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2:40 AM
I ain't pretending squat!
2:51 AM
@SoumikMukherjee Foundations of Differentiable Manifolds and Lie Groups
I'll most likely never have good exposition to the subject
@psie they are
3:30 AM
@psie well, just like Leslie said, Borel hierarchy for a perfect Polish space $X$ is a counter-example, say $\mathbb{R}$
maybe we can give an argument that $\bigcup_i E_i$ will never belong to $\mathcal{E}_\omega$ though
3:41 AM
@Jakobian we're the same GTM
we kind of have similar interests so maybe it should be something about Banach spaces or Banach algebras
would be more accurate
yeah i think there's only like 5 possible results in that quiz because they just did the classification by hand
there are any number of closer GTMs
4:00 AM
@leslietownes that was one extremely insightful interview.
4:23 AM
yeah it was cool
4:36 AM
😎
5:06 AM
@leslietownes there are at least 10, I counted:)
@Jakobian I think there are not any springer GTM topology book
GTM you mean
yes, corrected
@SoumikMukherjee Kelley General topology
and that's just for general topology
there are indeed books about topology in the springer GMT series
5:18 AM
the what series? :)
i think massey had one life as a gtm
springer would sometimes reprint stuff that initially came out elsewhere as gtm's
ah sorry. I made the same mistake as Soumik earlier. Of course GTM series, not geometric measure theory
the greenwich mean time series
hehe
I forgot about Kelley
 
4 hours later…
8:53 AM
@XanderHenderson ?
 
1 hour later…
10:21 AM
@PM2Ring @PM2Ring Yes, I'm working on the Quadratic Diophantine equation.
@PM2Ring @PM2Ring One more constraint is to be added & that is all of $a,b,c,d$ should have to be distinct from each other. As of, $(a,b,c,d)=(50,53,56,60)$ is an example.
I am reading on symmetric groups (chapter 1 section 3 on dummit and foote) and I have noticed that, given an $m-$cycle, say $(a_1,a_2 \cdots a_m)$, If I compose $\sigma$ with itself, suppose $i-$times, depending on $gcd(i,m)$, the cycle either stays an $m$-cycle, or splits. For eg, for $m=10$, and $i=3$, we see that the cycle is still an $10-$cycle, but if $i=2$, the cycle splits into 2, 5 sized disjoint cycles. If $i=4$, then still its 2* 5 sized cycles, but each one is a rearrangement away

from being the same as what we get for $\sigma^2$. How do I understand what is happening here?
 
2 hours later…
12:39 PM
I have never really studied ordinals by themselves. Is it correct to say that the countable ordinals are the initial segments of $\omega_1$, the set of countable ordinals?
Ok, thanks for confirming.
1:06 PM
I struggle with this idea that you can index sets by ordinal numbers when they are formally defined to be sets, e.g. writing $\mathcal E_\alpha$ where $\alpha$ is an ordinal number. To be more concrete, the sigma algebra generated by a collection of sets $\mathcal E$ is $\bigcup_{\alpha\in\Omega}\mathcal E_\alpha$, where $\mathcal E_\alpha$ are defined in a certain way. Here we have indexed the sets by ordinal numbers. This confuses me a lot.
why?
indexing just means "for each element of the indexing set I have an object"
@BenSteffan but $\mathcal E_\alpha$ here is indexed by a set. I've not seen anyone ever write $A_\mathbb{N}$ where $A_\mathbb{N}$ is a set. I'm not sure what that notation would even mean, but I feel like it is the same situation here with $\mathcal E_\alpha$.
You are in set theory
There is nothing except sets here
true :)
@psie But you have seen people write $A_n$, n'est-ce pas?
1:14 PM
@XanderHenderson indeed, that makes totally sense to me
$\alpha$ is an element of the indexing set $\Omega$.
So $\mathcal{E}_{\alpha}$ is to $\Omega$ as $A_n$ is to $\mathbb{N}$.
ah ok, that makes sense actually
For each element $\alpha \in \Omega$, there is some set in your sigma algebra which corresponds to that $\alpha$; call that set $\mathcal{E}_{\alpha}$
(or whatever your context is)
For example, consider the set of half-lines $\{(\alpha,\infty) : \alpha \in \mathbb{R}\}$. This is a set indexed by $\mathbb{R}$ (that is, $\Omega = \mathbb{R}$). Specifically, $\mathcal{E}_{\alpha} = (\alpha, \infty)$.
1:49 PM
Hi
2:02 PM
@Pizza hi
hi
2:28 PM
$\frac{1}{2}\left(\sum^{\infty}_{n=2} \frac{y^n}{n-1} - \sum^{\infty}_{n=2} \frac{y^n}{n+1}\right)$
I had thought the sum was $\frac{1}{2}\left(-y \log(1-y) - (-\frac{1}{y} \log(1-y))\right)$
But I think I made a mistake, because I have $n=2$ and the series of $-\log(1-y)$ starts from $n=1$
But I'm not sure about that
That is, if for example I had $\sum^{\infty}_{n=1} y^n$ the sum was $\frac{1}{1-y} - 1$ because i need the sum to start at 0
Right?
Hi everyone
I'm trying to solve the following differential $ydx - xdy = xydy - x^2dx$. Any views on how to solve it using exact differentials. I'm specifically trying to solve it with exact differentials.
2:50 PM
@Pizza correct
you have to be careful with the index
@SineoftheTime I didn't understand what to do in the above case
That is, I have $n=2$ but the Taylor series starts from $n=1$
you can shift or subtract the first terms
30 mins ago, by Pizza
I had thought the sum was $\frac{1}{2}\left(-y \log(1-y) - (-\frac{1}{y} \log(1-y))\right)$
I mean here
ok, so start with $\sum_{n\ge 2} \frac{y^n}{n-1}$
note that it's equal to $y\sum_{n\ge 2}\frac{y^{n-1}}{n-1}=y\sum_{k\ge 1}\frac{y^k}k$
I didn't understand why the index went from 2 to 1
3:04 PM
$n-1=k$
you computed the first one correctly
I had thought so $y \sum^{\infty}_{n=2} \frac{y^m}{m} = -y \log(1-x) - \frac{y^2}{2}$
for the second one?
No, practically at the first
no, the index starts from $1$
No but the series starts from $n=2$, I don't understand when it changes
3:08 PM
$n-1=k$, when $n=2$ then $k=1$
Ah ok
So we substituted $2-1 = m$ in my case
So $m=1$
I'll try to see the second one
$m = 3$
$\frac{1}{y} \sum^{\infty}_{m=3} \frac{y^m}{m}$
3:13 PM
correct
$-\frac{1}{y} \log(1-y) - \frac{y^3}{3} - \frac{y^2}{2}$?
The Taylor series starts from $n = 1$ , what should I do now?
$-\log(1-y)=\sum_{m\ge 1}y^m/m=y+y^2/2+\sum_{m\ge 3}\frac{y^m}m$
I am at $m = 3$
Oh ok I had to add
So I get to $n = 3$
3:18 PM
so $S(x)=-\log (1-y)-y-y^2/2$ and now divide all terms by $y$
Oh okay
So
$-\frac{1}{y} \left(\log(1-y) -y -\frac{y^2}{2}\right)$
It would be all ()
So right?
no, the signs are wrong
$-1/y( \dots+\dots +\dots )$
Right
Thank you! I continue with the exercises, I hope I understood enough
ok, if you have other questions you can ping me
4:05 PM
can someone help me:
0
Q: If $f$ is a strictly increasing function with $\lim_{x\rightarrow \infty}f(x)=\infty$ is it ture that $\lim_{y\rightarrow \infty}f^{-1}(y)=\infty$?

user123234 Let $f:[0,\infty)\rightarrow \mathbb{R}_+$ be a strictly increasing continuous function with $f(0)=0$ and such that $\lim_{x\rightarrow \infty}f(x)=\infty$. I want to check if $\lim_{y\rightarrow \infty}f^{-1}(y)=\infty$, where $f^{-1}$ is the inverse of $f$. My idea would be the following: Fir...

yes it looks fine
almost fine, with the correction stated in the comment it'll be fine
@SineoftheTime Why the last 3dots are lower than the first 6dots?
4:23 PM
@SoumikMukherjee that's strange, I've used \dots
$\dots +\dots$
also here the last three dots are lower
4:37 PM
I don't understand the following:
The text defines: More generally, let $\chi : M \to M_0$ be a smooth map such that $\chi(p) = p_0$.
Any tangent vector $v \in T_p M$ gives rise to a tangent vector $D_p \chi(v) \in T_{p_0} M_0$ defined by
$(D_p \chi(v))(f) := v(f \circ \chi)$.

And then a page later, we are talking about the tangent bundle, and a maximal atlas defined upon it,
which contains all charts of the form $D\psi : T U \to V \times \mathbb{R}^n : (p, v) := (\psi(p), D_p\psi(v))$.
But obviously the second element is not in $\mathbb{R}^n$, it is a map?
In matter of fact, they do not even explain what M_o is. another manifold?
5:25 PM
implicitly, M_0 is a smooth manifold (i.e., something capable of being the codomain of "a smooth map"), but, what is "the text"? is there a link to it? does everything have to be implicit inferences from things we aren't seeing and don't have access to? :)
presumably this is buried in what "charts" are in the reference. they permit identifications of all sorts of things with R^n (perhaps for varying n)
5:38 PM
An exact text fragment would be nice.
Is it proven at this point that the tangent bundle is a bundle, so we can identify $TU$ with $U \times \mathbb{R}^n$ for some opens?
I'd be willing to bet they do define what $M_0$ is
 
1 hour later…
6:53 PM
How to solve the Diophantine equation, $ab(a+b)=T$, for positive integers, where $a$ is any given Triangular number & $T$ are also certain Triangular numbers? As of, $(a,b,T)=(3,7,210), (3,82,20910)...$?
7:13 PM
@SineoftheTime Sorry I didn't notice the message, but thanks for the help before !
But in the first exercise can I calculate the potential? And then do $W = F(1) - F(0)$?
7:26 PM
@Pizza if the field is conservative, yes
@SineoftheTime Ok so can I calculate the rotor and set it = 0?
no
you don't set it =0
Yes, sorry I meant It has to be 0
yes, you see if rot F is 0
OK thank you
And what if it comes out ≠ 0?
7:31 PM
the field is not conservative, so you have to compute it using the definition
Oh ok, thanks, one last thing
So the field is liberalist/socialist?
But to evaluate $\int \sqrt{x^2+y^2} dx$, It is a good idea to use $x = y \tan u , \quad dx = y \sec^2 u du$?
yes, that's the standard sub
Then I should integrate $y^2 \int \sec^3 u du$
7:37 PM
yes, but maybe there is y somewhere
I arrived there
Can I proceed by parts, yes?
do you know the derivative of $\sec u$?
$\tan x \sec x$
do you know by heart $\int \sec u du$?
No
7:43 PM
:(
But in theory this is immediate, right?
it's not hard
$\ln\left(|\sec(x) + \tan(x)|\right)$
This would be the result
Anyways, this is how I'd solve it: $$I=\int \sec^3 u du=\int \sec u(1+\tan^2 u)du=\int \sec u du +\int (\sec u \tan u)\tan u du= \\ \int \sec u du+\sec u\tan u -\int \sec u (1+\tan^2 u)du=\int \sec u du+\sec u \tan u -I $$. Hence $2I=\sec u \tan u +\log(\sec u +\tan u)+k$
Thanks, then I will try to do as you did
7:49 PM
this exam is very hard if you you're not that fast and have strong basis in analysis 1. Maybe your professor should lower the standards or give at least 3 hours
the exercises are not difficult, they're all the same after all
Yes, I'm a bit lacking with these forms $\sec x , cot x , csc x $ etc
that's normal
but he could put in the exam simpler integrals
not that this one is hard
Yes but I found this integral online, it wasn't from an exam
ah ok
I thought it was from an exam
my bad
In the exam it came out, but thanks to the polar coordinates it became like $\int r$ :|
Instead with Gauss Green, It was asked to do it on $\iint_D x dx dy$
7:59 PM
that's nicer
Yes
But it's not always like this
If Gauss Green is to lengthy, I'd leave it to the end
you can do the double integral and then other exercises and at the end GG
but you have to evaluate
But in your opinion is analysis 1 or 2 more difficult?
Many say that analysis 1 is more difficult
it depends
analysis 1 is intuitive
analysis 2 is 30%-40% analysis 2, the rest is linear algebra and analysis 1
Can I show you the analysis 1 exam i did?
Or have I already shown it to you?
I mean the analysis 1 exam that I took
8:09 PM
@Pizza sure
I did this
But this was a different prof
that's standard
Instead this is the analysis 1 exam of the professor that I am taking now, because he teaches both analysis 1 and 2
In fact, the structure of the exam is the same as that of analysis 2
Which one do you think is more difficult? Of these 2 that I sent
I think this integral with sin is more difficult
@Pizza yes, maybe slightly difficult
In the other one, you can exploit the symmetry of the domain of integration
@SineoftheTime I had done the integration for simple fractions
8:23 PM
yes
if it was $\sin x -1$, it would have been simpler
But in the integral of the sin , is it useful to multiply by the conjugate?
@Pizza yes, $\int \frac{P(x)}{Q(x)}dx$ are easy but lengthy and annoying
Because I remember that $\int \frac{1}{1+\sin(x)} dx$ , It work multiply by the conjugate
the annoying thing is that $2$
multiplying by the conjugate may not be the fastest strategy
Yes, however in analysis 1 we only had the opportunity to see those in that form well, in fact I would have difficulty solving this
8:27 PM
do you know Weierstrass substitution?
Yes
$\sin t = \frac{2t}{1+t^2} dt$
you use it usually when you have $\int F(\sin x,\cos x)dx$
If I'm not mistaken?
yes, without the dt
Oh right.
$dx = \frac{2}{1+t^2} dt$
8:30 PM
yep
$t = \tan x/2$
Yes, I used it once for $\int \frac{1}{\sin(x)^4} dx$
Oh yes in the exercise I had posted on math Exchange, where you helped me with gauss green
I remeber it :(
Yes, then I stopped because I took the physics 1 exam
your next exam is physics 2 or analysis 2?
Analysis 2, then I will try to do either algebra and geometry or signal theory because there is the mid-course test
8:35 PM
mid course test are useful
However, while I was writing in that integral above I arrived at $\int \frac{2\sin(x)+1}{4\sin^2(x)-1} dx$
I multiplied by the conjugate :(
@SineoftheTime Yes, but you should follow the whole course and study gradually, I think
Maybe I should use $u = \tan(x/2)$
Yes, however it is better to use the Weiestrass substitution
8:40 PM
$u=\tan (x/2)$ is Weierstrass substitution
But does it change anything if I use it now, or would it have been better to use it from the beginning?
from the beginning it's better
Hi !!!!!!!!
@SineoftheTime Oh okay!
@Gian'sPizzeria Hi
8:47 PM
How can i graph $\sqrt{y^2 - x^4}$
$z=\sqrt{y^2-x^4}$?
you can use level curves
Oh... Sorry I meant how do I find the domain
do you have any idea?
8:51 PM
Maybe $(y-x^2)(y+x^2) \geq 0$
$y^2≥x^4$
$y≥x^2$
But I was wrong...
how do you solve $a^2\ge b$ for $b>0$?
$a≥±b$
are you sure?
8:55 PM
first of all, I think you meant to say $a\ge \pm \sqrt b$. Second: for example $x^2\ge 1$, so according to you $x\ge \pm 1$, i.e. $x\ge -1$. Do you think this is correct
@Pizza that's an idea
@SineoftheTime no
I solved it, I did this: $\begin{cases} y - x^2 \geq 0 \\ y + x^2 \geq 0 \end{cases}$ $\cup$ $\begin{cases} y - x^2 \leq 0 \\ y + x^2 \leq 0 \end{cases}$
it should be $a\ge \sqrt b$ or $a \le -\sqrt b$
I would do like this anyway
Maybe Gian wants to do it another way
9:00 PM
Ok, but there are 2 variables here
your method is fine
@Gian'sPizzeria so?
Why are external values ​​taken?
@SineoftheTime For example why here it wasn't -√b≤a≤√b
why should it be?
9:08 PM
It can't be
@SineoftheTime I was looking at the integrals in various analysis 1 exams and I found this $\int \sqrt{\frac{x-2}{x+2}} dx$
that's not easy
@Gian'sPizzeria Did you solve it?
@SineoftheTime The fact is that I wouldn't know what to do, but it seemed interesting
for example, $u=\frac{x-2}{x+2}$
or $u=\sqrt{\dots}$
9:19 PM
@SineoftheTime Oh yes, that's how it had to be done i think
In fact math df suggests this to me
$\def\arraystretch{2}\begin{array}{c|c}u=\dfrac{\sqrt{x-2}}{\sqrt{x+2}}&x=\dfrac{-2\,{u}^{2}-2}{{u}^{2}-1}\\&\mathrm{d}x=\dfrac{8\,u}{\left({{u}^{2}-1}\right)^{2}}\,\mathrm{d}u\end{array}$
$\int{\dfrac{8\,{u}^{2}}{\left({{u}^{2}-1}\right)^{2}}}{\;\mathrm{d}u}$
Then here It does it by parts
I quote Folland in the construction of a sigma algebra from a generating set $\mathcal E$:
> So one must start all over again. More precisely, one must define $\mathcal E_\alpha$ for every countable ordinal $\alpha$ by transfinite induction: If $\alpha$ has an immediate predecessor $\beta$, $\mathcal E_\alpha$ is the collection of sets that are countable unions of sets in $\mathcal E_\beta$ or complements of such; otherwise, $\mathcal E_\alpha=\bigcup_{\beta<\alpha}\mathcal E_\beta$.
I fail to see how he used transfinite induction to define $\mathcal E_\alpha$. How is transfinite induction used here?
Joe
Joe
@psie: Do you agree that he used transfinite recursion to define $\mathcal E_\alpha$?
$\ln^2(x) \geq y^2$
I solved this for internal values ​​but I don't get to the solution
@Joe well, the thing is, he hasn't introduced transfinite recursion 😅 he has only talked about transfinite induction so far. I actually don't know what transfinite recursion is, but I'd be happy to learn about it.
The statement of transfinite induction is as follows:
> 0.15 The Principle of Transfinite Induction. Let $X$ be a well ordered set. If $A$ is a subset of $X$ such that $x\in A$ whenever $I_x\subset A$, then $A=X$.
Here $I_x$ is the initial segment of $x$.
I think I'm missing the white part...
Joe
Joe
9:29 PM
Neither transfinite recursion nor transfinite recursion have precise definitions, similar to how "proof by contradiction" does not have a precise definition, even we can mostly agree what a proof by contradiction is. But... Transfinite Recursion usually refers to the principle by which we can define a function $f$ (or class function) on a well-ordered set (class) recursively, with $f(x)$ given in terms of $f(y)$ for $y<x$.
He is definitely not using transfinite induction in the sense that you wrote down.
@Gian'sPizzeria But it wouldn't be $|\ln(x)| \geq |y|$?
Ok.
Joe
Joe
He is instead "defining a function by induction" (although purists will insist that the word "recursion" should be used for definitions, and "induction" for proofs). The justification for why such a definition makes sense can be found in textbooks on set theory.
@Pizza why?
@Joe do you mean the function $f$ in your previous message when you say that he is defining a function by induction? I don't quite see yet which function he is defining.
Joe
Joe
9:34 PM
Sorry, I don't quite understand your comment, but let me try. The function that Folland is defining is $\mathcal E$. We are just writing down the values of $\mathcal E$ at inputs as $\mathcal E_{\alpha}$ rather than $\mathcal E(\alpha)$.
ah ok, now I understand :)
@Gian'sPizzeria $a^2 \geq b^2 \quad \Leftrightarrow \quad |a| \geq |b|$
I solved it thanks to everyone
Joe
Joe
@psie: See indexed family for an explanation of where this notation comes from.
Is there any consensus on how to pronounce "homotopy" in English? Do people say huh-MAH-tu-pee, or HOH-muh-toh-pee, or something else entirely?
9:41 PM
@Joe yeah, it seems to me he is defining $\mathcal E$ just inductively. We are dealing with ordinals, so maybe one has to call it transfinite induction then.
I don't know.
Joe
Joe
@psie: Yeah, transfinite is a kind of funny word that usually just refers to constructions that go beyond the finite, especially if they involve infinite ordinals.
By the way, although it should make intuitive sense to you that it is "legal" to define $\mathcal E$ recursively, if you would like to see how how recursive definitions are justified in axiomatic set theory, then you might want to look at this question, and my answer.
There are some additional subtleties that come up when we want to define a function on a well-ordered set recursively, but I think the main points in my answers remain. There are even more subtleties that come up if you have to deal with classes rather than sets.... I have to admit that I am quite scared of classes.
Ok :) cool answer
Joe
Joe
But, as a sort of philosophical aside, I should note that in "everday mathematics", people use recursive definitions all the time, even if they don't know the recursion theorem. The point is that axiomatic set theory has been designed to both formalise and emulate "everyday mathematics". If axiomatic set theory couldn't prove the recursion theorem, or something similar, then arguably we would consider it to be unsatisfactory as a foundation for mathematics.
@BrianTung: Sorry, I don't mean to be ignoring your comment; I just have no clue what the consensus is either...
@Joe: No worries! I don't expect you to know the consensus (or even if there is one), though if you did know, that'd have been great! However, I'm still curious how you pronounce it (if you have occasion to).
Joe
Joe
Well, I have always pronounced it the second way, but I can't say that it counts for much. But if we are to believe the Oxford English Dictionary, both pronounciations are acceptable.
10:01 PM
I've always pronounced it as HOH-muh-toh-pee. But also I say huh-MAH-luh-jee, so who knows.
10:34 PM
is there a badge for most reopened questions ? :p
11:23 PM
@mick wouldn't it be better to figure out how to ask questions that no one wants to close?
@BrianTung I pronounce it "that silly thing what moves lines around".
@BrianTung I've only ever heard it said with the first syllable stressed. Either HO-maw-to-pee, or HAW-maw-to-pee.
hi :3
im in a good mood
im gonna send an email to ted thanking him
im also gonna make cards for all the professors who have helped me when i graduate this year
11:39 PM
@Allie Don't send cards. Send whiskey. Lowland single malts are a good place to start.
For the teetotalers, try high quality chocolates.
yeah chocolates is a classic gift, but I don't think many people eat the chocolate but rather give it away in turn to someone else
That's their loss.

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