Mathematics - 2012-10-09

@Charlie when differentiating wrt $r,\theta$, we have that $x,y$ are functions of $r,\theta$. partial wrt $r$ holds $\theta$ constant, but not $x,y$; both of them are functions of both $r$ and $\theta$

Charlie

@robjohn hmmm

robjohn

Remember that $\displaystyle\frac{\partial}{\partial r}=\frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}$

Charlie

@robjohn Thanks, Robie!

robjohn

12:36 AM

and $\displaystyle\frac{\partial}{\partial \theta}=\frac{\partial x}{\partial \theta}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \theta}\frac{\partial}{\partial y}$

Then just apply those as many times as you need.

Charlie

Ok.Thanks again!

robjohn

I have to go, if you have an example that you want to work through, we can do that later.

Charlie

Ok.

Bye!

robjohn

2:37 AM

I'm all alone!

Austin Henley

hi

robjohn

3:06 AM

@AustinHenley what's up?

@JasperLoy: Hey there...

user19161

@robjohn Ah, there are three people now!

robjohn

@JasperLoy The third one is invisible until you hover over them.

I may not have been alone when I thought I was.

user19161

@robjohn Yes. At first I thought it was Austin above transformed, but it seems not.

robjohn

@JasperLoy They've been inactive for about 5 minutes, so they have been here for a while.

user19161

@agleaner Now that you have 20 rep, you can chat. I just upvoted your question so you have 21 rep.

user19161

3:23 AM

@FrankLen You need to know how to do matrix multiplication. This should be something covered in your class if you are asked to do it. In short, write the variables to form column vectors x and y. Write the coefficients of the kth equation as the kth row of the matrix A.

Jayesh Badwaik

The three ghosts in the chatroom..

Blue, Orange and Blackish-Red?

Mariano Suárez-Alvarez

6:59 AM

Hm, essentally no one here!

robjohn

7:46 AM

@MarianoSuárez-Alvarez we're essentially no one...

Old John

@robjohn I am nothing :)

Jayesh Badwaik

9:07 AM

Around Two and a half hours more for the chat to become more active.

robjohn

@JayeshBadwaik you ruined our wonderful idle time...

2

Zhen Lin

having achieved nothing in the last week feels about the same as being no-one...

robjohn

$$ $$ 1 hour earlier...

Jayesh Badwaik

@robjohn :-( You have been idling for over an hour now. ::P :P

Silly notation question. What does $\ell^p < \ell^{p+1}$ mean? Where $\ell^p$ is the normal notation for the $p$-norm space. Is the notation specific to only this or is it some kind of subspace order defined on spaces?

Zhen Lin

$<$ is frequently used to indicate a substructure.

Jayesh Badwaik

9:13 AM

@ZhenLin Okay, so in this case, since it is a vector space I am dealing with, it will be used to denote a subspace, right?

Zhen Lin

It is.

Jayesh Badwaik

Thanks.

robjohn

10:19 AM

@JayeshBadwaik To me it looks like inclusion.

skullpatrol

11:43 AM

$$ $$ 1 hour frozen in time...

infinitesimal

11:57 AM

@HenningMakholm Would you please make a quick comment on my answer here?

Henning Makholm

@infinitesimal I'm afraid I don't understand what it is you're trying to say.

Oh, you're suggesting an answer to the original question?

It seems rather random -- you add in words like "linear combination" that serve only to restrict what the "something" can be in "7+something", where actually to make the analogy work we'd be looking to name a concept where the "something" can be anything.

infinitesimal

@HenningMakholm Is there not a similar concept in the solution to indefinite integrals where the arbritrary constant C is added on?

Henning Makholm

12:14 PM

I don't think "7+something" is a particularly fruitful concept in integration. Of course addition itself is an important concept almost everywhere, but the concept of "something that is the result of adding 7 to something else" is not an interesting one.

Jayesh Badwaik

@infinitesimal And product need not be a multiple, if I multiply $7 $by $1/7$ then? Is $1$ a multiple of $7$? I think the question is farcial. If we restrict ourselves to natural numbers all the way, then set of product of a natural numbers by elements of $\mathbb{Z^{+}}$ is a set of multiples, and the corresponding set is a set of all numbers $a + n$ where $n = 1,2,3,\ldots$. I am not sure what name you want to give to it though. Since, it will cover all $b > a$ and hence is too general.

infinitesimal

@HenningMakholm OK fair enough, thanks for responding.

Jonas Teuwen

12:41 PM

@JayeshBadwaik It is just the fact that they are embedded (sequentially). Not only they are set-included, also the inclusion is continuous (iirc). That warrants another symbol. It is a common thing anyway, to define orders $\leqslant$ on spaces that are not necessarily numbers. See for example nets for an example!

Jonas Teuwen

1:01 PM

@JayeshBadwaik Actually Org-mode is our LaTeX cms.

infinitesimal

@JayeshBadwaik thank you also for your response.

Jayesh Badwaik

@JonasTeuwen Hmm, yup, I have seen order defined on variety of things now. But I was not sure about this, its clear now though. Thanks. I looked up embedding again. I guess you mean this for statement about continuous embeddings.

@JonasTeuwen Org-mode is just that. The Organizer. It can probably act as CMS, but then you are basically customizing its parser to make lists.

N3buchadnezzar

@JonasTeuwen Joonaaaaass

Jonas Teuwen

1:16 PM

Hi. And bye, need to go!

N3buchadnezzar

@JonasTeuwen Give me one responce, what is the best way to judge what to buy regarding liqour?

price of one litre / percent ?

Jonas Teuwen

If you want to pour it in one number, that ratio is the best!

N3buchadnezzar

:D Thanks!

Martin Sleziak

1:35 PM

If you're buying it as a present for someone, I would not go by that number....

N3buchadnezzar

It is a magical number of wonders.

Just imagine a bunch of guys shouting in the store "How low can you go!?" =)

Ragib Zaman

@MartinSleziak Do you know anything about algebraic curves/algebraic geometry?

Martin Sleziak

@RagibZaman Sorry, I do not know algebraic geometry :-(

Ragib Zaman

@MartinSleziak No need to be sorry lol. Just wondering if the complex analytic approach is worth learning or the "modern" approach via commutative algebra is sufficient.

I'm pretty sure Zhen would know heaps about this actually.

@ZhenLin I've browsed some introductory books and found one by Griffiths that gives an approach via complex analysis, and another by Fulton that develops the theory via commutative algebra. I am definitely going to learn the comm. alg approach but I'm wondering if learning the approach via complex analysis/riemann surfaces is something I should do? Is that what many other people do? What would you recommend?

Matt N.

2:39 PM

Hi

Anyone up for solving a puzzle with me?

Maybe @OldJohn?

Ok, looks like I'm the only one in here in the mood for a puzzle.

Old John

@MattN. Hi - I'm back :)

(but not for long ...)

Charlie

4:23 PM

@JayeshBadwaik Good evening!

Jayesh Badwaik

@Charlie Good afternoon!

Charlie

@JayeshBadwaik Yay!!!How are you?

Jayesh Badwaik

@Charlie I am fine. Wat about you?

@JasperLoy Hi! Good midnight.

user19161

@JayeshBadwaik Hi! The inner demons are very strong these few days.

Jayesh Badwaik

@JasperLoy Ohh. :-( I hope you are handling them well.

Charlie

4:30 PM

@JayeshBadwaik good !better now!

Jayesh Badwaik

Good!
@JasperLoy I sent you a mail.

Charlie

@JayeshBadwaik i read.

Jayesh Badwaik

@Charlie I sent Jasper a mail. :P

user19161

@Charlie Don't be jealous. Jayesh is mine!

Charlie

@JayeshBadwaik but i read your too

Jayesh Badwaik

4:34 PM

@Charlie Okay. :-)

user19161

@JayeshBadwaik So many secrets flying all over?

Jayesh Badwaik

@JasperLoy Some. Story telling is not going on for sure! :-) But yes, secrets are being discussed.

user19161

@JayeshBadwaik OMG! The secrets and people involved will form the nodes of a complicated graph.

Jayesh Badwaik

@JasperLoy It will be a directed graph.

user19161

@charlie did not say hi to me today. I wonder whether she is pissed with me.

Charlie

4:56 PM

@JasperLoy many secrets between Jay and I :P

$\huge\text {Hello Jasper!}$

user19161

@Charlie Any involving me? =)

Charlie

@JasperLoy no.

things between jayesh and i

user19161

@Charlie So exciting!

Jayesh Badwaik

@JasperLoy Your secrets stop inside my head, I don't let them get out.

Charlie

hmm

user19161

4:59 PM

@JayeshBadwaik Ah OK. I shared my secrets with Peter Jonas Ben Marilyn and you. The five of you can share among yourselves.

Charlie

ok

Jayesh Badwaik

@JasperLoy Okay.

Charlie

So ,@Jayesh...

what have you been doing?

Jayesh Badwaik

@Charlie I am currently reading this book "An Introduction to the Theory of Computation" -- Micheal Sipser. Right now, reading this.

Charlie

when i see" as many questions as you'd like"...

i go wild!

Jayesh Badwaik

5:08 PM

@Charlie :-) :-)

Charlie

once

Jayesh Badwaik

I had once gone to a lecture, at the end of which there were going to be questions. I did not know that it was necessary to give the questions beforehand, so after the lecture ended the organizer said "Now we will have questions." and I was a silly person with my hand in the air while all others were seated quietly and the organizer anyway went to some other kid. And then I raised my hand up again! :P :P Finally, I asked my question and I got an answer! :P :P
This was in my city's science center in high school and it was all formal and stuff. With around 200 students in the audience.

Charlie

when i was visiting a university, i went to the physics institute...then they did some interesting stuff to show the students,blahblah....so, There was a prof who was talking about Quantum...@jayesh He said:"you can make any questions..."
Ok !then , after"Questions " he said i could do...he was like"ok...no more questions,or she will keep asking me..."

@JayeshBadwaik :D

@JayeshBadwaik DID YOU SEE PHYSICS NOBEL PRIZE?

Jayesh Badwaik

@Charlie :P :P Which university you visit?

@Charlie Yes, I just read on my RSS feed, have opened it up in some tab, now will read it.

Charlie

@JayeshBadwaik local ones.my and another,called "Mackenzie"

@JayeshBadwaik I like the way you think my dear

user19161

5:13 PM

@Charlie Dear?

Charlie

@JasperLoy yes.why?

user19161

Oh OK, you can use whatever words you want I forgot.

Charlie

Do you mind, @Jayesh?

Jayesh Badwaik

@Charlie No. I was going to say "me too" as a reply to that. :-)

Charlie

@JayeshBadwaik :D

Do you see @jas... He's nice .

user19161

5:16 PM

@Charlie Hey using @jas... does not ping.

Charlie

@Jasper

user19161

Testing:@cha...

Charlie

what about now?

no,it didn't work

user19161

I think there must be a space, like @cha for example.

Charlie

@JasperLoy ok it did

OMG

Jayesh Badwaik

5:20 PM

@Charlie Nice it is!

Charlie

@JayeshBadwaik AWESOME, isn't it????

user19161

@Charlie OK, the two of you can get married then.

Jayesh Badwaik

@Charlie Yup. It is.

Charlie

:D

Jayesh Badwaik

This company says that it makes quantum computers (which is actually one of the application of that phenomena.) It recently built a 2 bit quantum computer, also IBM has been able to build a quantum factorizer that was able to factor 15 into 5 and 3. You can find it on google!

The company has shipped one such computer to USC and Lockheed Martin.

user19161

5:26 PM

@JayeshBadwaik I can factorise 15 as well. So what?

Jayesh Badwaik

@JasperLoy Well, its a start (for the computer). The success was in the proof of possibility of physical implementation of the shor's algorithm.

Right now, physicists are not able to maintain more than a 2-3 qubits for more than appreciable time, and hence the limit.

Nimza

Good evening!

user19161

@JayeshBadwaik Oh, Shor is on ELU!

Charlie

@Nimza Hi,Nimza!

Nimza

Maybe somebody knows, is there an easy way to convert ".tex" file to Word file with MathType?

@Charlie hello :)

Charlie

5:29 PM

@Nimza wassup?

user19161

@Nimza Have you tried google? I don't know.

Jayesh Badwaik

However, if it becomes possible to maintain more qubits in future, it makes it possible to factorize large prime numbers quickly. One bad effect is your Public Cryptography would be destroyed. (However, there are other techniques then.)

@JasperLoy Peter Shor?

user19161

@JayeshBadwaik Yes.

Nimza

@Charlie ok, how are you?)

Jayesh Badwaik

@JasperLoy Awesome. What is he doing there?

Charlie

5:30 PM

@Nimza Good!

Jayesh Badwaik

Or may be its just someone using his name as a screen name.

There is a Haskell Curry on MSE. @JasperLoy

Charlie

@JayeshBadwaik more likely

user19161

@JayeshBadwaik User. Just like us.

Nimza

@JasperLoy yes, I've found an "addon" to MathType that allows constructing formulas using tex. But... only "formulawise"

user19161

@Nimza So just paste the whole text file into word to start!

Nimza

5:32 PM

@JasperLoy yeah, yeah... this text is HUUUUGE

Jayesh Badwaik

@JasperLoy He does participate on SE. I have seen his answers somewhere on some community, I cannot remember.

user19161

@Nimza Well, then better start pasting!

user19161

@JayeshBadwaik Like I said, ELU!

Jayesh Badwaik

@JasperLoy no, probably MO and TCS.

Nimza

@JasperLoy problem is time... ok, I'll do something instead of speaking)))

Jayesh Badwaik

5:34 PM

@Nimza You have probably seen this but still. Also this.

Nimza

@JayeshBadwaik thank you for links!

Jayesh Badwaik

@Nimza Also, if you can do HTML -> Word. Then I would suggest using tex4ht which is a native TeX program (unlike latex2html) but produces HTML.

Charlie

@JayeshBadwaik that's something!

Jayesh Badwaik

@Charlie About your FB, is there a chance for being all of the three? ;-)

Progengineering Unicorn?

Charlie

@JayeshBadwaik awnnn

:D

@JayeshBadwaik you can be anything!

@OldJohn Hi Oldie!

Old John

5:49 PM

@Charlie Hi there

Charlie

:D

user19161

@jonas Have you eaten dinner?

Jonas Teuwen

No.

user19161

@JonasTeuwen How is your research coming along?

Jayesh Badwaik

I was going to post this. However, 9 minutes ago, the event was delayed due to high winds.
[An awesome event going on.](http://www.redbullstratos.com/live/) A man is jumping from around 30000 meters.

user19161

5:54 PM

Hmm, I am quite bored of the Eng and TeX sites now. I think I will just hang out on Math full time!

Charlie

@JasperLoy finally

Jayesh Badwaik

@JasperLoy Do not delete the accounts! People lose reps!

user19161

@JayeshBadwaik Yeah I think I will keep them, since I have more than 1k on each.

user19161

@JayeshBadwaik Are you still sleeping at odd hours?

Jayesh Badwaik

@JasperLoy No, I am improving. I generally sleep at 12 now, which I guess is okay. However, some days I just cannot sleep and then sleep at 2 or 3, even though I am very tired and practically useless. :-(

user19161

6:46 PM

@JayeshBadwaik Still there? Or preparing to sleep?

Jayesh Badwaik

@JasperLoy Preparing to sleep.

user19161

@JayeshBadwaik I will see you in your dreams.

user19161

Hello @anon. Is your headache gone?

anon

7:02 PM

hello. yes it was yesterday.

Charlie

Jasper loy.the man,the legend.

anon

7:22 PM

"The tensor product of two indecomposable representations $V_\lambda\otimes V_\mu$ (with diagonal action of $\mathfrak{S}_n$) decomposes into a direct sum of indecomposable representations of $\mathfrak{S}_n$, and in this case it is an open problem to describe the summands and their multiplicities. The problem naturally generalizes to any finite group." No wonder it's been so hard $-$ it's an open problem!

Old John

8:27 PM

@anon Yeah - some of those can be quite tough in my experience :)

user19161

8:37 PM

@OldJohn Hi!

user19161

Beaten to an answer again, damn.

user19161

@Charlie I am only a lunatic.

Old John

@JasperLoy happens to me all the time - I am not the fastest typist :(

Dan the Man

8:58 PM

How do I factor a polynomial?

Old John

@DantheMan depends on the degree - for large degree it can be tough

Dan the Man

$\frac{4x^2}{y^2}-\frac{36x}{y}+81$

Old John

I would first rewrite it as $\frac{1}{y^2}(4x^2-36xy+81y^2)$

or maybe replace $\frac{x}{y}$ by $z$

In the latter case, you end up trying to factorise $4z^2-36z+81$, which is much easier :)

Dan the Man

One sec.. my ChatJax isn't working

Woah... sorry epic fail...

What I meant to say is, "How do I find the square roots of a polynomial?"

Old John

@DantheMan In the case of the one you gave above, it factorises as a linear factor squared, so the square root is just the linear factor

Dan the Man

9:11 PM

@OldJohn Sorry, I can't understand that

user19161

@DantheMan Well, by observation that is a quadratic in x/y.

anon

@DantheMan Which part? Do you know what "linear factor" means? Do you know what "factor as a square of __" means?

Old John

$\frac{1}{y^2}(4x^2-36xy+81y^2) = \frac{1}{y^2}(2x-9y)^2$

so the square root is just $\frac{1}{y}(2x-9y)$

Dan the Man

@OldJohn Ah ok! I understand now.

user19161

@OldJohn Or rather, its absolute value.

Dan the Man

9:13 PM

thanks for the help

Old John

To check that you really understand, you ought to try one that I give you which is similar :)

Dan the Man

I have one

$4a^2-20ab+25b^2+4ac-10bc+c^2$

user19161

Haha.

Old John

OK - can you do that one?

Dan the Man

brain fart

don't know where to start.

user19161

9:16 PM

@DantheMan First factorise the first 3 terms into one product.

Old John

think of it as $(\lambda a + \mu b + \nu c)^2$, and try to work out what numbers $\lambda, \mu, \nu$ have to be

anon

It will be a square of something, that something will be of the form ([]a+[]b+[]c+[]), where the []'s are constants. (We see this by looking at degrees.) The last [] is zero (why?) and the other ones can be gleaned by looking at the coefficients of a^2,b^2,c^2 in the original polynomial and thinking about signs

bad idea using underscores

Old John

@anon :)

user19161

But students are not typically thought about squares of three terms added, so this is an extension.

user19161

I think this must be in the challenging section.

Old John

9:18 PM

@JasperLoy back in my day it would have been in the easy section :)

Dan the Man

$(2a-5b)^2+4ac-10bc+c^2$