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00:00
As I said, tell the prof what you know, and let the prof give advice.
I know very little harmonic analysis, but Pontryagin duality is sick
Actually, I have talked with her, and she asked me to try this and see if I like it. Interestingly, this course is open for registration both as a undegrad and grad course.
However, if it is taught in the version that you described to me then I really have to think about it. I knew nothing about Lie and let alone representation or so on.
though I admittedly have mostly used only the special case of Pontryagin duality between profinite and torsion groups to apply it in number theory
00:50
A pretty cool version of mean value theorem, the proof reminds me of gauge integration
I found a difficult problem
Does there exist a system of PDE's in $(n+m)$ dimensions that admits a unique solution(s) which remains invariant w.r.t. the system under any choice of $n$ and $m$ for $n,m>0$?
or something like that
 
2 hours later…
03:08
Can anyone give me an example of a non-commutative ring of order 4?
I found an example as follows: Let $R=\{0,a,b,c\}$ with a binary operation + defined on it.
The multiplication table is given by:
Does this suffice as an example? Or do we need to define what the operation + looks like explicitly.
@ThomasFinley There are none
@SoumikMukherjee why?
Least possible order of a non commutative ring is 8
@SoumikMukherjee Then is this page fake?!
Now, this seems to be a big confusion.
03:27
@ThomasFinley No. I was talking about non commutative ring with unity.
that article talks about not necessarily unital rings (heresy)
@SoumikMukherjee oh, so is my example valid then?
@Thorgott I am talking about the same thing as well
you do indeed have an example of a non-commutative, non-unital ring of order 4
@Thorgott do we need to define the addition as well in the example above?
03:51
of course
@NordineLotfi It's pretty unusual for an amateur to make a significant mathematical discovery, but it does happen occasionally, eg en.wikipedia.org/wiki/Marjorie_Rice
Australian scifi author Greg Egan occasionally publishes mathematical stuff, but he's not exactly an amateur. He has a Bachelor of Science degree in mathematics. (Incidentally, before he became a famous author he worked as a software engineer in biomedical science).
He has done interesting things in combinatorics en.wikipedia.org/wiki/Superpermutation geometry en.wikipedia.org/wiki/Egan_conjecture also arxiv.org/abs/gr-qc/0208010 and celestial mechanics arxiv.org/abs/1510.05345
Greg doesn't have a blog, but he has a lot of science and maths stuff on his website, and often posts on mathematical topics on Mastodon mathstodon.xyz/@gregeganSF
 
1 hour later…
05:29
$\int_{}^{}(\frac{3}{2}-(\frac{3}{2}-x^5)^5)^\frac{1}{5}dx$
I am not sure how to go further from here.
So, in this short paper I'm looking at, I find a mention that, for a prime ideal $P\subseteq R$ and an element $x\in R\setminus P$, then $(P+(x))^3=(P+(x))(P+(x^2))$.
This statement is just taken at face value at the start of a proof, as if it were obvious. However, it doesn't strike me as such, and I'm not sure how to try and verify it
Like, I could just expand both sides, but that seems to support them not being equal because one side has a $P^3$ term and the other has no such term (but, writing this out, I realize that $P^3\subseteq P^2$, so that means that $P^3+P^2=P^2$ . . . but there's no naked $P^2$ term on the LHS only a $3P^2(x)$)
05:52
I can't figure out the statement at the moment, so I'm gonna continue reading. If anyone has any insight into it, please ping me.
06:07
Is $R/P$ the quotient ring?
Or is it a set minus
Setminus
(Also, I think there's a typo in the paper. I kept reading and it started talking about $P^2+(x^2))$)
(Which I think resolves the issue. You get the same terms, but they're multiplied by constants)
I guess so I tried to expand the RHS and I saw no $P^3$
06:23
so this is sorta funny. i went trawling around looking for a certain type of question and found one with only one deleted answer
start looking through with interest, wonder why the person deleted it...and then notice the poster/date
... was it one of tomorrow's questions, asked and answered too early?
nope, wrong image
that bridge is gonna fall down.
"thou art the man!"
is the date of significance other than just being long enough ago that we shouldn't worry about your memory too much
06:27
other than it's a nine-year old deleted answer that i have absolutely no memory of ever making, no
nine years is long enough :)
it is
and also raises the question of why i deleted it
i would guess there's something wrong with it
the world wasn't ready for it
can you get some kind of badge for undeleting your own post more than 9 years later
i can think of several possible names, all of them potentially sacrilegious
06:29
necromancer, maybe?
necromancer is i think for reviving an old post with a new answer
true
mostly it's just bizarre to come across what feels like a previous life of mine
when my daughter was born i went through a period where i had about 6 hours of memory tape that i continuously cycled through, which made for some interesting times at work
"do you know if there's any source saying X?" "no, but someone wrote a memo about an issue like that, i think it's on the network somewhere" "uh, yeah, you wrote a memo about this about a month ago, which is why i'm asking you and not someone else"
@leslietownes "that was month-ago me, he still has my brain"
 
1 hour later…
07:43
Start with the bounds [1/4,7/2] and let e be the number we are interested in estimating. Now fit some family of polynomials P_n to the two numbers in the bound. Then we get the liouville criterion 1/q^2. Now use a program to half the interval each time towards e, then we can make this liouville crtierion arbitrarily small from 1/q^4, 1/q^6 , .... Thus we have a polynomial procedure to approach e at arbitrary accuracy, therefore the lower bound of transcendence degree of e is 1.
What does that say about NP and P I wonder?
Question for anyone who knows: If $R$ is a Prüfer domain that isn't almost Dedekind, do that require that $R$ have at least one idempotent maximal ideal?
(Bah, typos: "does that require that $R$ has")
08:08
Just a thought: should we have a poll sometime on MSE for the most beautiful equation? I think that Euler’s identity would still win, but we can get some new perspectives…
08:39
@SohamSaha No. "This question is likely to be answered with opinions rather than facts and citations. It should be updated so it will lead to fact-based answers."
@Secret Bisection gets you 1 extra bit of precision per loop. A linear increase in precision isn't very impressive. But if you can get quadratic convergence (i.e., double the precision on each loop), that's impressive. So you want 1/q^2, 1/q^4, 1/q^8, 1/q^16, ... .
There are algorithms that converge quadratically on pi. They are based on the arithmetic-geometric mean (AGM). Eg, the Gauss / Salamin / Brent algorithm, and a couple of similar algorithms devised by the Borwein brothers. I don't know of a direct quadratic algo for e, but there's an AGM algo for logarithm, and you can use that in Newton's method for finding e & its powers.
Those algorithms have some overhead, and aren't the fastest if you only need a small number of digits. But they're great if you want large numbers of digits (>100,000 or so). For a few thousand digits or fewer, the simplest algo is to use the factoradic representation of e, eg math.stackexchange.com/a/1295561/207316 which essentially converges linearly.
09:03
@PM2Ring I mean I agree it's not often this happens but it happened way more than just a few times. By the way I don't think "amateur" == "hobbyist" but that's maybe just semantics on my part
09:27
0
Q: Let $R$ be a ring with identity such that it has no zero divisors. If every subring of $R$ is an ideal of $R,$ then prove that $R$ is commutative.

Thomas FinleyLet $R$ be a ring with identity such that it has no zero divisors. If every subring of $R$ is an ideal of $R,$ then prove that $R$ is commutative. This means, that $R$ is an integral domain. By ideal, I think they mean a two-sided ideal. Now, a subring of a ring is a non-empty subset of a ring th...

Can someone please help me with this?
 
2 hours later…
11:07
Hi, I have a couple of questions on an exercise on Hilbert spaces. $H=L^2[-1,1]$ with the usual inner product, and consider the operator $Tf(x)=\int_{-1}^1(1+xy)f(y)dy$. Prove that $T(H)$ is finite dimentional and find an orthonormal basis
12:02
7 hours ago, by Kripke Platek
$\int_{}^{}(\frac{3}{2}-(\frac{3}{2}-x^5)^5)^\frac{1}{5}dx$
7 hours ago, by Kripke Platek
I am not sure how to go further from here.
@SineoftheTime we call such operators of finite rank
Closely related concept is compact operators
Note that T(f) is a linear function (as in $ax+b$)
I've never heard that term, what is the standard way to prove it?
12:19
Idk
Anyway
Here $Tf(x) = ax+b$ for some $a, b$ so you see its finite-dimensional
The coefficients depend on $f$
$Tf(x)=\int_{-1}^1 f(y)dy+x\int_{-1}^1 yf(y)dy$
It's like $\text{span}\{1,x\}$ (?)
Image is certainly contained in it
But it may be not equal to it
I'm trying to prove it's equal
Alright. Sounds good
At a first glance, try even and odd function $f$
$f(x) = 1, f(x) = x$
$T1=2$ and $Tx=\frac23x$
12:32
I wouldn't even calculate those
Coefficients. Just claim they're non-zero
I'm not understanding
wait a minute
Given $a+bx\in \text{span} \{1,x\}$, this element is the image of $f(x)=\frac a2+\frac32bx$
 
1 hour later…
13:52
I think that if I were to develop gauge integrals for functions $f:I\to F$ then this lemma would follow from $$\|f(b)-f(a)\| = \|\int f' \| \leq \int \|f'\| \leq \int \varphi' = \varphi(b)-\varphi(a)$$
Maybe thats why it feels very familiar to arguments from gauge integration
surely you can strengthen denumerable to measure $0$?
No
But if its like gauge integration then you should be able to assume that its enough that $f$ is differentiable outside of set $Z$ of measure $0$, on which it has negligible variation.
@Thorgott Counter-example is $f = $Cantor function, $\varphi = 0$
I'll try writing up a more general version of this in some pdf
By the way I really like the section about continuous differentiability in Dieudonne, the fact that its an iff statement, and that its happening in a more general way, I think those two things make it way more intuitive
It speaks more to me than "$\partial f/\partial x_i$ are continuous then $f$ is differentiable"
14:17
@Jakobian fair enough
14:43
So far, it seems tricky to prove that an indefinite integral of a gauge integrable function $f:I\to F$ would be differentiable a.e. with derivative $f$. The proof requires a corollary of Saks-Henstock lemma which doesn't seem to hold in general
But maybe there is some way to go around this
proof of continuity is fine though
 
1 hour later…
15:50
I have another questio. Is there a general method to find an explicit formula for the projection on a subspace? I'm in trouble with the following ex: consider $L^2[-1,1]$ with the usual inner product and the operator $Tf=\int_{-1}^1 f(x)e^x dx$, I have to find explicit formulas for the projection on the kernel of $T$ and on its orthogonal complement
I tried with the decomposition of $f$ in the sum of an even and odd function and I tried to multiply to $e^x$ and $e^{-x}$ to get the integral of an odd function
for example $f=f_1 +f_2$ with $f_1 \in \ker T$ and $f_2 \in (\ker T)^{\perp}$
16:20
@Jakobian I'm able to justify this using gauge integration.
and I think I even have an idea how to handle the general case
The argument is like this. If $f', \varphi'$ exist except for a countable set, then fundamental theorem of calculus holds for them, moreover $\|f'\|$ is measurable, which implies its integrable
Then the inequality like above
Proof of first sentence is like the standard proof, nothing wrong there, for the second one I've used linear functionals to justify that one
the inequalities also require justification, that I did
I can prove that if $f$ is gauge integrable and $F$ is its indefinite integral, then derivative of $h\circ F$ exists a.e. for linear functional $h$, and equals $h\circ f$
16:35
Hey there! We know per definition: M is globally hyperbolic if and only if for every $p,q\in M$ the 'causal diamonds' $J^+(p)\cap J^-(q)$ are compact.

Are the causal diamonds nonempty? Or what condition/constraint guarantees that the causal diamonds are nonempty?
Of course if $F'$ actually exists a.e. and equals $f$, then $h\circ F$ exists a.e. for all linear functionals $h$ and equals $h\circ f$. But here I've proven something weaker
16:47
Yes okay, I have it. If $f, \varphi$ are differentiable a.e. outside of a null set on which they are of negligible variation, and $\|f'\|\leq \varphi'$ a.e., then $\|f(\beta)-f(\alpha)\|\leq \varphi(\beta)-\varphi(\alpha).$
while I'm not sure if this implies that $f$ is gauge integrable, it implies the statement holds
@eigenvalue you'll probably have better luck asking that in the physics chat (the h-bar)
personally if I'm dealing with time it's always only mathematical time
Mad
Mad
how Can we expand this statement https://math.stackexchange.com/questions/1551065/t-compact-if-and-only-if-tt-is-compact
to there exists a natural number n such that (T*T)^n is compact
17:09
@Thorgott I've proven a more general version of that theorem if you're interested
17:26
This is strange. I'm working on finding the eigenvectors and generalized eigenvectors of $$A=\begin{pmatrix} 2&0&-2\\ 1&1&-1\\ 1&-1&-1\end{pmatrix}.$$ I have found the eigenvalues $0$ and $2$ with multiplicity $2$ and $1$ respectively. For the eigenvalue $0$, I have found the eigenvector $u=(1,0,1)$ and so to find the generalized eigenvector $v$, I put $$(A-0I)v=u.$$
Now, when I solve this for $v$, I get that it can be written as a linear combination of two vectors, $u$ and some other vector. This confuses me, since I expected to simply get $v$ in terms of a single vector. I am not sure if $v$ is even a generalized eigenvector.
18:20
What are you talking about? You solve and get lots of $v$'s (why?). Any one will work.
Hi! can we call the real numbers as "system of real numbers"? I thought we can only call it as set of real numbers and when we say "systems of numbers" it means decimal number system, binary number system, octal number system etc.?
I would never use that terminology, although I suppose for binary and octal it makes sense. But there is no such thing as the decimal number system.
You are representing real numbers by decimals, as opposed to some other way. But that doesn't make the representation a "number system."
Hi @TedShifrin Hmm okay. What exactly is a "number system"?
Good question. I don't use the term. What is your definition?
So it's wrong to say "system" to real numbers too right?
@TedShifrin I don't have a definition...I just came across these words that's it..
18:24
Now that I think of it, by binary you mean the binary representation of the integers. The integers is the set of numbers. You are just using a different representation. By decimal you meant base 10, still the integers? I thought you were referring to decimal representation of real numbers.
But when I was little, I've been taught that a system consisting of 0 to 9 is decimal ssytem, numbers made out of 0's and 1's are called binary system, etc. But in those days they never taught a definition. Only this categorization. But I came across that now again.
@TedShifrin Yes, I guess they mean base 10..
@TedShifrin hmm ok, so maybe I wasn't that wrong. But please, I'd be interested in hearing why we get lots of $v$'s? Because we have a lot of $u$'s right?
I would never call these a "number system," but I guess someone could. The underlying ring is the integers. I would talk about base 10, base 2 representation of the integers.
@TedShifrin Okay..Hmm... what is the connection with the system then? Why do they call as a system?
No, @psie. You have one $u$ up to scalar multiples. In general, what does the solution set of $Ax=b$ look like (if the system is consistent)?
18:29
@TedShifrin But then we will have to call it as a system of integers OR system of decimals,system of binary etc.?
@Buddhini People can use words any way they want, I guess. Googling just says a number system is a "way" to write numbers. Vague.
Not together with system of real numbers, system of complex numbers?
I don't say a system of anything. The underlying set is the integers, which I can represent in different bases.
I never say systems of real or complex numbers.
I don't know any mathematician who does.
We talk about them as fields.
Then, maybe I'd better say since it's not defined properly in that way, learn to say as sets of numbers right?
But there's nothing to stop people from using this language. As long as it's clear that you're talking about different representations of the same set (integers).
18:31
But then they start saying system of real numbers etc. I feel confused then.
Who are these "they"?
Calling system for system of decimal, system of binary system of octal etc. I can understand, as it can be said that we call "system" meaning that those are different ways to represent
@TedShifrin My colleagues
Are these students or professors?
Anyhow, I guess people can use words any way they want. One can guess from context what they mean. But you should ask them to define their terms.
We are writing a plan for some topics. In the introduction part they said write as system of numbers, but the sub topic they included was real numbers, integers, rationals etc. So actually so far I wrote it as "Introduction to sets of numbers" rather than saying systems
They are not professors
But graduate lecturers
There are such lot of variations in some other things they say too
I don't know what to do about it
So you all are math educators?
18:35
I'm junior to them new
Yes
In chapter 1 first page...
I suspect that the term is popular among math education folks. I could see making a list of different "number systems" — integers, real numbers, complex numbers. I still don't think of base $k$ representations as different number systems, but obviously some people do.
Rational numbers should definitely be on the list, too.
Hmm, I still don't get why we call it "system"
Joe
Joe
I think the main point is that "number" doesn't have a precise meaning in modern mathematics. Are ordinals "numbers"? What about the quaternions? But there is nothing wrong with talking about "numbers" or "systems of numbers" in informal, everyday speech.
@TedShifrin Oh, here you mean in the sense of doing "Math education".. No, we are just Mathematics graduate lecturers
Haven't done education type degree
@Buddhini So what is the audience of these lectures or this text you're writing?
@Joe I actually think we do often say ordinal numbers and cardinal numbers.
18:40
Software engineering undergraduates
Hi @Joe Thanks for the comment
Joe
Joe
@TedShifrin: Yes, that was my point – it seems like whether they should be considered "numbers" is a borderline case.
@TedShifrin Hmm yeah I've heard like this too
Interesting. So that's why you mentioned binary and octal. I think they should understand what I'm saying. We have different ways of representing/thinking about the integers.
I guess you can think of a number system as an underlying set (ring?) where typically we can do arithmetic together with a symbolic representation of them.
But when we move to sets of real, complex numbers etc. we are not talking only about integers?
As I say, I don't think I've ever used the term except perhaps in teaching future elementary school teachers, and I'm not sure I did there.
18:42
@TedShifrin Hmm okay @TedShifrin
Then, can I use this explanation if necessary or is it only if it's defined like that somewhere we can use it like that?
The underlying set does not need to be the integers, of course. But if you're talking about base 2,8,10 representations of integers as part of the number system structure, how are you representing real numbers? complex numbers?
I think you need to do your own research on this question. This is not something that mathematicians ordinarily fuss over.
Hmm I think we can't represent all the numbers there then?
@BuddhiniAngelika Even defining what the real numbers are is opening Pandora's box, and is a matter of frequent debate in here. But your students don't care what they are. They're used to drawing the number line and saying "here."
@TedShifrin Okay. My idea was that there's something wrong in saying them as sytem, as I referred...
I still say you should ask them to define the term.
18:48
Okay. I will try that. But I think sometimes, in the work place as some people have ego, they might say things to cover upeven if they make a mistake.. Some people are not sure whether all elements in the domain of an injective function is mapped to some value or not either...They said different things to the students
This is why most people choose to follow well-written textbooks instead of trying to create a text from the ground up.
There are, unfortunately, plenty of textbooks that are not well-written and that have flaws.
I'm still getting used to deal with people like this suddenly making absurd decisions, based on the thoughts they get, without checking whether it's really useful/applicable/etc. Just to show off like. So, I'm struggling to learn how to communicate my thoughts and usefulness of some ideas to juniors/seniors and peers despite the fact that they argue based on sudden instincts like this. It's more difficult as they are not aware about some facts too.
Are there books to learn these things too? How do you'll deal with such situations in academia?
@TedShifrin Yeah maybe the one I shared earlier could be having a fault too
No, probably need some industrial psychology course to learn things like this.
Hmm okay. :)
Thanks a million for all the advises @TedShifrin
Have a nice day!!!
You too. Sorry I couldn't be more definitive for you.
18:55
Okay 👍🏻
No, the question was about dealing with colleagues in the workplace.
@TedShifrin That's okay. It was very useful as I got a change to argue and see :)
@user85795 :) Thanks a lot!!
np
You could mix in social psychology
Personalities do clash.
Hmm okay
I didn't know about these aspects
Yeah I see clashes in practical scenario...
Try wikipedia for an overview
18:59
Okay, will do. Thanks a lot again @user85795 :)

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