« first day (5130 days earlier)      last day (185 days later) » 

06:12
@Shaun what's up man?
I further groupified twin primes if interested I'll share lin
*link it's group of units modulo a prime so a field, form
But prime-power equations ...
 
2 hours later…
08:18
0
Q: Find the expectation of the random variable $X^2$ when $X$ is a uniform random variable in $(0,1)$ following this particular way of reasoning.

Thomas FinleyFind the expectation of the random variable $X^2$ when $X$ is a uniform random variable in $(0,1).$ I tried solving the problem as follows: $X$ is a uniform random variable (URV) implies that, if $f_X$ is the probability density function of $X$ then, $f_X:\Bbb R\to \Bbb R$ such that $f_X(x)=1,x\i...

I need some help with this...
 
2 hours later…
10:09
Hi πŸ‘‹
11:08
@Debug Hi Daniel. I'm doing well. I hope things are okay with you. Yes, I'll take a look :)
11:29
I am reading a proof of the fact that if $E\subset\mathbb R^n$ is bounded, then it is totally bounded. It suffices here to prove that $T=[-b,b]^n\supset E$ is totally bounded, as any subspace of a totally bounded space is totally bounded. For this, let $\epsilon>0$. Then the balls $B(\epsilon j,\epsilon)$ cover $T$, where $j=(j_1,\ldots,j_n)$ ranges over all integral lattice points of $\mathbb R^n$ which satisfy $\epsilon |j_i|\leq 2b$, $1\leq i\leq n$.
(Recall, an integral lattice point is a point whose coordinates are integers.)
Since there are only finitely many such lattice points, $T$ is totally bounded, which completes the proof.
Question: How can I verify that the balls $B(\epsilon j,\epsilon)$ cover $T$? In particular, why the condition $\epsilon |j_i|\leq 2b$, $1\leq i\leq n$?
12:01
hi
is the trick here generalisable or specific to this problem
and is more of a language trick than math trick
is it*
13:46
Hi
@psie Here's my reasoning so far: if $x\in T$, then $x_i\in [-b,b]$ and $\epsilon j_i\in [-2b,2b]$, $1\leq i\leq n$. Now I want to show $d(x,\epsilon j)<\epsilon$, but I am not sure how to do that. I suspect one would like to consider a third point and use the triangle inequality somehow. Not sure.
14:04
@psie It'd be easier if you draw a picture. For example, consider [-5,5] in R. Say you want to cover it by balls of radius 1/3. Now if you just put such a ball at each of the points -5,-4,...,4,5 then it won't cover the whole interval. So you need more points where you can place a ball.
makes sense πŸ‘
And about the 1st question, suppose you have a point in R^2, and you place a ball of radius 1 at that point (the point is the center of the ball ). Can you claim that the ball contains an integral point?
@SoumikMukherjee yes, since the radius is $1$
Yes, though you have to argue that the nearest lattice point from any point in the plane is at a distance of at most $\sqrt{2}\over 2$
Anyway they are doing the same thing here, they are reducing the balls from radius 1 to $\epsilon$ but they are also adding more points
Another approach would be to consider the biggest cube that lies inside the ball of radius $\epsilon$ and then cover the space with such cubes and then replace each cube by a ball
@RyderRude This is a math trick
14:38
ok, hmm, I need to think about this some more probably. I still have trouble seeing how some $x\in T$ is contained in $B(\epsilon j,\epsilon)$ for some $j\in\mathbb R^2$ satisfying $\epsilon |j_i|\leq 2b$.
14:53
Try to draw a picture with a particular $\epsilon$ in mind
15:08
@SoumikMukherjee ok, for example, if $b=1$, $n=2$, $\epsilon=1.5$ and $x=(0.5,0.5)$, then it should be contained in the ball $B(\epsilon j,\epsilon)$ where $j=(0,1)$. Using the triangle inequality and $j$ as the third point (and the max metric) $$d(x,\epsilon j)\leq d(x,j)+d(\epsilon j,j)=\max\{|0.5-0|,|0.5-1|\}+\max\{|0-0|,|1.5-1|\}=1<\epsilon,$$but does this procedure generalize?
I don't see yet where I specifically used $\epsilon |j_i|\leq 2b$.
15:33
There are more than one points where you can place a ball
Wait
@SoumikMukherjee oh. i thought it was a language trick because we r kind of asking two questions by using a question within a question
@psie epsilon j is outside of your interval
@SoumikMukherjee are tricks like this studied, or does it only show up in this problem?
@SoumikMukherjee I don't think so: $\epsilon j=(0,1.5)$ and $1.5\leq 2$, or?
Your interval is [-1,1], no?
@RyderRude yeah but that's a math trick, ultimately they are asking something like if statement 1 is true then is statement 2 true?
15:40
@SoumikMukherjee yes, but is this a problem? I thought $j\in\mathbb R^2$ can be any point satisfying $\epsilon |j_i|\leq 2b$. I don't see the problem.
@SoumikMukherjee yeah... i find these tricks fascinating
@SoumikMukherjee probably, but to show that $T$ is included in $\bigcup_{j\in I} B(\epsilon j,\epsilon)$ where $I=\{j\in\mathbb R^n:\epsilon |j_i|\leq 2b\}$ we simply need to show that $x\in T$ is included in some ball
@RyderRude this is about logic so they just made an example of using logic
@user20458579510081670432 when you join, half of my screen is covered by your name. skull emoji
@SoumikMukherjee oh
15:43
☠️ @SineoftheTime ☠️
😎 πŸ™ˆ
@psie not a problem as [0,1] is a subspace of R
@RyderRude a trick that you see more than twice is a method.
@user20458579510081670432 i see similar tricks often
66
Q: What do named "tricks" share?

Joseph O'RourkeThere are a number of theorems or lemmas or mathematical ideas that come to be known as eponymous tricks, a term which in this context is in no sense derogatory. Here is a list of 11 such tricks (the last of which I learned at MO): the Whitney trick the DeTurck trick the Cayley trick the Rabinow...

15:49
@psie I will write things clearly once I am back to my laptop
@SoumikMukherjee thanks, that would be greatly appreciated
@SineoftheTime hi
I was doing an analysis exercise, could you tell me if this procedure is correct?
Calculate the absolute extrema of the function $g(x,y) = x - y$ at the unit center of the origin
the function is defined on a compact (the closed unit circle), therefore the Weierstrass theorem guarantees the existence of absolute extrema
just to be clear, by unit center do you mean $x^2+y^2\le 1$ or $x^2+y^2=1$?
15:56
<=
Since the function is linear It will have no local extremes within the domain
@RyderRude from the post: "The difference between a trick and a method is understanding."
So I can only analyze the part x^2 + y^2 = 1?
which would be the edge
And I parameterize it
Do you mean, since the gradient is constant you don't have max and min in the interior of the circle?
15:59
Yes
that's correct
the gradient is never $0$
So I would have parameterized the edge : $f(\theta) = -\sin(\theta) - \cos(\theta) \quad \theta \in [0,2\pi]$
$f'(\theta) = 0 \rightarrow \sin(\theta) = -\cos(\theta)$
Which is valid for $\theta = \frac{3\pi}{4} , \frac{7\pi}{4}$
Calculating the function at these points I find -√2 and √2
@SineoftheTime ah ok!
why $-\sin \theta$?
I wrote it wrong
$f(\theta) = \cos(\theta) - \sin(\theta)$
I used x = cosΞΈ , y = sinΞΈ
It should be correct right?
the method seems correct
in class did you prof parametrize when dealing with the edge?
16:11
@user20458579510081670432 sometimes tricks r just tricks
How one arrives at "understanding" is a different question.
There are also "tricks" to understand tricks.
@SineoftheTime Yes we did it
then ok, you can also use Lagrange multipliers
do you have other doubts?
@SineoftheTime yes this too
16:16
Knowing what to look for is the first step.
I would like to use the "quickest" methods where possible.
@psie or you can just say that $[-b, b]^n$ is compact, so totally bounded
@SineoftheTime if you are still free I could ask you something else
@Pizza sure, here it's good to parametrize
yes, you can always ask
Oh I forgot something here
16:18
if I'm not in chat and you have a question, just ping me
In this case I also had to check the values of $f(\theta)$ on the other notable points?
what are the other notable points?
As $\theta = 0, \pi/2, \pi$?
if the derivate is not $0$ in these points, you don't need to verify
@psie you need to choose $j_i$
16:21
@SineoftheTime ah ok
However I had this other doubt
Just a moment while I write the problem, anyway thanks for this help
@psie you don't want to use triangle inequality
I have this integral: $\int_D \frac{y \sin\sqrt{x^2+y^2}}{x^2+y^2} \text{dx dy}$ , on the domain D of the first quadrant delimited by the x axis and by the circle of radius 1 and centers O (0,0) and C (1,0)
Thanks:)
so D is the intersection of the two circles, with $y\ge 0$
if I understand correctly
16:28
the first circumference is $x^2+y^2 = 1$, the second one is $(x-1)^2 + y^2 = 1$
using x = r cosΞΈ, y = r sinΞΈ, x²+ y² = r², dx dy = r dr dΞΈ, I found at the first circumference $r = 1$ , to the second $r = 2\cos(\theta)$
@psie I don't think this actually covers $\mathbb{R}^n$
ΞΈ varies between Ο€ and 0?
let me grab a pencil
16:31
the problem is that I don't know what r varies between
@SineoftheTime ok
Given $\varepsilon > 0$, the vector $\frac{\varepsilon}{2}(1, 1, ..., 1)$ will be distance $\geq \sqrt{n}\frac{\varepsilon}{2}$ away from each vector of the form $\varepsilon j$ where $j$ is an point on the integer lattice.
so for $n\geq 4$ it won't belong to any ball of the form $B(\varepsilon j, \varepsilon)$
Yes you're right, I was thinking about R^2 only, my bad
We need to change the cover as to accommodate for the change in dimension
5 mins ago, by Jakobian
@psie I don't think this actually covers $\mathbb{R}^n$
I meant $T$ of course
@Jakobian things like these make me wonder if this has anything to do with the low-dimensional/high-dimensional topology division
@Pizza do you want to express theta as a function of r or viceversa?
4
A: An elementary way to show any bounded subset of $\Bbb{R}^k$ is totally bounded

PedroI'll be systematic here, I think it can help. D Let $S$ be any subset of $\Bbb R^n$. Given $\epsilon >0$, we say that $N$ is an $\epsilon$-net for $S$ if the set of open balls $$B_\epsilon(N)=\{B(x,\epsilon):x\in N\}$$ covers $S$. That is, the set of open balls of radius $\epsilon$ centered at...

16:43
@SineoftheTime ΞΈ = f(r)
Perhaps it would have been better to express r as a function of ΞΈ?
I'm pretty sure one just has to go for $B(\frac{\varepsilon j}{\sqrt{n}}, \varepsilon)$ instead, the inequality $\varepsilon |j_i|\leq 2b$ may be changed too
I'm making a silly mistake in the computation but I can't find it :(
unless we're talking about the maximum distance and psie didn't mention that
16:50
@SineoftheTime I think it's better to do it viceversa since r varies based on ΞΈ
Or not?
Maybe it's more complicated as I said before
5 minutes :
ok, figured it out. I'm stupid
Using polar coordinates, you should have $\theta \in [0,\pi/2]$ and $0\le r \le \min\{2\cos \theta,1\}$
you can split $\int_0^{\pi/2}=\int_0^{\pi/3}+\int_{\pi/3}^{\pi/2}$
to find the bounds, express $D$ in polar coordinates. $y\ge 0 \implies r\sin \theta \ge0$, i.e. $\theta \in [0,\pi]$. $x^2+y^2\le 1\implies 0\le r \le 1$
$(x-1)^2+y^2\le 1\implies r-2\cos \theta \le 0$ i.e $2\cos \theta\ge r$
since $r\ge 0$, $2\cos \theta \ge 0$, so that $\theta \in [0,\pi/2]$
$r\le 1$ and $r\le 2\cos \theta$ implies $r\le 1$ for $\theta \in [0,\pi/3]$ and $r\le 2\cos \theta $ for $\theta \in [\pi/3,\pi/2]$
this should be correct (?)
17:08
I'm trying to figure out the steps for a moment
let me know if there are mistakes and if it's clear
But $2\cos(\theta) \geq 0$ it's also true for -Ο€/2,0?
maybe it shouldn't be taken?
no
because you must also have $\sin \theta \ge 0$
all conditions must be fullfilled
$r\le 1$ and $r\le 2\cos \theta$ implies $r\le 1$ for $\theta \in [0,\pi/3]$ and $r\le 2\cos \theta $ for $\theta \in [\pi/3,\pi/2]$
This last thing is not clear to me
basically you have to solve $2\cos \theta \le 1$
if $\theta \in [0,\pi/3[$, then $2\cos \theta>1$
so $r\le 1$ and $r\le \text{something bigger than $1$}$
hence when $\theta\in [0,\pi/3]$ you choose $r\le 1$
is it clear?
17:27
But if r <= 1, shouldn't i solve 1 <= 2 cosΞΈ?
you can solve it
it doesn't matter
if you solve $1\le 2\cos \theta$, you find when you choose $r\le 1$
Oh ok, but so we expressed theta as a function of r?
you consider $\arccos$
but you have to be carefull with the domain
I prefer expressing $r=r(\theta)$
So the integral Is $\int_0^{\pi/3} \int_0^{min(1,2\cos\theta)} \sin \theta \sin r \ \ dr d\theta$
if $\theta \in [0,\pi/3]$, then $r\in [0,1]$
17:40
Ah ok!
$I=\int_0^{\pi/3}\int_0^1 +\int_{\pi/3}^{\pi/2}\int_0^{2\cos \theta}$
don't forget the jacobian :)
@SineoftheTime $$\int_0^{\pi/3}\int_0^1 \sin(\theta) \sin(r) dr d\theta+\int_{\pi/3}^{\pi/2}\int_0^{2\cos \theta} \sin(\theta) \sin(r) dr d\theta $$
@SineoftheTime I think it was simplified
oh right
$x^2+y^2=r^2$
$\frac{r \sin\theta \sin r}{r^2} r dr d\theta$
17:50
so it was simplified
So now I try to solve these integrals
18:04
I find rudin a bit "uninspiring" in that, I tend to look into books like, eg. Shirali & Vasudeva- Metric spaces, or Searcoid Metric spaces for intuition, which Rudin is devoid of. But that said, Rudin definitely does teach the techniques quite well (albeit in a few pages). I kind of like that about it
18:23
I hate Rudin
@Jakobian why? and what else would you prefer as an analysis text
Because it only gets done what it sets out to get done
18:40
Why would you hate that?
@nickbros123 I dislike baby Rudin because I think that it is overly terse and fails to be a pedagogical book---it is a reference, not something you are meant to learn out of. There is no intuition, and some of the stuff he does is, in my opinion, just backward. There are lots of better books for the material. Spivak and Apostal each have books, for example. I kind of like Tao's approach. Even Folland's undergrad analysis text is better (in my opinion---and I am not a huge fan of Folland).
Spivak's text is pretty good
@nickbros123 in particular this remark about teaching the techniques well, is misguided I think
it's a sort of Stockholm syndrome, where you're given so little about those techniques that you start appreciating them when they are present
@user20458579510081670432 hopefully that (and Xander's remark) explains why
How many books of one topic (say one variable calc) one should read to have the knowledge of a uni professor?
Probably just one
18:55
I'm facing a lack of self esteem due to the fact that I have high marks but deeply inside me I don't think I know how thing work
@SineoftheTime That question is nonsense, I think. University professors don't read books to be knowledgeable about what they teach (or, at least, that is not all that they do).
Teaching is a good way of learning
I talk about this with all the calc professors in my department and they say it's normal cuz now we tend to study all hastily just to graduate
@SineoftheTime I finished , thank you very much for the many helps :)
I would expect someone teaching calculus to at least be knowledgeable in real analysis. I would also expect them to know some diffy q and linear algebra (rigorously; beyond just being able to do computations). They should know where the ideas taught in calculus come from, how to prove them, and how to prepare students for future classes.
18:56
by Calculus what exactly do you mean
for example single variable analysis
Moreover, I would want a professor to have read several books on single variable calculus before teaching the class. I want them to have considered the approaches of different books, and made an active choice to choose a book which best reflects how they want to teach the class.
In Italy there's not difference between calculus and real analysis
(Or, at least, if they pick a book they don't like so much, there should be a good reason to make that choice.)
If you know where things come from, then you will be confident enough to see yourself as "knowing calculus"
@XanderHenderson I'd want them to read one, and browse the other ones
not necessarily full on read them
18:59
well, when doing theorems and proofs I understand almost everything but still I have the feeling of not being deep
@SineoftheTime you know, that should be considered in a case-by-case way, imo
Combinatorics is fun, I'm lovin' this class
@SineoftheTime Can you explain calculus to others and make them understand?
I think so
19:01
I think I could handle teaching a whole class of people, if I had some plan on how to teach them
there's nothing really in calculus that I'm struggling to understand
to my knowledge at least
@XanderHenderson Do you think is it bad "losing" one year (not graduating in time)?
@SineoftheTime What do you mean?
graduating in 4 years instead of 3
@SineoftheTime Already, that makes no sense to me. In the US, there are 2 year degrees (associates and masters), 4 year degrees (bachelors), and phds (which typically take at least 4 years, depending on field).
So I don't know of a program of study for which 3 years is normative.
In Europe a Bachelor usually takes 3 years
19:04
3 years is undergrad here
As such, I doubt the advice I could give you would apply to your situation, as you clearly don't come from the US, and so things are going to be very, very different.
3+2 plus PhD (depends on the university, but usally 3)
In India it takes 3 years, but they are probably going to make that 4 years in near future
in the past is was only 4 years
@SineoftheTime I don't understand this---what is the 3 and what is the 2?
19:06
3 bachelor and 2 master degree
3 years of Bachelor and 2 years of masters
4 years is if you want to get an engineering degree
math used to be 4 years and you were a math engineer before, but they changed it to bachelor
self studying maybe it's better
Ah. Sure. But it is also worth noting that European schools tend to start tracking students into college prep classes much earlier than in the US. So you miss out on some of the liberal arts in favor of specializing much earlier (high school students in a college preparatory track miss out on some gen ed, but get more specialized training younger).
Totally different philosophy from the US.
19:08
@XanderHenderson I don't really know what you mean by the college prep classes
@Jakobian Classes which are meant to filter into college programs. Things like calclulus.
Not a thing in here by default as far as I know
As opposed to, for example, metal shop.
@XanderHenderson Have you voted in the ongoing moderator election?
there is no real "college preparation" but for example the university I used to go did organize some classes for high school students to prepare them for university
but I didn't even participate in those classes since I didn't know they exist
so yeah, I don't know how it's in other countries but at least in here it's not like we have some kind of preparation class mentality
19:13
@SoumikMukherjee Why?
I mean, why do you ask?
(If you really want to know the answer to the question, you can look through the list of people who have earned a badge for voting in the last week, and determine whether or not my name is among them.)
tell your students to vote in this election and then give them one point bonus
that's be cool and all if only this election were available for everyone
@XanderHenderson Just curious, as you asked most of the questions that were asked to the candidates in the comment section, so whether any candidate seemed eligible to you( ofc I won't ask who you have voted for, if any)
I will ask who you have voted for
(I'm kidding)
The winner
@SineoftheTime Who do you think would win?
Gukesh or Ding?
19:24
Guky
@SoumikMukherjee "If you vote after the election is ended, you never lose" Sun Tzu
yesterday Ding had a better position
but it was hard to win
I don't consider Ding as a "real" WC
@SineoftheTime Run for an election and vote for the other party, either way you'll win
@SineoftheTime Is he an imaginary WC?
if you rotate him by $90^°$, then yes
@SineoftheTime ooh, have not seen that game yet
Imagine concentric spheres in $\Bbb R^{3}$ centered at the origin where you define a metric based on differences in volumes of pairs of spheres
19:40
but in all arctangents can the minus be taken out of the argument?
@BinkyMcSquigglebottom nice profile picture :)
Thanks
:⁠-⁠)
takeshi "gian" goda
@Sahaj Really popular character here from Doraemon
Is Gian the best character?
19:48
@BinkyMcSquigglebottom isn't there only one arctangent?
@BinkyMcSquigglebottom no
@SineoftheTime Just one
what do you mean by all arctangents
If like I have arctan(-(something)) can the minus always be taken out?
yes because it's an odd function
@ε†₯ηŽ‹Hades it's among the shows I used to watch on television as a kid about 10 years ago.. fond memories
19:52
@SineoftheTime and who is?
Guys Who Is the best doraemon character?
@SineoftheTime thanks
@BinkyMcSquigglebottom I'd say Dekisugi. Calm & composed, smart, helpful.
@BinkyMcSquigglebottom bro I don't even know who Gian is
f(x)=Arctan(-xe^(1-x)) if x<0 when Is f(x)>0?
When x<0?
20:20
@BinkyMcSquigglebottom Dekisugi
20:36
@Shaun no need to take a look. We found basic errors... -__- anyhow back on my ArrowGlue code for now, and I'll put twin primes on pause. It takes a lot of mental energy to work on it and doing it all day just isn't productive ^_^
I do however like the approach of prime-power equations, because there's a lot of group theory, just have to find the right framing of it, because there are actually many ways to frame it.
For example the minimal element > 1 in the group of units modulo $p_n\#$ is most definitely the residue $p_{n+1}$. But take some certain subgroups of the group of units modulo primorial, and how do you compute the "smallest non-trivial residue" in the group? That is essentially what I'm trying to answer. Or at least show that it can't be > $p_{n+1}^2 - 2$ and then we're done; twin primes would be proven.
20:51
@zetaspace Okay... why? It seems that you have constructed a space that is isometric to $[0,\infty)$...
Or have I missed something in your statement?
21:05
@Jakobian I think indeed you have found an error in the book. Thanks. Folland has another proof which is correct I think, but I'd really like to amend this proof; you suggest the balls $B(\frac{\varepsilon j}{\sqrt{n}}, \varepsilon)$. Under which condition on the components $j_i$ would they work? Grateful for any explanation.
21:17
Recall, we want to cover $[-b,b]^n$ with those balls for every $\epsilon>0$, where $j$ is a point on the integer lattice.
21:57
@Debug It happens to the best of us, don't worry.
 
1 hour later…
23:04
math.stackexchange.com/questions/4960643/… . My question : why is $m/m^3$ not a k vector space?
@jasmine because if you define $(a+\mathfrak{m})\cdot (b+\mathfrak{m}^3) := ab+\mathfrak{m}^3$ then this might not be well-defined since if $a'\in\mathfrak{m}, b'\in \mathfrak{m}^3$ then by the same definition $(a+a'+\mathfrak{m})(b+b'+\mathfrak{m}^3) = ab+a'b+\mathfrak{m}^3$ where the term $a'b$ isn't necessarily in $\mathfrak{m}^3$
2
@psie I think $\frac{\varepsilon}{2\sqrt{n}}|j_k|\leq b$ should work, but you can always go through the argument and check
23:30
@JakobianIf $a' \in m$ then $a'\in m^3$ because ideal $m$ contain $m^3$ .This implies $a'b \in m^3$. I don't understand why the term $a'b$ isn't necessarily in $m^3.$

« first day (5130 days earlier)      last day (185 days later) »