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01:13
@SillyGoose Not an expert, but I've come up with three main themes of analysis: 1) It's calculus done right, meaning rigorously 2) More philosophically, it's about grappling with infinity, and 3) It's the doorway to freaky calculus, like calculus with random quantities, or calculus on curved spaces.
01:35
@SillyGoose Estimation.
your name together with the preface of your message is funny :) @Novice
estimation :0 in what way @XanderHenderson
@Novice hm i was trying to think up like a road map. e.g. in Rudin at least, we start with sets, then introduce order, which gives us natural definitions like supremum and infimum, then move onto fields, which I guess is motivated by wanting to study the reals...
I've never read Rudin closely, but my feeling is that the material is not clearly motivated after calculus classes.
01:59
@SillyGoose I'm not even sure order is central to analysis. It's central to real analysis, but a lot of analysis can be done in Banach spaces, which have no inherent order
 
1 hour later…
03:18
@Thor That Georges question just reappeared!
04:09
i am trying to recall some basic group theory from last semester... I get thinking about quotient groups in terms of cosets, but I am having trouble motivating the perspective of looking at quotient groups G/N as the set of fibers with the usual group operations motivated from the original group of the natural homomorphism (homomorphism with kernal N)
is it because you want to guarentee an identity of the quotient group? and an easy way to do that is to use this fiber business
04:25
You have to think about why normality is crucial.
0
Q: Stein real analysis execise 1.18

one potato two potato Every measurable function is the limit a.e. of a sequence of continuous functions. I already proved the finite-valued case i.e. any measurable $f:A\to\Bbb R$ can be approximated by a sequence of continuous functions a.e. Now suppose $f:A\to\Bbb [-\infty,\infty]$: measurable and consider a trunc...

Just a minor question
 
1 hour later…
05:50
i'm sorry
06:09
Hope everyone is well, at peace and able-bodied. It would be interesting to note the topics that this room is interested in. For years, I've noticed real analysis is a trend. Yet, am unsure where to procure material to learn it or source from. Text and web-resource suggestions are most welcome.
06:24
Terry Tao’s text is highly respected. I do not know it personally. As with everything in math, deciding on a good text is dependent on your strength and learning style.
07:08
If F is a field such that for all a in F, $a^4=a$, how do I find characteristic of F?
0
Q: $F$ is a field such that for every $a\in F, a^4=a$, then what is the characteristic of $F$?

Koro$F$ is a field such that for every $a\in F, a^4=a$, then what is the characteristic of $F$? Take any $a,b \in F-\{0\}$. then $(a+b)^4=a+b\implies a+b +4a^3b+6a^2b^2+4ab^3=a+b\implies 4a+4b+6a^2b^2=0$. Multiplying throughout by $ab$ and using $a^3=b^3=1$, we get $4a^2+4b^2+6=0$. I am not sure how ...

 
1 hour later…
08:29
2
Q: Let $f$ be a monic polynomial in $\mathbb{Z}[x]$. Request to verify my thinking about a question on irreducibility of $f$.

Jun XuSuppose there are distinct primes $p,q$ such that $\bar{f}=gh\in F_p[x]$,$\bar{f}=uv\in F_q[x]$, where $\bar{f}$ is the image of $f$ under natural projection from $\mathbb{Z}[x]$ to $F_p[x], F_q[x]$ respectively. If $g,h,u,v$ are irreducible in their respective rings, and $\{deg(g), deg(h)\}$ as ...

Any hints for this one?
09:21
In what order should I study maths?
Like, I have completed high school pre-calc and stuff but where should I go from there?
I studied single variable calculus for a few months then jumped to multivariable calculus, however I found it difficult to remember my single variable stuff.
I tried my hand at abstract algebra but that didnt work out.
I also tried topology but that really messed me up as I had no solid set theory knowledge.
So i'm really all over the place right now
Anyone got any advice?
What order did you guys do it in?
09:45
I just wish I followed this ;_;
Thanks, looks great.
Well this is for university. You need to have good hs math background.
May be you are looking for this.
I don't know what is naive set theory.
My university has one course called fuzzy set theory.
May be related to that.
I recently found a book about Riemann surfaces an introduction to Riemann surfaces which is not very famous I think but it heavily uses analysis and differential geometry. Maybe @ted already aware of it.
Surprising thing is that the book is self-contained.
10:46
@JoeShmo "most likely" or "low confidence", choose one
went to dark side of MS again..
-1
Q: The new Calculus?

jake walshI read through some of John Gabriel's the new calculus last night, he does seem quite arrogant but some of the things he says seems to make sense. For example is his way of taking derivatives based of parallel secant lines right? To me it just seems like rolls theorem. Is this guy correct with h...

@SillyGoose The entire game of analysis is estimation. A limit expresses the idea that some error term in an estimate goes to zero; the derivative is an estimation (or approximation) of a function by a linear function; etc.
Many, many papers in analysis hinge on estimating some function, and then showing that the estimate vanishes, or that some error term can be made arbitrarily small, or something similar.
My algebra professor always said that if you know triangle inequality then you know analysis.
11:01
@onepotatotwopotato Boo!
(Of course a dirty algebraist would say such a thing.)
11:29
@Ajay my personal advice is: don't rush. Let it take time. Of course, this is not valid if you're enrolled at a university.
I rushed tensor products my last semester. What's the outcome? I'm still not confident with Tensor products. Do I remember even 40% of it. I don't think so.
And I don't remember alternating forms etc.
I guess that's what happens when concepts are rushed.
11:47
How is convergence defined in $L^p_{\text{loc}}$?
@robjohn sorry it again got frozen
Hey guys do any one of u know about a video which give some brief understand ing of natural language in AI work
12:11
@Koro Alternative explanation: you never actually learn topic [X] in a class named "[X]". You only really learn a topic when you need to in order to solve problems of interest. So, for example, students often master certain algebraic techniques (e.g. "rationalizing the numerator / denominator" in calculus classes, where those techniques become useful for differentiation.
If you needed to know something about tensor products in order to solve a problem of interest to you, it is nigh certain that you would pick it up fairly quickly. The goal of the "rushed" class was not to make you a master, but to give you enough knowledge to know where to Google for the things you really need.
2
@XanderHenderson excellent advice sir
12:29
Being PhD student seems to be so stressful.
I respect this guy. I hope one day I can achieve the level he is.
I think solving integral equation is graduate topic.
How would you write an isomorphism of lattices? $\cong$ or $\simeq$?
@shintuku what are you talking about
13:11
@JoeShmo you tagged me on that WSJ article
13:33
math.stackexchange.com/q/4648181/668308 The formula seems correct but can't understand why the formula makes sense.
@onepotatotwopotato In your question, you ask about the integrability of $(\zeta-z)^{-1}$. In what way could that function fail to be integrable?
@XanderHenderson Because as $\zeta\to z$, $(\zeta-z)^{-1}\to\infty$ so I thought an integral over any nbd near $z$ is not bounded.
13:50
Where in your problem are you assuming that $\zeta \to z$?
Your domain $\Omega$ seems to be fairly arbitrary...
Like, you know that $\zeta - z$ is going to have a singularity at $z$, but that isn't necessarily the end of the world.
I mean the statement is that $(\zeta-z)^{-1}$ is locally integrable on $\Bbb C$. So yes the integrand has a singularity at $z$. It seems it's not a big deal? hmm... mvp?
@onepotatotwopotato What is your definition of "locally integrable"? This should just mean that it is integrable on compact sets, right?
Each point has a nbd which is integrable on that nbd.
Ah, sure, right.
but anyway it's equivalent on $\Bbb R^d$.
14:42
Ah wait I think it's locally integrable using polar coordinate.
15:23
How to show "Arithmetic topology / Furstenberg topology" on $\Bbb{Z}$ homeomorphic to rational topology?
Given a field $K$ and a vector space $V$ over $K$, my lecturer defined affine spaces as a set $X$ with a translation operation $+_A:X\times V \to X$ defined as $(x,v)\mapsto x+_A v$ such that:

(i) for each $x\in X$, $x+_A 0_V=x$;

(ii) for each $x \in X$ and for each $v_1,v_2 \in V$, $(x+_A v_1)+_A v_2=x+_A (v_1+_V v_2)$;

(iii) for each $x_1,x_2 \in X$ there exists a unique $v \in V$ such that $x_1+_A v=x_2$.

He then proves that, actually, (i) can be deduced assuming only (ii) and (iii) using the uniqueness of $v$ from (iii).
I think I proved that (i) can be deduced from (ii) and (iii) without using uniqueness, but I am skeptical. Can someone check my proof, please? In the following, $+_V$ means the canonic addition for vector spaces.

Proof. Let $x \in X$. For (iii), is $x=x+_A v$. Hence, it is $x+_A 0_V =(x+_A v)+_A 0_V$. For (ii), it is $(x+_A v)+_A 0_V=x+_A (v+_V 0_V)$; since $0_V$ is the identity for $V$, it is $x+_A (v+_V 0_V)=x+_A v$. Again for (iii), it is $x+_A v=x$. Hence, $x+_A 0_V=x$.
I don't understand, you start with $x=x+_Av$, do a few steps to reach $x+_Av=x$, and then say "Hence $x+_A0_V=x$", but I don't see anything justifying the final "Hence"
Suppose $+_A\colon X\times V\to X$ satisfies (i),(ii),(iii) and let $w$ be any nonzero element of $V$. Define $+_B\colon X\times V\to X$ by $+_B(x,v)=x+_A(v+w)$. Doesn't $+_B$ satisfy (ii) and (iii) but not (i)?
15:41
@AlessandroCodenotti thanks for the answer. I did this: for (iii), it is $x=x+_A v$. Summing $0_V$ both sides respect to the operation $+_A$, it is $x=x+_A v \implies x+_A 0_V =(x+_A v)+_A 0_V$. Using (ii), it is $(x+_A v)+_A 0_V=x+_A (v+_V 0_V)=x+_A v=x$ (again using (iii) for the latter equality). Confronting first and last equality, it is $x+_A 0_V=x$. I hope it is clearer now.
@AlessandroCodenotti I will take some time to think about this!
Suggest any reference note on Golomb space.
@shintuku yes I tagged you on this article
IIRC you called it a conspiracy theory?
16:45
@JoeShmo yes, i also have low confidence that biden is a lizardman
17:08
perhaps a Sleestak?
@shintuku right, the WSJ and DOE are lizard men. Good job.
hm, well, they're probably not. actually, i'd like to check. oh wait, the report is not available to the public
cool, top tier news reporting
17:21
hi chat, hope you're all doing well :)
17:41
If a continued fraction for a Riemann zeta zero begins: 0, 1, 2, 4, 1, 16, 32, 1, does it mean anything?
ContinuedFraction[Re[Exp[-ZetaZero[1]/Im[ZetaZero[1]]*Pi/4]], 100]
Powers of 2:
1, 2, 4, 8, 16, 32, 64,...
18:21
@MatsGranvik Since it continues $0, 1, 2, 4, 1, 16, 32, 1, 3, 4, 5, 2, 1, 1, 1, 1, 3, 1, 3,\dots$, I don't think so.
18:54
When something is licensed by the Creative Commons license and it has the "Attribution" requirement, so you need to credit the author and indicate changes, does that also hold for just editing a private copy or is the whole "Attribution" requirement only regarding sharing the work?
@ILikeMathematics In general, you are allowed to do anything you like with a creative work, as long as you don't disseminate it.
Once you make it available to others, you have to start paying attention to the license.
yeah. and specifically to CC-attribution, while the short summaries don't make clear when the obligations supposedly attach, the long form i just read in english makes expressly clear that the attribution/indicating changes obligation only attaches if you "share" (a defined term, but one that roughly means what you'd think).
@leslietownes Thank you
so the language of the license you're asking about actually tracks logic, in this case. please ignore xander, who seems to basically be encouraging you to privately create unlicensed knockoffs of beloved characters.
which is objectively wrong, and horrible, even if nobody knows that you're doing it.
This is exactly why, by the way, animation companies take down YouTube videos involving their characters anywhere even on the thumbnail. Toei animation for example has done this multiple times.
looks at Beerus vs Sailor Galaxia still being down And I hate them for it
19:06
i've got a homemade adaptation of 'multivariable mathematics' where the definitions leave out important cases and half of the exercises are missing information or ask you to prove things that aren't true. what do we think, i'm thinking that when i distribute this, yes, attribution to ted, and no, no indication of changes.
19:21
Suppose that D is a unit disk.
Suppose that $g$ is holomorphic on $D\subset \mathbb C$ and has no zeroes in D.
i refuse. astyx doesn't speak for all of us
you obviously don't know how to capitalize
you have no say in this
ok, let's hear him out. it sounds like he's building to something, and we can decide then.
Then, it is to be shown that $2\pi \log |g(0)|=\int_0^{2\pi}\log|g(re^{i\theta})|d\theta$.
So, if instead of |,|, it were only g, complex function, then we could have used mean value theorem for complex functions and got the result.
But here, we have |,|. So how do we fix that?
log |g| is harmonic except at the zeros of g. similar principle.
19:34
hmmmm
Let me ponder over it for a while
again, how you tiptoe up to this with theorems may depend on your textbook's/instructor's route through the material.
So we have: $2\pi \log g(0)=\int_0^{2\pi} \log(g(0+re^{i\theta})d\theta$
I compare real parts on both sides.
using $\log (z)=\log|z|+iarg z$
and we are done :-).
thanks a lot, Leslie.
Right after the mean value theorem, there is a section about Harmonic functions, where they compared real parts on both sides.
So when you said harmonic, that occurred to me.
Do I need to worry about branch of log here?
well, yes, if you want to apply some result about analytic functions to something that you call "log g(z)" which you claim is an analytic function.
again, i'm not sure exactly how this would map out to a sequence of proved results in your material. some books, for whatever reason, do harmonic function stuff first.
Leslie, in Stein and Shakarchi, Harmonic stuff is after the mean value theorem.
log g(z) is analytic if g(z) is not on negative x axis (this is, if I take log branch to be $C-(-\infty, 0]$.
But here, I don't have any control on g(z).
ok, then i guess you need to do some work there. note that the original real-variable integral of log |g| does not involve a choice of log, but you presumably will need one if you want to talk about log g
19:46
Ohh, Harmonic is also in exercises in an earlier chapter in the book.
i suppose you don't expressly need "log g" here. you need to know that log |g| has a "harmonic conjugate," i.e. that there's some analytic function h with Re(h) = log |g| on the domain you're considering. stein and shakarchi have to discuss this sooner or later.
maybe around any discussion of branches of the logs, or when you can define log( ) on a domain. who knows. all roads lead to this one theme.
although somewhat vexingly it is sometimes put on page 10 of a book with 'complex analysis' in the title and it is sometimes put on page 200 of a book with 'complex analysis' in the title.
Ohh, I think I am familiar with this. u is harmonic means that u_xx+u_yy=0. v is its harmonic conjugate means u+iv is analytic.
So to cook up v: I apply Cauchy Riemann on u.
Here u=log |g(re^{i\theta})|.
1
Q: Mean value theorem for harmonic

vkubickiIn Problems and Solution in Mathematics by Ta-Tsien, exercise 5123, the mean value theorem is used as: \begin{equation} \text{log} |F(0)| = \frac{1}{2 \pi} \int_0^{2\pi} \text{log}|F(re^{i\theta})| d\theta \end{equation} Considering the expression provided by http://en.wikipedia.org/wiki/Harmon...

ahlfors has this on pages 161-163 of his book. it's a classical approach. he includes remarks like "in general, however, there is no single-valued conjugate function, and in these circumstances it is better not to use the notation dv" which are not the kinds of remarks you see in more modern books.
but the answer is not clear to me in the post.
@leslietownes so I need to prove that log|g| is harmonic, if g is analytic and non vanishing.
@shintuku interesting. In that case you also went (and indeed always make a point to go) to check every source of every news article that agreed with your preconceived points of view. In the name of good journalism.
20:03
Excuse me for this approach: deleted this spam math.stackexchange.com/questions/4648458/…
-5
Q: Urmăriți! Omul-Furnica si Viespea: Quantumania ( 2023 ) Film Online Subtitrat în Română 1080P

Irwin WhiteAnt-Man and the Wasp: Quantumania [2023] Film Online Subtitrat in Română HD 1080p Urmăriți filmul Ant-Man and the Wasp: Quantumania Online Gratis 𝐒ubtitrate in Romana, Filme Noi HD,Ant-Man and the Wasp: Quantumania Filme Online 𝐒ubtitrate Romana,Ant-Man and the Wasp: Quantumania Filme Online 𝐒...

So why are you spamming us here, @Sebastiano?
Leslie, I understood it now.
14
A: How do you prove that $\ln|f(z)|$ is harmonic?

Alex YoucisEDIT: As 5pm points out below, this is of course actually good enough since every domain in $\mathbb{C}$ is locally simply connected and harmonicity is a local property! This was too long for a comment, but I thought it might be nice to know. There is a much nicer, less computational way to pro...

@Koro i mean, when i say 'need to prove,' 'have to,' whatever, it's with the understanding that the important thing is more reaching the result in a way that makes sense and isn't circular. so yeah, that's definitely one way to do it.
disk D in my case is simply connected.
once you hear the magic words 'simply connected,' the clouds part and the sun comes out.
20:08
haha
I remember one day I was talking something about a limit of a function close to the limit point, Robjohn said -don't look directly into the Sun.
That was so funny 😅.
@leslietownes $dv$ is like $d\theta$. It’s a well-defined $1$-form even though $v$ is not globally defined. But Ahlfors introduces the Hodge star operator without calling it so.
Jan 15, 2022 at 18:18, by robjohn
@Koro look near $(0,0)$. Don't look directly at the sun!
yes, i'm not even really criticizing it. ahlfors absolutely knows what he's doing. but you don't see a lot of modern books reminding us of things that aren't "single valued", at least, without having defined a context where that distinction has formal meaning.
ahlfors even apologizes for using the term "single valued" in a footnote at the beginning. doesn't stop him from using it, though. he's not sorry.
Does anyone here know what’s going on here? I can do the question but have no idea what the OP is regurgitating from somewhere.
yes, i don't recognize the terminology or formulas at all. some musty old textbook strikes again, i bet.
20:14
@JoeShmo what are you on about
sigh
did you read the article, the one you tagged me in
you don't get to say, hey, here's what happened, most likely. and then two sentences later, say: but uhh... listen we're stating this with low confidence. the article itself states that the majority of intelligence agencies concluded against this hypothesis
dx.doi.org/10.1017/S1757748900001250 is an article from 1914 talking about generators of lambda-systems and mu-systems of a hyperboloid.
nice to know that some system of education somewhere is keeping the golden oldies alive.
@JoeShmo like, i bet what happened was "listen guys, we've got no confidence in this hypothesis, but our buddies over at the FBI are sounding like jingoist idiots being the only department to support the hypothesis, so i guess we can list them as a source and upgrade our statement to "low confidence"
i'd love to be wrong about this imaginary exchange, but since the report hasn't been made public....
and, (lol), the FBI's own statement states they've come to this conclusion with "moderate confidence"
shintuku: in order to prove that you don't just focus on media coverage that confirms your priors, you must now accept this reporting about a DOE hypothesis as the truth. this is the logical vise that you have been placed in.
although i have low confidence biden is a lizardman, i have moderate confidence in the case of the zucc
i don't like it any more than you do.
i can't tell, from the 1914 article, what a mu-system is supposed to be. people must have just lived and breathed this stuff.
20:26
@leslietownes oh no, not logic again!!
i wonder what stuff that is now published in 'mathematical notes' type sections of journals, for stuff of expository interest, will seem like this 100 years from now.
@leslietownes I call them $A$- and $B$-lines in my algebra book. :)
I’ll check Pedoe later.
now i kinda want to know, but i was born after hyperboloids went extinct, so i won't have much use for the information.
might as well tell me what passenger pigeons like to eat.
Suppose that a holomorphic function $f$ on the strip $S = \{|\Im z| < 1\}$ is bounded by $A > 0$. For any $0 < δ < 1$ define $S_δ = \{|\Im z| < δ\}$. Show that for any $n ∈ \mathbb N$, $f^{(n)}$is bounded on $S_δ$ by a constant which depends only on $A, n$ and $δ$.
I applied Cauchy inequality on $f^{(n)}$ to obtain: $f^{(n)}(z)\le A n!/ R^n$, where R is radius of circle centered at $z$.
special characters are not orthodox chatjax, koro
mr. chatjax is crying in heaven when he sees what you have done with his creation
20:38
The problem is how to get a bound on radius R.
One way is: $\delta- |\Im z|\le R$
But the problem with this is that it depends upon z.
And we want to bound f^(n) by a constant on S_\delta strip.
21:00
There is no reason why I can't take R=1-\delta. So this is done.
0
Q: fiber bundles and larger fiber bundle

MathematicallyInterestedGiven assumption: Given a fiber bundle $\pi:E\rightarrow B$ with fiber $F$. Let $\phi:U\times F\rightarrow \pi^{-1}(U)$ and $\psi:V\times F\rightarrow \pi^{-1}(V)$ be local trivializations of $\pi$. Suppose locally, so on $V\times F\rightarrow V$, for every local trivialization $\Psi$ and compact...

21:31
0
Q: $f: D\to \mathbb C$ is holomorphic such that $diam f(D)=d$, then $|f'(0)|=d/2\implies f$ is linear.

Koro$f: D\to \mathbb C$ is holomorphic such that $\sup_{u,v\in D}|f(v)-f(u)|=d$, then $|f'(0)|=d/2\implies f$ is linear. Here, let's suppose that $D$ is a closed unit disk. By Cauchy's integral formula, since $f(z)$ and $f(-z)$ are both holomorphic, we have for every circle $C_r$ centered at $0$ and ...

up to some normalizations, how close is this to the schwarz lemma?
i'm not suggesting that it just is that result, but surely you could translate f and scale it so that f(0) = 0 and d is 2, for example, and how far is that from the schwarz lemma. just remove away some of the 'noise' here that would appear in a proof with things subtracting off and then adding stuff like f(0) + f'(0)z.
ah, OK, some commenter linked to a post, a few arguments use the schwarz lemma.
21:47
Is it correct to write: $\emptyset=\emptyset \,\,\dot{\cup} \,\,\emptyset$? I saw it in a proof
the union of emptyset and emptyset is indeed emptyset, although i'm not sure if this dot over the symbol usually used for union is going to change things
it's used for disjoint union
But it feels like a trick
even if formally there are no elements in the intersection
if we have an ideal in the ring of formal power series, when is it closed under countable sum?
Is there a counterexample when it is not?
When I say "countable sum" I mean for each degree $d$ there are only finitely many coefficients to sum among all the series to sum
22:20
@leslietownes while I certainly placed him in a logical vise, that’s only a partial characterization of what I said. The point (as I’m sure you know) is that the lab leak hypothesis is NOT a conspiracy theory, unless the DOE now joined the FBI to purvey conspiracy theories and disseminate disinformation, via the WSJ. @shintuku will continue to read the article VERY selectively, and miss no opportunity to nitpick at the banal and irrelevant.
@shintuku, the jingoist idiots at the FBI — OK, Donald.
In fact, calling it a conspiracy theory was a CPC propaganda tactic to shut down inquiries. A tactic shintu will happily indulge, since deep down he thinks the CPC are the good guys ;-)
22:37
@leslietownes 😭
@leslietownes mr. chatjax DIED?!
No!!!!
When?!
HOW?!
suicide by cop. don't ask, you don't wanna know
Dang... sir... that went dark, fast.
just remember, you're not just putting unicode \in or \delta inside dollar signs, you're putting pressure on people who may already have been stressed out
@leslietownes $\unicode{x1f62d}$
22:51
Sadly, I was not put on a grand jury this morning. :(
They called in 52 people. I was number 49. They never got to me.
"i've been left out of grander juries than this," you said as you left and slammed the door behind you
3
Wish I'd thought of that. :D
23:09
If only you lived in ATL ….
23:26
but 49 is a nice number
@JoeShmo sick quotes from the little red book tho
23:45
I'm working through a proof, and I'm using a claim that I've just realized may be incorrect. If a < x < b, and c < y < d, then a + c < x + y < b + d (where a,b,c,d,x,y are real), right?
keith, sure.

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