma in my left arm, my orthopedist in Athens, GA, referred me to an orthopedic cancer specialist in Atlanta. It took me more than 2 months to get in to see him. Meanwhile, the damn tumor was growing and growing. I kept calling. Finally they gave me a 7 AM appointment. The guy was amazing and an amazing surgeon, but damn it took forever.
She has a leg that has been swelling since Friday when I took her in. We hadn't noticed it, but the vet did. Now, it is quite swollen.
17:46
Sounds like a return visit is in order.
it is obviously painful, as well
she is on tramadol and CBD oil
Poor thing.
@TedShifrin my wife had a similar experience trying to get a CT scan. She had to suggest things to the doctor before things moved.
Why do we pay these people so much when we have to do 2/3 of the work?
Well, I'm about to have a phone call with the NP at the pain clinic. So we'll see how that progresses.
good luck
17:50
You too!
18:08
isn't $\mathbb{Z}^2$ precisely $\mathbb{Z}\oplus \mathbb{Z}$? we have a question that sais find image of a map $\mathbb{Z}^2\rightarrow \mathbb{Z}^2$ and then on another question asks for the same map but with domain and codomain $\mathbb{Z}\oplus \mathbb{Z}$... this seems to be a silly question... no?
Yup.
Only in the case of infinite products do we have to distinguish between $\times$ and $\oplus$.
first $\mathbb{Z}/r$ now this
yup
Well, in fairness, $\Bbb Z/r$ is reasonably standard.
Is this a text or your prof?
18:11
there doesn't seem to be a lot of reflection on the notation being used. fine to use both but bad to mix and match without explanation.
Some teachers are more human than others.
some people do just forget what is and isn't known, or introduced, especially if it is below a threshold of 'difficulty'. but still.
I've not seen a serious algebraist write $\mathbb{Z}/r$
thorgott, are you one of those people who hates $\mathbb{Z}_r$
i'm trying to identify the most hated notation for that object
what if i defined $\mathbb{Z}_r$ to be the set $\{0, \dots, r-1\}$ with addition and multiplication defined in terms of remainders in the division algorithm
how bad can we make this? can someone think of worse?
18:28
The issue is that $\Bbb Z_p$ is used for the $p$-adics.
So one needs to know one's audience.
yes. that was the classic complaint. "but students might confuse this with something they haven't yet heard of and may never hear of!" :)
Damn, leslie, we agree again.
i don't mind what people use, although if there is a textbook, people should use what the textbook uses. if the textbook is that bad, choose a different textbook.
@leslietownes Terrible notation. You wanna have a $p$ there, 'cause it should be prime.
I agree. When I taught out of calculus books that were mandated, I adopted the book's notation. The prof is in a better position to make adjustments than the student.
When I taught out of my own books, I could complain about the idiot author.
@XanderHenderson Au contraire.
18:35
@TedShifrin $r$-adic numbers are nonsense. :P
how about we think of the cyclic group of order n as roots of unity, and write $\mathbb{C}^n$ for it? can you come up with worse?
Oh, that's pretty outstanding.
What idiot did that?
me, just now. it's a two-fer: C stands for cyclic, and to remind you that it's also complex numbers.
And the superscript for good measure.
yeah that's just a little extra style sauce on top
18:37
will it sell? if it sells i'm in
@leslietownes Yeah, but these roots can be expressed as $\mathrm{e}^{2k\theta/n}$ for some $k\in \{0,1,\dotsc,n-1\}$, so to specify a specific value, you want to use $C(k,n)$.
the operation on $\mathbb{C}^n$ isn't complex multiplication, by the way. that might confuse students who haven't taken complex analysis yet. so it's define in terms of representative angles in [0, 2pi) and remainders in the division algorithm and trigonometric identities which are established via triangles.
@leslietownes Seems totally reasonable.
Don't you have to take munchkin to the orthopedist?
you're not rid of me yet. that's at 4.
18:43
Grr.
olivia just came in here and jumped onto my desk for the sole purpose of swatting a pen off of the desk. then she jumped down and ran downstairs.
i'd love to know how the cat mind works.
Because of Screech, my pens and pencils are now all in a drawer.
we store ours under the couch or refrigerator. it's just more convenient that way.
Yes, not to mention the rubber bands that are everywhere under things.
Is it true that if a graph has ${3n\choose n,n,n}$ vertices, each with degree at most $3n-1$ then for every vertex you can find another vertex that is distant at least $3n^2/2$ from it?
18:47
@leslietownes My little guy, Shai, loves to bat just about anything off any table, counter, desk he can get his paws on!
He can also "catch", and he plays hockey in the bathtub using a soda cap!
user435118
Can the modulus of a complex number be negative?
no
what's the definition of the modulus?
user435118
It shouldn't be (I think) but my textbook uses a weird example.
oh, soda caps. i actually gave up bottled stuff because of the noise livvy would make with them in the middle of the night.
@leslietownes Yes. Shai also loves the plastic rings from milk jugs!!
18:50
i'm interested in this example. i wonder if 'modulus' is being given some new meaning.
these things must optimize some set of parameters for cats. light enough that a small swat sends them far away. they can make a whole lot of noise, but not too much noise. etc.
@Xnero What "weird example"?
my cat doesn't like things that roll very much, i think because the way they rebound off objects is unpredictable and scary.
user435118
Let me figure out the MathJax.
@leslietownes He doesn't roll that plastic rings; we works this paw in them to flip them. At times it looks like he uses his paw to "hool-la-hoop!"
user435118
@TedShifrin Write $-4(cos\frac{7pi}{6}+isin\frac{7pi}{6})$ in the form $x + iy$ but $-4$ is the modulus...
18:57
No, $-4$ is not the modulus.
$4$ is the modulus. The negative sign has to change the angle.
It becomes $\pi/6$ instead of $7\pi/6$.
user435118
@TedShifrin Makes sense, thanks.
good question. i could see a textbook goofing that up. "if you write a complex number as r (cos t + i sin t) [with r and t implicitly real], then r is the modulus." i bet that's out there somewhere.
I would bet not, but I don't have confidence in authors.
19:14
I was never clear if the $r$ could be negative in a polar equation :-).
Well, I always defined it to be nonnegative, but then you have equations like $r=\cos 2\theta$ ... Precalculus books typically allow it to have both signs. Ugh.
But modulus of a complex number is undeniably nonnegative.
It bugs me that elementary textbooks are sometimes not precise & explicit in such things.
mm yeah, i remember polar coordinates allowing for r negative in precalculus.
user435118
What is considered elementary level?
The first place you encounter a concept.
19:28
@leslietownes I prefer sticking to $\mathbb{Z}/r\mathbb{Z}$, but $\mathbb{Z}_r$ is alright if it's clear from the context
as long as you don't write $\mathbb{Z}_{(r)}$
haha. i agree with that.
@leslietownes The Earth is not flat; if it were, cats would have knocked everything off the edge.
@leslietownes one of our cats cannot resist the clear plastic wrappers that "secure" the tops of some bottles. Those get strewn about our house with wild abandon.
livvy will tear those apart and swallow portions of them, only to immediately vomit them up. she also does this with bits of ribbon.
19:45
hello, if $\lim_{x\to+\infty} f''(x)=0$ how to prove that $\lim_{x\to+\infty}\frac{f'(x)}{1+x}=0$ ?
maybe try with $g=f'$ first?
how ?
20:01
use a MVT type estimate?
for example on [a,b]: \exists $c\in ]a,b[$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$
i put for example $a=k b$ ?
20:20
you'll have to have a little more control over endpoints here but that's the kind of relation between f and f' i had in mind
Or did you intend the Cauchy (extended) MVT?
i don't understand you
if you know $|f'| < \epsilon$ on $[M, \infty)$, then $|f(x) - f(M)| < \epsilon (x - M)$ for all $x > M$
whatever gets you there is what i had in mind. i think it's something slightly weaker than the MVT because a version of it holds in higher dimensions. just, if f' is bounded, then you have an estimate like that
OK, this is beyond what's acceptable. Do NOT type oo. Geez.
bans leslie
stops rolling $\pi^5 + 6e$ eyes for the moment
i went t$\infty$ far with that one
20:24
Indeed. Unacceptable and worthy of jail time.
i have $\forall \varepsilon >0, \exists A>0, \forall x\in \mathbb{R}, |x|>A\Rightarrow |f''(x)|<\varepsilon$
oh sorry, i was starting via copper hats phrasing, where my "f" was copper's g, which was your "f'"
you apply MVT on [A,+\infty[ right ?
yeah, your A is my M.
all of the actors in today's play are wearing masks of other actors.
ok
so by MVT: |f'(x)-f'(A)|<\varepsilon |A-x|
when $x\to +\infty$ on a $|A-x|\to+\infty$
20:30
before taking limits, see what the above tells you about f'(x)/(1+x) on [A, $\infty$)
it's tell me that $\frac{\varepsilon(A-x)-f'(A)}{1+x}<\frac{f'(x)}{1+x}<\frac{\varepsilon(A-x)+f'(A)}{1+x}$
what i must do @leslietownes
this is where it's handy to have properties of limsup and liminf but i'm guessing that you don't. note that e(x - A)/(x + 1) = e(1 - A/x)/(1 - 1/x) goes to e as x gets large, and f'(A)/(1+x) goes to 0 as x gets large. so for sufficiently large x you can make both of those less than 2e (or whatever) in absolute value.
20:48
so we get that $-\varepsilon <\lim_{x\to+\infty}\frac{f'(x)}{1+x}\leq -\varepsilon, \forall \varepsilon >0$
so by letting $\varepsilon \to 0$ we get the answer
right ?
sort of. we're proving that the limit exists and is zero by establishing those inequalities for all e > 0. how you write this out in terms of quantifiers and definitions may depend on what tools you have access to. (we're not, for example, assuming that lim f'(x)/(1+x) exists and then using theorems about limits to evaluate it).
we have $\forall \varepsilon >0$ is the definition of the limit
how i must rwite it please ?
why don't you do:
1. Let $f''(x)$ be arbitrarily close to $0$
2. Prove that this implies $f'(x)$ is within $\delta$ of some constant $c$
3. use $\lim \limits_{x \to \infty} \frac{c}{1+x} = 0$?
i think that is what you do now
for #2 use the definition of local extremum
21:02
i don't understand
what is the difference with what you do before ?
you skip MVT and limsup/liminf
could someone elaborate on the following:


A matrix representation of a group $G$ is a homomorphism $R:G\rightarrow GL_n$.

If a group $G$ is given by $<x_1,......,x_n|r_1,.....,r_k>$ generators and relations and we have matrices $R_{x_1},......,R_{x_n}$ that satisfy the relations, Then we may obtain a matrix representation $R:G\rightarrow GL_n$.
no we use MVT
@shintuku
why is $R$ a homomorphism?
Definition of representation?
Too many letters R.
21:16
A matrix representation is just a homomorphism $G\rightarrow GL_n$
@TedShifrin I just wrote what artin has written
it seems like a universal property at play
Am I right?
yes, i can think of two relevant universal properties.
could you tell me what they are?
first, can you define a homomorphism from the free group into GL_n that sends x_j to R_{x_j}. one universal property might be relevant. second, does if you've got one of those, does that induce (or promote to or whatever you want to call it) a homomorphism on some quotient of the free group. another property there.
and i guess the definition of $\langle x_1, ..., x_n \mid r_1, ..., r_k \rangle$. that might come up too.
@Vrouvrou given $\lim \limits_{x \to \infty} f''(x) = 0$, acquire $|f'(x+h)-f'(x)+\varepsilon(h)| < \epsilon$
then, $\lim \limits_{x \to \infty} \frac{|f'(x+h)-f'(x) + \varepsilon(h)|}{1+x} = 0$
no need for mvt
where $\varepsilon(h)$ is the error function for $f''(x)$
21:36
can't believe leslie is explaining the universal properties
The world is upside-down?
i'm not leslie, i'm the guy who broke into his house. he left his computer on when he went to pick up his daughter from day care and i thought it would be fun to pop in and take some things.
blink a couple more times and I'll be doing spectral theorems
I have no idea what I'll be doing. Upside-down bumhood.
21:53
found something super cool today in a used bookstore; no idea if it's worth anything but i'm happy i found it - found chapter 7, divisors, from commutative algebra by bourbaki for $2, the french version
literally copyright says 1965, it's so neat having something that 'old' for lack of a better word
anyway time to piece together google translate and such to read it lmao
(to add the amusingness of this, found it in a revolutionary/communist bookstore on their used/cheap rack. only in berkeley!)
hah. bourbaki is the worker's weapon. look out, capitalists.
they also had a book on probability theory. really, a surprising number of science/math books
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