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00:17
Thank you for the advice :)
00:27
So the closed unit ball in $\ell^2$ isn't precompact, but it has the weaker property that every uniformly continuous function into $\mathbb{R}$ is bounded
and I am wondering, how is this property called, if anything. A name I could make is weakly precompact
maybe weakly precompact is a bad nomenclature considering how precompact is used in the context of vector spaces
apparently people did study this for metric spaces
apparently this is equivalent to, for each $\varepsilon > 0$, we can decompose $X$ into finite amount of sets $X_1, ..., X_n$ such that for every $x, y\in X_i$ there exists $x_1 = x, x_2, ..., x_n = y$ with $d(x_i, x_{i+1})\leq \varepsilon$ and we can do so while keeping $n$ bounded
sorry, replace $x_n$ by $x_m$ and last $n$ by $m$
so it is pretty much like totally bounded, but that one has $m = 1$
whereas here we allow longer steps
and by bounded $m$ I just mean, lets say, $m = m(\varepsilon)$ is a fixed natural depending on $\varepsilon$
@AlessandroCodenotti @Thorgott look at this pretty cool weakening of totally bounded metric spaces I've found
might generalize this result to uniform spaces
01:31
The new Disney movie, Wish, apparently was a big flop
plot, animation style, moreover it was 100th movie anniversary
apparently everything about the movie had a drop in quality in comparison to the other films
02:29
did any folks here ever read feller (An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition) for probability?
(Also if anyone gets the chance to look at my question above i would be very grateful :) )
ee18 feller is a classic (maybe not the best if you also want to learn the most modern mathematical frameworks as quickly as possible)
what do you mean by mathematical frameworks here Leslie?
as in measure-theoretic probability?
i assume no seeing as that's not modern :)
well actually i did mean anything to do with abstract measure theory, which as i recall feller either does not do or doesn't do until late
If you do have any other probability suggestions for someone after a first course in analysis (still months to even a year away to be fair) i'd also be all ears
or anything else that you'd need to do to rigorously develop basic theory of stochastic processes in continuous time
02:33
ah that is more problematic
i don't have other suggestions, i'm just saying, if you pick up feller i hope you actually want to learn probability theory and not measure theory
oh for sure, i definitely do
if you want to get to like black-scholes and stochastic DE as quickly as possible, there was some text that the financial math people i knew 15 years ago really liked for that that wasn't feller, but i don't remember what it was
Rosenthal?
maybe not for math math people actually
i might be thinking of evans
02:38
Probability and Statistics: The Science of Uncertainty?
evans 'introduction to stochastic differential equations.' it was a circulated set of notes before it was a book
it's kind of the opposite though, it assumed you know at least a little measure theory
ah ok, so it might be a feller followed by evans sort of thing?
you're saying evans might pick up what modern stuff feller misses?
if you want to do stochastic differential equations, maybe start with some book that has that more in mind than feller
feller's first few chapters are balls and urns and shit
i've heard it reads like a good story though ;)
i mean it all comes down to what do you want from a book
haha
02:44
fair play to that
hopefully i'll have a better answer to that once i have to decide on a book (months away)
the probability book i used five years ago was an engineering-ish one by papoulis, didn't love
i really like feller i just don't know that it's a great guide to what is going on right now in whatever is hot in mathematical applications of probability theory
i feel like a more modern book would get to abstract measure spaces sooner, even if they weren't emphasizing them
depends on the applications
I doubt there are people who do probability theory applications without any measure theory nowadays
physics? so i guess SDE probably matter there too?
well, SDE's are of interest in some physics, yes
as much as you can pick a field of math and say its of interest in some physics, I suppose
its probably almost all of them
better way would be to study math when you actually need it
if you want to absorb enough probability theory to address all potential applications, you might be in the market for more than one book
02:59
@EE18 whats your goal
is it just to learn some math
I assume not
This is all just for fun for me on the sides. To know more math and physics, i guess
I got fairly far with physics self-study over the course of 2-3 years but found myself not quite well-equipped enough with math from my engineering undergrad. ergo math time...and it turns out that's a ton of fun too. hence "wasting" time with logic and set theory ;)
this goes back to, you can spend ages learning just one field
for sure, and that's ok because in the end im not doing any of this professionally :)
learning logic and set theory is worthless in the first place for non-math applications
or at least for physics applications
unless you want to be professional goofball
03:03
LOL
definitely worthless for those applications, but a fun stretch of the mind. and maybe ill understand some of GEB next time i read it...
you want to start learning basic probability before even wondering about learning things like SDE
I've learned (and forgotten) basic probability theory to be fair
what that means to you and to me is probably different
I mean probability with measure theory background
they re-opened the pool near our house after several months, and my daughter and i went to swim in it, and this duck spent like 20 minutes next to the pool, just staring at us, clearly wondering when we were going to leave
ah that's fair jakobian
The duck was unimpressed with the company it seems
03:06
we used to swim with the ducks, i have a photo of my daughter with three ducks in the pool
this duck was clearly a newcomer who thought we were newcomers
I don't mean classical probability theory you learn in high school, like $P(A) = |A|/|\Omega|$. That's way more simpler
"you learn in high school" muffled laughter
they don't learn that in American high schools?
@leslietownes i've never had this said to me in lecture but i can only imagine the stereotype of the epic soviet math prof who...actually probably learned some of this stuff in high school
not in mine :)
jakobian, america does not have anything resembling a national educational system, high schools are controlled locally (~ 15000 districts) subject to state standards (which are variable among the 50 states)
there wasn't a whiff of probability at my high school but one county over in the rich area they may well have done all of it
it wouldn't surprise me if the rudiments of probability, as expressed in your example formula, are in at least some state standards
which isn't to say that local districts have ever been equipped to meet them
03:11
hmm. Weird. This is the simplest probability there is, and in Poland its usually what students score points on on their finals
i think my favorite activity is searching for a good book in field X and seeing the hilarious suggestions (made more hilarious by the fact they're not satire) for questions about "best intro to probability?"
Someone answers "Probability and measure, Billingsley". I've never read it but i know enough to know that is absurd
"and measure" tells you all you need to know
"billingsley" also sounds like the surname of someone one social class better than the reader
@EE18 well yeah thats a bit of a full introduction
@leslietownes LOL
AFAIK it's quite a famous book in probability circles?
at least, insofar as i had heard of it...
but if you're willing, you will learn a lot from it
if you're just trying to learn math, then why not?
03:16
i have about 6-8 months to go with my analysis, LA, set theory trifecta before getting to choose what's next but who knows, maybe billingsley will be it
okay, your majesty
would have made a better king than charles no doubt
well, which charles
the current monarch
@EE18 I suspect this won't go far. Learning three subjects at once like that usually doesn't make you progress. You end up lacking focus and thus compromising
03:18
is there another lol?
i wouldn't otherwise do it jakobian, and would otherwise drop set theory, but i am close enough to the end that ill push through. from there it will be just LA and analysis
ee18 he is charles the third, charles the first was executed for being kind of a dork
charles ii was OK in my personal ranking of monarchs
did not know we had an expert on the monarchy in our midst
i won't pretend to know anything more than the crown has taught me
the problem with ranking monarchs is they're all basically horrible
@Jakobian Ok, but thank you for your time!
absolute power does it historically, but i guess recently, without the power, there's less of an excuse. i do wonder if they'll still be around when i die or if the UK will become a republic...who knows
03:37
the problem with monarchy is that they keep it all in the family
you expect a genealogical tree? well how about a cycle instead
03:54
@EE18 my strategy would be to dump set theory, learn LA, then go for analysis
 
4 hours later…
07:53
3
Q: The intersection of any $r+1$ sets in a set $F$ is nonempty, then the intersection of all sets in $F$ is nonempty.

MathNoob420Let $F = \{E_{1}, E_{2}, \ldots, E_{s}\}$ be a family of subsets with $r$ elements of some set $X$. Show that if the intersection of any $r+1$ (not necessarily distinct) sets in $F$ is nonempty, then the intersection of all sets in $F$ is nonempty. Note that this has a solution but this statement...

Can someone please tell me whether this is a proof by contrapositive? I don't understand the contradiction here.
 
3 hours later…
10:28
is there a word for a structure that is an algebra but which has two multiplications?
10:55
@SillyGoose multiplication is too vague. For example Clifford algebras are equipped with a bilinear form
hm well the particular structure i had in mind was from quantum mechanics. we deal with an algebra of matrices equipped naturally with the structure of a vector space and then a multiplication via matrix multiplication. we further endow it with a lie bracket, the matrix commutator (another "multiplication"). so it can be viewed as either an algebra with matrix multiplication or a lie algebra with the lie bracket
to clarify, the lie bracket in this case is the matrix commutator
so i guess i mean to ask what to call a structure that is simultaneously two distinct algebras with the same underlying vector space
if that makes sense
11:21
@SillyGoose Poisson algebras are one way to generalize matrix Lie algebras
In mathematics, a Poisson algebra is an associative algebra together with a Lie bracket that also satisfies Leibniz's law; that is, the bracket is also a derivation. Poisson algebras appear naturally in Hamiltonian mechanics, and are also central in the study of quantum groups. Manifolds with a Poisson algebra structure are known as Poisson manifolds, of which the symplectic manifolds and the Poisson–Lie groups are a special case. The algebra is named in honour of Siméon Denis Poisson. == Definition == A Poisson algebra is a vector space over a field K equipped with two bilinear products, â‹… and...
Or any associative algebra with an induced Lie bracket
12:01
@Jakobian that's cute
12:13
I struggle with what probably is a basic set identity in theorem 1.18 in Folland's. Apparently $$\overline{E}\setminus U=E\setminus E\cap U$$ (the highlighted bit), where $U$ is such that $U\supset\overline{E}\setminus E$. Does anyone see why this "obviously" holds? If yes, I'd be grateful for a comment or two.
I also don't see where the fact that $E$ is bounded is used.
$\overline{E}\setminus E\subseteq U$ implies that $(\overline{E}\setminus E)\setminus U=\emptyset$, hence $\overline{E}\setminus U=E\setminus U=E\setminus E\cap U$
boundedness is used immediately in the next sense after being mentioned
12:31
Hi :) Suppose $G$ is a finite group with distinct primes $p,q\mid |G|$. That is, $G$ is finite and not a p-group. What does that say about the number of conjugacy classes of $G$?
I think I can answer this if I can show that, for at least one of $p,q$, the conjugacy classes of $x$ and $x^{-1}$ are distinct.
. . . where $|x|=p$ or $q$.
If $p=2$, then $x^G=(x^{-1})^G$.
Does the converse of the immediately above hold?
thanks Thorgott!
Cauchy's Theorem guarantees elements of orders $p,q$.
I think I'll ask about it on the main site. Thank you nonetheless :)
12:46
I've as yet only been introduced to $\Bbb C$'s rectangular decomposition, not polar. I have a few propositions with the basic properties of absolute values and complex conjugation, and I've seen that every real quadratic equation has two solutions in $\Bbb C$
But I'm not sure how to use any of those toward showing existence and uniqueness of $w$ as above.
13:24
what field in math do u hate/dislike?
field not to be taken to mean algebra
I feel like I dislike some approaches, where approach is independent of field
@EE18 use the quadratic formula
@Derivative what approaches
usually if you don't have lots of examples and counterexamples of every definition I dislike
yeah.. books without examples r unreadable
e.g. when they define continuous maps in topology, they give examples to demonstrate it aligns with the "draw the graph without lifting the pencil" thing
14:00
I disagree some things don't need an example
Moreover, sometimes examples only obscure the picture
example? :P
Let $M$ be a compact oriented manifold of dimension $\geq 3$. Let $[S_1] = [S_2]\in H_2(M;\Bbb Z)$ each are represented by compact oriented surfaces $S_1$ and $S_2$. Then is there a $3$-manifold $N$ in $M$ with $\partial N = S_1\coprod S_2$? (S_1 and S_2 are cobordant)
14:22
I'd be surprised if that were true
@Jakobian Are you talking about texts providing examples? because in that case, sure. The author needn't always provide an example.
But I think that if you want to understand something, you have to have some concrete example in mind.
My least favorite kind of talk is a speaker spending half an hour talking about all the properties of some really abstract object, without having any ability to give any kind of example at all.
I also kind of feel like examples have to exist in order to motivate the work. If you don't have an example or two, what are you doing? What is the goal of your research? What are you hoping to accomplish?
Should math be "intuitive"? I was reading this question and it seemed like the user wanted to understand intuitively what an exponential factor meant. I think we can intuit very simple mathematics but when it gets to a certain level, I'm not sure it's worth concerning oneself with "intuition"? Or maybe intuition develops overtime..
@Obliv I mean, once you become familiar with an area of mathematics, it becomes intuititive.
But "intuition" can only be brought down so far.
And when the gap between what you need to know in order to have intuition for a topic and the knowledge you actually have grows too large, it is very hard to provide meaningful intuition.
Does intuition vary from person to person? Like how they consider a concept intuitive is subjective? For example, I do like "fitting" math into my own understanding without just simply memorizing/following logic. I guess intuition to me then just means if the math relates to something I understand deeply?
People have all sorts of mechanisms for understanding concepts. Like I imagine some are more geometrically/visually inclined than others
@Thorgott the assumption does not imply there is a homotopy between $S_1$ and $S_2$ right?
14:37
@XanderHenderson So does it not make more sense to build intuition from the ground up, so that each step up isn't so dramatic? Building intuition from the very foundation of math via set theory, algebra, geometry etc so that when we derive more complicated results the person can rely on their intuition to move forward
@Obliv Yes. Intuition absolutely varies from person to person. That kind of follows from the idea that you build intuition over time.
I know it's not really feasible for a general education teacher but I feel like ideally that's how it'd go.
@Obliv I don't think that you should think about building intuition. It largely just happens and you learn and do more.
You build intuition by working through a lot of examples, and seeing how things get applied in other places.
@onepotatotwopotato I mean, they can be different surfaces
For example, I would guess that very few students taking an introduction to calculus have much intuition about epsilon-delta arguments. But by the time a student have gotten through real analysis, one would expect that they have a pretty good feeling for epsilon-delta proofs.
Honestly, this is a lot like Tao's breakdown of pre-rigorous / rigorous / post-rigorous mathematical thinking. Post-rigorous thinking is all about having the appropriate intuitions, and being able to kind of "skip over" certain steps.
Yeah, intuition and familiarity comes with time and effort. Can't really skip steps in the process to understand things on a deeper level.
can someone help me understand why when $\Gamma \subset PSL(2, \mathbb{R})$ is a fuchsian group , with limit set $L$ on $\partial \mathbb{H}$ not all of $\mathbb{R}$, for any point $z_0 \in \mathbb{R} \setminus L$, we can find a small half disk in which no two points are in the same orbit?
@XanderHenderson well if you say, have a hypothetical situation where you are trying to say, those objects are just those other objects in disguise and extra generality doesn't add anything
Would example help?
i can see that its impossible for any single orbit to come arbitrarily close to $z_0$, but i dont know how what I want follows from that
14:50
@Jakobian I don't like dealing with this hypothetical. Can you give me an actual example of a case where having an example is hurtful rather than helpful? I, personally, cannot imagine such a situation.
From the point of view of an author or speaking, my feeling is that if you cannot provide an elementary example, then you probably don't actually understand what you are talking about.
Note, again, that I am making a distinction between what is communicated to a student or reader, vs what the author themself should be able to do.
I also find it hard to imagine a case in which a student would be hurt by having an example handed to them, but that isn't really the argument that I am putting forward.
there is a theorem that says $\Gamma$ is of the second kind iff the dirichlet polygon has a free side, so i think this has to be true, but i dont understand that theorem at all and think there may be an easier way to see it
15:04
@XanderHenderson you seem to think there is a dichotomy here
@Jakobian I don't understand the point that you are making.
You wrote:
1 hour ago, by Jakobian
I disagree some things don't need an example
I don't think that is true.
I think that one should always have examples in mind, even for basic things.
The older word is sample:
Perhaps even especially for basic things.
But perhaps you would like me to respond more directly to
1 hour ago, by Jakobian
Moreover, sometimes examples only obscure the picture
c. 1300, saumple, "something which confirms a proposition or statement, an instance serving as an illustration" (a sense now obsolete in this word), from Anglo-French saumple, which is a shortening of Old French essample, from Latin exemplum "a sample," or a shortening of Middle English ensaumple (see example (n.)).The meaning "small quantity (of something) from which the general quality (of the whole) may be inferred" (later usually in a commercial sense) is recorded from early 15c.
Can you give an example of a situation in which an example "only obscures the picture"?
15:11
You're trying to argue that I've said examples can be hurtful, I imagine you think that not helpful and hurtful means the same thing
@Jakobian No, you argued that examples can be hurtful, in that they "obscure the picture". I am responding to that.
But even so, I am asking you to provide an example of the kind of situation which you have claimed exists, i.e. an example in which providing an example would "only obscure the picture" (whether or not we agree that this is equivalent to an example being hurtful).
There is still a difference between "can be" and "is" as examples are still subject to the reader
Okay, whatever. I disagree with your assertion that providing an example "can" obscure the picture. I am willing to be convinced otherwise, but you clearly have no desire to make any attempt at all to put forward an argument. I don't see any point in continuing this conversation if you can't back up your assertion with an example.
See definition of example :P
I wasn't trying to have a full out discussion about this, yes
15:20
@Jakobian They why bother replying? Now it just feels like you were trolling.
That's quite an accusation
Would you be trying to reply if someone started to shower you with words at random
I don't think you're being reasonable here
@Jakobian I have no idea what this is supposed to mean. You commented that examples might, in certain situations, only "obscure the picture." This is a controversial point, and when you were called on it, you quibbled over semantics, rather than either trying to defend the point or saying "You know, nevermind. I don't want to discuss this."
That feels like trolling (whether that was your intention or not).
We can all agree that without any examples things can become very "obscure" very quickly for the learner.
@user70432 That is certainly my feeling.
@XanderHenderson that's still an accusation
15:27
@Obliv I am not sure. But I do think that when mathematicians invent some new form of math or some new definition, they do it for a reason. Without understanding that backstory, math feels very obscure sometimes.
Whatever or not you feel rather than think, you've decided to voice it
Thus undermining my credibility
You undermine your own credibility when you get into these pointless arguments with people in room 36
I am simply explaining my actions from my point of view. That's not trolling but a simple reaction of someone who was suddenly "struck" by someone into a conversation they don't want to have, or rather don't feel like having right now
I'm polite enough to respond rather than outright try to ignore you
8 mins ago, by Xander Henderson
@Jakobian I have no idea what this is supposed to mean. You commented that examples might, in certain situations, only "obscure the picture." This is a controversial point, and when you were called on it, you quibbled over semantics, rather than either trying to defend the point or saying "You know, nevermind. I don't want to discuss this."
I didn't want to discuss from the start
15:32
If you didn't want to engage in the conversation, the polite thing to do would be to say "I don't want to engage in this conversation right now," not to tell your interlocutor that they have misunderstood your point.
(Telling another person that they have misunderstood is an invitation to further conversation.)
2
There is always a chance that if you explain that you didn't mean something in the way the other person understood then the conversation will end at that
@Jakobian If you don't want to engage in a conversation, the simplest way to not engage in that conversation is to simply stop engaging in that conversation. The next simplest way is to say "I don't want to engage in this conversation right now," and then stop engaging.
Comments like "you seem to think there is a dichotomy here" and "There is always a chance that if you explain that you didn't mean something in the way the other person understood then the conversation will end at that" are invitations to continue the discussion.
15:55
It's not like time didn't pass and my intentions didn't change throughout this time
9
Q: Which cognitive psychology findings are solid, that I can use to help my students?

Jim HefferonI read recently on this site that the growth mindset seems not to be real. I did not know that (I admit that I don't follow research into learning as closely as I would like). Can I turn that experience around and ask: which results useful to a working educator seem to be solid, in that they rep...

@XanderHenderson have you seen this^
@user70432 Yes? What about it?
@Thorgott OK, I will give this a go. My book only examined the case where all coefficients were real but I will spend some time examining that proof for the more general facts
@XanderHenderson just wondering if you were interested in it.
On second glance, I realize I was taking even that development for granted
Before I post a screenshot (gosh forbid!), it's correct to say that a real quadratic equations has precisely two solutions even in $\Bbb C$ right?
i.e. there are no "more"?
16:07
Are you counting with multiplicity?
yes, with multiplicity
Then yes, the statement is correct.
omg i think i forgot a key theorem: A nonconstant polynomial of degree m over a field has at most m zeros.
@EE18 Indeed. This is basically a corollary of a more generalized version of the fundamental theorem of algebra.
Understood, thank you Xander. Will keep at it
I was also pulling on another thread after reading an intro to sequences in my analysis text
i wrote the following and am wondering if it's correct, in particular around whether there are methods for algorithmically figuring out the limit given that we know we have convergence:
You'll notice in exercises like Exercises 3 and 5 to come that we are calculating limits by a very crude method: we are sort of "guessing" (based on our intuition) that the given sequence converges to some limit, and then we are using the definition of "converges to a limit at $a$" directly in order to verify this. However, this is only really working because we are dealing with simple sequences which are easily dealt with "by inspection".
What do we do in more complex cases? We will develop "algorithmic" methods applicable to certain classes of sequences that will allow us to check whether they converge. We will not though have an algorithmic method for determining the limit given that the sequence converges, though that should be clear "by inspection" in most cases.
I wrote those two bits in my notes based on what I know from undergrad engineering math courses, but wanted to check if that's roughly correct for big picture stuff
16:14
In general, you cannot necessarily look at some sequence and have any idea about whether or not it converges. The question of "does this sequence converge?" is at the heart of a lot of interesting mathematical problems. And even if you know that it does converge, determining the actual value of the limit might be very hard.
The Riemann hypothesis could, for example, be phrased as a question about the value to which certain sequences or functions converge.
Awesome, thank you Xander. I think that's roughly reflected in what I wrote but I will use your language in my notes because it's a lot more succinct (and gives context!)
But in the context of engineering, you are generally fine assuming that things which look like they converge actually do converge.
One can also prove existence of root of a polynomial over the complex field using degree of maps between smooth compact oriented manifolds.
@Koro Something about swatting gnats with nukes?
:)
16:18
@XanderHenderson I did want to engage but on energy-save mode
or using Liouiville's famous theorem on entire functions.
Switching gears one more time back to the earlier conversation, I've shown using the definitions etc. given in my book that $z \in \Bbb C$ being a zero of $aX^2 + bX + c \in \Bbb R[X]$ is equivalent (after using the homomorphism into the field of functions) to requiring $\left(z + \frac{b}{2a}\right)^2 = \frac{D}{4a^2}$ where $D = b^2 - 4ac$ of course. Now I can just use what I know the roots are from elementary school, verify that they fit the bill, and...
...then appeal to the max of 2 roots theorem I mentioned above.
@Koro Kincaide and Cheney have a very elementary proof. You don't even need to know much complex analysis. I like that proof.
However, that appears to have 2 issues to me: (1) it does not cover the case where our roots are degenerate (2) I would much prefer a proof that uses theorems I've seen in order to guide myself to the two roots, but I can't figure out what such an argument would look like
Elementary proofs tend to be lengthy imo.
16:20
@Koro Of course.
like pi is irrational.
But that is because more advanced proofs hide all of the details behind other results.
If your goal is to prove the fundamental theorem of algebra, and you have taken two semesters of calculus, you can either write a moderately long elementary proof, or you can write two or three chapters of a textbook on more advanced material until you get to the theorem that allows you to prove the fundamental theorem in one line.
I'm thinking something along the lines of if $D \geq 0$, then looking at $\left(z + \frac{b}{2a}\right)^2 = \frac{D}{4a^2}$ and using that $\frac{D}{4a^2} \in \Bbb R$ means, from a theorem I've seen, that $\frac{D}{4a^2}$ has only the roots $ \pm \frac{D}{4a^2}$ (this theorem covers the degenerate case too I guess, for if $D = 0$ then we must have $z + \frac{b}{2a} = 0$ because the complex field has no zero divisors), so (assuming for the moment $z \in \Bbb R$ so that $z + b/2a \in \Bbb R$) we
get that $z = -b/2a \pm \sqrt{D/4a^2}$. If $D \neq 0$ then $z \notin \Bbb R$ is impossible for otherwise we'd have more than 2 roots. For $D = 0$ we already argued that there can only be the one root.
I actually both didn't and did want to engage. I just didn't want to get into this annoying state of things where I'm being overburdened
Does the above seem like a reasonable argument (if poorly written)?
I need to still think about the $D < 0$ case of course
16:28
I guess there is actually no contradiction here
@Jakobian Dude... why are you still on this? If you don't want to engage, don't engage! No one here is going to judge you for stepping out. But these interjections of "I didn't want to engage!" are only encouraging more discussion.
Because its bothering me
You can both want and not want something and its confusing
Because your speech about me not wanting to engage and just not engaging didn't feel right
Am I even making sense
Guys I have solved an integral but the solution given to me is in evident disagreement with my result. Could someone quickly check the validity of my computations, which I believe to be correct obviously :P. It's only a 2 pages iPad pdf
Like lets say you want to eat a cake because tastes good but you don't want to eat a cake because makes you fat
16:41
What is a topology of the grassmann bundle $G_k(TM)$ of a smooth manifold $M$?
still don't know the formal definition of homotopic plane fields
That's why it feels so misplaced because your wants can contradict each other
How can a person knowingly contradict themselves though like its confusing me that this can even happen
the solution presents instead two gaussian distributions as the result, which is pretty surprising to be honest
I just want to sort of defend myself, you know? It isn't black or white type of situation, because I had feelings supporting both, so its not really appropriate to say I didn't know that I could just not engage
as to how my professor obtained these two Gaussian functions, that's something I haven't been able to decipher yet :P
I did know it, I chose to engage, but not full scale
16:54
@Koro lmao, NDT has just become a world-renowned meme at this point, hopefully he takes this fact lightheartedly
So if you're saying to me, lets say this is what you meant, next time just don't engage if you don't want to, I think its just oversimplified
So I'm sorry but you're wrong
@ClaudioMenchinelli I like ndt. Nonetheless, the video is too funny.
17:14
@ClaudioMenchinelli they use $\int_{-\infty}^\infty \delta (x-a) f(x) dx= f(a)$.
17:25
@Thorgott Sorry to ping you again, but would love to discuss this with you when you get the chance. I've just spent the last hour rigorously reexamining the theorem I was given regarding roots to the quadratic equation with real coefficients (I'll spare you the page I wrote up filling in the details lol) but I still can't see how this can be useful to me when the real coefficient hypothesis is relaxed-- i used numerous theorems about $\Bbb R$ (or, at least, its embedding in $\Bbb C$)
Somewhat unrelated: is it fair to $\Bbb C$ is partially ordered (we sort of tacitly do this), with the partial order applying only to the embedded reals?
@EE18 as in $x <_{\mathbb{C}} y$ iff $x, y\in\mathbb{R}$ and $x <_{\mathbb{R}} y$?
yes, to the extent we aren't being careful about distinguishing reals with their values under the given embedding
That partial order is often tacit right?
While this might agree with how people think about $\mathbb{C}$, no one treats $\mathbb{C}$ formally in this way
as far as I know
for sure, just figured i'd ask
I mean it is partially ordered like that
in what way its fair to say that?
when someone says $\mathbb{C}$ is partially ordered, they probably should put that into context
and context could be that the operations agree with partial order
but if you order $\mathbb{C}$ like that then $0 < 1$ would need to imply $i < 1+i$ for the operations to agree
it doesn't fit into that context for sure so maybe its a bit too vague of a statement
when I think of a partial order on $\mathbb{C}$, I think of $a+bi \leq c+di$ iff $a \leq c$ and $b \leq d$
because its just the most natural partial order on the product $\mathbb{R}^2$
and with this one I make $\mathbb{C}$ into a partially ordered additive group
17:42
Oh ya for sure i should be clearer about which specific partial order i mean
@Koro That's what I've done too : P, That's why I think the solution is wrong
but that still isn't good enough since multiplication isn't compatible with this partial order
17:53
@EE18 not my main point
its that saying partial order should usually imply more than just there being a partial order
there being a partial order is cool and all but maybe not essential
ah i see what you mean, it shouldbe interesting too
yeah
18:18
@EE18 there is something kinda like this generally in operator theory on hilbert space, where you define what it means to be 'positive' and say x < y if y - x is positive (often one also requires x and y to be at least self-adjoint before one writes this, but not always). so most operators aren't comparable like this, but some are
and if you do this for operators on a one dimensional complex hilbert space you do get exactly that ordering on C
where again, because of the ambient conventions you would tend not to write/use it if you weren't already comparing real numbers
Interesting, so its effectively "another way" of defining a given order in this context, sort of like how we can show that two definitions of a given set coincide
i was just thinking more, you had R, an ordered object, sitting inside C, an object that we tend not to think of as ordered (except sometimes when talking about elements of its ordered subset), and that is the n = 1 case of a more general and widely used thing in operator theory
the n = 1 case is not generally a very illuminating guide to operator theory, but the analogy is at least there :)
18:35
ah je comprende now, that makes sense
operator theory...one day
Then maybe i'll actually be able to make the QM I know rigorous :)
18:48
@EE18 thats only assuming you want to get a partially ordered ring
or vector space or whatever, group I suppose
and this is again mixing algebra with partial orders, its not just partial order
the positive elements satisfy conditions that make it so that the induced order has properties such as $x+z \leq y+z$ for $x\leq y$
so leslie's comment is more for orders such as $a+bi \leq c+di$ for $a\leq c, b\leq d$ rather than $x \leq y$ iff $x, y$ are reals and $x\leq y$ in the order of the reals
$k$ should depend on $n$ here right?
That is, we are describing a whole group of sequences (with $k:=(k_0,k_1,...)$ wherein each $k_i$ satisfies the given constraints)?
@EE18 impossible to tell from the phrasing
maybe the inequality leads to a unique $k$
Tis all I have to go with sadly. I am going to interpret it as such and then show that for each $n$ there is some $k_n$ satisfying the given constraints to show that this is "well-defined" I think
I don't think such a unique $k$ exists, because $k^2 + k - 2 \leq 2(0) = 0$ iff $k = 1$
But then $1^2 + 3(1) - 2 \geq n$ is certainly not true for all $n$
I mean unique $k = k(n)$ depending on $n$
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