12:00 AM
So what exactly is a PDE? Or a differential equation in general? I hate the nomenclature sometimes :\
So what exactly is a PDE? Or a differential equation in general? I hate the nomenclature sometimes :\

There's nothing wrong with the nomenclature. Do some basic reading.

Hi @TedShifrin Do you have any sources you recommend?
I did! I read the wiki page but it didn't really make sense to me

Simmons wrote a very nice book on differential equations. If the wiki page didn't make sense, it makes me wonder what your background is.

Okay i'll check that out, Thanks. I'm actually not a student I'm just curious

My comment had nothing to do with studentdom.

12:07 AM
its a differential equation, but with more differential

Oh, well I have a background in theatre and music

So, theater and music are wonderful, but if you have never learned calculus you can't expect to understand this.

I know it makes no sense, but I really like just skimming/understanding on a surface level

shrug

Just enough to satiate my curiosity! :D

12:10 AM
On what basis do you hate nomenclature if it's perfect descriptive?

@TedShifrin why do you like the Simmons book?

Simmons writes mathematics beautifully, and he has excellent taste. There are good applications and historical perspective.

Thanks

@TedShifrin I suppose I just don't like reading sentences about math over the math itself. something like this line: "An ordinary differential equation (ODE) is an equation containing an unknown function of one real or complex variable x, its derivatives, and some given functions of x" just bores/makes me lose interest
I'd rather just read the mathematical construction

There is no construction to read. I'm not continuing this discussion, though.

12:20 AM
not sure what kind of construction you were expecting
examples are abound and without looking, I'm gonna claim the wikipedia page has at least a couple

Yeah I guess diff eqs don't have a general form or whatever, and it's specific to the type
As I read through it more it seems to not just be a mathematical object like a group

i mean, it is a perfectly fine mathematical object, but not in any way similar to how groups are mathematical objects

Can anybody enlighten me how exactly the factoring of the morphism at the bottom of page 34 here works? https://www.math.uni-hamburg.de/home/dyckerhoff/higher/notes.pdf

The morphism itself is defined by the universal property of the pushout of the diagram Δ[n]x{0} <- dΔ[n] x {0} -> dΔ[n] x Δ[1], where {0} is the simplicial set with {0}(n) = {the unique function [n]->[0]}. The morphism to the left is (boundary inclusion,id) and the morphism to the right is (id,vertex inclusion).

The UP is applied to the extension of the diagram to a commutative diagram with Δ[n]xΔ[1] on the right corner.
And I wouldn't know what to search for on SE

1:01 AM
Oh no I now suspect this is just trivial, although I still can't fill in the details. Is it true that one would obtain Δ[n]xΔ[1] as the simplicial set generated by the (n+1)-simplices in any case?

5 hours later…
5:48 AM
@satan29

hi

write k+3 as integral (0 to 1 x^(k+2)dx)

@satan29 why

and bring the sum inside the integral, since the sum doesnt vary with "x"
@user69608 The factor k+3 is causing a lot of problems. So its better we re-write it.
the question becomes: (integral from(0 to 1) ( sum(k-0 to n) nCk*(x^k/n^k)*x^2dx.

5:54 AM
oh yeah done thanks

the" sum" simplifies to (1+(x/n)^n) using the binomial theorem
so the integral is (1+(x/n)^n)*x^2dx
and for large n, it becomes e^x*x^2 dx
which you can integrate by parts easily to get (e^x(x^2-2x+2)) evaluated at 0 and 1 to get e-2
so i believe the answer is 3?

yeah done :)

oh ok :)

method is correct. dont remember answer
do u know a good source for such type of question? @satan29

i had an entire matrix match on a similar question in the FIITJEE GMP

6:00 AM
oh

thats why kind of recognized that we can rewrite 1/k+3 as an integral

6:15 AM
@satan29 want to know about fog

6:28 AM
What's $[x]$ representing here?

@Drathora floor function

And what do you want to know about f o g?

differentiability

@Drathora yes

6:35 AM
But then think about what happens around the point $x = 1$
I.e. consider $f(g(1+\epsilon))$ for a small $\epsilon > 0$

oh, that wont be differentiable at integers

Well, it will be at $0$
But yeah, at the others it won't be

when $x^2$ is integers it wont be differentiable right?

Correct, since for a slightly lower $x$ fog(x) will jump to the previous integer

but what does c option tells , shouldnt it be$\sqrt{I}$ in place of I?

6:46 AM
What's "R ~ I" representing here?

real number,integer
@Drathora exclude integers from real numbers

ah
Well then I'd say no, since there are discontinuities at the square roots of integers also right?

@Drathora yes

2 hours later…
8:26 AM
Today I got question in numerical analysis
It is about something like decimals
too abstract to digest
related to machine number

8:56 AM
It's binary machine number and decimal machine number

9:25 AM
May be I will ask later. I really don't have energy to grasp any idea right now.

9:41 AM
@loch Maybe I can ask you about the content some time, depending on how much you got through
To understand the second section, I've gone off to read Paul Balmer's paper on Witt groups

@Alex the answer is not much esp. with the technical stuff :p

The first section is definitely rather technical :')
I think it's sections 2,4,10, and 11 that I mainly want from this paper, but I'll ask my adviser tomorrow about it

i just know that to see the stuff 'in action', you might want to look at the stuff that wickelgren did on counting things over non-algebraically closed fields

Do you have a specific paper/notes in mind?

yes
no there arent a lot of notes available
if any
(as far as im aware) - but i think the papers are pretty readable (definitely more readable than the levine paper lol)

9:49 AM
Well that goes without saying :')

Made a funny visualization of factors: desmos.com/calculator/0ssikt8mo7

@WeavingBird1917, taking too much time to load
Oh wait, that's cool, you can make an elevator gif xD

0

The objective is to: Evaluate $$\int_0^1 e^{\frac{1}{\log(\theta)}} ~d\theta$$ in polar coordinates. Using cartesian coordinates: The integral, where $K$ is the modified Bessel function of the second kind, $$\int_{0}^{1} e^{{\frac{1}{\log(x)}}} \, dx =2K_1(2),$$ Can be evaluated using the ...

@UmbQbify-Key20- Lol, I think it looks cooler backwards though.

10:04 AM
I don't actually understand what's happening, but it goes in reverse sometimes

The number being divided (a) is being varied across a range, when it reaches the end of the range it reverses. The full size versions are f3 and f4. The horizontal lines are actually meant to be points on the right side of them, which represent the remainder as a divides by x.

if a function is even
is it true that the integral from -00 to +00 is 2 times the limit from 0 to +00
?

10:20 AM
Try it for cosine

$\int_{-00}^{+00} cos(x) dx$
it is indeed 2 times the limiti to +00
right?

10:37 AM
@ManolisLyviakis It works - if the integral from 0 to +\infty exists. (So $\cos x$ is not an ideal example.)

ye
thats how i did it

A related question about odd functions: Can we use symmetry rules in improper integrals?

1 hour later…
11:43 AM
how many definite integrals yield the same result when swapping the variable from cartesian to polar? (not changing the coordinates)
math.stackexchange.com/a/3749735/460999 (relates to this answer...also did the answerer even use polar coordinates?)

What is the quotient of $S^\infty \times \Bbb R^2$ under diagonal action of $\Bbb Z/2$?
$\Bbb Z/2 = \{1, a\}$ acts on $\Bbb R^2$ by negation: $a\cdot (x, y) = (-x, -y)$
...and on $S^\infty$ by the usual anti-podal action.
Can I compute the fundamental group? I mean $S^\infty\times\Bbb R^2$ is simply-connected. I think the diagonal action is covering space action. So, $\pi_1((S^\infty\times\Bbb R^2)/\Bbb Z/2) = \Bbb Z/2$.
If $G$ acts on $X$ and $Y$, and one of them is a covering space action, then is the diagonal action on $X \times Y$ also a covering space action?

it only acts on the factors separately, so shouldn't the quotient be $\mathbb{R}P^{\infty}\times\mathbb{H}^2$

12:05 PM
What do you mean by 'its acts on the factor separately'?
What is the quotient of $S^1 \times \Bbb R$ under diagonal action of $\Bbb Z/2$?
I thinks its a disc not $S^1 \times \Bbb R_+$.

12:17 PM
oh, I see where I'm wrong
the actions on the factors aren't independent

@CalvinKhor @satan29

@feynhat Its not going to be a disc though. I was actually thinking of $S^1 \subset \Bbb R^2$, and applying the reflection action (not the anti-podal action).

what is this and why is the typesetting so bad?
also, hi?

ugh... no.

12:33 PM
$$\operatorname{\ell im}_{n\to\infty}\limits$$ lol
3

lol

> वास्तविक संख्याऒं
Does that mean real numbers?

how on earth lol

सीमा = limit... lol. अभिसारी = convergent... lmfao. What are these words?

1:13 PM
This is hindi language.

I think feynhat of all people could have guessed that. :P

I have learn basic hindi and I have no idea what you are writing
shima=limit?
abhishari=convergent?
It's time to go to bed. No questions for today

@robjohn hi sir
how are you?

1:31 PM
@feynhat This is false btw.

yes

@feynhat what are you up to?

Oh hi. Just trying to figure out the quotient $(S^\infty \times \Bbb R^2) / (\Bbb Z /2)$

As a Lie group?

No, just the topological space. (I don't even know how you put a smooth structure on $S^\infty$).

1:41 PM
Me either. Is it the antipodal action on S^\infty, and then what action on R^2?

$(x, y) \mapsto (-x, -y)$.

Oh so technically the same as S^\infty.

:54882438 $\sin^2(x)+2\cos(x)+1=3-(1-\cos(x))^2$ which has a range of $[-1,3]$

yea
@ABCD Yes, I know. I am a native speaker. Its just that I've never read math in that language. Those words seem funny being used in math.
@anakhro Btw, do you know more than half of people in the country do not speak Hindi? I am pretty sure Balarka doesn't (or at least, is not a native speaker, and speaks with an atrocious accent).

Yes, but I think significantly more than that can identify hindi.

1:55 PM
Oh yeah. That might be true, since Hindi appears in a lot official documents, bills etc.
@BalarkaSen Help me see this. Consider the finite cylinder $S^1 \times [-1,1]$. There is a $\Bbb Z/2$ diagonal action on this cylinder (it acts on both $S^1$ and $[-1,1]$ anti-podally). What is the quotient space?

identify but not understand? Even I can lol there are signs in hindi in my country

I think the fundamental domain will be $S^1 \times [0, 1]$.
What happens at $S^1 \times \{0\}$ ?

It's always ugly when you aren't gluing only on boundaries. :P
Are you able to think about doing it on the boundary to begin with, and then kind of zipping it up along the seams?
It already sucks imagining the boundary version so that's probably a bad suggestion, actually.

2:12 PM
@feynhat In what sense are you trying to identify the quotient?
It is a certain 2-plane bundle over $\Bbb{RP}^\infty$.

How do you conclude it is a plane bundle so quickly?

It will be great if we can figure out what exactly the space is, it would be great.
I mean some expression like $\mathbb{RP}^\infty \times \text{(something)}$.
Otherwise, I would like to compute $\pi_1$.

It is not a product, it is a fiber bundle. The most explicit you will get is identifying that fiber bundle. And honestly what you've described is more or less the most explicit description. If $\tau$ is the tautological line bundle (whose unit sphere bundle is $S^\infty$), then your plane bundle here is $\tau \oplus \tau$.
@anakhro Because $\Bbb Z/2$ acts freely on the first factor, one has an open cover of this space by things that look like $((U \times \Bbb Z/2) \times \Bbb R^2) / (\Bbb Z/2)$. That quotient is just $U \times \Bbb R^2$.
If $G$ is finite and acts freely on $X$ and in some arbitrary way on $Y$ then $(X \times Y)/G$ is a $Y$-bundle over $X/G$.
I will leave it to you to calculate $\pi_1$ from the fact that it's a fiber bundle over $\Bbb{RP}^\infty$ with fiber $\Bbb R^2$.
@feynhat The same fact tells you that this gives you a $[-1,1]$-bundle over $\Bbb{RP}^1$. Can you identify which one? You know a couple.

2:28 PM
Mobius bundle?

Maybe!

Nice, thanks for that elaboration.
@MikeMiller have you ever heard of something like the "discriminant" of a manifold?

Nope
Context?

There is this definition I read of "discriminant points" of a contact manifold $(M,\xi)$. A discriminant point of a contactomorphism $\phi\colon M\to M$ is a point $p\in M$ such that $\phi(p) = p$ and $(\phi^*\alpha)_p = \alpha_p$ for one (hence all) contact forms $\alpha$.
The discriminant of $M$ is then the set of all $\phi$ with discriminant points.
But I don't quite understand the intuition behind using the word "discriminant". Unless it's completely separate from every other use of discriminant in math. :P

What's the point/what's it used for?

2:37 PM
Well it's used with a certain type of Maslov index for a contact isotopy which is the intersection of a subspace of the discriminant and the contact isotopy.
<-- doesn't really understand the point but is trying to go through a paper.
This Maslov index has a bunch of applications, e.g. a special case of Arnold's conjecture for contact manifolds.
Also, contact non-squeezing re-proof.

Yeah, I don't know. Sorry about that.
I'm sure there's a good reason for calling it that but I'd have to be more immersed in the literature I'd guess.

It's fine, no worries. As long as I am not missing a painfully elementary comparison to a "discriminant" that I should of already.

@MikeMiller I only know how to do that when we have a covering space. But in that ^ case, the fiber is not discrete.

It might just be that they are more distinguished than regular fixed points.

@feynhat You don't know of the long exact sequence in homotopy groups of a fiber bundle? It's probably worth knowing.
@anakhro I would suspect it's only etymologically linked to "discriminant" as in $b^2-4ac$

2:43 PM
Yeah.

In the same way that the kernel of a linear map (== solution set to some linear equations) and the kernel of a differential operator (== something which generates the space of solutions by convolution) are both linked to the English word kernel (== the core of something).

20th century French colonialism. A friend also suckered me into a reading group for Hegel's phenomenology but it wouldn't be fair to say that I understand anything or that I'm spending a lot of time on it.

Is there a group structure on $\pi_0$?

$\pi_0$ of what?

2:50 PM
Fiber space, say (as in our case) $\Bbb R^2$.

Any finite set can be a fiber of something
You're basically asking (afaict) whether a random given set has a group structure
Which, yes, but not naturally so
If you want $\pi_0 G$ to be a group you probably want $G$ to be a group, in which case $\pi_0 G = G/G^e$, where $G^e$ is the path-component of the identity

Yeah. I meant, what does $\cdots \to \pi_1(E) \to \pi_1(B) \to \pi_0(F) \to \pi_0(E) \to 0$ being exact means.

If you're worried about applying the fiber sequence near the tail end here, when the fiber is connected you should just read it as ending $\pi_1 E \to \pi_1 B \to 1$, the last thing the trivial group (so that $\pi_1 E \to \pi_1 B$ is surjective).

Let $X$ be a topological space. Is it always true that if $A\subset X$ then $\operatorname{int}\bar A=\operatorname{int}A$?

When $F$ is disconnected and your fiber sequence isn't of the form $G \to X \to X/G$ then it's going to be more irritating to parse through the end of the sequence. But you're not in that situation, so I wouldn't worry.

2:54 PM
@JaakkoSeppälä no

What would be a counterexample?

$\Bbb Q\subseteq\Bbb R$

Consider a field $k$ and the $k$-algebra $k[x,xy,xy^2,...]\subset k[x,y]$. If this were finitely generated (as $k$-algebra, now and in the forthcoming) by elements $f_1,...,f_n$, then it would also be finitely generated by the monomial terms appearing in $f_1,...,f_n$, because any polynomial expression in $f_1,...,f_n$ is a polynomial expression in their monomial terms. So assume it is generated by some $xy^{a_1},...,xy^{a_k}$ and pick $n>\max a_i$. Looking at any polynomial expression in $xy^{a_1},...,xy^{a_k}$, we see that any monomial term containing the power $y^n$ necessarily contains

There's even counterexamples with $A$ open, that's why the concept of regular open set is interesting

Oh, okay. So, $1 \to \pi_1 E \to \pi_1 B \to 1$ is exact, where $E = (S^\infty \times \Bbb R^2)/(\Bbb Z/2)$ and $B = \mathbb{RP}^\infty$. So, $\pi_1 E = \Bbb Z /2$.

3:03 PM
@MikeMiller ask yourself: would a true friend sucker you into a reading group based on anything Hegel wrote?

Yes, just not the phenomenology.
The trick though is to read it without pretending you understand.
@feynhat Right. Another way to see your conclusion is that your space is a vector bundle over $\Bbb{RP}^\infty$, and every vector bundle def retracts onto the zero section.

oh duh. a vector bundle deformation retracts to the base space by by linear homotopy.

@user69608 the thing inside the bracket approches 1, while the power approches $\infty$. reminiscent of e.

@feynhat The really fancy observation is that if $Y$ is $G$-equivariantly homotopy equivalent to $Y'$ (and $G$ acts freely on $X$), then $(X \times Y)/G$ is homotopy equivalent to $(X \times Y')/G$. And if $G$ is any group of linear maps, the straight-line homotopy is an equivariant homotopy equivalence from your vector space to a point.
That's the fancy way of observing the vector bundle thing here. But it's unnecessarily fancy of course.

@MikeMiller Heh.

3:11 PM
@satan29 yeah how to do after that?

@user69608 if the thing inside the bracket is a, and the exponent is b, then your expression is : $(1+(a-1))^{\dfrac{1}{a-1}*{b*(a-1)}}$

@MikeMiller Bookmarked.

Well, you do you.

@user69608 so the limit boils down to e^ (limit of (b(a-1))

@Thorgott I'm starting to believe it just doesn't fail

3:20 PM
@user69608 now let x=1/t. so b(a-1) becomes (after writing 1 as n/n) :$\dfrac{n[(a_1^t-1)+(a_{2}^t-1).....]}{nt}$
which is a standard limit
$ln(a_{1})+ln(a_{2))...... =ln(a_{1}a_{2}.......$.
so e raised to this power is simply $a_{1}a_{2}..$
i.e D

@feynhat You can take the fundamental domain to be (the upper hemisphere of $S^1$) x [-1, 1].
Then simply observe the sides are getting glued like a Mobius strip :P

(balarka ninja)

Sniped. Also, my Hindi accent is fine.

nice :)

ok Bolorko

3:25 PM
Lmfao
@feynhat There's a cyclotron in Saha Institute in Calcutta. It doesn't work, do you know why?

umm no. Why?

Because V x B = 0 according to Bengalis

loooooooooooool

0

I've added an information to the Wikipedia article "Classification of discontinuities", but now I am in doubt if it's correct: "Intuitively, a function is continuous if (and only if) it has no breaks, jumps, vertical asymptotes, or wild oscillations." Is it correct? (ignoring the fact I give no p...

0

The two strings below describe the same English word: 7415963852 1475369963 What word is it? Just for fun! Credits: Hindemburg Melão Jr.

Anyone?

I'm thinking of $\mathbb R/\mathbb Z$ as the half-open interval $[0, 1)$. Is there any sense in which this $[0, 1)$ can be considered as a "circle" by joining its "endpoints"? I mean if the interval were $[0, 1]$ I suppose there is some terminology for gluing the endpoints but I'm not sure about $[0, 1)$.

3:40 PM
Can you make more precise what you're asking
$[0,1)$ only has one endpoint so it's hard to pin down what that would mean

@MikeMiller Umm, well first of all does it make sense to represent $\mathbb R/\mathbb Z$ as $[0, 1)$?

What does "to represent" mean
There's a natural set bijection between them

@MikeMiller That is representing the cosets of $\mathbb Z$ in $\mathbb R$ by taking a single representative element from each coset.
@MikeMiller I guess that's what I mean

The composite $[0,1) \hookrightarrow \Bbb R \to \Bbb R/\Bbb Z$ is a bijection
Because every number is (uniquely) of the form $n + r$ where $0 \leq r < 1$ and $n \in \Bbb Z$

However, in the construction of the Vitali set they're apparently representing $\mathbb R/\mathbb Z$ with the interval $[0, 1]$ so I'm a bit confused
From what I can understand, there is no reasonable bijection between $\mathbb R/\mathbb Z$ and $[0, 1]$. Is that right?

3:46 PM
I don't see anywhere that they use such a bijection. It also doesn't seem important to me, because this is about measurabiity, and chucking out a single point doesn't change whether or not a set is measurable

Isn't the Vitali set about picking a coset representative of each coset in $\Bbb R/\Bbb Q$?

@satan29 acha mistake milgaya ,thanks

Let us equip the space $X=\{1,2,3,4\}$ be the topology
$\{\emptyset,X,\{1\},\{2\},\{1,2\},\{3,4\},\{1,3,4\},\{2,3,4\}\}.$
Consider the partition $X=a\cup b\cup c$ where $a=\{1\}$, $b=\{2,3\}$, $c=\{4\}$. How can I compute the quotient space corresponding to this partition?

@BalarkaSen Yes, or rather the cosets of $\mathbb Q/\mathbb Z$ in $\mathbb R/\mathbb Z$

same thing

3:48 PM
Just phrase it like, picking a coset representative of each coset in $\Bbb R/\Bbb Q$ lying in $[0, 1]$

Oh, Wikipedia writes this as a note: The notation R/Z is somewhat ambiguous. If Z is understood to be a group acting on R via addition, then the quotient is the circle. However, if Z is thought of as a subspace of R, then the quotient is a countably infinite bouquet of circles joined at a single point.

That's irrelevant yeah forget about that

I'm basically trying to think of $\mathbb R/\mathbb Z$ as a circle

Ignore that note completely, they should not have written that

It is a circle.

3:49 PM
They're trying to clarify potential confusion but they've only created it lol

@BalarkaSen But how is $[0, 1)$ (which is in bijection with $\mathbb R/\mathbb Z$) a "circle" if one of its endpoints is missing?

$[0, 1)$ is also in bijection with the circle, $t \mapsto e^{2\pi i t}$.

a circle doesn't need endpoints

5

Possible Duplicate: Bijection between an open and a closed interval How do I define a bijection between $(0,1)$ and $(0,1]$? I wonder if I can cut the interval $(0,1)$ into three pieces: $(0, \frac{1}{3})\cup(\frac{1}{3},\frac{2}{3})\cup(\frac{2}{3},1)$, in which I'm able to map poin...

Oh, I think it makes sense now
We can always construct a bijection between $[0, 1)$ and $[0, 1]$

You don't want to be putting random bijections between different objects when doing math.
Then everything is a circle.
Clearly not true

3:52 PM
this doesn't matter
this how it be

@Thorgott Okay. Let's do this step by step. How do you say define or construct a circle given an interval like $(0, 1)$?

I wouldn't

By focusing so much on this you are obfuscating the point that the Vitali set is trying to make

@Thorgott Can this be made rigorous?

3 mins ago, by Balarka Sen
$[0, 1)$ is also in bijection with the circle, $t \mapsto e^{2\pi i t}$.

3:54 PM
A measurable set less one point is still measurable. A non-measurable set less one point is still non-measurable.

that's rigorous

You can write out the same construction as on that Wikipedia page with $[0,1)$ instead and you produce a non-measurable set
Or with $(0,1)$

missing a non- at the end, Mike

@MikeMiller The Vitali set isn't really my main point of concern here. I do understand that countable sets are measure zero :P

Well, it's pretty unclear what your main point of concern is.

3:56 PM
that said, it's suboptimal to think of $\mathbb{R}/\mathbb{Z}$ as $[0,1)$, because it obscures the topology of the space
if you don't want "too many representatives" for each equivalence class, think of it as $[0,1]/\sim$ where $\sim$ is the equivalence relation only identifying $0$ and $1$

@Thorgott Right, so even something like $(0, 1)$ is in bijection with $[0, 1]$. Then by gluing together $0$ and $1$ we can make a circle, yes?

$0$ and $1$ are not elements of $(0,1)$

Why do you want to do this? $\Bbb R^2$ is in bijection with $S^1$. What is to be gained from knowing that?

@Thorgott I guess that is one of my confusions. What exactly do you mean by it obscures the "topology of the space" ?

$\Bbb R/\Bbb Z$ is homeomorphic, as a topological space, to $S^1$.
$[0, 1)$ is not homeomorphic to $S^1$

3:59 PM
when you write $[0,1)$, you, I and everyone else thinks of this as subspace of $\mathbb{R}$ with the corresponding subspace topology
now refer to Balarka's comment