Mathematics - 2018-04-14 (page 1 of 2)

Nicholas Roberts

01:08

Can ayone explain how (x + y)^p = x^p + y^p in a field of characteristic p?

All the explanations I have seen are confusing :(

symmetric_cow

use binomial expansion --- (x+y)^p = x^p + \binom{p}{1} x^{p-1}y + \binom{p}{2} x^{p-2}y^2 + ... + y^p

then prove that \binom{p}{k} are divisible by p, for k = 1,...,p-1, , so in a field of characteristic p they are all 0

Nicholas Roberts

Ok so it boils down to showing that (p choose k) is divisible by p

Hmmm, ok. Thanks.

for k up to p-1

Ted Shifrin

Exactly correct.

@ALannister That's just wrong. :( What is correct is that $A^k = PJ^k P^{-1}$, where $A=PJP^{-1}$. And then $J^k = (\lambda I + N)^k$ becomes a finite sum because $N^s = 0$ for some $s$.

Nicholas Roberts

Does anyone here have math dreams?

Like when they sleep at night

Ted Shifrin

I don't think I ever have, Nicholas.

Nicholas Roberts

01:23

Thats interesting. When I started getting more serious into math, I would start having math dreams. Some good and some bad. The good ones are me finally solving something that I was working on in real life and the bad ones are me freaking out about not being able to do a question. So weird

Ted Shifrin

Or dreaming that you're sleeping through your comprehensive or oral exams. Friends of mine in grad school were tortured by that one.

Nicholas Roberts

Wow, that seems bad.

Btw can I ask you something? Is it any good to earn a master's degree in pure math? Not a PhD

Ted Shifrin

I should've kept my mouth shut.

A masters is enough to teach at some community colleges. Certainly in the job market, a masters is better than a bachelors, but I would still encourage you to have some applied strengths — definitely computer skills.

I see you're at StonyBrook. I have some old friends there.

Nicholas Roberts

Yes, currently doing a masters there. Oh really? Who?

Eric Silva

@TedShifrin i like just had a dream i overslept and missed my GRE

Ted Shifrin

01:27

Blaine Lawson was one of my favorite professors at Berkeley (before he left for Stony Brook). Fabulous guy. And I know Mike Anderson and Claude LeBrun and a few others.

Sorry, Eric. I didn't mean to jinx you. :)

Eric Silva

hopefully i dont atually oversleep tonight...

Nicholas Roberts

Wow! My friend currently has Lawson for complex and I will be taking his algebraic topology course in the fall! And Michael Anderson is phenomenal! I had him for Topology 1 just last fall.

Ted Shifrin

Borrow 5 of Demonark's alarms.

Nicholas Roberts

And Claude is the head the PhD program over here. So cool that you know them

Ted Shifrin

Say hi to them for me, Nicholas. Lawson is one of the top few teachers I've ever had.

Nicholas Roberts

01:29

Sure. I am very excited to have Lawson for Alg. Top.

Ted Shifrin

He actually read my thesis and wrote letters of rec for me (39 years ago).

Nicholas Roberts

Do you know Alexander Kirrilov Jr? He is my Algebra 2 professor

Eric Silva

@TedShifrin i actually have 3 alarms and i slept through all of them this morning so... does not inspire confidence

Ted Shifrin

I know his name, but don't know him personally.

Nicholas Roberts

Thats so cool. What was your thesis on?

Ted Shifrin

01:29

LOL, Eric. Should I wake up at 4 AM to call you?

Eric Silva

nah ill just make em super loud

Ted Shifrin

Chern form integrals and integral geometry for complex manifolds, Nicholas.

Eric Silva

my roommate has work at 6am so i told him to wake me up if i dont wake up by that point

Ted Shifrin

Oh, OK, Eric. Just for you, I would get up at 3 or 4 AM to call, if you needed me.

Nicholas Roberts

Sounds awesome lol

Doraemon

01:31

Excuse me. @TedShifrin And @EricSilva : Can u solve my Problem ?

Ted Shifrin

Um, what problem is that?

Eric Silva

ty but there will be 0 need

Doraemon

See my Question

Related to polynomials Tag

Ted Shifrin

That's absurd, @Doraemon. Give us a link.

And generally you'll get more attention on main than you will from 1 or 2 people here.

Doraemon

Ok

Eric Silva

01:33

main is blocked for me so im no help

Doraemon

Any reason to imagine it has a particularly sensible answer? For $a=2,3$ the answers look fairly random (though of course that might not mean much). — lulu yesterday

Ted Shifrin

LOL, Eric. You'll never release yourself from purgatory.

Eric Silva

ill release myself tomorrow after the hell-test is finally over

Ted Shifrin

@Doraemon: When you have something like $\sqrt a + \sqrt b = c$, the trick is always to move one of the square roots to the other side and then square. You'll have to play that game twice.

Doraemon

I had done that

Ted Shifrin

01:35

I see no evidence that you did that.

Doraemon

But there is no use after that as there will be root terms

Can u please solve that ?@TedShifrin

How can I get more help from main and what is main ? As I already posted the question.!

Ted Shifrin

It seems like an unbelievable mess. What makes you think there should be an elegant solution?

I don't see any reasonable route now that I work on it.

Doraemon

See graph of that expression and you will find that x is order of √a like term and this question is asked in some community and though being solvable.

Ted Shifrin

BTW, when you ask questions, please don't say "this is urgent." People will immediately ignore you when you do that.

I don't see a solution.

At best you're going to end up with a 4th degree equation, and in general those can be solved by radicals, but they will involve complex numbers.

Doraemon

Ok then do that way !

Ted Shifrin

01:47

I'm not interested in working on it any more.

Nicholas Roberts

Why is it urgent? Is it homework? Take-home exam?

Doraemon

Just to get solved it early

Ted Shifrin

By the way, $a\in\Bbb R$ is surely wrong. You need $a\ge 0$ for starters.

Daminark

@TedShifrin excellent idea!

Ted Shifrin

And then you need $a\ge \sqrt{a\pm x}$.

Doraemon

01:49

$a\ge0$ means ?

Ted Shifrin

I have those once in a while, Demonark.

Eric Silva

@Daminark hand em over

Ted Shifrin

\ge means greater than or equal to

Doraemon

Ok Then

Ted Shifrin

@EricSilva: Probably he needs to throw them (a distance) rather than handing them.

Daminark

01:51

Well, the issue is that my phone is my alarm

Ted Shifrin

@Doraemo: In fact, you need $a$ at least $1$ or negative for there to be any possible $x$.

I thought you had 6 of them, Demonark.

Doraemon

@TedShifrin but from here how can get solution as this is inequality ?

Ted Shifrin

I have no idea how to solve it. I'm just telling you that for certain $a$ there will clearly be no solution.

I'm too old to find these sorts of questions interesting. Sorry.

I'm gone for now.

Doraemon

a > = 1.3 approx there will be solution

Ted Shifrin

@EricSilva: Boa sorte !!!

Daminark

01:54

Well, when I talked about setting 6 alarms, I meant that I'd set like, an alarm every 10-15 minutes

Ted Shifrin

ohhh, Demonark. You need 6 simultaneously :P

Daminark

That's probably true, I am way too heavy of a sleeper

Ted Shifrin

It's all about series versus parallel (circuits).

Daminark

One time in high school I slept through a fire

Ted Shifrin

Oh, lovely.

Eric Silva

01:57

valeu @Ted!

@Daminark did u do this when you missed complex

Daminark

Yup

I had 6 alarms in succession but I ended up waking at 12:45

(I meant to wake up 9:30-10)

Eric Silva

wow what a catastrophe

this complex class is so weird dude

it's just a cornucopia of weird conformal map factoids

nitsua60

What's the common practice around these parts on handling problems whose solution is "OP didn't feel like doing work"? I'm looking at that one Doraemon dropped in here and told Eric and Ted they needed help urgently:

1

Q: Solving $\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x, \ a\in\Bbb{R}$

Given $$\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x$$ and $a\in\Bbb{R}$, express $x$ in terms of $a$. I rationalised the above expression and then again rationalised which gave me : $$\sqrt{a+\sqrt{a-x}}-\sqrt{a-\sqrt{a+x}}=2x+\frac{1}{\sqrt{a+x}-\sqrt{a-x}}$$ Now, what should I do? I...

polynomials

Daminark

Lol true, but it's quite a lot of fun I must say

nitsua60

There's a lot of rather straightforward work they show no work of having done.

Eric Silva

02:13

@Daminark i think it's great

Mike Miller

02:33

I love conformal maps

TraLaLa

Can someone help me on a function/symbol that i do not know about

?

It is stated right here awesomemath.org/wp-pdf-files/math-reflections/mr-2015-05/…

What does ω(P, C_1) mean?

Nvr mind I know it already

Daminark

02:51

@Mike what we did today was quite nice actually. So, we've been considering conformal maps where we renormalize the derivative and fix $0$, so we have $f(z) = z + a_2z^2 + \ldots$

(From the unit ball)

And it turns you can guarantee that $|a_2| \le 2$, and that the image of $f$ contains $B(0,\frac{1}{4})$. We're later gonna show that if $f$ is such that $f(z) \in \mathbb{R} \iff z\in \mathbb{R}$, that you have a general bound $|a_n| \le n$.

Eric Silva

03:07

@Daminark the general thing was proved in like the 80s, did she say she was actually gonna do it, i dont remember

Mike Miller

Oh that’s quite neat

Eric Silva

oh i see under a simpler condition

Daminark

Yeah she's proving it with the realness assumption

And yeah I think that's gonna be the end of the conformal maps part of the class, next up is gonna be entire functions

Eric Silva

03:23

i wish she had a syllabus

Daminark

I mean on the first day she said she wasn't even sure what she'd do at the end. Applications of complex analysis to something TBD

Though who knows, maybe she has a clearer picture now of what she'd like to do

Abcd

03:56

Hi @LeakyNun

Leaky Nun

?

Abcd

$k\left(1- \dfrac{x}{\sqrt{x^2+r^2}} \right)$

k, r are constants

@LeakyNun Its impossible to graph this?

Leaky Nun

@Abcd desmos.com

Abcd

Because at $x=\infty$ we get undefined form.

Leaky Nun

nobody told you to plot x=infty

Abcd

03:59

@LeakyNun My physics teacher said "you all will not be able to graph this"...

SO I wanted to discuss about this

@LeakyNun at $x \to \infty$?

Leaky Nun

1 min ago, by Leaky Nun

@Abcd desmos.com

Abcd

@LeakyNun I have already done that!

Leaky Nun

then you have plotted it and contradicted your physics teacher

Abcd

I want to know how to plot it myself at $x \to \infty$

Leaky Nun

well evaluate the limit

Abcd

04:00

A general physics graph is what I want to plot

Leaky Nun

well evaluate the limit

Abcd

@LeakyNun L Hospital right?

Leaky Nun

you don't need that

divide num and denom by x

Limits is $\Huge{\text{next}}$ chapter in maths

Abcd

@LeakyNun What do you mean?

Leaky Nun

?

nothing

Abcd

04:05

@LeakyNun then?

it becomes 0 using L Hospital?

Leaky Nun

whatever

Abcd

Why are you so angry?

Leaky Nun

i am not

Abcd

limit is 1 anyway according to WA

but it contradicts the graph: desmos.com/calculator/alywljlqg4

Leaky Nun

of what?

you forgot the square root in the graph

Abcd

04:14

Okay, done. Both results match.

Leaky Nun

ok

Silent

04:52

@LeakyNun, will you please have a look at this?

Leaky Nun

no idea

Rumplestillskin

Is anyone familiar with a reference that solves an IVP with an implicit Runge-Kutta ( of order at least 3 ) that includes all the steps? e.g. solving the implicit system of equations etc.

Balarka Sen

06:33

Echo

Anderson Felipe Viveiros

06:43

Does anybody know about a result like "If $X$ can be described by sequences then it is metrizable"? By "described by sequences" I mean that the topology of $X$ is given by a sequence convergence notion.

See my last comment in the accepted answer to this question math.stackexchange.com/questions/2493715/…

The problem is: if the sequence convergence notion does not have the limit uniqueness property, then $X$ cannot be metrizable, since metrizable implies Hausdorff which implies limit uniqueness property.

philmcole

09:01

Hi, if a polynomial $p(t) = q(t)s(t)$ and $p$ splits in linear factors, do we know then that $q$ and $s$ split too?

Astyx

Yes

philmcole

Can you tell me why?

Astyx

Write s and q as a product of irreducibles

philmcole

Yes, and then?

Astyx

write $s = \prod s_i$, $q = \prod q_i$, $p = \prod p_i$

Then $\prod p_i = \prod s_i \prod q_i$

All polynomials written are irreducible, then each $s_i$ is one of the $p_i$, and the same goes for the $q_i$s

philmcole

09:09

I understand. Thanks!

Astyx

It's polynomial arithmetics

philmcole

Yeah it's not hard but we only skimmed it in class back then

Need to do some resfreshing

Astyx

A more rigorous way of saying this is : Let $a$ be an irreducible factor of $s$, then $a|s|p$ so $a$ is an irreducible factor of $p$, and is therefor of degree 1

philmcole

yeah that's concise

YOUSEFY

Hi: Does the sum of 1/2+1/4+1/8 + .... for infinite is the same as for finite case?

It is known that 1/2+1/4+1/8+...=1 (called geometric series), now suppose we have finite case, then does this equality holds?

Dawood ibn Kareem

09:26

Are you asking whether there exists some $n$ for which $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = 1$ ?

YOUSEFY

10:03

@DawoodibnKareem yes!

Dawood ibn Kareem

Well, the left hand side can be written as $1-\frac{1}{2^n}$

And since there's no $n$ for which $\frac{1}{2^n}=0$, I think the answer to your question is no.

YOUSEFY

@DawoodibnKareem Wow, so from infinity to finite, the answer completely different!! I thought it would equal some numbers between 1 and 0.

Dawood ibn Kareem

Well, not completely different. As you add more and more terms, the $\frac{1}{2^n}$ part gets smaller and smaller. The infinite case is where it vanishes to zero.

John Rennie

10:33

@BalarkaSen you around?

Balarka Sen

Yup.

John Rennie

You'll have noticed that the PSE room owners/mods span a range of opinions.

Actually that's good because we end up with a mean that is mostly acceptable to most people.

I'm at the laissez faire end, which isn't always good because I've occasionally let things go too far and they get nasty.

When views at the other end of the spectrum are expressed I'd respect them and just move on.

Balarka Sen

I mean it's ok, it's just that almost 90% of the time it's not clear to me what the specific rule sets of the hbar are.

John Rennie

But isn't that always the case when you find yourself in any group of people? You have to kind of feel your way around.

Rumplestillskin

Is anyone familiar with implicit Gauss methods for IVP's of relatively high order?

Dawood ibn Kareem

10:40

I feel like I've stepped into a parallel universe.

Balarka Sen

@JohnRennie Well, there's obviously a difference because in an arbitrary group of people there isn't a stratum of people trying to manage the conversations in a specific way. It's a management issue, isn't it? The rules beyond which the moderators think certain messages are not appropriate is unclear, not the differing opinions of the users.

I mean I am part of no crusade to point out the inevitable inner contradictions in the moderation. It's just that it's never clear to me how broad they are.

skullpatrol

sorry about that pal

he pounces unexpectedly sometimes :P

John Rennie

The mods have a difficult balancing act to do. They have to consider the views of all the chatroom users, not just the noisy ones, and also the views of the SE management.

Balarka Sen

lol @Dawood I mean, I myself didn't think it was inappropriate, otherwise I wouldn't have said that. I have no problem with complying with the moderation rules, but adding one at a time just seems like an unplanned management issue

YOUSEFY

@DawoodibnKareem Thank you Dawood!

Balarka Sen

10:48

In this case I think the biggest irony is that the inside joke of hbar being ACM is a bot :p

@Dawood LOL

rip

John Rennie

@BalarkaSen I'm not sure I can claim to be consistent in my moderation of the room (though I'm consistently at the lax end :-)

Balarka Sen

@JohnRennie I am not even asking for consistency, which is probably unreasonable to some extent. It's simply uber-confusing to me given that I as a user should try to abide by the room ethics when the room ethics itself changes within such a short period of time

John Rennie

Trying to set hard and fast rules is impossible so the mods also have to feel their way to some extent. When eyebrows have been raised by the SE staff you'll find they are a bit more cautious.

Dawood ibn Kareem

Balarka, don't over-react. Think of it as a one-off ripple in space-time.

skullpatrol

lol

John Rennie

10:52

Yes, that would be my take. The weather changes all the time.

Dawood ibn Kareem

You're no worse off for having been told off by a grumpy moderator, right? I mean, you're not banned for 10 years or anything.

Balarka Sen

Lol

I'm not mad just vocalizing my confusion

John Rennie

And you weren't told off. David didn't criticise you personally he just said he didn't think the comment was appropriate.

Dawood ibn Kareem

OK. Humans are inconsistent. That's life. Get over it.

John Rennie

Life's like that ... then you die :-)

Dawood ibn Kareem

10:54

@JohnRennie Yeah, he was told off.

Leaky Nun

@BalarkaSen people are surely consonating

i'll show myself out

John Rennie

Anyway, I'm off to look at dmckee's question again and scratch my head some more ...

Balarka Sen

@LeakyNun yeah get outta here

go back to your algebraic hell

read the bible of Atiyah-MacDonald

Leaky Nun

:c

Dawood ibn Kareem

Who the heck is Atiyah MacDonald and why does he/she need a bible?

skullpatrol

11:01

*they

Balarka Sen

It's a very slim and terse commutative algebra textbook that every algebraist thinks of as their bible

Dawood ibn Kareem

I might ask how it compares with the real one, but I fear I might get censured.

Balarka Sen

lol

Maybe I should just say "religious textbook" than the bible in particular

Leaky Nun

I might say it has less violence and causes less deaths, but I fear I might get censored

Dawood ibn Kareem

I'd be worried if I found a commutative algebra textbook with as much bloodshed in it as the Bible.

Although it might make commutative algebra a more interesting field (no pun intended).

Balarka Sen

11:05

Eisenbud

Leaky Nun

certainly no talking snakes in atiyah-macdonald can be found

Balarka Sen

many a-people have died reading that text

Dawood ibn Kareem

@LeakyNun What, not even adders?

Leaky Nun

@DawoodibnKareem lol

Balarka Sen

Snake lemma

The snake lemma is a tool used in mathematics, particularly homological algebra, to construct long exact sequences. The snake lemma is valid in every abelian category and is a crucial tool in homological algebra and its applications, for instance in algebraic topology. Homomorphisms constructed with its help are generally called connecting homomorphisms. == Statement == In an abelian category (such as the category of abelian groups or the category of vector spaces over a given field), consider a commutative diagram: where the rows are exact sequences and 0 is the zero object. Then there is an...

obligatory

Leaky Nun

11:07

you win

Sha Vuklia

guys, how do we know that $s<1-1/2+1/3$?

Leaky Nun

@ShaVuklia the sum of the next two terms is negative

onwards

Dawood ibn Kareem

You can rewrite it as 5/6 - 1/20 - 1/42 - 1/72 - ...

Basically a whole lot of minus terms after the original 5/6.

Sha Vuklia

@LeakyNun yea I saw that 'intuitively', but I didn't know I could use that?

Leaky Nun

@ShaVuklia sure you can

lemma: if $a_n \ge 0$, then $\sum a_n \ge 0$, provided that the latter converges

proof: exercise

corollary: $s < \frac56$

done

Sha Vuklia

11:19

but we don't have $a_n\geq 0$? it's alternating

Dawood ibn Kareem

Usually, when you have to prove $ a \neq b$, there are loads of ways of doing it. You don't have to "spot" the same method that the book uses.

Leaky Nun

@ShaVuklia how about this

Sha Vuklia

yea but I would like to follow their argument

Leaky Nun

let $S_n$ be the partial sums

then $S_3 = \frac56$

and by induction, $S_{2n+1} < S_3$

for sufficiently large $n$

Dawood ibn Kareem

Following an argument isn't the same as being able to dream it up yourself.

Sha Vuklia

11:20

that's not my goal..

Leaky Nun

and subsequences converge to the same thing

so $S_{2n+1}$ goes to $s$ just fine

so $s \le S_3$

skullpatrol

@BalarkaSen the second biggest irony is him wearing a christmas hat in the middle of April and preaching about "appropriateness." :P

Sha Vuklia

right right

Leaky Nun

now use the same thing to show that $s \le S_5$ instead, and then it follows that $s<S_3$, since $S_5 < S_3$

so are you convinced lol

Sha Vuklia

haha, I am rereading

right, yea I have to really refresh my 1D real analysis course

actually, that is what I'm doing right now:p

thanks, I could follow the argument

Leaky Nun

11:23

me too

Sha Vuklia

you mean refreshing?

Leaky Nun

sure

Sha Vuklia

what book do you use?

I use Rudin now

but I learned real analysis from another book

Leaky Nun

just going through the lecture notes lol

Sha Vuklia

fair enough

skullpatrol

11:25

@DawoodibnKareem true, but some people like to get inside of the author's head...

Leaky Nun

well i learnt the stuff online before i even went into my uni

Sha Vuklia

true

Dawood ibn Kareem

Yes, but often authors of a mathematics book will go to great lengths to hide their thought processes. Like, here's the theorem and here's the proof, but how I thought of it is a secret!

Sha Vuklia

that's bs

Dawood ibn Kareem

You what?

skullpatrol

11:28

yup, "I have a secret and it is your job to try and find out what it is" :(

profs do it in lectures too

Sha Vuklia

lol if anything, mathematics books are the most transparent ones out there

Balarka Sen

@skullpatrol Bwahahah

That was funny

skullpatrol

:-D

that's why I ::ran for cover:: so fast

the "appropriateness" argument is a recurring theme with him

(just my 2 cents worth of observation :)

Balarka Sen

well, given there's a crowd which systematically trolls on the boundary of inappropriateness in the room, it's natural

Leaky Nun

@BalarkaSen closure of inappropriateness sans the interior thereof?

Balarka Sen

11:39

Accurate

the space of inappropriateness is just a manifold away from a discrete set of singularly inappropriate events anyway

skullpatrol

i'm still trying to find out what happened to Duffield?

specifically, who dropped the axe

a one year chat ban seems excessive, imo

Sha Vuklia

just out of curiosity; if you're banned (from the chat), can't you just set up a new account (temporarily) to join the chat? of will that one be banned automatically too?

Balarka Sen

if the mods notice that it has the same IP you get super-banned for evading an existing ban

they can IP ban you altogather

Sha Vuklia

11:54

makes sense

gotta learn hacking then

:p

I actually never really understood how IP addresses work, and I want to have an interview with Zucc the man for him to explain it to me

Balarka Sen

HACKERMAN

►

Sha Vuklia

that's my future

Zuckerberg explains the internet to Congress

►

this vid wasn't even that funny (well it was)

but the COMMENTS were so good

Balarka Sen

the Zucc was 10/10 on the congress

Sha Vuklia

I have a meme-crush on Zuckerberg

@BalarkaSen defo

Balarka Sen

@ShaVuklia we all do, Sha, we all do

ugh I broke my headphones again

Hey I fixed it.

Haxxx0rman

philmcole

12:37

he looks like a robot

Sha Vuklia

13:20

@Leaky have some time?

Leaky Nun

@ShaVuklia ?

Sha Vuklia

I have this question;

Let $A\subset R^n$ be a rectifiable set. I want to show that $v(A^0)=v(A)$, where $v$ stands for volume, and $A^0$ stands for the interior of $A$. I can show that $v(A^0)\leq v(A)$, because $A^0$ and $A$ have the same boundary, hence $A^0$ is also rectifiable, and for any rectangle $Q\supset A$ we have
$$
\int_{A^0}1=\int_Q1_{A^0}\leq\int_Q1_{A}=\int_A1.
$$
Now I want to show that $v(A^0)\geq v(A)$. Since the boundary has measure 0, we know that for each $\epsilon>0$, we can always find a covering of the boundary with rectangles $Q_i$, such that $\sum_iv(Q_i)<\epsilon$. So I was thinking of…

Leaky Nun

no idea

@BalarkaSen

Sha Vuklia

so I had this kind of idea in mind $v(A^0)\geq v(A)-v(A\cap(\bigcup_iQ_i))$, but I would have to think of a proper argument

Vrouvrou

If i take $O=\cup_{r\in L}\Omega_r=\{(x,y)\in\mathbb{R}^2, x^2+y^2=r^2\}$ and $S=\cup_{s\in S}\Omega_s=\{(x,y)\in\mathbb{R}^2, x^2+y^2=s^2\}$ what is $O\cap S$ please

Sha Vuklia

13:29

yea I think that will work

Vrouvrou

someone have an idea ?

Sha Vuklia

14:07

guys, can we say that manifolds have a tangent plane at each point?

Balarka Sen

Smooth manifolds do indeed

Sha Vuklia

oh, I thought that the condition about the rank for the coordinate patch somehow ensured the existence of a tangent plane

Balarka Sen

You are correct, it does. The image of the derivative of the inverse of the coordinate patch diffeomorphism is the tangent space.

Sha Vuklia

it shouldn't be the inverse right? because the coordinate patch maps onto $V\subset$manifold

Balarka Sen

Oh, your coordinate patches maps to the manifold? Then yeah. I thought we called those parametrizations, not coordinates.

I can never keep track of the terminologies anyway

:)

But same difference, yep

Sha Vuklia

14:13

ah okay, but if I am correct, then why didn't my initial question yield a 'yes'?

Balarka Sen

I said they "do indeed"!!!

That's a yes!!!!! I can't speak Russian m8

Leaky Nun

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Sha Vuklia

I thought you meant smooth manifolds do, but 'general' manifolds don't

maybe that's a terminology issue again

@BalarkaSen * laughs in Russian *\

Balarka Sen

@ShaVuklia What's your definition of a manifold? Subset of a Euclidean space which admits coordinate patches?

Sha Vuklia

I think yes

Balarka Sen

14:15

Yup, that's what we call a smooth manifold.

There are more general notions of manifolds :)

Sha Vuklia

oh, right, so we had confusion over terminology twice:p

alright, well I am glad all is settled

Balarka Sen

~~:thumbs:~~ :upside-down ok-hand:

Harry Evans

Anyone who can help? Suppose that for each $f \in L^p (\mathbb{R})$, the improper integral $$\int_{\mathbb{R}} f(x) \varphi(x) dx$$ converges. Show that $\varphi \in L^q (\mathbb{R})$.

I was about to use the Hölder inequality and "split up" the integral, then claim that since the integral converges, the "factors" converge, and so on

But then, I realize that it's possible for $$\bigg\{\int_{\mathbb{R}} \left| f(x) \right| ^p dx\bigg\}^{1/p} \bigg\{\int_{\mathbb{R}} \left| \varphi(x) \right| ^q dx\bigg\}^{1/q}$$

to converge even if one of the two factors doesn't.

Sha Vuklia

14:36

@Balarka hm, one question tho; a circle $M$ is a 1-manifold without boundary in $R^2$, but my book also says that $M$ must be a subspace of $R^2$, however, how is a circle a subspace?

Balarka Sen

Topological subspace, not vector subspace :)

Just consider the unit circle in $\Bbb R^2$ with the subspace topology

Sha Vuklia

ooh right, thanks

Harry Evans

15:06

@BalarkaSen Can you help? Thanks in advance!

Leaky Nun

A is a subspace of B iff there is a monomorphism A->B :)

change my mind

@Abcd what brings you here

Abcd

@LeakyNun I got the answer on physics chat. $\ln(r/0) \to \infty$ brought me here.

2

Q: Why is the force on the charge at the tip of a cone infinite?

Imagine a charge $q$ that is located at the top of a hollow cone with a surface charge density $\sigma.$ The slant height is $L$ and the charge $q$ sits at the vertex of angle $2\theta$. We are interested in the force acting on the charge $q$. Assume that there are shells inside this hollow co...

electrostatics singularities regularization

Harry Evans

@LeakyNun?

Leaky Nun

@HarryEvans?

Harry Evans

Can you help? Thanks

Charlz97

15:29

hello everyone

Leaky Nun

15:49

Let $V$ and $W$ be finite-dimensional vector spaces over a common ground field $F$. Let $T:V \to W$ be linear.

Let $T^\ast:W \to V$ be the transpose of $T$, i.e. the unique linear map $W \to V$ such that $\langle T(v),w \rangle_W = \langle v,T^\ast(w) \rangle_V$ for every $v \in V$ and $w \in W$

Claim: $\ker T \oplus_{\mathrm{int}} \operatorname{im} T^* = V$

Lembik

Suppose you have an odd number of white balls and the same number of black. How many partitions have an odd number of white balls and an odd number of black balls in each subset?

Leaky Nun

Let $v \in V$

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