The h Bar

Yashas

07:00

How?

user228700

@JohnRennie :-)

Anonymous

Whenever 0 to infinity is given in definite integration always check whether 0 to 1 cancels out 1 to infty

Anonymous

In 50 percent of cases, it does

Yashas

The area in $[0, 1]$ appears to be more than the area in $[1, \infty]$

@blue that wasn't the point

Anonymous

@Yashas 1 to infty is spread over a much much larger area

Yashas

07:01

this graph is not obvious

John Rennie

@Kaumudi.H They'll probably just leave it until Tuesday. It's configured as RAID6 so it can survive two disk failures.

Yashas

I know that in many cases with integrals with infinite limits usually have a zero part

Slereah

@0celouvsky any idea what this means?

The $A \star B $ part

Anonymous

@Yashas Cannot be determined =P

Anonymous

The question is incomplete

Anonymous

07:05

lol

user228700

@Yashas 25?

Yashas

@Kaumudi.H :O That is too huge.

user228700

Oh crap! >.< I'm an idiot.

user228700

5?

Yashas

25 is the max

1 is the min

user228700

07:08

How is 1 the min?

Anonymous

All the events are in past tense and the question is in present tense. Nonsense!

Yashas

I don't know the answer

@Kaumudi.H 1 is not the min, hmmm.

5 is the min

user228700

Initially, I thought that each elephant sees two different monkeys but all of them probably see the same two monkeys. Each monkey has one parrot in hand, which makes for 2 monkeys & 2 parrots. Let's not forget the rabbit and there, that's 5 animals.

user228700

If each elephant sees 2 different monkeys each time, that would make the number 25.

user228700

How can 1 be the minimum? There are at least 5 animals going to the river; one rabbit, two monkeys and two parrots.

John Rennie

07:12

God it's hard finding unused addresses on Gmail.

Yashas

The rabbit saw XXX while going to the river.

The rabbit might already be at the river.

Every elephant "saw" two monkeys.

The answer could be zero lol

user228700

"How many animals are going to the river?" "1 rabbit saw 6 elephants while going to the river"

Yashas

it is in the past

the question is in the present

Anonymous

I said it -_-

user228700

@Yashas -_- It's not a very good puzzle, then. Stupid wordplay is stupid.

Anonymous

07:15

@Kaumudi.H Almost all questions of this kind are stupid =P Be prepared to waste time when attempting them

Yashas

it made us think deep :P

user228700

@JohnRennie Sure is. Which is why I am so grateful to my past self for having made a decent id instead of something like "[email protected]" :-P

Yashas

I have 4 email ids :D

John Rennie

Boring gmail addresses are always good in the long run. Email addresses that seemed funny at the time tend to seem silly after a few months.

user228700

Yep yep yep.

John Rennie

07:17

I have to set up a new gmail address for my brother's mother in law (my aunt-in-law?) and it's proving troublesome.

Balarka Sen

@Slereah nah, not acid. cocaine

user228700

@JohnRennie Wasn't she the one for whom u fixed a computer? Man, you two are close :-P

Balarka Sen

@JohnRennie 'twas hip in the 70's

John Rennie

Her name isn't particularly unusual, so all the obvious choices are taken.

Balarka Sen

i mean, not snorting milk. you take bits down by drinking the milk

John Rennie

07:19

@Kaumudi.H I fix computers for my family, my friends, my family's friends, my family's pets, the local wildlife ... you get the idea.

user228700

:-P

John Rennie

My Mum tells me that one of her friends wants me to fix their iPad. Mind you that lady pays me in chocolate bars :-)

user228700

Aww :'-)

Balarka Sen

@Kaumudi.H relevant is the ever-present xkcd on them

user228700

Heil Randall Munroe!

Balarka Sen

07:23

he's a good guy

Anonymous

@Yashas Prove that composition of two differentiable functions is differentiable. How to?

Anonymous

Intuitively it seems correct

Yashas

did u try using first principle?

Anonymous

Trying

Anonymous

Any hints?

Anonymous

07:30

Say $h(x)=f(g(x))$

Anonymous

Let us apply chain rule.

Anonymous

$h'(x)=f'(g(x))g'(x)$

Balarka Sen

If you apply the chain rule you're, like, done. That's what you have to prove.

Anonymous

@BalarkaSen Is that sufficient?

Anonymous

I'm feeling like there's something missing

Balarka Sen

07:32

If you can prove the left and right derivatives are both $f'(g(x)) g'(x)$, you're done, sure. That's what $f, g$ being differentiable gives you.

Anonymous

@BalarkaSen So I need to prove using first using principle that $LHD=RHD=f'(g(x)) g'(x)$ or I guess only showing LHD=RHD will do?

Anonymous

g(x) is differentiable

Anonymous

So g'(x) must exist

Balarka Sen

LHD = RHD does it. But the easiest proof will probably go through showing both are equal to f'(g(x))g'(x).

Aka, "prove the chain rule".

Anonymous

Okay, I got it

Anonymous

07:37

Both LHD and RHD come out to be $f'(g(x))g'(x)$

Anonymous

@BalarkaSen Is only using the chain rule sufficient ? (I mean without using first principle) as g(x) is differentiable and so is f(x) And the terms we have are f'(g(x)) and g'(x) in the expression

Balarka Sen

I don't understand. You are proving chain rule. That composition of two diff functions is diff is part of the statement of the chain rule.

You can't just use chain rule to prove chain rule!

4

You have to use first principles, aka definition of derivative.

Anonymous

@BalarkaSen Okay. I get it now.

Anonymous

Thank you!

Balarka Sen

no problem

Yashas

07:45

lim f(g(x)) = f(lim g(x)) given that f(g(x)) at the approaching value is defined.

You can use that.

Anonymous

@Yashas Yes, I did use that

Anonymous

f is differentiable, so it must be continuous too

Anonymous

$$\lim_{x \to a}\frac{f(g(x)) - f(g(a))}{x-a}\\ = \lim_{x\to a}\frac{f(g(x)) - f(g(a))}{g(x) - g(a)}\cdot \frac{g(x) - g(a)}{x-a}=f'(g(a))\cdot g'(a)$$

user228700

@JohnR: Advice needed!

John Rennie

Advice?

user228700

07:50

Yep.

Balarka Sen

@blue Modulo rigor, that's correct. The point is $x \to a$ means $g(x) \to g(a)$ [by continuity of g, more importantly than continuity of f]. So that first limit becomes $f'(g(a))$ like you said.

John Rennie

@Kaumudi.H go on ...

Balarka Sen

But yeah, that's right.

Anonymous

@BalarkaSen Yep, got it now :-)

Anonymous

f and g are continuous by assumption

user228700

07:54

@JohnRennie I've already asked my mum about it and she's asked me to go with my gut but there is no gut feeling here. Hank & John are publishing a book of poetry and have asked for submissions. All proceeds from the book will go to The Foundation To Decrease World Suck (which may not sound like a real charity but I promise you that it is!) Now I plan on publishing some of the things I've written someday but it would also be amazing if my poem was published in this one and...yeah.

John Rennie

Submit it!

The worst that happens is that it won't be included.

user228700

If I submit it, I won't be able to re-publish it in the book I hope to publish someday, no?

John Rennie

But then there will be so may poems submitted that even really good poems won't make the cut, just down to the huge volume. So even if yur poem isn't included that's no shame.

@Kaumudi.H why not?

user228700

effyeahnerdfighters.com/post/159232938289/nerdfighter-poetry

John Rennie

Poems get anthlogised and re-anthologised all the time.

user228700

07:56

Wokay! I'll send it, then. I wonder if anybody will ever buy the book though :-P

John Rennie

I'm going to submit AFT's poem:
There was a young man from Madras
Whose bollocks were made from pure brass
In stormy weather
They rattled together
And sparks flew out of his ass!

Balarka Sen

that escalated quickly

John Rennie

:-)

Balarka Sen

also poetry is not charity, it's bourgeois endeavor

I can quote Trotsky on that

user228700

@JohnR: Oh, I only just saw ur e-mail. Yeah, no, I've had to re-install it a couple times :-P

John Rennie

08:00

Using phrases like bourgeois endeavor in a non-ironic sense is evidence of a deep detachment from reality.

@Kaumudi.H AVG? Oh well, as long as it's working.

Balarka Sen

well, i did use that ironically but i won't deny of my detachment from reality

user228700

ALSO. @JohnR: Should I or should I not make a meta post about the AMA vzn has been urging me to do?

user228700

Like I said:

user228700

3 hours ago, by Kaumudi. H

It's not so much modesty as a fear of sorts that...well, that nobody would want to listen about what I have to say because nobody else has shown any interest at all. Please don't get me wrong; I feel that it's an honor to have been asked by you and I'd gladly oblige but do u see where I'm coming from?

John Rennie

I not sure I should advise, even if I had useful advice to give. I don't think the AMAs have been a stunning success so my own view is to stand well back from them.

user228700

08:03

Ah, on that note; why did u decline from doing one?

John Rennie

Because I don't think the world needs me to ramble on for an hour. If people want to ask me stuff they can just ask in the regular chat.

Occaionally people do ask, and I do my best to make the answers entertaining, but I don't think a formal hour long session would be of benefit to anyone.

user228700

Ah, but the "world" does, right? That AMA would've been hugely successful for I think loads of people would've asked you loads of questions!

Anonymous

@BalarkaSen @Yashas Does any of you know the steps of the proof Circumcentre=$$(\frac{\sum x_i\sin(2A)}{\sum \sin(2A)},\frac{\sum y_i\sin(2A)}{\sum \sin(2A)})$$ ?

John Rennie

None of the previous AMAs have attracted an audience significantly different from the usual chat regulars.

Anonymous

I can't find the proof online

user228700

08:07

@JohnRennie Right, but fixing a place and time is a nice way to go about it, don't you think? (Even if it's just the chat regulars who show up)

John Rennie

If we had Stephen Hawking or Brian Cox on it would be a different matter.

@Kaumudi.H No, I honestly can't see that it would be educational or entertaining.

Anonymous

Circumscribed circle

In geometry, the circumscribed circle or circumcircle of a polygon is a circle which passes through all the vertices of the polygon. The center of this circle is called the circumcenter and its radius is called the circumradius. A polygon which has a circumscribed circle is called a cyclic polygon (sometimes a concyclic polygon, because the vertices are concyclic). All regular simple polygons, all isosceles trapezoids, all triangles and all rectangles are cyclic. A related notion is the one of a minimum bounding circle, which is the smallest circle that completely contains the polygon within it...

John Rennie

In any case a significant fraction of the chat regulars hate my guts.

user228700

Wtf...

John Rennie

Strange, but true.

Balarka Sen

08:12

@JohnRennie I don't find intestines appealing in general but sometimes aesthetically pleasing.

Maybe I shouldn't link that.

John Rennie

Videodrome?

Balarka Sen

Yup.

BAYMAX

Hi all!

John Rennie

I don't think there's any problem linking that. It's tame by modern standards.

Balarka Sen

@blue No, I don't.

Anonymous

08:15

@BalarkaSen Any link you know of ?

Balarka Sen

Maybe Yashas or the other JEE crew does though.

Nope. Can't help you on this one.

Anonymous

The proof isn't in JEE

Anonymous

I doubt they will

Anonymous

Anyhow, thanks

BAYMAX

I am new to Transmission Electron Microscope and tried googling to find out some references which has lucid description about various components used in TEM,like ETH Zurich sites , still any help in citing good reference for TEM components ?thanks.

Balarka Sen

08:16

@JohnRennie Well, fair. Interested readers can just look at the history of the message.

He shot that scene with actual sheep guts I think :P

Swapnil Das

I'm not from the JEE crew or whatsoever. But I don't understand why guys here don't like JEE.

Don't people discuss physics or math here? Yes. Then why not at the High school level?

Anonymous

@SwapnilDas Well there is only one such person. You need not worry :-).

Swapnil Das

@blue One such person who doesn't like JEE?

Anonymous

@SwapnilDas Yes.

Swapnil Das

Hmm. Yup.

Anonymous

08:23

Anyhow, I'm looking for a proof of the circumcenter formula :/

Anonymous

Tough luck

Anonymous

Got to ask in the main site perhaps

Swapnil Das

Oh, which one?

Anonymous

Scroll up a bit

Anonymous

That sin(2A) one

Swapnil Das

08:25

Is that the coordinates of the circumcenter?

Anonymous

Yes

Swapnil Das

Wait a lil

I perhaps had the proof

Anonymous

It is there in the Cengage book I think

Swapnil Das

yup!

Got it.

Anonymous

Found ?

Swapnil Das

08:28

yup

Anonymous

Can you upload a picture?

Swapnil Das

i'm trying

Nah, impossible with PC.

Anonymous

Ok np

Anonymous

Could you just tell the steps?

Swapnil Das

Sure

Anonymous

08:32

Anonymous

Let us label the vertices

Swapnil Das

A triangle has vertices A,B,c with coordinates (x1,y1),..(x3,y3)

Anonymous

Okay, then?

Swapnil Das

A line joining A and circumcenter O meets BC at E

Anonymous

Right

Swapnil Das

08:34

The foot of the perpendicular from O on BC is D

Anonymous

ok

Swapnil Das

BE/EC = 1/2XBEXOD/(1/2XECXOD)

= 1/2OBXOEX sinBOE /( 1/2OC X OEX sinCOE)

Anonymous

That's clever

Anonymous

=D

Anonymous

Then ?

Swapnil Das

08:37

Indeed it is.

= 1/2 R X OE X sin (pi- 2C) /( 1/2 R X OE X sin (pi - 2B)

= sin2C/sin2B

Anonymous

Wait. Angle BOE=pi-2C ?

Swapnil Das

Coordinates of D are

Anonymous

Lemme think a moment

Swapnil Das

Yup it writes so.

Anonymous

How?

Swapnil Das

08:40

Lemme see.

Anonymous

Got it

Swapnil Das

Yeah?

Anonymous

Angle subtended by AB at O is 2C

Anonymous

AOB + BOE = pi

Swapnil Das

Yup

So coordinates of D are

Anonymous

08:43

Coordinates of D or E? We found BE/EC, isn't it?

Anonymous

That should give coordinates of E

Swapnil Das

x2sin2B+ x3sin2C/(sin2B + sin2c) , y2sin2B+ y3sin2C/(sin2B + sin2c)

Anonymous

@SwapnilDas That should give the coordinate of E, right?

Anonymous

We found BE/EC= sin 2C/sin 2B

Swapnil Das

Hmm.

Anonymous

08:44

ok, so far it looks good

Anonymous

We found E

Anonymous

Now to find O

Swapnil Das

perhaps printing mistake

Anonymous

We need OE/OA

Swapnil Das

Yup. It's a bit long.

Like this:

sin2B + sin 2C / sin2A = 2sin(B+C) cos(B-c) / 2sinA cosA = cos(B-c)/cosA

in triangle ODE, OD = OB cosA = R cosA

Anonymous

08:47

I think I can do it from there. I will repeat the same steps on triangle AOB

Anonymous

Thanks @SwapnilDas

Swapnil Das

Yup.

Oh my pleasure.

Anonymous

You are in 11?

Swapnil Das

Just entered yup.

Classes not started though :P

Anonymous

Great. Enjoy your holidays. A storm is brewing ;).

Swapnil Das

08:49

hehe. I'm enjoying more of my prep.

Fortunately my prep^n isn't JEE centric. I perhaps got to study for KVPY

Anonymous

Don't have to worry about JEE in class 11. Make your maths and physics very strong.

Anonymous

Class 12 is sufficient to pick up chemistry

Swapnil Das

Yup. I need to make my Chem strong, actually :P

Anonymous

Inorganic ? =P

Anonymous

same problem, for everyone, hehe

Swapnil Das

08:51

In NTSE stage 1, which I wasn't able to qualify, I got Physics full, Maths 19/20 and Chem 0/12

Anonymous

@SwapnilDas That's a bit worrying then...

Anonymous

Chem maybe easy...but you need to devote time to it

Swapnil Das

Not really, I never studied.

Yup you're right.

Anonymous

Okay, see you around ;)

Anonymous

All the best!

Swapnil Das

08:53

Thanks. To you too!

user228700

09:26

@JohnR: Jon Hopkins?

John Rennie

@Kaumudi.H ???

user228700

The artist.

user228700

Jon Hopkins

Jonathan Julian "Jon" Hopkins (born 15 August 1979) is an English producer and musician who writes and performs his own melodic electronica and dance music. After starting his career playing keyboard for Imogen Heap, he has produced or contributed to albums by Brian Eno, Coldplay, David Holmes, and others. Hopkins composed the soundtrack for the 2010 film Monsters, which was nominated for an Ivor Novello Award for Best Original Score. His third solo album, Insides, reached no. 15 on the Dance/Electronic Album Chart in 2009. His collaborations on Small Craft on a Milk Sea with Brian Eno and Leo...

user228700

Do you listen to this sort of music?

John Rennie

I've never heard of him. I do listen to electronic music like Tangerine Dream, and I do have a couple of Imogen Heap albums, so yes his work looks interesting.

user228700

09:33

Jon Hopkins - "Inner Peace"

►

user228700

Jon Hopkins - Lost in thought (HQ)

►

Balarka Sen

Sounds like he's a collaborator of Brian Eno. Big fan of his.

John Rennie

Listening now ...

@Kaumudi.H: nice. If you haven't heard any Tangerine Dream try this:

TANGERINE DREAM - CLOUDBURST FLIGHT

►

user228700

@JohnRennie Actually no, I haven't. Will do immediately.

Slereah

there's a lot of definitions for faster than light spacetimes, which is slightly funny because there's pretty much only two of them

Up to some tweaking

it's more about the spacetimes you want to keep out

user228700

09:47

@JohnRennie It was OK :-)

John Rennie

Just OK?

Try:

Tangerine Dream - Stratosfear

►

user228700

OK...

John Rennie

Tangerine Dream is one of the greatest electronic bands of all time, and all people possessed of good taste in music love them :-)

user228700

Yep, the second one I like more :-) Not love but like a lot, yes.

Balarka Sen

10:05

@JohnRennie Pretty good!

As far as instrumentals go my favorite definitely has to be B-side of Low.

John Rennie

Their middle period albums are the most accessible. Albums like Force Majeure, Stratosfear, Tangram and Sorceror. Their earlier albums are on the ambient side and a bit trippy.

Balarka Sen

eg

John Rennie

The later albums don't improve much if any over Force Majeure etc.

Secret

11:18

Weird question about Fock space that I am most certain I don't have the required background to ask yet:

We knew a Fock space is defined to be the following construct:

$F_{\nu}(H)=\bigoplus_{n=0}^{\infty}S_{\nu}H^{\otimes n}=\Bbb{C}\oplus H \oplus S_{\nu}(H \otimes H) \oplus S_{\nu}(H \otimes H \otimes H) \oplus \cdots$, where $S_{\nu}$ antisymmetrize or symmetrize the n-particle hilbert space depending on whether there are fermions or bosons.

Suppose my hilbert space $H$ has 2 dimensions and I wrote some arbitrary element in $F_{\nu}(H)$ as a matrix, I should get the following:

$$\langle i\lvert F_{\nu}(H)\rvert j\rangle=\begin{pmatrix} * & \\
& * & * \\
& * & * \\
& & & * & * & * & *\\
& & & * & * & * & *\\
& & & * & * & * & *\\
& & & * & * & * & *\\
& & & & & & & * & * & * & * & * & * & * & *\\
& & & & & & & * & * & * & * & * & * & * & *\\
& & & & & & & * & * & * & * & * & * & * & *\\
& & & & & & & * & * & * & * & * & * & * & *\\
& & & & & & & * & * & * & * & * & * & * & *\\
& & & & & & & * & * & * & * & * & * & * & *\\
& & & & & & & * & * & * & * & * & * & * & *\\
& & & & & & & * & * & * & * & * & * & * & *\\

It is easy to see that each block diagonal corresponds to the $n$ particle states starting from the vacuum state (0 particles) all the way up, and the off block diagonal terms are all zero

But suppose some of the off block diagonal terms are nonzero, that is, there exists some linear map that maps from one of the $n$ particle state to a $m$ particle state, what will be the physical implication...?

(to be explored when I have enough background)

ACuriousMind

@Secret That obviously depends what the matrix/operator is supposed to represent. A matrix on its own has no physical meaning at all.

Secret

11:42

15

Q: What's the exact connection between bosonic Fock space and the quantum harmonic oscillator?

Let's suppose I have a Hilbert space $K = L^2(X)$ equipped with a Hamiltonian $H$ such that the Schrödinger equation with respect to $H$ on $K$ describes some boson I'm interested in, and I want to create and annihilate a bunch of these bosons. So I construct the bosonic Fock space $$S(K) = \bi...

Hmm, if I understood correctly, the harmonic oscillator is basically a Fock space where $H$ has dimensions = 1

Secret

@Qiaochu: I'll see what I can do. For the questions: 1. no, they can't be observables because they are not hermitian; 2. number is conserved only as long as you don't introduce interactions; but such a theory is not interesting because the Hilbert space is just a sum of non-interacting parts each of which is described by QM. To introduce an interaction you need to add Hamiltonian $H_I$ that has nontrivial matrix elements between states with differening particle numbers (in other words, it doesn't preserve the original Fock decomposition of the total Hilbert space). — Marek Jan 8 '11 at 14:41

Basically, my initial (probably naive) idea that I don't have background to investigate it yet, is that if the off block diagonal terms are nonzero, then we are effectively introducing an interaction. This is because a Fock space describes a non interacting field, thus two different n-particle states cannot interact with each other and you have this nice decomposition in terms of direct sums
(thus when all of this is wrote as a matrix, then one can observe the off block diagonal terms will all vanish)

(Unless direct sums of hilbert spaces are unlike the finite dimensional vector spaces in linear algebra in that they don't have a matrix representation(?) as a block diagonal matrix)

One reason I have this suspicion is because the off block diagonal terms are linear maps from the rows and columns that corresponds to the $n\times n$ blocks of the n particle states in the Fock space, to another rows and columns that corresponds to the $m \times m$ blocks of the m particle states.

Therefore if an n particle state of the fock space is fed into these linear operators, then it outputs an m particle state, suggesting it describes an n partcle state changing into an m particle one. And if we have change in particle number, then an interaction take places by definition

EDIT: Ok this is not enough, even if the above holds, it still cannot handle nonlinear interactions

You can write Taylor series for an operator, but that would look like $$ R(\mathbf x+\mathbf h) = \mathbf R(\mathbf x) + A(\mathbf x) \mathbf h + B(\mathbf x) (\mathbf h \otimes \mathbf h) + C(\mathbf x) (\mathbf h \otimes \mathbf h \otimes \mathbf h) + \dots $$ where $k$-th term is a tensor of $k$-th order, i.e. $k$-linear form. — uranix Jul 23 '15 at 13:25

This looks like fock space, hmm...

gertian

12:34

Hello :)
Why is this question getting downvoted ?
https://physics.stackexchange.com/questions/326544/why-does-the-differential-form-of-gauss-law-gives-div-vec-g-4-pi-g-rho-0-ou/326548#326548
The guy is clearly confused on something and we can help him ? Why downvote this ?

Thanks for the explenation :)

Ramanujan

Hello @yashas

Yashas

Hi @Ramanujan

Ramanujan

Whwhat's the unit of loudness?@yashas

Yashas

bells

You should learn to use Google.

Ramanujan

https://www.reference.com/science/unit-used-measure-loudness-sound-2079c250e72c7ab4?_e_pi_=7%2CPAGE_ID10%2C9283505196

@yashas.

Yashas

12:40

You can use intensity to express loudness too.

Phons is also a unit

Ramanujan

Which would be better to write in an exam?

@yashas

Yashas

whatever is given in your textbook

Ramanujan

I don't have a textbook. So

@yashas

Yashas

Phons and sones

Ramanujan

@yashas, oj

K

@yashas, If $abc(a+b+c)=3$, prove that $(a+b)(b+c)(c+a)\geq 8$

Yashas

12:55

expand (a+b)(b+c)(c+d)

you get 8 terms

take AM GM of that

It is a guess

I need to check if that'll give the right answer

Ramanujan

Where does d come.from?@yashas

Yashas

oops it must be a+c

yes

you get the answer

you can do it mentally

my first step is correct

$(a+b)(b+c)(c+a)$ can be written as the sum of 8 terms.

$AM \ge GM$

Ramanujan

OOn expanding I got, 2abc+ac^2+b^2c+a^2b+a^2c+ab^2

Yashas

@Ramanujan Take $2abc$ as $ abc + abc$

you have 8 terms if you count ^

the AM of that is (those 8 terms)/8

Ramanujan

Yeah i have 8 terms @yashas

Yashas

12:58

You know AM-GM inequality, right?

must be 8th or 9th grade syllabus

Ramanujan

II know that: $A.M\geq G.M$

But what's the gm? @Yashas

Transcript for