The h Bar - 2016-10-20 (page 3 of 5)

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2:00 AM

2. is a manifold a metric space?

@bl00, sadly, yes

er, wait, no on the second one

well...actually, I dunno what I'm supposed to ask here.

user218912

@heather can relate... my parents turn off the wifi at some point after 11.

user218912

but if they forget

user218912

my bedtime becomes undefined

user218912

or when I'm on campus I am free as well.

got to go, see you all later

2:03 AM

@heather huh?

@heather Yes!

It is!

the metric is not at all obvious

and it's not unique

but we will talk about that later

sleep

@bl00 Ok

Time to continue your training

@bl00 Compute the energy momentum tensor for Yang-mills

@bl00 You must get rid of the evil wifi radiation

It's dangerous at night

Agreed.

I got eye cancer from wifi

@bl00 how is GR going?

user218912

2:48 AM

@0celo7 didn't learn what yang mills is xD

user218912

@0celo7 on hold for now... giving condensed matter another try.

3:41 AM

@MAFIA36790 should I stack books?

user116211

Why people hate Feynman so much for beginners ?!

he's a crank

user116211

._.

user116211

@0celo7 You will go to see a doc; everything after that.

not at 11:43 I won't

user116211

3:43 AM

@0celo7 I know.

user116211

Well, I will have to go to see a doc too for mine is too much annoying.

user228700

Hey everyone :-)

user116211

Anyways, only one answer I can agree with:

user116211

1

A: Feynman Lectures for a newcomer

Depending on your personality, I think you would be very well placed to read the Feynman lectures and they would be an excellent beginning. Beware, though, of seeking "Bibles"- seek a few references and read them all at a relaxed rate at the same time. Seek more specialized, but broad, references...

user228700

@0celo7: I read about ur eye. Is it better now?

user116211

3:45 AM

@Kaumudi He is going to see a doc tomorrow.

user116211

Hope it's not cancer.

why is everyone so concerned about my eye??

user116211

Huh!

user228700

@0celo7 Dude, u made it sound horrible. We're just nice people (:-P)

4:02 AM

@Kaumudi Hi.

user116211

@DanielSank o/

@MAFIA36790 \o

@MAFIA36790 I can't say the same. vol I is where I think not for beginner (especially self study) vol II is okay (almost every book is same), vol III is one of the best quantum intro(you will be shocked by any quantum physics book anyway),

anyway, I am okay with ppl suggesting Feynman for a beginner. since it is just a matter of taste.

anyone with sense will pick a new book, if he is having a hard time.

however, I really do not think Landau is for a beginner...

it is even much deeper than Sommefield's ....

user116211

@Shing oh yes; i very much agree with that; although Mechanics can be readable but not the other books in the series at an undergrad level.

user218912

4:23 AM

i don't like landau

user218912

old book

user228700

5:35 AM

@JohnRennie: Morning! :-) U start work right after u wake up, these days?

@Kaumudi I turn on the computer as soon as I wake up, but the first few minutes are reading mail, eating breakfast etc :-)

user228700

Oh, OK :-)

At the moment I'm setting up a new laptop for my niece, which is much more fun than work :-)

@bl00 I think kip thorne is writing a new mechanics

user116211

Really?

5:48 AM

though not quite sure whether it is undergrad or grad text

@MAFIA36790 yeah

@MAFIA36790 here pmaweb.caltech.edu/Courses/ph136/yr2012

the draft

user116211

Whittaker wrote a very good book in Mechanics and it's extremely detailing.

user116211

It contains problems from Tripos' exams of Cambridge.

"This book is an introduction to the fundamentals and 21st-ce
ntury applications of all the
major branches of classical physics except classical mecha
nics, electromagnetic theory, and
elementary thermodynamics "

user116211

The notations might be outdated since it is 100 yrs old or more; but it's really a great book.

user116211

Not for freshman, though.

5:51 AM

probably I recalled wrongly

@MAFIA36790 this? amazon.com/Treatise-Analytical-Particles-Cambridge-Mathematical/…

user116211

yes.

user228700

6:02 AM

@JohnRennie I see :-) Do let me know when u will be relatively free...I have a few questions regarding the procedure to find the center of mass of a uniform mass distribution that I hope u'll help me with...

I'm happy to chat now ...

... for the 20 minutes anyway. In 20 minutes I have to go back to working (boo!).

user228700

Oh...

user228700

OK, I'll ask a smaller question now.

user228700

Dyou remember that bomb I told u about the other day?

Yes

user228700

6:14 AM

OK. Am I allowed to apply the conservation of momentum to a system defined to consist of he bomb alone?

Momentum is conserved when there is no external force acting on a system.

user228700

Aha. What about gravity?

user116211

@Kaumudi Take earth in your system.

@Kaumudi True, and obviously with gravity momentum isn't conserved because the bomb accelerates due to gravity.

However suppose we work in the rest frame of the bomb.

user228700

OK...

6:17 AM

In this frame the bomb is stationary at the origin.

And remains stationary at the origin until some force acts e.g. when the bomb explodes.

user228700

^ But that force is also internal, no?

Yes, that's an internal force, so it cannot change the total momentum of the bomb.

user228700

Why did u emphasize on the word total?

Because it can change the momentum of bits of the bomb e.g. one half can end up with a positive momentum while the other half ends up with a negative momentum.

But the changes to various bits of the bomb must balance out so the total momentum is unchanged.

user228700

Yeah OK...

user228700

6:22 AM

So, alright, what happens when the bomb bursts? The fragments fly off in different directions and then fall to the Earth under the influence of gravity, no?

user228700

Oh, u have to go work now, right? OK, I'll wait or something.

@Kaumudi yes, each fragment follows a parabola defined by the initial velocity of the fragments immediately after the explosion.

We should call this time of day the "Kaumudi and Sir John Physics Gallery".

Bring your friends, and learn something new, every day, same time, same place!

user228700

@JohnRennie Riight, so gravity does influence these particles. Now what can I say about momentum?

user228700

@DanielSank Sometimes featuring Sir Daniel Sank!

user116211

6:29 AM

@DanielSank, How is your book going?

@DanielSank It's surprising how often discussing apparently simple things forces you to think long and hard. For example this question about conservation of momentum of an exploding bomb falling under gravity.

user116211

@DanielSank h bar is confusion relieving center for most of us.

user228700

@MAFIA36790 True that.

user228700

@JohnRennie: Should I wait or..?

@Kaumudi well, we kind of fudge this and say that gravity affects all the bits of the bomb equally, so if we work in the rest frame of the bomb we can ignore gravity and just consider the relative motions of the bits.

But I do have to get to work now for about half an hour. Back soon ...

user228700

6:33 AM

@JohnRennie OK.

user228700

@JohnRennie Hm, so we can do this for literally any object in motion and conserve momentum, right?

@MAFIA36790 Well! I made real progress yesterday!

user116211

okay ;)

@JohnRennie But you're dropping knowledge bombs, so it's all ok.

::hides::

user228700

XD

6:37 AM

@Kaumudi yes, sort of, though it's actually a bit more complicated than that. For example gravity varies with height so a fragment that goes up experiences a slightly lower gravitational force than a fragment that goes down.

@JohnRennie Dang, we're getting technical up in here.

This difference is called a tidal force and it means that even in the rest frame of the bomb momentum is not conserved.

::ducks under shoe hurled by John::

user116211

Welcome in h bar @Bolaño.

But as long as it's a good approximation to say that gravity is everywhere constant then for a freely falling object we can ignore it.

user228700

6:38 AM

Yes, I was just gonna ask ^

If/when you learn general relativity you'll learn just how important this principle is :-)

user228700

OK, this is weird. I just learnt that I can conserve the momentum of most objects that I'll ever deal with b/w now and May 2017.

May 2017 is when you go to university?

user116211

@JohnRennie JEE.

Hi @MAFIA36790! I was just checking this chat (pretty cool!), in my country it's 4AM and I've got class at 10, so I won't be able to stay

user228700

6:40 AM

@JohnRennie No, my (last) exam's in the last week of May 2017.

I'll log in tomorrow. btw, great community guys!

user116211

@Bolaño You are welcome here at any time you feel comfortable.

@Kaumudi OK so up to May 17 you learn all this stuff because you have to. Once you get to university you'll be learning it because you want to, which is much more fun :-)

@Bolaño What country?

@JohnRennie That is SO true.

user116211

Now, I'm missing @0celo7 for I wanted to ask when an isotone function acts as an extension to linearly ordered chain and preserves the order.

user228700

6:48 AM

@JohnRennie I disagree just a little bit, 'cause this is fun too but I'm on a schedule so I don't get to have as much fun with these concepts as I'd like to :-)

7:01 AM

I'm back (for about half an hour) ...

user116211

Help me @JohnRennie :(

Is this answer answering the question?

0

A: Is a time change term (t or c squared) ALWAYS necessary to describe interactions between particles or objects?

The time-independent Schrodinger equation describes a particle at some instance $t$ in a state given by the wavefunction $\psi$. $$ \frac{-\hbar^2}{2m} \frac{\mathrm{d}^2\Psi}{\mathrm{d}x^2} + U(x)\Psi(x) = E\Psi(x) $$

@MAFIA36790 I think you've advanced past the point I can be of much use, but I'm willing to try ...

user116211

That's about Asaf's answer:

user116211

1

A: Why is restriction necessary on an order-preserving function for it to be an isotone extension?

Clearly $X_0$ has an upper bound in $\Bbb R$, for example $1$. But in $X_0$ it has no upper bound, because that would correspond to a maximal element strictly below $1$ in the real numbers. So how can we extend the identity function to remain monotone? It would have to map bounded sets to bounde...

7:06 AM

I don't even know what an isotone extension is, so I have no idea what Asaf's answer means!

user116211

Isotone means order preserving function; like monotone function.

user116211

7:17 AM

What I'm not getting is why $X_0$ in $X_0$ cannot have an upper bound.

user116211

He writes:

user116211

> that would correspond to a maximal element strictly below $1$ in the real numbers.

user116211

I didn't get this statement.

user116211

And... the maths room is empty and 0celot is not here.

user116211

So, @JohnRennie, anything that can shed light on this?

user228700

7:28 AM

@JohnRennie Aw man :-( I went to have lunch!

user228700

I've got a really quick question. If a man jumps off a cart moving with speed $u$ (w.r.t ground) with a speed $v$ (w.r.t cart) (both velocities are in the same direction) then his velocity w.r.t ground even after jumping off remains $u+v$?

user228700

See, this isn't the case with acceleration. Even if the cart had been accelerating, the man would accelerate with $g$, under the influence of only gravity after jumping off.

acceleration has to do with interaction

law III of Newton

user228700

So, him having the same velocity after jumping off is a consequence of the conservation of linear momentum but again, this momentum is conserved in the first place because..? Is there a qualitative way to understand this?

user228700

("this"=Conservation of linear momentum)

7:38 AM

what are the confusing part?

user228700

@Shing Yep, so it makes sense that he doesn't accelerate with the acceleration of the cart, even after jumping off.

the momentum of you after you jump off the cart?

user228700

No, no confusing part, I guess. Just wanted to know a qualitative way to understand the conservation of linear momentum.

user228700

Never mind, I'm gonna read about this some more.

@Kaumudi the two accelerations are at different directions... cart probably in x. gravity in y

user228700

7:45 AM

@Shing No, not after he actually jumps off...

@Kaumudi We can conveniently divide the velocity into horizontal and vertical components.

The horizontal component is unaffected by gravity because gravity acts vertically, so the horizontal component remains at $u+v$.

Obviously the vertical component changes due to gravity.

user228700

Yes, OK, that makes sense.

user228700

I have another question (All physics this time so at least there's that? :-P)

Yes ...

I've now officially finished work BTW. The only vitally important task remaining is to make more coffee.

user228700

Actually, no, let me retract that hand gesture. Idk what these things mean in different countries. Here anyway, 👌 means "Awesome!". Anyhoo, my question arises mostly 'cause I don't think I've understood the of principle of conservation of momentum properly.

user228700

7:54 AM

(And omg, u haven't had coffee yet?! :-o)

user228700

Basically, all that this principle is saying is that motion has to go somewhere, correct? (Sheesh, is that incorrect? :/)

@Kaumudi I've already drunk two pint (550 ml) mugs of coffee. The next mugful will be my third.

That happens when you start work at 05:00 :-)

user228700

Ah, I missed the "more coffee"

The rate of change of momentum is equal to the force: $$\frac{d\mathbf p}{dt} = \mathbf F$$

user228700

@JohnRennie Gosh.

user228700

7:57 AM

@JohnRennie Yes.

Or put another way the change of momentum is given by integrtaing the force (to get the *impulse): $$ \Delta \mathbf p = \int \mathbf F(t)\,dt $$

user228700

Yes, OK...

So conservation of momentum is the statement that there is no net force acting.

user228700

OK.

user228700

OK, say I throw something.

7:59 AM

This is all you need to work with momentum. I'm not sure what your statement: motion has to go somewhere means.

user228700

@JohnRennie Yeah, neither do I :/

user228700

Say I throw a rubber ball. I've defined the system to consist of my hand and the ball.

> I've defined the system to consist of my hand and the ball

Hmmm

user228700

Not OK?

I'd say either just consdier the ball, in which case momentum won't be conserved because an external force acts.

user228700

8:02 AM

@JohnRennie Yeah, that's why I thought I'd include my hand as well..?

Or consider the ball plus you plus the planet you're standing on, in which case momentum will be conserved because this is now a closed system.

The problem with including just your hand is that an external force is applied to your hand by your arm.

And to your arm by your shoulder and so on till ultimately the force is applied to you by the ground you're standing on.

user228700

Okay, I'll include my whole body.

user228700

One second.

user228700

Sorry about that :/ My neighbor needed me to fetch her something.

user228700

Anyway. Wth, "by the ground"?!

user228700

8:12 AM

Well, I mean, OK, I guess...

Suppose you're standing still on a set of scales. Those scales read your weight.

Now throw something, and you'll see the reading on the scales change as you throw.

That shows there is a change in the force between you and the ground.

So you can't treat the ball and your body as an isolated system. You could only do that if you and the ball were floating in space.

So when you threw the ball one way you would start moving in the other direction.

user228700

OK, I get this, but not really. Isn't my arm just converting it's (amazing :-P) muscle strength to apply a force on the ball?

Yes, your hand/arm apply a force to the ball, and Newton's third law tells us that in rteurn the ball applies an equal and opposite force to you.

So your mometum must change.

user228700

Sure...

user228700

Oh, dang it.

8:15 AM

But you could argue hang on I'm standing still

And that's true, but what it means is that the momentum of you plus the whole Earth changes.

user228700

It changes?

Though obviously the Earth is so massive that any change in its velocity due to you throwing a ball is immeasurably small.

user228700

Wait, I thought that it wouldn't change!

user228700

We're considering all the different components of the system so how is it changing? Won't it be actually conserved?

The momentum of the ball changes
The momentum of you and the earth changes
But if you add the two changes together they cancel out and the total momentum is unchanged.

user228700

8:19 AM

Phew. Okay. U almost triggered a breakdown!

user228700

Alright then. I'm floating in space. So are u! And I'm throwing this rubber ball to you.

OK

user228700

So now let's consider all the things happening. To conserve the momentum of the ball, I can pick any two arbitrary positions on its path, correct?

> To conserve the momentum of the ball, I can pick any two arbitrary positions on its path

Eh?

user228700

Yeah...what I meant was, I can pick any two points on the path of this ball and conserve the system's momentum at those two points, no?

8:23 AM

I'm not sure what you mean by that. The ball's momentum is constant because (after you've thrown it and before I catch it) there is no force acting on the ball.

user228700

OK, I think I've gone and confused my own self.

user116211

@JohnRennie, I want to share something funny ;P

@MAFIA36790 Yes ... :-)

user116211

Today, I saw a post where OP was asking whether Feynman is competent for beginners.

user116211

5

Q: Feynman Lectures for a newcomer

I'm a senior philosophy/economics undergrad and I'm recently finding myself very interested in physics. For several reasons, I'm not able to take physics courses in college, but I wanted to introduce myself to the "corpus" of the undergrad physics. I'm willing to invest time and have taken severa...

resource-recommendations education general-physics

user116211

8:27 AM

There were answers which said the Lectures are not suitable for complete beginners and all that.

user116211

Then there was also an answer of WetSavanna who wrote differently asserting they can be used by beginners.

user228700

OK, I think if we discuss just one case properly, my dumb questions will trouble me no more...how about me throwing a ball upward, on Earth and yes, I will include the planet in the system.

user116211

I left a comment there @JohnRennie:

user116211

Only one single answer I can agree with; thanks @WetSavannaAnimalakaRodVance; +1. I don't know why the other answers think Feynman is not good for a beginner. Well, here I am who learned physics literally from Feynman's Vol II and Vol III. While they are not extraordinary, they are different in the style of presentation; the lucid explanation and a innovative approach to deal with respective contents; he explained base states and operators with Stern-Garlech filters - that is insanely one of the brilliant things in his lectures. They are definitely accessible to undergrads. — MAFIA36790 5 hours ago

user228700

@JohnRennie: Is that OK?

8:29 AM

@Kaumudi suppose we leave you out and replace you by a cannon that fires the ball upwards.

user116211

and now I was checking the main page where I saw that someone edited the post of this question.

user228700

@JohnRennie OK, seriously, what difference does it make? I've included the Earth and everything!

So we start with the ball and the Earth floating in space. The Earth throws the ball upwards so the ball moves one way and the Earth moves the other way.

user116211

There I found this:

user228700

@MAFIA36790 Man, is this joke long af.

user228700

8:31 AM

@JohnRennie This is confusing but OK, the Earth moves a little bit, yeah?

user116211

> ... It probably arises from people wanting to say something about themselves ("look at me; I'm a very clever little chap, and I have read these difficult books on physics and claim them to be easy reading for beginners") rather than thinking what is best for the questioner.

user116211

> Someone who claims to have learnt physics solely from reading the Feynman lectures is trying to make you think they are an unnaturally intelligent and able person who, somehow or other, is able to piece together incomplete arguments, and is able to learn simply by reading, without doing any exercises...

user116211

> They're not trying to help you by recommending these books to you, they're trying to point out how brilliant their minds are.

user116211

That's clearly pointing at my comment.

user116211

I don't know what to say.

8:33 AM

@Kaumudi the point is that you throwing the ball and the Earth throwing the ball are basically the same. The only difference is that the Earth is a lot heavier than you are. However conservation of momentum works in exactly the same way in both cases.

user116211

And yes, this seemed to be a quite funny ranting @JohnRennie.

user228700

@MAFIA36790 I was expecting to laugh -_-

user228700

@JohnRennie For the momentum to be constant, the Earth moves a little, yeah?

user228700

LITTLE.

user116211

I'm really not in a mood to argue with the poster; and it would be futile, I know.

8:35 AM

@Kaumudi the Earth moves slowly rather tha a little. Remember that momentum is $mv$ so the velocity of the Earth changes by $\Delta v = \Delta p/M$.

user116211

But this is a little too much, imo;@JohnRennie.

user228700

@JohnRennie Well, I mean, what difference does it make, really? If it moves a little in the same time, it essentially moved very slowly, correct?

@MAFIA36790 Just ignore it. The language is unnecessarily provocative, but I have some sympathy with Hirose's view. I don't think the Feynmann lectures are suitable for complete beginners. You do need to have some knowledge of physics to understand them.

@Kaumudi The phrase moves a little suggests to me that the Earth moves a small distance then stops. I know that's not what you mean, but i think you need to be careful to phrase things in an unambiguous way.

user228700

@JohnRennie And then stops whilst the ball is still in motion, u mean?

user116211

@JohnRennie sure. Note I didn't say I solely learnt physics from his lectures. WetSavanna has explicitly mentioned to have some other reference books while reading the lectures.

8:40 AM

@Kaumudi the phrase moves a little suggests that the Earth moves a small distance then stops while the ball is still in motion. At least that's what it suggests to me.

user228700

OK. That's definitely not what I meant, sorry. Okay.

user116211

The answerer totally misinterpreted my comment and WetSavanna's answer; but I have other more important things to do; so yes, ignoring it.

@MAFIA36790 Well the OP says I literally don't know any physics

user116211

@JohnRennie He has the math background; he knows complex analysis.

@MAFIA36790 Yes, but that's not the same as having a basic grasp of physics.

user116211

8:42 AM

> I think you would be very well placed to read the Feynman lectures and they would be an excellent beginning. Beware, though, of seeking "Bibles"- seek a few references and read them all at a relaxed rate at the same time. Seek more specialized, but broad, references for the "pillar" subjects as well.

user116211

i think that sums it all.

user228700

OK, u know what? I think I sort of get it now. I'm just overanalyzing again. Someone go ahead and rename me. Henceforth, I shall be called Overanalyzer, sigh.

user228700

@JohnRennie: Thank you! 🙏

user116211

Thanks for the comment @JohnRennie.

@Kaumudi students often get confused about exactly what they should and shouldn't include when applying conservation of momentum. And to be fair it is sometimes hard.

We get lots and lots of questions here from students who have made just this mistake.

For example:

user116211

8:49 AM

I also many times could not comprehend Feynman at the first time; after re-reading between the lines or so, I could then get the explanation. I have also asked many questions on putting excerpt from the lectures; I remember @JohnRennie has also answered many of them; old memories :)

1

Q: Collisions between an object and a wall

Is momentum conserved when an object bounces back against a wall? The wall doesn’t move, but the object moves in the opposite direction. Assume this is an ideal, elastic collision. If, initially, the momentum of the system came from the ball, and after the collision, the momentum of the system w...

momentum conservation-laws collision

8:59 AM

@MAFIA36790 He got mad when someone downvoted him without a comment, and then he started being not quite polite...

user116211

@Shing Who?

who left a quite rude comment suggesting you

user116211

@Shing ohh.

user116211

@Shing John mentioned he downvoted.

user116211

@Shing And I didn't downvote.

9:05 AM

I know... you showed up quite late

user116211

I respect everyone's opinion. But I liked WetSavanna's and so I commented. I never said I learnt physics solely from the Lectures; he/she totally misinterpreted my comment.

I know. You are just passionate at Feynman's lecture, I can understand that.

btw

user116211

@Shing yeh ;))

there are actually problem-set for Feynman lecture... back to the days the exact homework of the course taught by Feynman (I recall)

user116211

@Shing yes, there was a separate book by Matthew Sands dedicated solely to exercises.

user116211

9:10 AM

And I have followed the Indian version of Feynman lectures where there are problem sets by Prof. V. Balakrishnan of IIT, Madras, at the end.

wow

that's nice

user116211

I must say though, I have not done that.

lol

user116211

@Shing The problems were too advanced for me then. Maybe I will do them in 2nd or third year.

I mostly do the problem set from MIT open course, and by David Morin. mostly for the detailed solution. that makes learning a lot more productive, when I finish my work, and have something in hand to compare with

user116211

9:12 AM

@yuggib o/

user116211

@Shing David Morin wrote a book on Mechanics and has modified Purcell, right?

yeah

what a great job

gotta do some Buddhism meditation, see you

user116211

@Shing o/

user228700

@JohnRennie: I am yet to check that post but in the meantime, do u think u could help me with some Math?

Yes, of course.

user228700

9:22 AM

You're so helpful :-D Nobody else says "Yes, of course".

I'm an English gentleman!

user116211

@JohnRennie sure you are, the Nerd, gentleman :))

user228700

Oh, damn, teacher's day is over! Well, I just wanted to let u know, BTW, that I've put u at the top 5 best teachers I've had in my whole life :-)

user228700

And I'm not just saying that, I promise.

Thanks, though you may be underestimating your other teachers. It's easy when you have a student who is hard working and enthusiastic. The hard bit is when your students are disillusioned and uninterested! That's when the really great teachers prove themselves.

user228700

9:26 AM

Of course, but seriously, u probably haven't come across the kind of teachers I've had all high school.

user228700

Anyhoo :-) Permutations and combinations is the most horrible chap. I have in Math.

user228700

And there's this thing where I'm supposed to find the coefficient of $x^k$ in an expression like this: $(1+x+x^2+x^3+...+x^p)(1+x+x^2+...+x^m)(1+x+x^2+...+x^n)$ and yeah, u get the point.

Is the $x$ deliberately missed out of the first bracket?

user228700

Oh, no. Let me fix that.

user228700

Damn, can't do it. But yeah, no, there's supposed to be an $x$ there.

user228700

9:32 AM

And I have no idea how to do this.

There's probably some cunning trick that mathematicians know, but I can't think of one. I would have to do it the hard way by multiplying out the terms.

user228700

Oh, alright...I've seen some posts about this but I wasn't able to understand the notations :/

Are $p$, $n$ and $m$ any three integers or are the contiguous i.e. differ by one from each other?

user228700

They can be any positive integers.

@Kaumudi if you link one of those posts I can have a look and maybe explain the notation.

user228700

9:34 AM

Searching...

user228700

OK, never mind. I think I found something that I actually do understand. Sorry for wasting ur time :/

user116211

I have now actual book of Herstein!! Just delivered by Amazon.

@MAFIA36790 \o

user116211

Could you help me @yuggib understanding one statement from an answer?

9:57 AM

@MAFIA36790 I can try

10:34 AM

@Kaumudi fixed it for you

@JohnRennie probably contour integrals :-P

user228700

@DavidZ Thank you :-)

11:20 AM

Hello

Hi peeps

user116211

Hello.

Anyone know a good, basic topology book?

user116211

@heather yes!!

Yay!! What?

user116211

11:27 AM

Well, there are many; I'm currently following Kelley. I have Munkres, Mendelson too.

Which one do you find the simplest?

user228700

11:44 AM

@heather: Wow, topology? Nice.

user116211

@heather Well, I'm loving Munkres; Mendelson is quite good; but am focussed in Kelley; so I'll say Kelley.

user116211

But take Munkres; the language is lucid.

user116211

He has many examples.

user116211

Kelley is bit vague sometimes and its definition of ordering is really weird.

user116211

Nevertheless, all are good books.

11:48 AM

@MAFIA36790, okay, thank you!

@Kaumudi, yeah, I've been conversing with 0celo

user228700

I figured XD

=P

@SirCumference, hello.

Any interesting revelations about black holes and mass limits and the end of the universe?

user116211

Afteernoon @JohnRennie.

Afternoon (12:55)

I'm just popping in for a quick look - I have to go shopping now (oh joy!)

user228700

Is that supposed to be sarcasm? :-P

11:56 AM

Ah, you spotted that :-)

user116211

Sarcasm ?! I never can spot one ;/

user228700

:-) Have fun sir! (:-P)

Relation vs function: is a function always a relation? Is a relation always a function? What would be a relation that is not a function?

Actually I don't mind shopping. I usually cycle to the shops so at least it's an opportunity to get some exercise, and the weather is nice and sunny in my bit of the UK today :-)

Thinking about this, it seems that a function must always be a relation

But a relation can not be a function if it doesn't follow the rules of a function

Like, for example, a relation that does not map each x to a unique y

But I'm not sure at the moment what an example of that would be.

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