Mathematics - 2023-05-20 (page 1 of 2)

03:13

Isom. thms for rings hold for nonunital rings right?

I looked up the proof but I think it doesn't matter if the ring is unital or not.

leslie townes

04:34

hard to say for sure without more specificity as to what you mean by "isomorphism theorems" and how you phrase them, but broadly speaking yes, it shouldn't make too big a difference, maybe no difference as long as you phrase the results the right way. "old enough" algebra books often include some results of this type.

shintuku

i have never encountered a ring without unity in my 2 weeks of ring theory

christ this profile pic is terrible

why would they do this

what happened to the old profile pictures

leslie townes

did yours change? the automatically generated ones sometimes do. koro's changed recently.

shintuku

yes it changed

it's no longer anthropomorphic and it's way greener

leslie townes

oh, you can switch back. you just need a copy of the image, or the hash that was used to generate it. gravatar.com/avatar/… in your case.

shintuku

WOAH!!!

thanks leslie!

leslie townes

04:44

i noticed koro's because it was accompanied by a change in color even more dramatic than your green-getting-greener.

shintuku

yeah he went to like grey

how did you find the hash btw?

test123

leslie townes

the internet archive keeps occasional copies of old profiles. web.archive.org/web/20220816162640/https://…

shintuku

ahh very cool

leslie townes

it might take a minute for the servers to catch up to your "new"/old image.

i used a gravatar for sufficiently long that i grew attached to it and kept a copy of it, but i like my current image.

shintuku

there was not enough water in the pool

or the pool was not adequately sized

D.C. the III

04:58

WHat have they done to my avatar???

Reading up a few lines I see I'm not the only on eperturbed

shintuku

find your profile page on internet archive

and save your pic

D.C. the III

And we shall return to our "old" tempered green in a few

Sourav Ghosh

05:20

1) Can a discontinuous function on a connected domain have connected graph?

2) Can a discontinuous function on a path connected domain have path connected graph?

I know 1) is true.

For an example, we can choose f:I\to\Bbb{R} such that f is discontinuous only at a single point which also an end point the interval and the corresponding point in the graph is a limit point of the rest of the graph.

f:[0, 1]\to\Bbb{R} defined by f(x) =\sin(\frac 1 x) ;x\in(0, 1] and 0 otherwise

Alessandro Codenotti

@SouravGhosh what is a domain here?

leslie townes

if you look at a portion of a helix such as {(cos(t), sin(t), t): t in [0, 2pi)} as the graph of a real-valued function on the unit circle, you get a nice example for both (1) and (2)

and yes what is a "domain" here, is my example even an example

the things that are often taken as branches of the complex logarithm on C \ {0} would also work, they will have discontinuities on whatever path you cut along, but this won't disconnect or path-disconnect the graph

Sourav Ghosh

06:15

@AlessandroCodenotti Any path connected space

Alessandro Codenotti

07:13

Then Leslie's examples work

leslie townes

you can change the answer to (2) if you impose more conditions on the domain. e.g. math.stackexchange.com/questions/2577156/… shows that if 'domain' means 'subset of R' the answer to (2) changes. it also links to examples like the one you gave for (1).

leslie townes

07:31

it makes sense that R would be somewhat special here. it is pretty restrictive in terms of how you can "get to" one point from another. not a lot of ways of sneaking in a discontinuity in a way that doesn't also disconnect a graph.

Sourav Ghosh

07:50

@leslietownes Thanks :)

user1176240

Can anyone cite the source, page # for the scan in math.stackexchange.com/q/4003576? Thanks!

PM 2Ring

For people wondering why their gravatar changed recently:

4

62

Q: We're regenerating all identicons

We've recently discovered a vulnerability in our identicon generation process. To remedy it, we have changed how we approach generating them and regenerated all identicons. We do not have any indication that any personally identifiable information (PII) was leaked as part of this vulnerability. S...

^ @shintuku @D.C.theIII @Semiclassical

user 85795

Thanks @PM2Ring 🙏

PM 2Ring

No worries.

leslie townes

08:06

@user1176240 it appears to be from some edition of steven lay's "analysis with introduction to proof." maybe the fifth edition. page 280 in the version excerpted here: math.colorado.edu/~walter/Math3001/section7.1pp275to283.pdf

PM 2Ring

I made an animated interactive version of Prince Rupert's cube in Sage.

sagecell.sagemath.org/…

user 85795

\o @copper.hat how's it going, pal?

Are you over the jet-lag yet.

Sourav Ghosh

I am going to show that every metric space is normal.

Let A, B be two non empty disjoint closed sets. f:X\to \Bbb{R} defined by f(x) =d(x, A) /{(d(x, A) +d(x, B) }

Continuous and well defined.

Koro

@shintuku you got it back :).

Sourav Ghosh

f(A) ={0}

f(B) =d(A, B) /d(A, B) +d(B, B)

A, B disjoint closed sets. Does this implies d(A, B) >0 ?

Koro

08:18

yes

Sourav Ghosh

Gr(exp) and x-axis in R^2

Alessandro Codenotti

@SouravGhosh no

Koro

ohh 😮😮

Sourav Ghosh

I guess so. One closed set must be compact.

Sourav Ghosh

08:35

f(x) =0 implies d(x, A) =0 implies x\in A

f(x) =1 implies d(x, B) =0 implies x\in B

Ignore what I have done above 😪. f(A) ={d(x, A) /(d(x, A) +d(x, B) : x\in A}

d(x, A) +d(x, B) >0 for all x\in X

Done!

f^{-1}{0}=A and f^{-1}{1}=B

We can choose two disjoint open sets containing 0 and 1 respectively (thanks to Hausdorff :).

Their pre images separates A and B

Hmm

10:50

Hey guys, I'd love to chat with a transdisciplinary, meta / holistic mathematician about a problem I'm trying to solve.

It has to do with sweeping the entirety of an N dimensional state space with a single variable, or finding a mapping from an N dimensional space to an X dimensional space (without any data, not at all talking about dimensionality reduction techniques like PCA or t-SNE or UMAP that require data)

But I want to do this hierarchically (like an autoencoder, or how I imagine a renormalization group works even though I barely know anything about it), and (hopefully) in some loc…

So imagine a second (44.1K samples / numbers) and we want to hear every possible second. This problem is a bit like using a hilbert space filling curve, but I've already tried implementing that and running into some problems programmatically. That solves my "many to one" dimension problem but not my "many to many"

or "many to fewer"

It's also a bit like a poincare disk, or any fractal

Any ideas, or concepts, or books you think I should look at would also be really helpful!

one potato two potato

12:12

For UFD $R$, if $a = bq$ for some $a,b,q\in R\setminus\{0\}$ then $R/(b)\simeq (q)/(a)$ as an $R$-submodule of $M$. Does this imply there exists a unique maximal ideal $M$ of $R$ with $M\simeq R/M$ as an $R$-module?

writing $a$ as a product of irreducibles,,, but it's PID that irred element generates maximal ideal

冥王 Hades

Hamburger

one potato two potato

Correction: there exists a unique maximal ideal $\mathfrak{m}$ of $R$ such that $M\simeq R/\mathfrak{m}$ as an $R$-module

Well actually, $(q)/(a) = \{r+(a)\in R/(a)\mid sr\in (a)\quad\forall s\in(b)\}$.

one potato two potato

12:52

what is the initial object of the full subcategory of unital rings with $1+\cdots+1$ invertible (in the ring)?

Thorgott

13:29

how many additions is that supposed to be

one potato two potato

Fixed $n$. Sorry

Thorgott

14:03

$\mathbb{Z}\left[\frac{1}{n}\right]$ then, no?

one potato two potato

ah

thanks. it factor through fraction ring

Koro

14:26

3

A: Existence of a set $A\subset\mathbb{R}$ such that $|G\setminus A|=\infty$ for every open set $G\supset A$

You are indeed right that your set $A$ need be non-measurable and uncountable (after all any countable set is measurable). Let $S \subseteq [0,1]$ be a non-measurable set. By the criteria that you stated there must exist $\epsilon_0 >0$ such that $|U \setminus S| > \epsilon_0$ for any open set $U...

why is $|U\setminus (S+n)|\ge \epsilon_0$?

Koro

15:21

I understand this now.

But why is $|U \setminus (S \cup (S+3))| \ge |(U\cap(-1,2)) \setminus S|+|(U \cap (2,5)) \setminus (S+3)|$?

Sourav Ghosh

15:45

"Möbius transformation is always a bijective holomorphic function from the Riemann sphere to the Riemann sphere"

Tz=(az+b) /(cz+d)

for c=0 , it's a polynomial. So infinity is a pole.

How we define holomorphicity at infinity?

Semiclassical

4

Q: Holomorphic at infinity (definition)

I struggle quite a bit with the usage of $\infty$ in complex analysis. In some cases, I can translate a definition involving infinity to equivalent statements using limits, or in the case of continuity I just make use of the topology defined on the Riemann sphere. However, what I don't understan...

complex-analysis

Sourav Ghosh

I haven't understood the answer.

Ted Shifrin

16:05

Set $w=1/z$ and decide if $f(w)$ is holo at $0$.

Sourav Ghosh

16:21

Can you explain with one example ?

f(z) =z has a pole at infinity

f(1/z) =1/z ohh

f(w) =w

f(1/z) =1/z ohh

Holomorphic at w=0

Ted Shifrin

Well, if $\infty$ is mapping to $\infty$, then you look at $w$ mapping to $1/f(w)$.

Otherwise, you just look at $f(w)$.

Koro

0

Q: Understanding a proof for existence of set $A\subset R$ such that $|G- A|=\infty$ for ever open set $G\supset A$.

I found an answer here https://math.stackexchange.com/a/4233925/266435. I'm pasting that here to avoid switching tabs and I'm highlighting the parts that I need clarification on: quote You are indeed right that your set $A$ need be non-measurable and uncountable (after all any countable set is me...

real-analysis measure-theory proof-explanation

Peter

Are $8$ and $144$ the only Fibonacci-numbers that are perfect powers ? Upto $n=10^5$ , no other perfect power $F_n$ exists.

Sourav Ghosh

(removed)

Koro

Measure theory and Algebraic topology imho are the perfect examples of subjects which one should do outside college on their own and never in a semester at a college.

noballpointpen

16:31

The link didn't work. Test italics link.

Weird

Sourav Ghosh

@Koro I am going to revise measure theory from tomorrow.

Koro

because in these subjects, one doesn't get get help so easily espacially when teachers at my current college are useless.

and another problem is: there are not many people who do this.

There is a high chance that the questions on these will go unanswered on mse.

on discord too

shintuku

what discord are you checking out Koro

Sourav Ghosh

@Koro Bad a.e

Koro

you shared the link once. The affinoid union.

Sourav Ghosh

16:34

@Koro No. Ask good question to get an immediate answer.

shintuku

oh cool yeah the people there are nice

Koro

@SouravGhosh you'll be surprised it is NOT true.

shintuku

koro maybe if it is too niche try mathoverflow?

Sourav Ghosh

@Koro Take your time. Try to improve your post.

Koro

otoh, a question on real analysis, complex analysis will almost never go unanswered.

@SouravGhosh ?? what is wrong with the post?

it is self contained.

shintuku

16:36

also up to now each time i added a bounty i've gotten a reply

Sourav Ghosh

@Koro False.

shintuku

koro you tried a bounty?

only issue is you gotta wait 3 days

Koro

@shintuku I tried it once at an answer few days ago. I was looking for an explanation for an answer, got none.

shintuku

that sucks

Koro

the bounty didn't give me the answer. It was an AT answer.

@SouravGhosh you are evading the question. What is wrong in the post?

What makes you think that it is not 'good'? Be specific.

Ted Shifrin

16:40

Instead of complaining and whining incessantly, you should appreciate the help you do get. It's remarkable that there is a place like this where professionals provide their services literally for free and out of the goodness of their own hearts.

8

Seriously.

Koro

I do appreciate that, seriously.

Ted Shifrin

You do nothing but complain when you don't receive instantaneous service that meets your (whatever) standards.

Koro

I'm just saying that there is 'shortage' in case of AT, measure theory etc.

Ted Shifrin

Just stop.

Ten years ago there was no site like this and yet students managed to study and learn mathematics somehow ... for centuries.

shintuku

but what if only the rich ones could

Ted Shifrin

16:42

What do you imagine that the rich ones did?

shintuku

the bastards lived off the rents of their tenants

Ted Shifrin

When I was a student, I worked with fellow students and we tried to answer one another's questions. If after a few days we still couldn't get anywhere, yes, we actually went and asked faculty at our universities.

Koro

that's nice but not applicable here at my college.

Ted Shifrin

We actually read books and articles.

I do not accept that, Koro, and I'm sick and tired of your complaints.

shintuku

no! not reading! my worst weakness

one potato two potato

16:47

5 hours ago, by one potato two potato

For UFD $R$, if $a = bq$ for some $a,b,q\in R\setminus\{0\}$ then $R/(b)\simeq (q)/(a)$ as an $R$-submodule of $M$. Does this imply there exists a unique maximal ideal $M$ of $R$ with $M\simeq R/M$ as an $R$-module?

Is this a true statement? I'm doubting.

I can't come up with a counterexample. For PID? maybe.

shintuku

will get back to you about that on monday

Sourav Ghosh

Don't try to be a robot! Take your time to think about your doubts. It may take 1 day or 7 days; it doesn't matter.

If you think your question is too advanced for MathSE, then post it on MO.
But asking for clarification of another's answer in a different post is not a good practise and should be avoided. Who cares about others' answers? Ask the original answerer in the comment and wait for his or her reply.

Ted Shifrin

Very rare that something that comes up in a class or textbook has any business being on MO.

@onepotatotwopotato Are you saying that the first sentence holds in every UFD? And what is $M$?

one potato two potato

That's what my professor said. $M = R/(a)$. (Oh, I didn't write that).

wait I think I typed incorrectly

Ted Shifrin

Then you use $M$ again in the second sentence. That isn't cool. Are those all the same $M$? ... I think I believe the statement for a PID, but I'm dubious in general.

one potato two potato

16:59

Let $R$ be a UFD, $a,b,q\in R\setminus\{0\}$ with $a = bq$. Let $M = R/(a)$ as an $R$-module. Then as an $R$-module, $R/(b)\simeq (q)/(a)$ by $1\mapsto\overline{q}$. Note that $(q)/(a) = \{r+(a)\in R/(a)\mid sr\in (a)\quad\forall s\in(b)\}$.

Now the claim is there exists a unique maximal ideal $\mathfrak{m}$ of $R$ such that $M\simeq R/\mathfrak{m}$ as an $R$-module.

The statement is believable if $R$ is PID as irred elements generate maximal ideal. buf for UFD...

So $(q)/(a)$ equals $\{m\in R/(a)\mid sm = 0\quad\forall s\in (b)\}$.

Ted Shifrin

What if we take something like $R=\Bbb Z[x,y]$ or even $\Bbb Z[i]$. Those are examples you should play with.

Koro

@SouravGhosh ? Did you even see the tags there? Proof explanation is one of the tags there. Did you see the answerer has not been online in a while (it's viewable by going to their profiles.)? 'Asking for clarification ... avoided'- is that a site policy (if yes, can you please share the related policy?) or you are enforcing it?

one potato two potato

@TedShifrin For counterexample or for intuition?

Koro

that was the whole point of me posting the question link here: just in case, anyone interested in that here could know about it.

I have shared that on discord as well so that the question gets more attention.

Sourav Ghosh

17:14

1

Q: Question on analytic continuation of the principal branch of logarithms on $B(1, 1) $

(Excercise $6$ ,Conway ,page $217$ ) Let $D_0=B(1, 1) $ and $f_0$ be the restriction to $D_0$ of the principal branch of logarithm. For $n\in \Bbb{Z}$ let $\gamma(t) =e^{2\pi i nt}$ for $0\le t\le 1$ . Find a continuation $\{(f_t,D_t):t\in[0,1]\}$ along $\gamma$ of $(f_0, D_0) $ and show that $[...

complex-analysis solution-verification proof-writing alternative-proof analytic-continuation

one potato two potato

If I just take $R = \Bbb Z$ and $a = 4$. Then the statement is false?

Sourav Ghosh

Unanswered ,12 month. No comment, no upvote.

Hi @AlessandroCodenotti

Ted Shifrin

@onepotatotwopotato I don't know, but they are concrete UFDs that are not PIDs :)

@onepotatotwopotato Which statement? Explain.

Sourav Ghosh

@Koro The above post is an example:)

noballpointpen

To Ted Shifrin: I hope I did the [thing](https://chat.stackexchange.com/transcript/message/63612014#63612014 "thing") how it was supposed to be done? I hope the link formatting works this time.
(You actually helped me to remember how the derivative is used to analyze functions. So thanks a lot.)
My reasoning was that since the partial derivative showed us for which values of $y$ the function stops changing, substitution of that value should yield us the values of the function where it peaks, and therefore taking $x \to 0$ shows us the behavior near $(0,0)$, from which we can analyze bounds…

I don't know why the chat doesn't format links to messages... uh...

Ted Shifrin

17:24

You are referring to this?

Don't put anything else inside the parentheses other than the URL.

noballpointpen

test

Koro

@SouravGhosh 'almost never'. Did I anywhere say 'never'? :)

noballpointpen

Thanks. In the help there was "optional title", and it seems to be working when it's not a link to the chat.

Ted Shifrin

At any rate, if you're actually getting into this question, you should try to solve my general question. What is the criterion in terms of $\alpha,\beta,\gamma,\delta$ for the function $$f(x,y) = \begin{cases} \frac{|x|^\alpha|y|^\beta}{|x|^\gamma + |y|^\delta}, & (x,y)\ne 0 \\ 0, & (x,y) = 0 \end{cases}$$ to be continuous?

Koro

anyways, I seem to have found one solution to my question. I'll post an answer to my question soon.

Ted Shifrin

17:27

The answer is quite nice.

noballpointpen

I will screenshot your question and think about it somewhere in time. Thank you.

Sourav Ghosh

@AlessandroCodenotti Please help me to prove that topologist's sine curve is not locally connected at (0, 0)?

Ted Shifrin

Where are you stuck, Sourav?

Sourav Ghosh

To prove it's not locally connected at (0, 0), we need to select an open ball in R^2

Ted Shifrin

OK

Sourav Ghosh

17:33

sorry open set in \Bbb{R}^2

Ted Shifrin

Containing the origin, I presume. A ball suffices.

Sourav Ghosh

Such that there doesn't exist any open disk containing (0, 0)

(0, 0) \in B(0,r) \cap T_s \subset U\cap T_s

Ted Shifrin

Soon we'll need dollar signs so I can read it. What is $T_s$?

Before you get buried in symbols, explain in words why the result is true and how the proof should proceed.

Sourav Ghosh

I know topologist's sine curve isn't path connected.

Ted Shifrin

Yes, but it is connected.

Sourav Ghosh

17:39

There is no path connecting (0, 0) to any points in the graph of f(x) =\sin(\frac 1 x) ; x\in (0, 1]

Ted Shifrin

So we do not want to be thinking about paths, do we? But what does the intersection of a (small) ball centered at $0$ with the TSC look like?

Sourav Ghosh

The curve has infinite oscillation near origin

Ted Shifrin

Yes, so answer my question.

shintuku

@PM2Ring sigh, i've decided to change back to the new gravatar in consideration of the potential security issues

thanks for the link

btw, gravatar isn't stack exchange's own terminology, but it seems to be some sort of company that makes avatars of something

are they related?

Sourav Ghosh

B(0, r) \cap T_s

shintuku

17:42

en.wikipedia.org/wiki/Gravatar (

Ted Shifrin

What is $T_s$? Don't just create notation.

Sourav Ghosh

TSC

Ted Shifrin

Good grief. Just call it $X$.

So describe $B(0,r)\cap X$ in words.

Sourav Ghosh

{(x, \sin (\frac 1 x)) : ||(x, \sin(1/x||)<r}

x^2+\sin^2(1/x) <r^2

Ted Shifrin

I don't want algebra or equations. I specifically said in words.

Draw a picture for yourself.

Then describe it to me.

Sourav Ghosh

17:48

Ok.

Ted Shifrin

17:58

@SouravGhosh Well?

This should have been one minute, not over ten.

Alessandro Codenotti

The pictures looks like this waves hand up and down really quick

Sourav Ghosh

Yes. Any open ball around origin contains infinitely many segments of f(x) =\sin(1/x)

Ted Shifrin

Roughly like little pieces of parallel line segments ... So ?

Sourav Ghosh

As f(x) =\sin(\frac{1}{x}) has infinite oscillation around origin.

@TedShifrin line segments are connected

As portion of continuous map on a connected domain.

Ted Shifrin

But what about all those infinitely many line segments?

@AlessandroCodenotti Precisely~

Sourav Ghosh

18:03

Those are maximal connected subsets contained in B(0, 1) and X

As Ball separates the graph with closed segments.

Ted Shifrin

Don't do $B(0,1)$. Choose $r$ arbitrarily small — you have to make the general argument, anyhow.

So are we done?

Sourav Ghosh

Yes.

Each line segments are connected components.

Ted Shifrin

Even if your definition requires that you choose an open set inside the given open set, the picture stays the same.

Yeah, so thinking about the origin as a special point is really misleading.

Sourav Ghosh

We can do same thing to any points on line segments joining (0, -1) and (0, 1).

Those are the limit points, so any open set must intersect the graph.

Ted Shifrin

Yes. But what would happen if we picked a point not on the y-axis?

Sourav Ghosh

18:08

Then the point will be on the graph.

Ted Shifrin

Yes, and ... ?

What if I intersect the graph with a small ball?

What if I intersect the graph with a larger ball?

Sourav Ghosh

Two points can be separated by two disjoint balls

Ted Shifrin

So what?

Sourav Ghosh

The portion of the graph inside the ball compact

Ted Shifrin

We were discussing local connectedness. Yes or no?

Sourav Ghosh

18:12

Must have positive distance

D.C. the III

Ted. Using the three noncollinear points to show the equation of the parabola is $0$ by taking the determinant. Is there any particular way I should think about it? Or is simply a matter of doing things by brute force through doing determinants with their minor or reducing the $4 \times 4$ matrix by row operations?

Sourav Ghosh

I mean we can choose the ball small enough to contain one component.

Ted Shifrin

Not brute force, no, @D.C.

@SouravGhosh Right. This is why the definition of local connectedness has to be phrased carefully.

D.C. the III

good...I'm glad I asked because it doesn't look appealing in the least

Ted Shifrin

What do you mean "to show the equation of the parabola is 0"???

Sourav Ghosh

18:17

Conclusion: Any nbd of origin contains infinitely many connected components; hence, no nbd of (0, 0) can contain a connected nbd.

Ted Shifrin

Correct.

Sourav Ghosh

So $T_s$ isn't locally connected at (0, 0)

Thanks 🙏

Ted Shifrin

Stop using that horrid notation.

It looks like some basis element of a topology.

I know what the question is. But you're not showing an equation is 0 !!!!

Sourav Ghosh

@TedShifrin T stands for topology and s stands for sine curve.

:)

Ted Shifrin

You have to show two things: Setting the determinant equal to 0 gives you such a quadratic equation. And the three points satisfy the equation.

D.C. the III

18:20

Oh......

for a determinant to equal $0$ means one row is equal to another

Ted Shifrin

GRRR.

D.C. the III

or at least a scalar multiple

Ted Shifrin

GRRRR.

D.C. the III

Lol...I noticed the 4th R this time

Oh....are you asking to literally set up the matrix that would come from the equation of the parabola by plugging in each of the individual points?

Ted Shifrin

I don't follow you.

Why does that determinant give you the quadratic equation you're looking for? (No, do not explicitly write out a $4\times 4$ determinant. Yet you can tell.) Why does $(x_1,y_1)$ satisfy that equation, etc.?

D.C. the III

18:28

So to save typing it would be $y = ax^2 + bx + c$, but I would have $y_1 = ax_1^2 + bx_1 + c$, etc but it would be four rows, then I would express that correctly in the matrix form. .....I just wanted to express what I was thinking even though you have said different above

Sourav Ghosh

If X is locally connected then every connected component of X open in X.

The component {0}×[-1, 1] of Topologist's sine curve is not open.

So TSC isn't locally connected.

{0}×[-1, 1] is the disturbing component!

Ted Shifrin

@D.C. I continue not to understand what you're saying.

@SouravGhosh Yes, but the reason comes from all the other stuff.

@D.C. So how do I look at that determinant and see instantaneously that it will give me an appropriate quadratic equation?

D.C. the III

Give me a second I'm working through what you asked of me but on a simpler one in #19 to see if what I'm thinking is the right idea.

Seeing instantaneously that it gives a quadratic I don't see, but is the first column of the determinant considered "variables" even though the other three columns are realized values?

I ask becasue if we want to say the determinant is $0$ if we take wny of those three points given and put it into the first column, I will have two identical columns

Ted Shifrin

Aha, so that answers the second question ... immediately.

D.C. the III

So that is a valid observation?

Ted Shifrin

18:42

So how can I calculate the determinant and see the equation fall out?

Somewhere we need to use the fact that the three points are non-collinear.

D.C. the III

Hmmm... pondering the noncollinear idea

Ted Shifrin

Worry about that last.

D.C. the III

As far as calculating the determinant you did say no brute force, so my idea of expanding along the first row is void for this question

Ted Shifrin

Yup.

D.C. the III

Dare I say Cramer?

Sourav Ghosh

18:52

Initial topologies are easier than final topology :)

Ted Shifrin

What could Cramer have to do with this?

D.C. the III

STill need to calculate deteminants tho

I was tyring to "think out the box"

Sourav Ghosh

Subspace topology( generated by the inclusion map) ,product topology (generated by the family of projections)

Subspace topology( initial topology generated by the inclusion map) ,product topology (initial topology generated by the family of projections)

Ted Shifrin

I think learning linear algebra out of FIS has made you row-obsessed.

D.C. the III

Lol...Indeed...I essentially do give short thrift to columns and almost think of them not existing

Ted Shifrin

18:55

If you've read my book, it's all columns. So wake up and pay attention.

Sourav Ghosh

Yes. Columns are nice.

D.C. the III

I haven't read your Lin-Alg book, but this is good to keep me sharp about the "column side" of life

Sourav Ghosh

Ax := linear combination of columns

Ted Shifrin

I'm talking about this book, DC. All vectors are columns.

Sourav Ghosh

Quotient topology is a final topology.

Ted Shifrin

18:59

I mean, even this problem makes that obvious.

Sourav Ghosh

I didn't understand the quotient topology.

D.C. the III

Do you want the determinant calculated through using column operations?

Ted Shifrin

I'm saying no more, DC.

D.C. the III

by getting it in an upper triangular form

fiiinnnnnnnneeee

Ted Shifrin

rolls $8\pi^{\sqrt e}$ eyes

geocalc33

19:59

Are steinmetz solids useful in research mathematics?

D.C. the III

hmmmmmmmm.

D.C. the III

20:18

Oh.... If I were to go through with calculating the determinant down the first column, Each cofactor is going to be a "number", so that times the respective entry from the first column will give me my equation. So in the case of this determinant it would be $1 c_{1,1} - x c_{2,1} + x^2 c_{3,1} - y c_{4,1}$, where $c_{i,j}$ is just the respective cofactor.

leslie townes

love seeing 'number' in scare quotes, wanna see this more often

D.C. the III

Then with some tidying up by isolating $y$ and dividing $y$ by $c_{4,1}$, I'll have the equation. Is this the reasoning you were looking for @TedShifrin

Ted Shifrin

Yes. Just have to justify why that coeff is nonzero.

And to see parabola, why is the coeff of $x^2$ nonzero?

D.C. the III

The coeff would be nonzero due to the noncollinearity

That was in response to the first question

Ted Shifrin

Guess again.

leslie townes

20:41

btphbpthbt

Ted Shifrin

Hi to Munchkin.

D.C. the III

I was looking at a triangle through $(0,0), (1,0), (0,1)$ to see the idea behind to justify the nonzero coefficient. I don't know how to phrase it properly, but if something such as the three noncollinear points I just stated were used, I wouldn't have a "proper" parabola

copper.hat

@user85795 thanks for asking! just getting over it now! 3 short europe trips in 6 mos is too much for me.

Ted Shifrin

That is needed for the coeff of $x^2$. Figure out why. The coeff of $y$ should be something you did in both FIS and my exercises.

At least no covid, copper!

copper.hat

@TedShifrin i'm a bit astounded i never got covid. both of my kids were vaxed yet got coide.

leslie townes

20:56

fuses to london, money to cork, transistors to belfast...

copper.hat

no longer needed it seems.

D.C. the III

@copper.hat Living the life of an Instagram Influencer

copper.hat

in fact, when i was returning last week through dublin where they have us preimmigration, the tsa green lighted me and waived my security check!

i have entered the real of true uselessness

Ted Shifrin

@copper.hat I finally got it two months ago.

copper.hat

@D.C.theIII my only real post on instagram (copper.hat surprisingly) is of a dung beetle

@TedShifrin sorry to hear it. i hope recovery was swift & complete

Ted Shifrin

20:59

Only A couple of slightly bad days, but more tired for quite a while,

copper.hat

i can imagine.

i do have a pic of broom bridge (quaternions) and an old one with my daughter on ig

D.C. the III

TEd I would've only done it in your book. FIS deal with quadratics in a special section in the inner product spaces chapter. One of the "starred" sections that I intend to return to........eventually

Ted Shifrin

No, the coeff of $y$ is a very famous determinant.

In every linear alg book.

D.C. the III

hmmm.

Do you rememebr where in your exercises I would've done it?

Ohhhhh....... Vandermode?

Ted Shifrin

Interesting, it appears several times in my plain lin alg book, but maybe not in this book.

Yes.

D.C. the III

21:05

Yes....I did the exercise looong ago.....clearly forgot it cause it fell under the "if I see this in the future I will have to refresh, until then I'll probably forget it"

Ted Shifrin

Not in this book, other than in this problem. Just too much stuff to cover.

D.C. the III

I don't thinki you would ask about Vandermode in this mbook. I may be wrong

Ted Shifrin

Nice interpretation with linear maps, rather than just grunging determinants.

D.C. the III

I do appreciate you placing this in here and unknowingly forcing me to recall old mateiral from previous subjects

Ted Shifrin

I should have put the conceptual version as an exercise in this section. My error.

D.C. the III

21:08

You probably thought about it and just left that concept up to the student to eventually encounter when they do more advanced work. Since it isn't a mainstream thing.

Ted Shifrin

Consider the linear map from the vector space of polys of degree $\le n$ to $\Bbb R^{n+1}$ given by evaluating at $n+1$ points. What is the matrix with respect to the obvious bases?

Now deduce the Vandermonde det is nonzero.

D.C. the III

Yea. I'm looking at the exercise now from FIS and going back to my notes where I did it

Ted Shifrin

The usual thing is inductive grunge with determinants and getting a formula. That too is useful, but all we need here is nonzero.

D.C. the III

Oh I was sneaky in showing the Vandermode determinant was nonzero. I used the fact that the linear map I had defined was an isomorphism and as such had an inverse.

Ted Shifrin

Read what I’ve been typing here.

leslie townes

21:25

there's two Ns in vandermonde

that's probably not what ted meant by 'read what i've been typing here,' but as the licensing representative of the vandermonde estate, i have to say this

Ted Shifrin

Dutch for worldly, I imagine.

leslie townes

21:39

the vandermode determinant is a regular determinant with a scoop of ice cream on top

geocalc33

Does Li(x)-f(x)=O(log x) essentially mean that the error grows on the order of log x?

Yes

leslie townes

h(x) = O(log x) usually means that there is some number M for which |h(x)| <= M log(x) holds for all x sufficiently large

it means no more, and no less than that

Ted Shifrin

It can be way less! 🤷‍♂️👹

leslie townes

for example, 0 = O(log x) but 0 does not "grow on the order of log(x)"

Ted Shifrin

You need to write $\sim$ to get what you want.

leslie townes

21:48

sometimes people talk about O( ) as though the thing inside is chosen 'best possible', but making useful sense of the thing i just wrote in quotes is not really meaningful, and that is not how the notation is defined

people on MSE/SE also ask stuff like "what is O( ) of [some function or algorithm]" which is not even a well-posed question, but could become one in a multiple choice environment

Ted Shifrin

Unfortunately, leslie and Ted agree yet again.

geocalc33

correction Li(x)-f(x) \sim log x

leslie townes

so while it is sadly quite standard to speak about O( ) as if it is more than it is, i would caution you against doing so

geocalc33

$\mathrm{Li}(x)-f(x) \sim \log x$

leslie townes

even with \sim (however you define that), we had something a while ago where some confusion resulted from someone thinking f(x) ~ g(x) allowed them to replace f(x) with g(x) anywhere in any subexpression without changing the limiting behavior at infinity

Ted Shifrin

21:50

Huh?

leslie townes

so, just be careful, and resist the impulse to boil down asymptotic descriptions to "oh, all this means is X"

ted: if you weren't there for it, think something like: x^2 + x ~ x^2 as x goes to infinity, but sqrt(x^2 + x) - x and sqrt(x^2) - x have different limits as x goes to infinity

D.C. the III

The linear map we are using is $T(f) = (f(c_0), f(c_1), \dots, f(c_n))$. So using the standard basis for polynomials of degree $\leq n$. I have as a matrix:

$$
\begin{pmatrix}
1 & c_1 & c_1^2 & \cdots & c_1^n \\
1 & c_2 & c_2^2 & \cdots & c_2^n \\
\vdots & \vdots & \ddots & \vdots \\
1 & c_n & c_n^2 \cdots & c_n^n
\end{pmatrix}
$$

Sorry I had to take a second to reacquaint myself with the Vandermonde ideas.

Ted Shifrin

@leslietownes Oh, right. Well, a symbol other than $=$ won’t be perfect,

You still haven’t read what I wrote above. This is the proof I suggested.

And the mapping is an iso because?

geocalc33

$\mathrm{Li}(x) \sim \pi(x)$ meaning the differences tend to increase as $x \to \infty$ but of course due to Skewe's number there are infinitely many times where $\mathrm{Li}(x)-\pi(x)$ changes sign. How should I think about this? $\mathrm{Li}(x)$ is getting further away and then at skewe's number just randomly spikes downward toward $\pi(x)$?

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