Mathematics - 2022-03-19

Ted Shifrin

00:07

The whole challenge in this subject is figuring out how to look at a problem in a reasonable and correct manner.

leslie townes

ted i'm glad that we share ignorance of the pseudocircle.

i think as a bottom line, for me, i need some connection to something. that at least means being a manifold, or something that might act on a manifold. satisfying the axioms for a topology is not enough.

somethign that might act on an infinite dimensional hilbert space is also OK, as a connection to something.

Koro

04:49

Prove the existence a splitting field for non constant f(x) over field F.
Proof: Let deg f(x)=n>0. If f(x) splits in F then we are done. So suppose that f(x) doesn't split in F. By fundamental theorem of field theory, there is an extension field $F_1$ of F in which f(x) has a zero. If f(x) splits in $F_1$ then there exist $a_i$'s in $F_1$ and $a$ in F such that $f(x)=a(x-a_1)...(x-a_n)$ and F$(a_1,a_2,...,a_n)$ is a splitting field for f(x) over F.

If $f(x)$ doesn't split in $F_1$ then we write $f(x)=(x-a_1)g(x)$, where $a_1$ is zero of f(x) in $F_1$. By FTFT, g(x) has a zero in an extension field $F_2$ of $F_1$. If f(x) splits in $F_2$ then we are done and if not we repeat this process at most $n$ time.

The finite collection of zeroes $A_1, A_2,...,A_n$ obtained at each step in the worst case when $n$ steps are executed gives a splitting field $F(A_1,...,A_n)$.

$F(a,b,c)$ stands for the intersection of all extension fields of F that contain a,b and c.

The proof doesn't look right, I think.

I think the steps are correct but not in order.

Alessandro Codenotti

09:04

@leslietownes Fun fact there are pseudo-circles in continuum theory as well

It was considered a good candidate as a fourth homogeneous planar continuum (after $S^1$, the pseudoarc and the circle of pseudoarcs) until Rogers showed in the 60s or 70s that it is not homogeneous

(Now we know that the only homogeneous planar continua are the three mentioned above thanks to the 2018 classification result of Hoehn and Oversteegen)

Matheus Sousa

11:21

Hi guys, how are you?

Please, can someone say why I need to use $e^{ln}$ tip for this limit? $\lim_{n \rightarrow \infty} (1 - 2\sin{\frac{1}{n}})^{3n}$

Leaky Nun

@MatheusSousa because it's the $1^\infty$ type

Matheus Sousa

Ok, but I don't understand the problem of $1^\infty$. For me, independant of what factor is in the power, $1^n$ will be always 1.

Jakobian

11:48

@AlessandroCodenotti oh, that's super interesting

Do you mean non-trivial continua?

Alessandro Codenotti

12:21

Yes, sorry, forgot to mention it since it is always assumed, singletons are not very interesting after all!

@Jakobian yes it was a big result, people had been thinking about it for decades

Hoehn and Oversteegen also classified the planar continua homeomorphic to all of their nontrivial subcontinua in 2020 (turns out that those are only the arc and the pseudoarc)

Silvinha

14:05

ok. I understood thanks for your attention.
Do you believe that the only refund I got after I posted this question here was a downvote? :( Sad, because I worked for more than a week in that question and in the answer I've provided.

copper.hat

14:24

Random stuff happens.

robjohn

@MatheusSousa You can use this answer

copper.hat

Good morning :-).

robjohn

@MatheusSousa It's because the $1$ is not actually $1$, just close to $1$ in those limits.

@copper.hat good morning!

@copper.hat: I am talking to ghosts again ;-)

afk for a while. BBL

Silvinha

@copper.hat Good Morning! :)

copper.hat

Just eating brekkie before a morning cycle :-)

Koro

15:06

The splitting field of $x^4+1$ over Q is $Q(\sqrt{i})$, where $i=\sqrt {-1}$.

leslie townes

you sure about that?

oh, i see, the sqrt. serves me right for not rendering chatjax.

it might help to actually see what that number is. simplifies the expression a little. none of this sqrt( ) sqrt( )

i think it could be rewritten as Q(i, sqrt(2))

Koro

$x^4+1=(x^2)^2-i^2=(x^2-i)(x^2+i)= (x\pm \sqrt i)(x\pm\sqrt i^3)$.

leslie townes

i cringe when i see \sqrt{ } on complex numbers, but you do you :D

it also factors over R as a product of two quadratics, of course

Koro

that’s how I concluded $Q(\sqrt i, \sqrt i^3)= Q(\sqrt i)$.

@leslietownes I’ll try write that in some different form.

leslie townes

if you wanna do the whole galois correspondence it helps to see what the number really is

Koro

15:19

$Q(\frac {1+i}{\sqrt 2})$

leslie townes

because of the primitive element theorem you can always write a finite extension of Q as Q(something), but seeing what the 'something' is can tease out information about intermediate subfields

robjohn

17:03

@Koro $e^{i\pi/4}=\frac{1+i}{\sqrt2}$

copper.hat

Beautiful cycling day, no hot sun to cope with.

leslie townes

my daughter experimented with picking the cat up again today. i have no idea why the cat allows this.

she picked the cat up and carried her into another room.

Koro

@robjohn yeah, that’s how I concluded this chat.stackexchange.com/transcript/message/60685015#60685015

@copper.hat Cycling is fun :-)

robjohn

@Koro I was just giving as many forms as I could at the time.

@leslietownes Practice for another stair run

leslie townes

you weren't kidding, she just tried to do that.

you must have ESP.

robjohn

17:15

how did it go?

leslie townes

olivia jumped down and ran away.

robjohn

same as last time. Cats are not stupid

leslie townes

i'm surprised that she doesn't attack. she can do this. she responds to aggressive petting with tooth and claw.

but is apparently fine with being picked up in a very uncomfortable way, providing no support whatsoever for her lower half.

Koro

One day a cat jumped on my head from a 2 storey house and as soon as I could do anything, the cat meowed and ran away.

The cat probably fell by accident.

robjohn

@Koro It's scary falling from a 2 story building when you are a foot high

leslie townes

17:18

i read somewhere that larger heights are better for cats than smaller ones. more time to adjust to the drop.

Koro

I too read that somewhere, there leg muscles are okay with such jumps.

robjohn

That's true, especially if they are not oriented properly

leslie townes

olivia sometimes jumps onto a ledge that looks down into our living room. it is a 15+ foot drop. i don't like it when she's on the ledge.

Koro

Leslie, does olivia respond to the call olivia?

PM 2Ring

High-rise syndrome

High-rise syndrome is a veterinary term for injuries sustained by a cat falling from a building, typically higher than two stories (7–9 m (23–30 ft)). == Injuries sustained by cats falling == Common injuries sustained in cats after a fall include: Broken bones, most likely the jawbone as the cat's chin hits the ground; a broken jawbone and shattered teeth are the classic signs of a cat having sustained injuries in a fall. Injuries to the legs: joint injury; ruptured tendons; ligament injury; broken legs. Internal injuries, especially to the lungsStudies done of cats that have fallen from 2 to...

In a study performed in 1987 it was reported that cats who fall from less than six stories, and are still alive, have greater injuries than cats who fall from higher than six stories.[6][7]

It has been proposed that this might happen because cats reach terminal velocity after righting themselves (see below) at about five stories, and after this point they are no longer accelerating, which causes them to relax, leading to less severe injuries than in cats who have fallen from less than six stories.

en.wikipedia.org/wiki/Cat_righting_reflex says "One cat survived a fall of 46 stories and landed with no injuries at all".

leslie townes

19:00

koro: yes. or "livvy"

Ted Shifrin

Wordle in 2 today :)

Jakobian

19:17

@PM2Ring "In a more recent study, it was observed that cats falling from higher places suffered more severe injuries than those experiencing shorter drops."

leslie townes

ted: i got it in 3. rare for me to dip below 4.

i'm assuming there are selection effects at work. most of the time, if an animal falls and dies it doesn't become a data point

Alessandro Codenotti

Took me 4 attemps today, I tried alloy as the third :/

Matías Matteini

In all the answers that I fount related to submersion of compact manifold in $R^{n}$ they always conclude using connection of $R^{n}$ (ie math.stackexchange.com/questions/1995022/…) Now, while of course it does not change much, would it be wrong arguing that f(M) must be compact and open but this is against Heine Borel?

leslie townes

19:34

matias: could you go one step further there? how is it contrary to heine borel? i do agree that using connectedness of R^n is maybe the most natural next step.

Matías Matteini

yeah that was dumb lol you are right

saying that being compact means closed and limited is a contraddiction to being open and compact only if i use that the only clopen set in Rn is Rn itself

leslie townes

that was my thinking, yeah. :)

Matías Matteini

Thank you

It's just that Heine Borel is such a way of life that as soon as i saw open and compact it had to be "him" haha

leslie townes

haha, i like that. Heine Borel is a way of life.

it is.

robjohn

20:00

@Ted: I don't know if you like geometry/trig problems, but I answered one, then noted that it was a PSQ, so I deleted my answer. It is a cyclic generalization of the Law of Tangents. It is sort of like Brahmagupta's Formula as a cyclic generalization of Heron's Formula.

Ted Shifrin

20:35

@robjohn Definitely not my expertise!

amWhy

@leslie Who is the orange user? He looks mean!

leslie townes

he looks like a mean square.

gosh, st patricks come and gone. i sorta forgot about it.

Ted Shifrin

20:53

How would you be mean in the $L^3$ norm?

leslie townes

someone asked a question on MSE a while ago about the L^4 norm. it was bananas.

like, nobody does anything in L^4. we interpolate through L^4, we don't stay there.

robjohn

21:10

@TedShifrin mean cube? In $L^4$, would it be the mean tesseract?

Koro

21:28

x-axis is closed in $R^2$
Proof: f:R->R defined as f(x)=0 is continuous and $f^{-1}(0)=x-$ axis and since {0} is a closed set so is x-axis.

to be more precise: f should be defined on {(a,0): a in R} such that (a,0) is mapped to (0,0).

Ted Shifrin

@robjohn in infinite dimensions, the mean Hilbert?

Alessandro Codenotti

f should be defined on R^2

Koro

not on {(a,0): a in R^2}?

leslie townes

what alessandro said

Ted Shifrin

I hate to agree …

Koro

21:33

But then $f^{-1}{(0,0)}=R^2$ as per my def. of f?

what is wrong with {(a,0): a in R}?

leslie townes

if f is a real valued function, there is no f^{-1} {(0,0)}.

i think the usual candidate here would be f from R^2 to R given by (x,y) mapsto x and then you look at the inverse image of {0}.

Koro

f is not real valued. f maps (a,0) to (0,0).

{(0,0)} is a closed set so f^{-1}{(0,0)} is closed.

Ted Shifrin

@leslietownes You mean $y$.

Koro

no, {(a,0): a in R}=S is not correct because the above shows that S is closed in S, which is trivial.

Alessandro Codenotti

So what does your function do on the rest of R^2?

leslie townes

21:42

i do mean y.

robjohn

@TedShifrin and I'll run a string of hotels

Koro

@leslietownes you're right as always. (x,y) to y. With this, now I have every n-1 dimensional subspace of $R^n$ is no-where dense.

:-)

@AlessandroCodenotti I realized the error in my argument. Thank you!

Ted Shifrin

@robjohn Mean rip-off?

robjohn

@TedShifrin Precisely, the Motel $6^\infty$

If they don't look close enough, it looks just like $6^{00}$ (you can check out any time you like, but you can never leave)

copper.hat

Tom Bodett

grinding through my depressing taxes

robjohn

21:54

He'll leave the light on for you

copper.hat

I stayed in one at Winemmuca

reddit.com/r/funny/comments/thx8hv/…

robjohn

@copper.hat I'd hate to get burned and pay full price

copper.hat

The price of enlightenment

Ted Shifrin

22:20

We have left enlightenment far in the dust!

robjohn

23:03

. o O ( dusty enlightenment, at least it's left enlightenment )

copper.hat

23:30

that's right

leslie townes

23:49

well, it's official. my daughter has carried the cat down the stairs.

copper.hat

baby steps

Ted Shifrin

@leslietownes Did Olivia survive to tell a tail?

leslie townes

yes. she seemed fine with it.

that's the part i don't understand about this. she could bring all of this experimentation to a stop very quickly.

Ted Shifrin

Screech would not be patient.

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$Mathematics$ Mathematics