The h Bar

8:00 AM

But does that mean it has no minimum/maximum on $[0, 1]$?

user228700

On the closed interval? No, the minimum value of the function would be given by $x=0$ and the maximum value by $x=1$, I think.

Balarka Sen

Exactly

So you have to study the function $f$ individually at the boundary of the set $[0, 1]$

That is exactly what your book does for a 2 variable function over a region on the plane

user228700

Riight. Analogously, for two-variable functions, we must examine the behavior of the function on the boundary curves of a given region, yeah?

Balarka Sen

Very correct

user228700

Cool, thanks! :-)

user228700

8:05 AM

I'mma eat lunch, then catch an auto to take me to the station to board my train. I'll be back once I'm comfortably seated in the train!

Balarka Sen

Safe journey

user228700

Zanks! :-)

user228700

(@JohnR: Lunch is quite appealing! Sambar and rice, as per usual, but with a side dish of sautéd cubes of brinjal!)

John Rennie

@Kaumudi.H Nice :-)

user228700

Most people I know hate brinjal. What about you guys?

John Rennie

8:07 AM

Though I have to say aubergine is not my favourite vegetable ...

user228700

Aha! :-)

Sid

isn't brinjal eggplant? Is aubergine the British name for that?

user228700

It can taste bitter and raw, if not cooked properly, that's true.

Anonymous

Any idea which equation they are talking about at 20:14?

Anonymous

Haynes Shockley Experiment on Drift and Diffusion

►

John Rennie

8:08 AM

@Sid yes

Anonymous

It sound's like "Bell's equation " or something

Anonymous

But on Googling I find nothing

Anonymous

Bell's equation for diffusion constant? What is that!!

John Rennie

@Sid it also turns to mush if cooked for a microsecond too long!

Fawad

Can anyone come to math chat and help me :(

user228700

8:09 AM

@JohnRennie That, too!

user228700

My mum/grandmum, however, happen to be experts at cooking brinjal perfectly because I like it so much! Hmm, that's a bit of a chicken-egg situation...

user228700

Ah, anyhoo, lunch awaits!

John Rennie

@Kaumudi.H apart from the hideous flavour and texture it's lovely :-)

Anonymous

@Fawad You can ask here...

John Rennie

Enjoy lunch :-)

Bernardo Meurer

8:10 AM

@JohnRennie Heya

Fawad

in Mathematics, 14 mins ago, by Fawad

how to solve for $\dfrac{1}{2!} +\dfrac{1+2}{3!} +\dfrac{1+2+2^2}{4!} +\cdots=\dfrac{1}{2}(e-1)^2$ ?

Bernardo Meurer

Sorry for the delay, I was out

John Rennie

@BernardoMeurer Morning :-)

Bernardo Meurer

Morning!

Fawad

in Mathematics, 12 mins ago, by Fawad

i figured out series is $\Sigma_{n=0}^\inf\dfrac{2^{n+1}-1}{(n+2)!}$

Bernardo Meurer

8:11 AM

@JohnRennie Can you lend me a server VM for a couple of days? I need to send some large files to Balarka and don't want to keep my computer on all the time (since I gotta go to class)

it's about 10 GB

Total

Anonymous

@Fawad What's the question ?

Fawad

@Blue I have to show lhs=rhs

starting from lhs

Anonymous

Just write the power series for $e^2$

Anonymous

And $e$

John Rennie

@BernardoMeurer I have loads of server space, but limited bandwidth

Anonymous

8:13 AM

Adjust them a bit and you're done

John Rennie

@BernardoMeurer If it's only 10GB just use a free Google Drive account. Free accounts allow up to 15GB.

Bernardo Meurer

@JohnRennie My GD is full :(

Anonymous

@BernardoMeurer Make a spare one :P

Bernardo Meurer

@JohnRennie @BalarkaSen's internet is really slow, 20Mbps, so bandwidth may not be a problem

and my US is not great either, It's mostly about space

John Rennie

Just create a new Google account ...

Rana

8:16 AM

Hi

Bernardo Meurer

::frowns::

I wanted to just rsync it all

Fawad

@Blue not helpful :(

Bernardo Meurer

@BalarkaSen Make a GD account and share a folder with my email

I'll put it the there

Anonymous

@Fawad It is.

Balarka Sen

@BernardoMeurer Alright, let me do that. I already have a GD account I think

Bernardo Meurer

8:18 AM

@BalarkaSen I just need like 6GB

Balarka Sen

I should def have that

lol I have used only about like 500 MB

out of 15

Bernardo Meurer

Alrighty

you know my email right, [last][first]@gmail.com

Sid

@Rana hello. Welcome to The h Bar. How can we be helpful?

Balarka Sen

yup i have it

@Bernardo Sent you a link

Bernardo Meurer

@BalarkaSen Looking

Fawad

8:25 AM

@blue what adjustments can I make?

Rana

Thank you @Sid, I am a newbie here

Anonymous

Find $\frac{1}{2}(e^2-2e+1)$ @Fawad

Fawad

😑

Sid

@Rana So.. what do you want to know?

Anonymous

@Fawad When someone is trying to help you out the minimum you can do is be respectful and not reply using emojis.

2

Rana

8:31 AM

Thanks for your attention @Sid, I am just thinking that for an event to happen both force and energy related concepts are there

Fawad

@Blue thanks for helping, my mistake I should have wrote rhs you dirty physicists :P

Sid

@Rana I don't quite get you. What?

Balarka Sen

@Bernardo I see things are there and coming along

How do I gauge the amount of time left for download to complete though?

Not sure how to do that in GD

Or, uh, maybe this is just uploading on your end

Rana

For example : let us consider gravitation scene of two bodies. We have both gravitation force to deal with the situation as well as gravitational energy and potential to deal with it as well. Hope I am not using confusing language

Bernardo Meurer

@BalarkaSen Yep, still uploading

Balarka Sen

8:36 AM

Right, I was being a n00b

Bernardo Meurer

Sending at about 2MBPS

Balarka Sen

'course you're uploading to the drive

@BernardoMeurer loool

Rana

But sometimes they don't cooperate well....

Fawad

@blue 😁🤤 thanks

Secret

First tidal locking, next (I forgot) and now stellar winds. Exoplanet habitability is SO conplicated

iopscience.iop.org/article/10.3847/2041-8213/aa6438/meta

enumaris

8:50 AM

hmmm, my cat dog classifier seems to be at about 88% accuracy...tho I only manually looked over 100 predictions to get this number lol...

Rana

Consider an experiment : take a solenoid in place free from other forces. And a magnet to bring flux changes. When magnet is brought near to solenoid, latter will repel (lenz) and magnet would move away, while moving away would decrease flux and make it to attract and so on.

Should I ping you @JohnRennie, @Sid for my above thought experiment

John Rennie

@Rana If your solenoid is connected to a circuit, so current can flow through it, then the magnet will decelerate as it passes through the solenoid. The kinetic energy of the magnet is converted to work done by the current. But the magnet will only be slowed.

That is, the magnet won't reverse direction and fly out again.

Rana

Not even at slightest push! @JohnRennie

Such that its kinetic energy is totally converted into potential energy just before the face of solenoid, moreover it will be repeling it too

Physics Meta

10:01 AM

0

Q: Chat option on Android app

The app is not showing direct option for chat. I have to open browser to get to chat. When I tap chat option, window comes to open it with app or with browser ; when I tap for app,, it opens browser.

feature-request

John Rennie

10:23 AM

@Rana I guess it depends on the mass and power of the magnet. If you simplify things by considering a single superconducting loop then the force on the magnet will be related to the rate of flux change as the magnet approaches the loop. With a powerful light magnet I think this could be great enough to repel the magnet and make it bounce back.

Anonymous

11:05 AM

@JohnRennie I'm having a bit of trouble understanding the concept of Landau quantization from Wikipedia. Basically I realize that for charged particles in a cyclotron have quantized energy levels. But for what size (type) of particles is this quantum effect observed ? Like electrons, protons or even large sized molecules? I guess as the size of particles increase the effect is less and less observable and the particles start following a spiral path (in the classical limit)

Anonymous

Any idea on what are the largest particles on which this quantization has been observed (in a cyclotron in the presence of magnetic field)?

John Rennie

@Blue I remember learning about Landau levels, but I have completely forgotten the details. Isn't it simply that the angular momentum is quantised?

Anonymous

@JohnRennie Well, it doesn't say anything directly about "angular momentum quantization".

Anonymous

In short: "Landau quantization in quantum mechanics is the quantization of the cyclotron orbits of charged particles in magnetic fields. As a result, the charged particles can only occupy orbits with discrete energy values, called Landau levels.

John Rennie

In a magnetic field particles move in a circle.

So they have a constant angular momentum.

Anonymous

11:08 AM

@JohnRennie Yeah, I got that

Anonymous

@JohnRennie Oh, okay. Yes

John Rennie

And the angular momentum is quantised in units of hbar

Anonymous

True. I was thinking about the size limit for this

Anonymous

What's the size range of particles for which this works?

John Rennie

In principle there is no upper limit, but in practice the energy spacing between levels quickly becomes so small that the quantisation can't be detected.

Anonymous

11:11 AM

@JohnRennie Mhm, I expect that it would reach the classical beyond a few hundred atom molecules and follow a spiral path

John Rennie

You can easily work out the energy spectrum for a particle of some mass m in a field B and you'll find the spacing becomes essentially zero for anything much above the molecular scale.

Why would it follow a spiral path?

Anonymous

@JohnRennie Ah, classical particles follow a spiral path in a cyclotron?

Anonymous

John Rennie

@Blue in a cyclotron the energy of the particle is increased twice an orbit by the electric field between the plates, so the radius of the orbit increases.

I can't see how that is related to Landau levels.

The particle orbits in a cyclotron are far too large to be quantised.

Anonymous

@JohnRennie The first sentence on Wikipedia (en.wikipedia.org/wiki/Landau_quantization) says "Landau quantization in quantum mechanics is the quantization of the cyclotron orbits of charged particles in magnetic fields"

Anonymous

11:14 AM

It's given a bit hazily on Wiki I agree

John Rennie

I think you're being confused by the temrinology

Anonymous

I guess so :/

John Rennie

A cyclotron orbit is just a circular orbit in magnetic fied.

The accelerator called a cyclotron is a different thing.

Anonymous

Ooooo....I see

Anonymous

I was thinking of a real cyclotron....that was pretty confusing

Anonymous

11:15 AM

Thanks!

Emilio Pisanty

11:34 AM

@JohnRennie @Blue Landau quantization is somewhat more complicated than just angular-momentum quantization

this is partly because the quantized levels generally have a preferred directionality

7

Q: For a constant magnetic field, is there a gauge with both canonical momenta conserved?

To describe a constant magnetic field $\mathbf B=(0,0,B)$ (ignoring the motion along the $z$ dimension) within hamiltonian (or quantum) mechanics, one needs to choose a gauge. One common gauge is the symmetric gauge, in which the vector potential is $\mathbf A=\frac12(-By,Bx,0)$ and the hamiltoni...

quantum-mechanics hamiltonian-formalism gauge-invariance gauge magnetostatics

John Rennie

@EmilioPisanty I last encountered it a long time ago :-)

Emilio Pisanty

I find it much more helpful to think of Landau levels as the quantized versions of cycloid-like orbits

i.e. a circular motion combined with a nonzero drift velocity

Anonymous

"Therefore, a gauge in which both canonical momenta are conserved is impossible."...ahh interesting

Anonymous

@EmilioPisanty Nice, that's a good way to visualize...gotta read more about this

Emilio Pisanty

@Blue yeah, Landau quantization is generally a problem with rotational symmetry (unless you're dealing with a finite rectangular strip that's long in one direction and confined in the other, which you often do) but there isn't normally a symmetric way to analyze it

i.e. you need to break the circular-orbit symmetry in some way, and then keep it in the back of your head that there is a 'hidden' symmetry that's no longer evident in your formalism

@Blue and yeah, this is a good thing to be learning. It's an incredibly important building block for stuff like this

nobelprize.org/nobel_prizes/physics/laureates/2016

(and btw, here's a nice tip: Nobel-prize lectures are very often an excellent place to learn about the topic)

Anonymous

11:43 AM

@EmilioPisanty Which lectures are you specifically talking about? (by nobel prize lectures)

Emilio Pisanty

nobelprize.org/mediaplayer/index.php?id=2690

Anonymous

nobelprize.org/nobel_organizations/nobelfoundation/publications/…

Anonymous

Oh, I see

Emilio Pisanty

nobelprize.org/mediaplayer/index.php?id=2692

Anonymous

Looks nice

Anonymous

11:44 AM

I'll see, thanks

Emilio Pisanty

those specific ones I hadn't seen

but e.g. the lecture text ("Read the Nobel lecture") by Roy Glauber, here nobelprize.org/nobel_prizes/physics/laureates/2005/…, is an excellent resource

Anonymous

Gotcha :)

Emilio Pisanty

ah, yes, this is the one I was looking for

Robert B. Laughlin's Nobel lecture on Fractional Quantization

featuring Landau quantization from about p.270 onwards

Qmechanic

0

Q: Categorizing condensed matter physics and statistical mechanics?

Categorizing condensed matter physics and statistical mechanics? How does one differentiate between these two? Can statistical mechanics be considered as a particular sub-field in condensed matter physics? Why is it so rare to see a research group on statistical physics/mechanics? But there wi...

statistical-mechanics condensed-matter terminology complex-numbers chaos-theory

Primarily opinion-based? Too broad?

Emilio Pisanty

12:04 PM

hey btw @Qmechanic, how come coherent-states synonymizes to hilbert-space?

no, wait, it doesn't

it's just a proposed synonym

@Qmechanic feels on the edge but still workable to me

no_choice99

12:27 PM

hey guys i've a question regarding partial derivatives. i really can't figure that one out

say we have sigma = sigma(T) where T depends on x. and I want to calculate the partial derivative of sigma w.r.t. x

sigma could be the conductivity, T the temperature and x the position

my doubt is whether this is worth 0 because sigma doesn't depend on x explicitely. It only does so implicitely

but if I blindly apply the chain rule, I get a non zero value

so which one is correct?

Anonymous

@no_choice99 Uh. No, it's not $0$, come to think of it.

Qmechanic

@EmilioPisanty : If there is a user that wants to be a custodian for the coherent-states tag then Ok, but if it is just applied only sometimes, then there seems to be little reason to have it. One is then better off synomizing to hilbert-space, and if needed, doing a direct search on the word 'coherent states' instead. Same with e.g. fock-space and quantum-states.

Anonymous

If $\sigma$ is a function of $T$ and $T$ depends on $x$, then chain rule is the way to go.

no_choice99

i mean if we were dealing with a lagrangian or thermodynamics functions, it would clearly be 0 since all the variables are kept constant except the one I vary, right?

Anonymous

$\sigma(T)$ can always be written as $\sigma(x)$ is that case.

no_choice99

12:30 PM

I agree

Anonymous

Try it with an example. Say $\sigma=aT$

Anonymous

And $T=x^2$

Anonymous

Then you can write $\sigma =ax^2$

no_choice99

I agree I can do that, but I can also do that for a lagrangian that depends on q(t)

Anonymous

Yes, it's always true.

no_choice99

12:31 PM

if t doesn't appear explicitely, then partial L / partial t = 0 despite that I can write it explcitely w.r.t. t

this is confusing me

Anonymous

Can you give an example where this doesn't work?

no_choice99

let me imagine a lagrangian, say L = 2(q(t))^2 where q is worth sin(t)

I can write it as L = 2 q^2

partial L / partial t = 0

Anonymous

@no_choice99 No. You can't write that. (Since $q$ is a function of $t$)

no_choice99

despite that I can also write L = 2(sin(t))^2 and in this case partial L / partial t is not 0

say we have a mass on a spring, its position is x(t)

lagrangian can be written in terms of x only (no explicit t)

yet I can replace x by an expression involving t

Physics Meta

0

Q: Why does the quantum-computing tag redirect to quantum-computer instead of the opposite direction?

I find this a confusing way to lay out our tag space. Lots of questions (say, this one as a recent example) are about quantum computing as a field, or about some specific aspect of a given quantum computation, without necessarily being about the specific hypothetical objects we call quantum compu...

discussion tags tag-synonyms

no_choice99

12:35 PM

and partail L / partial t is 0 when I wrote it using the notation involving x but not 0 anymore when i replace x by an expression involving t

Anonymous

@no_choice99 NOOOOOOO

Anonymous

"and partail L / partial t is 0 "

Anonymous

is wrong

Anonymous

Since $L$ consists of a function which is dependent on time $t$

Anonymous

This is basic differentiation. Check out Paul's notes or Wikipedia.

no_choice99

12:37 PM

mass on a spring L = 1/2 m (dot x)^2 + 1/2 k x^2 or something like that

Anonymous

Right. So $dL/dt=(1/2)m(2x)(dx/dt)+(1/2)(2)kxdx/dt$

no_choice99

what about the partial derivative

(not the total)

it's worth 0

Anonymous

No idea what you're saying. In this case partial derivative and total derivative means the same thing as $L$ is dependent on only $t$.

no_choice99

which implies energy is constant through time, right?

L depends on both x and t

Anonymous

No.

Anonymous

12:40 PM

It depends on $x$ which depends on $t$

Anonymous

It doesnt depend on $x$ AND $t$

no_choice99

ok I see

hmm great, I think I begin to understand

is it because x and t are not independent?

if x and t were independent then L would depend on both

Anonymous

Right

no_choice99

okay..... and in thermodynamics when we have S, V, N, etc. we assume they are independent, okay

uh wait maybe not S

but V and N are independent

well thanks a lot Blue

user228700

@Blue: I've reached my hostel ;_; (cc: @JohnR)

Anonymous

12:44 PM

Aaah, sympathies for you ;)

Anonymous

Good luck for the test tomorrow :)

user228700

Thanks, man :-) I will be around all night, as I continue to revise...

user228700

(I bought a can of Red Bull and everything, just in case :-P)

user228700

Don't you have an exam, too?

Anonymous

Me too..I've been procrastinating a lot ;_;

Anonymous

12:46 PM

Yeah, on Tuesday

Anonymous

I just want these tests to end fast

user228700

Ah, haha :-)

no_choice99

good luck guys and sorry blue to have taken some of yoru time. you did unstuck me though

user228700

@Blue Same. I have two more weeks to go, but I'm sure that they will fly by.

Felix.C

@JohnRennie Thx for explaining about the angular momentum...eventually I realised that when having a certain configuration, at least when having small numbers of $l,m$ and having let's say only 2 electrons it's kinda easy to see, why clebsch-cordans coefficients are how they are

user228700

12:48 PM

Oh, how I wish I could climb inside my laptop and render The h Bar a physical space at which to hang out.

user228700

Oh, well.

Anonymous

Hehe :P

Anonymous

We could actually do a hbar meet someday XD

Anonymous

Somewhere on this planet...lol

user228700

Yeah, right! x'D

Rana

12:58 PM

@JohnRennie, you might be being disturbed. I apologize in advance for that. How I analysed the situation : when magnet comes near the face of the solenoid, and the provided push is small enough to change into potential energy completely (come at rest) then it will be repeled and soon converted into kinetic energy and so on....continued

Initially it puzzled me but slight inspection told me that solenoid will be doing negative work on moving magnet (thereby not proving to be self perpetual) and cycle being completed. But if a bulb is connected in the circuit then it will radiate the transformed energy into light energy... (continued)

Then according to energy oriented concept energy is being radiated away but force of attraction and repulsion as per lenz law will continue to operate... Then these two approaches are not behaving in coordination. Will magnet stop moving...... Thanks for keeping patience

user228700

1:57 PM

Whoa, @Heather: Hello!! :-)

user228700

How's it going?

heather

@Kaumudi.H hello! pretty well; how are you?

user228700

Same :-) Although, my finals are starting t'row, so a little stressed, to the say the least, haha!

user228700

What have you been up to?

Sid

@Kaumudi.H it's tomorrow, isn't it?

user228700

1:59 PM

@Sid Yes...

Sid

Good luck for the big day

user228700

Lol, thanks! :-)

heather

@Kaumudi.H well, i've had quiz bowl competitions, robotics competition yesterday, all sorts of stuff - kinda a ton going on.

user228700

Wow, nice!

heather

@Kaumudi.H good luck! i'm sure you'll do great.

Sid

1:59 PM

@heather Quiz! SOunds interesting

user228700

Thanks, man! :-)

Sid

Quiz on what?

heather

quiz bowl is basically like really nerdy questions (literature, history, fine arts, science, etc) asked and you have to buzz in and answer.

an example question:

Sid

@heather Nerdy! Sounds great!

heather

> According to some traditions, the first day of this holiday commemorates the triumph of the son of King
Hima over Yama. It coincides with the Sikh holiday of Bandi Chhor Diwas, begins on Dhanteras and ends on
Bhai Dooj. In 2015, the Supreme Court of a certain nation refused to ban the use of (*) firecrackers during this
event, and it’s traditionally celebrated by the drawing of chalk rangoli designs and the lighting of lamps called
diyas. This autumnal holiday celebrates the return of Rama and Sita, and the birth of the goddess Lakshmi. For 10

the answer is of course Diwali.

since it's read aloud, you get the harder clues first, and the earlier you buzz in with an answer, the better.

(you can get extra points if you answer early enough)

Sid

2:02 PM

@heather that sounds too easy though..

(I would have got it during the firecrackers thing)

Anonymous

@Sid Well, an average American student wouldn't know that.

Anonymous

Requires knowledge of other cultures

heather

yeah, that was perhaps a poor example.

lemme find another...

> A passage in this poem paraphrases from Julius Caesar and describes “Between the idea / And the
reality.” It borrows from the Lord’s Prayer for the stuttering “For Thine is / Life is / For Thine is the,” and
images in this poem of (*) “Shape without form, shade without colour” and “Eyes I dare not meet in dreams” come
from The Inferno. Phrases in it like “A Penny for the Old Guy” and “headpiece filled with straw” reference Guy
Fawkes Day. This poem heavily references Heart of Darkness, as in the epigraph “Mistah Kurtz--he dead.” For 10

the answer is "The Hollow Men"

Sid

That is more interesting!

Any others?

user228700

@Balarka: You around, man?

user228700

2:11 PM

I have a silly question, with regard to Green's theorem. I have been asked to find the work done by a given force field given by $F$ on a particle that traverses counterclockwise along a given curve. Now, this work done would be the integral of $F.dr$, but, to make it so that it is in the form of the L.H.S of Green's theorem, my textbook has re-written it simply as F.(dx+dy) :-/ How does this make any sense?

heather

sure, you can read tons of packets of questions at quizbowlpackets.com @Sid

Anonymous

1

A: Find work done by a force field using Green's theorem?

The work done is $$ \oint{\bar{f} \cdot d\bar{s}} = \oint{ (f_x,f_y) \cdot (dx, dy)} =\\ \oint{f_x dx + f_y dy} $$ As per Green's theorem, this is $$ \oint{f_x dx + f_y dy} = \iint \limits_{D} \left ( \frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y} \right ) dx dy $$ Once y...

heather

or you can use a question reader like quinterest.org/quizbug.html

Anonymous

$\vec{F}.d\vec{r}=\vec{F}.(dx \hat{i} + dy \hat{j})$

user228700

@Blue Eyy, thanks!

user228700

2:15 PM

@Blue Ah, see, this is my primary question. How come this is true?

user228700

How is $dr=dxi=dyj$?

Anonymous

What are the $x$ and $y$ components of the displacement $d\vec{r}$ ?

user228700

:-/ When you put it that way...

user228700

I'm very, very dumb, it would seem. Thanks, man :-)

Anonymous

Hope you got it. :)

user228700

2:18 PM

I did :-D

Anonymous

If you want to convince yourself more, write the equation of the curve as $\vec{r}(t)=x(t) \hat{i} + y(t) \hat{j}$. Then differentiate both sides w.r.t the parameter $t$. $d\vec{r}$ by definition (of differentials) is actually $\frac{d\vec{r}}{dt}dt$. Probably Balarka can tell you more about 1-forms but I don't think you need those details at the moment.

user228700

:-D Yes, it makes sense. Thanks so much!

user228700

@Blue: I have one more question :-/ Dya have the time?

Anonymous

Maybe...just ask

Anonymous

:P

user228700

2:25 PM

'Kay :-P

user228700

Well, the statement of Green's theorem seems kinda...random to me.

user228700

What do those two functions $f(x,y)$ and $g(x,y)$...what, erm, why are they there, exactly? Is it just, hmm, do you see where I'm going with this?

user228700

It seems completely random to me, those two functions being there and then Green's theorem giving any easy way to compute that integral.

Anonymous

Green's theorem proof part 1 | Multivariable Calculus | Khan Academy

►

Anonymous

Green's theorem proof (part 2) | Multivariable Calculus | Khan Academy

►

Anonymous

2:28 PM

It'll take just 10-15 mins

Anonymous

Watch these two videos

user228700

Proof? Hmm...

ACuriousMind

Question to the Indians: Has "panchod" a non-offensive usage?

Anonymous

@ACuriousMind urbandictionary.com/define.php?term=panchod

Anonymous

It's offensive

Anonymous

2:31 PM

Where did you come across it? Someone in the comment section I guess...meh

ACuriousMind

@Blue I know the offensive usage. I'm asking whether there is a conceivably innocent meaning to it before I scold a user for using it :P

Anonymous

Not one that I know of...

Anonymous

If you could give me the context I might be able to say

ACuriousMind

@Blue It's "I edited it, that's all I got panchod" in response to a user requesting clarification of a question.

Anonymous

@ACuriousMind That's clearly an offensive usage (as far as I understand)

ACuriousMind

2:33 PM

Okay, thanks

user228700

@ACuriousMind (As far as I understand, too)

user228700

@Blue Hmm, although I haven't understood it fully just yet, I think I get that it's not random, lol.

ACuriousMind

I was mainly asking because I've had cases with spanish (I think? It's been a while) "curses" before that translate as very vulgar but aren't perceived as such by native speakers, apparently.

user228700

I see.

Anonymous

@Kaumudi.H It sure isn't random. :P

user228700

2:39 PM

My bad :-P

Anonymous

@ACuriousMind So, did you already join your new company? :) How are things going?

ACuriousMind

@Blue I'm starting in January

Anonymous

Oh, I see.

ACuriousMind

So currently I'm just chilling ;)

Anonymous

Hehe. Yeah, enjoy this time. After this you'll have only work, work and work!

ACuriousMind

2:47 PM

@Blue There's still vacations and holidays and weekends - I give them 8 hours of my workdays, not my life ;)

user228700

I have returned to pester you with yet another question!

user228700

So, when we evaluate line integrals along piecewise smooth curves, or, for that matter, when we evaluate line integrals period, we usually convert all of the things into parametric form, no?

Anonymous

Converting to parametric form normally makes things simpler

Anonymous

That's why we do that

user228700

Hmm, right. If we weren't to do that, say, we'd have to find the expression taken by the function over each piece of the curve and integrate that, yeah?

Anonymous

2:51 PM

Yeah, you could use a piecewise representation too (of $y$ as fn of $x$)

user228700

Right. Wokay, thanks! :-)

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