@JayeshBadwaik I'm sorry..i was interrupted...i needed your help..but i think your busy...do i disturb you?

@Charlie No, you do not (disturb me). Ask away.

@Charlie Maybe I can help?

Hi all

@OldJohn Hi!!

@JayeshBadwaik it's calculus thing.derivatives..some chain rule

@Charlie Okay.

@JayeshBadwaik just a sec andi will write the question

Who is andi? :-P

@PeterTamaroff It is quite likely that I will be utterly incompetent to help, and so, you will have to help anyhow.

@PeterTamaroff he can leap tall step functions with a single bound!

@robjohn like Jack the Ripper?

@JayeshBadwaik No, Superman leaps tall buildings with a single bound...

@JayeshBadwaik Just ask Jasper Loy!

@robjohn But Jonas is more human, more like bruce wayne!

@JayeshBadwaik you think that Jonas is a mere mortal?

afk for a while bbl

@robjohn How could I make that mistake! My basics must be slipping.

@PeterTamaroff Kinda!

@JonasTeuwen Kinda what?

This is all so confusing!

Kinda okay.

@Charlie This is simple transformation of laplacian from cylindrical to cartesian.

If $x=r\cos \theta$ and $y=r\sin \theta$ and $g(r,\theta)=f(x,y)$

i gotta show that equality

Do you know the formulas for divergence and curl in different co-oridinates?

@JayeshBadwaik yes..are you sure that that's is really necessairy?thought that chain rule from calculus 2 would give...

@JasperLoy Hi!!

@JayeshBadwaik I showed this:

@Charlie That is the shortest way. Just write a divergence of a gradient and you are done. But, I guess it can be done with chain rule, however, you might have to use the total differential. (I don't know what it is called in pure math, this is engineering terminology.)

@JayeshBadwaik hmm...

@JasperLoy Because people do not know what is the difference between 64-bit or 32-bit? Or for that matter between ARM and x86?

@JayeshBadwaik i did it then i found one part..but for the second derivative i got a little confused...

@Charlie First apply gradient and then apply divergence and you should be done. what exactly confused you?

@JasperLoy Hmm. I do not care, I get my software from pacman. I haven't visited the websites for over 9000 years.

@JayeshBadwaik i got too many derivatives...too many things goin' on..when i saw my notes,it was like:what the hell...

@Charlie Welcome! I am thinking that if you went into chain rule, things would be similarly clumsy.

@JayeshBadwaik but the objective is to use it :(

@Charlie Okay, is that given in the problem?

Hmm.

@JayeshBadwaik based upon what the prof gave to us...

@Charlie Okay!.

@argon Wie geht's?

So, now in chain rule, what's the problem?

@Charlie Good. I need to learn more German.

:)

@JayeshBadwaik Jay, when I make the second derivative...Jeez...what should I do?

I will write to you what I got.j a sec

@Argon :) Ich lerne...

@Charlie First simplify to remove/simplify terms like $\frac{\partial r}{\partial x}$ and then differentiate again and apply chain rule again.

Okay, now differntiate again wrt y and use the product rule to give an incredibly delightful mess.

@JayeshBadwaik :D

@Charlie good.

oh no..

@Charlie got ?

@JayeshBadwaik hmm...not yet..but i will

@Charlie Your first derivative seems to be wrong. (I just calculated it myself)

?

yes

@Jay it can't be!what's wrong?

@JayeshBadwaik hmm...

are..you...sure?

@Charlie and $$\frac{\partial f(x,y)}{\partial x}=\cos(\theta)\frac{\partial g}{\partial r}-\frac{sin(\theta)}{r}\frac{\partial g}{\partial \theta}$$

@Charlie No. Sorry. I think I am wrong.

@JayeshBadwaik I'm looking to my book...has something similar..it says i'm right...unless the book is wrong...

@Charlie No, you are right, I did something stupid.

@Charlie It all has to do with computing $$\frac{\partial(r,\theta)}{\partial(x,y)}=\left(\frac{\partial(x,y)}{\partial(r‌​,\theta)}\right)^{-1}$$

which is a simple $2\times2$ matrix inversion

@Jay

@Charlie I know. I said I made a mistake. A stupid one at that.

@JayeshBadwaik what mistake?tell me then i won't make the same

@Charlie Computation error. I calculated something wrongly in my head and missed some factor. I did not even try to write it down. Nothing conceptual.

@JayeshBadwaik Ok.

@robjohn Robie, could you explain that one better?think i didn't understand...

@Charlie http://en.wikipedia.org/wiki/Inverse_function_theorem

t2.gstatic.com/…

I'm off my rocker. Should go to sleep. It is a theorem from multivariable calculus. The matrix is a jacobian. and the reverse matrix is a reverse jacobian.

@Stefan Hallo!willkommen!wie gehts?

@JayeshBadwaik Thanks a lot , Jayesh!good night of sleep!

@Charlie Good night.,

@JayeshBadwaik :-D

good night,@jayesh!

daw

daw = observation of cuteness

oh...

@anon what is cute?

:>

@JasperLoy Jay seems to have fat cheeks good to pinch

We learn a lot here

@JasperLoy ?

@JasperLoy hmm...

@JasperLoy i see people who look like others all the time..in my uni there's a guy who looks like Pedro, i think,at least.then i stare him,thinking:hmm...no impossible.

@JasperLoy i don't know exactly how you look like...

@JasperLoy yes..pretty much,i think.

but i guess it's not the same as you look into one's eyes...

in person

@JasperLoy how do you look into someone's soul?

Jonas found me in FB!

...

@robjohn When you calculate the second derivative relative to $x$ , and how $x=r\cos \theta$, I'm sorry for the terrible question ... but $\cos\theta$... is it a constant when you differentiate it in terms of $x$?

@JasperLoy hey, look what I found makes sense...

@Jas Are you fine?

@SimonSheehan Jeez..only now i realized your gravatar is Darth Vader and Boba Fett...

@Charlie OK - I will :)

\o/

Oldie !!

@Charlie Hi - how are things?

@OldJohn cool.what about you?many holidays?

@Charlie Averaging one holiday a month at the moment - it is wonderful :)

@OldJohn that's how life should be lived, huh?

@Charlie Yep - for as long as the money (and health) holds out :)))

@OldJohn of course, of course!

- and still doing a bit of maths inbetween holidays :)

@OldJohn Sweet!

I want to be like Oldie John when i grow up

But now I have the freedom to do whatever maths I feel like doing

@Charlie Hmm - you want grey hair, wrinkles and a beard ???

@OldJohn I think that this is the best.We do what we wanna do.

Odd

Hello Pie boys.

@OldJohn i already have some gray hair... but no beard..

@JonasTeuwen Hi!!!

@JonasTeuwen Yo dude :)

@Charlie good to hear!

@OldJohn i never thought that with 20 years i would have some gray hair

@Charlie It happens - I have a nephew who has more grey than I do :)

@OldJohn wow

Have to go - back later

alright

@Charlie and @OldJohn Hi guys :))

@JonasTeuwen Hello, Jonas!How are you?

@Argon you are back!

@Charlie Hello again Charles

no jookes, i promise...

@Argon call me charlie, or marilyn

or marília

@Charlie Marilyn? :)

marília

I like that

@Argon yeah.i'm girl

@Argon :D

@Charlie How about that. Where did "Charlie" come from?

what about your true identity?

@Argon Chaplin.I'm a fan

@Charlie I see

@Charlie Aaron - hence, Argon

@Argon I like Aaron!

@Charlie Good. Pedro likes Arthemis.

@Argon what?

@Charlie I told him Argon was made from part of my name, so he thought it was Arthemis

Also, Argon comes from αργον meaning "lazy" or "the inactive one," according to Wikipedia... quite fitting, actually

@Argon wow...Aaron is much better

@Argon my name means "mary's daughter"

@Charlie Apparently, Aaron has unknown meaning from ancient Egyptian or Hebrew.

@Argon "From the Hebrew name אַהֲרֹן ('Aharon) which is most likely of unknown Egyptian origin. Other theories claim a Hebrew derivation, and suggest meanings such as "high mountain" or "exalted". In the Old Testament this name is borne by the older brother of Moses and the first high priest of the Israelites. He acted as a spokesman for his brother, and carried a miraculous rod. As an English name, Aaron has been in use since the Protestant Reformation."

Sure :)

@Argon Elvis presley was Aaron

@Charlie Really??

Where'd he get "Elvis" from?

@Argon Elvis Aaron

@Charlie Ah. Middle name

@Argon yep

@Charlie It seems "Marília" is a city in Brazil.

@Argon yes, it is

@Charlie Wait, where are you from again?

@Argon Brazil

@Charlie Ah. You should move there just so you can tell people you are from Marília.

@Argon it's a soccer team too, if i'm not wrong...

Hi, is this the right room to ask a math question?

@FrankLen Yes

@FrankLen think so

@FrankLen The math gods implore thee to ask thy query.

how does one know that a matrix becomes upper triangular?

very nice @peoplepower

*when is perhaps more correct than "that"

oh, its that complicated or was my question just very...broad?

I don't know linear algebra, unfortunately.

I think there is some inactivity right now. You could ask again later.

Impressive: We've doubled the number of 100k+ users since yesterday.

@HenningMakholm How many did we have yesterday? One?

Thats ok

@FrankLen just a sec ..i will check that...

@Argon Yes.

look at the last sentence

@HenningMakholm Who has 100k+?

no

ok

@Argon Arturo Magidin (on hiatus) and André Nicolas.

@HenningMakholm Wow, impressive!!

@Jasper Jassspeerrrr!!!!

@Charlie Could be better. I have this feeling of impending doom.

@JonasTeuwen oh, my dear... why?

I don't know! 8-)).

@JonasTeuwen Try to distract your mind...not to think about such things... keep these thoughts away from you...

@HenningMakholm LOL

Hmm, I am not really thinking, so I don't need distraction. It is a feeling.

@JasperLoy everything fine.What about you,honey?

@Charlie Woah, we're "honeying" each other now?

@JasperLoy Hey

@PeterTamaroff why?can't I?

Damn those downvoters

@Charlie You surely can. But how do you know Jasper is not a heartless bastard?

ok, jasper, I got it. Just assumed it would be a tad more complicated, thats why I asked.

@PeterTamaroff I don't know...i try to believe neither of you are...

..okay..don't call anyone honey, dear, love, or any cute name

okay

liebchen is cute too

@Charlie If you are speaking German...

@Argon yup

German does not usually sound like a cute language

to my ears

@Argon but i like liebchen...Mein liebchen...

@Argon what language is cute for you?

@Charlie I don't know. Definitely not German, however.

Excuse me, my dearest friends, but I will have my dinner.Have a good day, evening!

;-D

If this is something you have knowledge in

@JasperLoy i don't have neither..so i call honey whoever i want

don't take too seriously

i just think it's funny

"(2x−5y)⋅ey" if you derivate that based on f'y(x,y) as part of partial derivates, should one use the product rule or the 'core' rule or whatever its called in english?

@JayeshBadwaik Excellent. I found some good python bindings for Mendeley (local).

oops, the equation was wrong, it was: (x^2−5x⋅y)⋅e^y

anyone know what rule to use on this?

someone told me the product rule, but im not quite sure if thats right

derivative based on f'y(x,y)

nobody?

@skullpatrol Sorry for no reply earlier skullpatrol, didn't have enough points to enter the chat. It was indeed my first time here.