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00:14
what's there to doubt about the last line?
 
1 hour later…
01:29
If 1 column of a matrix $Q$ doesn't have length $1$, that automatically means that $Q$ is not orthogonal (orthonormal), correct?
01:52
Orthogonal. Correct.
02:09
Thanks again!
 
1 hour later…
03:27
How might I go about proving $||Qx||=||x||$ for any vector $x\in \mathbb{R}^n$ iff $Q$ is an orthogonal matrix? I don't know of a quantitative way to show this, it only seems true to me just because every row of $Q$ has length $1$ so it doesn't increase the length (Euclidean norm) of $x$
How do you actually calculate lengths of vectors?
Root of the sum of squares
Write that more conceptually.
$||Qx||=\sqrt{\langle Qx,Qx\rangle}$
03:43
Aha. Now use some famous formula to rewrite that inner product.
I forget which one that is lol
In my courses, I called it Ted’s favorite formula and no student ever forgot it.
$\langle Ax,y\rangle =$ ?
04:01
$=A\langle x,y\rangle$?
so then $\sqrt{\langle Qx,Qx\rangle}=\sqrt{Q^2\langle x,x\rangle}=Q\sqrt{\langle x,x\rangle}=Q||x||$?
@CottonHeadedNinnymuggins what you wrote is nonsense.
A scalar equals a matrix.
I'm unfamiliar with said formula and didn't know scalars equal matrices. If what I wrote nonsense, then what sense can I write?
@Goku have you watched DBZ? DBS?
$\langle Ax,y\rangle \ne A\langle x,y\rangle$?
yes, true. LHS is scalar and RHS is a matrix.
04:12
Try examples.
I believe your course/book has covered the required formula.
Go read about the transpose of a matrix.
@TedShifrin haha. The thing is that if I say "It's is I", some may think that I'm wrong because it should be "It is me" as per them.
One of my friends believed -'Is not it' is the full form of 'isn't it?'.
He won't believe that 'isn't it? = is it not?'. No one can convince him.
Oh, you have a typo in the second line.
I love grammar. :-)
What does he claim n’t means?
Doesn’r, aren’t, weren’t, ….
n't = not. No disagreement there.
04:18
So is not it. Does not, are not, etc.
doesn't he?= does he not?. But then according to him- 'does not he?', which is not a sentence.
The inversion of word order in a question is the issue.
Yes, I win.
@PM2Ring I understand that 'it is me' is more spoken. 'It is I' is what one writes in an exam.
Nah, write ‘tis I. But don’t ask your friend — he’ll need t’is or ‘t is.
yeah that's another way. 'tis -this is/it is. But is it formal ? (can one write it in exam?)
04:23
It’s poetic. In Shakespeare.
I remember losing points because of writing 'y'all' (for all of you) once. 😅
That’s Southern US only.
Although if other languages have you plural, why shouldn’t we!
@TedShifrin I think you'll be proud: $\sqrt{\langle Qx,Qx\rangle}=\sqrt{(Qx)^TQx}=\sqrt{x^TQ^TQx}=\sqrt{x^Tx}=\sqrt{\langle x,x\rangle}=||x||$
Good job! So what is Ted’s favorite formula?
$\langle Ax,y\rangle = (Ax)^Ty$?
04:31
Keep going. What inner product is this also?
$\langle Ax,y\rangle = (Ax)^Ty=\langle Ax,y\rangle$
Grr.
0 score on that.
maybe needs a $\ ^*$?
lol, I just know $Q^TQ$ went bye bye. Hold on one sec, I'm going to come up with a less sarcastic answer
No stars here, copper. Just $\top$.
04:39
$\langle Ax,y\rangle = (Ax)^Ty=x^TA^Ty=x^T\langle A,y\rangle$?
Loud spank.
Total garbage at the end. You’re not thinking.
I am thinking. You're overestimating my ability, sir. I don't know the end goal
Another inner product.
I would love to know which part of that equality is wrong too because I was convinced it was an equality. Is it not an equality? Or did I just not get what you're looking for?
What you wrote is worse than before.. Now you have an inner product of a matrix and vector, which is undefined, multiplied by a vector.
The key to many things in linear algebra is the associative law for matrix multiplication. $(x^\top A^\top)y = $?
04:47
3
Q: An exercise about countable basis

mcihakIn the book Munkres: Topology §30 I met the following problem: Show that if $X$ has a countable basis $\{B_n\}_{n \in \mathbb{Z}_+}$, then every basis $\mathscr{C}$ for $X$ contains a countable basis for $X$. [Hint: For every pair of indeces $n$, $m$ for which it is possible, choose $C_{n, m}...

I have confusion regarding the hint part in the OP. The hint says -"if possible". I think that it is always possible to find such $B_n, B_m$ (Contd.)
So the last step when I reintroduced the inner product was where I went wrong right?
To see this, fix an $m$. There exists a $C\in \scr C$ such that $C\subset B_m$. $\{B_n\}$ is a basis so there exists $B_{n_m}\subset C\subset B_m$. Let's rename one of such $C$'s as $C_{n_m,m}$. The subscript is as it is because we fixed $m$ so $n$ depends on $m$.
Now I claim that $\{C_{n_m, m}\}$ is the desired subcollection. Take any $x$ and a neighborhood V of x. There is a $B_m\subset V$ such that $x\in B_m$. There is a $C$ such that $C\subset B_m$ such that $x\in C$. There is a $B_n\subset C$ such that $x\in B_n$.
Now this C need not be $C_{n_m,m}$.
hmm, that's why I need all possible pairs (n,m) such that $C_{n,m}$ is...
I'm lost, sir. I don't know where to reintroduce the inner product again
Koro, the basis is numbered to start with. If one basis element isn’t a subset of the other, it’s impossible to find $C$.
@CottonHeadedNinnymuggins Just use the same facts you’ve already used. You might answer my associativity question.
05:04
yes, indeed.
05:15
$(x^\top A^\top)y = \langle(x^\top A^\top)^\top,y\rangle=\langle Ax,y\rangle$? What am I missing?
ASSOCIATIVITY means what?
It means (AB)C=A(BC)
Use that with the quantity I indicated.
$(x^\top A^\top)y = x^\top (A^\top y)=\langle x, A^Ty\rangle$?
Yippee!!
Never forget!
05:27
Well it took me long enough, it's pretty hard to forget at this point lol. Thank you, I'm still confused as to why you put the parenthesis around the $x^TA^T$
Because you had $(Ax)^\top y$.
Alright, I appreciate it. I'm glad I understand the formula rather than just having it fed to me
I've never actually seen that one in the textbook or class either :(
I am surprised. It’s very important. My book must be better :)
But, yes, understanding is the key!
If $D\subset\Bbb C$ is compact and simply connected, then $\partial D$ is simple closed curve?
05:38
Any counterexample?
Sure. Just take a figure 8 curve as the boundary.
I wouldn't be surprised if your book was better. And I actually just found your formula as a homework problem in the norms section but we never covered that in class and the homework answers aren't given necessarily
You’ll have to trust me that it’s highly useful and important as you go on. You are a step ahead.
I can tell that it would've instantly solved my original problem, that's for sure! I'll make sure to keep that one up my sleeve
05:47
Good :)
But the region surrounded by a simple closed curve is simply connected? (in $\Bbb C$)
Why does an orthogonal and orthonormal matrix mean the same thing? They called the $Q$ matrix in the $A=QR$ factorization an "orthogonal matrix" even though it's a collection of orthonormal vectors.
@onepotatotwopotato Yes, by Riemann mapping.
Yes, Cotton, matrices are orthogonal when their columns (or rows) form an orthonormal set. It’s just the way it is. No such thing as orthonormal matrices.
cotton, it's just a quirk of terminology.
@leslie This was a sorta interesting question. I don't see how to take my second derivative argument and get it to work.
05:56
"quirk of terminology"--just another way to confuse Cotton Headed Ninnymuggins
Yes, that's our total purpose in life :P
@Cotton Are you taking an applications sort of linear algebra course or a proof-oriented course?
Just name it after yourself at that point. It's like if I were to design a car and call the model "cheetah" only for people to find out it's not even fast.
It's called "Applied Linear Algebra". Our textbook is written by Peter J. Olver
it's one of those things in where, like in a lot of subjects with many users and a long history, it turns out to be more convenient to stick to quirky terminology that people do use, than invent perhaps more internally consistent terminology and then convince everybody to adopt it.
@leslietownes fair enough
Ah, it's an applied course, so the emphasis is not so much on proofs. I know of Olver. Haven't met him personally and don't know the book.
06:01
it wasn't done deliberately to inflict pain, that's all i'm saying. :)
the same could be said of standard units, or pre-decimalization currency in the UK.
@Cotton I don't know how much it might help, but perhaps some of my YouTube lectures may be of interest to you. There are ones that are explicitly linear algebra in a course that is a mixture of linear algebra and multivariable calculus/analysis.
Yes, why exactly did we not go metric?
i wouldn't expect an avowed francophile like yourself to understand, ted, but in some countries, we respect and cherish freedom, which includes the freedom to measure however we want.
Let's see what the MAGA folks do to your freedoms the next few years.
It doesn't seem like an emphasis on applications either. The closest application we had was data fitting using least squares. Absolutely nothing else has applied to real life. No indication of the usefulness of these things in the real world. Just: "find an orthonormal basis, prove this equality, find the kernel, show this is a subspace, for what b is this matrix positive-definite? Non-singular?, find LU factorization.
Don't get me started on the metric system. Physics makes me hate lbs, feet, inches, etc
Ah, too bad, @Cotton. Even in my relatively proof-oriented courses, I used to do some interesting applications.
06:06
Suppose $f$ is analytic on a simply connected domain $D$ and its boundary $\partial D = C$ which is a simple closed curve. If $f$ is injective on $C$ then $f$ is injective on $D$: By injectivity, $f(C)$ is also a simple closed curve. If $z_0\in D$ with $f(z_0)\notin f(C)$ then using argument principle along $C$, I can conclude $f(z) = f(z_0)$ on $D$ then $z = z_0$. But if there is a point $z\in C$ with $f(z) = f(z_0)$ then I can't use argument principle as the integrand is not well-defined.
Are you learning the concepts and then using technology to do the computational stuff?
I've never disliked math before. The professor is nice, It would just be cool if we saw the application. At some point I think we'll do some differential equations, maybe damping or something I don't know
We learn the concepts and do them ON PAPER. We aren't allowed to row-reduce using a dang calculator. Gotta write out the steps EVERY TIME.
Yes, linear algebra and ODE mix nicely. There is also a lot of interplay with computer graphics — not that I ever had time to cover that section, but I had fun writing it.
See. I think that's stupid. It's fine to give easy row reduction problems on a test (where calculators would allow all sorts of cheating), but once you've learned to do the mechanics, use technology for homework.
The prerequisite to this class was just "normal" linear algebra + diff eqs. It was nicer, we saw some applications because the class had a lab
Wait. You're doing this and you already took linear algebra?
This sounds like a messed up curriculum.
06:09
yeah, this was sounding a lot like the class you're describing as a prereq.
Totally.
where sometimes things are skipped because, OK, there's a more theoretical course later, and a more applications oriented course later.
and the first lin alg class I took, we did the first of the 3 midterms without a calculator then we got our "driver's license" to rref using a calculator on the other tests
This course is hardly doing anything you didn't see in the first course?
Makes sense if it's doing advanced applications (it isn't) or actual proofs of everything (it isn't).
No it is. 100% new stuff. The other class was like half linear algebra (matrices, rref, matrix powers, inverses, determinants) and half diff-eqs and systems of differential equations.
06:11
So no conceptual stuff at all.
Did you do eigenvalues/eigenvectors in first course? You certainly should have to tie in to DE.
Let's just say I understand matrices far better than before. Yes we did eigenvalues in the other one.
We never discussed kernel, image, LU decomposition, non-singular, positive definite, etc
Just took matrices for granted almost and used them to solve diff-eqs, etc
@onepotato Can that happen by the maximum principle?
Never mentioned vector spaces or subspaces in the first class.
06:15
Sounds like a crappy class :P
What would you expect it to be like instead?
Sorta like rote manipulation with no conceptual understanding. The way we teach calculus too often, sadly.
The second course sounds like what I taught as a first course.
Actually, I think I made students do more proofs.
I'm hoping to teach calculus 1 someday. That's my goal. About 1/3 to half the HW is proofs, it just feels like I'm spending way too much time computing orthonormal vectors on paper and all that garbage. Then they only grade 3 random problems so you don't even know what you get right
I loved calculus 1 WAY MORE. I have an introduction to proofs, sequences, series course and that one is awesome too. Hate this linear alg class
@TedShifrin I'm considering MMP or open mapping theorem but can't see the explicit contradiction.
In my multivariable math course (the linear/multivariable course with the videos), I gave all the computational homework on WebWork (so students knew immediately what they got right/wrong) and the written homework was all proofs (graded).
Yeah, I don't see it, either, potato. This question is reminding me of an issue 20 years ago with a qualifying exam question that someone wrote very much like this that turned out to be wrong. I'm too old to remember the question.
06:19
If $D$ is something like a disk then sure, a contradiction but in general I don't know how to use MMP.
Maybe I even wrote it. But I don't have all my files. They were destroyed when I retired.
Well, you may as well assume it's a disk. What difference does that make?
Sounds like a good way to do that. Now, I better go find the $QR$ factorization of 3 different matrices BY HAND
This is a problem appeared before riemann mapping theorem.
Just pretend.
@Cotton I don't blame you for being annoyed.
Anyhow, potato, I don't see how to prove it for a disk. That was my point.
Maybe we can use simple connectivity to define a log of something.
Then open mapping theorem + MMP gives a contradiction by taking a small ball centered at $z_0$ contained in the interior of $D$.
06:25
How so?
Oh, open mapping does it by itself, doesn't it? And I don't see why you need a disk.
I wish I could remember that false problem now.
Wait I was assuming $f:\Bbb D\to\Bbb D$. I can't see now either.
Ah, the question I wrote was for a simple closed curve contained in an (open) region on which $f$ is analytic. Otherwise the same as this question. Hmm.
I think this is actually correct. Are you assuming your function is analytic on the boundary (which means on a larger open set)? Or just continuous up to the boundary?
also analytic on its boundary.
OK, so analytic on an open set containing the boundary. Identical question to the one I wrote back in 1995. I think this is correct. I've forgotten what the false problem was.
The one you wrote on qual?
06:46
Does anyone happen to have a moment to answer a question i have about the fourth isomorphism theorem for rings :o
they might need to see the question to make that determination.
e.g. some folks who maybe know what the 'fourth isomorphism theorem' is, without remembering that it's called that.
the main problem i am trying to solve is to determine the ideals of a particular quotient group of the integers
And i am thinking that the fourth iso theorem: "Let I be an ideal of R. Let A be an element of the set of subrings of R containing I. A is an ideal of R iff A/I is an ideal of R/I". paraphrased from dummit and foote
can be applied here since it is easy (unless i am understanding incorrectly) to find the ideals of the integers up to a certain point
Well, let me pose the problem I am trying to solve more clearly. I want to find all homomorphisms from $\mathbb{Z} \rightarrow \mathbb{Z/nZ}$ where $n$ is some small (for simplicity) non-prime integer. Since each homomorphism has a kernal which is necessarily an ideal, my approach is to find the number of ideals. Then, the homomorphisms will just be the natural projections corresponding to those ideals, which I am thinking will be all the homomorphisms
well i see a flaw--i can't consider the natural projections. but is my reasoning for finding the ideals along the right track?
07:01
mm, i don't know that that framework is the most helpful. you know what ideals in Z look like, presumably, irrespective of this. here you want to count maps from Z into something else, with a certain property.
i'd focus on something special about Z, namely, an additive map on Z is determined by where 1 goes.
when you say 'homomorphism' here do you mean in the sense of groups or rings? and [if the latter] is a "ring homomorphism" required to send 1 to 1 when the rings are unital (as they are here)?
ring homomorphism yes
ah mayn
the fact that you say n is not prime suggests to me that maybe your ring homomorphisms don't have to be unital, which makes the problem more complicated but also arguably more interesting.
if it's the problem you want to solve.
i am confused about that--so in group theory group homomorphisms necessarily send identity of the domain to the identity of the codomain
but that isn't the case in ring theory?
err wait
i guess that is the case for the additive group structure, but it isn't the case for the identity of the multiplicative structure?
yes. 0 is going to 0 here.
there is no restriction requiring 1 to be sent to 1 to my understanding
i feel quite lost trying to attempt this problem xD
07:06
the question is, does 1 have to go to 1, or just anything consistent with being multiplicative.
ok, so maybe not. a "ring homomorphism" here is just a map f with f(x+y) = f(x)+f(y) and f(xy) = f(x)f(y) for all x, y, with no requirement that f(1) = 1. that seems plausible.
i think 1 does not have to go to 1; does the condition that $1 \neq 0$ imply that 1 does not have to go to 1?
where 0 is the additive identity
it's a little more complicated than that, although depending on where you're coming from, you might be thinking along the right lines.
i'll give an example. if you define f from Z to Z_6 by f(x) = 3x. then 1 goes to 3. this turns out to be an additive map, the interesting part is the multiplication. f(xy) = 3xy is the same thing as f(x) f(y) = 3x 3y = 9xy because in Z_6, 3 and 9 happen to be the same thing.
ack i will be back in a little bit thank you for the help so far :)
or to put it more suggestively, 3^2 = 3 in Z_6 and in particular 0 and 1 aren't the only things k in that ring for which k^2 = k.
food for thought.
@TedShifrin You should get more than 2 digits per round in Newton's method, once you're close to a root. "Under suitable hypotheses" it converges quadratically, so the number of correct digits should double at each step, assuming you're calculating with sufficiently high precision.
11
Q: Proof that Newton Raphson method has quadratic convergence

MosesAI've googled this and I've seen different types of proofs but they all use notations that I don't understand. But first of all, I need to understand what quadratic convergence means, I read that it has to do with the speed of an algorithm. Is this correct? Ok, so I know that this is the Newto...

As Wikipedia notes, "Newton's method is a powerful technique [...] However, there are some difficulties with the method". In particular, it blows up when it encounters a stationary point, and it's slow at finding multiple roots.
07:21
if $fv$ and $w$ are vector fields where $f$ is a smooth function, $g(fv,w) = fg(v,w)$ right?
ah nvm, it is true. Because I can just break both vector fields down by its coordinate basis representation, say for $n=2$, $f(v^1 \partial_1 + v^2\partial_2)$ and $g(fv,w) = \sum fv^iw^i g_{ij}$
how do you go about starting to think about solving such a problem @leslietownes. Because initially, since what one is trying to find seems so particular, i didn't even think about considering the property of the homomorphism itself.
mm, i dunno if i had a deeper thought than "Z is generated as an abelian group by 1, so if you want to know what a group homomorphism f does on Z, then you only need to know what f(1) is." then you add in the other stuff, okay, can i declare "f(1)" to be anything and actually get a group homomorphism, and what happens if i also want ring homomorphism.
for more complicated rings, it's a more complicated problem.
so maybe the brainstorm is just thinking of Z as a specific kind of object, as opposed to just as one example of a ring that has ideals/etc. as all rings do.
ah okay that is helpful. to think about the group structure in addition to all that comes with talking about rings
07:38
yeah that is also a useful idea. looking at the groups problem first. (we didn't fully do that above, but you could, it turns out to be significantly simpler and not to capture the subtlety of the ring problem. but it's a good problem.)
so you've established more or less that we need to look at idempotents in the rings $\Bbb Z/N\Bbb Z$. You can decompose $\Bbb Z/N \Bbb Z$ by the chinese remainder theorem. If $N$ is a prime power, you have a local ring, which implies the only idempotents are $0$ and $1$. If you have a product $R \times S$, idempotents are of the form $(e,f)$ where $e \in R, f \in S$ are idempotent.
In particular, you get that the number of non-unital homomorphisms $\Bbb Z \to \Bbb Z/N\Bbb Z$ is $2^{\omega(N)}$, where $\omega$ denotes the number of distinct prime factors of $N$
we haven't learned the chinese remainder theorem yet :-)
it's natural if you want to solve this problem
it's not difficult to prove
i speculate that this problem is one of the problems which you solve before building up the machinery which makes solving it much easier
if you have a concrete small value for $n$, then you can solve this without Chinese remainder theorem
just solve the equation $x^2\equiv x\pmod{N}$
you can try all values $0, \dots, N-1$ and just check which are solutions
07:46
hm okay
leslie gave you the correct idea. I think you should be able to work out that every homomorphism $\Bbb Z \to \Bbb Z/N\Bbb Z$ is of the form $x \mapsto \overline{ax}$ for some $a$ with $a^2\equiv a \pmod{N}$. And conversely, given an $a$ with this property, the map defines a homomorphism
that kind of classifies the homomorphisms
now, can applying the fourth iso theorem as stated above serve as a check that i have obtained the correct number of such elements (which solve $x^2 \equiv x \pmod{N}$)?
I only know three isomorphism theorems
hm i think it is just coincidence in this case. i mean to reference the lattice iso theorem (for rings)
ah I see
I don't think that helps, actually
07:49
is it because the homomorphisms corresponding to those ideals (from the lattice iso theorem) have difference codomains?
I don't know how to determine the number of solutions to $x^2 \equiv x \pmod{N}$ without the Chinese remainder theorem
@SillyGoose the problem is when you pass from $\Bbb Z/N\Bbb Z$ to $\Bbb Z$, you lose the property $x^2 = x$. For example, $3^2 \equiv 3 \pmod{6}$, but $3^2 \neq 3$ in $\Bbb Z$
sorry, what is that map referring to?
I mean when you take some preimages under the map $\Bbb Z/N\Bbb Z \to \Bbb Z$
anyway, what's actually the problem you want to solve? You said: "the main problem i am trying to solve is to determine the ideals of a particular quotient group of the integers" this is different (and easier!) from classifying nonunital homomorphisms $\Bbb Z \to \Bbb Z/N\Bbb Z$
the problem i am trying to solve is the latter of the two you stated
oh okay
07:55
i just thought i could solve it by finding the number of ideals of $Z/nZ$ which seems to be a poor route to go down
or a non-helpful
hmm yeah. I think the fourth iso theorem isn't going to help you much for that
maybe I'm blind because I know the machinery and can't do without it, but I really don't see a way to do this without CRT
do you know Bezout's theorem? Or the fact that every ideal in $\Bbb Z$ is principal? it's quite straightforward to show the relevant case of CRT from either of those
No to both of those xD
seems pretty easy to see that a nontrivial factorization of n into things that are coprime, if it exists, would give you an idempotent that isn't 0 or 1, but this doesn't count them
yeah true
actually even that is maybe using at least a case of CRT
hm
but not the 'full' CRT
08:00
are you sure that your text doesn't require the convention that $f(1)=1$? If it does, that makes the whole problem kind of trivial
it does not make that requirement
hmm
maybe the best idea is to return to this exercise when you have more tools available
08:59
@Koro yes. I've watched DB, DBZ, DBS and even DBGT
I bought their hard copy manga volumes when j was in Japan
I have also watched them.
one piece is my favourite anime now.
What's the point of introducing the term univalent? It's just an injective holomorphic map.
 
1 hour later…
10:35
@TedShifrin I posted the question on MSE a few hours ago and someone left a comment and deleted it almost immediately which I think works: Since $f$ is holomorphic on $\overline{D}$, $f$ is holomorphic on an open set $U$ containing $\overline{D}$. If $f(z_0)\in f(C)$ then by the compactness of $C$, I can find a curve $C'$ enclosing $C$ but contained in $U$. Since $f$ is injective on $C$, I can choose $C'$ close enough to $C$ so that $f$ remains injective on $C'$
($f$ is injective on $C$ so $f'\neq 0$ on $C$ so choosing $C'$ close, $f'\neq 0$ on $C'$.... oh this is the reason why he deleted the comment. Only ensures local injectivity).
Then applying argument principle on $C'$ was the plan...
11:11
4
Q: If an analytic $f$ is injective on $\partial D$ then $f$ is injective on $D$

one potato two potato Let $f(z)$ be analytic in a simple closed domain $D$ and on its boundary, the simple closed contour $C$. If $f(z)$ is injective on $C$, then $f(z)$ is injective on $D$. If $z_0\in D$ is a point such that $f(z_0)\notin f(C)$ then we can use an argument principle to conclude $f(z_0) = f(z)$ then ...

11:21
@Koro My problem with One Piece is that, its all in just one piece unfortunately
1000+ episodes, no seasons. That's not good
11:35
@TedShifrin I want in here when you start talking about the Oxford Comma. I will rip everyone up (then make more people to talk with using Banach-Tarski)
11:46
@Goku 1k+ episodes is too much. I had watched only canon episodes and skipped fillers.
@Koro I've thought of just reading the manga and watch the big fights for fun as the quickest possible way to catch up with the story. Funnt enough One Piece is really big in Japan, almost everyone had seen some of it
Even older anime and one of my favorites, Saint Seiya, was still alive and well in Japan
@Goku I have read some manga. But I'm not used to it. Like imagine Jiren vs UI goku without that background music.
Suppose that T and T' are topologies on X, $T\subset T'$. Suppose that T is regular (saying so for brevity; actually I should say the space (X,T) is regular), can it be concluded that T' is regular?
I think -yes. First an observation- every closed set $A\subset X$ w.r.t. T is closed w.r.t T' (X-A is open in T, hence X-A is in T' and hence complement of that, which is A is closed in T').
Given $x\in X$ and a closed set B not containing A. There exist disjoint open sets $U\ni x$ and $V\supset B$ in T. T is a subset of $T'$ so U and V are in $T'$ too. It follows that T' is regular.
But the converse may not be true as open sets in T' need not be open in T.
Is my understanding correct? Thanks.
I mean-"Given $x\in X$ and a closed set B not containing x here-
4 mins ago, by Koro
Given $x\in X$ and a closed set B not containing A. There exist disjoint open sets $U\ni x$ and $V\supset B$ in T. T is a subset of $T'$ so U and V are in $T'$ too. It follows that T' is regular.
12:02
@Koro Agreed that is why I'll watch most of the big fights. Plus the DBS anime is "more canon" than the manga so I watched it first
One old anime that I watched long time back is Claymore. Since they cancelled the anime, I read some of its manga to see how it ends.
So, I was just browsing geometry problems and answers on here as usual, including my own, and I came across an interesting pattern regarding the people that answer these questions and sometimes even those that ask these questions too.
What I've noticed is that, there's two groups that people can be generalized into. There's the "euclidean geometry' guys like myself and some others that usually solve problem via conventional synthetic geometry. And then there's the second group that uses various other methods, the most common of which is trigonometry
However that isn't the interesting part, the interesting part is, the second group usually tends to have this notion that they can "never find such elegant geometric proofs" referring to the first group of people, and I find that statement weird. Why is that they imply they could "never" do this? What drives that conclusion?
Surely anyone can do it with some practice, I did it, what makes them any different?
 
1 hour later…
13:15
@Goku You need to be good at visualisation to do traditional synthetic geometry proofs. And you need to be aware of a whole bunch of standard geometry theorems. It also helps if you learned (and practised) synthetic geometry when you were young.
Trig-based proofs mostly rely on symbol manipulation skill, although visualisation is also helpful, especially at the start of the proving process, when you're trying to figure out how to attack the problem. Many people who are good at algebra & symbol manipulation aren't so good at visualising, and didn't get much geometry training in school.
I'm old enough that I was introduced to traditional Euclidean geometry in the last couple of years of primary school and the first couple of years of high school. Our geometry textbook closely followed Euclid's Elements. I enjoyed it, but I'm not great at visualisation. It was a great relief when I learned how to solve geometry problems trigonometrically. :)
@PM2Ring Huh, I see. So the problem isn't just "not enough practice" like I naively thought it was? Its a real thing that means, not just an artificial problem signifying a lack of practice?
13:31
Practice certainly helps, but you also need good visualisation. If you don't have the innate visualisation ability, it takes a lot of practice to make a little bit of progress. And if you find it a lot easier to do stuff algebraically it can be hard to get motivated to put in the work to practice doing synthetic geometry.
@PM2Ring I know this might be silly but perhaps playing video games such as Minecraft and more when I was young, which does require some visualization, gave me an "edge"?
That sounds reasonable.
I'm calling it. Mandatory Minecraft classes in schools now!
13:47
@Goku There's a 3D structure that Conway et al called Tetrastix.
> it is possible to fill 3/4 of the volume of three-dimensional Euclidean space by three sets of infinitely-long square prisms aligned with the three coordinate axes, leaving cubical voids
I made an interactive 3D version in Sage you might enjoy exploring.
14:16
@PM2Ring This is the kind of stuff that "tickles" a part of my brain that enjoys problem solving. Thank you!
My pleasure! BTW, you can pass the extend parameter negative values.
@Goku Here's another 3D geometry thing that may tickle your fancy:
May 30, 2021 at 20:09, by PM 2Ring
But you can get unwieldy proofs with a seemingly small problem space, too. Eg, that problem about 7 mutually touching cylinders of equal radius that I linked here a few weeks ago. https://puzzling.stackexchange.com/q/109154/36040 From the solution paper: "The first real solution was found after tracking 80 million paths, and the second one was found after tracking another 25 million paths."
14:33
How can you right away tell the degree of a polynomial function? Counting the roots/multiple roots? But that alone won't always be enough, for example for $x^3 + 1$ you would only have one root with multiplicity 1, so you also need to count the maximas/minimas?
You need to count the complex roots too (and their multiplicity). The stationary points may also have multiplicity, so counting maxima/minima doesn't really simplify things.
@PM2Ring So you can't directly read off the degree with just a graph of a polynomial function, right?
14:56
Take a look at $y=x^{2n}$, for any positive integer $n$. They all look pretty similar in shape.
15:07
@PM2Ring need the exponent necessarily be even?
Well, odd exponents give you odd functions, which look different to even functions. ;)
Speaking of functions, I noticed that while solving some complicated homogenous equations involving lots of radicals and stuff, there can usually be a lot of extraneous roots with only one actual solution and sometimes no real solution at all
While I can sometimes explain away the extraneous roots, I can't always explain them always. Is there a "universal" reasons as to why these "fake solutions" pop up?
A common reason for the fake solutions is that you've done an exponentiation on the equation.
A very simple example: $y=\sqrt5$ has one solution, $y^2=5$ has two.
 
1 hour later…
16:28
Alright, thanks
And there is also no way to accurately read off turning/inflection points, right?
Given something like that for example
The inflection point is at -0.66.. in that graph, how would you read that off?
We know that it's at least a cubic, with a repeated root at x=0. But it could have a higher degree.
There is a maxima at $x = 0$ and a minima at $x = -1.33..$ and $\frac{-1.33 + 0}{2} = -0.66..$, but is the middle value always the inflection point?
nein, it depends on what your polynomial is
@ILikeMathematics It's hard to judge it by eye. But if you put a ruler on it tangentially, and slide the ruler along the curve you can roughly estimate where the slope changes from increasing to decreasing.
@PM2Ring Alright, thank you
16:45
Hi folks! I'm struggling with calculation of this expected value: imgur.com/VmGXS7d p ~ Beta(x+1, n-x+1) Could you give me some hints?
17:40
looking for someone with probability brain:
with equal probability i pick between two coins, one which is fair and one that's heads on both sides
i flip the coin k times and find that it's heads every time. what's the probability that it's nevertheless fair?
the text i'm referring to says: if it's fair, the probability to get k heads in a row is $2^{-(k-1)}$. on the basis of Bayes rule, it concludes that the probaility of being fair given that you get k heads is $1/(2^{k-1}+1)$
which i do follow
but it then somehow concludes that the probability of it being fair after k flips is $1-\dfrac{1}{2^{k-1}}\dfrac{1}{2^{k-1}+1}$
oh. i mucked it up slightly: the coin is either fair, or it's the same side (i.e., could be tails/tails). so if you get heads once, that doesn't actually tell you anything
so i should replace every $k-1$ with $k$ in the above
17:59
Yes. I was going to complain about the first sentence. I haven’t gone further.
the paragraph i'm stealing from is unfortunately not that precise. it's very tossed off
oh. igi
So the prob of $k$ heads is $\frac12(1+2^{-k})$ by Bayes. So prob you picked fair coin given $k$ heads is $2^{-k}$ over that.
right. derp
Could anybody weigh in this task?
0
Q: $\mathcal P((-1,1))$ and the set of all functions $f:\Bbb R\to\Bbb R$ which attain every value in $\Bbb Z$ uncountably many times are equinumerous?

Invisible Prove that $\mathcal P((-1,1))$ and the set of all functions $f:\Bbb R\to\Bbb R$ which attain every value in $\Bbb Z$ uncountably many times are equinumerous. My attempt: My first claim: Let $S=\{A\subseteq\Bbb R\mid A\text{ is uncountable}\}.\operatorname{card}(S)=2^{\mathfrak c}.$ $\boxed{\l...

What is $\mathcal P$?
18:12
I found one question on probability that may interest some (based on geometric probability):math.stackexchange.com/…
Power set? Why $(-1,1)$ rather than $\Bbb R$?
@Koro No link?
I don’t consider that geometric probability. It’s just areas by standard calculus.
oh I thought of that as geometric probability.
Nah, not really. Geometric probability is what I call integral geometry (in which I wrote my thesis). But who knows …
You’ve been perverted by Leslie: Anything with a picture is de facto geometric.
18:24
I thought that if we have infinitely many outcomes in a sample space then we call it geometric probability - for example: finding probability of randomly chosen point to lie below x axis etc.
But I am not sure. I never really thought much about this.
Any time you have a continuous prob distribution it’s geometric? Good grief.
Actually, infinitely many … can still be discrete — like Poisson. Forget that notion.
yeah, right :-). Discreteness.
Even wiki says there should be a group action involved in the geometry.
I thought everything that involved uniform distribution is considered geometric probability?
18:35
I went to the ELU chat. Probably shouldn't have
Whatever that is.
19:17
Yes, I won't say anything, just to keep everyone else safe
19:35
Sounds like you're making some "bad decisions" of late.
@TedShifrin man, this just hasn't been my year
Suppose X is connected $T_4$ space with more than 1 element, then X is uncountable.
can this statement be proven with Uryshon's lemma?
I never know what the $T_i$ mean. Is that normal?
If so, yes, and that's a standard exercise.
yes, normal.
Is it not obvious to you?
19:42
There is one $T-3\frac12$ too. (trennungsaxiom)
with Uryshon's lemma, yes.
But without Uryshon's lemma, no.
Oh I see. I wrote with earlier. I meant to write without.
Oh. Well, you need to use normality somehow. You can regurgitate the proof of Urysohn, but why bother?
take any two elements a and b in X. {a} and {b} are closed disjoint. So by UL, there is continuous real valued f on X s.t. f(a)=0 and f(b)=1. By IVT, all values btw 0 and 1 should be attained too. That is, range f is uncountable, which is possible only if X is uncountable.
I was thinking if it could be done without UL.
but nvm
hmm, does this even need T_4
that's part (b) of the exercise @Thorgott.
i would think T_3 suffices
19:49
Let's consider T_3.
I think you need at least complete regularity.
But meh.
suppose on the contrary that X is countable. $\color{red} {\text{Suppose that X has a countable basis.}}$. I know that $T_3+$ countable basis = normal. So by the previous part, we get a contradiction. Hence, X must be uncountable.
I believe I have an argument that T_3 suffices, but I'll let Koro dwell on it as I go and clean the bathroom
If the red part is true, then I am done.
But that's trivial as X is countable.
Is my understanding correct?
Oh. Thor wins. It's an exercise in Munkres (I think it was added to the second edition, and I taught out of the first edition).
Challenge (without looking at Steen-Seebach, which I no longer possess): Find a connected Hausdorff space that is countably infinite.
19:56
That's a daggered remark in Munkres on the same page where the above exercise was taken from.
That's where I am, obviously
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