« first day (4495 days earlier)      last day (822 days later) » 

00:02
I’d say it’s the kernel of the orthogonal projection onto W.
WAs just going to say you need a linear map to have the existence of a kernal
But rob john is a lot more precise than me
@CottonHeadedNinnymuggins you talk about kernel of a linear map, not of a set.
Hi @robjohn … Hope you’re feeling better.
confused about $hf$......$hf(t) = h(t)f(t)$? How wouldn't it be $h(f(t))$?
Yes?
Multiplication of functions, yes.
oh.....I thought it was implying composition
00:11
Then you need the $\circ$.
true......there was short circuting of thinking of matrices while in truth they are functions.
Ah.
Matrix multiplication is still multiplication! 🤷‍♂️😻
true, but I went to the jump of matrix multiplication --> function composition.....went too far.
How should I prove that the set of all vectors orthogonal to a subspace forms a subspace? I cannot mention kernel (which is a subspace itself) since that implies a linear map?
WHat does it mean to form a subspace?
or rather I should say what does it mean to be in a subsapce?
 
1 hour later…
01:29
the word subspace is commonly used to mean (i) a linear (sub) space and (ii) more in a topology context, a subspace of a larger space. in the above comment it is (i).
@CottonHeadedNinnymuggins just the usual show that multiples and sums of orthogonal vectors are them selves orthogonal.
 
2 hours later…
03:12
How do you study analysis?
Should you either just read the proof and practice the practice problem or you attempt to do it yourself then read the proof?
I think the latter takes a lot of time. But I want to develop the skill to write clever proofs.
I wonder how do you develop the creativity.
I wonder how did those great mathematician study.
Planning takes a lot of time.
Is there a biography where you can learn how well known mathematician study?
04:01
Planning on avoiding my family this thanksgiving/winter break by instead watching/studying linear algebra Zoom lectures that professors published online similar to this format. My university gave me keycard access to classrooms so I am able to post-shop to mimic a distance-learning classroom so I can try to make the most of it. However, I don't do online classes or lectures, so I have no idea which ones are decent
The Schwartz reflection principle on the unit circle really comes from the reflection principle on a half plane with conformal equivalences
I just checked them all.
one potato: what something "really comes from" is a matter of opinion. disc v. half plane is a coke or pepsi kind of thing. (the disc is coke, and it's the objectively right choice.)
Many arguments on the half plane can be boiled down to the unit disk by the conformal equivalence map. I recently saw two examples including reflection principle.
btw leslie, do you remember the annulus problem I asked?: If $r_1<|z|<R_1$ and $r_2<|z|<R_2$ are conformally equivalent then $R_1/r_1 = R_2/r_2$. I said I have a proof using reflection principle. It assumes a slightly stronger condition that the reflection function is continuous on the boundary but in RCA, there's a proof without assuming that. Completely different argument.
04:32
nvm
04:44
complex analysis is weird.
04:57
@TedShifrin thanks. Getting another test on Tuesday. Whether insurance does what it should tells whether I have to pay a huge amount or just a painful amount.
👀🙄🥲🤦
@MissMae Depending on your interest and topics, I also have linear algebra lectures posted. See my profile for a link.
good luck, rob.
ted: we saw an osprey carry a fish away from the koi pond again today. the fish was still wriggling. it's all munchkin could talk about.
She’s jealous.
@TedShifrin The smiling face with tear is not everywhere, yet, it seems.
If a, b ∈ R satisfy a < b, then there is a q ∈ Q such that a < q < b.
05:09
@TedShifrin That's what I'm looking for, thank you! Also, thank you for being a cool and making education more accessible by posting that for free online
My students wanted to do it, @MissMae. They get the credit.
I wonder why you take n>max{1/a,1/(b-a)}
I think choosing a such that k_0/n is already enough where 1/(b-a) is consequence of that.
Why bother with n>1/a?
How do get $E$ bounded above?
E is already bounded above by a*n
by definition
now we need to choose n
but isn't choosing n>1/(b-a) enough using Archimedes theorem?
You need the other term to get $na>1$.
you need to read the proofs like a detective,
05:17
@TedShifrin I think I don't follow.
My guess is to make sure E is nonempty
You said $E$ is bounded above by $na$. You need $na>1$ for that to be possible.
You also need it for non-emptiness, yes.
@TedShifrin k is natural so yes
my be using pen and paper will be better to connect the ideas
oh i get it I see the sequence trying to adjust itself so that it is between a b
05:35
my professor said well ordering principle is axiom because its proof is proved by an axiom but isn't all theorem proved by axiom then we call every theorem axiom?
Every nonempty subset of N has a smallest element. In fact, we cannot prove the principle of well-ordering with just the familiar properties that the natural numbers satisfy under addition and multiplication. Hence, we shall regard the principle of well-ordering as an axiom
strange
well, whether it's an axiom or a theorem depends on the particulars of what you assume about N. a lot of books aren't in the business of building N from any particular foundation, so it might be one, or the other, or both.
i really don't like this book, haha
all of the proofs you've pasted from it hint at a pretty unusual pathway being taken through the material.
I think well-ordering is equivalent to induction.
i might have to agree with Leslie, yet again.
what book is this?
if my professor didn't mention about equivalence to induction I wouldn't know that
how do you get the idea of induction?
@TedShifrin wade william
analysis
oh, interesting. I actually like the later parts of his book.
The beginning seems a bit torturous.
@TedShifrin why would that be herr shifrin?
05:43
Because it’s more natural to say tHat induction characterizes the natural numbers and well-ordering is a consequence. No lub axiom needed.
But he must have had his reasons.
i would have only thought about being consequence not converse...
yeah it seems to be better to show the equivalence
> Alcohol and calculus don’t mix so don’t drink and derive.
no i don't agree .the proof is gonna be bit longer for equivalence
05:57
Beginnings of books can be problematic. @leslie That property is essential for students to understand. Whether it should be a theorem, I dunno.
oh, i meant the style of proof. it's an unnecessary use of contradiction, and i don't know how much students get out of a proof by contradiction that ends "but epsilon being less than or equal to 0 contradicts it being positive!"
strikes me as more of a thing you'd put in a book to illustrate 'proof techniques,' or something.
in theory i suppose you could rewrite most analysis proofs to end with epsilon \leq 0 as the contradiction. that would be a good idea for a prank textbook.
that's my get rich quick scheme. i will write one and advertise it on tiktok.
I guess some contradiction is unavoidable. If not, we’d have a smaller upper bound. End.
proof by authority
I never taught out of Wade, but I looked at it as a reference for students.
what book you use to teach student?
06:03
@shin I often told my students they were indulging in proof by intimidation.
@NotTfue I never taught real analysis per se. I taught Spivak’s Calculus 14 times.
Multivariable analysis is contained in my multivariable math book.
@TedShifrin that is more of analysis than calculus
ted's marketing division told him that putting 'analysis' in the title hurts sales.
makes sense lol
let's call it mathematical analytics
LOL, no, but true
@NotTfue It has everything one needs from calculus in it.
06:13
Happy Thanksgiving to those that celebrate it!
2
06:26
Happy Thanksgiving!
for everyone else, happy thursday!
for thursday, happy everyone else!
well, that's covered just about everything.
@TedShifrin then why not teach algebra I using abstract algebra book :) it has everything one needs from algebra too
jk
06:46
Happy
for x all, let x be happy
today, chat is wholesome
somehow xxx does not seem wholesome
just looking at my dvds, i mean
some selected titles, fields medalists do pde in the dark, push forwards on gigantic manifolds & endless one point compactions
today, there was an attempt to steer chat clear of grotesque reference to acts of impudence
oops, i didn't get the memo.
my favourite, smooth manifolds with two point discontinuities
07:37
Does anyone know whether $n!\mod m$ can be calculated efficiently for large $n,m$ assuming that the prime factorization of $m$ is known ? And if yes, how ?
 
3 hours later…
10:36
Anyone know some cool Fibonacci facts?
oh btw, happy Fibonacci day!
10:57
is it right to argue that there isn't a matrix $M$ such that $M^3 \ne 0$ and $M^4=0$ because, assuming by contradicion that this exist, it would be from $M^3 \ne 0$ that $M^3 \cdot M \ne 0 \cdot M$ and so $M^4 \ne 0$. Hence, it would be $M^4 \ne 0$ and $M^4=0$, a contradiction
11:08
3
A: Finding the correct contour to evaluate an integral with finite bounds of integration

Marko RiedelI would like to add some commentary to the excellent answer by @DrMV, showing how to compute the residues involved. We will use $$f(z) = \frac{1}{1+z^2} \exp(1/2 \times\mathrm{LogA}(1+z)) \exp(1/2 \times\mathrm{LogB}(1-z)).$$ Here $\mathrm{LogA}$ denotes the branch of the logarithm where $-...

So if there are three components $\sqrt{(1+z)(2+z)(3+z)}$ what branch can be chosen?
11:39
@Sonozaki no
matrices have something called zero divisors
which means the implication $x \neq y \implies xz\neq yz$ doesn't always hold
12:08
@Jakobian thanks for help, what if I add the hypothesis that $\text{rank}(M) \ge 1$?
12:25
Hi Folks, could someone have look at my question about the delta method? math.stackexchange.com/questions/4584128/…
 
3 hours later…
15:30
In a macbook, suppose that one wants to look at collection of photos. Is there a way to click one photo and then clicking forward in the photo app to get to the other photos?
In Windows, there is an option like that.
But in mac, it seems to me that to switch to another photo, has to first exit the photo app and then go to the desired image and then click it.
15:47
wdym @Koro?
Like using the pre-installed photos app?
Just use the arrow keys if that's the case.
yes in the pre-installed app. arrow keys don't take me to the next photo in the list.
@Ajay
What OS are you using?
you mean version of mac OS?
Yes.
it's ventura 13.0.1.
15:53
hmmm... that's strange.
I'll switch back to windows/androids in next few years in a phase wise manner. I am not quite satisfied with Apple products. I'm finding lot of issues in Apple devices.
Try exporting the photos into a folder on your mac.
They are in the same folder already.
See here for example. I am not surprised.
I've been using my mac for more than 3 years and haven't had any issues.
I'm not familiar with windows. Juts use what you're most comfortable with.
Not just issues/glitches. The problem is lack of convenience in using it. For example-typing something on iphone, pointing cursor is difficult, in mac this photos issue ad nauseam.
anyways, nvm.
15:59
Well, for the cursor you can adjust the sensitivity. For the iphone you can make the keyboard bigger so it's easier to type.
For the photos, have you tried looking at the photo user guide?
no I have not.
I didn't know until now it even existed.
Well take a look, maybe mac isn't so bad.
it is the worst for sure. At least, for me.
is $e^n\times e^m $ an m+n cell?
As an iPhone user, I will concede that pointing cursors is annoyingly difficult
16:06
$e^{m+n}$
It used to be easier on 3D touch enabled iPhone but not anymore
@PNDas Yes.
how should I see this?
$(e^m)^n$ is a multiplication cell.
I am talking about cell complex in algebraic topology
are you talking about the same thing? Ajay?
16:09
yeeessss......
nah i'm cappin
I thought u were talking about regular indices and stuff
$X\times Y$ has cells of the form $e_\alpha^n\times e_\beta^m$ where $e_\alpha^n$ is a cell in $X$ and $e_\beta^m$ is a cell in $Y$.
idk anything about AT, but here.
Actually, just read this
@PNDas
Is $\sqrt{x^2}=|x|$ a definition or can it be proven?
@Ajay, I came here from that page. Ha ha
Anyway $S^1\times S^1$ has cell structures $\{e^0\times e^0,e^1\times e^0,e^0\times e^1,e^1\times e^1\}$ also $\{e^0,e_1^1,e_2^1,e^2\}$
So I was thinking whether $e^n\times e^m=e^{n+m}$?
I would guess so. However, i'm not sure and I don't want to give you wrong info. I think it would be best that you ask someone else.
I can answer questions relating to analysis of some sorts. However, I can't do geometry, topology or anything similar.
16:25
Actually $e^n$ is just the open n-disk. also there is a attached map $\varphi:S^{n-1}\to X^{n-1}$.
16:50
yes, $D^n\times D^m\cong D^{n+m}$
note $I^n\cong D^n$ for all $n$ and $I^n\times I^m=I^{n+m}$
 
1 hour later…
18:09
@Thorgott Yeah but do we need to consider the attaching maps?
18:34
Also $\pi_{n+1}(X)$ is trivial implies $\pi_{n}(X)$ is trivial. It was only states in my notes. No proofs were given.
So I need to show that a map from $S^n\to X$ can be extended to $S^{n+1}\to X$.
How to do that?
@PNDas This certainly looks wrong.
@schn No, not definition. The definition of $|x|$ is $$|x|=\begin{cases} x, & x\ge 0 \\ -x, & x<0\,.\end{cases}$$
@TedShifrin any counter examples?
hmm got it $S^1$
Yup. That's the easiest one.
I don't know may be I wrote it wrong.
the converse is also not true.
Very much not.
18:42
$S^2$ is a counter example
With what $n$?
You might look here for more interesting ones.
@PNDas for what? to construct a CW structure on a product? yes. to see that a product of cells is a cell? no.
@Thorgott okay.
where else does the chinese remainder theorem show in mathematics up as a theorem of great importance?
if that's not a too open question to ask
18:59
Where else besides what?
besides in an intro absract algebra book
It's not a deep theorem I'd say. Certainly shows up in elementary NT and in the ring-theoretic generalisation it's useful in commutative algebra and algebraic NT
im trying to determine whether or not to skip reading the section basically
It's a basic theorem, I'd say don't skip it
there's a lot of theory that doesn't use it, certainly. anywhere that might end up drawing from the ring structure of Z, it might end up coming up.
19:02
oh okay
it strikes me as one of those fairly fundamental things that you can skip on a first go-round, but, maybe don't skip if you might be in a class or other environment where people expect you to know it.
It tells you, at the basic level, that $\Bbb Z_{mn} \cong \Bbb Z_m\times\Bbb Z_n$ as rings when $m,n$ are relatively prime. This is not obvious.
here's an example application in commutative algebra: any Artinian commutative ring is a product of local Artinian rings. That's certainly useful if you want to study Artinian rings
Certainly, my algebra students couldn't skip the result.
our class is skipping it and going straight into euclidean ideal domains, PIDs, and unique factorization domains, which are our final topics for the semester :o
19:03
it came up the other day when someone was trying to count not-necessarily-unital ring homomorphisms from Z to Z/nZ.
it's also useful when you study modules over a PID
if your class is skipping it, that's a sign that you might be able to skip it. but it couldn't hurt to know it. and any later class that assumes your class as a prerequisite, might assume that you know it.
this is the general problem with instructors skipping stuff. :D
wasn't that someone exactly @SillyGoose ?
hahah i was about to say, i looked it up.
it came up when you asked that question the other day. hahaha.
oh XD
oh yes!
i remember
19:06
six long days ago.
lukas was there too. we're having a reunion.
Count me out!
ted: phbbhtht
@SillyGoose if you plan to take other courses in the algebraic or number-theoretic direction, then you probably should study the CRT at some point. If your course is officially skipping it, you don't have to do it now
ted's just sad because when he went to his high school reunion, his best friend eudoxus couldn't make it.
but also studying it now is an option, because I'd say that the CRT is not that hard to grasp or prove
19:13
hm okay i think i will learn it ! it's break anyways
i just looked at my notes from my first algebra course. we skipped it.
don't turn out like me. learn it now
we learned it in our linear algebra course (which to be fair, covered module theory over PIDs, so it was anatural inclusion)
in a linear algebra course 0.0 nice
yeah, i landed in a linear algebra class which assumed we knew it already.
wait is it usual for an intro linear algebra course to cover module theory?
19:18
@SillyGoose in our university, it is. We have freshman linear algebra courses, the second semester of which covers module theory over a PID
the average first course would not do that. this was a second course
yeah second course, here, too
but linear algebra I and II are required courses for freshmen math majors here
yeah, we also did structure theory of f.g. modules over PIDs in 2nd semester linear algebra
yeah I think that's standard in Germany
ah my school has very...um <strike>bad<\strike> questionable major requirements xD
you can get a physics major without taking one of the core courses (of electricity and mag, stat mech, quantum, mech)
hm maybe i will take advanced linear algebra as one of my electives though--though would you think self studying hoffman and kunze would suffice to have decent linear knowledge? it's hard to know whether to delve deeper into the same subject or get more exposure
19:24
linear algebra is essential in a lot of different fields. Certainly couldn't hurt to take an advanced linera algebra course
@SillyGoose it doesn't seem to cover the STFGMPID
what is STFGMPID?
Suppose that T is a path connected space.
Is the sum of two distinct positive irrational numbers irrational?
silly there is a rough division between applications-oriented linear algebra and 'abstract algebra'/theory oriented linear algebra. hoffman and kunze is closer to the first half of that.
Then fundamental group at any point is isomorphic to that at another.
19:29
@SillyGoose structure theorem for finitely generated modules over a principal ideal domain
at my undergrad instructors had a lot of freedom to choose what the second course would cover, some would get very algebraic, some would get very into numerics and matrix factorizations and things that only mattered over the real and complex numbers.
So we could identify the fundamental group of T as fundamental group of T w.r.t. some x in T.
seeker: sometimes yes, sometimes no.
But apparently this could lead to error. Why,how?
oh really @leslietownes what text would treat the latter well?
19:30
koro i don't know that it would 'lead to error' all by itself. but sometimes you do need to keep track of basepoints and forgetting about them requires you to do work later on.
If X is path connected, all the groups $\pi_1(X, x)$ are isomorphic, so it is tempting
to try to “identify” all these groups with one another and to speak simply of the fundamental
group of the space X, without reference to base point. The difficulty with
this approach is that there is no natural way of identifying $\pi_1(X, x0) $with $\pi_1(X, x1)$;
different paths α and β from x0 to x1 may give rise to different isomorphisms between
these groups. For this reason, omitting the base point can lead to error.
@leslietownes Yeah I was thinking the same. I thought I remembered reading somewhere that it was always true. But then I realised that then it would be trivial to prove that pi + e is irrational. But thanks!
@LukasHeger I dont think so 0.0
seeker: stuff like sqrt(2) + (2 - sqrt(2)) = 2. or more generally a + (b - a) for your favorite irrational a and rational b. :)
silly: friedberg/insel/spence is a popular book in this area. it isn't super numerical, but has at least the basis for some stuff that can get numerical.
oh wait are you saying FIS is closer to the applications side?
19:33
for modules over pid, you could also study TS Blyth's.
@leslietownes Got it!
@leslietownes I think "the latter" from silly refered to "abstract algebra'/theory oriented linear algebra"
I don't understand why the last line in the para I posted above says 'can lead to error'.
Why would there be an error?
@SillyGoose if you want modules over a PID and the abstract/theoretic side of linear algebra, you could study chapter 3 of Jacobson's "Basic Algebra I"
it simply says there's no canonical way of identifying $\pi_1$ at different points
19:40
and maybe supplement that with something that deals with inner products etc. these are special to real/complex numbers, but pretty important, too
if you choose a fixed path between any pair of points in $X$ and identify the $\pi_1$ along that, it won't lead to error, but that's a silly thing to do
just work with the fundamental groupoid
essentially, if you just pretend the $\pi_1$ at different points are equal and do calculations, you unknowingly end up working in the set of conjugacy classes of $\pi_1$ instead
conjugacy classes?
you mean equivalence classes?
no, I mean conjugacy classes
if you identify $\pi_1(X,x_0)$ with $\pi_1(X,x_1)$ via two distinct paths, the difference of going back and forth with each is an automorphism of $\pi_1(X,x_0)$ that is given as conjugating with the element represented by the loop given as the difference of those two paths
19:54
ok. that's not been defined in the chapter.
:(
it's a general concept from group theory
koro: it's probably also buried in the proof that pi_1(X,x_0) and pi_1(X,x_1) "are isomorphic." there's got to be a map in there somewhere.
20:21
yes, isomorphic if there exists a path starting at x_0 and ending at x_1.
so holds in particular in a path connected space.
rather than worrying about the precise wording or what can be said more precisely, I think the point you should understand is simply that whilst this isomorphism exist, it does in general depend on the choice of that path
and there's no reason for why any one path should be preferred over any other in doing as such, so there is no "natural" way of identifying $\pi_1(X,x_0)$ and $\pi_1(X,x_1)$ and hence we don't do it usually
and if you were to suddenly identify these two groups in multiple ways at once, that's when error can happen
Suppose group G is isomorphic to group G'. Then it does not matter what isomorphism we take between G and G'. So why is it the concern here?
i.e., in this case why do distinct paths matter? The groups $\pi_1(X.x_0) $ and $\pi_2(X.x_1)$ are isomorphic.
20:36
please don't call it $\pi_2$ :o
it doesn't matter to the claim that they are isomorphic
koro if you have groups G and G' whose elements are different things, and there is more than one isomorphism from G to G', and you're in the "identify everything" mode, you might wind up in a situation where f and h are isomorphisms from G to G' and g is in G and you're regarding f(g) and h(g) as "the same" when they are in fact different elements of the same group G'
e.g. your 'the same' can wind up blurring what equality is
in a way that you aren't necessarily keeping track of
my high level comment would be, don't worry about it and wait until it happens, then you'll have a reason to care
i'm being entirely serious
I understand. So basically, the 'error' part was put there thinking that the reader might not be familiar with abstract algebra.
because yes f(g) and h(g) need not be the same.
fun fact: there is always more than one isomorphism between two isomorphic groups unless they're both isomorphic to $0$ or $\mathbb{Z}/2\mathbb{Z}$
thanks a lot @leslietownes and @Thorgott.
I can now go back to trying to understand coproducts of modules.
:-)
20:46
ahhhhh
nice
@Thorgott this is a fun exercise to toy with btw
20:59
i agree
21:24
Great note to self.
 
3 hours later…
23:57
i agree

« first day (4495 days earlier)      last day (822 days later) »