Mathematics - 2020-05-02

geocalc33

00:04

took an L today friends :)

my answer got downvoted :/. oh well

BAYMAX

00:21

@user714630 hmm thank you for looking

I think somewhere the question is incomplete as I am given only that much info.

Real Dumb foggy brain

01:08

Does Lie group theory differs from group theory?

What is pigeonhole theory btw.

Pigeonhole Principle. Can anyone explain?

Knight

01:25

@robjohn Can you please explain how you got that?

@Thorgott Would you please help me?

KReiser

01:39

Here's a (potentially silly) question: must a local homomorphism of noetherian local rings $A\to B$ which is finite type also make $B$ a finite $A$-module? It seems like either there should be a well-known proof or a well-known counterexample, but my google skills (and my head) are coming up empty right now.

BAYMAX

04:07

Hi
Let $F$ be a field and $K$ an extension of $F$. the element $a$ in $K$ is algebraic over $F$ if and only if $F (a)$ is a finite extension of $K(a)$
I think this is true??

KReiser

04:24

@BAYMAX Yes, this is true. Think about linear independence of the powers of $a$.

BAYMAX

Cool! thank u!

skullpatrol

04:55

Has the mean square taken up the tango? @robjohn

Balarka Sen

05:12

@KReiser I am not an expert, but can it be argued as follows? Suppose the local rings are $(A, \mathfrak{m}), (B, \mathfrak{n})$. If $B$ is a finite type $A$-algebra, $B/\mathfrak{n}$ is a finite type $A/\mathfrak{m}$-algebra. By Nullstellensatz field extensions which are finite type are finite, hence $A/\mathfrak{m}$ is a finitely generated $B/\mathfrak{n}$-module. By Nakayama's lemma, this basis gives rise to a generating set of $B$ as an $A$-module, not?

Ah, no, ignore me. I need to assume $B$ is a finitely generated $A$-module in the first place for Nakayama.

skullpatrol

::ignoring Balarka: A Geometric Approach::

Balarka Sen

05:27

Haha, there are actually two issues with what I am saying. (1) I am confusing $B/\mathfrak{n}$ with $B/\mathfrak{m}B$; there's no reason $\mathfrak{m}B = \mathfrak{n}$, unless the original map of local rings was unramified. (2) The form of Nakayama I want to apply is $M$ is an $A$-module and $M/\mathfrak{m}M$ is a finitely generated $A/\mathfrak{m}$-vector space, and to get that $M$ is a finitely generated $A$-module out of there. I think this is true if $A$ is Artinian

Maybe these will help to find a counterexample

Knight

@robjohn Hey Robbie! Are you here?

robjohn

05:46

@skullpatrol amWhy suggested I use the rose as my gravatar for MayDay. I didn't want to do that, but I figured I would Tango with it.

@Knight yes

Real Dumb foggy brain

youtu.be/GJDNkVDGM_s Just relax.

skullpatrol

@robjohn I see.

robjohn

@skullpatrol Of course, amWhy never even came to this chat since then.

skullpatrol

robjohn ♦, West Hills, CA

292k 29 351 686

see^ what you've done @amWhy

Knight

06:01

@robjohn Sir please see this

@skullpatrol Hi Pal!

skullpatrol

hi pal, how are you?

robjohn

@Knight what about that?

the answer looks good.

Knight

@robjohn Please see my comment

I wanted to know if the argument can itself be a function

@skullpatrol I’m fine pal, are you chilling? :-)

skullpatrol

@robjohn just chalk it up to one of your several unusual deviations: (with one standard deviation and several unusual ones)

@Knight yup

:-)

Knight

Pal! You know the word “quarantine” stems from Venetian usage which means “forty days”

skullpatrol

06:07

yeah, that's interesting

robjohn

@skullpatrol since that is a link to my profile, if I change my gravatar back, that image will change as well.

@Knight: what star is that in your gravatar?

skullpatrol

Knight

@robjohn I think it is some galaxy

Robbie did you reject someone’s proposal, that rose shows something

(Sorry for the joke sir)

robjohn

@Knight it's so small, it is only a point.

Knight

@robjohn It could be a star also :-)

robjohn

06:12

@Knight That is what I asked... what star is it?

amWhy

@skullpatrol I love it!!

Knight

@robjohn I don’t know sir, I just used it (stole it from bing)

I am Why

robjohn

@Knight ah... know what you steal!

5

Knight

@robjohn I will surely follow that advice from next time.

skullpatrol

@amWhy you love the tango?

Knight

06:14

Because there are chances I may steal artificial jewel from the original. I must know what I Rob

Robbie sir please help me with that question.

Balarka Sen

07:35

@KReiser @loch Consider $k[t^2]_{(t^2 -1)} \subset k[t]_{(t-1)}$ where $k$ is algebraically closed. This is finite type, because anything in the codomain can be written as $f(t)/\prod(t - c_i)$ where none of the $c_i$'s are $1$. Multiply top and bottom by $t + c_i$ whenever $c_i \neq -1$ to get something of the form $P(t)/(Q(t^2) \cdot (t + 1)^n)$ where $t^2-1$ does not divide $Q(t^2)$. So the latter can be obtained from the former by adjoining $t$ and $1/(t + 1)$

2

Doesn't look like the latter is finite over the former though. I think $t + 1$ in the latter ring is not integral over the former

Finite extensions are integral, so that can't happen

After my observation on the unramified thing I was trying $k[t^2]_{(t^2)} \subset k[t]_{(t)}$ for a long time but that's actually finite :)

KReiser

Turns out the answer to this problem is yes, any local ring homomorphism of noetherian local rings of finite type actually makes the target a finite module over the source.

Balarka Sen

What's wrong with my example though

I don't see an issue

KReiser

getting there! writing up the proof now

(sorry, any flat local morphism)

Let our rings be $(A,\mathfrak{m})$ and $(B,\mathfrak{n})$ with $f:A\to B$. Then $B\cong A[x_1,\cdots,x_n]/(f_1,\cdots,f_m)$, and base-changing by $A\to A/\mathfrak{m}$, we see that $B/\mathfrak{m}B\cong (A/\mathfrak{m})[x_1,\cdots,x_n]/(\overline{f_i})$. But the quotient of a local ring is local, so this ring has one maximal ideal. By Noether normalization, this means it's a finite module over $A/\mathfrak{m}$, and by fpqc descent, this means $B$ is module-finite over $A$.

Balarka Sen

Oh, sure, my thing isn't flat, I am sure

KReiser

That would be my first guess, but I need to take a second to read it

Balarka Sen

07:49

@KReiser This is cool! Can you explain why $B/\mathfrak{m}B$ is a finite $A/\mathfrak{m}$-module implies $B$ is a finite $A$-module? I wanted to do something like this earlier using Nakayama but thought I needed $\mathfrak{m}$ to be nilpotent in $A$ for this (aka $A$ is Artinian). I don't know fqpc descent though :)

I guess that is where you're using flatness

KReiser

Yeah, that's where flatness happens.

Oh crap did I verify faithfully flat? Uhh gotta check that

@BalarkaSen Anyways your example is a localization of $k[t^2,t^{-2}]\to k[t,t^{-1}]$, which is a finite flat map of rings (as long as characteristic isn't 2), I think.

@BalarkaSen Descent is basically the converse of base-change: if you have a base change square (arrows point down and right) with the bottom map nice enough and the left hand map with some property, then the right hand map must have that property sometimes too.

The reason it's the converse is that usually we make a conclusion about the left arrow from knowing something about the right arrow.

Balarka Sen

Hm, I don't quite see why it should be finite though, because I think integral closure of $k[t^2]_{(t^2-1)}$ in $k(t)$ is essentially germs of functions vanishing at both the preimages of $1$ by $t \mapsto t^2$, so $t+1$ is there in particular

KReiser

@BalarkaSen Did you take an integral closure? I must have missed that.

Balarka Sen

Oh no I am arguing that it can't be finite because finite maps are integral

@KReiser I see

KReiser

I mean, I think it's finite because we have a basis: $\{1,t\}$.

Sorry let me explain better: $\{1,t\}$ is a basis of $k[t,t^{-1}]$ as a $k[t^2,t^{-2}]$-module, so it's finite free and thus flat. These things are preserved by localization, which is what you've done, right?

Balarka Sen

08:02

Let me see if I follow. How are you realizing $k[t]_{(t-1)}$ as a localization of $k[t, t^{-1}]$?

Ok, just localize at $(t-1)$

Ok, so you're arguing it's flat. That seems right to me. I am still not convinced that $k[t^2]_{(t^2-1)} \to k[t]_{(t-1)}$ is finite though; there must be something wrong

KReiser

Oh dang it I applied fpqc descent wrong and my conclusion is wrong :(. I need to try again tomorrow on this when I've slept more.

@BalarkaSen Can you demonstrate an infinite collection of elements which are linearly independent?

Balarka Sen

Yeah, I think $\{1/(t+1)^n:n \in \Bbb N\}$ does the trick.

Note that $\{t, 1/(t+1)\}$ is a generating set for $k[t]_{(t-1)}$ as a $k[t^2]_{(t^2-1)}$-algebra.

KReiser

Huh, something is bugging me about this but I can't quite put my finger on it right now.

Balarka Sen

Alright, let me know if you see an issue with this.

KReiser

08:17

Sure, thanks for the discussion.

KReiser

08:40

@BalarkaSen I've convinced myself you're correct. We can write the map as a composition $k[t^2]_{(t^2-1)}\to S^{-1}k[t] \to k[t]_{(t-1)}$ where $S$ is the multiplicatively closed set of polynomials in $t^2$ not vanishing at $t=\pm1$. Then this first map gives the middle term as a finite free module over the first with basis $\{1,t\}$, and so it's finite type and flat. The second map comes from inverting $t+1$, which is flat and finite-type, so we win. Thanks again for the help!

Balarka Sen

09:18

@KReiser That's a concise way to say it! Thanks for the discussion!

I meant $1/(t+1)$ isn't an integral element throughout what I said, kept saying $t+1$ for some reason.

I guess scheme theoretically you'd say that Spec of that map isn't proper? It's like picking out a branch of the square root at $1$

IIRC proper + finite-type + finite fibers are exactly the finite morphisms, that's a theorem of Deligne or something

northerner

Any tips on how not to make simple mistakes or catch them quickly? I just spent half an hour making sure all the calculations were correct and thinking the problem had to do with how the modulus was taken. Turned out I just swapped the value of two variables.

Leaky Nun

09:44

@BalarkaSen help where do I learn about Euler class, Stiefel-Whitney class, Thom class, and Chern class

Alessandro Codenotti

You take a class on classes

EnjoysMath

0

Q: A generalization of homological chain maps, reference?

Let $R$ be an integral domain with $1$ that is multiplicatively generated by the set $G$ and $Q = \text{Frac}(R)$ its field of fractions. Suppose that there exists a group homomorphism $d : (Q, \cdot) \twoheadrightarrow (R, +)$ from the multiplicative structure of the field onto the additive str...

abstract-algebra prime-numbers homology-cohomology homological-algebra rational-numbers

CosmicChalk

10:22

So what do we discuss here?

Can anyone explain me how to intutively understand countable and uncountable sets or infinity?

Mike Miller

10:43

A countable set is more or less like a sequence. If your set is countable, you can list off the elements --- $x_1, x_2$, and so on --- so that for any element of your set, it gets listed off in finite time. (It won't take forever to see it, it might just take a long time.)

An uncountable set is a set that's so large you can't list off the elements.

It's sort of famously hard to see that uncountable sets exist; the argument is a little tricky.

Real Dumb foggy brain

11:06

Guys how do I read full description of chat. I use android tablet.

@CosmicChalk Well a set is called countable if it's either countable infinite or is finite. Else it's uncountable.

A set S is said to be Countably Infinite if there exists a bijection f:N→S.

I think it's easy to understand it intutively.

I am just repeating what Mike Miller said 😂😂😂.

robjohn

11:25

@amWhy back to the same old, boring, mean square.

Knight

Robbie do you help with physics also?

@Astyx What is mass balance ?

Astyx

Something that measures mass ?

Knight

11:42

@Astyx But this other part of the question

If mass balance is just a sale of some sort then when ball hits it should show 3.35 kg but it’s showing something else.

Is it due to the impact of the ball on it?

Balarka Sen

11:57

@LeakyNun I barely know these things

Edward Evans

Just a yes/no question; are all norms on a finite dimensional space equivalent to the $\ell^1$ norm?

Thorgott

yes

Edward Evans

The hint given in my exercise was to consider an appropriate $\ell^p$ norm and compare an arbitrary norm to that guy

okay nice, I'll try and show that then

seems to work with triangle inequality and a normalised basis, at least for one direction

Thanks @Thorgott :P

Alessandro Codenotti

Hint: the unit ball is compact if the space is finite dim

Thorgott

heh

that's the fancy way

robjohn

12:12

@Knight It depends on exactly how the "mass balance" works. Most scales actually measure force, not mass. Force times time is momentum, and the added $6.7$ kg at $9.81\text{ m}/\text{s}^2$ times $90\text{ ms}$ would provide $6.7\cdot9.81\cdot0.09/0.35=16.9\ \text{m}/\text{s}$. This means to me that the ball bounced back at $6.9\text{ m}$/$s$

Thorgott

actually, how do you get compactness of the unit ball without equivalence of norms?

hmm, you need continuity of linear maps, I guess that can be done by computation with transformation matrices explicitly

Alessandro Codenotti

@Thorgott By Banach-Alaoglu since finite dim spaces are reflexive and isomorphic to their dual, obviously

Jokes apart I don't think you need the equivalence of norms things to show that finite dim iff the unit ball is compact

Edward Evans

@Alessandro Functional analysis is nice atm, although all we're doing is building up structure from topological spaces up to Hilbert spaces

and getting a feel for how they work

Thorgott

I don't know how "if" goes, actually, but for "only if", I think you want to know it's homeomorphic to $\mathbb{R}^n$ and use Heine-Borel

so you want to know that linear maps from finite dim spaces are continuous, which follow from equivalence of norms by looking at the graph norm

but I think you can just show it directly too

Alessandro Codenotti

I'm pretty sure we showed that linear maps are continuous explicitely in my first analysis course, but that was too long ago

@EdwardEvans nice

Edward Evans

12:24

Algebra 2 has just descended into masses of diagram chasing

the proof of the snake lemma was truly vile

Alessandro Codenotti

Oh that's one of the proofs that you should do once and never again

Edward Evans

yeah I should hope so

Thorgott

ok yeah, something like this is what I had in mind math.stackexchange.com/questions/112985/…

Edward Evans

we proved the snake lemma, horseshoe lemma, and five lemma in one lecture

and it was basically just a wall of arrows

Alessandro Codenotti

horseshoe?

Edward Evans

12:26

yeah en.wikipedia.org/wiki/Horseshoe_lemma

Knight

@robjohn Yep! I too think that mass balance measures the downward force and then it divides it by 10 to give us the mass

robjohn

@Knight The weird thing is the square shape of the curve, but whatever...

Alessandro Codenotti

@EdwardEvans Oh ok, I had never seen this one before

Knight

12:57

@robjohn Sir but I’m getting a different answer. $$\text{Initial momentum of ball}= 3.5 \\ \text{Final momentum of ball}= p \\ \text{Time elapsed}= 90 \times 10^{-3} \\ \text{Force applied = Force experienced} = 6.7 \times 9.8 $$

$$F =\frac{p- 3.5}{90 \times 10^{-3}}\\ p = 9.4094$$ and therefore the final velocity is $9.4304/0.35= 26.875$

Stupid Kid

My brain is empty at this moment. What should I start with?

Don't u think it is wierd that formal definition of curl differs from the formula?

May be this is also boring to start conversation. It's too trivial.

Or talk about navier-strokes?

What am I talkin about fluid Mechanics in Mathematics anyway.

Can somebody explain me what actually Reimann hypothesis states?

r9m

13:12

@skullpatrol LOL XD

Stupid Kid

Mike testing mike testing 123... .123 can somebody see me in chat?

Knight

@StupidKid Yes

@nitsua60 Hi

Edward Evans

PLEASE spell names properly

Stupid Kid

@Knight Do you understand what Reimann hypothesis states?

Knight

@StupidKid No

Balarka Sen

13:19

@EdwardEvans Loool

Stupid Kid

@Knight 😭

nitsua60

@Knight hi.

Stupid Kid

I think the chat is now kinda normal. Because you guys start to use Lol ,XD...etc. I was think everyone here are sensitive and serious. Also you will get flagged easily 😂😂😂.

Alessandro Codenotti

Only @Balarka gets flagged easily

Knight

@nitsua60 My niece (the other one) slept from 7:30 PM up to 8:00 AM :-)

Is it fine ?

Thorgott

13:34

and Ted when he calls you demonic

Balarka Sen

thanks for roasting him for me

Alessandro Codenotti

@BalarkaSen flagged

Leaky Nun

@AlessandroCodenotti in italian songs do they lengthen geminated consonants?

Balarka Sen

@Alessandro gonna report you too

Leaky Nun

@BalarkaSen Why do you take this so seriously ? it's the internet, no one cares except for you

Alessandro Codenotti

13:37

@LeakyNun Sure

Leaky Nun

@AlessandroCodenotti how?

Alessandro Codenotti

It's an important part of Italian pronunciation. Not doing it would be like not reading umlauts in German

Leaky Nun

yeah but I mean in terms of the music

like if "mattina" spans 3 beats

on which beat do you pronounce the t

Alessandro Codenotti

Uh not sure

Leaky Nun

1.5? 1.75? 2?

Alessandro Codenotti

13:41

I guess you arrange the lyrics in a way such that there are no issues lol

Balarka Sen

musik

to me that means screaming at a potato stuck on top of the mic stand

Leaky Nun

poor potato

Balarka Sen

yes, i have been listening to burzum

Alessandro Codenotti

@BalarkaSen Ah, I see you listen to grindcore

Balarka Sen

lol

Leaky Nun

13:44

@BalarkaSen how about transgression

Balarka Sen

little to nothing

Leaky Nun

:c

Balarka Sen

i dunno algebraic topology dude

shits hard

Alessandro Codenotti

Wait since when are you doing honest math instead of higher topos theory? @Leaky

Leaky Nun

since I have a problem set due two days later

Alessandro Codenotti

13:47

Fair enough

Leaky Nun

@BalarkaSen if I apply SSS to a vector bundle $E \to B$ do I get $H_\ast(B) = H_\ast(E)$

Balarka Sen

yeah the fibers are contractible, the $E^2$ page only consists of the bottom row, all differentials zero, so $E^2 = E^\infty$

Leaky Nun

(I guess the joke is that $E$ def. ret. to $B$)

Balarka Sen

did you intend it to be a troll question or did you just realize that lmao

i wont be surprised by the latter

Leaky Nun

the latter

Balarka Sen

13:53

boy o boy

thats actually the best i can compute with SSS

Leaky Nun

ouch

Balarka Sen

not even joking

Leaky Nun

what does this mean, element of $H^0(B; H^{n-1}(F))$ that evaluates to $1$ on every orientation class

$F \to E \to B$ is a spherical fibration

$F = S^{n-1}$ I think

like, what is an orientation class, and why does such an element exist

Balarka Sen

$H^0(B; H^{n-1}(F)) = H^{n-1}(F)$, they're talking about the dual of the fundamental class of $H_{n-1}(F)$

Leaky Nun

wait is the base the sphere or the fibre the sphere

Balarka Sen

13:57

spherical fibration means spherical fibers yeah

altho you're prolly doing gysin sequence which also applies with spherical base instead

different story

Leaky Nun

but then $H^0(B;H^{n-1}(F)) = H^0(B)$

so I guess it means 1 on every connected component

but then what is orientation class

Balarka Sen

$H^0(B; G)$ is $G$ my mate, that's my identification

Leaky Nun

hi @loch

Balarka Sen

there's no canonical isomorphism $H^{n-1}(F) \to \Bbb Z$ a-priori, the choice of the isomorphism is given by choice of orientation of $F$

Leaky Nun

ah

Balarka Sen

14:00

the fibers vary point-to-point on the base, so you need choice of an orientation of every fiber compatible with each other, is all

Leaky Nun

an oritentation class is a choice of generator of $H^{n-1}(F)$?

Balarka Sen

yeah

well, $H_{n-1}(F)$, rather, but whatever, dualize

Leaky Nun

how does this work

how do I choose it across fibres

"continuously"

Balarka Sen

You can't in general; it's a restriction on the bundle. You need it to be an oriented spherical bundle

Eg $S^1 \to K \to S^1$ where $K$ is Klein bottle is a nonorientable spherical bundle

Leaky Nun

as in, how do you define it

Balarka Sen

14:02

You're reading the Thom class or what

Leaky Nun

Euler class

Balarka Sen

1 sec

3

A: Boil down the formal definition of Euler class

A choice of a local orientation of $\Bbb R^n$ at the origin is equivalent to choosing a vector space orientation of $T_0\Bbb R^n\simeq \Bbb R^n$ which is in turn equivalent to choosing a basis $(\mathbf{e}_1, \mathbf{e}_2, \cdots, \mathbf{e}_n)$ and remarking that every basis $\mathbf{b}$ of $\Bb...

This is specseq free tho

Leaky Nun

oh wow thanks

also it then says that the Euler class is the transgression of this class

8 mins ago, by Leaky Nun

what does this mean, element of $H^0(B; H^{n-1}(F))$ that evaluates to $1$ on every orientation class

Balarka Sen

Yeah this is something I do not understand. There's a transgression map $E^{0, n-1} \to E^{n, 0}$ or something, annoying af

Comes from doing a stupid snake lemma

I don't know why this is related to Euler class but it makes sense

Leaky Nun

so what sense does it make to you

Balarka Sen

14:07

Lmao just that they make the same long exact sequence (Gysin)

i have no idea why they are the same maps; i know how to interpret the weird map in Gysin as "cup product with Euler class" and i know how to compute it separately from specseq as induced from the transgression

i never figured out why they match

it just makes sense because most long exact sequences irl are natural

so it should be the same maps lol - bullshit reason

Stupid Kid

Why are math people so serious and get offended so easily lol. I got flagged lmao and couldn't talk for 30 min. BTW I am math people myself with asperger's syndrome.

Leaky Nun

14:34

@BalarkaSen can you apply Poincaré duality to CP^\infty

loch

i think apriori this doesn't make sense, since you don't have a fundamental class

but for any specific purpose my guess is you can probably get aroudn it by working with CP^{N} for large enough N

EnjoysMath

15:10

0

Q: The infinite intersection $\bigcap_{p=2 \\ \text{prime}}^{\infty} (K - \{1/p^2\})$ where $K$ is a multiplicative (not additive) group, equals $K$?

Necessary background: prime $\Omega$ extended to all of $\Bbb{Q}^{\times}$. The twin prime conjecture is that $\Omega^{-1}(2) \cap (K - 1) \cap \Bbb{Z}$ is an infinite set where $K = \{ x^2 : x \in \Bbb{Q}^{\times}\}$ is the subgroup. This is true since if $\Omega(x) = 2$ then $x = pq$ and if $...

abstract-algebra group-theory elementary-number-theory group-homomorphism prime-twins

Mats Granvik

15:21

@StupidKid oeis.org/A003418 "An assertion equivalent to the Riemann hypothesis is: | log(a(n)) - n | < sqrt(n) * log(n)^2. - Lekraj Beedassy, Aug 27 2006." (due to L. Schoenfeld)

A003418(n) = a(n)

Knight

15:46

@StupidQuestionsInc Hello! Seeing you after a long time :-)

Stupid Questions Inc

I have been bugging my head for a while now, does anyone know how to show that 3 is a primitive root modulo 7^2 (i wonder how one can do that efficiently)

@Knight thanks for the welcome! indeed i've been extremely busy with life and inefficient with my time management

Knight

@StupidQuestionsInc I understand

Tobias Kildetoft

@StupidQuestionsInc Hmm, probably not much better way that to test the smaller possible orders it could have

Knight

How’s everything going @StupidQuestionsInc?

Stupid Questions Inc

@Knight confinement is doing wonders to my mental and physical health /s + we have so much more work for college, sitting in front of a computer for a longer time is causing me back problems again

@TobiasKildetoft i got one idea, there's this theorem that if $s$ is a primitive root mod $p$ then either $s$ or $s+p$ is a primitive root mod $p^2$, maybe if i show that $s+p$ is not a primitive root it may turn out to be easier

@Knight what about you? what did you learn during this time? :)

Tobias Kildetoft

15:51

@StupidQuestionsInc Sure, if you can somehow quickly see that. Though you might be unlucky and it turns out they are both primitive

Stupid Questions Inc

@TobiasKildetoft actually 10 is also a primitive root mod 7^2 :(

Knight

@StupidQuestionsInc Do you teach in college or study in a college?

Stupid Questions Inc

@Knight study

Knight

@StupidQuestionsInc I’m doing okay :-)

@StupidQuestionsInc Are you an undergrad ?

Stupid Questions Inc

@Knight yes

Knight

15:54

Wow!

Stupid Questions Inc

@Knight why are u that surprised x) ?

Knight

@StupidQuestionsInc Becasue I thought the man in your avatar was you, and to me that man doesn’t look like an undergrad

Stupid Questions Inc

@Knight oh, no that's definitely not me :) he's just a person who turned into a famous meme

Knight

Hahahahaha

And one more thing due to our talks previously it seemed to me that you were a self learner

That’s why knowing you as an undergrad surprised me :-)

Stupid Questions Inc

@Knight yes, I don't like the atmosphere surrounding courses, it stresses you out and doesn't give you enough time to digest the material unlike if you were instead a self learner

Knight

16:01

@StupidQuestionsInc I totally agree with that

topologicalorientablesurface

17:08

0

Q: Separability of countable product my attempt

Let $X_n$ be separable spaces for each $n$. Then, $X=\prod_nX_n$ is separable. (n is a natural number) Note: I am considering the product topology. My attempt: Let $D_n$ denote the countable dense subset of $X_n$. For each $n$, choose $\textbf{x}_n\in X$ such that $x_n\in D_n$ . Put $D=\bigcup_...

general-topology solution-verification

Victor

19:08

What are some of the characteristics of the real-numbered quadratic formula challenge problem for root finding even for mathematican by hand?

Thorgott

19:22

I'm not sure what you are asking

Victor

20:11

0

Q: A critique for po shen loh's quadratic formula solving way?

Is there some characteristics exist for one to critique this simple new way of solving qudratic equation without looking or comparing to the old way? Is there any challenging features on those problem exist to solving those quadratic equation even for mathmatician given only pen and paper? Elabor...

abstract-algebra algebra-precalculus number-theory

Thorgott

what kind of "characteristics" do you want?

what does it mean to critique a way of solving a problem? mathematically speaking, it works and that's about it. is your question of didactical nature?

Victor

20:34

Hi, does that formula able to work for the quaduatic formula of the form ax^2+bx+c=0, if not, is there any challenging feature to solve these equation by hand except factoring assuming both root are real also rational.

Hi, anybody can make good suggestion will be greatly appreciate!

geocalc33

20:57

@EnjoysMath what's good

user76284

22:53

Both $\mathrm{d} \mu(t)$ or $\mu(\mathrm{d} t)$ are used for Lesbesgue integrals, but which notation is more "technically correct"?

Thorgott

$\mathrm{d}\mu$

user76284

I mean when you have to show the variable.

Otherwise, I do prefer $\int f \,\mathrm{d}\mu$.

Thorgott

then I'm not sure if one is more correct than the other

they're both conventions

Transcript for

$Mathematics$ Mathematics