Mathematics - 2023-02-24

冥王 Hades

01:31

Dead as a morgue

leslie townes

pbhbht

冥王 Hades

We should charge money from people from other SE sites visiting Mathematics chat

And then throw a pizza party

Sine of the Time

@冥王Hades seems a good idea

冥王 Hades

02:29

@SineoftheTime "authentic" Italian Pizza

Jack Rod

03:07

@robjohn chat room sir

Romm where I used to ask question shall I share the link of the room?

copper.hat

03:39

just had a nice deep dish pizza

The Real Masochist

08:04

Aha, n-multisubsets of x is the answer.

robjohn

15:47

@JackRod that was already unfrozen.

@冥王Hades ?

leslie townes

16:48

when i see a solar eclipse, i think, is the moon eating the sun? i don't know. i'm just a caveman. but there's one thing i do know...

冥王 Hades

17:08

@robjohn what's up?

Xander Henderson

17:25

@leslietownes NOT GUILTY!

(Man, Phil Hartman was so very good... I miss him.)

leslie townes

same.

robjohn

17:44

@JackRod this was the most recently used inactive room. It was reactivated over a day ago.

@leslietownes Wait... it's not a dragon? That's what I was told.

Koro

17:56

An example of an algebraic extension that is not a finite extension

leslie townes

the field of all algebraic numbers would be one example.

Koro

Consider $\mathbb Q$ and its extension $\mathbb C$. Let $K$ be subfield of $\mathbb C$ consisting of all algebraic nos. over $\mathbb Q$.

leslie townes

can probably find smaller stuff with, i dunno, adjoin every nth root of 2 to Q.

Koro

@leslietownes ohh, same example :-).

leslie townes

or all square roots of primes. proving those things are at least linearly, if not algebraically independent, is a common exercise.

but yeah, A:Q is the example i think most people would think of first.

Koro

18:02

Then, $[K:Q]$ is not finite: suppose on the contrary that it is n. Consider f(x)= x^{n+1}-3 in Q[x], which is irreducible by Eisenstein's criterion. Let a be a root of f(x) in C, then a is algebraic. [Q(a): Q] =n+1. Moreover, [K:Q]=[K:Q(a)] (n+1), whence we get [K:Q(a)]= n/(n+1), which is disaster.

@leslietownes Ohh

I was wondering if the following is true? Suppose that K is a finite extension of F. Then there exists a in K such that K=F(a).

If it were true, then Q(√2,√3)= Q(p) for some p in Q(√2,√3).

Consequence would be that we'll be then able to speak of 'minimal polynomial of p' over Q.

leslie townes

Primitive element theorem

In field theory, the primitive element theorem is a result characterizing the finite degree field extensions that can be generated by a single element. Such a generating element is called a primitive element of the field extension, and the extension is called a simple extension in this case. The theorem states that a finite extension is simple if and only if there are only finitely many intermediate fields. An older result, also often called "primitive element theorem", states that every finite separable extension is simple; it can be seen as a consequence of the former theorem. These theorems...

the wikipedia page even discusses the example of Q(sqrt(2), sqrt(3)) :)

Koro

ahh, so these are called simple extensions.

leslie townes

yeah, so buried within that wiki entry is the fact that a finite extension of a field of characteristic 0 (e.g. Q or R) is simple

robjohn

@leslietownes so, complex is simple

insipidintegrator

18:17

hi, can someone help me with a hint: consider a point P in a triangle ABC. Let the feet of perpendiculars from P on AB, BC, CA be M, N, Q. If PM^2-AM.MB=PN^2-BN.NC=PQ^2-CQ.QA then prove/disprove that P is the orthocenter of ABC.

I actually need it as a lemma for solving a math.se problem

anak

I have a homogeneous DE of the form $\dot v^2 \dddot v - 2\ddot v\dot v + A\dot v^5 - B\ddot v^2 = 0$...does anyone know of a technique for solving something like this?

leslie townes

robjohn: galaxy brain

robjohn

lots of black holes...

Koro

and theoretic white holes

robjohn

spewing out unformed ideas?

Koro

18:21

yeah, weird stuff happening

robjohn

Be glad I didn't mention the civilized element theorem ;-)

user 726941

Would that've been an element of western civilization.

Koro

footfall in chat has reduced over last few days.

user 726941

18:39

As in, talk about soccer? ⚽

🙊🙉🙈

anak

🏈🏈🏈

user 726941

🙏

March madness will soon be upon us, grasshopper.

Koro

Keith Conrad's articles are so amazing :-).

Hi @TedShifrin! :-)

anak

They are excellent for basing tutorials around, @Koro

leslie townes

ted, short time no see but hi.

user 726941

18:43

👋👋👋

Ted Shifrin

Hi Koro, anak

Howdy.

anak

Hi Ted.

Koro

@anak I was having so much difficulty in understanding why topological sine curve is not path connected until I saw Keith Conrad's

Ted Shifrin

How can a path leap? 🤷‍♂️

anak

With a bridge.

Ted Shifrin

18:45

Not stilts?

anak

I've never tried jumping with stilts, but it might be difficult.

Ted Shifrin

Just put one stilt where you need to end up?

anak

From this point forward, all functions will be outfitted with a pair of stilts.

Ted Shifrin

Might be a convenient fad.

Actually, it’s the oscillation with non-shrinking amplitude that’s the killer. Stilts could help there, too.

冥王 Hades

I tried speaking Latin and accidentally summoned an ancient demon instead

Ted Shifrin

19:00

You summoned me?

anak

I thought that was Demonark

Ted Shifrin

Demonark has been scarce.

anak

Hope he is well.

user 726941

Perhaps he changed his username again.

anak

speaking of people who should change their _user_name.

stares @ user 726941.

user 726941

19:13

:(

anak

It's okay. We love you anyway.

user 726941

👍

Ted Shifrin

I can’t tell one user from another.

anak

This one makes emotive use of emojis, so they are highly approved.

Sha Vuklia

heyo Ted, just wanted to share that I passed my diffgeo exam <xD

anak

19:17

Congrats!

Sha Vuklia

thanks:D

user 726941

🎇🎇🎇

^celebration fireworks 🎆

Ted Shifrin

Congratulations, Sha. I’m far from surprised.

Sha Vuklia

@user726941 so kind, thanks:D

leslie townes

differential geometry is the wrong path. it leads to becoming ted shifrin.

there's still a chance to turn away. it's never too late to change your ways.

user 726941

19:22

Come back to algebra?

Sha Vuklia

@TedShifrin let's say it wasn't my first try :') we had lectures notes that weren't always very complete/detailed/clear, so I had to dive into alternative literature because I was unable to connect the dots based on these notes, which made the whole process a whole less efficient - but hey, in the end I got there

and it also didn't help that I had never grasped basic smooth manifold theory during my undergraduate years, so it was a looong journey x)

Ted Shifrin

You’re ahead of most of the denizens of chat.

leslie townes

i am?! thanks, ted!

Ted Shifrin

Typical lawyer. Takes after Rudy.

Koro

19:40

algebraic top. teacher here is like: how much time is left for the class to end? '5 mins', say students. Then he says: ok, we can do 1 more theorem. Then he writes the theorem and proof, then leaves.

'But how does that step follow?', ask students. He says, 'I have left some steps here and there, complete them on your own.'

leslie townes

i'm split between being annoyed by that, and wanting a job at your university

Koro

so people skip his classes.

shintuku

he did it

he found the loophole

leslie townes

👑👑👑

Koro

we have midsem and the syllabus for alg. top. is -Chain complexes, CW complex, fundamental group, homotopy, s -delta sets etc.

Quotient topology was removed from the midsem because then people would score marks.

I shared one image the other day here, wherein he had cut my marks in a test. He was telling me that mine had a mistake. I told him that he was wrong and that my answer was correct. He realized his mistake and said-oh marks don't matter, understanding matters etc.

😅

user2236

19:48

Did you get the marks back?

Koro

no

user2236

Then he needs to understand his mistake.

Koro

probably because then he would have to have the trouble of changing the marks in his databse.

(too much task to power on laptop and do it, ughh leave it)

shintuku

or maybe office politics

but that sucks

why wouldn't he give you your marks back

Koro

because 'marks don't matter, understanding matters'

😅

user2236

19:52

probably just lazy.

Koro

yeah, or full of ....

user2236

that too

shintuku

key his car

i didn't mean that

Koro

I don't understand how one becomes like that: don't they introspect what we did today?

user2236

nah, just live and learn

Koro

19:54

Did my teaching help anyone?

I guess they don't ask these questions to themself.

user2236

don't you have a rate my professor site?

Koro

unfortunately, no.

and there is no feedback system also here.

user2236

start one

leslie townes

yeah, i need to get a job at this place

user2236

be the change you want to see

Koro

19:59

The college where I did my UG from had a feedback form system: at the end of a semester, feedback forms will be distributed among students who will then fill it up and give to the teacher. It's anonymous. Then the teacher submits it to the administration.

shintuku

@user2236 -lenin

Koro

last semester, one of the teachers who taught us is said be 'one of the best teachers' in the country. Is it strange that I politely disagree with that?

I feel that if I just listened to what he said in the class, I won't be able to even touch an exercise problem in the book.

@leslietownes come here Leslie. You will be a great help here. :-)

Xander Henderson

@Koro I strongly disbelieve in the notion of "the" best teacher, or any objective measure of "best".

What works for some people will fail utterly for others.

And the people that I, personally, feel are the best instructors are also often the most polarizing.

Koro

indeed. But I think it's like -'best actor award, best song award', not really welldefined.

Xander Henderson

@Koro Well, the awards are, in fact, very well defined (in general).

You get a bunch of nominations, and there is a vote of some sort.

But whether or not that relates to any "objective" measure of talent, skill, or ability is anyone's guess.

D.C. the III

21:01

I just did a first pass on reading on Jordan Canonical Forms......seems OK, but I need clarification on a concept. Anybody available?

We have this matrix and after doing all the necessary work we get to a basis of the solution space for $K_{\lambda_2}$. In the book they chose the vector $(-1,2,0)$ to begin their cycle. I tried using the vector $(1,-3,-1)$ initially. I was expecting to be able to get the vector $(-1,2,0)$ after I multiplied by the correct matrix, but I didn't.

Isn't it the case that I should be able to get every vector in that solution space by cycling through it?

When I did it the other way and started with $(-1,2,0)$ I got the second vector, but it didn't work the other way around

Astyx

21:54

One thing to notice is that solutions to $(A-2I)v$ are solutions to $(A-2I)^2v$

i.e. you have $E_{\lambda_2}\subset K_{\lambda_2}$

If you have a basis of the former you can complete it into a basis of the latter

Since $\dim E_{\lambda_2} =1$, it is sufficient to find one nonzero vector in this space to get a basis

To do this, one thing you may do is pick a vector of $K_{\lambda_2}$ and apply $A-2I$ to it – if you're lucky you will find a nonzero vector, which has to be in $E_{\lambda_2}$

D.C. the III

So even though they didn't show the computation, then $(-1,2,0)$ was an eigenvector?

and that is what allowed for the creation of the "cycle" per se?

Astyx

If you do this with $(1,-3,-1)$ it works. If you do it with $(0,-1,-1)$ it works. If you do it with $(-1,2,0)$ it does not work.

Here you can take any vector in $K_{\lambda_2}$ that is not colinear to $(-1, 2, 0)$ and it will work (can you prove that?)

Right, they did not actually do the computation, they just claimed it works (and it does)

D.C. the III

Yea. As I finished the section they said the next section will show the ways to compute these things because I one o fthe things I was wracking my head about was how they were getting some of these results

So when they solved for the basis of the solution space and got the vectors $(1,-3,-1), (-1,2,0)$, they "found" the acceptable vector to be $(-1,2,0)$. I take it won't always be the case that a basis of such a homogenous system will provide me with a cycle of generalized eigenvectors?

Astyx

22:18

It depends what you mean

if your generalized eigenspaces staired well, you will always be able to find such a thing

more precisely, here $\dim E_{\lambda_2} = \dim K_{\lambda_2} - \dim E_{\lambda_2} = 1$

if there were three independent vectors in K and only one in E, then you should be able to convince yourself that there is never a cycle which yields a basis

however if it is straired (I don't know the actual word) nicely, then you alway will be, by taking one in the upmost space that is not in the one below it

D.C. the III

Yea, they showed that idea so it makes sense to me.

Hmmm. I'll pause the rest of my inquiries until a little later when I've read the next section but I do have a bit of better understanding. Thanks for the help

Astyx

Glad to help

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$Mathematics$ Mathematics