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00:01
@XanderHenderson I meant send it to you via email
but alternatively I could just link the arxiv link here in a few days
 
12 hours later…
12:21
Normalize dimension theory
@Jakobian You do silly topological dimension theory stuff. Ew.
WHERE'S YOUR METRIC?!
dimension theory is not my favourite part of commutative algebra honestly
but then what is
That's another dumb version of dimension theory.
:P
12:39
@XanderHenderson I wouldn't say I do it. It was my interest for a while, sure, but now I focus on other things. What just strikes me as odd is how people are not accepting of using dimension theory for separable metrizable spaces in their arguments
 
1 hour later…
pie
pie
14:04
I always wonder: if $1+1=2$ is an axiom, does that mean someone could create a whole new 'system' of mathematics if they made $1+1=3$ (whatever that might mean)? Or are there some axioms that cannot form a 'system'?
If that is true is there a 'math field' where mathematicians just change the rules and try to build a new 'system'?
In what system is $1 + 1 = 2$ an axiom, exactly?
@pie yes, this would be logic :)
pie
pie
Hmmm Maybe I misunderstood, but as I recall one can't prove $1+1=2$ so it is an axiom, right?
one big part of mathematical logic is studying different axiomatic systems
@pie it's a definition
pie
pie
@BenSteffan There is a math field called "logic" or you mean the logic itself?
the mathematical field of "logic"
14:08
there is no mathematical field of logic, logic and mathematics are two different, albeit overlapping fields
there is mathematical logic however
pie
pie
@BenSteffan ngl that is a terrible name for a field, as an outsider like me can't just know such thing exist because if I hear about it I will say they mean the general logic
weird because I know several people at the math department who are working in logic
pie
pie
@Jakobian wait what
@pie it's a perfectly fine and appropriate name for the field
pie
pie
14:09
there is a whole separate field called logic?
yes.
or "mathematical logic" so as not to upset Jakobian
yes, logic is not just mathematics, its also lawyering and other uninteresting things
pie
pie
@BenSteffan It will create confusion can't you see?
@pie eh
everybody knows about this
we don't generally cater to outsiders :)
pie
pie
not if you aren't a math major
14:11
neither do other academic disciplines
pie
pie
@Jakobian So how to study it? is there a separate degree to it other than math? is there a road map for it
logic is also in philosophy
logic in the larger sense is multidisciplinary I guess
pie
pie
@BenSteffan I really wish I wasn't an outsider:(
@pie I don't know, but you can specialize in model theory and this will be largely overlapping with logic
pie
pie
14:13
don't tell me there is a math field called that?
I guess studying logic more generally would be something you'd do in a philosophy course, potentially
pie
pie
"model theory"? pure math?
model theory is a subfield of logic though (?)
at least I believe people generally consider it as such, even if it overlaps with things like AG
model theory would be pure math, yes
is model theory a subfield of logic... I don't want to think about that lol. Maybe
pie
pie
14:15
Why can't "they" choose names that aren't confusing to people like me?
it studies what kind of mathematical structures ("models") satisfy a given axiomatic system
@pie pie why should they care?
again, these names are fine
they are descriptive
I'd focus on the content, not on the name
pie
pie
@BenSteffan I thought It would be something like how to model 3d stuff for 3d artist or modelling using MAtLAb or something
@SineoftheTime If you are a self learner you wouldn't guess or know that this is a pure math topic
That's why
then you go google it, look at wikipedia and learn what it is
the names aren't confusing, you just need to learn what the field is instead of assuming what it is
if you encounter a term like that in a mathematical context, it would make sense to think that it's something mathematical, no?
if you encounter it in a non-mathematical context, then there's a good chance it refers to something else anyways
pie
pie
14:20
@Jakobian I would assume it is an applied math thing, and I kinda hate applied math so I just decide not to check, That turns out to be a mistake
if you encounter "surgery theory" or "deformation theory" or something like this in a mathematical text, you wouldn't assume that refers to medicine, e.g.
@pie that doesn't sound like something that anyone would have a reason to care about
if it's pure maths and you want to study pure maths this does not mean you should study model theory
pie
pie
@BenSteffan Yeah because medicine don't use math, but engineers and programmers do use math (or abuse it) so I might suspect that "surgery theory" is a pure math thing, but model theory seems like a name for a course in engineering
well there's your problem :)
14:23
the divide between applied and pure math is a pretty outdated take anyway
pie
pie
@Jakobian You are right, but I am just surprised
you let things like these to hold you back, its only going to be worse for you
I wouldn't want to not study something because of some arbitrary boundaries that someone decided for me
pie
pie
@Jakobian I am in engineering school right now, I hate the way they never show a proof for anything and just "take my word for it, it works", so I generally assume applied math don't care about proofs but how to use the theorems, maybe I am wrong?
and that is why I 'hate' applied math
same things happen at those courses that are considered to be pure
perhaps less often, but it does happen. And you are in school, so its not like they prove a lot of things
I'll admit that when it comes to academia, a lot of things were dumbed down, and I probably wouldn't like our applied mathematics courses
but this has more to do with academia itself, than with the fields
@pie that sounds like a local experience to me. our applied mathematicians are quite rigorous
14:36
I feel like this is precisely the kind of cliche that makes people turn away from applied mathematics (however vague what this means is; and thus from a lot of mathematics)
pie
pie
@BenSteffan WOAH
@Jakobian It made me hate any form of applied math like physics (although I liked it in high school), in the first year we took thermodynamics and Schrodinger equation without taking linear algebra or ode or multi variable calculus or pde,
I remember struggling to memorise the solution of second degree PDE without knowing what the hell was going in any step
pie
pie
that made me hate anything that is not pure math.
but I hear PDEs are kind of just like that :)
pie
pie
@BenSteffan applied math books has proofs? I learned a lot of things today ngl
14:47
@Jakobian Mathematics is a subfield of philosophy.
@XanderHenderson >:(
@XanderHenderson subfield or subring?
@pie at least some of them should
@SineoftheTime I SEE WHAT YOU DID THERE!
@BenSteffan The truth hurts...
in the words of Gilles Deleuze: pee pee poo poo
pie
pie
14:48
@XanderHenderson is philosophy a sub-field of anything?
@pie "Applied math" is a spectrum. At my phd institution, there was an applied mathematics research group in the mathematics department. They were a rigorous bunch who also had industry ties. They cared about proof, or at least about having a strong understanding of the theoretical underpinnings of what they were doing.
@BenSteffan "poop (no context needed)"- Xander
There are also people who are even more rigorous, e.g. people who focus on numerical analysis.
But there are also folk who really don't need to know the proofs.
In any event, my brother-in-law is off, and I need to go sit on a nibblet, so later.
15:01
@pie calling physics applied math is ... uh
I guess a lot of it consists of what you might call applied math
But calling physics math is something that a physicist wouldn't allow
neither a mathematician :)
@XanderHenderson I'm not going to argue with that, but definitely, philosophy is not a subfield of math
@pie again, the dichotomy between "applied" and "pure" math is really more of an issue with the philosophy with which you do math with. This only shows that you dislike that particular exposition to physics. Concluding more would be a stretch.
But you can always study PDE theory and see that, it really isn't that bad after all if you learn it properly (or just hate PDEs)
15:30
@BenSteffan even the numerical analysts, they are rigorous. If you write dog as dog=cat +(dog-cat), they make sure (dog-cat) is small for large enough values of cat
oh numerics is a very rigorous business
Yeah, and they need to be, or else you'll encounter failed rocket launches and stuff
@pie i wouldn't classify physics as applied math, really, though there is a lot of math applied in it. Applied mathematicians still must be good at theory, physicists needn't even touch analysis textbooks
@BenSteffan ofcourse model (category) theory is pure math :)
@SineoftheTime have you watched the match today?
16:04
@SoumikMukherjee yeah, embarassing
This WCC is a delusion
both sides are disappointing
16:25
Yeah
@BenSteffan what's your opinion on taking theoretische Physik 1, 2, 3 and be done with the Nebenfach with that? Can't get more mathematical than that, huh haha
Experimentalphysik is a recommended pre-req though
And I won't have that
16:40
I took Theoretische 1, 2 and 4 as Nebenfach in my Bachelor's and then CS as Nebenfach in my Master's and I preferred the latter
Physics is the most mathematical, but the mathematics are kinda poor and, in terms of the exercises, it's just doing a lot of standard computations
but of course your mileage may vary, and so may the courses from university to university
16:56
@Jakobian I mean, math is a subfield of philosophy. If philosophy were a subfield of math, then the two would be equal, and that doesn't seem to be the case...
17:06
i have completed sheldon axler's Linear algebra done right, what do I do now with this knowledge
the sequence $f_k(x) = e^{-|x-k|}, k \in \mathbb{N}, x \in \mathbb{R}$ can be uniformly bounded by $M = 1$ a.e. in $\mathbb{R}$ and $$\lim_{k \to \infty}f_k(x) = 0:=f(x), \forall x \in \mathbb{R}$$ but $$\lim_{k \to \infty} \int_{\mathbb{R}} |f_k(x)-f(x)|dx = \int_{\mathbb{R}} e^{-|x-k|}dx = 2 \nrightarrow 0$$
@nickbros123 Move on to Hoffman and Kunze?
Or Hungerford?
what am I getting wrong here, this is what Wikipedia calls Bounded convergence theorem
17:22
@Claudio that theorem you cite is for finite measure spaces
but the Lebesgue measure of $\mathbb{R}$ is infinite, so it can't be applied
oh :p
well forget everything I wrote ahahah
@XanderHenderson hmmmmm
Generally speaking, introductory texts like Lin Alg Done Right are meant to give you a baseline, which is supposed to help you solve problems later on.
The tools taught in lin alg are often of practical use in, for example, 3d graphics. They also show up in many numerical schemes for solving problems (e.g. linear optimization problems (in the form of "linear programming"), root finding algorithms, etc). Lin alg shows up all over the place in computer and data science.
time to chase the bag then with all this lin alg knowledge \$
@nickbros123 Not really. You have a long way to go before you get to a useful a place from where you are. Again, Axler is just building a foundation.
17:35
true true. I was just joking
Hoffman Kunze seems like a good book to spend my time in
Aug 28 at 17:14, by nickbros123
My course forcing me to let go of the love of my life Hoffman & Kunze- Linear algebra to pick up Sheldon Axler -Linear Algebra. At my lowest rn :'(
Weird cuz i kinda became a LADR fanboy after a few days
moral of the story- dont trust me
@nickbros123 now read linear algebra done wrong
@SoumikMukherjee Or Linderholm's book... :D
(Several years ago, I managed to get copy of Linderholm's book for less than \$100. I feel special.)
(It was on my watchlist for almost a decade.)
@XanderHenderson ive seen some parts of that book lol. I intended on reading its pdf every night as a bedtime reading thing, at the start of the semester
but LADR ended up being my bedtime reading book
I can't imagine any mathematics text as "bedtime reading". Math is read with a pen in hand.
u can actually read LADR as a bedtime thing
17:48
Not if you are doing it right.
most of the arguments can be constructed in ur head itself
Again, math is read with pen in hand.
Until it is written down, you can't tell that it is wrong.
@XanderHenderson me neither. When I'm in bed, I want to sleep
sure you can read one quick proof before going to sleep, but you know
If the proof is easy enough, I think you can read it in bed
some proofs you do need to grab a piece of paper for though
some problems ofc need pen and paper, like for eg: Proving that for any 3 vector spaces V1, V2, V3 over a field with more than 2 members, the union of the 3 is a vector space if and only if one of them contains the other two. This was a notorious exercise for me, cant imagine solving without pen n paper. but say for eg, proving that for finite dim vector spaces, the only two sided ideals of $L(V,V)$ are (0) and the space itself, argument can be constructed without pen n paper ofc
for me, its not that I can't tell if a proof in my head is wrong, but rather, when looking at a proof, I can't tell if the steps are correct or not without writing them down explicitly sometimes
17:55
most of the exercises in LADR are just imaginable
@Jakobian well at the level of math u do I wouldnt expect to be able to
depends on how technical and well-written something is, doesn't really have to do with "level" of it
how technical is isomorphic to "level"
computations you definitely need paper for, if something is terse and not well-written, that also demands paper
but if someone wrote it for you in a comprehensible manner, and its not computations, that doesn't necessarily require you to write it out
but if you have multiple concepts introduced, that again, maybe not require paper, but paper helps to get you oriented among all the definitions
this helps with conceptual understanding and memorization
if I were reading books and papers on analysis, say, I think I would have to grab a piece of paper for every argument
I would not be able to do analysis arguments with just my head
because those type of fields are computation heavy
similarly, abstract algebra has either computation heavy, or just very conceptual approach, and I would need paper to either get through computations or orient myself among the definitions
in general topology there is much less of this in comparison, and proofs are relatively tame
18:05
fair points, I am not saying all problems are air solvable
I'm not even talking about "problems" because constructing an argument in your head is another thing entirely
I mean it of course has elements of "verifying a proof" in itself, since you do so with your own proof, but it also has the additional step of having to make an argument for why something is true
But I do agree that a lot of non-computational type exercises are picked to be easy enough so that they can often be "done in your head"
this conceptual kind of thinking, if its not super long, can often be done without the aid of paper
@Jakobian u mean like actually learning a field? that I havent obviously. I did linear algebra throughout the day / week, with pen n paper ofcourse, and atop of that, I did a bit of bedtime reading of LADR. Reasons are similar to the points you raised, memorization of definitions, getting into the context of things, making notes, etc.
@nickbros123 I don't really know what 'learning a field' means, but sure, lets go with that
neither do i. its the thing u do in a course, or something like that
I mean reading mathematical papers, books, parts of books etc.
whatever that is for the purpose of getting acquinted with a given subfield of mathematics or not
basically anything that a mathematician would have to do
18:30
@BenSteffan do you think the tag choice in this post is correct? In particular the general topology tag
hmm... I suppose it is "topological" enough for it to be considered general topology
19:34
Hi
$\int_{0}^{\infty} \frac{x}{2x^2 + 3} \sin\left(\frac{π}{3}x \right) \, dx$
Can anyone help me with this integral?
looks like something you can easily solve with complex analysis
I.e. what's the fastest way to evaluate this
19:39
residue theorem seems the fastest method
Can I tell you what I did?
$f(z) = \frac{z e^{i \frac{\pi}{3} z}}{2z^2 + 3}$
Breaking it down $f(x) = \frac{x e^{i \frac{\pi}{3} x}}{2x^2 + 3} = \frac{x \cos\left(\frac{\pi}{3} x\right)}{2x^2 + 3} + i \frac{x \sin\left(\frac{\pi}{3} x\right)}{2x^2 + 3}$
I define $\int_{0}^{\infty} \frac{x \cos\left(\frac{\pi}{3} x\right)}{2x^2 + 3} dx = a$
$\int_{0}^{\infty} \frac{x \sin\left(\frac{\pi}{3} x\right)}{2x^2 + 3} dx = b$
For singular points: $2z^2 + 3 = 0 \implies z^2 = -\frac{3}{2}$
$|z|^2 = \frac{3}{2}, \quad \text{Arg } z^2 = \pi$
Exponential form: $z^2 = \frac{3}{2} \left(\cos\pi + i \sin\pi \right)$
The roots are: $z = \pm \sqrt{\frac{3}{2}} \left(\cos\frac{\pi}{2} + i \sin\frac{\pi}{2}\right)$
$z = \left\{\sqrt{\frac{3}{2}} i, -\sqrt{\frac{3}{2}} i\right\}$
So far Is okay ?
why you need the exponential form? $z^2=-\frac 32 \implies z=\pm i \sqrt{\frac 32}$
19:55
@SineoftheTime Oh okay, this way we avoid a superfluous step and simplify the solution
This is Ramanujans lost formula for the zeta function $$I(s)=\zeta(s)+F(s)$$
@SineoftheTime is the rest okay?
Ok, now for the Cauchy residue theorem
$\int_{\partial D} f(z) dz = 0$
And i take $\partial D = \{x, |z| = R\} \cup \{-R, -R + i\} \cup \{R, R - i\}$
@Jakobian yeah. I never quite know what to do with questions like that either, but I wouldn't edit to remove the tag
@ILikeMathematics Don't really have an opinion on that :)
20:01
@Pizza I don't understand the path
I know very little about physics as a nebenfach
@SineoftheTime $\partial D = \Gamma_R \cup [-R, R]$
Ok so $\Gamma_R$ is the semicirle or radius $R$?
semicircular arc
And [-R,R] segment along the real axis.
@Pizza in the upper or lower semiplane?
20:08
the semicircular arc in the upper half-plane, defined by $z = Re^{i\theta}$ for $\theta \in [0, \pi]$
By the residue theorem, since $f(z)$ is holomorphic on $\partial D$ except at isolated points:

$\int_{\partial D} f(z) \, dz = 2 \pi i \sum \text{Residues of } f(z) \text{inside} \partial D$
I write the integral along the contour as: $\int_{\partial D} f(z) \, dz = \int_{-R}^{R} f(z) \, dz + \int_{\Gamma_R} f(z) \, dz$
When $R \to \infty$, the integral along $\Gamma_R$ tends to zero, under the assumption that the function $f(z)$ decreases quite rapidly on the semicircular arc
yes, but you have to be careful
$\lim_{R \to \infty} \int_{\Gamma_R} f(z) \, dz = 0$
@Pizza did you verify this? Or did you use a particular theorem?
@SineoftheTime Jordan's Lemma
if $f(z) = g(z) e^{iaz}$ and $g(z)$ is analytic and bounded on $\Gamma_R$, with $a > 0$, then:
$\int_{\Gamma_R} f(z) \, dz \to 0 \quad \text{as } R \to \infty$
Right ?
20:25
that's correct, but I won't check in this specific case
I'll leave the computation to you
Ok
so now you have to compute the residue
$\text{Res}\left(f(z), z = i \sqrt{\frac{3}{2}}\right) = \lim_{z \to i \sqrt{\frac{3}{2}}} (z - i \sqrt{\frac{3}{2}}) f(z)$
$f(z) = \frac{z e^{i \frac{\pi}{3} z}}{2z^2 + 3},$ and substituting $z = i \sqrt{\frac{3}{2}}:$
just write the result, instead of every step so that you don't waste time
the procedure here is easy, nothing to invent
$\text{Res}\left(f(z), z = i \sqrt{\frac{3}{2}}\right) = \frac{i \sqrt{\frac{3}{2}} e^{i \frac{\pi}{3} i \sqrt{\frac{3}{2}}}}{4i \sqrt{\frac{3}{2}}}$
20:32
looks good
$\int_{0}^{\infty} \frac{x \sin\left(\frac{\pi}{3} x\right)}{2x^2 + 3} dx = \text{Im}\left( \pi i \cdot \text{Res}\left(f(z), z = i \sqrt{\frac{3}{2}}\right)\right)$
ok that's correct
:)
Can I check on wolfram if I did it right?
did you check, before choosing this path, that the integrand is even?
@Pizza yeah, no one is preventing you :)
you should get $I=\frac{\pi}4 \exp(-\pi/\sqrt{6})$
which matches with your computation
for a similar integal, take a look at problem 7.17 :)
well done!
@SineoftheTime I didn't check :(
20:41
you have to check
otherwise you're computing $\int_{-\infty}^{\infty}$ and if the function is not even this contour is not useful
Ah ok
Thanks for the help and advice
But are there other things I wrote that I can avoid?
Bros are cooking
@Pizza this part here
@Jakobian HI!
just introduce $f(z)$ and say $I=\text{Im} \int f(z)$
20:56
@mo-_- hi
plus, if I recall correctly, there's a faster method to compute the residue at $z_0$ of $\frac{h(z)}{g(z)}$ when $h(z)$ is holomorphic at $z_0$
it should be something like $h(z_0)/g'(z_0)$
@SineoftheTime Oh ok, thank you very much!
Let $S_1 = \{ z \in \mathbb{C} : |z| > 1 \}$ ; verify that $(S_1, \cdot)$ is a semigroup but not a monoid.
semigroup = algebraic structure with binary operation that satisfies associativity
monoid = semigroup with identity element
@Jakobian Can you by any chance help me with this?
+ sine, pizza etc
where are you stuck?
21:07
@SineoftheTime I have to add one more thing : (if S is a set and * : SxS --> S is a binary operation, then for all s there exists x such that x * s = s * x = s)
@SineoftheTime ↓
For the first part i'm trying to prove that associativity holds using z1, z2, z3, each written in trigonometric form but i get stuck in the calculations and im not sure it's the right approach. For the second part, S1 is a subset of Z, the identity element in (Z, *) is 1 and for every z in Z 1 is the only element such that 1 * z = z * 1 = 1. 1 is not in S1 since abs(1) is not greater than 1, therefore, S1 cannot be a monoid
do you
do you not know that multiplication in $\mathbb{C}$ is associative?!
what is Z?
what do you mean by trigonometric form?
@not_a_chatbot I guess polar form
@SineoftheTime set of integers
21:11
polar form, presumably
Then I don't see how $S_1$ is a subset of $\Bbb Z$
Hm, I would use polar form to prove that multiplication is associative, if I did not have that fact already
@BenSteffan yes i know
then why are you spending time proving something you already know?
@not_a_chatbot I wouldn't, because your proof will likely be circular :)
I'm not in the least convinced that you can prove the polar form exists without making use of associativity even once
@SineoftheTime i meant to write C
21:14
I definitely need associativity of R, and a stronger statement that rz = zr for complex z and real r.
Not sure where I would need associativity of complex multiplication. Point taken, though, that this approach is not the best way to show complex multiplication is associative.
@BenSteffan what you mean ?
ponder this
why is the multiplication on $S_1$ associative, given that you already know the multiplication on $\mathbb{C}$ is associative?
21:39
@Jakobian This, more or less. But also: you need to know it in your soul, and the only way to get there is to write it down yourself. Probably five or six times, definitely at least twice without constantly looking at the book.
The point, @nickbros123, is that the goal is to really internalize the mathematics, and that doesn't happen from reading alone.
2
the nlab has a great article on internalization
good for the nlab, good for the nlab
If I eat the pages of a book, am I internalizing it?
@BenSteffan That's a pretty low bar. :P
@SineoftheTime Only if you swallow.
21:47
indeed
eating pages of a book as a pastime activity
> 1-8 Theorem. If $A \subset \mathbb{R}^n$, a function $f: A \to \mathbb{R}^m$ is continuous if and only if for every open set $U\subset \mathbb{R}^m$ there is some open set $V\subset \mathbb{R}^n$ such that $f^{-1}(U) = V\cap A$.
I have been thinking some time about the "if" direction and I feel like I'm going in circles.
Attempt. Let $\epsilon > 0$ and $x_0\in A$. Now $U=\{y: |y-f(x_0)|<\epsilon\}$ is open in $\mathbb{R}^m$. By hypothesis, there is a $V$ open in $\mathbb {R}^n$ such that $f^{-1}(U)=V\cap A$. Notice $x_0 \in V$, therefore there exists $\delta > 0$ such that $B(x_0,\delta) \subset V$. Then $$\forall x \in A, |x-x_0| < \delta \implies x \in V\cap A \implies |f(x)-f(x_0)| < \epsilon.$$
Any mistakes in the above?
@psie That kind of seems like the definition of continuity?
What is your definition of continuity?
Not in Spivak's book :)
(i.e. the definition I know best is "preimage of an open set is open", which is exactly what your theorem is saying).
@XanderHenderson I feel like he's a bit sloppy actually, but he just defines it in passing to be $\lim_{x\to a} f(x)=f(a)$.
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