Hi @inf.

Hi my friend @abe

Sorry to bother but I can't find what the symbol δ means? Its used in thermodynamics and other areas. Is it just another symbol for the inexact differential?

The symbol means what it is supposed to mean. Look up the definition in the text.

I haven't found it, and its been long a time now. I thought its meaning was as concrete as a plus sign, but apparently not...

All I can say is the symbol is delta, usually representing change.

In thermodynamics you are right @David

Alright thanks to both

@infinitesimal How do you know?

Check Wikipedia

Oh OK. I usually don't answer if I don't know clearly myself. Because I may misread Wiki.

As you said it depends on context

Like all symbols ;-)

My previous chat suspension happened so quickly. I wonder what happened. Maybe a mod suspended me.

When?

When I mentioned boobs, lol. Anyway, nvm.

My appointment is in 10 days time.

Are you going to the same doc @abe

Yes. The doc is not too important, just for meds. I need to choose the therapist more carefully.

I don't like the previous one.

Find someone you like @abe

I am trying to get completely well by the end of next year. I hope I make it this time.

Or you are wasting your time

And also money.

Get a referral to as many as you need until you find one you like.

Hi @kaj.

hey hey

@infinitesimal These few days, I am scared I may never get well, which is why I have been feeling very anxious.

You will be fine. Just keep working at it my friend @abe

23 is such a strange number. Hawking's house number, Nash's favourite prime, and my date of birth.

It does have a certain beauty to it.

Hi @kaj

Hey @infinitesimal

All numbers are beautiful. If there was an ugly number, that in itself would be beautiful.

@infinitesimal What is your favourite movie?

A Beautiful Mind, obviously

Hello @ᴇʏᴇs. You are my favourite user in this chat.

Honestly, I don't think I have a favourite.

Sometimes, the anxiety gets so bad that I think I am losing my mind, even though I have already lost it.

Please stop trying to convince yourself that you are "crazy."

No, I am not doing that. I am just trying to describe how bad it is.

I know it's painful.

Survey: Do you use cotton buds to clean out ear wax?

I do, and sometimes, it gets pushed in further, not good.

Q-tips?

What is Q tips?

Q-tips are what we call "cotton buds" in the US.

Any new girlfriend yet? @kaj

Working on it @ABeautifulMind

Life is full of surprises. I did not expect I would get so sick. Maybe one day I would look back and say I did not expect myself to get well.

:)

Until that day comes, I cannot really smile.

Did you watch John Nash's son interview @abe

@infinitesimal Yes. I don't think they were acting.

Hi, random question: Does the existence of a well-ordering on $\mathbb R$ necessarily depend on the axiom of choice?

@user Who are you?

I cannot remember. Please tell your name.

Are you John?

Hi @andres nice beard.

Apparently, the answer to my question is yes.

@user Please don't say such frightening things to me, thanks.

Just to be clear: You're @ABeautifulMind's worst nightmare, but not necessarily my worst nightmare. Right?

Don't try and scare my friend pal or I'll call skullpatrol to come here and lay a beating on you @user

>8(

Well, you know what they say: You're either a user or a loser. Or something like that.

skullpatrol is my pal

Orly?

what's that?

It either means "Oh really" or a female given name. (EDIT: That's "Orli," sorry.)

Is user=skullpatrol?

Or a city in France.

Cool, cool. Wanna do some math?

@user Not funny to me.

@user Please don't frighten me anymore.

"If someone makes a big show and tell about 'coming for someone,' chances are that he will never actually get around to coming for that person." - Some rule I just made up

Wait, what the hell?

I don't believe in the crucifix.

What? That's like saying "I don't believe in the gun."

There still are people being crucified in the middle east.

ISIS sucks, by the way.

Oh, there is no need to suspend user.

@ACuriousMind Nice avatar.

No need to flag him either.

> Oh, there is no need to suspend you, sir.

Why, thank you!

@ABeautifulMind - Multiple flags raised.

@MikeMiller Heh, thanks.

In any case... Anyone have any interesting maths to talk about?

@columbus8myhw - I heard that 1+1 = 3. Discuss

Yes. Is there anyone going to solve another Millennium problem soon?

@Richard Totally true. Once I took one man and one woman, and later I ended up with three people.

bye

That was a fast bye.

Hehe. If there are any other issues, please don't hesitate to click the 'flag' button.

Peace out, Maths dudes.

A 4 hour suspension is too serious.

Back. By the way, you know how "infinity" isn't a number, but it's considered to be bigger than anything else?

Go on.

Is there a similar idea for cardinals? Like, $\aleph_{\mathbb{Ord}}$ is bigger than any cardinal.

And $\mathbb{Ord}$ is bigger than any ordinal. Does that make them like an "infinity" for the cardinals and ordinals?

I am not too sure how chat flags work now.

Can I flag myself?

EDIT: Yeah, but it might annoy the hell out of moderators.

I think.

I don't know. There are so many things that can be flagged in this chat that I reserve it only for the extreme cases.

I only flag when there is a clear pattern of abuse by one user against another. Which almost never happens.

What if I start cursing at you, but it's in another language so you don't know that I'm cursing at you?

(Suka blyat.)

I think the one time I did it was against a mod who probably unsuspended himself.

bye again

Why are you running in and out like that?

@MikeMiller I thought you were talking to me. Curious mind.

I am going to shower, bbl.

@ABeautifulMind Mods can't be autobanned from flags in the first place.

If they get approved, nothing happens.

@GnomeSlice Hi, why did you come here?

nice to meet you too

I can't remember your name, Jasper here, the guy who always deleted accounts.

@GnomeSlice Is your name Shawn? I forgot.

Oh I think it's Michael.

@MaryStar I think you should ask your instructor instead.

@ᴇʏᴇs Still here?

Quick question you lot, with linear isometries, do the norms have to be the same?

Like it's a map from one vector space to another, where the norm of the thing is the same as the norm of the image, but what if you can't apply a norm to the image vector space?

@AlecTeal Oh, I get what you mean.

@PedroTamaroff How is school?

@ABeautifulMind Still on vacations.

@AlecTeal Say you have a map $f:X\to Y$.

And you have norms (or distances) in $X$ and $Y$.

Then $f$ is said to be an isometry if $f$ is distance preserving.

So you want $d_X(x,y)=d_Y(f(x),f(y))$ for all $x,y\in X$.

Or the same with norms.

So you can use different norms. I thought as much, just wanted to be sure.

anyone know if this equation $a^3+3a^2(\lambda)-4(\lambda)^2=0$

is factorable!

unless that $3a^2$ is the trace but the trace is the diagonal of the matrixx.x

So how are things you lot?

?

Are you unbanned @AlecTeal

No?

@I'mGettingThere What

Sorry

Eyes without a face?

what's an Elasticity Matrix

@usukidoll this nasty tensor thing you wont like.

Very similar to viscosity matrix

wait whaattt

I got the roots and the sensitivity ... just need elasticity if only I know what to do

Question on module theory. I came across a question I was about to answer: "Show that $\langle x \rangle = \{rx + nx : r \in R, n \in \mathbb Z\}$. My initial thought was to explain how we can pull the $n$ out as follows $nx = 1 \cdot (nx) = (1 \cdot n)x$ by looking at the action of $R$ on $M$, but I realized my reasoning presupposes $R$ has identity and the question doesn't mention this at all. Is this correct? If not, how would one show this?

So $nx$ looks at addition in the additive strucutre of $M$ and $1 \cdot n$ looks at the additive structure of $R$.

I assumed that $\langle x \rangle = \{rx : r \in R\}$ was the original definition they were working with.

@RobertCardona there would no point in writing rx+nx at all if R had unity

show any submod containing x contains all elts of the form rx+nx, then show this set of elts is a submod

(note <x> means the smallest submod containing x)

okay!

thanks!

I'm used to seeing $\langle x \rangle = \{rx : r \in R\}$.

or maybe I'm confused.

that's equivalent if R has unity

if R doesn't have unity then we say Rx={rx:r in R}, and <x> is the smallest submod containing x

Thanks! this clears it up for me!

You've got mail, @MikeMiller.

Not very interesting mail, but mail nonetheless. ;)

sigh :/

Can I have a hint on why matrix $A$ having $A^n = \operatorname{Id}$ means that $A$ is diagonalisable?

@Committingtoachallenge Over what field?

That's not true always.

Over $\Bbb R$ (Apparently not true over $\Bbb C$ said another student)

The other way around. =)

Really?

The lecturer agreed with him and limited it to $\Bbb R$ lol

He's also mistaken then. =/

I think it was being put on the spot(although you were put on the spot)

But over a field that's algebraically closed, like $\Bbb C$, you can use that $x^n-1$ has distinct roots, and since the minimal polynomial of $A$ must divide $x^n-1$, this minimal polynomial has distinct roots, i.e. no repeated roots.

Ahhh I see, I will see what I can do, thanks!

@Committingtoachallenge For a counterexample in $\Bbb R$, think about rotation matrices.

@PedroTamaroff Doing a full rotation, but being unable to get rid of the $\sin$ terms off of diagonal?

@Committingtoachallenge Come again?

Don't worry, I will think more about the counter example

Do you know that say a rotation matrix over $\Bbb R$ of angle $2\pi/3$ has no nonzero eigenvectors at all?

You can see that geometrically.

Hi guys :-)

hey guys

How are you?

Yup im doing well

U?

Fine thanks.

Do u enjoy learning mathmatics?

Im enjoy it

I

Sometimes.

I want to train and improve my anlaytical and critical thinking skills

Thats why i learn mtahs

Good choice.

How aout u?

ame fr me

Same for me

I enjoy learning it

Also beacuse i need it for physics

Hi maybe someone could help me. I was using Conway's book to learn complex analysis and in one question en the third chapter he said: can you map the unit disk conformally to the punctured disk. I almost sure that the answer is no but I can't prove it...

@JoseAntonio Do you know what it means for a region to be homologically simply connected?

Show that for the unit disk, every holomorphic function has a primitive.

Now show this fails for the punctured unit disk.

Finally show that if two regions are biholomorphic, and if one has the property that every holomorphic function has a primitive, so does the other.

I know what is simply connected the problem is that he didn't defined it yet. Neither that every holomorphic function has primite I suppose using Goursat's lemma as other sources.

@JoseAntonio The point is that for the unit disk, every holomorphic function integrates to zero over a closed curve. This fails for the punctured unit disk.

The canonical example being $z^{-1}$.

The chapter uses very elementary machinery. Integral are not defined yet.

Integrals are elementary!

it's a chapter before of the definition of integrals.

only conformality theorem is mentioned (the proof is left but is really easy) and a quickly explanation of the linear fractional and nothing else.

yes are elementary but Conway uses more elementary tools. Really I'm not sure how proceed.

@JoseAntonio What's the "conformality theorem"?

holomorphic preserves angle

@PedroTamaroff

@JoseAntonio OK.

@Committingtoachallenge You probably should not be trying to write a math paper without faculty guidance.

@MikeMiller I'm not at all, just looking at how to format long homework

It's fine, the words I were looking for are lemma and corollary mainly

Oh :) I've never done anything more than just writing down the numbers and then a proof.

It's very rare that I feel the need to prove a lemma, even in something long, because to think of subdividing in that manner usually requires some structure, and I guess I do most of my homework ad hoc.

@MikeMiller Yeah haha, it usually just gets a little messy for my liking

Sure, sure.

If you want to see what the average math paper looks like, just grab a section on the arXiv...

I just checked this out xD

@MikeMiller Good night.

Goodbye people. Please behave! >:)

Good night

@PedroTamaroff Reported for angry happy face

(e.g. mischievous face)

Morning Pedro.

maybe someone can help me with some hints. In the chapter 3 in Conway's complex analysis book (chapter before integrals): Can the unit disk can be mapped conformally onto the punctured unit disk. I don't know how proceed, there is not even definition of simply connected and not ntegral the only result is the definition of conformal and some easy result that the holomorphic functions are conformal when the derivative is non-zero.

Hi guys, can someone help me understand a set of vectors spanning a subspace? I did a couple of practice questions and got them right. But does the result imply anything? What's the significance.

hi

anyone have any ideas about math.stackexchange.com/questions/1179943/… ?

@robjohn Good morning, Did you have time to look at my problem ?

@howcan, The result shows us that a particular point in space can or can't be a linear combination of the vectors. It also helps us in getting minimum distance between a point and the subspace spanned by vectors,i.e. column space, If I understood your question correctly.

Hello,@robjohn ,this is old problem can't solution, can you see it,?text[math.stackexchange.com/questions/314548/…

Hello!! Is someone familiar with Ackermann's Function??

How can I prove that if $F$ is a field, then $F[x_1, \ldots, x_n]$ is not a PID (for $n>1$)?

I'm not really sure where I'm supposed to start.

@user112495 Look at the ideal $(x_1, \cdots , x_n)$. Is it a principle ideal?

Hi @Bal

Hi, @ᴇʏᴇs

@BalarkaSen I'm not really sure what the next step from here would be.

@user112495 Well, you have $(f) = (x_1, \cdots, x_n)$.

@BalarkaSen Isn't that what I wrote above?

Note that $x_i$ is irreducible in $F(x_1, \cdots, x_n)$.

@user112495 Yes. But you needn't be too explicit here.

@BalarkaSen Why is $x_i$ irreducible?

@ᴇʏᴇs Eyes = bad at math = nabla?

@user112495 Think about what irreducibility means.

@Committingtoachallenge Yes

Can you factor $x_i$ nontrivially in your ring?

@BalarkaSen Oh, no. Okay then.

Think about what being a generator means and somehow contradict/use this irreducibility of $x_i$.

Where $M(T,b_1,b_2)$ is the matrix of transformation from basis $b_1$ to basis $b_2$

Is that the correct interpretation?

Hello everyone, I apologize for leaving on such a sour note last night :L

@teadawg1337 Sorry if we offended you

@BalarkaSen We know that $f|x_i$. Since the $x_i$ are irreducible, f must be a unit, so $(x_1, \cdots , x_n) = F[x_1, \cdots , x_n]$. However, $1 \notin (x_1, \cdots , x_n)$. If it did, then there would exist $h_i \in F[x_i]$ such that $1 = \sum_{i=0}^n h_ix_i$. However, all the $h_ix_i$ have zero constant term, so the sum has zero constant term. Contradiction.

Well done!

@ᴇʏᴇs In retrospect, you guys didn't say anything offensive. I had a rough day yesterday, and I overreacted as a result.

@BalarkaSen Me?

The gist of this problem is really that $x_i$s are all $F$-linearly independent.

@user112495 Yes :)

I especially owe an apology to Mike... :L

@teadawg1337 I just read what I believed to be all of the text in the transcript and I must have missed this conflict(or it is being exaggerated here)

I don't see a conflict, to be frank. Some messages been deleted?

@Committingtoachallenge @Bal About scoring on the GRE

Something Mike said was removed, which is what I overreacted to.

Oh okay, I just see heaps of bragging about being great at math

lol

Also something I need to work on: modesty

If anyone should be bragging it's Balarka I must say(but he doesn't[well not directly ;P])

no, no, i used to have an enormously inflated head, @Committingtoachallenge.

I don't mind if people brag if they're really nice people

but i eventually stopped doing it seeing how little i know

:P

@Bal At least you'll know more than me by the time I finish college

@BalarkaSen There is a slang term for this in some circles of the internet: being a fedora.

Most people grow out of being a fedora and cringe at their past :P

I really have no reason to brag, I have so little experience and talent compared to many who frequent this chatroom

@ᴇʏᴇs you never know. my knowledge of mathematics is based entirely upon algebra and topology, and there is a lot more mathematics other than those. i can assure you i know nothing of analysis.

I had amazing scores in Math in highschool and I feel below average in university

Speaking of talent, has anyone seen beginner around lately?

@Bal I don't know any mathematics :(

I know like a little bit from each introductory course, but that's it

there's loads of time, @ᴇʏᴇs. study topology thoroughly.

I spent the first 2 years doing all my gen. ed. classes (like art, music, history, etc.) so I started taking math classes last semester

@Balarka Could you answer my question posted in the chat room 25min ago if you have time please

I am not familiar with linear algebra. I plan to study it this summer.

Oh okay, thanks anyway

Going to use Axler?

Well, I am thinking of Hoffman-Kunze. But maybe I'll get better books.

Didn't Axler sue someone for using one of the exercises from his book

Wouldn't know and also wouldn't change my opinion on the book(which is good :) )

H-K seems the tersest out there, so I guess I will use it.

@BalarkaSen You prefer it that way? Although I have to say Zorich's MA is the opposite and I dislike that

I like really long books that explain everything

@Committingtoachallenge I'm just excited to get to multivariable calculus. :P

Personally, I like huge books.

@BalarkaSen I can tell with you liking D&F :P

D&F has good exercises.

@teadawg1337 heya, any ideas how to figure out the brown/redish/maroon colored area in the figure? =)

@N3buchadnezzar Area of the sector minus the area of the blue triangle

$a \theta$, @N3buchadnezzar. You already know $\tan(\theta)$.

Oh, the area.

Well, it still shouldn't be hard knowing $\theta$

The sector is $\pi a^2/\theta$, and you already know the area of the blue triangle...

Ugh, I should be studying Mayer-Vietoris right now.

What's a Mayer-Vietoris

Sounds like a star or sometihng