The h Bar

user228700

01:15

Yello, everyone :-)

heather

@Kaumudi.H, hello =)

user228700

@DanielS: This is what I meant:

user228700

in Mathematics, Nov 7 at 3:42, by Brody

@Kaumudi Range cannot be determined analytically, for now you need the graph of the function, which will tell all

user228700

And I slept very well last night! :-)

user228700

So apparently, I have to find the maxima, minima and then the horizontal asymptotes (to eliminate)

DanielSank

01:38

@Kaumudi.H well sure, in many cases you cannot solve for x in terms of y.

user228700

@DanielSank Yeah, the example that I brought up was unfortunately one such function.

user228700

I don't remember if I've asked this question before but I have a very troubling question that I've been struggling with for weeks.

user228700

This is what my textbook says:

user228700

> "If two functions, $f$ and $g$ are both one-one, then $g \circ f$ and $f \circ g$ would also be one-one (if they exist)."

user228700

But I've been trying to prove this using these diagrams:

user228700

01:49

user228700

And I'm not getting anywhere :-/ (Well, that is a lie. I got to the point where I disproved it.)

Balarka Sen

First off, what does one-to-one mean?

user228700

One-one means that every element in the range has but a single pre-image in the domain.

user228700

That every $f(x)$ is associated with only a single $x$ value for a function $f$.

Balarka Sen

aka, $f$ is one-to-one if $f(a) = f(b)$ implies $a = b$. Agreed?

user228700

01:53

Yep.

Balarka Sen

Alright. So you're given $f$ and $g$ are both one-to-one. To prove $f \circ g$ is one-to-one too, all you need to do is to prove is that $(f \circ g)(a) = (f \circ g)(b)$ implies $a = b$.

user228700

Uh, yes.

Balarka Sen

What is $(f \circ g)(a)$? Can you "expand that out"?

user228700

Um, that would be $f[g(a)]$, yes?

Balarka Sen

Right, that's it. Similarly, $(f \circ g)(b)$ is $f(g(b))$.

user228700

01:56

Right...

Balarka Sen

Then $(f \circ g)(a) = (f \circ g)(b)$ translates to $f(g(a)) = f(g(b))$. Since $f$ is one-to-one, what do you infer from that?

user228700

Ohh, since $f$ is one-one, we have that $g(a) = g(b)$ and then since $g$ is one-one, we have $a=b$.

Balarka Sen

Da. Quod erat demonstrandum for you.

user228700

Okay, I understand this but please bear with me as I share a picture of some diagrams that I drew to try and prove (/disprove) this:

user228700

Never mind the diagrams, they look very childish, I'll write it out. Oh, um, how to write the arrow pointing to the right in LaTeX?

Balarka Sen

02:01

\rightarrow

user228700

(Y)

DanielSank

@BalarkaSen вы русский?

user228700

Say we have two functions, $f(x)$ and $g(x)$ defined as $f : A \rightarrow B$ and $g: A \rightarrow C$. Is this correct so far? (I'm concerned about my definitions)

Balarka Sen

@DanielSank I am guessing you're asking if I speak Russian? Nah. I just like Russian literature/films (translated and subtitled in English, the only language I know :P)

@Kaumudi.H Well, no. If $f \circ g$ has to make sense, the codomain/range of $g$ must be the same as the domain of $f$.

Frotaur

I think you should define you g going from B to C if you want to compose the functions. So that gof goes from $A\rightarrow B \rightarrow C$

DanielSank

02:06

Ahh, good old gof and fog.

2

Frotaur

By the way guys, just quick question if anyone knows some particle physics, can a quark and antiquark annihilate into a Z boson ? By that is there a vertex quark antiquark Z boson ?

user228700

@BalarkaSen Dang it, okay, gimme a moment...

user228700

Forgive me for being so dumb (I've just started all this) but if we're given two functions $f(x)$ and $g(x)$, doesn't that mean that their domains are the same?

DanielSank

@Kaumudi.H Please don't call yourself dumb.

It is not helpful in the slightest way.

If nothing else, it's wasted words. At worst, it actually begins to convince your brain that you are less intelligent than you are.

So just don't do it.

As a student of math and science, we must be our own psychologist. Be nice to your brain and to your self-esteem!

2

There used to be a sticker on the keyboard at the main experimental apparatus in my grad school lab: "90% of education is encouragement"

Balarka Sen

@Kaumudi Given two arbitrary functions $f$ and $g$, $f \circ g$ and $g \circ f$ need not always make sense, no. This is where the little note in your book, "(if they exist)", is useful. Let me explain:

user228700

02:11

@DanielSank :-) Okay, I understand. Will not repeat, thanks.

Balarka Sen

Recall, by definition, $(f \circ g)(a) = f(g(a))$. But this only makes sense if $f$ can eat $g(a)$. What can $f$ eat? Stuff in $f$'s domain. Where does $g(a)$ live? In $g$'s range. So $f \circ g$ only makes sense if domain of $f$ agrees with the range of $g$.

DanielSank

@BalarkaSen a diagram of a case where fog or gof doesn't exist would help.

(but that's a great explanation, so feel free to ignore me)

I like thinking of functions as eating things and pooping out a result.

Apologies if that's gross.

user228700

Oh, no, no, I understand how/why/when $f \circ g$ and $g \circ f$ might not exist.

DanielSank

I also like Kaumudi's diagrams a lot (the two circles with arrows).

Balarka Sen

@DanielSank That's how I always ever think of 'em.

DanielSank

02:13

That's also how I think of tensors, for whatever that's worth.

Balarka Sen

In abstract contexts, I mean. Sometimes different pictures are useful.

DanielSank

@heather have you learned about partially applied functions?

@BalarkaSen Absolutely.

Balarka Sen

@Kaumudi.H So, what's you're question?

user228700

My question is the silliest of all-since we write $f(x)$ and $g(x)$, this indicates that both functions take different $x$ values from the same domain, no? (I'm actually kinda embarrassed to be asking this but if I don't ask now, my brain will screw me over during the exam)

Balarka Sen

Nah. $x$ is something abstract, which can live anywhere it wants. Might as well be different things living in different domains.

Nothing is specified unless you specify the domain/range.

It's just a notation, is all.

user228700

02:17

(Y) Wokay, so back to my example...let me recreate it so that the range of $g$ is equal to the domain of $f$...

DanielSank

What is (Y)?

Balarka Sen

it's a facebook thing. never got that.

apparently it's supposed to denote "thumbs up"

DanielSank

I don't know Facebook things.

Thumbs up? I don't see that.

user228700

Oh, um, that's s'posed to be the thumbs-up sign :-P I dunno "facebook things" either but my friends who are on fb keep using it in places like hangouts where it doesn't automatically translate to the sign so yeah...

Balarka Sen

@DanielSank me neither. heard it the first time in my life in the chat

user228700

02:20

(Working on the example...)

user228700

@Balarka: For how much longer will u be around?

Balarka Sen

For a finite amount of time. I am postponing the quickish nap I should be getting though.

DanielSank

@heather here's something you should know about:

user228700

Um, that's not very helpful :-P Thing is, I'm still working on the example and I have like, two more questions (that I've been struggling with for weeks >.<) that I'd like to ask...and also breakfast because I'm starving :-(

DanielSank

import functools
def sum(x, y):
    return x + y
add_3 = functools.partial(sum, 3)
seven = add_3(4)

Balarka Sen

02:27

Feel free to ask the questions. Better to ask on the MSE chat - If I am not around, someone else would help.

DanielSank

Here, add_3 is a "partially applied" version of sum, where the first argument is set to 3.

So, add_3 takes one argument and returns 3 + that argument.

user228700

The MSE chat is very messy and there are too many people (for me to keep track of) online and having lengthy (and probably interesting if I could understand anything :-P) conversations at the same time. OK, I'm not risking it, I'll ask them all now itself. I'll go back to working on the example...

DanielSank

@Kaumudi.H Use the "ignore" feature.

Or start a new room once you pick up someone to chat with :)

That's why I made Linear Algebra Happy Fun Time in order to have linear algebra happy fun time.

Balarka Sen

That's fair. It has gotten a bit messed up lately.

I suppose I (along with many others) am partially to blame to make all the conversations about topology and geometry...

user228700

That's absolutely fine-it's the MSE room, after all. I gave up on my example because it was stupid. I'll ask the second question now, then...

Balarka Sen

02:34

sure

user228700

OK, this is what my textbook says:

heather

@DanielSank nope

DanielSank

@heather Read on...

heather

(sorry, i'll have to go in a few minutes, and i was gone doing chores for my mom)

@DanielSank will do

okay

user228700

> "The composite of two bijections is a bijection iff $f$ and $g$ are two bijections such that $g \circ f$ is defined, then $g \circ f$ is also a bijection only when co-domain of $f$ is equal to the domain of $g$"

Sir Cumference

02:38

I was literally about to say it was suspicious that @Pissedofflayman and @0celo7 were never on at the same time, but here they are...

heather

hmm...so add_3 takes in an argument and then adds three to it using sum?

DanielSank

add_3 is a function. It is sum but with the first argument always set to 3.

So yes.

heather

huh, that's cool.

DanielSank

It's cool and useful in programming, and this idea comes up in mathematics all the time.

Balarka Sen

That's so poorly phrased

DanielSank

02:39

I can give you a very simple example but I'd rather do it when you have time.

user228700

Nope, nvm that.

Pissed off layman

@SirCumference we are of the same blood ;-)

Sir Cumference

@Pissedofflayman Oh, so you're his brother that he spoke about

Pissed off layman

nope

user228700

@BalarkaSen :-| How would u phrase it..?

Sir Cumference

02:41

@Pissedofflayman Quit confusing us

Pissed off layman

:D

Sir Cumference

You his relative?

His girlfriend?

Pissed off layman

ghost?

Sir Cumference

GRAHHHH

heather

@DanielSank I'll have plenty of time tomorrow - wouldn't want to interrupt another conversation with gtg =)

DanielSank

02:43

@heather no problem.

Having fun with your arduino?

Balarka Sen

@Kaumudi $g\circ f$ being defined, and codom $f$ = dom $g$ are really the same thing, aren't they?

heather

@DanielSank indeed =)

user228700

@BalarkaSen Exactly! Why is that a condition for the function to be bijective?

Balarka Sen

I want to say, "because your book is dumb, like most things in high schools are", but maybe I shoudn't.

It seems like poor phrasing on their part.

Pissed off layman

Is high school algebra dumb?

user228700

02:46

@BalarkaSen Nah, my book is almost useless without the help of you people. So OK, what is the condition for it to be surjective (since we've just proved that they will be injective) then?

Sir Cumference

@Pissedofflayman Most people in my high school said so

Pissed off layman

true dat

Sir Cumference

I was the only one who liked it :/

Balarka Sen

@Kaumudi So you want $g \circ f$ to be surjective? Hm. Well, $g$ being surjective suffices, nope? :)

user228700

:-| Let me have a think...

Pissed off layman

02:49

It is the first barrier to college @sir

Balarka Sen

@Kaumudi Nah, you need $f$ to be surjective too. Sorry about that.

Sir Cumference

@Pissedofflayman People who don't know algebra are freaking out in my classes

They assumed Astronomy would be easy

user228700

@BalarkaSen Oh, um, okay, now I'm trying to actually prove this...

Balarka Sen

$A \stackrel{f}{\to} B \stackrel{g}{\to} C$ is your sequence of arrows. If image of $f$ fills up $B$, and image of $g$ fills up $C$, then the image of the whole thing fills up $C$.

Sir Cumference

Even those who don't know calc are having trouble

Pissed off layman

02:52

Astronomy is the first science

Sir Cumference

"The first science"?

Pissed off layman

Yup

Sir Cumference

I think it was astrology back then

If you mean "first science chronologically"

user228700

@BalarkaSen I don't understand what you're trying to say...

Sir Cumference

@Pissedofflayman Say, are you in college?

Pissed off layman

02:54

Perhaps

Sir Cumference

What you majoring in?

Balarka Sen

@Kaumudi Surjective literally means the image fills up the codomain.

Every point is in the image, that is to say

Try to cook up a proof, you'll understand

Sir Cumference

Did I piss @Pissedofflayman off?

Dunno why he left

user228700

When you say "Every point is in the image", dyou mean how every element of the codomain is also in the range?

Balarka Sen

yup

user228700

02:56

Okay, so when you say "Image of the whole thing fills up $C$", what does that mean?

Sir Cumference

@Kaumudi.H You got me to start re-reading Harry Potter

I hope you're happy D:

heather

good night everyone

DanielSank

@EmilioPisanty I'm tempted to take a whack at this but what exactly are you looking for?

user228700

@SirCumference :-P I'm sorry..? Dude, it was your choice.

Sir Cumference

@heather 'Night

user228700

02:57

@heather Bye :-)

DanielSank

You just want a more mathematically and physically complete answer?

Balarka Sen

@Kaumudi Whole thing is $g \circ f$. As I said, start writing a proof and you'll understand

DanielSank

Goodnight, @heather.

user228700

@BalarkaSen "Start writing a proof". I dunno how to. I'm trying but I don't really know how to prove stuff in sets...I mean, extremely trivial stuff yes, but other stuff, not really...

Balarka Sen

Start with definitions. You need to prove $g \circ f$ is surjective. What does it mean to say, "$g \circ f$ is surjective"?

user228700

03:01

It means that the codomain of $g$ is the range of $g$...wait, that's what u said before, but then you corrected yourself for some reason...

Balarka Sen

Not $g$.

You mean $g \circ f$ instead.

user228700

But what's the codomain of $ g \circ f$ supposed to be..?

user228700

Isn't it just the codomain of $g$?

Balarka Sen

@Kaumudi Right

user228700

:-| So..?

Balarka Sen

03:07

and what's the image of $g \circ f$?

user228700

Image=Range?

Balarka Sen

yes

user228700

Um, wouldn't that just be the range of $g$?

Balarka Sen

yes; why?

user228700

Well, uh, because $g$ is surjective..?

Ramanujan

03:09

Good morning :D

user228700

Huh, I didn't think of that fact before asking if the $g$ would be the range or not tho :-| It was just an intuitive guess but I see that it wouldn't work without $g$ being surjective...yeah?

user228700

@Ramanujan G'morning :-)

Balarka Sen

think about that again. you need the surjectivity of both $f$ and $g$. Why don't you do this formally? take a $y \in C$ (look in the arrow diagram above), goal being to find a $y \in A$ such that $g(f(x)) = y$.

in fact let me just tell you. since $g$ is surjective, there is an $b \in B$ such that $g(b) = y$. since $f$ is surjective, there is an $x \in A$ such that $f(x) = b$.

so there's your desired $x$.

user228700

Huh. And this is us proving that $g \circ f$ is surjective..?

Balarka Sen

yes

we just proved that for all $y \in C$, there is an $x \in A$ such that $(g\circ f)(x) = y$.

that's the defn of surjectivity

user228700

03:14

Riight. Alright, those are all my questions. I think I'll be OK on my own now. Thanks ever so much! _/\ _ :-)

Balarka Sen

it's fine. good luck.

user228700

04:30

Let me rephrase (and correct) that: Is the cis isomer of compounds such as 2,3 dimethyl but-2-ene less stable than its trans isomer due to steric repulsion?

user228700

Gah, nvm.

user228700

Alright, I have a different question that is related to hyperconjugation. I fear that I have asked this question once before but I am still just a little bit confused. It is a lot of trouble to write down the whole thing so please ping if you're interested to read/answer, thanks.

user228700

04:46

Ah, @Sec: Hello! :-) We have, in fact, discussed the question once before.

Secret

For simple enough compounds, usually sterics have a large effect on stability. Thus making a tert butyl substituted ethene is not going to be easy

and so are cis bulky ethenes

user228700

Ah, I see.

Secret

One of my physical organic professor always reminded us not to underestimate sterics

user228700

Well, that wasn't the question, actually. Dyou remember this:

user228700

04:57

Secret

yup, though we discuss that in a more overview kind of fashion

user228700

Yeah...

user228700

If we take hyperconjugation into account, the reaction rates is in the order 4>3>2>1. On the website where I found this, it was written that at a time when this effect wasn't known, this wasn't the order since only inductive effect was taken into account...

user228700

If we take only inductive effect into account, what's the order..?

DanielSank

20

Q: What does a blackbody sound like?

Update: According to this wikipedia article, blackbody radiation is just thermal noise (Johnson–Nyquist noise); if that's what I'm looking for, what does it sound like? If a blackbody has a temperature such that its peak frequency was well within our audible range, for example 1 KHz, what would ...

I am drawing attention to that question because it is awesome.

Secret

05:01

actually, what you wrote out is the inductive effect only result. Hyperconjugation reverses that to 1>2>3>4 as this reaction is facilitated by electron releasing groups and hyperconjugation delocalise the electrons in the C-C bonds thus less electron is released into the ring than expected

user228700

> "This reaction is facilitated by electron releasing groups"

user228700

What, really?

user228700

What dyou mean by that, in the first place..?

Secret

Normally, the more electron releasing the groups are, the faster this reaction will go (e.g. -OH < -NH2). However, for the alkyls, hyperconjugation compete with the electron releasing ability of the alkyls and as seen from the experimental data, seemed to dominate the inductive effect and instead made them more electorn withdrawing

hence the reversed order compared to the inductive effect's case

user228700

I urge you to please have a look at this, if you have the time:

user228700

05:10

acadblock.com/organic-chemistry/baker-nathan-effect-f0q9

user228700

Also, @Kenshin: Afternoon :-)

user228700

@Sec: U there..?

Secret

I am still reading. That link is confusing as both users there seemed to have opposite claims on what is the observed reaction sequence. I am currently looking up journal articles to find the actual observed order, else I think I will be just confusing myself

user228700

Oh, cool :-) Okay...

user228700

05:58

@JohnR: Morning :-)

John Rennie

morning :-)

Pissed off layman

Top of the morning to you :-)

user228700

@Sec: Still looking? :-o

Secret

yes (especially mum told me to go downstaris to deal with a spider a few minutes ago). However I do find a journal article article that describe the reaction in detail

scs.illinois.edu/~mainzv/HIST/bulletin_open_access/v37-2/…

user228700

Oh, wow, thanks!

Sir Cumference

06:08

0

A: Effect of test charge on electric field intensity

What is the meaning taken out with underlined paragraph?

Sigh...

Not sure why people post that stuff as an answer to an unrelated question

koolman

Morning @JohnRennie

John Rennie

Morning :-)

Secret

@SirCumference well, that textbook does contain a partial answer to the question, but the answerer instead hijacked the post and ask his question

A question is not an answer. If you are going to ask a question, please check the site to see if it already had the answer to your question, if there is not, then post a question onto the site (make sure enough prior research is done). — Secret 8 secs ago

Sir Cumference

@Secret Oh by the way, want to see a magic trick?

Secret

?

Sir Cumference

06:15

Well, not so much a magic trick as a cool thingy in math

user228700

Are u gonna tell us or what?

user228700

Somebody please ban @SirC for not keeping up his promise. Where the heck is the "trick"?

John Rennie

@SirCumference Cough :-)

Pissed off layman

0/0 = 1

End of trick

user228700

@JohnRennie @SircC: Ahem.

Sir Cumference

06:23

Oh crud

Sorry, I'm not hearing the ping notification :/

Ok, @Kaumudi.H, pick a number

It could be a decimal, imaginary, infinity, etc.

Any number whatsoever

...uh, you there?

user228700

Oh, crap, um, yep, I'm here but I'm suddenly a little busy.

Sir Cumference

@JohnRennie Oh crud...something must be wrong with my head XD

user228700

I need to figure out why this:

Sir Cumference

I just realized there's also David D...

user228700

$A \stackrel{f}{\to} B \stackrel{g}{\to} C$

Sir Cumference

06:27

There's quite a lot of people actually...

John Rennie

@SirCumference And dmckee

user228700

Why ^ that represents $g(f(x))$

Sir Cumference

@Kaumudi.H Finally, someone using $g(f(x))$ instead of $g \circ f$

user228700

:-P

user228700

Oh, yes of course.

Sir Cumference

06:29

So, uh, who wants to try the magic trick?

._.

user228700

:-(

Sir Cumference

You were all so eager a second ago...

Swapnil Das

Morning all.

Sir Cumference

Uh, @JohnRennie, want to try?

Or @SwapnilDas?

Swapnil Das

@SirCumference Yes what?

Sir Cumference

06:32

A magic number trick

1. Ok, pick any number

Swapnil Das

Sure, though mostly it will reduce to simple algebra :P

Sir Cumference

It can be a cardinal, ordinal, an imaginary number, real number, infinity, decimal, etc.

Swapnil Das

Ok done.

Sir Cumference

2. Ok, now write out your number in english

Swapnil Das

Wait, my pen.

yes.

Sir Cumference

06:34

3. Count the number of letters in your number

Swapnil Das

fine.

Sir Cumference

Ok, now take the number of letters as your new number. Repeat steps 2 and 3.

Swapnil Das

yes

Sir Cumference

If you keep doing this, you will always find yourself at 4, with an infinite loop, since "four" has 4 in it.

Swapnil Das

yeah!

Sir Cumference

06:36

Cool?

Swapnil Das

Cool.

Sir Cumference

.–.

Swapnil Das

my loop was small. i chose 5 :P

Sir Cumference

It works with any number, real or not, finite or not, cardinal or ordinal, etc.

Swapnil Das

yeah.

Sir Cumference

06:37

@Kaumudi.H Was it worth it?

Swapnil Das

Hey, i use my real name here.

Sir Cumference

Then again, most people here might...

Swapnil Das

Ohk.

Sir Cumference

Yeah it was my stupid realization. XD

Swapnil Das

My name isn't any kinda fancy. Simple Indian name.

:P

Sir Cumference

06:41

Most people here seem to be eastern hemispherers.

Swapnil Das

I'm northener, perhaps?

Secret

@Kaumudi.H You act f on any element in A to send it to B, then you act g on the result of f(A) to send it to C. When you write it as $g(f(x))$, it is the same as $(g \circ f) (x)$ because you always evaluate from the innermost bracket first

Swapnil Das

@SirCumference You're an American?

user228700

@Secret Yeah, I was able to figure it out, thanks :-)

Sir Cumference

@SwapnilDas Yep :)

user228700

06:47

@SirCumference Haven't read it yet :-P

Sir Cumference

@Kaumudi.H ;-;

I thought people wanted to hear it...

Swapnil Das

@SirCumference Oh. Don't switch your field to Condensed Matter Physics :P

user228700

:-P A little busy, dude.

Sir Cumference

@Kaumudi.H I'm kidding

@SwapnilDas What?

I'm into astronomy :P

Swapnil Das

@SirCumference Astronomy or Astrophysics?

Sir Cumference

06:48

They're basically the same thing.

Nowadays, at least, all astronomy is astrophysics

Swapnil Das

Ohk. Why different names, then?

Sir Cumference

Well historically Astronomy was about cataloguing objects, while Astrophysics was about the physics in space

Then the latter basically became the basis of all space research and the two terms took a synonymous meaning

Swapnil Das

Ah. Nice.

Sir Cumference

So technically I'm more into the physics part :)

How bout you?

Swapnil Das

I just was gonna ask. Great ;)

@SirCumference I'm into GR, if it sounds well from my side.

Sir Cumference

06:51

@SwapnilDas Oh neat!

Swapnil Das

I don't of course understand it, but can feel its beauty and elegance.

Sir Cumference

I'm pretty new to GR

Swapnil Das

It's cool?

Sir Cumference

Well yeah :)

user228700

Quick question. Range of polynomials = Real numbers, correct?

Sir Cumference

06:52

But I end up confusing myself since I'm so used to Newtonian mechanics

@SwapnilDas Oh wait, you're in 10th grade, right?

Swapnil Das

@SirCumference That of course of happen if beginners.

@SirCumference Lol yes.

Sir Cumference

Huh, you're like me

Swapnil Das

How so?

Sir Cumference

I started looking into QM qualitatively around then

It's a lot nicer when you understand the math though :)

Swapnil Das

Of course. Actual fun begins.

user228700

06:54

Aaaanybody?

Swapnil Das

@Kaumudi.H Why not complex?

user228700

The exponents of the variables can only ever be whole numbers in case of polynomials. There's my answer, I guess.

Sir Cumference

Sigh...can someone remind me what a Fourier transform is?

Wait, nvm

Secret

:34409360

in Mathematics, 2 mins ago, by Secret

Given the string length function $sl(n)$ for any english name of any mathematical object $n$. It is known that $n=4$ is a fixed point. Prove or give cnounterexample that $n=4$ is an attractive fixed point for all $n$ for this algorithm.

John Rennie

@Kaumudi.H are you assuming that the domain is the real numbers?

Obviously is the domain is the complex numbers then the range will be the complex numbers as well.

Secret

07:07

If your polynomial have complex coefficients, the range can be complex as well

user228700

@JohnRennie Yes, of course.

user228700

I asked the same question at the MSE chat and it seems to have sparked a big decision, so I quietly left the room :-P

user228700

@JohnRennie Right, yes, thanks :-)

Secret

This is normal. We mathematicians get very predentic on the precise definition because otherwise there are just too many possibilities

Swapnil Das

@Secret Bound by the chains of rigor :P

user228700

07:11

$Mathematics$ The JEE LaunchPad

We love Physics, Chemistry and Mathematics. And we are prepari...

user228700

^ Pretty sure this room is eventually going to be frozen due to inactivity, now that Doraemon has been banned.

David Z

07:43

@Kaumudi.H Deleted. Not banned.

2

There's a big difference.

John Rennie

08:20

@Kaumudi.H: you can answer this now! :-)

0

Q: At the freezing point, why is the vapor pressure of the liquid phase equal to the vapor pressure of the solid phase?

At the freezing point of a substance, the solid phase is in dynamic equilibrium with the liquid phase. My textbook defines the freezing point of a substance as "the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase". Why...

freezing

VIP

08:42

@JohnRennie maybe because in the PVT surface, the triple state is a straight line perpendicular to the pressure axis, or is this just another way to see the problem ?

John Rennie

@VIP: see my answer. It's a straightforward result of both phases being in equilibrium with the same vapour.

Secret

ugh, that's the only thing that none of those hyperconjugation journal articles are explaining:

2

Q: Why don't we have hyperconjugation with C-C bonds?

C-C bonds are weaker than C-H bonds.So why is hyperconjugation with C-C bonds not possible?

organic-chemistry hyperconjugation

while drawing resonance structures do suggest a possible reason in that e.g. for $^tBu-C-C^{\delta +}$ I will end up with methyl carbocations as resonance contributors which are highly unstable. I still felt there is a better explanation than that

meanwhile...

pubs.rsc.org/en/Content/ArticleLanding/2013/OB/…

So in order to perfectly answer kaumudi's question, we need to understand why for common systems such as that Baker Nathan reaction, CC hyperconjugations are so weak

Osheen Sachdev

09:10

Hey people

koolman

Hii

Welcome @user140525

Jonathan Richard Lombardy

Hi everyone, I have completed high school and was looking for a book that teaches mathematics necessary for physics, at undergraduate level. I want to find a book similar to Mathematical Methods for physical sciences by Mary Boas, can anyone recommend similar books, or give a review of this one. Thank you.

Osheen Sachdev

Please go through question 60

okay so her's how I approach the quastion

*here's

now the acceleration of the block m is gsin(a)cos(a) and that of mass M should simply be (m/M)gsinacosa

I don't know wherre I'm going wrong. it would be great if you could just help

user228700

09:30

@DavidZ What's the difference?

user228700

@JohnRennie Ah, yes, but I see that you've given an excellent answer, so I'll pass :-)

user228700

@Secret I'll look into this in a bit, thanks!

Secret

09:57

0

Q: On baker nathan hyperconjugation

Recently I have been reading about the history of the Baker Nathan reaction. In addition to the large rate increase for Me vs H, all the other alkyl groups showed decreases with respect to Me in a regular manner as would be predicted by reduced hyperconjugative ability. By drawing ...

organic-chemistry molecular-orbital-theory hyperconjugation

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