3:58 AM
@Obliv How do you have a $t$ in the limit of integration when $t$ is the variable of integration?

it's notational, was trying to get the indefinite integral idk why I wrote it like that @robjohn
some FTC jazz like

3 hours later…
6:42 AM

7:35 AM
I just heard a legendary tale of a PDE professor from her student: So one day, one of the students in her analysis class wanted to ask some questions about her work on PDE after the class. But she refused by saying "You'd better not ask me those questions. My first impression of you will get bad."

7:51 AM
I wish I could say the same in my interviews

8:30 AM
soumik: ... i mean, you could say the same in your interviews

Hello

8:47 AM
can anyone help me with this:
https://math.stackexchange.com/questions/4407959/power-series-identity-proof

9:24 AM
@AsishMohanty hi

9:42 AM
> Definition 4.8 If $f:X\to\mathbb{\overline{R}}$ is a measurable function, then $$\int f \ d\mu=\int f^+ \ d\mu-\int f^- \ d\mu,$$ provided that at least one of the integrals $\int f^+ \ d\mu$,$\int f^- \ d\mu$ is finite. The function $f$ is integrable if both $\int f^+ \ d\mu$,$\int f^- \ d\mu$ are finite, which is the case if and only if $$\int |f| \ d\mu < \infty.$$
> Remark Note that, according to this definition, the integral may take the values $-\infty$ or $\infty$, but it is not defined if both $\int f^+ \ d\mu$,$\int f^- \ d\mu$ are infinite.
I'm confused about this definition. The integral can only take the values positive or negative infinity if one of $\int f^+ \ d\mu$,$\int f^- \ d\mu$ takes one of those values, but then it isn't defined. Shouldn't something that obtains a value be defined?

it is

> but then it isn't defined
it is defined, as those values of $\infty$ or $-\infty$
but we don't call $f$ to be (Lebesgue) integrable in that case
its a difference

right ok, maybe my confusion mislead me, subtle difference there

subtle in the sense that integrable doesn't mean can be integrated, but it means can be integrated with a finite value (i.e. a real number)

9:58 AM
@leslietownes ðŸ˜†ðŸ˜†

2 hours later…
11:31 AM
anyone want to play this? lichess.org/fLcQScN2

3 hours later…
2:04 PM
Hello @Thorgott , hope all is well. Do you recall a while back we were discussing intervals of R and how they were uniquely identified by their endpoints and whether they contain them?
This is very much not important, but now that I look back at it it seems the claim was only made for nonempty intervals. But why not empty? The empty set would be the unique interval for which $\inf A = \infty$ and $\sup A = - \infty$

2:38 PM
Take a disk with a given metric $g,$ $(D^2, g)$ and identify points on $\partial D^2$ by considering the space of lines with the same slope. Choose an element in this space. Ex. take vertical lines through $D^2$. This quotient space is homeomorphic to $S^2$. But what is the quotient metric on this quotient space?

This is not an adjunction space, as is the topological sphere as the quotient space of a disk

Anyone game to help me with this one? Subtract the polynomials. (4t^3-8t+19)-(-16^3+t^2+11t)
wonders if @pizza has anchovies and spinach in addition to the giant pinapple on the pizza in his picture

3:01 PM
is there a way to solve this without taylor/hopital?
$$\lim_{x \to 0}\frac{1}{x^3}\left(\log(\sin x+\cos x) - \frac{x}{1-x}\right)$$
@YourLordJoyBoy yes!

@Pizza Question is, do you know where I got that from XD

2 hours later…
4:35 PM
I killed it, didn't I?

1 hour later…
5:36 PM
@EE18 your claim is correct, but one would not call these the "endpoints" of the empty interval
simply because doing so does not make any sense as far as our geometric intuition is concerned
you also have to be careful of what pairs of endpoints are admissible in your classification
a non-empty interval $I$ has $\inf I\le\sup I$ and this fails for the empty interval
the case of equality is also dangerous: for any real number $a$, there is a unique closed interval with both endpoints equal $a$. it is $[a,a]=\{a\}$. however, we also have the corresponding open interval $(a,a)=\emptyset$, but this representation is not unique anymore!
if I were to state the classification of intervals, I would provide the following list:
- the empty interval $\emptyset$,
- the degenerate intervals $\{a\}$ for any real number $a$,
- the non-degenerate, bounded intervals $(a,b)$ (open), $[a,b]$ (closed), $[a,b)$ (right half-open) and $(a,b]$ (left half-open) for any real numbers $a<b$
- the unbounded below, bounded above intervals $(-\infty,a)$ (open) and $(-\infty,a]$ (closed) for any real number $a$
- the bounded below, unbounded above intervals $(a,\infty)$ (open) and $[a,\infty)$ (closed) for any real number $a$
you can arrange them in other ways, too, though

@Pizza Why do you presume a limit exists?

@Pizza As $x \to 0$, then $\sin x \to x$ and $\cos x \to 1$. So $(\sin x+\cos x) \to (x + 1)$. And as $x \to 0$, $\log(1 + x) \to x$. So that simplifies things a fair bit.

6:25 PM
hi all. I am wondering if there is a well known operation on a matrix which returns the form of the following sum
where this is $1 - \text{Operation on Matrix}$
I see that the sum over $R_{ii} R_{ii}$ and $R_{ij}R_{ji}$ can be condensed into the trace of the $R$ matrix squared (redefining some coefficients), $\text{Tr}(R^2)$, but that leaves these $R_{ii}R_{jj}$ terms

@SillyGoose is $R_{ij}$ short for, matrix $R$ but with $i$th row and $j$th column crossed out?

sorry $R_{ij}$ is the $i,j$ entry of the $R$ matrix. the entries of this $R$ matrix are themselves matrices.

the fact that this is so awfully complicated already makes me think that probably there isn't a well known operation
what makes you think that there is one?

well (maybe I misunderstand) but i think if the operation exists for a matrix with scalar entries it should work fine for this case (or i can just notationally define it or something)

so you want to assume $R_{ij}$ are scalars for now?

6:31 PM
yeah let's do that

What are $D_i$ and $C_{ij}$?

and $D_i$ and $C_{ij}$ are actually scalars. they are just coefficients
so maybe i can simplify this by asking about $\sum_{i = 0}^n D_i R_{ii}R_{ii} + \sum_{i \neq j = 0}^n C_{ij}(R_{ij}R_{ji} + R_{ii}R_{jj})$
and treating $R_{ij}$ as scalars and $D_i$ and $C_{ij}$ are also scalars

the two terms are sort of independent

where $i \neq j = 0$ means summing over all $i \neq j$ starting at $i = 0$ and $j = 0$. where this indexing came about because i pulled out one summation over the "diagonal terms" and if i were to allow $i = j$, then i would get two extra such summations from the second term

The only thing I see to potentially make this work is to write this in matrix notation. First sum is product of $D = [D_0, ..., D_n]$ with $[R_{00}^2, ..., R_{nn}^2]^T$ so you just need to find expression for $[R_{00}^2, ..., R_{nn}^2]^T$ in terms of $U$

6:37 PM
sorry what is $U$?

Sorry, I assumed the matrix of interest is called $U$ from above screenshot

ohh sorry $S(U)$ is just my naming of the RHS and the RHS is secretly a function of a unitary matrix $U$, but I have hidden the RHS's dependence on $U$ because it doesn't actually change the form as presented on the RHS
these $R_{ij} \equiv \text{Tr}_E(U e_{ij} U^\dagger)$ where $\text{Tr}_E$ is the partial trace over all but one tensor factor, $e_{ij}$ is the matrix with $1$ in the $i,j$ entry and $0$ elsewhere and $U$ is a unitary matrix. but the $U$ dependence doesn't change how i would manipulate the actual form of the RHS of my screenshot

Well I stated what I would do as a strategy here, I'm not a big linear algebra person
hopefully you reach some conclusion on this problem, good luck

thanks for the help

@SillyGoose also I believe that this can be stated as just one sum and then you might have a chance to multiply some expression with $R$ together with a matrix $C = (C_{ij})$

6:44 PM
yes i think i should just state it as one sum
i think it can be rewritten as $\text{Tr}(R^2) + \text{Tr}(R)^2$ with some modifications to the coefficients of $R$
because $\text{Tr}(R^2)$ gives sum over all $R_{ij}R_{ji}$ terms and $\text{Tr}(R)^2$ gives $R_{ii}R_{jj}$ terms

7:41 PM
@PM2Ring If I do this I can't get to the solution [$-1/3$]
I think because the denominator is at degree 3... But I don't want to say anything stupid
And so I think this process doesn't work

THank you for all the comments @Thorgott !

8:14 PM
Can $\Bbb{R}^2$ be covered triangles such that no two are similar? also they can intersect only at the boundary.
Covered by* triangles

@SoumikMukherjee Yes, I think so.

I also feel so but i can't think of a construction

Tile $\mathbb{R}^2$ with equilateral triangles. In each triangle, choose a point, and join each vertex of the triangle to that point in order to divide the triangle into three sub-triangles. It is possible to choose points in such a way that (1) no two sub-triangles in any one triangle are similar, and (2) each subdivision is unique.
Or do the same with squares or regular hexagons.

@SoumikMukherjee tessellated is a proper term, I believe

@XanderHenderson Nice!

8:22 PM
@Jakobian Yes, "tile" or "tessellate".

@Jakobian Thanks, i didn't know this term

Penrose made this term famous, I believe.

@XanderHenderson smart

8:54 PM
I am confused by terminology again. Number of distinct right cosets of a normal subgroup $N$ of $G$, means $|G|$ right

Anyone wanna give this problem a go? Divide the polynomials.
8x^5 + 9x^4 - 5x3/
x2

since a coset is a congruence class of an element $a \in G$ modulo $N$ so $Na$ for all $a \in G$
@YourLordJoyBoy Can't read that last term, is it $5x^3/x^2 = 5x$?

@obliv It is divided by x2
OH! I see the issue. Thats supposed to be 5x^3

is x2 supposed to be x^2? or 2x just backwards? or x_2

Nope

8:59 PM
@Obliv You only wrote |G|, which denotes the order of G

@Obliv Wait, maybe it is.
These study guides tend to have errors for some reason.
What I know is that on the sg it's stated as x2 but perhaps they meant x^2

9:25 PM
@Obliv Lets suppose it is x^2. I'm thinking we can't solve it unless we do.

is it always possible to write a diagonal matrix $D = A \otimes I + I \otimes B$ where $D$ is $2n \times 2n$, $A$ is $2 \times 2$ and $B$ is $n \times n$ and $\otimes$ is the kronecker product?
this equation is a linear system of $2n$ equations in $2 + n$ variables (if I understand correctly), but it is unclear how many of the $2n$ equations are linearly independent. And it seems unlikely to have a general solution because it seems to heavily depend on the elements of $D$.

Wondering if @SillyGoose went to Silly Goose University XD

h o n k

i think i have an approach
i think i can represent this system as an augmented matrix and it will probably tell me if it is consistent or not

9:48 PM

10:01 PM
And class starting now

10:35 PM
how do advanced papers in math get to be so long? Don't we develop a ton of theorems overtime, so proving something can be greatly condensed by referencing theorems?
like the classification of finite simple groups or fermat's last theorem for example are really long :O

@Obliv The "scope" of interesting things that can be proven with a given result is quite small.
As a general observation, that is. The exceptional theorems for which this does not hold are usually just the standard theorems that you learn in classes.

I feel like the standard theorems in class are "common sense" in the more advanced proofs so probably don't even need to be mentioned explicitly except to be formal.

But taking a random lemma L from some paper? It's unlikely to be ever applied anywhere else ever again due to the specific nature of most research topics.
@Obliv heh, I hope to never read one of your papers then. :P

I haven't actually sifted thru long proofs/papers but similar to an abstract, is there a "solution statement" that is a condensed version of the paper that academics can read and get the gist of

Just read the abstract, introduction, or theorem statements.
Usually the abstract tells you the major result. If that's not detailed enough, usually the introduction states the theorems and needed context. If that's not detailed enough, just skim through the paper and read definitions and theorem statements. If that's not detailed enough, just read the paper in full.

10:42 PM
Right, but the fact that the paper itself could be longer than 100 pages is incredible to me.
Like, wouldn't it make more sense to just publish sections of that proof or something lol

BACK!
Class ended quicker than normal, what is with these short lessons I wonder

In my own research, papers in journals are usually significantly shorter than 100 pages.

It can't be helped though, I guess. Even if it's in a trilogy, the necessary reading is the same :P

@Obliv Hope I'm not being a bother but you um able to help with that problem I mentioned earlier?

Can you post the problem in its entirety?
I don't know what "divide the polynomial" means

10:45 PM
I did. I suppose you can consider it as a fraction.
3 terms, I think, on top of one term.

Try writing it with LaTeX
So that it's unambiguously stated.

I have tried :)
:(

do u have chatjax enabled

Have tried that too. I'm not sure why it's not working.

He has Chrome, but cannot get it to work

10:47 PM
oh dang I just realized what the problem is asking.. it's asking you to divide $8x^5+9x^4-5x^3$ by $x^2$
I thought you were saying divide the polynomial: $8x^5+9x^4-5x^3/x^2$

@YourLordJoyBoy when you drag the link to the bookmark bar, does something appear there?

Yes.
It looks like an equation

when you click "start chatjax" it should render the symbols on this site into latex

I think you've dragged the wrong thing. Did you drag the text "start ChatJax"?

Yes, I did.
I dragged start chatjax into the bookmark bar.

10:50 PM
And "start ChatJax" does not appear on the bookmark bar?

Yes it does.

click on it, does it change the text here into symbols

When I click it it takes me to where it's talking about the bookmarking

$\Sigma$

I see $\Sigma$
Wish there was a way to send a snippit of the problem.

10:52 PM
do u see the symbols or \sigma between two dollar signs

Hmmm

I see sigma between two dollar signs :(

when ur on the page and click start chatjax do u see the symbols on there?

how about u right click and bookmark the link "start chatjax" manually
idk if that'll do anything
I feel like I had this issue a long time ago I forgot how I fixed it

10:54 PM
It gives me the option to save the link and open it in another tab, neither of which will do me any good.
Or will saving the link help? Cause I'm unsure.

@YourLordJoyBoy when you try to save the link, what options does it give you?

It doesn't o.O

wait I just realized they might be on mobile

No they're not

I'm not
This is on a laptop
But it is on chrome. Hmm....

10:58 PM
maybe try this
I think this might be the issue

Hmm
Divide the polynomials.
8x5 + 9x4 - 5x3
x2

how many times does $x^2$ go into $8x^5$?
@YourLordJoyBoy ?

It says its set to sites can use javascript

its ok don't worry about it for now

@Obliv I'm not sure.

11:06 PM
Is it possible for an functional equation (automorphism) of a function to carry over to a functional equation (automorphism) through Fourier transform, in the complex space, absent of any summation (meaning Poisson summation is not used)?

if you have $ax$ times $bx$ whats the product equal to? @YourLordJoyBoy
where a,b are constants

@Obliv trying to think of that myself. Ugh....its like my brain's fried

is the $x$ part tripping you up?

@Obliv Unfortunately yes

consider then 3*4 * 2*4 where * denotes the usual multiplication. Note that 3*4*2*4 is the same as 3*2*4*4 which is 3*2*4^2
here we set x=4. In general ax*bx = abx^2 does that make sense

11:15 PM
@Obliv Yes, I believe so yes.

so then x^2 times what is 8x^5 (if x*x = x^2)?
brb

8x^5 would be 8*xxxxx correct?
And actually in a bit I'm off for the night.
8*xxxxx
Ugh it keeps bringing them together :(
Hmm

yes

Thought so.

so to get to 8xxxxx from xx you wanna multiply xx by 8xxx = 8x^3

11:24 PM
@Obliv Yes
And thinking perhaps Thursday or maybe tomorrow, my dad might be able to help with getting MathJax working.

another way to think of it is you're missing the 8 and 3 factors of x so x^2*8x^3 gives u 8*x^(2+3)

@Obliv Yes that seems right so far.

I am interested in differential geometry

That'll make things easier to present for sure. So now are you familiar with long division of polynomials?

@Obliv Yeah. And more than likely I'm going to be tackling it tomorrow.
I starred the bits you said so I can remember because I doubt I will once I log off.

11:27 PM
Please don't star everything at once!

This is probably the last time I'm doing that.
I prefer not to but with chats you can't recall everything said. Since thousands of people are chating.

Okay, well just remember that when dividing a polynomial by another polynomial (or in this case x^2 is called a monomial), you want to "fit" the smaller polynomial into the larger one. It's done in steps using long division. i.imgur.com/3LP827B.png This is my crudely drawn depiction on mspaint :P

And since this is important stuff, it needed to be done.

star trolling

Alright, will do @Obliv. Thank you so much for the help.

11:29 PM
star recycling

each term on the top I multiplied by x^2 and then subtracted it from the base
trying to whittle it down

@Obliv Got it. And good night guys

take care

You too :D

11:41 PM
how can I prove $G/N$ is abelian iff $(ab)(ba)^{-1}\in N$ for all $a,b \in G$
$N$ is a normal sg of $G$
I get that I have to prove $Nab = Nba$
tfw the proof of a theorem ur reading references a theorem in a later section
like i checked 8.21, that had to be a typo because it didn't really help or make any sense. I also checked 7.21 which was cayley's theorem so that's a no-no
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