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Bob
Bob
12:00 AM
I have posted several problems and my answers to then to be checked by the group. They all deal with the same topic. Is that a problem?
 
Ted Shifrin
Ted Shifrin
Well, Fargle, I'm always happy to chat diff geo with you.
 
0celo7
0celo7
@AlessandroCodenotti The undergrads here don't know what PDEs are.
They also don't know that there are seminars.
 
Alessandro Codenotti
Alessandro Codenotti
Here we have an undegrad PDE course so everybody can dislike them
 
0celo7
0celo7
We do too, but what you get from that class has little to do with PDEs.
 
Ted Shifrin
Ted Shifrin
We had an undergrad PDE course at UGA, but it was mostly engineering-style and often badly taught.
I never taught that, although I taught a lot of the material in the year-long applied math class I taught in 1986-7.
 
Bob
Bob
12:01 AM
PDE can be very important in engineering
 
Ted Shifrin
Ted Shifrin
they're important in tons of things, including mathematics.
 
Fargle
Fargle
@Ted I don't know how I missed this in earlier reads, but I like how the Frenet frame kinda just builds itself. It almost feels like cheating. That constant length iff orthogonal to derivative theorem is insanely powerful once you see you can just normalize stuff.
 
Alessandro Codenotti
Alessandro Codenotti
True, our undergrad PDE course wasn't very good, mostly because we don't have the prerequisits to actually do a PDE course
I skimmed the PDE part of Brezis when studying functional analysis, that looked much more interesting but still not my cup of tea :P
 
0celo7
0celo7
I think that's the worst part of Brezis, and I'm a PDE person.
 
Daminark
Daminark
In my functional class I liked Hille-Yosida
 
Alessandro Codenotti
Alessandro Codenotti
12:04 AM
Interesting, I really liked the first 5 or 6 chapters (and I tried quite a few texts before deciding to study from Brezis)
 
0celo7
0celo7
I still don't understand the dedication of the book. Something about viewing PDEs through the eyes of a functional analyst, but the treatment uses virtually no functional analysis.
 
Bob
Bob
what is functional analysis?
 
0celo7
0celo7
@AlessandroCodenotti Brezis and Yosida are my go-to books for functional analysis.
@Bob The theory of infinite-dimensional topological vector spaces.
 
Daminark
Daminark
Sobolev spaces were a bit iffy though, we didn't talk weak derivatives and instead described them as the completion of the compactly supported smooth functions, but then it wasn't clear when things were defined everywhere
 
Alessandro Codenotti
Alessandro Codenotti
That's the second recommendation for Yosida in two minutes, I'll see if our library has a copy
 
Daminark
Daminark
12:08 AM
(e.g. Laplacian is unbounded so we weren't sure how to extend it from smooth functions to H^2)
 
0celo7
0celo7
@Daminark You extend it as a map $H^2\to L^2$.
Then it's bounded.
 
Alessandro Codenotti
Alessandro Codenotti
@Dami you have a new quarter starting soon, right?
 
Daminark
Daminark
Stuff like that made it more confusing, also it didn't feel like functional as much as a long drawn out example that used functional at the end. So yeah I dunno
@Alessandro yeah I've got this week of break and then spring quarter starts
 
0celo7
0celo7
@Daminark That's what people in geometry do because it's easy and defining weak derivatives on manifolds is a pain in the ass. A careful student then has to spend a lot of time checking that everything makes sense.
It's definitely an issue when working with vector bundles.
 
Ted Shifrin
Ted Shifrin
@Fargle: Yes, the orthogonal group is very powerful. (I was fixing a martini and getting some appetizers ...)
 
0celo7
0celo7
12:10 AM
@AlessandroCodenotti It's on Springer.
 
Alessandro Codenotti
Alessandro Codenotti
Which courses are you going to take? @Dami
 
Daminark
Daminark
Right now the plan is complex analysis, algebraic topology, Galois theory, and combinatorics
 
Ted Shifrin
Ted Shifrin
I'm glad to see you taking things besides mathematics.
 
Daminark
Daminark
Very much so
 
Fargle
Fargle
sweats
 
Alessandro Codenotti
Alessandro Codenotti
12:12 AM
@TedShifrin that's not a nice thing to say about combinatorics
 
Fargle
Fargle
Sorry, that was just a really hot take.
 
Ted Shifrin
Ted Shifrin
Huh? @Fargle
 
Fargle
Fargle
Never mind.
 
Ted Shifrin
Ted Shifrin
@Alessandro: I appreciate your sarcastic retort to my sarcasm.
 
Alessandro Codenotti
Alessandro Codenotti
:P
 
0celo7
0celo7
12:14 AM
The oldest trick in the book -- don't define what you're talking about so any claims you make cannot be falsified.
 
Alessandro Codenotti
Alessandro Codenotti
I'm actually going to sleep now though, bye!
 
Ted Shifrin
Ted Shifrin
That's how I taught for 40 years, 0celo.
Night, @Alessandro.
 
0celo7
0celo7
It's how certain geometers write.
 
Daminark
Daminark
But yeah this quarter is the first where I'm doing this. We'll see how it goes. If it's too much then later I'll modify the strategy.
 
Ted Shifrin
Ted Shifrin
I was being sarcastic, yet again.
Demonark: My personal advice is to take advantage of your last bit of undergraduateness before becoming a grad student.
Demonark: Grad complex?
 
Daminark
Daminark
12:16 AM
I was considering linguistics but one section conflicts with Galois and the other with complex. That and econ are probably the main non-math things I'm interested in trying
Yeah grad
 
Ted Shifrin
Ted Shifrin
Probably a bit math-heavy, then.
 
0celo7
0celo7
@TedShifrin Did you get the feeling at UGA that seminar/colloquium attendance by grad students was low? Alternatively, that there was a lack of curiosity?
 
Daminark
Daminark
Perhaps. If I were to let go of anything it'd be AT since that's likely auditable (I do wanna take grad AT next fall so I'm gonna have to learn the undergrad stuff somehow by then)
 
Ted Shifrin
Ted Shifrin
Colloquium attendance was generally pitiful on the parts of both faculty and grad students. On the other hand, when I was a Berkeley grad student, I think I missed a total of 5 colloquia (when I wasn't traveling), and I was one of the only grad students (out of 400) to attend regularly.
Demonark: Really? Don't they teach grad AT for incoming grad students who don't know anything from undergrad?
 
0celo7
0celo7
400 grad students??
 
Ted Shifrin
Ted Shifrin
12:19 AM
Yes, Berkeley has now shrunk. Down around 200 PhD students, I think.
The attrition rate was humongous in the 70's.
 
Daminark
Daminark
Uh, depends on who teaches. Danny Calegari taught last year and he said he was gonna go very quickly and assume that people have had undergrad AT
 
Ted Shifrin
Ted Shifrin
That's not fair to all the first-years from elsewhere (which is most of 'em), Demonark.
What's in undergrad?
 
Daminark
Daminark
Depends a lot on who teaches, course description says "Topics include the fundamental group of a space; Van Kampen's theorem; covering spaces and groups of covering transformation; existence of universal covering spaces built up out of cells; and theorems of Gauss, Brouwer, and Borsuk-Ulam."
 
0celo7
0celo7
What is the Gauss theorem
 
Daminark
Daminark
But this rarely is what the actual class looks like I think
 
Ted Shifrin
Ted Shifrin
12:22 AM
Definitely not fair to assume all beginning grad students have had fundamental group and covering spaces, van Kampen, etc. Ridiculous.
 
Daminark
Daminark
And I'm not sure exactly what background the grad students have. I think you can still follow even Danny's class without prior background, you'll just have to work damn hard. Also next year Weinberger is teaching and I'm not sure what he assumes
 
Ted Shifrin
Ted Shifrin
Well, in principle, one can learn homology and cohomology to a large extent without knowing the fundamental group, but that's not appropriate for, nor fair to, the grad students.
 
Daminark
Daminark
Maybe they only take students who know some basic AT?
I'm not really sure. I'm guessing that not everyone has the background but they just make it work very quickly
 
Ted Shifrin
Ted Shifrin
Demonark, that would be ludicrous, although I'm sure some of the grad students have in fact taken AT as undergrads.
 
Daminark
Daminark
Maybe. Also I'm not sure if he necessarily assumes what's in the course description, because that's not really a good representation of the class. Someone took it with Peter a year ago and then grad with Danny
She said undergrad covered 85% of grad but that the latter was more in depth, so while he did go through all of Hatcher, start to finish, it was hard to absorb it in its full depth if it was your first time, but possible if you devote a ton of time
Whatever the system is, it seems to work out alright, and it's one in place, so you just have to play the cards you've been dealt
 
Tanuj
Tanuj
1:00 AM
Guys, can a parabola have a center (point where a chord is bisected)?
On partially differentiating the general equation of parabola wrt x and then y and then solving them together gets me to the conclusion that a parabola cannot have a center. Is it true? If yes then how?
 
Maneesh Narayanan
Maneesh Narayanan
1:14 AM
How do you define centre as a point where chord bisected taken?
consider the circle.
Do you able to get the centre of the circle according to the definition?
@Tanuj
 
Niing
Niing
Is the reverse of Symmetry of second derivatives, or Schwarz's theorem, also correct?
Larson calculus p.895
 
Tanuj
Tanuj
1:50 AM
@ManeeshNarayanan Well it's not the other way round. Every chord is not bisected at the center, but only those that pass through it.
@Secret Hi! Help needed.
 
Secret
Secret
I need a picture to see which chords are we talking about
 
Niing
Niing
if $f_{xy} \not= f_{yx}$ then $f$ will have two value at a point?
 
orbit-stabilizer
orbit-stabilizer
If $f$ is a function, it cannot map one thing in the domain to two things in the co-domain.
 
Niing
Niing
Yes, so it will be a surface which has vertical line, so it's not even a function.
I'm imagining the picture of the function when the theorem is false.
Like you've said, if the theorem were false then it will lead to a contradiction.
 
Secret
Secret
2:10 AM
Is Schwartz theorem an iff?
If not, what is an example of f that has commuting mixed derivatives but is discontinuous?
 
Tanuj
Tanuj
@Secret well, I don't know if they even exist, so I can't draw the picture, but tell me this, can a parabola have a center?
 
Secret
Secret
@Tanuj I don't know what the centre will be like if you don't tell me what the chords on the parabola like
Because I will expect all horizontal chords to be bisected by the line of symmetry of the parabola
 
Niing
Niing
it's not an iff, I was just imaging the case when the conclusion is false...
 
Tanuj
Tanuj
@Secret nvm leave it. I don't think there is a center for parabola.
 
orbit-stabilizer
orbit-stabilizer
 
Secret
Secret
2:28 AM
No, I am thinking about the counterexample to the converse of Schwartz theorem, where a function where mixed partial diffs are equal but they are discontinuous on the open disk
 
Niing
Niing
3:07 AM
it's impossible...?
since the those diffs exist means they're continuous?
 
orbit-stabilizer
orbit-stabilizer
No
 
Eric Silva
Eric Silva
@TedShifrin they go through the same material the grad course is just harder and faster, so it works as a first course I think just a hard one
Danny definitely didn't assume we knew what the fundamental group was day one but I guess the pace would've probably been much harder without kind of already knowing what was going on
 
Secret
Secret
what's an example of a function with equal but discontinuous mixed partial derivatives?
 
Niing
Niing
3:32 AM
So a function $f(x,y)$ both diffs exist and the same doesn't promise that the function is continuous?
 
Secret
Secret
response does not exist
 
Niing
Niing
So I should not ask this question?
Is this a bad question?
 
Secret
Secret
No its just the chat is currently too empty and nobody is responding to our questions
 
user76284
user76284
3:47 AM
Let $S$ generate a group $G$. Then $x \in G$ is in the center of $G$ if and only if it commutes with all the elements of $S$, right?
 
Secret
Secret
Room is currently empty, responses does not exist, no questions will be answered
 
Niing
Niing
Life is too short to be rigorous
 
Secret
Secret
> Don't turn the room into a graveyard of deleted messages.
lol
Consider the following:
Define a lugotheropen clastopheromen be the defino $X: Y \to Y{X \to E^{e^x}}$ . Now let westerine $W$ be the squaoshmorin $s = \int _S^{s} s^{s^{gib}}df^{dsdt}$ such that $G = M\circ s_kc$. Now, it follows that the qwe of the 45768 is casnsurate to the following
$$\int^{\int^{\int}}D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr}$$
 
Niing
Niing
I can prove it
Use horseshoe
I'm giving you the link about horseshoe
wait a minute
@Secret: Take a look
 
Secret
Secret
4:06 AM
You cannot use the horseshoe lemma because the above pseudointegral is irreflexive :P
also typo:
$X: Y \to Y^{X \to E^{e^x}}$
Let $S$ be a sextent. The the following holds:
$S^{S^{S(ubetj \mathop{E}(Putin)+\text{Rus}\frac{@}{3})}}$
 
Niing
Niing
maybe you can try differentiation, when fail, try twice or more
 
Secret
Secret
ok, let's see :P
$$\frac{d}{d\tau}\int^{\int^{\int}}D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr} = \int^{\int^{|}} D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr-1}$$
$$\frac{d^2}{d\tau^2}\int^{\int^{\int}}D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr} = \frac{d}{d\tau}\int^{\int^{|}} D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr-1} = \int^{|} D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr-2ux}$$
 
Akiva Weinberger
Akiva Weinberger
\heartsuit $\heartsuit$
 
Secret
Secret
Ah finally, we are bored to death here
 
Akiva Weinberger
Akiva Weinberger
\varheartsuit $\varheartsuit$
Aw dangit
 
Niing
Niing
4:21 AM
I'm sad, but they look beautiful.
And still, they're easier than Riemann Hypothesis.
 
Secret
Secret
O btw, we did formalise "$dx^{dx}$" somewhat so "$\int^{\int^{\int}}$" is not very far fetched
 
mowwwalker
mowwwalker
Heya, I'm a comp sci grad looking to start studying math recreationally starting from the top and trying to understand everything intuitively. I'm planning to start with geometry / trigonometry. Anyone have any recommendations for resources?
 
Secret
Secret
I knew nothing about reference materials, I pretty much learnt most of it from the chat
 
Secret
Secret
$$\int^{\int^{\int}}d\tau = \int ^{e^{\int \ln \int}}d\tau = e^{\int \ln (\int \ln \int)} d\tau$$
 
Niing
Niing
@Secret: by the way, what do you meant by the Figure?:????
 
Akiva Weinberger
Akiva Weinberger
4:29 AM
Eli Maor has a book called Trigonometric Delights
I don't remember if it's any good though
@mowwwalker
 
Secret
Secret
@Niing o that. That's just some worldbuilding stuff taken from h bar
also typo:
$$\int^{\int^{\int}}d\tau = \int ^{e^{\int \ln \int}}d\tau = e^{e^{\int \ln \int} \ln \int} d\tau$$
and then we can expand $\ln (id+\int)$ with the Taylor series (too lazy to write down), integrate that, expand the result in Taylor series again, and then integrate that, and then expand the exponential twice.
The result, I believe, is an infinite series of integral operators which has the general form:
$$\sum_{k=0}^{\infty} a_k \prod_{\ell=0}^{\infty}(\int^{(-k)})^{\ell}$$
which I think convergence will be a big issue
Actually on a related note, does:
$$\lim_{n\to \infty}\int \underbrace{\cdots}_{\text{n}} \int f(x) dx^n$$ always converge for all $f$ integrable functions?
We knew it does not for e.g. $\sin x, \cos x$, and it converges for $e^x$
A more interesting question will be. Is the set of all integrable function always closed under $\int$ ?
Recall an integrable function is something where $\int f(x) dx < \infty$ for all $f$
we also have complications like $x^m$ for integers $m > 0$ because its iterated integral can grow without bound, yet each member is a polynomial and hence integrable
 
Semiclassical
Semiclassical
If I want the Gamma function $\Gamma(x)$, I do $\int e^{-t}t^{x-1}\,dt$ with an appropriate integration contour
 
0celo7
0celo7
@Semiclassical preach
 
Secret
Secret
4:45 AM
@0celo7 0celo, are the counterexamples to the converse of Schwartz theorem, i.e. a continuous function with equal mixed derivatives $f_{xy}=f_{yx}$ but $f_{xy}$, $f_{yx}$ are discontinuous?
 
Semiclassical
Semiclassical
If I sub $t=xe^z$, this becomes $x^x \int e^{-x(e^z-z)}\,dx$
With saddle points at $2\pi i \mathbb{Z}$
 
0celo7
0celo7
@Secret Presumably, yes
 
Semiclassical
Semiclassical
So if for some perverse reason I wanted to have $ze^z$ inside that exponent instead
 
Secret
Secret
Semi: You seemed to want to do some Lambert W stuff, hmm...
 
Semiclassical
Semiclassical
Yuuup
I’m trying to see the algebra on mobile
If I back-sub to get that modified integral in terms of t, that’s...
$(e^{-t})^z=(e^{z})^{-t}=(x/t)^t$ instead of $e^{-t}$
 
Secret
Secret
4:55 AM
hmm... $ze^z = e^{-t} \implies \ln z + z = -t \implies -\ln z - z = t \implies (-\frac{1}{z} - 1)dz = dt$, ok that does not work as $t^{x-1}$ becomes umanagable... hmm...
 
Semiclassical
Semiclassical
So $\int (x/t)^t t^{x-1}\,dt$ I think?
Which, oof
 
Secret
Secret
$\frac{1}{t^t}$ does not looks like a nice thing to integrate. In fact, its special function has no name as far I remember
 
Semiclassical
Semiclassical
That just looks grotesque
Yeah. Do not want
 
0celo7
0celo7
tfw you discover federer has elliptic regularity too
I think this book contains all math
 
Silent
Silent
how to write bold vector in math.se sites?
 
orbit-stabilizer
orbit-stabilizer
5:09 AM
what book
 
Secret
Secret
what happens if $e^{-t}t^{x-1} = ze^z \implies e^{-\frac{t}{x-1}}t = (ze^z)^{\frac{1}{x-1}} \implies e^{-\frac{t}{x-1}}(-\frac{t}{x-1}) = (-\frac{1}{x-1})(ze^z)^{\frac{1}{x-1}} \implies -\frac{t}{x-1} = W((-\frac{1}{x-1})(ze^z)^{\frac{1}{x-1}})$
ok that's nasty, nvm
 
Silent
Silent
Suppose $C\subset \Bbb R^k$ is closed in $\Bbb R^k$ and let $\pmb z\in \Bbb R^k$. If $F=\pmb z-C$ where $\pmb z -C$ is the set of all $\pmb{z-y}$ with $\pmb y\in C$. How to show that $F$ is closed?
 
orbit-stabilizer
orbit-stabilizer
So, basically a translation.
 
Silent
Silent
5:35 AM
Is this argument correct for my above question? Since $\pmb y\to \pmb {z-y}$ and $\pmb {z-y}\to \pmb y$ continuous, we get a homeomorphism, hence $F$ closed too. @Secret
 
Secret
Secret
I think it looks fine, though I am still very shaky on my background on continuous functions in topology
in particular, I am not sure how to show that the preimage of z-y (which I suspect to be closed in the usual topology) is also closed
 
Silent
Silent
ok
 
Zee
Zee
5:54 AM
Whose here
 
Secret
Secret
No one only none persons
 
Zee
Zee
All persons are none persons
 
 
2 hours later…
Niing
Niing
7:35 AM
Can a single point be called a region? If it is, the point is interior or boundary point?
 
Niing
Niing
8:03 AM
It's not a interior point since every $\delta-neighborhood$ will contain points outside. And it's not a boundary point, since there is no interior point so it cannot contain points inside and points outside
 
Alessandro Codenotti
Alessandro Codenotti
That's not how boundary points are defined
 
hungryWolf
hungryWolf
8:41 AM
@Secret /o
o/
sorry :p
Guys if I have a stackoverflow account wih say 5000, will I get 100 if I signup for math.stackexchange?
 
Tobias Kildetoft
Tobias Kildetoft
@hungryWolf Yes
 
hungryWolf
hungryWolf
@TobiasKildetoft thanks for the info man
 
Secret
Secret
9:31 AM
Fun fact (not): A coliation does not work because it is incomplete, not dense ans does not converge to all points of view
Consider the following:
Take a piece of Tol, wrap it into a Dyson cloth
Compute its generaloicseries which is given as follows:
$$\int^{\int^{\int}} e^{x^{\heartsuit}}gskfdx$$
where the tetration integral is defined as follows:
$\int^{\int^{\int}}=e^{e^{\int \ln \int}\ln \int}$
 
abenthy
abenthy
If $F$ is a number field and $p$ a prime in $\mathbb{Q}$, are there any embeddings of $F$ into the $p$-adic completion $\mathbb{Q}_p$ and if so, how many?
 
Secret
Secret
In particular, the chat is currently empty for the above question to receive the answer it deserves
 
TheNotMe
TheNotMe
9:55 AM
How do I show that $\epsilon_n = \frac{1}{2} - (\frac{1}{2})^n \sum_{k=0}^n \binom{n}{k}max\{\frac{k}{n},1-\frac{k}{n}\} \approx \frac{-1}{\sqrt{2\pi n}}$?
 
hungryWolf
hungryWolf
10:10 AM
my friend has created a chatroom, but it does not look like this one
it is more kind of orange
can I post the link here?
 
Shobhit
Shobhit
10:29 AM
On wikipedia, under the topic second countable space, its given "lower limit topology on the real line is not metrizable". We have not studied metrizability yet, but i found its definition. I wanted to know how can we know if a space is metrizable or not?
 
Secret
Secret
I don't think that question has a first principle answer:
Metrization theorem
In topology and related areas of mathematics, a metrizable space is a topological space that is homeomorphic to a metric space. That is, a topological space ( X , T ) {\displaystyle (X,{\mathcal {T}})} is said to be metrizable if there is a metric d : X × X → [ 0 , ∞ ) {\displaystyle d\colon X\times X\to [0,\infty )} such that the topology induced by d is ...
though it seems second countable is a requirement for metrizablility
 
Shobhit
Shobhit
this is what i got when i googled, i have been reading this. But there are many technical terms, separable, second countable,..., so i am stuck in a web. Tbh, these are very cool. @Secret
 
TheNotMe
TheNotMe
10:46 AM
asking again
50 mins ago, by TheNotMe
How do I show that $\epsilon_n = \frac{1}{2} - (\frac{1}{2})^n \sum_{k=0}^n \binom{n}{k}max\{\frac{k}{n},1-\frac{k}{n}\} \approx \frac{-1}{\sqrt{2\pi n}}$?
 
Secret
Secret
@Shobhit yeah, you just have to go slow, topology has a lot of terms wrapped in them. But if you are clear what limit points and open sets are, then you are halfway there
for me, I like to find the bases directly, which is one reason why I spend a lot of time in uncountable stuff
 
Shobhit
Shobhit
how do you find the basis directly? @Secret
 
Secret
Secret
Well, Leaky trained myself somewhat. Basically say for the lower limit topology, you knew what the open sets are, you then try to find a family of these open sets such that they generate the topology For example $\{[x,\infty)\}$ will give open sets that are half open intervals of the form $[a,\infty)$ or a disjoint union of them
Feb 26 '17 at 8:03, by Secret
@DHMO $(-\infty,a)\cup [a,\infty) =\Bbb{R}$ is clopen. $(a,b)\cup [b,-\infty)=(a,\infty)$ is open. $[a,b] \cup (b,\infty)=[a,\infty)$ is closed.
 
Shobhit
Shobhit
i thought you had a general method :P @Secret
 
Secret
Secret
I just try to identify the different kinds of open sets in the topology ,and then attack from there
 
Shobhit
Shobhit
10:57 AM
yes, thats basically it
 
Secret
Secret
For example $[a,\infty)$ and $[b,\infty)$ are of the same kind
maybe I am thinking in terms of equivalence classes of open sets in a topology?
Btw, my PhD started just when I reached the chapter on function continuity in munkres, thus I won't say I am very confident on function continuity proofs
 
Shobhit
Shobhit
you were helpful, ty
 
Silent
Silent
Suppose $0<x<1$. Let $k$ is any integer, and $n$ any positive integer. Fix $k,n$. If $q$ is such integer that $qn\le k<(q+1)n$, then how to show that there exists positive integer $p$ such that $k<px+qn<k+1$?
 
Silent
Silent
11:32 AM
@AkivaWeinberger, will you please look at this (above post)?
 
Shobhit
Shobhit
11:43 AM
@Silent first see that $k$ and $q$ must have same signs ($q \ne 1$), then from the second inequality $k = qn + \alpha$ where $\alpha$ is $0,1,..,n-1$ (for the case $k>0$), substitute this in the inequality which you want to prove, you will end up with $\alpha < px < \alpha +1$ so now you just need to prove that there exists $p$ for all these $\alpha$ 's such that this inequality is true. I leave that to you.
This is just over my head, i have exam tommorow, you can try if this works, until someone smarter comes along.
 
Dattier
Dattier
12:05 PM
@Lozansky : the path you choose is very difficult for me so here with an another path is more easy : fr.wikipedia.org/wiki/…
but there exists an intersting hint, maybe that can give the solution
with Rodrigues formula
$R(x)$ polynomial then $D^n[x\times R(x)]=xD^n[R(x)]+nD^{n-1}[R(x)]$
@Lozansky : ok ?
A topology enigma :
$\mathbb T$ all the connex by arc of $\mathbb R^2$ and, with $\forall A\in\mathbb T,\forall a\in A,A-\{a\}$ isn't convex by arc.
Is it true that $card(\mathbb E)=card(\mathbb R^{\mathbb R})$ ?
 
Silent
Silent
12:21 PM
@Shobhit Thanks for help. Best wishes for exam.
 
Shobhit
Shobhit
@Silent ty :)
 
Secret
Secret
what is "convex by arc", convex curve?
 
Dattier
Dattier
sorry it's a mistake, "connex by arc"
 
Silent
Silent
@Shobhit I am sorry, but can't show that there is $p$ such that $\alpha < px < \alpha +1$.
 
Shobhit
Shobhit
12:43 PM
@Silent this inequality is equivalent to $p \in (\frac{\alpha}{x}, \frac{\alpha +1}{x})$, so all you need to show is that there is always an integer in this interval. Can you do now?
 
Silent
Silent
@Shobhit No! suppose $(1/5,2/5)$ does not have an integer in between them, can't do!
 
Shobhit
Shobhit
you are forgetting the range of x @Silent
 
Silent
Silent
@Shobhit Oh! $p$ exists because $\frac{\alpha} x<\frac{\alpha} x+1<\frac{\alpha} x+\frac1x$, because $0<x<1$, right?
Thank yo.
 
Shobhit
Shobhit
@Silent perfect. Np :)
@Silent I think you should argue that since the length of the interval in which $p$ belongs is $\frac{1}{x} >1$ so it contains an integer.
 
Silent
Silent
Ok!, i will do that.
@Shobhit
 
Lozansky
Lozansky
12:58 PM
@Dattier Yes, that's what I got so far, namely $(n+1)P_{n+1} = \dfrac{n+1}{2^n n!}[xD^n((x^2-1)^n)+nD^{n-1}((x^2-1)^n)]$. So you see it's that last term that needs fixing
 
Dattier
Dattier
@Lozansky : the path you choose is difficult, do without the Rodrigues fromula
 
Lozansky
Lozansky
Via fonction génératrice?
 
Dattier
Dattier
yes
in this link the solution is give
do you speak french ?
@Lozansky
 
Akiva Weinberger
Akiva Weinberger
@Vrouvrou speaks French (@Dattier)
 
Secret
Secret
55 mins ago, by Dattier
$\mathbb T$ all the connex by arc of $\mathbb R^2$ and, with $\forall A\in\mathbb T,\forall a\in A,A-\{a\}$ isn't convex by arc.
 
Lozansky
Lozansky
1:10 PM
@Dattier Très peu
 
Secret
Secret
$\Bbb{T}$ is the topology where all open sets are connected curves in $\Bbb{R}^2$ except for a point?
 
Dattier
Dattier
@Lozansky : it does not matter, did you understand the proposed solution?
@Secret : no, the topology is fixed it's the canonical one of R^2
 
Secret
Secret
ok, usual topology of \Bbb{R}^2$ and you are considering all the connected components on it except for a point?
 
Lozansky
Lozansky
@Dattier Yes, I got it. But without a GF, it would be difficult no?
 
Dattier
Dattier
@Secret : no, T is a set of part A connex by arc of R^2, and A-{a} is not connex by arc (forall a in A)
@Lozansky : GF ?
 
Lozansky
Lozansky
1:19 PM
@Dattier Generating function
 
Dattier
Dattier
ok
Maybe you can do that with the definition of Legendre Polynomial
@Lozansky
Show by induction : $\int_{-1}^1 P_m(x)P_n(x)dx=\frac{2}{2n+1}\delta_{m,n}$
 
Secret
Secret
this set is so weird
o wait what I am doing, a curve in $\Bbb{R}^2$ already fits the criteria of $A$
 
Dattier
Dattier
@Secret : this set contains the open segment so card(T)\geq card(R)
@Secret an open curve. (for exemple [0,1]\times {0} is not in T)
but ]0,1[\times {0} is in T
 
Secret
Secret
Yeah, since $A$ is connected, but $A-\{a\}$ is not for all $a \in A$, it follows $A$ has to be an open curve in $\Bbb{R}^2$
So I think $T$ is like this:
So that means, we are counting the number of open curves in $\Bbb{R}^2$
(computing...)
 
Dattier
Dattier
I think he misses you a lot
what about the open cross ?
@Secret
 
Secret
Secret
1:34 PM
ah right, cause A-{a} only need to be disconnected, not disconnected with 2 components
So T has to be all star shaped sets in $\Bbb{R}^2$ I think...?
(no, wrong terminology, let me think again)
 
Air Conditioner
Air Conditioner
Lately, I've seen some posts that aren't in English, and they almost immediately get comments saying something along the lines of "We use English only here". The top answers to this meta post (from 2011) seem to mostly agree that posting in a different language is okay. Has something changed in the attitude toward these questions in recent years?
 
ÍgjøgnumMeg
ÍgjøgnumMeg
@AirConditioner I don't think so, there are just a lot of comments from people who never look at meta
 
Dattier
Dattier
It's a good news (for me, my english is very bad)
 
Silent
Silent
Let $A\subset B\subset G$, where $G$ is group and $A$ nonempty, then centralizers satisfy this relation: $C_G(B)\subset C_G(A)$. Can something be said about relation between normalizers $N_G(A)$ and $N_G(B)?$
 
Akiva Weinberger
Akiva Weinberger
*It's good news
 
Dattier
Dattier
1:42 PM
that's the proof of "my english is very bad" lol
 
Akiva Weinberger
Akiva Weinberger
@Silent Remind me what a normalizer is?
 
Silent
Silent
ok
 
Secret
Secret
T is the set of all star domains and open curves and their union in $\Bbb{R}^2$
 
Akiva Weinberger
Akiva Weinberger
Oh, it's the set of things where $xA=Ax$?
 
Silent
Silent
@AkivaWeinberger $N_G(A)=\{g\in G: gAg^{-1}=A\}$
 
Dattier
Dattier
1:45 PM
@Secret : what about the open disc is a star domains, no ?
 
Secret
Secret
but open disk won't become disconnected if you remove any point, is it?
 
Silent
Silent
@AkivaWeinberger yes!
 
Dattier
Dattier
@Secret : so T is not the set of all star domains and open curves and their union in $\Bbb{R}^2$
 
user193319
user193319
Dumb question: if $\varphi : E \to \Bbb{R}$ is a simple function that is equal to $0$ almost everywhere on $E$, then isn't $\varphi = 0$ everywhere on $E$?
 
Akiva Weinberger
Akiva Weinberger
In $S_3$, is $(23)$ in $N_{S_3}\{(12),(13)\}$?
 
Secret
Secret
1:47 PM
ugh, how do i describe this shape?
 
Dattier
Dattier
@Secret The question is about cardinality, just have enough to answer the question
it is not necessary to describe everything
 
Akiva Weinberger
Akiva Weinberger
I think $N_{S_3}\{(12)\}=\{e,(12)\}$ and $N_{S_3}\{(12),(13)\}=\{e,(23)\}$ @Silent
 
Secret
Secret
Well, T is gonna to have something to do with some union of $\gamma : [0,1] \to \Bbb{R}^2$. That means we need to compute $\text{card}((\Bbb{R}^2)^{[0,1]})$. Let's see...
 
Akiva Weinberger
Akiva Weinberger
$\{(12)\}\subset\{(12),(13)\}$ but $N_G\{(12)\}\not\subseteq,\not\supseteq N_G\{(12),(13)\}$
(No idea what's the best way to notate that)
Neither is a subset of the other
 
Silent
Silent
I think $(23)$ is in $N_{S_3}\{(12),(13)\}$. @AkivaWeinberger
 
Dattier
Dattier
1:52 PM
@Secret : I think your idea of ​​working with starred ensembles is good, but you have to restrict yourself or only good together
 
user193319
user193319
Any care to answer the dumb question I posed above? I'm pretty certain it's true.
 
Silent
Silent
Oh, you already gave more info@AkivaWeinberger
Thank u!!!
 
Akiva Weinberger
Akiva Weinberger
@user193319 What about $1_{\Bbb Q}$?
 
Secret
Secret
$\Bbb{R}^2 = \mathfrak{c}$, $[0,1]=\mathfrak{c}$ thus we have $\text{card}(T) = \mathfrak{c}^{\mathfrak{c}} = 2^{\mathfrak{c}} = \text{card}(\Bbb{R}^{\Bbb{R}})$
 
Akiva Weinberger
Akiva Weinberger
I don't know what notation you use; I mean the function that is $1$ on the rationals and $0$ everywhere else @user193319
 
user193319
user193319
1:57 PM
@AkivaWeinberger Hmm...That seems to be a counter-example...I am trying to show that if $\varphi$ is a simple function that is $0$ almost everywhere on $E$, then $\int_E \varphi = 0$.
 
Dattier
Dattier
@Secret : why card(T)=c^c ?
 
Akiva Weinberger
Akiva Weinberger
@user193319 Well, $\int_E(1_{\Bbb Q})$ does equal $0$
It equals the Lebeg measure of $\Bbb Q$ which is 0
 
user193319
user193319
@AkivaWeinberger Wait! $1_{\Bbb{Q}}$ isn't $0$ almost everywhere, is it?
It's equal to $1$ almost everywhere, right?
 
Akiva Weinberger
Akiva Weinberger
It's 0 on all the irrationals, which is almost everywhere
It's only 1 on the rationals, which is almost nowhere
Very few numbers are rational
 
user193319
user193319
Whoops...I was thinking things backwards.
I still don't see how to prove that $\int_E \varphi = 0$ in general.
 
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