Mathematics - 2022-12-04 (page 1 of 2)

12:10 AM

Got the result for my Real Analysis 2 exam

98 out of 100

And the reason I lost those 2 points, most likely, is because I messed up the OMR sheet and you can't really change it once you've marked it. So I was stuck with having marked one answer wrong knowingly

Ted Shifrin

Are you telling me a Real Analysis 2 course had a multiple choice exam?

Goku

@TedShifrin I was surprised too. Though to be fair it was only 10 really easy questions

Ted Shifrin

@leslie @robjohn Please check me if you have a chance.

Goku

The rest was just a normal exam

Ted Shifrin

Oh, so you had to write proofs for the rest. OK.

I don't mind some T/F or multiple choice; they can be very thought-provoking, actually.

Balarka Sen

12:18 AM

omr sheet for 10 multiple choice qs seems overkill

Goku

Yes the rest of it was just proofs. Some improper integrals here and there too

@BalarkaSen all of our MCQs are conducted on an OMR

Balarka Sen

interesting

Ted Shifrin

@Goku Check out the question/answer I just linked.

Goku

@TedShifrin I was just reading that.

leslie townes

i would use multiple choice and t/f mainly as a way of giving weaker students a randomized chance at free points and a quick way of differentiating solid A students from the rest (which is hard to do with questions individually worth a lot that you expect everyone to answer)

Goku

12:27 AM

@leslietownes Thats exactly why I think we have multiple choice. Not sure if "weeding out weaker students" is still necessary at Real Analysis 2 but most of us suspect that's why it was there

Ted Shifrin

The second analysis course at MIT when I was there covered Lebesgue integration and multivariable analysis/differential forms. I opted not to take it from Warren Ambrose (quite famous) because he gave only take-home true/false exams. The questions were sooooo tricky.

Goku

I personally prefer more "traditional" style questions than multiple choice or true/false

That's why I find contest problems interesting. You're given a question statement and you just need to solve/prove it.

leslie townes

oh, i wouldn't use it for weeding, as much as reducing score variance by boosting lower scoring students a little (e.g. many could get 3/10 on T/F by random chance) while adding enough static at the top end to differentiate A from A-.

Ted Shifrin

I typically threw in a "prove or give a counterexample" question once in a while.

leslie townes

haha, i've heard of ambrose.

Ted Shifrin

12:33 AM

Ambrose-Singer may be the most famous theorem, but he did plenty of PDE analysis.

Goku

Did anyone here ever have the unfortunate experience of walking through the humanities department?

Ted Shifrin

You talking about MIT?

Balarka Sen

I have had the unfortunate experience of being a graduate student in the mathematics department

Closely comparable

leslie townes

i've never heard of a department called 'humanities' although you sometimes find that in the names of higher academic divisions.

Ted Shifrin

I had plenty of wonderful humanities experiences, but there were lots of locations (at MIT).

leslie townes

12:35 AM

sometimes even divisions that include math.

Balarka Sen

I am switching to the Collapse of Western Civilization Studies department

Goku

I think one of the history professors here was suspended a few days ago. Why? He referred to some old historical figure using the "incorrect" pronoun (according to a student who brought attention to this), even though that's exactly what's written in the book? Yeah I'm confused too

Ted Shifrin

That doesn't merit suspension. Something's very fishy.

leslie townes

i'd need more details to even begin to form an opinion about that.

that's very close to, "Mrs. Krabappel and Principal Skinner were in the closet making babies and I saw one of the babies and the baby looked at me"

Goku

It absolutely doesn't merit suspension. The student also accused the same professor of racism right after that. Its making no sense

Jakobian

1:11 AM

@leslietownes that's from Simpsons right

leslie townes

yes.

shintuku

1:22 AM

HUMANITIES QUEEN OF SCIENCES

GO HUMANITIES

HUMANITIES REPRESENT

Koro

6:54 AM

Suppose that f is a 0 form. $dx_I$ represents elementary k form. Then consider the k form $w=f dx_I$. Let d be the differential operator. $dw= d(f dx_I)= d(f\wedge dx_I)=df\wedge dx_I+ f\wedge d(dx_I)=df\wedge dx_I$.

I don't quite understand why the last inequality is true. Why is $f\wedge d(dx_I)=0$?

one potato two potato

@Koro $d^2= 0$.

leslie townes

Koro

oh nvm. I got it- $dx_I:= dx_{i_1}\wedge \cdots dx_{i_k}$. Now I know the result that $d(df_1\wedge ...\wedge df_k)=0$, where $f_i$'s are k forms.

@leslietownes: I thought that the groups upto order 5 were all abelian.

leslie townes

koro, this is me interrogating the chatGPT chatbot, which is trained on text alone. hence the mixture of truth and gibberish.

its group multiplication table is correct (it's missing the left hand column telling you what the rows correspond to multiplication by), and some of what it says about the cyclic group of order 3 is correct. but obviously not all of it.

Koro

ah, I got some news notification about that chatbot. Why is it famous/trending?

I didn't read through the news.

leslie townes

7:01 AM

it's pretty good at prompts with lesser precision.

that's all.

Koro

oh

leslie townes

if you say "compare and contrast the positions of [BS artist 1] and [BS artist 2] on [BS topic]," it will generate something that would get at least a C, maybe a B, maybe an A, in the kind of university-level class where that kind of exercise is homework.

but it's based on learning texts, and text prediction, so it gets a lot of math spectacularly wrong.

Koro

oh I see.

@onepotatotwopotato that is true but I don't think this is the correct explanation.

$dx_I$ is the notation for elementary $k-$ form. d(form ) need not be zero but d(d(form))=0 is correct, I think.

leslie townes

if you ask it to factor an integer of moderate size, it will correctly state the brute force method that proceeds by trying each of the primes in succession, but (for example) it will test the "divisibility" of a floating point N/2 by the integer 3 even when N/2 is not an integer, and generate a list of nonsense factors on that basis, based on how close text representations of floating point numbers come to integers, up to the number of primes that you ask it to spend CPU time on considering.

it's a mess, but a funny mess.

one potato two potato

@Koro $dx_i$ is a differential of $i$th coordinate function $x_i:\Bbb R^n\to\Bbb R$.

Koro

7:11 AM

yes, indeed. That equals the elementary k form.

one potato two potato

7:23 AM

@Koro So anyway you can say $d^2 =0$ so $d(fdx_I) = df\wedge dx_I$. But doesn't the textbook define that as a definition? exterior derivative of $n$-form.

Koro

actually, the doubt that I asked earlier arose from a theorem showing existence and uniqueness of differentials.

In the proof of the theorem, the book shows that $d(fdx_I)=df \wedge dx_I$.

based on the following:

33 mins ago, by Koro

oh nvm. I got it- $dx_I:= dx_{i_1}\wedge \cdots dx_{i_k}$. Now I know the result that $d(df_1\wedge ...\wedge df_k)=0$, where $f_i$'s are k forms.

the result is also proven by induction in the proof of theorem.

the proof of the result assumes that d(d(form))=0.

Koro

8:39 AM

what is a simplex?

and how to compute its boundary?

leslie townes

need more context, koro. lots of definitions out there, who knows what you're interacting with?

rudin's PMA has one definition, is that it? or is it another?

hatcher has a definition, too.

one potato two potato

6

Q: Smooth lifting criteria of smooth vector fields given smooth surjective submersion whose fibers are connected.

I am working on an exercise: Suppose $F : M \to N$ is a smooth submersion, where $M$ and $N$ are positive-dimensional smooth manifolds. Given $X \in \mathfrak{X}(M)$ and $Y \in \mathfrak{X}(N)$, we say that $X$ is a lift of $Y$ if $X$ and $Y$ are $F$-related. A vector field $V \in \mathfrak{X}(M)...

leslie townes

same with "boundary."

one potato two potato

How can I show this without using flow? It's a problem that appeared before the notion of flow is defined.

Koro

9:18 AM

@leslietownes a function defined on [0,1]*[0,1] with values in R^3 was given. It was asked to find boundary of this.

Mad

10:03 AM

https://chat.stackexchange.com/transcript/message/62490157#62490157
could someone look at this

Nick

11:55 AM

Hi,

I found something of interest: youtu.be/uMMQbcisins

Do we have her visiting here sometime?

one potato two potato

1:02 PM

If $f$ is an entire function and $f(a+bi)\to 0$ as $b\to\infty$ then $f\equiv 0$?

Oh $f$ can be $e^{iz}$

robjohn

@Nick I should start one of those: "idiot answers stupid questions"

Magnus Alexander

1:33 PM

Hello everyone! I'm struggling with a task about Dirichlet kernel: I need to prove that $|D_n (t)| \leq \dfrac{\pi}{2t}, \ t\in [-\pi, \pi], t\neq 0$. Can anyone give me a hint with this one? I know, as a fact, that the Dirichlet kernel can be represented as $\dfrac{\sin{(n+0.5)t}}{2\sin{\dfrac{t}{2}}}$, but still can't really get the needed inequality.

Jakobian

2:41 PM

@MagnusAlexander $|\sin(x)| \geq 2|x|/\pi$ for $x\in [-\pi/2, \pi/2]$

this is one of those standard inequalities you have

geocalc33

3:14 PM

$$g_{jk}(\theta)= \int_X
\frac{\partial \log p(x,\theta)}{\partial \theta_j}
\frac{\partial \log p(x,\theta)}{\partial \theta_k}
p(x,\theta) \, dx.$$

do you include the normalization constant in this calculation? It makes the calculation messy because the logarithm doesn't let you take out this constant outside of the integral

I did an example without the normalisation constant, just the pdf with variable and parameter, and got a simple result

schn

3:33 PM

In the general definition of the limit, for a function $f$ continuous at $a$, we have $$|x-a|<\delta\implies |f(x)-f(a)|<\epsilon.$$ For right-continuity at $a$, however, $|x-a|<\delta$ reduces to $a\le x < a+\delta$. I am confused about this $\le$. Why $a\le x < a+\delta$ and not $a< x < a+\delta$?

user123234

4:15 PM

We have discussed the following theorem:

>Let $X$ be a Hilbert space and $T:X\rightarrow X$ a bounded self-adjoint linear operator. Then there exists a unique map $$F:C(\sigma(T))\rightarrow B(X)$$ ($B(X)$ is the set of bounded linear operators) such that
>
> 1. for polynomials $p$, $F(p)=\sum_{n=0}^N a_n T^n$
>
> 2. $F$ is a $*$-homomorphism i.e. $F(\phi+\psi)=F(\phi)+F(\psi)$, $F(\phi\cdot \psi)=F(\phi)\cdot F(\psi)$, $F(\lambda \phi)=\lambda F(\phi)$, $F(\bar \phi)=F(\phi)^*$
>
> 3. $F$ is an isometry: $||F(\phi)||=||\phi||_\infty$

Obliv

for $\int_{e^y}^{y} y\frac{\ln x}{x}\text{d} x , y>0$ aren't these limits backwords? Shouldn't e^y be the upper limit?

Magnus Alexander

4:46 PM

@Jakobian Thanks! I already got it myself though but still I greatly appreciate your help!

Magnus Alexander

5:05 PM

Here I am with another question :) Given the task: Consider n-dimensional representation $\rho$ of the group $S_n$ set by the formula: $\rho (\sigma)(e_i)=e_{\sigma(i)}$. Decompose this representation into the sum of irreducible representation.

To start with, I don't really get what $e_i$ is. It's a base element of $GL_n (\mathbb{F}$ space?

Obliv

if e^y > y for all y>0 shouldn't the limits be swapped.. I got the answer that my textbook gave but I don't get why it's correct

Magnus Alexander

@Obliv Does it matter after all?

You can swap limits of integration by changing the sign, I don't think it's a big deal, maybe they mistyped it

Obliv

I thought it mattered, but it's been a while lol. Thank you.

I mean it does matter if you're using the result in some way, because it changes the sign but for the purposes of these exercises it does not matter very much.

Lukas Heger

5:22 PM

@MagnusAlexander $e_i$ is the standard basis of $\Bbb F^n$

so the action is just permuting the components of the vectors in $\Bbb F^n$

you have one irreducible component given by the span of $(1,1, \dots, 1)$ one can show that the complement of this is actually irreducible

you can prove this e.g. by using character theory

more generally if a finite group $G$ acts 2-transitively on a set $X$, then the corresponding permutation representation has two irreducible components, one corresponding to a trivial subrepresentation and another irreducible one of dimension $|X|-1$

Ted Shifrin

5:44 PM

@Obliv The answer had better be negative unless $y<0$.

Obliv

twas not, but I at least confirmed here that the textbook missed that :P

I'm running into so many walls doing these integrals because of stuff I learned years ago that I've forgotten :(

Ted Shifrin

It’s easy to forget lots if you don’t use it!

Obliv

$\int_0^{x^3}ye^{\frac{-y}{x}}\text{d}y$ for example: I had to look up integration techniques and figure out that this is an integration by parts situation

leslie townes

it's like running. can't skip 10 years of running and then run a mile without stopping and probably feeling like s--t

Obliv

yeah exactly

Xander Henderson

5:47 PM

@leslietownes Flagged for implied offensive language!

leslie townes

and you're like "wtf this is one mile"

uh oh i did it again

Xander Henderson

@leslietownes FLAGGED AGAIN!

leslie townes

flagged doubly for willful violations

Obliv

but the part I chose for $\text{d}v = e^{\frac{-y}{x}}$ I have to look up now to figure out how to integrate lol

I suspect it's something to do with a chain rule..

Xander Henderson

@leslietownes I'm willing to give you the benefit of the doubt. It might have been unintentional. :P

Obliv

5:48 PM

lots of exponential function properties that I do not remember.

leslie townes

i was feeling like silt, and asking "why this failure"

Xander Henderson

@leslietownes Took a while to come up with that, eh?

:P

leslie townes

obliv: now also with the fun of keeping track of which variable you're doing the integration with respect to.

Obliv

oh yeah so since it's w.r.t. y, I treat x as a constant and it's a simple u sub

wait

Magnus Alexander

@LukasHeger I guess I need to specify. $\Bbb F^n$ is an n-dimensional vector space? I'm kinda lost in the notation

Lukas Heger

5:53 PM

yes

Obliv

yes so $\int e^{\frac{-y}{x}}$ becomes $-xe^{\frac{-y}{x}}$..?

Lukas Heger

if $\Bbb F=\Bbb R$, then surely you know the notation $\Bbb R^n$

same notation here, for a general field

leslie townes

obliv: life will be easier for you if you remember the dx or dy part of integration by parts. think if IBP as involving a "term you differentiate" (often called u - but it has a variable that goes with it that you differentiate with respect to) and a "term you integrate" (often called dv - but it has a variable that goes with it, written "d[variable]" as part of what dv is)

Magnus Alexander

@leslietownes There is actually an interesting method called "DI-method", i guess it was invented by blackpenredpen, nevertheless, he is the one to explain this on his YT channel, be sure to check out

@LukasHeger So e_i is a base vector of the field, not matrix? Meaning our action works like this: it takes element of S_n and changes the given base vector on another one, by changing it's index correspondingly to the chosen permutation?

Lukas Heger

yes

and this action extends linearly

you just permute the coordinates of vectors

Obliv

6:00 PM

@leslietownes ah okay yeah it's coming back to me, I would pick the terms that get simpler by differentiation and integration.

Magnus Alexander

Hm. I just know that representation is a homomorphism from the group into general linear group of corresponding order, meaning there has to be matrices at some point and I don't really get where they will come up here

Let's go down for $\Bbb R^n$ then. Base elements are described like this: $e_i = (0.....1...0)$, where $1$ stands on $i$'th place. Let's take the permutation $\sigma \in S_n$, doesn't really matter how it works exactly at this point. So, what this representation does? It takes this $\sigma$ and acts on the index, so we get another base vector. But as far as I know, within this notation given in the task, $e_{\sigma(i)}$ should be an $n\times n$ matrix, not vector

Obliv

6:18 PM

ooh didn't realize integration by parts evaluates the whole expression over the definite integral limits..so I'm guessing if there is ever a situation where $\int u\text{d}v = uv - \int v\text{d}u$ the indefinite integral on the right can't be found, then this method doesn't work for a definite integral

because you have to evaluate the indefinite integral to get something to plug in the definite integral limits with

leslie townes

obliv, but bear in mind sometimes there are situations where one wants to apply IBP multiple times. in these situations it's helpful to ignore any bounds that might be relevant to you and compute an antiderivative first. and then only when you have it, work with bounds.

put slightly more vaguely, if you don't yet have something that you can plug numbers into, forget about the fact that you're taking a definite integral and keep trying other integration techniques as if you are only taking an indefinite integral.

Obliv

so you do the IBP the first time for a definite integral, the bounds to plug in remain the same, you can then do IBP for the indefinite integral as many times as you wish without worrying about the bounds until the end?

leslie townes

this goes for other techniques too, e.g. substitution. it's often a lot simpler to first find an antiderivative, and when you're done with that, plug in bounds.

instead of using the formulations of those 'rules' that keep track of how the bounds change.

there are exceptions where you very much want to pay attention to that, but i think a useful default rule is, don't worry about bounds until you're "done integrating"

Obliv

oh yeah changing bounds for the substitution seems unhelpful

leslie townes

as in, do not write them in the integral at all. act as if you don't know what they are.

Magnus Alexander

6:22 PM

@Obliv Keep in mind that there are integrals that are, strictly speaking, won't be calculated via IBP method in a sense that $\int vdu$ part won't be a simpler integral, but still there is a good idea to use IBP. Example: $\int e^{\alpha x}\cdot \sin{\beta x}dx$

Obliv

yeah so that's an example of when IBP can't solve the definite integral

Lukas Heger

@MagnusAlexander do you know what a group action is?

a representation is a group action on a vector space that is linear

I'm very certain that $e_{i}$ and $e_{\sigma(i)}$ are vectors, not matrices

if you want to describe the homomorphism $S_n \to \mathrm{GL}_n(\Bbb R)$ that corresponds to this representation, you can do that. It sends an element of $S_n$ to the corresponding permutation matrix

there's not just one way to write down a representation

if you have a homomorphism $\rho:G \to \mathrm{GL}(V)$ you get an action of $G$ on $V$ via $g \cdot v := \rho(g)v$

this is a group action with the property that every for every $g \in G$, the map $v \mapsto gv$ is linear

and if you have such an action, you can build a group homomorphism $G \to \mathrm{GL}(V)$ from that

Magnus Alexander

@LukasHeger We had the definition of the group $G$ acting on the set $X$, that this is a map $G\times X \rightarrow X$, that satisfies certain conditions like (not sure that's how you say it on english) outer associativity. And we had a definition of linear representation: it's a homomorphism $G \rightarrow GL(V)$, where $V$ -- is a vector space above the field $\Bbb F$

Lukas Heger

there are different equivalent ways to write down the datum of a group action or a representation

for example maybe you know that a group action of $G$ on $X$ is the same as a homomorphism $G \to \mathrm{Sym}(X)$, where the latter is the group of all bijections from $X$ to itself

Magnus Alexander

Yeah, I think so, I'm just giving you the definitions that I've been given, thank you for your patience by the way :)

Lukas Heger

6:30 PM

have my explanations made things clear?

Magnus Alexander

Yes, we had this statement as a theorem

Lukas Heger

ah I see

Magnus Alexander

Not really, yet. But I'm trying to understand

Lukas Heger

so for (non-linear) group actions you have two different ways of presenting them: one as a homomorphism and one as a map $G \times X \to X$

there are similarly two different equivalent ways to present a group representation

so you have an analogous theorem in the linear world, if you wish

Magnus Alexander

I'm confused about the fact that $e_i$ is a vector, not a matrix, I don't get it ;(

It doesn't correspond to the definition of representation that I have

Lukas Heger

6:33 PM

a matrix is just a linear representation

you can specify a linear representation by saying where each basis vector goes

Magnus Alexander

Oh

Lukas Heger

and they are giving you that information

Magnus Alexander

Now I get it ahahah

Lukas Heger

in addition, I'm saying that there's a similar theorem to the one on group actions that you know

a group homomorphism $G \to \mathrm{GL}(V)$ is "equivalent" to a map $G \times V \to V$ that satisfies some properties

Magnus Alexander

They described linear representation by describing how it works on base elemtns

And to get the matrix, they just stack those vectors, right?

Lukas Heger

6:35 PM

yes

Magnus Alexander

Such a relief gosh

Lukas Heger

the properties are $1 \cdot v = v$, $(gh) \cdot v= g \cdot (h \cdot v)$ (these two say that you have a group action). and in addition you have a linearity condition, that is, $g \cdot (v+\lambda w)= g \cdot v+ \lambda(g\cdot w)$

so representations are a special case of group actions

because $\mathrm{GL}(V) \subset \mathrm{Sym}(V)$

Magnus Alexander

I see! Thank you very much !!
So, another problem is, how do I decompose my representation into irreducible ones?

Lukas Heger

do you know character theory?

Magnus Alexander

The point here is we had a definition of "irreducible representation": it's a representation that doesn't have any invariant subspace

I'm familiar with it, we had a few lectures, but I never had to deal them in terms of solving problems

I'm pretty sure this task has to be solved without characters, but I would love to hear how it works anyway

(Because we have two different problem sheets, one on representations, another one on characters)

Lukas Heger

6:39 PM

I'm pretty sure this question has been answered before on MSE, also without characters

Magnus Alexander

Well, I haven't found it by so far

Lukas Heger

14

Q: Permutation module of $S_n$

Let $G=S_n$ and let $V$ be the permutation module of $G$ with basis $\{x_1,\ldots,x_n\}.$ Let $\lambda, \mu \in \mathbb{C}$ to allow one to define a $\mathbb{C}G$-homomorphism $\rho:V \to V$ by $$\rho(x_j):=\lambda x_j+\mu\sum_{i \neq j}x_i.$$ By using the above fact or otherwise, how can we pr...

abstract-algebra representation-theory modules

this works over $\Bbb C$

but the proof over $\Bbb R$ is the same

Magnus Alexander

Module is something a bit cooler than the field right?

Thanks, I'll try to work it through!

Lukas Heger

so there's a third way to describe reprentations. You can build a ring from a field $k$ and a group $G$, the group algebra $k[G]$ and modules over that thing are representations

Magnus Alexander

Sadly, I'm unfamiliar with these terms ahah

Obliv

6:44 PM

integrating a partial derivative of a function nets you the part of the function with that variable, and also leaves you with a constant function of integration, correct?

Lukas Heger

to understand the answer, you can just replace the term $FG$-submodule with subrepresentation, and you will be fine

Obliv

what's a common way to write that? $f(x,y) = \int f_x(x,y)dx = f(x) + C(y)$? feel like that's not right...

Magnus Alexander

@LukasHeger Do you mind giving me an answer to my original problem? I need to decompose this representation into a direct sum of irreducible ones. I'll try to play with the given solution that you've sent, but I'm afraid that I might make a mistake somewhere so I want to have an answer to check myself

Lukas Heger

@MagnusAlexander the decomposition is $V_1:=\langle (1,1,\dots, 1)\rangle$ and $V_2:=\{(a_1, \dots, a_n) \mid \sum_{i=1}^n a_i = 0 \}$

Magnus Alexander

@LukasHeger Thank you very much!!

Lukas Heger

6:51 PM

of course you need to prove that these are irreducible

that's what is done in the linked answer

Magnus Alexander

Yes, I understand

Koro

is there an algorithm to write a diffeomorphism as a composition of primitive diffeormorphisms?

say, I want to write $(x,y)\mapsto (x+y^2+2x^3y+x^6, y+x^3)$ as a composition of primitive diffeomorphisms.

Thorgott

of what

Koro

for the map I wrote, for example.

Thorgott

no I'm asking what a primitive diffeomorphism is

never heard of it

Koro

7:11 PM

Suppose that $f:U\to R^n, U\subset R^n$ is $C^1$ and is of the form $f(x_1,x_2,...,x_n)= (x_1,...,x_i+\alpha (x), x_{i+1},...,x_n)$.

where $\alpha: U\to R$ is a $C^1$ map.

So f changes in atmost one coordinate. Such f is called primitive diffeomorphism.

@Thorgott.

Obliv

7:25 PM

Do you need information of the functions if asked to find the area of the region bounded by $g(x)$ and $f(x)$ between $a \leq x \leq b$?

or can I just set up the integral like $\int_a^b \int_{g(x)}^{f(x)} dy dx$

leslie townes

two functions of one variable bounding a 'region' with an 'area' on an interval? how is that a multivariable calculation?

if you can do it via calc 1 methods, just do that

Obliv

iterated integrals, i've learned these years ago too I'm just relearning it for this section

leslie townes

oh, so you want to set the thing up just to fit it into the multivar framework

sure, what you've done will work, (if f(x) >= g(x) on the interval of x)

can you give an example of something you'd like to do, but can't, because of this perceived obstacle? certain restrictions on 'variables' are simply necessary for the iterated integral to make sense.

Obliv

well in the textbook it says i.imgur.com/Q2gB46G.png

the idea is that the inner integral gives you something standard for the outer but I was wondering why in general you'd need the outer integral's bounds be independent to the inner bounds.

so like given $\int_{f_1(x)}^{f_2(x)} [\int_{x_1}^{x_2} f(x)dx]dy$ the inner integral evaluates to $F(x_2)-F(x_1)$ then you take the integral of the outer

Koro

7:46 PM

if w= P dx+Qdy is the 1-form on $R^2$, then why is dw= $(D_1Q-D_2P)dx\wedge dy$?

$dw = (D_1P dx + D_2P dy) \wedge dx + (D_1 Q dx + D_2Q dy) \wedge dy$, but from here what happens to $D_1P$ and $D_2 Q$?

Thorgott

@Koro hmm, interesting

surely that's not something explicit

I would guess it's of a similar strength as the implicit function theorem

should correspond to the linear algebra fact that GL(n) is generated by shears (and dilations)

@Koro yes

Koro

ahh, nvm. $dx\wedge dx=0$.

I'm very new to this. :D

$\color{red}{dx_i}\wedge dx_i= (-1)^1 dx_i\wedge \color{red}{dx_i}$.

Obliv

does this make any sense: $y=x, \int_{y_1}^{y_2}\int_{x_1}^{x_2} y dxdy = \int_{y_1}^{y_2}[yx]_{x_1}^{x_2}dy = \int_{y_1}^{y_2}(yx_2 - yx_1) dy = [\frac{y^2x_2}{2}-\frac{y^2x_1}{2}]_{y_1}^{y_2} = \frac{{y_2}^2 x_2}{2} - \frac{{y_2}^2 x_1}{2}-\frac{{y_1}^2 x_2}{2}+\frac{{y_1}^2 x_1}{2}$

like this is why you evaluate $\int_{y_1}^{y_2}ydy$ first then $\int_{x_1}^{x_2}dx$ of the thing you get from the first one

otherwise you just get a bunch of interrelated terms

wow I actually had such a dejavu I literally tried this back in calc 1 to see why the order mattered.

Ted Shifrin

8:04 PM

What's the $y=x$ at the beginning? Recommendation: Keep things factored to avoid mistakes. Write $y(x_2-x_1)$ for the first integtration.

Obliv

oops meant to say y=f(x)

oh yeah..

Ted Shifrin

But what does $y=f(x)$ mean? What's the actual problem? I have a feeling this is wrong.

Obliv

so it should actually be $\int_{y_1}^{y_2}F(x_2)-F(x_1)dy$ i was trying to make it look simpler with y and x but it seems to make no sense lol

Ted Shifrin

You might want to review setting up iterated integrals correctly (see a couple of my videos, perhaps).

Obliv

I was trying to see what happens when you set them up incorrectly

Ted Shifrin

8:07 PM

What is the actual problem?

Obliv

@TedShifrin i.imgur.com/Q2gB46G.png my textbook mentions this fact and I was wondering what happens if the integral bounds were the other way around.

Ted Shifrin

It can get very complicated. Practice on these — Rewrite in the other order of integration: $\int_0^2\int_0^x dy\,dx$, $\int_0^2\int_0^{x^2} dy\,dx$, $\int_0^2\int_x^{x^2} dy\,dx$. Be very careful. Draw pictures.

Don't do any integrals. Just set up limits of integration in the other order.

Obliv

so instead of $\int_{a}^{b}\int_{g_1(x)}^{g_2(x)}f(x,y)dydx$ i wanted to see what happens with $\int_{g_1(x)}^{g_2(x)}\int_{a}^{b}f(x,y)dxdy$

okay I'll try those.

Ted Shifrin

What you just wrote canNOT be written.

You must have constants as the outer limits, ALWAYS.

I suggest you watch my video on this if you're not going to read your textbook very carefully.

Koro

8:46 PM

what is a singular n-chain?

n-chain is linear combination (coefficients should be integer) of singular n-cubes (i.e., continuous map from [0,1]^n to R^n).

Xander Henderson

@Koro It is a type of $n$-chain which happens to not have a date for prom.

Koro

But then every n-chain is also singular.

say $c_1+3c_2$, where c_i is a singular n-cube. Take c= $c_1+3c_2$, then c=1c.

So every n-chain is singular n chain?

Thorgott

oh wow, Ronnie Brown is channeling his spirit through Koro

Magnus Alexander

9:09 PM

Question (Representation theory): can a group have only two two-dimensional and two five-dimensional complex irreducible representations?

Ted Shifrin

@Koro "Singular" is there to contrast with, say, "simplicial" or "cellular." Singular chains are built out of continuous (or smooth) mappings from a simplex to your space.

Magnus Alexander

I have a problem with this one, since the problem that we've solved in class was about three one-dimensional and four two-dimensional. Therefore, we got that $|G|=1+1+1+4\cdot 2^2 =19 $. This is a prime number, and the group has to be cyclic. So we only have $\Bbb Z_{19}$ which is abelian. So all its irreducible representations are one-dimensional then. But in this task we have the group of order 52.

Koro

@leslietownes I understood what it is now. Suppose a:[0,1]^k--> U, U is open in R^n, is continuous, then a is called a k-simplex in U.

Ted Shifrin

A singular $k$-simplex. The name comes from the fact that the image of your mapping may be very degenerate and hardly "$k$-dimensional" at all.

Koro

Its boundary is defined similar to how boundary is defined for an n-chain.

n-chains are continuous maps from I^n to R^n, but k-simplex in U (open in R^n) are continuous maps from I^k to U. So we may say that k-simplex is generalization of n-chain.

Lukas Heger

9:16 PM

you can work with cubical singular homology instead of "simplicial" singular homology if you want

this makes some things easier and other things more difficult

Ted Shifrin

Totally NOT your last sentence, Koro.

Koro

oh, what a symbol $\xrightarrow[]{n\to \infty}$. :-)

Lukas Heger

@MagnusAlexander every group has a one-dimensional irreducible complex representation

Koro

oops, it was looking like a tilted arrow on mac. But I checked in other device, it's the usual arrow.

Ted Shifrin

Not tilted here.

Magnus Alexander

9:22 PM

oh

@LukasHeger Gosh Im so stupid

Koro

Ted Shifrin

That's not tilted. It's a slight pixel error. I bet if you quit and reopen the browser, it'll be back to normal.

Thorgott

or it's an issue with the zoom, I think we had something similar in the past

Ted Shifrin

Is anyone else having issues with MSE continuing to show old notifications, even after one checks "mark all read"? It's relentless and annoying.

Magnus Alexander

@LukasHeger Ok. What about this one: 5 one-dimensional and one 5-dimensional? The order of this group is 30, and the order of the quotient by its commutator subgroup gotta be 5. I have no idea where to go from here

Koro

9:31 PM

@TedShifrin it's very annoying.

The notification keeps showing unless you click 'mark all read'.

Ted Shifrin

The last time I pointed out something on Meta, I got skewered for not being all-knowing.

Even after I check "mark all read" several times, it continues to show the same one(s).

Koro

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Ted Shifrin

I'm gonna go look and see if anyone's complained.

Koro

probably related.

Ted Shifrin

Indeed. It's like all "improvements." It makes things worse for the common man.

I don't see any responses complaining that clicking the "mark all read" doesn't do its job.

Lukas Heger

9:35 PM

@MagnusAlexander now that question is more interesting. it's not possible

every group of order $30$ has a subgroup of order $15$, which is necessarily normal and has abelian quotient of order $2$

hence commutator subgroup needs to have even index

Koro

I don't understand one thing:-

Consider w= $\frac {-y}{x^2+y^2} dx+\frac{x}{x^2+y^2} dy$. I know that w is a closed form on $R^2-\{0\}$.

Magnus Alexander

@LukasHeger Can you explain the last statement? I agree that a subgroup of order $15$ is abelian, and that the quotient by this subgroup has an order $2$, but it's not obvious for me why it means that the commutator subgroup needs to have an even index.

Oh!! Because this abelian group is the subgroup of the commutator subgroup, so the commutator subgroup should have index of $2\cdot C$

Koro

w is apparently not an exact form. I don't understand why without using Stokes theorem.

Ted Shifrin

You can't use Stokes's Theorem.

Wrong statement, Lukas.

Lukas Heger

just check that intergration of a closed path is not zero

Ted Shifrin

9:42 PM

Right. This is just what people learn in standard multivariable calculus.

Lukas Heger

so people learn in multivariable calc that $H^1_{dR}(\Bbb R^2\setminus \{0\},\Bbb R) \neq 0$? interesting

Ted Shifrin

Not in that terminology, but, yes, absolutely.

Koro

@TedShifrin why not?

Ted Shifrin

We learn that curl = 0 does not imply conservative and this is the counterexample.

Lukas Heger

the fact that this form is not exact makes complex analysis interesting

Ted Shifrin

9:44 PM

Well, it's topology that complex analysis steals.

How do you intend to use Stokes, Koro?

Lukas Heger

@MagnusAlexander more generally every abelian quotient of a group is quotient of the abelianization

the important fact is not if your subgroup is abelian, the question is if it's normal and has abelian quotient. If it does, then it contains the commutator subgroup

@Ted I meant if that form was exact, the residue theorem would get a lot less interesting

Koro

I integrate w on 1-simplex $t\mapsto (\cos 2\pi t, \sin 2\pi t)$ to get 2\pi. Assuming that w= df for some $\int_\gamma w=\int_\gamma df=\int_{\delta \gamma}f$.

The second equality is due to Stokes. This evaluates to 0 giving contradiction.

Ted Shifrin

Oh, I don't consider the fundamental theorem of calculus in one dimension to be Stokes's Theorem. It's far more basic.

@Lukas He's not doing that.

@Koro You should prove once and for all, independent of Stokes's Theorem, that $\int_\gamma df = f(Q)-f(P)$ for any path ($1$-chain, if you want) $\gamma$ from $P$ to $Q$.

"Every calculus student" should know this result.

Lukas Heger

@TedShifrin yeah I agree I don't really think of that as an application of Stoke'ś's theorem

Ted Shifrin

@LukasHeger LOL, the name is not Stoke :D

Lukas Heger

9:53 PM

I know

Koro

Strokes :D

Lukas Heger

I was joking

Ted Shifrin

Any more of this and I'll have a stroke!

Koro

@TedShifrin I know that result but never proved it before. I'll try proving this.

It looks like fundamental theorem of calculus.

Ted Shifrin

Exactly. If you use the definition of a smooth $1$-chain, you'll see that it is :P

Lukas Heger

9:59 PM

I remember we defined path integrals via pullbacks of differential forms and almost did no examples, so we were lost at the beginning of complex analysis when actually computing path integrals became important

Ted Shifrin

My largest complaint — from what I've observed for years — about the European math education system is that it is too abstract and that there aren't enough computations/examples.

But the same might be said of certain professors in the US, too.

Lukas Heger

I agree that examples are important. But you shouldn't wait until year 3 with starting to do proofs! I heard that there are some US colleges which almost have no proof-based courses for math majors, which sounds absurd

Ted Shifrin

Oh, I'm not praising the US system. The point, though, is that with our one-size-fits-all viewpoint, at most schools we have a calculus course for science, math, engineering and a separate one for business, economics, perhaps biology. And since our high school education is generally unrigorous, expecting all but the most motivated/talented students to take a proof-based calculus (such as I taught 15 times at the first- and second-year levels) is asking for disaster.

The "advanced placement" calculus courses that most of our high school students take are very much lacking in rigor. Indeed, when I took the exam in 1970, I had to write a $\delta$-$\epsilon$ proof. Now students don't even have to justify the hypotheses of the intermediate value theorem to get full credit.

What do you mean by abstraction?

Lukas Heger

our analysis ("calculus") courses were taken by math, physics majors and some CS majors. Same for linear algebra. All entirely proof-based, abstract and rigorous

Ted Shifrin

I don't see the point for an engineering or physics student of doing proofs that give absolutely no insight. The proofs that do give insight, I usually did in classes.

Lukas Heger

10:11 PM

note that 50%-60% fail rates for first year courses here are standard

Ted Shifrin

@Lukas Yes, it's a different culture. And your high schools track from early on. Ours do not.

Yeah, and a failure rate of more than 20% in calculus here gets the department in trouble with the administration in many cases.

Lukas Heger

the university doesn't make money from each student, which I think also makes a difference

Ted Shifrin

Also education is essentially free in the EU. Here it costs up to $30,000 a year some places.

Lukas Heger

not all of the EU

but in Germany it's basically free yeah

in some EU countries you have to get a scholarship for your undergrad or else pay a lot

Ted Shifrin

OK, I was wrong. But $60,000 for four years is extremely common.

Lukas Heger

10:13 PM

that's quite a lot

even our algorithms class from the CS department was proof-based

Ted Shifrin

Well, that course should be more proof-y than most CS.

Lukas Heger

we proved all lemmas we needed on Landau notation rigorously and then proved correctness and complexity of all algorithms either in class or in exercises

user4539917

Do they still have grade 13 in most of the EU?

Ted Shifrin

I thought only Canada had that.

Lukas Heger

in Germany some states have it and some states don't

not sure about other countries

Ted Shifrin

10:15 PM

Just shows how ignorant I am.

Lukas Heger

but if you're not in the university-track part of the school system, then the school ends after grade 10

Ted Shifrin

And in many countries (France and Germany among them), university-track bifurcates so that students interested in science/engineering and students interested in social science or humanities are in separate tracks, no?

Lukas Heger

well, in Germany you can't enroll without choosing a major

and the vast majority of your required courses will be from your major

Ted Shifrin

But I thought that even the high school education was different ...

Lukas Heger

no that's not true

if you want to major in say, philosophy or French, you still need to learn calculus in high school

Ted Shifrin

10:19 PM

Oh, OK. I thought I had heard this was so 30-40 years ago. Maybe I'm wrong, or maybe it changed.

Lukas Heger

not sure about France

but you can choose majors/A-levels/whatever you want to call them in high-school too

Ted Shifrin

I remember being astonished when my one PhD student, who was educated in Poland, told me that as a student majoring in elementary education in Poland she had to take a Rudin analysis course. I was stunned. In the US we're lucky if we can get these students to understand basic arithmetic to teach it.

Lukas Heger

but everyone will still need to take math and language classes and history classes for example

Ted Shifrin

That's good, Lukas.

Lukas Heger

I meant in high school

Ted Shifrin

10:22 PM

Yeah, here in college a broad basic background is required of everyone. I actually approve of that. The European system definitely is more a graduate school mentality for undergraduate.

Lukas Heger

@TedShifrin it's not that extreme for elementary education in germany, but if you want to do university-track high school education, you need to take classes with the pure math majors and that includes stuff like measure theory

my high school physics teacher took courses in algebraic number theory and algebraic geometry

Ted Shifrin

Crazy :)

No, she swore that she had to do that to teach little kids. She was, indeed, and still is a very gifted teacher.

But she's been teaching college level instead :)

Lukas Heger

we have special colleges of education for elementary school level education

so they don't share the courses with the usual math majors (and the university-track high school education majors)

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