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12:00 AM
@PeterTamaroff eat well!
@robjohn Hehe, the last exchange I read between you two was the same thing about food.
@WillHunting you always thought we were just a couple of bots...
where is everybody?
@MarianoSuárez-Alvarez The site was down for a while, It may take a while for people to trickle back.
@MarianoSuárez-Alvarez You mean in chat or on main?
12:09 AM
Even without maintenance, there are times when it is quiet.
@WillHunting Oh, be quiet!
@meand You are a big fan of Charlie?
@WillHunting yep!
@robjohn I'm wondering the same thing. Nothing has been even edited in 6hrs. Is the site read-only right now?
@AlexBecker Well for one it is Sat night!
12:24 AM
@AlexBecker You should be able to edit questions, it is just that the main page seems to be unchanged for over 2 hours now.
Oh, huh. I cleared my cache and now the active page has been updated
Alex looks so green.
@AlexBecker I answered this question since the freeze, can you see my answer?
@robjohn Yes, I can.
12:26 AM
@WillHunting What, my gravatar?
@AlexBecker Yup.
@robjohn I forgot, does the Muppet Show have anything to do with Sesame Street?
Guess it's just a slow night...
@AlexBecker Wow, you are here and not partying?
@WillHunting Perhaps
12:29 AM
Math is a party :)
(so lonely)
My university doesn't start for a couple more weeks
so I'm stuck at home in the meantime
Well, you can just read on your own.
@AlexBecker Meantime? I like meantime!
@robjohn No, robjohn, not your kind of meantime. I'm not stealing candy from small children.
@AlexBecker Aww! >8(
@AlexBecker Wait, is there a show with the mean square stealing candy?
12:31 AM
We are like this in the chat!
@WillHunting There should be.
One of my friends dogs just got xylitol toxicity, we think from eating some sugar free candy.
Can someone tell me what $\mathbb{G}$ usually means?
or $\mathbb{G}_m$
xylitol and dogs don't mix
which appears in the first new paragraph of the second page of this: math.uga.edu/~pete/SC5-AlgebraicGroups.pdf
12:35 AM
@QualFighter Can be anything.
that's not a nice thing to say
@QualFighter I've never seen that before. If you find out, I would be interested to learn.
it is the target space of the set of characters on an algebraic torus, i think.
characters that i know are sent to circles
what you mean
Pragmatics is a subfield of linguistics which studies the ways in which context contributes to meaning. Pragmatics encompasses speech act theory, conversational implicature, talk in interaction and other approaches to language behavior in philosophy, sociology, and linguistics. Unlike semantics, which examines meaning that is conventional or "coded" in a given language, pragmatics studies how the transmission of meaning depends not only on structural and linguistic knowledge (e.g., grammar, lexicon, etc.) of the speaker and listener, but also on the context of the utterance, any preexis...
@QualFighter from that reference, it seems that $\mathbb{G}_a$ is "the affine line endowed with its natural addition law"
12:40 AM
@QualFighter Look, I was just trying to say G can mean anything. There's no other intention.
Haven't found $\mathbb{G}_m$
@robjohn do you know if characters on algebraic tori go to affine lines?
that would make sense to me, from the various sources i've been reading
@QualFighter I have no idea. All I know about that is what I just read :-)
ok, thanks anyhow @robjohn
@robjohn you saw this link i posted, yes? martinorr.name/blog/2010/01/24/…
that is representative of the various other places i have seen this inexplicable $\mathbb{G}_m$
sorry to harp if you did check this out
@QualFighter Can't say that I've seen that link. Even if I did, I wouldn't have understood much on it :-)
12:44 AM
how frustrating that notation is consistently not defined!
a conspiracy to keep lesser mortals out of the game
@MeAndMath Yep.
This is weird... I just saw a change on the main site, but only for something 31 minutes ago.
@QualFighter Maybe you can comment there and ask about it.
@robjohn I think things are behaving randomly for the time being. Intermittent disruption.
1:00 AM
You can exprress any wave with exponential function. Does it mean that you can express $cos(t)$ with $Re(e^{it})$ where $Re$ is the real part?
@hhh $e^{it}=\cos t+i\sin t$ and $e^{-it}=\cos t-i\sin t$. Add the two equations and see what you get.
@hhh that is the real part of $e^{it}$, yes
@hhh So $\cos t=\frac{1}{2}(e^{it}+e^{-it})$.
So the word is that ALL the home and Q&A pages are frozen everywhere except SO.
@robjohn Is that an official statement?
1:07 AM
@robjohn What do you mean?
@WillHunting Thank you for your help, got the point...
@WillHunting I asked on the mountain and that is the word I was given
(so unclear)
I want to understand why you can express any wave with the below?

$$\bar f(z,t)=\int_{-\infty}^{\infty} \bar A(k) e^{i(kz-wt)} dk$$
It is like

$$\bar f(z,t)=\sum_{-\infty}^{\infty} \bar A(k) e^{i(kz-wt)} dk$$
Wait! 'allelujah! What is $\bar A(k)$? Vector amplitude?
I can understand it if it is $-1$ or $1$ that depends each case of $k$ -value but I cannot yet see how it is vector...$e^{i(kz-wt)}$ is a complex number so consisting of Real and Complex part, the crux point here?
(studying page 370 in Griffits, Electrodynamism book that tries to explain things such as fourier-transform and this equation, I stopped here because not yet understand $\bar A(k)$...ideas?)

Book notation $\tilde A= A e^{i\gamma}$ where the bar is a "tilde", actually.
Yes it actually means just a phase-shift, not vector, roger.
Where should I learn Fourier-Transform? Some book suggestion?
2:11 AM
I cannot understand the minus sign in the second equation of 9.76.
(Griffiths p.384)
Magnetic field got a minus sign but electric field not.
@hhh A-bar(k) is a complex number, If A-bar(k) = complex conjugate of A-bar(-k) then $f(z)$ is a real valued function, which is the case we are interested. I'd first look up Wikipedia page on Fourier Transform and Fourier series of real valued functions which gives some clarity. Good Luck
I don't see any new questions on the front page. Is there a problem?
@Thomas Someone should try posting a question and see, I don't have one.
The questions are not ordered chronologically either.
@RajeshD But $k$ has no bar, $\bar A$ has a bar and the book explains that the bar (actually tilde) means a phase -shift, not complex-conjuage.
Or did I misunderstood something?
2:26 AM
@hhh I don't mean Bar is complex conjugate either
I am not talking about the bar at all
But the notation is $\bar A(k)$, not $A(\bar k)$
"A-bar(k) is a complex number"?
k is a real number hence no bar
that is correct
k is scalar
A-bar(k) is complex
Yes $\tilde A$ is complex, it is a phase-change -operation, $\tilde A:=Ae^{i\beta}$
2:28 AM
I don't think it means phase shift here
it only means it is a phasor
the quatity is a phasor
@hhh I am not sure on these things, so don't take things from me seriosly....But my advice is turn some pages back in the Griffith's book and read clearly what is the notation he is following, it should solve the problem
1 hour later…
3:39 AM
I see the Q&A page is back to updating
Most of the people who posted answers during the freeze will not ever be on the front page.
4:19 AM
It's quiet... too quiet.
4:59 AM
Hi, Ravi.
$$\overrightarrow{a}\cdot \overrightarrow{b} = ||\overrightarrow{a}|| \times ||\overrightarrow{b}||\cos \theta $$
Hmm, well, very well!
6:07 AM
Hey @Matt can I ask you something about this question you asked?
6:36 AM
@robjohn I am here!
6:47 AM
Hello everybody! Is there anybody who deals with visualisation of surfaces?
@DavidK. Sure, ask.
@Matt Hey did you change your avatar? I noticed!
@WillHunting It changes automatically, I have no email address in my profile.
@Matt But isn't it filled in once you log back in?
@DavidK. I'll edit my answer to that, maybe it helps.
@WillHunting No.
6:54 AM
@Matt You may find it filled in the next time you log out and then in. At least that is what happened to me a few months ago.
@Matt Well I'm really trying to finish off part (b). I'm at the point were the existence of a retraction implies that $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ is in injection. But, why is that a contradiction?
7:06 AM
@DavidK. I think the fundamental group of the torus is generated by two circles, one around the middle hole and one orthogonal to that, around the "side" of the torus. The retract will map one of these two generators, namely the second one, to zero. So I'm not sure how you can have an injection.
@WillHunting Ok.
You're not free to choose the induced map on the groups.
@matt Right, the generators. That makes sense. I knew it was something simple.
@Matt Thanks!
@DavidK. No problem : ) So it's all clear?
It's been a while since I looked at fundamental groups...
I think the retract shrinks the torus to $S^1$.
Which maps one generator to a point while leaving the other one alone.
@Matt Yeah I think so. I mean, $\mathbb{Z}$ is generated by $1$, so the generators of $\mathbb{Z}\times\mathbb{Z}$ are $(1,0)$ and $(0,1)$. Both of those must map somewhere.
Yes. Well, I think of the generators as actual loops on $S^1 \times S^1$.
So $(1,0)\mapsto 1$ is fine, but $(0,1)\mapsto 0$ is a contradiction.
@Matt Right, I see what you mean there.
7:15 AM
@DavidK. If I ignore the geometry behind it, I get confused. For example in your argument (just using groups): we can find a bijection between $Z \times Z$ and $Z$, right? So how do we get a contradiction?
Could you explain the contradiction you are talking about there in more detail, please?
@Matt Right. That has been an issue for me here. But how can we get a bijection? I've been trying to work out an actual bijection. I can get an injection, but NOT a homomorphism
@Matt Yes:
Let $i_{\star}:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}\times\{0\}$. Then, for any integers $m,n$ $i_{\star}(m,n)=m(i_{\star}(1,0))+n(i_{\star}(0,1))$.
@DavidK. Yes, ok. There cannot be a homomorphism that is bijective since then the two groups would be isomorphic which they are not since on is cyclic and one is not.
So, the map $i_{\star}$ is completely determined by $i_{\star}(1,0)$ and $i_{\star}(0,1)$. So if $(0,1)\mapsto 0$, then so does any element $(0,n)$ in $\mathbb{Z}\times\mathbb{Z}$.
I spent a while trying to show that a homomorphism couldn't exist, but I couldn't figure out how. But as soon as you mentioned generators I figured it out.
7:24 AM
That's good. : )
@Matt Thanks again!
@DavidK. My pleasure : )
If there is anything else at any time just ping me.
@Matt Oh I will!
does XnA , where X and A are sets could mean anything other than $X\cap A$
@RaviUpadhyay What is XnA? Never seen this notation.
7:41 AM
Morning @jonas! How was your beauty sleep?
@RaviUpadhyay It is probably a way to refer to the intersection if the author was too lazy to use the actual symbol. That book probably sucks typographically then, and cannot contain too much math.
Again I ask, what is the best resource for learning Mathematics?
Khan Academy covers only basic topics, so not him.
Paul's Notes doesn't cover topics so well for a layman like me.
Anyone can look this?
Q: Is it possible to perform polynomial composition on polynomials of $deg=0$?

Gustavo BandeiraI'm reading Barbeau's polynomials, and he states: For the polynomial $a_nt^n+a_{n-1}t^{n-1}+...+a_1t+a_0$, with $a_n \neq0 $, the numbers $a_i$ $(0 \leq i \leq n)$ are called coefficients. Some pages later, there's a question: Is $deg(p \circ q)$ related in any way to $deg(q \circ p...

Hey, @WillHunting again!
8:00 AM
@ParthKohli Do you know Coursera?
How do we make hyperlinks in this chat?
Thank you.
Coursera? Yes, heard about it.
@GustavoBandeira I am now answering your question...
@ParthKohli There's a course on mathematical thinking
@WillHunting Thanks. =)
8:03 AM
Thank you, I am looking for it.
I don't want to join a course as such. Is there anything else?
@ParthKohli There are only a light calculus course - You may find something useful on OpenCourseWare
Well, I just need some tutorials and practice problems. Nothing else :)
The good thing on coursera is that you can do it on your pace.
There's one guy on YT
Let' me search it.
Erm, can you please explain how they teach on Coursera?
Are there any live lessons?
@ParthKohli Nope, recorded lessons. And there's a forum.
8:09 AM
Well, signed up already :)
I'm watching algorithms classes now.
Thanks, @GustavoBandeira.
@ParthKohli If you can't stream, you can download the lessons too.
@GustavoBandeira that YouTube channel is useful. Accept my gratitude :-)
@ParthKohli Someone here recommended me this channel.
8:15 AM
@Gustavo can you give the link for the algorithms classes. Can't find it.
This one - There are two parts.
You're welcome. =)
@WillHunting It was beauty decreasing!
@ParthKohli It depends on what topics. Your question is too vague.
8:24 AM
I'd like to start from Calculus II.
@ParthKohli Well, I don't know what "Calc 2" treats, but I think Paul Dawkin's notes are good.
Yes, of course.
I figured out that there is something missing in his notes and that makes me ask for some other resource.
@ParthKohli What is missing that you need?
It doesn't just work out for me.
It seems that there is a dimension less.(just like in textbooks)
@ParthKohli What do you mean by a dimension less?
8:36 AM
@gustavo Answered.
@WillHunting ty, I'm gonna look it in a minute.
Reading a bunch of text isn't my cup of tea. @WillHunting
@ParthKohli So you think one should not have math books at all?
One should, but I, personally, need more resources.
Oh well, I think Paul's notes are fairly complete.
8:38 AM
Maybe you should not study calculus at such a young age!
Yes, it's good enough for genius minds. :-)
I always end up lecturing on how age doesn't matter.
@ParthKohli More resources?
@GustavoBandeira Yes. I like audio-visual learning... it's much better than reading a boring book.
@ParthKohli Got it. =)
Then the channel I gave you will help a lot.
@ParthKohli Then take a look at the MIT OCW on youtube.
8:45 AM
Did u see opencourseware?
MIT OCW? Yes, I've seen it a number of times.
You can watch lessons there.
BUT, textbooks will help a lot when you're lost.
Yesterday I've watched a calculus class that left me lost - textbook helped me.
I don't think Courseera or khan's academy can ever beat books. But then again every one has their own preferences.
Yes, I just need something more than textbooks. I'm not saying anything against textbooks. :-)
I like what you're doing.
Trying to learn advanced stuff and all that.
9:06 AM
@IanWilliams Hello. =)
@Noah Hello. =)
Hey, @WillHunting
Hello @GustavoBandeira
Whats up
Not much, you?
No majro complains
9:14 AM
What ya doing here? Math problems?
@ParthKohli buy an iPad
I think they have got some interative books on there
@GustavoBandeira Nope, just hanging around
Good morning
Plus I used to be good at math
@Noah Why did you quit?
I didnt quit...
9:17 AM
@Noah But you stopped being good at it?
I just dont feel like I am the same as I used to be
Thats right
@GustavoBandeira I more into arts
Painting, writing, etc.
Do you do math?
I'm studying math now, but I used to make music. I also study classic piano.
Oh, that's cool
What kind of music did you make?
9:22 AM
I had "musical periods" since I was 15.
Now I'm compilling some piano etudes and learning math to understand some 20th century msic theories.
This is from the time I was young and less insane.
Oh, cool, thanks for the link. Do you consider yourself a madman :)
My recent etudes:

Yeah, iPad, but it's not the best thing to spend money for textbooks. :)
9:25 AM
And the other ones are the stuff I do now.
@Noah It's because today, I make music no one wants to hear. =D
@ParthKohli I dont think you get only textbooks
Yes, you get apps as well.
The thing is, I already have an iPad. ;)
@GustavoBandeira I am listining to some of the melodies
they seem to be pretty cool
@ParthKohli Then why not give it a try
Some of the etudes have a easy listening sounding, because they're intended for basic piano study,
@GustavoBandeira omm
9:27 AM
I wouldn't like to spend dollars for textbooks.
@ParthKohli Pirate bay FTW.
@ParthKohli Why dont you try borrowing books from your school library?
In India, you get textbooks for 1 dollar.
@ParthKohli then what seems to the be problem?
Good idea, but I mentioned above that I don't like textbooks much, and interactive textbooks on iPad cost a lot!
9:29 AM
@ParthKohli Oh, always remember of sending gifts to your friends in Brazil....
@ParthKohli I dont know much about prices in India, but I guess iPad books are way cheeper than hardcopy books
Wait, yeah, 5 dollars for a book is cool enough.
@ParthKohli Plus, I dont really like the idea of carrying bundles of hard copy books on your back...
@Noah I don't really like the idea of buying a tablet for 650 dollars ;)
@ParthKohli You said you had one
So I dont think taht's an issue
9:33 AM
I do... thank God somebody gifted me one.
@Noah In soviet Russia, the books carry you on their back.
Yeah, that isn't. :-)
@GustavoBandeira How's that?
@Noah That is a Soviet Russia joke.
9:35 AM
Can combine them with gay.
We already have cat-themed jokes.
I don't need an extra "s" with cat-jokes.
@GustavoBandeira that was interesting
@JonasTeuwen looks like you have a special love for the bat man?
9:49 AM
@Noah Yes, he is Batman and I am Superman.
@GustavoBandeira No, that is not cool.
@ParthKohli Hmm, maybe you should get used to reading then, because that is the only source of higher information.
@ParthKohli All the more reason for you to get them. iPads are not for serious reading, fiction yes, math no.
@WillHunting I disagree
I love reading, but just reading Math doesn't work.
It involves practicing as well, which may be done through audio-visual learning.
Is there a completing the square formula for a polynomial with (deg = 3)?
10:26 AM
@GustavoBandeira You still have not looked at my answer?
@WillHunting Gonna do it now.
@GustavoBandeira Yes, I usually remove my answers if they have no votes after a few hours.
@WillHunting Yep, it was what I was looking for.
@GustavoBandeira Do you understand it now?
@WillHunting Yep.
10:35 AM
Ah, now I can create tag synonyms with 2.5k.
Need 500 more to closevote questions.
My question has almost no views
@GustavoBandeira It's Sunday and the kids are partying.
When I posted it, I thought:
-Well, I guess people is going to answer it because it's so elementary
@GustavoBandeira No, sometimes people don't know what to answer because it is too simple. But I know what the doubt is so I answer.
10:39 AM
I usually don't go to parties.
I love to stay at home. =)
@gustavo Why did you put a number for your age in your profile that is not the actual number? You can always leave it blank!
@WillHunting Oh, I wanted to earn that badge.
@GustavoBandeira Then why don't you just put the real number?
@WillHunting It had the real number, then I changed because I was feeling ashamed of knowing little math for my age.
@GustavoBandeira No need to be ashamed! Age is not relevant at all.
10:49 AM
You can call me stupid now.
@WillHunting Thanks for your words.
Everytime I start something new, I feel this. It was the same with the piano.
One only needs to be ashamed if one has done evil deeds to cause others suffering.
Kids with 1/4 of my age could play 4x what I could play.
@WillHunting I have this feeling too.
When I asked "what do girls think about you?" and you suddenly disapeared, I entered in a micro-panic.
@GustavoBandeira No need to compare with others. People are born different and go through different life experiences.
@GustavoBandeira I am always disappearing like I said.
@WillHunting Yep. I started to ignore others performance after some time - and started to focus more in the problems I was finding.
10:55 AM
@GustavoBandeira Yes, just be the best you can be.
Ask yourself what the most important things in life are, and focus on them.
Thanks for your words.
1 hour later…
12:17 PM
@JayeshBadwaik Hello, I try to understand how an EM wave behaves when it hits a physical object. I can find something like the reflection depends on the impedance of the materials and reflection is different with reactive (capacitive/inductive) material, something here. A question asks what happens to EM wave when it hits a cloud and I don't any source of information. Ideas?
can someone help me understand the greedy algorithm in childs terms
@hhh So, basically you must ask what happens to EM waves at the boundaries of surfaces.
Wait a minute, I will be back.
Probably I am reading the page 331 in Griffiths but it is different scope, it covers the Maxwell equations in boundaries and some differential-equation analysis.
@hhh That is exactly what the concept of reflection of EM waves is based on. If you can understand this, you would understand the further thing.
The boundary on which the wave is incident can be of basically two types "dielectric" and "conducting" duh . And then you must see what happens to the electric field and the magnetic field at the boundary. There is then a transmission co-efficient and a reflection co-efficient which add up to 1 duh again and then based on this information, you can calculate stuff about normal incidence and oblique incidence. And finally use this information to solve your problem about the clouds.
@hhh If you want to "bookmark" a chat message for future reference, use the bookmark feature of the chat which can be obtained by using the drop down arrow besides the link "room" on the top right.
(found it, thank you for the tip -- have to read this chapter now again, not sure yet whether things work similarly to material waves -- atached wave or like freely moving wave? I think it can act the both waves depending on its impedance -profile, right?)
12:33 PM
@hhh yes, the principle is same. the things get complicated by the fact that magnetic and electric waves behave differently.
And also because there is transmission + reflection instead of pure reflection .
Well I have to rush now. See you later.
12:48 PM
Wadafuq, my account was "automatically suspended"? Who are these sour people in this room?
@JonasTeuwen That's what happens when you tell the logo to "shut up."
1:00 PM
If $\det A=0$, does anyone know any example of a non-zero $B$ such that $AB=0$ ?
I first thought of the adjugate, but its entries are the determinants of the minors of A, and if the rank of A is less than n-1, all these minors have determinant $0$ so the adjugate is the zero matrix.
By Cayley-Hamilton we know that $A$ satisfies something like $A^n + a A^{n-1} + ... + z A = 0$ so $B= A^{n-1} + a A^{n-2} + ... + z $ would work if we knew that $B$ was non-zero.
I'm working with entries over an arbitrary commutative ring by the way, otherwise if I had a field I could rig something up with the minimal polynomial of $A.$
@JonasTeuwen They do that? What is the charge?
@RagibZaman If you are talking about matrices, then one good method would be to use the fact that $A$ can be transformed into a column-triangular form with one of the columns completely zero by only right multiplication, once this is done, take the transpose of the result and right-multiply it and you will get a zero matrix.
@JayeshBadwaik That sounds perfect! But why would one of the columns be completely zero?
@JohnJunior Sure. Bye.
1:16 PM
@RagibZaman Determinant is zero, so rank is less than n (where the matrix is nxn). Which kind of actual triangular form will be needed I am not sure though.
@JayeshBadwaik I can't see why rank A < n implies a zero column will form we when put A into column-echelon form.
Would it help to assume the rank is $n-2$ or lower?
1:33 PM
@JayeshBadwaik Also, I can't see why right multiplying by the transpose gives the 0 matrix?
If a matrix in Column echelon form is given by the rows (a,0,0), (b,0,0), (d,e,0) then the 1,1 entry in the product with its transpose is $a^2.$

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